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NDA/NA, National Defence Academy / Naval Academy, , ABOUT THE EXAMINATION, Education is the glorious route through which anyone can attain the goal of success. And if education, has been acquired through a renowned institution, it leads to achieve glorious heights in career., National Defence Academy (NDA) is one such institution which propels the students in the arena of life, and contributes to a very successful and fulfilled career. But to get enrolled in this institution means, goal directed study for passing in a competitive examination which is conducted by Union Public, Service Commission nationwide. For recruitment to Army, Navy and Air Force wings of Indian Army,, there is prestigious National Defence Academy Entrance Examination. The examination is conducted, twice a year, and the duration of training is three years., Though the candidate may give his preference for a particular wing of the Armed Forces, the final, selection depends upon his performance and place in the merit list., , EDUCATIONAL QUALIFICATIONS, (i) For Army wing of National Defence Academy, Class XII pass of the 10+2 pattern of school, education or equivalent examination conducted by a State Education Board or a University., (ii) For Air Force and Naval wings of National Defence Academy and for the 10+2 (Executive Branch), Course at the Naval Academy, Class XII pass of the 10+2 pattern of school education or equivalent, with Physics and Mathematics conducted by a State Education Board or a University., , SYLLABUS, Algebra Concept of a set, operations on sets, Venn diagrams. De-Morgan laws. Cartesian product,, relation, equivalence relation. Representation of real numbers on a line. Complex numbers – basic, properties, modulus, argument, cube roots of unity. Binary system of numbers. Conversion of a, number in decimal system to binary system and vice-versa. Arithmetic, Geometric and Harmonic, progressions. Quadratic equations with real coefficients. Solution of linear inequations of two variables, by graphs. Permutation and Combination. Binomial theorem and its application., Logarithms and their applications., Matrices and Determinants Types of matrices, operations on matrices Determinant of a matrix, basic, properties of determinant. Adjoint and inverse of a square matrix, Applications – Solution of a system of, linear equations in two or three unknowns by Cramer's rule and by Matrix Method., Trigonometry Angles and their measures in degrees and in radians. Trigonometrical ratios., Trigonometric identities Sum and difference formulae. Multiple and Sub-multiple angles. Inverse, trigonometric functions. Applications – Height and distance, properties of triangles.
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Analytical Geometry of Two and Three Dimensions Rectangular Cartesian Coordinate system., Distance formula. Equation of a line in various forms. Angle between two lines. Distance of a point, from a line. Equation of a circle in standard and in general form. Standard forms of parabola, ellipse and, hyperbola. Eccentricity and axis of a conic. Point in a three dimensional space, distance between two, points. Direction Cosines and direction ratios. Equation of a plane and a line in various forms. Angle, between two lines and angle between two planes. Equation of a sphere., Differential Calculus Concept of a real valued function – domain, range and graph of a function., Composite functions, one to one, onto and inverse functions. Notion of limit, Standard limits –, examples. Continuity of functions – examples, algebraic operations on continuous functions. Derivative, of a function at a point, geometrical and physical interpretation of a derivative – applications., Derivatives of sum, product and quotient of functions, derivative of a function with respect of another, function, derivative of a composite function. Second order derivatives. Increasing and decreasing, functions. Application of derivatives in problems of maxima and minima., Integral Calculus and Differential Equations Integration as inverse of differentiation, Integration by, substitution and by parts, Standard integrals involving algebraic expressions, trigonometric,, exponential and hyperbolic functions. Evaluation of definite integrals – determination of areas of plane, regions bounded by curves – applications. Definition of order and degree of a differential equation,, formation of a differential equation by examples. General and particular solution of a differential, equation, solution of first order and first degree differential equations of various types – examples., Application in problems of growth and decay., Vector Algebra Vectors in two and three dimensions, magnitude and direction of a vector. Unit and, null vectors, addition of vectors, scalar multiplication of vector, scalar product or dot product of twovectors. Vector product and cross product of two vectors. Applications-work done by a force and, moment of a force and in geometrical problems., Statistics and Probability Statistics: Classification of data, Frequency distribution, Cumulative, frequency distribution – examples Graphical representation – Histogram, Pie Chart, Frequency Polygon, – examples. Measures of Central tendency – Mean, Median and Mode. Variance and standard deviation, – determination and comparison. Correlation and regression., Probability Random experiment, outcomes and associated sample space, events, mutually exclusive, and exhaustive events, impossible and certain events. Union and Intersection of events., Complementary, elementary and composite events. Definition of probability – classical and statistical –, examples. Elementary theorems on probability – simple problems. Conditional probability, Bayes', theorem – simple problems. Random variable as function on a sample space. Binomial distribution,, examples of random experiments giving rise to Binomial distribution.
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CONTENTS, NDA / NA Solved Paper 2019 (II/I), , 1-30, , NDA / NA Solved Paper 2018 (II/I), , 1-44, , NDA / NA Solved Paper 2017 (II/I), , 1-44, , 1. Sets, Relations and Functions, 2. Complex Numbers, , 1-19, 20-40, , 16. Functions, , 297-316, , 17. Limits, Continuity, and Differentiability, , 317-339, , 3. Quadratic Equations, and Inequalities, , 41-63, , 18. Differentiation, , 340-362, , 4. Sequence and Series, , 64-87, , 19. Application of Derivative, , 363-388, , 5. Logarithms, , 88-97, , 20. Indefinite Integration, , 389-411, , 21. Definite Integration, , 412-429, , 6. Matrices, , 98-123, , 7. Determinant, , 124-149, , 22. Area Bounded by Region, , 430-445, , 8. Binomial Theorem, , 150-165, , 23. Differential Equations, , 446-468, , 9. Permutations and, Combinations, , 166-181, , 24. Rectangular, Cartesian System, , 469-489, , 10. Probability, , 182-202, , 25. The Straight Line, , 490-512, , 11. Binary Numbers, , 203-212, , 26. The Circle, , 513-535, , 12. Trigonometric Ratios, and Equations, , 27. Conic Sections, , 536-567, , 213-246, , 28. Vector Algebra, , 568-605, , 13. Properties of Triangles, , 247-264, , 14. Height and Distance, , 265-280, , 29. Three Dimensional, Geometry, , 606-635, , 15. Inverse Trigonometric, Functions, , 30. Statistics, , 636-654, , 281-296, , 31. Correlation and Regression, , 655-663
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NDA /NA, , National Defence Academy/Naval Academy, , SOLVED PAPER 2019 (II), PAPER I : Mathematics, 1., , If both p and q belong to the set, {1, 2, 3, 4 }, then how many equations, of the form, px 2 + qx + 1 = 0 will have real roots?, (a) 12, (c) 7, , 3., , 2, Ê (d) Equation px + qx + 1 = 0, has real, , subset of U., We know that, A ∪ ( A ∩ B) = A,, So option (a) is not correct., A ∩ ( A ∪ B) = A, so option (b) is correct., ( A ∩ B) ∪ C = ( A ∪ C ) ∩ ( B ∪ C ),, so option (c) is correct., and ( A ∪ B) ∩ C = ( A ∩ C ) ∪ ( B ∩ C ), so option (d) is correct., , [Q for real roots of a quadratic equation, b 2 − 4ac ≥ 0], , What, is, the, value, 1 − 2 + 3 − 4 + 5 − ...... + 101?, (a) 51, (c) 110, , 4., , of, , (b) 2, (d) 4, , Ê (a) Sum of first n term of a series = n + 12, , ⇒ a1 + a2 + a3 + .........+ an = n + 12, Put n = 1, a1 = 1 + 12 = 13, Put n = 2, a1 + a2 = 2 + 12 ⇒ a1 + a2 = 14, ⇒ 13 + a2 = 14 ⇒ a2 = 1, Put n = 3, a1 + a2 + a3 = 3 + 12, ⇒ 13 + 1 + a3 = 15, ⇒ a3 = 15 − 14 = 1, , Ê (a) Given series,, , 5., , k+1, 2, = k 2 + 4 − 4k + k + 1, = ( k − 2 )2 + 2 ×, , = k 2 − 3k + 5, 9 9, = k 2 − 3k + − + 5, 4 4, 2, 3, 11, = k − +, , 2, 4, 3, 2, 2, , α + β is minimum, if k − = 0, , 2, 3, k=, ⇒, 2, , If the sum of first n terms of a series, is (n + 12), then what is its third, term?, (a) 1, (c) 3, , (b) 55, (d) 111, = 1 − 2 + 3 − 4 + 5 − ...... + 101, = (1 + 3 + 5 + ..... + 101), − (2 + 4 + 6 + ..... + 100), = (1 + 3 + 5 + .... 51 terms), − (2 + 4 + 6 + ...... 50 terms), 51, =, [2 + ( 51 − 1) × 2 ], 2, 50, −, [4 + ( 50 − 1) × 2 ], 2, [Q both series are AP and, n, S n = [2 a + ( n − 1)d ], 2, 51, 50, =, × 102 −, × 102, 2, 2, = 2601 − 2550 = 51, , 2( k − 2 ), = k − 2,, 2, − ( k + 1), αβ =, 2, We know that, α 2 + β 2 = (α + β )2 − 2αβ, , Qα + β =, , Ê (a) Let U be the set and A, B andC are the, , roots, where p and q belong to the set, {1, 2, 3, 4}., ∴, q 2 − 4p ≥ 0, , 2., , 2 x 2 − 2( k − 2 ) x − ( k + 1) = 0, , (a) A ∪ ( A ∩ B) = A ∪ B, (b) A ∩ ( A ∪ B) = A, (c) ( A ∩ B) ∪ C = ( A ∪ C ) ∩ ( B ∪ C ), (d) ( A ∪ B) ∩ C = ( A ∩ C ) ∪ ( B ∩ C ), , (b) 10, (d) 6, , It, is, possible, if, value, of, ( p, q ) = (1, 2 ), (1, 3), (1, 4), (2, 3), (2, 4), and ( 3, 4), Hence, the number of equations are 6., , Ê (c) Let α, β be the roots of equation., , If A , B and C are subsets of a given, set, then which one of the following, relations is not correct?, , What is the value of k for which the, sum of the squares of the roots of, is, 2x 2 − 2(k − 2)x − (k + 1) = 0, minimum?, (a) − 1, 3, (c), 2, , (b) 1, (d) 2, , 6., , If the roots of the equation, a( b − c )x 2 + b(c − a )x + c( a − b ) = 0, , are equal, then which one of the, following is correct?, (a) a, b and c are in AP, (b) a, b and c are in GP, (c) a, b and c are in HP, (d) a, b and c do not follow any regular, pattern, , Ê (c) The roots of the equation, , a( b − c ) x 2 + b(c − a ) x + c( a − b ) = 0, , are equal., ∴b 2(c − a )2 − 4a( b − c ). c( a − b ) = 0, [Q ax 2 + bx + c = 0 of roots are real if, b 2 − 4ac ≥ 0, ⇒ b 2(c 2 + a 2 − 2ca ) − 4ac ( ab − b 2, − ac + bc ) = 0
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2, , NDA/NA, ⇒ b 2c 2 + a 2b 2 – 2 ab 2c − 4a 2bc, , 9., , + 4ab 2c + 4a 2c 2 − 4abc 2 = 0, ⇒ b 2c 2 + a 2b 2 + 2 ab 2c, − 4a 2bc − 4abc 2 + 4a 2c 2 = 0, ⇒ b 2(c 2 + a 2 + 2 ac ) − 4abc ( a + c ), + 4a 2c 2 = 0, ⇒ b (c + a ) − 4abc ( a + c ) + (2 ac ) = 0, 2, , 2, , 2, , ⇒ [b(c + a ) − 2 ac ]2 = 0, ⇒ b (c + a ) − 2 ac = 0, 2 ac, ⇒ b(c + a ) = 2 ac ⇒ b =, c+ a, , 7., , (a) 3, (c) 8, , (a) 48, (b) 40, (c) 28, (d) 20, , 2, 2, Ê (c)| x − 3 x + 2| > x − 3 x + 2, , ⇒ − ( x2 − 3 x + 2 ) > x2 − 3 x + 2, [if x 2 − 3 x + 2 < 0, and x 2 − 3 x + 2 > 0, not possible], − 2( x 2 − 3 x + 2 ) > 0, , ⇒, , x2 − 2 x − x + 2 > 0, , ⇒, ( x − 2 )( x − 1) > 0, ∴ 1 < x < 2 is correct., , A geometric progression (GP), consists of 200 terms. If the sum of, odd terms of the GP is m, and the, sum of even terms of the GP is n,, then what is its common ratio?, (a) m / n, (c) m + ( n / m), , (b) n / m, (d) n + ( m / n), 2, , Ê (b) Let a, ar, ar ...... 200 terms be a, geometric progression., Where, a is the first terms and r be the, common ratio., GP of odd terms a, ar 2, ar 4 , ..... 100 terms., GP of even terms ar, ar 3, ar 5, …… 100, terms., ∴Sum of odd terms of the GP = m, a{ r 200 − 1}, …(i), ⇒, =m, r−1, Sum of even terms of the GP = n, ar( r 200 − 1}, =n, ⇒, r −1, , …(ii), , Dividing of Eq. (i) by Eq. (ii),, 1 m, n, =, ⇒r =, ⇒, r, n, m, n, Hence, the common ratio of the GP is ., m, , Ê (d) The number of vertices of an octagon, , points], 8!, 8×7, =, = 28, 2 ! 6!, 2, ∴ The number of diagonals of an octagon, = Total number, of straight line form by 8 points − number, of sides of octagon, = 28 − 8 = 20, =, , = 2 ! 3! 4! = 2, , 6, , 3! 4! 5!, , 0, , 24 = 2, , 2, , 6, , 6 12, , 48, , 12. What are the values of x that satisfy, 0 2, , 2x 2 1 + x 2 2 1 = 0 ?, 1 1 1, 0 1 1, (a) −2 ± 3, (b) −1 ± 3, (c) −1 ± 6, (d) −2 ± 6, , 2, −4 ± 24 −4 ± 2 6, =, =, 2, 2, = −2 ± 6, , a, a, , b, , c, , x +b, c ?, b, x +c, (b) ( a + b + c )2, (d) a + b + c − 2, , x+ a, , b, , a, , x+ b, , c, , a, , b, , x+c, , x+ a+ b+c, = x+ a+ b+c, x+ a+ b+c, , c, , b, , c, , x+ b, , c, , b, , x+c, , [byC1 → C1 + C 2 + C 3], 1, b, c, = ( x + a + b + c) 1 x + b, c, 1, b, x+c, , 14., 0, , [by C 2 → C 2 − 2C1, C 3 → C 3 − 3C 2], = 1( 96 − 72 ) − 0 + 0, [expression w.r.t. first row], = 24, , the equation, x 0 2, 3x, , 16 − 4 (1) ( −2 ), , [ x + a + b + c common from C1] = 0, [Q x + a + b + c = 0], 1, , 6 24 120, , −4 ±, , Ê (a) Given, x + a + b + c = 0, , (b) 12, (d) 36, 6, , x=, , (a) 0, (c) a 2 + b 2 + c 2, , 4! ?, 3! 4 ! 5!, , 2, , x2 + 4 x − 2 = 0, , value of, , 2! 3!, , 1, , ⇒, , x +a, , 3!, , 1! 2 ! 3 !, , 2 x2 + 8 x − 4 = 0, , 13. If x + a + b + c = 0, then what is the, , 11. What is the value of the determinant, , Ê (c) Given determinant, , ⇒, , ⇒, , [Q 1 straight line form by 2, , (a) 0, (c) 24, , 3x 0 2, 1 + x2 2 1 = 0, 1 1, 0 1 1, , [expression w.r.t. first row], ⇒ x + 4 x − 4 + 3 x + 2 x2 = 0, , =8, ∴The number of points in a plane = 8, ∴ Total number of straight line form by 8, points = 8C 2, , 1! 2!, , 0 2, , ⇒ x(2 − 1) − 0 + 2(2 x − 2 ) + 3 x(2 − 1), − 0 + 2 ( x 2 − 0) = 0, , ∴The minimum number of elements in, A∪B= 6, i.e, n( A ∪ B) = 6, (because max n( A ∩ B) = 3, , If | x 2 − 3x + 2| > x 2 − 3x + 2, then, which one of the following is, correct?, , x2 − 3 x + 2 > 0, , 1, , Ê (b) n( A ) = 3, n( B) = 6, , an octagon?, , ⇒, , x, , 2x 2, , (b) 6, (d) 9, , So, a, b and c are is HP., , ⇒, , Ê (d) Given equation,, , 10. What is the number of diagonals of, , (a) x ≤ 1 or x ≥ 2 (b) 1 ≤ x ≤ 2, (c) 1 < x < 2, (d) x is any real value except 3 and 4, , 8., , If a set A contains 3 elements and, another set B contains 6 elements,, then what is the minimum number, of elements that ( A ∪ B ) can have?, , Solved Paper 2019 (II), , 1 −1, If A = , , then the expression, −1 1 , A 3 − 2A 2 is, , (a) a null matrix, (c) equal to A, 1 −1, Ê (a) A = −1 1 , , , , (b) an identity matrix, (d) equal to − A, , 1 −1 1 −1, ∴ A2 = A ⋅ A = , ⋅, , −1 1 −1 1 , 1 + 1 −1 − 1 2 −2 , =, =, , −1 − 1 1 + 1 −2 2 , 2 −2 1 −1, and A 3 = A 2 ⋅ A = , ., , −2 2 −1 1 , 2 + 2 −2 − 2 4 −4, =, =, , −2 − 2 2 + 2 −4 4
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NDA/NA, , 3, , Solved Paper 2019 (II), , Now,, , 4 −4, A3 − 2 A2 = , −2, −4 4 , 4 −4 −4, =, + , −4 4 4, , 2, −2, , 4, −4, 4 − 4 −4 + 4 0, =, =, −4 + 4 4 − 4 0, , 17. What is the value of, , −2 , 2 , , i + 3 , , , 2 , , 15. Let m and n(m < n ) be the roots of the, , (b) 30, (d) 35, , m + n = 16, mn = 39 …(ii), , …(i), , We know that, n − m = ( m + n)2 − 4mn, (Q m < n), = 256 − 156 = 100, …(iii), n − m = 10, Solving the Eqs. (ii) and (iii), n = 13, m = 3, Four terms p, q , r and s are inserted, between m and n to form an AP., ∴ AP is 3, p, q , r, s, 13, Here, a = 3, l = 13, n = 6, ∴, l = a + ( n − 1)d, 13 = 3 + ( 6 − 1)d, ⇒, d =2, ∴, p = a + d = 3 + 2 = 5,, q = a + 2d = 3 + 4 = 7, r = a + 3 d = 3 + 6 = 9,, d = a + 4 d = 3 + 8 = 11, Now, p + q + r + s = 5 + 7 + 9 + 11, = 32, , 16. Under which one of the following, conditions will the quadratic, equation, x 2 + mx + 2 = 0 always have real, roots?, (a) 2 3 ≤ m2 < 8, , (b) 3 ≤ m2 < 4, , (c) m2 ≥ 8, , (d) m2 ≤, , 3, , Ê (c) The quadratic equation, x 2 + mx + 2 = 0,, , have real roots., ∴, m2 − 4(1)(2 ) ≥ 0, equation, , ax 2 + bx + c = 0, , have real roots if b − 4ac ≥ 0], 2, , ⇒, , m2 − 8 ≥ 0, , ⇒, , m2 ≥ 8, , and 40% play football. If 10% of, students play both the games, then, what per cent of students play, neither cricket nor football?, , ?, , (a) 10%, , i + 3, , 2 , , 2019, , i − 3, + , , 2 , , 3 1 , =, + i, 2 , 2, , 2019, , 2019, , 3 1 , −, − i, 2 , 2, , π, π, = cos, + i sin , , 6, 6 , , 2019, , 2019, , 2019, , Ê (c)2 m and n be the roots of the equation, x − 16 x + 39 = 0 ( m < n)., , [quadratic, , i − 3 , +, , 2 , , Ê (c) , , equation x 2 − 16x + 39 = 0. If four, terms p , q , r and s are inserted, between m and n to form an AP, then, what is the value of p + q + r + s ?, , ∴, and, , 19. In a school, 50% students play cricket, 2019, , (a) 1, (b) − 1, (c) 2 i, (d) − 2 i, , 0, 0, , = a null matrix, , (a) 29, (c) 32, , 2019, , π, π, – cos, − i sin , , 6, 6 , 2019 π, 2019 π, = cos, + i sin, 6, 6, 2019 π, 2019 π, − cos, + i sin, 6, 6, [De-moivre’s theorem, (cos θ ± i sin θ)n = cos nθ ± i sin n θ], 2019 π, = 2 i sin, 6, 3π , , = 2 i sin 168 × 2 π +, , , 6 , 3π, = 2 i sin, 6, [Q sin (2 nπ + θ) = sin θ, n is an integer], π, = 2 i sin = 2 i, 2, , 18. If α and β are the roots of, x 2 + x + 1 = 0,, 3, , ∑ (α j, , then, , what, , is, , + β j ) equal to?, , j =0, , (a) 8, (c) 4, , (b) 6, (d) 2, x2 + x + 1 = 0, , α + β = −1, αβ = 1, 3, , Now,, , ∑ (α, , j, , + β j ) = (α 0 + β 0 ), , j=0, 1, , + (α1 + β ) + (α 2 + β 2 ) + (α 3 + β 3 ), = (1 + 1) + ( −1) + {α 2 + β 2 + 2αβ − 2αβ}, + (α + β ) (α + β − αβ ), 2, , 2, , = 2 − 1 + {(α + β )2 − 2αβ} + ( −1), {α 2 + β 2 + 2αβ − 3αβ}, = 1 + {( −1)2 − 2(1)} − {(α + β )2 − 3(1)}, = 1 − 1 − {( −1)2 − 3}, = − (1 − 3) = 2, , (d) 25%, , 20. If A = { x : 0 ≤ x ≤ 2} and B = {y : y is, a prime number}, then what is, A ∩ B equal to?, (a) φ, , (b) {1}, , (c) {2}, , (d) {1, 2}, , Ê (c) A = { x : 0 ≤ x ≤ 2} = { 0, 1, 2}, B = { y : y is a prime number}, = {2, 3, 5, 7, 11, ..... }, ∴ A ∩ B = { 0, 1, 2} ∩ {2, 3, 5, 7, 11, ......}, = {2}, and, , 21. If x = 1 + i , then what is the value of, x 6 + x 4 + x 2 + 1?, (a) 6i − 3, (c) −6i − 3, , (b) −6i + 3, (d) 6i + 3, , Ê (c) Given, x = 1 + i, , 1, i , = 2 , +, , 2, 2, π, π, = 2 cos, + i sin , , 4, 4, Now, x 6 + x 4 + x 2 + 1, = x 4 ( x 2 + 1) + 1( x 2 + 1), , Ê (d) α and β are the roots of the equation, ∴, and, , (b) 15% (c) 20%, , Ê (c) Students, who play cricket = 50%, Students, who play football = 40%, Students who play both games = 10%, Students who play only cricket, = 50 − 10 = 40%, Students who play only football, = 40 − 10 = 30%, ∴Total students who play any game, = 40 + 30 + 10 = 80%, ∴ Students who play neither cricket nor, football = 100 − 80 = 20%, , = ( x 2 + 1) ( x 4 + 1), 2, , π, π, = ( 2 )2 cos + i sin +, , 4, 4, , 4, , π, π, 4, ( 2 ) cos + i sin +, , 4, 4, , 2π, 2π, , , = 2 cos, + i sin, + 1, , 4, 4 , , , , 1, , , 1, , , 4π, 4π , , , 4 cos, + i sin, + 1, , 4, 4 , , [Q (cos θ + i sinθ)n = cos nθ + i sin nθ], π, π, , , = 2 cos + i sin + 1, , , 2, 2, [4(cos π + i sin π ) + 1], = [2( 0 + i ) + 1] [4( −1 + 0) + 1], = (2 i + 1) ( −4 + 1) = − 6i − 3
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4, , NDA/NA, , 22. What, 2+, , is, 1, , 2+, , the, , of, , x=, , (1 + 2 x + x 2 )5 + (1 + 4 y + 4 y 2 )5, = [(1 + x )2 ]5 + [(1 + 2 y)2 ]5, , 1, 2 + ... ∞, , = (1 + x )10 + (1 + 2 y)10, ∴ Total number of terms in given, expansion., = (10 + 1) + (10 + 1) = 22, [Q total number of terms in expansion of, (1 + x )n = n + 1], , (a) 2 − 1 (b) 2 + 1 (c) 3 (d) 4, 1, Ê (b) Let, x = 2 +, 1, 2 +, 1, 2 +, 2 + ... ∞, 1, x=2 +, ⇒, ⇒ x2 = 2 x + 1, x, ⇒, x2 − 2 x − 1 = 0, ⇒, , Ê (d) Given expansion,, , ?, , 1, 2+, , value, , 2±, , 25. If the middle term in the expansion, 2n, , 1, , is 184756x 10 , then, of x 2 + , , x, what is the value of n?, , ( −2 )2 − 4(1) ( −1), , 2, 2 ± 8 2 ±2 2, =, = 1± 2, 2, 2, = 2+1, (Q x > 2 ), , (a) 10, (c) 5, , =, , term in the expansion of, Ê (a) The middle, 2n, x2 + 1 , , , , x, 2n, = , + 1 th term [Q 2 n is even], 2, , = ( n + 1)th term., According to the question,, Value of middle term = 184756 x10, , 23. If P (n, r ) = 2520 and C (n, r ) = 21,, then what is, C (n + 1, r + 1) ?, (a) 7, (c) 28, , the, , value, , of, , (b) 14, (d) 56, , Ê (c) If P( n, r ) = 2520 and C ( n, r ) = 21,, ∴, , n, , ⇒, and, , pr = 2520, n!, = 2520, ( n − r )!, , Cn ( x ), , [Q Tr, , C r = 21, n!, = 21, r ! ( n − r )!, , ⇒, , n( n − 1) ( n − 2 ) ( n − 3)( n − 4), = 7 × 6× 5× 4× 3, ∴, n=7, Now, C ( n + 1, r + 1) = n + 1C r + 1, =, , 7 +1, , C 5 + 1 = 8C 6, , 8!, 8×7, =, 6 !2 !, 2, = 28, =, , 24. How many terms are there in the, expansion of, ( 1 + 2x + x 2 ) 5 + ( 1 + 4y + 4y 2 ) 5 ?, (b) 20, (d) 22, , +1, , n, , 1 = 184756 x10, , x, , = nC r x n − r a r in expansion, of ( x + a )n], , …(ii), , From Eqs. (i) and (ii), we get, 2520, = 21, r!, 2520, r! =, = 120, ⇒, 21, ⇒, r! = 5 !, ∴, r=5, Putting the value of r in Eq. (i),, n!, = 2520, ( n − 5)!, , (a) 12, (c) 21, , ⇒, , …(i), , 2 2n − n, , 2n, , n, , ⇒, , (b) 8, (d) 4, , ⇒, ⇒, , 2n, , 4 n − 2n − n, , C n( x ), , = 184756 x, , 10, , C n( x ) = 184756 x10, , 2n, , n, , Comparing the power of x both sides, n = 10, , 26., , 1 2, , , 1 2, If A = 2 3 and B = , , then, 2 1, 3 4, , , which one of the following is, correct?, , (a) Both AB and BA exist, (b) Neither AB nor BA exists, (c) AB exists but BA does not exist, (d) AB does not exist but BA exists, 1 2, 1 2, , , Ê (c)We have, A = 2 3 and B = 2 1, , , 3 4, , , order of A = 3 × 2 and order of B = 2 × 2, Q Number of column of A = Number of, row of B, ∴ AB exists., and number of column of B ≠ Number of, raw of A, ∴BA does not exist., Hence, AB exists but BA does not exist., , Solved Paper 2019 (II), , 27. If n ! has 17 zeros, then what is the, value of n?, (a) 95, (b) 85, (c) 80, (d) No such value of n exists, , Ê (b) We know that each interval of 5!is one, zero., i.e. 5! has one zero., 10! has two zeros., ∴85! has 17 zeros., Hence, the value of n is 85., , 28. Let A ∪ B = { x |( x − a )( x − b ) > 0,, where a < b }. What are A and B equal, to?, , (a) A = { x| x >, (b) A = { x| x <, (c) A = { x| x <, (d) A = { x| x >, , a} and B = { x| x >, a} and B = { x| x >, a} and B = { x| x <, a} and B = { x| x <, , b}, b}, b}, b}, , Ê (c) Let A ∪ B = { x :( x − a)( x − b ) > 0,, where a < b}., It is possible if x − a < 0 and x − b < 0, or x < a and x < b, ∴ A = { x : x < a} and B = { x : x < b}, , 29. If the constant term in the expansion, 10, , k , , of x − 2 is 405, then what can, , x , be the values of k?, (a) ±2, (c) ±5, , (b) ±3, (d) ±9, , Ê (b) Let ( r + 1)th term in the expansion of, 10, , x − k is constant., , , , x2 , ∴, , Tr, , +1, , [Q Tr, , −k, = 10C r ( x )10 − r 2 , x , +1, , r, , = nC r x n − r a r in expansion, of ( x + a )n], 10 − r, , ⇒ 405 = 10C r ( x ), , 2, , − 2r, , ⋅ ( − k )r, , 10 − 5r, , ⇒ 405 =, , 10, , C r ( x), , 2, , ⋅ ( − k )r, , For constant term, 10 − 5r, = 0 ⇒ 10 − 5r = 0, 2, ∴, r=2, Putting the value of r, in Eq. (i),, 405 = 10C 2.( − k )2, ⇒, ⇒, ⇒, ⇒, ⇒, , 10 !, × k2, 2 ! 8!, 10 × 9 2, 405 =, .k, 2, 405, k2 =, 45, k2 = 9, 405 =, , k=± 3, , …(i)
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NDA/NA, , 5, , Solved Paper 2019 (II), , 30. What is C ( 47, 4 ) + C (51, 3) + C (50, 3), , 33. Let Sn be the sum of the first n terms, , + C ( 49, 3) + C ( 48, 3) + C ( 47, 3) equal, to?, , of an AP. If S 2n = 3n + 14n 2 , then, what is the common difference?, , (a) C( 47, 4), (c) C( 52, 4), , (b) C( 52, 5), (d) C( 47, 5), , (a) 5, (c) 7, , Ê (c) C ( 47, 4) + C ( 51, 3) + C ( 50, 3), , =, , 51, , C3 +, , 50, , =, , 51, , C3 +, , 49, , C3 +, , 50, , C3 +, , C3 +, , 49, , C3 +, , 48, , C3 +, , =, , 51, , C3 +, , 50, , C3 +, , 47, , C3 +, , 48, , =, , 51, , C3 +, , 50, , =, , 51, , C3 +, , 51, , =, , 52, , n, , C3 +, , 49, , C3 +, , C3 +, , 50, , C3, , C3, , [Q C r + C r − 1 =, n, , 47, , 48, , +, , 47, , C4, , 48, , C4, , n +1, , Cr ], , 49, , C4, , C4, , C4, , C 4 = C( 52, 4), , 31. Let a, b, c be in AP and k ≠ 0 be a real, number. Which of the following are, correct?, 1. ka, kb, kc are in AP, 2. k − a, k − b, k − c are in AP, a b c, 3. , , are in AP, k k k, Select the correct answer using the, code given below., (a) 1 and 2 only, (c) 1 and 3 only, , Ê (c) S 2n = 3n + 14n (S n be the sum of first, 2, , + C ( 49, 3) + C ( 48, 3) + C ( 47, 3), = 47C 4 + 51C 3 + 50C 3 + 49C 3, +, , (b) 2 and 3 only, (d) 1, 2 and 3, , n terms of an AP), 3, 7, S 2n = .(2 n) + (2 n)2, 2, 2, Put 2 n = n, 3n 7 n 2, we get, S n =, +, 2, 2, ∴, Tn = S n − S n −1, ⇒, , We know that equal number addition,, subtraction and multiply, divide, by equal, number of each term of an AP, the, resultent, series be an AP., ∴ka, kb, kc are in AP (multiplying by k)., k − a, k − b, k − c are in AP (subtraction, a b c, from k) and , , are in AP (divide by k), k k k, Hence, option (d) is correct answer., , Put n = 1, 2, ...., T1 = 7(1) − 2 = 5, , 34. If 3rd, 8th and 13th terms of a GP are, p , q and r respectively, then which, one of the following is correct?, (b) r 2 = pq, (a) q 2 = pr, (c) pqr = 1, (d) 2q = p + r, Ê (a) Let first term and common ratio of a, GP be a and R., …(i), ∴, T3 = aR 2 = p, , 32. How many two-digit numbers are, divisible by 4?, (b) 22, (d) 25, , Ê (b) Series of two-digit number that, divisible by 4 is, 12, 16, 20, ........., 96, This series is an AP, Here, A = 12,d = 4, l = 96, Let total number of terms be n., ∴, l = a + ( n − 1)d, ⇒, 96 = 12 + ( n − 1) 4, ⇒, 84 = ( n − 1)4, ⇒, n − 1 = 21, ⇒, n = 21 + 1 = 22, , T8 = aR 7 = q, , …(ii), , T13 = aR, , …(iii), , =r, , Multiplying of Eqs. (i) and (iii), ( aR 2 ) ( aR12 ) = pr, ( aR 7 )2 = pr, , ⇒, , q 2 = pr [from Eq. (ii)], , 35. What is the solution of x ≤ 4, y ≥ 0, and x ≤ − 4, y ≤ 0?, (a) x ≥ − 4, y ≤ 0, (c) x ≤ − 4, y = 0, , (b) x ≤ 4, y ≥ 0, (d) x ≥ − 4, y = 0, , Ê (c) Given inequalities, , log 7 x > ( ± 1), x > 71 ⇒ x > 7, x < 7 −1 ⇒ x <, , 1, 7, , 1, Hence, x ∈ , 7 , 7 , , equation x 2 + 3| x | + 2 = 0 have?, , d = T2 − T1 = 12 − 5 = 7, , ⇒, , and, , [Q log a a = 1], , 37. How many real roots does the, , T2 = 7(2 ) − 2 = 12, , a 2R14 = pr, , (log 7 x )2 > 1, , ⇒, ∴, , 3, 7, 3, 3 7, 7 7, n + n2 − n + − n2− + .2 n, 2, 2, 2, 2 2, 2 2, Tn = 7 n − 2, , ⇒, , 1, (b) x ∈ , 7 , 7 , , 1, (c) x ∈ 0, ∪ (7, ∞ ), 7, 1, (d) x ∈ , ∞ , 7, , , ⇒, , =, , 12, , (a) x ∈ ( 0, ∞ ), , Taking log on base 7 both sides, log 7 x. log 7 x > log 7 7, [Q log a mn = nlog a m], , 3, 7n, 3, 7, = n+, − ( n − 1) − ( n − 1)2, 2, 2, 2, 2, , ∴, , one of the following is correct?, , log x, Ê (b) x 7 > 7 where x > 0., , 2, , Ê (d) a, b,c are in AP., , (a) 21, (c) 24, , (b) 6, (d) 9, , 36. If x log 7 x > 7 where x > 0, then which, , …(i), x ≤ 4, y ≥ 0, and, …(ii), x ≤ − 4, y ≤ 0, Possible value of x and y., x = { 4, 3, 2, 1, 0, − 1, − 2, − 3, − 4, − 5, ...}, …(i), y = { 0, 1, 2, 3, 4 .....}, and x = { −4, − 5, − 6, − 7, ...},, y = { 0, − 1, − 2, − 3, − 4 ...} …(ii), Take combine (i) and (ii),, x = { −4, − 5, − 6, − 7 ... }, y = 0, or, x ≤ − 4, y = 0., , (a) Zero, (c) Two, , (b) One, (d) Four, , 2, Ê (a) Given equation, x + 3| x| + 2 = 0, , Case I x 2 + 3 x + 2 = 0 (when x > 0), ⇒, , x2 + x + 2 x + 2 = 0, , ⇒ x( x + 1) + 2( x + 1) = 0, ⇒, ( x + 1) ( x + 2 ) = 0, ∴, x = − 1, − 2, Hence, no real roots because x > 0., Case II x 2 − 3 x + 2 = 0 (when x < 0), ⇒, , x2 − 2 x − x + 2 = 0, , ⇒, x( x − 2 ) − 1( x − 2 ) = 0, ⇒, ( x − 2 )( x − 1) = 0, ∴, x = 1, 2, Hence, no real roots because x < 0., ∴ The number of real roots of given, equation is zero., , 38. Consider the following statements in, respect of the quadratic equation, 4( x − p )( x − q ) − r 2 = 0,, where p , q and r are real numbers., 1. The roots are real., 2. The roots are equal, if p = q and, r = 0., Which of the above statements is/are, correct?, (a) Only 1, (c) Both 1 and 2, , (b) Only 2, (d) Neither 1 nor 2, , Ê (c) Given quadratic equation,, 2, , 4( x − p) ( x − q ) − r = 0, , ⇒, , 4 x 2 − ( 4q + 4 p) x + 4 pq − r 2 = 0, , Comparing it Eq. by ax 2 + bx + c = 0
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6, , NDA/NA, a = 4, b = − 4 ( p + q ), c = 4 pq − r 2, b − 4ac = 16 ( p + q ) − 4 × 4 ( 4 pq − r ), 2, , 2, , 2, , = 16 p2 + 16 q 2 + 32 pq − 64 pq + 16r 2, = 16 p2 + 16 q 2 − 32 pq + 16r 2, = 16( p − q )2 + 16r 2, Q b 2 − 4ac will be positive., , 42. Consider the following statements, , So, the roots are real., If p = q and r = 0, then b 2 – 4ac = 0, So, the roots are equal., Hence, the statements both 1 and 2 are, correct., , 39. Let S = {2, 4, 6, 8, ......,20}., What are the maximum number of, subsets of S?, (a) 10, (c) 512, , (b) 20, (d) 1024, , Ê (d) S = {2, 4, 6, 8, ......., 20}, Here, number of elements of set, S = 10 ( n), ∴Maximum number of subsets of set, S = 2 n = 2 10 = 1024, , 40. A binary number is represented by, (cdccddcccddd )2 , where c > d . What, is its decimal equivalent?, (a) 1848, (c) 2842, , (b) 2048, (d) 2872, , where, c > d . We know that only two bit, (digits) 0 and 1 be any binary number., ∴Given binary number, = (101100111000)2, = (1 × 2 11 + 0 × 2 10 + 1 × 2 9 + 1 × 2 8 +, + 0 × 27 + 0× 26 + 1 × 25 + 1 × 24, + 1 × 2 3 + 0 × 2 2 + 0 × 2 1 + 0 × 2 0 )10, = (2048 + 512 + 256 + 32 + 16 + 8)10, = (2872 )10, , 41. If cosec θ = 29 , where 0 < θ < 90°,, 21, , (a) 5, , the, , value, , (b) 10, , (c) 15, (d) 20, 29, Ê (b) Given, cosec θ =, 21, where, 0 < θ < 90°, H 29, =, = k (let), Qcosec θ =, P 21, ∴ H = 29k, P = 21 k, ∴ B = ( H )2 − ( P )2 = (29k )2 – (21k )2, =, , 841k 2 − 441k 2, , =, , 400k = 20k, , ∴sec θ =, , 2, , H 29k 29, =, =, B 20k 20, , 1. cos θ + sec θ can never be equal to, 1.5., 2. tan θ + cot θ can never be less, than 2., Which of the above statements is/are, correct?, (a) Only 1, (c) Both 1 and 2, , (b) Only 2, (d) Neither 1 nor 2, , of, , Ê (c) Given, radius of circle = 1 unit, Angle subtends at the centre of circle by, chord = θ, We know that, length of chord, θ, θ, θ, = 2 r sin = 2 × 1 sin = 2 sin, 2, 2, 2, , 45., , , What is tan 2 tan −1, , 2, 3, 1, (d), 9, , (b), , (a), , − 1 ≤ sec θ ≤ ∞ but cos θ = sec θ ifθ = 0, and θ = 180°, ∴, − 2 ≤ cos θ + sec θ ≤ ∞, So, cos θ + sec θ = 15, . is possible., and again 0 ≤ tan θ ≤ ∞ and, 0 ≤ cot θ ≤ ∞ , but tan θ = cot θ, if, θ = 45°, ∴ 2 ≤ tanθ + cot θ ≤ ∞, So, tan θ + cot θ can never be less than 2., Hence, only the Statement 2 is correct., , below the top of a vertical flagstaff., From the foot of the ladder, the, elevation of the flagstaff is 60°. What, is the height of the flagstaff?, (a) 9 m, (c) 13.5 m, , (b) 10.5 m, (d) 15 m, , Ê (*) Let AP be a ladder and QR be a vertical, , flagstaff. P is a point 9 m below the top on, flagstaff. A is the foot of ladder and h is the, height of point P from the ground., ∴ AP = 9 m, PR = 9 m, PQ = hm, PQ, In ∆ AQP,, sin θ =, AP, h, 3 h, =, ⇒ sin 60° =, ⇒, 9, 2, 9, 9 3, ⇒ 9 3 = 2h ⇒ h =, 2, 9 × 173, ., 15.57, = 7.7 m, =, =, 2, 2, ∴Height of flagstaff, = h + 9 = 7.7 + 9, = 167, . m, , 44. What is the length of the chord of a, , unit circle which subtends an angle θ, at the centre?, , θ, (a) sin , 2, θ, (c) 2 sin , 2, , θ, (b) cos , 2, θ, (d) 2 cos , 2, , 1 , equal to?, 3 , , 3, 4, , (c), , 3, 8, , , −1 1 , Ê (b) tan 2 tan , 3 , , , 1 , 2 ×, , , −1, 3 , = tan tan, 2, 1, , 1 − , , 3 , , −1, −1 2 x , , Q 2 tan x = tan, 1 − x2 , , 2, 2 ×9 3, −1, = tan tan 3 =, =, 8, 3×8 4, 9, , Ê (b) We know that, − 1 ≤ cos θ ≤ 1 and, , 43. A ladder 9 m long reaches a point 9 m, , Ê (d) Binary number = (cdccddcccddd )2, , then what is, 4 sec θ + 4 tan θ?, , P 21k 21, =, =, B 20k 20, Now, 4 sec θ + 4 tan θ, 29, 21, =4×, + 4×, 20, 20, 50, =4×, = 10, 20, and tanθ =, , Solved Paper 2019 (II), , 46. What is the scalar projection of, a = $i − 2$j + k$ on b = 4 i$ − 4 $j + 7 k$ ?, 9, 6, (d), 19, 19, $, $ $, $, $, $, Ê (b) a = i − 2 j + k, b = 4 i − 4 j + 7k, (a), , 6, 9, , (b), , 19, 9, , (c), , Projection of a on b, a ⋅b, =, |b|, ( $i − 2 $j + k$ ) . ( 4$i − 4$j + 7k$ ), =, 16 + 16 + 49, =, , 4+ 8+7, 81, , =, , 19, 9, , 47. If the magnitude of the sum of two, non-zero vectors is equal to the, magnitude of their difference, then, which one of the following is, correct?, (a) The vectors are parallel, (b) The vectors are perpendicular, (c) The vectors are anti-parallel, (d) The vectors must be unit vectors, , Ê (b) Let a and b are the two non-zero, vectors., According to the question,, |a + b| = |a − b|, ⇒ a 2 + b 2 + 2a ⋅ b = a 2 + b 2 − 2a ⋅ b, , ⇒, 4a ⋅ b = 0 ⇒ a ⋅ b = 0, So, a and b are perpendicular.
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NDA/NA, , 48. Consider the following equations for, two vectors a and b., 1. (a + b) . (a − b) = | a| 2 − | b| 2, 2. (| a + b| ) (| a − b| ) = | a| 2 − | b| 2, 3. | a ⋅ b| + | a × b| = | a 2 | b| 2, Which of the above statements are, correct?, (a) 1, 2 and 3, (c) Only 1 and 3, , (b) Only 1 and 2, (d) Only 2 and 3, , Ê (c) I : (a + b ) ⋅ (a − b ), = a ⋅ a − a ⋅ b + b⋅ a − b⋅ b, |a|2 − a . b + a . b − |b|2, [Q a ⋅ b = b ⋅ a ], = |a|2 − |b|2, So, Statement 1 is correct., 2. (|a + b|) (|a − b|) = |a + b||a − b|, ≠ |a|2 − |b|2, So, Statement 2 is not correct., 3.|a ⋅ b|2 + |a × b|2 = ||a||b|cos θ|2, + ||a||b|sin θ|2, = |a |2 |b|2 cos 2 θ + |a|2 |b|2 sin2 θ, = |a | |b| (cos θ + sin θ), 2, , 7, , Solved Paper 2019 (II), , 2, , 2, , = |a |2 |b|2, , 2, , [Q cos 2 θ + sin2 θ = 1], , So, statement 3 is correct., Hence, only Statements 1 and 3 are correct., , 49. Consider the following statements., 1. The magnitude of a × b is same as, the area of a triangle with sides a, and b, 2. If a × b = 0, where a ≠ 0 ,b ≠ 0,, then a = λ b., Which of the above statements is/are, correct?, (a) Only 1, (c) Both 1 and 2, , (b) Only 2, (d) Neither 1 nor 2, , Ê (b) 1. We know that,, Area of triangle with sides a and b, 1, = |a × b|, 2, So, statement 1 is not correct., 2. a × b = 0, where a ≠ 0, b = 0,, So, a and b are parallel., ⇒, a = λb, So, Statement 2 is correct., Hence, only statement 2 is correct., , 50. If a and b are unit vectors and θ is the, angle between them, then what is, θ, sin 2 equal to?, 2, (a), , |a + b|2, 4, , (b), , |a − b|2, 4, , (c), , |a + b|2, 2, , (d), , |a − b|2, 2, , Ê (b) Given,|a| = 1,|b| = 1, , We know that,, |a − b|2 = |a|2 + |b|2 − 2a ⋅ b, , [Q Distance between two points ( x1, y1 ), and ( x2, y2 ), = | ( x2 − x1 )2 + ( y2 − y1 )2|, , ⇒|a − b|2 = 1 + 1 − 2 |a||b|cos θ, ⇒|a − b|2 = 2 − 2 cos θ, ⇒|a − b|2 = 2 (1 − cos θ), θ, ⇒|a − b| = 2 1 − 1 + 2 sin2 , , 2, θ, ⇒|a − b|2 = 2 ⋅ 2 sin2 , , 2, , m2 cos 2 2β + m2 cos 2 2α, = | − 2 m2 cos 2β cos 2α + m2 sin2 2β, + m2 sin2 2α − 2 m2 sin 2β sin2α|, , 2, , ⇒, , sin2, , θ |a − b|2, =, 2, 4, , equation, ax + by + c = 0, represents a straight line, , 51. The, , (a) for all real numbers, a, b and c, (b) only when a ≠ 0, (c) only when b ≠ 0, (d) only when at least one of a and b is, non-zero., , Ê (d) The equation ax + by + c = 0, represents a straight line only when at, least one of a and b is non zero., , 52. What is the angle between the lines, x cos α + y sin α = a and, x sin β − y cos β = a ?, , (a) β − α, (b) π + β − α, ( π + 2β + 2α ), ( π − 2β + 2α ), (d), (c), 2, 2, , Ê (d) The equations of given lines, , (cos 2 2β + sin2 2β ), 2, 2, = | m + (cos 2α + sin 2α ), − 2 (cos 2β cos 2α, , + sin 2β sin 2α )|, = | m 1 + 1 − 2 cos (2α − 2β )|, = | m 2 [1 − cos 2(α − β )]|, 2(α − β ), = | m 2 [1 − 1 + 2 sin2, |, 2, = | m 2 × 2 sin2 (α − β )|, = | 2 m sin (α − β )|, , 54. An equilateral triangle has one, , vertex at ( −1, − 1) and another vertex, at( − 3, 3 ). The third vertex may lie, on, , (a) ( − 2 , 2 ), (c) (1, 1), , (b) ( 2 , − 2 ), (d) (1, − 1), , Ê (c) Consider two vertices of an equilateral, , triangle are A( −1, − 1) and B( − 3, 3 )., Let third vertex x be C ( x, y)., Q ∆ABC is equilateral, AC = AB ( x + 1)2 + ( y + 1)2, , x cos α + y sin α = a … (i), and x sin β − y cos β = a, … (ii), − cos α, Slope of Eq. (i), m1 =, = − cot α, sin α, π, = tan + α , 2, , − sin β, Slope of Eq. (ii), m2 =, = tan β, − cos β, , ∴, , Let θ be the angle between the lines, then, m − m2, tan θ = 1, 1 + m1m2, , From option only point (1, 1) is satisfying, of it equation. Hence, the third vertex may, lie on (1, 1)., , π, tan + α − tan β, 2, , =, 1 + tan ( π − α ) tan β, π, tan θ = tan + α − β , 2, , π + 2α − 2β, π, θ=, + α −β=, 2, 2, , 53. What is the distance between the, , points P (m cos 2α, m sin 2α ) and, Q (m cos 2β, m sin 2 β )?, (a)|2 m sin (α − β )| (b)|2 m cos (α − β )|, (c)| m sin (2α − 2β )|, (d)| m sin (2α − 2β )|, , Ê (a) Given points, p( m cos 2α, m sin 2α ), and Q ( m cos 2β, m sin 2β ), ( m cos 2β − mcos 2α )2 +, ∴PQ = |, ( m sin 2β − m sin 2α )2|, , = ( − 3 + 1)2 + ( 3 + 1)2, ⇒, , x 2 + 1 + 2 x + y2 + 1 + 2 y, , ⇒, , = 3 + 1− 2 3 + 3 + 1 + 2 3, x 2 + y2 + 2 x + 2 y + 2 = 8, , ⇒, , x 2 + y2 + 2 x + 2 y = 6, , 55. If the angle between the lines joining, the end points of minor axis of the, x2 y2, ellipse 2 + 2 = 1 with one of the, a, b, π, its foci is , then what is the, 2, eccentricity of the ellipse?, 1, 2 2, x2, y2, Ê (b) Equation of the ellipse, 2 + 2 = 1, a, b, End points of minor axis are, ( 0, b ), ( 0, − b ) and one foci is ( ae, 0), (a), , 1, 2, , (b), , 1, 2, , (c), , 3, 2, , (d), , Slope of line BS =, , 0−b, b, ( m1 ), =−, ae − 0, ae, , Slope of line B′ S =, , 0+ b, b, ( m2 ), =, ae − 0 ae
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8, , NDA/NA, According to the question, angle between, π, BS and B′ S is ., 2, i.e. BS and B′ S are perpendicular,, ∴, m1m2 = − 1, −b, b, ×, = − 1 ⇒ b 2 = a 2e 2 … (i), ae, ae, b2, We know that, e = 1 − 2, a, ⇒, ⇒, ⇒, ⇒, , ae =a −b, 2 2, , 2, , 2, , a 2e 2 = a 2 − a 2e 2 [from Eq. (i)], 1, 2 a 2e 2 = a 2 ⇒ e 2 =, 2, 1, e =, 2, , 56. A point on a line has coordinates, ( p + 1, p − 3, 2p ) where p is any, real number. What are the direction, cosines of the line?, 1 1 1, 1 1 1, (a) , ,, (b), , ,, 2 2 2, 2 2 2, 1 1 1, (c), , ,−, 2 2 2, (d) Cannot be determined due to, insufficient data, , Ê (d) Coordinate of a point on a line is, , ( p + 1, p − 3, 2 p), p is any real, number., Equation of a line, whose direction ratios, are a, b and c and passing through the, point ( x1, y1, z1 ), x − x1, y − y1, z − z1, =, =, =r, a, b, c, ∴ ( ar + x1, br + y1, cr + z1 ) any point on, the line., According to the questions,, ( ar + x1, br + y1, cr + z1 ), = ( p + 1, p − 3, 2 p), … (i), ∴, ar = p + 1 − x1, br = p − 3 − y1 … (ii), cr = 2 p − z1 …(iii), Squaring and adding of (i), (ii) and (iii), ( a 2 + b 2 + c 2 )r 2 = ( p + 1 − x1 )2, + ( p − 3 − y1 ) + ( 2 p − z1 ), 2, , 2, , We can not find the values of a, b and c., Hence, the direction cosines of the line, can not be determined due to insufficient, data., , 57. A point on the line, , x −1 y −3 z +2, =, =, 1, 2, 7, has coordinates, (a) (3, 5, 4), (c) ( − 1, − 1, 5), , (b) (2, 5, 5), (d) (2, − 1, 0), , Ê (b) Equation of the line, , x−1 y− 3 z+ 2, =, =, 1, 2, 7, , From option, point (2, 5, 5) is satisfying, the given equation of line., Q 2 − 1 = 5 − 3 = 5 + 2 ⇒ 1 = 1 = 1, , , 1, 2, 7, Hence, the coordinates of required point, (2, 5, 5)., , 58. If the line x − 4 = y − 2 = z − k lies, , 3, 3, 1, , m1 = , n1 =, 4, 4, 2, 3, 3, 1, and l 2 = −, , m2 = − , n2 =, 4, 4, 2, We know that,, cos θ = | l1l 2 + m1m2 + n1n2|, 3, 9, 1, ⇒ cos θ = −, −, +, 16 16 4, , (b) 3, (d) 7, , Ê (d) Equation of line, x− 4 y−2, z−k, =, =, =, =r, 1, 1, 2, ∴ ( r + 4, r + 2, 2 r + k ) point lies on the, line., This line lies on the plane, 2 x − 4y + z = 7, Then, the point ( r + 4, r + 2, 2 r + k ) lies, on the plane, we get, 2( r + 4) − 4 ( r + 2 ) + (2 r + k ) = 7, ⇒, 2 r + 8 − 4r − 8 + 2 r + k = 7, ⇒, k=7, Hence, the value of k is 7., , 59. A straight line passes through the, point (1, 1, 1) makes an angle 60°, with the positive direction of Z -axis,, and the cosine of the angles made by, it with the positive directions of the, Y -axis and the X -axis are in the ratio, 3 : 1. What is the acute angle, between the two possible positions, of the line?, (a) 90°, (c) 45°, , (b) 60°, (d) 30°, , Ê (b) Let the straight line makes the angle, with X-axis, Y-axis and Z-axis be α, β, and γ., cos β, 3, =, ∴γ = 60° and, cos α, 1, , If l, m and n are the direction cosines of, the lines, then, 1, n = cos γ = cos 60° =, 2, m cos β, 3, and, =, =, l, cos α, 1, m, 3, =, = k (Let), l, 1, ∴ m = 3k, l = k, We know that, l 2 + m2 + n2 = 1, ⇒, , 1, k + 3k + = 1, 4, 2, , ⇒, ⇒, , 2, , 4k 2 = 1 −, k2 =, , 1 3, =, 4 4, , 3, 3, ⇒k=±, 16, 4, , l1 =, , ∴, , 1, 1, 2, on the plane 2x − 4y + z = 7, then, what is the value of k?, , (a) 2, (c) 5, , Solved Paper 2019 (II), , =, , −3 − 9 + 4, −8, =, 16, 16, , 1, = cos 60°, 2, θ = 60°, , cos θ =, ∴, , 60. If the points ( x , y , − 3), (2, 0, − 1) and, C ( 4, 2, 3) lie on a straight line, then, what are the values of x and y, respectively?, (a) 1, − 1, (c) 0, 2, , (b) −1, 1, (d) 3, 4, , Ê (a) Points, A( x, y, − 3), B (2, 0, − 1) and, ( 4, 2, 3). These points lie on a straight, line, then direction ratios of AB = λ, (direction ratios of BC ), ∴(2 − x, 0 − y, − 1 + 3), = ( 4 − 2, 2 − 0, 3 + 1), ⇒ (2 − x, − y, 2 ) = (2, 2, 4), ⇒ (2 − x, − y, 2 ) = 2(1, 1, 2 ), Comparing both sides,, 2 − x =1 ⇒ x = 1, and − y = 1 ⇒ y = − 1, , 61. What is the minimum value of, a2, cos 2 x, b > 0?, , +, , b2, , where a > 0 and, , sin 2 x, , (b) ( a − b )2, (d)| a 2 + b 2|, a2, b2, +, Ê (*) Let p =, 2, cos x sin2 x, = a 2 sec 2 x + b 2 cosec 2 x, (a) ( a + b )2, (c) a 2 + b 2, , − 2 ab sec x cosec x, + 2 ab sec x cosec x, = ( a sec x − b cos ec x )2, + 2 ab sec x cosec x, For minimum value of p,, a sec x − b cos ec x = 0, ⇒, a sec x = b cos ec x, sec x, b, ⇒, =, cos ec x a, ⇒, ∴ sin x =, , tan x =, , b, a, , b, a2 + b 2, , , cos x =, , a, a2 + b 2, , ∴Minimum value of p, a 2 ( a 2 + b 2 ) b 2( a 2 + b 2 ), =, +, a2, b2, = 2( a 2 + b 2 )
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NDA/NA, , 9, , Solved Paper 2019 (II), 65. If 2 tan A = 3 tan B = 1, then what is, , 62. If the angles of a triangle ABC, are in AP and b : c = 3 : 2, then, , tan ( A − B ) equal to?, , what is the measure of angle A?, , (a), , 1, 5, , (b), , 1, 6, , (a) 30°, (c) 60°, , (c), , 1, 7, , (d), , 1, 9, , (b) 45°, (d) 75°, , Ê (d) Angles of a triangle ABC are in AP,, then 2 B = A + C, We know that, A + B + C = 180°, ⇒, 3B = 180° ⇒ B = 60°, sin A sin B sin C, By sine rule,, =, =, a, b, c, sin A sin 60° sin C, ⇒, =, =, a, b, c, sin 60° sin C, Take II and III,, =, b, c, sin 60° b, 3 /2, 3, =, ⇒, = ⇒, sin C, c, sin C, 2, , ⇒, ⇒, ∴, , Ê (c) Given, 2 tan A = 3 tan B = 1, 1, 1, ∴tan A = , tan B =, 2, 3, Now, tan ( A − B), tan A − tan B, =, 1 + tan A tan B, , sin C = sin 45° ⇒ C = 45°, A = 180° − ( B + C ), = 180° − ( 60° + 45° ) = 75°, , 66. What is cos 80° + cos 40° − cos 20°, equal to?, (a) 2, (c) 0, , (b) 1, (d) −19, , Ê (c) cos 80° + cos 40° − cos 20°, 80° + 40°, 80° − 40°, . cos, 2, 2, − cos20°, = 2 cos 60° cos 20° − cos 20°, 1, = 2 × cos 20° − cos 20°, 2, = cos 20° − cos 20° = 0, , = 2 cos, , and, tan A − tan B = x, cot B − cot A = y , then what is the, value of cot ( A − B )?, , 63. If, , 1, 1, +, x, y, xy, (c), x + y, (a), , 1 1, −, y x, 1, (d) 1 +, xy, , (b), , Ê (a) Given, tan A − tan B = x, andcot B − cot A = y, From Eq. (i), tan A − tan B = x, 1, 1, −, = x, ⇒, cot A cot B, , Now,, , … (ii), , cot B − cot A, = x, cot A cot B, , ⇒, ⇒, , 67. If angle C of a triangle ABC is a right, … (i), , y, [from Eq. (ii)], cot A cot B =, x, cot A cot B + 1, cot ( A − B) =, cot B − cot A, y, +1, y+ x 1 1, = x, =, = +, y, xy, x, y, , 64. What is, , sin (α + β ) − 2 sin α cos β +, sin (α − β ) equal to?, (a) 0, (c) 2 sin β, , (b) 2 sin α, (d) sin α + sin β, , Ê (a) sin (α + β ) − 2 sin α cos β, , + sin (α − β ), = sin α cos β + cos α sin β, − 2 sin α cos β + sin α cos β, − cos α sin β = 0, , angle, then what is tan A + tan B, equal to ?, , a −b, ab, b2, (c), ca, 2, , (a), , 2, , 2, , a, bc, c2, (d), ab, , (b), , A, (a) tan , 2, A, (c) 2 tan , 2, , A, (b) cot , 2, A, (d) 2 cot , 2, , =, , cos A, cos A + 1, 1, +, =, sin A, sin A, sin A, , A, − 1+ 1, 2, =, A, A, 2 sin cos, 2, 2, A, A, 2 cos 2, cos, 2 = cot A , 2, =, =, A, A, A, 2, 2 sin cos, sin, 2, 2, 2, 2 cos 2, , 70. What is tan 25° tan 15° + tan 15°, tan 50° + tan 25° tan 50° equal to?, , (a) 0, (c) 2, , (b) 1, (d) 4, , Ê (b)Q tan 50° = tan( 90° − 40° ), ⇒ tan 50° = cot 40°, 1, ⇒ tan 50° =, tan 40°, 1, ⇒ tan 50° =, tan (25° + 15° ), ⇒ tan 50° =, , 1 − tan25° tan15°, tan25° + tan15°, , ⇒ tan25° tan 50°+ tan15° tan 50°, = 1 − tan25° tan15°, ⇒ tan25° tan15° + tan15° tan 50°, + tan25° tan 50° = 1, , Ê (d) In ∆ABC , ∠C = 90°, , ∴ c 2 = a2 + b 2, [by Pythagoras theorem] … (i), a, b, tan A = , tan B =, b, a, a, b, Now, tan A + tan B = +, b, a, a2 + b 2 c 2, [from Eq. (i)], =, =, ab, ab, , 68. What is cot A − tan A equal, 2, , 69. What is cot A + cosec A equal to?, , Ê (b) cot A + cos ec A, , 3−2, 1 1, −, 1, 2 3, =, = 6 =, 6+1 7, 1 1, , 1+ , 2 3, 6, , [Q Given, b : c = 3 : 2 ], 3, 2, 1, sin C =, ×, =, 2, 3, 2, , A, A, − sin2, 2 cos A, 2, 2, =, =, A, A, A, A, 2 sin cos, sin cos, 2, 2, 2, 2, 2 cos A, =, = 2 cot A, sin A, cos 2, , 2, , to?, (a) tan A, (b) cot A, (c) 2 tan A, (d) 2 cot A, A, A, Ê (d) cot − tan, 2, 2, A, A, cos, sin, 2 −, 2, =, A, A, sin, cos, 2, 2, , 71. What is the area of the region, bounded by | x | < 5, y = 0 and y = 8?, , (a) 40 sq units, (c) 120 sq units, , (b) 80 sq units, (d) 160 sq units, , Ê (b) Given curve y = 0 and y = 8 and, | x| < 5, Case I When x < 0, then, area of the region bounded, =, , 0, , 0, , ∫−5 0 dx − ∫−5 8 dx = 0 − 8 [ x]−5, 0, , = − 8 [0 + 5] = − 40, = 40 sq units, [Qarea will not be negative], Case II when x > 0, then, Area of the region bounded, =, , 5, , 5, , ∫0 0 dx − ∫0 8 dx = 0 − 8 [ x]0, 5, , = − 8[5 − 0] = − 40 = 40 sq units
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10, , NDA/NA, ∴Required area = 40 + 40 = 80 sq units, Y, y=8, , Case I, , Case II, , 74. What is the degree of the differential, , y=0, , X¢, , X, , 72. Consider the following statements in, respect, , of, the, function, 1, f ( x ) = sin for x ≠ 0 andf (0) = 0 :, x, 1. lim f ( x ) exists, x→ 0, , 2. f ( x ) is continuous at x = 0, Which of the above statement is/are, correct?, (a) Only 1, (c) Both 1 and 2, , (b) Only 2, (d) Neither 1 nor 2, sin 1 , x ≠ 0, , Ê (d) Given, f( x) = x, 0, ,x=0, 1 , 1, LHL = lim sin = lim sin, , x h→ 0 0 − h , x→ 0−, 1, = lim − sin = − sin ∞, h→ 0, h, = − (a rational number), [Q sin θ lies between −1to1], 1 , 1, RHL = lim sin = lim sin, , x h→ 0 0 + h, x → 0+, 1, = lim sin = sin ∞, h→ 0, h, = a rational number, [Q sin θ lies between − 1to1], Q LHL ≠ RHL, So, f( x ) does not exists., Q f( x ) = 0 at x = 0, ∴LHL ≠ RHL ≠ f( 0), So, f( x ) is not continuous., Hence, the statements neither 1 nor 2, correct., , 73. What is the value of lim sin x ° ?, x→ 0, , 1, (a), 4, , tan 3x °, , 1, (b), 3, , 1, (c), (d) 1, 2, x , sin, x×, , , sin x °, x, = lim , , Ê (b) xlim, → 0 tan 3 x °, x→ 0, 3x, tan, 3x ×, , , 3x , sin x , , , x , 1, 1, = lim, =, 3 x→ 0 tan 3 x , 3, , , 3x , Q lim sin θ = 1 and lim tanθ = 1, , θ→0 θ, θ → 0 θ, , QCurve intersects Y-axis at a point P, then, x=0, ∴, y = me 0 ⇒ y = m, ∴Point P ( 0, m), Now, differentiation w.r.t x of given curve,, dy, = m.e mx . m, dx, dy, = m2e mx, dx, dy, at point P( 0, m),, = m2e 0 = m2, dx, , Ê (a) Given differential equation,, 2, , d 4 y, dy, + − x 2 4 = 0, 3, , , dx, dx, dx , , x=5, Y¢, , Given curve y = me mx where m > 0, , (b) 2, (d) 4, d 3y, , x=–5, , Ê Solutions (Q. Nos. 76-78), , equation, 2, d 4y , d 3y dy , − x 2 4 = 0?, +, , , 3, dx , dx, dx , (a) 1, (c) 3, , ⇒, , d4y, dx 4, , 2, , −, , 1 d 3y, 1 dy , − 2 =0, , x 2 dx 3 , x dx , , We know that power of the heighest order, of differentiation is the degree of, differential equation., So, the degree of it equation is 1., , 75. Which one of the following is the, second degree polynomial function, f ( x ) where, f (0) = 5, f ( −1) = 10 and, f (1) = 6?, (a) 5x 2 − 2 x + 5, (c) 3x 2 − 2 x + 5, , Ê 76. (b) Slope of the curve at the point, P( 0, m), dy, = at point P ( 0, m) = m2, dx , , Ê 77. (c) Let the tangent makes the angle, with X-axis be θ, then, dy, tan θ = at P ( 0, m), dx , , (b) 3x 2 − 2 x − 5, (d) 3x 2 − 10x + 5, , tan θ = m2 ⇒ θ = tan−1 m2, , ⇒, , Now, the tangent will make the angle with, Y-axis, π, π, =, −θ =, − tan−1 m2, 2, 2, π, = cot −1 m2 Q tan−1 x + cot −1 x = , , 2 , , Ê (c) From the option (c),, , f( x ) = 3 x − 2 x + 5, 2, , f( 0) = 3( 0)2 − 2( 0) + 5, =5, f( −1) = 3( −1)2 − 2( −1) + 5, = 3 + 2 + 5 = 10, and f(1) = 3(1)2 − 2(1) + 5, , , 1, = sin−1 , 1 + m4, , , = 3−2 + 5= 6, Hence, the required polynomial, f( x ) = 3 x 2 − 2 x + 5., , Directions (Q. Nos. 76-78) Read the, , point of intersection P?, (a) m, (c) 2 m, , (b) m2, (d) 2 m2, , 77. How much angle does the tangent at, P make with y-axis?, (a) tan−1 m2, (b) cot −1(1 + m2 ), , 1, (c) sin−1 , 1 + m4, , , , , , , , (d) sec −1 1 + m4, , 78. What is the equation of tangent to, the curve at P ?, (a) y = mx + m, (b) y = − mx + 2 m, (c) y = m2x + 2 m (d) y = m2x + m, , , , , , , , Q cot −1 x = sin, , , , , , 1, , , 1 + m2 , , , , Ê 78. (d) Equation of tangent to curve at P is, dy, y − y1 = , dx ( x, , following information and answer the, three items that follow ., A curve y = memx where m > 0 intersects, Y-axis at a point P., , 76. What is the slope of the curve at the, , Solved Paper 2019 (II), , ( x − x1 ), , 1 , y1 ), , ⇒ y − m = m2 ( x − 0), ⇒, , y = m2 x + m, , Directions (Q. Nos. 79 and 80) Read, the following information and answer, the two items that follow., Let f ( x ) = x 2, g( x ) = tan x and, h( x ) = log x., 79., , π, , what is the value of, 2, [ho( gof )]( x )?, For x =, , (a) 0, , (b) 1, , (b), , π, 4, , (d), , π, 2, , 80. What is [ fo( fof )](2) equal to ?, (a) 2, (c) 16, , (b) 8, (d) 256, , (Q. Nos. 79 and 80) Given,, Ê Solutions, 2, f( x ) = x , g ( x ) = tan x and h( x ) = log x
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NDA/NA, , 11, , Solved Paper 2019 (II), , 2, Ê 79. (a) ( gof ) ( x) = g { f( x)} = tan x, , Ê (d) Suppose,, , Now, [ho( gof )] ( x ) = h {( gof )( x )}, = log (tan x 2 ), π, 2, π, π, [ho( gof )] , = log tan , 4, 2 , for x =, , = log 1 = 0, , Ê 80.(d) ( fof ) ( x) = f{ f( x)}, = (x ) = x, 2 2, , I=, , =−, , ∴[fo( fof )] (2 ) = 2 = 256, , 2x 2 − 2x + 1, , equal to?, , tan−1(2 x − 1), +c, 2, −1, (b) 2 tan (2 x − 1) + c, tan−1 (2 x + 1), (c), +c, 2, −1, (d) tan (2 x − 1) + c, dx, Ê (d) Let I = ∫ 2, 2x −2x + 1, 1, dx, = ∫, 2 x2 − x + 1, 2, 1, dx, = ∫, 2 x2 − x + 1 − 1 + 1, 4 4 2, 1, dx, = ∫, 2, 2 , 1, 1, x− +, , 2, 4, 1, dx, = ∫, 2, 2, 2 , 1, 1, x− + , , 2, 2, 1 , , x − , , 1, 2, −1 , = .2 tan , +c, 1, 2, , , 2, , , , x, dx, 1, = tan−1 , Q ∫ 2, 2, a, a, a, x, +, , , = tan−1(2 x − 1) + c, , dx, , ∫ x (1 + ln x )n, , (n ≠ 1)?, 1, , (where n ≠ 1), , ( n − 1) t n −1, , log x, dy, =, dx (1 + log x )2, dy, Ê 84. (a)Q =, dx, , +c, , 1, ( n − 1) (1 + log x )n − 1, , +c, , +c, , ( n − 1) (1 + ln x )n − 1, 1− n, (b), +c, (1 + ln x )1 − n, n+1, (c), +c, (1 + ln x )n + 1, 1, (d) −, +c, ( n − 1) (1 + ln x )n − 1, , equal to, , dy, dy 1, = 4 xy 2 (b), =, dx, dx, y, dy, dy, 2, (c), = x y (d), = − 4 xy 2, dx, dx, , Ê (d) The differential equation of family of, 1, , 2 x2 − C, , … (i), , (where, C is any arbitrary constant), Differentiation w.r.t. x of Eq. (i), d, dy, 1, =−, (2 x 2 − C ), dx, (2 x 2 − C )2 dx, dy, = − y 2. ( 4 x − 0), dx, dy, = − 4 xy 2, it is required differential, ⇒, dx, equation., ⇒, , Directions (Q. Nos. 84 and 85) Read, the following information and answer, the two items that follow., Consider the equation x y = ex − y, (a) 0, (c) 2, , 85. What is, , dy, at x = 1 equal to?, dx, d 2y, 2, , log 1, dy, =, dx (1 + log 1)2, , 0, =0, [Q log 1 = 0], 1, log x, dy, Ê 85. (b)Q =, dx (1 + log x )2, =, , (a), , 84. What is, , log x, (1 + log x )2, , At x = 1,, , differential equation that represents, 1, the family of curves y = 2, ,, 2x − C, where C is an arbitrary constant?, , curves y =, , Differentiation w.r.t. x, we get, 1, (1 + log x ).1 − x 0 + , , dy, x, =, dx, (1 + log x )2, dy 1 + log x − 1, =, dx (1 + log x )2, , 83. Which one of the following is the, , dx, , (a), , (a), , 1, , =−, , 4, , 8, , 82. What is, , ∫ x (1 + ln x)n, , Let 1 + ln x = t, Diff. w.r.t. x, we get, dx, 1 dt, = dt, 0+ =, ⇒, x, x dx, dt, t −n + 1, I=∫ n=, ∴, +c, −n+1, t, , Now, [fo( fof )] ( x ) = f{( fof ) ( x )}, = ( x 4 )2 = x 8, , 81. What is, ∫, , dx, , (b) 1, (d) 4, , at x = 1 equal to?, , dx, (a) 0, (b) 1, (c) 2, (d) 4, Solutions (Q. Nos 84 and 85), Given equation, x y = e x − y, , On taking log both sides, we get, y log x = ( x − y) log e, ⇒, y log x = x − y[Q loge e = 1], x, ⇒ (1 + log x ) y = x ⇒ y =, (1 + log x ), , Differentiation. w.r.t. x, we get, 1, (1 + log x )2. − (log x )., x, 1, 2 (1 + log x ) 0 + , , d 2y, x, =, (1 + log x )4, dx 2, 1, (1 + log x ) (1 + log x − 2 log x ), = x, (1 + log x )4, 1, (1 + log x ) (1 − log x ), = x, (1 + log x )4, At x = 1, , d 2y, dx 2, , =, , 11, ( + 0) (1 − 0), (1 + 0)4, , =1, , Directions (Q.Nos. 86-88) Read the, following information and answer the, three items that follow., Consider the function, f ( x ) = g( x ) + h( x ), x, where, g( x ) = sin , 4, 4x, and, h( x ) = cos , 5, 86. What is the period of the function, g ( x )?, (a) π, (c) 4 π, , (b) 2 π, (d) 8 π, , 87. What is the period of the function, h ( x )?, (a) π, (c), , 5π, 2, , 4π, 5, 3π, (d), 2, (b), , 88. What is the period of the function, f (x ) ?, (a) 10 π, (c) 40 π, , (b) 20 π, (d) 80 π
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12, , NDA/NA, , Ê Solutions (Q. Nos. 86-88), Given, f( x ) = g ( x ) + h( x ),, x, where, g ( x ) = sin and, 4, 4x, h( x ) = cos , 5, , x, g ( x ) = sin , 4, x + 8π , g ( x + 8 π ) = sin , , , 4 , x, = sin 2 π + , , 4, x, = sin = g ( x ), 4, , Ê 86. (d), , ∴ Period of the function g ( x ) = 8 π, 4x, Ê 87. (c) h( x) = cos , 5, 5π , 4, 5π , h x +, = cos x +, , , 2 , 5, 2 , 4x, = cos 2 π +, , , 5, 4x, = cos = h( x ), 5, 5π, ∴ Pperiod of the function h( x ) =, 2, , Ê 88. (c) f( x) = g ( x) + h( x), , x, 4x, = sin + cos , 4, 5, x + 40 π , 4, f( x + 40 π ) = sin , + cos, , , 4, 5, ( x + 40 π ), 4x, x, = sin 10 π + + cos 32 π +, , , , 4, 5, 4x, x, = sin 5 × 2 π + + cos 16 × 2 π +, , , , 4, 5, x, 4x, = sin + cos = f( x ), 4, 5, ∴Period of the function f( x ) = 40 π, , Directions (Q. Nos. 89 and 90) Read, the following information and answer, the two items that follow., consider the function, f ( x ) = 3x 4 − 20 x 3 − 12x 2 + 288 x + 1, , 89. In which one of the following, intervals is the function increasing?, (a) ( −2, 3), (c) ( −3, − 2 ), , (b) ( 3, 4), (d) ( −4, − 3), , 90. In which one of the following, intervals is the function decreasing?, (a) ( −2, 3), (c) (4, 6), , (b) (3, 4), (d) (6, 9), , Ê Solutions (Q. Nos. 89 and 90), , Given function,, f( x ) = 3 x 4 − 20 x 3 − 12 x 2 + 288 x + 1, Differentiation w.r.t. x, we get, f ′ ( x ) = 12 x 3 − 60 x 2 − 24 x + 288, , Ê 89. (b) f( x) is increasing, if f ′( x) ≥ 0, , 12 x − 60 x − 24 x + 288 ≥ 0, 3, , 2, , ⇒, , x 3 − 5 x 2 − 2 x + 24 ≥ 0, , ⇒, , ( x + 2 )( x 2 − 7 x + 12 ) ≥ 0, , ⇒ ( x + 2 ) ( x − 3) ( x − 4) ≥ 0, ∴, x ≤ − 2, x ≥ 3, x ≥ 4, Hence, f( x ) is increasing the interval, ( 3, 4)., , Ê 90. (a) f( x) is decreasing, if f ′( x) ≤ 0, 12 x 3 − 60 x 2 − 24 x + 288 ≤ 0, ⇒, , x − 5 x − 2 x + 24 ≤ 0, , ⇒, , ( x + 2 ) ( x − 7 x + 12 ) ≤ 0, , 3, , 2, Ê 92. (d) 1. f[g ( x)] = 25 x + 310 x + 955, , f[g ( x )] is a polynomial of degree 2., So, Statement 1 is not correct., 2. g [g ( x )] = 25 x + 180, g [g ( x )] is a polynomial of degree 1., So, Statement 2 is not correct., Hence, the Statement neither 1 nor 2, correct., , Ê 93. (b) Given, h( x) = 5f( x) − xg ( x), , = 5( x 2 + 2 x − 5) − x ( 5 x + 30), , 2, , 2, , ⇒ ( x + 2 ) ( x − 3) ( x − 4) ≤ 0, ∴, x ≥ − 2, x ≤ 3, x ≤ 4, Hence f( x ) is decreasing the interval, ( −2, 3)., , Directions (Q.Nos. 91-93) Read the, following information and answer the, two items that follow ., Let, f ( x ) = x 2 + 2x − 5, and, g( x ) = 5x + 30, , 91. What are the roots of the equation, g [( f ( x )] = 0?, , (a) 1, − 1, (c) 1, 1, , (b) −1, − 1, (d) 0, 1, , 92. Consider the following statements., 1. f [ g ( x )] is a polynomial of degree, 3., 2. g [ g ( x )] is a polynomial of degree, 2., Which of the above statements is/are, correct?, (a) Only 1, (c) Both 1 and 2, , = 5 x 2 + 10 x − 25 − 5 x 2 − 30 x, = − 20 x − 25, Differentiation w.r.t. x, we get, h′ ( x ) = − 20, Hence, derivative of h( x ) is −20., , Directions (Q.Nos. 94 and 95) Read, the following information and answer, the questions given below., Consider the integrals, π, xdx, and, I1 = ∫, 0 1 + sin x, π, (π − x )dx, I2 = ∫, 0 1 − sin (π + x ), 94. What is the value of I 1?, (a) 0, , (a) 2 π, , ⇒, , ( x + 1)2 = 0, , ∴, x = − 1, − 1, Hence, the roots of this equation are, −1, − 1., , π, , ( π − x ) dx, , ∫0 1 − sin ( π +, π, , xdx, , π, , ( π − x ) dx, , I1 =, , x), …(i), , ∫0 1 + sin ( π −, , Q f( x ) dx =, ∫0, , f[g ( x )] = ( 5 x + 30)2 + 2( 5 x + 30) − 5, , x + 2x + 1= 0, , (d) 0, , xdx, , a, , = 5 x 2 + 10 x + 5, , ⇒, , π, 2, , (c), , ∫0 1 + sin x, , =, , ∴g [f( x )] = 5( x 2 + 2 x − 5) + 30, , 2, , π, , I1 =, , Nos. 91-93) Given,, Ê Solutions (Q., f( x ) = x 2 + 2 x − 5, g ( x ) = 5 x + 30, , Ê 91. (b) The equation, g [f( x)] = 0, 5 x 2 + 10 x + 5 = 0, , (d) 2 π, , ∫0 1 + sin x,, , I2 =, , (b) −20, (d) 0, , and g [g ( x )] = 5( 5 x + 30) + 30, = 25 x + 180, , (b) π, , Given, I1 =, , is the derivative of h ( x )?, , = 25 x 2 + 310 x + 955, , (c) π, , Ê Solutions (Q.Nos. 94 and 95), , (b) Only 2, (d) Neither 1 nor 2, , = 25 x 2 + 900 + 300 x + 10 x + 60 − 5, , π, 2, , (b), , 95. What is the value of I 1 + I 2 ?, , 93. If h( x ) = 5 f ( x ) − xg ( x ), then what, (a) −40, (c) −10, , Solved Paper 2019 (II), , π, , ∫0, , x), , ∫0 f(a − x) dx, a, , ( π − x ) dx, 1 + sin x, , … (ii), , Adding Eqs. (i) and (ii), we get, π ( x + π − x ) dx, π, π dx, ∫0 1 + sin x = ∫0 1 + sin x, , 2 I1 =, , π, , (1 − sin x ) dx, , = π, , ∫0, , = π, , ∫0, , = π, , ∫0 (sec, , π, , π, , (1 − sin2 x ), (1 − sin x ) dx, cos 2 x, 2, , x − sec x tan x ) dx, , = π [tan x − sec x ]π0, = π [(tan π − sec π ), − (tan 0 − sec 0)
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NDA/NA, , = π [( 0 + 1) − ( 0 − 1)], 2 I1 = π (1 + 1) = 2 π, ⇒ I1 = π, π, ( π − x ) dx, Now, I2 = ∫, 0 1 − sin ( π + x ), =, , 98. For what value of k is the function, , 1, 2x + , x < 0, 4, , f ( x ) = k, , x = 0 continuous?, , 2, x + 1 , x > 0, , 2, , ( π − π + x ) dx, , π, , ∫0 1 − sin ( π − π −, , Q af( x ) dx =, ∫0, π, xdx, =∫, 0 1 − sin ( − x ), =, , 13, , Solved Paper 2019 (II), , π, , x), , a, , ∫0 f ( a −, , x ) dx , , , (a), , xdx, , = I1 = π, π, , xdx, , ∫0 1 + sin x =, , π, , (a) e tan ydx + (1 − e ) dy = 0, x, , (b) e x tan ydx + (1 − e x ) sec 2 ydy = 0, (c) e (1 − e ) dx + tan ydy = 0, x, , (d) e x tan ydy + (1 − e x ) dx = 0, , Ê (b) The equation of the family of curves., tan y = C (1 − e x ), , … (i), , Differentiation w.r.t. x, we get, dy, sec 2 y ., = C(0 − e x ), dx, dy, … (ii), ⇒ sec 2 y ., = − Ce x, dx, EleminatingC from Eqs. (i) and (ii), we get, tan y, dy, sec 2 y. = − e x, dx, (1 − e x ), ⇒ (1 − e ) sec y dy = − e tan y. dx, 2, , x, , 97. What is the derivative of 2(sin x ), , 2, , with respect to sin x?, 2, , (sin x )2, , with respect to, sin x, 2, d, d (sin x )2, 2, (sin x )2, 2 (sin x ) log 2, dx, dx, =, =, d, cos x, (sin x ), dx, 2, 2 (sin x ) log 2. 2 sin x.cos x, =, cos x, 2, , = 2 log 2. (sin x ). 2 (sin x ), 2, , = sin x. 2 (sin x ) .log 4, , 99. What is the area of the region, enclosed between the curve y 2 = 2x, and the straight line y = x ?, 2, sq unit, 3, 1, (c) sq unit, 3, , 4, sq unit, 3, (d) 1 sq unit, , (b), , (a), , y = 0, 2, 2, , y2, , , − y dy, , 2, , 1 y, y , = ., −, 2 0, 2 3, 3, , x2 − 3 x − 2 x + 6 ≥ 0, , ⇒, x( x − 3) − 2 ( x − 3) ≥ 0, ⇒ ( x − 3) ( x − 2 ) ≥ 0 ⇒ x ≤ 2, x ≥ 3, ∴, T = ( − ∞,2 ) ∪ ( 3, ∞ ), Again, f( x ) is decreases in interval S., ∴, f ′( x) ≤ 0, ⇒, x2 − 5 x + 6 ≤ 0, ⇒ x2 − 3 x − 2 x + 6 ≤ 0, ⇒ x( x − 3) − 2( x − 3) ≤ 0, ⇒, ( x − 3) ( x − 2 ) ≤ 0, ⇒, 2 < x < 3 ⇒ x ∈ (2, 3), ∴, S = (2, 3), , 101. A coin is biased so that heads comes, up thrice as likely as tails. For three, independent tosses of a coin, what is, the probability of getting at most, two tails?, (b) 0.48, (d) 0.98, , Ê (d) Let X be a random variable, it, … (i), , ∫0 2, , (a) T = ( −∞, 2 ), ∪ ( 3, ∞ ) and S = (2, 3), (b) T = φ and S = ( −∞, ∞ ), (c) T = ( −∞, ∞ ) and S = φ, (d) T = (2, 3) and S = ( −∞, 2 ) ∪ ( 3, ∞ ), x3 5 x2, + 6x + 7, Ê (a) Given, f( x) = −, 3, 2, Differentiating w.r.t x, we get, 3 x2 5, f ′( x) =, − . 2 x + 6 = x2 − 5 x + 6, 3, 2, Q f( x ) is increases in interval T,, ∴, f ′( x) ≥ 0, ⇒, x2 − 5 x + 6 ≥ 0, , (a) 0.16, (c) 0.58, , and, … (ii), y= x, From Eqs. (i) and (ii),, y 2 = 2 y ⇒ y( y − 2 ) = 0, , ∴Required area =, , (sin x )2, , Ê (a) Derivative of 2, , and f( 0) = k, Q The function f( x ) is continuous at x = 0, LHL = f( 0) = RHL, ∴, 1, LHL = f( 0) ⇒, =k, ⇒, 4, 1, Hence, k =, 4, , ∴, , 2, , (b) 2 sin x 2 (sin x ) ln 4, , (sin x )2, , 1, 4, , y2 = 2 x, , (a) sin x 2 (sin x ) ln 4, , (d) 2 sin x cos x 2, , 1, 4 , =, , x 3 5x 2, −, + 6x + 7 increases, 3, 2, in the interval T and decreases in the, interval S, then which one of the, following is correct?, , If f ( x ) =, , ⇒, , Ê (a) Equations of curves,, , ⇒ e x tan y dx + (1 − e x ) sec 2 y dy = 0, , (c) ln (sin x ) 2, , (d) 2, , = lim 2( 0 − h) +, h→ 0 , , 1, , = lim −2 h + , h → 0, 4, , represents the family of curves given, by tan y = C (1 − e x ) is, , x, , (c) 1, , , 1, 2 x + , x < 0, 4, , f( x ) = k, , x = 0 continuous., 2, , x + 1 , x > 0, , 2, , 96. The differential equation which, , x, , 1, 2, , 1, LHL = lim f( x ) = lim 2 x + , x → 0−, h → 0− , 4, , Ê 95. (a) I1 + I2 = π + π = 2 π, , x, , (b), , Ê (a) Given,, , ∫0 1 + sin x [Q sin (− θ) = sin θ], , Ê 94. (c) I1 =, , 1, 4, , 100., , 2, , 1, (2 ) , = (2 )3 −, −0, 2 , 6, 8 4 8 − 12, = − =, 6 2, 6, 4 −2 2, =− =, = sq unit, 6, 3, 3, (∴area will not be negative), 2, , represents of the number tail comes of, three tosses of a coin., ∴Possible value of X are 0, 1, 2, 3., According to the question, the coin is, biased in which the probability to comes, head is thrice as likely as tails., 3, 1, ∴P(H) = and P(T) =, 4, 4, 3, , 3, 27, P(X = 0) = P({HHH}) = =, 4, 64, P( X = 1) = P (2 heads and 1 tail), = P({HHT}) + P ({HTH}) + P({THH}), 3 3 1 3 1 3 1 3 3, = ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅, 4 4 4 4 4 4 4 4 4, 27, =, 64, P( X = 2 ) = P (1 head and 2 tails), = P({HTT}) + P({THT}) + P({TTH})
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14, , NDA/NA, 3 1 1 1 3 1 1 1 3, ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅, 4 4 4 4 4 4 4 4 4, 9, =, 64, ∴Required probability, = P( X = 0) + P( X = 1) + P( X = 2 ), 27, 9, 27, =, +, +, 64 64 64, 63, =, = 0.98, 64, =, , 102. A bag contains 20 books out of, which 5 are defective. If 3 of the, books are selected at random and, removed from the bag in succession, without replacement, then what is, the probability that all three books, are defective?, (a) 0.009, (c) 0.026, , (b) 0.016, (d) 0.047, , 104. Arithmetic mean of 10 observations, is 60 and sum of squares of, deviations from 50 is 5000. What is, the standard deviation of the, observations?, (a) 20, (c) 22.36, , Ê (a) Arithmetic mean of 10 observations, = 60, , Q x = Σxi , , n , If, A = 50, then Σd i2 = 5000, , ∴Σxi = 60 × 10 = 600, , Q d i = xi − A, ∴ Σd i = Σ( xi − A ) = Σxi − AΣ1, = 600 − 50 × n, [Q Σ1 = n], = 600 − 50 × 10 = 100, , 103. The median of the observations 22,, 24, 33, 37, x + 1, x + 3, 46, 47, 57, 58 in, ascending order is 42. What are the, values of 5th and 6th observations, respectively?, , (a) 42, 45, (c) 43, 46, , Σd i2 Σd i , −, , n , n, , 5000 100 , −, , 10 , 10, , 2, , =, =, , 500 − 100 =, , 400 = 20, , 105. If p and q are the roots of the, equation x 2 − 30x + 221 = 0, what is, the value of p 3 + q 3 ?, (a) 7010, (c) 7210, , equation, x 2 − 30 x + 221 = 0, , ∴ p + q = 30 and pq = 221, Now, p3 + q 3 = ( p + q ) ( p2 + q 2 − pq ), = 30 [ p + q + 2 pq − 3 pq ], 2, , Value of 6th observation, 2, ⇒ 84 = x + 1 + x + 3, ⇒ 2 x = 84 − 4, 80, = 40, ⇒ x=, 2, ∴ 5th observation = x + 1 = 40 + 1 = 41, and 6th observation, = x + 3 = 40 + 3 = 43, , 2, , = 30 [( p + q )2 − 3 pq ], , 106. For the variables x and y, the two, , regression lines are 6x + y = 30 and, 3x + 2y = 25. What are the values of, x , y and r respectively?, 20 35, ,, , − 0.5, 3 9, 35 20, (c), , , − 0.5, 9 3, , 20 35, ,, , 0.5, 3 9, 35 20, (d), , , 0.5, 9 3, (b), , Ê (c) Given lines, 6 x + y = 30, and, 3 x + 2 y = 25 …(ii), where, x and y are two variables., Solving these equations,, 35, 20, x = , and y =, 9, 3, These lines are regression,, 35, 20, Then, x = , y =, 9, 3, , 107. The class marks in a frequency table, are given to be 5, 10, 15, 20, 25, 30, 35,, 40, 45, 50. The class limits of the first, five classes are, (a) 3-7, 7-13, 13-17, 17-23, 23-27, (b) 2.5-7.5, 7.5-12.5, 12.5-17.5,, 17.5-22.5, 22.5-27.5, (c) 1.5-8.5, 8.5-11.5, 11.5,-18.5,, 18.5-21.5, 21.5-28.5, (d) 2-8, 8-12, 12-18, 18-22, 22-28, , Ê (b) Given, class marks in a frequency, , table are, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50., Let L1 and L2 be the lower limit and upper, limit of first interval., L + L2, L + L2, 5= 1, Q Class mark = 1, 2, 2, … (i), ⇒ L1 + L2 = 10, and L2 − L1 = Class interval, … (ii), or L2 − L1 = 5, Solving Eq. (i) and (ii),, L2 = 7.5 and L1 = 2.5, ∴Class limit of first classes is 2.5 − 7.5, Similarly find class limit of other classes., Hence, class limits of the first five classes, are, 2.5 − 7.5, 7.5 − 12.5, 12.5 − 17.5,, 17.5 − 22.5, 22.5 − 27.5., , 108. The mean of 5 observations is 4.4 and, variance is 8.24. If three of the five, observations are 1, 2 and 6, then what, are the other two observations?, (b) 9, 4, (d) 81, 4, , Ê (b) Let x1, x2, x3, x4 and x5 are five, , = 30 [900 − 663], = 30 × 237 = 7110, , (a), , 3, 1, or −, 6, 2, = − 0.5, , r=−, , and, , (a) 9, 16, (c) 81, 16, , = 30 [( 30)2 − 663], , Ê (b) The observations in ascending order, , ⇒ 42 =, , (b) 7110, (d) 7240, , Ê (b) Since, p and q are the roots of the, , (b) 41, 43, (d) 40, 40, , are, 22, 24, 33, 37, x + 1, x + 3, 46, 47, 57, 58, Here, n = 10, ∴Median, N, Value of the observations +, 2, N, Values of + 1 th observations, 2, , =, 2, Value of 5th observations +, , 2, , Now, SD =, , Ê (a) Total books in bag = 20, Defective books = 5, ∴Undefective books = 20 − 5 = 15, ∴ Probability to selected three books are, defective without replacement, 5, 4, 3, =, ×, ×, 20 19 18, 6, =, 684, = 0.0087 = 0.009, , (b) 21, (d) 24.70, , Solved Paper 2019 (II), , …(i), , observations., ∴ x1 = 1, x2 = 2 , and x3 = 6, x + x2 + x3 + x4 + x5, ∴ x= 1, 5, x + x2 + x3 + x4 + x5, ⇒ 4.4 = 1, 5, ⇒ x1 + x2 + x3 + x4 + x5 = 22, ⇒, 1 + 2 + 6 + x4 + x5 = 22, ⇒, x4 + x5 = 22 − 9, …(i), ⇒, x4 + x5 = 13, and variance,, ( x1 − x )2 + ( x2 − x )2 + ( x3 − x )2 +, σ2 =, , ( x4 − x )2 + ( x5 − x )2, 5, (1 − 4.4)2 + (2 − 4.4)2 + ( 6 − 4.4)2 +, , . =, ⇒ 824, , ( x4 − 4.4)2 + ( x5 − 4.4)2, 5
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NDA/NA, , ⇒ 824, . × 5 = 1156, . + 576, . + 2.56, + ( x4 − 4.4)2 + (13 − x4 − 4.4)2, , 111. Two, , [from Eq. (i)], . = 19.88 + ( x4 − 4.4)2 + ( 8.6 − x4 )2, ⇒ 4120, ⇒ 4120, . − 19.88 = x42 + 19.36 − 8.8 x4, + 73.96 + x42 − 17.2 x, ⇒ 2132, . = 2 x42 − 26 x4 + 93.32, ⇒, , 2 x42 − 26 x4 + 72 = 0, , ⇒, , x42 − 13 x4 + 36 = 0, , ⇒, , x42 − 9 x4 − 4 x4 + 36 = 0, , 109. If a coin is tossed till the first head, appears, then what will be the, sample space?, (a) {H}, (b) {TH}, (c) {T, HT, HHT, HHHT, ………}, (d) {H, TH, TTH, TTTH, ………}, , Ê (a) A coin is tossed till the first head, appears, then the sample space will be, = {H}, , 110. Consider the following discrete, frequency distribution., x, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , f, , 3, , 15, , 45 57, , 5, 0, , 36 25, , 9, , What is the value of median of the, distribution?, (b) 5, , (c) 6, , (d) 7, , Ê (b), x, 1, 2, 3, 4, 5, 6, 7, 8, , f, 3, 15, 45, 57, 50, 36, 25, 9, N = 270, , Here, N = 270, ∴Median, N, Value of th term + value of, 2, N + 1 th term, , , 2, , =, 2, Value of 135th term + Value of, =, =, , 136th term, 5+ 5, =5, 2, , 2, , dice, are, thrown, simultaneously. What is the, probability that the sum of the, numbers appearing on them is a, prime number?, , 5, 12, 7, (c), 12, (a), , 1, 2, 2, (d), 3, , (b), , Ê (a) Total number of sample space of two, , ⇒ x4 ( x4 − 9) − 4 ( x4 − 9) = 0, ⇒, ( x4 − 9) ( x4 − 4) = 0, ∴, x4 = 4, 9, From Eq. (i), x5 = 9, 4, Hence, other two observations are 9, and 4., , (a) 4, , 15, , Solved Paper 2019 (II), , C, 3, 18, 63, 120, 170, 236, 261, 270, , dice are thrown, n( s ) = 6 × 6 = 36, Total number of favourable outcomes the, sum of numbers appearing on them is a, prime number., (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3),, (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2),, (5, 6), (6, 1), (6, 5), ∴, n( E ) = 15, n( E ), ∴Required probability =, n(S ), 15, 5, =, =, 36 12, , 112. If 5 of a Company’s 10 delivery, trucks do not meet emission, standards and 3 of them are chosen, for inspection, then what is the, probability that none of the trucks, chosen, will, meet, emission, standards?, 1, 8, 1, (c), 12, , (a), , 3, 8, 1, (d), 4, , (b), , Ê (c) Total trucks of a company’s = 10, Number of trucks that do not meet, emission standards = 5, Number of trucks that are chosen for, inspection = 3, 5, C, ∴Required probability = 10 3, C3, 5!, 5 !7 !, = 3 !2 ! =, 10 ! 10 !2 !, 3 !7 !, 5⋅ 4⋅ 3, 1, =, =, 10 ⋅ 9 ⋅ 8 12, , 113. There are 3 coins in a box. One is a, two-headed coin; another is a fair, coin; and third is biased coin that, comes up heads 75% of time. When, one of the three coins is selected at, random and flipped, it shows heads., What is the probability that it was, the two-headed coin?, 2, 9, 4, (c), 9, , (a), , 1, 3, 5, (d), 9, (b), , Ê (c) Let E1, E 2 and E 3 represent the events, of two-headed coin, a fair coin and, biased coin respectively., 1, 1, 1, ∴P( E1 ) = , P( E 2 ) = , P( E 3 ) =, 2, 2, 4, E 1 E 1 E 1, P = , P = , P =, E1 2 E 2 2 E 3 4, Apply Baye’s theorem,, E, P 1 =, E, , E, P( E1 ) ⋅ P , E1 , E, E, P( E1 ) ⋅ P + P( E 2 ) ⋅ P , E2 , E1 , E, + P( E 3 ) ⋅ P , E3 , , 1 1, 1, ., 2 2, 4, =, 1 1 1 1 1 1 1 1, 1, + +, . + . + ., 2 2 2 2 4 4 4 4 16, 1, 4, 4, =, =, 4+ 4+1 9, 16, =, , 114. Consider the following statements:, 1. If A and B are mutually exclusive, events, then it is possible that, P ( A ) = P ( B ) = 06., ., 2. If A and B are any two events, such that P ( A / B ) = 1, then, P ( B / A ) = 1., Which of the above statement is/are, correct?, (a) Only 1, (c) Both 1 and 2, , (b) Only 2, (d) Neither 1 nor 2, , Ê (b) Statement 1 : A and B are mutually, exclusive events, then P( A ∩ B) = 0, ∴P( A ∪ B) = P( A ) + P( B), = 0.6 + 0.6, = 1.2 , it is not possible, So, Statement 1 is not correct., Statement 2 : A and B are any two events, such that, A, P = 1, B, ⇒, , P( A ∩ B), = 1 ⇒ P( A ∩ B) = P( B) … (i), P( B), , ∴P( A ∪ B) = P( A ) + P( B) − P( A ∩ B), P( A ∪ B) = P( A ) + P( B) − P( B), [from Eq. (i)], … (ii), ⇒ P( A ∪ B) = P( A ), B P( B ∩ A ) P( A ∪ B), Now, P =, =, P( A ), P( A ), A, =, , 1 − P ( A ∪ B) 1 − P( A ), =, =1, 1 − P( A ), 1 − P( A ), , So, Statement 2 is correct., Hence, only the Statement 2 is correct.
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16, , NDA/NA, , 115. If a fair die is rolled 4 times, then, what is the probability that there are, exactly 2 sixes?, 5, 216, 125, (c), 216, , 25, 216, 175, (d), 216, (b), , (a), , [By Bernoulli distribution], , 4!, 1, ×, ×, 2 !2 ! 36, 4⋅ 3, 1, =, ×, ×, 2 ⋅ 1 36, =, , 116. Mean of 100 observations is 50 and, standard deviation is 10. If 5 is added, to each observation, then what will, be the new mean and new standard, deviation respectively?, (b) 50, 15, (d) 55, 15, , Ê (c) Mean of 100 observations = 50, and standard deviation = 10, We know that, if k is added to each, observation, then new mean will be more, than k and standard deviation no change., ∴After 5 added to each observation., mean = 50 + 5 = 55, and standard deviation = 10, , 117. If the range of a set of observations, on a variable X is known to be 25, and if Y = 40 + 3X , then what is the, range of the set of corresponding, observations on Y ?, (a) 25, (c) 75, , (b) 40, (d) 115, , Ê (c) Range of set of observations on a, variable, X = 25, We know that, Range, R X = Xmax = Xmin, ⇒, 25 = Xmax − 0, ⇒, , Xmax = 25, , [Q Xmin = 0], , Ymax = 40 + 3 Xmax, = 40 + 3(25) [Q Xmax = 25], , Now,, , R Y = Ymax − Ymin, = 115 − 40 = 75, , 118. If V is the variance and M is the mean, of first 15 natural numbers, then, what is V + M 2 equal to?, 124, 3, 248, (c), 3, , (a), , 148, 3, 124, (d), 9, (b), , Ê (c) Mean of first 15 natural numbers, M, 1+ 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9, + 10 + 11 + 12 + 13 + 14 + 15, =, 15, 15(15 + 1), =, 2 × 15, , 2, , 25, 36, 25, 25, =, 36 216, , (a) 50, 10, (c) 55, 10, , 119. A car travels first 60 km at a speed of, , = 40 + 75 = 115, , Here, n = 4, r = 2, ∴Required probability = nC r prq n − r, 1, 5, = 4C 2 , 6 6, , Y = 40 + 3 X, Ymin = 40 + 3 Xmin, = 40 + 3( 0), = 40, , and, , Ê (b) Let X be a random variable that, represents to appearing 6 of rolled a die., Probability of to get 6 to rolled a die,, 1, p=, 6, ∴Probability of not get 6 to rolled a die,, 1, 1 5, q = 1− = 1− =, p, 6 6, , 2, , Q, ∴, , [Q Xmin = 0], , Solved Paper 2019 (II), , Q 1 + 2 + 3 + ... + n = n( n + 1), , 2, , =8, Variance of first 15 natural numbers, V, 1, =, [(1 − 8)2 + (2 − 8)2 + ( 3 − 8)2, 15, + ( 4 − 8)2 + ( 5 − 8)2 + ( 6 − 8)2, + (7 − 8)2 + ( 8 − 8)2 + ( 9 − 8)2, + (10 − 8)2 + (11 − 8)2 + (12 − 8)2, + (13 − 8)2 + (14 − 8)2 + (15 − 8)2 ], 1, =, [( − 7 )2 + ( − 6)2 + ( − 5)2 + ( − 4)2 + ( − 3)2, 15, + ( − 2 )2 + ( − 1)2 + 0 + (1)2 + (2 )2 + ( 3)2, + ( 4)2 + ( 5)2 + ( 6)2 + (7 )2 ], 2 2, =, [1 + 2 2 + 3 2 + 4 2 + 5 2 + 6 2 + 7 2 ], 15, 7(7 + 1) (14 + 1), 2, =, ×, 15, 6, Q 12 + 2 2 + 3 2 + ... + n2 , , , = n ( n + 1) (2 n + 1), , , , 6, 7 × 8 × 15 56, 2, =, ×, =, 15, 6, 3, 56, Now, V + M 2 =, + 64, 3, 56 + 192, =, 3, 248, =, 3, , 3 v km/h and travels next 60 km at 2, v km/h. What is the average speed of, the car?, (a) 2.5 v km/h, (b) 2.4 v km/h, (c) 2.2 v km/h, (d) 2.1 v km/h, , Ê (b) Time taken for first 60 km with speed 3, v km/h, 60 20, h, =, =, 3v, v, , Distance , , Q Time = Speed , , , , Time taken for next 60 km with speed, 2v km/h, 60, 30, h, =, =, 2v, v, Total distance, ∴Average speed =, Total time, 60 + 60, =, 20 30, +, v, v, 120 v, =, 50, = 2.4 v km/h, , 120. The mean weight of 150 students in a, certain class is 60 kg. The mean, weight of boys is 70 kg and that of, girls is 55 kg. What are the number, of boys and girls respectively in the, class?, (a) 75 and 75, (b) 50 and 100, (c) 70 and 80, (d) 100 and 50, , Ê (b) Let number of boys and girls be x and, y respectively., … (i), ∴, x + y = 150, Mean weight of 150 students = 60 kg, ∴Total weight of 150 students, = 60 × 150, = 9000kg., Mean weight of boys = 70 kg, ∴Total weight of boys = 70 x kg, and mean weight of girls = 55 kg, ∴Total weight of girls = 55 y kg, ∴Total weight of 150 students = 9000 kg, ⇒, , 70 x + 55 y = 9000, , ⇒ 14 x + 11y = 1800 … (ii), Solving Eqs. (i) and (ii), we get, x = 50, y = 100, Hence, the number of boys and girls are, 50 and 100 respectively.
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NDA /NA, , National Defence Academy/Naval Academy, , SOLVED PAPER 2019 (I), PAPER I : Mathematics, 1., , (a) ( −5)2n−1, (c) ( −1)2n−1 5 n +1, , (b) ( −1)2n 5 n +1, (d) ( −1)n−1 5 n +1, , Ê (d), , Given,, sequence, −125, 625, − 3125, T, T, Here, 2 = 3 = ..........., T1 T2, , 3., 25,, , So, this sequence in GP whose common, ratio is −5., then a = 25, r = −5, ∴n th term of sequence = ar n−1, = 25( − 5)n − 1, = ( − 1)n − 1 5 2 × 5 n − 1 = ( − 1)n − 1 5 n + 1, , 2., , ∴R is symmetric., Hence, R is neither reflexive nor, transitive but symmetric., , What is the nth term of the sequence, 25, − 125, 625, − 3125, …?, , Suppose X = {1, 2, 3, 4 } and R is a, relation, on, If, X., R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1),, (2,3), (3, 2)}, then which one of the, following is correct?, (a) R is reflexive and symmetric, but not, transitive, (b) R is symmetric and transitive, but not, reflexive, (c) R is reflexive and transitive, but not, symmetric, (d) R is neither reflexive nor transitive, but, symmetric, , Ê (d) We have, X = {1, 2, 3, 4}, R = { (1,1), (2, 2), (3, 3),, (1, 2), (2, 1), (2, 3), (3, 2)}, Since, ( 4, 4) ∉ R,, Hence, R is not reflexive., Since,, but, (1, 2 ) ∈ R, (2, 3) ∈ R, (1, 3) ∉ R, R is not transitive., (1, 2), (2, 3) ∈ R, and also (2, 1), (3, 2) ∈ R, , 4., , A relation R is defined on the set N of, natural, numbers, as, Then,, xRy ⇒ x 2 − 4 xy + 3y 2 = 0., which one of the following is, correct?, , (a) Null set, −1 + 3i −1 − 3i , (b) , ,, , 2, 2, , , −1 + 3i −1 − 3i , (c) , ,, , 4, 4, , , 1 + 3i 1 − 3i , (d) , ,, , 2 , 2, , (a) R is reflexive and symmetric, but not, transitive, (b) R is reflexive and transitive, but not, symmetric, (c) R is reflexive, symmetric and transitive, (d) R is reflexive, but neither symmetric, nor transitive, , 3, Ê (b) We have, A = { x ∈ Z : x − 1 = 0}, , and, , 2, 2, Ê (d) Given, xRy ⇒ x − 4 xy + 3 y = 0, , For reflexive, xRx ⇒ x 2 − 4 x 2 + 3 x 2 = 0, So, ( x, x ) ∈ R, ∀ x ∈ N, Hence, R is reflexive., For symmetric, yRx ⇒ y 2 − 4 xy + 3 x 2, , It is not clear, that y − 4 xy + 3 x is equal, 2, , 2, , to zero or not., i.e. ( x, y)∈ R but ( y, x ) ∉R ⋅ ∀ x, y ∈ N, Hence, R is not symmetric., For transitive, xRy ⇒ x 2 − 4xy + 3 y 2 = 0, yRz ⇒ y 2 − 4 yz + 3 z2 = 0 (let), xRz ⇒ x 2 − 4xz + 3 z2, It is not clear, that x 2 − 4xz + 3 z2 is, equal to zero or not., So, ( x , y) ∈ R ,( y, z)∈ R, ⇒( x , z) ∉ R ∀ x , y, z ∈ N, Hence, R is not transitive., , B = { x ∈ Z : x 2 + x + 1 = 0}, − 1 + 3i − 1 − 3i , A = 1,, ,, , 2, 2, , , − 1 + 3i − 1 − 3i , B=, ,, , 2, 2, , , , − 1 +, A∩B= , 2, , , xRy ⇒ x 2 − 4 xy + 3 y 2 = 0, ∴, , If, and, A = { x ∈ Z : x 3 − 1 = 0}, B = { x ∈ Z : x 2 + x + 1 = 0}, where, Z, is set of complex numbers, then, what is A ∩ B equal to?, , 5., , 3i − 1 − 3i , ,, , 2, , , Consider the following statements, for the two non-empty sets A and B., 1. ( A ∩ B ) ∪ ( A ∩ B ) ∪ ( A ∩ B ), =A∪B, 2. ( A ∪ ( A ∩ B )) = A ∪ B, Which of the above statements, is/are correct?, (a) Only 1, (c) Both 1 and 2, , Ê (a) We have,, , (b) Only 2, (d) Neither 1 nor 2, , 1. ( A ∩ B) ∪ ( A ∩ B ) ∪ ( A ∩ B) = A ∪ B, LHS ≡ ( A ∩ B) ∪ ( A ∩ B ) ∪ ( A ∩ B), = { A ∩ ( B ∪ B )} ∪ ( A ∩ B), [by distributive property]
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18, , NDA/NA, , 2., , 6., , = ( A ∩ U ) ∪ ( A ∩ B), [Q B ∪ B = U ], = A ∪ ( A ∩ B), = ( A ∪ A ) ∩ ( A ∪ B), = U ∩ ( A ∪ B)= A ∪ B = RHS, Hence, 1 is correct., A ∪ ( A ∩ B) = A ∪ B, LHS ≡ A ∪ ( A ∩ B ), = ( A ∪ A) ∩ ( A ∪ B), = U ∩ ( A ∪ B), = A∪B ≠ A ∪ B, Hence, 2 is false., ∴Only 1 is correct., , B31 = 0, B32 = 0, B33 = 8, B11, adj B = B21, , B31, , 8., , 3 2 0, If B = 2 4 0, then what is adjoint, , , 1 1 0, of B equal to?, 0 0 0, (a) 0 0 0, , , −2 −1 8, 0 0 2 , (c) 0 0 1, , , 0 0 0, , 0 0 −2 , (b) 0 0 −1, , , 0 0 8 , (d) It does not exist, , 3 2 0, 2 4 0, (a), We, have,, B, =, Ê, , , 1 1 0 , Co-factor of B,, B11 = 0, B12 = 0, B13 = − 2, B21 = 0, B22 = 0, B23 = − 1, , 10., , B13 ′, B23 , , B33 , 0, 0, , − 2, , 0, 0 0, , − 1 8 , 0, , What are the roots of the equation, | x 2 − x − 6| = x + 2?, (b) 0, 2, 4, (d) −2, 2, 4, , Ê (d) We have,, , | x 2 − x − 6| = x + 2, , ⇒, | ( x − 3) ( x − 2 )| = x + 2, Case I x < 2, x2 − x − 6 = x + 2, x2 − 2 x − 8 = 0, x2 − 4 x + 2 x − 8 = 0 , , , x ( x − 4) + 2 ( x − 4) = 0, ( x − 4) ( x + 2 ) = 0, x = − 2 but x ≠ 4 [Q x < 2], Case II 2 ≤ x < 3, x2 − x − 6 = − ( x + 2 ), , Ê (d) Let X = {1,2, 3, 4}, , 7., , B32, , (a) −2, 1, 4, (c) 0, 1, 4, , (b) Only 2 and 3, (d) 1, 2 and 3, , A = {1, 2}, B = {2, 3, 4}, C = {1, 2, 3}, A⊂C, A ∩ B = {2}, C ∩ B = {2, 3}, Clearly, ( A ∩ B) ⊂ (C ∩ B), A ∪ B = {1, 2, 3, 4}, (C ∪ B) = {1, 2, 3, 4}, ( A ∪ B) ⊂ (C ∩ B), Hence, Statement 1 is correct., 2.( A ∩ B) ⊂ (C ∩ B) for all sets B ⇒ A ⊂ C, Hence, Statement 2 is also correct., 3. ( A ∪ B) ⊂ (C ∪ B) for all sets, B⇒A⊂ C, Hence, Statement 3 is also correct., , B22, , 0 0 − 2 ′, = 0 0 −1 =, , , 0 0 8 , , Let X be a non-empty set and let, A , B, C be subsets of X . Consider the, following statements., 1. A ⊂ C ⇒ ( A ∩ B ) ⊂ (C ∩ B ),, ( A ∪ B ) ⊂ (C ∪ B ), 2. ( A ∩ B ) ⊂ (C ∩ B ) for all sets, B ⇒A ⊂C, 3. ( A ∪ B ) ⊂ (C ∪ B ) for all sets, B ⇒A ⊂C, Which of the above statements are, correct?, (a) Only 1 and 2, (c) Only 1 and 3, , B12, , x − 3i 1 , If y, i = 6 + 11i , then what, 1, , , 0 2i −i , are the values of x and y, respectively?, (a) −3, 4, (c) 3, − 4, , (b) 3, 4, (d) −3, −4, x − 3i 1, i = 6 + 11 i, Ê (a) We have, y 1, 0 2i, −i, ⇒ x ( − i + 2 ) − y ( − 3 − 2 i ) = 6 + 11 i, ⇒ 2 x + 3 y + ( − x + 2 y) i = 6 + 11 i, On equating real and imaginary parts, on, both sides,, we get 2 x + 3 y = 6, ...(i), and − x + 2 y = 11, ...(ii), On solving Eqs. (i) and (ii), we get, x = − 3 and y = 4, , 11. The common roots of the equations, z 3 + 2z 2 + 2z + 1 = 0, and z 2017 + z 2018 + 1 = 0 are, (a) −1, ω, (c) −1, ω 2, , x −4=0, x=±2, x = 2 but x ≠ − 2 [Q x ∈ (2, 3)], Case III x ≥ 3, x2 − x − 6 = x + 2, x2 − 2 x − 8 = 0, ( x + 2 ) ( x − 4) = 0, x = 4 but x ≠ − 2 [Q x ≥ 3], x = − 2, 2, 4, , ∴, , 9., , 0 1, If A = , , then the matrix A is, 1 0, a/an, (a) singular matrix, (b) involutory matrix, (c) nilpotent matrix, (d) idempotent matrix, 0 1, Ê (b) We have, A = 1 0 , , , | A| = − 1, Since,| A | ≠ 0, Hence, A is not singular., 0 1 0 1, A2 = A ⋅ A = , ⋅, , 1 0 1 0, 1 0, =, , 0 1, A2 = I, Hence, A is involutory matrix., , (b) 1,ω 2, (d) ω, ω 2, , 3, 2, Ê (d) We have, z + 2 z + 2 z + 1 = 0, , ( z + 1) ( z2 + z + 1) = 0, , x2 − x − 6 + x + 2 = 0, 2, , Solved Paper 2019 (I), , ⇒, , z + 1 = 0 or z2 + z + 1 = 0, , z = −1, −1 ± 1 − 4, or, z=, 2, −1 + i 3 −1 − i 3, =, ,, = ω, ω 2, 2, 2, Now, z2017 + z2018 + 1 = 0, Put z = − 1,, LHS = ( −1)2017 + ( −1)2018 + 1, = − 1+ 1+ 1, = 1 ≠ 0 (RHS), ∴, z = − 1 is not a root of equation., Put z = ω,, LHS = (ω )2017 + (ω )2018 + 1, = (ω 3 )672.ω + (ω 3 )672.ω 2 + 1, = ω + ω2 + 1, , [Q ω 3 = 1], [Q1 + ω + ω 2 = 0], , = 0 = RHS, ∴ z = ω is a root of equation., put z = ω 2,, LHS = (ω 2 )2017 + (ω 2 )2018 + 1, = ω 4034 + ω 4036 + 1, = (ω 3 )1344 .ω 2 + (ω 3 )1345.ω + 1, = ω 2 + ω + 1 = 0 RHS, ∴, , z = ω 2 is a root of equation., , Hence, ω, ω 2 are the common roots of, these equations.
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NDA/NA, , 19, , Solved Paper 2019 (I), , 12. If C (20, n + 2) = C (20, n − 2), then, what is n equal to, (a) 8, (c) 12, , (b) 10, (d) 16, , Ê (b) We have, C (20, n + 2 ) = C (20, n − 2 ), Cn + 2 =, , ⇒, , 20, , ⇒, , 20, , Cn − 2, , n + 2 + n − 2 = 20, [Q nC x = nC y ⇒ x + y = n], , ∴, , n = 10, , Directions (Q. Nos. 16 and 17) Read, the information carefully and answer, the given questions., In a school, all the students play atleast one, of three indoor games– chess, carrom and, table tennis. 60 play chess, 50 play table, tennis, 48 play carrom, 12 play chess and, carrom, 15 play carrom and table tennis, 20, play table tennis and chess., 16. What can be the minimum number of, students in the school?, , 13. There are 10 points in a plane. No, , (a) 123, , three of these points are in a straight, line. What is the total number of, straight lines which can be formed, by joining the points?, , Ê (b) Let, , (a) 90, (c) 40, , (b) 45, (d) 30, , Ê (b) Given, 10 points in a plane where no, , three of these points are in straight line., Total number of straight line formed from, 10 points is, 10 × 9, 10 !, 10, C2 =, =, = 45, 2 ! 8!, 2, , equation px 2 + qx + r = 0, (where p , q , r , all are positive) has, distinct real roots a and b. Which one, of the following is correct?, , 14. The, , (a) a >, (b) a <, (c) a >, (d) a <, , 0, b >, 0, b <, 0, b <, 0, b >, , Ê (b) Given,, , 0, 0, 0, 0, px 2 + qx + r = 0,, , where, , Now,, , r> 0, ab > 0, a > 0, b > 0, a < 0, b < 0, −q, < 0 q, p > 0, p, , 17. What can be the maximum number of, students in the school?, (b) 123, (d) 135, , Ê (b) For maximum number of students in, school n( A ∩ B ∩ C ) must be 12., ∴, n( A ∪ B ∪ C ), = 60 + 50 + 48 − 20 − 15 − 12 + 12, = 123, , 18. If A is an identity matrix of order 3,, then its inverse ( A −1 ), , ...(i), ...(ii), , ∴, , a+ b< 0, ...(iii), a < 0, b < 0, From Eqs. (i), (ii) and (iii), we get, ∴, a < 0 and b < 0, , 15. If A = { λ , { λ ,µ }}, then the power set, of A is, (a) { φ, { φ}, { λ}, { λ,µ }}, (b) {φ, { λ},{ λ,µ },{ λ,{ λ,µ }}}, (c) { φ, { λ},{ λ,µ },{ λ,{ λ,µ }}}, (d) {{ λ},{ λ,µ },{ λ,{ λ ,µ }}}, , Ê (b) We have, A = { λ, { λ, µ }}, , (d) 63, , A = Student play chess, B = Student play table tennis, C = Student play carrom, Given, n( A ) = 60, n( B) = 50, n(C ) = 48, n( A ∩ B) = 20, n( B ∩ C ) = 15, n( A ∩ C ) = 12, For minimum number of students in, school, n ( A ∩ B ∩ C ) must be zero., ∴n( A ∪ B ∪ C ) = n( A ) + n( B) + n(C ), − n( A ∩ B) − n( B ∩ C ), − n( A ∩ C ) + n( A ∩ B ∩ C ), = 60 + 50 + 48 − 20 − 15 − 12 + 0 = 111, , (a) 111, (c) 125, , p, q , r > 0 and a and b are distinct roots., −q, and ab = r, a+ b=, ∴, p, Now,, ∴, ⇒, or, , (b) 111 (c) 95, , P( A ) = { φ, { λ}, {{ λ, µ }}, { λ, { λ, µ }}}, , (a) is equal to null matrix, (b) is equal to A, (c) is equal to 3 A (d) does not exist, 1 0 0, 0 1 0, (b), Given,, A, =, Ê, , , 0 0 1 , 1 0 0, ∴, A −1 = 0 1 0 = A, , , 0 0 1 , , 19. A is a square matrix of order 3 such, that its determinant is 4. What is the, determinant of its transpose?, (a) 64, (c) 32, , (b) 36, (d) 4, , Ê (d) Given, | A | = 4, ∴, , | A′ | = 4 [Q | A | = | A ′ |], , 20. From 6 programmers and 4 typists,, an office wants to recruit 5 people., What is the number of ways this can, be done so as to recruit atleast one, typist?, (a) 209, (c) 246, , (b) 210, (d) 242, , Ê (c) We have,, , 6 programmers and 4 typists, Number of ways of 5 recruit people such, that atleast one typist, = 4C1 6C 4 + 4C 2 6C 3 + 4C 3 6C 2, + 4C 4 6C1, , = 4 × 15 + 6 × 20 + 4 × 15 + 1 × 6, = 60 + 120 + 60 + 6 = 246, , 21. What is the number of terms in the, expansion of [(2x − 3y )2 (2x + 3y )2 ]2 ?, (a) 4, (c) 8, , (b) 5, (d) 16, , 2, 2 2, Ê (b) Given, [(2 x − 3 y) (2 x + 3 y) ], , = [4x 2 − 9 y 2 ]4, ∴ Total number of terms = 4 + 1 = 5, , 22. In the expansion of (1 +ax )n , the first, three terms are respectively 1, 12x, and 64 x 2 . What is n equal to?, (a) 6, (c) 10, , (b) 9, (d) 12, , first three terms of expansion, Ê (b) Given,, n, 2, (1 + ax ) is 1, 12 x, 64 x ,, , Now,, , n( n − 1) 2 2, a x +K, 2, On equating first three terms, we get, n ( n − 1) 2, na = 12 and, a = 64, 2, On putting the value of a in, n ( n − 1) 2, a = 64, we get, 2, 2, n ( n − 1) 12 , = 64, n, 2, 144 ( n − 1), ⇒, = 64, 2n, ∴, n=9, (1 + ax )n = 1 + nax +, , 23. The numbers 1, 5 and 25 can be three, terms, (not, consecutive)of, , necessarily, , (a) only one AP, (b) more than one but finite numbers of, APs, (c) infinite number of APs, (d) finite number of GPs, , Ê (d) We have, 1, 5, 25 be three terms., Clearly, 1, 5, 25 are finite number of GPs.
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20, , NDA/NA, , 24. The sum of ( p + q )th and ( p − q )th, terms of an AP is equal to, (a) (2 p) th term, (b) (2q ) th term, (c) twice the p th term, (d) twice the q th term, , Ê (c) Let a is first term and d is common, difference of AP., ap + q = a + ( p + q − 1) d, and ap − q = a + ( p − q − 1) d, Sum of ( p + q ) th and ( p − q ) th terms, = ap + q + ap − q = 2 a + (2 p − 2 ) d, = 2 ( a + ( p − 1) d ) = 2 ap, = twice of p th term, , 25. If A is a square matrix of order n > 1,, then which one of the following is, correct?, (a) det ( − A ) = det A, (b) det ( − A ) = ( −1)n det A, (c) det ( − A ) = − det A, (d) det ( − A ) = n det A, , Ê Sol. (b) We know that if A is a square, matrix of order n > 1, then, det( − A ) = ( −1)n det A, , 2 3, For example If A = , ,, 4 5, −2 −3, then − A = , , −4 −5, 2 3, = 10 − 12 = − 2, …(i), ∴det A = , 4 5, −2 −3, = 10 − 12 = − 2, and det( − A ) = , −4 −5, = ( −1)2( −2 ) [Q here n = 2], = ( −1)2 det A [from Eq. (i)], 1 2 3 , if A = 3 1 0 , , , 4 3 −2 , −1 −2 −3, Then, − A = −3 −1 0 , , , −4 −3 2 , 1 2 3, ∴ det A = 3 1 0, 4 3 −2, = 1( −2 − 0) − 2( −6 − 0) +, = − 2 + 12 + 15 = 25, −1 −2, and det ( − A ) = −3 −1, , −4 −3, , 3( 9 − 4), −3, 0, , 2, , = − 1( −2 − 0) + 2( −6 − 0) − 3( 9 − 4), = 2 − 12 − 15 = − 25, [here n = 3], = ( −1)325, = ( −1)3det A [from Eq. (i)], , 26. What is the least value of, 25 cosec2 x + 36sec2 x ?, (a) 1, (b) 11, (c) 120, (d) 121, , Directions (Q. Nos. 29 and 30) Read, the information carefully and answer, the given questions., A complex number is given by, z=, , 2, 2, Ê (d) Given, 25 cosec x + 36 sec x, , = 25 (1 + cot 2 x ) + 36 (1 + tan2 x ), , Solved Paper 2019 (I), , 1 + 2i, ., 1 − (1 − i )2, , 29. What is the modulus of z?, (a) 4, , (b) 2, , (c) 1, , = 25 + 25 cot x + 36 + 36 tan x, 2, , (d), , 2, , = 25 + 36 + 25 cot 2 x + 36 tan2 x, , 1 − (1 − i )2, 1 + 2i, z=, 1 − (1 − 1 − 2 i ), 1 + 2i, =, =1, 1 + 2i, , = 61 + ( 5 cot x − 6 tan x )2 + 2 × 5 × 6, ≥ 61 + 60 = 121, , [Qminimum value of, ( 5 cot x − 6 tan x )2 = 0], , ∴ Minimum value of, 25 cosec 2 x + 36 sec 2 x = 121, , Directions (Q. Nos. 27 and 28) Read, the information carefully and answer, the given questions., Let A and B be 3 × 3 matrices with det A = 4, and det B = 3 ., , ∴, , | z| = 1, , 30. What is the principal argument of z?, (a) 0, , (b), , π, 4, , (c), , π, 2, , (d) π, , lm( z) , , Re ( z), , −1, Ê (a) arg ( z) = tan , , 0, = tan−1 = tan−1 0 = 0, 1, , 27. What is det (2AB ) equal to?, (a) 96, (b) 72, (c) 48, (d) 36, , 1 + 2i, , Ê (c) We have, z =, , 1, 2, , 31. What is the value of, , Ê (a) A and B be ( 3 × 3) matrices with, det A = 4 and det B = 3, We know that,, det ( KAB) = K ndet( A ) × det( B), , where, n is the order or A and B, K is a real, number., ∴det(2 AB) = (2 )3detA × detB, [Q n = 3 and k = 2], =8×4×3, = 96, , 28. What is det (3AB −1 ) equal to?, (a) 12, (c) 36, , (b) 18, (d) 48, , Ê (c) A and B be ( 3 × 3) matrices with, det A = 4 and det B = 3, , (a) −2, , (b) −1, , (c) 2, , (d) 1, , Ê (a) We have,, , sin 34° cos 236° − sin 56° sin 124°, cos 28° cos 88° + cos 178° sin 208°, sin 34° cos (180° + 56° ), − sin 56° sin ( 90° + 34° ), cos 28° cos 88° + cos ( 90° + 88° ), sin (180° + 28° ), − sin 34° cos 56° − sin 56° cos 34°, =, cos 28° cos 88° + sin 88° sin 28°, =, , =, =, , − sin ( 56° + 34° ) − sin 90°, =, cos ( 88° − 28° ), cos 60°, −1, = −2, 1, 2, , 32. tan54° can be expressed as, , We know that,, 1, ,, det( KAB ) = K det( A ) ×, det( B), −1, , sin 34 ° cos 236° − sin 56° sin 124 °, ?, cos 28° cos 88° + cos 178° sin 208°, , n, , where n is the order of A and B, K is a, real number], 1, −1, 3, ∴det ( 3 AB ) = ( 3) det( A ) ×, det B, 1, = 27 × 4 ×, 3, = 36, , sin 9° + cos 9°, sin 9° − cos 9°, (b), sin 9° − cos 9°, sin 9° + cos 9°, cos 9° + sin 9°, sin 36°, (d), (c), cos 9° − sin 9°, cos 36°, (a), , Ê (c) We have, tan 54° = tan ( 45° + 9° ), =, , tan 45° + tan 9°, 1 + tan 9°, =, 1 − tan 45° tan 9° 1 − tan 9°, =, , cos 9° + sin 9°, cos 9° − sin 9°
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22, , NDA/NA, 3, 4, , 42. If cos A = , then what is the value, A 3A , of sin sin ?, 2 2 , (a), , 5, 8, , (b), , 5, 16, , (c), , 5, 24, , (d), , 7, 32, , 3, Ê (b) Given, cos A =, , Ê (b) Given, sin 2θ = cos 3θ, 2 sin θ cos θ = 4 cos 3 θ − 3 cos θ, , ⇒, , 2 sin θ = 4 cos 2 θ − 3, , ⇒, , [Qcosθ ≠ 0], 2 sin θ = 4(1 − sin2 θ) − 3, , ⇒, , 4 sin2 θ + 2 sinθ − 1 = 0, sinθ =, , ⇒, , A, 3A, sin, 2, 2, 1, 3A, A, = 2 sin sin, , 2 , 2, 2 , , ⇒, , sin θ =, , ∴, , sin θ =, , Now, sin, , =, , 1, 2, , 3A , , A 3A, A, cos 2 − 2 − cos 2 + 2 , , , , 1, = [cos A − cos 2 A ], 2, 1, = (cos A − 2 cos 2 A + 1), 2, 1 3, 9, = − 2 ×, + 1, 2 4, 16, , =, , 1 3 9, 1 6 − 9 + 8, 5, − + 1 = , = 16, 2 4 8, 8, 2, , , 43. What is the value of tan 75° + cot 75° ?, (a) 2, (c) 2 3, , (b) 4, (d) 4 3, , Ê (b) We have, tan 75° + cot 75°, = tan ( 90° − 15° ) + cot ( 90° − 15° ), = cot 15° + tan 15°, cos 15°, sin 15°, =, +, sin 15°, cos 15°, =, , cos 2 15° + sin2 15°, sin15° cos 15°, , 44. What is the value of, , cos 46° cos 47 ° cos 48° cos 49 °, cos 50° .....cos 135°?, (b) 0, (d) Greater than 1, , Ê (b) We have,, , cos 46° cos 47 ° cos 48°, cos 49° cos 50° … cos 90° … cos 135°, [Qcos 90° = 0], =0, , π, 45. If sin 2θ = cos 3θ, where 0 < θ < ,, 2, then what is sinθ equal to ?, (a), (c), , , π , Q θ ∈ 0, 2 , , , , , x + px + q = 0 are tan19° and, tan26°, then which one of the, following is correct ?, 2, , (a) q − p = 1, (c) p + q = 2, , (b) p − q = 1, (d) p + q = 3, , Ê (a) Given, tan 19° and tan 26° are roots of, x 2 + px + q = 0., ∴tan 19° + tan 26° = − p, tan 19° ⋅ tan 26° = q, tan 19° + tan 26°, tan (19° + 26° ) =, 1 − tan 19° tan 26°, ⇒, , tan 45° =, , ⇒, ∴, , −p, −p, ⇒1 =, 1− q, 1− q, , 1− q = − p, q − p=1, , 47. What is the fourth term of an AP of n, (b) 8, (d) 20, , Ê (b) Given,, , 2, 2, =, =, =4, 1, sin 30°, 2, , 5 +1, 4, 5 +1, 16, , 5, , terms whose sum is n (n + 1)?, , 2, 2 sin 15° cos 15°, , (a) −1, (c) 1, , ⇒, , 2 ×4, , 5 −1, 4, , (b), (d), , 5 −1, 4, 5 −1, 16, , Sum of n terms of an AP, i.e., S n = n ( n + 1), a4 = S 4 − S 3 [Q an = S n − S n−1], a4 = 4( 4 + 1) − 3 ( 3 + 1), a4 = 20 − 12 = 8, ∴ Fourth term of an AP = 8, , 48. What is − sec 2 α sec 2 β equal to?, (a) 0, (c) 2, , (b) 1, (d) 4, , Ê (a) We have,, , (1 + tan α tan β )2 + (tanα − tanβ )2, − sec 2α sec 2β, , = 1 + tan α tan2 β + 2 tanα tanβ, 2, , + tan2 α + tan2 β, − 2 tanα tanβ − sec 2α sec 2β, = 1 + tan2 α tan2 β + tan2 α + tan2 β, − sec 2α sec 2 β, = (1 + tan2 α ) (1 + tan2 β ) − sec 2α sec 2β, = sec 2αsec 2β − sec 2αsec 2β = 0, , (b) p = q, (d) p + q = 0, , q = (cosec θ + cot θ)−1, , 4 + 16, , − 2 ± 2 5 − 1±, =, 2 ×4, 4, , (a) pq =1, (c) p + q = 1, , Ê (b) Given, p = cosec θ − cot θ, , 46. If the roots of the equation, , (a) 6, (c) 12, , 1, =, sin 15° cos 15°, =, , −2 ±, , and, p = cosecθ − cot θ, q = (cosecθ + cot θ )−1, then which, one of the following is correct?, , 49. If, , ⇒, , 4, , Solved Paper 2019 (I), , , cosec θ − cot θ , 1, q =, , , cosec θ + cot θ cosec θ − cot θ , ⇒, ∴, , q = cosec θ − cot θ, q = p, , 50. If the angles of a triangle ABC are in, the ratio 1 : 2 : 3, then the, corresponding sides are in the ratio, (a) 1 : 2 : 3, (c) 1 : 3 : 2, , (b) 3 : 2 : 1, (d) 1 : 3 : 2, , Ê (c) We have, angle of triangle ABC are in, the ratio 1 : 2 : 3, ∴, x + 2 x + 3 x = 180°, ⇒, x = 30°, ∴ Angles of triangle are 30° , 60° , 90°., We know that, sine rule, a, b, c, =, =, sin A sin B sinC, a, b, c, =, =, sin 30° sin 60° sin 90°, a, b, c, =, =, ⇒, 1, 1, 3, 2, 2, ∴ a : b : c = 1: 3 : 2, , 51. Consider the following statements, 1. For an equation of a line,, x cos θ + y sin θ = p , in normal, form, the length of the, perpendicular from the point, to, the, line, is, ( α, β ), | α cos θ + β sin θ + p |., The length of the perpendicular, from the point (α,β ) to the line, aα + bβ − ab, x y, + = 1 is, ., 2, 2, a b, a +b , Which of the above statements, is/are correct?, (a) 1 only, (c) Both 1 and 2, , (b) 2 only, (d) Neither 1 nor 2, , Ê (d) 1. Equation of line, , x cos θ + y sin θ = p, Perpendicular distance from (α,β ) to the, given line is, α cos θ + β sin θ − p, cos 2 θ + sin2 θ, = (α cos θ + β sinθ − p), Hence, statement 1 is incorrect.
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NDA/NA, , 2. Length of the perpendicular from the, x, y, point (α,β ) to the line +, = 1 is, a, b, α, , β, − 1 , +, αb + aβ − ab, a, b, , , = , a2 + b 2 , 1 + 1 , b 2, a2, Hence, statement 2 is incorrect., ∴ Neither 1 nor 2., , 52. A circle is drawn on the chord of a, circle x 2 + y 2 = a 2 as diameter. The, chord lies on the line x + y = a., What is the equation of the circle?, (a) x 2, (b) x 2, (c) x 2, (d) x 2, , 23, , Solved Paper 2019 (I), , + y2, + y2, + y2, + y2, , − ax − ay + a2 = 0, − ax − ay = 0, + ax + ay = 0, + ax + ay − 2 a2 = 0, , Ê (b)2, , Given, equation, x + y 2 = a 2., , of, , circle, , 54. The equation 2x 2 − 3y 2 − 6 = 0, represents, (a) a circle, (b) a parabola, (c) an ellipse, (d) a hyperbola, 2, 2, Ê (d) Given, 2 x − 3 y − 6 = 0, , ⇒, , 2 x 2 − 3 y2 = 6, , x2, y2, −, =1, 3, 2, Which represents the equation of a, hyperbola., ⇒, , 55. The two parabolas y 2 = 4ax and, x = 4ay intersect, 2, , is, , x + y = a is chord of a circle., (0, a), , (a) at two points on the line y = x, (b) only at the origin, (c) at three points one of which lies on, y+ x =0, (d) only at (4a, 4a), 2, Ê (a) Given, y = 4ax, , and, , x 2 = 4ay, , The graph of given curve is clearly from, graph the given curve is intersect at two, points on the line y = x, y=x, , Ê (c) The lines 3 y + 4 x = 1, y = x + 5 and, 5 y + bx = 3 are concurrent., 3 4 − 1, 1 − 1 − 5 = 0, ∴, , , 5 b − 3, , ⇒ 3 ( 3 + 5b ) − 4( − 3 + 25) −1( b + 5) = 0, ⇒, ⇒, ∴, , 9 + 15b + 12 − 100 − b − 5 = 0, 14b = 84, b=6, , 58. What is the equation of the straight, line which is perpendicular to y = x, and passes through (3, 2)?, , (a) x, (b) x, (c) x, (d) x, , −, +, +, −, , y=5, y=5, y=1, y=1, , Ê (b) Equation of line perpendicular to y = x, is x + y = λ., Since, this line is passes through (3, 2), ∴, 3 + 2 = λ⇒λ = 5, Hence, equation of required line is, x + y = 5., , 59. The straight lines x + y − 4 = 0,, 3x + y − 4 = 0 and x + 3y − 4 = 0, form a triangle, which is, , ,, (a, 0), , (a) isosceles, (c) equilateral, , (b) right angled, (d) scalene, , Ê (a) Given, equation of line, , 53. The sum of the focal distances of a, point on an ellipse is constant and, equal to, (a) length of minor axis, (b) length of major axis, (c) length of latusrectum, (d) sum of the lengths of semi major, and semi minor axes, , Ê (b) The sum of the focal distance of a, , point on a ellipse is constant and equal, to the length of major axis., We know that, PS + PS ′ = 2 a, P, , x+ y=4, 3x + y = 4, x + 3y = 4, , 56. The points (1, 3) and (5, 1) are two, opposite vertices of a rectangle. The, other two vertices lie on the line, y = 2x + c . What is the value of c?, (a) 2, (c) 4, , Ê (d) The points (1, 3) and (5, 1) are two, , opposite vertex of rectangle. The other, two vertices lie on the line y = 2 x + c., ∴ The mid point of vertices lie on the line, 1 + 5 3 + 1, i.e. , ,, ≡ ( 3, 2 ) lie on the line, 2, 2 , ∴, ⇒, , S′, , y = 2x + c, 2 = 2( 3) + c, c=−4, , 5y + bx = 3 are concurrent, then, what is the value of b?, , (a) 1, (c) 6, , 3x+y=4, , (b) −2, (d) −4, , 57. If the lines 3y + 4 x = 1, y = x + 5 and, S, , ...(i), ...(ii), ...(iii), , A (0, 4), , x+, y=, 4, , ∴End points of diameter of required circle, is (a,0) and ( 0, a )., ∴Equation of circle is, x ( x − a) + y ( y − a) = 0, ⇒ x 2 + y 2 − ax − ay = 0, , (b) 3, 1, (d), 2, , B (4, 0), , x+3y=4, , C (1, 1), , On solving Eqs. (i) and (ii), we get, x = 0, y = 4, A = ( 0, 4), On solving Eqs. (i) and (iii), we get, x = 4, y = 0, B = ( 4, 0), On solving Eqs. (ii) and (iii), we get, x = 1, y = 1, C = (1, 1), Clearly, AC = BC, ∴Triangle is an isosceles., , 60. The circle x 2 + y 2 + 4 x − 7y + 12 = 0,, cuts an intercept on Y -axis equal to, (a) 1, (c) 4, , (b) 3, (d) 7
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24, , NDA/NA, , 2, 2, Ê (a) Given, x + y + 4 x − 7 y + 12 = 0, , For intercept on Y-axis put x = 0, we get, y 2 − 7 y + 12 = 0, ( y − 4) ( y − 3) = 0, y = 3, y = 4, Length of intercept on Y-axis, = | y2 − y1 | = | 3 − 4 | = 1, , 61. The centroid of the triangle with, , vertices A(2, − 3, 3), B(5, − 3, − 4 ) and, C(2, − 3, − 2) is the point, , (a) (−3, 3, − 1), (b) (3, − 3, − 1), (c) (3, 1, − 3), (d) (−3, − 1, − 3), , 9 −9 −3 , = ,, ,, = ( 3, − 3, − 1), 3 3 3 , , 10 x − 2 y + 2 z = 28, , and B are 3$i − 2$j + k$ and 2$i + 4 $j − 3k$, respectively, then what is the length, of AB ?, , (b) 29, (d) 53, (d), We, have, ,, =, 3$i − 2 $j + k$, OA, Ê, OB = 2 $i + 4$j − 3k$, (a) 14, (c) 43, , ∴| AB | = | OB − OA |, = | (2 $i + 4$j − 3k$ ) − ( 3$i − 2 $j + k$ )|, = | ( − $i + 6$j − 4k$ |, , 1, 4, 3, (c), 2, , = 1 + 36 + 16 =, , 1, 2, 7, (d), 4, , (b), , (a), , (b) 2, (d) 3, , Ê (c) Given, equation of sphere, , x 2 + y 2 + z2 − 6 x + 8 y − 10 z + 1 = 0, , On comparing with, x 2 + y 2 + z2 + 2u x + 2 vy + 2 wz, , hypotenuse AC = p , then what is, AB ⋅ AC + BC ⋅ BA + CA ⋅ CB equal, to?, , ...(i), , 8 x − 4 y + 8 z + 21 = 0, 21, 4x − 2 y + 4z +, = 0 ...(ii), ⇒, 2, Distance between parallel planes (i), and (ii), 21 − 9 , , 2, =, , 49 = 7, , 63. The equation of the plane passing, through the intersection of the, planes 2x + y + 2z = 9,, 4 x − 5y − 4z = 1 and the point, (3, 2, 1) is, , Z -axis?, (b) <1, 0, 0>, (d) <0, 0, 1>, , Ê (d) Direction cosines of Z-axis are, < cos 90° , cos 90° , cos 0°> < 0, 0,1 >, , then what is ( b − a ) ⋅ (3a + b) equal, to ?, (b) −106, (d) −53, $, $, $, Ê (b) We have, a = i − 2 j + 5k, $, $, $, b = 2 i + j − 3k, (a) 106, (c) 53, , Ê (a) Equation of the plane passing through, the intersection of plane, 2 x + y + 2 z = 9, 4 x − 5 y − 4 z = 1 is, (2 x + y + 2 z − 9), ...(i), , Since, plane (i) passes through the point, (3, 2, 1), ∴ (2 × 3 + 2 + 2 × 1 − 9), + λ ( 4 × 3 − 5 × 2 − 4 × 1 − 1) = 0, , (b) 2 p2, (d) p, , In right angled ∆ABC, we have, ∠ABC = 90°, C, , Ê (a), , p, , 2, , 66. If a = $i − 2$j + 5k$ and b = 2$i + $j − 3k$ ,, , − 2 y + 2 z = 28, + 2 y + 2 z = 28, + 2 y − 2 z = 28, − 2 y − 2 z = 24, , + λ ( 4 x − 5 y − 4 z − 1) = 0, , 2, , 3, 3, 1, 2, 2, =, =, 16 + 4 + 16 6 4, , (a) <1, 1, 1>, (c) <0, 1, 0>, , = ( − 3)2 + ( 4)2 + ( − 5)2 − 1, , (a) p2, p2, (c), 2, , ( 4) + ( − 2 ) + ( 4), 2, , 65. What are the direction cosines of, , ∴Radius of sphere = u 2 + v 2 + w 2 − d, , 53, , 68. If in a right angled triangle ABC,, , Ê (a) Given equation of planes, , =, , + d = 0, we get, 2 u = − 6, 2 v = 8, 2 w = − 10, d = 1, ⇒ u = − 3, v = 4, w = − 5, d = 1, , = ( − 1)2 + ( 6)2 + ( − 4)2, , and, , and, , x 2 + y 2 + z 2 − 6x + 8y − 10z + 1 = 0?, , 9 + 16 + 25 − 1 =, , planes 4 x − 2y + 4z + 9 = 0, 8x − 4y + 8z + 21 = 0, , 4x − 2 y + 4z + 9 = 0, , 62. What is the radius of the sphere, , (a) 10x, (b) 10x, (c) 10x, (d) 10x, , ⇒ 10 x − 2 y + 2 z − 28 = 0, , 67. If the position vectors of points A, , 64. The distance between the parallel, , A(2, − 3, 3), B( 5, − 3, − 4), and C(2, − 3, − 2 ), Centroid, of, ∴, ∆ABC, 2 + 5 + 2 − 3 − 3 − 3 3 − 4 − 2, =, ,, ,, , , , 3, 3, 3, , =, , 1 + λ ( − 3) = 0, 1, λ=, ⇒, 3, 1, On putting λ = in Eq. (i), we get, 3, (2 x + y + 2 z − 9), 1, + ( 4 x − 5 y − 4 z − 1) = 0, 3, ⇒ 6 x + 3 y + 6 z − 27 + 4 x − 5 y − 4 z, − 1= 0, ∴, , Ê (b) Given vertices of triangle ABC are, , (a) 5, (c) 7, , ⇒, , Solved Paper 2019 (I), , ∴b − a = (2 $i + $j − 3k$ ) − ( $i − 2 $j + 5k$ ), = $i + 3$j − 8k$, and 3 a + b = 3( $i − 2 $j + 5k$ ), + (2 $i + $j − 3k$ ), $, $, = 5 i − 5 j + 12k$, ∴(b − a ) ⋅ (3a + b ), = ( $i + 3$j − 8k$ ) ⋅ ( 5$i − 5$j + 12k$ ), = 5 − 15 − 96 = − 106, , θ, B, A, Let, ∠BAC = θ, Then, ∠ACB = ( 90° − θ), ∴AB ⋅ AC + BC ⋅ BA + CA ⋅ CB, = | AB || AC | cos θ + | BC || BA |, cos 90°+ | CA || CB | cos ( 90° − θ), = | AB || AB | + 0 + | CB || CB |, = | AB |2 + | CB |2, , = | AC |2 = p2, , 69. The sine of the angle between, vectors, a = 2 $i − 6$j − 3k$ and b = 4$i + 3$j − k$ is, , 1, 26, 5, (c), 26, , 5, 26, 1, (d), 26, , (a), , (b), , Ê (b) Let θ be the angle between vectors a, and b, , ∴, , cos θ =, , a ⋅b, | a || b |, , Since,, a ⋅ b = (2 $i − 6$j − 3k$ ) ⋅ ( 4$i + 3$j − k$ ), = 8 − 18 + 3 = − 7, | a | = 2 2 + ( − 6)2 + ( − 3)2, =, , 49 = 7
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NDA/NA, , 25, , Solved Paper 2019 (I), , | b| =, , 4 2 + 3 2 + ( − 1)2 = 26, , ∴, , −7, 1, cos θ =, =−, 7 × 26, 26, , ∴, , sin θ = 1 − cos 2 θ =, , 73. A function f defined by, f ( x ) = ln( x 2 + 1 − x ) is, , 5, 26, , 70. What is the value of λ for which the, , vectors, and, 3$i + 4 $j − k$, −2i$ + λ$j + 10k$ are perpendicular?, , (a) 1, (c) 3, , (b) 2, (d) 4, , Ê (d) Since, given vectors are, , perpendicular., ∴ ( 3 $i + 4$j − k$ ) ⋅ ( − 2 $i + λ $j + 10 k$ ) = 0, , ⇒ − 6 + 4λ − 10 = 0, ⇒, 4λ − 16 = 0, ∴, λ=4, , is the derivative, sec 2 (tan −1 x ) with respect, to x?, , of, , f( − x ) = log ( x 2 + 1 + x ), ( x 2 + 1 + x) ( x 2 + 1 − x), , = log , , , x2 + 1 − x, , , x2 + 1 − x2 , , = log , x2 + 1 − x , , , , On differentiating both sides w.r.t x, we get, dy d, sec 2 (tan−1 x ), =, dx dx, = 2 sec (tan−1 x ) ⋅ sec (tan−1 x ), d, (tan−1 x ), dx, 1, , tan (tan−1 x ), = 2 sec 2(tan−1 x ) ⋅ x ⋅, , 1+ x, , = 2 (1 + tan2 (tan−1 x )) ⋅, x, 1 + x2, , , 1, , x 2 + 1 − x , , 2, , x, 1 + x2, , = 2x, , 72. If f ( x ) = log10 (1 + x ), then what is, 4 f ( 4 ) + 5 f (1) − log10 2 equal to?, , (a) 0, (b) 1, (c) 2, (d) 4, , Ê (d) We have, f( x) = log10 (1 + x), ∴ 4f( 4) = 4 log10 (1 + 4) = 4 log10 5, 5f(1) = 5 log10 (1 + 1) = 5 log10 2, ∴ 4f( 4) + 5f(1) − log10 2, = 4 log10 5 + 5 log10 2 − log10 2, = 4 log10 5 + 4 log10 2, = 4 log10 ( 5 × 2 ), = 4 log10 10 = 4 × 1 = 4, , (a) 1, (c) 3, , (b) 2, (d) 4, , Ê (c) We have,, , Perimeter of a circle with radius r, Area of a circle with radius r, 2 πr 2, =, f( r ) =, ⇒, r, πr 2, 2, 2, f(1) = = 2 ⇒ f(2 ) = = 1, ∴, 1, 2, ∴ f(1) + f(2 ) = 2 + 1 = 3, f( r )=, , 77. If f ( x ) = 31 + x , then f ( x ) f (y ) f (z ), is equal to, (a) f( x + y + z), (b) f( x + y + z + 1), (c) f( x + y + z + 2 ) (d) f( x + y + z + 3), 1 +x, Ê (c) We have, f( x) = 3, , = − log ( x + 1 − x ), , Similarly, f( y) = 31 + y, , = − f( x ), , and, , 2, , f( z) = 31 + z, , ∴ f( x ) f( y) f( z) = 31 + x, , (a) x > 10, (b) x > 0 excluding x = 10, (c) x ≥ 10, (d) x > 0 excluding x = 1, , Ê (d) We have, f( x) = log x 10, =, , log 10, 1, =, log x, log x, , ∴ f( x ) is define when x > 0 and x ≠ 1., , 75. lim, , 1 − cos 3 4 x, , x→ 0, , x, , (a) 0, (c) 24, , 2, , is equal to, (b) 12, (d) 36, , Ê (c) xlim, →0, , 1 − cos 3 4 x 0, , 0 form, x2, , On apply L′ Hospital rule we get, − 3 cos 2 ( 4 x ) ( − sin 4 x ) ( 4), lim, x→ 0, 2x, 12 cos 2 4 x sin 4 x 0, , 0 form, x→ 0, 2x, , Again, apply’s L′ Hospital rule, we get, 12 [2 cos ( 4 x ) ( − sin 4 x ) ( 4), sin 4 x + cos 2 4 x (cos 4 x ) ( 4)], x→ 0, 2, lim, , 12 [− 8 cos 4 x sin2 4 x, = lim, , x→ 0, , +1 + y +1 + z, , 1 +2 +x +y +z, , =3, , = f(2 + x + y + z), , = lim, , = 4 (log10 5 × log10 2 ), , perimeter to area of a circle of radius, r . Then, f (1) + f (2) is equal to, , defined by, f ( x ) = log x 10 is, , 2, −1, Ê (a) Let y = sec (tan x), , = 2 (1 + x ) ⋅, , ∴, , 74. The domain of the function f, , (b) x 2 + 1, (d) x 2, , 2, , 2, Ê (b) We have, f( x) = log ( x + 1 − x), , , = log , , , , 71. What, , (a) 2x, (c) x + 1, , (a) an even function, (b) an odd function, (c) both even and odd function, (d) neither even nor odd function, , 76. For r > 0, f (r ) is the ratio of, , + 4 cos 3 4 x ], 2, , = 6 ( − 8 × 0 + 4), = 24, , 78. The number of real roots for the, equation x 2 + 9| x | + 20 = 0 is, (a) zero, (b) one, (c) two, (d) three, 2, Ê (a) Given, x + 9| x | + 20 = 0, , ⇒, , x 2 + 9 x + 20 = 0, , or, , x 2 − 9 x + 20 = 0, , ⇒, , x + 4 x + 5 x + 20 = 0, , or, , 2, , x 2 − 4 x − 5 x + 20 = 0, , ⇒, x ( x + 4) + 5 ( x + 4) = 0, or, x ( x − 4) − 5 ( x − 4) = 0, ⇒, ( x + 4) ( x + 5) = 0, or, ( x − 4) ( x − 5) = 0, ⇒, x = − 4, − 5, or 4, 5, But these values of x does not satisfy the, given equation., Hence, number of real roots of the given, equation is zero., , 79. If f ( x ) = sin(cos x ), then f ′( x ) is, equal to, (a) cos(cos x ), (b) sin( − sin x ), (c) (sin x )cos(cos x ), (d) (− sin x )cos(cos x ), , Ê (d) Given, f( x) = sin (cos x), ⇒ f ′ ( x ) = cos (cos x ) ( − sin x )
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26, , NDA/NA, π, , 80. The domain of the function, f ( x ) = (2 − x )( x − 3) is, (a) ( 0, ∞ ), (c) [2, 3], , 4, , (b) [0, ∞ ), (d) (2, 3), , f( x ) will be define if (2 − x ) ( x − 3) ≥ 0, ⇒, ( x − 2 ) ( x − 3) ≤ 0, ∴, 2 ≤ x≤ 3, , 81. The solution of the differential, equation, dy, = cos(y − x ) + 1 is, dx, , ⇒, sec t dt = dx, On integrating both sides, we get, , ∫ sec t dt = ∫ dx, , log (sec t + tan t ) = x + a, ⇒, , sec t + tan t = e x ⋅ e a, ex, = e −a, sec t + tan t, , ⇒, ⇒, , +a, , e x (sec t − tan t ), = e −a, (sec t + tan t ) (sec t − tan t ), e x (sec t − tan t ), , ⇒, , sec t − tan t, 2, , 2, , = e −a, , −2, , 83. If y = a cos 2x + b sin 2x , then, , (c), , d 2y, 2, , dx, d 2y, dx 2, , + y=0, , (b), , − 4y = 0, , (d), , d 2y, dx 2, d 2y, dx 2, , + 2y = 0, + 4y = 0, , Ê (d) Given, y = a cos 2 x + b sin 2 x ...(i), dy, = − 2 a sin 2 x + 2 b cos 2 x, dx, 2, d y, = − 4a cos 2 x − 4b sin 2 x, dx 2, 2, d y, = − 4 ( a cos 2 x + b sin 2 x ), dx 2, 2, d y, [using Eq. (i)], = − 4y, dx 2, , ⇒, ⇒, ⇒, ⇒, ∴, , d 2y, dx 2, , + 4y = 0, , cast into a half cylinder, (i.e. with a rectangular base and, semicircular ends). If the total, surface area is to be minimum, then, the ratio of the height of the half, cylinder to the diameter of the, semicircular ends is, (b) ( π + 2 ) : π, (d) None of these, , Ê (a) Let r be the radius and h be the height, of the half cylinder,, Then, surface area, S = πrh + πr 2 + 2 rh, , e x (sec t − tan t ) = e − a, , ⇒, , ∴ e x [sec ( y − x ) − tan ( y − x )] = c,, [where, c = e − a], , h, , π, , ∫02 |sin x − cos x |dx is equal to, (b) 2( 2 − 1), (d) 2( 2 + 1), , (a) 0, (c) 2 2, π, , Ê (b) ∫02| sin x − cos x | dx, =, , π, , ∫04 (cos x − sin x) dx, +, , π, , ∫ π2(sin x − cos x) dx, 4, , 85., , π, 2 e sin x, 0, , ∫, , cos xdx is equal to, , (a) e + 1, (c) e + 2, , (b) e − 1, (d) e, I=, , Ê (b) Let, , π, , ∫02 e, , sin x, , cos x dx, , sin x = t, cos x dx = dt, π, When x = , t = 1, 2, x = 0, t = 0, , Let, ⇒, , I=, , ∴, , 1 t, , ∫0 e, , dt = [et ]10, , = e1 − e 0 = e − 1, , 86. If f ( x ) = x − 2 , x ≠ −2, then what is, f, , x +2, ( x ) equal to ?, , −1, , 4( x + 2 ), x −2, x +2, (c), x −2, , x+2, 4( x − 2 ), 2(1 + x ), (d), 1−x, , (b), , Ê (d) Given, f( x) =, ⇒, ⇒, ⇒, ⇒, , x−2, x−2, ⇒ y=, x+2, x+2, , x − 2 = xy + 2 y, x − xy = 2 y + 2, x (1 − y) = 2 y + 2, 2 ( y + 1), x=, 1− y, , ⇒, , f −1( y) =, , 2 ( y + 1), 1− y, , ∴, , f −1( x ) =, , 2 ( x + 1), 1− x, , 87. What is ∫ ln( x 2 )dx equal to?, , ⇒ e x [sec ( y − x ) − tan ( y − x )] = e − a, , 82., , Neglecting − sign as r and h can not be, negative., h, π, ∴, =, 2r, π+2, , (a), , 84. A given quantity of metal is to be, , (a) π :( π + 2 ), (c) 1 : 1, , 2r =, , ⇒, , 1, 1 , , −, −, , , 2, 2 , , = 2 2 − 2 = 2 ( 2 − 1), , (a), , (a) e x [sec( y − x ) − tan( y − x )] = c, (b) e x [sec( y − x ) + tan( y − x )] = c, (c) e x sec( y − x ) tan( y − x ) = c, (d) e x = c sec( y − x ) tan( y − x ), dy, ...(i), Ê (a) Given, = cos ( y − x) + 1, dx, Let, y− x = t, dy, dt, dy, dt, ⇒, − 1=, ⇒, = 1+, dx, dx, dx, dx, dt, From Eq. (i), 1 +, = cos t + 1, dx, dt, = cos t, ⇒, dx, , sec t + tan t = e x, , , + ( − 0 − 1) −, , 2, 4, 2, =, − 1− 1+, =, 2, 2, 2, , − h (π + 2), π, − π, 2r − (π + 2), h, =, =, ⇒, h, 2r, π, π+2, , ⇒, , 1, , 1 , = , +, − ( 0 + 1), , , 2, 2, , , , Ê (c) We have, f( x) = (2 − x) ( x − 3), , ⇒, , π, , = [sin x + cos x ]04 + [− cos x − sin x ]π2, , Solved Paper 2019 (I), , (a) 2 x ln( x ) − 2 x + C, 2, +C, x, (c) 2 x ln( x ) + C, 2 ln( x ), (d), −2x + C, x, (b), , Ê (a) Let, dS, = πh + 2 πr + 2 h, dr, dS, On putting, =0, dr, ( πh + 2 h), 2r = −, ⇒, π, , ∴, , I = ∫ ln ( x 2 ) dx = ∫ 2 ln x dx, = ln x ∫ 2 dx, d, − ∫ (ln x ) ∫ (2 dx ) dx, dx, 1, = ln x ⋅ 2 x − ∫ ⋅ 2 x dx, x, = 2 x ln x − 2 x + C
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NDA/NA, , 88. The minimum distance from the, point (4, 2) to y = 8x is equal to, 2, , y 2 = 8 x., , Then, the distance between ( x, y) and, ( 4, 2 ) is, D 2 = ( x − 4)2 + ( y − 2 )2, 2, , , y2, − 4 + ( y − 2 )2, D2 = , , 8, , ...(i), , [Q y 2 = 8 x], 2y, y2, dD 2, =2 , − 4 + 2 ( y − 2 ), dy, 8, 8, 3, 2 y, 2 y, =2 , −4×, + 2 (y − 2), 8, 64, , y3, y3, − 2y + 2y − 4=, −4, 16, 16, d 2D 2 3 y 2, ⇒ 2 =, 16, d y, dD 2, y3, On putting, −4=0, =0 ⇒, dy, 16, ⇒, y 3 = 64 ⇒ y = 4, =, , y = 4,, , At, , d 2D 2, , >0, , d 2y, So, it is point of minima., ∴Minimum D = (2 − 4)2 + ( 4 − 2 )2, =, , 4+ 4=, , 8 =2 2, , 89. The differential equation of the, system of circles touching theY -axis, at the origin is, dy, =0, dx, dy, (b) x 2 + y 2 + 2 xy, =0, dx, dy, (c) x 2 − y 2 + 2 xy, =0, dx, dy, (d) x 2 − y 2 − 2 xy, =0, dx, (a) x 2 + y 2 − 2 xy, , ⇒, , x + a − 2 ax + y = a, , ⇒, , x − 2 ax + y = 0, , 2, , 2, , 2, , ...(i), , On differentiating Eq. (i) w.r.t. x, we get, 2 x −`2 a + 2 yy′ = 0, ⇒, x + yy′ = a, Put value of a in Eq. (i), we get, x 2 − 2 ( x + yy′ ) x + y 2 = 0, ⇒, , x 2 − 2 x 2 − 2 xyy′ + y 2 = 0, , ⇒, , − x 2 − 2 xyy′ + y 2 = 0, , ∴, , dy, x − y + 2 xy, =0, dx, 2, , 2, , c2 , z = a2 x + b 2 , x, , ⇒, , occuring in the differential equation is 2, and its degree is 1., , 91. What is the general solution of the, dy x, differential equation, + = 0?, dx y, (a) x 2 + y 2 = C, (b) x 2 − y 2 = C, 2, 2, (c) x + y = Cxy (d) x + y = C, , d 2z, , ⇒, , dx, , ∴, , a2 −, , At x =, , − bc d 2 z, 2 a3, , 2 =−, <0, a, bc, dx, Gives maximum value, At x =, , ax, +C, ln(a), ex, (c), +C, ln(ae ), , ex, +C, ln(a), ae x, (d), +C, ln(a), (b), , x, , =, , ∫a, , x, , dx =, , ax, +C, ln a, , 95. What is the area of one of the loops, , between the curve y = c sin x and, X -axis ?, (b) 2c, (d) 4c, π, , Y, X′, , a 2 x + b 2y where xy = c 2 ?, , O, , π, 2, , y = csin x, X, π, , Y′, , = 2c, , (b) 2 abc, (d) 4abc, , On putting y =, , ...(i), , π, , ∫0, , sin x dx = 2c[− cos x ]π0, , = 2c [− (cos π − cos 0)], = 2c (2 ) = 4c sq units, , 96. If sin θ + cos θ = 2 cos θ, then what, is (cos θ − sin θ ) equal to ?, , c, in Eq. (i), we get, x, , is, , Ê (d)∴ Required area = 2 ∫0 c sin x dx, , lim sin x = k, k=0, , 2, , bc, a, , x ln a, dx = ∫ elna dx, Ê (a) Let I = ∫ e, , 93. What is the minimum value of, , ⇒, , bc, a, , ⇒ Gives minimum value, , (a) c, (c) 3c, , x→ 0, , c2, y=, x, , = a2, , bc d 2 z 2 a 3, ,, =, >0, a dx 2, bc, , lim f( x ) = f( 0), , Since, xy = c 2, , =0, , x=±, , ⇒, , x→ 0, , Ê (b) Let z = a x + b y, , x2, b 2c 2, x2, , (a), , Ê (d) Given, f( x) is continuous at x = 0., , 2, , b 2c 2, , dz, =0, dx, , 94. What is ∫ e x ln(a )dx equal to?, , (b) 1, (d) 0, , 2, , ...(iv), , x3, , For maxima and minima, we put, , 92. The value of k which makes, , (c) 3abc, , 2 b 2c 2, , abc + abc = 2 abc., , ⇒, ydy = − xdx, Integrating both sides, we get, y2 − x 2, =, + C1, 2, 2, [where,C = 2C1], ∴, x 2 + y2 = C, , (a) abc, , =, , 2, , ∴ Minimum value of z at x =, , Ê (a) Given differential equation,, dy, x, + =0, dx, y, dy − x, =, ⇒, dx, y, , ∴, , ...(ii), , On differentiability Eq. (ii) both sides, we, get, dz, b 2c 2, ...(iii), = a2 −, dx, x2, , ⇒, , Ê (c) The order of highest order derivative, , ⇒, ∴, , ( x − a )2 + y 2 = a 2, 2, , (a) Only 1, (b) Only 2, (c) Both 1 and 2, (d) Neither 1 nor 2, , (a) 2, (c) −1, , Y-axis at the origin is, 2, , dy , + 2 + 9 x = x, 2, dx , dx, 1. The degree of the differential, equation is 1., 2. The order of the differential, equation is 2., Which of the above statements, is/are correct ?, , sin x , x ≠ 0, continuous at, f (x ) = , x =0, k,, x = 0, is, , Ê (c) The system of circles touching the, , 2, , 2, , d 2y, , Ê (b) Let ( x, y) be any point on the curve, , ⇒, , 90. Consider the following in respect of, the differential equation :, , (b) 2 2, (d) 3 2, , (a) 2, (c) 2, , ⇒, , 27, , Solved Paper 2019 (I), , (a) − 2 cosθ, , (b) − 2 sinθ, , (c) 2 sinθ, , (d) 2sinθ
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28, , NDA/NA, , Ê (c) Given, sin θ + cos θ = 2 cos θ, , ...(i), sin θ = ( 2 − 1) cos θ, ⇒, Now, cos θ − sin θ, [from Eq. (i)], = cos θ − ( 2 − 1) cos θ, = cos θ [1 − ( 2 − 1)] = cos θ [2 − 2 ], = cos θ ⋅ 2 ( 2 − 1), [using Eq. (i)], = 2 sin θ, , 97. In a circle of diameter 44 cm, the, length of a chord is 22 cm. What is, the length of minor arc of the chord?, 484, cm, (a), 21, 121, (c), cm, 21, , 242, (b), cm, 21, 44, (d), cm, 7, , 100. The angle of elevation of a tower of, , 103. A problem is given to three students, , height h from a point A due South of, it is x and from a point B due East of, A is y. If AB = z , then which one of, the following is correct ?, , A , B and C whose probabilities of, solving the, 1 3, 1, problem are , and respectively., 2 4, 4, What is the probability that the, problem will be solved if they all, solve the problem independently ?, , (a) h2 (cot 2 y − cot 2 x ) = z2, (b) z2 (cot 2 y − cot 2 x ) = h2, (c) h2 (tan2 y − tan2 x ) = z2, 2, , 2, , 2, , Ê (a) P, h, , O, , O, , B, , ⇒, 2 r = 44, ⇒, r = 22, ⇒ ∆OAB is an equilateral triangle., ⇒, ∠AOB = 60°, ∴ Length of minor arc, 60° , = , × 2 π × 22, 360° , 1, 22, = ×2 ×, × 22, 6, 7, 484, cm, =, 21, , 1, 1, , then in, 2, 3, which quadrant does θ lie?, , 98. If sinθ = − and tanθ =, (a) First, (b) Second, (c) Third, (d) Fourth, , ∴ Required probability, , A, Here, OP be the tower,, OA = h cot x, OB = h cot y, In right-angled ∆OAB,, h2 cot 2 y = z2 + h2 cot 2 x, , ∴, , z2 = h2 (cot 2 y − cot 2 x ), , 101. From a deck of cards, cards are taken, out with replacement. What is the, probability that the fourteenth card, taken out is an ace?, 1, (a), 51, , 4, (b), 51, , 1, (c), 52, , 1, (d), 13, , Ê (d) Total number of possible outcomes, = 52, , And number of favourable outcomes = 4, 4, 1, =, ∴Required probability =, 52 13, , (c) We know that, if θ lies in third quadrant, then, sinθ < 0 and tanθ > 0., , 99. How many three digit even numbers, can be formed using the digits 1, 2, 3,, 4 and 5 when repetition of digits is not, allowed?, (a) 36, (c) 24, , 2, 2, 3, 1, P( B) = , P( B ) =, 4, 4, 1, 3, P(C ) = , P(C ) =, 4, 4, , = 1 − P( A ) P( B ) P(C ), 1 1 3 29, = 1− × × =, 2 4 4 32, , cm, , 22 cm, , and, , B, , x, , 22, , A, , 27, 32, 23, (d), 32, (b), , 1, 1, Ê (a) We have, P( A ) = , P( A ) =, , y, cm, , 29, 32, 25, (c), 32, (a), , (d) z (tan y − tan x ) = h, 2, , Ê (a) Given, diameter of a circle be 44 cm., , 22, , Solved Paper 2019 (I), , (b) 30, (d) 12, , Ê (c) Here, unit digit can be filled by 2 or 4., so number of ways is 2. Since repetition, is not allowed therefore hundred place, and ten place can be fill in 4 C 2 × 2 ways, ∴ Total number of three digits even, number = 4 × 3 × 2 = 24, , 102. If A and B are two events such that, P ( A ) = 0.5, P ( B ) = 06, . and, P ( A ∩ B ) =0. 4, then what is, P ( A ∪ B ) equal to ?, (a) 0.9, (c) 0.5, , (b) 0.7, (d) 0.3, , Ê (d) P( A ∪ B) = 1 − P( A ∪ B), We have,, P( A ∪ B) = P( A ) + P( B) − P( A ∩ B), = 0.5 + 0.6 − 0.4, = 11, . − 0.4, = 07, ., ∴ P( A ∪ B) = 1 − 07, . = 0.3, , 104. A pair of fair dice is rolled. What is, the probability that the second dice, lands on a higher value than does the, first?, 1, 4, 5, (c), 12, , 1, 6, 5, (d), 18, , (a), , (b), , Ê (c) Total number of possible outcomes, , = 36, Favourable outcomes, = (1, 2 ), (1, 3), (1, 4), (1, 5), (1, 6),, (2, 3), (2, 4), (2, 5),(2, 6) ( 3, 4),, ( 3, 5),, ( 3, 6), ( 4, 5), ( 4, 6), ( 5, 6), ∴Total number of favourable outcomes, = 15, 15, 5, =, ∴Required probability =, 36 12, , 105. A fair coin is tossed and an unbiased, dice is rolled together. What is the, probability of getting a 2 or 4 or 6, along with head?, (a), , 1, 2, , (b), , 1, 3, , (c), , 1, 4, , (d), , 1, 6, , Ê (c) Total number of possible outcomes, , = 2 × 6 = 12, And favourable outcomes, = (H, 2), (H, 4), (H, 6), ∴Total number of possible outcomes = 3, 3, 1, =, ∴Required probability =, 12 4
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NDA/NA, , 29, , Solved Paper 2019 (I), , 106. If A , B and C are three events, then, , 109. The mean of 100 observations is 50, , 111. Two symmetric dice flipped with, , what is the probability that atleast, two of these events occur together ?, , and the standard deviation is 10. If 5, is subtracted from each observation, and then it is divided by 4, then what, will be the new mean and the new, standard deviation respectively ?, , each dice having two sides painted, red, two painted black, one painted, yellow and the other painted, white. What is the probability that, both land on the same colour ?, , (a) 45, 5, (b) 11.25, 1.25, (c) 11.25, 2.5, (d) 12.5, 2.5, , (a), , (a) P( A ∩ B) + P( B ∩ C ) + P(C ∩ A ), (b) P( A ∩ B) + P( B ∩ C ) + P(C ∩ A ), − P( A ∩ B ∩ C ), (c) P( A ∩ B) + P( B ∩ C ) + P(C ∩ A ), − 2 P( A ∩ B ∩ C ), (d) P( A ∩ B) + P( B ∩ C ) + P(C ∩ A ), − 3P( A ∩ B ∩ C ), , Ê (c) If A, B and C are three events, then, atleast two events occur i.e., ( A ∩ B ∩ C ′ ) ∪ ( A ∩ B′ ∩ C ), ∪ ( A′ ∩ B ∩ C ) ∪ ( A ∩ B ∩ C ), ∴Required probability, = P( A ∩ B) + P( B ∩ C ) + P(C ∩ A ), −2 P( A ∩ B ∩ C ), , 107. If two variables X and Y are, independent, then what is the, correlation coefficient between them?, (b) −1, (d) None of these, , (a) 1, (c) 0, , Ê (c) Correlation coefficient between two, independent variables is zero., , 108. Two independent events A and B are, such, , that, , P( A ∪ B ) =, , 2, 3, , and, , 1, P ( A ∩ B ) = . If P ( B ) < P ( A ), then, 6, what is P ( B ) equal to ?, 1, 4, 1, (c), 2, (a), , 1, 3, 1, (d), 6, (b), , 2, Ê (b) Given, P( A ∪ B) =, , 3, , 1, 6, ⇒ P( A ∪ B) = P( A ) + P( B) − P( A ∩ B), 2, 1, = P( A ) + P( B) −, ⇒, 3, 6, 2, 1, P( A ) + P( B) = +, ⇒, 3 6, 5, ...(i), P( A ) + P( B) =, ⇒, 6, 1, And also, P( A ∩ B) =, 6, 1, ...(ii), P( A ) P( B) =, ⇒, 6, From Eqs. (i) and (ii), we get, 1, 1, P( A ) or P( B) = or, 2, 3, Also, given P( B) < P( A ), 1, ∴, P( B) =, 3, and P( A ∩ B) =, , Ê (c) Given, mean ( x ) = 50, , 50 − 5, The new mean =, 4, 45, =, = 11.25, 4, And standard deviation ( σ ) = 10, ∴The new standard deviation, 10, =, 4, = 2.5, Since, addition and subtraction does not, effect standard deviation., , 110. If two fair dice are rolled, then what, is the conditional probability that, the first dice lands on 6, given, that the sum of numbers on the, dice is 8?, 1, 3, 1, (b), 4, 1, (c), 5, 1, (d), 6, (a), , 3, 18, 5, (c), 18, , 2, 9, 1, (d), 3, (b), , 2 2, Ê (c) P (two sides painted red) = ×, , 6 6, 2 2, P (two sides painted black) = ×, 6 6, 1 1, P (one side painted yellow) = ×, 6 6, 1 1, and P (other side painted white) = ×, 6 6, , ∴ Required probability that both land on, the same colour, 2 2 2 2, 1 1 1 1, = × + × + × + ×, 6 6 6 6 6 6 6 6, 4 + 4 + 1+ 1, =, 36, 10, 5, =, =, 36 18, , 112. There are n socks in a drawer, of, which 3 socks are red. If 2 of the, socks are chosen randomly and the, probability that both selected socks, 1, are red is , then what is the value, 2, of n ?, , Ê (c) Let E1 = Event of first dice on 6, E 2 = Event of the sum of numbers on, dices 8, ∴ Total number of sample space of two, dices are rolled, n( s ) = 36, Possible outcomes of E1 (6, 2), Possible outcomes of E 2 (2, 6) (3, 5), (4, 4) (5, 3) (6, 2), 1, P( E1 ∩ E 2 ) =, ∴, 36, 5, and, P( E 2 ) =, 36, E , ∴Required probability = P 1 , E2 , =, , P( E1 ∩ E 2 ), ,, P( E 2 ), , when P( E 2 ≠ 0), 1, 1, = 36 =, 5 / 36 5, , (a) 3, (b) 4, (c) 5, (d) 6, , Ê (b) Total number of socks = n, P (first socks is red) =, P (second socks is red) =, , 3, n, , 2, n−1, , According to the question,, 3, 2, 1, ×, =, n n−1 2, n2 − n = 12, , ⇒, ⇒, , n2 − n − 12 = 0, , ⇒, , n − 4n + 3n − 12 = 0, , ⇒, , n ( n − 4) + 3 ( n − 4) = 0, , ⇒, ∴, , 2, , ( n − 4) ( n + 3) = 0, n = 4, − 3
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30, , NDA/NA, , 113. Two cards are chosen at random, from a deck of 52 playing cards., What is the probability that both of, them have the same value ?, 1, 17, 5, (c), 17, , (a), , (b), , 3, 17, , (d), , 4, , C 2 × 13, 52, , C2, , 114. In eight throws of a die, 5 or 6 is, considered a success. The mean, and standard deviation of total, number of successes is respectively, given by, 8 16, (a) ,, 3 9, 4 4, (c) ,, 3 3, , 8 4, (b) ,, 3 3, 4 16, (d) ,, 3 9, 1 1 1, (b), We, have,, (success), p, = + =, Ê, 6 6 3, 2, ∴, q = 1− p=, 3, Given, n = 8, 1 8, ∴ Mean = np = 8 × =, 3 3, Standard deviation = npq, =, , 1 2, 8× ×, 3 3, , =, , 16 4, =, 9, 3, , 115. A and B are two events such that A, and B are mutually exclusive. If, P ( A ) = 0.5 and P ( B ) = 06,, . then what, is the value of P ( A / B ) ?, 1, 5, 2, (c), 5, , 1, 6, 1, (d), 3, (b), , P( A ∪ B) = 0, , ⇒, , 1 − P( A ∪ B) = 0, , ⇒, , P( A ∪ B) = 1, , P( A ∪ B) = P( A ) + P( B) − P( A ∩ B), , 4 × 3 × 13, =, 52 × 51, 1, =, 17, , (a), , ⇒, , We know that,, , 7, 17, , Ê (a)∴ Required probability =, , Ê (b) Given, P( A ∩ B ) = 0, , ⇒, , 1 = 0.5 + 0.6 − P( A ∩ B), , ⇒, ∴, , P( A ∩ B) = 01, ., A P( A ∩ B), , P =, B, P( B), =, , 01, ., 1, =, 0.6 6, , 116. Consider the following statements, 1. The algebraic sum of deviations, of a set of values from their, arithmetic mean is always zero., 2. Arithmetic mean > Median >, Mode, for, a, symmetric, distribution., Which of the above statements, is/are correct?, (a) Only 1, (b) Only 2, (c) Both 1 and 2, (d) Neither 1 nor 2, , 118. If all the natural numbers between 1, and 20 are multiplied by 3, then what, is the variance of the resulting, series?, (a) 99.75, (c) 299.25, , (b) 199.75, (d) 399.25, , Ê (c) Variance of first n natural number, , n2 − 1 20 2 − 1, =, 12, 12, 399, =, = 33 .25, 12, If all the natural number between 1 and 20, multiplied by 3, then, ∴Required variance = 9 × 33 .25, = 299 .25, =, , 119. What is the probability that an, interior point in a circle is closer to, the, centre, than, to, the, circumference?, 1, 1, (b), 4, 2, 3, (c), 4, (d) It cannot be determined, , (a), , Ê (a) Let radius of circle be r, then the points, , Ê (a) We know that, the algebraic sum of, deviations of a set of values from their, arithmetic mean is always zero., , 117. Let, , the correlation coefficient, between X and Y be 0.6. Random, variables Z and W are defined as, Y, Z = X + 5 and W = . What is the, 3, correlation coefficient between Z, andW ?, , (a) 0.1, (c) 0.36, , Solved Paper 2019 (I), , (b) 0.2, (d) 0.6, , Ê (d) Since, the correlation coefficient is, , independent of change of origin and, scale. It is given that correlation, coefficient between X and Y be 0.6., So, correlation coefficient between Z, and W be 0.6., , closer to centre if circumference will lie, r, within radius of ., 2, So, the favourable outcome would be the, points inside the area of circle with radius, r, whereas the total possible outcomes, 2, could be all the points inside the area of, circle with radius r., , ∴ Required probability =, , r, π , 2, πr 2, , 2, , =, , 1, 4, , 120. If A and B are two events, then what, is the probability of occurrence of, either event A or event B ?, (a) P( A ) + P( B), (c) P( A ∩ B), , (b) P( A ∪ B), (d) P( A ) P( B), , Ê (b) If A and B are two events, then the, probability of occurrence of either event, A or event B is P( A ∪ B).
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NDA / NA, National Defence Academy/Naval Academy, , Solved Paper, , 2018 (II), , Paper 1 (Mathematics), 3. Consider the following expressions, , 1. What is the value of log 7 log 7 7 7 7 equal to ?, (a) 3 log 2 7, , (b) 1 − 3 log 2 7 (c) 1 − 3 log 7 2 (d), , 7, 8, , 1. x + x 2 −, , 7 7 7 =, , 1, 72, , Now, log 7 log 7, , ⋅, , 1, 74, , ⋅, , 1, 78, , =, , 7, 78, , 5., , 7 7 7, , = log 7 7 − log 7 8, , m, [Q log, = log m − log n], n, , = log 7 7 − log 7 23 = 1 − 3 log 7 2, [Q log b a n = n log b a ], , 2. If an infinite GP has the first term x and the sum 5, then, which one of the following is correct?, (a) x < −10, (c) 0 < x < 10, , x, x, =1 −r ⇒ r =1 −, 5, 5, x, x, < 1 ⇒− 2< − < 0, 5, 5, , −1< −, , ⇒, , − 10 < − x < 0, , ⇒, , 10 > x > 0, , (a) 1, 4 and 5, , (b) 1, 3, 4 and 5, , (c) 2, 4 and 5, , (d) 1 and 2, , Ê (a) We know that, rational expressions are those expression, which can be write in the form of, , p (x ), , q(x) ≠ 0, q(x), , So, 1, 4, 5 are rational expressions, (b) A ′ = A −1, , (c) A = A −1, , (d) A = A′, , where A′ is the transpose of A, , Ê (b) A square matrix is called an orthogonal matrix if AA′ = I, multiply. by A− 1, , [Q sum of infinity GP =, , ⇒, , 1, 2, −, x x +5, , (a) A = A 2, , Ê (c) Given that first term of an infinity GP is x and sum = 5, , Where, |r |< 1, , 2, x 2 − ax + b 3, , 4. A square matrix A is called orthogonal if, , (b) −10 < x < 0, (d) x > 10, , x, =5, 1 −r, , 4., , d, e, − 2, x x, , Which of the above are rational expressions?, , 7, , 7, = log 7 log 7 (7 ) 8 = log 7 , 8, , ⇒, , 2. ax 2 + bx + x − c +, , 3. 3x 2 − 5x + ab, , Ê (c) We have,, , ∴, , 1, x, , a, ], 1 −r, , ∴, ⇒, , A− 1 ( AA′ ) = A− 1 I ⇒ IA′ = A− 1, A′ = A− 1, , 5. IfA, B and C are subsets of a universal set, then which one, of the following is not correct?, (a) A ∪ ( B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C ), (b) A ′∪( A ∪ B) = ( B′∩ A )′ ∪ A, (c) A ′∪( B ∪ C ) = (C ′∩ B)′ ∩ A ′, (d) ( A ∩ B) ∪ C = ( A ∪ C ) ∩ ( B ∪ C ), , where A′ is the complement of A
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2, , NDA/NA Solved Paper 2018 (II), , Ê (c) Let A, B and C are subsets of a universal set., , We have,, n ( A ∪ B ∪ C ) = 300, , Let A = { 1 } , B = { 2 } , C = { 3 }, , n ( A) = 125 ⇒ n (B) = 145, , ∪ = {1, 2, 3 } , A′ = {2, 3 }, B′ = {1, 3 }, C′ = {1, 2}, , n (C ) = 90, , by checking options, we get, , n ( A ∪ B ∪ C ) = n( A) + n (B) + n (C ), , LHS = A′∪ (B ∪ C ) = { 2, 3 } ∪ { } = {2, 3 }, , − [n ( A ∩ B) + n (B ∩ C ) + n (C ∩ A)], , RHS = (C ′ ∩ B)′ ∩ A′, , + n ( A ∩ B ∩ C), , = ( { 1, 2 } ∩ {1, 3 } ) ∩ { 2, 3 }, = ( { 2 } )′ ∩ { 2, 3 } = { 1, 3 } ∩ { 2, 3 } = { 3 }, LHS ≠ RHS, , 6. Let x be the number of integers lying between 2999 and, 8001 which have at least two digits equal.Then x is equal, to, (b) 2481, , − [n ( A ∩ B) + n (B ∩ C ) + n (C ∩ A)] + n ( A ∩ B ∩ C ), ⇒ n ( A ∩ B) + n (B ∩ C ) + n (C ∩ A), , So, option (c) is wrong, , (a) 2480, , ⇒ 300 = 125 + 145 + 90, , (c) 2482, , (d) 2483, , = 60 + n ( A ∩ B ∩ C ), n ( A ∩ B) + n (B ∩ C ) + n (C ∩ A), , − 3 n ( A ∩ B ∩ C ) = 32, ⇒ n ( A ∩ B) + n (B ∩ C ) + n (C ∩ A), = 32 + 3n ( A ∩ B ∩ C ), , Ê (b) We have, x be the number lying between 2999 and 8001, , ...(ii), , if repetition allowed, , From Eqs. (i) and (ii), we get, , total numbers = 5 × 10 × 10 × 10 = 5000, , 60 + n ( A ∩ B ∩ C ) = 32 + 3n ( A ∩ B ∩ C ), , if repetition not allowed, , ⇒, , ∴ total numbers = 5 × 9 × 8 × 7 = 2520, , (a) 196, , = 5000 − 2520 + 1 = 2481, [Q add 1 because of number 8000], , 1 1, 3 9, , 7. The sum of the series 3 − 1 + − +.... is equal to, 20, 9, , (b), , Ê (c) Given series, , 3 −1 +, , ∴, , r=, , 9, 20, , (c), , 9, 4, , (d), , 4, 9, , 3, , a , Q Sn =, , − r, 1, , , (d) 268, , Ê (c) Number of students like to play exactly one, , game = n ( A) + n (B) + n (C ), − 2 [n ( A ∩ B) + n (B ∩ C ) + n (C ∩ A)], + 3n ( A ∩ B ∩ C ), = 125 + 145 + 90 − 2 [32 + 3 × 14] + 3 × 14, = 360 − 106 = 254, , (a) Least value −, , 1, 4, , (c) Greatest value, , quadratic, , (b) Least value −, 1, 4, , expression, 9, 4, , (d) Greatest value, , 9, 4, , Ê (d) α and β are the roots of quadratic equation., , (Q. Nos. 8 and 9) Consider the information given, below and answer the two items that follow., , A survey was conducted among 300 students. It was found that, 125 students like to play cricket, 145 students like to play football, and 90 students like to play tennis. 32 students like to play exactly, two games out of the three games., , 8. How many stdudents like to play all the three games ?, (b) 21, , (c) 254, , x 2 + αx − β = 0, then the, − x 2 + αx + β, where x ∈ R has, , Directions, , (a) 14, , (b) 228, , 10. If α and β ( ≠ 0) are the roots of the quadratic equation, , 1 1, − + ..... are in GP, 3 9, −1, , 3, Sn =, 1, 1 − − , 3, 3 9, = =, 4 4, 3, , 2n ( A ∩ B ∩ C ) = 28 ⇒ n ( A ∩ B ∩ C ) = 14, , 9. How many students like to play exactly only one game?, , So, x = atleast two digit repeated, , (a), , ...(i), , Again,, , (c) 28, , (d) 35, , Ê (a) Let,, A be the set of students like to play cricket, B be the set of students like to play football., C be the set of students like to play tennis., , x2 + αx − β = 0, So, (αβ = − β ) ⇒αβ + β = 0, ⇒, β (α + 1 ) = 0, [Qβ ≠ 0], α = −1, α + β = − α ⇒ 2α + β = 0 ⇒ β = 2, [Qα = − 1, β = 2], ∴, − x2 + αx + β, = − x2 − x + 2, (− 1 ) 2 , Greatest value = 2 − , , 4 (− 1 ) , 2, , b2 , [Q Greatest value = c − ], 4a , 1 , 1 9, =2− , =2+ =, 4 4, − 4
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3, , NDA/NA Solved Paper 2018 (II), 11. What is the coefficient of the middle term in the binomial, expansion of (2 + 3x ) 4 ?, (a) 6, , (b) 12, , Ê (d) We have, (2 + 3 x), , (c) 108, , cos θ, (a) , − sin θ, , Here, n = 4, so middle term is, th, , = nCr ar bn − r ] 1 =, , 4×3, 2×1, , × 4 × 9 x2, , Ê (a) We have,, , T3 = 216 x2, Hence, coefficient of middle term is 216., , 12. For a square matrix A, which of the following properties, , Now,, , hold?, 1, det A, , 3. ( λA )−1 = λA −1, where λ is a scalar, (b) 2 and 3, , (c) 1 and 3, , (d) 1, 2 and 3, , ), , Statement 2, det ( A, , 3, , Ê (a) We have,, , (d) 0, , 3n, , , − 1 + i 3 2 − 1 − i 3 , ,ω = , , Q ω =, 2, 2, , , , [Qω3 = 1], , (b) C (12, 8), (d) C ( 5, 3) × C (12, 8), , Ê (d) There are 17 cricket players, out of which 5 players can, , 9 contain ?, , (b) x − y, , −1 − i 3, + , , 2, , , , (a) C (17, 11), (c) C (17, 5) × C ( 5, 3), , bowl., , x 3 10y 5 27, (a) x − 3, , (c) 1, , bowl. In how many ways can a team of 11 players be, selected so as to include 3 bowlers?, , 13. Which one of the following factors does the expansion of, , 5y, , 3n, , 16. There are 17 cricket players, out of which 5 players can, , So, Statement 3 is correct., , x, , −1 − i 3 , +, , 2 , , , =1 + 1=2, , (λA)− 1 = λA− 1 , where λ is a scalar., , 3, , 3a, , = (ω3 )n + (ω3 )2 n (1 )n + (1 )2 n, , Statement 3, , 2, , 3n, , = (ω)3 n + (ω2 )3 n, , 1, )=, det A, , Statement 2 is correct, , the determinant, x, y, , (b) 2, , −1 + i 3, , , 2, , , , = A, , Statement 1 is true, −1, , −1 + i 3 , , , 2 , , , where i = −1 ?, , Ê (b) We have,, , Statement 1, (A, , cos (− θ) − sin (− θ) , − sin (− θ) cos (− θ) , cos θ sin θ , sin θ cos θ , , C11 = cos θ, C12 = − sin θ, C21 = − sin θ, , (a) 3, , Ê (d) For a square matrix A, −1 −1, , , A=, , , A=, , , 15. What is the value of, , Select the correct answer using the code given below., (a) 1 and 2, , cos θ − sin θ, (d) , , sin θ cos θ , , C22 = cos θ, T, cos θ − sin θ , cos θ − sin θ, adj A = , = − sin θ cos θ , sin, θ, cos, θ, −, , , , , , 1. ( A −1 )−1 = A, 2. det ( A −1 ) =, , cos θ sin θ , (b) , , sin θ cos θ, , − sin θ , , cos θ, , cos θ sin θ , (c) , , − sin θ cos θ, , T3 = 4C2 × 22 × (3 x)2, +1, , cos ( −θ) − sin ( −θ), , ?, − sin ( −θ) cos ( −θ), , (d) 216, , 4, , 4 + 1 = 3 rd term, , , 2, , [Tr, , 14. What is the adjoint of the matrix, , required number of ways =, (c) y − 3, , C8 × 5C3, , 12, , = C (12, 8 ) × C (5, 3 ), , (d) x − 3 y, , 17. What is the value of log 9 27 + log 8 32 ?, y, 3 , x, x2 5 y3, 9 , , 3, 5, x 10 y 27 , , y, x−3, = x2 − 9 5 y3, 3, 5, x − 27 10 y, 1, , x+3, = (x − 3 ) , 2, x + 9 +, , [C1 → C1 − C3 mes,], 3 , 9 , , 27 , , (a), , 7, 2, , 19, (b), 6, , (c) 4, , (d) 7, , Ê (b) We have, log 9 27 + log 8 32, = log, , y, 5 y3, , 3 x 10 y, , 5, , 3 , 9 , , 27 , , =, , 32, , 3 3 + log 2 3 25 =, , 3 5 19, + =, 2 3 6, , 5, 3, log 3 3 + log 2 2, 3, 2, n, Q log, log a, bn =, am, , m, , b,
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4, , NDA/NA Solved Paper 2018 (II), | A| = 2 [− 2 (− 1 ) − 2 (− 3 )] − 1 [3 (− 1 ) − 2 (5)], + (− 3 ) [3 (− 3 ) − (− 2)(5)], , 18. If A and B are two invertible square matrices of same, order, then what is ( AB ), −1 −1, , −1, , equal to?, , −1 −1, , (a) B A, , (c) B−1 A, , (b) A B, , Ê (a) If, , A and B are two invertible square matrices of same, order, then, ( AB)− 1 = B− 1 A− 1, , c, , b, , c, , b−x, , a, , b, , a, , c −x, , (d) None of the above, , Ê (c), , (b) x =, 2 (a2 + b 2 + c 2 ), 3, , (c) x =, , cube roots of unity?, (b) They lie on a circle of radius 3, (c) They form an equilateral triangle, , = 0 is, , (a) x = a, , 22. Which one of the following is correct in respect of the, (a) They are collinear, , 19. If a + b + c = 0, then one of the solutions of, a−x, , = 2 (8 ) − 1 (− 13 ) − 3 (1 ) = 16 + 13 − 3 = 26 ≠ 0, So, System is consistent with unique solution., , (d) A −1B, , 3 (a2 + b 2 + c 2 ), 2, , We know that, cube roots of unity is 1, ω, ω2 , where, −1 + i 3, −1 − i 3, and ω2 =, ω=, 2, 2, , (d) x = 0, , B, , Ê (d) We have,, , R1 → R1 + R2 + R3, , a + b + c − x a + b + c − x a + b + c − x, = 0, c, b− x, a, , b, a, c−x, , , , ω1, (–½, √3 ) C, 2, , They form an equilateral triangle., , 1, 1 , −x , 1, − x − x, a = 0, a = 0 ⇒ (−x) c b − x, ⇒ c b− x, , , , , b, a, c, −, x, b, a, c, x, −, , , , ⇒ x=0, Hence, x = 0 is a solution, , 2 4, , −8 x , , 20. What should be the value of x, so that the matrix , does not have an inverse?, (b) −16, (c) 8, (d) −8, 2 4, (b) Let, A = , , −8 x, Matrix does not have any solution if, | A| = 0, 2x + 32 = 0, 32, ⇒ x = − 16, 2x = − 32 ⇒ x = −, 2, (a) 16, , 21. The system of equations, 2x + y − 3z = 5, 3x − 2y + 2z = 5 and 5x − 3y − z = 16, (a) is inconsistent, (b) is consistent, with a unique solution, (c) is consistent, with infinitely many solutions, (d) has its solution lying along X-axis in three-dimensional space, , 23. If u, v and w (all positive) are the p th , q th and r th terms of, ln u p 1, , , a GP, then the determinant of the matrix ln v q 1 is, , , ln w r 1, (a) 0, , (b) 1, , (c) ( p − q )(q − r ) ( r − p), , (d) ln u × ln v × ln w, , Ê (a) Given that u, v and w are the pth ,qth, , and r th term of GP, [Q an = aRn −1 ], , ∴ u = aR p − 1 , v = aRq − 1, r −1, , w = aR, ln u p 1 , We have, ln v q 1 , , , ln w r 1 , and, , ln a R p − 1 p 1 ln a +, = ln a R q − 1 q 1 = ln a +, , , , , ln a Rr − 1 r 1 ln a +, , , ln a p 1 ( p − 1 ) ln R, = ln a q 1 + (q − 1 ) ln R, , , ln a r 1 (r − 1 ) ln R, p−1, 1 p 1 , , , = ln a 1 q 1 + ln R q − 1, , , , r −1, 1 r 1 , , y − 3z = 5, , 3 x − 2 y + 2 z = 5 and 5x − 3 y − z = 0, 2 1 −3 , A= 3 −2 2 , , , 5 − 3 −1 , , X, A (1,2), , Y´, , ⇒ , , Ê (b) The system of equations 2x +, , 120º, 120º, , 120º, , X´, , c, b , a − x, c, b− x, a = 0, , , a, c − x, b, , Ê, , Y, , ω, (–½, √3 ), 2, , p−1 p−1, = 0 + ln R q − 1 q − 1, , r −1 r −1, , 1, 1= 0, , 1, , p − 1 ln R p 1 , q − 1 ln R q 1 , , r − 1 ln R r 1 , p 1, q 1, , r 1, p 1, q 1, , r 1, C2 → C2 − C3
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5, , NDA/NA Solved Paper 2018 (II), 1, n, 1, Tn =, m, 1, a + (n − 1 ) d =, m, , a + (m − 1 ) d =, , 24. Let the coefficient of the middle term of the binomial, expansion of (1 + x ) be α and those of two middle terms, of the binomial expansion of (1 + x ) 2n −1 be β and γ. Which, one of the following relations is correct?, 2n, , (a) α > β + γ, , (b) α < β + γ, , (c) α = β + γ, , Ê (c) We have,(1 + x)2n, , 2n, Middle term = , + 1 , 2, , , (d) α = βγ, , term = (n + 1 )th term, 2n, , α=, , 2n, , Cn, , Put in Eq. (i),, , Cn, , Again, we have binomial expansion of, coefficient of middle terms are,, β=, , 2n − 1, , β+ γ=, , 2n − 1, , ∴, Now,, , Cn and, Cn +, , γ=, , 2n, , ⇒, , Cn − 1, , Cn − 1, , Now,, , [Q nCr + nCr, =, , −1, , =, , n+1, , Cr ], , Cn, , 1, n, 1, 1, ⇒ a=, n, mn, , Tmn = a + (mn − 1 ) d, 1, 1, 1, 1, =, + (mn − 1 ), =, +1−, mn, mn mn, mn, , Tmn = 1, , 25. Let A = [x ∈ R :−1 ≤ x ≤ 1] ,, B = [y ∈ R : − 1 ≤ y ≤ 1] and S be the subset of A × B,, defined by, S = [( x , y ) ∈ A × B : x 2 + y 2 = 1]., , 27. Suppose f ( x ) is such a quadratic expression that it is, positive for all real x., If g ( x ) = f ( x ) + f ′( x ) + f ′′ ( x ), then for any real x, (a) g (x ) < 0, , Which one of the following is correct ?, , (b) g (x ) > 0, , (c) g (x ) = 0, , (d) g (x ) ≥ 0, , Ê (b) Given that f (x) is a quadratic expression, , (a) S is a one-one function from A into B, (b) S is a many-one function from A into B, (c) S is a bijective mapping from A into B, (d) S is not a function, , Ê (d) Given that,, , 1, =, mn, 1, 1, a+ −, =, n mn, , a + (m − 1 ), , (1 + x)2 n − 1, , 2n − 1, , 2n − 1, , ...(ii), , Subtracting Eq. (ii) from Eq. (i), we get, 1 1, (m − 1 ) d − (n − 1 ) d = −, n m, m−n, 1, (m − n) d =, ⇒ d=, ⇒, mn, mn, , th, , Coefficient of (n + 1 )th term =, , and, , ...(i), , Let f (x) = ax2 + bx + c, a > 0, , ∴, , b2 − 4ac < 0, , ⇒, , A = { x ∈R: − 1 ≤ x ≤ 1 } ,, , f ′ (x) = 2ax + b, , Now,, and, , B = { y ∈R: − 1 ≤ y < 1 }, , [∴ f (x) > 0], , b2 < 4ac, f ′ ′ (x) = 2a, , We have,, , and S = { (x, y) ∈ A × B : x2 + y2 = 1 }, , g (x ) = f (x ) + f ′ (x ) + f ′ ′ (x ), , Y, , = ax2 + bx + c + 2ax + b + 2a, , 1, , = ax2 + (b + 2a ) x + 2a + b + c, Now, (b + 2a ) − 4a (2a + b + c), 2, , X′, , 1, , -1, , = b2 + 4ab + 4a 2 − 8 a 2 − 4ab − 4ac, , X, , = b2 − 4ac − 8 a 2 < 0, ⇒, , -1, , same order., , By vertical line test. when we draw a vertical line, then, line cuts the circle in two points. Hence, S is not a, function., , 26. Let T r be the r th term of an AP for r = 1, 2, 3, .... If for some, , distinct positive integers m and n we have T m = 1 / n and, T n = 1 / m, then what is T mn equal to ?, (b) m−1 + n−1, , (c) 1, , (d) 0, , Ê (c) Let first term of an AP is a and common difference is d, Given that,, Tm =, , g (x ) > 0, , 28. Consider the following in respect of matrices A, B and C of, , Y′, , (a) (mn)−1, , [Q b2 − 4ac < 0], , 1, n, , 1. ( A + B + C )′ = A ′ + B ′ + C ′, 3. ( ABC )′ = C ′ B ′ A ′, , 2. ( AB )′ = A ′ B ′, , Where A′ is the transpose of the matrix A. Which of the, above are correct?, (a) 1 and 2, (c) 1 and 3, , (b) 2 and 3, (d) 1, 2 and 3, , Ê (c) Given that A, B and C are matrices of same order, Statement 1, ( A′ + B + C )′ = A′ + B′ + C ′, So, Statement 1 is correct, , [Q ( A + B)′ = A′ + B′]
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6, , NDA/NA Solved Paper 2018 (II), Statement 2, , 33. What is the determinant of the matrix, , We know that, ( AB)′ = B′ A′, , x y, , z x, , y z, , Hence, Statement 2 is incorrect, Statement 3, ( ABC )′ = C ′ B′ A′, , [Q ( AB)′ = B′ A′], , Hence, Statement 3 is correct., , 29. The sum of the binary numbers (11011) 2 , (10110110) 2 and, (10011x 0y ) 2 is the binary numbers (101101101) 2 . What, are the values of x and y ?, , (a) ( x − y) ( y − z) ( z − x ), (c) ( y − z) ( z − x ), , and, , (10011 xoy)2 is (101101101 )2, So,, , (101101101), − 10110110, 10110111, − 11011, 10011100, , Compare with (10011 x o y)2, , x y y + z, z x z + x, , , y z x + y, , 30. Let matrix B be the adjoint of a square matrix A, I be the, identity matrix of same order as A. If k( ≠ 0) is the, determinant of the matrix A, then what is AB equal to ?, , Ê (b), , (c)k 2 l, , (b) kl, , (d) (1 / k )l, , 31. If (02, , then what is the value of x, . ) x = 2 and log10 2 = 03010, ., to the nearest tenth?, , Ê (c) We have,, , (c) −0.4, , (d) −0.2, , (0.2) = 2, , ⇒ x [0.3010 − 1] = 0.3010, 0.3010, ⇒, x=−, ≈ − 0.43, 0.6990, , [Q log a a = 1], , (b) 30240, , (c) 27216, , (d) 15120, , Ê (c) 5-digit number that can be composed by distinct digits, from 0 to 9 is given as, , 8, , 7, , 1, 1, 1, , , =0, 1 + sin A, 1 + sin B, 1 + sin C, , , 2, 2, 2, sin A + sin A sin B + sin B Sin C + Sin C , R1 → R1 − R2 , R3 → R3 − R2, , composed of distinct digits from 0 to 9 is, , 9, , then which one of the following is correct?, , Ê (a) We have,, , 32. The total number of 5-digit numbers that can be, , 9, , 1, 1, = 0,, 1 + sin B, 1 + sin C, SinA + sin 2 A sin B + sin 2 B sin C + sin 2 C, , (d) No conclusion can be drawn with regard to the nature of the, triangle, , x log10 0.2 = log10 2, 2, x log10 = log10 2, 10 , , ⇒ x [log10 2 − log10 10] = log10 2, , (a) 45360, , 1, I + sin A, , (a) The triangle ABC is isosceles, (b) The triangle ABC is equilateral, (c) The triangle ABC is scalene, , x, , taking log10 both side, ⇒, , = (x + y + z ) [1 (x − z ) (x − y) − (x − z ) ( z − y)], , 34. If A, B and C are the angles of a triangle and, , AB = A (adjA) = ( A)l = kl, , (b) −0.5, , 0, 0 , 1, = (x + y + z ) z x − z x − z , , , y z − y x − y, , = (x + y + z ) (x − z ) (x − z ) = (x + y + z ) ( z − x ) 2, , ( A) = K, , (a) −10.0, , R1 + R2 + R3, y + z x + y + z 2 (x + y + z ) , , z, x, z+ x, , y, z, x+ y, , 2 , 1 1, = (x + y + z ) z x z + x , , , y z x + y, , = (x + y + z ) [(x − z ) (x − y − z + y)], , Given, B = adjA, l = identity Matrix, ∴, , R1 →, x +, , , , , C2 → C2 − C1 and C3 → C3 − 2C1, , We get, x = 1 and y = 0, , (a) l, , (b) ( x − y) ( y − z), (d) ( z − x )2 ( x + y + z), , Ê (d) We have,, , (a) x = 1, y = 1 (b) x = 1, y = 0 (c) x = 0, y = 1 (d) x = 0, y = 0, , Ê (b) Sum of the binary number (11011 )2, (10110110 )2, , y + z, , z + x ?, , x + y, , 6, , required number = 9 × 9 × 8 × 7 × 6 = 27216, , − sin B, − sin C , − sin A, 1 + sin A 1 + sin B 1 + sin C = 0, , , 2, 2, 2, sin A − 1 sin B − 1 sin C − 1 , − sin A − sin B − sin C , , = 0, 1, 1, 1, , , 2, 2, 2, − cos A − cos B − cos C , , R2 → R2 + R1, , R3 → R3 + R2
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7, , NDA/NA Solved Paper 2018 (II), sin B, sin C , sin A, = 0, ⇒ , 1, 1, 1, , , 2, 2, 2, 1 − cos A 1 − cos B 1 − cos C , sin A sin B sin C , ⇒ 1, 1, 1 = 0, , , 2, 2, 2, sin, A, sin, B, sin, C, , , 37. If sec (θ − α ), sec θ and sec (θ + α ) are in AP, where, cos α ≠ 1, then what is the value of sin 2 θ + cos α ?, (a) 0, , ⇒ (sin A − sin B) (sin B − sin C ), sin C , 1, 1, , , 0, 0, 0 , , , 2, sin A + sin C sin B + sin C sin C , , A = B and B = C, , 35. Consider the following in respect of matrices A and B of, , θ + α + θ − α cos θ + α − θ + α , 2 cos , , , , , , , , 2, 2, 2, 2, cos θ − sin α, , C + D, C − D , Q cos C + cos D = 2 cos 2 cos 2 , , , , 2, =, cos θ, , ⇒ sin 2 α = cos 2 θ (1 − cos α ) ⇒ cos 2 θ =, , same order, 1. A 2 − B 2 = ( A + B ) ( A − B ), , =, , 2. ( A − I ) ( I + A ) = O ⇔ A 2 = I, , where I is the identity matrix and O is the null matrix., Which of the above is/are correct?, (b) 2 only, (d) Neither 1 nor 2, , [Q AB ≠ BA], , ≠ A2 − B2, Statement 2, ( A − I ) (I + A) = 0, AI + A − I − IA = 0, , ⇒, , A + A2 − I 2 − A = 0, , ⇒, , A =I, 2, , [Q AI = IA = A], [Q I = I], 2, , Statement 2 is correct, , 36. What is, , 2 tan θ, 1 + tan θ, , (a) cos2θ, , 2, , Ê (c) We have,, 2 tan θ, 1 + tan 2 θ, =, , equal to ?, , (b) tan2θ, , (c) sin2θ, , (d) cosec 2θ, , 2 sin θ cos θ, , cos 2 θ, , = 2 sin θ cos θ = sin 2θ, , (b) −4cos A sin B cos C, (d) −4sin A cos Bcos C, , Ê (d) We have,, , Q Sin C + Sin D = 2 Sin C + D Cos, , , , 2 , , C − D , , , 2 , , = 2 sin A cos A − [2 sin (B + C ) cos (B − C )], [Q Sin 2 A = 2 Sin A cos A], = 2 sin A cos A − [2 sin (180 ° − A) cos (B − C )], [Q A + B + C = 180], = 2 sin A cos A − 2 sin A cos (B − C ), [Q sin (180° − θ) = sin θ], = 2 sin A [cos A − cos (B − C )], , = − 2 sin A [cos (B + C ) + cos (B − C )], , sin 2 θ, , cos 2 θ + sin 2 θ, , cos θ = 1 + cos α, 1 − sin 2 θ = 1 + cos α, [Q sin 2 θ + cos 2 θ = 1], 1 − sin 2 θ = 1 + cos α ⇒ sin 2 θ + cos α = 0, , = 2 sin A [− cos (B + C ) − cos (B − C )], , cos θ, 1+, , 1 − cos α, , = 2 sin A [cos (180 − (B + C ) − cos (B − C )], , 2 sin θ, =, , (1 + cos α ) (1 − cos α ), , sin 2 A − sin 2B − sin 2 C = sin 2 A − [sin 2B + sin 2C], 2B + 2C , 2B − 2C , = sin 2 A − [2 sin , cos , , , , , , 2, 2, , Statement 1 is not correct, , ⇒, , =, , 2, , (a) −4sin A sin Bsin C, (c) −4 cos A cos Bsin C, , 2, , 1 − cos α, , equal to?, , RHS = ( A + B) ( A − B), , 2, , ⇒, ⇒, ⇒, , 1 − cos 2 α, , sin 2 α, 1 − cos α, , 38. If A + B + C = 180º, then what is sin 2A − sin 2B − sin 2C, , Statement 1, = A2 − AB + BA − B2, , are in AP, then, , ⇒ sin 2 α = cos 2 θ − cos 2 θ cos α, , So, ABC is an isosceles triangle, , Ê (b), , 1, 2, , [and cos ( A + B) cos ( A − B) = cos 2 A − sin 2 B], 2 cos θ cos α, 2, ⇒, =, ⇒cos 2 θ cos α = cos 2 θ − sin 2 α, cos θ cos 2 θ − sin 2 α, , sin A = sin B and sin B = sin C, , (a) 1 only, (c) Both 1 and 2, , (d), , 2 sec θ = sec (θ − α ) + sec (θ + α ), 2, 1, 1, ⇒, =, +, cos θ cos (θ − α ) cos (θ + α ), cos (θ + α ) + cos (θ − α ), 2, =, ⇒, cos θ, cos (θ − α ) (cos (θ + α ), ⇒, , ∴ S in A − sin B = 0 or sin B − sin C = 0, ⇒, , (c) –1, , Ê (a) If sec (θ − α ), sec θ and sec (θ + α ), , [C1 → C1 − C2 C2 → C2 − C3 ], sin B − sin C, sin C , sin A − sin B, ⇒, 0, 0, 1 = 0, , , 2, 2, 2, 2, 2, sin, A, −, sin, B, sin, B, −, sin, C, sin, C, , , ⇒, , (b) 1, , [Q sin 2 θ + cos 2 θ = 1], , = − 2 sin A [2 cos B cos C], , C + D, C − D , Q cos C + cos D = 2 cos 2 cos 2 , , , = − 4 sin A cos B cos C
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8, , NDA/NA Solved Paper 2018 (II), , of depression of the other end of the bridge from the, ballon is 48º. If the height of the balloon above the bridge, is 122 m, then what is the length of the bridge?, , ⇒, , (a) 122 sin 48º m, (c) 122 cos 48º m, , ⇒, , C, , H −h, x, , H −h, 1, =, 3, x, , ⇒, , x = 3 (H − h), H, = 3 (H − h) ⇒ H = 3 (H − h), 3, 3, H = 3 H − 3 h ⇒ 2H = 3 h ⇒ H = h, 2, , ⇒, , (b) 122 tan 42 º m, (d) 122 tan 48º m, , Ê (b), , tan 30° =, , In ∆BDE, , 39. A balloon is directly above one end of a bridge. The angle, , 42. What is/are the solution (s) of the trigonometric equation, , 48º, , cosec x + cot x = 3 where 0 < x < 2π ?, , 122 m, , (a), 48º, A, , B, , [Q cot (90° − θ) = tan θ], , ⇒, , (c) 330º, , Ê (a) We have, 3 (3 − tan 2, , ⇒, , A − cot A)2 = 1, , (a) 0, , π, π, and, 6, 3, , respectively. What is the height of the hill?, 3h, 2, , (c) h, , (d), , cosec 2 x − cot 2 x, cosec x − cot x, , cosecx − cot x =, , Ê, , (b) 1, , (c) 2, , (2 cos θ + 1 )10 (2 cos 2θ − 1 )10, , ⇒ (4 cos 2 θ − 1 )10 (2 cos 2 θ − 1 )10, ⇒ [2 (2 cos θ) − 1], 2, , 10, , ⇒ (2 cos 2 θ +, π, ⇒ 2 cos +, 4, , , h, , (2 cos 2θ − 1 ), , In ∆ACD, , ⇒, , 1), , 10, , 1, , , 1, + 1 , ⇒ 2 ×, , , 2, ...(i), , ⇒, ⇒, , 10, , (2 cos 4θ − 1 )10, , (2 cos 2 θ − 1 ), , 10, , (2 cos 2 θ − 1 ), 10, 2 cos π − 1 , , , 4, 10, , 2 cos π − 1 , , , 2, , C, X, , H, H, tan 60° =, ⇒ 3 =, x, x, H, x=, 3, , 10, , (2 cos 4θ − 1 )10, 10, , ⇒ [2 (1 + cos 2 θ) − 1], , E, , 60º, , (d) 4, , (2 cos 2θ − 1 )10 (2 cos 4θ − 1 )10, , H, , A, , ...(ii), , π, (b) If θ = , then, 8, , 10, , h, , 1, 3, , ⇒ (2 cos θ + 1 )10 (2 cos θ − 1 )10, , D, , 30º, , 1, = 3, cosecx − cot x, , (2 cos θ − 1 )10 (2 cos 4 θ − 1 )10, , h, 2, , Ê (b) height of building be h and let height of hill is H, B, , = 3 ⇒, , π, 8, (2 cos θ + 1)10 (2 cos 2θ − 1)10 (2 cos θ − 1)10 (2 cos 4θ − 1)10 ?, , 41. The top of a hill observed from the top and bottom of a, building of height h is at angles of elevation, , ...(i), , 43. If θ = , then what is the value of, , So, option (a) is correct, , (b), , π 5π, ,, 3 3, , Adding Eqs. (i) and (ii), we get, 1, 4, 2, 2 cosecx = 3 +, ⇒ 2 cosecx =, ⇒ cosecx =, 3, 3, 3, 3, ⇒, sin x =, 2, π 2π, ∴, x= ,, 3 3, , (d) 345º, , Checking through options A = 300, So,, 3 [3 − tan 2 300 ° − cot (300 )° ]2, = 3 [3 − tan 2 (360 ° − 60 ° ) − cot (360 ° − 60 )° ]2, = 3 [3 − tan 2 60 ° + cot 60 ° ]2, 2, 1 , 1, = 3 3 − 3 +, =3 × =1, , 3 , 3, , (a) 2h, , (d) π,, , (c) π only, , cosecx + cot x = 3, , trigonometric equation 3(3 − tan 2 A − cot A ) 2 = 1. Which, one of the following is a value of A ?, (b) 315º, , π, only, 3, , Now,, (cosec x + cot x) (cosec x − cot x), = 3, ⇒, cosec x − cot x, , 40. A is an angle in the fourth quadrant it satisfies the, , (a) 300º, , (b), , Ê (b) We have,, , BC, 122, ⇒ tan 48 ° =, AB, AB, 122, AB =, = 122 cot 48 °, tan 48 °, , tan 48° =, , = 122 cot (90 ° − 42° ), = 122 tan 42°, , 5π, only, 3, , (2 cos 4 θ − 1 )10, (2 cos 4 θ − 1 )10, , 10, , 2 ⋅ 1 − 1, , , , , 2, , Q θ = π , , 8 , 10, , (0 − 1 )10, , ( 2 + 1 )10 ( 2 − 1 )10 × 1, (( 2 )2 − (1 )2 )10 × 1 ⇒ (2 − 1 )10 ⇒1
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9, , NDA/NA Solved Paper 2018 (II), 44. If cos α and cos β (0 < α < β < π) are the roots of the, quadratic equation 4 x 2 − 3 = 0, then what is the value of, sec α × sec β ?, 4, (a) −, 3, , 4, (b), 3, , Ê (c) Let first term of a GP is a and common ratio is r, , 3, (d) −, 4, , 3, (c), 4, , ∴, , a2 = 2, , ⇒, , ar = 2, , and, , Sn = 8 ⇒, , ...(i), a, =8, 1 −r, , From Eq. (i), , Ê (a) Given,, , 2. −4, 1, 4. −, 4, , Which of the above values of x is/are the solution (s) of, the equation, π, tan −1 (2x ) + tan −1 (3x ) = ?, 4, (a) 3 only, , (b) 2 and 3, , Ê (a) We have,, , (c) 1 and 4, , (d) 4 only, , π, 4, π, − 1 2x + 3 x , tan , =, 1 − 2x ⋅ 3 x 4, , tan − 1 (2x) + tan − 1 (3 x) =, , ⇒, , 4r − 4r + 1 = 0, , ⇒, , 4r 2 − 2r − 2r + 1 = 0, , 47. If a, b, c are in AP or GP or HP, then, b, b, or 1 or, a, c, a, a, (c) 1 or or, b c, , (b), , ⇒, , −1, , a− b= b− c, a−b, =1, b− c, , If a, b, c are in GP, then, a b a−b, = =, ∴, b c b− c, , and tan a + tan b = π + tan, a + b, , , ab > 1, 1 − ab , π, 5x, 5x, = tan, ⇒, =1, ⇒, 4, 1 − 6 x2, 1 − 6 x2, , If a, b, c are in HP, then, 2ac, b=, ∴, a+ c, ⇒, , ab − ac = ac − bc, , ⇒, , ⇒, , a (b − c) = c (a − b), a−b a, =, b− c c, , ⇒, ⇒, , ⇒, , 1 − 6 x2 = 5x ⇒6 x2 + 5x − 1 = 0, 6 x2 + 6 x − x − 1 = 0, (6 x − 1 ) (x + 1 ) = 0 ⇒ x = − 1,, , 1, 6, , x = − 1 is not possible, 1, So,, x=, 6, , ⇒, , ab + bc = 2ac, , 48. What is the sum of all three digit numbers that can be, formed using all the digits 3, 4 and 5, when repetition of, digits is not allowed?, , 46. If the second term of a GP is 2 and the sum of its infinite, terms is 8, then the GP is, 1 1, (a) 8, 2, , , …, 2 8, 1 1, (c) 4, 2, 1, , 2 …, 2 2, , a −b, is equal to, b −c, , c, c, or or 1, a, b, a c, (d) 1 or or, b, a, , (a), , ∴, , ab ≤ 1, −1, , 8r − 8r 2 = 2, 2, , Ê (c) a, b, c are in AP, then, , a + b, [Q tan − 1 a + tan −1 b = tan − 1 , ,, 1 − ab , −1, , ⇒, , ⇒ 2r (2r − 1 ) − 1 (2r − 1 ) = 0 ⇒ (2r − 1 ) (2r − 1 ) = 0, 1, ⇒, r=, 2, ∴Put in Eq. (i), 1, a = 2 ⇒ a = 4, 2, 1, ∴GP is 4, 2, 1, ..., 2, , 45. Consider the following values of x, 1. 8, 1, 3., 6, , 2, =8, r (1 − r ), , ⇒, , 4x2 − 3 = 0 ⇒ 4x2 = 3, 3, 3, ⇒, x2 = ⇒ x = ±, 2, 4, 3, − 3, and cosβ =, So, cosα =, 2, 2, − 2 − 4, 2, , Now, secα ⋅ secβ =, ×, =, 3 3, 3, , 2 2, (b) 10, 2, , ,…, 5 25, 3. 3, (d) 6, 3, , ,…, 2 4, , (a) 2664, , (b) 3382, , (c) 4044, , (d) 4444, , Ê (a) Three digit number. that can be formed using 3, 4 and 5, , when repetition not allowed are 543, 534, 453, 435, 354,, 345, Sum = 543 + 534 + 453 + 435 + 354 + 345, = 2664
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11, , NDA/NA Solved Paper 2018 (II), 54. Consider the following statements, y = mx + c 1 and, , 1. The distance between the lines, | c − c2 |, y = mx + c 2 is 1, 1 + m2, , Ê (a) Given lines,, , x, y, x, y, + = 1 and + = 1, 2 3, 3 2, l1=0, , 2. The distance between the linesax + by + c 1 = 0 and, | c − c2 |, ., ax + by + c 2 = 0 is 1, a2 + b2, , l2=0, l1+λl2=0, , 3. The distance between the lines x = c 1 and x = c 2 is, | c 1 − c 2 |., , Equation of line passing through the point of, intersection are l1 + λl2 = 0, , Which of the above statements are correct?, , (3 x + 2 y − 6 ) + λ (2x + 3 y − 6 ) = 0 ...(i), , (a) 1 and 2, , (b) 2 and 3, , (c) 1 and 3, , (d) 1, 2 and 3, , Slope of line, − (3 + 2λ ), 4, =−, ∴, 2 + 3λ, 5, , Ê (b) Statement 1 Given lines,, , ⇒, ⇒, , + 15 + 10 λ = 8 + 12λ ⇒ 7 = 2λ, 7, λ=, 2, 7, (3 x + 2 y − 6 ) + (2x + 3 y − 6 ) = 0, 2, ⇒2 (3 x + 2 y − 6 ) + 14x + 21 y − 42 = 0, , y = mx + c1, , y = mx + c2, , and, , both lines are parallel, Q, , ⇒ 6 x + 4 y − 12 + 14x + 21 y − 42 = 0, P, , ⇒, , c −c, 1, 2, PQ = , , 2, 1, m, +, , , 20 x + 25 y − 54 = 0, , 56. What is the distance of the point (2, 3, 4) from the plane, 3x − 6y + 2z + 11 = 0 ?, , So, Statement 1 is correct, , (a) 1 unit, , Statement 2 for ax + by + c1 = 0, , Ê (a), , and, ax + by + c2 = 0, both lines are parallel so,, c −c , 1, 2, D =, ∴, , , , 2, a, b2, +, , , (b) 2 units, , Distance, , of, , point, , (x1 , y1 , z1 ), , ax + by + cz + d = 0 is, ax + by + cz +, 1, 1, 1, d =, , 2, 2, a, b, c2, +, +, , , (d) 4 units, from, , plane, , d, , , , , P (2, 3, 4), , x = c1, , Statement 3, , (c) 3 units, , x = c2, D = | c1 − c2 |, Statement 3 is correct, M, , Y, , X′, , 3 (2) − 6 (3 ) + 2 (4) + 1, PM = , , , 2, 2, 2, (3 ) + (− 6 ) + (2) , , X, , Y′, , x=c1, , 3x–6y+2z+11=0, , 6 − 18 + 8 + 11, =, =, 9 `+ 36 + 4 , , 7, 49, , PM = 1 unit, , x=c2, , 57. Coordinates of the points O, P, Q and R respectively:, 55. What is the equation of straight line passing through the, x y, x y, + = 1 and + = 1, 2 3, 3 2, and parallel to the line 4 x + 5y − 6 = 0 ?, point of intersection of the lines, (a) 20x + 25 y − 54 = 0, , (b) 25x + 20 y − 54 = 0, , (c) 4x + 5 y − 54 = 0, , (d) 4x + 5 y − 45 = 0, , (0, 0, 0), (4, 6, 2m), (2, 0, 2n) and (2, 4, 6) L, M, N and K, OR, OP, PQ and QR respectively such that LMNK is a, parallelogram whose two adjacent sides LK and LM are, each of length 2 ?, (a) 6, 2, (c) 3, 1, , (b) 1, 3, (d) None of these
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12, , NDA/NA Solved Paper 2018 (II), , Ê (c), , LMNK is a parallelogram we know that, if we join, mid-point of any quadrilateral we get a parallelogram, R, (2, 4, 6), , Q, (2, 0, 3n), , K, , L, , N, , O, (0, 0, 0), , 1. The angle between the planes 2x − y + z = 1 and, π, x + y + 2z = 3 is, 3, 2. The distance between the planes 6x − 3y + 6z + 2 = 0 and, 10, 2x − y + 2z + 4 = 0 is, 9, , Which of the above statement is/are correct?, , P, (4, 6, 2m), , M, , 59. Consider the following statements, , (a) 1 only, , So. M, N, K , L are mid- points of OP, PQ, QR and RO, respectively, 0 + 4 0 + 6 0 + 2m , ,, ,, ∴Coordinate of M = , = (2, 3, m), 2, 2, 2 , [Q Coordinate of mid-point, x + x2 y1 + y2 , ,, = 1, ], 2, 2 , , (b) 2 only, , Ê (c) Statements 1 Given,, , 2x − y + z = 1 and x + y + 2 z = 3, Here, a1 = 1, b1 = −1, c1 = 1, and a2 = 1, b2 = 1, c2 = 2, a1 a2 + b1 b2 + c1 c2, cos θ =, 2, a1 + b12 + c12 a22 + b22 + c22, =, , Coordinate of L = (1, 2 , 3 ), Coordinate of K = (2 , 2 , 3 + n), , =, , LM = 2, , Now,, , ⇒ (2 − 1 )2 + (3 − 2)2 + (m − 3 )2 = 2, 1 + 1 + (m − 3 ) 2 = 2, , ⇒, , ∴, , Squaring both side,, ⇒, , 2 + (m − 3 ) 2 = 2 ⇒ (m − 3 ) 2 = 0, , ⇒, , m=3, , ⇒ (2 − 1 ) + (2 − 2) + (3 + n − 3 ) = 2, 2, , 2, , 2, , 1 + 0 + n2 = 2, , ⇒, Squaring both side,, ⇒, , 58. The line, , 1 + n2 = 2 ⇒ n2 = 1 ⇒ n = 1, , x −1 y −2 z −3, is given by, =, =, 2, 2, 4, , (a) x + y + z = 6, x + 2 y − 3 z = − 4, (b) x + 2 y − 2 x = − 1, 4x + 4 y − 5 z − 3 = 0, (c) 3x + 2 y − 3 z = 0, 3x − 6 y + 3 z = − 2, (d) 3x + 2 y − 3 z = − 2, 3x − 6 y + 3 z = 0, , Ê (d) Given lines,, , x−1, , ∴, , and, , x−1, , 2, y−2, z −3, 4, , 3, , 2, , =, , y−2, 3, , =, , z −3, 4, , = λ (let), , = λ ⇒ x = 2λ + 1, = λ ⇒ y = 3λ + 2, , = λ ⇒ z = 4λ + 3, , by checking options if 3 x + 2 y − 3 z, then, 3 (2λ + 1 ) + 2 (3 λ + 2) − 3 (4λ + 3 ), = 6 λ + 3 + 6 λ + 4 − 12λ − 9 = − 2, 3 (2λ + 1 ) − 6 (3 λ + 2) + 3 (4λ + 3 ), ⇒6 λ + 3 − 18 λ − 12 + 12λ + 9 = 0, So, option (d) is correct, , 2 × 1 + (− 1 × 1 ) + 1 × 2, 22 + (− 1 )2 + (1 )2, , 1 2 + 1 2 + 22, , 4 −1, 4+ 1 + 1, , 4+ 1 + 1, , 3, 3 1, =, = =, 6 6 6 2, π, θ=, 3, , So, Statement 1 is correct., Statement 2 Distance between two planes, , LK = 2, , Again,, , (c) Both 1 and 2 (d) Neither 1 nor 2, , ax + by + cz + d1 = 0, ax + by + cz + d2 = 0, , , d1 − d2, distance S = , , , 2, 2, 2, a + b + c , 2, d1 = , d2 = 4, 3, 2, 10, , 4−, = 3 = 10 = 10, , 3, distance =, 22 + 1 2 + 22 9 3 × 3 9, , , , 60. Consider the following statements, Statement I If the line segment joining the points, P(m, n ) and Q (r , s ) subtends an angle α at the origin, then, ms − nr, ., cos α =, 2, (m + n 2 ) (r 2 + s 2 ), Statement II In any triangle ABC, it is true that, a 2 = b 2 + c 2 − 2bc cos A., Which one of the following is correct in respect of the, above two statements?, (a) Both Statement I and Statement II are true and Statement II is the, correct explanation of statement I., (b) Both Statement I and Statement II are true, but Statement II is not, the correct explanation II is not the correct explanation of, Statement I, (c) Statement I is true, but Statement II is false, (d) Statement I is false, but Statement II is true
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13, , NDA/NA Solved Paper 2018 (II), Ê (d), , The equation of circle is x2 + y2 + gx + fy +, , Statement I If the line segment joining the point, P(m, n) and Q(r, s ) subtends angle α at origin, then, mr + ns, cosα =, 2, m + n2 r 2 + s 2, , y −f , Q centre = − ,, and radius =, 2 2 , , So, Statement I is not correct, , Since, circle touches Y-axis, then, , In any triangle ABC, 2, , 2, , ∴circle touches Y-axis at, 2, f, f2, −f, y2 + fy +, = 0 ⇒ y + = 0 ⇒ y =, , 2, 2, 4, −f , , A 0,, ∴, , 2 , , 61. What is the area of the triangle with vertices, , 1 , 1 , 1, x 1, , x 2 , , x 3 , ?, x1 , x2 , x3, , , ∴Normal at A will pass through centre C and intersect, circle again at B., −f , ∴Coordinates of B are − g,, , , 2 , , (a)| ( x 1 − x 2 ) ( x 2 − x 3 ) ( x 3 − x 1 )|, (b) 0, (x − x 2 ) (x 2 − x 3 ) (x 3 − x1 ) , , (c) 1, x 1x 2x 3, , , (x1 − x 2 ) (x 2 − x 3 ) (x 3 − x1 ) , , , (d), 2 x 1x 2x 3, , , , r, , r, , r, , r, , r, , and only if, , r, r, (a) a and b are perpendicular, , r, r, (b) a and rb are parallel, r, (c) a and b are inclined at an angle of 45º, r, r, (d) a and b are anti-parallel, , , 1 , 1 , 1, A x1 , , B x2 , , C x3 , , x2 , x3 , x1 , , x1, 1, x2, 2, , x3, , , , 1, , 1, x1, 1, x2, 1, x3, , x1, 1, A = x2, 2, x3, , y2, y3, , 1, 1 =, , 1, , , x1, 1, = x2, 2, , x3, , , 1, x1, 1, x2, 1, x3, , , 1, x12 1 x1 , , 1, x2 1 x , 1 ⇒, 2, 2, , x, x, x, 2, 1 2 3 x2 1 x , , 3, 3, , , , 1, , , y1, , , 1, , 1, , , , (x − x2 ) (x2 − x3 ) (x3 − x1 ) , = 1, , 2x1 x2 x3, , , , r, , r, r, r, r, r r, (a + b) ⋅ (a + b) = | a |2 + | b|2 + 2a ⋅ b, r, r, r, r, a ⋅ b = b⋅ a, r r, r r, a⋅ b=0⇒ a ⊥ b, , r, r, 64. If r = xi$ + yj$ + zk$, then what is r ⋅ (i$ + $j + k$ ) equal to ?, , Ê, , (a) x, (b) x + y, (c) − (x + y + z) (d) (x + y + z), r, $, $ + yj$ + zk, (d) r = xi, r, Now, r ⋅ (i$ + $j + k$ ), i$. i$ = 1, = (xi$ + yj$ + zk$ ) ⋅ (i$ + $j + k$ ) = ( x + y + z ), , 2i$ − $j + k$ and 3i$ − 4 $j − k$ is, 1 $, 1 $, 1 $, i +, j −, k, 3, 3, 3, 1 $, 1 $, 1 $, (c), i −, j −, k, 3, 3, 3, (a), , c, = 0, then the normal at this point, 4, intersects the circle at the point., x 2 + y 2 + gx + fy +, , g, f, (b) − g , − (c) − , f , , 2 , 2, , (d) (− g , − f ), , -g , -f, ( 2 2(, , x′, , a× b, r, n = r r, | a × b|, , B, , x, y′, , 1 $, 1 $, 1, i +, j + k$, 2, 2, 2, 1 $, 1 $, 1 $, (d), i +, j +, k, 3, 3, 3, (b), , Ê (a) We know that,, r r, , y, , -1, ( 0, 2 ( A, , r, , Ê (a) Given,r | a | ≠ 0r,| b| ≠ 0, , 65. A unit vector perpendicular to each of the vectors, , 62 If Y-axis touches the circle, , O, , r, , r, , 63. Let | a | ≠ 0, | b | ≠ 0. ( a + b) ⋅ ( a + b) = | a | 2 + | →b | 2 holds if, , Ê (d) Area of ∆ABC, , Ê (b), , g, g2 f 2 − c, f2 c, =, =, +, ⇒, 2, 4, 4 4, 4, 4, , ⇒, , a = b + c − 2bc cos A, Statement II is correct, , g f, (a) − , , 2 2, , g2 f 2 c, +, −, 4, 4 4, , AC = radious, , Statement II, 2, , c, =0, 4, , ∴, , $j, i$, k$ , r r , a × b= 2 −1 1 , , , , , 3 − 4 −1, , , r, r, a ×r b = 5i$ + 5 $j − 5k$, r, | a × b| = 5 3, r 5 (i$ + $j − k$ ) i$ + $j − k$, n =, =, 5 3, 3
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14, , NDA/NA Solved Paper 2018 (II), r, , r, , r, , r, , 66. If | a | = 3, | b | = 4 and | a − b | = 5, then what is the value of, r r, | a + b| ?, , 70. In a triangle ABC, if taken in order, consider the following, statements, , (a) 8, , (b) 6, , (c) 5 2, , (d) 5, , r, |a|= 3, r, r r, | b| = 4 and | a − b| = 5, r, r, a⋅ b=0, r, r r 2, r, | a + br| = | a |2 | + | b|2 + 0 = 25, r, | a + b| = 5, , r, r, r, (a)| A | > | B | > | C |, r, r, r, (c)| A | = | B | =| C |, , −→, , −→, , −→, , −→, , −→, , −→, , −→, , −→, , −→, , −→, , −→, , −→, , How many of the above statement are correct?, , r, , r, , r, , 68. What is ( a − b) × ( a + b) equal to?, r r, (b) a × b, , r r, (c) 2 (a × b), , r, r, (d)| a |2 − | b|2, , Ê (c) Wer have,, r, , r r, (a − b) × (a + b), r r r r r r r r, = a × a + a × b− b× a − b× b, r r, r r r r, r r, = a × b− b× a, [Q a × a = 0, b × b = 0], r, r, r, r, r, r, r, r, r r, = a × b + a × b [Q a × b = − b × a] = 2 (a × b), , 69. A spacecraft located at i$ + 2 $j + 3k$ is subjected to a force, λk$ by firing a rocket. The spacecraft is subjected to a, moment of magnitude, (b) 3λ, , (c) 5λ, , (d) None of these, , Ê (c) We have, r, , r, r = i$ + 2 $j + 3 k$ and F = λk$, We know that,r, r, Moment = r × F = (i$ + 2 $j + 3 k$ ) × (λk$ ), ∴ i$ × k$ = − $j , , , = − λ$j + 2λi$ $j × k$ = i$ , k$ × k$ = 0 , , , , Magnitude of moment =, =, , (a) One, (c) Three, , (− λ )2 + (2λ )2, , λ2 + 4λ2 = 5 λ2 = 5 λ, , (b) Two, (d) Four, , Ê (a) We know that,, In a triangle ABC, −→, , r, r, r, (b)| A | = | B | ≠ | C |, r, r, r, (d)| A | ≠ | B | ≠| C |, , r r r, A = a + b+ c, r r r, B= a − b+ c, r r r r, C = a − b− c, r, r, r, | a | = | b| = | c | = 1, r, r, r, r, r r, | a ⋅ b| = | b ⋅ c | = c ⋅ a = 0, r, Now, | A | = a 2 + b2 + c2 = 1 + 1 + 1 = 3, r, | B | = a 2 + b2 + c2 = 1 + 1 + 1 = 3, r, |C | = a 2 + b2 + c2 = 1 + 1 + 1 = 3, r, r, r, ⇒| A | = | B | = |C |, , (a) λ, , −→, , 4. BA − BC + CA = 0, , Ê (c) Given that,r, , r, (a) 0, , −→, , 3. AB − BC + CA = 0, , r, r r, 67. Let a, b and c be three mutually perpendicular vectors, r r r r r r r r, each of unit magnitude. IfA = a + b + c, B = a − b + c and, r r r r, C = a − b − c , then which one of the following is correct?, , r, , −→, , 2. AB + BC − CA = 0, , Ê (d) Given that,, ∴, , −→, , 1. AB + BC + CA = 0, , −→, , −→, , −→, , AB + BC + CA = O [Q by triangle law], So, only first statement is correct., , 71. Let the slope of the curve y = cos − 1 (sin x ) be tan θ. Then, the value of θ in the interval (0, π ) is, (a), , π, 6, , 3π, 4, , (b), , (c), , π, 4, , (d), , π, 2, , Ê (b) We have,, , y = cos − 1 (sin x), differentiation w.r.t. x, we get, −1, dy, =, cos x, dx, 1 − sin 2 x, − cos x, =, [Q sin 2 x + cos 2 x = 1], cos x, dy, = −1, dx, Slope of the curve = tan θ, ∴, tan θ = − 1, π, tan θ = − tan, ⇒, 4, π, tan θ = tan π − , ⇒, [Q θ ∈ (0, π )], , 4, 3π, ⇒, θ=, 4, , 72. If f ( x ) =, , x −1, x−4, , , defines a function on R, then what is its, , domain ?, (a) ( − ∞, 4) ∪ ( 4, ∞ ), (c) (1, 4) ∪ ( 4, ∞ ), , Ê (d) We have,, f (x ) =, ∴, and, ⇒, So,, , (b) [4, ∞ ), (d) [1, 4) ∪ ( 4, ∞ ), , x−1, x−4, , x−1≥0, x− 4≠0, x≥1, x≠4, x ∈ [1, 4) ∪ (4, ∞ ), Domain = [1, 4) ∪ (4, ∞ )
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15, , NDA/NA Solved Paper 2018 (II), , f ( x ) − f (1), equal to?, x →1, x −1, , 73. Consider the function, , 76. If f ( x ) = 25 − x 2 , then what is lim, , sin 2x , if x ≠ 0, , f ( x ) = 5x, 2, if x = 0, ,, , 15, , (a) −, , Which one of the following is correct in respect of the, function?, (a) It is not continuous at x = 0, (c) It is not continuous at x = π, , (b) It is continuous at every x, , (b), , 1, 4 3, , 1, , (d), , 4 3, , f (x) = 25 − x2, , Now,, , lim, , f (x) − f (1 ), , x→ 1, , x−1, − 2x, , 2 25 − x2, 25 − x2 − 24, =, = lim, x→ 1, x→1, 1 −0, x −1, =−, , −0, , 5 − 2 tan x , dy, equal to?, , then what is, dx, 2 + 5 tan x , , (a) −, , 74. For the function f ( x ) = | x − 3,| which one of the following, is not correct?, , 1, 2 x, , Ê (a) We have,, , (a) The function is not continuous at x = −3, (b) The function is continuous at x = 3, (c) The function is differentiable at x = 0, (d) The function is differentiable at x = −3, , Ê (a) We have, f (x) =| x − 3 |, We know that, modulus function is continuous in R., , Let,, , So, option (a) is incorrect., , 75. If the function f ( x ) =, , 2x − sin − 1 x, 2x + tan, , −1, , 1, 3, , (b), , 1, 3, , (c), , x, , 2, 3, , (b) 1, , (c) − 1, , (d), , 1, 2 x, , 5 − 2 tan x , y = tan − 1 , , 2 + 5 tan x , 5 − tan x , , −1 2, y = tan , , 5, 1 + tan x , , , 2, , tan, A − tan x , 5, = tan A = tan − 1 , , 2, 1 + tan A tan x , = tan − 1 [tan( A −, 5, y = tan − 1 − x, 2, , is continuous at each, , point in its domain, then what is the value of f (0)?, , [by L-Hosptial], , 1, 24, , 77. If y = tan − 1 , , lim f (x) ≠ f (0 ), , x→ 0, , at x = 0 function is discontinuous., , differentiation w.r.t x, we get, , (d) 2, , x)] = A −, , x, , 1, dy, =−, 2 x, dx, , 78. Which one of the following is correct in respect of the, function, , Ê (b) We have,, f (x ) =, , 2x − sin, , −1, , 2x + tan, , −1, , x, , f ( x ) = x sin x + cos x +, , x, , Function is continuous at each point, ∴, , f (0 ) = lim f (x), x→ 0, , Now, lim f (x) = lim, x→ 0, , ∴, , (c) −, , lim, , sin 2x, ,x ≠ 0, , f (x) = 5x, ,x = 0, 2, 15, lim f (x), sin 2x × 2 2, 2, at x = 0, = ⇒ f (0 ) =, = lim, x→ 0, 5, 5, x→0, 5x × 2, , (a) −, , 1, 24, , Ê (a) We have,, , (d) It is continuous at x = 0, , Ê (a) We have,, , ∴, , 1, 24, , 2x − sin − 1 x, , 2x + tan − 1 x, 1, 2−, 1 − x2, = lim, 1, x→ 0, 2+, 1 + x2, 2−1 1, =, =, 2+ 1 3, 1, f (0 ) =, 3, x→ 0, , [by L-Hospital], , 1, cos 2 x ?, 2, , π, (a) It is increasing in the interval 0, , 2, π, (b) It remains constant in the interval 0, , 2, π, (c) It is decreasing in the interval 0, , 2, π π, (d) It is decreasing in the interval , , 4, 2 , , Ê (a) We have,, , f (x) = x sin x + cos x +, , 1, cos 2 x, 2, , f ′ (x) = x(cos x) + sin x − sin x +, f ′ (x) = x cos x − sin x cos x, , 1, ⋅ 2 cos x (− sin x), 2
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16, , NDA/NA Solved Paper 2018 (II), By checking options, we put, π, x=, 4, π, π, π, π, f ′ (x) = cos − sin cos, 4, 4, 4, 4, π 1, 1, 1, π, 1, = ⋅, −, ×, =, − >0, 4, 2, 2, 2 4 2 2, π, So, f (x) is increasing in the interval 0, , 2, , 79. What is lim, , 1 − cos θ, θ, , θ→ 0, , equal to?, 1, (d) −, 2 2, , 80. A function f : A → R is defined by the equation, f ( x ) = x 2 − 4 x + 5, where A = (1, 4 ). What is the range of, the function?, (b) (1, 5), , (c) [1, 5), , (d) [1, 5], , Ê (c) We have,, , A function f : A → R is defined by, f (x) = x2 − 4x + 5,, Where, A = (1, 4), Let,, y = x2 − 4x + 5, dy, = 2x − 4, dx, dy, Now,, = 0 ⇒ 2x − 4 = 0, dx, , (b) 3, , 5, , 8, , =−, , 5, , 8, , ∫2 (x − 5) dx + ∫5 (x − 5) dx, 5, , 83. What is ∫ sin 3 x cos x dx equal to ?, (a) cos 4 x + C, (1 − sin2 x )2, (c), +C, 4, , (b) sin4 x + C, (1 − cos 2 x )2, (d), +C, 4, , Where C is the constant of integration., , Ê (d) We have,, 3, ∫ sin x cos x dx, Let,, , sin x = t, , cos x dx = dt =, , ∫, , t 3 dt, , t4, sin 4 x, +C =, +C, 4, 4, (sin 2 x)2, =, +C, 4, (1 − cos 2 x)2, =, +C, 4, =, , 2, 1, 1, , 2, , [Q sin 2 x + cos 2 x = 1], , 4, , 84. What is ∫ e, , x = 2, y = 1, , At x = 1 , y = (1 ) − 4 (1 ) + 5 = 2, 2, , y ∈[1, 5), b, , a, , a, , 81. What is ∫ [x ] dx + ∫ [− x ] dx equal to, where [,] is the, greatest integer function?, (b) a − b, , dx equal to, (b) ln| sec x | + C, (d) e tan x + C, , Where C is the constant of integration., , Range = [1, 5), b, , ln (tan x ), , (a) ln| tan x | + C, (c) tan x + C, , At x = 4 y = (4)2 − 4 (4) + 5 = 5, , (a) b − a, , 8, , x2, , , x2, − 5x, =−, − 5x + , 2 2, 2, 5, 25, 25, , , , , =− , − 25 − (2 − 10 ) + (32 − 40 ) − , − 25 , 2, , , , 2, 25, 25, =9, =, −8−8+, 2, 2, , 5, , So,, , (d) 9, , ∫2 | x − 5| dx|+ ∫5 | x − 5| dx, b, c, b, [Q ∫ f (x) dx = ∫ f (x) dx + ∫ f (x) = dx], a, a, c, , x=2, , At, , (c) 4, , Ê (d) We8 have,, ∫2 | x − 5| dx, =, , θ, 1 − 1 − 2 sin 2 , , 2, = lim, θ→ 0, θ, θ, 2θ, 2 sin, 2 sin, 2 = lim, 2= 1, = lim, θ, θ→ 0, θ→ 0, θ, 2, ×2, 2, , ⇒, , = − (x)ba = − (b − a ) = a − b, , (a) 2, 1, (c), 2, , θ, , (a) (2, 5), , ∫a, , [Q[x] + [− x] = − 1, if x ∈, / z], (− 1 ) dx, , 2, , 1 − cos θ, , θ→ 0, , b, , 82. What is ∫ | x − 5 | dx equal to?, , Ê (c) We have,, lim, , =, 8, , (b) 2 2, , (a) 2, , Ê (b) Web have, b, ∫a[x] dx + ∫a [− x] dx, b, = ∫ ([x] + [− x]) dx, a, , (c) 0, , (d) 2(b − a), , Ê (b) We have, ln (tan x ), dx, ∫e, [Q eln (x ) = x], = ∫ tan x dx = log| sec x| + C
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17, , NDA/NA Solved Paper 2018 (II), 1, , d, dx, − 1, , 85. What is ∫ , , tan − 1 1 dx equal to ?, , , , x , , (b) −, , (a) 0, , π, 4, , (c) −, , π, 2, , Ê (b) We have,, (d), , π, 2, , have,, Ê (d) We, 1, d , −1 1, ∫ dx tan x dx, −1, , = tan − 1, , π, π, = +, 4, 4, , 1, , 1 = tan − 1 (1 ) − tan − 1 (− 1 ), , x −1, π, =, 2, , 86. In which one of the following intervals is the function, f ( x ) = x 2 − 5x + 6 decreasing?, , (a) (− ∞, 2 ], , Ê (a) We have,, , (b) [3, ∞ ), , (c) (− ∞, ∞ ), , (d) (2, 3), , f (x) = x2 − 5x + 6, f ′ (x) = 2x − 5, , For decreasing, f ′ (x) < 0, 2x − 5 < 0, 5, x<, 2, x < 2.5, x ∈ (− ∞, 2.5), , (c), , d y, dx 2, d 2y, dx 2, , (b), , + ay = 0, , Ê (d) We have,, , (d), , d y, dx 2, d 2y, dx 2, , are arbitrary, , − ay = 0, + a y=0, 2, , y = p cos (ax) + q sin (ax), , differentiation. w. r. t. x, we get, dy, = − p a sin ax + q a cos ax, dx, Again, differentiation. w. r. t. x, we get, d2 y, = − pa 2 cos ax − qa 2 sin ax, dx2, d2 y, = − a 2 ( p cos ax + q sin ax), ⇒, dx2, d2 y, d2 y, ⇒, = − a2 y ⇒, + a2 y = 0, 2, dx, dx2, , 88. The equation of the curve passing through the point, ( −1, − 2), which satisfies, , dy, 1, = − x 2 − 3 , is, dx, x, , (a) 17 x 2 y − 6x 2 + 3x 5 − 2 = 0, (b) 6x 2 y + 17 x 2 + 2 x 5 − 3 = 0, (c) 6xy − 2 x 2 + 17 x 5 + 3 = 0, (d) 17 x 2 y + 6xy − 3x 5 + 5 = 0, , y=, , ⇒, , − x3, 3, , +, , 1, 2x2, , +C, , ...(i), , given curve is passing through (− 1, − 2), 1 1, 5, − 2 = + + C ⇒− 2 = + C, 3 2, 6, − 17, −5, C = −2, ⇒C=, ⇒, 6, 6, Put in Eq. (i), − x3, − 2x5 + 3 − 17 x2, 1, 17, ⇒y=, + 2 −, y=, 3, 6, 2x, 6 x2, ⇒, , 6 x2 y = − 2x5 + 3 − 17 x2, , ⇒, , 6 x2 y + 2x5 + 17 x2 − 3 = 0, , (a) 1, , 2, , − a2 y = 0, , Integrating both sides, 1, 2, , ∫ dy = ∫ − x − x3 dx, x3 x− 2 , ⇒, y=−, −, + c, 3, −2, , solution is y = a cos x + b sin x + ce − x + d , where a, b, c, and d are arbitrary constants?, , y = p cos (ax ) + q sin (ax ), where p , q, constants, is, (a), , − x2 − 1 dx, , , , x3 , , 89. What is the order of the differential equation whose, , 87. The differential equation of the family of curves, 2, , 1, dy, = − x2 − 3 ⇒ dy =, dx, x, , (b) 2, , (c) 3, , (d) 4, , Ê (d) We have,, , y = a cos x + b sin x + ce− x + d,, , a, b, c and d are arbitrary constants., We know that,, order = number of arbitrary constant, So, order = 4, , 90. What is the solution of the differential equation, dy , ln = ax + by ?, dx , (a) ae ax + be by = C, (c) ae ax + be − by = C, , 1, 1, (b) e ax + e by = C, a, b, 1, 1, (d) e ax + e − by = C, a, b, , Ê (d) We have,, , dy, ln = ax + by, dx , dy, dy, = eax + by ⇒, = eax ⋅ eby, dx, dx, dy, = eax dx, eby, , Integrating both sides, − by, , ∫e, ⇒, ⇒, , dy = ∫ eax dx, e, eax, =, +C, a, −b, − by, , eax, e− by, +, + C =0, a, b
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18, , NDA/NA Solved Paper 2018 (II), , 91. If u = e ax sin bx, , and v = e ax cos bx , then what is, , du, dv, equal to?, u, +v, dx, dx, , θ., , r, , (b) ( a 2 + b 2 )e ax, (d) ( a + b ) e ax, , (a) ae 2ax, (c) ab e, , Ê (c) Let radius of sector be r and angle subtended at centre be, , 2ax, , θ, , Ê (a) We have,, , ∴, , u = eax sin bx, , ...(i), , = eax (− b sin bx + a cos bx), Now we have,, dv, du, + v, dx, dx, , 1, , 94. What is the minimum value of [x ( x − 1) + 1] 3 , where, a≤ x ≤1?, , = eax sin bx [eax (b cos bx + a sin bx)], + eax cos bx [eax (− b sin bx + a cos bx)], [b sin bx cos bx + a sin 2 bx] + e2 ax, (− b cos bx sin bx + a cos 2 bx), , 1, , 3 3, (a) , 4, , 92. If y = sin (ln x ), then which one of the following is, , Ê (a) We have,, , dx, , (c) x 2, , 2, , + y=0, , d 2y, dx, , 2, , +x, , dy, + y=0, dx, , (b), , d 2y, dn2, , (d) x, , =0, , d 2y, dx 2, , −x, , dy, + y=0, dx, , differentiation. w. r. t. x, we get, dy, dy cos (log e x), = cos (log e x), =, ⇒x, dx, dx, x, Again, differentiation. w. r. t. x, we get, d2 y dy − sin (log e x), =, x 2 +, dx, x, dx, 2, dy, 2 d y, + x, = − sin (log e x), x, ⇒, dx, dx2, dy, d2 y, ⇒, =− y, x2 2 + x, dx, dx, dy, d2 y, + y=0, ⇒ x2 2 + x, dx, dx, , 93. A flower-bed in the form of a sector has been fenced by a, wire of 40 m length. If the flower-bed has the greatest, possible area, then what is the radius of the sector?, (b) 20 m, , (c) 10 m, , (d) 5 m, , 1/ 3, , y = (x 2 − x + 1 ) 3, 1, , 1, at x = , y is minimum y =, 2, , 3 3, , 4, , 95. If y = | sin x || x | , then what is the value of, (a), , y = sin (log e x), , (a) 25 m, , 3, (d) , 8, , 1, 2, , 1, , Where C is the constant of integration, , Ê (c) We have,, , (c), , 1 1, 3, y = x2 − x + − + 1 , , , 4 4, , correct?, d 2y, , (b) 1, , 1, , = e2 ax [a sin 2 bx + a cos 2 bx] = ae2 ax, , (a), , 40 − 2r, , r, 1 2, r θ, 2, 1, 1 2 40 − 2r , A= r , ⇒ A = r (40 − 2r ), , 2, r, 2 , 1, 2, A = (40r − 2r ), 2, dA 1, differentiating. w. r. t. r,, = (40 − 4r ), dr 2, dA, Now,, = 0 ⇒ 40 − 4r = 0 ⇒ r = 10 m, dr, , differentiation. w. r. t. x, we get, dv, = eax (− b sin bx) + cos bx (aeax ), dx, , = e2 ax, , 2r + rθ = 40 ⇒ θ =, , Area of the sector =, , Now, v = eax cos bx, , u, , r, , r + r + rθ = 40, , ⇒, , differentiation. w. r. t. x, we get, du, = eax (b cos bx) + sin bx (aeax ), dx, du, = eax (b cos bx + a sin bx), dx, , rθ, , (c), , 2, , 2, , −, , π, 6, , −, , π, 6, , dy, π, at x = − ?, dx, 6, , π, , ( 6 ln 2 −, 6, , 3π), , ( 6 ln 2 +, 6, , 3π), , 2 6 ( 6 ln 2 +, (b), 6, (d), , 2, , −, , π, 6, , ( 6 ln 2 −, 6, , 3π), , 3π), , Ê (a) We have, y =|sin x|| x |, ∴, , x=−, , π, , sin x < 0, x < 0, 6, , y = (− sin x)− x, Q, y = f (x) g (x ) , So, dy, g (x ), , = f (x ) g ( x ) , f ′ (x) + log| f (x)|⋅ g ′ (x), dx, f (x ), , dy, = (− sin x)− x, dx, (− x ), , ⋅ (− cos x) + log|(− sin x)|(− 1 )|, , (, sin, ), −, x, , , π, , − π6, dy, x = − π / 6 = − sin , , , , 6 , dx , ,
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19, , NDA/NA Solved Paper 2018 (II), , π, , π, = sin 6, , 6, π, 1 6, , = , 2, −, , =2, , 96. What is, , π, 6, , , π /6, π , − cos − , , 6, −, −, 6, sin, (, π, /, ), , , π, , − log| − sin − , 6 , , − π, 6 3, 1, log, −, , 1 2 , 2, , , , 2, , − 3π, 1, − log = 2− π / 6, , 2, 6, 6 log 2 − 3 π , , , 6, , , , d 1 − sin 2x, dx, , 2, 3π, log − 6 , , , , equal to, where, , (a) cos x + sin x, (c) ± (cos x + sin x ), , (b) − (cos x + sin x ), (d) None of these, , 1, a tan x , tan− 1 , , b , ab, , (b) C −, , (c) C +, , 1, b tan x , tan− 1 , , , ab, b , , (d) None of these, , , , x, , 1, a tan x , tan− 1 , , b , ab, , Where C is the constant of integration., , Ê (a) We have,, dx, ∫ a2 sin 2 x + b2 cos 2 x, Divide numerator and denominator by cos 2 x, sec2 x dx, I=∫ 2, a tan 2 x + b2, Let, , f (x + y ) = f (x ) f ( y ), f (x) = 1 + x g (x) ⋅ φ(x), , and, , Thus, f ( y) = 1 + yg ( y) φ ( y), Where lim g (x) = a and lim φ(x) = b, x→ 0, , y→ 0, , equal to ?, , (a) C +, , (d) ab f (x ), , y→ 0, , = f (x) ⋅ a ⋅ b = ab f (x), , d, |cos x − sin x|, dx, d, Q π < x < π ∴ sin x > cos x], (sin x − cos x), =, 4, dx, 2, = cos x − (− sin x) = cos x + sin x, , a 2 sin 2 x + b 2 cos 2 x, , Ê (d) Let,, , (c) ab, , = f (x) lim g ( y) lim φ( y), , =, , dx, , (a) 1 + abf (x ) (b) 1 + ab, , using first principal, f (x + y ) − f (x ), f ′ (x) = lim, y→ 0, y, f (x ) ⋅ f ( y ) − f (x ), = lim, y→ 0, y, f (x)[ f ( y) − 1], = lim, y→ 0, y, 1 + yg ( y)φ( y) − 1 , lim f (x) , , y→ 0, y, , , , Ê (a) We have,, , 97. What is ∫, , andf ( x ) = 1 + xg ( x ) φ ( x ),, , where lim g ( x ) = a and lim φ ( x ) = b. Then, what is f ′ ( x ), x→ 0, x→ 0, equal to?, , x→ 0, , π, π, <x< ?, 4, 2, , π, π, d, 1 − sin 2 x, < x <, dx, 4, 2, d, cos 2 x + sin 2 x − 2 sin x cos x, =, dx, Q sin 2 x + cos 2 x = 1, , and sin 2 x = 2 sin x cos, , f ( x + y ) = f ( x ) f (y ), , 98. Let, , a tan x = t, a sec2 x dx = dt, dt, sec 2 x dx =, a, 1, 1 1, dt, t, I= ∫ 2, = × tan − 1 + C, b, a b, a t + b2, , , 1, 1, x, dx = tan − 1 + C , Q ∫ 2, 2, a, a, x + a, , , 1, − 1 a tan x , =, tan , +C, , ab, b , , 99. What is the solution of the differential equation, dx x + y + 1, ?, =, dy x + y − 1, , (a) y − x + 4 ln ( x + y) = C, (c) y − x + ln ( x + y) = C, , (b) y + x + 2 ln ( x + y) = C, (d) y + x + 2 ln ( x + y) = C, , Where C is an arbitrary constant., dx, , x+ y+1, , Ê (c) We have, dy = x +, Let, , ⇒, ⇒, ⇒, ⇒, ⇒, , y−1, , x+ y=u, dx, du, +1=, dy, dy, dx du, =, −1, dy dy, u+1, du, −1 =, dy, u−1, du u + 1, =, +1, dy u − 1, u+ 1 + u−1, du, =, dy, u−1, u − 1, dy, 2u, =, ⇒ , du = 2 dy, dy u − 1 u , , Integrating both sides,, 1, , ∫ 1 − u du =, , ∫ 2 dy, , ⇒, , u − log u = 2 y + C, , ⇒, ⇒, ⇒, , x + y − log (x + y) = 2 y + C, x − log (x + y) = y + C, y − x + log (x + y) = C, , [Q u = x + y]
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20, , NDA/NA Solved Paper 2018 (II), , 100. What is lim, , π, x→, 6, , (a) −, , 1, 2, , 2 sin 2 x + sin x − 1, 2 sin x − 3 sin x + 1, 2, , (b) −, , 1, 3, , Ê (d) In group of men, let number of men = a, , equal to?, , X1 = 26 yr and n1 = a (let), , and in group of women, number of women = b (let), , (c) − 2, , 2 sin 2 x + 2 sin x − sin x − 1, , and combined mean X = 25, n X1 + n2 X 2, Now,, X= 1, n1 + n2, 26 a + 21 b, ⇒, 25 =, ⇒25a + 25b = 26 a + 21 b, a+ b, , 2 sin 2 x − 2 sin x − sin x + 1, , ⇒, , 2 sin x (sin x + 1 ) − (sin x + 1 ), , ⇒, , Ê (d) We have,, lim, x→, , π, 6, , 2 sin 2 x + sin x − 1, , 2 sin 2 x − 3 sin x + 1, , = lim, x→, , π, 6, , = lim, x→, , π, 6, , = lim, x→, , π, 6, , 2 sin x (sin x − 1 ) − 1 (sin x − 1), , 104. If sin β is the harmonic mean of sin α and cos α and sin θ is, , (sin x − 1 ) (2 sin x − 1 ), , the arithmetic mean of sin α and cos α, then which of the, following is/are correct?, π, , 1. 2 sin α + sin β = sin 2α, , 4, , = −3, , 101. If two dice are thrown and atleast one of the dice shows 5,, then the probability that the sum is 10 or more is, 1, 6, , (b), , 4, 11, , (c), , 3, 11, , (d), , 2, 11, , Ê (c) Let A be event of dice shows 5 and B be the event that the, sum is 10 or more, , n(S) = 36, , Here,, , n( A) = { (1, 5), (2, 5), (3, 5), (4, 5), (5, 5),, (6, 5), (5, 1 ), (5, 2), (5, 3 ), (5, 4), (5, 6 ) }, n(B) = { (5, 5), (6, 4), (4, 6 ), (6, 5), (5, 6 ), (6, 6 ) }, n( A ∩ B) = { (5, 5), (6, 5), (5, 6 ), 3, B 36, , P =, A 11, 36, 3, = ., 11, , 4b = a, a 4, =, b 1, , By checking options (d) is correct., , (sin x + 1 ) (2 sin x − 1 ), , 3, 1, +1, sin x + 1 2, = lim, =, = 2, π sin x −, 1 1 −1 − 1, x→, 6, 2, 2, , (a), , X 2 = 21 yr, , (d) − 3, , P (B ∩ A) , B, , Q P =, , A, P( A) , , , π, , 2. 2 sin θ = cos α − , , 4, , Select the correct answer using the code given below., (a) 1only, , (b) 2 only, , (c) Both 1 and 2, , (d) Neither 1 nor 2, , Ê (c) Given that,, sinβ is HM of sinα and cosα, 2 sin α cos α, So,, sin β =, sin α + cos α, ⇒, , sin β (sin α + cos α ) = sin 2α, 1, 1, sin α +, cos α , ⇒ sin β × 2 , 2, , 2, , π, π, ⇒ sin β × 2 cos sin α + sin cos α , , , 4, 4, , observations is 0.8. Then the percentage of variation not, explained by linear regression is, , = sin 2α, π, ⇒ 2 sin β sin α + = sin 2α, , 4 , , π, , ⇒ 2 sin α + sin β = sin 2α, , 4, , (a) 80%, , Statement I is true., , 102. The correlation coefficient computed from a set of 30, (b) 20%, , (c) 64%, , (d) 36%, , Ê (b) Given that,, correlation coefficient = 0.8 = 80%, if the relation is 80%, explained, then 20% of variation will not explained by, near regression., , Now, sinθ is AM of sin α and cosα, sin α + cos α, sin θ =, 2, , (a) 20, 80, , (b) 40, 60, , 2 sin θ = sin α + cos α, 1, 1, ⇒2 sin θ = 2 , sin α +, cos α , 2, , 2, π, π, , ⇒2 sin θ = 2 sin sin α + cos cos α , , , 4, 4, π, 2 sin θ = cos α − , ⇒, , 4, , (c) 60, 40, , (d) 80, 20, , Hence, Statement II is also correct., , 103. The average age of a combined group of men and women, is 25 yr. If the average age of the group of men is 26 yr and, that of the group of women is 21 yr, then the percentage, of men and women in the group is respectively, , ⇒, , = sin 2α
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21, , NDA/NA Solved Paper 2018 (II), 25, 2, ×, 100, 100, =, 25, 2, 35, 4, 40, 5, ×, +, ×, +, ×, 100 100 100 100 100 100, 25 × 2, =, 25 × 2 + 35 × 4 + 40 × 5, 50, 50, 5, =, =, =, 50 + 140 + 200 390 39, , 105. Let A, B and C be three mutually exclusive and exhaustive, events associated with a random experiment. If, P( B ) = 15, . P( A ) and P(C ) = 05, . P( B ), then P( A ) is equal to, (a), , 3, 4, , (b), , 4, 13, , (c), , 2, 3, , (d), , 1, 2, , Ê (b) We have,, , 3, P( A), 2, 1, P(C ) = 0.5 P(B) = P(B), 2, 1 3, 3, = × P( A) = P( A), 2 2, 4, P(B) = 1.5 P( A) =, , and, , 107. 8 coins are tossed simultaneously. The probability of, getting atleast 6 heads is, (a), , Now, A, B and C are mutually exclusive and exhaustive, events, P( A) + P(B) + P(C ) = 1, 3, 3, P( A) + P( A) = 1, 2, 4, 3 3, , =1, P( A) 1 + +, , 2 4 , 13, 4, P( A) = 1 ⇒ P( A) =, 4, 13, , 106. In a bolt factory, machines X ,Y , Z manufacture bolts that, are respectively 25%, 35% and 40% of the factory’s total, output. The machines X , Y , Z respectively produce 2%, 4%, and 5% defective bolts. A bolt is drawn at random from, the product and is found to be defective. What is the, probability that it was manufactured by machine X ?, (a), , 5, 39, , (b), , 14, 39, , (c), , 20, 39, , (d), , 34, 39, , (c), , 37, 256, , (d), , 229, 256, , 1, = , 2, , 8, , 1, = , 2, , 8, , 8 probability of, Probability of, , 1, , 2, , 8, , (8 C6 + 8 C7 + 8 C8 ), (28 + 8 + 1 ) =, , 37, 256, , 108. Three groups of children contain 3 girls and 1 boy; 2 girls, and 2 boys; 1 girl and 3 boys. One child is selected at, random from each group. The probability that the three, selected consist of 1 girl and 2 boys is, (a), , 13, 32, , (b), , 9, 32, , (c), , 3, 32, , (d), , 1, 32, , Ê (a) In first group, 3 girls and 1 boy, , Ê (a) Let, , P(G ) =, , A : bolt manufactured from machine, B : bolt manufactured from machine, C : bolt maufactured from machine, and E : bolt is defective, ∴, , 57, 64, , 8 coins are tossed simultaneously i.e n =, 1, 1, getting head p = , so q = 1 − p = ., 2, 2, getting atleast 6 heads., 1, 6, 2, 7, 1, 1, 1, 1, = 8 C6 + 8 C7 + 8 C8, 2 2, 2 2, , ⇒ P( A) +, , ⇒, , (b), , Ê (c) We have,, , So,, , ⇒, , 7, 64, , 25, 100, 35, P(B) = 35% =, 100, 2, 40, E, and P = 2% =, P(C ) = 40% =, A, 100, 100, 4, E, P = 4% =, B, 100, 5, E, P = 5% =, C, 100, P( A) = 25% =, , Probability of defective bulb that was manufactured by, machine X,, E, P( A) ⋅ P , A, A, , P =, E, E, E, P( A) ⋅ P + P(B) ⋅ P , A, B, E, + P(C ) ⋅ P , C, , 3, 1, and P(B) =, 4, 4, , In second group, 2 girls and 2 boys, 2 1, 2 1, P(G ) = = and P(B) = =, 4 2, 4 2, One child is selected at random from each group and, consists of 1 girl and 2 boys are GBB or BGB or BBG., So, required probability, 3 1 3 1 1 3 1 1 1, = × × + × × + × ×, 4 2 4 4 2 4 4 2 4, 9, 3, 1, 13, =, +, +, =, 32 32 32 32, , 109. Consider the following statements, 1. If 10 is added to each entry on a list, then the average, increases by 10., 2. If 10 is added to each entry on a list, then the standard, deviation increases by 10., 3. If each entry on a list is doubled then the average doubles., , Which of the above statements are correct?, (a) 1, 2 and 3, , (b) 1 and 2, , (c) 1 and 3, , (d) 2 and 3
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22, , NDA/NA Solved Paper 2018 (II), Statement 1 is correct, , Ê (c) Statement 1 The average is affected by the change of the, origin., , For Statement 2, P( A ∩ B ) = 1 − P( A ∪ B), = 1 − [P( A) + P(B) − P( A ∩ B)], = 1 − [P( A) + P(B) − 2 P( A ∩ B) + P( A ∩ B)], = 1 − [q + p] = 1 − q − p, , So, if 10 is added to each entry on list then average, increase by 10. Statement 1 is correct., Statement 2 Standard deviation is independent on, change in origin. Statement 2 is incorrect., Statement 3 The average is affected by change in scale, in same ratio as each entry is changes., So, Statement 3 is correct., , 110. The variance of 25 observations is 4. If 2 is added to each, observation, then the new variance of the resulting, observations is, (a) 2, , (b) 4, , (c) 6, , (d) 8, , Ê (b) We know that,, , Statement 2 is also correct., , 113. If the regression coefficient of Y on X is −6 and the, 1, correlation coefficient between X and Y is − , then the, 2, regression coefficient of X on Y would be, (a), , So, if 2 is added to each observation, then variance is, remain same, , variables X and Y with geometric means P and Q, X, is, respectively, then the geometric mean of, Y, P, (b) antilog , Q, (d) n (log P + log Q ), , 1, 2, , (i = 1, 2, 3,...., n) ⇒ P = (x1 × x2 × x3 × ..... xn, Q = ( y1 × y2 × y3 × ......× yn, x, Now, geometric mean of, y, and, , 1, )n, , and B is p and the probability that exactly one of A, B, occurs is q, then which of the following is/are correct?, 1. P ( A ) + P ( B ) = 2 − 2p − q, , 1, 6, , 1, =, 2, , − 6 × bxy, , 1, 1, = − 6 × bxy ⇒ bxy = −, 4, 24, , set, of, bivariate, observations ( x 1 , y 1 ),, ( x 2 , y 2 )....( x n , y n ) are such that all the values are distinct, and all the observations fall on a straight line with, non-zero slope. Then the possible values of the, correlation coefficient between x and y are, (b) 0 and −1, , (c) 0, 1 and −1 (d) −1and 1, , Ê (d) Given that,, All the observations fall on a straight line with, non-zero, slope then if slope is positive then r = 1, and if slope is negative then r = −1, So, values of the correlation coefficient between x and y, are −1 and 1., , 1, , 112. If probability of simultaneous occurrence of two events A, , (d), , 114. The, , 1, )n, , x, x, x n P, = 1 × 2 × ....× n =, y, y, yn , Q, 1, 2, , byx × bxy ⇒ −, , Squaring both sides, ⇒, , (a) 0 and 1, , Ê (a) We have, xi > 0, yi > 0,, , 115. Two integers x and y are chosen with replacement from, the set [0, 1, 2, ...., 10]. The probability that | x − y| > 5 is, (a), , 6, 11, , (b), , Ê (c) Given that,, , 35, 121, , (c), , 30, 121, , (d), , 25, 121, , S = { 0, 1, 2, ..., 10 }, , n(S) = 11 × 11 = 121, , 2. P ( A ∩ B ) = 1 − p − q, , Select the correct answer using the code given below., (a) 1 only, (c) Both 1 and 2, , byx = − 6 and rxy = −, , rxy =, , 111. If x i > 0, y i > 0 (i = 1, 2, 3,..., n ) are the values of two, , (c) n (log P − log Q ), , 1, 24, , We know that,, , Hence, variance = 4, , P, Q, , 1, 6, , (b) −, , Ê (b) Given that,, , Variance is independent on change in origin., , (a), , (c) −, , 1, 24, , (b) 2 only, (d) Neither 1 nor 2, , Ê (c) Given that,, , P( A ∩ B) = P and P( A) + P(B) − 2P ( A ∩ B) = q, , Now, Statement 1, Given, P( A) + P(B) − 2P ( A ∩ B) = q, ⇒, , 1 − P ( A ) + 1 − P (B ) − 2 p = q, , ⇒, , P ( A ) + P (B ) = 2 − 2 p − q, , Now,, , x − y>5, , Now, E is the set of element such that| x − y|> 5, E = { (6, 0 ), (0, 6 ), (7, 1 ), (1, 7 ), (8, 2),, (2, 8 ), (9, 3 ), (3, 9 ), (10, 4), (4, 10 ), (7, 0 ),, (0, 7 ), (8,1 ), (1, 8 ), (9, 2), (2, 9 ), (10, 3 ),, (3, 10 ), (8, 0 ), (0, 8 ), (9, 1 ),, (1, 9 ), (10, 2), (2, 10 ), (9, 0 ), (0, 9 ),, (10, 1 ), (1, 10 ) , (10, 0 ), (0, 10 ) }, n(E ) = 30, 30, So, required probability =, 121
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23, , NDA/NA Solved Paper 2018 (II), Possible ways, , 116. An analysis of monthly wages paid to the workers in two, , = { (6, 5, 1 ), (6, 5, 2), (6, 5, 3 ), (6, 5, 4), , firms A and B belonging to the same industry gives the, following result, Firm A, , Firm B, , Number of workers, , 500, , 600, , Average monthly wage, , ` 1860, , ` 1750, , Variance of distribution, of wages, , 81, , 100, , (6, 4, 1 ), (6, 4, 2), (6, 4, 3 ), (6, 3, 2), (6, 3, 1 ) (6, 2, 1 ), So, possible ways = 10, Case II If x = 5, Then, possible ways = { (5, 4, 3 ), (5, 4, 2), (5, 4, 1 ), (5, 3, 2), (5, 3, 1 ), (5, 2, 1 )}, So, possible ways = 6, Case III If x = 4, , The average of monthly wage and variance of, distribution of wages of all the workers in the firms A and, B taken together are, (a) ` 1860, 100, (c) ` 1800, 81, , Then, possible ways = { (4, 3, 2), (4, 3, 1 ), (4, 2, 1 ), So, possible ways = 3, , (b) ` 1750, 100, (d) None of these, , Case IV If x = 3, Then, possible ways = 1 (3, 2, 1 ), , Ê (d) For firm A, , So, required possible outcomes, , n1 = 500, X1 = 1860, and variance = σ12 = 81, So,, S.D. = σ1 = 9, For firm B, n2 = 600, X 2 = 1750, and variance σ 22 = 100, So,, S.D. = σ = 10, n X1 + n2 X 2, Now, combined mean X = 1, n1 + n2, 500 × 1860 + 600 × 1750, =, 500 + 600, , = 10 + 6 + 3 + 1 = 20, , 118. Which one of the following can be obtained from an, ogive?, , =, =, , (d) Mode, , than or more than type ogive curve where both curve, cuts each other that point median., , 119. In any discrete series (when all values are not same), if x, represents mean deviation about mean and y represents, standard deviation, then which one of the following is, correct?, , X = 1800, , Q, , (b) Median, , (c) Geometric Mean, , Ê (b) We know that, median is always calculated from less, , Combined variance, n (σ 2 + d12 ) + n2 (σ 22 + d22 ), = 1 1, n1 + n2, =, , (a) Mean, , (a) y ≥ x, (c) x = y, , 500 [81 + (− 60 )2 ] + 600 [100 + (50 )2 ], , Ê (d) We know that,, , 500 + 600, , MD =, , d1 = X − X1 = 1800 − 1860 = − 60, , ⇒, , and d2 = 1800 − 1750 = 50, 500 (81 + 3600 ) + 600 (100 + 2500 ), 1100, 500 (3681 ) + 600 (2600 ), 1100, , = 3091.36, , marked I, II, and III and rolled. Let x, y and z represent the, number on die-I, die-II and die-III, respectively. What is, the number of possible outcomes such that x > y > z ?, (b) 16, , (c) 18, , (d) 20, , Ê (d) Three dice having digit 1, 2, 3, 4, 5 and 6 and given that, x > y > z., , So, possiblities are, Case I If x = 6, x> y> z, , 4, S.D., 5, , 5MD = 4.S.D., , ⇒, , 5x = 4 y, , ∴, , x< y, , [Q MD = x and SD = y], , 120. In which one of the following cases would you expect to, , 117. Three dice having digits 1, 2, 3, 4, 5 and 6 on their faces are, , (a) 14, , (b) y ≤ x, (d) x < y, , get a negative correlation?, (a) The ages of husbands and wifes, (b) Shoe size and intelligence, (c) Insurance companies profits and the number of claims they have, to pay, (d) Amount of rainfall and yield of crop, , Ê (c), , In negative correlation, if x is increases then y is, decreases by checking options Insurance companies, profits and the number of claims they have to pay are, negatively correlated.
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NDA / NA, National Defence Academy/Naval Academy, , Solved Paper, , 2018 (I), , Paper 1 (Mathematics), 1. If n ∈ N , then 121n − 25n + 190n − ( − 4 )n is divisible by, which one of the following?, (a) 1904, , (b) 2000, , (c) 2002, , (d) 2006, , 3. In the expansion of (1 + x ) 43 , if the coefficients of (2r + 1), th and (r + 2) th terms are equal, then what is the values of, r (r ≠ 1) ?, (a) 5, , Ê (b) We have,, , On putting n = 1, we get, , ∴ General term, Tr, , (121 )1 − (25)1 + (1900 )1 − (−4)1, , Now,, , = 121 − 25 + 1900 + 4 = 2000, , (d) 22, , Tr, , +1, +1, + 2, , =, , 43, , =, , 43, , =, , 43, , Cr xr, C2r x2r, Cr, , +1, , xr, , +1, , Now, according to the question, Coefficients of (2r + 1 ) th and (r + 2) th terms are equal, , 1, 1, 1, 2. If n = (2017 )! , then what is, +, +, log 2 n log 3 n log 4 n, 1, equal to ?, +…+, log 2017 n, , ∴, , 43, , ⇒, , (b) 1, , C2r =, , 43, , Cr, , +1, , 2r + r + 1 = 43 [Q if nCx = nC y ⇒ x + y = n], , ⇒, , 3r = 42, , ⇒, , r = 14, , 4. What is the principal argument of ( − 1 − i ) , where, , (d) n, , i = − 1?, , Ê (b) We have,, , 1, 1, 1, 1, +, +, + .... +, log 2 n log 3 n log 4 n, log 2017 n, , (a), , = log n 2 + log n 3 + log n 4 + .... + log n 2017, , 1 , Q log b a = log b , a , , [Q log a + log b = log ab], , = log n (1 ⋅ 2 ⋅ 3 ⋅ 4......2017 ), = log n (2017 )!, , T2r, , and, , Which is divisible by 2000., , = log n (2 ⋅ 3 ⋅ 4......2017 ), , (c) 21, , Ê (b) We have, (1 + x)43, , 121 n − 25n + 1900 n − ( − 4)n, , (a) 0, n, (c), 2, , (b) 14, , [Q n(n − 1 )(n − 2) K 3 ⋅2⋅1 = n!], , = log (2017 )! (2017 )!, , [Q n = 2017 !], , =1, , [Q log a a = 1], , π, 4, , (b) −, , Ê (c) Let z − 1 − i, Now, tan α =, ∴, , π, 4, , (c) −, , |b| | − 1 |, =, |a| | − 1 |, , α = tan −1 (1 ) =, , arg( z ) = α − π, =, , (d), , 3π, 4, , [Q a = − 1, b = − 1], π, 4, , Since a, b both are negative,, ∴, , 3π, 4, , π, −3 π, − π=, 4, 4
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25, , NDA/NA Solved Paper 2018 (I), 5. Let α and β be real number and z be a complex number. If, z 2 + αz + β = 0 has two distinct non-real roots with Re, (z ) = 1, then it is necessary that, (a) β ∈ ( −1, 0), (c) β ∈ (1, ∞ ), , (b) |β| = 1, (d) β ∈ ( 0, 1), , ⇒, , z 2 + αz + β = 0, , (x + iy) + α (x + iy) + β = 0, , ⇒, , x2 − y2 + 2ixy + αx + iαy + β = 0, , ⇒, , x2 − y2 + αx + β = 0 and 2xy + αy = 0, , 9. If the ratio of AM of GM of two positive numbers a and b is, , ⇒ x2 − y2 + αx + β = 0 and (2x + α ) y = 0, , 5 : 3, then a : b is equal to, [Q y ≠ 0], , ⇒ x − y + αx + β = 0 and 2x + α = 0, 2, , ⇒, , x − y + αx + β = 0 and α = − 2x, , ⇒, , x2 − y2 + αx + β = 0 and α = − 2, , 2, , (a) 3 : 5, , (b) 2 : 9, , ⇒, , β = y2 + 1, , ⇒, , β ∈ (1, ∞ ), , 2, , 6. Let A and B be subsets of X and C = ( A ∩ B ′ ) ∪ ( A ′ ∩ B ),, , ⇒, , (a) ( A ∪ B′ ) − ( A ∩ B′ ), (c) ( A ∪ B) − ( A ∩ B), , ⇒, , C = ( A ∩ B′ ) ∪ ( A′ ∩ B), [Q X ∩ Y ′ = x − y], , = ( A ∪ B) − ( A ∩ B), , [from venn diagram], B, , (A–B), , ab, , (B–A), U, , with the digits 5, 6, 7, 8, 9, if the repetition of digits is not, allowed?, (c) 120, , (d) 60, , Ê (d) Number lying between 100 and 1000 are of three digit., Since the numbers are to be formed with 5, 6, 7, 8, 9 and, repetition is not allowed, so total number of numbers, = 5 × 4 × 3 = 60, , 8. The number of non-zero integral solution of the equation, | 1 − 2i | x = 5 x is, , (a) Zero (no solution), (b) One, (c) Two, (d) Three, , (a + b)2 10 2, 10, ⇒, = , 3 , ab, 3, , 100, =, 9, ab, 100, a b, + + 2=, 9, b a, 1, 100, t + + 2=, t, 9, 2, t + 1 + 2t 100, =, 9, t, , ⇒, ⇒, , 9 t 2 − 82t + 9 = 0, , ⇒, , (t − 9 ) (9 t − 1 ) = 0 ⇒, , ∴, , a, a 1, = 9 or =, b, b 9, , ⇒, , 7. How many numbers between 100 and 1000 can be formed, , =, , ab , , , a 2 + b2 + 2ab, , ⇒, , = ( A − B) ∪ (B − A), , A, , a+ b, , ⇒, , where A′ and B′ are complements of A and B respectively, in X . What is C equal to ?, (b) ( A ′ ∪ B) − ( A ′ ∩ B), (d) ( A ′ ∪ B′ ) − ( A ′ ∩ B′ ), , 2 =5, 3, ab, , Q A :G = 5 : 3, A = a + b , G =, , 2, , [Q y ≥ 0 ⇒ y + 1 ≥ 1], 2, , (d) 5 : 3, , a+ b, , According to the question,, , [Q x = 1, α = − 2], , 1 − y2 − 2 + β = 0, , (b) 5 3, , (c) 9 : 1, , Ê (c) Let a and b be two numbers., , [QRe( z ) = 1 = x], , Ê (c) We have,, , x =0, , ∴ Given equation has no solution., , On comparing,, , 2, , ⇒, , But x is non-zero integral., , ⇒ (x2 − y2 + αx + β ) + (2xy + αy) i = 0, , 2, , a 2 + b2 ], , 5x / 2 = 5x, x, =x, [Q a m = a n ⇒ m = n], 2, x, x, x− =0 ⇒, =0, 2, 2, , 2, , [Q i2 = −1], , [Q a + ib =, , ( 1 + 4 )x = 5x ⇒ ( 5 )x = 5x, , ⇒, , ⇒, , (a) 3 5, , ( (1 ) + (−2)2 )x = 5x, , ⇒, , Now, we have, , |1 − 2i|x = 5x, 2, , ⇒, , Ê (c) Let z = x + iy, , ⇒, , Ê (a) We have,, , Q a = t , , b, , t = 9,, , 1, 9, Q t = a , , b , , a : b = 9 : 1 or 1 : 9, , 10. If the coefficients of a m and a n is the expansion of, , (1 + a )m + n are α and β, then which one of the following is, correct?, (a) α = 2β, (c) 2α = β, , Ê (b) We have, , (b) α = β, (d) α = ( m + n)β, , (1 + a )m +, , n, , =, , m+ n, , =, , m+ n, , and coefficient of a =, , m+ n, , ∴, , Tr, , +1, m, , ∴ Coefficient of a, , n, , ∴, and, ∴, , Cr ar, , Cm, , [Qr = m], , Cn, , [Qr = n], , α = m + nCm, β = m + nCn, = m + nCm + n − n, = m + nCm = α, α =β, , [Q n Cr = nCn − r ]
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26, , NDA/NA Solved Paper 2018 (I), , 11. If x + log15 (1 + 3 x ) = x log15 5 + log15 12, where x is an, integer, then what is x equal to?, (a) − 3, , (b) 2, , (c) 1, , (d) 3, , Ê (c) We have, x + log15 (1 + 3 x ) = x log15 5 + log15 12, , ⇒ log15 15 + log15 (1 + 3 ) = log15 5 + log15 12, [Q log a a = 1 and log b a m = m log b a], x, , x, , 15. If α and β are different complex numbers with | α | = 1,, then what is, (a)|β|, , Ê (c) We have,, , α −β, , 1 − αβ, , [Q log a + log b = log ab], , ⇒, ⇒, ⇒, ⇒, , [where y = 3 x ], , y (1 + y) = 12, , =, , y + y − 12 = 0, ( y + 4)( y − 3 ) = 0 ⇒ y = − 4, 3, 3 x = − 4, 3, 3x = 3, , =, , 12. How many four-digit numbers divisible by 10 can be, formed using 1, 5, 0, 6, 7 without repetition of digits?, (c) 44, , (d) 64, , Ê (a) We have to form four digit numbers which are divisible, , by 10 and using 1, 5, 0, 6, 7. Since numbers must be, divisible by 10, so unit place must be zero., , ∴ Total number of such numbers = Permutations of, three digits using 1, 5, 6, 7, 4!, = 4 P3 =, = 4! = 24, (4 − 3 )!, , [Q|z | = |z |], [Q|α| =1], , [Q x < 1 ⇒|1 − x| = 1 − x], , x2 − x − 4 = 0, 1±, x=, x=, , 1 − 4 (1 ) (− 4), 2, , 1 − 17, 2, , − (1 − x) + x2 = 5, , =, , 1±, , 17, 2, [Qx < 1], , [Q x ≥ 1 ⇒|1 − x| = − (1 − x)], , ⇒ −1 + x+ x =5, 2, , Hindi or Mathematics but not English?, , ⇒, , (d) 130, , x2 + x − 6 = 0, , ⇒ (x + 3 ) (x − 2) = 0 ⇒ x = − 3, 2, , 14. What is the number of students who are good in Hindi, , ⇒, , and Mathematics but not in English?, , [Qx ≥ 1], , x=2, , ∴Given equation has a rational root and an irrational, root., , (d) 8, , 17. The binary number expression of the decimal number 31, , H, , M, 54, , |α − β|, |α ||α − β|, |α − β|, , Case II When x ≥ 1, , 13. What is the number of students who are good in either, , Solution (Q. Nos. 13-14), , 1 α −β, |α| α − β, , 1 − x + x2 = 5, , ⇒, , (c) 10, , =, , Case I When x < 1, , In a class, 54 students are good in Hindi only, 63 students are good, in Mathematics only and 41 students are good in English only., There are 18 students who are good in both Hindi and, Mathematics. 10 students are good in all three subjects., , (b) 12, , α −β, α (α − β ), , |1 − x| + x2 = 5, , ⇒, , (a) 18, , [Q|α| = 1 ⇒|α|2 = 1 ⇒ α ⋅ α = 1], , Ê (a) We have,, , (Q. Nos. 13-14) Consider the information given, below and answer the two items that follow, , (c) 125, , α −β, αα − αβ, , (a) a rational root and an irrational root, (b) two rational roots, (c) two irrational roots, (d) no real roots, , Directions, , (b) 107, , (d) 0, , 16. The equation |1 − x | + x 2 = 5 has, , ⇒, , (a) 99, , (c) 1, , |α ||α − β|, 1, =, =1, |α|, , [Q3 x ≠ − 4], , x=1, , (b) 36, , =, =, , 2, , ⇒, , (a) 24, , (b) 2, , 15x (1 + 3 x ) = 12 ⋅ 5x ⇒ 3 x (1 + 3 x ) = 12, , ⇒, , equal to ?, , x, , ⇒ log15 [15x (1 + 3 x )] = log15 (5x × 12), ⇒, , α −β, 1 − αβ, , 8, 10, , 63, , 41, E, , Ê (c) From given Venn-diagram, , is, (a) 1111, , (b) 10111, , Ê (d), 2, , 31, , 2, , 15, , 1, , 2, , 7, , 1, , 2, , 3, , 1, , 2, , 1, , 1, , n (H ∪ M ∪ E ′ ) = 54 + 8 + 63 = 125, , Ê (d) From given Venn-diagram n (H ∩ M ∩ E ′ ) = 18 − 10 = 8, , (c) 11011, , 0, , ∴( 31)10 = (11111)2, , 1, , (d) 11111
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27, , NDA/NA Solved Paper 2018 (I), 18. What is i 1000 + i 1001 + i 1002 + i 1003 equal to (where, , n + 2 , if n is even, , = 2, n+1, , if n is odd, , 2, ∴ Number of terms in, , i = −1) ?, (a) 0, , (c) − i, , (b) i, , (d) 1, , Ê (a) We have,, , i1000 + i1001 + i1002 + i1003, = i1000 [1 + i + i2 + i3 ], [Q i2 = − 1, i3 = − i], , = i1000 [1 + i − 1 − i], =0, , 1, 1, 1, 1, 19. What is, +, +, +…+, log100 N, log 2 N log 3 N log 4 N, equal to ( N ≠ 1) ?, (a), , 1, , (b), , log100! N, , 1, , log 99! N, , 1, , Ê (a) We have, log, , 2, , N, , +, , 99, (c), log100! N, , 99, (d), log 99! N, , [Q log a + log b = log ab], [Q n! = n (n − 1 ) (n − 2) ... 2 ⋅ 1], , π, π, (a) 2 cos + i sin , , 3, 3, π, π, , (c) 4 cos + i sin , , 3, 3, , π, π, (b) 2 cos + i sin , , 6, 6, π, π, , (d) 4 cos + i sin , , 6, 6, , (b) 8, , (c) 11, , (d) 12, , ∴55 + 7 5 is divisible by 5 + 7 = 12 as m = 5 is odd., , 23. If x = 1 − y + y 2 − y 3 … up to infinite terms, where, | y | < 1, then which one of the following is correct?, 1, 1+ y, y, (c) x =, 1+ y, , 1, 1− y, y, (d) x =, 1− y, , (b) x =, , ( 3 )2 + (1 )2, , 4=2, b, Now, amp ( z ) = tan −1 , a, π, −1 1 , = tan , =, 3 6, , a 2 + b2 ], , = 3+1=, , Q tan π = 1 , 6, 3 , , , z = r (cos θ + i sin θ), π, π, = 2 cos + i sin , , 6, 6, , [Q r = | z | = 2 and θ = amp ( z ) =, , π, ], 6, , (b) 5, (d) 11, , a, , r < 1], 1 −r, , 1, 1+ y, , cos θ sin θ 0, , , 24. What is the inverse of the matrix A = − sin θ cos θ 0 ?, 0, 0, 1, , cos θ − sinθ 0, , , (a) sinθ cos θ 0, 0, 0, 1, , , cos θ 0 − sinθ, , , (b) 0, 1, 0 , sinθ 0 cos θ , , , , 0, 0 , 1, , , (c) 0 cos θ − sinθ, 0 sinθ cos θ , , , , cos θ sinθ 0, , , (d) − sinθ cos θ 0, 0, 0, 1, , , Ê (a) We have,cos θ, , 21. What is the number of non-zero terms in the expansion of, (1 + 2 3x )11 + (1 − 2 3x )11 (after simplification)?, , x = 1 − y + y2 − y3 + ... ∞,|y|< 1, 1, =, 1 − (− y ), [Q a + ar + ar 2 + K ∞ =, =, , 3 + i, , [Q z = a + ib ⇒|z | =, , (a) 4, (c) 6, , which the number 5 5 + 7 5 is divisible?, , Ê (a) We have,, , 20. The modulus-amplitude form of 3 + i, where i = − 1 is, , ∴, , 22. What is the greatest integer among the following, by, , (a) x =, , = log N (100 )!, 1, =, log (100 )! N, , |z | =, , 12, =6, 2, , (xm + ym ) is divisible by (x + y)., , = log N 1 ⋅ 2 ⋅ 3 ⋅ 4...100, , ∴, , =, , [Q n =11, is odd], , Ê (d) We know that when m is odd then, , = log N 2 + log N 3 + log N 4 + ... + log N 100, , 1 , Q log b a = log b , a , , , Ê (b) Let z =, , (1 + 2 3 x)11 + (1 − 2 3 x)11, 11 + 1, =, 2, , (a) 6, , 1, 1, 1, +, + .... +, log 3 N log 4 N, log100 N, , = log N 2 ⋅ 3 ⋅ 4...100, , Ê (c) In (a + b)n + (a − b)n , number of terms, , sin θ 0, , , A = − sin θ cos θ 0, , , 0, 1, 0, |A| = 1 [cos 2 θ − (− sin 2 θ)] = 1 ≠ 0, cos θ 0, = cos θ, C11 =, 0, 1, − sin θ 0, C12 = −, = sin θ, 0, 1, − sin θ cos θ, C13 =, =0, 0, 0
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28, , NDA/NA Solved Paper 2018 (I), sin θ 0, = − sin θ, 0, 1, cos θ 0, = cos θ, 0, 1, cos θ sin θ, −, =0, 0, 0, sin θ 0, =0, cos θ 0, cos θ 0, −, =0, − sin θ 0, , C21 = −, C22 =, C23 =, C31 =, C32 =, C33 =, , cos θ, , sin θ, , − sin θ cos θ, , 27. What is the number of triangles that can be formed by, choosing the vertices from a set of 12 points in a plane,, seven of which lie on the same straight line?, (a) 185, , =, , (a) C ( n + 1, r ), , cos θ − sin θ 0, 1 , sin θ cos θ 0, , 1 , 0, 1, 0, , T, , cos θ − sin θ 0, = sin θ cos θ 0 , , , 0, 1 , 0, , 25. If A is a 2 × 3 matrix and AB is a 2 × 5 matrix, then B must, (b) 5 × 3 matrix, (d) 5 × 2 matrix, , Ê (d) We have,, , C (n1r ) + 2C (n, r − 1 ) + C (n, r − 2), = nCr + 2 ⋅ nCr, = ( Cr + Cr, n, , n, , =, , n+1, , =, , n+1+1, , Cr +, , −1, , + nCr, , n, − 1 ) + ( Cr, , n+1, , Cr =, , m = 3 and n = 5, , Cr, , −1, n+ 2, Cr, , (c) 8, , (d) − 8, , Ê (a) We have,, , − 2), , [Q n Cr + nCr, , = C ( n + 2, r ), , −1, , (c) Two, , =, , n+1, , Cr ], , (d) Three, , ∴, , x2 − 4x + 0 = 0, , ⇒, , x2 − 4x = 0, , [Q x∈[0, 1] ⇒[x] = 0], , x (x − 4) = 0 ⇒ x = 0, 4, x=0, , [Q x∈[0, 1]], , Case II x∈[1, 2], ∴, ⇒, , x2 − 4x + 1 = 0, 4±, x=, , ⇒, , x=2±, , 16 − 4, 2, , [Q x∈[1, 2] ⇒ x = 1], 4± 2 3, ⇒ x=, 2, , 3 ⇒ x = 0 .268, 3.732, , No solution, , 1 2, A=, , 2 3 , , [Q x∈[1, 2] ], , ∴Given equation has only one solution i.e. x = 0., 1 2, 2 3 , , , 1 ⋅ 2 + 2 ⋅ 3 5 8 , =, 2 ⋅ 2 + 3 ⋅ 3 8 13 , , 2, 3 , , Now, it is given that, A2 − kA − I 2 = 0, 5 8 k 2k 1 0 , ⇒, 8 13 − 2k 3 k − 0 1 = 0, , , , , 4 8 k 2k, ⇒, 8 12 = 2k 3 k, , , , k= 4, , + nCr, , number of solutions of the equation x 2 − 4 x + [x ] = 0 in, the interval [0, 2] ?, , ⇒, , 1 2, 2, 26. If A = , and A − kA − I 2 = O, where I 2 is the 2 × 2, 2 3, identity matrix, then what is the value of k ?, , ⇒, , −1, , 29. Let | x | denote the greatest integer function. What is the, , ⇒, , ∴ Order of B is 3 × 5., , 1, A2 = A ⋅ A = , 2, 1, 1, 2, ⋅, +, ⋅2, , =, 2 ⋅ 1 + 3 ⋅ 2, , −2, , Case I x∈[0, 1], , A2 × 3 × Bm × n = ( AB)2 × 5, , ∴, , 7 × 6 ×5, 3 ×2×1, , x2 − 4x + [x] = 0, , Now, according to the question, , (b) − 4, , −, , Ê (b) We have,, , (a) Let order of B is m × n., , (a) 4, , 3 ×2×1, , (b) C ( n − 1, r + 1) (c) C ( n, r + 1) (d) C ( n + 2, r ), , (a) Zero (no solution) (b) One, , be a, , ∴, , 12 × 11 × 10, , 28. What is C (n, r ) + 2C (n, r − 1) + C (n, r − 2) equal to ?, , = cos 2 θ + sin 2 θ = 1, , (a) 3 × 5 matrix, (c) 3 × 2 matrix, , (d) 105, , = 220 − 35 = 185, , cos θ − sin θ 0, cos θ sin θ 0, ∴ adj A = − sin θ cos θ 0 = sin θ cos θ 0 , , , , , 0, 1 , 0, 1, 0, 0, 1, adj, ∴ A− 1 =, | A|, , Ê, , (c) 115, , Ê (a) Required number of triangle = 12C3 − 7C3, , T, , A=, , (b) 175, , 30. A survey of 850 students in a University yields that 680, students like music and 215 like dance. What is the least, number of students who like both music and dance?, (a) 40, , (b) 45, , (c) 50, , (d) 55, , Ê (b) Let A be the set of students who like music and B be the, set of students whose like dance., , ∴n( A) = 680, n(B) = 215 and n(U ) = 850, We know that,, n( A ∩ B) = n( A) + n(B) − n( A ∪ B), ⇒, , n( A ∩ B)min = n( A) + n(B) − n( A ∪ B)max, , ⇒ n ( A ∩ B)min = 680 + 215 − 850, [Q n( A ∪ B)max = n(∪)], = 45
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29, , NDA/NA Solved Paper 2018 (I), 31. What is the sum of all two-digit numbers, which when, divided by 3 leave 2 as the remainder ?, (a) 1565, , Ê, , (b) 1585, , (c) 1635, , n =2, , (d) 1655, , (a) i, , (c) Required numbers are 11, 14, 17, ... 98 which is an AP., We know that,, 98 = 11 + 3 n − 3 ⇒98 = 3 n + 8, , ⇒, , 90 = 3 n ⇒ n = 30, , ∴, , = (1 + i) Σ in = (1 + i) [i2 + i3 + i4 + ... + i11 ], , Q S = n (a + l), n, , , 2, , (a) Negative power of 10 is less than 1, (b) Negative power of 10 is between 0 and 1, (c) Negative power of 10 is positive, (d) Negative power of 10 is negative, , =, , a = 10 x, , =, , 0 < 10 < 1 ⇒ x must be negative, x, , ∴ If 0 < a < 1, the value of log10 a is negative implies that, negative power of 10 is between 0 and 1., , 33. The third term of a GP is 3. What is the product of the first, five terms?, (a) 216, (b) 226, (c) 243, (d) Cannot be determined due to insufficient data, , a3 = 3, , ⇒, , ar 2 = 3, , [Qan = ar n − 1 ], , … (i), , Required product = a1 ⋅ a2 ⋅ a3 ⋅ a4 ⋅ a5, = (a )(ar )(ar 2 )(ar 3 )(ar 4 ) = a 5r10 = (ar 2 )5, = (3 )5, , [from Eq. (i)], , = 243, , =, , − (1 + i) (i2 − 1 ), (i − 1 ), − (1 + i) (− 1 − 1 ), (i − 1 ), 2 (1 + i), (i − 1 ), , 3, 2, following will be in HP?, (b) x , 4, z, , (c) x , 2, z, , (d) x , 1, z, , Ê (a) We have,, x,, ⇒, , x+ z 3, 3, , z are in AP. ⇒, =, 2, 2, 2, x+ z =3, , Also, x, 3, z are in GP ⇒ x z = 3, ⇒, , … (i), 2, , x z =9, , Now, from Eqs. (i) and (ii), we have, 2x z, 2×9, 2x z, =, ⇒, =6, x+ y, 3, x+ z, ⇒ x, 6, z are in HP., , … (ii), , i+1, (i + 1 ), , −1 − 1, , [Q i2 = − 1], =, =, , 2 (1 + i), (i − 1 ), 2 (i + 1 + i 2 + i), i2 − 1, , = − 2i, , π, π, , where 0 < x < , 0 < y < , then, 2, 2, 5, 10, what is ( x + y ) equal to ?, , 36. If sin x =, , 1, , , sin y =, , (b), , 1, , π, 2, , (c), , π, 4, , (d) 0, , 1, 1, and sin y =, 5, 10, −1 1, −1 1, and y = sin, ⇒ x = sin, 5, 10, −1 1, −1 1, Now, x + y = sin, + sin, 5, 10, 2, 2, 1, 1, 1 , 1, = sin −1 , 1 − , 1 − , +, 10 , 5 , 10, 5, , , Ê (c) We have, sin x =, , [Q sin −1 x + sin −1 y = sin −1 [x 1 − y2 + y 1 − x2 ], , 34. If x , , z are in AP; x , 3, z are in GP; then which one of the, (a) x , 6, z, , ×, , 2 (i + 1 − 1 + i), , (a) π, , Let a and r be the first term and common ratio of of the, GP., ∴, , i10 − 1 , = (1 + i) i2 , , i −1 , , r n − 1 , n −1, 2, =a, Q a + ar + ar + K + ar, , , r − 1 , (1 + i)i2 (i2 × 4 + 2 − 1 ), =, (i − 1 ), =, , It is given that 0 < a < 1, , Ê (c), , n=2, , 11, , 32. If0 > a < 1, the value of log10 a is negative. This is justified by, , ⇒, , 11, , n=2, , = 15 × 109 = 1635, , ⇒, , Ê (c) We11 have,, , (d) 1 + i, , n=2, , Sum = 11 + 14 + 17 + .... + 98, 30, =, [11 + 98], 2, , Ê (b) Let log10 a = x, , (c) − 2 i, , (b) 2 i, , Σ (in + in + 1 ) = Σ in (1 + i), , an = a + (n − 1 ) d ⇒98 = 11 + (n − 1 ) (3 ), ⇒, , 11, , 35. What is the value of the sum ∑ (i n + i n+1 ), wherei = − 1 ?, , , = sin −1 , , −1 , = sin, , , = sin −1 , , π, =, 4, , 37. What is, (a) sin x, , 1, 1, +, 1−, 10, 5, 1, 3, ×, +, 5, 10, , 1, 10, , 1−, , 1, 5 , , 1, 2 , ×, 10, 5 , , , 5, −1 1 , = sin , 5 × 10 , 2, , Q sin π = 1 , , 4, 2 , , sin 5x − sin 3x, equal to ?, cos 5x + cos 3x, (b) cos x, , (c) tan x, , (d) cot x
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30, , NDA/NA Solved Paper 2018 (I), sin 5x − sin 3 x, , Ê (c) Given, cos 5x + cos 3 x =, , 2 cos, 2 cos, , 5x + 3 x, 2, 5x + 3 x, , . sin, . cos, , 5x − 3 x, 2, 5x − 3 x, , 2, 2, C + D, , C − D, , Q sin C − sin D = 2 cos 2 ⋅ sin 2 and, , C + D, C − D , cos C + cos D = 2 cos , , cos , 2 , 2 , 2 cos 4x sin x sin x, =, = tan x, =, 2 cos 4x cos x cos x, , Ê, , (b) cos 50°, , (c), , (not ane quilateral triangle) such that tan( x − y ), tan x, and tan( x + y ) are in GP, then what is x equal to ?, (a), , π, 4, , 1, 2, , (d) 0, , (c) We have, sin 105° + cos 105° = sin (90 ° + 15° ) + cos 105°, [Q sin(90 + θ) = cos θ], = cos 15° + cos 105°, 105° + 15° , 105° − 15° , , , = 2 cos , , cos , , , , , 2, 2, , C + D , Q cos C + cos D = 2 cos 2 , , , C − D, cos , , , 2 , , , 1, 1, = 2 cos 60 ° cos 45° = 2 × ×, 2, 2, Q cos 60 ° = 1 , cos 45° = 1 , , 2 , 2, 1, =, 2, , (b), , ∴, , Ê, , π, (b), 2, , π, (c), 3, , π, (d), 6, , 2, (b) We have, a = 2, b = 3 and sin A =, 3, 2, sin B, sin A sin B, 3, Now, from sine formula⇒, =, ⇒, =, b, 2, a, b, ⇒, ⇒, , sin B = 1, π, B=, 2, , Q sin π = 1 , , , 2, , 40. What is the principal value of sin −1 sin, , , π, (a), 4, π, (c), 3, , Ê, , 2π , ?, 3, , (d), , π, 2, , 42. ABC is a triangle inscribed in a circle with centre O. Let, , α = ∠ BAC, where 45° < α < 90°. Let β = ∠ BOC. Which, one of the following is correct?, , (a) cos β =, (c) cos β =, , 1 − tan2 α, , (b) cos β =, , 1 + tan α, 2 tanα, 2, , 1 + tan2 α, 1 − tan2 α, , (d) sinβ = 2 sin2 α, , 1 + tan2 α, , Ê (a) We know that angle subtended by a chord at centre is, , always double the angle subtended by it at any other, part of the circle., A, α, O, β, B, , ∴, ⇒, ⇒, , C, , β = 2α, cos β = cos 2α, 1 − tan 2 α, cos β =, 1 + tan 2 α, , , 1 + tan 2 θ , Q cos 2θ =, , 1 + tan 2 θ , , , 43. If a flage-staff of 6 m height placed on the top of a tower, throws a shadow of 2 3 m along the ground, then what is, the angle that the sun makes with the ground ?, (a) 60°, , (b) 45°, , (c) 30°, , (d) 15°, , Ê (a) Let OB and BD be the tower and flag-staff respectively., , OA and AC be the shadow of tower and flag-staff, respectively., D, , π, (b), 2, 2π, (d), 3, , π, 2π, (c) We have, sin −1 sin = sin −1 sin π − , , , , 3 , 3 , π, [Q sin ( π − θ) = sin θ], = sin −1 sin, 3, π, Q sin −1 sin θ = θ, if θ ∈ − π , π , =, , 2 2 , 3, , π, 6, , x+ x− y+ x+ y= π, π, 3x = π ⇒ x =, 3, , 2, 3, , π, (a), 4, , (c), , x, x − y, x + y are the angles of a triangle. Since, sum of, angles of a triangle = π, , 39. In a ∆ABC, if a = 2 , b = 3 and sin A = , then what is ∠ B, equal to ?, , π, 3, , Ê (b) We have,, , ⇒, , 38. What is sin 105° + cos 105° equal to ?, (a) sin 50°, , 41. If x , x − y and x + y are the angles of a triangle, , 6m, B, xm, α, O, , α, A, , ym, , C, 2√3 m, , Again let α be the angle that sun makes with the, ground., ∴, , ∠OAB = ∠OCD = α
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31, , NDA/NA Solved Paper 2018 (I), Now, in ∆OAB, tanα =, and in ∆OCD, tanα =, , x, y, , In ∆OAP, α OA, sin =, 2 OP, α, r, α, ⇒ sin =, ⇒OP = r cosec, 2 OP, z, OL, In ∆OPL, sinβ =, ⇒ OL = OP sinβ, OP, α, ⇒, OL = r cosec . sin β, 2, r sin β, ∴, OL =, α, sin , 2, , … (i), x+6, , … (ii), , y+2 3, , From Eqs. (i) and (ii), we get, x+6, x, =, y y+2 3, ⇒, ⇒, , xy + 2 3 x = xy + 6 x, x, = 3, y, , ⇒, , tanα = 3, , ⇒, , α = 60 °, , 44. What is tan, , −1, , (b), , π, 4, , (c), , tan, , + tan, 4, , −1 3 , , = tan, 5, , π, 3, , (d), , 1 + 3 , 4 5 , , 1 3, 1 − × , 4 5, , , [Q tan −1 x + tan −1 y = tan −1, , = tan, π, =, 4, , 5 + 12 , −1 17 , −1, 20 , 20 − 3 = tan 17 = tan 1, , , 20 , , x+ y, 1 − xy, , , xy < 1], , Q tan −1 1 = π , , 4 , , eye of an observer, while the angle of elevation of its, centre is β. What is the height of the centre of the balloon, (neglecting the height of the observer)?, β, r sin , 2, (c), sinα, , r sinα, (d), β, sin , 2, , Ê (a) Let O bet the centre of the balloon, P be the eye of the, observer and ∠APB be the angle subtendd by the, balloon at the eye of the observer. ∠APB = α, A, , α, , sin x cos y, cos x sin y, , =, , a−b, a+ b, , a+ b, a−b, , tan x a, a, =, ⇒, tan y b, b, , 47. If sin α + sin β = 0 = cos α + cos β , where 0 < β < α < 2π,, then which one of the following is correct?, (a) α = π − β, (c) α = 2 π − β, , (b) α = π + β, (d) 2α = π + 2β, , Ê (b) We have,, , sin α + sin β = 0 = cos α + cos β, , ∴(sin α + sin β )2 + (cos α + cos β )2 = 0, ⇒ sin 2 α + sin 2 β + 2 sin α sin β + cos 2 α, + cos 2 β + 2 cos α cos β = 0, ⇒ (sin 2 α + cos 2 α ) + (sin 2 β + cos 2 β ), + 2 (cos α cos β + sin α sin β ) = 0, ⇒, , 1 + 1 + 2 cos(α − β ) = 0, , ⇒, , 2 cos (α − β ) = − 2 ⇒ cos (α − β ) = − 1, , ⇒, , α −β = π⇒ α =β + π, , (a) An odd multiple of 90°, (c) An odd multiple of 180°, , B, , ∠APO = ∠BPO =, , (d), , possible, then A must be, (b) A multiple of 90°, (d) A multiple of 180°, , A, , β, , ∴, , =, , a+ b, a−b, , 2, , r, , P, , sin(x + y), , (c), , A, 48. Suppose cos A is given. If only one value of cos is, O, , α/2, , b, a, , On using componendo and dividendo rule, we get, sin(x + y) + sin (x − y) a + b + a − b, =, sin(x + y) − sin(x − y) a + b − a + b, x+ y+ x− y, x+ y− x+ y, 2 sin , , cos , , , , , 2a, 2, 2, =, ⇒, 2b, x+ y+ x− y, +, −, +, x, y, x, y, , 2 cos , sin , , , , 2, , , ⇒, , 45. A spherical balloon of radius r subtends an angle α at the, , r sinβ, (b), α, sin , 4, , (b), , sin (x − y), , −1 , , r sinβ, (a), α, sin , 2, , a, b, , π, 2, , −1 , , [from Eqs. (i)], , sin( x + y ) a + b, tan x, equal to ?, , then what is, =, tan y, sin( x − y ) a − b, , Ê (a) We have,, , Ê (b) We have, −1 1 , , 46. If, , (a), , 1, −1 3, + tan , 4, 5, , equal to ?, (a) 0, , [from Eq. (i)], , … (i), , L, , α, 2, , X, , Ê (c) We know that, cos A = 2 cos 2 2, , −1, , A, has only one solution. So,, 2, A must be odd multiple of 180°., , Since, cos A is given and cos
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33, , NDA/NA Solved Paper 2018 (I), , y1 ) from the line, , Ê (c) We know that x1 distance of a point (x1, , Ê (d) From the given figure, it is clear that, , Ax + By + C = 0 is given as, Ax + By1 + C , Distance = 1, , A2 + B2 , , , 30º, , Statement I, 2, , (0, 0), p, , Slope of line = tan30 ° =, , 1, and line passes through the, 3, , ax+by–c=0, , point (0, − 2)., ∴Equation of line is, 1, y − (− 2) =, (x − 0 ), 3, 1, ⇒, y+ 2=, x⇒ 3 y + 2 3 = x, 3, ⇒, , ∴, , ⇒ p2 =, , x − 3y − 2 3 = 0, , a ⋅ 0 + b⋅ 0 − c, , p=, , a + b, 2, , 2, , c2, a + b2, , Statement II, (0, 0), p, , (b) 5x − 7 y + 18 = 0, (d) x − y + 5 = 0, , x y, –1=0, a +b, , Ê (c) Equation of line passing through intersection point of, lines x + 2 y − 3 = 0 and 2x − y + 5 = 0 is, x + 2 y − 3 + λ (2x − y + 5) = 0, , … (i), , ⇒ (1 + 2λ ) x + (2 − λ ) y + 5λ − 3 = 0, (1 + 2λ ), ∴ Slope of above line = −, (2 − λ ), , ∴, , Since line is parallel to y − x + 10 = 0, − (1 + 2λ ) − ( − 1 ), =, =, (2 − λ ), 1, , ⇒, , 0 0, + −1, a b, , p=, , 2, , 1 + 1, , , a, b, 1, 1, 1, =, + 2, p2 a2, b, , ⇒p=, , 1, 1, a2, , +, , 1, b2, , Statement III, , − λ =3⇒ λ = −3, , (0, 0), , Putting λ = − 3 in Eq. (i), we get, x + 2 y − 3 − 3 (2x − y + 5) = 0, , p, , 5x − 5 y + 18 = 0, , Which is equation of required line., , y–mx–C=0, , 55. Consider the following statements, I. The length p of the perpendicular from the origin to the, c2, ., line ax + by = c satisfies the relation p 2 = 2, a + b2, II. The length p of the perpendicular from the origin to the, x y, 1, 1, 1, line + = 1 satisfied the relation 2 = 2 + 2 ., a b, p, a, b, III. The length p of the perpendicular from the origin to the, 1, 1 + m2 + c 2, ., line y = mx + c satisfies the relation 2 =, p, c2, , Which of the above is/are correct?, (a) I, II and III, , 2, , It is true., , ⇒ − (1 + 2λ) = 2 − λ ⇒ − 1 − 2λ = 2 − λ, , ⇒, , a + b2, , 2, , of intersection of the lines x + 2y − 3 = 0 and 2x − y + 5 = 0, and parallel to the line y − x + 10 = 0 ?, , ⇒, , c, 2, , It is true., , 54. What is the equation of the line passing through the point, (a) 7 x − 7 y + 18 = 0, (c) 5x − 5 y + 18 = 0, , ⇒ p=, , (b) I only, , (c) I and II, , (d) II only, , ∴, ⇒, , p=, 1, p2, , =, , 0 − m×0 − c, (− m) + (1 ), 2, , 2, , ⇒p=, , c, m +1, 2, , 1 +m, , 2, , c2, , It is false., , 56. What is the equation of the ellipse whose vertices are, ( ± 5, 0) and foci are at ( ± 4, 0) ?, , x2, y2, +, =1, 25, 9, 2, 2, x, y, (c), +, =1, 25 16, (a), , x2, y2, +, =1, 16, 9, 2, 2, x, y, (d), +, =1, 9, 25, (b)
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34, , NDA/NA Solved Paper 2018 (I), A (1, 8, 4), , Ê (a) We have,, , Vertices = (± 5, 0 ) and Foci = (± 4, 0 ), , ∴ a = 5 and ae = 4, [Q vertex = (± a, 0 ) and focus (± ae, 0 ), 4, ⇒e =, 5, e= 1−, , Now,, ⇒, ⇒, , a, , 2, , = < 2λ − 1, 8 λ − 19, − 3 λ >, , 2, , ∴ 2 (2λ − 1 ) + 8 (8 λ − 19 ) − 3 (− 3 λ ) = 0, ⇒, , = (4, 5, − 2), , 59. What is the equation of the plane passing through the, points ( −2, 6, − 6), ( −3, 10, − 9 ) and ( −5, 0, − 6) ?, , the point (2, 3) and making an intercept on the positive, Y -axis equal to twice its intercept on the positive X -axis?, , Ê (b) Let the equation of line be, , (c) x + 2 y = 7 (d) 2 x − y = 1, , x2, y, + =1, a, b, , ⇒, , 7 = 2a, 7, a=, 2, , Equation of the plane passing through three points, (x1 , y1 , z1 ), (x2 , y2 , z2 ) and (x3 , y3 , z3 ) is, x − x1, y − y1, z − z1, x2 − x1 y2 − y1 z2 − z1 = 0, z3 − z1, , Equation of plane is, x − ( − 2), y−6, z − (− 6 ), − 3 − (− 2) 10 − 6 − 9 − (− 6 ) = 0, − 5 − (− 2) 0 − 6 − 6 − (− 6 ), x+ 2 y−6, −1, 4, −6, , z+6, −3 =0, 0, , ⇒ (x + 2) ( − 18 ) − ( y − 6 ) (− 9 ) + ( z + 6 ) (18 ) = 0, ⇒ 2 (x + 2) − ( y − 6 ) − 2 ( z + 6 ) = 0, , be, (1, 8, 4 ), (0, − 11, 4 ) and (2, − 3, 1) respectively. What are the, coordinates of the point D which is the foot of the, perpendicular from A on BC ?, (c) ( 4, 5, − 2 ), , y3 − y1, , ⇒ (x + 2) [4 × 0 − (− 6 ) (− 3 )] − ( y − 6 ), [(− 1 ) (0 ) − (− 3 ) (− 3 )] + ( z + 6 ) [(− 1 ) ( − 6 ) − (−3 ) (4)] = 0, , 2x + y = 7, , (b) ( 4, − 2, 5), , (b) 2 x + y + 3 z = 3, (d) x − y − z = 3, , −3, , 58. Let the coordinates of the points A, B, C, , (a) ( 3, 4, − 2 ), , Ê (a), , ⇒, , 7, b = 2a = 2 × = 7, ⇒, 2, x, y, ∴Equation of line is, + =1, 7 /2 7, ⇒, , (a) 2 x − y − 2 z = 2, (c) x + y + z = 6, , x3 − x1, , It is given that, b =2a and line passes through the point, (2, 3)., 4+ 3, 2, 3, ∴, +, =1⇒, =1, a 2a, 2a, ⇒, , 77 λ = 154 ⇒ λ = 2, , ∴Coordinates of D = (2 × 2, 8 × 2 − 11, − 3 × 2 + 4), , 57. What is the equation of the straight line passing through, (b) 2 x + y = 7, , 4λ − 2 + 64λ − 152 + 9 λ = 0, , ⇒, , ∴ Equation of ellipse is, x2, y2, +, =1, a2, b2, 2, 2, x, y, ⇒, +, =1, 25, 9, , (d) (2, 4, 5), , A (1, 8, 4), B (0, − 11, 4) and C (2, − 3, 1), , ∴Equation of BC is, x−0, y + 11, z−4, =, =, 2 − 0 − 3 + 11 1 − 4, x y + 11 z − 4, ⇒, =, =, =λ, 2, 8, −3, x = 2λ, y = 8 λ − 11, z = − 3 λ + 4, , ⇒, , 2x − y − 2 z − 2 = 0, , ⇒, , 2x − y − 2 z = 2, , 60. A sphere of constant radius r through the origin, intersects the coordinate axes in A, B and C. What is the, locus of the centroid of the ∆ ABC?, (a) x 2 + y 2 + z2 = r 2, (c) 9 ( x 2 + y 2 + z2 ) = 4r 2, , Ê (c) We have,, , ⇒, , Since, AD ⊥ BC, , [Q a = 5], , b2 = 9 ⇒ b = 3, , (a) 2 x + y = 5, , C, , Now, DR’s of, AD =< 2λ − 1, 8 λ − 11 − 8, − 3 λ + 4 − 4 >, , b2, , 2, 4 = 1 − b, , 5, (5)2, b2, 16, =1 −, ⇒ 16 = 25 − b2, 25, 25, , ⇒, , D (x, y, z), , B, , (b) x 2 + y 2 + z2 = 4r 2, (d) 3 ( x 2 + y 2 + z2 ) = 2 r 2, , Ê (c) Let A (a, 0, 0), B (0, b, 0) and C(0, 0, c), , ∴Equation of sphere passing through A, B, C and origin, is, x2 + y2 + z 2 − ax − by − cz = 0, , [say], ∴, ⇒, , Radius = r =, , a2, b2, c2, +, +, 4, 4, 4, , 4r 2 = a 2 + b2 + c2, , … (i)
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35, , NDA/NA Solved Paper 2018 (I), Let (α, β, γ ) be the centroid of triangle., a+0+0, 0 + b+ 0, 0+0+ c, ,γ =, ,β =, ∴ α=, 3, 3, 3, ⇒ a = 3α, β = 3β, c + 3 γ, , 63. The equation of the line, when the portion of it, intercepted between the axes is divided by the point (2, 3), in the ratio of 3 : 2, is, … (ii), , From Eqs. (i) and (ii), we have, (3α )2 + (3β )2 + (3 γ )2 = 4r 2, ⇒, , 9 (α 2 + β 2 + γ 2 ) = 4r 2, , (a) Either x, (b) Either x, (c) Either x, (d) Either x, , 9 (x + y + z ) = 4r z, 2, , 2, , y=, y=, y=, y=, , 4 or 9x + y = 12, 5 or 4x + 9 y = 30, 4 or x + 9 y = 12, 5 or 9x + 4 y = 30, , Ê (d) Case I, , ∴Locus of the centroid of ∆ABC is, 2, , +, +, +, +, , 2, , (0, b), , 61. The coordinates of the vertices P, Q and R of a triangle, , b, , (2, 3), 2:3, a (a, 0), (0, 0), , PQR are (1, − 1, 1), (3, − 2, 2) and (0, 2, 6) respectively. If, ∠ RQP = 9, then what is ∠ PRQ equal to ?, (a) 30° + θ, , (b) 45° − θ, , Ê (d), , (c) 60° − θ, , (d) 90° − θ, , From above figure,, 2a + 3 × 0, 2 × 0 + 3b, = 2 and, =3, 2+ 3, 2+ 3, , P (1, –1, 1), , ⇒, θ, Q (3, –2, 2), , Case II, , < a1 , b1 , c1 > = < 2, − 1, 1 >, , (0, b), , and DR’s of PR = < 0 − 1, 2 − (− 1 ), 6 − 1 >, , b, , < a2 , b2 , c2 > = < − 1, 3, 5 >, , (2, 3), 3:2, a (a, 0), (0, 0), , Now, a1 a2 + b1 b2 + c1 c2 = 2 × (− 1 ) + (− 1 ) × 3 + 1 × 5, = −2−3 + 5 =0, ∴ PQ ⊥ PR ⇒ ∠QPR = 90 °, , From above figure,, 3a + 2 × 0, 3 × 0 + 2b, = 2 and, =3, 3+2, 3+2, , Now, by angle sum property, ∠PQR + ∠QPR + ∠PRQ = 180 °, ⇒, , θ + 90 ° + ∠PRQ = 180 °, , ⇒, , ∠PRQ = 90 ° − θ, , line 2x + 11y = 5 upon the two straight lines 24 x + 7y = 20, and 4 x − 3y = 2 are, (b) 11 and 5 respectively, (d) Not equal to each other, , Ê (c) Let (− 3, 1 ) be a point on 2x + 11 y = 5, Now, perpendicular from (− 3, 1 ) on 24x + 7 y = 20, =, =, , 24 (− 3 ) + 7 (1 ) − 20, , 576 + 49, , =, , − 85, 25, , =, , 17, 5, , =, , 4(− 3 ) − 3 (1 ) − 2, 4 + (− 3 ), 2, , − 12 − 3 − 2, 16 + 9, , x, y, +, =1, 10 / 3 15 / 2, , 9 x + 4 y = 30, , 64. What is the distance between the straight lines, 3x + 4y = 9 and 6x + 8y = 15 ?, (a), , 3, 2, , (b), , 3, 10, , (c) 6, , (d) 5, , Ê (b) Given equation of straight lines are, and, , Again, perpendicular from (− 3, 1 ) on 4x − 3 y = 2, =, , ⇒, , 10, 15, and b =, 3, 2, , 3x + 4y = 9, , (24)2 + (7 )2, − 72 + 7 − 20, , 3 a = 10 and 2b = 15 ⇒ a =, , ∴ Equation of line is, , 62. The perpendiculars that fall from any point of the straight, (a) 12 and 4 respectively, (c) Equal to each other, , a = 5 and b = 5, x, y, + =1⇒ x+ y=5, 5 5, , ∴ Equation of line is, , R (0, 2, 0), , DR’s of PQ = < 3 − 1, − 2 − (− 1 ), 2 − 1 >, , ⇒, , 2a = 10 and 3 b = 15, , ⇒, , 2, , 17, =, 5, , ∴Both perpendicular are equal to each other., , ⇒, , … (i), , 6 x + 8 y = 15, 15, 3x + 4y =, 2, , … (ii), 9−, , ∴ Required distance =, , 15, 2, , 3 2 + 42, , [Q distance between two lines ax + by = c1 and, ax + by = c2 is given, , by, , 3, c2 − c1 2, 3, =, =, 2, 2, 5, 10, a + b
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36, , NDA/NA Solved Paper 2018 (I), , 65. What is the equation of the sphere whose centre is at, ( −2, 3,4 ) and radius is 6 units?, +, +, +, +, , 2, , (a) x, (b) x 2, (c) x 2, (d) x 2, , 2, , y, y2, y2, y2, , +, +, +, +, , +, +, +, +, , 2, , z, z2, z2, z2, , 4x, 6x, 4x, 4x, , −, −, −, +, , So, equation of sphere, ⇒, {x − (− 2)}2 + ( y − 3 )2 + ( z − 4)2 = 6 2, ⇒ (x + 2)2 + ( y − 3 )2 + ( z − 4)2 = 36, ⇒, x2 + 4x + 4 + y2 + 9 − 6 y + z 2 + 16 − 8 z = 36, ⇒ x2 + y2 + z 2 + 4x − 6 y − 8 z + 29 − 36 = 0, ⇒, x2 + y2 + z 2 + 4x − 6 y − 8 z = 7, , r, r, r, r, 66. If a and b are vectors such that | a | = 2, | b | = 7 and, r r, $ then what is the acute angle between, a × b = 3$i + 2$j + 6k,, a and b ?, (c) 60°, , →, , →, , →, , 9|p|2 = 4|q|2, , ⇒, , 9 p 2 = 4q2, , →, a × b = 3 $i + 2$j + 6k$, , →, , →, , |a × b |, →, , →, , =, , |3 $i + 2$j + 6k$|, 2× 7, , |a||b|, 3 +2 +6, 2, , =, , 2, , 2, , =, , 14, , represented by 3i$ + k$ acting through the point 2i$ − $j + 3k$, ?, (a) − 3$i + 11$j + 9k$, (c) 3$i + 4$j + 9k$, , (d), , →, , r = (2$i − $j + 3k$ ) − ($i + 2$j − k$ ), , $i, , θ = 30 °, , (b) 4 p = 9 q, 2, , 2, , (c) 9 p = 4 q, , (d) 4 p = 9 q, , (a) The points Rand S divider s PQ internally and externally, respectively in the ratio 2 : 3. The position vectors of R, and S are, →, , →, , →, , →, , →, , OR =, , ⇒, , →, , →, , →, , →, , →, , →, , →, , →, , →, , →, , →, , →, , →, , →, , →, , →, , →, , a × b = 3 (b × c ) [b × b = 0], →, , →, , →, , →, , →, , →, , →, , [Qc × b = − b × c ], , a × b + 2 b × b = 3 (b × c ), , … (i), →, , →, , →, , →, , →, , 3 c + a = − 2 b ⇒ (3 c + a ) × a = − 2 b × a, →, , →, , →, , →, , 3 c × a + a × a = 2 (a × b ), →, , →, , →, , →, , →, , →, , ⇒, , 3 (c × a ) = 2 (a × b ), , ⇒, , 3 (c × a ) = 6 (b × c ), , ⇒, , →, , →, , →, , →, , →, , [from Eq. (i)], , →, , c × a = 2 (b × c ), , →, , →, , →, , →, , →, , →, , →, , →, , … (ii), →, , →, , Now, a × b + b × c + c × a = λ (b × c ), →, , →, , →, , →, , →, , →, , = 3 (b × c ) + (b × c ) + 2 (b × c ), [from Eqs. (i) and (ii)], , r, , r, , →, , →, , ⇒, , 3 p + 2q, →, . (3 →, , p − 2q ) = 0, , , 5, , , , →, , →, , →, , →, , →, , 70. If the vectors, r rK and A are parallel to each other, then, , OR ⋅ OS = 0, , →, , →, , →, , on comparing, we get λ = 6, , →, , →, , →, , (d) 6, , a + 2b + 3 c = 0 ⇒ a + 2b = − 3 c, , = 6 (b × c ), , ⇒, , ⇒, , →, , →, , Now, OR ⊥ OS, , →, , Again, ⇒, , (c) 4, , (a + 2 b ) × b = − 3 c × b, , 3 p − 2q, , 5, →, →, →, OS = 3 p − 2 q, , →, , 1, , (b) 3, , and 3 p − 2 q respectively., , 5, , 0, , = $i (− 3 − 0 ) − $j (1 − 12) + k$ (0 + 9 ) = − 3 $i + 11 $j + 9k$, , ⇒, , →, , k$, , r r, r r r, r r r r r r, r, 69. If a + 2b + 3c = 0 and a × b + b × c + c × a = λ ( b × c),, then what is the value of λ ?, , respectively with respect to origin O. The points R and S, divide PQ internally and externally respectively in the, ratio 2 : 3. If, →, →, OR and OS are perpendicular, then which one of the, following is correct?, , 3 p + 2q, , $j, , 3, , ⇒, , Ê, , F = 3 $i + k$, , →, →, →, ∴ Moment τ = r + F = ($i − 3 $j + 4k$ ) × (3 $i + k$ ), , Ê (d) Given that,, , r, , (a) 9 p = 4 q, , →, , = $i − 3 $j + 4k$ and, , 67. Let p and q be the position vectors of the points P and Q, , 2, , 3$i + 2 $j + 9k$, $i + $j + k$, , (b), , Ê (a) Given that,, , (a) 2, , 49, 14, , 7 1, = ⇒ sin θ = sin 30 ° ⇒, ⇒ sinθ =, 14 2, , 2, , → →, , 68. What is the moment about the point $i + 2$j − k$ of a force, , →, , →, , →, , sin θ =, , r, , → →, , →, , →, , ⇒, , (d) 90°, , |a × b| = |a||b|sin θ, , ⇒, , →, , = 1 −3 4, , |b| = 7 and, →, , →, , 9|p|2 − 4|q|2 = 0, , →, , Ê (a) Given,|a| = 2, Q, , →→, , →→, , and radius = 6 units, , (b) 45°, , →→, , [Qa . a =a2 and a ⋅ b = b ⋅ a ], , Equation of the sphere having centre at (α, β, γ ) and, radius r is (x − α )2 + ( y − β )2 + ( z − γ )2 = r 2, , (a) 30°, , →→, , ⇒, , 6y − 8z = 7, 4y − 8z = 7, 6y − 8z = 4, 6y + 8z = 4, , Ê (a) Given, centre = (− 2, 3, 4), , →→, , ⇒ 9 p. p − 6 p. q + 6 q. p − 4 q. q = 0, , →, , →, , (3 p + 2 q ) ⋅ (3 p − 2 q ) = 0, , what is kK × A equal to ?, r, (a) k 2A, , →, , Ê (b) Since, a, →, , r, r, (c) − k 2A (d) A, , r, (b) 0, , →, , →, , →, , →, , × b = 0, if a and b are parallel., →, , →, , So, k K × A = 0 if K and A are parallel to each other.
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37, , NDA/NA Solved Paper 2018 (I), 71. When one of the following is correct in respect of the, +, , function f : R → R defined as f ( x ) = | x + 1| ?, (a) f( x 2 ) = [f( x )]2, (c) f( x + y) = f( x ) + f( y), , (b) f(| x|) = | f( x )|, (d) None of these, , is f ′ (0) equal to ?, (a) 0, (c) − 1, , Ê (d) Given, f (x) = |x + 1|, , x ln|x|, x ≠ 0, f (x ) = , x=0, 0,, f (h) − f (0 ), f ′ (0 ) = lim, ∴, h→ 0, h, h2 log h, = lim h log h = 0, = lim, h→ 0, h→ 0, h, , (a) f (x2 ) = |x2 + 1| { f (x)}2 = (x + 1 )2, Which implies that f (x2 ) ≠ { f (x)}2, (b) f (|x|) = ||x| + 1| |f (x)| = ||x + 1|= |x + 1|, which implies that f (|x|) ≠ |f (x)|, (c) f (x + y) = |x + y + 1|, f (x) + f ( y) = |x + 1| +|y + 1|, , 75. What is the area of the region bounded by the parabolas, , which implies that f (x + y) ≠ f (x) + f ( y), , y 2 = 6( x − 1) and y 2 = 3x ?, , So, option (d) is correct., , x2, 1+ x, , . What is, 2, , the range of the function?, (a) [0, 1), , (b) [0, 1], , (c) (0, 1], , (b) 1, (d) It does not exist, , is, Ê (a) Given function, 2, , By checking the options, we get, , 72. Suppose f : R → R is defined by f ( x ) =, , x 2 In | x | x ≠ 0, What, x =0, 0, , , 74. Consider the function f ( x ) = , , (a), , 6, 3, , (b), , 2 6, 3, , (c), , 4 6, 3, , (d), , 5 6, 3, , Ê (c) Given,, , y 2 = 6 (x − 1 ), and, y2 = 3 x, on solving Eqs. (i) and (ii), we get, , (d) (0, 1), , Ê (a) Let f (x) = y, , x = 2 and y = ±, , Then, y ≥ 0 and f (x) = y, x2 + 1 1, x2, ∴, = for y > 0, = y⇒, 2, y, x2, x +1, 1− y, y, 1, ⇒, =, ⇒ x=, 1− y, y, x2, , 2, , x=y /3, , … (i), … (ii), , 6, y2=6(x–1), , y=√6, , x=1, , y, y, Now,, is real ⇒, ≥0, 1− y, 1− y, , y=–√6, , ⇒ 0≤ y<1, So, Range of f (x) is [0, 1)., , 73. If f ( x ) = | x | + | x − 1 |, then which one of the following is, , −, , correct?, (a), (b), (c), (d), , ⇒, , ⇒, , f (x) = |x| + |x − 1|, x<0, − 2x + 1,, , f (x) = x − x + 1, 0 ≤ x < 1, x + x − 1,, x≥1, , x<0, − 2x + 1,, , 0≤ x<1, f (x) = 1,, 2x − 1,, x≥1, , , Clearly, lim f (x) = 1 = lim+ f (x), x → 0 −1, , x→ 0, , and lim− f (x) = lim+ f (x)., x→1, , x −1, , So, f (x) is continuous at x = 0, 1., , ∫, , 6, , , y2 , y3 , = 2 ∫ 1 −, dy = 2 y −, 6 , 18 0, , 0 , , f (x) is continuous at x = 0 at x =1, f (x) is continuous at x = 0 but not at x = 1, f (x) is continuous at x = 1 but not at x = 0, f (x) is neither continuous at x = 0 nor at x = 1, , Ê (a) We have,, , , y2, y2 , −, dy, 1 +, 6, 3 , , 6, , 6, , ∴Required area =, , 6, , 6, , 18 y − y3 , 18 6 − 6 6 , =2× , =2× , , 18, 18, , , , 0, 12 6 4 6, =, =, 9, 3, , Directions (Q. Nos. 76-78) Consider the following information, for the next three items that follow Three sides of a trapezium are, each equal to 6 cm. Let αε 0, π be the angle between a pair of, 2, adjacent sides., 76. If the area of the trapezium is the maximum possible, then, what is α equal to ?, π, 6, π, (c), 3, , (a), , π, 4, 2π, (d), 5, (b)
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38, , NDA/NA Solved Paper 2018 (I), π, , Ê (c), , D, , 79. What is ∫ e x sin x dx equal to ?, , C, , 6, , 0, , 6, , eπ + 1, (a), 2, , 6, , √36 – x2, , α, x, , A, , ∴ Area = A =, , (c) e π + 1, , x, , 6, , B, , π, , 1, (6 + 6 + 2x) 36 − x2, 2, , Ê (a) Let I = ∫ ex sin x dx, 0, , = (6 + x) 36 − x2, ∴, , d( A), d, =, [(6 + x) 36 − x2 ], dx, dx, , , −2x, +, = (6 + x) , 2 36 − x2 , , , = 36 − x2 −, , x (6 + x), 36 − x2, , =, , 0, π, , = [sin x ⋅ ex ]π0 − ∫ cos x ⋅ ex dx, , 36 − x2, , 0, , π, , , = 0 − [cos x ⋅ ex ]π0 + ∫ sin x ⋅ ex dx, 0, , , , 36 − 6 x − 2x2, 36 − x2, , dA, =0, dx, ⇒, , 2x2 + 6 x − 36 = 0, , ⇒, , x + 3 x − 18 = 0, , ⇒, , x2 + 6 x − 3 x − 18 = 0, , ⇒, , x (x + 6 ) − 3 (x + 6 ) = 0, , At x = 3,, ∴, , dx, d2 A, , (− 6 − 4x) 36 − x2 −, =, , (36 − 6 x − 2x2 )(−2x), , (36 − x ), , e, , 81. What is ∫ x In x dx equal to ?, 1, , length of the fourth side?, , Ê (d) So, fourth side = x + 6 + x, , (c) 10 cm, , (d) 12 cm, , = 3 + 6 + 3 = 12, , 78. What is the maximum area of the trapezium?, (a) 36 3 cm, , 2, , (b) 30 3 cm, , (c) 27 3 cm 2, , 2, , (d) 24 3 cm 2, , Ê (c) Maximum area = (6 + x), , x − 2x − 3, 2, , Applying L′ Hospital rule, d 2, (x − 9 ), dx, f (3 ) = lim, x→ 3 d, (x2 − 2x − 3 ), dx, 2x, 2⋅ 3, 6, = lim, =, = = 1 .5, x → 3 2x − 2 2 ⋅ 3 − 2, 4, , π, π, ⇒α =, 3, 3, , (b) 9 cm, , x2 − 9, , x→ 3, , 2 36 − x2, 2, , 77. If the area of the trapezium is maximum, what is the, (a) 8 cm, , Therefore, f (3 ) = lim, , = − 6 − 12 = − 18, dx2, d2 A, <0, dx2, , cos α = cos, , (b) f( 3) = 15, ., (d) f( 3) = − 15, ., , Ê (b) Since, f (x) is continuous at x = 3, , So, at x = 3 is maximum., x 3 1, Now, cosα = = =, 6 6 2, ⇒, , 2, , (a) f( 3) = 0, (c) f( 3) = 3, , (x + 6 ) (x − 3 ) = 0 ⇒ x = 3, − 6, , 2, , x −9, , , x ≠ 3 is continuous at x = 3, then, x − 2x − 3, which one of the following is correct?, , 80. If f ( x ) =, , Again, on differentiating it, d2 A, , I + I = eπ + 1 ⇒ 2I = eπ + 1, eπ + 1, I=, 2, , ⇒, , 2, , ⇒, , I = − [− eπ − 1] − I, , ⇒, ⇒, , 36 − 6 x − 2x2 = 0, , π, , d, = [sin x ⋅ ex ]π0 − ∫ {sin x} ⋅ ex dx, dx, , , For maximum area,, , ⇒, , eπ − 1, 2, eπ + 1, (d), 4, , (b), , 36 − x2, , = (6 + 3 ) 36 − 9, = 9 × 27 = 9 × 27, = 9 × 3 3 = 27 3 cm 2, , e +1, (a), 4, e −1, (c), 4, , Ê (b), , e2 + 1, 4, e2 − 1, (d), 4, (b), , e, , Let 1 = ∫ x log xdx, 1, , ], , e d, = [log x⋅ ∫ xdx] 1e − ∫ {log x}. ∫ x dx dx, 1, dx, e, , , x2 , = log x ⋅ −, 2 1, , , e1, , ∫1 x ., , x2, dx, 2, , 2, e2 1 1 2 e e2 [e − 1], − × [x ] 1 =, −, 2 2 2, 2, 4, 2e2 − e2 + 1 e2 + 1, =, =, 4, 4, , =
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39, , NDA/NA Solved Paper 2018 (I), 2, , ∫, , 82. What is, , [x 2 ] dx equal to (where [.] is the greatest, , 0, , (b) 1 − 2, 2, , x dy − y dx = 0 ?, (a) xy = c, , integer function) ?, (a) 2 − 1, , 86. What is the solution of the differential equation, , (c) 2 ( 2 − 1), , 1, , (d) 3 − 1, , 0, , 0, , = ∫ 0 dx +, 0, , , Q [x] =, , , ∫ 1 dx, 1, , Variable seprate on both sides, dy dx, =, y, x, , ⇒, , 0 0 ≤ x < 1 , 1 1 ≤ x < 2 , , , , On integration both sides, we get, dy, dx, ∫ y = ∫ x ⇒ log y = log x + log c, , = 0 + [x]1 = 2 − 1, 2, , ⇒, , 83. What is the maximum value of 16 sin θ − 12 sin 2 θ ?, (a), , 3, 4, , (b), , 4, 3, , (c), , 16, 3, , [where log c is integrating constant], , (d) 4, , ⇒, , (a), , 42 + (− 3 )2, a + b ], , S = Range of f = [− 4, 6], , 85. For f to be a function, what is the domain of f , if, |x| − x, , (a) ( − ∞, 0), , Ê, , (c) ( − ∞, ∞ ), , (d) ( − ∞, 0), , 1, (a) We have, f (x) =, |x| − x, f (x) is defined, if |x| − x > 0 ⇒ |x|> x, Case I x > 0, ∴, , (a), , d 2x, 2, , + µx = 0, , d 2x, , − µx = 0, dt 2, dx, (d), + µxt = 0, dt, (b), , dt, dx, (c) x, + µt = 0, dt, , d2 x, , + µx = 0 is the differential equation of simple, dt 2, harmonic motion, which has a periodic solution., , 89. What is the period of the the function f ( x ) = sin x ?, (a), , π, 4, , (b), , π, 2, , (c) π, , (d) 2 π, , Ê (d) We have, f (x) = sin x, , x > x [Q|x| = x, x > 0], , which is not possible, Case II, ∴, , [Qa log a b = b], , 88. Which one of the following differential equations has a, , Ê (a), , ?, , (b) ( 0, ∞ ), , (d) 4e, , periodic solution?, , ∴ f (x) ∈ [− 4, 6] since, f (x) is onto., , f (x ) =, , (c) 2e, , function, , 2, , ⇒ − 5 + 1 ≤ 4 sin x − 3 cos x + 1 ≤ 5 + 1 ⇒ − 4 ≤ f (x) ≤ 6, , 1, , the, π, at x = ?, 4, , = e1 ( 2 )2 + 1 − 1 = e⋅2 + 1 − 1 = 2e, 2, , ⇒ − 5 ≤ 4 sin x − 3 cos x ≤ 5, , ∴, , (b) e, , of, , = sec2 x etan x + tan x − 1, π, tan, π, π, ∴[ f ′ (x)] π = e 4 ⋅ sec2 ⋅ + tan − 1, x=, 4, 4, 4, , We know that,, [Q − a + b ≤ a sin x + b cos x ≤, , + ln ( sec x ) − e, , ln x, , On differentiating with respect to x both the sides, we, get, 1, f ′ (x) = etan x ⋅ sec2 x +, ⋅ sec x tan x − 1, sec x, , (d) [− 4, 6], , 2, , derivative, , f (x) = etan x + log(sec x) − x, , Ê (d) We have, f (x) = 4 sin x − 3 cos x + 1, 2, , e, 2, , the, , F (x) = etan x + log(sec x) − elog x, , onto, then what is S equal to ?, , − 42 + (− 3 )2 ≤ 4 sin x − 3 cos x ≤, , tan x, , Ê (c) We have,, , 84. If f : R → S defined by f ( x ) = 4 sin x − 3 cos x + 1 is, (c) ( −4, 6), , is, , f (x ) = e, , 16, 4, sin θ = − 12 sin 2 θ − sin θ, = − 12 sin 2 θ −, , , , , 12, 3, 2, 2, , 2, 4, 2, 16, = − 12 sinθ − − = − 12 sinθ − +, , 3, 9 , 3, 3, , 16, ∴, f (x ) ≤, 3, 16, ∴ Maximum value of f (x) =, 3, , (b) ( −5, 5), , y = xc ⇒ y = cx, , 87. What, , Ê (c) Let f (x) = 16 sin θ − 12 sin 2 θ, , (a) [− 5, 5], , (d) x − y = c, , x dy − y dx = 0 ⇒ x dy = y dx, , 1, , 2, , (c) x + y = c, , Ê (b) Given differentiation equation, , 2, , Ê (a) Let I = ∫ [x2] dx = ∫ [x2] dx + ∫ [x2] dx, 1, , (b) y = cx, , x<0, , –2π, , –π, , O, , π, , 2π, , 3π, , 4π, , − x > x [Q|x| = − x, x < 0], , ⇒ 0 > 2x ⇒ x < 0, Which is possible ∴Domain of f (x) = (− ∞, 0 ), , f (x + 2 π ) = sin (x + 2 π ) = sin x = f (x), ∴ Period of f (x) is 2π.
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40, , NDA/NA Solved Paper 2018 (I), , 90. What is ∫, , dx, 2x − 1, , (a) ln (2 − 1) + C, , ln (1 − 2 −x ), (b), +C, ln 2, , ln (2 − x − 1), (c), +C, 2 ln 2, , ln (1 + 2 −x ), (d), +C, ln 2, , x, , Ê, , b, , b, , a, , a, , of a and b respectively?, (a) − 1, 1, , (b) 1, 1, , Ê (a) We have,, , Put, ⇒, , −x, , 2, , ∴, , 2− x log 2, 1, dx, ∫, log 2 1 − 2− x, , b, , ⇒, , 1 − 2−x = t, , ⇒, , log 2 dx = dt, 1, I=, log 2, , But ∫ x dx = 0 and x is an odd function., , =, , b= ± a, 3, , dt, 1, ∫ t = log 2 . log t + C, , log (1 − 2− x ), , a, , ∴, , a=−b, b, , +C, , log 2, , 3, , Again, ∫ x2 dx =, a, , ⇒, , y 2 = 4a ( x − a ), where ‘a’ is an arbitrary constant, are, respectively, (a) 1, 2, , ⇒, , (b) 2, 1, , (c) 2, 2, , y = 4a (x − a ), 2, , … (i), , dy, dy, − y2 , dx , dx, , b3 − a 3 = 2, [Qfrom Eq. (ii)], , ⇒, , b =1, , ⇒, , a= −1, , [from Eq. (ii)], , 0, , … (ii), , (b), , (c), , 1, , 1, , 0, , 0, , (b), , (c) 0, , (d) 2, , 1, 2, , a, Q ∫ f (x) dx =, 0, , 1, 240, , a, , , , 0, , , , ∫ f (a − x) dx, , 1, , 95. What is lim, , x→0, , tan x, equal to, sin 2x, , 1, 2, (c) 2, , (a), , (b) 1, (d) Limit does not exist, , Ê (a) We have,, , π /4, , Ê (c) Let I = ∫− π / 4 (sin x − tan x) dx, , lim, , x→ 0, , f (x) = sin x − tan x, f (− x) = sin (− x) − tan (− x) = − sin x + tan x, , [Q sin ( − θ) = − sin θ, tan (− θ) = − tan θ], = − (sin x − tan x) = − f (x), ∴ f (x) is odd function., π /4, , ∫− π / 4 (sin x − tan x) dx = 0, a, , (d), , x10 x11 , 1 , 1, 1, = ∫ (x9 − x10 ) dx = , −, = 10 − 11 =, 110, 10, 11, , , 0, 0, , −π / 4, , −a, , 1, 148, , Ê (a) Let I = ∫ x (1 − x)9 dx = ∫ (1 − x) x9 dx, , 1, , ∫ ( sin x − tan x ) dx ?, , [Q ∫, , 1, 132, , 2, , 1, 1 , (a) −, + ln , , 2, 2, , ∴ I=, , 2, ⇒, 3, , 2b3 = 2 ⇒ b3 = 1, , 1, (a), 110, , π /4, , Let, , x , 2, 2, ⇒ =, 3, 3, 3, a, , 94. What is ∫ x (1 − x ) 9 dx equal to ?, , ∴Order = 1 and degree = 2, , ∴, , b, , 1, , On putting the values of a from Eq. (ii) in Eq. (i), we get, 1 dy , 1 dy , y2 = 4 × y, x − y , , 2 dx, 2 dx , , 92. What is the value of, , =, , 3, , ⇒, , y2 = 4ax − 4a 2, , y2 = 2xy, , b3 − a 3, , … (ii), 3, , ⇒b3 − (− b)3 = 2, , (d) 1, 1, , On differentiating both sides, we get, dy, = 4a, 2y, dx, 1 dy, a= y, ⇒, 2 dx, , ⇒, , … (i), , b, , 91. The order and degree of the differential equation, , Ê (a) We have,, , (d) 2, − 2, , x4 , 3, ∫ x dx = 0 ⇒ 4 = 0, a, a, b4 − a 4, = 0 ⇒ b4 = a 4, 4, , 2, (b) Let I = ∫ 2, dx, =∫, 1, − 2− x, 2x − 1, =, , (c) 0, 0, , b, , −x, , dx, , 2, 3, , 93. If ∫ x 3 dx = 0 and ∫ x 2 dx = , then what are the values, , equal to ?, , f (x) dx = 0, if f (x) is odd], , tan x, sin 2x, , By using L’ Hospital rule, we have, sec2 x, = lim, x → 0 2 cos 2x, 1, = lim, x → 0 2 cos 2x ⋅ cos 2 x, 1, 1, =, =, 2 cos 0 ° cos 2 0 ° 2 × 1 × 1, 1, =, 2
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41, , NDA/NA Solved Paper 2018 (I), 96. What is lim, , 2x + 3h − 2x, , x→0, , (a), , Ê, , 1, 2 2x, , 2h, 1, 2x, , (b), , (d) lim, , equal to ?, , (c), , 3, 2 2x, , (d), , 3, 4 2x, , log(1 + 2x) + log (1 − 2 y) = − 2C ′, , ⇒, , log (1 + 2x) (1 − 2 y) = − 2C ′, , ⇒, , (1 + 2x) (1 − 2 y) = e − 2 C′, , ⇒, , 1 − 2 y + 2x − 4xy = e − 2 C′, , 2x + 3 h − 2x, , ⇒, , 2h, , ⇒, , x→ 0, , By using, L’ Hospital rule, we get, 1, .3 − 0, 2 2x + 3 h, 3, = lim, = lim, h→ 0, h → 0 4 2x + 3 h, 2, =, , 3, 1, 3, ., =, 4 2x + 0 4 2x, , one of the following is correct?, (a) f ′ (x) is an even function, (b) f ′ (x) is an odd function, (c) f ′ (x) may be an even or odd function depending on the, type of function, (d) f ′ (x) is a constant function, , Ê (b) We have,, ∴, , 2, , 2, , 100. What are the order and degree, respectively, of the, 5, d 3y , dy , differential equation 3 = y 4 + ?, dx , dx , , (a) 4, 5, , (b) 2 π e π, , 2, , dy, at x = π equal to ?, dx, , (c) 2e π, , 2, , (d) e π, , 5, d3 y , 4, dy , 3 = y + , dx , dx , , Here, highest order derivative is, , 2, , + 2 πeπ (0 ) = 2eπ, , 2, , 80, 243, 20, (c), 243, , dx3, , ., , 40, 243, 10, (d), 243, (b), , 2, , Ê (a) According to the question,, ⇒, ⇒, Q, ⇒, ∴, , 99. What is the solution of, , (1 + 2x ) dy − (1 − 2y ) dx = 0 ?, , np = 3 npq, 1, q=, 3, p+ q=1⇒p+, p=, , 1, =1, 3, , 2, 3, 3, , =, , 3, , 2, , 2, , 80, 5!, 2, 1, × × =, 3, 243, 3 ! 2! 3 , , 102. Consider the following statements, , (1 + 2x) dy − (1 − 2 y) dx = 0, , ⇒ (1 + 2x) dy = (1 − 2 y) dx ⇒, , [where n = number, of trials], , 2, 1, p ( X = 3 ) = 5C3 × , 3, 3, , (b) y − x − 2 xy = c, (d) x + y + 2 xy = c, , Ê (a) We have,, , dy, dx, =, 1 − 2 y 1 + 2x, , On integrating both the sides, we get, dy, dy, ∫ 1 − 2 y = ∫ 1 + 2x, 1, 1, ⇒, − log (1 − 2 y) = log(1 + 2x) + C ′, 2, 2, ⇒, , d3 y, , Mean = 3 (Variance), , On differentiating both the sides, we get, 2, 2, dy, = 2 cos 2x ex + 2x ex sin 2x, dx, 2, 2, dy, ∴ , = 2 cos 2 πeπ + 2 πeπ sin 2 π, dx x = π, , (a) x − y − 2 xy = c, (c) y + x − 2 xy = c, , (d) 5, 4, , equation is, Ê (c) Given differential, 2, , (a), , 2, , 2, , (c) 3, 2, , variance. What is the probability of exactly 3 successes, out of 5 trials ?, , f ′ (− x ) = − f ′ (x ), , (c) We have, y = ex sin 2x, , = 2(1 ) eπ, , (b) 2, 3, , 101. In a Binomial distribution, the mean is three times its, , ∴ f ′ (x) is an odd function., , 98. If y = e x sin 2x , then what is, , x − y − 2xy = C, 1, [where C = (e − 2 C′ − 1 )], 2, , So, order = 3, and degree = 2, , f (− x ) = f (x ), , On differentiating both the sides, we have, , (a) (1 + π ) e π, , ⇒, , f (x) is an even function., , − f ′ (− x ) = f ′ (x ) ⇒, , 2x − 2 y − 4xy = e − 2 C′ − 1, 1, x − y − 2xy = (e − 2 C′ − 1 ), 2, , 2, , 97. If f ( x ) is an even function, where f ( x ) ≠ 0, then which, , Ê, , ⇒, , − log (1 − 2 y) = log(1 + 2x) + 2C ′, , I. P ( A ∪ B ) = P ( A ) = P ( B ) − P ( A ∩ B ), II. P ( A ∩ B ) = P ( B ) − P ( A ∩ B ), III. P ( A ∩ B ) = P ( B ) P ( A | B ), , Which of the above statements are correct?, (a) I and II, , (b) I and III, , (c) II and III, , (d) I, II and III, , Ê (b) Here,, Statement I, P( A ∪ B) = P( A ) + P(B) − P( A ∩ B) is correct.
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42, , NDA/NA Solved Paper 2018 (I), Statement II, , Ê (d) According to correlation condition,, , P ( A ∩ B ) = P(B) − P( A ∩ B) is wrong as, , If correlation coefficient rr ⋅ y = 0, then lines of regression, are parpendicular, , P ( A ∩ B ) = P( A) − P( A ∩ B)., Statement III, , And if rry = 1, then lines of regression are parallel., , A, P( A ∩ B) = P(B) × P is correct. [by conditional, B, theorem], , So, both statements are wrong., , Hence, Statements I and III are correct., , 103. If the correlation coefficient between x and y is 0.6,, covariance is 27 and variance of y is 25, then what is the, variance of x?, (a), , 9, 5, , 81, 25, , (b), , (c) 9, , 107. If 4 x − 5y + 33 = 0 and 20x − 9y = 107 are two lines of, regression,, then, are the values of x and y respectively?, (a) 12 and 18, , 0.6 =, , (d) 81, , and, , 4x − 5 y + 33 = 0, , … (i), … (ii), , on multiplying Eq. (i) by 5 and subtract Eq. (ii) from it,, we get, , C⋅V, σx ⋅ σ y, , 20 x − 25 y + 165 = 0, 20 x − 9 y − 107 = 0, , 27, 27, 27, ⇒ σx =, =, =9, σx × 5, 0.6 × 5 3, , −, , Ê (b) Given that, P( A) = 0.4 and, , Here, X = 13 and Y = 17, , 108. Consider the following statements, , = 1 − (0.3 ) = 0 . 7, , 105. Let x be the mean of x 1 , x 2 , x 3 ,… x n . If x i = a + cy i for, some constants a and c, then what will be the mean of, y 1 , y 2 , y 3 , …, y n ?, 1, x, c, , (c), , 4x − 85 + 33 = 0, 4x = 52 ⇒ x = 13, , The mean of two regression lines are the solution set at, given regression lines,, , P(B) = 0.5, , = 1 − [(1 − 0.4) × (1 − 0.5)] = 1 − (0.6 )(0.5), , (b) a −, , ⇒, , (d) 0.9, , Q P( A ∪ B) = 1 − P ( A′ ∩ B′ ), , (a) a + c x, , +, , on putting the value of y in Eq. (i), we get, , Questions B are 0.4 and 0.5 respectively. What is the, probability that he solves atleast one of the two, questions?, (c) 0.8, , +, , − 16 y = − 272, y = 17, , 104. The probabilities that a student will solve Question A and, , (b) 0.7, , (d) 17 and 13, , 20 x + 9 y − 107 = 0, , ∴ Variance of x = σ 2 (x) = (9 )2 = 81, , (a) 0.6, , (c) 13 and 17, , Ê (c) Given lines of regression are, , Ê (d) Given, σ 2 ( y) = 25 ⇒ σ( y) = 5, Correlation coefficient =, , (b) 18 and 12, , what, , 1, x −a, c, , (d), , x −a, c, , Ê (d) Given that,, , Mean of x1 , x2 , x3 , .... xn i.e. x = x, , I. Mean is independent of change in scale and change in, origin., II. Variance is independent of change in scale but not in, origin., , Which of the above statements is/are correct?, (a) I only, (c) Both I and II, , Ê (d), , (b) II only, (d) Neither I nor II, , Since, mean changes with changes in origin. So,, Statement I is wrong., And variance is independent to the choice of origin. So,, Statement II is also wrong., , Hence, both statements are wrong., , Now, we have, 1, xi = a + cyi ⇒ yi = (x − a ), c, 1, ⇒ y = (x − a ), c, , 106. Consider the following statements, I. If the correlation coefficient, rxy = 0, then the two lines of regression are parallel to, each other., II. If the correlation coefficient rxy = 1, then the two lines, of regression are perpendicular to each other., , 109. Consider the following statements, I. The sum of deviations from mean is always zero., II. The sum of absolute deviations is minimum when taken, around median., , Which of the above statements is/are correct., (a) I only, (c) Both I and II, , Ê (c) By the property of deviation both statement are correct., 110. What is the median of the numbers 4.6, 0, 9.3, −4.8, 7.6 2.3,, , Which of the above statements is/are correct?, , 12.7, 3.5, 8.2, 6.1, 3.9, 5.2 ?, , (a) I only, (c) Both I and II, , (a) 3.8, (c) 5.7, , (b) II only, (d) Neither I nor II, , (b) II only, (d) Neither I nor II, , (b) 4.9, (d) 6.0
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43, , NDA/NA Solved Paper 2018 (I), Ê (b) On arranging the given number is ascending order, we, have, , − 4.8, 0, 2.3, 3.5, 3.9, 4.6, 5.2, 6.1, 7.6, 8.2, 9.3, 12.7, Here, n = 12, So, median, 12, Value of th number, 2, 12, + Value of + 1 th number, , 2, =, 2, . + 52, ., Value of 6th number + Value 7th number 46, =, =, = 49, ., 2, 2, , 115. If A and B are two events such that 2P( A ) = 3P( B ), where, 0 < P( A ) < P( B ) < 1, then which one of the following is, correct?, (a) P( A| B) < P( B| A ) < P( A ∩ B), (b) P( A ∩ B) < P( B| A ) < P( A | B), (c) P( B| A ) < P ( A | B) < P( A ∩ B), (d) P( A ∩ B) < P( A | B) < P( B| A ), , 2P( A) = 3 P(B), P( A), 3 P(B), ⇒2, =, P( A ∩ B) P( A ∩ B), , Ê (b) Given that,, , 1 P ( A ∩ B) 1, ×, =, 2, 3, P( A), 1, B, , ⇒, × P =, A, 2, B, P <, ⇒, A, , 111. In a test in Mathematics, 20% of the students obtained, , ⇒, , ‘‘first class’’. If the data are represented by a pie chart,, what is the central angle corresponding to ‘‘first class’’?, (a) 20°, , (b) 36°, , (c) 72°, , (d) 144°, , Ê (c) Pie chart contains total angle equal to 360°., So, central angle corresponding to “First class”, 20, × 360 ° = 72°, = 20% of 360° =, 100, , 112. The mean and standard deviation of a set of values are 5, and 2 respectively. If 5 is added to each value, then what is, the coefficient of variation for the new set of values?, (a) 10, , (b) 20, , (c) 40, , (d) 70, , Ê (b) Given, mean = 5, , is drawn at random and kept aside. From the remaining, a, second chit is drawn at random. What is the probability, that the second chit drawn is ‘‘9’’ ?, 1, 10, 1, (c), 90, , E2 be the event of drawing second chit bearing number, 9., 9, 1, C, C, 9, 1, and P(E2 ) = 9 1 =, ∴, P(E1 ) = 10 1 =, 10, C1, C1 9, , But standard deviation will remain same., Hence, coefficient of variation, σ, 2, =, × 100 =, × 100 = 20, mean, 10, , ∴ Required probability = P(E1 ) ⋅ P (E2 ), 9, 1 1, =, × =, 10 9 10, , 113. A train covers the first 5 km of its journey at a speed of 30, km/h and the next 15 km at a speed of 45 km/h. What is, the average speed of the train?, (d) 40 km/h, , Total distance, Total time, 5 + 15, 20 × 6, 20, 20, =, =, =, =, = 40 km/h, 5, 15 1 1 1 + 2, 3, +, +, 30 45 6 3, 6, , Ê (d) Average speed =, , 114. Two fair dice are rolled. What is the probability of getting, a sum of 7?, (a), , 1, 36, , Ê (b) Here,, , (b), , 1, 6, , (c), , 7, 12, , (d), , 5, 12, , n(S) = 36 and E be the event of getting a sum of 7, , on two fair dice., = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}, ∴ n(E ) = 6, So, required probability =, , n( E ) 6, 1, =, =, n(S ) 36 6, , 1, 9, (d) None of these, , (b), , (a), , So, new mean = 5 + 5 = 10, , (c) 39.5 km/h, , P(B), 1 A, P , 3 B, A, P , B, , 116. A box has ten chits numbered 0, 1, 2, 3, …, 9. First, one chit, , Since, 5 is added to each value., , (b) 37.5 km/h, , [dividing both sides by P( A ∪ B)], P ( A ∩ B), , Ê (a) Let E1 be the event at drawing a chit which is not 9 and, , and standard deviation (σ ) = 2, , (a) 35 km/h, , ×, , 117. One bag contains 3 white and 2 black balls, another bag, contains 5 white and 3 black balls. If a bag is chosen at, random and a ball is drawn from it, what is the change, that it is white?, (a), , 3, 8, , Ê (b) Let, , (b), , 49, 80, , (c), , 8, 13, , (d), , 1, 2, , E1 be the event of selecting the first bag and E2 be, , the event of selecting the second bag. Let A be the event, of drawing white ball., So, by theorem at total probability., A, A, P( A) = P(E1 ) × P + P(E2 ) × P, , E2 , E1 , 3, 5, C, C, 1, 1, ×5 1 + × 8 1, 2, C1 2, C1, 1 3 5, 1 24 + 25, = + = ×, , , 2 5 8 2, 40, , =, , =, , 1 49 49, ×, =, 2 40 80
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44, , NDA/NA Solved Paper 2018 (I), , 118. Consider the following in respect of two events A and B, I. P ( A occurs but not B ) = P ( A ) − P ( B ) if B ⊂ A, II. P ( A alone or B alone occurs) = P ( A ) = P ( B ) − P ( A ∩ B ), III. P ( A ∪ B ) = P ( A ) + P ( B ) if A and B are mutually, exclusive, , Which of the above is/are correct?, (a) I only, , (b) I and III, , (c) II and III, , (d) I and II, , Ê (b) If B ⊂ A, then P( A − B), = P( A) − P( A ∩ B) = P( A) − P(B), [QB ⊂ A ⇒ A ∩ B = B], So, Statement I is correct., P (A alone or B alone), = P( A) − P( A ∩ B) + P(B) − P( A ∩ B), = P( A) + P(B) − 2 P( A ∩ B), So, Statement II is wrong., If A and B are mutually exclusive,, then P( A ∩ B) = 0, ⇒ P( A ∪ B) = P( A) + P(B), So, Statement III is correct., Here, Statement I and III are correct., , Ê (a) Total number of selecting three members = 9C3, Favourable numbers of selecting two members as men, = 4C2 × 5C1, 4, C2 × 5C1, So, required probability =, 9, C3, 4 ×3 5, ×, 2×1 1 2×3 ×5, 5, =, =, =, 9 × 8 × 7 3 × 4 × 7 14, 3 ×2×1, , 120. The standard deviation σ of the first N natural numbers, can be obtained using which one of the following, formulae?, N2 − 1, 12, N−1, (c) σ =, 12, (a) σ =, , 5, 14, 3, (c), 14, (a), , 1, 21, 8, (d), 21, (b), , (d) σ =, , N2 − 1, 12, N2 − 1, 6N, , 1, ΣXi2 − ( X )2, N, 1 2, =, (1 + 22 + ... + N 2 ), N, , Ê (b) Q σ 2 =, , 1, − (1 + 2 + 3 + ... + N ), , N, , 119. A committee of three has to be chosen from a group of 4, men and 5 women. If the selection is made at random,, what is the probability that exactly two members are, men?, , (b) σ =, , =, =, σ=, , N ( N + 1 ) (2 N + 1 ) ( N + 1 ) , 1, ×, −, , , N, 6, 2, 2, N −1, 12, N2 − 1, 12, , 2, , 2
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NDA / NA, National Defence Academy/Naval Academy, , Solved Paper, , 2017 (II), , Paper 1 (Mathematics), 1. If x + log10 (1 + 2 x ) = x log10 5 + log10 6 then x is equal to, (a) 2, -3, , (b) 2 only, , (c) 1, , 1 1 0, , (d) 3, , x + log10 (1 + 2 ) = x log10 5 + log10 6, x, , Ê (c), Þ, , 0 × 20 = 0, 1 × 21 = 2, 1 × 22 = 4, 6, , x(1 - log10 5) = log10 6 - log10 (1 + 2x ), , Þ x(log10 10 - log10 5) = log10 6 - log10 (1 + 2 ), x, , Þ, , Þ, , 10 ö, 6, xæç log10, ÷ = log10, è, 5ø, 1 + 2x, , [Qlog10 10 = 1], , éQ log a - log b = log a ù, 10, 10, 10, ëê, b ûú, 6, x log10 2 = log10, 1 + 2x, , \ (101110 ) = (46 )10, \, , (110 )2 = (6 )10, , Now,, , 6 ) 46 ( 7, 42, 4, , \ Quotient = 7,, , For x = 1 , above equation is satisfied, , 2, , 7, , \, , 2, , 3, , 1, , 2, , 1, , 1, , 0, , 1, , x = 1 is only solution., , 2. The remainder and the quotient of the binary division, (101110) 2 ¸ (110) 2 are respectively, (a) (111)2 and (100)2, (b) (100)2 and (111)2, , \, , (c) (101)2 and (101)2, (d) (100)2 and (100)2, , Ê (b), , (7 )10 = (111 )2, , \ Quotient = (111 )2, Remainder = 4, , 1 0 1 1 1 0, , 0 × 20 = 0, 1 × 21 = 2, 1 × 22 = 4, 1 × 23 = 8, 0 × 24 = 0, 1 × 25 = 32, 46, , \, , 2, , 4, , 2, , 2, , 0, , 2, , 1, , 0, , 0, , 1, , (4)10 = (100 )2, , \ remainder = (100 )2
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2, , NDA/NA Solved Paper 2017 (II), 3. The matrix A has x rows and x + 5columns. The matrix B, has y rows and 11 - y columns. Both AB and BA exist., What are the values of x and y respectively?, (a) 8 and 3, , (b) 3 and 4, , (c) 3 and 8, , (d) 8 and 8, , Ê (c), , (a) B¢ È C ¢, , (b) B È C, , \ E - A = A¢}, = E - (E - (E - A)), , [Q E - A¢ = A], , = B¢ Ç C ¢ [By Demorgan’s Theorem], , x = 3 and y = 8 ., , n (n - 1)Q, , where S n denotes the sum of the, 4. If S n = nP +, 2, first n terms of an AP, then the common difference is, (a) P + Q, , (b) 2 P + 3Q, , (c) 2Q, , (d) Q, , 7. If A = {x : x is a multiple of 2}, B = {x : x is a multiple of 5}, and C = {x : x is a multiple of 10}, then A Ç ( B Ç C ) is, equal to, (a) A, , (b) B, , (c) C, , (d) {x : x is a multiple of 100}, , Ê (c) From questions,, , A = {2, 4, 6, 8, ¼¼ }, , n(n - 1 )Q, , …(i), , 2, , We know, sum of n terms of AP, n, = [2a + (n - 1 )d], 2, n(n - 1 ), = na +, d, 2, [Here a = first term and d = common difference], By comparing (i) and (ii), we get, common difference, d = Q, , 5. The roots of the equation, (q - r )x 2 + (r - p )x + ( p - q ) = 0 are, , (q - r ), ,1, (p - q), , [Q E is universal set, , = E - A = A¢ = (B È C )¢ [Q A = B È C], , 11 - y = x, , (r - p) 1, (a), ,, (q - r ) 2, , = E - (E - (E - (E - A¢ )))), , = E - (E - A¢ ), , By solving above two equations,, , Sn = np +, , (d) B Ç C, , \ E - (E - (E - (E - (E - A)))), , \ For AB and BA both to be exist, , (d) We have,, , (c) B¢ Ç C ¢, , Ê (c) Given, E is the universal set and A = B È C, , x+ 5= y, , (c), , E - ( E - ( E - ( E - ( E - A )))) is same as the set, , We know, two matrices P and Q can be multiplied, (i.e., PQ exist) only when number of columns of, P (pre-multiplier) is equal to the number of rows of, q (post multiplier)., , and, , Ê, , 6. If E is the universal set and A = B È C , then the set, , (p - q), (b), ,1, (q - r ), (d), , (r - p) 1, ,, (p - q) 2, , Ê (b) We have,, , (q - r )x2 + (r - p )x + ( p - q) = 0, , Here, sum of coefficients, (q - r ) + (r - p ) + ( p - q) = 0, \ One roots of given equation must be 1., p-q, Here, Product of roots =, q-r, [Q In quadratic eq n ax2 + bx + c = 0, c, Product of roots = ], a, p-q, \ Another root ´ 1 =, q-r, p-q, ,1, \ Roots of the given equation are, q-r, , B = {5, 10, 15, 20, ¼¼ }, C = {10, 20, 30, 40, ¼¼ }, Here, C is subset of A. i.e C Î A and C is subset of B, i.e., C Ì B, , …(ii), , Now, ( A Ç (B Ç C ) = A Ç C = C, , 8. If a and b are the roots of the equation 1 + x + x 2 = 0,, é 1 b ù éa b ù, then the matrix product ê, ú is equal to, úê, ëa a û ë 1 b û, é1 1ù, (a) ê, ú, ë1 2 û, é 1 -1ù, (c) ê, ú, ë -1 2 û, , é -1, (b) ê, ë -1, é -1, (d) ê, ë -1, , -1ù, 2 úû, -1ù, -2 úû, , Ê (b) Q a and b are the roots of the equations., \ a = w and b = w2, , [Here, w is cube roots of unity], Now,, b + b2 ù, é1 b ù é a b ù é a + b, ú, ê a aú ê1 b ú = ê 2, û ë a + a ab + ab û, ûë, ë, é w + w2 w2 + w4 ù, =ê 2, ú, 2w3 û, ëw + w, é w + w 2 w 2 + w(w 3 )ù, =ê 2, ú, 2w3, ëw + w, û, é w + w 2 w 2 + wù, [Q w3 = 1], =ê 2, ú, 2, +, w, w, w, ë, û, é- 1 - 1ù, [Q1 + w + w2 = 0], =ê, ú, ë- 1 2 û, \ option (b) is correct.
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3, , NDA/NA Solved Paper 2017 (II), 9. If | a | denotes the absolute value of an integer, then which, of the following are correct?, 1. | ab | = | a || b | 2. | a + b | £ | a | + | b |, 3. | a - b | ³ | a | - | b |, , 12. The sum of all real roots of the equation, | x - 3| 2 + | x - 3| - 2 = 0 is, (a) 2, , (b) 3, , (c) 4, , Ê (d) We have, |x - 3|, , 2, , Select the correct answer using the code given below., , + |x - 3| = - 2 = 0, |x - 3| = t, , Let, , (a) 1 and 2 only, , (b) 2 and 3 only, , \, , (c) 1 and 3 only, , (d) 1, 2 and 3, , Þ, , t 2 + 2t - t - 2 = 0, , Þ, , t (t + 2) - 1 (t + 2) = 0, , Ê (d) For asbsolute value, , (d) 6, , t2 + t - 2 = 0, , 1. |ab| = |a||b|is true, , \, , 2. |a + b| £ |a| + |b|is true, , [Here, t = - 2is not possible because t = |x - 3|is always, + ve], \, t =1, \, |x - 3| = 1, Þ, x - 3 = 1 or x - 3 = - 1, x = 4 or x = 2, \ Sum of roots of given equation = 4 + 2 = 6, , 3. |a - b| ³ ||a| - |b||is true, \ For absolute value all the given conditions are true., , 10. How many different permutation can be made out of the, letters of the word ‘PERMUTATION’?, (a) 19958400, , (b) 19954800 (c) 19952400, , (d) 39916800, , Ê (a) In word ‘PERMUTATION’ we have to arrange 11 letters, in which number of letter ‘I’ is 2., 11 !, \ Number of permutations =, 2!, 11 ´ 10 ´ 9 ´ 8 ´ 7 ´ 6 ´ 5 ´ 4 ´ 3 ´ 2 ´ 1, =, 2´1, , = 11 ´ 10 ´ 9 ´ 8 ´ 7 ´ 6 ´ 5 ´ 4 ´ 3 = 19958400, , 1, é 4i - 6 10i ù, and k = , where i = -1, then kA, ú, 2i, ë 14i 6 + 4i û, is equal to, , 13. It is given that the roots of the equation, , x 2 - 4 x - log 3 P = 0 are real. For this the minimum, value of P is, (a), , (b), , 1, 64, , (c), , 1, 81, , (d) 1, , x2 - 4x - log 3 P = 0, , …(i), , It is given that the roots of the given equation are real, \ Discriminant, D ³ 0, \, , 5 ù, é2 - 3i, (b) ê, 2 + 3i úû, ë 7, 5 ù, é2 + 3i, (d) ê, 2 + 3i úû, ë 7, , (-4)2 - 4(1 ) (- log 3 P) ³ 0, [Q D = b2 - 4ac], , Þ, , 16 + 4 log 3 P ³ 0, , Þ, , é 4 i - 6 10 i ù, ú, ë 14 i 6 + 4 iû, , Ê (a) Given, A = ê, , -1, 1 i i, and k = ´ = 2 =, i [Q i2 = - 1], 2i i 2i, 2, é 4i - 6 10 i ù, Now, kA = k ê, ú, ë 14i 6 + 4iû, k(10 i) ù, é k(4i - 6 ), =ê, k, (, 14, i, ), k, (, 6 + 4i)úû, ë, 1, é - 1 i(4i - 6 ), - (10 i) ù, ê 2, ú, 2, =ê, ú, 1, 1, +, i, (, 14, i, ), i, (, 6, 4, i, ), ê, ú, ë 2, û, 2, 2, 2, é - 2i + 3 i, ù, - 5i, =ê, ú, 2, - 3 i - 2i2 û, ë - 7i, - 5 (- 1 ) ù, é - 2 (- 1 ) + 3 i, =ê, - 3 i - 2 (- 1 )úû, ë - 7 (- 1 ), 5 ù, é2 + 3 i, =ê, 7, 2, 3 iúû, ë, , 1, 27, , Ê (c) We have,, , 11. If A = ê, , 5 ù, é2 + 3i, (a) ê, 2 - 3i úû, ë 7, 7ù, é2 - 3i, (c) ê, 2 + 3i úû, ë 5, , t = 1 or t = - 2, , Þ, , 4 log 3 P ³ - 16, 16, log 3 P ³ 4, , Þ, , log 3 P ³ - 4, P ³ 3 -4, 1, P³, 81, , Þ, Þ, , \ Minimum value of, , P=, , 1, 81, , 14. If A is a square matrix, then the value of, adj A T - (adj A )T is equal to, (a), (b), (c), (d), , A, 2| A|I, where I is the identity matrix, null matrix whose order is same as that of A, unit matrix whose order is same as that of A, , Ê (c) We know, in square matrix A, , adj AT = (adj A)T, , Þ, , adj AT - (adj A)T = 0, , \ adj A - (adj A)T is null matrix whose order is same, as that of A., T
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4, , NDA/NA Solved Paper 2017 (II), , 15. The value of the product, 1, 2, , 1, 4, , 1, 8, , 19. If a, b, c are non-zero real numbers, then the inverse of, , 1, 16, , the matrix, , 6 ´ 6 ´ 6 ´ 6 ´ … up to infinite terms is, (a) 6, , Ê (a) 6, , (b) 36, 1, 2, , 1, , 1, , (c) 216, , éa 0 0ù, A = ê 0 b 0ú is equal to, ú, ê, êë 0 0 c úû, , (d) 512, , 1, , ´ 6 4 ´ 6 8 ´ 616 ´ ¼¼ up infinite terms, =6, , 1, 1, 1, 1, +, +, +, + ¼¼ ¥, 2, 4, 8 16, , =6, , 1/2, 1, 2, , [Q we know, sum of an infinite GP with first terms a, and d common ratio r is, a, , where - 1 < r < 1] = 6, S¥ =, 1 -r, , ½cos 2 q, 2, 16. The value of the determinant ½, ½sin 2 q, ½, 2, for all values of q, is, , q, sin 2 ½, 2½, 2 q½, cos, 2½, , Ê, , number of terms in the expansion, ( x + a )100 + ( x - a )100 after simplification is, (b) 101, , (c) 51, , \ Inverse fo matrix A, i.e.,, , of, , (d) 50, , Number of terms in the expansion of (x + a )100 will be, 101 and number of terms in the expansion of (x - a )100, also will be 101., , \ In the expansion of, (x + a )100 + (x - a )100 50 terms will be cancel out and 51, terms will be added up, \ Number of terms in the expansion of, (x + a )100 + (x - a )100 will 51, , 18. In the expansion of (1 + x ) 50 , the sum of the coefficients, of odd powers of x is, (b) 2 49, , (c) 2 50, , (d) 2 51, , Ê (b) We know, the sum of the coefficients of (1 + x), , n, , A-1 = diag (a -1 b-1 c-1 ), é a -1 0, 0 ù, ú, ê, = ê 0 b-1 0 ú, -1, ê 0, 0 c ú, û, ë, , 20. A person is to count 4500 notes. Let an denote the, number of notes he counts in the nth minute. If, a1 = a 2 = a 3 = … = a10 = 150, and a10 , a11 , a12 , … are in AP, with the common difference -2, then the time taken by, him to count all the notes is, (a) 24 minutes (b) 34 minutes (c) 125 minutes (d)135 minutes, , Ê (c) We know,, , (a) 2 26, , A = diag (a b c), , = diag (a1-1 , a2-1 a3-1 ¼¼ an-1 ), , = cos q × 1 = cos q, , 17. The, , é a -1 0, 0ù, ú, 1 ê, -1, (b), 0 b, 0ú, abc ê, ê 0, 0 c -1 ú, û, ë, é a 0 0ù, 1 ê, (d), 0 b 0ú, ú, abc ê, êë 0 0 c úû, , We know that, the inverge of a diagonal matrix diag, (a1 a2 a3 ¼¼ an ), , [Q a 2 - b2 = (a + b) (a - b)], , (a) 202, , 0, , \, , (b) cos q, (c) sinq, (d) cos2q, é cos 2 q sin 2 q ù, ê, 2, 2ú, (b) ê, ú, 2 q, 2 q, cos, ú, ê sin, ë, 2, 2û, 2, q, q, = æç cos 2 ö÷ - æç sin 2 ö÷, è, 2ø è, 2ø, q, q, æ, 2 q, 2 qö æ, = ç cos - sin ÷ - ç cos 2 + sin 2 ö÷, è, 2, 2ø è, 2, 2ø, (a)1, , Ê, , 0ù, ú, -1, 0ú, b, 0 c -1 ú, û, é 1 0 0ù, 1 ê, (c), 0 1 0ú, ú, abc ê, êë 0 0 1úû, éa 0 0ù, (a) We have, A = ê0 b 0 ú, ú, ê, êë0 0 c úû, é a -1, ê, (a) ê 0, ê 0, ë, , 1-, , ., , i.e., C0 + C1 + C2 + ¼¼ + Cn = 2n, , \ sum of the coefficients of odd powers of (1 + x)n ., 2n, i.e., C1 + C3 + C5 + ¼¼ =, 2, \ sum of the coefficients of odd powers of, 250, (1 + x)50 =, = 249, 2, , Ê (b) Given, a, , 1, , = a2 = a3 = ¼¼ = a10 = 150, , \ From question, Number of notes counted by the person in 9 minutes, = 9 ´ 150 = 1350, Let total minutes to count all the 4500 notes by the, person = t, \ Remaining minutes after 9 minutes = t - 9, We know sum of n terms of an AP whose first term ‘a’, and common d is given by, n, Sn = [2a + (n - 1 )d], 2, Now, From question,[2 ´ 150 + {(t - 9 ) - 1 } (-2)] = 4500, [QTotal number of notes = 4500 and common difference, = - 2], t -9, 1350+, [300 - 2t + 20] = 4500, 2, t 2 - 169 t + 4950 = 0 Þ (t - 135) (t - 34) = 0, , \, , t = 135 or + = 34, , Here + = 135 (impossible), \, , t = 34 Minutes
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5, , NDA/NA Solved Paper 2017 (II), n, , æ1 + i ö, 21. The smallest positive integer n for which ç, ÷ = 1, is, è1 - i ø, (a) 1, , (b) 4, , (c) 8, , n, , Ê (b), , (d) 16, , n, , é (1 + i)2 ù, æ1 + i ö, æ1 + i 1 + i ö, ´, ç, ÷ =ç, ÷ =ê 2, 2 ú, è1 - i ø, è1 - i 1 + iø, ë (1 ) - (i) û, , n, , 24. If a and b are the roots of the equation 3x 2 + 2x + 1 = 0,, then the equation whose roots are a + b -1 andb + a -1 is, (a) 3x 2 + 8x + 16 = 0, , (b) 3x 2 - 8x - 16 = 0, , (c) 3x 2 + 8x - 16 = 0, , (d) x 2 + 8x + 16 = 0, , Ê (a) Given a and b are the roots of equations, , 3 x2 + 2x + 1 = 0, 2, …(i), a+ b = \, 3, 1, and, …(ii), ab =, 3, We have to find the equation whose roots are a + b -1, and b + a-1 ., \ Sum of roots of required equation, , [Q a - b = (a + b) (a - b)], 2, , 2, , n, , é 1 + i2 + 2i ù, =ê, ú, ë 1 - (- 1 ) û, n, é 1 - 1 + 2i ù, = (i) n, =ê, ú, 2, û, ë, , [Qi2 = - 1], , \ n = 4 is the smallest value for which (i)n = 1., , = (a + b -1 ) + (b + a-1 ), a+ b, 1 1, = a+ b + + = a+ b +, a b, ab, - 2 - 2/ 3, [from (i) and (ii)], =, +, 3, 1 /3, 8, =3, and product of roots of required equation, = (a + b -1 ) (b + a-1 ) = ab + a × a-1 + b -1 × b + b -1 a-1, 1, = ab + 1 + 1 +, ab, 1, 1, [From (ii)], = + 2+, 1, 3, 3, 16, =, 3, \ Required equation will be, x2 - (sum of roots) x + product of roots = 0, 8, 16, \, = 0 Þ 3 x2 + 8 x + 16 = 0, x2 - æç - ö÷ x +, è 3ø, 3, , 22. If we define a relation R on the set N ´ N as (a, b ) R (c , d ), Û a + d = b + c for all (a, b ), (c , d ) Î N ´ N, then the, relation is, (a) symmetric only, , (b) symmetric and transitive only, , (c) equivalence relation, , (d) reflexive only, , Ê (c), , (a, b)R(a, b)" (a, b) Î N ´ N, Since, a + b = b + a,, \ R is Reflexive., for R is to Symmetric, (a, b) R (c, d), Þ a + d = b+ cÞd+ a = c+ b, Þ c + d = d + a Þ (c, d) R (a, b), \ R is Symmetric., Now, By definition at R,, a + d = b + c and c + f = d + c, equality both sides, a + d + c + f = b + c + d + e as, a + f = b+ e, Now, (a, b) R (c, d) and (c, d) R (c, f ), Þ, , 25. The value of, , (a, b) R (e, f ), , 1, 1, 1, +, +, + …up to infinite terms is, log 3 e log 3 e 2 log 3 e 4, , \ R is transitive., Since R is reflexive, symmetric and transitive, , (a) loge 9, , \ R is an equivalance relation., , 23. If y = x + x 2 + x 3 + … up to infinite terms where x < 1,, then which one of the following is correct?, (a) x =, , y, 1+ y, , Ê (a) Given, y = x + x, where,, , y, 1- y, , (b) x =, 2, , (c) x =, , 1+ y, y, , (d) x =, , 1- y, y, , + x + ¼¼ upto infinite terms, 3, , x<1, , Which is sum of infinite term of GP whose first term is x, and common ratio is x when x < 1., x, \, y=, 1-x, [Q Sum of infinite terms of GP whose first term is a, a, and common ratio r =, where, r < 1], 1 -r, y, x=, Þ, 1+ y, , Ê (a), , (b) 0, , (c) 1, , (d) loge 3, , 1, 1, 1, Let, y=, +, +, + …¥, log 3 e log 3 e2, log 3 e4, 1, 1, 1, =, +, +, + …¥, log 3 e 2log 3 e 4log 3 e, ù, é, 1 ê 1 ú, 1 é, 1 1, 1 + + + .... ¥ù =, ú, ê, úû, 2 4, log 3 e ëê, log 3 e ê 1 - 1 ú, ë, 2û, 1, =, × 2 = 2 log e 3 = 2 log e 3 2 = log e 9, log 3 e, =, , 26. A tea party is arranged for 16 people along two sides of a, long table with eight chairs on each side. Four particular, men wish to sit on one particular side and two particular, men on the other side. The number of ways they can be, seated is, (a) 24 ´ 8! ´ 8! (b) (8!)3, , (c) 210 ´ 8! ´ 8!, , (d) 16!
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6, , NDA/NA Solved Paper 2017 (II), , Ê (c), , Since four particular men want to sit on a particular, side A (say) and two other porticular men on the other, side B. So, we are left with 10 guests out of which we, choose 4 for side A and 6 four side B., Hence, the number of selection for the two sides, = 10 C4 ´ 6 C6, Now, 8 persons on each side of the table can be, arranged among themselves in 8! ways., Hence, the total number of arrangements, 10 !, = 10 C4 ´ 6 C6 ´ 8 ! ´ 8 ! =, ´ 1 ´ 8! ´ 8!, 4!6 !, 10 ´ 9 ´ 8 ´ 7, =, ´ (8 !)2 = 210 ´ (8 !)2, 4´3 ´2´1, , Q, p, P, 29. In DPQR, ÐR = . If tan æç ö÷ and tan æç ö÷ are the roots of, è2ø, è2ø, 2, the equation ax 2 + bx + c = 0, then which one of the, following is correct?, (a) a = b + c, , (b) b = c + a, , (c) c = a + b, , (d) b = c, p, , Ê (c) We have, D PQR in which Ð R = 2 ., , p, 1, 1, p, Þ ÐP + ÐQ =, 2, 2, 4, 2, p, P Q, Þ, tan æç + ö÷ = tan, è2, 2ø, 4, P, Q, tan + tan, 2, 2 =1, Þ, P, Q, 1 - tan tan, 2, 2, P, Q, P, q, …(i), Þ tan + tan = 1 - tan tan, 2, 2, 2, 2, P, Q, Since tan and tan are the roots of the equation, 2, 2, , \, , 27. The system of equations kx + y + z = 1, x + ky + z = k, and x + y + kz = k 2 has no solution if k equals., (a) 0, , (c) -1, , (b) 1, , (d) -2, , kx + y + z = 1, , Ê (d) We have,, , x + ky + z = k, , x+ y+, ½k, For no solution, ½1, ½, ½1, , kz = k2, 1 1½, k 1½ = 0, ½, 1 k½, , Þ, , k(k2 - 1 ) - 1 (k - 1 ) + 1 (1 - k) = 0, , Þ, , k - k- k+ 1 + 1 - k=0, , Þ, , k3 - 3 k + 2 = 0, , ax2 + bx + c = 0, P, Q, b, \ tan + tan = 2, 2, a, P, Q c, and tan × tan =, 2, 2 a, , 3, , Þ, , 4, 30. If½z - ½ = 2, Then the maximum value of | z | is equal to, , k= -2, , 28. If 13, . + 23, . + 33, . + ¼ + n.3, , (a) 1 +, , n, , (2 n - 1)3 + b, then a and b are respectively, 4, (a) n, 2, (b) n, 3, (c) n + 1, 2, (d) n + 1, 3, a, , =, , Ê (d) Given, series is, , 1 × 3 + 2 × 3 + 3 × 3 + ¼¼ + n × 3, 2, , 3, , …(i), , - 2Sn = 1 × 3 + [3 + 3 + ¼¼ + 3 ] - n × 3, , Þ, Þ, Þ, , n, , |z|2 - 2|z| - 4 £ 0, , Þ (|z| - 1 +, , 5 ) (|z| - 1 - 5 ) £ 0, , (d) 5 - 1, , 5, , Thus, the maximum value of|z|is 1 +, n +1, , -2Sn = (3 + 3 2 + 3 3 + ¼¼ + 3 n ) - n × 3 n + 1, -2Sn = (3 + 3 2 + 3 3 + ¼¼ + 3 n ) - n × 3 n + 1, é3 n - 1, ù, - 2Sn = 3 ê, - n3 n + 1 ú, ë 3 -1, û, n n+1 3 n, Sn = × 3, - (3 - 1 ), 4, 2, 2n × 3 n + 1 - 3 n + 1 + 3, Sn =, 4, (2n - 1 ) 3 n + 1 + 3, Sn =, 4, , on comparing a = n + 1, b = 3, , Þ, , Þ 1 - 5 £ |z| £ 1 +, , …(ii), , On subtracting eq (ii) from eq (i), we get, , Þ, , 5, , [using triangle inequality], 4, 4, [Q½z - ½ = 2], |z| £ 2 +, |z|, z½, ½, , n, , 3, , (c) 1 -, , 5, , Þ, , Þ 3Sn =1 × 3 2 + 2 × 3 2 + ¼¼, + (n - 1 )3 n n × 3 n + 1, 2, , (b) 1 +, , 4 4, 4, 4, |z| = ½ z - + ½ £ ½z - ½ + ½ ½, z, z½ ½, z½ ½z½, ½, , n, , Let, Sn = 1 × 3 + 2 × 3 2 + 3 × 3 3 + ¼¼ + n × 3 n, , n, , 3, , Ê (b) We have,, , The above series is AGP, , Þ, , z½, , ½, , 3, , …(iii), , b, c, =1 a, a, - b = a - cÞc = a + b, , Þ, , (k + 2) (k - 1 ) (k - 1 ) = 0 Þ k = 1, - 2, , 2, , …(ii), , \ From eq n (i), (ii) and (iii), we have -, , But for k = 1, first two equations will be identical, \, , ÐP + ÐQ =, , 5, , 31. The angle of elevation of a stationary cloud from a point, 25 m above a lake is 15° and the angle of depression of its, image in the lake is 45°. The height of the cloud above the, lake level is, (a) 25 m, , Ê (b), , (b) 25 3 m, , (c) 50 m, , (d) 50 3 m, , Let AB be the surface of the lake and P be the point of, observation such that AP = 25 m. Let C be the position, of the cloud and c¢ be its reflection in the lake. The, CB = C ¢ B. Let PM be perpendicular from P on CB. Then,, ÐCPM = 15° and ÐC ¢ PM = 45°. Let CM = h, then, CB = h + 25 consequently, C ¢ B = h + 25., In DCMP, we have
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7, , NDA/NA Solved Paper 2017 (II), CM, h, Þ 2- 3 =, PM, PM, h, …(i), PM =, Þ, 2- 3, h + 50, C¢ M, In D PMC¢, we have tan 45° =, Þ1=, PM, PM, tan15° =, , Þ, , PM = h + 50, , 33. The value of 3 cosec 20° - sec 20° is equal to, (a) 4, , 15º, 45º, , P, , …(ii), h, h+25, , 25 m, , A, , B, h+25, , 34. Angle a is divided into two parts A and B such that, , A - B = x and tan A : tan B = p : q . The value of sin x is, equal to, , C′, , From (i) and (ii), , h, = h + 50, 2- 3, , (a), , h = h(2 - 3 ) + 50 (2 - 3 ), , Þ, Þ, , h - 2h +, , 3 h = 50 (2 - 3 ), , Þ, , \, , =, , 75 - 25 3, ( 3 - 1), , =, , 25(3 - 3 ), 3 -1, , 25(3 3 + 3 - 3 - 3 ), 3 -1, , =, , =, , 25(3 - 3 ), 3 -1, , ´, , 3 +1, 3 +1, , 25(2 3 ), = 25 3 m, 2, , 32. The value of, tan 9° - tan 27° - tan 63° + tan 81° is equal to, (a) -1, , (b) 0, , (c) 1, , (d) 4, , Ê (d) We have,, , tan 9° - tan 27° - tan 63° + tan 81°, , psina, p-q, , (d), , ( p - q )sin a, p+ q, , A + B = a and A - B = x, a+ x, a- x, and B =, A=, 2, 2, , on applying componendo and dividendo , we get, sin a + sin x + sin a - sin x p + q, =, sin a + sin x - sin a + sin x p - q, sin a p + q, =, sin x, p-q, p-q, Þ, sin x =, sin a, p+ q, , = tan 9° - tan 27° - tan (90° - 27° ) + tan (90° - 9° ), = tan 9° - tan 27° - cot 27° + cot 9°, = (tan 9° + cot 9° ) - (tan 27° + cot 27° ), sin 9° cos 9° ù é sin 27° cos 27° ù, +, +, =é, êë cos 9° sin 9° úû êë cos 27° sin27° úû, é sin 2 9° + cos 2 9° ù é sin 2 27° + cos 2 27° ù, =ê, ú-ê, ú, ë sin 9° cos 9° û ë sin 27° cos 27° û, 1, 1, =, sin 9° cos 9° sin 27° cos 27°, =, , (c), , Now, according to question, a + xö, tan æç, ÷, è, tan A p, 2 ø p, =, Þ, =, a - xö q, tan B q, tan æç, ÷, è 2 ø, a + xö, a - xö, sin æç, cos æç, è 2 ÷ø, è 2 ÷ø p, Þ, =, a - xö, a + xö, q, sin æç, cos æç, ÷, ÷, è 2 ø, è 2 ø, a + xö, a - xö, cos æç, 2 sin æç, è 2 ÷ø, è 2 ÷ø p, =, Þ, a - xö, a + xö, q, cos æç, 2 sin æç, ÷, ÷, è 2 ø, è 2 ø, sin a + sin x p, Þ, =, sin a - sin x q, , \ Height of the cloud above the lake level = h + 25, 100 - 50 3 + 25 3 - 25, 100 - 50 3, =, + 25 =, 3 -1, 3 -1, =, , ( p + q )sin a, psina, (b), p+ q, p-q, , Ê (d) We have,, , h( 3 - 1 ) = 50 (2 - 3 ), 50 (2 - 3 ), h=, ( 3 - 1), , Þ, , (d) -4, , 3 cosec 20° - sec 20, 3 cos 20° - sin 20°, 3, 1, =, =, sin 20° cos 20°, sin 20° cos 20°, ö, æ 3, 1, 4ç, cos 20° - sin 20° ÷, 2, ø, è 2, =, 2 sin 20° cos 20°, æ sin 60° cos 20° - cos 60° sin 20° ö, = 4ç, ÷, è, ø, sin 40°, 4 sin (60° - 20° ) 4 sin 40°, =, =, =4, sin 40°, sin 40°, , M, , 25 m, , (c) 1, , Ê (a) We have,, , C, h, , (b) 2, , é sin 54° - sin 18° ù, 2, 2, = 2ê, ú, sin 18° sin 54°, ë sin 18° sin 54° û, , é 2 cos 36° sin 18° ù 4 cos 36° sin 18°, = 2ê, =4, ú=, sin 18° cos 36°, ë sin 18° sin(90° - 36° ) û, , 35. The value of, æ3ö, æ1ö, sin -1 ç ÷ + tan -1 ç ÷ is equal to, è5ø, è7 ø, (a) 0, (c), , p, 3, , p, 4, p, (d), 2, (b)
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9, , NDA/NA Solved Paper 2017 (II), ì, ï, =í, ïî, , A, A, + cos ,, 2, 2, æ sin A + cos A ö ,, ç, ÷, è, 2, 2ø, , 41. The points (a, b ), (0, 0), ( -a, - b ) and (ab, b 2 ) are, , sin, , (a) the vertices of a parallelogram, 2np -, , p, 3p, £ A £ 2np +, 4, 4, otherwise, , A, A, 1 + sin A = - æç sin + cos ö÷, è, 2, 2ø, 3 p A 5p, <, <, 4, 2, 4, 3p, 5p, < A<, Þ, 2, 2, so,, , is, , true, , when, , cos 2 A + cos 2 B + cos 2 C, , (c) the vertices of a square, (d) collinear, , Ê (d) All given point lie on the line having equation ay = bx., \ All point are collinear, , 42. The length of the normal from origin to the plane, x + 2y - 2z = 9 is equal to, (a) 2 units, , 39. In triangle ABC, if, sin 2 A + sin 2 B + sin 2 C, , (b) the vertices of a rectangle, , =, , Ê (a) We have,, , cos 2 A + cos 2 B + cos 2 C, , =2, , = 2 cos A + 2 cos B + 2 cos C, 2, , = 2 cos 2 A + 2 cos 2 B + 2 cos 2 C, Þ 3 = 3 cos 2 A + 3 cos 2 B + 3 cos 2 C, Þ cos 2 A + cos 2 B + cos 2 C = 1, 1 + cos 2 A 1 + cos 2B 1 + cos 2C, +, +, =1, Þ, 2, 2, 2, Þ 3 + cos 2 A + cos 2B + cos 2C = 2, Þ cos 2 A + cos 2B + cos 2C = - 1, cos 2 A + cos 2B = - (1 + cos 2C ), 2 cos( A + B) cos( A - B) = - 2 cos 2 C, , Þ2 cos(180° - C ) cos( A - B) = - 2 cos 2 C, - cos C cos( A - B) = - cos 2 C, , Þ, , cos( A - B) = CosC, , Þ, , A- B=C, , Þ, , A= B+ C, A+ B+ C = p, , Again,, Þ, Þ, , A + A = p Þ 2A = p, p, A=, 2, , \ D ABC is a right angled triangle., , 40. The principal value of sin -1 x lies in the interval, , Ê, , -9, , =, , 1 + 4+ 4, , =, , 9, = 3 units, 3, , 2, , Þ1 - cos 2 A + 1 - cos 2 B + 1 - cos 2 C, , Þ, , 2, , 3. sin 2 a + sin 2 b + sin 2 g = 2, 2, , Þ, , 2, , the origin) makes with positive direction of the, coordinate axes, then which of the following are correct?, 1. cos 2 a + cos 2 b = sin 2 g 2. sin 2 a + sin 2 b = cos 2 g, , Þ sin 2 A + sin 2 B + sin 2 C, , Þ, , (1 ) + (2) + (-2), 2, , ¾®, , (d) obtuse-angled, sin 2 A + sin 2 B + sin 2 C, , 0 + 2 .0 - 2 .0 - 9, , (d) 5 units, , 43. If a , b and g are the angles which the vector OP (O being, , (b) equilateral, , (c) isosceles, , (c) 4 units, , Ê (b) Length of normal, , =2, , then the triangle is, (a) right-angled, , (b) 3 units, , Select the correct answer using the code given below., (a) 1 and 2 only, , (b) 2 and 3 only, , (c) 1 and 3 only, , (d) 1, 2 and 3, , Ê (c) We know that, cos, , 2, , a + cos 2 b + cos 2 g = 1, , Þ cos 2 a + cos 2 b = 1 - cos 2 g, Þ cos 2 a + cos 2 b = sin 2 g, so, statement 1 is correct, Again, 1 - sin 2 a + 1 - sin 2 b + cos 2 g = 1, Þ, , 1 + cos 2 g = sin 2 a + sin 2 b, , so, statement 2 is incorrect, Again, 1 - sin 2 a + 1 - sin 2 b + 1 - sin 2 g = 1, Þ, , sin 2 a + sin 2 b + sin 2 g = 2, , \ Statements 3 is correct, , 44. The angle between the lines x + y - 3 = 0 and, x - y + 3 = 0 is a and the acute angle between the lines, x - 3y + 2 3 = 0 and 3x - y + 1 = 0 is b. Which one of, the following is correct?, (a) a = b, , (b) a > b, , (c) a < b, , (d) a = 2b, , Ê (b) We have, equation of lines, , x + y - 3 = 0 and x - y + 3 = 0, , p p, (a) æç - , ö÷, è 2 2ø, , p p, (b) é - , ù, êë 2 2 úû, , \ slopes of the above lines are - 1 and 1 respectively., , p, (c) é 0, ù, êë 2 úû, , (d) [0, p ], , \ angle between these lines is 90°., , p p, (b) Range of sin x is é - , ù, êë 2 2 úû, p p, \, sin x Î é - , ù, ëê 2 2 ûú, -1, , Since product of their slope is - 1, \, , a = 90°, , Also, angle between the lines x - 3 y + 2 3 = 0 and, 3 x - y + 1 = 0 i,e,. b is acute, \, , a>b
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10, , NDA/NA Solved Paper 2017 (II), ®, , ®, , ®, , 45. Let a = i$ + 2 $j - k$, b = 2i$ - $j + 3k$ and g = 2i$ + $j + 6k$ be, ®, , Ê (b), , ®, Now, a ´ $i = (xi$ + yj$ + zk$ ) ´ i$, , ®, , three vectors. If a and b are both perpendicular to the, vector d and d × g = 10, then what is the magnitude of d?, , Ê, , 3, (b) 2 3 units (c), unit, (a) 3 units, 2, r, (b) Let d = xi + yi + zk, r r, r, It is given that, r r d 1r arand d ^ b, \, d× a = d×b = 0, \, x + 2y - z = 0, and, , 2x - y + 3 z = 0, , Again,, \, , r r, d × g = 10, 2x + y + 6 z = 10, , (d), , = - yk$ + zj$, r $2, \ |a ´ i| = ( ( z )2 + (- y))2 = z 2 + y2, ®, ®, Similarly, |a ´ $j|2 = x2 + z 2 and|a ´ k|2 = x2 + y2, , 1, unit, 3, , ®, ®, ®, \ |a ´ i$|2 + |a ´ $j|2 + |a ´ k$|2, , = ( z 2 + y 2 ) + (x 2 + z 2 ) + (x 2 + y 2 ), = 2(x2 + y2 + z 2 ), , …(i), r, [Q a = i$ + 2 $j - k$], …(ii), r, $, $, [Qb = 2i - j + 3 k$, …(iii), r, $, $, $, [Q g = 2i + j + 6 k], , on solving (i), (ii), (iii), we get, x = - 2, y = 2, rz = 2, r, \, d = 2i$ + 2 $j + 2k$ Þ |d| = (- 2)2 + (2)2 + (2)2, = 4 + 4 + 4 = 12 = 2 3, , ®, , =, =, =, , (a) 18p square metres, , (b) 15p square metres, , (c) 12p square metres, , (d) 8p square metres, , Ê (b) Since the shape of a racecourse is of the form of ellipse., Let the equation of racecourse be, x2, y2, + 2 =1, 2, a, b, Now, according to the question, , (c) (2a$ - b$ ), , (d) (2a$ + b$ ), , PF1 + PF2 = 10 and F1 F2 = 8, \, , 2a = 10 and 2ae = 8, 4, Þ, a = 5 and e =, 5, b2, b2, Again,, e = 1 - 2 Þ e2 = 1 - 2, a, a, 2, b2, 4ö, 16, b2, æ, Þ, =1 Þ, ç ÷ =1 è5ø, (5)2, 25, 25, 9, b2, 2, =, Þb = 9 Þ b = 3, Þ, 25 25, Now, area of an ellipse = pab, Required area = p (5) (3 ), \, = 15 p sq unit, , (a$ + b$ ) ´ (a$ + b$ ), a$ ´ (a$ ´ b$ ) + b$ ´ (a$ ´ b$ ), (a$ × b$ )a$ - (a$ × a$ ) b$ + (b$ × b$ )a$ - (b$ × a$ )b$, [Q a$ × a$ = b$ × b$ = 1], (a$ × b$ ) a$ - b$ + a$ - (b$ × a$ )b$, (a$ × b$ ) a$ - (b$ × a$ ) b$ + a$ - b$, [Q a$ × b$ = b$ × a$ ], , = (a$ × b$ ) a$ - (a$ × b$ ) b$ + a$ + b$, = (a$ × b$ ) (a$ - b$ ) + (a$ - b$ ), = (a$ × b$ + 1 ) (a$ - b$ ), \ (a$ + b$ ) ´ (a$ ´ b$ )||(a$ - b$ ), as = (a$ × b$ + 1 ) is a scalar quantity., , 50. The distance of the point (1, 3) from the line 2x + 3y = 6,, , ®, , 47. A force F = i$ + 3 $j + 2k$ acts on a particle to displace it, from the point A(i$ + 2 $j - 3k$ ) to the point B (3i$ - $j + 5k$ )., The work done by the force will be, (a) 5 units, , x2 + y2 + z 2, , the distances of two flag-posts from him is always 10 m, and the distance between the flag-posts is 8 m. The area, of the path he encloses is, , Ê (a) We have,, =, , [Q|a| =, , 49. A man running round a racecourse notes that the sum of, , (a$ + b$ ) ´ (a$ ´ b$ ) is parallel to, (b) (a$ + b$ ), , ®, , = 2|a|2, , 46. If a$ and b$ are two unit vectors, then the vector, (a) (a$ - b$ ), , ®, Let a = xi$ + yi$ + zk$, , (b) 7 units, , (c) 9 units, , (d) 10 units, , Ê (c) We have,, ®, , measured parallel to the line 4 x + y = 4, is, (a), , 5, units, 13, , (b), , 3, units, 17, , Ê (d) The slope of the line 4x +, , (c) 17 units, , (d), , 17, units, 2, , y = 4 is - 4., , \ Equation of a line passing through (1, 3 ) with slope, - 4 is, , F = i$ + 3 $j + 2k$, A(i$ + 2 $j - 3 k$ ) and B(3 i$ - $j + 5k$ ), , y - 3 = - 4(x - 1 ), , \ AB = (3 i$ - $j + 5k$ ) - (i$ + 2 $j - 3 k$ ) = 2i$ - 3 $j + 8 k$, , Þ, , Now, work done = F × AB, , The coordinate of the point of intersection of the lines, 3, 4x + y = 7 and 2x + 3 y = 6 is æç , 1 ö÷., è2 ø, , ®, , = (i$ + 3 $j + 2k$ ) × (2i$ - 3 $j + 8 k$ ), = 1 ´ 2 + 3 ´ (-3 ) + 2 ´ 8 = 2 - 9 + 16 = 9 units, ®, , ®, , ®, ®, 48. For any vector a | a ´ i$| 2 + | a ´ $j | 2 + | a ´ k$| 2, , y - 3 = - 4x + 4 Þ 4x + y = 7, , \ Required distance, 2, , =, , æ 3 - 1 ö + (1 - 3 )2, ç, ÷, è2, ø, , =, , 1, 17, units, + 4=, 4, 2, , is equal to, ®, , (a)|a|2, , ®, , (b) 2|a|2, , ®, , (c) 3|a|2, , ®, , (d) 4|a|2
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11, , NDA/NA Solved Paper 2017 (II), 51. If the vectors ai$ + $j + k$, i$ + bj$ + k$ and i$ + $j + ck$, (a, b, c ¹ 1) are coplanar, then the value of, 1, 1, 1, +, +, 1-a 1-b 1-c, , 53. If the angle between the lines whose direction ratios are, p, (2, - 1, 2) and x, 3, 5 is , then the smaller value of x is, 4, (a) 52, , Ê (b), , is equal to, (a) 0, (c) a + b + c, , Ê (b) Since ai$ +, \, , (b) 1, (d) abc, $j + k$, i$ + bj$ + k$ and i$ + $j + ck$ are coplanar, ½a 1 1½, ½1 b 1½ = 0, ½, ½, ½1 1 c½, , 7, 8, (a) æç , - , 0ö÷, è3 3 ø, 7 8, (c) æç - , , 0ö÷, è 3 3 ø, , 7, 8, (b) æç - , - , 0ö÷, è 3 3 ø, 7 8, (d) æç , , 0ö÷, è3 3 ø, , If two lines have direction ratios < a1 , b1 , c1 > and, , Þ, , Þ a (b - 1 ) (c - 1 ) - (1 - a ) (c - 1 ) - (1 - a ) (b - 1 ) = 0, , ( -3, 4, - 8) and (5, - 6, 4 ) with XY -plane is, , (d) 1, , Now, according to the question, 2x - 3 + 10, p, cos =, 4, (2)2 + (-1 )2 + (2)2 x2 + (3 )2 + (5)2, , ½a 1 - a 1 - a½, ½1 b - 1, 0 ½= 0, ½, ½, 0, c - 1½, ½1, , 52. The point of intersection of the line joining the points, , (c) 2, , < a2 , b2 , c2 > then the angle between them i.e., q is, given by, a1 a2 + b1 b2 + c1 c2, cos q=, a12 + b12 + c12 a22 + b22 + c22, , on applying c2 ® c2 - c1 , c3 ® c3 - c1 , we get, , on dividing by (1 - a ) (1 - b) (1 - c), we get, a, 1, 1, +, +, =0, 1-a 1-b 1-c, 1 -1 + a, 1, 1, +, +, =0, Þ, 1-a, 1-b 1-c, 1, 1, 1, -1 +, +, =0, Þ, 1-b 1-c, 1-a, 1, 1, 1, Þ, +, +, =1, 1-a 1-b 1-c, , (b) 4, , 2x + 7, 1, =, 2 3 34 + x2, , Þ, , 3 34 + x2 = 2 (2x + 7 ), , Þ, , 9 (34 + x2 ) = 2(2x + 7 )2, , Þ, , 306 + 9 x2 = 8 x2 + 56 x + 98, , Þ, , x - 56 x + 208 = 0, , Þ, , (x - 4) (x - 52) = 0 Þ x = 4, 52, , 2, , \ smaller value of x is 4., , 54. The position of the point (1, 2) relative to the ellipse, 2x 2 + 7y 2 = 20 is, (a) outside the ellipse, (b) inside the ellipse but not at the focus, (c) on the ellipse, (d) at the focus, , Ê (a) Let S : 2x, , 2, , + 7 y2 - 20, , \ S1 = 2(1 )2 + 7 (2)2 - 20, , [Put x = 1, y = 2], , Þ, , S1 = 2 + 28 - 20 Þ S1 = 10, , \, , S1 > 0, , Hence, point (1, 2) lies outside the ellipse., , Ê (a) Equation of line joining (-3, 4, - 8 ) and (5, - 6, 4) is, x+3, , =, , y-3, , =, , z+ 8, , =l, , 5 + 3 -6 - 4 4 + 8, x+ 3 y-3 z + 8, =, =, =l, Þ, - 10, 8, 12, , Þ x = 8 l - 3, y = - 10 l + 4,, z = 12l - 8, , 55. The equation of straight line which cuts off an intercept, of 5 units on negative direction ofY -axis and makes and, angle 120° with positive direction of X -axis is, (a) y +, , 3x + 5 = 0, , (b) y -, , 3x + 5 = 0, , (c) y +, , 3x - 5 = 0, , (d) y -, , 3x - 5 = 0, , y, , Ê (a), , since line intersect XY plane, \, Þ, Þ, , z =0, , 120º, x′, , 12l - 8 = 0, 2, l=, 3, , \ Point of intersection, 2, 2, = æç 8 ´ - 3, - 10 ´ + 4, 0 ö÷, è, ø, 3, 3, æ7 -8 ö, , 0÷, =ç ,, è3 3, ø, , O, , x, , (0, –5), y′, , We have, Slope = m = tan120° = - 3, and, , y-intercept = c = - 5, , \ Equation of line is y = mx + c Þ y = - 3 x - 5, or, , 3x + y + 5 = 0
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12, , NDA/NA Solved Paper 2017 (II), , 56. The equation of the line passing through the point (2, 3), , and the point of intersection of lines 2x - 3y + 7 = 0 and, 7 x + 4y + 2 = 0 is, , 58. The equation of the circle which passes through the, points (1, 0), (0, -6) and (3, 4) is, , (a) 4x 2 + 4 y2 + 142 x + 47 y + 140 = 0, , (a) 21x + 46 y - 180 = 0, , (b) 21x - 46 y + 96 = 0, , (b) 4x 2 + 4 y2 - 142 x - 47 y + 138 = 0, , (c) 46x + 21y - 155 = 0, , (d) 46x - 21y - 29 = 0, , (c) 4x 2 + 4 y2 - 142 x + 47 y + 138 = 0, , Ê (b) Since the line passes through the intersection point of, the lines, , (d) 4x 2 + 4 y2 + 150x - 49 y + 138 = 0, , Ê (c) Let the equation of the circle be x, , 2x - 3 y + 7 = 0 and 7 x + 4 y + 2 = 0, , \, , Now, required line passes through (2,3 ), So, put (2,3) in equation of line given in Eq. (i), , or, , ( 4 - 9 + 7 ) + l (14 + 12 + 2) = 0, , 28 x - 42 y + 98 - 7 x - 4 y - 2 = 0, , Þ, , 21 x - 46 y + 96 = 0, , Þ, , 3, and latus, 4, , rectum 4 units is, 7 y2, x2, +, =1, 1024 64, 7x 2, 49 y2, (c), +, =1, 1024, 64, (a), , …(iii), , 4x2 + 4 y2 - 142x + 47 + 138 = 0, , 59. A variable plane passes through a fixed point (a, b, c ) and, , 57. The equation of the ellipse whose centre is at origin,, major axis is along X -axis with eccentricity, , 6 g + 8 f + c + 25 = 0, , \ Equation of circle will be, 47, 69, 71, =0, x2 + y2 + 2 æç - ö÷ x + 2 æç ö÷ y +, è 4ø, è 8 ø, 2, , \ Equation of the required line is, 1, 2x - 3 y + 7 + æç - ö÷ (7 x + 4 y + 2) = 0, è 14 ø, Þ, , ...(i), ...(ii), , On solving (i), (ii) and (iii), we get, 71, 47, 69, g = - , f = , c=, 4, 8, 2, , 2 + l (28 ) = 0, -1, l=, 14, , Þ, , 1 + 2g + c = 0, 36 - 12 f + c = 0, , and 9 + 16 + 6 g + 8 f + c = 0, , (2 ´ 2 - 3 ´ 3 + 7 ) + l (7 ´ 2 + 4 ´ 3 + 2) = 0, Þ, , 49x 2 7 y2, +, =1, 1024 64, x2, y2, (d), +, =1, 1024 64, , (b), , cuts the axes in A, B and C respectively. The locus of the, centre of the sphere OABC, O being the origin, is, x, y, z, + + =1, a b c, a b c, (c) + + = 2, x, y, z, , a b c, + + =1, x, y, z, x, y, z, (d) + + = 2, a b c, , (b), , (a), , Ê (c) Equation of plane passing through A, B, C is, x y z, + + =1, p q r, y, , Ê (b) Given that,, , 3, eccentricity of ellipse = e =, 4, , B (0, q, 0), , …(i), , It is given that major axis is along x-axis., 2b2, \, =4, a, Þ, , + y2 + 2 gx + 2 fy + c = 0, , Since above circle passes through (1, 0), (0, - 6 ) and (3, 4), , \ Equation of required line is, , Þ, , 2, , b2 = 2a, , (0, 0, 0), , …(ii), , Since, eccentricity of an ellipse, b2 = a 2 (1 - e2 ), 9, Þ, 2a = a 2 (1 - (3 / 4)2 ) Þ 2a = a 2 æç1 - ö÷, è 16 ø, 7, 2a = a 2 Þ 32a = 7 a 2, Þ, 16, 32, [Qa ¹0], Þ, a=, 7, 32, Put the value of a in eqn (ii), we get b2 = 2 ´, 7, 64, b2 =, Þ, 7, x2 y2, \ Equation of the ellipse is, +, =1, a 2 b2, x2, y2, 49 2 +7 y2, x, =1, +, =1 Þ, 2, 64, 1024, 64, æ 32 ö, ç ÷, 7, è7ø, , z, , x, A (p, 0, 0), , C (0, 0, r), , Since above plane passes through (a, b, c), a b c, + + =1, \, p q r, , ……(i), , Now, equation of sphere passing through A, B, C, O is, \, \, , x2 + y2 + z 2 - px - qy - rz = 0, p q r, Centre = æç , , ö÷ Let the locus of centre be (x, y, z ), è 2 2 2ø, q, P, r, x = , y = , z = Þ P =2x, q =2 y, r =2 z, 2, 2, 2, , Putting these value in (i), we get, a, b, c, a b c, +, +, = 1 Þ + + =2, 2x 2 y 2 z, x y z
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13, , NDA/NA Solved Paper 2017 (II), 60. The equation of the plane passing through the line of, intersection of the planes x + y + z = 1, 2x + 3y + 4z = 7,, and perpendicular to the plane x - 5y + 3z = 5is given by, , (a) x + 2 y + 3 z - 6 = 0, , (b) x + 2 y + 3 z + 6 = 0, , (c) 3x + 4 y + 5 z - 8 = 0, , (d) 3x + 4 y + 5 z + 8 = 0, , 63. If y = (cos x ) (cos x ), (a) (c), , Let P1 : x + y + z - 1 = 0 and P2 : 2x + 3 y + 4 z - 7 = 0, , Ê (a), , \ Equation of plane passing through line of, intersection of P1 and P2 is given by, P1 + lP2 = 0, , y2 tan x, 1 - y ln(sin x ), , Þ, , (1 - 2)x + (1 - 3 ) y + (1 - 4) z - 1 + 7 = 0, , (a) x = y, , 1, ln 5, , (c) x = y, , (b) x = y , y > 0, , ,y<0, , (d) x = 5ln y, y > 0, , Þ, Þ, , y =5ln x, ln y = ln 5ln x Þ ln y = ln x ln 5, 1, ln y, Þ ln x = ln y ln 5, ln x =, ln 5, x = y ln 5 , y >0, , [Q5ln x > 0], , x, ì, , x ¹0, ïf (x ) = í, x2, ïî 0 ,, x =0, , - y2 tan x, dy, =, dx 1 - y log cos x, , (b) 2 and 3 only, , (c) 1 and 3 only, , (d) 1, 2 and 3, , Ê (a) Since x + x, , f(x ) is continuous at x = 0 but not differentiable at x = 0, f(x ) is continuous as well as differentiable at x = 0, f(x ) is discontinuous at x = 0, None of the above, , Now,, and, \, Hence,, , ì -x , x >0, -x, -x, , x ¹ 0 ìï , x ¹ 0 ïï x, =í x, = í 0, x = 0 =, x2, 0, x = 0, ï -x , x <0, îï 0, x = 0, ïî - x, LHL (x = 0 ) = 1, RHL (x = 0 ) = - 1, LHL ¹ RHL, f (x) is discontinuous at x =0, , 2, , is a polynomial function, therefore it will be, , continuous everywhere., 1, 1, Again cos is not continuous at x =0 as cos æç ö÷ Î[-1, 1], è0 ø, x, i.e., an oscillating value., \ statements 1 and 2 are correct., , 65. Consider the following statements, , Which one of the following is correct in respect of the, above function?, , ì, ï, f (x ) = í, ïî, , (cos) ¼¥, , (a) 1 and 2 only, , 62. A function is defined as follows, , Ê (c) We have,, , y2 sin x, 1 + y ln(sin x ), , 1. x + x 2 is continuous at x = 0, 1, 2. x + cos is discontinuous at x = 0, x, 1, 3. x 2 + cos is continuous at x = 0, x, Which of the above are correct?, , ln 5, , 1, , Þ, , (d), , 64. Consider the following, , ,y>0, , Ê (a) We have,, , y2 tan x, 1 + y ln(cos x ), , ù, dy é 1, - log cos xú = - y tan x, dx êë y, û, , x + 2y + 3 z - 6 = 0, , 1, ln 5, , (b), , (cos x ) ¥, , Þ, , - x - 2y - 3 z + 6 = 0, , 61. The inverse of the function y = 5 ln x, , (a), (b), (c), (d), , y = (cos x)(cos x ), , dy, is equal to, dx, , Let, y = (cos x), \, y = (cos x) y, Taking log both the sides, we get, log y = y log cos x, Differentiating both sides, w.r.t x, we get, 1 dy dy, 1, =, log cos x + y×, (- sin x), y dx dx, cos x, , Þ (1 + 2l )x + (1 + 3 l ) y + (1 + 4l ) z - 1 - 7 l = 0, This plane is perpendicular to the plane x - 5 y + 3 z = 5, \ 1 (1 + 2l ) - 5 (1 + 3 l ) + 3 (1 + 4l ) = 0, Þ, 1 + 2l - 5 - 15l + 3 + 12l = 0, Þ, - l - 1 = 0 Þ l = -1, Equation of required plane is, , Þ, , , then, , y2 tan x, 1 - y ln(cos x ), , Ê (a) We have,, , Þ (x + y + z - 1 ) + l (2x + 3 y + 4 z - 7 ) = 0, , Þ, , (cos x ) ¥, , dy, at a point on the curve gives slope of the tangent at, dx, that point., 2. If a(t ) denotes acceleration of a particle, then, ò a(t )dt + c gives velocity of the particle., 1., , 3. If s(t ) gives displacement of a particle at time t, then, ì-1, x > 0, ï, í 0, x = 0, ï 1, x < 0, î, , gives its acceleration at that instant., Which of the above statements is/are correct?, (a) 1 and 2 only, (b) 2 only, (c) 1 only, (d) 1, 2 and 3, , ds, dt
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17, , NDA/NA Solved Paper 2017 (II), 83. The set of all points, where the function f ( x ) = 1 - e - x, , \ f (x) is continuous at x =0, f (h) - f (0 ), Now, LHD, (at x =0) = lim, h® 0, -h, f (- h) - f (0 ), = lim, h® 0, -h, -h ( -h - h + 1 ) -0, = lim, = lim ( h - h + 1 ) = - 1, h® 0, h® 0, -h, f (0 + h) - f (0 ), RHD, (at x =0) = lim, h® 0, h, f ( h - h + 1 ) -0, f (h) - f (0 ), = lim, = lim, h® 0, h® 0, h, h, , 2, , is differentiable, is, (a) (0, ¥), , (b) (-¥, ¥), 1 - e- x, , Ê (c) We have, f (x) =, f ¢ (x ) =, , Þ, , (c) (-¥, 0) È (0, ¥) (d) (-1, ¥), 2, , 1, 2 1 - e- x, , (- e- x ) (-2x) =, 2, , 2, , xe- x, , 2, , 1 - e- x, , 2, , which is defined " x ÎR except x =0, \ f (x) is differentiable on (- ¥, 0 ) È (0, ¥), , 84. Match List-I with List-II and select the correct answer, , = lim ( h - h + 1 ) = - 1, h® 0, , using the code given below the lists :, List-I, (Function), , LHD = RHD, Q, \ f (x) is differentiable at x =0., , List-II, (Maximum value), , A. sin x + cos x, , 1., , 10, , B. 3sin x + 4cos x, , 2., , 2, , C. 2sin x + cos x, , 3. 5, , D. sin x + 3cos x, , 4., , 86. Which one of the following graph represents the, x, function f ( x ) = , x ¹ 0?, x, (a), , y, , 5, +1, , Code, A, , B, , C, , D, , A, , B, , C, , D, , (a) 2, , 3, , 1, , 4, , (b) 2, , 3, , 4, , 1, , (c) 3, , 2, , 1, , 4, , (d) 3, , 2, , 4, , 1, , Ê (b), , x, , O, , (b), , y, +1, , \ - a 2 + b2 £ f (x) £ a 2 + b2, \ - 1 2 + 1 2 £ sin x + cos x £ 1 2 + 1 2, , x, , O, , If f (x) = a sin x + b cos x, , –1, , (c), , y, +1, , Þ - 2 £ sin x + cos x £ 2, - 3 2 + 42 £ 3 sin x + 4 cos x £ 3 2 + 42, Þ - 5 £ 3 sin x + 4 cos x £ 5, - 22 + 1 2 £ 2 sin x + cos x £ 22 + 1 2, Þ - 5 £ 2 sin x + cos x £ 5, - 1 2 + 3 2 £ sin x + 3 cos x £ 1 2 + 3 2, Þ - 10 £ sin x + 3 cos x £ 10, , 85. If f ( x ) = x ( x - x + 1 ), then f ( x ) is, (a), (b), (c), (d), , continuous but not differentiable at x = 0, differentiable at x = 0, not continuous at x = 0, None of the above, , Ê (b) We have,, , f (x ) = x ( x - x + 1 ), , \ LHL (at x =0) = lim- x ( x - x + 1 ), x® 0, , = lim (0 - h) ( 0 - h - 0 - h + 1 ), h® 0, , = lim - h ( - h - 1 - h ) = 0, h® 0, , RHL (at x =0) = lim+ x ( x - x + 1 ), x® 0, , = lim (0 + h) ( 0 + h - 0 + h + 1 ), h® 0, , = lim h ( h - 1 + h ) = 0, h® 0, , Q, , f (0 ) = 0 ( 0 - 0 + 1 ) = 0, LHL = RHL = f (0 ), , x´, , x, , O, , (d) None of the above, y, +1, , Ê (c) We have, , x, f (x) = , x ¹0 =1, x ¹0, x, x´, , \ Graph of f (x) will be, , 87. Let f (n ) = éê +, 1, ë4, , O, , x, , n ù, , where [x ] denot the integral part, 1000 úû, 1000, , of x. Then the value of, , å f (n ) is, , n =1, , (a) 251, , (b) 250, , (c) 1, , (d) 0, , 1, n ù, f (n) = é +, êë 4 1000 úû, 1000, 1, 1 ù é1, 2 ù, \ å f (x ) = é +, + +, êë 4 1000 úû êë 4 1000 úû, n =1, 1 749 ù é 1 750 ù, + ¼+ é +, + +, êë 4 1000 úû êë 4 1000 úû, 1 751 ù, 1 1000 ù, +é +, + ¼+ ...+ é +, ëê 4 1000 ûú, ëê 4 1000 ûú, , Ê (a) We have,, , 1 750 ù é 1 751 ù, 1 1000 ù, + +, 0 + 0 + ¼+ ...0 + é +, + ¼+ é +, ëê 4 1000 ûú ëê 4 1000 ûú, ëê 4 1000 ûú, = 1 + 1 + 1 + ¼+ 1 (251 times) = 251
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21, , NDA/NA Solved Paper 2017 (II), 103. Let the sample space consist of non- negative integers up, , 106. Given that the arithmetic mean and standard deviation, , to 50, X denote the numbers which are multiples of 3 and, Y denote the odd numbers. Which of the following is/are, correct?, 8, 1, 1. P( X ) =, 2. P(Y ) =, 25, 2, , of a sample of 15 observations are 24 and 0 respectively., Then which one of the following is the arithmetic mean, of the smallest five observations in the data?, , Select the correct answer using the code given below., (a) 1 only, , (b) 2 only, , (c) Both 1 and 2, , (d) Neither 1 nor 2, , (a) 0, , Þ, Þ, , S = {0, 1, 2, 3, ¼, 49, 50 }, , Þ, , n(Y ) = 25, n( X ) 17 1, \, P( X ) =, =, =, n(s ) 51 3, n(Y ) 25, and P(Y ) =, =, n(s ) 51, , Þ S (xi - x )2 = 0, , N, , xi = x Þ xi =24, , 107.Which one of the following can be considered as, appropriate pair of values of regression coefficient of y, on x and regression coefficient of x on y?, (a) (1, 1), , 1 10, (d) æç , ö÷, è3 3 ø, , 1, (c) æç - , 2 ö÷, è 2 ø, , (b) (-1, 1), , Ê (a) We know that, , 104. For two events A and B,, 1, 2, 1, let P( A ) = , P( A È B ) = and P( A Ç B ) = . What is, 2, 3, 6, P( A Ç B ) equal to?, , Ê, , S (xi - x )2, , \ Mean of any five observation will be 24., , g = Numbers which are odd = {1,3,5¼49 }, , 1, 4, , N, , xi - x =0, , Þ, , = {3,6,9¼, 48 }, , (b), , 0=, , S(xi - x )2, , \ Each observation will be 24., , n( X ) =17, , 1, 6, , (d) 24, , [Q (xi - x )2 is a positive value], , X = Number which are multiple of 3., , (a), , s=, , Q, , Þ n(s ) =51, , Þ, , (c) 16, , Ê (d) Since standard deviation is 0., , Ê (d) Let S = set of all non negative integers upto 50., \, , (b) 8, , (c), , 1, 3, , (d), , 1, 2, , (a) We have,, 1, 2, 1, P( A) = ,P( A È B) = , P( A Ç B) =, 2, 3, 6, We know that,, P( A È B) = P( A) + P(B) - P( A Ç B), 2 1, 1, 1, = + P(B) - Þ P(B) =, 3 2, 6, 3, 1 1 1, Again, P( A Ç B) = P(B) - P( A Ç B) = - =, 3 6 6, , regression coefficient of x on y and regression, coefficient of y on x are always equal., \ Correct option is (a)., , 1, 3, , 108.Let A and B be two events with P( A ) = , P( B ) =, P( A Ç B ) =, (a), , (a) 1 and 2 only, , (b) 2 and 3 only, , (c) 1 and 3 only, , (d) 1, 2 and 3, , Ê (a) Both statement 1 and 2 are fundamental concept and, are correct., , But, Mean deviation is least when measured about, mean., \ Statement 3 is incorrect., , 1, . What is P( B| A ) equal to?, 12, (b), , 1, 7, , (c), , 1, 8, , (d), , 1, 10, , 1, 3, 1, 1, P(B) = , P( A Ç B) =, 6, 12, P(B Ç A ) P(B) - P( A Ç B), Now, P(B / A ) =, =, P (A), 1 - P( A), , Ê (c) We have,, , 105. Consider the following statements, 1. Coefficient of variation depends on the unit of, measurement of the variable., 2. Range is a measure of dispersion., 3. Mean deviation is least when measured about, median., Which of the above statements are correct?, , 1, 5, , 1, and, 6, , P( A) =, , [Q P(B Ç A ) = P(B) - P( A Ç B) and P( A ) = 1 - P( A)], 1 1, 1, 1, 6, 12, 12, =, =, =, 1, 2 8, 13, 3, , 2, 3, , 109. In a binomial distribution, the mean is and the variance, 5, is . What is the probability that X = 2 ?, 9, (a), , 5, 36, , (b), , 25, 36, 2, , Ê (c) We have, mean = 3, \, , (c), , 25, 216, , and variance =, , np =, , 2, 5, and npq=, 3, 9, , (d), 5, 9, , 25, 54
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22, , NDA/NA Solved Paper 2017 (II), 5/9 5, 5 1, = Þ P =1 - q=1 - =, 6 6, 2/3 6, 2/3, n=, =4, 1 /6, , \, , 3, 1, 1, and P( A Ç B Ç C ) = ,, 4, 3, 3, then what is P( B Ç C ) equal to?, , q=, , Þ, , p (x = 2) = C2 P, n, , Now,, , n -2, , q = C2 p, 2, , 4, , 4-2, , 113. If P( B ) = , P( A Ç B Ç C ) =, (a), 2, , q, , = 4 C2 p 2 q2, 2, 2, 4 ´ 3 æ1 ö æ 5 ö, 25, =, ´ç ÷ ç ÷ =, 216, 2 ´ 1 è6 ø è6 ø, , 1, 3, probability that out of 5 ships, at least 4 ships would, arrive safely is, 1, 243, , (b), , 10, 243, , (c), , 11, 243, , (d), , 13, 243, , 1, , Ê (c) According to question, n=5, P = 3 ,, q=1 - P =, , 17, 72, , (c), , 1, 144, , (d), , 2, 9, , 12 11 10, , =, , Ê (b) P (None born in same month) = 12 ´ 12 ´ 12, , 110, 144, , \ P (at least two person born in same month), = 1 - P (none born in same month), 110 144 - 110 34 17, =1=, =, =, 144, 144 72, 144, , 112. It is given that X = 10, Y = 90,, s X = 3, s Y = 12 and r XY = 08, . . The regression equation of, X on Y is, (a) Y = 32, . X + 58, , (b) X = 32, . Y + 58, , (c) X = - 8 + 02, . Y, , (d) Y = - 8 + 02, . X, , Ê (c) We have, X =10, Y =90,s, , x, , =3, , s y =12, rxy =0.8, Now, we know that, Regression equation of X and Y is given by, s, X - X = rx y × x ( y - y ), sy, 3, x - 10 = 0.8 ´ ( y - 90 ), Þ, 12, Þ, Þ, , X - 10 = 0.2 (Y - 90 ) Þ X = 0.2Y - 18 + 10, X = - 8 + 0.2Y, , 1, 15, , (d), , 1, 9, , Also, P(B) = P(B Ç C ) + P(B ÇC ), 3, 2, 3, 2, [Q P(B) = and P (B Ç C ) = ], Þ, = P (B Ç C ) +, 4, 3, 4, 3, 3 2 1, Þ P (B Ç C ) = - =, 4 3 12, , 114. The following table gives the monthly expenditure of, Expenditure (in `), , group of three persons were born in the same month, (disregard year)?, (b), , (c), , P (B Ç C ) = P ( A Ç B Ç C ) + P ( A Ç B Ç C ), 1 1 2, = + = (given), 3 3 3, , 2, 3, , 111. What is the probability that at least two persons out of a, 33, 144, , 3, 4, , two families, , \ Required probability = P( x ³ 4) = P(x = 4) + P (x = 5), 5, 0, 4, 1, 5 ´ 2 1 ´1, 1, 2, 1, 2, = 5 C4 æç ö÷ æç ö÷ + 5 C5 æç ö÷ æç ö÷ = 5 + 5, è3 ø è3 ø, è3 ø è3 ø, 3, 3, 10, 1, 11, =, +, =, 243 243 243, , (a), , (b), , Ê (a) We know that,, , 110. The probability that a ship safely reaches a port is . The, , (a), , 1, 12, , Items, , Family A, , Family B, , Foods, , 3,500, , 2,700, , Clothing, , 500, , 800, , Rent, , 1,500, , 1,000, , Education, , 2,000, , 1,800, , Miscellaneous, , 2,500, , 1,800, , In construction a pie diagram to the above data, the radii, of the circles are to be chosen by which one of the, following ratios?, (a) 1 : 1, (c) 100 : 91, , (b) 10 : 9, (d) 5 : 4, , Ê (b) According to the given data,, Total expenditure of Family A, = 3500 + 500 + 1500 + 2000 + 2500, = 10,000, Total expenditure of Family B, = 2700 + 800 + 1000 + 1800 + 1800 = 8100, Q The ratio of area of circle for pie diagram of both, family will be equal to ratio of their total expenditure., \ Area of A : Area of B = 10,000 : 8100 = 100 : 81, Þ Radius of circle for A : Radius of circle for B, = 100 : 81 = 10 :9, , 115. If a variable takes values 0, 1, 2, 3, …, n with frequencies, 1, C (n, 1), C (n, 2), C (n, 3), … , C (n, n ), respectively, then the arithmetic mean is, (a) 2n, (c) n, , (b) n + 1, n, 2, , (d)
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23, , NDA/NA Solved Paper 2017 (II), Ê (d) Let X denote, , n, , the required mean. Then, X =, , år ×, , r=0, n, , å, , n, , n, , cr, , (a) x1 + x 2 > 2 x1 x 2, , (b) x1 +, , (c)| x1 -, , (d) x1 + x 2 < 2( x1 x 2 + 1), , x 2| > 2, , Ê (c) We have,, , cr, , r=0, n, , =, , =, , n n -1, år × r cr -1, r=0, n, , å, , n, , r=0, n -1, , n´2, 2n, , cr, , n, , =, , å, , n -1, , r=0, , n, , å, , n, , cr - 1, , Þ ( x1 )2 + ( x2 )2 - 2 x1, , x2 > 2, , Þ ( x1 - x2 ) > 2, 2, , cr, n, , n, 2, , [Q å n - 1 cr - 1 = 2n - 1 ], r=0, , 116. In a multiple-choice test, an examinee either knows the, correct answer with probability p, or guesses with, probability 1 - p. The probability of answering a, 1, question correctly is , if he or she merely guesses. If the, m, examinee answers a question correctly, the probability, that he or she really knows the answer is, , Ê, , AM - GM >1, , Þ x1 + x2 - 2 x1 x2 > 2, , r=0, , =, , x2 > 2, , (a), , mp, 1 + mp, , (b), , mp, 1 + (m - 1)p, , (c), , (m - 1)p, 1 + (m - 1)p, , (d), , (m - 1)p, 1 + mp, , (b) Let E1 be the event that examinee knows the answer., E2 be the event the examine guesses the answer., A be the event that the answer is correct., Now, According to the question, P(E1 ) = p, P (E2 ) = 1 - p, P ( A / E1 ) = 1, 1, P ( A / E2 ) =, m, Required probability = P (E1 / A), P (E1 ) P ( A / E1 ), =, P (E1 ) P( A / E1 ) + P(E2 ) P ( A / E2 ), P ´1, mp, mp, =, =, =, 1, pm, +, 1, p, 1, +, (, m - 1) p, p ´ 1 + (1 - p ) ´, m, , 117. If x 1 and x 2 are positive quantities, then the condition for, the difference between the arithmetic mean and the, geometric mean to be greater than 1 is, , Þ, , x1 - x2 > 2, , 118. Consider the following statements, 1. Variance is unaffected by change of origin and, change of scale., 2. Coefficient of variance is independent of the unit of, observations., Which of the statements given above is/are correct?, (a) 1 only, , (b) 2 only, , (c) Both 1 and 2, , (d) Neither 1 nor 2, , Ê (b) Variance is independent of change of origin but is not, independent of change of scale., So, statement 1 is correct., Coefficient of variance is independent of the unit of, observations., So, statement 2 is correct., , 119. Five sticks of length 1, 3, 5, 7 and 9 feet are given. Three, of these sticks are selected at random. What is the, probability that the selected sticks can form a triangle?, (a) 0.5, , (b) 0.4, , (c) 0.3, , (d) 0, , Ê (c) Let S be the event of selecting three sticks., \, , n(s ) = 5 c3 = 10, , E be the event that selected sticks forms a triangle., \, , n(E ) = 4 c3 - 1 = 3, n (E ) 3, P (E ) =, =, = 0.3, n (S) 10, , 120. The coefficient of correlation when coefficients of, regression are 0.2 and 1.8 is, (a) 0.36, , (b) 0.2, , Ê (c) Coefficient of Correlation, , (c) 0.6, , = 0.2 ´ 1.8 = 0.36 = 0.6, , (d) 0.9
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NDA / NA, National Defence Academy/Naval Academy, , Solved Paper, , 2017 (I), , Paper 1 (Mathematics), 1. Let S be the set of all persons living in Delhi. We say that, x , y in S are related if they were born in Delhi on the same, day. Which one of the following is correct?, (a) The relation is an equivalent relation, (b) The relation is not reflexive but it is symmetric and, transitive, (c) The relation is not symmetric but it is reflexive and, transitive, (d) The relation is not transitive but it is reflexive and, symmetric, , Ê (a) Given set S is the set of all persons living in Delhi, and x, y in S are related such that they were born in, Delhi on the same day., Let R be a relation on S defined as x and y were born in, Delhi on same day., For reflexive Here (x, x) Î R as x and x was born on the, same day., \ R is reflexive, For symmetric Let (x, y) Î R, Þ x and y were born in Delhi on same day, Þ y and x were born in Delhi on same day, Þ ( y, x) Î R, \ R is symmetric., For Transitive Now let (x, y) Î R and ( y, z ) Î R, , 2. Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Then the number of, subsets of A containing two or three elements is, (a) 45, , Ê (c), , (b) 120, , Þ (x, z ) Î R, \ R is transitive., Thus, R is reflexive symmetric and transitive., \ R is an equivalence relation., , (d) 330, , Given set, A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }, The number of subsets of A containing two or three, elements, 10 ´ 9 10 ´ 9 ´ 8, +, = 10 C2 + 10 C3 =, = 45 + 120 = 165, 2´1, 3 ´2´1, , 3. The value ofi 2n + i 2n + 1 + i 2n + 2 + i 2n + 3 , wherei = -1, is, (a) 0, , Ê (a) i, , (b) 1, , 2n, , + i, , =i, , 2n, , 2n + 1, , + i, , (d) -i, , (c) i, , 2n + 2, , + i, , 2n + 3, , (1 + i + i + i3 ) = i2 n (1 + i - 1 - i) = i2 n ´ 0 = 0, 2, , 4. If the difference between the roots of the equation, , x 2 + kx + 1 = 0 is strictly less than 5, where | k | ³ 2 ,, then k can be any element of the interval, (a) (-3, - 2)] È[(2, 3), , (b) (-3, 3), , (c) [-3, - 2] È [2, 3], , (d) None of these, x2 + kx + 1 = 0, , Ê (b) Given equation,, , Let roots of equation be a and b., According to the question, we get a - b < 5, 2, , = a-b < 5, , Þ x and y were born in Delhi on same day and y and z, were born in Delhi on same day., Þ x and z were born in Delhi on same day, , (c) 165, , ...(i), [Squaring both sides], , Now,, , a + b = - k and ab = 1, (a - b )2 = (a + b )2 - 4ab Þ (a - b )2 = k2 - 4 ×1, , Þ, (a - b )2 = k2 - 4, On substituting (a - b )2 = k2 - 4 in Eq. (i), we get, Þ k2 - 4 < 5 Þ k2 < 9 Þ - 3 < k < 3
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26, , NDA/NA Solved Paper 2017 (I), Modulus of z is | z | =, , (1 )2 + (0 )2 = 1 + 0 = 1, , (where b and c are non-zero) is equal to the sum of the, 1 c, reciprocals of their squares. Then, , b, are in, c, b, , Þ tan q = tan 0° Þ q = 0°, , (a) AP, (c) HP, , 12. If the graph of a quadratic polynomial lies entirely above, X -axis, then which one of the following is correct?, (a), (b), (c), (d), , 16. The sum of the roots of the equation x 2 + bx + c = 0, , and argument,, imaginary part, 0, tan q =, =, =0, real part, 1, , Both the roots are real, One root is real and the other is complex, Both the roots are complex, Cannot say, , Ê (c) Let a and b be roots of given equation x, , Þ, , (b) 4, (d) 10, £ | z + 4| + | - 3 |, , Þ | z + 1| £ 3 + 3 Þ | z + 1| £ 6, , 14. The number of roots of the equation z = 2z is, 2, , (b) 3, , (c) 4, , (d) zero, , z 2 = 2z, , Ê (c), Þ, , (x + iy)2 = 2(x - iy), , Þ, , x2 - y2 + 2xy. i = 2x - 2iy, , 17. The sum of the roots of the equation ax 2 + x + c = 0, (where a and c are non-zero) is equal to the sum of the, reciprocals of their squares. Then a, ca 2 , c 2 are in, (a) AP, (c) HP, , ax2 + x + c = 0, , Then,, , 2xy + 2 y = 0 Þ xy + y = 0, y(x + 1 ) = 0 Þ y = 0 or x = - 1, y = 0,, , When,, , 2, , æ - 1 ö - 2(c / a ), ç, ÷, 2, 1 1 / a - 2c / a, 1 è aø, Þ, Þ, - =, =, (c / a ) 2, a, a, c2 / a 2, 2, (, 1, 2, ac, ), /, a, (, 1, 2, ac, ), 1, 1, - =, Þ- =, Þ, 2, 2, 2, a, c /a, a, c, Þ, - c2 = a (1 - 2ac) Þ - c2 = a + 2a 2 c, a + c2, Þ, 2a 2 c = a + c2 Þ a 2 c =, 2, Þ a, ca 2 , c2 are in AP., , x2 - y2 - 2x = 0, , then, , x2 - 2x = 0, , Þ, Þ, , x(x - 2) = 0 Þ, , x = 0, 2, , When, x = - 1, then x - y - 2x = 0, 2, , 2, , Þ (- 1 )2 - y2 - 2(- 1 ) = 0 Þ 1 - y2 + 2 = 0, y2 = 3 Þ y = ±, , Þ, , 3, , Therefore, four roots exist., , 15. If cot a and cot b are the roots of the equation, x 2 + bx + c = 0 with b ¹ 0, then the value of cot(a + b ) is, (a), , c -1, b, , (b), , 1- c, b, , (c), , b, c -1, , a + b = - 1 / a and ab = c / a, , According to question,, 1, 1, a+ b = 2 + 2, a, b, 2, a2 + b 2, 1 (a + b ) - 2ab, Þ, a+ b =, Þ - =, 2 2, ab, a, (ab )2, , x2 - y2 - 2x = 0 and 2xy + 2 y = 0, , Þ, , (b) GP, (d) None of these, , Ê (a) Let a and b are roots of the equation, , On comparing real and imaginary parts, we have, Now,, , b, +c, 1 c, 2c, 2, =, Þ, b, =, Þ, 2, b, 2, b+ c, b/ c + c, , Þ b / c, 1 / b, c are in AP Þ c / b, b, 1 / c are in HP., , Þ x2 - y2 - 2x + i(2xy + 2 y) = 0, Þ, , b(b + c2 ) - 2c = 0, , Þ b=, , [by Triangle Inequality], , (a) 2, , b2 - 2c = - bc2 Þ b2 + bc2 - 2c = 0, , Þ, , 13. If | z + 4 | £ 3, then the maximum value of | z + 1 | is, , | z + 1| =| z + 4 - 3|, , + bx + c = 0, , According to question,, a2 + b 2, 1, 1, a + b = 2 + 2 Þa + b =, a, b, a2 b 2, 2, (a + b ) - 2ab, (- b)2 - 2(c), Þ - b=, Þ - b=, 2, (ab ), (c)2, , the X-axis, then real roots does not exists because for, the existence of real roots the graph of polynomial, should cut X-axis at at least one point., , Ê (c), , 2, , \ a + b = - b and ab = c, , Ê (c) If the graph of a quadratic polynomial lies entirely above, , (a) 0, (c) 6, , (b) GP, (d) None of these, , (d), , b, 1- c, , Ê (b) Here, cot a + cot b = - b / 1 = - b, and cot a × cot b = c / 1 = c, cot a × cot b - 1, c-1 1 - c, \, cot(a + b ) =, =, =, cot a + cot b, -b, b, , value, of [C (7, 0) + C (7, 1)] + [C (7, 1) + C (7, 2)], + ¼ + [C (7, 6) + C (7, 7 )] is, , 18. The, , (a) 254, , Ê (a), , (b) 255, , (c) 256, , (d) 257, , [C (7, 0 ) + C (7, 1 )] + [C (7, 1 ) + C (7, 2)], + ... + [C (7, 6 ) + C (7, 7 )], = C (8, 1 ) + C (8, 2) + C (8, 3 ) + C (8, 4), + C (8, 5) + C (8, 6 ) + C (8, 7 ), (Q n Cr + n Cr - 1 =, , n +1, , Cr )
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27, , NDA/NA Solved Paper 2017 (I), =, , 8, , C1 + 8 C2 + 8 C3 + 8 C4 + 8 C3 + 8 C2 + 8 C1, , = 2 C1 + 28 C2 + 28 C3 + 8 C4, 8´7, 8 ´ 7 ´6 8 ´ 7 ´6 ´5, + 2´, +, = 2´ 8 + 2´, 2´1, 3 ´2´1, 4´3 ´2´1, 8, , = 16 + 56 + 2 ´ 56 + 2 ´ 7 ´ 5, , ending and beginning with a consonant which can be, made out of the letters of the word ‘EQUATION’ is, (c) 3000, , (d) 2160, , Ê (b), Vowel, , Consonant, , Q There are, ‘EQUATION’., , (c) 2 - n + n - 1 (d) 2 n - 1, , 3 7 15, + +, +K, 4 8 16, 1, 1, 1, 1, Þ æç1 - ö÷ + æç1 - ö÷ + æç1 - ö÷ + K + æç1 - n ö÷, è, è, 8ø, 4ø è, 2ø è, 2 ø, 1 1 1, 1, = n - æç + + + K + n ö÷, è2 4 8, 2 ø, n, 1 æ 1 - (1 / 2) ö, = n - çç, ÷ = n + 2- n - 1, 2 è 1 - 1 / 2 ÷ø, , 23. Consider the following in respect of sets A and B, , Consonant, , three, , (a) 2 n - n - 1 (b) 1 - 2 - n, 1, , 19. The number of different words (eight-letter words), (b) 4320, , 1 3 7 15, + + +, + ¼ is equal to, 2 4 8 16, , Ê (c) 2 +, , = 16 + 56 + 112 + 70 = 254, , (a) 5200, , 22. The sum of the first n terms of the series, , consonants, , in, , the, , word, , \ First and last place of required word can be filled in, 3, P2 ways, 3!, =, =6, (3 - 2)!, and there are 6 vowels in the word ‘EQUATION’., , 1. ( A - B ) È B = A, 3. ( A - B ) Ç B = f, , Which of the above are correct?, (a) 1, 2 and 3 (b) 2, 3 and 4 (c) 1, 3 and 4 (d) 1, 2 and 4, , Ê (b) 1., , A, , \ 6 places in between first and last placed can be filled, by 6 vowels in 6 P6 ways = 6 ! = 720, , Here ( A - B) È B = A È B, , 20. The fifth term of an AP of n terms, whose sum is n 2 - 2n,, , So, 1 is incorrect, 2., , (a) 5, , (b) 7, , (c) 8, , B, , A–B, , \ Required number of different words = 6 ´ 720 = 4320, , is, , 2. ( A - B ) È A = A, 4. A Í B Þ A È B = B, , A, , B, , (d) 15, , = n2 - 2n, , Ê (b) Given, S, , n, , A–B, , On replacing n by (n - 1 ), we get, , Here ( A - B) È A = A, So, 2 is incorrect, , Sn - 1 = (n - 1 )2 - 2(n - 1 )= n2 + 1 - 2n - 2n + 2, = n - 4n + 3, 2, , 3., , Now, Tn = Sn - Sn - 1 = (n2 - 2n) - (n2 - 4n + 3 ), , A, , B, , = n - 2n - n + 4n - 3, 2, , 2, , = 2n - 3, , … (i), , On putting n = 5 in Eq. (i), we get, T5 = 2 ´ 5 - 3 = 10 - 3 = 7, , 21. The sum of all the two-digit odd numbers is, (a) 2475, (c) 4905, , Ê, , (b) 2530, (d) 5049, , (a) Two-digit odd numbers are 11, 13, 15, ... , 99, Here, , a = 11, d = 2, an = 99, , \, , an = a + (n - 1 ) d, , Þ, , 99 = 11 + (n - 1 ) (2), , Þ 88 = (n - 1 ) (2), Þ, , n - 1 = 44 Þ n = 45, , \ Sum of all the two-digit odd numbers, 45, n, S45 =, (11 + 99 ) éQSn = (a + l)ù, 2, 2, ëê, ûú, 45, =, ´ 110 = 45 ´ 55 = 2475, 2, , A–B, , Here ( A - B) Ç B = f. So, 3 is correct, 4. A Í B Þ A È B = B. So, 4 is also correct, \ option (b) is correct, , 24. In the binary equation (1p 101) 2 + (10q 1) 2 = (100r 00) 2, where p , q and r are binary digits, what are the possible, values of p , q and r respectively?, (a) 0, 1, 0, , (b) 1, 1, 0, , (c) 0, 0, 1, , Ê (a) Given binary equation,, , (d) 1, 0, 1, , (1 p101 )2 + (10 q1 )2 = (100r00 )2, 1 p 1 0 1, æQ 0 + 0 = 0 ö, ç, ÷, + 1 0 q 1, ç1 + 0 = 1 ÷, ç1 + 1 = 10 ÷, è, ø, 1 0 0 r 0 0, , So, Here, p = 0, q = 1 and r = 0
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34, , NDA/NA Solved Paper 2017 (I), , 60. If the centroid of a triangle formed by (7, x ), (y, - 6) and, , RS =, =, , (9, 10) is (6, 3), then the values of x andy are respectively, (a) 5, 2, , (b) 2, 5, , (c) 1, 0, , Ê (a) Given,, , Centroid of a triangle formed by (7, x), ( y, - 6 ) and, (9, 10 ) is (6, 3), æ 7 + y + 9 x + (- 6 ) + 10 ö, ,, \ ç, ÷ º (6, 3 ), ø, è, 3, 3, 7+ y+9, x - 6 + 10, \, =3, = 6 and, 3, 3, y = 2 and x = 5, , (2 - 3 )2 + (- 1 - 2)2 + (6 - 4)2, , = 1 + 9 - 4 = 14, Since, PQ ¹ RS, Þ Given points are not vertices of a rhombus., and also,, RS = RQ + QP + PS, Þ, , 126 = 14 + 14 + 14 = 126, , Þ collinear points, Now, direction ratios of PQ are, , 61. A straight line with direction cosines 0, 1, 0 is, , 4 - 3, 5 - 2, 2 - 4 i.e., 1, 3, - 2, , (a) parallel to X-axis, , Ê, , (- 3 )2 + (- 9 )2 + (6 )2, , = 9 + 81 + 36 = 126, , (d) 0, 0, , and PS =, , Þ, , (2 - 5)2 + (- 1 - 8 )2 + (6 - 0 )2, , direction ratios of PR are, , (b) parallel toY-axis, , 5 - 3, 8 - 2, 0 - 4 i.e., 2, 6, - 4, , (c) parallel to Z-axis, , direction ratios of PS are, , (d) equally inclined to all the axes, , 2 - 3, - 1 - 2, 6 - 4 i.e., - 1, - 3, 2, , points. What are the coordinates of the point which is, equidistant from the four points?, , ® ®, Given, points P, Q, R and S are coplanar if PQ, PR and, ®, PS are coplanar, ® ® ®, i.e.,, [PQ PR PS] = 0, , a+b +c a+b +c a+b +cö, (a) æç, ,, ,, ÷, è, ø, 3, 3, 3, , Here, , (b) We know that direction cosines of Y-axis are (0, 1, 0)., , 62. (0, 0, 0), (a, 0, 0), (0, b, 0) and (0, 0, c) are four distinct, , (b) (a, b, c), a b c, (c) æç , , ö÷, è2 2 2ø, , Let, , Þ P, Q, R and S are coplanar., , the, , required, , point, , be, , P(x, y, z )Also,, , let, , 0 (0,0,0 ), A (0,0,0 ) B (0, b,0 ) and C (0,0, C ) be four distinct, points According to the equations, we have, PO = PA = PB = PC, 2, , 2, , 2, , 2, , Þ x2 + y2 + z 2 = (x - a ) 2 + y2 + z 2, = x + ( y - b) + z, 2, , 2, , 2, , = x2 + y2 + ( z - c)2, Þ a 2 - 2ax = 0, b2 - 2by = 0 and c2 - 2cz = 0, Þ x = a / 2, y = b / 2, z = c / 2, \ The required point is P(a / 2, b / 2, c / 2), , 63. The points P(3, 2, 4 ), Q( 4, 5, 2), R(5, 8, 0) and S (2, - 1, 6) are, (a), (b), (c), (d), , Ê (c), , vertices of a rhombus which is not a square, non-coplanar, collinear, coplanar but not collinear, PQ =, =, QR =, , 3, 6, , = 0 (since R1 and R3 are identical), ® ®, ®, ÞPR , PS and PS are coplanar, , a b c, (d) æç , , ö÷, è3 3 3ø, , Ê (c), , -2, -1 -3 2, - 4 = (- 1 ) 2, 6 -4, -1 -3 2, -1 -3 2, 1, 2, , (4 - 3 )2 + (5 - 2)2 (2 - 4)2, (1 )2 + (3 )2 + (- 2)2, , = 1 + 9 + 4 = 14, , (5 - 4)2 + (8 - 5)2 + (0 - 2)2, , = (1 )2 + (3 )2 + (- 2)2 = 1 + 9 + 4 = 14, , 64. The line passing through the points (1, 2, -1) and (3, - 1, 2), meets the yz-plane at which one of the following points?, 7 5, (a) æç 0, - , ö÷, è, 2 2ø, , 7 1, (b) æç 0, , ö÷, è 2 2ø, , 7, 5, (c) æç 0, - , - ö÷, è, 2, 2ø, , 7, 5, (d) æç 0, , - ö÷, è 2, 2ø, , Ê (d) Equation of line passing through the points, (1, 2, - 1 ) and (3, - 1, 2) is, x-1, y-2, z+1, =, =, 3 -1 -1 -2 2+ 1, x -1 y -2 z + 1, =, =, = l(say), Þ, 2, 3, 3, General point on line (i) is, x = 2l + 1, y = - 3 l + 2 and z = 3 l - 1, Since, line (i) meets the yz-plane, \ x-coordinate will be zero, Þ, , 2l + 1 = 0 Þ l = - 1 / 2, , \, , y = - 3 (- 1 / 2) + 2 = 3 / 2 + 2 = 7 / 2, -3 -2, and z = 3 (- 1 / 2) - 1 =, = -5 / 2, 2, required point is (0, 7 / 2 , - 5 / 2)., , … (i)
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35, , NDA/NA Solved Paper 2017 (I), 65. Under which one of the following conditions are the, lines x = ay + b, z = cy + d and x = ey + f , z = gy + h, perpendicular?, , (a) ae + cg - 1 = 0, (c) ae + cg + 1 = 0, , (b) ae + bf - 1 = 0, (d) ag + ce + 1 = 0, , Ê (c) For the first line, x - b = ay, z - d = cy, Þ, , x-b, a, , 68. ABCD is a quadrilateral whose diagonals are AC and BD., Which one of the following is correct?, ®, ®, (a) BA + CD =, ®, ®, (c) BA + CD =, , Ê (b), , y z-d, = =, 1, c, , We have, In quadrilateral ABCD, AC and BD are, diagonals., D, , C, , and for the second line, x - f = ey, z - h = gy, x- f, y z-h, = =, Þ, e, 1, g, , P, , Q these are perpendicular, , A, , So a × e + 1 × 1 + cg = 0 Þ ae + cg + 1 = 0, ®, $ ®b = 2i$ + 3j$ + 2k$ and ®c = i$ + mj$ + nk$ are, 66. If a = i$ - $j + k,, ®, , three coplanar vectors and | c | = 6, then which one of, the following is correct?, (a) m = 2 and n = ± 1, , Ê, , (b) m = ± 2 and n = - 1, , (c) m = 2 and n = - 1, (d) m = ± 2 and n = 1, ® ®, ®, ® ® ®, (d) a , b and c are coplanar if [ a b c ] = 0, 1 -1 1, Þ, , 2, 1, , 3, m, , ®, , ®, , ®, , ®, , ®, , ® ® ® ®, at P and let O be the origin. What is OA + OB + OC + OD, equal to?, ®, (c) 6OP, , ®, , ®, (d) 8OP, , ®, , ® ® ®, ® ® ®, , ®, , ® ® ®, , ®, , ®, , (c) a , b, c are orthogonal in pairs but | a | ¹ | c |, ® ® ®, c, , Ê (a) Given, a ´ b =, ®, , … (i), ®, , ®, , \ vector c is perpendicular to both vectors a and b ., ® ® ®, … (ii), and,, b´ c =a, ®, , ®, , ®, , ®, , ®, , \ a , b and c are orthogonal in pairs., ® ® ®, ® ®, ®, Now,Q a ´ b = c Þ | a ´ b | = | c |, ® ®, ®, Þ | a ||, × b|sin 90° = | c |, ®, ®, [Q a and b are perpendicular to each other], ®, ®, ®, … (iii), Þ | a | | b| = | c |, , C, , P, , B, , We known that, the diagonals of a parallelogram, bisects each other,, , From (i) and (ii), ® ® ® ®, ®, Þ OA + OB + OC + OD = 4OP, , ®, , (b) a , b, c are non-orthogonal to each other, , ®, , therefore P is midpoint of AC and BD both, ® ®, ®, \ OA + OC = 2OP, ®, ®, ®, OB + OD = 2OP, , ®, , \ vector a is perpendicular to both vectors b and c, , Ê (b) Given, ABCD is a parallelogram., D, , ®, , (a) a , b, c are orthogonal in pairs and | a | = | c | and | b | = 1, , (d) a , b, c are orthogonal in pairs but | b | ¹ 1, , 67. Let ABCD be a parallelogram whose diagonals intersect, , A, , ®, , following is correct, , m2 + 2 = 6 Þ m2 = 6 - 2 = 4 Þ m = ± 2, , ®, (b) 4OP, , adding (i) and (ii), we get, ® ®, ® ®, ® ®, BA + CD = (BP + PA) + (CP + PD), ® ®, ® ®, = (BP + PD) + (CP + PA), ® ® ® ®, Þ BA + CD = BD + CA, , 69. If a ´ b = c and b ´ c = a , then which one of the, , Þ 5n - 5 = 0 Þ n = 1, ®, ®, Now, c = i$ + mj$ + k$ Þ | c | = 1 2 + m2 + (1 )2 = 6, , ®, (a) 2OP, , B, , Let diagonals AC and BD are intersects each other at P., ® ® ®, … (i), then, BA = BP + PA, ® ® ®, and, … (ii), CD = CP + PD, , ®, , 2 =0, n, , Þ 1 (3 n - 2m) + 1 (2n - 2) + 1 (2m - 3 ) = 0, , Þ, , ®, ® ®, ®, (b) BA + CD = BD + CA, ®, ® ®, ®, (d) BA + CD = BC + AD, , ®, ®, AC + DB, ®, ®, AC + BD, , … (i), … (ii), , [Q sin 90° = 1], ® ® ®, ® ®, ®, and b ´ c = a Þ | b ´ c | = | a |, ®, ®, ®, Þ, | b | | c | sin90° = | a |, ®, ®, [Q b and c are perpendicular to each other], ®, ®, ®, … (iv), Þ | b | | c |= | a |, ®, ®, ®, ®, From (iii) and (iv) | b | | a | | b | = | a |, ®, ®, ®, Þ, | b | = 1 and | a | = | c |
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40, , NDA/NA Solved Paper 2017 (I), , 93. Consider the following statements :, , 96. Let f (a ) =, , 1. If lim f ( x ) and lim g ( x ) both exist, then lim{ f ( x )g ( x )}, x ®a, x ®a, x ®a, exists., , a -1, Consider the following, a +1, , 1, 1. f (2a ) = f (a ) + 1 2. f æç ö÷ = - f (a ), èa ø, , 2. If lim{ f ( x )g ( x )} exists, then both lim f ( x ) and, x ®a, x ®a, lim g ( x ) must exist., , Which of the above is/are correct?, , Which of the above statement is/are correct?, , (a) Only 1, , (a) Only 1, , (b) Only 2, , (c) Both 1 and 2, , (c) Both 1 and 2, , (d) Neither 1 nor 2, , x ®a, , Ê (b) We have,, , Ê (a) We know,, When, , lim f (x), , x® a, , and, , lim g (x), , x® a, , both, , exist,, x® a, , x® a, , lim g (x) both exist., , x® a, , 94. Which one of the following functions is neither even nor, odd?, 3, (b) x +, x, , (a) x - 1, , Ê (d) For option (a), , (c) | x |, , 2., , (d) x (x - 3), 2, , f (x ) = x 2 - 1, , f (- x ) = (- x ) 2 - 1 = x 2 - 1, \ even function., For option (b) f (x) = x +, , 3, 3, = - æç x + ö÷, è, (- x ), xø, , 97. What is the maximum area of a triangle that can be, inscribed in a circle of radius a?, (a), , f (- x ) = - f (x ), , Here,, , æa - 1ö 1 - a, - f (a ) = - çç, ÷÷ =, èa + 1ø 1 + a, , So, statement 2 is true., , 3, x, , f (- x ) = - x +, , a+1, 2a - 1, , So, statement 1 is not true., 1/a-1, f (1 / a ) =, 1/a+ 1, (1 - a ) / a, 1-a, =, =, (1 + a ) / a 1 + a, , and, , f (x ) = f (- x ), , Here,, , f (a ) =, , a-1, , 2a + 1, a-1, f (a ) + 1 =, +1, a+1, (a - 1 ) + (a + 1 ), 2a, =, =, a+1, a+1, , exists, then it is not necessary that lim f (x) and, , 2, , (d) Neither 1 nor 2, , 1. Now, f (2a ) =, , then, , lim{ f (x) g (x)} must exist. But when lim{ f (x) g (x)}, , x® a, , (b) Only 2, , \ odd function., , 3a 2, 4, , (b), , a2, 2, , (c), , 3 3a 2, 4, , (d), , 3a 2, 4, , Ê (c), , For option (c) f (x) = | x|, f (- x ) = | - x | = | x |, \, , f (x ) = f (- x ), , \ even function., , a, , For option (d) f (x) = x2 (x - 3 ), f (- x) = (- x)2 [(- x) - 3] = - x2 (x + 3 ), Here, f (x) ¹ f (- x) and f (x) ¹ - f (- x), , For area to be maximum triangle inscribed in the circle, must be equilateral triangle, , \ Neither even nor odd., , 95. What is the derivative of log10 (5x 2 + 3) with respect to x?, (a), , x log10 e, 5x 2 + 3, , (b), , 2 x log10 e, 5x 2 + 3, , (c), , 10x log10 e, 5x 2 + 3, , (d), , x loge 10, 5x 2 + 3, , Ê (c) let y = log, , 10, , \ Area of the equilateral triangle, 3, 3, 3 3a2, =, (side)2 =, ( 3 a )2 =, 4, 4, 4, , On differentiating both sides w.r.t. x, we get, 1, d, dy, (log10 e), (5x2 + 3 ), =, dx, dx 5x2 + 3, [by chain rule], =, , 5x + 3, 2, , (10 x) =, , 1, x, following is correct?, , 98. Let f ( x ) = x + , where x Î(0, 1). Then which one of the, , (5x2 + 3 ), , log10 e, , Given, radius of the circle = a, \ side of the equilateral triangle = 3a, , 10 x log10 e, 5x + 3, 2, , (a) f (x) fluctuates in the interval, (b) f (x) increases in the interval, (c) f (x) decreases in the interval, (d) None of the above
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41, , NDA/NA Solved Paper 2017 (I), 1, x, x 2 - 1 (x + 1 ) (x - 1 ), 1, f ¢ (x ) = 1 - 2 =, =, x, x2, x2, f (x ) = x +, , Ê (c) Given,, \, , [Q a 2 - b2 = (a + b) (a - b)], From question x Î(0, 1 ) and in the interval (0, 1), f ¢ (x ) < 0, , 102. The mean of a group of 100 observations was found to be, 20. Later it was found that four observations were, incorrect, which were recorded as 21, 21, 18 and 20. What, is the mean if the incorrect observation are omitted?, (a) 18, , i, , 99. Suppose the function f ( x ) = x , n ¹ 0is differentiable for, , \, , all x. Then n can be any element of the interval, , Þ, , 1, (c) æç , ¥ö÷ (d) None of these, è2 ø, , \, , 2, , ½ x ½, , ò, , I=, , Ê (b) Let, , e2, , e-, , =-, , (c) 3, , ln x, dx =, x, , 1, , ò, , 5, 2, , e0, , e-, , 1, , ò, , e0, , e-, , 1, , ln x d(ln x) +, e, , (d) 4, , æ - ln x ö, ç, ÷ dx +, è x ø, , ò, , e2, , e0, , ò, , e2, , e0, , ln x, dx, x, , e, , is multiplied by 3, then what is the new variance of the, resulting observations?, (c) 15, , (d) 45, , Ê (d) Let x , x , … x be the given observations we have,, 1, å (x - x ) = 5, 20, 1, 20, , 2, , 20, , 2, , i, , i=1, , then we have to find variance of 3 x1 , 3 x2 , … , 3 x20, Let X denotes the mean of new observation then,, 20, , 20, , å3 x, , i, , X =, , i=1, , =, , 20, , 3 å xi, i=1, , 20, , = 3x, , \ variance of new observation, 1 20, 1 20, 2, =, (3 xi - 3 x )2 =, å, å (9 )(xi - x ), 20 i = 1, 20 i = 1, =, , 9, 20, , 20, , å (x, , i, , i=1, , - x ) = 9 ´ 5 = 45, , 1, 6, , (b), , 1, 3, , (c), , 1, 2, , (d), , 2, 3, , Ê (a) Number of ways of forming a committee of two persons, from two men and two women = 4 C2, , and number of ways of forming a committee of two, persons from two men and two women in which only, women are there, , 2, , 101. The variance of 20 observations is 5. If each observation, (b) 10, , (a), , ln x d(ln x), , 0, , é (ln x)2 ù, é (ln x)2 ù, + ê, =-ê, ú, ú, ë 2 û e0, ë 2 û e- 1, -1, 1, =, [(ln e0 )2 - (ln e- 1 )2 ] + [(ln e2 )2 - (ln e0 )2 ], 2, 2, -1, 1, 5, 1, 2, 2, 2, =, [(0 ) - (- 1 ) ] + [(2) - (0 )2 ] = + 2 =, 2, 2, 2, 2, , (a) 5, , i, , and two women. What is the probability that the, committee will have only women?, , -1, , (b), , 96, , å x = 2000 - (21 + 21 + 18 + 20 ), , 103. A committee of two persons is constituted from two men, , e ln x½, ½ ½dx equal to?, 100. What is ò ½, , 3, 2, , = 100 x = 100 ´ 20 = 2000, , i, , = 2000 - 80 = 1920, 1920, Hence, new mean =, = 20, 96, , n³1, , \ For f (x) to be differentiable n can be any element of, the interval [1, ¥)., , (a), , åx, , 100, , i=1, , For f (x) to be differentiable n - 1 ³ 0, , e, , 100, , i=1, , Four observation 21, 21, 18, 20 are incorrect and, omitted, , Here, f ¢ (x) = nxn - 1, \, , x=, , i=1, , f (x ) = x n , n ¹ 0, , Ê (a) Given,, , (d) 22, , Ê (b) We have x = 20, åx, , n, , (b) (0, ¥), , (c) 21, , 100, , \ f (x) decreases when x Î(0, 1 )., , (a) [1, ¥), , (b) 20, , = 2 C0 ´ 2 C2, 2, C0 ´ 2 C2 1 ´ 1 1, \ Required probability =, =, =, 4, 6, 6, C2, , 104. A question is given to three students A, B and C whose, 1 1, 1, chances of solving it are , and respectively. What is, 2 3, 4, the probability that the question will be solved?, (a), , 1, 24, , (b), , 1, 4, , (c), , 3, 4, , (d), , 23, 24, , Ê (c) We have Chances of solving the questions A, B and C are, 1 1, 1, , , and respectively, 2 3, 4, 1, 1, 1, \, P( A) = , P(B) = , P(C ) =, 2, 3, 4, 1, 2, 3, P( A¢ ) = , P(B¢ ) = P(C ¢ ) =, 2, 3, 4, Required, , probability, , = 1 - probability of problem not solved, = 1 - P( A¢ ) ´ P(B¢ ) ´ P(C ¢ ), 1 2 3, =1 - ´ ´, 2 3 4, 1 3, =1 - =, 4 4
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42, , NDA/NA Solved Paper 2017 (I), \ Required Probability, , 105. The mean weight of 150 students in a certain class is, , a, 2, =, Area of circle whose radius is a, 2, a, p æç ö÷, è2ø, 1, =, =, 2, 4, pa, Area of circle whose radius is, , 60 kg. The mean weight of boys in the class is 70 kg and, that of girls is 55 kg. What is the number of boys in the, class?, (a) 50, , (b) 55, , (c) 60, , (d) 100, , Ê (a) We have, , Total number of students = 150, , 108. If two regression lines between height ( x ) and weight (y ), , Let the number of boys = x, , are 4y - 15x + 410 = 0 and 30x - 2y - 825 = 0, then what, will be the correlation coefficient between height and, weight?, , So, the number of girls = 150 - x, Mean weight of boys = 70, , 1, 3, 2, (c), 3, , Mean weight of girls = 55, Mean weight of students = 60, \ Mean weight of students, (70 ´ x + 55 ´ (150 - x), =, 150, 70 x + 55 ´ 150 - 55x, Þ 60 =, 150, Þ 60 ´ 150 - 55 ´ 150 = 15x Þ, Þ, , 150 ´ 5, 15, , \, , P( A ) = 02, . and P( B ) = 05., . If A Í B. Then the values of, conditional probabilities P( A| B ) and P( B| A ) are, respectively, 2, ,1, 5, , (d) Information is insufficient, , Now, correlation coefficient between height and weight, 15 1, 1, = bxy ´ byx =, ´, =, 4 15 2, , AÍB, \, , A Ç B = A Þ P( A Ç B) = P( A), P( A Ç B) P( A) 0.2 2, P( A / B) =, =, =, =, P(B), P(B) 0.5 5, , 109. In an examination, 40% of candidates got second class., When the data are represented by a pie chart, what is the, angle corresponding to second class?, , P( A Ç B) P( A), =, =1, P( A), P( A), , (a) 40°, (c) 144°, , 107.A point is chosen at random inside a circle. What is the, probability that the point is closer to the centre of the, circle than to its boundary?, (a), , 1, 5, , (b), , 1, 4, , (c), , 1, 3, , (d), , byx, , = - 410 + 15x, - 410 15, y=, +, x, 4, 4, 15, x=, 4, , From (ii) 30 x = 2 y + 825, 2 y 825, x=, +, Þ, 30, 30, 825, 1, x=, +, y, Þ, 30, 15, 1, \, bxy =, 15, , Ê (b) We have P( A) = 0 × 2 P(B) = 0.5, , P(B / A) =, , … (i), … (ii), , Þ, , 106. For two dependent events A and B, it is given that, , (b), , 4 y - 15x + 410 = 0, , and 30 x - 2 y - 825 = 0, From (i) 4 y = 15x - 410, , =x, , \ Number of boys in the class = 50, , 2 3, ,, 5 5, 2, (c) 1,, 5, , (b), , Ê (b) Given, regression lines, , x = 50, , (a), , 1, 2, 3, (d), 4, , (a), , 1, 2, , Ê (c), , (b) 90°, (d) 320°, Second Class, 40%, , a, , Ê (b)QAny point which lies interior to circle whose radius is 2, is close to the centre than circumference., a, a/2, , Q In the examination candidates who got second class, = 40%, \ Angle corresponding to second class, 40, =, ´ 360°, 100, 2, = ´ 360° = 2 ´ 72° = 144°, 5
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43, , NDA/NA Solved Paper 2017 (I), 110. Consider the following statements, Statement 1 Range is not a good measure of dispersion., Statement 2 Range is highly affected by the existence of, extreme values., Which one of the following is correct in respect of the, above statements?, (a) Both Statement 1 and Statement 2 are correct and, Statement 2 is the correct explanation of Statement 1, (b) Both Statement 1 and Statement 2 are correct but, Statement 2 is not the correct explanation of Statement 1, (c) Statement 1 is correct but Statement 2 is not correct, (d) Statement 2 is correct but Statement 1 is not correct, , 114. For given statistical data, the graphs for less than ogive, and more than ogive are drawn. If the point at which the, two curves intersect is P, then abscissa of point P gives, the value of which one of the following measures of, central tendency?, (a) Median, , (c) Mode (d) Geometric mean, , Ê (a) Median of grouped data is the x-coordinate of the point, , of intersection of ‘less than’ and ‘more than’ ogive, curves., , 115. Consider the following statements, , of extreme values. So range is not a good measure of, dispersion., , 1. Two events are mutually exclusive if the occurrence, of one event prevents the occurrence of the other., 2. The probability of the union of two mutually, exclusive events is the sum of their individual, probabilities., , \ option (a) is correct., , Which of the above statement is/are correct?, , Ê (a) We know that, range is highly affected by the existence, , 111. A card is drawn from a well-shuffled ordinary deck of 52, cards. What is the probability that it is an ace?, 1, 13, 3, (c), 13, (a), , 2, 13, 1, (d), 52, (b), , (a) Only 1, , (b) Only 2, , (c) Both 1 and 2, , (d) Neither 1 nor 2, , Ê (c) We know,, Two events are mutually exclusive if the occurance of, one event prevents the occurance of the other and the, probability of the union of two mutually exclusive, events is the sum of their individual probabilities., , Ê (a) Q Number of ace in a deck of 52 cards = 4, and Total number of cards = 52, 4, 1, \ Required probability =, =, 52 13, , 112. If the data are moderately non-symmetrical, then which, one of the following empirical relationships is correct?, (a), (b), (c), (d), , (b) Mean, , 2 ´ Standard deviation = 5, 5 ´ Standard deviation = 2, 4 ´ Standard deviation = 5, 5 ´ Standard deviation = 4, , ´ Mean deviation, ´ Mean deviation, ´ Mean deviation, ´ Mean deviation, , Ê (c) We know, If the data are moderately non-symmetrical,, then the following empirical relationship holds., 5, Standard deviation = ´ Mean deviation., 4, , \ Both statement (1) and (2) are correct., , 116. If the regression coefficient of x on y and y on x are -, , 1, respectively then what is the correlation, 8, coefficient between x and y?, , and 1, 4, 1, (c), 16, (a) -, , Ê (d) Given,, , (a) 1 and 2 Only, (b) 2 and 3 Only, (c) 1 and 3 Only, (d) 1, 2 and 3 only, , Ê (b) We know, Data can be represent in Tabular form and, Graphical form., , (d), , 1, 4, , 1, 16, , Regression coefficient of x on y =, -1, , -1, 2, , and, , 8, , We know coefficient of correlation is the geometric, mean between the regression coefficients., , 113. Data can be represented in which of the following, , Select the correct answer using the code given below., , (b) -, , Regression coefficient of y on x =, , Þ 4 ´ Standard deviation = 5 ´ Mean deviation, , forms?, 1. Textual form, 2. Tabular form, 3. Graphical form, , 1, 2, , \ Correlation coefficient between x and y, 1/2, éæ - 1 ö æ - 1 öù, 1, 1, = êç, =, =, ÷ú, ÷´ç, ø, è, ø, è, 4, 16, 2, 8, ë, û, , 117. A sample of 5 observations has mean 32 and median 33., Later it is found that an observation was recorded, incorrectly as 40 instead of 35. If we correct the data,, then which one of the following is correct?, (a), (b), (c), (d), , The mean and median remain the same, The median remains the same but the mean will decrease, The mean and median both will decrease, The mean remains the same but median will decrease
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44, , NDA/NA Solved Paper 2017 (I), , Ê (b) From question for correcting the data we have to replace, 40 by 35., , P( A) =, , 1, 1, and P(B) =, 3, 4, , \ Median of corrected data remain the same i.e., 33, , Q A and B are two mutually exclusive events, , and, as 35 is less than 40, so the mean of correct data, will decrease., , We know,, , 118. If two fair dice are thrown, then what is the probability, that the sum is neither 8 nor 9?, 1, 6, 3, (c), 4, (a), , 1, 4, 5, (d), 6, (b), , Ê (c) Total number of outcome when two fair dice are thrown, = 6 ´ 6 = 36, , Outcome in which sum is either 8 or 9 will be (2, 6),, (3, 5), (3, 6), (4, 4), (4, 5), (5, 3), (5, 4), (6, 2), (6, 3)., \ Total number of outcome after throwing two dice in, which sum is either 8 or 9 = 9, 9, \ Probability that the sum is either 8 or 9 =, 36, \ Probability that the sum is neither 8 or 9, 9 27 3, =1 =, =, 36 36 4, , 119. Let A and B are two mutually exclusive events with, 1, 1, P( A ) = and P( B ) = . What is the value of P( A Ç B )?, 3, 4, 1, 6, 1, (c), 3, , (a), , Ê (d) Given,, , 1, 4, 5, (d), 12, (b), , \, , P( A Ç B) = 0, , P( A È B) = P( A) + P(B) - P( A Ç B), 7, 1 1, = + -0 =, 12, 3 4, Now, \ P( A Ç B ) = P( A È B), [by Demorgan’s Theorem], = 1 - P( A È B) = 1 -, , 7, 5, =, 12 12, , 120. The mean and standard deviation of a binomial, distribution are 12 and 2 respectively. What is the, number of trials?, (a) 2, , (b) 12, , (c)18, , Ê (c)Q mean = 12 and standard deviation = 2, then,, , np = 12 and npq = 4, , Þ, , 12 ´ q = 4, 1, Þ, q=, 3, 1 2, \p=1 - q=1 - =, 3 3, and also,, Þ, Þ, , np = 12, 2, = 12, 3, , n´, , n = 12 ´, , 3, = 18, 2, , (d) 24
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Sets, Relations, and Functions, Sets, In our mathematical language, everything in this, universe whether living or non-living is called an object., Any collection of well defined objects is called a set. By, ‘well defined objects’, we mean that given a set and an, object, it must be possible to decide whether or not the, object belongs to the set. The objects in set are called its, members or elements., Sets are denoted by capital letters A, B, C etc., while, the elements are denoted in general by small letters a, b, c,, etc., Following collections are sets, (i) The collection of all positive integers., (ii) The collection of all capitals of states of India., Let A be any set of objects and let ‘a’ be a member of A,, then we write a ∈ A and read it as ‘a belong to A’ or ‘a is an, element of A’ or ‘a is member of A’. If a is not an object of A,, then we write a ∉ A and read as ‘a does not belong to A’ or ‘a, is not an element of A’ or ‘a is not member of A’., Some standard notations for some special points., (i) The set of all natural numbers i.e., the set of all, positive integers is denoted by N., (ii) The set of all integers is denoted by Z or I., (iii) The set of all rational numbers is denoted by Q., (iv) The set of all real numbers is denoted by R., (v) The set of all positive real numbers is denoted by R + ., (vi) The set of all complex numbers is denoted by C., (vii) The set of all positive rational numbers is denoted, by Q + ., , (i) Listing method In this method, the set is, represented by writing elements in a bracket and, separated by comma (,)., e.g., A = Set of vowels of English alphabet, = { a , e, i , o , u }., This method is also known as Tabular method or, Roster method., (ii) Set builder method In this method, instead, of listing all elements of a set, we write the set by some, special property or properties satisfied by all elements and, write it as,, A = { x : P ( x )}, where P ( x ) is a property which is satisfied by x., e.g.,, A = { 1, 2, 3, 4} = { x : x ∈ N and x < 5}, This method is also known as Rule method or Property, method., %, , There are two methods to represent sets, (i) Listing method (ii) Set builder method, , In writing the elements of any set there is no consideration of, sequence, e.g.,{ a, e, i , o, u } and { e, i , o, u, a } are two same sets., , Types of Sets, (i) Empty set A set consisting of no element is, called an empty set or null set or void set and is denoted by, symbol φ or { }., e.g., A = { x : x ∈ N and 3 < x < 4} = φ ., A set which is not empty is called non-empty set or, non-void set., (ii) Singleton set A set consisting of only one, element is called a singleton set., e.g., {2}, {0}, {φ}, %, %, , Representation of Sets, , 1, , The set {0} is not an empty set as it contains one element 0., The set {φ} is not an empty set as it contains one element φ., , (iii) Finite set A set having finite number of, elements is called finite set., e.g., A = {1, 2, 3} is a finite set.
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2, , NDA/NA Mathematics, , (iv) Infinite set A set which is not finite is called, an infinite set., e.g., A = Set of points lie in a plane is an infinite set., (v) Cardinal number of a finite set The, number of elements of a finite set A is called its cardinal, number and it is denoted by n(A) or o(A)., (vi) Equivalent sets Two finite sets A and B are, said to be equivalent if they have the same cardinal, number. Thus, sets A and B are equivalent if n( A ) = n( B )., (vii) Subset and super set The set B is said to, be subset of set A, if every element of set B is also an, element of set A. Symbolically we write it as, B ⊆ A or, A ⊇ B, where A is super set of B., (a) B ⊆ A is read as B is contained in A or B is subset, of A or A is super set of B., (b) A ⊇ B is read as A contains B or B is a subset of A., Evidently, if A and B are two sets such that, x ∈ B ⇒ x ∈ A, then B is subset of A. The symbol ‘⇒’ stands, for ‘implies’, we read it as x belongs to B implies that x, belongs to A., , %, , Whenever, we have to show that two sets A and B are equal i.e.,, A = B ⇔ A ⊆ B and B ⊆ A., , (x) Universal set In any discussion in set theory,, we need a set such that all sets under consideration in that, discussion are its subsets. Such a set is called the universal, set and is denoted by U., (xi) Power set The set of all the subsets of a given, set A is said to be the power set of A and is denoted by P( A )., e.g., If A = { 1, 2, 3}, then, P ( A) = { φ ,{ 1},{ 2},{ 3},{ 1, 2},{ 2, 3}, { 3, 1},{ 1, 2, 3}}., %, %, %, , Elements of power set are the subset of A., The power set of each given set is always non-empty., If A is a finite set of n elements, then number of elements in P(A), will be 2n ., , Venn Diagram, To express the relationship among sets in a, perspective way, we represent them pictorially by means of, diagrams, known as Venn diagrams., , e.g., Let A = { 1, 2, 3, 4}; B = { 1, 2, 4}, , U, , Here, B is a subset of A., , (viii) Proper subset The set B is said to be a, proper subset of set A, if every element of set B is an, element of A whereas every element of A is not an element, of B., We write it as B ⊂ A and read it as ‘B is a proper subset, of A’. Thus, B is a proper subset of A, if every element of B is, an element of A and there is atleast one element in A which, is not in B., Observe that A ⊆ A i. e., every set is a subset of itself,, but not a proper subset., e.g., Let A = { 1, 2, 3}; B = { 1, 2}, then B ⊂ A., %, %, , The universal set is usually represented by a, rectangular region and its subsets by circle or closed, bounded regions inside this rectangular region., , Operations on Sets, (i) Union of sets Let A and B are two sets, then, union of A and B is denoted by A ∪ B and it consists of each, one of which is either in A or in B or in both A and B., U, , Null set is a subset of every set and each set is subset of itself., Number of subset of a finite set of n elements is 2n ., A, , B, , (ix) Equal sets Two sets A and B are said to be, equal, if each element of A is an element of B and each, element of B is an element of A., Thus, two sets A and B are equal, if they have exactly, the same elements but the order in which the elements in, the two sets have been written down is immaterial., Thus, if, , x ∈ A ⇒ x ∈B, , and, , y ∈ B ⇒ y ∈ A,, , then A and B are equal, e.g.,, , {4, 8, 10} = {8, 4, 10}, , [The order in which the elements of a set is also, immaterial], , Thus, A ∪ B = { x : x ∈ A or x ∈ B}, Clearly, x ∈ A ∪ B ⇔ x ∈ A or x ∈ B, and x ∉ A ∪ B ⇔ x ∉ A and x ∉B, In the figure, the shaded part represents A ∪ B., It is evident that A ⊆ A ∪ B, B ⊆ A ∪ B., , Example 1. If A = {1, 2, 3} and B = {1, 3, 5, 7}, then the, value of A ∪ B is, (a) {4, 5, 7}, (c) {1, 2, 3, 5}, , (b) {1, 2, 3, 5, 7}, (d) {6, 3, 5, 7}, , Solution (b) Q A = {1, 2, 3} and B = {1, 3, 5, 7}, ∴, , A ∪ B = {1, 2, 3, 5, 7}
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3, , Sets, Relations and Functions, , (ii) Intersection of sets The intersection of two, sets A and B, denoted by A ∩ B is the set of all elements,, common to both A and B., , (v) Symmetric difference of two sets The, symmetric difference of two sets A and B, denoted by A ∆ B, is the set ( A − B) ∪ ( B − A)., U, , U, , A–B, A, , B, , Example 2. If A = {1, 2, 3, 4} and B = {2, 4, 6}, then the, (b) {6, 5}, (d) {1, 2, 3}, , Solution (a) Q A = {1, 2, 3, 4} and B = {2, 4, 6}, ∴, , B, , A, , Thus,, A ∩ B = { x : x ∈ A and x ∈ B}, Clearly, x ∈ A ∩ B ⇔ x ∈ A and x ∈ B, and x ∉ A ∩ B ⇔ x ∉ A or x ∉B, In the figure, the shaded part represents A ∩ B., It is evident that A ∩ B ⊆ A, A ∩ B ⊆ B., value of A ∩ B is, (a) {2, 4}, (c) {2, 3, 4}, , B–A, , Thus,, A ∆ B = ( A − B) ∪ ( B − A) = { x : x ∉ A ∩ B}, The shaded part represents A ∆ B., , Example 3. If A = {1, 3, 5, 7, 9} and B = {2, 3, 5, 7, 11},, then find the value of A ∆ B is, (a) {5, 7, 11}, (c) {3, 5, 7}, , Solution (b) Q A = {1, 3, 5, 7, 9} and B = {2, 3, 5, 7,11}, ∴, ∴, , A ∩ B = {2, 4 }, , (iii) Disjoint sets Two sets, U, A and B are said to be disjoint sets,, B, A, if they have no common element, i.e., A ∩ B = φ., The disjoint sets can be, represented by Venn diagram as shown in the figure, e.g., let A = { 1, 2, 3} and B = { 4, 6}, Here, A and B are disjoint sets because A ∩ B = φ ., , A − B = {1, 9 } and B − A = {2,11}, A ∆ B = ( A − B) ∪ (B − A), = {1, 9 } ∪ {2,11} = {1, 2, 9,11}, , Complement of a Set, If U is a universal set and A ⊂ U ,then complement set, of A is denoted by A′ or U − A., U, A, , (iv) Difference of sets If A and B are two sets,, then their difference A − B is the set of all those elements, of A which do not belong to B., U, , (b) {1, 2, 9, 11}, (d) {1, 3, 5, 11}, , B, , Thus, A′ = U − A = { x : x ∈U , but x ∉ A}, It is clear that x ∈ A′ ⇔ x ∉ A, %, %, , φ = U′, A ∪ A′ = U, , %, %, , φ′ = U, A ∩ A′ = φ, , %, , (A′ )′ = A, , A–B, A, , Laws of Algebra of Sets, , B, , Thus,, A − B = { x : x ∈ A and x ∉B}, Clearly, x ∈ A − B ⇔ x ∈ A and x ∉B., In the figure, the shaded part represents A − B., Similarly, the difference B − A is the set of all those, elements of B that do not belong to A i.e.,, B − A = { x : x ∈ B and x ∉ A}, U, B–A, A, , B, , In the figure, the shaded part represents B − A., e.g., if A = { 1, 3, 5, 7, 9} and B = { 2, 3, 5, 7, 11}, then, A − B = { 1, 9} and B − A = { 2, 11}., , If A, B and C are three sets, then, 1. Idempotent laws, (a) A ∪ A = A, (b) A ∩ A = A, 2. Identity laws, (a) A ∪ φ = A, (b) A ∩ U = A, 3. Commutative laws, (a) A ∪ B = B ∪ A, (b) A ∩ B = B ∩ A, 4. Associative laws, (a) ( A ∪ B) ∪ C = ( A ∪ B) ∪ C, (b) A ∩ ( B ∩ C ) = ( A ∩ B) ∩ C, 5. Distributive laws, (a) A ∪ ( B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C ), (b) A ∩ ( B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C ), 6. De-Morgan’s laws, (a) ( A ∪ B)′ = A′ ∩ B′, (b) ( A ∩ B)′ = A′ ∪ B′
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4, , NDA/NA Mathematics, 7. (a) A − B = A ∩ B′, (b) B − A = B ∩ A′, (c) A − B = A ⇔ A ∩ B = φ (d) ( A − B ) ∪ B = A ∪ B, (e) ( A − B) ∩ B = φ, (f) ( A − B) ∪ ( B − A) = ( A ∪ B) − ( A ∩ B), 8. (a) A − ( B ∪ C ) = ( A − B) ∪ ( A − C ), (b) A − ( B ∩ C ) = ( A − B) ∩ ( A − C ), (c) A ∩ ( B − C ) = ( A ∩ B) − ( A ∩ C ), (d) A ∩ ( B ∆ C ) = ( A ∩ B) ∆ ( A ∩ C ), , Important Results, If A, B and C are any three finite sets, then, 1. n( A ∪ B ) = n( A) + n( B ) − n( A ∩ B ), 2. n( A ∪ B ) = n( A) + n( B ), if and only if A ∩ B = φ, 3. n( A − B ) = n( A) − n( A ∩ B ), 4. n( A ∆ B ) = n( A − B ) + n( B − A) = n( A) + n( B ) − 2n( A ∩ B ), 5. n( A ∪ B ∪ C ) = n( A) + n( B ) + n(C ) − n( A ∩ B ), − n( B ∩ C ) − n( A ∩ C ) + n( A ∩ B ∩ C ), 6. n( A′ ∪ B′ ) = n(U ) − n( A ∩ B ), 7. n( A′ ∩ B′ ) = n(U ) − n( A ∪ B ), , Example 4. If in a group of 850 persons, 600 can speak, Hindi and 340 can speak Tamil. Then, the number of persons, can speak both Hindi and Tamil is, (a) 40, (b) 90, (c) 85, (d) 120, Solution (b) Let A and B denote the sets of persons who can, speak Hindi and Tamil, respectively., Then, n( A) = 600 ; n(B) = 340 and n( A ∪ B) = 850, A–B, A, , A∩B, , B–A, B, , n( A ∩ B) = n( A) + n(B) − n( A ∪ B), = 600 + 340 − 850 = 90, Thus, 90 persons can speak both Hindi and Tamil., , Ordered Pair, Two elements a and b listed in a particular order, is, called an ordered pair and it is denoted by ( a , b). In an, ordered pair ( a , b); ‘a’ is regarded as the first element and, ‘b’ the second element., It is evident from the definition that, (i) ( a , b) ≠ ( b, a ) ⇔ a ≠ b, (ii) ( a , b) = ( c, d ) ⇔ a = c, b = d, , Cartesian Product, Let A and B be two non-empty sets. The cartesian, product of A and B, denoted by A × B is defined as the set of, all ordered pairs ( a , b), where a ∈ A and b ∈ B., Symbolically, A × B = {( a , b); a ∈ A and b ∈ B}, Thus, (a, b) ∈ A × B ⇔ a ∈ A and b ∈ B, , Facts Related to Cartesian Product, If A, B and C are non-empty sets, then, 1. (a) A × ( B ∪ C ) = ( A × B ) ∪ ( A × C ), (b) A × ( B ∩ C ) = ( A × B ) ∩ ( A × C ), 2. A × ( B − C ) = ( A × B ) − ( A × C ), 3. A × B = B × A ⇔ A = B, 4. If A ⊆ B ⇒ A × B ⊆ ( A × B ) ∩ ( B × A), 5. If A ⊆ B ⇒ A × C ⊆ B × C, 6. A ⊆ B and C ⊆ D ⇒ A × C ⊆ B × D, 7. ( A × B ) ∩ (C × D ) = ( A ∩ C ) × ( B ∩ D ), 8. A × ( B′ ∪ C ′ )′ = ( A × B ) ∩ ( A × C ), 9. A × ( B′ ∩ C ′ )′ = ( A × B ) ∪ ( A × C ), 10. If n elements are common in A and B, then in A × B and, B × A, n 2 elements will be common., , Relation, Let A and B are two non-empty sets, then a relation R, from A to B is a subset of A × B., Equivalently, any subset of A × B is relation from A to B., Thus, R is a relation from A to B ⇔ R ⊆ A × B, ⇔ R ⊆ {( a , b): a ∈ A, b ∈ B}, If ( a , b) ∈ R , then we write a R, / b which is read as ‘a is, related to b by the relation R’. If ( a , b) ∉R , then we write ab, and we say that a is not related to b by the relation R., , Domain and Range of Relation, If R is a relation from set A to set B, then the set of all, first coordinates of elements of R is called the domain of R,, while the set of all second coordinates of elements of R is, called the range of R,, ∴ Domain ( R ) = { a ∈ A :( a , b) ∈ R for some b ∈ B}, and Range ( R ) = { b ∈ B :( a , b) ∈ R for some a ∈ A}, , Codomain of a relation If R be a relation from, A to B, then B is called the codomain of relation R., , Types of Relation, (i) Empty relation Since, φ ⊂ A × A, it follows, that φ is a relation on A, called the empty or void relation., e.g., Let A = {1, 2}, B = {1, 3}, Let, R = {(1, 1), (1, 3), (2, 1), (2, 3)}, Here, R = A × B, Hence, R is the universal relation from A to B., (ii) Universal relation Since, A × A ⊆ A × A, it, follows that A × A is a relation on A, is called the universal, relation., (iii) Identity relation The relation I A = {( a , a );, a ∈ A} is called the identity relation to A., e.g., If A = { 1, 2, 3}, then the identity relation on A is, given by IA = {(1, 1),( 2, 2),( 3, 3)}.
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5, , Sets, Relations and Functions, , (iv) Inverse relation If R is a relation on A, then, the relation R −1 on A, defined by R −1 = {( b, a ):( a , b) ∈ R } is, called an inverse relation to A., Clearly, domain ( R −1 ) = range (R), and, , range ( R −1 ) = domain (R), , Example 5. Let A = {1, 2, 3} and let R = {(1, 2), (2, 2), (3, 1),, (3, 2)}, then the domain and range of R−1 is, (a) {2, 1}, {1, 2, 3}, (b) {1, 2}, {1, 2, 3}, (c) {1, 2, 3} {2, 3}, (d) {2, 3} {1, 2}, Solution (a) Q A = {1, 2, 3} and R = {(1, 2),(2, 2),(3, 1), (3, 2)}., Then, R being a subset of A × A, it is a relation on A., Clearly, 1R2 ; 2R2; 3R1 and 3R2., ∴ Domain (R) = {1, 2, 3 } and range (R) = {2, 1}, Also,, R −1 = {(2, 1),(2, 2),(1, 3),(2, 3)}, Domain (R −1) = {2, 1} and range (R −1) = {1, 2, 3 }, , Various Types of Relations, (i) Reflexive relations A relation R on a set A is, said to be a reflexive relation, if ( a , a ) ∈ R , ∀ a ∈ R. It should, be noted, if ∃ any a ∈ A such that ( a , a ) ∉R , then R is not, reflexive., e.g., let A = { 1, 2, 3} and R = {(1, 1),( 2, 2)}, Then, R is not reflexive, since 3∈ A but ( 3, 3) ∉R., (ii) Symmetric relations A relation R on a set, A is said to be a symmetric relation on A, if, ( x , y ) ∈ R ⇒ ( y , x ) ∈ R , ∀ x , y ∈ R. i.e., if R related to y, then y, is also R related to x. It should be noted that R is, symmetric, iff R −1 = R., Let A = {1, 2, 3}, Let R1 = {(1, 2),( 2, 1)}, R2 = {(1, 2),( 2, 1),(1, 3),( 3, 1)}, Here, R1 and R2 are symmetric relations on A., (iii) Anti-symmetric relations A relation R on a, set A is said to be an anti-symmetric relation, if ( a , b) ∈ R, and ( b, a ) ∈ R ⇒ a = b. Thus, if a ≠ b,then a may be related to, b or b may be related to a, but never both., e.g., let N be the set of natural numbers. A relation, R ⊆ N × N is defined by xRy iff x divides y ( i. e. , x / y ), Then, xR y , yRx ⇒ x divides y , y divides x., ⇒, x= y, e.g., let A = {1, 2, 3}, Let, R1 = {(1, 2), (1, 3), (1, 1)}, R2 = {(1, 2)}, Here, R1 and R2 are anti-symmetric relations on A., , (iv) Transitive relations A relation R on a set A, , is said to be a transitive relation, if ( a , b) ∈ R ,, ( b, c) ∈ R ⇒ ( a , c) ∈ R., In other words, if a is related to b, b is related to c, then, a is related to c., , Transitivity fails only when there exists a , b and c such, that aRb, bRc but a R/ c., e.g., let A = { 1, 2, 3} and the relation, R = {(1, 2),( 2, 1),(1, 1)}., Then, R is not transitive, since R,, ( 2, 1) ∈ R ,(1, 2) ∈ R but ( 2, 2) ∉R., , Equivalence Relation, A relation R on a set A is said to be an equivalence, relation, if, (i) R is reflexive i.e., ( a , a ) ∈ R , ∀ a ∈ A, (ii) R is symmetric i.e.,( a , b) ∈ R ⇒ ( b, a ) ∈ R , ∀ a , b ∈ A, (iii) R is transitive i.e., ( a , b),( b, c) ∈ R ⇒ ( a , c) ∈ R, , Example 6. The relation ‘>’ on the set R of all real, numbers is, (a) transitive, (b) reflexive, (c) symmetry, (d) anti-symmetric, , Solution (a) The relation ‘>’ on R is transitive, since a > b and, b > c ⇒ a > c., Since, no real number can be greater than itself, the relation, is not reflexive., Also,, a>b ⇒, / b > a,, e.g.,, 3 > 2 but 2 >, / 3., So, it is not symmetric., , Composition of Relations, If R and S are relations from A to B and B to, C respectively, then SoR is a relation from A to C, which is, defined as follows, ( a , c) ∈SoR ⇔ ∃ b ∈ B s.t. ( a , b) ∈ R and ( b, c) ∈S ., This relation is known as composition of R and S., , Function, Let A and B are two non-empty sets, then subset of, A × B i. e. , f is known as a function from A to B iff ∀ a ∈ A ∃, a unique element in B such that ( a , b) ∈ f ., Function is also known as mapping, transformations,, operators., f, Function is denoted by f : A → B or A , → B., , Domain, Codomain and Range of, a Function, Let f : A → B. Then, the set A is known as the domain of, f and the set B is known as the codomain of f. The set of all, f-images of elements of A is known as the image of f or, image set of A under f and is denoted by f ( A)., Thus,, f ( A) = { f ( x ) : x ∈ A} = Range of f., Clearly, f ( A) ⊆ B.
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6, , NDA/NA Mathematics, (b) it is onto i.e., ∀ y ∈ B, there exists y ∈ A such that, f ( x ) = y., , Various Types of Functions, (i) Many-one function Let f : A → B. If two or, more than two elements have the same image in B, then f is, said to be many-one function., e.g., the function f : A → B given by f ( x ) = x 2 is a, many-one function., A, 1, –1, 2, , f, , B, 1, 4, 7, , (ii) One-one function (injective) Let f : A → B., Then, f is said to be one-one function or an injective, if, different elements of A have different images in B., Thus, f : A → B is one-one, ⇔ a ≠ b ⇒ f ( a ) ≠ f ( b),, ∀ a, b ∈ A, ⇔ f ( a ) = f ( b) ⇒ a = b,, ∀ a, b ∈ A, e.g., the function f : A → B given by f ( x ) = 2x is an, one-one function., 1, –1, 3, , f, , 2, –2, 6, 7, , (iii) Onto function (surjective) Let f : A → B. If, every element in B has atleast one preimage in A, then f is, said to be an onto function., Thus, f : A → B is a surjective, iff for each b ∈ B, ∃ a ∈ A, such that f ( a ) = b clearly, f is onto ⇔ range ( f ) = B., 1, –1, 2, , f, , (iv) Into function Let f : A → B. If there exists, even a single element in B having no preimage in A, then f, is said to be an into function., e.g., the function f : A → B given by f ( x ) = x 2 is an into, function., f, , (vii) Identity function Let A be a non-empty, set. Then, the function, defined by IA : A → A. IA ( x ) = x, ∀, x ∈ A, is called an identity function on A., This is clearly a one-one onto function with domain A, and range A., (viii) Equal functions Two functions f and g are, said to be equal, written as f = g, if they have the same, domain and they satisfy the condition f ( x ) = g( x ), ∀ x., (ix) Even and odd functions A function, f : A → B is said to be an even or odd function according as, f ( − x ) = f ( x ), ∀ x ∈ A and f ( − x ) = − f ( x ), ∀ x ∈ A,, respectively., Example 7. Let N be the set of all natural numbers. If, f : N → N defined as f ( x) = 2 x, ∀ x ∈ N, then f is, (a) one-one onto, (b) many-one onto, (c) one-one into, (d) many-one into, , Solution (a) Clearly, each x ∈ N has its unique image 2x ∈ N., So, f is onto., Also, f is one-one, since, f ( x1) = f ( x2) ⇒ 2x1 = 2x2 ⇒ x1 = x2, Hence, f is one-one and onto., , 3, –3, 6, , e.g., the function f : A → B is given by f ( x ) = 3x is an, onto function., , A, 1, –1, 2, , (vi) Constant function Let f : A → B is defined, in such a way that all the elements in A have the same, image in B, then f is said to be a constant function., Thus, f ( x ) = c for every x ∈ A, where c is a fixed number, Clearly, domain of f = A and range of f = { c}., Thus, a function f : A → B is a constant function, if, range of f is a singleton set., e.g., A = {1, 2, 3} and B = {5, 7}, let f : A → B defined by, f ( x ) = 5, ∀ x ∈ A., Then, all the elements in A have the same image in B., So, f is a constant function., , B, 1, 4, 7, , (v) Bijective function A one-one and onto, function is said to be bijective., A bijective function is also known as a one-to-one, correspondence., In other words, a function f : A → B is a bijection, if, (a) it is one-one i.e., f ( x ) = f ( y ) ⇒ x = y, ∀ x , y ∈ A., , (x) Inverse function Let f be a one-one onto, function from A to B., f, x, x = f –1 (y), A, , y, f −1, , y = f(x), B, , Let y be an arbitrary element of B. Then, f being onto,, there exists an element x ∈ A, such that f ( x ) = y., Also, f being one-one, this x must be unique., Thus, for each y ∈ B, there exists a unique element, x ∈ A, such that f ( x ) = y., So, we may define a function, denoted by f − 1 as, f − 1 : B → A, such that f − 1 ( y ) = x ⇔ f ( x ) = y., The above function f − 1 is called the inverse of f., %, , A function f is invertible if and only if f is one-one onto.
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7, , Sets, Relations and Functions, , Example 8. The inverse of the function f : R → R defined by, , Composition of Functions, , f ( x) = 4x − 7 is, x+4, (a), 2, x−4, (c), 5, , Let A, B and C be three non-empty sets. Let f : A → B, and g : B → C. Since, f : A → B, for each x ∈ A,there exists a, unique element g[ f ( x )] of C. Thus, for each x ∈ A, there is, associated a unique element g[ f ( x )] of C. Thus, from f and, g, we can define a new function from A to C. This function is, called the product or composite of f and g, denoted by gof, and defined by, ( gof ) : A → C such that ( gof ) ( x ) = g[ f ( x )] for all x ∈ A., , x+7, (b), 4, (d) does not exist, , Solution (b) We have, f ( x) = 4x − 7, x ∈R, f is one-one Let x1, x2 ∈R, Now,, f ( x1) = f ( x2), ⇒, 4x1 − 7 = 4x2 − 7, ⇒, 4x1 = 4x2 ⇒ x1 = x2, ∴f is one-one., f is onto Let y ∈codomain R, Let, f ( x) = y ⇒ 4x − 7 = y, y+7, ⇒, x=, ∈R, 4, ∴ f is onto., Hence, f is invertible i.e., f −1 exists., , A, x, , ⇒, ⇒, , y+7, 4, y+7, −1, f (y) =, 4, x+ 7, −1, f ( x) =, 4, , B, f(x), , g, , C, g(f(x)), , gof, , Properties of Composition of Functions, , To find f −1 : f ( x) = y ⇒ 4x − 7 = y, ⇒, , f, , x=, , [Q f ( x) = 4 ⇔ x = f −1(y)], , 1. The product of any function with the identity, function is the function itself., 2. The product of any invertible function f with its, inverse function f − 1 is an identity function., 3. Composite of functions is associative., 4. Let f : A → B and g : B → A, such that gof is an, identity function on A and fog is an identity function, on B. Then, g = f − 1., 5. Let f : A → B and g : B → C be a one-one onto, functions., Then, gof is also one-one onto and ( gof )− 1 = f − 1og− 1, , Comprehensive Approach, n, , n, , n, , n, n, n, , n, n, n, , n, , n, n, , A natural number p is a prime number, if ‘p’ is greater than one and, its factors are one and ‘p’ only., Finite sets are equivalent sets only when they have equal numbers, of elements., Equal sets are equivalent sets, but equivalent sets may not be equal, sets., Number of proper subsets of a set containing ‘n’ elements is2 n − 1., If A ⊆ B, we may haveB ⊆ A but, if A ⊂ B,we cannot haveB ⊂ A., {x, y} and {y , x} are equals sets but ( x, y) and (y , x) are not equal, ordered pair., If ‘A’ is any set, then A ⊆ A is true but A ⊂ A is false., Total number of relations from set A to set B is equal to 2 n ( A )⋅n (B)., The universal relation on a non-empty set is always reflexive,, symmetric and transitive., The identity relation on a non-empty set is always reflexive,, symmetric and transitive., The identity relation on a non-empty set is always anti-symmetric., If R is relation from A to B and S is a relation from B to C, then, (RoS) − 1 = S − 1 oR− 1., , n, , n, n, n, , n, , n, , n, , n, , n, , n, , For two relations R and S, the composite relation RoS, SoR may be, void relation., The product of two even or odd function is an even function., The product of an even and an odd function is an odd function., Every function f(x) can be expressed as the sum of an even and an, odd function., If A and B have n and m distinct elements respectively, then the, number of mappings from A to B is equal to mn ., If A and B have n equal number of distinct elements, then the, number of mappings from A to B is equal to nn ., If A and B have n equal number of distinct elements, then the number, of objective functions from A to B is equal to n !., The number of one-one functions that can be defined from a finite set, A into a finite set B is n (B) Pn ( A ), if n (B) ≥ n ( A) and zero, otherwise., The number of onto functions that can be defined from a finite set, A containing n elements on finite set B containing 2 elements, = 2 n − 2., If two curves do not intersect each other, then intersection of two, sets is a null set.
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Exercise, Level I, 1. If A × B = {(1, 1), (1, 2), (1, 3), ( 2, 1),( 2, 2), ( 2, 3)}, then A, is equal to, (a) {1, 2}, (b) {1, 2, 3}, (c) {2, 3}, (d) None of these, 2. If A = {1, 2, 3} and B = {3, 4},then ( A ∪ B) × ( A ∩ B) is, (a) {3, 3}, (b) {(1, 3), (2, 3), (3, 3), (1, 4), (2, 4), (3, 4)}, (c) {(1, 3), (2, 3), (3, 3)}, (d) {(1, 3), (2, 3), (3, 3), (4, 3)}, 3. Which of the following is correct?, (a) A ∩ B ⊂ A ∪ B, (b) A ∩ B ⊆ A ∪ B, (c) A ∪ B ⊂ A ∩ B, (d) None of these, 4. If n( A) = 8, n( A ∩ B) = 2, then n( A − B) is equal to, (a) 8, (b) 2, (c) 6, (d) 9, 5. A set contains n elements. Then, the power set, contains, (b) n elements, (a) n 2 elements, (d) 2n elements, (c) ( 2n − 1) elements, 6. If A = { 1, 2, 3, 4} and B = { 5, 6, 7}, then number of, relations from A to B is equal to, (a) 24, (b) 23, (c) 27, (d) 212, 7. If φ is a null set, then which one of the following is, correct?, (NDA 2011 II), (a) φ = 0, (b) φ = { 0}, (c) φ = { φ }, (d) φ = { }, 8. If A = {a, b, c}, then what is the number of proper, subsets of A?, (NDA 2011 II), (a) 5, (b) 6, (c) 7, (d) 8, 9. If A = { 1, 2, 5, 6} and B = { 1, 2, 3}, then what is, (NDA 2011 I), ( A × B) ∩ ( B × A) equal to?, (a) {(1, 1), (2, 1), (6, 1), (3, 2)}, (b) {(1, 1), (1, 2), (2, 1), (2, 2)}, (c) {(1, 1), (2, 2)}, (d) {(1, 1), (1, 2), (2, 5) (2, 6)}, 10. If E is the universal set and A = B ∪ C , then the set, ( E − ( E − ( E − ( E − ( E − A)))) is the same as the set, (b) B ∪ C, (a) Bc ∪ C c, (c) Bc ∩ C c, (d) B ∩ C, 11. If a set A contains 3 elements and another set B, contains 6 elements, then the number of elements in, A ∪ B would be, (a) 9, (b) either 8 or 9, (c) either 7 or 8 or 9, (d) either 6 or 7 or 8 or 9, 12. In a class of 100 students, 70 have taken Science, 60, have taken Mathematics, 40 have taken both Science, and Mathematics. The number of students who have, , not taken Science or Mathematics or both Science, and Mathematics, is equal to, (a) 90, (b) 10, (c) 30, (d) 20, 13. The relation R on a set A = { 1, 2, 3, 4} is defined as, {(1, 1), (1, 3), ( 2, 2), ( 2, 3), ( 3, 1) ( 3, 2)}. Then, R is, (a) reflexive, (b) symmetric, (c) anti-symmetric, (d) transitive, 14. If A, B and C are non-empty sets such that A ∩ C = φ,, then what is ( A × B) ∩ (C × B) equal to? (NDA 2011 I), (a) A × C, (b) A × B, (c) B × C, (d) φ, 15. If A = { 4n + 2|n is a natural number} and B = { 3n|n, is a natural number}, then what is ( A ∩ B) equal to?, (a) { 12n 2 + 6n|n is a natural number} (NDA 2011 I), (b) { 24n − 12|n is a natural number}, (c) { 12n 2 − 6n|n is a natural number}, (d) { 12n − 6|n is a natural number}, 16. If A and B are two disjoint sets, then which one of the, following is correct?, (NDA 2010 II), (a) A − B = A − ( A ∩ B) (b) B − A′ = B ∩ A, (c) A ∩ B = ( A − B) ∩ B (d) All of these, 17. If the cardinality of a set A is 4 and that of a set B is 3,, then what is the cardinality of the set A∆B?, (a) 1, (NDA 2010 II), (b) 5, (c) 7, (d) Cannot be determined as the sets A and B are not, given., 18. If A and B are finite sets, then which one of the, following is the correct equation?, (a) n( A − B) = n( A) − n( B), (b) n( A − B) = n( B − A), (c) n( A − B) = n( A) − n( A ∩ B), (d) n( A − B) = n( B) − n( A ∩ B), [n( A) denotes the number of elements in A], 19. Which one of the following is correct?, The relation R = {(1, 1), ( 2, 2), ( 3, 3)} on, A = { 1, 2, 3} is, (a) only reflexive, (b) only symmetrics, (c) only transitive, (d) reflexive, symmetric and transitive, , a, , set, , 20. If X = {multiples of 2}, Y = {multiples of 5},, Z = {multiples of 10}, then X ∩ (Y ∩ Z ) is equal to, (a) {multiples of 10}, (b) {multiples of 5}, (c) {multiples of 2}, (d) {multiples of 20}
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9, , Sets, Relations and Functions, 21. Let X be any non-empty set containing n elements., Then, what is the number of relations on X?, (a) 2n, , 2, , (b) 2n, , (c) 22n, , (d) n 2, , 22. Which of the following is a null set?, (a) { x :|x|< 1, x ∈ N }, (b) { x :|x|= 5, x ∈ N }, (c) { x : x 2 = 1, x ∈ Z }, (d) { x : x 2 + 2x + 1 = 0, x ∈ R }, 23. The order of a set A is 3 and that of a set B is 2. What, is the number of relations from A to B? (NDA 2010 I), (a) 4, (b) 6, (c) 32, (d) 64, 24. The set of intelligent students in a class is, (a) a null set, (b) a singleton set, (c) a finite set, (d) not a well defined collection, 25. If A = P {1, 2}, where P denotes the power set, then, which one of the following is correct?, (NDA 2010 I), (a) { 1, 2} ⊂ A, (b) 1 ∈ A, (c) φ ∉ A, (d) { 1, 2} ∈ A, 26. If A and B are any two sets, then what is the value of, (NDA 2012 I), A ∩ ( A ∪ B) ?, (a) Complement of A, (b) Complement of B, (c) B, (d) A, 27. Let A = { x : x is a digit in the number 3591}, B = { x : x ∈ N , x < 10}, which of the following is not, correct?, (a) A ∩ B = { 1, 3, 5, 9}, (b) A − B = φ, (c) B − A = { 2, 4, 6, 7, 8}, (d) A ∪ B = { 1, 2, 3, 5, 9}, 28. If A = {( x , y ) : y = ex , x ∈ R }, and B = {( x , y ) : y = e− x , x ∈ R }, then A ∩ B is, (a) empty set, (b) singleton set, (c) not a set, (d) None of these, 29. The relation R defined on set A = { x :| x|< 3, x ∈ Z } by, R = {( x , y ) : y =| x|} is, (a) {( − 2, 2), ( − 1, 1), ( 0, 0), (1, 1), ( 2, 2)}, (b) {( − 2, − 2), ( − 2, 2), ( − 1, 1),( −1, − 1), ( 0, 0),, (1, − 2),(1, 2), ( 2, − 1), ( 2, − 2)}, (c) {( 0, 0),(1, 1), ( 2, 2)}, (d) None of the above, 30. Let A = { 2, 3, 4, 5} and R = {( 2, 2),( 3, 3), ( 4, 4),( 5, 5)}, be a relation in A. Then, R is, (a) reflexive, (b) symmetric, (c) transitive, (d) None of these, 31. For non-empty subsets A, B and C of a set X such that, A ∪ B = B ∩ C, which one of the following is the, strongest inference that can be derived? (NDA 2007 I), (a) A = B = C, (b) A ⊆ B = C, (c) A = B ⊆ C, (d) A ⊆ B ⊆ C, , 32. The function f ( x ) =, , x, x +1, 2, , from R to R is, (NDA 2010 I), , (a) one-one as well as onto, (b) onto but not one-one, (c) neither one-one nor onto, (d) one-one but not onto, 33. Let A = { −1, 2, 5, 8}, B = { 0, 1, 3, 6, 7} and R be the, relation ‘is one less than’ from A to B, then how many, elements will R contain?, (NDA 2009 II), (a) 2, (b) 3, (c) 5, (d) 9, 34. If n( A) = 115, n( B) = 326, n( A − B) = 47, then what is, (NDA 2009 II), n( A ∪ B) equal to?, (a) 373, (b) 165, (c) 370, (d) 394, 35. If P ( A) denotes the power set of A and A is the void, set, then what is number of elements in, (NDA 2009 II), P { P { P { P ( A)}}}?, (a) 0, (b) 1, (c) 4, (d) 16, 36. If N a = { ax| x ∈ N }, then what is N 12 ∩ N 8 equal to?, (NDA 2009 II), , (a) N 12, , (b) N 20, , (c) N 24, , (d) N 48, , 37. Consider the following Venn diagram, E, A, , B, , If| E| = 42,| A| = 15,| B| = 12 and| A ∪ B| = 22, then, the area represented by the shaded portion in the, above Venn diagram is, (a) 25, (b) 27, (c) 32, (d) 37, 38. If A, B and C are three sets and U is the universal set, such that n (U ) = 700, n ( A) = 200 , n( B) = 300 and, n( A ∩ B) = 100, then what is the value of ( A′ ∩ B′ ) ?, (NDA 2007 II), , (a) 100, , (b) 200, , (c) 300, , (d) 400, , 39. Let A = { x : x is a square of a natural number and x is, less than 100} and B is a set of even natural numbers., What is the cardinality of A ∩ B?, (NDA 2012 I), (a) 4, (b) 5, (c) 9, (d) None of these, 40. If X = {( 4n − 3n − 1)| n ∈ N } and, Y = { 9( n − 1)|n ∈ N }, then X ∪ Y equals to, (NDA 2009 II), , (a) X, (c) N, , (b) Y, (d) a null set, , 41. Sets A and B have n elements in common. How many, elements will ( A × B) and ( B × A) have in common?, (NDA 2009 II), , (a) 0, , (b) 1, , (c) n, , (d) n 2
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10, , NDA/NA Mathematics, , 42. What is the number of proper subsets of a given finite, set with n elements?, (NDA 2009 I), (a) 2n − 1, (b) 2n − 2, (c) 2n − 1, (d) 2n − 2, 43. If A, B and C are three sets, such that A ∪ B = A ∪ C, and A ∩ B = A ∩ C, then which one of the following is, correct?, (NDA 2008 II), (a) A = B only, (b) B = C only, (c) A = C only, (d) A = B = C, 44. If A, B and C are three sets, then A − ( B − C )equals to, (NDA 2008 II), , (a) A − ( B ∩ C ), (c) ( A − B) ∪ ( A ∩ C ), , (b) ( A − B) ∪ C, (d) ( A − B) ∪ ( A − C ), , 45. In a class containing 120 students, 65 students drink, tea and 84 students drink coffee. If x students drink, both tea and coffee, then what is the value of x?, (a) 39, (b) 65, (c) 29 ≤ x ≤ 65, (d) 29 ≤ x ≤ 84, 46. If A and B are two sets satisfying A − B = B − A, then, which one of the following is correct?, (NDA 2007 II), (a) A = φ, (b) A ∩ B = φ, (c) A = B, (d) None of these, 47. Two finite sets have m and n elements, respectively., The total number of subsets of the first set is 56 more, than the total number of subsets of the second set., What are the values of m and n, respectively?, (a) 7, 6, (b) 6, 3, (c) 5, 1, (d) 8, 7, , Level II, 1. Consider the following in respect of subsets A and B, of X, I. A ⊆ B ⇔ A ∪ B = A, II. ( A ∪ B)c = Ac ∩ Bc, III. A / B = B / A, IV. A ∪ B = A ∩ B, iff A = B, Which of these are correct?, (a) I and III (b) I and IV (c) II and III(d) II and IV, , 5. What is the range of f ( x ) = cos 2x − sin 2x ?, (a) [2, 4], (b) [− 1, 1], (NDA 2011 I), (d) ( − 2 , 2), (c) [− 2 , 2 ], , 2. Consider the following statements, I. All poets (P) are learned people (L)., II. All learned people (L) are happy people (H)., , 7. Q = (sin θ + cos θ ). Which one of the following is the, correct range of Q?, (a) − 2 ≤ Q < 2, (b) − 2 < Q < 2, (d) None of these, (c) − 2 ≤ Q ≤ 2, , Which one of the following Venn diagrams correctly, represents both the above statements taken, together?, H, , (a), , P, , L, , (b), , 6. Which one of the following is the union of the closed, 1, 1, , sets 2 + , 10 − , n = 1, 2, ... ?, n, n, , (a) [2, 10], (b) (2, 10), (c) [2, 10), (d) (2, 10], , 8. What does the shaded portion of the Venn diagram, given above represent?, , L, , P, , Q, , P, , H, , R, , (c), , P, , L, , (d), , P LH, , H, , 3. If R be a relation on N × N defined by ( a , b) R ( c, d ) if, and only if ad = bc; then R is, (a) an equivalence relation, (b) symmetric and transitive but not reflexive, (c) reflexive and transitive but not symmetric, (d) reflexive and symmetric but not transitive, 4. If A = { 1, 2, 3, 4} and B = { 2, 3, 5}, then identify the, correct relation, among the following from A to B, given by xRy, if and only if x < y, (a) R = {(1, 2), (1, 3), ( 2, 2), ( 2, 3)}, (b) R = {( 3, 2), ( 3, 3), ( 3, 4), ( 3, 5)}, (c) R = {(1, 2), (1, 3), ( 2, 3), ( 2, 5)}, (d) R = {(1, 3), (1, 5), ( 3, 2), ( 4, 2)}, , (a), (b), (c), (d), , (P ∩ Q ) ∩ (P ∩ R), (( P ∩ Q ) − R ) ∪ (( P ∩ R ) − Q ), (( P ∪ Q ) − R ) ∩ (( P ∩ R ) − Q ), (( P ∩ Q ) ∪ R ) ∩ (( P ∪ Q ) − R ), , 9. Consider the following with regard to a relation R on, a set of real numbers defined by xRy if and only if, (NDA 2011 I), 3x + 4 y = 5, I. 0R1, 1, II. 1R, 2, 2 3, III. R, 3 4, , Which of the above are correct?, (a) I and II, (c) II and III, , (b) I and III, (d) I, II and III
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11, , Sets, Relations and Functions, 10. Let M be the set of men and R is a relation ‘is son of’, (NDA 2011 I), defined on M. Then, R is, (a) an equivalence relation, (b) a symmetric relation only, (c) a transitive relation only, (d) None of the above, 11. Let N denote the set of natural numbers and, A = { n 2 : n ∈ N } and B = { n3/ 2 : n ∈ N }. Which one of, the following is correct?, (NDA 2010 II), (a) A ∪ B = N, (b) The complement of ( A ∪ B) is an infinite set, (c) A ∩ B must be a finite set, (d) A ∩ B must be a proper subset of { m 6 : m ∈ N }, 12. The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3),, (1, 3)} on a set A = {1, 2, 3} is, (NDA 2010 II), (a) reflexive, transitive but not symmetric, (b) reflexive, symmetric but not transitive, (c) symmetric, transitive but not reflexive, (d) reflexive but neither symmetric nor transitive, 13. Let R be a set of real numbers and let S be a relation, defined on R as follows, x Sy, if and only if, x 2 + y 2 = 1, Which one of the following statements is correct?, (a) S is a reflexive relation, (b) S is a symmetric relation, (c) S is a transitive relation, (d) S is an anti-symmetric relation, 14. If F ( n ) denotes the set of all divisors of a natural, number n, what is the least value of y satisfying, [F ( 20) ∩ F (16)] ⊂ F ( y )] ?, (a) 2, (b) 1, (c) 4, (d) 6, 15., H, , E, T, , K, , The Venn diagram shown above represents four sets, of people who can speak Telugu (T ), English (E),, Hindi (H ) and Kannada (K ). What does the marked, region represent?, (a) People, who can speak Hindi and Kannada only, (b) People, who can speak English, Telugu and, Kannada only, (c) People, who can speak Hindi and English only, (d) People, who can speak Hindi, English and, Kannada only, 16. The function f ( x ) = ex , x ∈ R is, (a) onto but not one-one, (b) one-one onto, (c) one-one but not onto, (d) neither one-one nor onto, , (NDA 2010 II), , 17. Consider the function f : R → { 0, 1} such that, 1,, f (x) = , 0,, , if x is rational, if x is irrational, , Which one of the following is correct?, (a) The function is one-one into, (b) The function is many-one into, (c) The function is one-one onto, (d) The function is many-one onto, , (NDA 2010 II), , 18. Which one of the following functions f : R → R is, injective?, (NDA 2009 II), 2, (a) f ( x ) =| x|, ∀ x ∈ R, (b) f ( x ) = x , ∀ x ∈ R, (c) f ( x ) = 11, ∀ x ∈ R, (d) f ( x ) = − x , ∀ x ∈ R, 19. If A = { a , b, c} and R = {( a , a ), ( a , b), ( b, c), ( b, b),, ( c, c), ( c, a )} is a binary relation on A, then which one, of the following is correct?, (NDA 2009 I), (a) R is reflexive and symmetric but not transitive, (b) R is reflexive and transitive but not symmetric, (c) R is reflexive but neither symmetric nor, transitive, (d) R is reflexive symmetric and transitive, 20. If α , β , ξ and η are non-empty sets, then, (a) (α × β ) ∪ ( ξ × η) = (α × β ) ∩ ( ξ × η), (b) (α × β ) ∩ ( ξ × η) = (α × ξ ) ∩ (β × η), (c) (α ∩ β ) × ( ξ ∩ η) = (α × ξ ) ∪ (β × η), (d) (α ∩ β ) × ( ξ ∩ η)= (α × η) ∩ (β × ξ ), 21. In a Euclidean plane, which one of the following is, not an equivalence relation?, (a) Parallelism of lines (a line being deemed parallel, to itself), (b) Congruence of triangles, (c) Similarity of triangles, (d) Orthogonality of lines, 22. Which one of the following is correct?, (a) The relation R0 defined on the set of real, numbers as R0 = {( a , b) such that a 2 + b2 = 1 for, all a , b ∈ R } is an equivalence relation, (b) The relation R0 defined on the set of real, 1, numbers as R0 = {( a , b) such that| a − b|≤ for, 3, all a , b ∈ R } is an equivalence relation, (c) The relation I 0 defined on the set of integers as, 2, 2, I1lo I 2 : I1 − 3I1I 2 + 2I 2 = 0 for all I1 , I 2 ∈ I is an, equivalence relation, (d) We define AEcB by the open sentence : A is, cardinally equivalent to B on the family of sets, Then, the relation Ec on family of sets is an, equivalence relation, 23. Let g : R → R be a function such that, g( x ) = 2x + 5., Then, what is the value of g−1( x )?, (NDA 2008 II), x−5, (b) 2x − 5, (a), 2, 5, x 5, (d), (c) x −, +, 2, 2 2
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12, , NDA/NA Mathematics, , 24. If A = { 1, 2, 3, 4} and R = {(1, 1), (1, 3), ( 2, 2), ( 3, 1),, ( 3, 4), ( 4, 3), ( 4, 4)} is a relation on A × A, then which, one of the following is correct?, (NDA 2008 II), (a) R is reflexive, (b) R is symmetric and transitive, (c) R is transitive, but not reflexive, (d) R is neither reflexive nor transitive, 25. The function f : R → R defined by f ( x ) = ( x 2 + 1)35 for, all x ∈ R is, (NDA 2008 II), (a) one-one but not onto, (b) onto but not one-one, (c) Neither one-one nor onto, (d) Both one-one and onto, 26. Let N be the set of integers. A relation R on N is, defined as R = {( x , y )|xy > 0, x , y , ∈ N }. Then, which, one of the following is correct?, (NDA 2007 II), (a) R is symmetric but not reflexive, (b) R is reflexive but not symmetric, (c) R is symmetric and reflexive but not transitive, (d) R is an equivalence relation, 27. Let R be the relation defined on the set of natural, number N as aRb; a, b ∈ N , if a divides b. Then,, which one of the following is correct?, (NDA 2008 I), (a) R is reflexive only, (b) R is symmetric only, (c) R is transitive only, (d) R is reflexive and transitive, 28. Consider the following statements, I. φ ∈{ φ }, II. { φ } ⊆ φ, , (NDA 2008 I), , Which of the statements given above is/are correct?, (a) I only, (b) II only, (c) Both I and II, (d) Neither I nor II, 29. Which one of the following is correct?, (NDA 2008 I), (a) A ∪ P ( A) = P ( A), (b) A ∩ P ( A) = A, (c) A − P ( A) = A, (d) P ( A) − { A} = P ( A), Here, P (A) denotes the power set of a set A., 30. If A, B and C are three finite sets, then [( A ∪ B) ∩ C ]′, equals to, (NDA 2009 I), (a) A′ ∪ B′ ∩ C ′, (b) A′ ∩ B′ ∩ C ′, (c) A′ ∩ B′ ∪ C ′, (d) A ∩ B ∩ C, 31. If A and B are two non-empty sets having n elements, in common, then what is the number of common, elements in the sets A × B and B × A?, (NDA 2012 I), (a) n, (b) n 2, (c) 2n, (d) Zero, 32. If A = { x : x 2 = 1} and B = { x : x 4 = 1}, then A∆B is, equal to, (a) { i , − i }, (b) { − 1, 1}, (c) { − 1, 1, i , − i }, (d) None of these, 33. There are 100 families in a society, 40 families buy, newspaper A, 30 families buy newspaper B, 30, families buy newspaper C, 10 families buy, newspapers A and B, 8 families buy newspapers B, and C, 5 families buy newspaper A and C, 3 families, , buy newspapers A, B and C, then the number of, families, who do not buy any newspaper is, (a) 20, (b) 80, (c) 0, (d) None of these, 34. Two finite sets A and B having m and n elements., The total number of relation from A to B is 64, then, possible values of m and n are, (a) 2 and 4 (b) 2 and 3 (c) 2 and 1 (d) 64 and 1, 35. Suppose A1 , A2 ,... , A30 are thirty sets each having 5, elements and B1 , B2 , ... , Bn are n sets each having 3, 30, , n, , i =1, , j =1, , elements. Let ∪ Ai = ∪ Bj = S and each elements, of S belongs to exactly 10 of Ai ’s and exactly 9 of Bj$ ’ s ., The value of n is equal to, (a) 15, (b) 3, (c) 45, (d) None of these, 36. Universal set,, U = { x x5 − 6x4 + 11x3 − 6x2 = 0}, A = { x x2 − 5 x + 6 = 0 }, B = { x x2 − 3 x + 2 = 0 }, What is the value of ( A ∩ B)′ ?, , (a) {1, 3}, , (b) {1, 2, 3} (c) {0, 1, 3} (d) {0, 1, 2, 3}, , 37. A relation R is defined on the set Z of integers as, follows mRn ⇔ m + n is odd., Which of the following statements is/are true for R?, I. R is reflexive., II. R is symmetric., III. R is transitive., Select the correct answer using the code given below, (a) II only (b) II and III (c) I and II (d) I and III, 38. Which of the following statements is not correct for, the relation R defined by aRb if and only if b lives, within 1 km from a?, (a) R is reflexive, (b) R is symmetric, (c) R is not anti-symmetric, (d) None of the above, 39. What is the region that represents A ∩ B, if, A = {( x , y ) x + y ≤ 4 } and B = {( x , y ) x + y ≤ 0 } ?, (b) {( x , y ) 2x + y ≤ 4 }, (a) {( x , y ) x + y ≤ 2 }, (c) {( x , y ) x + y ≤ 0 }, , (d) {( x , y ) x + y ≤ 4 }, , 40. Let X and Y be two non-empty sets and let R1 and R2, be two relations from X into Y . Then, which one of, the following is correct?, (a), (b), (c), (d), , −1, , −1, , ( R1 ∩ R2 )−1 ⊂ R1 ∩ R2, −1, −1, ( R1 ∩ R2 )−1 ⊃ R1 ∩ R2, −1, −1, ( R1 ∩ R2 )−1 = R1 ∩ R2, −1, −1, −1, ( R1 ∩ R2 ) = R1 ∪ R2, , 41. If a set A contains 4 elements, then what is the, number of elements in A × P ( A)?, (NDA 2008 II), (a) 16, (b) 32, (c) 64, (d) 128
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13, , Sets, Relations and Functions, 42. Let f : R → R be a function defined as f ( x ) = x| x|; for, each x ∈ R, R being the set of real numbers. Which, one of the following is correct?, (NDA 2009 I), (a) f is one-one but not onto, (b) f is onto but not one-one, (c) f is both one-one and onto, (d) f is neither one-one nor onto, 43. If A and B are subsets of a set X, then what is the, value of { A ∩ ( X − B)} ∪ B ?, (NDA 2009 I), (a) A ∪ B (b) A ∩ B (c) A, (d) B, 44. Let U = { x ∈ N : 1 ≤ x ≤ 10} be the universal set, N, being the set of natural numbers. If A = { 1, 2, 3, 4}, and B = { 2, 3, 6, 10}, then what is the complement of, (NDA 2012 I), ( A − B)?, (a) { 6, 10}, (b) { 1, 4}, (c) {2, 3, 5, 6, 7, 8, 9, 10 (b) {5, 6, 7, 8, 9, 10}, , Directions (Q. Nos. 45-47), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 45. Assertion (A) { x ∈ R|x 2 < 0} is not a set. Here, R is, the set of real numbers., Reason (R) For every real number x, x 2 ≥ 0., (NDA 2008 I), , 46. Assertion (A) If A = {1, 2, 3, 4}, then the relation R, defined on A is {(1, 1), (2, 2), (3, 3), (4, 4)} is transitive., Reason (R) A relation R defined on A is transitive iff, aRb, bRc ⇔ cRa ∀ a, b, c ∈ A, 47. Assertion (A) If A = {1, 2, 3}, B = {2, 4}, then the, number of relation from A to B is equal to 26., Reason (R) The total number of relation from set A, to set B is equal to { 2n ( A )⋅ n ( B ) }., , Directions (Q. Nos. 48-51), , Consider a relation R, is defined from a set A = {2, 3, 4, 5} to a set B = {3, 6, 7, 10}, as follows (x, y) ∈R ⇔ x divides y., 48. Express R, as a set of ordered pairs is, (a) {(2, 4), (2, 3)}, (b) {(3, 2), (3, 7), (3, 9)}, (c) {(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)}, (d) None of the above, 49. The domain of R is, (a) {2, 3, 5}, (c) {2, 3}, , (b) {1, 2}, (d) None of these, , 50. The range of R is, (a) {3, 6, 10} (b) {1, 2}, , (c) {2, 3}, , (d) {1, 3}, , 51. The inverse relation R −1 is, (a) {(6, 2), (10, 2), (3, 3), (6, 3), (10, 5)}, (b) {2, 4}, (c) {(3, 2), (1, 3), (4, 5)}, (d) None of the above, , Directions, , (Q. Nos. 52-55) Consider the, universal set S = {0, 1,2, 3, 4, 5, 6, 7, 8, 9} and, A = { 1, 2, 3, 4}, B = {2, 3, 5,6}, C = {2, 3, 7}, then, , 52. The value of A′ is, (a) {0, 5, 6, 7, 8, 9}, (c) {5, 6}, , (b) {2, 3}, (d) {7, 8}, , 53. The value of ( A − B)′ is, (a) {0, 2, 3, 5, 6, 7, 8, 9}, (c) {2, 6, 7, 8, 9}, , (b) {1, 2, 5}, (d) None of these, , 54. The value of A′∩ B is, (a) {2, 3}, (b) {5, 6}, , (c) {1, 3}, , (d) {1, 4}, , 55. The value of B′ − A′ is, (a) {1, 4}, (b) {2, 3}, , (c) {6, 5}, , (d) {4, 2}, , Directions (Q. Nos. 56-60) Read the following, passage and give answer., (NDA 2011 I), The students of a class are offered three languages, (Hindi, English and French). 15 students learn all the, three languages whereas 28 students do not learn any, language. The number of students learning Hindi and, English but not French is twice the number of students, learning Hindi and French but not English. The, number of students learning English and French but, not Hindi is thrice the number of students learning, Hindi and French but not English. 23 students learn, only Hindi and 17 students learn only English. The, total number of students learning French is 46 and the, total number of students learning only French is 11., 56. How many students learn precisely two languages?, (a) 55, (b) 40, (c) 30, (d) 13, 57. How many students learn atleast two languages?, (a) 15, (b) 30, (c) 45, (d) 55, 58. What is the total strength of the class?, (a) 124, (b) 100, (c) 96, (d) 66, 59. How many students learn English and French?, (a) 30, (b) 43, (c) 45, (d) 73, 60. How many students learn atleast one language?, (a) 45, (b) 51, (c) 96, (d) None of these, 61. For a set A, consider the following statements, I. A ∪ P ( A) = P ( A), (NDA 2010 I), II. { A} ∩ P ( A) = A, III. P ( A) − { A} = P ( A), Where P denotes power set., Which of the statements given above is/ are correct?, (a) I only, (b) II only, (c) III only, (d) I, II and III
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Answers, Level I, 1., 11., 21., 31., 41., , (a), (d), (a), (d), (d), , 2., 12., 22., 32., 42., , (d), (b), (a), (d), (c), , 3., 13., 23., 33., 43., , (b), (b), (d), (b), (b), , 4., 14., 24., 34., 44., , (c), (d), (d), (a), (c), , 5., 15., 25., 35., 45., , (d), (d), (d), (d), (c), , 6., 16., 26., 36., 46., , (d), (d), (d), (c), (c), , 7., 17., 27., 37., 47., , (d), (d), (d), (a), (b), , 8., 18., 28., 38., , (c), (c), (b), (c), , 9., 19., 29., 39., , (b), (a), (a), (a), , 10., 20., 30., 40., , (c), (a), (a), (b), , 2., 12., 22., 32., 42., 52., , (d), (a), (d), (a), (c), (a), , 3., 13., 23., 33., 43., 53., , (a), (b), (a), (a), (a), (a), , 4., 14., 24., 34., 44., 54., , (c), (c), (d), (b), (c), (b), , 5., 15., 25., 35., 45., 55., , (c), (d), (c), (c), (a), (a), , 6., 16., 26., 36., 46., 56., , (b), (c), (d), (c), (a), (c), , 7., 17., 27., 37., 47., 57., , (c), (d), (d), (a), (a), (c), , 8., 18., 28., 38., 48., 58., , (b), (d), (d), (c), (c), (a), , 9., 19., 29., 39., 49., 59., , (c), (c), (c), (c), (a), (a), , 10., 20., 30., 40., 50., 60., , (d), (d), (c), (d), (a), (c), , Level II, 1., 11., 21., 31., 41., 51., 61., , (d), (d), (d), (b), (c), (a), (b), , Hints & Solutions, Level I, 1. If (a , b) ∈ A × B, then a ∈ A , b ∈ B, so a = 1, 2, A = {1, 2}, 2. ( A ∪ B) = {1, 2, 3, 4} and A ∩ B = { 3}, ∴ ( A ∪ B) × ( A ∩ B) = {(1, 3), (2, 3), (3, 3), (4, 3)}, 3. If A = B, then A ∩ B = A ∪ B, so A ∩ B ⊆ A ∪ B., 4. n ( A − B) = n ( A ) − n ( A ∩ B) = 8 − 2 = 6, 5. The power set is a set of all subsets. So, it contain 2n, elements., 6. Given, n ( A ) = 4, n (B) = 3, We know that, the total number of relations from two, finite sets A to B is given by = 2n( A)⋅n(B ) = 24 × 3 = 212., 7. If φ is a null set, then its other representation is { }., i.e.,, φ = { }., 8. Given, A = { a , b, c}, Number of subsets of A = 2n = 23 = 8, {where, n = number of elements in A, n = 3}, Proper subset of A = 2n − 1 = 8 − 1 = 7, ∴, 9. Q A = {1, 2, 5, 6} and B = {1, 2, 3}, ∴ A × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3),, (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}, and B × A = {(1, 1), (1, 2), (1, 5), (1, 6), (2, 1), (2, 2), (2, 5), (2, 6), (3, 1), (3, 2), (3, 5), (3, 6)}, ⇒ ( A × B) ∩ (B × A ) = {(1, 1), (1, 2), (2, 1), (2, 2)}, 10. We have, A = B ∪ C, ∴ E − (E − (E − (E − (E − A )))), = E − (E − (E − (E − A c ))), = E − (E − (E − A )), = E − (E − A c ) = E − A = A c, ⇒, A c = (B ∪ C )c, ∴, A c = Bc ∩ C c, , …(i), , [Q E − A c = A ], [from Eq. (i)], , 11. Given that, there are two sets A and B, which contains 3, and 6 elements each., ∴ We have, n ( A ∪ B) = n ( A ) + n (B) − n ( A ∩ B), = 3 + 6 − 0 = 9 [Q n ( A ) = 3, n (B) = 6], or, n ( A ∪ B) = 3 + 6 − 1 = 8, or, n ( A ∪ B) = 3 + 6 − 2 = 7, or, n ( A ∪ B) = 3 + 6 − 3 = 6, [Q These are 4 possibilities of n ( A ∩ B)], 12. Total number of students = 100, Number of students having Science = 70, Number of students having Mathematics = 60, Number of students having both Science and, Mathematics = 40, Now, number of students having Science only, = 70 − 40 = 30, Number of students having Mathematics only, = 60 − 40 = 20, Thus, number of students having Science only,, Mathematics only and both subjects = 30 + 20 + 40 = 90, Number of students, who have not taken Science, Mathematics both of the subjects, = 100 − 90 = 10, 13. We have, A = {1, 2, 3, 4}, and, R = {(1, 1), (1, 3), (2, 2), (2, 3), (3, 1), (3, 2)}, (i) R is not reflexive, since (3, 3) ∈ R., (ii) R is symmetric, since for each (a , b) ∈ R, we have (b, a ) ∈ R e.g., (2, 3) ∈ R ⇒ (3, 2) ∈ R,, (iii) R is not transitive, since for any (a , b) ∈ R and, (b, c) ∈ R, we do not find (a , c) ∈ R e.g., (1, 3) ∈ R,, (3, 2) ∈ R but (1, 2) ∉ R., (iv) R is not anti-symmetric, since for any (a , b) ∈ R and, (b, a ) ∈ R, we do not have a = b., e.g., (1, 3) ∈ R, (3, 1) ∈ R but 1 ≠ 3., Hence, R is only symmetric.
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15, , Sets, Relations and Functions, 14. Q A, B and C are non-empty sets, such that A ∩ C = φ., ∴ ( A × B) ∩ (C × B) = ( A ∩ C ) × B = φ × B = φ, 15. Q A = {4n + 2 : n ∈ N }, = {6, 10, 14, 18, 22, 26, 30, K }, and, B = {3n : n ∈ N }, = {3, 6, 9, 12, 15, 18, 21, 24, K }, ∴, A ∩ B = {6, 18, 30, K }, = {6 + (n − 1)12|n ∈ N }, = {12n − 6|n ∈ N }, 16. Q A ∩ B = φ, ∴ A − B = A − ( A ∩ B), Now,, B − A′ = φ, and, B∩ A=φ, or, B − A′ = B ∩ A, and, ( A − B) ∩ B = A ∩ B, , (given), (Q A − B = A ), , 17. Since, the sets A and B are not known, then cardinality, of the set A∆B cannot be determined., 18. If A and B are finite sets, then, n ( A − B) = n ( A ) − n ( A ∩ B), 19. Q A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3)}, This relation is reflexive, since 1 R1, 2 R2 and 3 R3., 20. We have, X = {multiples of 2}, Y = {multiples of 5} and Z = {multiples of 10}, ∴ X ∩ (Y ∩ Z ), = X ∩ {multiples of 5} ∩ {multiples of 10}, = X ∩ {multiples of 10}, = {multiples of 2} ∩ {multiples of 10}, = {multiples of 10} = Z, ∴ X ∩ (Y ∩ Z ) = Z, , 29. A = { − 2, − 1, 0, 1, 2},, Since, R = {(x, y ): y =|x|}, ∴ − 2R 2, − 1R 1, 0R 0, 1R 1 and 2R2 satisfies relation, y =| x|on A., 30. A = {2, 3, 4, 5}, if a ∈ A ,then (a , a ) ∈ R for every a.So, R is, reflexive., 31. Q A ∪ B = B ∩ C, From above strong inference is A ⊆ B ⊆ C., e.g., A = { a }, B = { a , b},C = { a , b, c}, A ∪ B = { a , b}, B ∩ C = { a , b}, x, 32. Given function, f (x) = 2, x +1, For the function f (x) is one-one, f (x1 ) = f (x2), x1, x2, ⇒, =, x12 + 1 x22 + 1, ⇒, x1x22 + x1 = x12x2 + x2, ⇒, x1x2(x1 − x2) − (x1 − x2) = 0, ⇒, (x1 − x2)(x1x2 − 1) = 0, ⇒, x1 = x2 and x1x2 ≠ 1, So, the function is one-one., x, Let, ⇒ yx2 + y = x, y= 2, x +1, ⇒, ⇒, , 21. Since, number of elements in X be n , then the number, 2, of relations on X and 2n ., 22. | x| < 1 ⇒ − 1 < x < 1 (In this interval no natural number, will satisfy it). So, it is a null set., 23. Q n ( A ) = 3 and n (B) = 2, ∴ Number of relations from A to B = 2[ n( A) × n(B )], = 2(3 × 2) = 26 = 64, 24. Since, intelligency is not defined for students in a class., i.e., Not a well defined collection., 25. A = P ({1, 2}) = { φ , {1}, {2}, {1, 2}}, From above, it is clear that, {1, 2} ∈ A, , B, , +, –∞, , –, –1/2, , 1/2, , +, +∞, , 1 1, y∈ − ,, ~ {0}, 2 2 , 1 1, Here, the range of f (x) = − ,, ~ {0}, 2 2 , and, codomain = R, Range ≠ Codomain, ⇒, ∴ f (x) is not onto., Hence, f (x) is one-one but not onto., 33. Given, A = { −1, 2, 5, 8} and B = {0, 1, 3, 6, 7}, , 26. Here, A and B are any two sets and U be the universal, set., A, , x2 y − x + y = 0, 1 ± 1 − 4 y2, x=, ; 1 − 4 y2 ≥ 0, 2y, (1 − 2 y)(1 + 2 y) ≥ 0, (2 y − 1)(2 y + 1) ≤ 0, , U, , A ∩ (A ∪B), , 27. A = {3, 5, 9, 1}, B = {1, 2, 3, 4, 5, 6, 7, 8, 9}, , Clearly, option (d) is not correct., 28. y = ex and y = e− x only common solution at (0, 1), so, A ∩ B is singleton set., , ∴, , R = {( −1, 0), (2, 3), (5, 6)}, , (Q R = A is one less than from B), Hence, total number of elements in R is 3., , 34. Now, n ( A − B) = n ( A ) − n ( A ∩ B), ⇒, 47 = 115 − n ( A ∩ B), ⇒, n ( A ∩ B) = 68, ∴, n ( A ∪ B) = n ( A ) + n (B) − n ( A ∩ B), = 115 + 326 − 68 = 373, , 35. The number of elements in power set of A is 1., , [Q P ( A ) = 20 = 1], , ∴, ⇒, ⇒, , P{ P ( A )} = 21 = 2, P{ P{ P ( A )}} = 22 = 4, P{ P{ P{ P ( A )}}} = 24 = 16
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16, , NDA/NA Mathematics, , 36. Given, N a = { ax| x ∈ N }, ∴, N 12 = {12, 24, 36, 48, K }, and, N 8 = {8, 16, 24, K }, ∴, N 8 ∩ N 12 = {24, 48, K }, = N 24, 37. From the given figure clearly,, Area of shaded portion = ( A ∪ B)c + ( A ∩ B), We are given, | E | = 42,| A | = 15,| B| = 12 and |( A ∪ B)| = 22, Now,|( A ∪ B)|c = | E | − |( A ∪ B)|= 42 − 22 = 20, Also, we know that, |( A ∪ B)| = | A | + | B| − |( A ∩ B)|, ⇒|( A ∩ B)| = | A | + | B| − |( A ∪ B)|= 15 + 12 − 22 = 5, ∴ Area of shaded portion = 20 + 5 = 25, 38. Given that, n (U ) = 700, n ( A ) = 200, n (B) = 300, and, n ( A ∩ B) = 100, We know that,, n ( A ∪ B) = n ( A ) + n (B) − n ( A ∩ B), = 200 + 300 − 100 = 400, Now, n ( A′ ∩ B′ ) = n (U ) − n ( A ∪ B), = 700 − 400 = 300, 39. Given,, A = {1, 4, 9, 16, 25, 36, 49, 64, 81}, and, B = {2, 4, 6, ...}, Now,, A ∩ B = {4, 16, 36, 64}, ∴ The cardinality of (A ∩ B), = Number of elements in (A ∩ B) = 4, 40. X = {(4n − 3n − 1)| n ∈ N }, and, Y = {9 (n − 1)| n ∈ N }, ⇒, X = {0, 9, 54, K }, and, Y = {0, 9, 18, 27, 36, 54, K }, ∴, X ∪ Y = {0, 9, 18, 27, 36, 54, ... } = Y, 41. The total number of elements common in ( A × B) and, (by property), (B × A ) is n 2., 42. Total number of proper subsets of a finite set with n, elements = 2n − 1, (by property), 43. ∴ Given that, A ∪ B = A ∪ C, and, A∩ B= A∩C, Then, let A = { a , b}, B = { a , c}, and, C = { a , c}, , ⇒ A ∪ B = { a , b, c}, A ∪ C = { a , b, c}, A ∩ B = { a }, A ∩ C = { a },, ∴ For B = C, ⇒, A∪ B= A∪C, and, A∩ B= A∩C, 44., , A, , B, , C, , In the above Venn diagram, shaded area shows, A − (B − C ) ., A, , B, , C, , In the above Venn diagram, horizontal lines mean, ( A − B) and vertical lines mean ( A ∩ C )., Total shaded portion = ( A − B) ∪ ( A ∩ C ), ∴ ( A − B) ∪ ( A ∩ C ) = A − (B − C ), 45. n (T ) = 65, n (C ) = 84,, n (T ∪ C ) = 120 and n (T ∩ C ) = x, Q, n (T ∪ C ) = n (T ) + n (C ) − n (T ∩ C ), ⇒, 120 ≥ 65 + 84 − x, x ≥ 149 − 120 ≥ 29, and, x ≤ n (T ) ⇒ x ≤ 65, ∴, 29 ≤ x ≤ 65, 46. Q A and B are two sets satisfying A − B = B − A , which, is possible only, if A = B., 47. Let n ( A ) = m, n (B) = n, The total possible subsets of A and B are 2m and 2n ,, respectively., According to the given condition,, 2m − 2n = 56, n, m− n, ⇒, 2 (2, − 1) = 23 (23 − 1), ⇒, n = 3, m − n = 3 ⇒ m = 6, n = 3, , Level II, 1. We know that by the definition of subsets, if A = B, Then, A ∪ B = A ∩ B, ∀ A , B ∈ X, [Q n ( A ∪ B) = n ( A ) + n (B) − n ( A ∩ B)], and by De-Morgan’s law ( A ∪ B)c = A c ∩ Bc, 2. For first statement i.e., all poets (P) are learned people, (L) and from second statement all poets who learned, people (L) are happy (H)., P L, , L, H, , P LH, , ∴Venn diagram represents both in (d)., 3. R is relation defined on N × N by (a , b) R (c, d ) if and, only if ad = bc, then R is, (i) Reflexive Since, for any a , b ∈ N, ⇒, ab = ba ⇒ (a , b) R (a , b), (ii) Symmetric Let (a , b) R (c, d ) ⇒ ad = bc ⇒ bc = ad, (Q commutativity of natural numbers), ⇒ cb = da, ⇒, (c, d ) R (a , b), (iii) Transitive Let (a , b) R (c, d ) and (c, d ) R (e, f ), ⇒, ad = bc and cf = de, ⇒, ade = bce ⇒ acf = bce
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17, , Sets, Relations and Functions, (cancellation law), ⇒, af = be, ⇒, (a , b) R (e, f ), Hence, from above observations, we conclude that R is, an equivalence relation on N × N ., 4. A = {1, 2, 3, 4} and B = {2, 3, 5}, Now, (x, y) = ordered pair of A × B, where x ∈ A and, y ∈ B., The ordered pairs (x, y) such that x < y are, (1, 2), (1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)., 5. Q f (x) = cos 2x − sin 2x, [Q f (x) = a cos x + b sin x − a + b ≤ f (x) ≤ a + b ], 2, , 2, , 2, , 2, , − 1 + 1 ≤ cos 2x − sin 2x ≤ 1 + 1, − 2 ≤ cos 2x − sin 2x ≤ 2, So, Range of f (x) is [− 2 , 2 ]., 1, 1, , 6. Union of the closed sets 2 + , 10 −, , n = 1, 2, ... is, n, n , , (2, 10)., 7. Q = sin θ + cos θ, Which is of the form Q = a sin θ + b cos θ, then, ⇒, ∴, , Symmetric, Q, 1R2 but 2R1, ∴ R is not a symmetric relation., Transitive, Q, 1R2, 2R3 ⇒ 1R3, ∴ R is a transitive relation., 13. Given that, xSy is defined relation,, if x2 + y2 = 1 and yS x is defined relation, if, y2 + x 2 = 1, ⇒, x2 + y2 = 1, Hence, xS y ⇒ yS x, ∴The relation S is symmetric relation., 14. F (n ) = Set of all divisor of a natural number n., F (20) = {1, 2, 4, 5, 10}, F (16) = {1, 2, 4, 8}, Now,, ∴, ∴, , F (20) ∩ F (16) = {1, 2, 4}, F ( y) = {1, 2, 4}, = set of all divisor of a natural number y, y=4, , − a 2 + b2 ≤ Q ≤ a 2 + b2, , 15. In the given figure marked region represents the people, who can speak Hindi, English and Kannada only., , − 1 + 1 ≤Q ≤ 1 + 1, − 2 ≤Q≤ 2, , 16. It is clear from the graph that f (x) = ex , ∀ x ∈ R is, one-one but not onto. Since, range ≠ Codomain, so f (x) is, into., y, , 8. The shaded portion represents in the given figure is, ((P ∩ Q ) − R) ∪ ((P ∩ R) − Q )., 9. Since, on the set of real numbers, R is a relation defined, 1, 2 3, by xRy if and only if 3x + 4 y = 5 for which 1R and R ., 2, 3 4, 1, 1, i.e.,, 1R, ⇒ 3 ⋅ 1 + 4 ⋅ = 5,, 2, 2, 2 3, 2 3, and, R ⇒ 3 × + ×4 =5, 3 4, 3 4, Hence, both the statements II and III are correct., 10. Q M = Set of men and R is a relation is son of defined on, M., Reflexive relation aRa., Since, a cannot be a son of a., Symmetric relation, aRa ⇒ bRa, which is also not possible., Transitive relation aRb, aRc ⇒ cRa, which is not possible., , y = ex, x, , x′, , y′, , 17. Since, on taking a straight line parallel to x-axis, the, group of given function intersect it at many points., ∴ f (x) is many-one., And as range of f (x) = Codomain, ∴ f (x) is onto., Hence, f (x) is many-one onto., 18. An injective function means one-one., In option (d), f (x) = − x., For every values of x, we get a different value of f., Hence, it is injective., , 11. Q A = { n 2 : n ∈ N } and B = { n3 : n ∈ N }, A = {1, 4, 9, 16, 25, 49, 64, 81, K }, B = {1, 8, 27, 64, 125, K }, A ∩ B = {1, 64, ... }, ∴ A ∩ B must be a proper subset of {m6 : m ∈ N }, , 19. Q (a , a ), (b, b), (c, c) ∈ R, ∴ R is a reflexive relation., But (a , b) ∈ R and (b, a ) ∉ R., ∴ R is not a symmetric relation., Also, (a , b), (b, c) ∈ R, ⇒, (a , c) ∉ R, ∴ R is not a transitive relation., , 12. (a) Q R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}, Reflexive, Q, 1R1, 2R2, 3R3, ∴ R is a reflexive relation., , 20. Let us consider a and b ∈ (α ∩ β ) × (ξ ∩ η), ⇒ a ∈ (α ∩ β ) and b ∈ (ξ ∩ η), ⇒ (a ∈α ) and (a ∈β ) and (b ∈ ξ ) and (b ∈ η), ⇒ (a ∈ α and b ∈ η) and (a ∈ β and b ∈ ξ )
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18, , NDA/NA Mathematics, ⇒ ab ∈ (α × η) and ab ∈ (β × ξ ), ∴ ab ∈ (α × η) ∩ (β × ξ ), ⇒ (α ∩ β ) × (ξ ∩ η) = (α × η) ∩ (β × ξ ), , 21. In a Euclidean plane, orthogonality of lines is not an, equivalence relation., , 28., 29., , 22. We define AE c B by the open sentence : A is cardinally, equivalent to B on the family of sets. Then, the relation, E c on family of set is an equivalence relation, which is, true., , 30., , 23. Let y = 2x + 5, , 31., , ⇒, ∴, , y − 5 = 2x, g −1 (x) =, , ⇒ x=, , x−5, 2, , y−5, = g −1 ( y), 2, , 24. Since, 3 ∈ A, But (3, 3) ∉ R, So, it is not reflexive., and (3, 4) ∈ R and (4, 3) ∈ R, But (3, 3) ∉ R, So, it is also not transitive., Hence, R is neither reflexive nor transitive., 25. Since, f (−1) = f (1) = 235, i.e., two real number 1 and –1 have the same image., So, the function is not one-one and let, y = (x2 + 1)35 ⇒ x = ( y)1/35 − 1, Thus, every real number has no pre image. So, the, function is not onto., Hence, the function is neither one-one nor onto., 26. Q R = {(x, y)| xy > 0, x, y ∈ N }, Reflexive, Q, x, y ∈ N, ∴, x, x ∈ N ⇒ x2 > 0, ∴ R is reflexive., Symmetric, Q, x, y ∈ N, and, xy > 0 ⇒ yx > 0, ∴ R is also symmetric., Transitive, Q, x, y, z ∈ N ⇒ xy > 0, yz > 0 ⇒ xz > 0, ∴ R is also transitive., Thus, R is an equivalence relation., 27. For reflexive, aRa ⇒ a divides a, ∴ R is reflexive., For symmetric, aRb ⇒ a divides b, bRa ⇒ b divides a, which may not be possible., ∴ R is not symmetric., For transitive, aRb ⇒ a divides b ⇒ b = ka, bRc ⇒ b divides c ⇒ c = lb, Now, c = lka, , 32., , 33., , ⇒ a divides c ⇒ aRc ⇒ aRb, bRc ⇒ cRa, Thus, R is transitive., Both the statements are incorrect., A − P ( A) = A, Which is correct as A and P ( A ) are disjoint sets., We know that,, [( A ∪ B) ∩ C ]′ = ( A ∪ B)′ ∪ C′ = ( A′ ∩ B′ ) ∪ C′, = A′ ∩ B′ ∪ C′ (by De-Morgan’s law), Let us consider an example., Let A = { a , b, c} and B = { a , b, c, d }, Here, 3 elements are common in A and B., Now,, A × B = {(a , a ), (a , b), (a , c), (a , d ), (b, a ), (b, b),, (b, c), (b, d ), (c, a ), (c, b), (c, c), (c, d )}, B × A = {(a , a ), (a , b), (a , c), (b, a ), (b, b), (b, c),, (c, a ), (c, b), (c, c), (d , a ), (d , b), (d , c)}, here common element in ( A × B) and (B × A ) is 9 i.e.,, (3) 2., So, in general, if A and B are two non-empty sets having, ‘n’ elements in common, then (n) 2 is the common, elements in the sets A × B and B × A., A = { − 1, 1}, B = { − 1, 1, − i , i }, A − B = φ, B − A = { − i , i }, ∴, ( A − B) ∪ (B − A ) = { − i , i }, ∴n ( A ∪ B ∪ C ) = n ( A ) + n (B) + n (C ) − n ( AB) − n (BC ), , − n(CA) + n( ABC ), = 80 (number of families reading atleast, one newspapers A, B and C), ∴ Total number of families = 100, So, 20 families do not read any newspaper., 34. 2mn = 64 ⇒ mn = 6, ∴ The possible value of m and n are 2 and 3., 35. If elements are not repeated, then number of elements, in A1 ∪ A2 ∪ ...∪ An is 30 × 5. But each element is used, 30 × 5, 10 times so, S =, = 15. Similarly, if elements in, 10, B1 , B2, ... , Bn are not repeated, then total number of, elements is 3n but each element is repeated 9 times so,, 3n, S=, = 15 ⇒ n = 45., 9, 36. U = { x : x5 − 6x4 + 11x3 − 6x2 = 0} = {0, 1, 2 , 3}, A = { x : x2 − 5x + 6 = 0} = {2 , 3}, and, B = { x : x2 − 3x + 2 = 0} = {2 , 1}, A ∩ B = {2 }, ∴ ( A ∩ B)′ = U − ( A ∩ B), = {0, 1, 2 , 3} − {2 } = {0, 1, 3}, 37. Q R is a relation defined on the set 2 of integers as, follows, mRn ⇔ m + n is odd., (I) We know that the sum of two odd and even numbers, is an even number. Thus, it may be reflexive or not., (II) If m and n are numbers, such that, mRn ⇔ m + n is odd.
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19, , Sets, Relations and Functions, Thus, nRm ⇔ n + m is odd., ∴ This relation is symmetric, (III) This relation is not transitive., Therefore, only (II) statement is correct., 38. R is not anti-symmetric., 39. Q A = {(x, y) x + y ≤ 4} and B = {(x, y) x + y ≤ 0}, ∴ A ∩ B = {(x, y) x + y ≤ 0}, , 40. The correct relation is (R1 ∩ R2)−1 = R1−1 ∪ R2−1., 41. Since, the number of elements in set A is 4., ∴ Number of elements in P ( A ) = 24 = 16., So, the number of elements in A × P ( A ) = 4 × 16 = 64., , 42. Q, f (x) = x| x|, If, f (x1 ) = f (x2), ⇒, x1 | x1 | = x2| x2|, ⇒, x1 = x2, ∴ f (x) is one-one., Also, range of f (x) = codomain of f (x), ∴ f (x) is onto., Hence, f (x) is both one-one and onto., , ∴ (2, 6) ∈ R, (2, 10) ∈ R, (3, 3) ∈ R, (3, 6) ∈ R, (5, 10) ∈ R, Thus, R = {(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)}, , and, , 49. Clearly, domain (R) = {2, 3, 5}, 50. Range (R) = {3, 6, 10}, 51. By definition, R−1 = {(6, 2), (10, 2), (3, 3), (6, 3), (10, 5)}, 52. A′ = { x : x ∈S and x ∉ A } = {0, 5, 6, 7, 8, 9}, 53. A − B = { x : x ∈ A and x ∉ B} = {1, 4}, ( A − B)′ = S − ( A − B) = S − {1, 4}, = {0, 2, 3, 5, 6, 7, 8, 9}, 54. A ′ ∩ B = {0, 5, 6, 7, 8, 9} ∩ {2, 3, 5, 6} = {5, 6}, 55. B′ = S − B = S − {2, 3, 5, 6} = {0, 1, 4, 7, 8, 9}, A′ = S − A = S − {1, 2, 3, 4} = {0, 5, 6, 7, 8, 9}, ∴ B′ − A′ = {0, 1, 4, 7, 8, 9 } − {0, 5, 6, 7, 8, 9} = {1, 4}, , Solutions (Q. Nos. 56-60), The total number of students learning French = 46, H, , 43. Q A ⊆ X and B ⊆ X, ∴, {( A ∩ (X − B))} ∪ B, (Q X − B = B′ ), = ( A ∩ B′ ) ∪ B = ( A ∪ B) ∩ (B′ ∪ B), = ( A ∪ B) ∩ X = A ∪ B, , 23, , 2x, , 17, , E, , 15, , x, , 3x, 11, F, , 28, , 44. Given that, U = { x ∈ N : 1 ≤ x ≤ 10}, A = {1, 2, 3, 4} and B = {2, 3, 6, 10}, Now, A − B = {1, 4}, ∴ Complement of (A − B), = ( A − B)′ = U − ( A − B), = {2, 3, 5, 6, 7, 8, 9, 10}, , 56. The number of students learning precisely two, languages, = x + 2x + 3x, = 6x = 6 × 5 = 30, , 45. Both A and R are true and R is the correct explanation, of A., Q x2 is never negative but it can be positive or zero., , 57. The number of students learning atleast two languages, = x + 2x + 3x + 15, = 6x + 15 = 30 + 15 = 45, , 46. Since, {(1, 1), (2, 2), (3, 3), (4, 4)} is identity relation and, identity relation is always equivalence relation. So, it is, transitive., By definition of transitive relation., aRb, bRc ⇔ cRa, ∀ a , b, c ∈ A, ∴ A and R are both correct and R is the correct, explanation of A., , 58. The total strength of the class, = 28 + 23 + 17 + 11 + x + 2x + 3x + 15, = 79 + 6x + 15 = 79 + 30 + 15 = 124, , 47. We know by the property of relation, the total number of, relation from set A to set B is 2n( A)⋅ n(B )., So, Both A and R are true and R is the correct, explanation of A., 48. Recall that a /b stands for ‘a divides b’. For the elements, of the given sets A and B, we find that 2/6, 2/10, 3/3, 3/6, and 5/10., , ⇒, ⇒, , 15 + 11 + x + 3x = 46, 4x = 20 ⇒ x = 5, , 59. The number of students learn English and French, = 15 + 3x = 15 + 15 = 30, 60. The number of students learn atleast one of the, languages, = 124 − 28 = 96, 61. Let A = {1, 2} and { A } = {{1, 2}}, ⇒, P ( A ) = {{1}, {2}, {1, 2}, φ },, Then,, A ∪ P ( A) ≠ P ( A), and, { A } ∩ P ( A ) = {{1, 2}} ∩ {{1}, {2}, {1, 2}, φ }, = {1, 2} = A
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Complex Numbers, Imaginary Numbers, Square root of a negative real number is imaginary, number, while solving equation x 2 + 1 = 0, we get, x = ± − 1 which is imaginary. So, the quantity − 1 is, denoted by ‘i’ called 'iota'. Thus, i = − 1. e.g., −3 , − 5, are imaginary numbers. Each imaginary number can be, expressed as a product of a real number and i., %, , i =, , − 1 , i 2 = − 1, i 3 = − i and i 4 = 1., , %, , If m is an integer, then, i 4 m = 1, i 4 m + 1 = i , i 4 m + 2 = − 1 and i 4 m + 3 = − i ., , %, , i 4m + i 4m + 1 + i 4m + 2 + i 4m + 3 = 0, , %, , 'i' is neither positive, zero nor negative., , Example 1. If a < 0, b > 0, then the value of a ⋅ b is, (a) | a | b i, (c) ab i, , (b) | b | a i, (d) | ab | i, , Solution (a) As we can only multiply the positive values in, square root., ∴, a ⋅ b = −| a| b , as a < 0 and b > 0., = − 1 ⋅ | a| b = i | a| b = | a| b i, , Complex Numbers, A number of the form ( x + iy ), where x and y are real, numbers and i = − 1 is known as complex number. It is, denoted by z, ∴, z = x + iy, x is known as real part of number z and it is denoted by, Re(z), y is known as imaginary part and it is denoted by, Im (z)., %, , %, , The set of all complex numbers is denoted by C. The order, relation ‘ greater than’ and ‘less than’ are not defined in C. Thus,, the inequality like 7i > 3i or i<0 have no meaning., A complex number z is purely real, if Im(z ) = 0 and it is said to be, purely imaginary, if Re(z ) = 0. The complex number 0 = 0 + i 0, is both purely real and purely imaginary., , 2, , Equality of Complex Numbers, Two complex numbers x1 + iy1 and x2 + iy2 are said to, be equal if and only if x1 = x2 and y1 = y2., i.e., z1 = z 2 ⇔ Re ( z1 ) = Re ( z 2 ) and Im ( z1 ) = Im ( z 2 ), , Example 2. The real part of, (a), , 1, 2, , (b), , 1, 3, , 1, is, 1− cos θ + i sin θ, (c) 2, , (d) 3, , 1, 1 − cos θ + i sin θ, 1, =, 2, 2 sin (θ /2) + 2 i sin (θ /2) cos (θ /2), 1, =, 2 sin (θ /2) [(sin (θ /2) + i cos (θ /2)], 1, =, 2 i sin (θ /2) [(cos (θ /2) − i sin (θ /2)], −i, =, [cos (θ /2) + i sin (θ /2)], 2 sin (θ /2), 1, sin (θ /2), 1, is, Thus, real part of, =, 1 − cos θ + i sin θ 2 sin (θ /2) 2, , Solution (a), , Operation of Complex Numbers, Addition of Complex Numbers, Let z1 = x1 + iy1 and z 2 = x2 + iy2 are two complex, numbers, then, z1 + z 2 = x1 + iy1 + x2 + iy2 = ( x1 + x2 ) + i ( y1 + y2 ), ⇒, Re( z1 + z 2 ) = Re( z1 ) + Re( z 2 ), and Im( z1 + z 2 ) = Im( z1 ) + Im( z 2 ), , Properties of Addition of Complex Numbers, 1. Closure property If z1, z 2 ∈C , then z1 + z 2 ∈ C., 2. Commutative law If z1 , z 2 ∈ C, then z1 + z 2 = z 2 + z1., 3. Associative, law, If, then, z1 , z 2 , z3 ∈ C,, ( z1 + z 2 ) + z3 = z1 + ( z 2 + z3 )
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22, , NDA/NA Mathematics, , Solution (a) If − 3 + i x2y and x2 + y + 4 i are conjugate complex, , 11. | z1 ± z 2|2 = ( z1 ± z 2 ) ( z1 ± z 2 ), =| z1|2 +| z 2|2 ± ( z1z 2 + z1z 2 ), , numbers, then, − 3 + ix y = x + y + 4 i, 2, , ⇒, , 2, , =| z1|2 +| z 2|2 ± 2 Re( z1z 2 ), , − 3 + ix y = x + y − 4 i, 2, , 2, , 12. | z1 + z 2| =| z1|2 +| z 2|2, 2, , On comparing real and imaginary parts, we get, − 3 = x2 +y, −4, and, x2y = − 4 or y = 2, x, On putting the value of y in Eq. (i), we get, 4, − 3 = x2 − 2 ⇒ x4 + 3x2 − 4 = 0, x, ⇒, ( x2 + 4) ( x2 − 1) = 0 ⇒ x2 − 1 = 0, , ...(i), ...(ii), , (Q x2 ≠ − 4), , ⇒, x= ±1, From Eq (ii), we get y = − 4 when x = ± 1, ∴, x = 1, y = − 4 or x = − 1, y = − 4, , Example 6. If z is a complex number satisfying the, relation| z + 1| = z + 2 (1 + i ) , then the value of z is, 1, 1, (b) (1 − 4 i), (a) i, 3, 2, 1, 1, (c) (1 + 4 i), (d) i, 2, 2, , Solution (b) Let z = x + iy, , Modulus of a Complex, Number, Let z = x + iy be any complex number. Then,, | z| = ( x 2 + y 2 ) is called the modulus of the complex, number z, where modulus|z| represents distance of z from, origin., Y, , ∴, , | x + iy + 1| = x + iy + 2 (1 + i), , ⇒, , ( x + 1) 2 + y 2 = ( x + 2) + i(y + 2), , ⇒, , ( x + 1) 2 + y 2 = x + 2 and y + 2 = 0, , ⇒, , ( x + 1) 2 + 4 = ( x + 2) 2 and y = − 2, , ⇒, , 2x + 5 = 4x + 4 and y = − 2, 1, x = and y = − 2, 2, 1, z = (1 − 4 i), 2, , ⇒, ∴, , Argument of a Complex Number, , Imaginary axis, , z = (x + iy), y, , x, , Real axis, , X, , Let z = x + iy is a complex number, then argument of, complex number is denoted by arg (z)., y, arg( z ) = tan− 1, ∴, x, Argument of z is not unique. General value of, argument of z is 2nπ + θ., , |z| = x 2 + y 2 = {Re( z )} 2+ {Im( z )} 2, , Y, , Properties of Modulus of Complex Numbers, , Imaginary axis, , O, , 13. | z1 + z 2|2 +|z1 − z 2|2 = 2 {|z1|2 +|z 2|2 }, , If z , z1 , and z 2 are complex numbers, then, 1. | z|≥ 0 ⇒| z| = 0, iff z = 0 and| z|> 0 ,iff| z| ≠ 0, 2. −| z|≤ Re( z ) ≤| z| and −| z|≤ Im( z ) ≤| z|, 3. | z| =| z | =| − z| =| − z |, , O, , z = (x + iy), , θ, Real axis, , X, , 4. zz =| z|2, 5. | z1z 2| =| z1|| z 2|, , Principal Value of Argument, , |z |, z, 6. 1 = 1 , z 2 ≠ 0, z 2 | z 2|, , The value of θ of the argument, which satisfies, inequality − π < θ < π is called the principal value of, argument., Principal, values, of, the, argument, θ , π − θ , − π + θ , − θ according as the complex number, on the 1st, 2nd, 3rd or 4th quadrant., | y|, Here,, θ = tan− 1, | x|, , 7. | z1 ± z 2|≤| z1| +| z 2|, 8. | z1 ± z 2|≥|| z1| −| z 2||, 9. | z n | =| z|n, 10. || z1| −| z 2||≤| z1 + z 2|≤| z1| +| z 2|, , the, the, are, lies
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23, , Complex Numbers, , (arg z = π – θ), , Geometrical Representation, , (arg z = θ), , Let X ′OX and YOY ′ be the coordinate axes and O be, the origin. Then, any complex number z = a + ib = ( a , b), may be represented by a unique point P whose coordinates, are (a, b)., The representation of complex number as points in a, plane forms an Argand diagram., , θ, θ, , θ, θ, (arg z = –π + θ), , (arg z = – θ), , 1 π, =, 1 4, π, e.g., arg ( − 1 − i ) = − π +, 4, π, e.g., arg (1 − i ) = −, 4, π, e.g., arg ( − 1 + i ) = π −, 4, e.g.,, , %, %, %, %, , Y, , arg (1 + i ) = tan− 1, , P (x, y ), y, X´, , X, , x, , O, , Y´, , Argument of 0 is not defined., If z1 = z 2 , then | z1 | = | z 2 | and arg (z1 ) = arg (z 2 ) ., π, π, Argument of purely imaginary number is or − ., 2, 2, Argument of purely real number is 0 or π., , Properties of Argument of Complex Numbers, If z , z1 and z 2 are three complex numbers, then, 1. arg( z ) = − arg( z ), 2. arg( z1z 2 ) = arg( z1 ) + arg( z 2 ), 3. arg( z1 z 2 ) = arg( z1 ) − arg( z 2 ), z , 4. arg 1 = arg( z1 ) − arg( z 2 ), z2 , 5. | z1 + z 2|2 =| z1|2 +| z 2|2 + 2| z1|| z 2| cos (θ1 − θ 2 ), where θ1 = arg( z1 ) and θ 2 = arg( z 2 ), or | z1 + z 2|2 =| z1|2 +| z 2|2 + 2 Re( z1 z 2 ), 6. | z1 − z 2|2 =| z1|2 +| z 2|2 − 2| z1|| z 2| cos (θ1 − θ 2 ), where θ1 = arg( z1 ) and θ 2 = arg( z 2 ), or| z1 − z 2|2 =| z1|2 +| z 2|2 − 2 Re( z1 z 2 ), 7. | z1 + z 2| =| z1 − z 2| ⇔ arg( z1 ) − arg( z 2 ) =, , π, 2, , 8. | z1 + z 2| =| z1| +| z 2| ⇔ arg( z1 ) = arg( z 2 ), z, 9. | z1 + z 2|2 =| z1|2 +| z 2|2 ⇔ 1 is purely imaginary., z2, , The plane on which complex numbers are represented, is known as the Complex plane or Argand’s plane or, Gaussian plane., Let a be a real number. Then, we can write, a = a + i ⋅ 0 = ( a , 0), Clearly, ( a , 0) lies on x-axis., Thus, a purely real number will be represented by, some point on x-axis, therefore x-axis is called the real axis., Similarly, purely imaginary numbers lie on y-axis., So, y-axis is called the imaginary axis., The complex number z = a + ib is known as the affix of, the point ( a , b) which it represents., , Polar Form of a Complex, Number, Let X ′OX and YOY ′ be the coordinate axes. Let, z = a + ib be represented by a point P ( a , b)., Draw PM ⊥ OX . Then, OM = a and PM = b. Join OP., Let OP = r and ∠ XOP = θ. Then, a = r cos θ and, b = r sin θ., ∴, z = a + ib = r (cos θ + i sin θ ), On comparing real and imaginary parts, we get, Y, , Example 7. If z = x + iy satisfies arg( z − 1) = arg( z + 3i),, then the value of ( x − 1) : y is, 1, 1, (b), (a), 4, 2, , (c), , 1, 3, , (d), , P(a, b), r, , 1, 6, , X´, , b, , θ, O a, , M, , X, , Solution (c) arg [( x − 1) + iy ] = arg [( x + i (y + 3)], ⇒, ⇒, ∴, , tan − 1, , y, y+3, = tan − 1, x −1, x, xy = ( x − 1) ( y + 3) ⇒ 3 ( x − 1) = y, x −1 1, =, y, 3, , Y´, , and, , a = r cos θ, b = r sin θ, , ...(i), ...(ii)
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25, , Complex Numbers, If b is negative, then from Eq. (ii) x and y are of, different sign., 1, i.e.,, a + ib = ± [ a 2 + b2 + a ], 2, 1, − i [ a 2 + b2, 2, , , − a ], , , Example 10. If 1, ω and ω 2 are the cube roots of unity, then, the value of (2 − ω) (2 − ω 2) (2 − ω10) (2 − ω11) is, (a) 15, (b) 25, (c) 49, (d) 40, , Solution (c) Now, (2 − ω) (2 − ω 2) (2 − ω10) (2 − ω11), = (2 − ω) (2 − ω 2) (2 − ω 9 ⋅ ω) (2 − ω 9 ⋅ ω 2), = (2 − ω) (2 − ω 2) (2 − ω) (2 − ω 2), = [(2 − ω) (2 − ω 2)]2 = [ 4 − 2 (ω + ω 2) + ω3 ]2, , Example 9. The square root of (3 − 4 i) is, (a) ± (2 − i), , Solution (a) Let, , (b) ± 3i, , (c) ± (3 − i), , 3 − 4 i = x − iy, then, 3 − 4i = ( x − iy), , (3 − 4 i) = ( x − y ) − (2xy) i, , ⇒, ∴, , 2, , Important Results, , 2, , ⇒, and, , = ( 4 + 2 + 1) 2 = 7 2 = 49, , (d) ± ( 4 − i), , 1. x + x + 1 = ( x − ω ) ( x − ω 2 ), 2, , 2, , 2. x 2 − x + 1 = ( x + ω ) ( x + ω 2 ), , 2, , x −y =3, , ...(i), , 3. x 2 + xy + y 2 = ( x − ωy ) ( x − ω 2 y ), , 2xy = 4, , ...(ii), , 4. x 2 − xy + y 2 = ( x + ωy ) ( x + ω 2 y ), , 2, , 5. x 2 + y 2 = ( x + iy ) ( x − iy ), , x2 + y 2 = ( x2 − y 2) 2 + 4x2 y 2, , 6. x 3 + y 3 = ( x + y ) ( x + ωy ) ( x + ω 2 y ), , = 9 + 16 = 5, x2 + y 2 = 5, , ...(iii), , On solving Eqs. (i) and (iii), we get, x2 = 4 and y 2 = 1, ∴, x = ± 2 and y = ± 1, Since, xy > 0 , it follows that x and y are of the same sign., ∴, x = 2 ,y =1, or, x = − 2 ,y = −1, Hence,, 3 − 4 i = ± (2 − i), , Cube Roots of Unity, x = 3 1 ⇒ x3 − 1 = 0, ( x − 1) ( x 2 + x + 1) = 0, −1+ i 3 −1− i 3, Therefore,, x = 1,, ,, 2, 2, If second root be represented by ω , then third root will, be ω 2., ∴ Cube roots of unity are 1, ω , ω 2 and ω , ω 2 are, called the imaginary cube roots of unity., Let, ⇒, , Properties of Cube Roots of Unity, 1. 1 + ω r + ω 2r = 0, if ‘ r ’ is not a multiple of ‘3’, = 3, if ‘ r ’ is a multiple of ‘3’, 2. ω3 = 1 or ω3 r = 1, 3. ω3 r + 1 = ω , ω3 r + 2 = ω 2, 4. It always forms an equilateral triangle, 5. z3 − 1 = ( z − 1) ( z − ω ) ( z − ω 2 ), 6. Roots of equation z 2 + z + 1 = 0 are ω and ω 2, 7. Cube roots of –1 are − 1, − ω , − ω 2, 8. ω = ω 2 and ω 2 = ω, 9. a + bω + cω 2 = 0 ⇒ a = b = c, if a , b and c are real, numbers., , 7. x 3 − y 3 = ( x − y ) ( x − ωy ) ( x − ω 2 y ), 8. x 2 + y 2 + z 2 − xy − yz + zx, = ( x + ωy + ω 2 z ) ( x + ω 2 y + ωz ), = (ωx + ω 2 y + z ) (ω 2 x + ωy + z ), = (ωx + y + ω 2 z ) (ω 2 x + y + ωz ), 9. x 3 + y 3 + z 3 − 3 xyz, = ( x + y + z ) ( x + ωy + ω 2 z ) ( x + ω 2 y + ωz ), , nth Roots of Unity, Let z = 11/ n . Then,, z = (cos 0° + i sin 0° )1/ n, ⇒, ⇒, , ⇒, ⇒, ⇒, , z = (cos 2rπ + i sin 2rπ )1/ n , r ∈ Z, 2rπ, 2rπ, + i sin, , r = 0 , 1 , 2 , ... , ( n − 1), z = cos, n, n, [Using De-Moivre’s theorem], z=e, z=, , i 2rπ, n, , , r = 0, 1, 2, ... , ( n − 1), , i 2π, { e n }r , r, , z = αr , α =, , = 0, 1, 2, ... , ( n − 1), , i 2π, e n, , , r = 0 , 1 , 2 , ... , ( n − 1), , Thus, nth roots of unity are, 1, α , α 2 , ... , α n − 1 , where α = e, , i 2π, n, , = cos, , 2π, 2π, + i sin, n, n, , Properties of nth Roots of Unity, , i 2π, nth roots of unity form a GP with common ratio e n, , 1., ., 2. Sum of nth roots of unity is always zero., 3. Sum of pth powers of nth roots of unity is zero, if p is, not a multiple of n.
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26, , NDA/NA Mathematics, , 4. Sum of pth powers of nth roots of unity is n, if p is a, multiple of n., 5. Product of nth roots of unity is ( − 1)n − 1., , z=, %, , 1 + α + α 2 + ... + α n − 1, 1 − α n 1 − (cos 2 π + i sin 2 π), =, =, =0, 1−α, 1−α, , %, , %, , %, , 1 ⋅ α ⋅ α 2 K α n −1 = (−1) n − 1, %, , Example, , 11. If, , the fourth roots of unity, z1, z 2, z3 and z 4 , then the value of z12 + z 22 + z32 + z 42 is, (a) 4, (b) 0, (c) 1, (d) 5, 1/ 4, , Solution (b) 1, , ⇒, ∴, , 11/ 4, z12, , = (cos 2πr + i sin 2πr), πr, πr, ; r = 0 ,1, 2 , 3, = cos, + i sin, 2, 2, = 1, i , − 1, − i., , Distance between Two Points, Distance between two points P ( z1 ) and Q( z 2 ) = PQ, =| z 2 − z1|, , %, , z1, If z1 , z 2 and z3 will be collinear, then z 2, z3, , Q (z2), , Distance of a point P(z ) from origin = | z |., Three points A(z1 ), B(z 2 ) and C(z3 ) will be collinear, if, AB + BC = AC, i.e.,, | z1 − z 2 | + | z 2 − z3 | = | z1 − z3 |, A (z1), , B (z2), , C (z3), , Section Formulae, If R( z ) divides the joining of P ( z1 ) and Q( z 2 ) in the ratio, m : n., (a) If R( z ) divides the line segment PQ internally,, mz 2 + nz1, then, z=, m+n, (b) If R( z ) divides the line segment PQ externally,, then, , z1, , 1, , z2 1, z3, , =0, , 1, , Equation of Circle, (a) Equation of a circle with centreC( z 0 ) and radius, r is, | z − z 0| = r, , r, C (z0), , + z 22 + z32 + z 42 = 12 + i 2 + ( −1 ) 2 + ( − i) 2 = 0, , Application in Coordinate, Geometry, , %, , z1 + z 2, ⋅, 2, If z1 , z 2 and z3 are the vertices of a triangle, then centroid of the, z + z 2 + z3, triangle = 1, ., 3, If R(z ) is mid–point of PQ, then z =, , are, , 1/ 4, , P (z1), , mz 2 − nz1, m−n, , %, , P (z), , If the centre of circle is at the origin, then equation of circle is, | z | = r., , General Equation of Circle, The general equation of a circle is zz + az + az + b = 0,, where a ∈ C and b ∈ R., Centre of this circle is – a and radius is | a|2 − b., , Example 12. If| z + 1| = 2 | z − 1,, | then the locus described, by the point z in the argand diagram is a, (a) straight line, (b) circle, (c) parabola, (d) None of these, , Solution (b) Given, | z + 1| = 2| z − 1|, Put, ∴, , z = x + iy, | x + iy + 1| = 2| x + iy − 1|, , ⇒, , |( x + 1) + iy| = 2|( x − 1) + iy|, , ⇒, , ( x + 1) 2 + y 2 = 2 [( x − 1) 2 + y 2], x2 + y 2 − 6x + 1 = 0, , Hence, it represents a equation of circle.
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27, , Complex Numbers, , Comprehensive Approach, n, , n, n, , π , if amp( z) > 0, amp( z) − amp( − z) = , − π , if amp( z) < 0, log e( z) = log e| z | + i amp( z), Triangle of the vertices P( z1), Q( z 2) and R( z3 ) will be an equilateral, triangle, if, 1, 1, 1, +, +, =0, z1 − z 2 z 2 − z3, z3 − z1, or, , n, , If z +, , n, , n, , n, , 1., , a + a2 + 4, 1, = a, then maximum value of | z | =, z, 2, −a+, , a2 + 4, , 2, (iz) = − iz , Re(iz) = − Im( z) and Im(iz) = Re( z), If z1 and z 2 are two complex numbers, such that z1 + z 2 or z1z 2 is a, real number, then it is not necessary that z1 and z 2 are conjugate to, each other., The order relation is not defined on the set C of all complex, numbers as it is not a complete ordered field. Thus, the statements, , a + ib +, , 2. a + ib −, , z12 + z 22 + z32 = z1z 2 + z 2z3 + z3 z1, , and minimum value of | z | =, , n, , n, , z1 > z 2 and z1 < z 2 have no meaning unless z1 and z 2 both are, purely real., Since, | z |2 = [Re( z)]2 + [Im( z)]2 , therefore Re ( z) ≤ | z |,, Im( z) ≤ | z |., For any a, b ∈ R, , n, , n, , n, , n, , a − ib = 2 { a 2 + b 2 + a}, a − ib = i 2 { a 2 + b 2 − a}, , The one and only one case in which, | z1 | + |z 2 | + ... + | z n | = | z1 + z 2 + ... + z n |, is that the numbers z1 , z 2 , ...., z n have the same amplitude., The sum and product of two complex numbers are real, simultaneously if and only if they are conjugate to each other., If three points z1 , z 2 and z3 are connected by relation, az1 + bz 2 + cz3 = 0, where a + b + c = 0, then the three points, are collinear., If three complex numbers are in AP, then they lie on a straight line, in the complex plane.
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Exercise, Level I, 1. The value of (1 + i )5 × (1 − i )5 is, (a) − 8, (b) 8i, (c) 8, (d) 32, 2. Inequality a + ib > c + id can be explained only when, (a) b = 0, c = 0, (b) b = 0, d = 0, (c) a = 0, c = 0, (d) a = 0, d = 0, 3. For the complex number z, are from z + z and z z is, (a) a real number, (b) an imaginary number, (c) both are real, (d) both are imaginary numbers, 4. The product of two complex number each of unit, , modulus is also a complex number, of, (a) unit modulus, (b) less than unit modulus, (c) greater than unit number, (d) None of the above, 5. The values of x and y satisfying the equation, (1 + i ) x − 2 i ( 2 − 3i ) y + i, = i are, +, 3+ i, 3− i, (a) x = − 1, y = 3, (b) x = 3, y = − 1, (c) x = 0, y = 1, (d) x = 1, y = 0, 6. The complex number z = x + iy, which satisfy the, z − 5i, equation, = 1 lie on, z + 5i, (a) real axis, (b) the line y = 5, (c) the line y = 3, (d) None of these, 3 + 2 i sin θ, will be real, if θ is, 7., 1 − 2 i sin θ, π, (a) 2nπ, (b) nπ +, 2, (c) nπ, (d) None of these, 8. If α and β are imaginary cube roots of unity, then, 1, is equal to, α4 + β4 +, αβ, (a) 3, (b) 0, (c) 1, (d) 2, 9. The points represented by the complex numbers, 5, 1 + i , − 2 + 3i , i on the Argand diagram are, 3, (a) vertices of an equilateral triangle, (b) vertices of an isosceles triangle, (c) collinear, (d) None of the above, 10. The amplitude of sin, (a), , π, 5, , (b), , 2π, 5, , π, π, , + i 1 − cos is, , 5, 5, π, π, (d), (c), 10, 15, , 11. What is the argument of (1 − sin θ ) + i cos θ? ( i = −1 ), (NDA 2011 II), , π θ, (a), −, 2 2, π θ, (c), −, 4 2, , π θ, (b), +, 2 2, π θ, (d), +, 4 2, , 12. If α and β are the complex cube roots of unity, then, what is the value of (1 + α )(1 + β )(1 + α 2 )(1 + β 2 )?, (NDA 2011 II), , (a) –1, (c) 1, , (b) 0, (d) 4, , 13. If z = 1 + cos, (a) 2 cos, , π, 5, , π, π, + i sin , then| z| is equal to, (NDA 2011 I), 5, 5, π, π, π, (b) 2 sin, (c) 2 cos, (d) 2 sin, 5, 10, 10, , 14. If ω is the imaginary cube root of unity, then, ( 2 − ω + 2 ω 2 )27 is equal to, (NDA 2011 I), 27, 27, 2, (b) − 3 ω, (a) 3 ω, (c) 327, (d) −327, 6, , 3 + i, 15. What is the value of , ?, 3 − i, (a) –1, , (b) 0, , 6i, 16. If x + iy = 4, 20, , −3i, 3i, 3, , ( x − iy ) ?, (a) 3 + i, (c) 3i, 17. If A + iB =, value of A?, (a) – 8, , (c) 1, , (NDA 2010 II), , (d) 2, , 1, −1 , then what is the value of, i, (NDA 2010 II), , (b) 1 + 3i, (d) 0, 4 + 2i, , where i = −1, then what is the, 1 − 2i, (NDA 2012 I), , (b) 0, , (c) 4, , (d) 8, , π, π, 18. If xr = cos r + i sin r , then x1 ⋅ x2 ... ∞ is, 2 , 2 , (a) –3, (b) –2, (c) –1, (d) 0, 19. If x = a + b, y = aα + bβ and z = aβ + bα,where α and, β are complex cube roots of unity, then xyz is equal to, (b) a3 + b3, (a) a 2 + b2, 3 3, (d) a3 − b3, (c) a b, 20. Given that the equation z 2 + ( p + iq ) z + r + is = 0,, where p, q , r and s are real and non-zero root, then, (a) pqr = r 2 + p2s, (b) prs = q 2 + r 2 p, (c) qrs = p2 + s2q, (d) pqs = s2 + q 2r
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29, , Complex Numbers, 21. The cube roots of unity, when represented on the, Argand plane from the vertices of a /an, (a) equilateral triangle, (b) isosceles triangle, (c) right angled triangle (d) None of these, 22. What is the value of ( −1 + i 3 )48?, (a) 1, (c) 224, , (b) 2, (d) 248, ω, , 1, 23. What is the value of 2, 3, , (b) x + iy, , 2ω 2, , 2ω 2, , 4 ω3 , where ω is the, , 3ω, , 6ω, , cube roots of unity?, (a) 0, (c) 2, 1+ x+, 24. If x 2 + y 2 = 1, then, 1+ x−, (a) x − iy, , (NDA 2010 II), , 3, , 4, , (NDA 2010 II), , (b) 1, (d) 3, iy, is equal to, iy, (NDA 2010 I), (c) 2x, (d) − 2iy, , 25. What is the least positive integer n for which, n, 1 + i, , = 1?, 1 − i, (NDA 2010 I), (a) 16, (b) 12, (c) 8, (d) 4, 26. If α is a complex number, such that α 2 + α + 1 = 0,, then what is the value of α31?, (NDA 2009 II), (c) 0, (d) 1, (a) α, (b) α 2, , 32. If ω is the cube root of unity, then what is the, conjugate of 2ω 2 + 3i ?, (NDA 2009 II), (a) 2 ω − 3i, (b) 3 ω + 2i, (c) 2 ω + 3i, (d) 3 ω − 2i, 33. If z is a complex number, such that z + z −1 = 1, then, what is the value of z 99 + z −99 ?, (NDA 2009 II), (a) 1, (b) –1, (c) 2, (d) –2, 34. What is the value of, (a) 1 + i, , ( 3 + i), (1 + 3i ), , (b) 1 − i, , 35. If, 2x = 3 + 5i , then, 2x3 + 2x 2 − 7x + 72 ?, (a) 4, , (b) – 4, , ?, (NDA 2009 I), , 3(1 − i ), ( 3 − i), (d), 2, 2, , (c), what, , is, , the, , value, , of, , (NDA 2009 I), , (c) 8, , (d) –8, , 36. The principal argument of −1 − i is, π, π, 3π, (b) −, (c) −, (a), 4, 4, 4, , (d), , 5π, 4, , 37. The modulus and amplitude of 1 + cos θ + i sin θ ,, respectively are, θ θ, θ θ, (a) 2 sin ,, (b) 2 cos ,, 2 2, 2 2, θ θ+π, θ π −θ, (d) 2 sin ,, (c) 2 cos ,, 2, 2, 2, 2, , 27. If α , β and γ are the cube roots of p ( p < 0), then for, xα + γβ + zγ, is equal to, any x , y and z,, xβ + yγ + zα, 1, 1, (b) (1 + i 3 ), (a) ( − 1 + i 3 ), 2, 2, 1, (c) (1 − i 3 ), (d) None of these, 2, , 38. If i = − 1, what is the value of i ?, , 28. If a and b are real numbers between 0 and 1, such, that the points z1 = a + i , z 2 = 1 + ib and z3 = 0 form, an equilateral triangle, then, (a) a = b = 2 + 3, (b) a = b = 2 − 3, (c) a = 2 − 3 , b = 2 + 3 (d) None of these, , 40. What is the value of ( − − 1 )8n + 1 + ( − − 1 )8n + 3 ,, where n is a natural number?, (a) 0, (b) 1, (c) 2 − 1, (d) − 2 − 1, , 29. Let z1 and z 2 be two complex numbers with α and β as, their principal arguments, such that α + β > π , then, principal arg( z1z 2 ) is given by, (a) α + β + π, (b) α + β − π, (c) α + β − 2π, (d) α + β, 30. If ω is a complex cube root of unity, then the value of, ω 99 + ω100 + ω101 is, (a) 1, , (b) –1, , 31. What is the modulus of, (a) 5, (c) 3, , (c) 3, 1 + 2i, 1 − (1 − i )2, (b) 4, (d) 1, , (d) 0, ?, , (NDA 2009 II), , (a), , 1− i, 2, , (b), , 1+ i, 2, , (c), , 2+ i, , (d), , 2, , 2− i, 2, , 39. If ω is a complex cube root of unity and, 1 + ω n + ω 2n = 0, what is the value of n?, (a) 3, , (b) 5, , (c) 6, , (d) 9, , 41. If ω, is, a, cube, root, of, unity, and, p = a + b, q = aω + bω 2 , and r = aω 2 + bω ;then what, is the value of pqr?, (a) a 2b + ab2, (b) a 2 + ab + b2, 3, 3, (d) ( a + b)3, (c) a + b, 42. If ω is a complex non-real cube root of unity, then ω, satisfies which one of the following equations?, (NDA 2008 I), , (a) x − x + 1 = 0, (c) x 2 + x − 1 = 0, 2, , (b) x + x + 1 = 0, (d) x 2 − x − 1 = 0, 2, , 43. What is the value of i + − i ?, (a), (c) ±, , 2, 1+ i, 2, , (b) 0, (d) ±, , (NDA 2007 I), , 1− i, 2
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30, , NDA/NA Mathematics, , Level II, 1. If z is a complex number, such that, imaginary, then, (a)| z| = 0, (c)| z|> 1, , z −1, is purely, z +1, , (b)| z| = 1, (d)| z|< 1, , 2. The points z1 , z 2 , z3 and z 4 in the complex plane are, the vertices of a parallelogram taken in order, if and, only if, (a) z1 + z 4 = z 2 + z3, (b) z1 + z3 = z 2 + z 4, (d) None of these, (c) z1 + z 2 = z3 + z 4, 3. If z be the conjugate of the complex number z, then, which of the following relations is false?, (b) z ⋅ z =| z |2, (a)| z| =| z |, (c) z1 + z 2 = z1 + z 2, , (d) arg ( z ) = arg ( z ), , 4. For any two complex numbers z1 and z 2 and any real, numbers a and b; |( az1 − bz 2 )|2 +|( bz1 + az 2 )|2 is, equal to, (a) ( a 2 + b2 ) (| z1| +| z 2|), (b) ( a 2 + b2 ) (| z1|2 +| z 2|2 ), (c) ( a 2 + b2 ) (| z1|2 −| z 2|2 ), (d) None of the above, 5. Consider the following statements, 1 − 3i , I. The value of , , 2 , , 36, , − 1 − 3i , +, , 2, , , , 36, , is 2., , II. The modulus of 2 i − − 2 i is 2., Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) None of these, 1 + 2i, 2− i, 6. If z =, , then what is the value of, −, 2− i, 1 + 2i, z 2 + zz ? ( i = −1 ), (NDA 2011 II), (a) 0, , (b) –1, , (c) 1, , (d) 8, , 7. The smallest positive integral value of n for which, n, 1 − i, , is purely imaginary with positive imaginary, 1 + i, part is, (NDA 2011 II), (a) 1, (b) 3, (c) 4, (d) 5, 8. What is the value of (1 + i ) + (1 − i ) , where i = −1 ?, (a) – 8, (b) 8, (NDA 2011 II), (c) 8 i, (d) − 8 i, 5, , 5, , 9. The moduli of two complex numbers are less than, unity, then the modulus of the sum of these complex, numbers is, (a) less than unity, (b) greater than unity, (c) equal to unity, (d) any of (a), (b) and (c), , 10. The maximum value of | z|, where z satisfies the, 2, condition z +, = 2 is, z, (b) 3 + 1 (c) 3, (d) 2 + 3, (a) 3 − 1, 11. If 1, ω , ω 2 , ω3 ,... , ω n − 1 are the n , nth roots of unity,, then (1 − ω ) (1 − ω 2 ) K (1 − ω n − 1 ) equals to, (a) 0, (b) 1, (c) n, (d) n 2, 12. If z1 and z 2 be complex numbers, such that z1 ≠ z 2 and, | z1| =| z 2|. If z1 has positive real part and z 2 has, ( z + z2 ), negative imaginary part, then 1, may be, ( z1 − z 2 ), (a) purely imaginary, (b) real and positive, (c) real and negative, (d) None of these, 13. The region of the complex plane for which, z −a, = 1 [Re( a ) ≠ 0] is, z+a, (a) x-axis, (b) y-axis, (c) the straight line x = a, (d) None of the above, 14. Common roots of the equations z3 + 2z 2 + 2z + 1 = 0, and z1985 + z100 + 1 = 0 are, (a) ω , ω 2, (b) ω , ω3, 2, 3, (c) ω , ω, (d) None of these, 15. If complex numbers z1 , z 2 and z3 represent the, vertices A, B and C, respectively of an isosceles ∆, ABC of which ∠C is right angled, then correct, statement is, (a) z12 + z 22 + z32 = z1z 2z3, (b) ( z3 − z1 )2 = z3 − z 2, (c) ( z1 − z 2 )2 = ( z1 − z3 ) ( z3 − z 2 ), (d) ( z1 − z 2 )2 = 2 ( z1 − z3 ) ( z3 − z 2 ), 16. If ω is a complex cube root of unity and x = ω 2 − ω − 2,, then what is the value of x 2 + 4x + 7 ?, (NDA 2010 I), (a) –2, , (b) –1, , (c) 0, , 17. What is the modulus of, (a) 1, , (b), , 1 + 2i, 1 − (1 − i )2, (c), , 5, , − 1 + i 3, 18. The value of , , 2, , , 3, (a) 3, (b), 2, , (d) 1, , 3n, , ?, , 3, , (NDA 2010 I), , (d) 5, , − 1 − i 3, +, , 2, , , , 3n, , (c) 0, , (d) 2, , is equal to, , 19. The complex number z, satisfying the equation, i−z, = 1 lies on, i+z
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31, , Complex Numbers, (a) a circle with the centre (0, 0) and radius 1, (b) the x-axis, (c) the y-axis, (d) the line y = x + 1, 20. Match List I with List II and select the correct, answer using the code given below the lists, (NDA 2008 II), , List I, A., B., C., D., , List II, , A cube root of unity, A square root of –1, Cube of 1 − i, Square of 1 + i, , Codes, A B, (a) 4 1, (c) 4 3, , C, 3, 1, , D, 2, 2, , 1., 2., 3., 4., , A, (b) 2, (d) 2, , −2(1 + i), 2i, i, 1, − (1 + i 3), 2, , B, 1, 3, , C, 3, 1, , D, 4, 4, , 21. For positive whole number of n, what is the value of, (NDA 2008 II), i 4n + 1?, (a) 1, (b) –1, (c) i, (d) − i, 22. If ω is complex cube root, then what is the value of, 1, 1, 1−, −, ?, (1 + ω ) (1 + ω 2 ), (NDA 2008 II), (a) 1, (b) 0, (c) ω, (d) ω 2, 23. A straight line is passing through the points, represented by the complex numbers a + ib and, 1, , where ( a , b) ≠ ( 0, 0)., (NDA 2008 I), − a + ib, Which one of the following is correct?, (a) It passes through the origin, (b) It is parallel to the x-axis, (c) It is parallel to the y-axis, (d) It passes through ( 0, b), 24. What is the value of, integer?, (a) 1, , [1 + ( i5 )4n − 1 ]4n + 1, [1 + ( i5 )4n + 1 ]4n − 1, , (b) 2, , 25. What is the real part of, 1, (a), 1 + cos θ, 1, (c) −, 2, , (c) 4, , (c) A = { x + iy x 2 ∈ R x , y ∈ R }, (d) A = { iy y ∈ R }, 28. Which one of the following is correct? If z and w are, complex numbers and w denotes the conjugate of w,, then| z + w| =| z − w| holds only, if, (NDA 2008 I), (a) z = 0 or w = 0, (b) z = 0 and w = 0, (c) z ⋅ w is purely real, (d) z ⋅ w is purely imaginary, 29. What is the square root of, 3 i, (a) ± , + , 2, 2, 1, 3, (c) ± + i, , 2, 2, , 1, 3, −i, ?, (NDA 2008 I), 2, 2, 3 i, (b) ± , − , 2, 2, 1, 3, (d) ± − i, , 2, 2, , 30. Let C be the set of complex number and z1 ,z 2 are in C., I. arg ( z1 ) = arg ( z 2 ) ⇒ z1 = z 2, II. | z1| =| z 2| ⇒ z1 = z 2, Which of the statements given above is/are correct?, (NDA 2008 I), , (a) Only I, (c) Both I and II, , (b) Only II, (d) Neither I nor II, , 31. If 1, ω and ω 2 are the three cube roots of unity, then, ( aω 6 + bω 4 + cω 2 ), ?, what is the value of, ( b + cω10 + aω 8 ) (NDA 2007 II), a, (a), (b) b, b, (c) ω, (d) ω 2, 32. If z = − z , then which one of the following is correct?, (NDA 2012 I), , , n is a positive, (d) 5, , 1, ?, 1 + cos θ + i sin θ, 1, (b), 1 − sin θ, 1, (d), 2, , 26. If a is a real number and z is a complex number, then, what is the value of ( z + a ) ( z + a ) ?, (a) z 2 + a 2 (b) a 2 − z 2 (c)| z − a|2 (d)| z + a|2, 27. Suppose that A denotes the collection of all complex, numbers whose square is a negative real number., Which one of the following statements is correct?, (a) A ⊆ R, (b) A ⊇ R, , (a) The real part of z is zero., (b) The imaginary part of z is zero., (c) The real part of z is equal to imaginary part of z., (d) The sum of real and imaginary parts of z is z., 33. If the complex numbers z1 and z 2 and the origin form, an equilateral triangle, then z12 + z 22 is equal to, (a) z1z 2, (c) z 2z1, , (b) z1z 2, (d)| z1|2 =| z 2|2, , 34. If z1 and z 2 are any two complex numbers, such that, |z1 + z 2| =|z1| +|z 2|, which one of the following is, correct?, (b) z1 ≥ 0 or z 2 ≥ 0, (a) z1 = αz 2 with α ∈ R, (c) z1 = αz 2 with α > 0, (d)|z1| =|z 2|, α + iβ, ?, 35. If α and β are real, what is the value of , β + iα, 1, (a) 0, (b), 2, (c) 1, (d) 2
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32, , NDA/NA Mathematics, , 36. What is the value of, [( −1 + i 3 )/ 2], , 10, , (a) 1, , + [( −1 − i 3 )/ 2] ?, 10, , (b) –1, , (c) 2, , (NDA 2007 I), , (d) 0, , 37. If ω denotes the cube root of unity, then what is the, real root of the equation x3 − 27 = 0 ?, (NDA 2007 I), (b) 3 ω, , (a) 3ω, , (c) −3ω, , 2, , (d) 3 ω, , 3, , 38. Let O be the origin and point A be represented by z. If, π, in the anti, OA is rotated through an angle, 2, clockwise direction keeping the length of OA same,, then what represents the new point?, NDA 2007 I), (a) − iz, (b) | z| i, (c) iz, (d) z, , Directions (Q. Nos. 39-40), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., , 39. Assertion (A) If z1 = 3 + −4 and z 2 = 3 + −25 ,, z1/ z 2 is a complex number., Reason (R) If z1 and z 2 are complex numbers, then, z1/ z 2 is always a complex number., − 1 + − 3, 40. Assertion (A) , , 2, , , , 29, , − 1 − − 3, +, , 2, , , , 29, , = −1, , Reason (R) ω 2 = − 1, , Directions (Q. Nos. 41-43) Consider the cubic, equation x3 = 1 whose roots are 1, ω and ω 2 where, −1+ i 3, ω=, , then, 2, 41. Find the modulii and arguments of complex number, ω 2, respectively are, 4π, π, (a) 1,, (b) 2,, 3, 3, 4π, 2π, (d) 2,, (c) 3,, 3, 3, 42. The polar form of complex numbers ω 2 is, 4π, 4π, 2π, 2π, (a) cos, (b) cos, + i sin, + i sin, 3, 3, 3, 3, π, π, (d) None of these, (c) cos + i sin, 3, 3, 43. The value of ω17 is, (a) ω 2, (b) 1, , (c) 2, , (d) 3, , Directions (Q. Nos. 44-46), , Consider the, statement 1 + ω + ω 2 = 0, where ω is a cube roots of, unity, then, 44. The value of (1 − ω + ω 2 ) (1 − ω 2 + ω 4 ) (1 − ω 4 + ω 8 ) ..., 2n factors, (a) 22n, (b) 22, (c) 23 n, (d) 2n, 45. Find the value of, (a) − 1, (c) 3, , a + bω + cω 2, c + aω + bω 2, (b) 2, (d) 0, , +, , a + bω + cω 2, b + cω + aω 2, , ., , 46. The equation ( x − 1)3 + 8 = 0 in the set C of all, complex numbers are, (b) 1, 2ω + 1, 1 − 2ω 2, (a) − 1, 1 − 2ω , 1 − 2ω 2, 2, (c) 1, 2ω , ω, (d) 1, ω , ω 2, , Answers, Level I, 1., 11., 21., 31., 41., , (d), (d), (a), (d), (c), , 2., 12., 22., 32., 42., , (b), (c), (d), (a), (b), , 3., 13., 23., 33., 43., , (c), (c), (a), (d), (a), , 4., 14., 24., 34., , (a), (d), (b), (d), , 5., 15., 25., 35., , (b), (c), (d), (a), , 6., 16., 26., 36., , (a), (d), (a), (c), , 7., 17., 27., 37., , (c), (b), (d), (b), , 8., 18., 28., 38., , (b), (c), (b), (b), , 9., 19., 29., 39., , (c), (b), (c), (b), , 10., 20., 30., 40., , (c), (d), (d), (a), , 2., 12., 22., 32., 42., , (b), (a), (b), (a), (a), , 3., 13., 23., 33., 43., , (d), (b), (a), (a), (a), , 4., 14., 24., 34., 44., , (b), (a), (b), (a), (a), , 5., 15., 25., 35., 45., , (a), (d), (d), (c), (a), , 6., 16., 26., 36., 46., , (d), (c), (d), (b), (a), , 7., 17., 27., 37., , (b), (a), (d), (d), , 8., 18., 28., 38., , (d), (d), (a), (c), , 9., 19., 29., 39., , (d), (b), (b), (a), , 10., 20., 30., 40., , (b), (c), (d), (c), , Level II, 1., 11., 21., 31., 41., , (b), (c), (c), (c), (a)
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3, Quadratic Equations, and Inequalities, Polynomial, , Quadratic Identity, , Let a0 , a1 , a2 , ... , an are real numbers, then, f ( x ) = a0 + a1x + a2x 2 + ... + an x n is known as polynomial., , If two quadratic expression in x are equal for all values, of x. This statement of equality between the two, expressions is called an identity., , Polynomial a0 + a1x + a2x 2 + K + an x n is known as a, polynomial of degree n., , Polynomial Equation, If f ( x ) is a real or complex polynomial, then f ( x ) = 0 is, known as a polynomial equation., If f ( x )is a polynomial of degree 2, then f ( x ) = 0 is known, as quadratic equation. General equation of quadratic, equation is ax 2 + bx + c = 0 ,where a , b, c ∈ R or C., The quadratic equation of the form ax 2 + c = 0 is, known as pure quadratic equation., , Quadratic Equation, An equation of the form ax 2 + bx + c = 0, where a, b and, c are certain numbers and a ≠ 0 is called a quadratic, equation., The numbers a , b, and c are called the coefficients of, the quadratic equation and the number b2 − 4ac is called, its discriminant., Discriminant of a quadratic equation is usually denoted, by D., , Quadratic Expression, An expression of the form ax 2 + bx + c,where a , b and c, are some numbers and a ≠ 0 is called a quadratic, expression., , Identical Equations, Two equations are said to be identical, if they have, same roots., , Roots of an Equation, The values of the variables satisfying the given, equation are called its roots i. e. , if f ( x ) = 0 is a polynomial, equation and f ( a ) = 0, then a is known as root of the, polynomial equation f ( x ) = 0., %, , An equation of degree n has n roots real or imaginary., , %, , If an equation has a root a +, be a −, , b or a + ib, then another root will, , b or a − ib, respectively., , %, , An odd degree equation has atleast one real root whose sign is, opposite to that of its last term, provided that the coefficient of, the highest degree term is positive., , %, , Every equation of an even degree whose constant term is, negative and the coefficient of the highest degree term is positive,, has atleast two real roots, one positive and one negative., , %, , If all the terms of an equation are positive and the equation, involves no odd power of x, then all its roots are complex., , Roots of a Quadratic Equation, Roots of a quadratic equation ax 2 + bx + c = 0 are, − b + b2 − 4ac, − b − b2 − 4ac, and, , where b2 − 4ac is, 2a, 2a, known as discriminant and is denoted by D.
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42, , NDA/NA Mathematics, , Example 1. The roots of the equation x2 + | x| − 6 = 0 is, (a) ± 9, (c) ± 3, , (b) ± 2, (d) ± 4, , Solution (b)| x|2 + | x| − 6 = 0, ⇒ (| x| + 3) (| x| − 2) = 0 ⇒| x| = 2 ⇒ x = ± 2, , Relation between Coefficient, and Roots of an Equation, (i) Quadratic equation Let α and β be the roots, of the quadratic equation, ax 2 + bx + c = 0, then, b, α +β = −, a, c, and, αβ =, a, (ii) Cubic equation Let α , β and γ be the roots of, the cubic equation ax3 + bx 2 + cx + d = 0, then, b, c, d, and αβγ = −, α + β + γ = − , αβ + βγ + γα =, a, a, a, %, , If α and β be the roots of the quadratic equation, ax 2 + bx + c = 0, then ax 2 + bx + c = a (x − α) (x − β)., , π, P, Q, . If tan and tan, are, 2, 2, 2, 2, the roots of the equation ax + bx + c = 0, then which of the, following is correct?, (a) a + b = 2c, (b) a + b = c, (c) a + b + c = 0, (d) 2 a = b − c, , Example 2. In a ∆ PQR, ∠ R =, , Solution (b) Q tan, , P, Q, and tan, are the roots of the equation, 2, 2, , ax2 + bx + c = 0, then, P, Q, b, P, Q c, tan + tan, = − and tan tan, =, 2, 2, a, 2, 2 a, π, π, Also,, ∠R =, ⇒ ∠P + ∠Q =, 2, 2, P, Q, tan + tan, π, P Q, 2, 2, ⇒ 1 = tan = tan + ⇒ 1 =, P, Q, 2 2, 4, 1 − tan tan, 2, 2, −b / a, −b, ⇒ 1=, ⇒ 1=, ⇒ a+ b=c, 1− c / a, a−c, , Equation of Given Roots, (i) Quadratic equation If the roots of a, quadratic equation are α and β, then equation will be, x 2 − (α + β ) x + αβ = 0., (ii) Cubic equation If α, β and γ are the roots of, the cubic equation, then the equation will be, x3 − (α + β + γ )x 2 + (αβ + βγ + γα ) x − αβγ = 0, , Example 3. If the product of the roots of the equation, mx 2 + 6 x + (2m − 1) = 0 is − 1, then the value of m is, 1, 1, 2, (a) m =, (b) m =, (c) m =, (d) m = 2, 3, 2, 3, Solution (a) Let α and β be the roots of the equation, mx2 + 6x + (2m − 1) = 0 ., 6, ...(i), ∴, α+β=−, m, 2m − 1, ... (ii), and, αβ =, m, According to the question,, 2m − 1, 1, −1=, ⇒ − m = 2m − 1⇒ 3m = 1 ⇒ m =, m, 3, , Nature of Roots, Let the equation is ax 2 + bx + c = 0, then, 1. if D = b2 − 4ac > 0, then the roots of equation are real, and distinct., 2. if D = b2 − 4ac = 0, then the roots of equation are real, and coincident., 3. if D = b2 − 4ac < 0, then the roots of equation are, imaginary., 4. if D = b2 − 4ac > 0 and a perfect square, then the, roots of the equation are rational., 5. if D = b2 − 4ac < 0 and not a perfect square, then the, roots of the equation are irrational., , Example, 4. If, the, roots, of, the, equation, x 2 − 8 x + a 2 − 6 a = 0 are real and distinct, then all possible, values of a is, (a) − 3 < a < 2 (b) − 2 < a < 8 (c) − 3 < a < 4 (d) − 4 < a ≤ 1, Solution (b) Since, the roots of the given equation are real and, distinct, we must have, D > 0 ⇒ 64 − 4 ( a2 − 6a) > 0 ⇒ 4 [16 − a2 + 6a] > 0, ⇒, , −4 ( a2 − 6a − 16) > 0 ⇒ a2 − 6a − 16 < 0, , ⇒, ( a − 8) ( a + 2) < 0 ⇒ −2 < a < 8, Hence, the roots of the given equation are real, if a lies, between −2 and 8., , Symmetric Functions, The algebraic expressions in α and β which remain, unchanged when α and β are interchanged, are known as, symmetric functions in α and β., An important property of such functions is that they, can always be expressed in terms of (α + β) and αβ. So, they, can be evaluated for a given quadratic equation., The following relations serve the purpose of useful, tools for the same., (i) (α 2 + β 2 ) = [(α + β )2 − 2αβ ], (ii) ( 2 − β )2 = [( 2 + β )2 − 4αβ ]
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44, , NDA/NA Mathematics, , 6. If a1 , a2 ,... , an and b1 , b2 ,... , bn are positive real, numbers such that, a1 > b1 , a2 > b2 ,... , an > bn ,, then, , a1a2a3 ... an > b1b2b3 K bn ., , Arithmetico-Geometric, Mean Inequality, 1. If a , b > 0 and a ≠ b, then, , 7. If a and b are positive real numbers such that a < b, and if n is any positive rational number, then, (a) a n < bn, , (b) a − n > b− n, , (c) a1/ n < b1/ n, 8. If 0 < a < 1 and n is any positive rational number,, then, (a) 0 < a n < 1, , (b) a − n > 1, , 9. If a > 1 and n is any positive rational number, then, (a) a n > 1, , (a) m > n ⇒ a m < a n, , (b) m < n ⇒ a m > a n, , 11. If a > 1 and m , n are positive rational numbers, then, (a) m > n ⇒ a m > a n, , Example 8. If a, b and c are three distinct positive real, numbers, then, 1 1 1, ( a + b + c) + + is, a b c, , (a) >9, ∴, , and, , (b) (5, 9), , ⇒, , 2, , (d) − 8, , , 3, , and, , Solution (c) Given,| 2x − 3| < | x + 5|, , (b) <4, , (c) >5, , (d) <6, , Solution (a) We have AM > GM, , (b) m < n ⇒ a m < a n, , Example 7. If| 2 x − 3| < | x + 5 |, then x belongs to, (a) ( − 3, 5), − 2 , (c) , , 8, , 3, , 2. If ai > 0, where i = 1, 2, 3, ... , n , then, n, a1 + a2 + ... + an, ≥ ( a1 ⋅ a2K an )1/ n ≥, 1, 1, 1, n, +, + ...+, a1 a2, an, , (b) 0 < a − n < 1, , 10. If 0 < a < 1 and m , n are positive rational numbers,, then, , 2, a+b, ., > ab >, 1, 1, 2, +, a b, , ⇒, , a+ b+ c, > ( abc)1/3, 3, 1 1 1, + +, 1/3, a b c, 1 , >, , abc, 3, , a + b + c > 3 ( abc)1/3, 3, 1 1 1, + + >, a b c ( abc)1/3, 1 1 1, ( a + b + c) + + > 9, a b c, , ⇒, , | 2x − 3| −| x + 5| < 0, , ⇒, , 3 − 2x + x + 5 < 0 , x ≤ − 5, , Example 9. For real positive numbers a, b and c the, , ⇒, , 3, 3 − 2x − x − 5 < 0 , − 5 < x ≤, 2, 3, 2x − 3 − x − 5 < 0 , x >, 2, , minimum value of, , ⇒, ⇒, ⇒, ⇒, ⇒, , x > 8, x ≤ − 5, −2, 3, , −5 < x≤, x>, 3, 2, 3, x < 8, x >, 2, − 2 3 3 , −2 , ,, , 8, ∪ , 8 ⇒ x ∈ , x ∈, , 3 2 2 , 3, , (a) 5, , b +c c + a a+ b, is, +, +, a, b, c, (b) 4, (c) 6, , (d) 3, , 1 a b b c c a, Solution (c) Q + + + + + , 6 b, , a, , c, , b, , a, , c, , a b b c c, ≥ ⋅ ⋅ ⋅ ⋅ ⋅, b a c b a, a b b c c a, ⇒, + + + + + ≥6, b a c b a c, a+ b c+ a b+ c, +, +, ⇒ Minimum value of, c, b, a, , a, , c, , 1/ 6, , =6
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45, , Quadratic Equations and Inequalities, , Comprehensive Approach, n, , n, , n, , If f(α) = 0 and f ′(α) = 0, then α is a repeated root of the quadratic, equation f ( x) = 0 and f ( x) = a ( x − α) 2, b, In fact α = −, 2a, For the quadratic equation ax2 + bx + c = 0, 1. One root will be reciprocal of the other, if a = c, 2. One root is zero, if c = 0, 3. Roots are equal in magnitude but opposite in sign, if, b = 0., 4. Both roots are zero, if b = c = 0, 5. Roots are positive, if a and c are of the same sign and b is, of the opposite sign., 6. Roots are of opposite sign, if a and c are of opposite sign., 7. Roots are negative, if a, b and c are of the same sign., If the ratio of roots of the quadratic equation ax2 + bx + c = 0 be, p : q, then pqb 2 = ( p + q) 2 ac, If one root of the quadratic equation ax2 + bx + c = 0 is equal to, the n th power of the other, then, 1, , (b) whose roots are Aα and Aβ, is, ax2 + Abx + A 2c = 0, α, β, (c) whose roots are and , is, A, A, aA 2x2 + bAx + c = 0, 1, 1, (d) whose roots are and , is, α, β, n, , n, , n, , n, , 1, , ( ac n ) n + 1 + ( a n c) n + 1 + b = 0, n, , n, , n, , n, , n, , n, , n, , n, , If one root of the equation ax2 + bx + c = 0 be n times the other, root, then nb 2 = ac (n + 1) 2, k+1, If the roots of the equation ax2 + bx + c = 0 are of the form, k, k+2, and, , then ( a + b + c) 2 = b 2 − 4 ac, k+1, If the roots of ax2 + bx + c = 0 are α and β, then the roots of, 1, 1, cx2 + bx + a = 0 will be and ., α, β, The roots of the equation ax2 + bx + c = 0 are reciprocal to, a′ x2 + b ′ x + c ′ = 0, if, ( cc ′ − aa′) 2 = ( ba′ − cb ′) ( ab ′ − bc ′), Let f ( x) = ax2 + bx + c, where a > 0. Then,, 1. conditions for both the roots of f ( x) = 0to be greater than a given, −b, number K are b 2 − 4 ac ≥ 0; f (K) > 0 ;, > K., 2a, 2. the number K lies between the roots of f ( x) = 0, if f (K) < 0., 3. condition for exactly one root of f ( x) = 0 to lie between d and e,, is f (d ) f ( e) < 0., If the common root of quadratic equation a1x2 + b1x + c1 = 0 and, a2x2 + b2x + c2 = 0 is α, then, c a − c2a1, b c − b2c1, or 1 2, α= 1 2, a1b2 − a2b1, c1a2 − c2a1, If both the roots of the quadratic equations a1x2 + b1x + c1 = 0 and, a2x2 + b2x + c2 = 0 are common, then, a1 b1 c1, =, =, a2 b2 c2, Let α and β be the roots of the equation ax2 + bx + c = 0, then that, equation, (a) whose roots are α ± A , β ± A, is, a ( x m A) 2 + b ( x m A) + c = 0, , n, , n, n, , cx2 + bx + a = 0, If in ax2 + bx + c, a > 0 and b 2 − 4 ac < 0, then ax2 + bx + c will, always be positive., If in ax2 + bx + c , a < 0 and b 2 − 4 ac < 0, then ax2 + bx + c will, always be negative., If the sum of coefficients of the polynomial equation, a0 + a1x + a2x2 + K + an xn = 0 is zero, then x = 1will be the root, of that equation., If the equation is ax2 + bx + c = 0 such that, a + b + c = 0, then, c, roots of equation ax2 + bx + c = 0 will be1, and if a − b + c = 0,, a, c, then roots of the equation will be −1, − ., a, If sum of roots of the equation ax2 + bx + c = 0 is equal to the sum, b a, of their reciprocal, then ab 2 , bc 2 and ca 2 will be in AP or , and, c b, c, will be in AP., a, If a1 = a2 = K = an , then AM = GM, If a, b > 0 and a ≠ b, then, m, , (a), , am + b m a + b , >, , if m < 0 or m > 1, 2 , 2, , (b), , am + b m a + b , <, , if 0 < m < 1, 2 , 2, , m, , n, , n, , n, , n, , n, , f a > 1and n is any positive rational number, then, (a) a n > 1, (b) 0 < a − n < 1, If 0 < a < 1 and m,n are positive numbers, then, (a) m > n ⇒ a m < a n, (b) m < n ⇒ a m > a n, (a) if a > 1and x > y > 0, then log a x > log a y, (b) if 0 < a < 1 and x > y > 0, then log a x < log a y, If a and b are positive real numbers, then, a+ c a, (a) a < b ⇒, > , ∀c > 0, b+ c b, a+ c a, (b) a > b ⇒, < , ∀c> 0, b+ c b, 1, If a is a positive real number, then a + ≥ 2 and if a is a negative, a, 1, real number, then a + ≤ −2., a, | a + b | ≤ | a | + | b | , in general, | a1 + a2 + K + an |≤| a1| + | a2| + ... + | an|
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Exercise, Level I, 1. If α and β, , are the, 1, 4x + 3x + 7 = 0, then +, α, 3, 3, (a) −, (b), 7, 7, 2, , roots of the equation, 1, is equal to, β, 3, 3, (c) −, (d), 5, 5, , 2. If the roots of the equation 3x 2 − 5x + q = 0 are equal,, then what is the value of q?, (NDA 2011 II), (a) 2, (b) 5 / 12, (c) 12 / 25, (d) 25 / 12, 3. If the product of the roots of the equation, ( a + 1) x 2 + ( 2a + 3) x + ( 3a + 4) = 0 be 2, then the, sum of roots is, (a) 1, (b) −1, (c) 2, (d) −2, 4. If the roots of the equation ax 2 + bx + c = 0 be α and β,, then the roots of the equation cx 2 + bx + a = 0 are, 1, (a) −α , − β, (b) α,, β, 1 1, (d) None of these, (c), ,, α β, 5. If α and β are the roots of the equation, ax 2 + bx + c = 0, then the equation, whose roots are, 1, 1, α + and β + , is, β, α, (a) acx 2 + ( a + c) bx + ( a + c)2 = 0, (b) abx 2 + ( a + c) bx + ( a + c)2 = 0, (c) acx 2 + ( a + b) cx + ( a + c)2 = 0, (d) None of the above, 6. If the equations x 2 − px + q = 0 and x 2 − ax + b = 0, have a common root and the roots of the second, equation are equal, then which one of the following is, correct?, (NDA 2011 II), (a) aq = 2( b + p), (b) aq = b + p, (c) ap = 2( b + q ), (d) ap = b + q, 7. The equation x 2 − 4x + 29 = 0 has one root 2 + 5i., What is the other root? ( i = −1 ), (a) 2, , (b) 5, , (c) 2 + 5i, , 8. If, one, of, the, roots, a ( b − c) x 2 + b ( c − a ) x + c ( a −, second root?, b( c − a ), (a) −, (b), a( b − c), c( a − b), (c), (d), a( b − c), , 9. What are the roots of the equation, 2( y + 2)2 − 5( y + 2) = 12? (NDA 2011 II), (a) − 7 / 2 , 2, (b) − 3 / 2 , 4, (c) − 5 / 3, 3, (d) 3 / 2, 4, 10. Let α and β be the roots of the equation x 2 + x + 1 = 0., The equation, whose roots are α19, and β7 is, (NDA 2011 II), , (a) x − x − 1 = 0, (c) x 2 + x − 1 = 0, 2, , (b) x − x + 1 = 0, (d) x 2 + x + 1 = 0, 2, , ( x + 1) ( x − 3), , then all real values of x for, ( x − 2), which y takes real values, are, (a) −1 ≤ x < 2 or x ≥ 3, (b) −1 ≤ x < 3 or x > 2, (c) 1 ≤ x < 2 or x ≥ 3, (d) None of these, , 11. Let y =, , 12. If 2 + i 3 is a root of the equation x 2 + px + q = 0,, where p and q are real, then ( p, q ) is equal to, (a) ( −4, 7), (b) ( 4, − 7), (c) ( 4, 7), (d) ( −4, − 7), 13. The coefficient of x in the equation x 2 + px + q = 0, was taken as 17 in place of 13, its roots were found to, be −2 and −15. The roots of the original equation are, (a) 3, 10, (b) −3, − 10, (c) −5, − 8, (d) None of these, 14. If p, q and r are real and p ≠ q, then the roots of the, equation, ( p − q ) x 2 + 5 ( p + q ) x − 2 ( p − q ) = r are, (a) real and equal, (b) unequal and rational, (c) unequal and irrational, (d) nothing can be said, 15. If the roots of the equation ( p2 + q 2 )x 2 − 2q ( p + r ) x, + ( q 2 + r 2 ) = 0 be real and equal, then p, q and r will, be in, (a) AP, (b) GP, (c) HP, (d) None of these, 16. What is the value of, , (NDA 2011 II), , (d) 2 − 5i, , of, the, equation, b) = 0 is 1, what is the, (NDA 2011 II), , b( c − a ), a( b − c), c( a − b), −, a( b − c), , 8 + 2 8 + 2 8 + 2 8 +K∞ ?, (a) 10, (c) 6, , (NDA 2011 II), , (b) 8, (d) 4, , 17. If α and β are the roots of the equation, x 2 − q (1 + x ) − r = 0, then what is the value of, (NDA 2012 I), (1 + α ) (1 + β )?, (a) 1 − r, (b) q − r, (c) 1 + r, (d) q + r
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47, , Quadratic Equations and Inequalities, 18. What is the solution, x 4 − 26x 2 + 25 = 0 ?, (a) { − 5, − 1, 1, 5}, (c) { 1, 5}, , set, , for, , the, , equation, , (NDA 2011 I), , (b) { − 5, − 1}, (d) { − 5, 0, 1, 5}, , 19. If the equations x 2 + kx + 64 = 0 and x 2 − 8x + k = 0, have real roots, then what is the value of k?, (NDA 2010 II), , (a) 4, , (b) 8, , (c) 12, , (d) 16, , 20. If the product of the roots of the equation, x 2 − 5x + k = 15 is − 3, then what is the value of k?, (NDA 2010 I), , (a) 12, , (b) 15, , (c) 16, (d) 18, α, β, 21. If the roots of the equation, +, = 1 be equal, x−α x−β, in magnitude but opposite in sign, then α + β is equal, to, (a) 0, (b) 1, (c) 2, (d) None of these, 22. If the equation x 2 − bx + 1 = 0 does not possess real, roots, then which one of the following is correct?, (NDA 2010 I), , (a) − 3 < b < 3, (c) b > 2, , (b) − 2 < b < 2, (d) b < − 2, , 24. If α and β are the roots of ax 2 + bx + b = 0, then what, β, b, α, equal to?, is, +, +, (NDA 2009 II), a, β, α, (b) 1, , (c) 2, , (d) 3, , 25. If the roots of ax 2 + bx + c = 0 are sin α and cos α for, some α, then which one of the following is correct?, (NDA 2009 II), , (a) a + b = 2ac, (c) b2 − a 2 = 2ac, 2, , 2, , (b) b − c = 2ab, (d) b2 + c2 = 2ab, 2, , 2, , 26. If x = 2 + 21/3 + 22/ 3 , then what is the value of, x3 − 6x 2 + 6x?, (NDA 2009 II), (a) 1, , (b) 2, , (c) 3, , (b) Only 1, (d) − 2 and − 1, , 29. If α and β are the roots of the equation, 4x 2 + 3x + 7 = 0,then what is the value of (α −2 + β −2 )?, (NDA 2011 I), , (a) 47 / 49, (c) − 47 / 49, , (b) 49 / 47, (d) − 49 / 47, , 30. If the roots of the equations px 2 + 2qx + r = 0 and, qx 2 − 2 pr x + q = 0 be real, then, (a) p = q, (b) q 2 = pr, 2, (d) r 2 = pq, (c) p = qr, 31. If a < b, then, a, b, (a), <, ( −2) ( −2), 1 1, (c), <, a b, , a b, >, 2 2, a, b, (d), >, −2 −2, , (b), , 32. If 2x + 3 y = 17 and 2x + 2 − 3 y + 1 = 5, then what is the, value of x?, (NDA 2009 I), (a) 3, (b) 2, (c) 1, (d) 0, 33. What is the value of x satisfying the equation, 3, , 23. If p and q are the roots of the equation x 2 − px + q = 0,, then what are the values of p and q, respectively?, (a) 1, 0, (b) 0, 1, (c) − 2, 0, (d) − 2, 1, , (a) 0, , (a) Only − 2, (c) − 2 and 1, , (d) −2, , 27. The roots of the equation ( x − p)( x − q ) = r 2, where p,, q and r are real, are, (NDA 2009 II), (a) always complex, (b) always real, (c) always purely imaginary, (d) None of the above, 28. For the two equations x 2 + mx + 1 = 0 and, x 2 + x + m = 0, what is/are the value/values of m for, which these equations have atleast one common, root?, (NDA 2009 II), , a − x, a+x, 16 , ?, =, a + x, a−x, a, a, (b), (a), 2, 3, , (NDA 2009 I), , (c), , a, 4, , (d) 0, , 34. The roots of Ax 2 + Bx + C = 0 are r and s. For the, roots of x 2 + px + q = 0 to be r 2 and s2 , what must be, the value of p?, (NDA 2009 I), ( B2 − 4 AC ), ( B2 − 4 AC ), (a), (b), A2, A2, 2, ( 2 AC − B ), (d) B2 − 2 C, (c), A2, 35. If the roots of the equation x 2 − bx + c = 0 are two, consecutive integers, then what is the value of, b2 − 4c ?, (NDA 2008 II), (a) 1, (c) –2, , (b) 2, (d) 3, , 36. If r and s are roots of x 2 + px + q = 0, then what is the, value of (1/ r 2 ) + (1/ s2 )?, (NDA 2008 II), (a) p2 − 4q, (c), , p2 − 4q, q2, , (b), (d), , p2 − 4q, 2, p2 − 2q, q2, , 37. If α and β are the roots of x 2 + 4x + 6 = 0, then what is, the value of α3 + β3 ?, (NDA 2008 II), 2, 2, (b), (c) 4, (d) 8, (a) −, 3, 3
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48, , NDA/NA Mathematics, , 38. The number of, x 2 − 3|x| + 2 = 0, is, (a) 2, (b) 3, , solutions, , of, , (c) 4, , the, , equation, , (d) 5, , 39. What is the sum of the squares of the roots of the, equation x 2 + 2x − 143 = 0 ?, (NDA 2012 I), (a) 170, (b) 180, (c) 190, (d) 290, 40. If the roots of the equation λ + 8λ + µ + 6 µ = 0, are real, then µ lies between, (a) −2 and 8, (b) −3 and 6, (c) −8 and 2, (d) −6 and 3, 2, , 41. What is the value of 5 5 5 K ∞ ?, (a) 5, (c) 1, , 2, , (NDA 2008 I), , (b) 5, (d) ( 5)1/ 4, , 42. One root of the equation x 2 = px + q is reciprocal of, the other and p ≠ ± 1. What is the value of q?, (NDA 2008 I), , (a) q = − 1, , (b) q = 1, 1, 2, , (d) q =, , (c) q = 0, , 43. If α and β are the roots of the equation, lx 2 − mx + m = 0, l ≠ m , l ≠ 0, then which one of the, following statements is correct?, (NDA 2007 II), (a), , α, β, +, −, β, α, , m, =0, l, , (b), , α, β, +, +, β, α, , m, =0, l, , (c), , α +β, −, αβ, , 46. If| x|> 5, then, (a) 0 < x < 5, (c) −5 < x < 5, , (b) x < − 5 or x > 5, (d) x > 5, , 47. What is the solution set for the, x 2/ 3 + x1/ 3 − 2 = 0 ?, (a) { −8, 1}, (b) { 8, 1}, (c) { −8, − 1}, (d) { 8, − 1}, , 49. If α and β are the roots of the equation, ax 2 + bx + c = 0, then what is the value of, ( aα + b)−1 + ( aβ + b)−1 ?, (NDA 2007 I), a, b, (b), (a), ( bc), ( ac), −b, −a, (d), (c), ( ac), ( bc), 50. If α and β are the roots of the equation x 2 − 2x − 1 = 0,, then what is the value of α 2β −2 + α −2β 2 ? (NDA 2007 I), , 2, , (d) The arithmetic mean of α and β is the same as, their geometric mean., 44. For what value of k, are the roots of the quadratic, equation ( k + 1)x 2 − 2( k − 1)x + 1 = 0 real and equal?, (NDA 2007 II), , (b) Only k = − 3, (d) k = 0 or k = − 3, , equation, , 48. What is the polynomial, whose zero is 2 ?, (a) x 2 − 2x + 2, (b) ( x 4 − 2) ( 3 − 4x + 3x3 ), 4, 3, (c) x − 2x + x − 4, (d) x 4 − 3x3 + 3x 2 − 3x + 2, , (a) –2, (b) 0, a c, 51. If < , then, b d, a+b c+d, (a), <, a−b c−d, , m, =0, l, , (a) Only k = 0, (c) k = 0 or k = 3, , 45. If the equation x 2 + k2 = 2( k + 1)x has equal roots,, then what is the value of k?, (NDA 2007 I), 1, 1, (b) −, (c) 0, (d) 1, (a) −, 3, 2, , c, a, (c) < , d, b, , (c) 30, , (b), , (d) 34, , a−b c−d, <, a+b c+d, , 2, , 52. If a > b, then, (a) a + 5 > b + 5, (c) a + b < b + 5, , (d) None of these, , (b) a − 5 < b − 5, (d) depends on a and b, , 53. If one of the roots of the equation x 2 + ax − b = 0 is 1,, then what is the value of ( a − b)?, (NDA 2012 I), (a) −1, (b) 1, (c) 2, (d) − 2, , Level II, 1. Let α and β be the roots of the equation, ( x − a ) ( x − b) = c, c ≠ 0. Then, the roots of the, equation ( x − α ) ( x − β ) + c = 0 are, (NDA 2011 II), (a) a, c, (b) b, c, (c) a, b, (d) a + b, a + c, 2. If p, q and r are rational numbers, then the roots of, the equation x 2 − 2 px + p2 − q 2 + 2qr − r 2 = 0 are, (NDA 2011 I), , (a) complex, (c) irrational, , (b) pure imaginary, (d) rational, , 3. If, the, roots, of, the, equation, ( a 2 + b2 ) x 2 − 2b ( a + c) x + ( b2 + c2 ) = 0 are equal,, then which one of the following is correct?, (NDA 2010 II), , (a) 2b = a + c, (c) b + c = 2a, , (b) b2 = ac, (d) b = ac, , 4. Which of the following are the two roots of the, equation ( x 2 + 2)2 + 8x 2 = 6x ( x 2 + 2)?, (NDA 2010 II), (a) 1 ± i, (b) 2 ± i, (c) 1 ± 2, (d) 2 ± i 2
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49, , Quadratic Equations and Inequalities, 5. If α and β are the roots of the equation x 2 + x + 1 = 0,, then which of the following are the roots of the, equation x 2 − x + 1 = 0 ?, (a) α7 and β13, (b) α13 and β7, 20, 20, (c) α and β, (d) None of these, 6. If α and β be the roots of the equation, 2x 2 + 2 ( a + b) x + a 2 + b2 = 0, then the equation,, whose roots are (α + β )2 and (α − β )2, is, (a) x 2 − 2abx − ( a 2 − b2 )2 = 0, (b) x 2 − 4abx − ( a 2 − b2 )2 = 0, (c) x 2 − 4abx + ( a 2 − b2 )2 = 0, (d) None of the above, 7. If x 2 − 3x + 2 be a factor of x 4 − px 2 + q, then ( p, q ) is, equal to, (a) ( 3, 4), (b) ( 4, 5), (c) ( 4, 3), (d) ( 5, 4), 8. If α and β be the roots of x 2 − px + q = 0 and α ′ , β ′ be, the roots of x 2 − p′ x + q ′ = 0, then the value of, (α − α ′ )2 + (β − α ′ )2 + (α − β ′ )2 + (β − β ′ )2 is, (a) 2 { p2 − 2q + p′ 2 − 2q ′ − pp′ }, (b) 2 { p2 − 2q + p′ 2 − 2q ′ − qq ′ }, (c) 2 { p2 − 2q − p′ 2 − 2q ′ − pp′ }, (d) 2{ p2 − 2q − p′ 2 − 2q ′ − qq ′ }, 9. If one root of the quadratic equation ax 2 + bx + c = 0, is equal to the nth power of the other root, then the, 1, n n +1, , value of ( ac ), , + ( a c), n, , 1, n +1, , is equal to, 1, , (a) b, , (b) −b, , (c) b n + 1, , 1, , (d) − b n + 1, , 10. If the difference between the roots of ax 2 + bx + c = 0, is 1, then which one of the following is correct?, (NDA 2012 I), , (a) b2 = a ( a + 4c), (c) a 2 = c( a + 4c), , (b) a 2 = b ( b + 4c), (d) b2 = a ( b + 4c), , 1, is one of the roots of ax 2 + bx + c = 0,, 2 − −2, where, a, b and c are real, then what are the values of, a, b and c, respectively?, (NDA 2010 I), (a) 6, − 4, 1 (b) 4, 6, − 1 (c) 3, − 2, 1 (d) 6, 4,1, , 11. If, , 12. If α and β are the roots of the quadratic equation, x 2 − x + 1 = 0, then which one of the following is, correct?, (NDA 2010 I), 4, 4, 5, 5, (a) (α − β ) is real, (b) 2(α + β ) = (αβ )5, 6, 6, (c) (α − β ) = 0, (d) (α 8 + β 8 ) = (αβ )8, 13. If p and q are positive integers, then which one of the, following equations has p − q as one of its roots?, (NDA 2009 I), (a) x 2 − 2 px − ( p2 − q ) = 0, 2, 2, (b) x − 2 px + ( p − q ) = 0, (c) x 2 + 2 px − ( p2 − q ) = 0, (d) x 2 + 2 px + ( p2 − q ) = 0, , 14. The equation x − 2( x − 1)−1 = 1 − 2( x − 1)−1 has, (a) no roots, (NDA 2009 I), (b) one root, (c) two equal roots, (d) infinite roots, 15. If a, b and c are real numbers, then the roots of the, equation, ( x − a )( x − b) + ( x − b)( x − c) + ( x − c)( x − a ) = 0 are, always, (NDA 2009 II), (a) real, (b) imaginary, (c) positive, (d) negative, 2, , 16. The solution set of the equation x log x (1 − x ) = 9 is, (a) { −2, 4}, (b) { 4}, (c) { 0, − 2, 4}, (d) None of these, 2x, 1, , then, 17. If 2, >, 2x + 5x + 2 x + 1, (a) −2 > x > − 1, (b) −2 ≥ x ≥ − 1, (c) −2 < x < − 1, (d) −2 < x ≤ − 1, 1, 1, 1, 18. If the roots of the equation, +, = are, x+ p x+q r, equal in magnitude but opposite in sign, then the, product of the roots will be, ( p2 + q 2 ), p2 + q 2, (a), (b) −, 2, 2, ( p2 − q 2 ), p2 − q 2, (d) −, (c), 2, 2, x+2, takes all, 19. If x is real, the expression, 2x 2 + 3x + 6, values in the interval, 1 1, 1 1, (b) −, (a) , , ,, 13 3, 13 3, 1 1, (c) − , , (d) None of these, 3 13, 20. If a < b < c < d, then the roots of the equation, ( x − a ) ( x − c) + 2 ( x − b) ( x − d ) = 0 are, (a) real and distinct, (b) real and equal, (c) imaginary, (d) None of these, 21. If x is an integer and satisfies 9 < 4x − 1 ≤ 19, then x is, an element of which one of the following sets?, (a) { 3, 4}, (b) { 2, 3, 4} (NDA 2008 II), (c) { 3, 4, 5}, (d) { 2, 3, 4, 5}, 22. If x is real and x 2 − 3x + 2 > 0, x 2 − 3x − 4 ≤ 0, then, which one of the following is correct?, (NDA 2008 I), (a) −1 ≤ x ≤ 4, (b) 2 ≤ x ≤ 4, (c) −1 < x ≤ 1, (d) −1 ≤ x < 1 or 2 < x ≤ 4, 23. The numerical value of the perimeter of a square, exceeds that of its area by 4. What is the side of the, square?, (NDA 2008 I), (a) 1 unit, (b) 2 units, (c) 3 units, (d) 4 units
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50, , NDA/NA Mathematics, , 24. If the equation x 2 + kx + 1 = 0 has the roots α and β,, then what is the value of (α + β ) × (α −1 + β −1 )?, (NDA 2008 I), 2, , (a) k, , (b), , 1, 2, , k, , 2, , (c) 2k, , (d), , 1, ( 2k2 ), , 25. If α and β are the roots of the equation x 2 + x + 1 = 0,, then what is the equation whose roots are α19 and β7 ?, (NDA 2007 II), , (a) x 2 − x − 1 = 0, (c) x 2 + x − 1 = 0, , (b) x 2 − x + 1 = 0, (d) x 2 + x + 1 = 0, , 26. The number of real solution of the equation, |x|2 − 3|x| + 2 = 0 are, (a) 1, (b) 2, (c) 3, (d) 4, 27. Let α and γ be the roots of Ax 2 − 4x + 1 = 0 and β , δ be, the roots of Bx 2 − 6x + 1 = 0. If α, β, γ and δ are in HP,, then what are the values of A and B, respectively?, (NDA 2009 I), , (a) 3, 8, (c) 3, –8, , (b) –3, –8, (d) –3, 8, , 28. If x is real, then the maximum and minimum values, x 2 − 3x + 4, of the expression 2, will be, x + 3x + 4, 1, (a) 2, 1, (b) 5,, 5, 1, (d) None of these, (c) 7,, 7, 29. If l , m and n are real and l ≠ m, then the roots of the, equation ( l − m ) x 2 − 5 ( l + m ) x − 2 ( l − m ) = 0 are, (a) complex, (b) real and distinct, (c) real and equal, (d) None of these, 30. The number of rows in a lecture hall equals to the, number of seats in a row. If the number of rows is, doubled and the number of seats in every row is, reduced by 10, the number of seats is increased by, 300. If x denotes the number of rows in the lecture, hall, then what is the value of x?, (NDA 2007 II), (a) 10, (b) 15, (c) 20, (d) 30, 31. If α and β are the roots of ax 2 + 2bx + c = 0 and α + δ ,, β + δ are the roots of Ax 2 + 2Bx + C = 0, then what is, ( b2 − ac)/( B2 − AC ) equal to, (NDA 2007 I), 2, , b, (a) , B, ( a 2b2 ), (c), ( A2B2 ), , 2, , a, (b) , A, ( ab), (d), ( AB), , 32. How many real values of x satisfy the equation, (NDA 2007 I), | x| +| x − 1| = 1 ?, (a) 1, (b) 2, (c) Infinite, (d) No value of x, , 33. If the roots of the equations x 2 − ( a − 1)x + ( a + b) = 0, and ax 2 − 2x + b = 0 are identical, then what are the, values of a and b?, (NDA 2007 I), (a) a = 2, b = 4, (b) a = 2, b = − 4, 1, 1, (d) a = − 1, b = −, (c) a = 1, b =, 2, 2, 34. If − x 2 + 3x + 4 > 0, then which one of the following is, correct?, (NDA 2007 I), (a) x ∈ ( −1, 4), (b) x ∈ [−1, 4], (c) x ∈ ( −∞ , − 1) ∪ ( 4, ∞ ) (d) x ∈ ( −∞ , 1] ∪ [4, ∞ ), 35. If the two quadratic equations x 2 − bx + c = 0 and, x 2 − b′ x + c′ = 0 have a common root, what is the, value of the common root ?, b − b′, c − c′, (a), (b), c − c′, b − b′, b − b′, c − c′, (d), (c), c′ − c, b′ − b, 36. f ( x ) = x 2 + 2ax + 1 and α is a root of the equation, f ( x ) = 0, where a is real., Which one of the following is correct ?, (a) f(α ) = 0 and f(1 / α ) ≠ 0, (b) f(α ) = 0 and f(1 / α ) = 0, (c) f(α ) ≠ 0 and f(1 / α ) = 0, (d) f(α ) ≠ 0 and f(1 / α ) ≠ 0, 37. The roots of the quadratic x 2 + 4a = 8x − 12a 2 are real, and unequal. Which one of the following is correct?, (a) 4 / 3 < a < 2, (b) − 4 / 3 < a < − 1, (c) −4 / 3 < a < 2, (d) −4 / 3 < a < 1, 38. The sum of the two roots of a quadratic equation is, 3, λ and the sum of their squares is 5 µ 2 . Which one of, the following is that equation ?, 3, , 5, , (a) x 2 − 3 λ x + ( λ 2 − λ 2 ) = 0, 3, , (b) x 2 − 3 λ x + ( λ 2 + 5 µ 2 ) = 0, 3, , (c) 2x 2 − 2 3 λ x + ( λ 2 − 5 µ 2 ) = 0, 3, , (d) 2x 2 − 2 3 λ x + ( λ 2 + 5 µ 2 ) = 0, 39. The roots of the equation x 2 + px + q = 0 both real, and greater than 1. If r = p + q + 1, then which one of, the following is correct?, (a) r must be greater than 0, (b) r must be less than 0, (c) r must be equal to 0, (d) r may be equal to 0, 40. Which one of the following is correct?, 7 , 7 , The equation x − , = 3− , , x − 3, x − 3, (a) has only one integral root, (b) has no roots, (c) has two equal integral roots, (d) has two unequal integral roots
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51, , Quadratic Equations and Inequalities, 41. One of the roots of a quadratic equation with real, 1, . Which of the following, coefficients is, (2 − 3 i), implications is/are true?, 1, ., I. The second root of the equation will be, (3 − 2 i), II. The equation has no real root., III. The equation is 13x 2 − 4x + 1 = 0., Which of the above is/are correct ?, (a) Only I and II, (b) Only III, (c) Only II and III, (d) I, II and III, , A. 6 x2 + 7 x − 10 = 0, , C. 3 x2 + 5 x + 3 = 0, , Select the correct answer using the code given below, (a) Only I and II, (b) I, II and III, (c) Only II, (d) Only I, 43. Which one of the following sets has all elements as, odd positive integers?, (a) S = { x ∈ R x3 − 8x 2 + 19x − 12 = 0}, (b) S = { x ∈ R x3 − 9x 2 + 23x − 15 = 0}, (c) S = { x ∈ R x3 − 7x 2 + 14x − 8 = 0}, (d) S = { x ∈ R x3 − 12x 2 + 44x − 48 = 0}, solutions, , of, , 45. Consider the following statements, I. If the quadratic equation is ax 2 + bx + c = 0 such, that a + b + c = 0, then roots of the equation, c, ax 2 + bx + c = 0 will be 1, ., a, II. If ax 2 + bx + c = 0 is a quadratic equation such, that a − b + c = 0, then roots of the equation will, c, be −1, ., a, Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, , List II, (Their roots), , List I, (Equations), , B. x + 1 = 5, x 2, , 42. Given, 4a − 2 b + c = 0, where a , b, c ∈ R, which of the, following statements is/are not true in general?, I. ( x + 2) will always be a factor of the expression, ax 2 + bx + c., II. ( x − 2) will always be a factor of the expression, ax 2 + bx + c., III. There will be a factor of the expression, ax 2 + bx + c different from ( x + 2)., , 44. What is the number of real, |x 2 − x − 6| = x + 2 ?, (a) 4, (b) 3, (c) 2, (d) 1, , 46. Match List I (equations ) with List II (their roots ), and select the correct answer using the codes given, below the lists., , Codes, A B, (a) 3 4, (c) 1 4, , C, 2, 2, , 1. 6, 40, 13, 2. − 5 + i 11 −5 − i 11, ,, 6, 6, 3. −2, 5, 6, 4. 2, 1, 2, , A, (b) 3, (d) 1, , B, 1, 3, , C, 2, 2, , 47. Let a , b ∈ {1, 2, 3}. What is the number of equations of, the form ax 2 + bx + 1 = 0 having real roots?, (a) 1, (b) 2, (c) 5, (d) 3, 48. If 3 <| x|< 6, then x belongs to, (a) ( −6, − 3) ∪ ( 3, 6), (b) ( −6, 6), (c) ( −3, − 3) ∪ ( 3, 6), (d) None of these, 1 1, 49. If < , then, a b, (a) | a|>| b|, (b) a < b, (c) a > b, (d) None of these, 2, 50. x + < 3, then x belongs to, x, , (a) ( −2, − 1) ∪ (1, 2), (b) ( −∞ , − 2) ∪ ( −1, 1) ∪ ( 2, ∞ ), (c) ( −2, 2), (d) ( −3, 3), 51. x 2 − 3|x| + 2 < 0, then x belongs to, (a) (1, 2), (b) ( −2, − 1), (c) ( −2, − 1) ∪ (1, 2), (d) ( −3, 5), 52. The set of values of x satisfying the inequalities, ( x − 1) ( x − 2) < 0 and ( 3x − 7) ( 2x − 3) > 0 is, 7, (a) (1, 2), (b) 2 , , 3, 7, 3, (d) 1, , (c) 1, , 3, 2, 53. log2 x > 4, then x belongs to, (a) x > 4, (b) x > 16, (c) x > 8, (d) None of these
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52, , NDA/NA Mathematics, , Directions (Q. Nos. 54-58), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true and R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 54. Assertion (A) One root of the equation, 7x 2 + 5x + 7 = 0 is reciprocal of the other., Reason, (R) For, the, quadratic, equation, ax 2 + bx + c = 0 one root will be reciprocal of the, other, if a = c., 55. Assertion (A) The common root of the quadratic, equation 2x 2 − 3x − 5 = 0 and x 2 − 3x − 4 = 0 is −1., Reason (R) If the common root of the quadratic, equation a1x 2 + b1x + c1 = 0 and a2x 2 + b2x + c2 = 0 is, α, then, c a − c2a1, ., α= 1 2, a1b2 − a2b1, 56. Assertion (A) The real quadratic equation whose, one root is 2 − 3 is x 2 − 4x + 1 = 0., Reason (R) If an equation has a root 2 − 3, then, another root will be 2 + 3., 57. Assertion (A) For b = − 5, x + 3 is a factor of, x3 + 2x 2 + bx − 6 ., Reason (R) If f ( x ) is a polynomial and f ( a ) = 0,, then x − a is a factor of f ( x )., 58. Assertion (A) The equation x 2 + 2ax − b2 = 0 can, have repeated roots, where a and b are real numbers., Reason (R) The equation Ax 2 + Bx + C = 0 will, have repeated roots when the discriminant becomes, zero., , Directions (Q. Nos. 59-61), , Consider the, quadratic equation 2x 2 − 8x + 3 = 0, whose roots are α, and β, then, 59. The sum and product of the roots are, respectively is, 3, (a) 3, 4, (b) 4,, 2, 3, (d) 4, 7, (c) 5,, 2, 60. The value of α3 + β3 is, (a) 46, (b) 40, (c) 30, (d) − 18, 1, 1, 61. The quadratic equation, if roots are and , is, α, β, (b) x 2 − 4x + 3 = 0, (a) 3x 2 − 8x + 2 = 0, (c) 3x 2 − 4x + 1 = 0, (d) None of these, , Directions (Q. Nos. 62-64), , Consider, equation ax 2 + bx + c = 0, then condition that, , the, , 62. One roots is the reciprocal of the other roots is, c, (a) a = c, (b) a = −, 2, (c) 2b = a, (d) b = a, 63. One roots is n times the other root is, (b) ab2 ( n + 1)2, (a) ac ( n + 1)2 = b2n, 2, 2, (d) 4a 2 = b2, (c) ac ( n + 2) = b, 64. One root is the square of the other root, (a) ca 2 + c2a − 3abc = − b3 (b) a + b + c = 0, (d) None of these, (c) a3 + b3 + c3 = 0, , Directions (Q. Nos. 65-66) The equation formed, by multiplying each root of ax 2 + bx + c = 0 by 2 is, x 2 + 36x + 24 = 0., (NDA 2012 I), 65. What is the value of b : c?, (a) 3 : 1, (b) 1 : 2, (c) 1 : 3, , (d) 3 : 2, , 66. Which one of the following is correct?, (b) bc = 36a 2, (a) bc = a 2, 2, (c) bc = 72a, (d) bc = 108a 2, , Answers, Level I, 1., 11., 21., 31., 41., 51., , (a), (a), (a), (d), (a), (d), , 2., 12., 22., 32., 42., 52., , (d), (a), (b), (a), (a), (a), , 3., 13., 23., 33., 43., 53., , (b), (b), (a), (b), (a), (a), , 4., 14., 24., 34., 44., , (c), (d), (a), (c), (c), , 5., 15., 25., 35., 45., , (a), (b), (c), (a), (b), , 6., 16., 26., 36., 46., , (c), (d), (b), (d), (b), , 7., 17., 27., 37., 47., , (d), (a), (b), (d), (a), , 8., 18., 28., 38., 48., , (c), (a), (c), (c), (d), , 9., 19., 29., 39., 49., , (a), (d), (c), (d), (b), , 10., 20., 30., 40., 50., , (d), (a), (b), (c), (d), , 2., 12., 22., 32., 42., 52., 62., , (d), (c), (d), (b), (c), (d), (a), , 3., 13., 23., 33., 43., 53., 63., , (b), (b), (b), (b), (b), (b), (a), , 4., 14., 24., 34., 44., 54., 64., , (a), (a), (a), (a), (b), (a), (a), , 5., 15., 25., 35., 45., 55., 65., , (d), (a), (d), (b), (c), (a), (a), , 6., 16., 26., 36., 46., 56., 66., , (b), (b), (d), (b), (a), (a), (d), , 7., 17., 27., 37., 47., 57., , (d), (c), (a), (d), (d), (a), , 8., 18., 28., 38., 48., 58., , (a), (b), (c), (c), (a), (a), , 9., 19., 29., 39., 49., 59., , (b), (b), (b), (a), (d), (b), , 10., 20., 30., 40., 50., 60., , (a), (a), (d), (b), (a), (a), , Level II, 1., 11., 21., 31., 41., 51., 61., , (c), (a), (c), (b), (c), (c), (a)
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54, , NDA/NA Mathematics, 3, ,4, 2, 3, ⇒, y + 2 = − ,4, 2, 3, y = − 2 − , 4 –2, ⇒, 2, 7, ⇒, y = − ,2, 2, 7, ∴ Required roots are − , 2., 2, ⇒, , So,, , z=−, , [from Eq. (i)], , 10. The given quadratic equation is x2 + x + 1 = 0, −1 ± 1 − 4 −1 ± i 3, x=, =, 2, 2, −1 + i 3 −1 − i 3, x=, ,, 2, 2, or, x = ω, ω2, i.e.,, α = ω and β = ω 2, 19, 19, ⇒, α = ω = (ω3 )6 ⋅ ω = (1)6 ⋅ ω = ω, ⇒, β7 = (ω 2)7 = ω14 = (ω3 )4 ⋅ ω 2 = (1)4 ⋅ ω 2 = ω 2, Now, sum of roots, α 19 + β7 = ω + ω 2 = − 1, (Q 1 + ω + ω 2 = 0), and product of roots, α 19 ⋅ β7 = ω ⋅ ω 2 = ω3 = 1, (Q ω3 = 1), So, the required quadratic equation, whose roots are α 19, and β7, is, x2 − (α 19 + β7 )x + (α 19 ⋅ β7 ) = 0, ⇒, x2 − (−1)x + (1) = 0, ⇒, x2 + x + 1 = 0, (x + 1) (x − 3), . Here, x cannot be 2., 11. We have y =, (x − 2), ⇒ Either both N r and D rare positive., ∴, x ≥ − 1, x ≥ 3 and x > 2 ⇒ x ≥ 3, or N r is negative and D r is negative, x ≥ −1 and x < 2 ⇒ −1 ≤ x < 2, From Eqs. (i) and (ii), we get, −1 ≤ x < 2 or x ≥ 3, , ...(i), ...(ii), , 12. Since, 2 + i 3 is a root of the equation x2 + px + q = 0., Therefore, 2 − i 3 will be other root., Now, sum of the roots, (2 + i 3 ) + (2 − i 3 ) = − p ⇒ 4 = − p, and product of roots (2 + i 3 ) (2 − i 3 ) = q, ⇒, 7=q, Hence, ( p, q) = (−4, 7), 13. Let the equation (in correctly written form) be, x2 + 17x + q = 0, Roots are −2 and −15. Then, its equation, (x + 2) (x + 15) = 0, ⇒, x2 + 17x + 30 = 0, On comparing Eqs. (i) and (ii), we get q = 30, So, correct equation is x2 + 13x + 30 = 0., Hence, roots are −3 and − 10., 14. Given equation is,, ( p − q) x2 + 5 ( p + q) x − 2 ( p − q) − r = 0, Let, A = ( p − q), B = 5 ( p + q), and, C = − 2 p + 2q − r = − (2 p − 2q + r ), , …(i), , …(ii), , B2 − 4 AC = 25 ( p + q)2 + 4 ( p − q) (2 p − 2q + r ), = 25 ( p + q)2 + 4{2 ( p − q)2 + rp − rq}, , = 25 ( p + q)2 + 8 ( p − q)2 + 4r ( p − q), Now, it depends on the value of p, q and r., 15. Given equation is,, ( p2 + q2)x2 − 2q ( p + r ) x + (q2 + r 2) = 0, Since, roots are real and equal, then, b2 − 4ac = 0, ⇒, 4q2 ( p + r )2 − 4( p2 + q2) (q2 + r 2) = 0, 2, 2, 2, ⇒ q ( p + r + 2 pr ) − ( p2q2 + p2r 2 + q4 + q2r 2) = 0, ⇒ q2p2 + q2r 2 + 2 pq2r − p2q2 − p2r 2 − q4 − q2r 2 = 0, ⇒, 2 pq2r − p2r 2 − q4 = 0 ⇒ (q2 − pr )2 = 0, Hence, q2 = pr. Thus, p, q and r are in GP., 16. Let y = 8 + 2 8 + 2 8 + 2 8 + K ∞, y = 8 + 2 y ⇒ y2 = 8 + 2 y, ⇒, ⇒, y2 − 2 y − 8 = 0, 2, ⇒, y − 4y + 2y − 8 = 0, ⇒ ( y − 4)( y + 2) = 0 ⇒ y = − 2, 4, ∴ Required value of expression is 4., 17. Given that, α and β be the roots of the equation, x2 − q(1 + x) − r = 0, ⇒, x2 − qx − (q + r ) = 0, Then sum of the roots = q, i.e.,, α+β=q, and product of the roots = − (q + r ), i. e.,, αβ = − (q + r ), Now, (1 + α ) (1 + β ) = 1 + (α + β ) + αβ, = 1 + q − (q + r ), =1 − r, , …(i), …(ii), , 18. x4 − 26x2 + 25 = 0 ⇒ x4 − 25x2 − x2 + 25 = 0, ⇒, (x2 − 25)(x2 − 1) = 0, ⇒, (x − 5)(x + 5)(x − 1)(x + 1) = 0, ⇒, x = 5, − 5, 1, − 1, ∴ Solution set for the given equation is { −5, 5, − 1, 1}., 19. Since, the equations x2 + kx + 64 = 0 and x2 − 8x + k = 0, have real roots., ∴, k2 ≥ 4 × 64, (Q B2 − 4 AC ≥ 0), …(i), ⇒, k ≥ 16, and, 64 ≥ 4k, …(ii), ⇒, k ≤ 16, From Eqs. (i) and (ii), we get, k = 16, 20. Let α and β be the roots of the equation, x2 − 5x + k − 15 = 0, ∴, αβ = k − 15, But, (given), αβ = − 3, ⇒, −3 = k − 15, ⇒, k = 15 − 3 = 12, α, β, 21. Given equation, +, = 1 can be rewritten as, x−α x−β, ⇒, x2 − 2 (α + β )x + 3αβ = 0, Let roots be α′ and − α ′., ∴, α ′ + (− α ′ ) = 2 (α + β ) ⇒ 0 = 2 (α + β ), ⇒, α + β =0
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55, , Quadratic Equations and Inequalities, 22. Q Equation x2 − bx + 1 = 0 has no real roots i.e., it has, imaginary roots which is possible only, if, b2 − 4 < 0, (Q B2 − 4 AC < 0), 2, ⇒, b <4, ⇒, −2 < b < 2, , Now,, , 1, 1 α 2 + β2, + 2=, 2, (αβ )2, α, β, 9, 7, −, (α + β )2 − 2αβ 16 2, =, =, 49, (αβ )2, 16, 9 − 56, −47 16, 47, = 16 =, ×, =−, 49, 16, 49, 49, 16, , α −2 + β −2 =, , 23. Q p and q are the roots of x2 − px + q = 0, ∴, p + q = p, pq = q, ⇒, q( p − 1) = 0, ⇒, q = 0, p = 1, 24. Since, α and β are the roots of ax2 + bx + b = 0, b, b, and αβ =, ∴, α+β=−, a, a, b α+β, b, α, β, ∴, +, +, =, +, a, a, β, α, αβ, − b /a, b, b, b, =0, =, +, =−, +, a, a, a, b /a, , 30. Given, equations, qx2 − 2 pr x + q = 0., , 2x + 3y = 17, 2, − 3y + 1 = 5, x, 4 ⋅ 2 − 3 ⋅ 3y = 5, From Eqs. (i) and (ii), we get, 2x = 8 and 3y = 9, ⇒, x = 3 and y = 2, , 32. Q, and, , …(i), , x+ 2, , …(ii), , 3, , On taking cube of both the sides, we get, (x − 2)3 = 2(1 + 21/3 )3, x3 − 8 − 6x2 + 12x = 2(1 + 2 + 3 ⋅ 21/3 + 3 ⋅ 22/3 ), x3 − 6x2 + 6x = 14 + 6 ⋅ 21/3 + 6 ⋅ 22/3 − 6x, 14 + 6 ⋅ 21/3 + 6 ⋅ 22 /3 − 6(2 + 21/3 + 22/3 ) = 2, , (Q D = B2 − 4 AC ), , = ( p − q )2 + 4 r 2 ≥ 0, Hence, the roots are always real., 28. Let given equations have common root α., Then, α 2 + mα + 1 = 0 and α 2 + α + m = 0, 1, α2, α, ⇒, =, =, m2 − 1 1 − m 1 − m, α, 1, =, ⇒ α =1, ⇒, 1 −m 1 −m, α2, 1, Also,, =, m2 − 1 1 − m, ⇒, ⇒, ⇒, ⇒, , and, , ...(i), ∴, 4q2 − 4 pr ≥ 0 ⇒ q2 ≥ pr, and from second 4 ( pr ) − 4q2 ≥ 0, ...(ii), ⇒, pr ≥ q2, From Eqs. (i) and (ii), we get q2 = pr, 2, 3, 31. Since, 2 < 3 ⇒, >, (−2) (−2), On multiplying or dividing an inequality by a negative, number on both sides its sign changes., , 26. Given, x − 2 = 21/3 (1 + 21/3 ), , 27. Given, (x − p) (x − q) = r 2, x2 − ( p + q) x + pq − r 2 = 0, Now, D = ( p + q)2 − 4( pq − r 2), , px2 + 2qx + r = 0, , They have real roots., , 25. Since, sin α and cos α are the roots of ax2 + bx + c = 0,, −b, …(i), ∴, sin α + cos α =, ,, a, c, …(ii), and, sin α cos α =, a, 2, b, [from Eq. (i)], ⇒ sin 2 α + cos 2 α + 2 sin α cos α = 2, a, 2c b2, [from Eq. (ii)], =, 1+, ⇒, a a2, ⇒, b2 − a 2 = 2ac, , ⇒, ⇒, ⇒, , are, , 1 − m = m2 − 1, m2 + m − 2 = 0, (m + 2)(m − 1) = 0, m = 1 and –2, , 29. Q α and β are the roots of the equation, 4 x2 + 3 x + 7 = 0, 3, 7, and αβ =, α+β=−, ∴, 4, 4, , (Q α = 1), , a − x, a+x, 16 , =, a + x, a−x, , 33., , 4, , ⇒, , 4, a − x, 1, , = , 2, a + x, , ⇒, , a−x 1, =, a+x 2, , ⇒, , 2a − 2x = a + x, , ⇒, , a = 3x ⇒ x =, , a, 3, , 34. Since, r and s are the roots of Ax2 + Bx + C = 0, then, B, C, and rs =, r + s=−, A, A, Now, the roots of x2 + px + q = 0 be r 2 and s2., ∴, r 2 + s2 = − p and r 2s2 = q, 2, ⇒, (r + s) − 2rs = − p, B2 2C, −, =−p, ⇒, A, A2, B2 − 2 AC, ⇒, =−p, A2, 2 AC − B2, ⇒, p=, A2, 35. Let the roots of the equation x2 − bx + c = 0 be α and, α + 1., ∴, α + (α + 1) = b, ⇒, 2α + 1 = b
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58, , NDA/NA Mathematics, Eliminating α, we get, c, , a, ⇒, , a⋅a, , −, , 1, n+ 1, , ⇒, , c, , 1, n+ 1, , 1, n+ 1, , c, + , a, , + a⋅a, , (a nc), , 1, n+1, , −, , n, n+ 1, , n, n, n+ 1 n + 1, , c, , 1, n n+1, , + (ac ), , =−, , b, a, , = −b, = −b, , 10. Let the roots of the equation ax + bx + c = 0 are α and, (α − 1) by given condition., b, Then, sum of the roots = −, a, b, ⇒, α + (α − 1) = −, a, b, a−b, 2α = 1 −, ⇒ α=, ⇒, a, 2a, c, and product of the roots =, a, c, α (α − 1) =, ⇒, a, (a − b) a − b, c, ⇒, − 1 =, , 2a 2a, a, (a − b) (− b − a ) c, ⋅, =, ⇒, 2a, 2a, a, ⇒, − (a 2 − b2) = 4ac, ⇒, b2 − a 2 = 4ac, ⇒, b2 = a (a + 4c), 1, 1, 2 + 2i, 11. ∴ Given, root =, =, ×, 2 − −2 2 − 2 i 2 + 2 i, 2 + 2i 2 + 2i, =, =, 4+2, 6, 2 − 2i, ∴ Another root will be, ., 6, 2 + 2i 2 − 2 2i 4, Thus, sum of roots =, =, +, 6, 6, 6, 2 + 2i 2 − 2i , and product of roots = , , , 6 6 , 4+2 1, =, =, 36, 6, ∴ Required equation is, x2 − (sum of roots) x + (product of roots) = 0, 4, 1, x2 − x + = 0, 6, 6, ⇒, 6 x2 − 4 x + 1 = 0, Thus, the values of a , b and c are 6, – 4 and 1,, respectively., 2, , 12. Q α and β be the roots of the equation x2 − x + 1 = 0, ∴, α + β = 1 and αβ = 1, Now, α − β = (α + β )2 − 4αβ = 3i, ⇒, Now,, and, , 1+ i 3, 1−i 3, and β =, 2, 2, π, π, α = cos + i sin, 3, 3, π, π, β = cos − i sin, 3, 3, α=, , 4π, 4π, 4π, 4π, + i sin, − cos, + i sin, 3, 3, 3, 3, (by de-Moivre’s theorem), 4π, = 2i sin, 3, ⇒ α 4 − β 4 is not real., 5π, 5π, 5π, 5π , , (b) 2(α 5 + β5 ) = 2 cos, + i sin, + cos, − i sin, , , 3, 3, 3, 3, 5π, 1, = 2 ⋅ 2 cos, = 4⋅ = 2, 3, 2, Now,, (αβ )5 = 1, ⇒ 2(α 5 + β5 ) ≠ (αβ )5, 6π, 6π, 6π, 6π, (c) α 6 − β 6 = cos, + i sin, − cos, + i sin, 3, 3, 3, 3, = 2i sin 2π = 0, 8π, 8π, 8π, 8π, (d)α 8 + β 8 = cos, + i sin, + cos, − i sin, 3, 3, 3, 3, 8π, 1, = 2 cos, = 2 − = −1, 2, 3, 8, Now,, (αβ ) = (1)8 = 1, 8, ⇒, (α + β 8 ) ≠ (αβ )8, , (a) α 4 − β 4 = cos, , 13. If any equation has p − q as a root, then the another, root will be p + q., Sum of roots = p − q + p + q = 2 p,, and product of roots = ( p − q )( p + q ), = p2 − q, Now, the required equation is, x2 − (sum of roots)x + (product of roots)= 0, ⇒, x2 − 2 px + ( p2 − q) = 0, 14. Given, x − 2(x − 1)−1 = 1 − 2(x − 1)−1, 2, 2, x−, =1 −, ⇒, ⇒x=1, x−1, x−1, But x = 1 doesn’t satisfy the given equation., Hence, no roots exist., 15. Given,, (x − a )(x − b) + (x − b)(x − c) + (x − c)(x − a ) = 0, ⇒, 3x2 − 2(b + a + c)x + ab + bc + ca = 0, Now,, D = 4(a + b + c)2 − 12(ab + bc + ca ), = 2 a 2 + b2 + c2 − ab − bc − ca, 1, {(a − b)2 + (b − c)2 + (c − a )2} > 0, 2, Hence, roots are always real., =2, , 2, , 16. We have xlog x (1 − x ) = 9, Taking log on both sides, we get, log x 9 = log x (1 − x)2, (Q a x = N ⇒ log a N = x), 2, 2, ⇒, 9 = (1 − x) ⇒ 1 + x − 2x − 9 = 0, ⇒, x2 − 2 x − 8 = 0, ⇒, x = − 2, 4, ⇒, x=4, (Q x ≠ −2), 2x, 1, 17. Given that,, >, 2 x2 + 5 x + 2 x + 1, 2x, 1, ⇒, >, (2x + 1) (x + 2) (x + 1)
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59, , Quadratic Equations and Inequalities, , ⇒, ⇒, ⇒, ⇒, , 2x, 1, −, >0, (2x + 1) (x + 2) (x + 1), 2x (x + 1) − (2x + 1) (x + 2), >0, (x + 1) (x + 2) (2x + 1), 2 x 2 + 2 x − 2 x2 − 4 x − x − 2, >0, (x + 1) (x + 2) (2x + 1), −3 x − 2, >0, (x + 1) (x + 2) (2x + 1), , 22. Q, ⇒, , +, –∞, , ⇒, and, ⇒, , Equating each factor equal to 0, we have, 2, 1, x = − 2, − 1, − , −, 3, 2, 2, 1, It is clear that − < x < − or −2 < x < − 1 ., 3, 2, 1, 1, 1, 18. Given equation, +, = can be rewritten as, x+ p x+q r, x2 + x ( p + q − 2r ) + pq − pr − qr = 0, Let roots are α and −α, then the product of roots, , −α 2 = pq − pr − qr = pq − r ( p + q ), , …(i), , ...(ii), , and sum of roots, 0 = − ( p + q − 2r ), p+ q, ...(iii), r=, ⇒, 2, Solving Eqs. (ii) and (iii), we get, p+ q, 1, ( p + q) = − {( p + q)2 − 2 pq}, − α 2 = pq −, 2, 2, =−, , ( p2 + q 2 ), 2, , 19. Let the given expression be y, i.e., y =, , x2 − 3x + 2 > 0, (x − 1)(x − 2) > 0, , x+2, 2x + 3x + 6, 2, , ⇒, 2x2y + (3 y − 1) x + (6 y − 2) = 0, If, y ≠ 0,, then ∆ ≥ 0 for real x, i. e. ,, B2 − 4 AC ≥ 0, ⇒ (3 y − 1)2 − 8 y (6 y − 2) ≥ 0 ⇒ −39 y2 + 10 y + 1 ≥ 0, 1, 1, ≤ y≤, ⇒, (13 y + 1) (3 y − 1) ≤ 0 ⇒ −, 13, 3, If y = 0, then x = − 2 which is real and this value of y is, included in the above range., 20. Given equation can be rewritten as, 3x2 − (a + c + 2b + 2d ) x + (ac + 2bd ) = 0, Its discriminant, D = (a + c + 2 b + 2 d )2 − 4 ⋅ 3 (ac + 2 bd ), = {(a + 2 d ) + (c + 2 b)}2 − 12 (ac + 2 bd ), = {(a + 2 d ) − (c + 2 b)}2 + 4 (a + 2 d ) (c + 2 b), − 12 (ac + 2 bd ), = {(a + 2 d ) − (c + 2 b)}2 − 8ac + 8ab + 8dc − 8bd, = {(a + 2 d ) − (c + 2 b)}2 + 8 (c − b) (d − a ), Which is positive, since a < b < c < d. Hence, roots are, real and distinct., 21. Given, 9 < 4x − 1 ≤ 19, ⇒, 9 < 4x − 1, and 4x − 1 ≤ 19, and 4x ≤ 19 + 1, ⇒, 9 + 1 < 4x, 5, and, ⇒, x>, x≤ 5, 2, ⇒, 5 /2 < x ≤ 5, ∴, x ∈{3, 4, 5}, , {Q x ∈ Z }, , +, +∞, , 2, , x < 1 or x > 2, x2 − 3 x − 4 ≤ 0, (x − 4)(x + 1) ≤ 0, , –∞, , ∴, ⇒, , –, 1, , +, –1, , …(i), , –, , +, 4, , −1 ≤ x ≤ 4, −1 ≤ x < 1 or 2 < x ≤ 4, , +∞, , …(ii), [from Eqs. (i) and (ii)], , 23. Let the side of the square = x units, Area of square = x2 units,, ∴, and perimeter of the square = 4x units, According to the question,, x2 + 4 = 4 x, 2, ⇒, x − 4x + 4 = 0, ⇒, (x − 2)2 = 0, ⇒, x=2, Side of square = 2 units, ∴, 24. Q The roots of the equations x2 + kx + 1 = 0 are α and β., ∴, α + β = − k and αβ = 1, 1 1, Now, (α + β )(α −1 + β −1 ) = (α + β ) + , α β, α + β, = (α + β ) , , αβ , =, , (α + β )2 (− k)2, =, = k2, 1, αβ, , 25. Since, α and β are the roots of the equation x2 + x + 1 = 0, ∴, α = ω and β = ω 2, ∴, α 19 + β7 = ω19 + ω14 = ω + ω 2, (Q ω 2 + ω + 1 = 0 and ω3 = 1), = −1, and, α 19 ⋅ β7 = ω19 ⋅ ω14 = ω33 = 1, ∴ The required equation whose roots are α 19 and β7 , is, x2 − (α 19 + β7 ) x + α 19 ⋅ β7 = 0, ⇒, x2 + x + 1 = 0, 26. We have |x|2 − 3|x| + 2 = 0, ⇒ |x| − 2|x| − |x| + 2 = 0, ⇒, (|x| − 2) (|x| − 1) = 0, ⇒, |x| = 1 and|x| = 2 ⇒ x = ± 1, ± 2, 27. Since, α and γ be the roots of Ax2 − 4x + 1 = 0, 4, 1, and αγ =, α+γ=, ∴, A, A, And β and δ be the roots of Bx2 − 6x + 1 = 0., 6, 1, and βδ =, ∴, β+δ=, B, B, Also, α , β , γ and δ are in HP., 1 1 1, 1, and are in AP., , ,, ∴, α β γ, δ, 1 1 1 1, − = −, ⇒, β α δ γ
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60, , NDA/NA Mathematics, 1 1 1 1, − = −, β δ α γ, δ −β γ −α, =, βδ, αγ, , ⇒, ⇒, , and, ⇒, , (δ + β )2 − 4βδ, (γ + α )2 − 4αγ, =, βδ, αγ, 36 4, 16 4, −, −, B2 B = A 2 A, ⇒, 1, 1, A, B, 36 − 4B = 16 − 4 A, ⇒, 36 − 4B = 16 − 4 A, ⇒, 4 A − 4B = − 20, It is possible, when A = 3 and B = 8., ⇒, , 28. Let y =, , c, a, , ⇒, , c, a, , 2, , ⇒, ⇒, ⇒, , x2 − 3 x + 4, x2 + 3 x + 4, , ⇒, ( y − 1)x2 + 3( y + 1)x + 4( y − 1) = 0, For x is real, D ≥ 0, ⇒ 9 ( y + 1)2 − 16 ( y − 1)2 ≥ 0 ⇒ −7 y2 + 50 y − 7 ≥ 0, ⇒ 7 y2 − 50 y + 7 ≤ 0 ⇒ ( y − 7) (7 y − 1) ≤ 0, 1, 1, ⇒, y ≤ 7 and y ≥ ⇒ ≤ y ≤ 7, 7, 7, , ⇒, , C, [from Eq. (iii)], A, C, αβ + (α + β ) δ + δ 2 =, A, 2, C, 2b b B b B , −, − + − =, a a A a A, A, [from Eqs. (i) and (iv)], 2, 2, 2b2 2bB b , 2bB C, B, =, − 2 +, + + −, , , , , aA, a, A, aA, A, a, (α + δ )(β + δ ) =, , ⇒, , ...(i), , 1, Hence, maximum value is 7 and minimum value is ., 7, 29. Given equation is (l − m)x2 − 5 (l + m)x − 2 (l − m) = 0, Its discriminant, D = 25 (l + m)2 + 8 (l − m)2 > 0, Since, l, m and n are real and l ≠ m., Hence, roots are real and distinct., 30. Q Number of rows = x, ∴ Number of seats in each row = x, Thus, total number of seats = x2, Now, new number of rows = 2x, and new number in each row = x − 10, Thus, total number of seats = 2x(x − 10) = 2x2 − 20x, According to the question, 2x2 − 20x − x2 = 300, ⇒, x2 − 20x − 300 = 0, 2, ⇒, x − 30x + 10x − 300 = 0, ⇒, (x − 30)(x + 10) = 0, ⇒, x = 30, (Q x ≠ − 10), 31. Since, α and β are the roots of ax2 + 2bx + c = 0., 2b, α+β=−, ∴, a, c, and, …(i), αβ =, a, Also, α + δ and β + δ are the roots of, Ax2 + 2Bx + C = 0., 2B, …(ii), (α + β ) + 2δ = −, A, C, …(iii), and, (α + δ )(β + δ ) =, A, 2b, 2B, [from Eqs. (ii) and (i)], ⇒, −, + 2δ = −, a, A, b B, …(iv), δ= −, ⇒, a A, , 2, , c b, C, B, − + =, A, a a, A, 2, B, C b2 c, −, =, −, 2, A a2 a, A, B2 − AC b2 − ac, =, A2, a2, 2, 2, b − ac, a, = , 2, B − AC A , , 32. The given equation is, | x| + | x − 1| = 1, If x > 0, then given equation becomes, x + x − 1 = 1 ⇒ 2x = 2 ⇒ x = 1, If x = 0, then given equation becomes, 0 + 1 =1, and if x < 0, then the given equation becomes, − x− x+ 1 =1, ⇒, −2 x = 0 ⇒ x = 0, Thus, only two real values of x satisfy the given, equation., 33. Let α and β be the roots of, …(i), x2 − (a − 1)x + (a + b) = 0, and, …(ii), ax2 − 2x + b = 0, From Eq. (i), we get, α + β = (a − 1) and αβ = (a + b), and from Eq. (ii), we get, 2, b, and αβ =, α+β=, a, a, 2, (according to the question), ∴, a −1 =, a, ⇒, a 2 − a − 2 = 0 ⇒ a = − 1, 2, b, and, (according to the question), a+ b=, a, 1, If, a = − 1, b = and if a = 2, b = − 4, 2, 34. − x2 + 3x + 4 > 0, ⇒, x2 − 3 x − 4 < 0, ⇒, (x − 4)(x + 1) < 0, , –∞, , ⇒, , +, –1, , –, , +, 4, , +∞, , x ∈ (−1, 4), , 35. If α is the common root of the equation, then, α2 − b α + c = 0, and, α 2 − b′ α + c′ = 0, , ...(i), ...(ii)
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61, , Quadratic Equations and Inequalities, , 1 i.e., 2 + 3 i = 2 + 3 i, then another, 4 + 9 13 13, 2 −3i, 1, 2 3i, ., i.e.,, root will be, −, 2 + 3i, 13 13, , On subtracting, we get, , ⇒, , c − c′, α (b′ − b) + (c − c′ ) = 0 ⇒ α = −, b′ − b, c − c′ , α=, , b − b′ , , 41. If one root is, , 36. Given that f (x) = x2 + 2ax + 1, Qα is the root of the equation f (x) = 0, ∴, f (α ) = α 2 + 2aα + 1 = 0, 1, 2a, 1, 1, and, f = 2+, + 1 ⇒ f = 1 + 2aα + α 2 = 0, α, α α, α, 1, ∴, f (α ) = 0 and f = 0, α, 37. The given equation is, x2 + 4a = 8x − 12a 2 or x2 − 8x + 4a + 12a 2 = 0, On comparing with Ax2 + Bx + C = 0, Here,, A = 1, B = − 8, C = 4a + 12a 2, Since, the roots are real and unequal., So,, B2 − 4 AC > 0 ⇒ (−8)2 − 4 × 1 × (4a + 12a 2) > 0, ⇒ 64 − 4 (4a + 12a 2) > 0 ⇒ 64 − 4 × 4 (a + 3a 2) > 0, ⇒, 4 − a − 3a 2 > 0 ⇒ 3a 2 + a − 4 < 0, 2, ⇒ 3a + 4a − 3a − 4 < 0 ⇒ a (3a + 4) − 1 (3a + 4) < 0, ⇒, (a − 1) (3a + 4) < 0, 4, ∴, − < a <1, 3, 38. Let the roots of quadratic equation be α and β., According to the given condition,, α + β =3 λ, and, α 2 + β2 = 5 µ 2, , ...(i), , (α + β )2 − 2αβ = 5 µ 2 ⇒ 2αβ = 3 λ 2 − 5 µ 2, 1, ⇒, αβ = (3 λ 2 − 5 µ 2 ), 2, ∴ Required quadratic equation is, x2 − (α + β ) x + αβ = 0, 1, x2 − 3 λ x + (3 λ 2 − 5 µ 2 ) = 0, ⇒, 2, 2x2 − 2 3 λ x + (3 λ 2 − 5 µ 2 ) = 0, ⇒, ⇒, , 39. We know that, if the roots of the equation x2 + px + q = 0, are greater than 1, then, …(i), p2 > 4 q, ...(ii), f (1) > 0, p, ...(iii), and, − >1, 2, From Eq.(ii), f (1) > 0, ⇒, p+ q + 1 >0, But, r= p+ q+1, ∴, r >0, So, r must be greater than 0., 7 , 7 , 40. We have x − , ...(i), =3 − , , x − 3, x − 3, ⇒, x (x − 3) − 7 = 3 (x − 3) − 7, ⇒, (x − 3) (x − 3) = 0, ⇒, (x − 3)2 = 0 ⇒ x = 3, But x = 3 is not satisfied the equation., ∴The given equation has no roots., , Also, a quadratic equation has two roots, thus this, equation has only imaginary roots., ∴ The equation is, x2 − (Sum of roots) x + (Product of roots) = 0, 4, 1, x2 −, x+, ⇒, = 0 ⇒ 13x2 − 4x + 1 = 0, 13, 13, Thus, II and III statements are correct., 42. Given that, 4a − 2 b + c = 0, where a , b, c ∈ R, ∴ (x + 2) is a factor of ax2 + bx + c and there cannot be a, factor of the expression ax2 + bx + c of the form (x − 2)., Thus, II statement is not correct., 43. (a) x3 − 8x2 + 19x − 12 = 0 ⇒ (x − 1) (x2 − 7x + 12) = 0, ⇒ (x − 1) (x − 3) (x − 4) = 0 ⇒ x = 1, 3 , 4, , Thus, it is not a set of elements as odd positive, integers., (b) x3 − 9x2 + 23x − 15 = 0 ⇒ (x − 1) (x2 − 8 x + 15) = 0, ⇒ (x − 1) (x − 3) (x − 5) = 0 ⇒ x = 1, 3 , 5, Thus, S will be a set of elements as odd positive integers., 44. Q |x2 − x − 6| = x + 2 ⇒ ± (x2 − x − 6) = x + 2, Take positive sign,, ⇒ x2 − 2x − 8 = 0 ⇒ (x − 4) (x + 2) = 0 ⇒ x = 4, − 2, Take negative sign, − x2 + x + 6 = x + 2, ⇒, x2 = 4 ⇒ x = ± 2, Thus, the number of real relations of, |x2 − x − 6| = x + 2 are 3., 45. I. Given equation is, ax2 + bx + c = 0, put, x =1, ∴, a + b + c = 0 ⇒ 0 = 0 (a + b + c = 0 given ), Let another root be α,, c, c, 1 ⋅α =, ⇒ α=, ∴, a, a, II. Given, ax2 + bx + c = 0, On putting, x = − 1, we get, a − b + c = 0 ⇒ 0 = 0 (Q a − b + c = 0, given ), Let another root be α, c, c, α=, ⇒ 2=, ∴ I., a, a, 46. A. 6x2 + 7x − 10 = 0, − 7 ± 49 + 240 − 7 ± 289, x=, =, ∴, 2 ×6, 12, − 7 ± 17 10 − 24 5, =, =, = ,−2, ,, 12, 12 12 6, 1, 5, B. x +, =, ⇒ 2 (x2 + 1) = 5x, x 2, ⇒, 2x2 − 5x + 2 = 0, 2, ⇒, 2x − 4x − x + 2 = 0, ⇒, 2x(x − 2) − 1(x − 2) = 0, 1, (x − 2) (2x − 1) = 0 ⇒ x = , 2, ⇒, 2
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62, , NDA/NA Mathematics, (−5) (1) − (−4) (2), (2) (−3) − (1) (−3), −5 + 8 3, =, =, = −1, − 6 + 3 −3, , C. 3x2 + 5x − 3 = 0, − 5 ± 25 − 36 − 5 ± − 11, x=, =, ∴, 2(3), 6, − 5 + 11i − 5 − 11i, =, ,, 6, 6, , α=, , 47. Given equation,, ax2 + bx + 1 = 0, Since, the roots of given equation be real., ∴, b2 − 4a ≥ 0 ⇒ b2 ≥ 4a, It is possible, if (a , b) is (1, 2), (1, 3) and (2 , 3) ., ∴The number of equations of the form ax2 + bx + 1 = 0, will be 3., 48. 3 < |x| < 6 ⇒ −6 < x < − 3 or 3 < x < 6, ∴, x ∈ (−6, − 3) ∪ (3, 6), 1 1, (if a and b are of same sign), 49., <, ⇒ a>b, a b, 3, 1 2, e.g.,, (inequality changes), <, ⇒ 2>, 2, 2 3, −1, 1, or, (inequality remain same), < 3 ⇒ −2 <, 2, 3, 1, 1, If, <, ⇒ 5 < | − 2| ⇒ which is not true., (−2) 5, 2, 50. −3 < x + < 3, x, (x2 + 2) x, ⇒, −3 <, <3, x2, ⇒, −3x2 < (x2 + 2) x < 3x2, (x ≠ 0), 2, ⇒ x (x + 3x + 2) > 0 and x (x2 − 3x + 2) < 0 (x ≠ 0), ⇒, x (x + 1) (x + 2) > 0 and x (x − 1) (x − 2) < 0, …(i), ⇒, x ∈ (−2, − 1) ∪ (0, ∞ ), …(ii), x ∈ (−∞ , 0) ∪ (1, 2), From Eqs. (i) and (ii), x ∈ (−2, − 1) ∪ (1, 2), 51. x2 − 3| x| + 2 < 0, ⇒, (| x| − 1) (|x| − 2) < 0, ⇒, 1 < | x| < 2, ⇒, −2 < x < − 1 or 1 < x < 2, ∴, x ∈ (−2, − 1) ∪ (1, 2), 52. (x − 1) (x − 2) < 0 ⇒ 1 < x < 2, , 3, 7, and, (3x − 7) (2x − 3) > 0 ⇒ x < or x >, 2, 3, From Eqs. (i) and (ii), we get, 3, 1<x<, 2, , ... (i), …(ii), , 53. logb a > c ⇒ a > b , if b > 1, So,, log 2 x > 4 ⇒ x > 24 ⇒ x > 16, c, , 54. We know that the roots of quadratic equation, ax2 + bx + c = 0 will be reciprocal to each other, if a = c., , Thus, the roots of the equastion 7x 2 + 5x + 7 = 0 will, be reciprocal to each other., 55. Let α be the common root of the quadratic equation, 2x2 − 3x − 5 = 0 and x2 − 3x − 4 = 0, then, c a − c2a1, α= 1 2, a1b2 − a 2b1, , Thus, both A and R are true and R is the correct, explanation of A., 56. Since, one root is 2 − 3, then the another root will be, 2 + 3., Now, sum of roots = 2 − 3 + 2 + 3 = 4, and product of roots = (2 − 3 ) (2 + 3 ) = 4 − 3 = 1, 57. Assertion f (x) = x3 + 2x2 + bx − 6, when b = − 5, then, f (x) = x3 + 2x2 − 5x − 6, when x = − 3, then f (−3) = − 27 + 18 + 15 − 6 = 0, ∴ (x + 3) is one factor of x3 + 2x2 − 5x − 6., ∴Both A and R are individually true and R is the correct, explanation of A., 58. Both A and R are individually true and R is the correct, explanation of A., (− 8), 59. The sum of roots = −, =4, 2, 3, and product of roots =, 2, 60. α 3 + β3 = (α + β )3 − 3αβ (α + β ), 3, = (4)3 − 3 × (4) = 64 − 18 = 46, 2, 1 1, 1 1, 61. x2 − + x + ⋅ = 0, α β, α β, 1, α + β, ⇒, x2 − , =0, x+, αβ , αβ, , , 1, 4, x2 − x + = 0, ⇒, 3, 3, 2, 2, 8, 2, 2, ⇒, x − x+ =0, 3, 3, ⇒, 3x2 − 8x + 2 = 0, , Solutions (Q. Nos. 62-64), We have, ax2 + bx + c = 0, a ≠ 0, Let α and β be the roots of this equation. Then,, −b, α+β=, a, c, and, αβ =, a, 1, c, 62. Let β =, , then αβ = 1 ⇒ = 1, α, a, ⇒, c=a, , ... (i), ... (ii), , 63. Let β = n α , putting β = nα in Eqs. (i) and (ii), we get, −b, c, and nα 2 =, α + nα =, a, a, c, −b, and α 2 =, α (n + 1), ⇒, a, na
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4, , Sequence and, Series, Definitions, , Arithmetic Progression, , Sequence, , A sequence is called an arithmetic progression, if the, difference of any two consecutive terms is constant. The, constant difference of terms is known as common, difference., , A succession of numbers formed and arranged in a, definite order according to a certain definite rule is called a, sequence. A sequence is said to be finite or infinite according, of the number of distinct terms in it is finite or infinite., The number 3, 5, 7, 9, 11, ... form a sequence, given by, the rule tn = ( 2n + 1)., , The difference of any two consecutive terms of a, a + d ,, a + 2d, … is constant, then this series is known as, arithmetic progression., Its first term is a and common difference is d., , Series, By adding the terms of a sequence we obtain a series., A series is finite or infinite according as the number of, terms added is finite or infinite., , Progressions, Sequences following certain patterns are called, progressions., , nth term of a series, Tn = a + ( n − 1) d, Last term of a series, l = a + ( n − 1) d, Sum of n terms of a series,, n, S n = [2a + ( n − 1) d ], 2, n, = (a + l), 2, , Important Results, 1. If a constant is added to or subtracted from each term of an AP,, then the resulting sequence is also an AP with the same, common difference., 2. If each term of a given AP is multiplied or divided by a non-zero, constant k, then the resulting sequence is also an AP with, common difference kd or d/k , where d is the common difference, of the given AP., 3. In a finite AP, the sum of the terms equidistant from the, beginning and end is always same and is equal to the sum of, first and last term i. e. , ak + an − (k − 1 ) = a1 + an for all, k = 1, 2, 3, K , ( n − 1), 4. Three numbers a, b and c are in AP iff 2b = a + c., 5. If the sum of three consecutive terms of an AP is given, it is, convenient to assume them as a − d, a and a + d, where the, common difference is d., 6. If the sum of four consecutive terms of an AP is given, it is, convenient to assume them as a − 3d, a − d, a + d and a + 3d,, where the common difference is 2d ., , 7. Tn = S n − S n − 1 ( n ≥ 2)., 8. If Tn = pn + q, then it will form an AP of common difference p, and first term is p + q ( n = 1)., 9. If the terms of an AP are chosen at regular intervals, then they, form an AP., 10. A sequence is an AP iff its nth term is of the form An + B i. e. , a, linear expression in n. The common difference in such a case is, A i. e. , the coefficient of n., 11. If the pth term of an AP is q and the qth term is p, then its, ( p + q)th term is zero and nth term is ( p + q − n )., 12. If the pth term of an AP is q and the qth term is p, then its, ( p + q)th term is zero and nth term is ( p + q − n )., 13. If in an AP, S p = q and S q = p, then, S p + q = − ( p + q).
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65, , Sequence and Series, , Example 1. If the ratio of the sums of m and n terms of an, AP is m 2: n2, then the ratio of mth and nth terms is, (a) (2m − 1) : (2n − 1), (b) (2m + 1) : (2n − 1), (c) (2n − 1) : (2m − 4), (d) None of these, Solution (a) Let a be the first term and d be the common, difference, then, , ⇒, ⇒, ⇒, ∴, , m, [2a + (m − 1) d ], m2, S m m2, = 2, = 2 ⇒ 2, n, Sn n, n, [2a + (n − 1) d ], 2, 2a + (m − 1) d m, =, 2a + (n − 1) d, n, 2an + n (m − 1) d = 2am + m (n − 1) d, 2a (n − m) = (n − m) d ⇒ d = 2a, Tm a + (m − 1) d 2m − 1, =, =, Tn, a + (n − 1) d, 2n − 1, , Arithmetic Mean of Two, Quantities, Let a and b be two quantities, then arithmetic mean of, a+b, ., a and b is A =, 2, %, , If A1 , A2 , A3 ,..., An are arithmetic mean between a and b, then, an + b, a + nb, a (n − 1) + 2b, A1 =, , A2 =, ,...., An =, n+1, n+1, n+1, , %, , The sum of n arithmetic mean between two quantities a and b is, n times of arithmetic mean of a and b, i.e.,, a + b , = nA, A1 + A2 + ... + An = n , , 2 , , , Example 2. There are n arithmetic means between 3 and, 17. If the ratio of the last mean to the first mean is 3 : 1, then, the value of n is, (a) 5, (b) 4, (c) 6, (d) 1, Solution (c) Let A1, A2, A3 ,... , An be n arithmetic means, between 3 and 17, then, 3, A1, A2 ... , An , 17 are in AP., Let d be the common difference., 14, ∴, 17 = 3 + (n + 2 − 1) d ⇒ d =, n +1, ∴, , A1 = a + d = 3 +, , and, , An = 3 +, , Given that,, ⇒, ⇒, , 14, 3n + 17, =, n +1, n +1, , 14n 17n + 3, =, n +1, n +1, , An 3, =, A1 1, 17n + 3, =3, 3n + 17, n =6, , Geometric Progression, A sequence is known as geometric progression, if the, ratio of any term to its previous term is constant., If a1 , a2 , a3 , ... , an are in GP., a2 a3, a, Then,, =, =K= n = r, a1 a2, an − 1, where, r is known as common ratio of GP., nth term of GP, Tn = ar n − 1, Last term of GP, l = ar n − 1, Sum of n terms of GP,, a (r n − 1), a (1 − r n ), , when r > 1 =, , when r < 1, Sn =, r −1, 1−r, Sum of infinite terms of GP,, a, , where|r|< 1, S∞ =, 1−r, , Important Results, 1. If all the terms of a GP be multiplied or divided by the same, non-zero constant, then it remains a GP with the same, common ratio., 2. The reciprocals of the terms of a given GP form a GP., 3. If each terms of a GP be raised to the same power, the resulting, sequence also forms a GP., 4. In a finite GP the product of the terms equidistant from the, beginning and the end is always same and is equal to the, product of the first and the last term., 5. Three non-zero numbers, a, b and c are in GP, if b2 = ac., 6. If the terms of a given GP are chosen at regular intervals, then, the new sequence so formed also forms a GP., 7. If a1 , a2 , a3 , K , an , . . . is a GP of non-zero, non-negative, terms, then log a1 , log a2 , . . . , log an , . . . is an AP and, vice-versa., 8. The geometric mean G of two non-zero numbers a and b is, given by ab. It is to be noted that a, G and b are in GP. If, a1 , a2 , . . . , an are n non-zero numbers, then their geometric, mean is given by G = ( a1 a2 . . . an )1 / n ., 1, is 1., 9. The product of n geometric mean between a and, a, 1 / (n + 1 ), b, 10. If n GM’s are inserted between a and b, then r = , ., a, 11. The odd number of terms in a GP should be taken as, a a a, . . . , ar 3 , ar 2 , ar, a, , 2 , 3 ,…, while the even number of, r r r, terms in a GP should be taken as,, a a a, . . . ar5 , ar 3 , ar, , 3 , 5 , . . ., r r r, 12. Let the first term of a GP be positive, then if r > 1, then it is an, increasing GP, but if r is positive and less than 1 i. e. ,0 < r < 1,, then it is a decreasing GP., 13. Let the first term of GP be negative, then if r > 1, then it is a, decreasing GP, but if 0 < r < 1, then it is an increasing GP.
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66, , NDA/NA Mathematics, , Example 3. The number which should be added to the, numbers 2, 14, 62, so that the resulting numbers may be in, GP is, (a) 1, (b) 2, (c) 3, (d) 4, Solution (b) Let the number x be added, then, x + 2, x + 14, x + 62 be in GP, ( x + 14) 2 = ( x + 2)( x + 62), , ∴, ⇒, , Recurring Decimals, If in a decimal fraction, a set of figures is repeated or a, figure is repeated, then the fraction is called recurring, 2, decimal. e.g., = 0.66666 ... = 0.6, 3, , Example 6. Find the value of 0.14189, 17, 148, 13, (c), 148, , x + 196 + 28x = x + 64x + 124, 2, , ⇒, , (a), , 36x = 72 ⇒ x = 2, , Example 4. If the ratio of the sum of first three terms and, the sum of the first six terms of a GP be 125 : 152, then the, common ratio r is, 3, 5, 2, 3, (a), (b), (c), (d), 5, 3, 3, 2, , Solution (a) According to the given condition,, ∴, , (r3 − 1) 152 = 125 (r 6 − 1), , ⇒, , (r3 − 1) 152 = 125 (r3 − 1) (r3 + 1), 3, 5, , Geometric Mean of Two, Quantities, If three numbers a, G,and b are in GP, we say that G is, the geometric mean between a and b., Thus, G is the GM between a and b., ⇔ a, G and b are in GP., G b, ⇔, =, ⇔ G 2 = ab i.e., G = ab, a G, %, , If G1 , G2 ,K , Gn are n geometric means between a and b, then, b, G1 = a , a, , %, , 1 / (n + 1 ), , b, , G2 = a , a, , 2 / (n + 1 ), , b, ,...,Gn = a , a, , n / (n + 1 ), , (d) None of these, , = 0.14 + 0.00189 + 0.000000189 + ..., 14, 1, , 1, =, + 189 5 + 8 + ..., , 10, 100, 10, , S3 125, =, S 6 152, , 152 = 125 (r3 + 1) ⇒125 r3 = 27 ⇒ r =, , ⇒, , 1 / 10 5 7, 1 10 3 , 7, + 189 , =, + 189 5 ×, 3 , 50, 999 , 1 − (1 / 10 ) 50, 10, 7, 189, 7, 7, =, +, =, +, 50 999 × 100 50 3700, 7, 7, 21, =, +, =, 50 25 × 148 148, =, , Harmonic Progression, A sequence a1 , a2 ,K , an of non-zero numbers is called, a, harmonic, progression,, if, the, sequence, 1 1 1, 1, ,, ,, ,K ,, ,... is an arithmetic progression., a1 a2 a3, an, 1 1 1, The sequence 1, , , ,… is a HP, because the, 3 5 7, sequence 1, 3, 5, 7,… is an AP., 1, nth term of HP, Tn =, 1, 1, 1, + ( n − 1) , − , a1, a2 a1 , a1a2, =, a2 + ( n − 1) ( a1 − a2 ), nth term from the end of HP,, , Product of n geometric means between a and b is equal to the, nth power of geometric mean of a and b., i.e.,, G1 , G2 , K , Gn = (ab) n / 2 = G n ., , 1, 1, 1, − ( n − 1) , − , an, a2 a1 , a1a2an, =, a1a2 − an ( n − 1) ( a1 − a2 ), , (d) 18, 4, , Solution (b) Let the numbers be a and b., Then,, ⇒, ∴, , a+ b, = 34 and ab =16, 2, a + b = 68 and ab = 256, , ( a − b) = ( a + b) 2 − 4ab = (68) 2 − 4 × 256 = 60, , On solving, a + b = 68 and a − b = 60, we get, a = 64 and b = 4., Hence, the required numbers are 64 and 4., , 1, , T ′n =, , Example 5. Two numbers whose arithmetic mean is 34, and the geometric mean is 16 are, (a) 60, 3, (b) 64, 4, (c) 20, 4, , 21, 148, , Solution (b) Let S = 0.14189, , a(r3 − 1) / (r − 1) 125, , where r is the common ratio., =, a(r 6 − 1) / (r − 1) 152, , ∴, , (b), , %, , −, , 1, 1, 1, 1, +, =, +, Tn, T ′n, a1, an, , %, , Any term of HP cannot be zero., , Example 7. The second term of HP is, 1, . The 10th term of the HP is, 5, 3, (a) 25, (b), 25, , (c), , 25, 3, , 1, and its 5th term is, 3, , (d) 40
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67, , Sequence and Series, , Solution (b) In the given HP,, 1, 1, and 5th term =, 3, 5, The corresponding AP has, Second term = 3 and 5th term = 5, Let a and d be the first and common difference of this AP, then, ...(i), a+ d =3, and, ...(ii), a + 4d = 5, 7, 2, On solving Eqs. (i) and (ii), we get a = and d =, 3, 3, 2 25, 7, 10th term of the AP = a + 9d = + 9 × =, ∴, 3, 3, 3, 3, Hence, 10th term of the given HP =, 25, second term =, , Harmonic Mean of Two, Quantities, Let two quantities are a and b, 2ab, harmonic mean of a and b =, a+b, , respectively, then, , H1 , H 2 ,K , H n are n harmonic mean between a and b, then, bn + a, an + b, , K, H n =, H1 =, (n + 1) ab, (n + 1) ab, , %, , a, , Example 8 The value of n for which , , , n+1, , +b, , a +b, , harmonic mean between a and b is, (a) 4, (b) −1, (c) 5, , n, , n + 1, n, , is the, , , A=, , ⇒, , A > G > H and G 2 = AH ., , Arithmetico-Geometric Series, A series of the form a, ( a + d ) r ,( a + 2d ) r 2 ,... is called, an arithmetico-geometric series., Sum of n terms of an arithmetico-geometric series,, Sn =, , a, dr (1 − r n − 1 ) [a + ( n − 1) d ] r n, +, −, 1−r, 1−r, (1 − r )2, , Sum of infinite terms of an arithmetico-geometric, series,, a, dr, ,|r|< 1, S∞ =, +, 1 − r (1 − r )2, , Example 9. Find the sum of infinity of the series, 1 − 3x + 5 x 2 − 7x3 + .... ∞, when| x|<1, 1− x, 1+ 2 x, (a), (b), 2, (1+ x), (1+ x) 2, 1− 2x, (c), (d) None of these, (1+ x) 2, , Solution (a) Let S ∞ =1 + 3( − x ) + 5 ( − x) 2 + 7( − x)3 + ... ∞ ...(i), ⇒ ( − x) S ∞ = ( − x ) + 3 ( − x) 2 + 5( − x)3 + ... ∞, (1 + x) S ∞ = 1 + [ − 2x + 2x2 − 2x3 + ... ∞ ], − 2x , 2x, 1− x, =1+ , = 1 − (1 + x ) = 1 + x, 1 − ( − x) , , 2 ab, Solution (b) We know that, HM between a and b =, ., a+b, , ∴, ⇒, ⇒, ⇒, ⇒, ⇒, ⇒, , an + 1 + b n + 1, 2ab, =, n, n, a+ b, a +b, ( an + 1 + b n + 1) ( a + b) = 2ab ( an + b n), an + 2 + ab n + 1 + ban + 1 + b n + 2 = 2ban + 1 + 2ab n + 1, an + 2 − ban + 1 = ab n + 1 − b n + 2, an + 1( a − b) = b n + 1 ( a − b), an + 1 = b n + 1, [Q a − b ≠ 0 ], a, , b, , n+1, , 0, , a, = 1 = ∴n + 1 = 0 ⇒ n = −1, b, , ...(ii), , Subtracting Eq. (ii) from Eq. (i), we get, , (d) 3, , an + 1 + b n + 1 , But , is HM between a and b., an + b n , , a+b, 2ab, , G = ab and H =, a+b, 2, , ∴, , ⇒, , S∞ =, , 1− x, (1 + x) 2, , Method of Difference, Sometimes the nth term of a series cannot be, determined by the methods discussed so far. If a series is, such that the difference between successive terms are, either in AP or in GP, then we determine its nth term by, the method of difference and then find the sum of the series, by using the formulae for Σn , Σn 2 and Σn3 . The method of, difference is illustrated in the following examples., , Example 10. The sum to n terms of the series, , Relation among Arithmetic, Mean, Geometric Mean and, Harmonic Mean, If a and b are two real numbers and A, G and H are, arithmetic mean, geometric mean and harmonic mean,, respectively., , 3 + 15 + 35 + 63 + K is, n, (a) ( 4n2 + 6n − 1), 3, (c) 2n − 4, , (b), , 1 2, ( 4n + 5n − 4), 2, , (d) n2 − 5n + 6, , Solution (a) The difference between the successive terms are, 15 − 3 = 12, 35 − 15 = 20 , 63 − 35 = 28, ..., Clearly, these difference are in AP.
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68, , NDA/NA Mathematics, , Let Tn be the nth term and S n denote the sum to n terms of, the given series., Then,, …(i), S n = 3 + 15 + 35 + 63 + ... + Tn − 1 + Tn, Also,, …(ii), S n = 3 + 15 + 35 + K + Tn − 1 + Tn, Subtracting Eq. (ii) from Eq. (i), we get, 0 = 3 + [12 + 20 + 28 + K + (Tn − Tn − 1)] − Tn, (n − 1), ⇒, Tn = 3 +, [2 × 12 + (n − 1 − 1) × 8], 2, = 3 + (n − 1 ) [12 + 4n − 8], ⇒, Tn = 3 + (n − 1) ( 4n + 4) = 4n 2 − 1, ∴, , n, , n, , n, , n, , k =1, , k =1, , k =1, , k =1, , S n = Σ Tk = Σ ( 4k2 − 1) = 4 Σ k2 − Σ 1, n, n (n + 1) (2n + 1) , 2, =4, − n = ( 4n + 6n − 1), 6, 3, , , , %, , %, , %, , %, , Example 11. The sum of serise 12 ⋅ 2 + 2 2 ⋅ 3 + 32 ⋅ 4 + ... to n, terms is, , n3(n + 1)3 (2n + 1), 24, n (n + 1) (3n2 + 7n + 2), (b), 12, n(n + 1), (c), [n(n + 1) + (2n + 1)], 6, (d) None of the above, (a), , Sum to n terms of Special, Series, 1. Sum of first n natural numbers, n ( n + 1), 2, 2. Sum of squares of n natural numbers, n ( n + 1) ( 2n + 1), = 12 + 22 + 32 + 42 + K + n 2= Σn 2 =, 6, 3. Sum of cubes of n natural numbers, = 13 + 23 + 33 + 43 + K + n3, = 1 + 2 + 3 + 4 + K + n = Σn =, , Sum of cubes of n natural numbers is divisible by sum of natural, numbers., 1, 1, 1, 1, n, +, +, +K+, =, 1⋅ 2 2⋅ 3 3⋅ 4, n (n + 1) n + 1, 1, 1, 1, +, +K+, 1⋅ 2⋅ 3 2⋅ 3⋅ 4, n (n + 1) (n + 2), 1, 1, = −, 4 2 (n + 1) (n + 2), n (n + 1) (n + 2), 1 ⋅ 2 + 2 ⋅ 3 + ... n (n + 1) =, 3, , Solution (b) Hence nth term of the given series is, Tn = n 2 (n + 1) = n3 + n 2, ∴, , S n = ΣTn = Σn3 + Σn 2, 2, , n(n + 1) n(n + 1)(2n + 1), +, 2 , 6, n(n + 1) n(n + 1 ) 2n + 1, =, +, 2, 2, 3 , =, , 2, , = Σn 3 =, , n 2( n + 1)2 n ( n + 1), 2, =, = [Σn ], 4, 2, , , =, , n (n + 1)(3n 2 + 7n + 2), 12, , Comprehensive Approach, n, , n, n, , n, n, , n, , If a, b , c and d ,... are in GP, they are also in continued proportion, a b c, 1, i. e., = = = K = (say)., b c d, r, If a, b and c are in AP, then xa , xb and xc will be in GP ( x ≠ ± 1)., If an AP consists of n(odd terms) and its middle term is m, then the, sum of the AP is mn., If pth , qth andrth terms of a GP are in GP, then p , q and r are in AP., If first term of a GP of n terms is a and last term is l. Then, the, product of all the terms of the GP, is ( al) n / 2., If a, b and c are in AP as well as in GP, then a = b = c., , n, , n, , n, , n, n, , a, b and c are in AP, GP or HP according as,, , a−b a a, a, = , or ., b−c a b, c, , If n th term of a series is Tn = an2 + bn + c , then sum of its n terms, is given by Sn = a Σn2 + b Σn + cn. In general Sn = ΣTn ., The sum of all possible products of the first n natural numbers, 1, taken two at a time is, n (n2 − 1) (3n + 2)., 24, 1 + 3 + 5 + K to n terms = n2., 2 + 4 + 6 + .... to n terms = n (n + 1)
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Exercise, Level I, 1, 1, , − , 1 , K is −128?, (NDA 2011 I), 4, 2, (a) 9th, (b) 10th, (c) 11th, (d) 12th, 1, 1, 2. What is the sum of 3 +, +, + K? (NDA 2011 I), 3 3 3, 1. Which term of a series, , (a), , 3, 2, , (b), , 3 3, 2, , (c), , 2 3, 3, , (d), , 3, , . ., , 3. The value of 0.4 2 3 is, 419, (a), 990, 417, (c), 990, , 419, 999, 417, (d), 999, , (b), , 4. If b2 , a 2 and c2 are in AP, then a + b, b + c and c + a, will be in, (a) AP, (b) GP, (c) HP, (d) None of these, 5. If the ratio of the sum of n terms of two AP’s be, ( 7n + 1) : ( 4n + 27), then the ratio of their 11th terms, will be, (a) 2 : 3, (b) 3 : 4, (c) 4 : 3, (d) 5 : 6, , 1, 1, 1, be consecutive terms of an, ,, and, b− c c− a, a−b, AP, then ( b − c)2 , ( c − a )2 and ( a − b)2 will be in, (a) GP, (b) AP, (c) HP, (d) None of these, , 11. If, , 12. If the sum of the roots of the equation ax 2 + bx + c = 0, be equal to the sum of the reciprocals of their, squares, then bc2 , ca 2 and ab2 will be in, (a) AP, (b) GP, (c) HP, (d) None of these, 13. α and β are the roots of the equation x 2 − 3x + a = 0, and γ and δ are the roots of the equation, x 2 − 12x + b = 0. If α , β , γ and δ form an increasing, GP, then ( a , b) is equal to, (a) ( 3, 12), (b) (12, 3), (c) ( 2, 32), (d) ( 4, 16), 14. If A be an arithmetic mean between two numbers, and S be the sum of n arithmetic means between the, same numbers, then, (a) S = nA, (b) A = nS, (c) A = S, (d) None of these, , 6. If the pth term of an AP be q and qth term be p, then, its rth term of an AP will be, (a) p + q + r, (b) p + q − r, (c) p + r − q, (d) p − q − r, , 15. The 59th term of an AP is 449 and the 449th term is, 59. Which term is equal to 0 (zero)?, (NDA 2010 I), , 7. The interior angles of a polygon are in AP. If the, smallest angle be 120° and the common difference be, 5, then the number of side is, (a) 8, (b) 10, (c) 9, (d) 6, , 16. If x 2 , y 2 and z 2 are in AP, then y + z , z + x and x + y, are in, (NDA 2009 II), (a) AP, (b) HP, (c) GP, (d) None of these, , 8. If the sum of first 10 terms of an arithmetic, progression with first term p and common difference, q , is 4 times the sum of the first 5 terms, then what is, the ratio of p : q?, (a) 1 : 2, (b) 1 : 4, (c) 2 : 1, (d) 4 : 1, 9. The sum of an infinite geometric progression is 6. If, the sum of the first two terms is 9/2, then what is the, first term?, (NDA 2010 II), (a) 1, (b) 5/2, (c) 3 or 3/2, (d) 3 or 9, 10. The sum of n terms of three AP’s is whose first term is, 1 and common differences are 1, 2 and 3 are, S1 , S 2 and S3 , respectively. Then, the true relation is, (b) S1 + S3 = 2S 2, (a) S1 + S3 = S 2, (c) S1 + S 2 = 2S3, (d) S1 + S 2 = S3, , (a) 501st term, (c) 508th term, , (b) 502nd term, (d) 509th term, , 17. If x, 2x + 2 and 3x + 3 are the first three terms of a GP,, then what is its fourth term?, (NDA 2009 II), (a) − 27/2 (b) 27/2, (c) − 33/2, (d) 33/2, 1, 1, 3, 18. Which term of the sequence 20, 19 ,18 ,17 ,…, is, 4, 2, 4, the first negative term?, (NDA 2009 II), (a) 27th, (b) 28th, (c) 29th, (d) No such term exists, 19. In an AP, the mth term is 1/n and nth term is 1/m., What is its (mn)th term?, (NDA 2009 II), (a) 1/(mn), (b) m/n, (c) n/m, (d) 1, 20. If G be the geometric mean of x and y, then, 1, 1, is equal to, + 2, 2, 2, G −x, G − y2, 1, 2, (c) 2, (d) 3 G 2, (b) 2, (a) G 2, G, G
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70, , NDA/NA Mathematics, , 21. If n geometric means be inserted between a and b,, then the nth geometric mean will be, n, , b, (b) a , a, , b n − 1, (a) a , a, , n −1, n, 1, , n, , b n, (d) a , a, , b n + 1, (c) a , a, , 31. If the mth term of HP be n and nth term be m, then, the rth term will be, r, mn, (a), (b), mn, r +1, mn, mn, (d), (c), r, r −1, , 22. If |x|< 1, then the sum of the series, 1 + 2x + 3x 2 + 4x3 + K ∞ will be, 1, 1, 1, 1, (d), (b), (c), (a), 2, 1− x, 1+ x, (1 + x ), (1 − x )2, , 32. The geometric mean and harmonic mean of two, non-negative observations are 10 and 8, respectively., Then, what is the arithmetic mean of the, observations?, (NDA 2012 I), (a) 4, (b) 9, (c) 12.5, (d) 25, , 23. If a1/ x = b1/ y = c1/ z and a , b and c are in GP, then, x , y and z will be in, (a) AP, (b) GP, (c) HP, (d) None of these, , 33. If loga x , logb x and logc x be in HP, then a , b and c are, in, (a) AP, (b) HP, (c) GP, (d) None of these, , 24. If the arithmetic mean of two numbers be A and, geometric mean be G, then the numbers will be, (a) A ± ( A2 − G 2 ), (b) A ± A2 − G 2, A ± ( A + G) ( A − G), (c) A ± ( A + G ) ( A − G ) (d), 2, 25. The nth term of the series, 13 13 + 23 13 + 23 + 33, +, +, + ... will be, 1, 1+ 3, 1+ 3+ 5, n 2 + 2n + 1, (b), (a) n 2 + 2n + 1, 8, n 2 + 2n + 1, n 2 − 2n + 1, (c), (d), 4, 4, 26. What is the 15th term of the series 3, 7, 13, 21, 31,, 43, …, (NDA 2008 II), (a) 205, (b) 225, (c) 238, (d) 241, 27. If the AM and GM of two numbers are 5 and, 4 respectively, then what is the HM of those, numbers?, (NDA 2008 II), 5, 16, 9, (b), (c), (d) 9, (a), 4, 5, 2, 28. Which one of the following is correct?, 1, 1, 1 1, If, +, = + , then a, b and c are in, b− c b− a a c, (NDA 2008 I), (a) AP, , (b) HP, , (c) GP, , (d) None of these, , 29. If a , b and c be in GP, then log a n , log bn and log cn, will be in, (a) AP, (b) GP, (c) HP, (d) None of these, 30. If, , a n + 1 + bn + 1, , be the harmonic mean between a and, , a +b, b, then the value of n is, (a) 1, (b) −1, n, , n, , (c) 0, , (d) 2, , 34. If 1, x, y, z and 16 are in geometric progression, then, what is the value of x + y + z?, (NDA 2007 II), (a) 8, (b) 12, (c) 14, (d) 16, 35. If the nth term of an arithmetic progression is 3n + 7,, then what is the sum of its first 50 terms?, (NDA 2007 II), , (a) 3925, (c) 4175, , (b) 4100, (d) 8200, , 36. The sum of integers from 1 to 100 that are divisible, by 2 or 5 is, (a) 3000, (b) 3050, (c) 4050, (d) None of these, 37. If the geometric mean between a and b is, a n + 1 + bn + 1, , then the value of n is, a n + bn, 1, (a) 1, (b) −, 2, 1, (c), (d) 2, 2, 38. If a , b and c are in AP and|a|,|b|,|c|< 1, and, x = 1 + a + a 2 + ... ∞, y = 1 + b + b2 + ... ∞, z = 1 + c + c2 + K ∞, Then, x , y and z shall be in, (a) AP, (b) GP, (c) HP, (d) None of these, 39. If a1 , a2 , a3 ,... , an are in AP, where ai > 0 for all i,, then the value of, 1, 1, 1, is, +, + ... +, a1 + a2, a2 + a3, an − 1 + an, (a), (c), , n −1, a1 + an, n −1, a1 − an, , (b), (d), , n +1, a1 + an, n +1, a1 − an
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71, , Sequence and Series, 40. If A1 , A2 ; G1 , G2 and H 1 , H 2 be two AM’s, GM’s and, HM’s between two quantities, then the value of, G1G2, is, H 1H 2, A1 + A2, H1 + H 2, A1 + A2, (c), H1 − H 2, (a), , A1, H1, A1, (d), H1, , (b), , −, +, −, −, , A2, H2, A2, H2, , 41. The sum to n terms of the infinite series, 1 ⋅ 32 + 2 ⋅ 52 + 3 ⋅ 72 + ... ∞ is, n, (a), ( n + 1) ( 6n 2 + 14n + 7), 6, n, (b), ( n + 1) ( 2n + 1) ( 3n + 1), 6, (c) 4n3 + 4n 2 + n, (d) None of the above, 42. If a , b and c are in GP and, log a − log 2b, log 2b − log 3c and log 3c − log a are in, AP, then a , b and c are the length of the sides of a, triangle which is, (a) acute angled, (b) obtuse angled, (c) right angled, (d) equilateral, 43. The sum of the first and third term of an arithmetic, progression is 12 and the product of first and second, term is 24, then first term is, (a) 1, (b) 8, (c) 4, (d) 6, 44. If 1 / 4, 1 / x , 1 / 10 are in HP, then what is the value of, x?, (NDA 2012 I), (a) 5, (b) 6, (c) 7, (d) 8, 45. If y = x − x 2 + x3 − x 4 + ... ∞, then value of x will be, 1, y, (b), (a) y +, y, 1+ y, y, 1, (d), (c) y −, 1− y, y, 46. Which of the following statement is correct?, (a) If to each term of an AP a number is added or, subtracted, then the series so obtained is also, an AP., (b) The nth term of geometric series whose first term, is a and common ratio r, is ar n − 1., (c) If each term of a GP be raised to the same power, the resulting terms are in GP., (d) All of the above, 1, 47. The sum of the first n terms of the series, 2, 3 7 15, + + +, + Kis, 4 8 16, (a) 2n − n − 1, (b) 1 − 2n, −n, (c) n + 2 − 1, (d) 2n − 1, 48. If a x = b y = cz , where a , b and c are in GP and, a , b, c, x , y , z ≠ 0; then x , y and z are in, (a) AP, (b) GP, (c) HP, (d) None of these, , 49. If the number of terms in an AP is 2n + 1, then the, ratio of the sum of the odd terms to the sum of even, terms is, n +1, n, n2, n +1, (a), (c), (b), (d), n +1, n, n +1, 2n, 50. The sum of the series, is equal to, 1, 1, (b), (a), 6, 3, , 1, 1, 1, +, +, +K∞, ( 3 × 5) ( 5 × 7) ( 7 × 9), (c) −, , 51. If a , b and c are in GP, then, (a), (c), , 1, 3, , (d), , 1, , +, , 1, , a −b, b2, 1, (b) 2, b − c2, 1, (d) 2, b − a2, , 1, c2 − b2, 1, c2 − a 2, , 2, , 2, , is, , −1, , 52. What, , does, , represent?, (a) AP, (c) HP, , the, , series, , 5, 6, , 1+ 32 + 3+, , 1, 3 3, , +........, , (NDA 2012 I), , (b) GP, (d) None of these, , 21π , 11π , 283π , 53. The numbers tan −, and cot , , tan , , 4 , , 6 , 6 , are in, (a) AP, (b) GP, (c) HP, (d) None of these, 54. If the pth , qth and rth terms of a GP are again in GP,, then which one of the following is correct?, (a) p, q and r are in AP., (b) p, q and r are in GP., (c) p, q and r are in HP., (d) p, q and r are neither in AP nor in GP nor in HP., 55. The geometric mean of two numbers is 6. Their, arithmetic mean A and harmonic mean H satisfy the, equation, 90 A + 5H = 918, Which one of the following is correct?, (a) A = 10, (b) A = 1 / 5; A = 10, (c) A = 5 ; A = 10, (d) A = 1 / 5; A = 5, 56. An even number of AM’s are inserted between two, numbers whose sum is 13/6. If the sum of means, exceeds their number by 1, what is the number of, means?, (a) 8, (b) 18, (c) 12, (d) 6, 57. A man saves ` 135 in the first year ` 150 in the second, year and in this way he increases his savings by ` 15, every year. In what time will his total savings be `, 5550 ?, (a) 20 yr, (b) 25 yr, (c) 30 yr, (d) 35 yr, 58. If the AM and HM of two numbers are 9 and 4, respectively, then what is their GM?, (a) 13/2, (b) 6, (c) 3, (d) 2
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72, , NDA/NA Mathematics, , Level II, 1. If S n = nP +, , n ( n − 1), Q, where S n denotes the sum of, 2, the first n terms of AP, then the common difference is, (a) P + Q, (b) 2P + 3Q, (c) 2Q, (d) Q, , 12. If p, q and r are in an AP, then which one of the, following points shall the straight line px + qy + r = 0, always pass through?, (a) (1, 2), (b) ( 2, 1), (c) (1, − 2), (d) ( 2, − 1), , 2. If x > 1, y > 1 and z > 1 are in GP, then 1 / (1 + log x ),, 1 / (1 + log y ) and 1 / (1 + log z ) are in, (a) AP, (b) HP, (c) GP, (d) None of these, , 13. The sum of the first nine terms of an arithmetic, progression is 171. Which one of the following, statements is nor correct about this AP?, (a) The sum of the first and the ninth terms can be, determined, (b) No terms of the AP can be determined, (c) The first term of the AP cannot be determined, (d) The common difference cannot be determined, , 3. If the sequence { sn } is a geometric progression and, s2s11 = s p s8 , then what is the value of p? (NDA 2012 I), (a) 1, (b) 3, (c) 5, (d) Cannot be determined, 4. If p, q and r are in AP as well as GP, then which one, of the following is correct?, (NDA 2012 I), (a) p = q ≠ r, (b) p ≠ q ≠ r, (c) p ≠ q = r, (d) p = q = r, 5. If the sum of the first ten terms of an arithmetic, progression is four times the sum of the first five, terms, then the ratio of the first term to the common, difference is, (a) 1 : 2, (b) 2 : 1, (c) 1 : 4, (d) 4 : 1, 6. If n !, 3 × ( n !) and ( n + 1) ! are in GP, then the value of, n will be, (NDA 2011 II), (a) 3, (b) 4, (c) 8, (d) 10, 7. What is the 10th common term between the series, (NDA 2011 II), 2 + 6 + 10 + ... and 1 + 6 + 11 + ...?, (a) 180, (b) 186, (c) 196, (d) 206, 8. If the 10th term of a GP is 9 and 4th term is 4, then, what is its 7th term?, (NDA 2011 II), (a) 6, (b) 14, (c) 27/14, (d) 56/15, 9. If a, b, c, d, e and f are in AP, then ( e − c) is equal to, which one of the following?, (NDA 2011 II), (a) 2( c − a ), (b) 2( d − c), (c) 2( f − d ), (d) ( d − c), 10. If the arithmetic and geometric means of two, numbers are 10 and 8 respectively, then one number, exceeds the other number by, (NDA 2011 II), (a) 8, (b) 10, (c) 12, (d) 16, 11. What is the sum of numbers lying between 107 and, 253, which are divisible by 5?, (a) 5220, (b) 5210, (c) 5200, (d) 5000, , 14. If 5 ( 3a − 1 + 1), 6 ( 22a − 3 + 2) and 7 ( 5a − 2 + 5) are in, AP, then what is the value of a ?, (a) 7, (b) 6, (c) 5, (d) 3, 15. Two numbers x and y are given. Let A denote the, single AM between them and S denote the sum of n, AM’s between them. On which one of the following, does S / A depend ?, (a) n, (b) n , x, (c) n , y, (d) n , x , y, 16. Let x be one AM and g1 and g2 be two GM’s between y, and z. Then, what is the value of g13 + g32 ?, (a) xyz, (b) xy 2z, 2, (c) xyz, (d) 2xyz, 17. If A, G and H denote AM, GM and HM, respectively, between two numbers and G is thrice as great as H ,, ( A + H ) (G + H ), ?, then what is the value of, ( A − H ) (G − H ), (a) 3/2, (b) 5/2, (c) 3, (d) 5, 18. In a GP of positive terms, any term is equal to, one-third of the sum of next two terms. What is the, common ratio of the GP?, (NDA 2011 I), (a), (c), , 13 + 1, 2, 13 + 1, 3, , (b), , 13 − 1, 2, , (d), , 13, , 19. Which one of the following option is correct?, (NDA 2011 I), , (a), (b), (c), (d), , sin2 30°, sin2 45° and sin2 60° are in GP., cos2 30°, cos2 45° and cos2 60° are in GP., cot2 30°, cot2 45° and cot2 60° are in GP., tan2 30°, tan2 45° and tan2 60° are in GP., , 20. The arithmetic mean of two numbers exceeds their, geometric mean by 2 and the geometric mean exceeds, their harmonic mean by 1.6. What are the two, numbers?, (NDA 2010 II), (a) 16, 4, (b) 81, 9, (c) 256, 16, (d) 625, 25
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73, , Sequence and Series, 21. A square is drawn by joining mid-points of the sides, of a square. Another square is drawn inside the, second square in the same way and the process is, continued indefinitely. If the side of the first square, is 16 cm, then what is the sum of the areas of all the, squares?, (NDA 2010 II), (a) 256 sq cm, (b) 512 sq cm, (c) 1024 sq cm, (d) 512/3 sq cm, 22. If the AM and GM between two numbers are in the, ratio m : n, then what is the ratio between the two, numbers?, (NDA 2010 II), (a), , (c), , m + m2 − n 2, m − m2 − n 2, m2 − n 2, , m+n, (b), m−n, m 2 + n 2 − mn, , (d), , m2 + n 2, , m 2 + n 2 + mn, , 23. If the sum of n terms of a series is a quadratic, expression in n , then the series is in, (a) GP, (b) HP, (c) AP, (d) neither in GP nor in HP nor in AP, 24. What is the sum of the first 50 terms of the series, (1 × 3) + ( 3 × 5) + ( 5 × 7) + ... ?, (a) 171650 (b) 26600 (c) 26650, (d) 26900, 25. Which one of the following statement is correct?, The numbers log6 7, log42 7 and log294 7 are in, (a) AP, (b) GP, (c) HP, (d) None of these, 2, , 3, , 26. What is the value of 76/ 7 ⋅ 76/ 7 ⋅ 76/ 7 ... upto ∞?, (a) log7 ( 6 / 7), (b) 6, (c) 6 / 7, (d) 7, 27. What is the sum of the integers from 1 to 100 that are, divisible by 2 or 5?, (a) 3600, (b) 3550, (c) 3050, (d) 2550, 28. If the AM and HM of two numbers be 27 and 12, respectively, then what is their GM equal to?, (NDA 2010 I), , (a) 12, , (b) 18, , (c) 24, , (d) 27, , 29. What is the sum of all natural numbers between 200, and 400 which are divisible by 7?, (NDA 2010 I), (a) 6729, (b) 8712, (c) 8729, (d) 9276, 30. Let a, b and c be in AP., Consider the following statements, 1 1, 1, I., are in AP., ,, and, ab ca, bc, 1, 1, 1, II., and, are in AP., ,, b+ c, c+ a, a+ b, Which of the statements given above is/are correct?, (a) I only, (b) II only, (NDA 2010 I), (c) Both I and II, (d) Neither I nor II, , 31. If p times the pth term of an AP is q times the qth, term, then what is the ( p + q )th term equal to?, (a) p + q, (b) pq, (NDA 2010 I), (c) 1, (d) 0, 2, , 3, , y y, y, + + + ... , where| y|< 2 , what is, 2, , , 2, 2, y equal to?, x−1, x−1, 2x − 2, 2x + 1, (c), (b), (d), (a), x, 2x, x, 2x, , 32. If x = 1 +, , 33. What is the product of first 2n + 1 terms of a, geometric progression?, (a) The ( n + 1) th power of the nth term of the GP, (b) The ( 2n + 1) th power of the nth term of the GP, (c) The ( 2n + 1) th power of the ( n + 1) th term of, the GP, (d) The nth power of the ( n + 1) th term of the GP, 34. Natural numbers are divided into groups as (1),, (2, 3), (4, 5, 6), (7, 8, 9, 10) and so on. What is the sum, of the numbers in the 11th group?, (NDA 2009 II), (a) 605, (b) 615, (c) 671, (d) 693, 35. In a geometric progression with first term a and, common ratio r, what is the arithmetic mean of first, five terms?, (NDA 2009 I), (a) a + 2r, (b) ar 2, (c) a(r5 − 1) / (r − 1) 5, (d) a(r5 − 1) / [6(r − 1)], 36. If the number of terms of an AP is ( 2n + 1), then what, is the ratio of the sum of the odd terms to the sum of, even terms?, (NDA 2008 II), 2, n, n, n +1, n +1, (b), (d), (a), (c), n +1, n +1, n, 2n, 37. If the sum of ‘n’ terms of an arithmetic progression is, n 2 − 2n, then what is the nth term?, (NDA 2008 II), (a) 3n − n 2, , (b) 2n − 3, , (c) 2n + 3, , (d) 2n − 5, , 38. If a, 2a + 2 and 3a + 3 are in GP, then what is the 4th, term of that series?, (NDA 2008 II), (a) − 13.5, (b) 13.5, (c) − 27, (d) 27, 39. If the nth term of an arithmetic progression is 2n − 1,, then what is the sum upto n terms?, (NDA 2008 II), 2, 2, (b) n − 1, (a) n, 1, 2, (d), (c) n + 1, n( n + 1), 2, 40. The arithmetic mean of 4 numbers is 15. The, arithmetic mean of another 6 numbers is 12. What is, the arithmetic mean of the combined 10 numbers?, (a) 12.2, (b) 12.8, (NDA 2008 II), (c) 13.2, (d) 13.8
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74, , NDA/NA Mathematics, , 41. The product of first nine terms of a GP is, in general,, equal to which one of the following?, (NDA 2008 I), (a) The 9th power of the 4th term, (b) The 4th power of the 9th term, (c) The 5th power of the 9th term, (d) The 9th power of the 5th term, 42. If for positive real numbers x, y and z the numbers, x + y, 2y and y + z are in harmonic progression, then, which one of the following is correct?, (NDA 2007 II), (a) x, y and z are in geometric progression., (b) x, y and z are in arithmetic progression., (c) x, y and z are in harmonic progression., (d) None of the above, 43. What is the sum of the series, (NDA 2007 II), 1 1⋅ 3, 1⋅ 3⋅ 5, + ... ∞ ?, +, 1+ +, 8 8 ⋅ 16 8 ⋅ 16 ⋅ 24, 2, 3, 1, (a), (d), (b) 2 3, (c), 2, 3, 2 3, 44. What is the geometric mean of the ratio of, corresponding terms of two series, where G1 and G2, are geometric means of the two series? (NDA 2007 II), (a) log G1 − log G2, (b) log G1 + log G2, G1, (d) G1G2, (c), G2, , Directions (Q. Nos. 45-47), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 1, 45. Assertion (A) 0.3 + 0.03 + 0.003 + ... = ., 3, (NDA 2007 II), , Reason (R) For each positive integer n, let, an = a + nd, a and d are real numbers. Then,, n, a1 + ... + an = [2a + ( n + 1)d ]., 2, 46. Assertion (A) If mth term of an AP is n and nth, term is m, then its ( m + n )th term is zero., Reason (R) If n th term of an AP is q and qth term, is p, then its ( p + q )th term is zero and nth term is, ( p + q − n )., 47. Assertion (A) The product of n geometric means, 1, between 3 and is 1., 3, Reason (R) The product of n geometric means, 1, between a and is 1., a, , 48. Which one of the following is correct? If the positive, numbers a, b, c and d are in AP, then bcd, cda, dab, and abc are in, (NDA 2007 II), (a) AP, (b) GP, (c) HP, (d) None of the above progression, 49. What is the value of 91/3 ⋅ 91/19 ⋅ 91/ 27 . .. ∞? (NDA 2007 II), (a) 9, , (b) 3, , (c) 91/3, , (d) 1, , 50. If a, b, c and d are in harmonic progression such that, a > b, then which one of the following is correct?, (NDA 2007 II), , (a) a + c = b + d, (c) ac = bd, , (b) a + c > b + d, (d) ab = cd, , 51. Consider the following statements, 2355, I. The value of 2 . 357 is, ., 999, II. If 1 + cos α + cos2 α + ... ∞ = 2 − 2, then α is equal, 3π, to, ., 4, Which of the statements given above is/ are correct?, (a) I only, (b) II only, (c) Both I and II, (d) Neither I nor II, 52. If b1, b2 and b3 are three consecutive terms of an, arithmetic progression with common difference d > 0,, then what is the value of d for which, b32 = b2b3 + b1d + 2 ?, (NDA 2007 I), 1, (b) 0, (c) 1, (d) 2, (a), 2, , Directions (Q. Nos. 53-56), , Consider the series, , A1 = 12 + 22 x + 32 x 2 + 42 x3 + ... ∞, and A2 = 1 +, , 4 9 16, +, +, + ... ∞, 3 32 33, , 53. The sum of the series A1 is, (1 + x ), (1 − x ), (a), (b), (1 + x ), (1 − x )3, 2, (1 − x ), (d) None of these, (c), 1+ x, 54. The nth term of the series A1 is, (b) n3 x n, (c) n 2x n, (a) n 2x n −1, 55. The sum of the series A2 is, 1, (b) 12, (c) 13, (a) 4, 2, , (d) nx, (d) 14, , 56. If S1 and S 2 is the sum of the series A1 and A2, then, A1| A2 is, 2 (1 + x ), 1+ x, (a), (b), 3, 9 (1 − x ), 1− x, 1− x, (1 − x )2, (d), (c), (1 + x )2, (1 + x )2
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Answers, Level I, 1., 11., 21., 31., 41., 51., , (b), (b), (c), (c), (a), (b), , 2., 12., 22., 32., 42., 52., , (b), (a), (d), (c), (b), (d), , 3., 13., 23., 33., 43., 53., , (a), (c), (a), (c), (c), (b), , 4., 14., 24., 34., 44., 54., , (c), (a), (c), (c), (c), (a), , 5., 15., 25., 35., 45., 55., , (c), (c), (c), (c), (d), (b), , 6., 16., 26., 36., 46., 56., , (b), (b), (d), (b), (d), (d), , 7., 17., 27., 37., 47., 57., , (c), (a), (b), (b), (c), (a), , 8., 18., 28., 38., 48., 58., , (a), (b), (b), (c), (c), (b), , 9., 19., 29., 39., 49., , (d), (d), (a), (a), (a), , 10., 20., 30., 40., 50., , (b), (b), (b), (a), (a), , 2., 12., 22., 32., 42., 52., , (b), (c), (a), (c), (a), (c), , 3., 13., 23., 33., 43., 53., , (c), (b), (c), (c), (a), (a), , 4., 14., 24., 34., 44., 54., , (d), (d), (a), (c), (c), (a), , 5., 15., 25., 35., 45., 55., , (a), (a), (c), (c), (d), (a), , 6., 16., 26., 36., 46., 56., , (c), (d), (d), (c), (a), (a), , 7., 17., 27., 37., 47., , (b), (b), (c), (b), (a), , 8., 18., 28., 38., 48., , (a), (b), (b), (a), (c), , 9., 19., 29., 39., 49., , (b), (d), (c), (a), (b), , 10., 20., 30., 40., 50., , (c), (a), (c), (c), (a), , Level II, 1., 11., 21., 31., 41., 51., , (d), (a), (b), (d), (d), (c), , Hints & Solutions, Level I, 1, 1 1, = −2, ,r = −, 4, 2 4, So, the given series forms a GP., ∴, Tn = ar n − 1, 1, −128 = (−2)n − 1, ⇒, 4, ⇒, (−2)9 = (−2)n − 1, ⇒, 9 = n −1, ⇒, n = 10, , ⇒, , 1. Q a =, , 2., , 3+, , 1, 1, +, +K, 3 3 3, , ⇒, ⇒, or, , a , Q S ∞ =, , , 1 − r, , 3 3, 3, =, =, 1, 2, 1−, 3, ⋅ ⋅, , 3. We have, 0.4 2 3 = 0.4232323 ..., = 0.4 + 0.023 + 0.00023 + 0.0000023 + ... ∞, 4, 23, 23, 23, =, + 3 + 5 + 7 +K∞, 10 10, 10, 10, =, , 1, 1, 4, 23 , , 1+, +, +, +K∞, 10 103 , 102 104, , , , , 4, 23 419, 4, 23 1 , +, =, =, +, =, , 10 1000 1 − 1 10 990 990, , 102 , 4. Given that, b2, a 2and c2 are in AP., ∴, a 2 − b2 = c2 − a 2, , ⇒, , (a − b) (a + b) = (c − a ) (c + a ) ⇒, , a −b c−a, =, c+ a a + b, , b+ c−a −c, a+ b−b−c, =, (c + a ) (b + c) (b + c)(a + b), 1, 1, 1, 1, −, =, −, c+ a b+ c b+ c a + b, 1, 1, 1, 1, −, =, −, b+ c a + b c+ a b+ c, 1, 1, 1, are in AP., ,, and, a+ b b+ c, c+ a, , Therefore, (a + b), (b + c) and (c + a ) are in HP., 5. Let S n and S′n be the sums of n terms of two AP’s and T11, and T′11 be the respective 11th terms, then, n, [2a + (n − 1) d ], 7n + 1, Sn, (given), = 2, =, n, S′n, [2a′ + (n − 1) d′ ] 4n + 27, 2, (n − 1), a+, d, 7n + 1, 2, ⇒, =, (n − 1), n + 27, 4, a′ +, d′, 2, Now, put n = 21, we get, a + 10d, T, 148 4, = 11 =, =, a′ + 10d ′ T ′11 111 3, 6. Given that, T p = a + ( p − 1) d = q, and, Tq = a + (q − 1) d = p, From Eqs. (i) and (ii), we get, ( p − q), d=−, = −1, ( p − q), , ...(i), ...(ii), …(iii)
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76, , NDA/NA Mathematics, Putting the value of d in Eq. (i), then, a + ( p − 1) (−1) = q, …(iv), ⇒, a = p+ q −1, Now, rth term is given by an AP, Tr = a + (r − 1) d = ( p + q − 1) + (r − 1) (−1), [from Eqs. (iii) and (iv)], = p+ q−r, , 7. Let the number of sides of the polygon be n. Then, the, sum of interior angles of the polygon, π, = (2n − 4) = (n − 2) π, 2, Since, the angles are in AP and a = 120° , d = 5, n, Therefore,, S n = [2a + (n − 1) d ], 2, n, [2 × 120 + (n − 1) 5 ] = (n − 2) 180, ⇒, 2, ⇒, n 2 − 25n + 144 = 0, ⇒, (n − 9) (n − 16) = 0, ⇒, n = 9, 16, Take, n = 16, ∴ T16 = a + 15d = 120° + 15 ⋅ 5° = 195°, which is, impossible as an interior angle cannot be greater than, 180°. Hence, n = 9., 8. Since, first term = p, and common difference = q, According to question,, 10, 5, [2 p + 9q] = 4 × [2 p + 4q], 2, 2, ⇒, 2 p + 9q = 4 p + 8q, ⇒, 2p = q ⇒ p : q = 1 :2, a, 9. Q, = 6 ⇒ a = 6(1 − r ), 1−r, 9, and, a + ar =, 2, 9, ⇒, 6(1 − r ) + 6r (1 − r ) =, 2, ⇒, 12 − 12r + 12r − 12r 2 = 9, 3 1, r2 =, =, ⇒, 12 4, 1, 1, or −, ⇒, r=, 2, 2, ⇒, a = 3 or 9, , …(i), (given), [from Eq. (i)], , 10. Let a1 , a 2, a3 and d1 , d2, d3 are the first terms and, common differences of the three AP’s., We have, a1 = a 2 = a3 = 1 and d1 = 1, d2 = 2, d3 = 3,, therefore, n, ...(i), S1 = (n + 1), 2, n, ...(ii), S 2 = [2n ], 2, n, ...(iii), S3 = [3n − 1], 2, On adding Eqs. (i) and (iii), we get, n, n, , (2n ) = 2S 2, S1 + S3 = [(n + 1) + (3n − 1)] = 2, 2, 2, , Hence, correct relation is S1 + S3 = 2S 2., , 11. Now, we assume, (b − c)2, (c − a )2 and (a − b)2 are in AP, then we have, (c − a )2 − (b − c)2 = (a − b)2 − (c − a )2, ...(i), ⇒ (b − a ) (2c − a − b) = (c − b) (2a − b − c), 1, 1, 1, Also, if, are in AP, then, ,, and, b−c c−a, a−b, 1, 1, 1, 1, −, =, −, c−a b−c a −b c−a, b + a − 2c, c + b − 2a, ⇒, =, (c − a ) (b − c) (a − b) (c − a ), ⇒, (a − b) (b + a − 2c) = (b − c) (c + b − 2a ), ⇒, (b − a ) (2c − a − b) = (c − b) (2a − b − c), Which is equal to Eq. (i)., So, our hypothesis is true., 12. Given equation is ax2 + bx + c = 0 and let the roots are, b, c, α and β. So, α + β = − and αβ =, a, a, b2 2c, Now,, α 2 + β 2 = (α + β )2 − 2αβ = 2 −, a, a, b2 2c, −, 2, b2 − 2ac, 1, 1 α 2 + β2, Now,, = a 2 a =, + 2=, 2, 2 2, c, c2, α, β, αβ, a2, According to the given condition,, 1, 1, b b2 − 2ac, α + β = 2 + 2 ⇒− =, a, α, β, c2, 2, 2, 2, ⇒, − bc = ab − 2a c, Hence,, 2a 2c = ab2 + bc2 ⇒ ab2, ca 2 and bc2 or, bc2, ca 2 and ab2 be in AP., 13. Since, α , β , γ and δ form an increasing GP, so αδ = βγ, where,, α <β < γ <δ, We have,, x2 − 3 x + a = 0, 1, …(i), ⇒, x = (3 ± 9 − 4a ), 2, 1, Also, α < β . Hence, α = (3 − 9 − 4a ) ,, 2, 1, β = (3 + 9 − 4a ), 2, 12 ± 144 − 4b, 2, Similarly, from x − 12x + b = 0 ⇒ x =, 2, Since, γ < δ, 1, 1, ∴, γ = (12 − 144 − 4b ), δ = (12 + 144 − 4b ), 2, 2, Substituting the values of α , β , γ and δ in Eq. (i), we, get, 1, 1, (3 − 9 − 4a ) ⋅ (12 + 144 − 4b ), 2, 2, 1, 1, = (3 + 9 − 4a ) ⋅ (12 − 144 − 4b ), 2, 2, Only the option (c) i.e., (2, 32) satisfy the equation., 14. Let the two quantities be a and b and let A1 , A2, K , An, be the n, AM’s between them. Then, a , A1 , A2, K , An, b, are in AP and let d be the common difference., b−a, Now, Tn + 2 = b = a + (n + 2 − 1) d ⇒ d =, n+1, Also,, , A1 + A2 + ...+ An = S n + 1 − a
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77, , Sequence and Series, , 1, (b − a ) , −a, (n + 1) 2a + (n + 1 − 1), 2, (n + 1) , , n, n, a + b, = [2 a + (b − a )] = [a + b] = n , = nA, 2 , 2, 2, =, , 15. Let a and d be the first term and common difference of, the AP., …(i), ∴, a + 58d = 449, and, …(ii), a + 448d = 59, On solving Eqs. (i) and (ii), we get, a = 507 and d = − 1, Now, assume that nth term will be zero., ∴, 0 = 507 + (n − 1)(−1), ⇒, 507 = n − 1, ⇒, n = 508, 16. Since, 2 y2 = x2 + z 2, (Q x2, y2 and z 2 are in AP), Let y + z , z + x and x + y are in HP., 2 ( y + z )(x + y), z+ x=, ∴, y+ z+ x+ y, 2 ( y + z )(x + y), z+ x=, ⇒, 2y + z + x, ⇒, , 2 yz + z 2 + zx + 2xy + xz + x2, = 2 yx + 2 y2 + 2zx + 2 yz, 2, 2, ⇒, z + x = 2 y2, 2, 2, ⇒ x , y and z 2 are in AP., 17. Since, (2x + 2)2 = x(3x + 3) ⇒ x2 + 5x + 4 = 0, ⇒, x = − 1, − 4, Now, first term, a = x, 2(x + 1), Second term,, ar = 2(x + 1) ⇒ r =, x, 2(x + 1), ∴ Fourth term, ar3 = x , , , x , On putting, , 3, , x = − 4, we get, 3, , 3, , 27, 3, 2(−4 + 1), T4 = − 4 , = −4 × = −, 2, , 2, −4 , 18. Given, series can be rewritten as, 77 37 71, 20,, ,, ,, ,K, 4 2 4, This is an AP series., 3, Here,, a = 20, d = −, 4, ∴, Tn = a + (n − 1) d, 3, = 20 + (n − 1) − , 4, 83 3, =, − n, 4 4, For first negative term, Tn < 0, 83 3, − n <0, ⇒, 4 4, ⇒, 83 < 3n, 83, ⇒, n>, 3, So, n should be 28., Hence, 28th term is the first negative term., , Tm = a + (m − 1) d, 1, = a + (m − 1) d, n, and, Tn = a + (n − 1) d, 1, = a + (n − 1) d, ⇒, m, On solving Eqs. (i) and (ii), we get, 1, 1, 1, a=d=, ∴ Tmn =, + (mn − 1), =1, mn, mn, mn, , 19. Given,, ⇒, , …(i), , …(ii), , 20. Since, G is the geometric mean between x and y., ∴, G = xy, 1, 1, 1, 1, Now,, +, =, +, G 2 − x2 G 2 − y2 xy − x2 xy − y2, 1, 1, 1, 1 1, =, =, − + =, (x − y) x y xy G 2, 21. If n geometric means G1 , G2, K , Gn are to be inserted, between two positive real numbers a and b, then, a , G1 , G2, ... , Gn , b are in GP, then, G1 = ar , G2 = ar 2, K , Gn = ar n, So,, , b, b = ar n + 1 ⇒ r = , a, , 1/( n + 1 ), , b, Now, nth geometric mean (Gn ) = ar n = a , a, , n/ n+ 1, , S = 1 + 2x + 3x2 + 4x3 + K ∞, xS = x + 2x2 + 3x3 + K ∞, Subtracting Eq. (ii) from Eq.(i), we get, (1 − x) S = 1 + x + x2 + x3 + K ∞, 1, 1 1 , S=, =, ⇒, (1 − x) 1 − x , (1 − x)2, , 22. Let, , 23. Let a1/ x = b1/ y = c1/ z = k, ⇒, a = kx , b = ky and c = kz, Now, a , b and c are in GP., ⇒ b2 = ac ⇒ k2y = kx ⋅ kz = kx + z ⇒ 2 y = x + z, ⇒, x, y and z are in AP., a+b, 24. Since,, AM =, =A, 2, and, GM = ab, ⇒, G 2 = ab, Now,, (a − b)2 = (a + b)2 − 4ab, = (2 A )2 − 4G 2 = 4 ( A 2 − G 2), ⇒, a − b = ± 2 ( A − G) ( A + G), On solving Eqs. (i) and (iii), we get, a = A ± ( A − G) ( A + G), and, b = A ± ( A − G) ( A + G), 25. We have,, ∴, , 13 13 + 23 13 + 23 + 33, +, +, +…, 1, 1+3, 1+3+5, 13 + 23 + 33 + K + n3, Tn =, 1 + 3 + 5 + K upto n terms, Σn3, 1 n 2 (n + 1)2, =, = ⋅, n, n2, [2 + (n − 1) 2] 4, 2, 1, = (n 2 + 2n + 1), 4, , ...(i), ...(ii), , …(i), …(ii), , ...(iii)
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78, , NDA/NA Mathematics, S = 3 + 7 + 13 + 21 + 31 + K + a n, − S = 3 + 7 + 13 + 21 + K +, an − 1 + an, – – –, –, –, –, –, 0 = (3 + 4 + 6 + 8 + 10 + 12 + K + n terms) − a n, ⇒ a n = 3 + {4 + 6 + 8 + K + (n − 1) terms }, (n − 1), an = 3 +, {8 + (n − 1 − 1) 2}, 2, (n − 1), =3 +, ⋅ 2 {4 + n − 2}, 2, = 3 + (n − 1)(n + 2), ∴ 15th terms = a15 = 3 + (15 − 1)(15 + 2), = 3 + 14 ⋅ 17, = 3 + 238 = 241, , 26. Let, , (GM)2, AM, 16, HM =, 5, , 27. We know that, HM =, ∴, 28., , Let a, b and c are in HP., 2ac, ∴, b=, a+c, 1, 1, 1, 1, LHS =, +, =, +, 2ac, 2ac, b−c b−a, −c, −a, a+c, a+c, a+c, a+c, +, =, 2, 2ac − ac − c, 2ac − a 2 − ac, a+c, a+c, +, =, ac − c2 ac − a 2, , (a + c)(2ac − a 2 − c2), − ac (c − a )2, , (a + c)(c − a )2 a + c, =, ac, ac(c − a )2, a, c 1 1, =, +, = + = RHS, ac ac c a, Hence, a, b and c are in HP., 29. If a , b and c are in GP, then b2 = ac, Taking log on both sides, we get, 2 log e b = log e a + log e c, ⇒, 2n log e b = n log e a + n log e c, ⇒, 2 log e bn = log e a n + log e cn, ⇒, log e a n , log e bn and log e cn be in AP., 2ab, ., 30. Since, harmonic mean between a and b is, a+b, a n + 1 + bn + 1, 2ab, =, n, n, a, +b, a +b, , ⇒, ⇒, , (given), , a n + 2 + abn + 1 + ba n + 1 + bn + 2 = 2a n + 1b + 2bn + 1a, a n + 1 (a − b) = bn + 1 (a − b), , ⇒, Hence,, , a, , b, , n+ 1, , a, = (1) = , b, n = −1, , 0, , for, , HP, therefore the, 1, 1, , nth term =, n, m, Let a and d be the first term and common difference of, this AP, then, 1, ...(i), a + (m − 1) d =, n, 1, ...(ii), a + (n − 1) d =, m, On solving Eqs. (i) and (ii), we get, 1, 1, a=, ,d =, mn, mn, Now, rth term of AP = a + (r − 1) d, 1, 1, =, + (r − 1), mn, mn, 1 + r −1, r, =, =, mn, mn, mn, ∴, rth term of HP =, r, , 33. Since, the given series log a x, logb x and log c x are in, HP., log x log x, log x, are in HP., ,, and, ⇒, log a log b, log c, log a log b, log c, are in AP., ⇒, ,, and, log x log x, log x, ⇒, log x a , log x b and log x c are in AP, ⇒, a , b and c are in GP., , =, , ⇒, , Tm = n , Tn = m, , corresponding AP of mth term =, , 32. Given that, geometric mean (G ) = 10, and harmonic mean (H ) = 8, Let A be the arithmetic mean, G2, Then,, G 2 = AH ⇒ A =, H, (10)2 100, ⇒, A=, =, = 12.5, 8, 8, , ac − a 2 + ac − c2, = (a + c) , , ac(a − c)(c − a ) , =, , 31. Given,, , 34. Q 1, x, y and z, 16 are in geometric progression., Here,, a = 1, l = 16, n = 5, l = ar n − 1, 16 = 1 ⋅ r 4 ⇒ r = 2, ∴, x = 1⋅ r = 2, y = 1 ⋅ r 2 = 4,, z = 1 ⋅ r3 = 8, ∴, x + y + z = 2 + 4 + 8 = 14, 35. Given that, Tn = 3n + 7, S n = ΣTn, = Σ (3n + 7) = 3Σn + 7 Σ1, 3n (n + 1), 3n + 3 + 14 , =, + 7n = n, , , 2, 2, 3n + 17 , =n, 2, , 3 × 50 + 17 , Now, sum of 50 terms = S50 = 50, , , 2, 167 , = 50, = 25 × 167 = 4175, 2 , 36. The sum of integers from 1 to 100 that are divisible by 2, or 5 = sum of series divisible by 2 + sum of series divisible, by 5 − sum of series divisible by 2 and 5
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79, , Sequence and Series, = (2 + 4 + 6 + K + 100) + (5 + 10 + 15 + ... + 100), − (10 + 20 + 30 + K + 100), 50, 20, =, {2 × 2 + (50 − 1) 2} +, {2 × 5 + (20 − 1) 5}, 2, 2, 10, −, [10 × 2 + (10 − 1) 10], 2, = 25 (102) + 10 (105) − 5 (110), = 2550 + 1050 − 550 = 3050, 37. The geometric mean between a and b = ab, a n + 1 + bn + 1, ⇒, = (ab)1/ 2, a n + bn, a n + 1 − a n + 1/ 2b1/ 2 + bn + 1 − a1/ 2bn + 1/ 2 = 0, (a n + 1/ 2 − bn + 1/ 2) (a1/ 2 − b1/ 2) = 0, n + 1/ 2, a, − bn + 1/ 2 = 0, (Q a ≠ b ⇒ a1/ 2 ≠ b1/ 2), n + 1/ 2, 0, 1, 1, a, a, =1 = ⇒n + =0⇒n = −, , b, b, 2, 2, , ⇒, ⇒, ⇒, ⇒, , ∴, ⇒, ⇒, , 41. Given, series is 1 ⋅ 32 + 2 ⋅ 52 + 3 ⋅ 72 + ... ∞, This is an arithmetic-geometric series whose nth term is, equal to, Tn = n (2n + 1)2 = 4n3 + 4n 2 + n, , ⇒, , z = 1 + c + c2 + K ∞, 1, z=, 1−c, , and, ⇒, , Since, a , b and c are in AP., ⇒, 1 − a , 1 − b, 1 − c are in AP., 1, 1, 1, are in HP, ,, and, ⇒, 1−a 1−b, 1−c, ∴ x, y and z are in HP., 39. Since, a1 , a 2, a3 , K , a n are in AP., Then , a 2 − a1 = a3 − a 2 = ... = a n − a n − 1 = d, where, d is the common difference of the given AP., Also,, a n = a1 + (n − 1) d, Then, by rationalizing each term, 1, 1, 1, + ... +, +, an + an − 1, a 2 + a1, a3 + a 2, =, , a 2 − a1, +, a 2 − a1, , a3 − a 2, + ... +, a3 − a 2, , an − an − 1, an − an − 1, , 1, ( a 2 − a1 + a3 − a 2 + ... + a n − a n − 1 ), d, a n + a1, 1, = ( a n − a1 ) ×, d, a n + a1, , , 1, 1 (n − 1) d , a n − a1, n −1, = , =, = , d a n + a1 d a n + a1 , a n + a1, =, , 40. Let the two quantities be a and b. Then, a , A1 , A2 and b, are in AP., …(i), ∴, A1 − a = b − A2 ⇒ A1 + A2 = a + b, Again, a , G1 , G2 and b are in GP., G1, b, ...(ii), =, ⇒ G1G2 = ab, ∴, a G2, Also, a , H 1 , H 2 and b are in HP., , n, , n, , 1, n, , 1, , ∴ S n = Σ Tn = Σ (4n3 + 4n 2 + n ), n, , n, , = 4 Σ n3 + 4 Σ n 2 + Σ n, 1, , 1, , 1, , 2, , 4, n, , n, (n + 1) + n (n + 1) (2n + 1) + (n + 1), =4, 6, 2, , 2, 4, 1, 2, = n (n + 1) n + n + (2n + 1) +, , 6, 2 , n, 2, = (n + 1) (6n + 14n + 7), 6, , 38. We have, x = 1 + a + a 2 + K ∞, 1, ⇒, x=, 1−a, y = 1 + b + b2 + ... ∞, 1, y=, 1−b, , 1, 1 1, 1, 1, 1, 1 1, − = −, ⇒, +, = +, H1 a b H 2, H1 H 2 a b, H 1 + H 2 a + b A1 + A2, [from Eqs. (i) and (ii)], =, =, H 1H 2, ab, G1G2, G1G2, A + A2, = 1, H 1H 2 H 1 + H 2, , 42. Since, a , b and c are in GP., ⇒, b2 = ac, and log a − log 2b, log 2b − log 3c and log 3c − log a are, in AP., ⇒ 2 (log 2b − log 3c) = log a − log 2b + log 3c − log a, ⇒, b2 = ac and 2b = 3c ⇒ b = 2a / 3 and c = 4a / 9, 5a, 10a, 13a, Since, a + b =, > c, b + c =, > a, c + a =, >b, 3, 9, 9, It implies that, a , b and c form a triangle with a as the, greatest side., Now, let us find the greatest ∠ A of ∆ABC by using the, cosine formula,, 29, b2 + c2 − a 2, cos A =, =−, <0, 2bc, 48, ∴The ∠ A is obtuse., 43. Let first 3 terms be a − d , a and a + d., Now, (a − d ) + (a + d ) = 12 ⇒ 2a = 12 ⇒ a = 6, and, , (a − d )a = 24 ⇒ 6(6 − d ) = 24, , ⇒, , 6 − d =4 ⇒ d =2, , ∴ First term,, , a − d =6 −2 =4, , 44. Given that, 1 / 4, 1 / x and 1 / 10 are in HP., ⇒ 4, x and 10 are in AP., ∴ Arithmetic mean,, 4 + 10 14, x=, =, = 7., 2, 2, 45. Given , y = x − x2 + x3 − x4 + K, x, y=, ∴, 1 − (− x), ⇒, ⇒, ⇒, , y + yx = x, x(1 − y) = y, x=, , y, 1− y, , 46. All hypothesis are based on fundamental concepts of, progressions.Therefore, option (d) is correct.
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80, , NDA/NA Mathematics, , 47. Suppose, S n be the sum of first n terms of given series., 1 , 1 , 1, 1, , , S n = 1 − + 1 − + 1 − + K+ 1 − n , , , 2 , 4 , 8, 2 , 1, 1, 1, , , = n − + 2 + K + n, 2 2, 2 , 1 1, 1, 1, Here, + 2 + 3 + K + n is a GP series., 2 2, 2, 2, 1, 1, Which has first term a = and common ratio =, 2, 2, 1, , 1 − n , 1 , 2 = n − 1 + 2−n, ∴, Sn = n −, 1, 2 , 1 − , , 2, 48. We have, a x = by = cz, or, a x = by = cz = k, ⇒, a = k1/ x , b = k1/ y and c = k1/ z, Given that, a , b and c are in GP., ∴, b2 = a ⋅ c, ⇒, k2/ y = k1/ x ⋅ k1/ z, 2 1 1, ⇒, = +, y x z, ∴ x, y and z are in HP., , (say), , 49. Number of terms in an AP = 2n + 1. We know that, the, ratio of the sum of the odd terms to the sum of even, terms is given by, n + 1 sum of odd terms, =, n, sum of even terms, 50. We have, sum of series is, 1, 1, 1, +, +, +K∞, 3 ×5 5 × 7 7 ×9, 1 1 1 1 1 1 1, 1, Let S =, − + − + − + .. =, 6, 2 3 5 5 7 7 9, 51. Since, a , b and c are in GP., ⇒, b2 = ac, 1, 1, 1, 1, +, ∴, = 2, +, a 2 − b2 b2, a − ac ac, a, c+ a −c, =, =, ac (a − c), ac (a − c), 1, 1, =, = 2, 2, ac − c, b − c2, , …(i), [from Eq. (i)], , 52. Given, series is, 1, 1, +3+, + ..., 3, 3 3, Here, Between each two consecutive terms, no common, difference and common ratio are form., Hence, the given series does not form any series., 11 × 180°, 11π , 53. Now, tan −, = − tan, , 6 , 6, = − tan 330°, = − tan (360° − 30° ), 1, = tan 30° =, 3, 21 × 180°, 21 π , tan , = tan, 4 , 4, 1+, , = tan 945°, = tan (3 × 360° − 135° ), = tan (−135° ), = − tan (180° − 45° ), = tan 45°, =1, 283 × 180°, 283 π , cot , , = cot , , , 6 , 6, = cot 8490° = cot (24 × 360° − 150° ) = − cot 150°, = − cot (180° − 30° ) = cot 30° = 3, 1, , 1 and 3 are in GP., Q, 3, ∴ Given numbers are in GP., 54. Let pth term is aR p − 1, qth term is aRq − 1 and rth term is, aRr − 1. These are in GP, then, (aRq − 1 )2 = aR p − 1aRr − 1, ⇒, a 2R2q − 2 = a 2R p + r − 2, ⇒, R2q − 2 = R p+ r− 2, ⇒, 2q − 2 = p + r − 2, ⇒, 2q = p + r, ⇒, p, q and r in AP., 55. Given that, GM = 6, AM =A, HM = H, ...(i), ∴, 90 A + 5H = 918, We know that, (GM)2 = (AM) × (HM), ⇒, (6)2 = A × H, ⇒, AH = 36, 36, ⇒, H =, A, 36, in Eq. (i), we get, On putting H =, A, 36, 90 A + 5 ×, = 918, A, ⇒, 90 A 2 − 918 A + 180 = 0, ⇒, 5 A 2 − 51 A + 10 = 0, 2, ⇒, 5 A − 50 A − A + 10 = 0, ⇒, 5 A ( A − 10) − 1 ( A − 10) = 0, ⇒, (5 A − 1) ( A − 10) = 0, 1, ⇒, A = , 10, 5, 13, , 56. Let the two numbers are a and , − a in which 2n AM, 6, , are inserted, then the series AP will be, , 13, − a ., a , A1 , A2,... , A2n , , , 6, 13, Here, first term = a, last term =, −a, 6, Total number of terms = 2n + 2, Using the relation, Tn = a + (n − 1) d, 13, ∴, − a = a + (2n + 1) d, 6, 13 − 12a, ⇒, d=, 6 (2n + 1), 13 − 12a, A1 = a + d = a +, ∴, 6 (2n + 1)
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81, , Sequence and Series, 12na + 6a + 13 − 12a, 6 (2n + 1), 12na − 6a + 13, =, 6 (2n + 1), (13 − 12a ), and A2n = a + 2nd = a + 2n, 6 (2n + 1), 12na + 6a + 26n − 24na, =, 6 (2n + 1), 26n − 12na + 6a, =, 6 (2n + 1), , 57. Given a = ` 135, d = ` 15 and S n = ` 5550, Let in n years his total savings be ` 5550, n, ∴, S n = [2a + (n − 1) d ], 2, n, 5550 = [2 × 135 + (n − 1) 15], ⇒, 2, , =, , Now,, ⇒, ⇒, , (given), A1 + A2 + K + A2n = 2n + 1, 2n, [ A1 + A2n ] = 2n + 1, 2, 12na − 6a + 13 + 26n − 12na + 6a , n, = 2n + 1, , , 6 (2n + 1), , ⇒, ⇒, , 13n = 12n + 6, n =6, , ⇒, ⇒, , 11100 = n [270 + 15n − 15], 15 n 2 + 255n − 11100 = 0, , ⇒, ⇒, , n 2 + 17n − 740 = 0, n 2 + 37n − 20n − 740 = 0, , ⇒, ⇒, , (n + 37) (n − 20) = 0, n = 20 yr, , (Q n ≠ − 37), , 58. Q AM = 9 and HM = 4, Then,, GM = AM × HM, = 9 × 4 = 36 = 6, , Level II, n (n − 1), ...(i), Q, 2, Let us put n = n − 1, then Eq. (i) becomes, (n − 1) (n − 2), S n − 1 = (n − 1)P +, Q, 2, We know, the nth term is given by, Tn = S n − S n −1, n (n − 1) , , Tn = nP +, Q, ⇒, , , 2, (, n − 1) (n − 2) , , Q, − (n − 1) P +, , , 2, n (n − 1), Tn = nP +, Q − (n − 1)P, ⇒, 2, (n − 1) (n − 2), −, Q, 2, ⇒, Tn = P (n − n + 1), n (n − 1) (n − 1) (n − 2) , +Q, −, , , 2, 2, 2, −, +, n, n, , , ⇒, Tn = P + Q (n − 1), , , 2, ...(ii), ⇒, Tn = P + Q (n − 1), ...(iii), ⇒, Tn − 1 = P + Q (n − 2), From Eqs. (ii) and (iii), we get, d = Tn − Tn − 1 = P + Q (n − 1) − P − Q (n − 2), =Q, , 1. Given that, S n = nP +, , 2. Given that, x > 1, y > 1 and z > 1 are in GP or x, y and z, are in GP., ∴, y2 = xz, Taking log on both sides, we get, 2 log y = log x + log z, ⇒, 2 (1 + log y) = (1 + log x) + (1 + log z ), Which shows that, 1 + log x, 1 + log y and 1 + log z are in, AP., 1, 1, 1, are in HP., ∴, ,, and, 1 + log x 1 + log y, 1 + log z, , 3. We know that, in a GP the product of two terms, equidistant from the beginning and end is a constant, and is equal to the product of first term and last term,, i.e., if a1 , a 2, a3 , K a ( n − 2) a n − 1 are in GP then, a1a n = a 2a n = a3 a n − 2 = K, Given that,, S 2 S11 = S pS 8, ⇒, (P + 8) = (2 + 11), ⇒, p = 13 − 8 = 5, 4. Given that, p, q and r are in AP., ∴, 2q = p + r, As well as are in GP., ∴, q2 = pr, From Eqs. (i) and (ii), we get, p + r = 2 pr, ( p )2 − 2 pr ⋅ r + ( r )2 = 0, ⇒, ⇒, ( p − r )2 = 0, p − r =0, ⇒, p= r, ⇒, ⇒, p=r, From Eq. (ii), we get, q2 = r ⋅ r = r 2 ⇒ q = r, Now, from Eqs. (iii) and (iv), we get, p=q=r, 5. Sum of n terms of an AP is given by, n, S n = [2a + (n − 1) d ], 2, ∴ S10 = 5 [2a + 9d ], 5, and S5 = [2a + 4d ], 2, According to question,, 5, 5 [2a + 9d ] = 4 × [2a + 4d ], 2, ⇒, 2a + 9d = 4a + 8d ⇒ d = 2a, a 1, ⇒, = ⇒ a : d = 1 :2, d 2, , …(i), …(ii), , …(iii), …(iv)
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82, 6. Given that,, n !,3 × (n !) and (n + 1) ! are in GP., Then, {3 × (n !)}2 = (n !) × (n + 1) !, ⇒, (3)2 (n !) = (n + 1) !, ⇒, 9(n !) = (n + 1) ⋅ (n !), ⇒, 9=n+1, ⇒, n =8, 7. Let the series;, Ist, IInd, S1 = 2 + 6 + 10 + 14 + 18 + 22 + 26 + 30 + 34 + 38 + 42, IIIrd, + 46 + K, and S 2 = 1 + 6 + 11 + 16 + 21 + 26 + 31 + 36 + 41, + 46 + K, The number sequence of common terms in S1,, S1′ = 2 + 7 + 12 + K, and number sequence of common terms in S 2,, S 2′ = 2 + 6 + 10 + K, Now, we find the 10th term in both S1′ and S 2′, for S1′,, T10 = 2 + (10 − 1) ⋅ 5 = 2 + 45 = 47, and for S 2′, T10 = 2 + (10 − 1) ⋅ 4 = 2 + 36 = 38, So, the 47th term in S1 and 38th term in S 2 are the, 10th common term in both series., For S1, T47 = 2 + (47 − 1) × 4 = 2 + 46 × 4 = 186, and for S 2, T38 = 1 + (38 − 1) × 5 = 1 + 37 × 5 = 186, 8. Given, 10th term of GP = 9, T10 = ar(10 − 1) = 9, …(i), ⇒, ar 9 = 9, (where, a = first term; r = common difference), and, 4th term of GP = 4, …(ii), T4 = ar( 4 − 1) = ar3 = 4, On divide Eq. (i) by Eq. (ii), we get, ar 9 9, 9, …(iii), =, ⇒ r6 =, 4, ar3 4, Now, 7th term of GP is, 9a, …(iv), T7 = ar 6 =, 4, On multiplying Eqs. (i) and (ii), we get, (ar 9 )(ar3 ) = 9 ⋅ 4, ⇒, a 2r12 = 36, 2 6 2, ⇒, a (r ) = 36, 2, 36 × 16, 9, a 2 = 36 ⇒ a 2 =, ⇒, 4, 81, 6 ×4, 8, a=, ⇒ a=, ⇒, 9, 3, From Eq. (iv), we get, 9 8, T7 = × = 2 × 3 = 6, 4 3, 9. Given, a, b, c, d, e and f are in AP., b−a =c−b, …(i), ⇒, b−c=a −b, and, …(ii), e−d =d −c, On adding both the equations,, (e − c) + (b − d ) = (a − c) + (d − b), …(iii), ⇒, (e − c) = (d − c) + (a + d ) − 2b, (Q a, b and c are in AP.), ⇒, 2b = a + c, , NDA/NA Mathematics, From Eq. (iii), we get, ⇒, (e − c) = (d − c) + (a + d ) − (a + c), = (d − c) + (d − c), ⇒, (e − c) = 2(d − c), a+b, 10. Arithmetic mean of a and b is, 2, a+b, = 10, ⇒, 2, ⇒, a + b = 20, Geometric mean of a and b is ab, ⇒, ab = 8, ⇒, ab = 64, Now, we have,, (a − b)2 = (a + b)2 − 4ab, = (20)2 − 4(64), = 400 − 256 = 144, ⇒, a − b = 12, From Eqs. (i) and (iii), we get, a = 16 and b = 4, So, one number exceeds the other by 12., , (given), …(i), (given), …(ii), , …(iii), , 11. Number lying between 107 and 253 are 108, 109, 110,, 252 and the number which are divisible by 5 are 110,, 115, 120,…, 250, which form an AP., ∴, a = 110, d = 5, l = 250 and n = 29, 29, n, ∴, S = [a + l] =, [110 + 250], 2, 2, 29 × 360, =, = 29 × 180 = 5220, 2, 12. Since, p, q and r are in AP, then 2q = p + r, ⇒, p − 2q + r = 0, Here, the option (c) satisfy the above condition., 13. Let a be the first term of an AP and d be the common, difference,, n, S n = [2a + (n − 1) d ], ∴, 2, Where, n be the number of terms., 9, 9, ⇒ 171 = [2a + (9 − 1) d ] ⇒ 171 = (2a + 8d ), 2, 2, ⇒ 38 = 2a + 8d, From this we conclude that no terms of the AP can be, determined., 14. Put a = 3 in given term, the term satisfied the condition, of AP., x+ y, n (x + y), 15. Given that, A =, ,S =, 2, 2, (x + y), n, S, 2, Now,, =, =n, x+ y, A, 2, y+ z, 16. Let y, x and z are in AP, then x =, 2, …(i), ⇒, 2x = y + z, and y, g1 g2 and z are in GP
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83, , Sequence and Series, , Then,, , g1 g2, z, =, =, ⇒ g12 = g2y, y, g1 g2, , ⇒, , g13 = g1 g2 y, , and, ⇒, , g22, g32, , and, , yz = g1 g2, , On solving Eqs. (i) and (iii), we get, a = 16 and b = 4, ...(ii), , 21. Required sum, , = g1z, , G, , = g1 g2z, , ...(iii), , K, , L, , ...(iv), , On adding Eqs. (ii) and (iii), we get, g13 + g32 = yg1 g2 + zg1 g2 = g1 g2 ( y + z ) = yz ⋅2x, [from Eqs. (i) and (iv)], = 2xyz, 17. Let two numbers are x and y,, x+ y, 2xy, then, A=, , G = xy , H =, 2, x+ y, G = 3H, 2xy, ⇒, xy = 3 ×, ⇒ x + y = 6 xy, x+ y, , D, , A, , F, , H, I, B, , J, E, , C, , 1, 1, (16)2 + (16)2 + K ∞, 2, 4, 1 1, , 2, = (16) 1 + + + K ∞ , 2 4, , , , , 1 , = 162 , = 256 × 2 = 512 sq cm, 1, 1 − , , 2, = (16)2 +, , (given), , ∴, , x+ y, 2xy , +, , (3H + H ), x + y, ( A + H ) (G + H ) 2, =, ( A − H ) (G − H ), x + y 2xy , −, , (3H − H ), 2, x + y, , ⇒, , 6 xy, 2xy , +, , , 6 xy 4H 5, 2, ×, =, 6 xy 2xy 2H 2, −, , , 6 xy , 2, , ...(i), , 18. If a , ar , ar 2, K are in GP, then, 1, According to question, a = (ar + ar 2) ⇒ 3 = r + r 2, 3, −1 ± 1 + 4 × 3, 2, ⇒, r + r −3 =0⇒ r =, 2, −1 ± 13, 13 − 1, (Q r > 0), r=, =, ⇒, 2, 2, 1, 19. Q, tan 2 30° =, 3, 2, tan 45° = 1 and tan 2 60° = 3, 2, ∴ tan 30° , tan 2 45° and tan 2 60° are in GP., Because its common ratio is same, i.e., 3., 20. Let H be the harmonic mean of two numbers., ∴, A = G + 2 and G = H + 1.6, and, A = H + 1.6 + 2 = H + 3.6, We know that,, AH = G 2, (H + 3.6)H = (H + 1.6)2, ⇒, H 2 + 3.6H = H 2 + 2.56 + 3.2H, 2.56, H =, = 6.4, ⇒, 0.4, ∴, A = 6.4 + 3.6 = 10 and G = 6.4 + 1.6 = 8, Let two numbers are a and b., …(i), ∴, a + b = 20, and, …(ii), ab = 64, We know that,, (a − b)2 = (a + b)2 − 4ab = 400 − 256 = 144, …(iii), ⇒, a − b = 12, , 22. A : G = m : n, ⇒, , a+b m, (a + b)2 m2, = 2, =, ⇒, 4ab, n, 2 ab, n, , …(i), , (a + b)2 − 4ab m2 − n 2, (by componendo), =, 4ab, n2, (a − b)2 m2 − n 2, …(ii), ⇒, =, 4ab, n2, From Eqs. (i) and (ii), we get, (a + b)2, m2, (a + b), m, ⇒, =, = 2, 2, 2, (a − b), (a − b), m − n2, m − n2, and, , ⇒, , 2, 2, (a + b) + (a − b) m + m − n, =, (a + b) − (a − b) m − m2 − n 2, , (using componendo and dividendo rule), ⇒, , 2, 2, a m+ m −n, =, b m − m2 − n 2, , 23. Let S n be the sum of n terms of a sequence a1 , a 2, K , a n, of the form An 2 + Bn., We have,, S n = An 2 + Bn, ⇒, S n − 1 = A (n − 1)2 + B (n − 1), Now, a n = S n − S n − 1, = An 2 + Bn − { A (n − 1)2 + B(n − 1)}, = An 2 + Bn − { An 2 − 2 An + A + Bn − B}, = 2 An − ( A − B), ⇒ a n + 1 = 2 A (n + 1) − ( A − B), ∴ a n + 1 − a n = 2 A (n + 1) − ( A − B) − 2 An + ( A − B) = 2 A, Since, a n + 1 − a n = 2 A, So, the sequence is an AP with common difference 2A., 24. The given series is, (1 × 3) + (3 × 5) + (5 × 7) + ..., Tn = (2 n − 1) × (2 n + 1) = 4n 2 − 1, Sum of series = 4Σn 2 − Σ1
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84, , NDA/NA Mathematics, 4 n (n + 1) (2 n + 1), −n, 6, 2, , 2 (2 n + 3n + 1), − 1, Sn = n , 3, , , , = 2550 + 1050 − 550, = 3050, , Sn =, , ∴, , 4n 2 + 6n + 2 − 3 , Sn = n , , 3, , , 2, n (4n + 6n − 1), Sn =, 3, 50 [4(50)2 + 6(50) − 1], S50 =, 3, 50 × (10000 + 300 − 1) 50 × 10299, =, =, = 171650, 3, 3, , 25. Given series is, log 6 7, log 42 7 and log 294 7, or it can be rewritten as, 1, 1, 1, ,, and, log7 6 log7 42, log7 294, Let us assume the series log7 6, log7 42, log7 294 are in, AP., ∴, 2 log7 42 = log7 6 + log7 294, ⇒, log7 422 = log7 (6 ⋅ 294), ⇒, (42)2 = 6 ⋅ 294, ⇒, 1764 = 1764, ∴Our assumption is true., If the series log7 6, log7 42 and log7 294 are in AP., 1, 1, 1, are in HP., ⇒ The series, ,, ,, log7 6 log7 42 log7 294, Hence, log 6 7, log 42 7 and log 294 7 are in HP., 6, , 6, , 6, , 7, , 2, , 3, , 26. Let S = 7 , 77 , 77 , K ∞, 6 [1 + 1 + 1 + K ∞ ], 7 72, , = 77, =, , 6 7 , , 77 6 , , =, , , , 1 , , 6, 1 − 1 , 77 7 , , =7, , 27. Let the sum of series which are divisible by 2 is, P = 2 + 4 + 6 + K + 100, = 2 (1 + 2 + 3 + K + 50), 50 (50 + 1), = 2⋅, = 2550, 2, Again, let the sum of series which are divisible by 5 is, Q = 5 + 10 + 15 + K + 100, = 5 (1 + 2 + 3 + K + 20), 20 (20 + 1), = 5⋅, = 50 ⋅ 21, 2, = 1050, and the sum of series which are divisible by both 2, and 5 (i.e., 10) is, R = 10 + 20 + 30 + ... + 100, = 10 (1 + 2 + 3 + K + 10), 10 (10 + 1), = 10 ⋅, = 50 ⋅ 11, 2, = 550, ∴ Required sum of integers, =P+Q−R, , 28. Q AM = 27 and HM = 12,, and we know that,, (GM)2 = (AM) (HM), = 27 × 12, ⇒, GM = 3 × 3 × 2, = 18, 29. The numbers between 200 and 400 which are divisible, by 7, are, 203, 210, 217, …, 399, Now, let the number of terms be n., ∴, 399 = 203 + (n − 1) 7, 196, = (n − 1) ⇒ n = 29, ⇒, 7, 29, Thus, required sum =, [203 + 399], 2, 29 × 602, =, 2, = 8729, 1 1, 1, 30. Let, are in AP., ,, and, ab ca, bc, 1, 1, 1, 1, b−c a−b, =, −, =, −, ⇒, ⇒, ca ab bc ca, abc, abc, ⇒, b − c = a − b ⇒ 2b = a + c, ⇒ a, b and c are in AP., 1, 1, 1, and, are in AP., Now,, ,, b+ c, c+ a, a + b, 2, 1, 1, ∴, =, +, c+ a, b+ c, a + b, ⇒ 2( b + c )( a + b ) = ( c + a )( a + 2 b + c ), ⇒ 2( ab + b + ac + bc ) = ac + 2 bc + c + a, + 2 ab + ac, ⇒ 2 ab + 2b + 2 ac + 2 bc = 2 ac + 2 bc, + 2 ab + c + a, ⇒, 2b = a + c, ⇒ a, b and c are in AP., Hence, both the statements are correct., 31. Let a and d be the first term and common difference of, an AP., According to question, p ⋅ T p = q ⋅ Tq, ⇒, p[a + ( p − 1)d ] = q[a + (q − 1)d ], ⇒, pa + ( p2 − p)d = qa + (q2 − q) d, ⇒, ( p − q)a = (q2 − p2 + p − q) d, ⇒, ( p − q)a = ( p − q)(− p − q + 1) d, ⇒, a = − ( p + q − 1) d, Now,, T p + q = a + ( p + q − 1) d, = − ( p + q − 1) d + ( p + q − 1) d, =0, 2, , 32. x = 1 +, ⇒, ⇒, , 3, , y y, y, + + + ..., 2, , , 2, 2, 2, 1, x=, ⇒x=, y, 2, −y, 1−, 2, 2x−2, 2 x − xy = 2 ⇒ y =, x
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85, , Sequence and Series, 33. The GP is a , ar , ar 2, ... , ar 2n, ∴ P = a ⋅ ar ⋅ ar 2 ⋅ ar3 ⋅ ... ⋅ ar 2n, = a 2n + 1 r(1 + 2 + ... + 2n), = a( 2n + 1) r, , 2n ( 2n + 1 ), 2, , = a 2n + 1 r n( 2n + 1) = (ar n )( 2n + 1), = (2n + 1)th power of the (n + 1)th term of GP., 34. Given that, the group of natural numbers, (1), (2, 3), (4, 5, 6), (7, 8, 9, 10), K, whose number of terms in each group are, 1, 2, 4, 7, ..., Let S = 1 + 2 + 4 + 7 + K + tn,, and S =, 1 + 2 + 4 + K + tn − 1 + tn, –, – – –, –, –, –, 0 = 1 + 1 + 2 + 3 + K + tn − tn − 1 + tn, tn = 1 + (1 + 2 + 3 + K + (n − 1) terms ), (n − 1), tn = 1 +, {2 + (n − 2)}, 2, n (n − 1), tn = 1 +, 2, Therefore, the first term in 11th group of natural, number is, 11 ⋅ 10, t11 = 1 +, = 56, 2, ∴ The 11th group is (56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66)., Sum of the terms of 11th group, 11, S=, {2 × 56 + (11 − 1) 1}, 2, 11, =, (112 + 10) = 11 × 61, 2, = 671, 35. First five terms of a geometric progression are a, ar,, ar 2, ar3 and ar 4., a + ar + ar 2 + ar3 + ar 4, Mean =, ∴, 5, a (1 + r + r 2 + r3 + r 4 ), =, 5, r5 − 1, a, , r −1 , =, 5, a (r5 − 1), =, 5(r − 1), 36. Let the AP is, a , a + d , a + 2d , ... , a + (2n − 1)d , a + 2nd, Series of odd terms, a + d , a + 3d , ... , a + (2n − 1) d , has n terms., n, Sum = [(a + d ) + { a + (2n − 1)d }], ∴, 2, n, = [2a + 2nd ], 2, = n [a + nd ], Series of even terms, a , a + 2d , a + 4d , ... , a + 2nd, has (n + 1) terms., n+1, Sum =, ∴, [a + (a + 2nd )], 2, , n+1, (2a + 2nd ), 2, = (n + 1)(a + nd ), n+1, So, the ratio =, n, =, , 37. Given,, ∴, , S n = n 2 − 2n, a n = Sn − Sn − 1, = n 2 − 2n − [(n − 1)2 − 2(n − 1)], = n 2 − 2n − [n 2 + 1 − 2n − 2n + 2], = 2n − 3, , 38. Q a , 2a + 2 and 3a + 3 are in GP., ∴, (2a + 2)2 = a (3a + 3), 2, ⇒, 4a + 4 + 8a = 3a 2 + 3a, ⇒, a 2 + 5a + 4 = 0, ⇒, (a + 4)(a + 1) = 0, ⇒, a = −1 or – 4, Let the 4th term is x., a, 3a + 3, =, ∴, 2a + 2, x, (3a + 3)(2a + 2), ⇒, x=, a, When,, a = − 4, x = − 13.5, and, when a = − 1, x = 0, So, the 4th term is –13.5., 39. Given, a n = 2n − 1, ∴, , n, , n, , k =1, n, , k =1, , S n = Σ a k = Σ (2n − 1), = 2 Σ (n − n ), k =1, , n (n + 1), −n, 2, = n2 + n − n = n2, , =2⋅, , 40. Arithmetic mean of the combined, , 4 × 15 + 6 × 12, x n + x2n2, 10 numbers =, Q x12 = 1 1, , , 10, n1 + n2 , where, n1 = 4, x1 = 15 and n2 = 6, x2 = 12, 60 + 72, =, = 13.2, 10, 41. First nine terms of a GP are a , ar , ar 2, ... , ar 8., ∴, P = a ⋅ ar ⋅ ar 2 K ar 8, 8, , = a 9 ⋅ r1 + 2 + ... + 8 = a 9 ⋅ r 2, = a9 ⋅ r, , 8 ⋅9, 2, , (1 + 8 ), , = a 9r36, , = (ar 4 )9 = (T5 )9, = 9th power of the 5th term, 42. Q x + y, 2 y and y + z are in harmonic progression., 2(x + y)( y + z ), 2y =, x+ y+ y+ z, ⇒, 2 y(x + 2 y + z ) = 2(xy + xz + y2 + yz ), ⇒, 2xy + 4 y2 + 2 yz = 2xy + 2xz + 2 y2 + 2 yz, ⇒, 2 y2 = 2xz, ⇒, y2 = xz, ⇒ x, y and z are in geometric progression.
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86, , NDA/NA Mathematics, , 43. The given series is, , 1, 1 ⋅3, 1 ⋅3 ⋅5, +, +, +K∞, 8 8 ⋅ 16 8 ⋅ 16 ⋅ 24, On comparing this series with, n (n − 1) 2, S = 1 + nx +, x + K ∞ , we get, 2!, 1, …(i), nx =, 8, n (n − 1) 2 1 ⋅ 3, …(ii), and, x =, 2!, 8 ⋅ 16, S =1 +, , From Eqs. (i) and (ii), we get, n (n − 1) 2, 1 ⋅3, x, 2!, 8, = ⋅ 16, 1 1, n 2x2, ⋅, 8 8, n −1 3, =, ⇒ n − 1 = 3n, ⇒, 2n, 2, 1, n=−, ⇒, 2, On putting this value in Eq. (i), we get, 1, 1, 1, ⇒ x=−, − x =, 2, 8, 4, But, , 1, , S = (1 + x)n = 1 − , , 4, 3, = , 4, , −1/ 2, , =, , 47. We know, if n geometrical means be inserted between, two quantities a and b, then their product will be (ab)x|2., Both A and R are individually true and R is the correct, explanation of A., −1/ 2, , 2, 3, , 44. The geometric mean of the ratio of corresponding terms, of two series, where G1 and G2 are geometric means of, G, two series is 1 ., G2, 45. 0.3 + 0.03 + 0.003 +K, = 3 [0.1 + 0.01 + 0.001 + ...], 3, =, [0.9 + 0.09 + 0.009 + K ], 9, 1, =, [(1 − 0.1) + (1 − 0.01) + (1 − 0.001) + ...], 3, 1, =, [n − {0.1 + 0.01 + 0.001 + ...}], 3, 1, 0.1(1 − (0.1n ) , = n −, , 3, 0.9, , n 1, n , =, −, (1 − (0.1) ), 3 27, , Also, a n = a + nd, ∴, a1 + a 2 + K + a n, = a + d + a + 2d + K + a + nd, = na + (1 + 2 + K + n ) d, n (n + 1) d, = na +, 2, n, = [2a + (n + 1) d ], 2, Hence, A is false but R is true., 46. Let the first term of AP = a and cd = d, Then, mth term = a + (m − 1)d, and nth term = a + (n − 1)d, , According to question,, a + (m − 1)d = n ⇒ m{ a + (m − 1)d } = mn, a + (n − 1)d = m, ⇒, n{ a + (n − 1)d } = mn, ∴, m{ a + (m − 1)d } = n { a + (n − 1)d }, ⇒, ma + m(m − 1)d = na + n (n − 1)d, (ma − na ) = n (n − 1)d − m(m − 1)d, (m − n )a = (n 2 − n − m2 + m)d, = {(n 2 − m2) − (n − m)}d, = (n − m)(n + m − 1)d, ⇒ (m − n )a + (m − n )(m + n − 1)d = 0, ⇒, (m − n ){ a + (m + n − 1)}d = 0, ...(i), ⇒, a + (m + n − 1)d = 0, m ≠ n, Now,, (m + n )th term = a + (m + n − 1)d, =0, Both A and R are individually true and R is the correct, explanation of A., , 48. Q a , b, c and d are in AP., 1 1 1, 1, ⇒, , , and are in HP., a b c, d, abcd abcd abcd, abcd, are in HP., ,, ,, and, ⇒, a, b, c, d, ⇒, bcd , acd , abd and abc are in HP., , a , 49. 91/3 91/ 9 91/ 27 K ∞, Q (GP)∞ = 1 − r , , , 1, , = 93, , +, , 1, 1, +, + K∞, 9 27, , 1/3, , 1, , = 91 − 1/3 = 93, , 2, 3, , = 91/ 2 = (3), , 2×, , 1, 2, , =3, , 50. If a, b, c and d are in HP. Such that a > d , then the only, possible condition is a + c = b + d., 51. I. 2.357 = 2 + 0.357 + 0.000357 + K, 257 357, =2 + 3 +, + ..., 10, 106, 357, 3, 357 103, = 2 + 10, =2 + 3 ×, 1, 999, 10, 1− 3, 10, 2355, =, 999, II. Given, 1 + cos α + cos 2 α + K = 2 − 2, 1, 1, 2+ 2, ∴, =, ×, 1 − cos α 2 − 2 2 + 2, 2+ 2 2+ 2, =, =, 4 −2, 2, 1, 1 − cos α = 1 +, ⇒, 2, 1, 3π, ⇒, ⇒α =, cos α =, 4, 2, 52. Since, b1 , b2 and b3 are in AP with common difference d,, then b2 = b1 + d and b3 = b1 + 2d
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87, , Sequence and Series, Also,, ⇒, ⇒, ⇒, ⇒, , b32 = b2b3 + b1d + 2, , 55., , (b1 + 2d ) = (b1 + d )(b1 + 2d ) + b1d + 2, 2, , b12, , + 4d 2 + 4b1d = b12 + 2b1d + b1d + 2d 2 + b1d + 2, 2d 2 = 2 ⇒ d 2 = 1, d = ± 1 or, , d =1, , 53. The given series is not an arithmetic geometric series,, since 12, 22 ,32 ,42... are not in AP., However, their successive differences (22 − 12), (32 − 22),, (42 − 32) , ... i.e., 3, 5, 7, ... are in AP., So, the above process will be repeated twice, S ∞ = 1 + 4x + 9x2 + 16x3 + ... ∞, xS ∞ = x + 4x2 + 9x3 + ... ∞, On subtraction,, , =1+, , (1 − x)S ∞ = 1 + 3x + 5x2 + 7x3 + ... ∞, x(1 − x)S ∞ = x + 3x2 + 5x3 + ... ∞, On subtracting Eq. (iv) from Eq. (iii), we get, (1 − x)2S ∞ = 1 + 2x + 2x2 + 2x3 + ... ∞, = 1 + 2x(1 + x + x2+ ... ∞ ), 1, 1+ x, = 1 + 2x, =, 1−x 1−x, (1 + x), S∞ =, (1 − x)3, 2 n−1, , 54. The nth term of the series A1 is tn = n x, , ...(i), …(ii), , 4 9 16, +, +, + ... ∞, 3 32 33, 1, 1 4, 9, S ∞ = + 2 + 3 + ... ∞, 3, 3 3, 3, On subtraction, we get, 2, 5 5, 7, S∞ = 1 + + 2 + 3 + K ∞, 3, 3 3, 3, 2, 1 3, 5, S ∞ = + 2 + 3 + ... ∞, 9, 3 3, 3, On subtracting the two series, we get, 2 2, 2, 4, ⋅ S ∞ = 1 + + 2 + 3 + ... ∞, 9, 3 3, 3, S∞ = 1 +, , ...(iii), …(iv), , =1+, , 2, 1 1, , 1 + + 2 + ... ∞, , 3, 3 3, 2, 1, =2, ×, 1, 3 , 1 − , , 3, , 9, 4, 1, =4, 2, , S∞ = 2 ×, , ., , (1 + x), S1 (1 − x)3 2(1 + x), 56., =, =, 9, S2, 9(1 − x)3, 2
Page 215 : 5, , Logarithms, Logarithm, , 9. a > 1, then, , If a is a positive real number, other than 1 and a x = m,, then x is called the logarithm of m to the base a. It is, denoted by loga m. Thus, a x = m ⇔ x = loga m, 23 = 8 ⇔ log2 8 = 3, , e.g.,, %, %, %, , If x < 0, then loga x is imaginary., If x = 0, then loga x is meaningless., loga x exists, if and only if x > 0, a > 0 and a ≠ 1., , Properties of Logarithms, , (a) loga x > p ⇒ x > a p, (b) 0 < loga x < p ⇒ 1 < x < a p, 10. If 0 < a < 1, then, (a) loga x > p ⇒ 0 < x < a p, (b) 0 < loga x < p ⇒ a p < x < 1, , Graph of y = log a x, y, , 1. a log a x = x ; a ≠ 0, ≠ 1, x > 0, 2. a logb x = x logb a ; a > 0, b > 0, ≠ 1, x > 0, , O, , 3. loga a = 1, loga 1 = 0 ; a > 0, ≠ 1, 4. loga x =, , y, y = logax, a>1, , 0<a<1, (1, 0), x, , O, , x, , y = loga x, , 1, ; x , a > 0, ≠ 1, logx a, , 5. loga x = logb x ⋅ loga b =, , logb x, ; a , b > 0, ≠ 1, x > 0, logb a, , 6. For m , n > 0, a > 0 and ≠ 1, (a) loga ( m ⋅ n ) = loga m + loga n, m, (b) loga = loga m − loga n, n, (c) loga ( m n ) = n loga ( m ), , Example 1. If a = log 2 3, b = log 2 5, c = log7 2, then the, value of log140 63 in terms of a, b, c is, 2 ac + 1, a +1, (a), (b), 2c + bc + 1, b +1, ac + 1, (d) None of these, (c), 2b +1, , Solution (a) log140 63 =, , 7. For x > 0, a > 0, ≠ 1, (a) loga n x =, , 1, loga x, n, , m, (b) loga n x m = loga x, n, , (b) loga x < loga y , if 0 < a < 1, , =, , 2 log 2 3 + log 2 7, 2 log 2 2 + log 2 5 + log 2 7, , =, , 2a + 1/ c, 2ac + 1, =, 2 + b + 1 / c 2c + bc + 1, , Example 2. If log10 1+ x + 3 log10 1 − x, , 8. For x > y > 0, (a) loga x > loga y , if a > 1, , log 2 63 log 2 (3 × 3 × 7), =, log 2 140 log 2(2 2 × 5 × 7), , = log10 1 − x2 + 2, then the value of x is, , (a) 4, (c) 5, , (b) −99, (d) None of these
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89, , Logarithms, , =, , Solution (d) log10 ( x + 1)1/ 2 + log10 (1 − x)3/ 2, = log10 (1 − x2)1/ 2 + log10 100, , log a + log b + log c, =1, log a + log b + log c, , 4. If x = 9 is a root of the equation, , ⇒, , ( x + 1)1/ 2 (1 − x)3 / 2 = (1 − x2)1/ 2 × 100, , Example, , ⇒, , 1 − x (1 − x − 100) = 0, , 8 ax , log π ( x 2 + 15 a 2) − log π ( a − 2) = log π , , then the other, a − 2, root is, (a) 12, (b) 15, (c) 13, (d) 10, , 2, , ⇒, x = ± 1, x = − 99, but, x ≠ ±1, − 99, Thus, there is no value of x which satisfy the given equation., , 1, +, 1+ log b a + log b c, 1, 1, is, +, 1+ log c a + log c b 1+ log a b + log a c, (b) 2, (d) 4, , Example 3. The value of, , (a) 1, (c) 3, , 1, 1, Solution (a), +, 1 + log b a + log b c 1 + log c a + log c b, 1, +, 1 + log a b + log a c, log b, log c, =, +, log a + log b + log c log a + log b + log c, log a, +, log a + log b + log c, , Solution (b) log π ( x 2 + 15 a 2) − log π ( a − 2), 8ax , = log π , , a − 2, ⇒, ⇒, ⇒, , x2 + 15a2 , 8ax , log π , , = log π , a, −, 2, a − 2, , , x2 + 15a2 = 8ax, x2 − 8ax + 15a2 = 0, , ⇒, ( x − 5a) ( x − 3a) = 0, ⇒, x = 3a = 9 ⇒ a = 3, Another root x = 5a = 5 × 3 = 15
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Exercise, Level I, 1. If log2 x = 4, then x is equal to, (a) 16, (b) 4, (c) 2, , (d) 64, , 2. What is the value of, 3, 27, 9, log10 − log10 + log10 ?, 4, 32, 8, (a) 3, (c) 1, , (NDA 2011 I), , (b) 2, (d) 0, , 13. If a , b and c are the pth, qth and rth terms, respectively, of a GP, then ( q − r ) log a + (r − p) log b + ( p − q ) log c, is equal to, (a) 0, (b) 1, (c) −1, (d) abc, (log27 9)(log16 64), 14. What is the value of, ?, log4 2, (NDA 2009 II), (a) 1, , 3. If log3 log4 x > 0, then, (a) x > 1, (c) x > 64, , (b) x > 4, (d) None of these, , 4. If log10 x < 1, then, (a) 0 < x < 10, (c) x < 10, , (b) x > 10, (d) 0 ≤ x < 10, , (NDA 2011 II), , 7. If log3 [log3 (log3 x )] = log3 3, then what is the value of, x?, (NDA 2010 II), (a) 3, (b) 27, (c) 39, (d) 327, 8. If a , b and c are in GP, then logax x , logbx x , logcx x, are in, (a) GP, (b) HP, (c) AP, (d) None of these, 9. The value of x satisfying, log10 1 + x + 3 log10 1 − x = log10 1 − x 2 + 2 is, (a) −1 < x < 1, (b) x = 0, (c) No solution, (d) None of these, a + b 1, 10. If loge , = (loge a + loge b), then, 2 2, b, (a) a = b, (b) a =, 2, b, (c) 2a = b, (d) a =, 3, 11. If logax x , logbx x, logcx x are in AP,, a , b, c, x ∈ (1, ∞ ), then a , b and c are in, (a) AP, (b) GP, (c) HP, (d) None of these, , (c) 4, , (d) 8, , 16. The number of solutions of log2 ( x − 1) = 2 log2 ( x − 3), is, (a) 2, (b) 1, (c) 6, (d) 7, , 5. If y = 21/ log x ( 8) , then x is equal to, (a) y, (b) y 2, 3, (c) y, (d) None of these, 6. What is the value of log2 (log3 81)?, (a) 2, (b) 3, (c) 4, (d) 9, , (b) 2, , 15. If (logx x )(log3 2x )(log2x y ) = logx x 2 , then what is the, value of y?, (NDA 2009 II), 9, (b) 9, (c) 18, (d) 27, (a), 2, , where, , 12. If 2 log( x + 1) − log( x 2 − 1) = log 2 , then x equals to, (a) 1, (b) 0, (c) 2, (d) 3, , 17. If log2 x + log2 y ≥ 6, then the least value of ( x + y ) is, (a) 4, (b) 8, (c) 16, (d) 32, 18. If 4log 9 3 + 9log 2 4 = 10log x 83 , then x is, (a) 4, (b) 9, (c) 10, (d) None of these, 19. The least value of 2 log10 x − logx ( 0.01) for x > 1 is, (a) 1, (b) 2, (c) 4, (d) 6, log3 9, ?, 20. What is the value of 2 log8 2 −, 3, (NDA 2012 I), (a) 0, (b) 1, (c) 2, (d) 1 / 3, 21. If x > 1 and log2 x , log3 x , logx 16 are in GP, then what, is the value of x?, (NDA 2009 I), (a) 9, (b) 8, (c) 4, (d) 2, 22. If log10 ( x + 1) + log10 5 = 3, then what is the value of x?, (a) 199, (b) 200, (NDA 2009 I), (c) 299, (d) 300, 23. Solution of the equations x log x 2 = log3 ( x + y ) and, x 2 + y 2 = 65 is, (a) x = 8, y = 1, (b) x = 1, y = 8, (c) ( x = 8, y = 1); ( x = 1, y = 8), (d) None of the above, 24. The identity, loga n logb n + logb n logc n + logc n loga n is, loga n logb n logc n, logabc n, (a), (b), logabc n, loga n, logb n, (d) None of these, (c), logabc n
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91, , Logarithms, 25. If log4 7 = x , then log7 16 is equal to, (a) 2 / x, (b) x 2, (c) x, (d) 2x, 26. The value of e(log10 tan 1° + log10 tan 2° + .... + log10 tan 89° ) is, equal to, (a) 0, (b) 1, 1, (c) e, (d), e, , 27. If the logarithm of a number of the base 8 is 6, then, the number is, 8, (a) 48, (c) 6 8, (b), (d) 512, 6, 28. Find the value of log5, 2, 1, (b), (a), 3, 3, , 5, , 5., , (NDA 2007 I), , 1, (c), 2, , (d) 2, , Level II, 1. If x < 0, y < 0, then log ( xy ) is equal to, (a) log x + log y, (b) log ( − x ) + log( − y ), (c) − log x − log y, , 2. If y = a, , 1, 1 − log a x, , (d) None of these, 1, , and z = a 1 − log a y , then x is equal to, 1, , 1, , (b) a z + log a z, , (a) a 1 + log a z, 1, , (c) a 1 − log a z, , (d) None of these, , 3. If log10 2 , log10 ( 2 − 1), log10 ( 2x + 3) are three, consecutive terms of an AP, then which one of the, following is correct?, (NDA 2011 II), (a) x = 0, (b) x = 1, (c) x = log2 5, (d) x = log5 2, x, , , , 1, equals, 4. What is log ( a + a 2 + 1 ) + log , a + a2 + 1 , , , to?, (NDA 2011 I), (a) 1, (b) 0, 1, (c) 2, (d), 2, 5. The least value of n in order that the sum of first n, 2, 3, 3 3, 3, terms of the infinite series 1 + + + + .... ,, 4, 4 4, should differ from the sum of the series by less than, 10−6 is, (Given, log10 2 = 0.30103, log10 3 = 0.47712), (a) 14, (b) 27, (c) 53, (d) 57, 6. Which of the following is not true?, 1, 1, (a), +, >2, log3 π log4 π, (b) log3 5 is an irrational number, 10, (c) log 8 x =, ⇒ x = 16, 3, (d) If logx ( a 2 + 1) < 0, ( a ≠ 0), then 0 < x < 1, 7. The number of solution (s) of the equation, log2( x 2 − 1) = log1/ 2 ( x − 1) = a real number, is, (a) 0, (b) 1, (c) 2, (d) 3, , 8. The value of 22 − log 2 5 is equal to, (a) 4/5, (b) 5/4, (c) 2/5, (d) 5/2, r, 9. If logr 6 = m and logr 3 = n , then logr is equal to, 2, log2 r, (b) 1 − logr 2, (a), 2, (c) 1 − m − n, (d) 1 − m + n, 1 2, 10. If log8 m + log8 = , then m is equal to, 6 3, (a) 24, (b) 18, (c) 12, (d) 4, 11. If a x = b y = cz and logb a = logc b, then which one of, the following will hold true?, (b) x 2 = yz, (a) y 2 = xz, (d) y = xz, (c) z 2 = xy, log, 12. What is the value of, , log, , (a) logαβ (α ), (c) logαβ (αβγ ), , αβ, αβγ, , (H ), (H ), , ?, , (NDA 2010 I), , (b) logαβγ (αβ ), (d) logαβ (β ), , 13. If logk x log5 k = 3, then what is the value of x?, (a) k5, (c) 243, , (b) 5k3, (d) 125, , 14. For what value(s) of x is, , (NDA 2009 II), , (NDA 2007 II), , log10 ( 999 + x − 3x + 3 ) = 3 ?, (a) 0, (b) Only 1, (c) Only 2, (d) 1, 2, 2, , 15. The positive solution of the equation, logx + 3 ( x 2 + 6x + 9) + log5 x + 2 ( 6x 2 − 6x ), = log2x − 1 ( 8x3 − 12x 2 + 6x − 1) is, (a) 9, (c) 5, , (b) 6, (d) 2, , 16. How many number of digits are there in 298?, (Given that log10 2 = 0.30103), (a) 98, (b) 99, (c) 30, (d) 29
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92, , NDA/NA Mathematics, , 17. If x , y and z are distinct positive numbers different, from, 1, such, that (log y x logz x − logx x ) +, (logx y logz y − log y y ) + (logx z log y z − logz z ) = 0., What is the value of xyz ?, (a) 2, (b) 1, (c) − 1, (d) 0, 18. If log30 8 = a and log30 5 = b, what is the value of, log30 8?, (a) 3 (1 + a + b), (b) 3 (1 − a + b), (c) 3 (1 + a − b), (d) 3 (1 − a − b), 19. What, is, the, least, integral, value, 2 log10 x − logx ( 0.01)?, (a) 0, (b) 2, (c) 4, (d) 3, 20. What is the value of log, , b, , of, , a log( c)1/3 b log( a )1/ 4 ?, , (a) 12, (b) 24, (c) 1/12, (d) 1/24, log x log 36 log 64, 21. If, =, =, , what are the values of x, log 5, log 6, log y, and y, respectively?, (a) 8, 25, (b) 25, 8, (c) 8, 8, (d) 25, 25, 22. If x (log10 | x |) = 2, what is the value of x?, (a) Only 2, (c) 2 or –2, , (NDA 2008 I), , (b) Only –2, (d) 1 or –1, , Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 25. Assertion (A) If x , y , z > 0 and x ≠ 1, y ≠ 1, z ≠ 1 such, that, log x log y log z, , then x x y y z z = 1, =, =, y−z z−x x− y, Reason (R) If log12 18 = a and log24 54 = b, then, ab + 5( a − b) = 1, 26. Assertion (A) If x is a positive real number different, from unity such that loga x , logb x and logc x are in, AP, then c2 = ( ca )log a b, Reason (R) If x = 2002! ,, 1, 1, 1, 1, then, =4, +, +, + ...... +, log2002 x, log2 x log3 x log4 x, 27. Assertion (A) If f ( x ) = log x , then f ( x ) > 0, ∀ x > 0., Reason (R) f ( x ) = log x is defined for all x > 0., (NDA 2008 I), , 23. What is the number of digits in the numeral form of, 817 ?, (NDA 2007 I), (a) 51, (b) 16, (c) 15, (d) 14, 24. If (log3 x )2 + (log3 x ) < 2 , then which one of the, following is correct?, 1, 1, (a) 0 < x <, (b) < x < 3, 9, 9, 1, (c) 3 < x < ∞, (d) ≤ x ≤ 3, 9, , Directions (Q. Nos. 25-27), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., , Directions (Q. Nos. 28-30) Let us consider, log 2 = 0.3010, log 3 = 0.4771, 81, 25, 16, is, 28. The value of 7 log, + 5 log, + 3 log, 80, 24, 15, (a) 0.3010, (b) 0.3512, (c) 0.412, (d) None of the above, 70, 22, 7, is, 29. The value of log, + log, − log, 33, 135, 18, (a) − 0.512, (b) 0.4213, (c) 0.3010, (d) None of these, 30. The value of 32 − log3 ( 4 × log 36) is, (a) 1.45, (b) 0.69, (c) 1.2, (d) 3.2, , Answers, Level I, 1. (a), 11. (b), 21. (a), , 2. (d), 12. (d), 22. (a), , 3. (b), 13. (a), 23. (a), , 4. (a), 14. (c), 24. (a), , 5. (c), 15. (b), 25. (a), , 6. (a), 16. (b), 26. (b), , 7. (d), 17. (c), 27. (d), , 8. (b), 18. (c), 28. (a), , 9. (c), 19. (c), , 10. (a), 20. (a), , 2. (c), 12. (c), 22. (c), , 3. (c), 13. (d), 23. (b), , 4. (b), 14. (d), 24. (b), , 5. (c), 15. (d), 25. (b), , 6. (c), 16. (c), 26. (c), , 7. (a), 17. (b), 27. (d), , 8. (a), 18. (d), 28. (a), , 9. (d), 19. (c), 29. (a), , 10. (a), 20. (b), 30. (a), , Level II, 1. (b), 11. (a), 21. (b)
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Hints & Solutions, Level I, 1. Q log 2 x = 4, ∴ x = 24 = 16, 9, 27, 3, 2. log10 − log10, + log10, 8, 32, 4, 9 32 3, = log10 ×, × = log10 1 = 0, 8 27 4, 3. log3 log 4 x > 0, log 4 x > 1 ⇒ x > 41 ⇒ x > 4, y = 21/log x ( 8), ⇒, , y = 2log 8 x ⇒ y = 2, , ⇒, , x = y3, , log 3 x, 2, , 6. Given, log 2 (log3 81), = log 2 {log3 (3)4 }, = log 2 {4 log3 3}, = log 2 (4), = log 2 (2)2, = 2 log 2 2, =2× 1= 2, , ⇒, ⇒, ⇒, , (Q log3 3 = 1), (Q log 2 2 = 1), , log3 [log3 [log3 x]] = log3 3, ⇒, log3 [log3 x] = 3, ⇒, log3 x = 33, ⇒, log3 x = 27, ⇒, x = 327, 1, 1, 8. log ax x =, ⇒ log ax x =, log x ax, 1 + log x a, , 7., , 13. a = αβ p − 1 , b = αβ q − 1 , c = αβ r − 1, (where α is first term of GP and β is common ratio ), Take logarithm, ...(i), log a = log α + ( p − 1) log β, ...(ii), log b = log α + (q − 1) log β, ...(iii), log c = log α + (r − 1) log β, Put values of log a , log b and log c from Eqs. (i), (ii), and (iii) in (q − r ) log a + (r − p) log b + ( p − q) log c, we, get 0., (log 27 9)(log16 64), 14., log 4 2, , If a , b and c are in GP., ⇒, log x a , log x b, log x c are in AP., ⇒, 1 + log x a , 1 + log x b, 1 + log x c are in AP., 1, 1, 1, are in HP., ,, ,, ⇒, 1 + log x a 1 + log x b 1 + log x c, ⇒, , ⇒, , =, , ( 1 − x2 ) (1 − x) = 100 1 − x2, 1 − x (1 − x − 100) = 0, 2, , ⇒, x = ± 1, x = − 99, But x = − 99, ± 1 do not satisfy domain., So, no solution exist., a + b 1, 10. log e , = (log e a + log e b), 2 2, a+b, ⇒, = ab, 2, a + b − 2 ab = 0, ⇒, ⇒, a = b, ⇒, a=b, , log3 3 (32) log 42 (4)3, log 22 (21/ 2), , 2, 3, log3 3 × log 4 4, 1, 3, 2, =, = =4, 1, 1, log 2 2, 2 ×2, 4, , log ax x, logbx x, log cx x are in HP., , 9. log10 ( 1 + x (1 − x)3/ 2) = log10 (100 1 − x2 ), ⇒, , 1 + log x a , 1 + log x b, 1 + log x c are in AP., log x a , log x b, log x c are in AP., a , b, c are in GP., , 12. By definition, x ≠ 1, − 1, Given equation is, 2 log (x + 1) − log (x2 − 1) = log 2, (x + 1)2, = log 2, log 2, ⇒, x −1, x+1, =2 ⇒ x=3, ⇒, x−1, , 4. log10 x < 1 ⇒ 0 < x < 10, 5., , 11. log ax x, logbx x and log cx x are in AP., 1, 1, 1, are in HP., ⇒, ,, ,, 1 + log x a 1 + log x b 1 + log x c, , 15. (log x x)(log3 2x)(log 2x y) = log x x2, ⇒, 1 (log3 2x)(log 2x y) = 2, ⇒, log3 y = 2, ⇒, y = 32 ⇒ y = 9, 16., , (Q log x x = 1), , log 2 (x − 1) = 2 log 2 (x − 3), ⇒ log 2 (x − 1) = log 2 (x − 3)2, ⇒, (x − 1) = (x − 3)2 ⇒ x2 − 7x + 10 = 0 ⇒ x = 2, 5, Since, x = 2 does not satisfy the equation., Hence, x = 5 is only solution., , 17., Now,, ⇒, , log 2 xy ≥ 6 ⇒ xy ≥ 26, x+ y, ≥ xy, 2, x+ y, ≥ 23 ⇒ x + y ≥ 24 = 16, 2, , [Q AM ≥ GM ]
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94, , NDA/NA Mathematics, , 18. 4, , log 2 3, 3, , 2, , + 9log 2 2 = 10log x 83 ⇒ 41/ 2 + 92 = 10log x 83, , ⇒, , 2 + 81 = 10, ⇒ 83 = 10, ⇒ x = 10, 1, 19. 2 log10 x − log x, = 2 log10 x + 2 log x 10, 100, 1, , = 2 t + ≥ 4, where t = log10 x, , t, log x 83, , log x 83, , 1, 1, log3 9 = 2 log( 23 ) 2 − log3 (3)2, 3, 3, 2, 1, = log 2 2 − ⋅ 2 log3 3, 3, 3, 1, b = log a b, log a bn = n log a b, log a a = 1), n, 2, 2, 2 2, = (1) − (1) = − = 0, 3, 3, 3 3, , 20. 2 log 8 2 −, , (Q log an, , 21. Q log 2 x, log3 x, log x 16 are in GP., ∴, (log3 x)2 = log 2 x ⋅ log x 16, ⇒, (log3 x)2 = log 2 16, ⇒, (log3 x)2 = 4 log 2 2, ⇒, log3 x = 2 ⇒ x = 32 = 9, , ⇒, ⇒, ∴, , xlog x 2 = 2 = log3 (x + y), x + y = 9, x2 + y2 = 65, x = 8, y = 1 or x = 1, y = 8, x = 8, y = 1, , [Q logm n =, log n c + log n a + log n b, log n a log n b log n c, log a n logb n log c n, log n (abc), =, =, log abc n, log n a log n b log n c, , (but x ≠ 1), , 1, ], log n m, , =, , 25. We have, log 4 7 = x, ∴, log7 16 = log7 4 × 4, = log7 42, , , 1 , Q log e a = log e , a , , 1, 1 2, =2 ×, =2 × =, log 4 7, x x, , = 2 log7 4, , 26. e(log10 tan 1° +, , 22. Given, log10 (x + 1) + log10 5 = 3, ⇒, log10 5(x + 1) = 3, ⇒, 5(x + 1) = (10)3 ⇒ 5x + 5 = 1000, ⇒, 5x = 995 ⇒ x = 199, 23., , 24. log a n logb n + logb n log c n + log c n log a n, 1, 1, 1, =, +, +, log n a log n b log n b log n c, log n c log n a, , log10 tan 2° + ... + log10 tan 89 °), , = elog10 (tan 1° tan 2° ... tan 89° ), = elog10 {tan (90° − 89° ) ⋅ tan (90° − 88° ) .... tan 89° }, = elog10 (cot 89° ⋅ cot 88° .... tan 89° ), = elog10 1 = e0 = 1, 27. Suppose the number be x., ∴, log 8 x = 6, ⇒, x = ( 8 )6 = (81/ 2)6 = 83 = 512, log 5, log 5, log 5, 2, 28. log5 5 5 =, =, =, =, log 5 5 log 5 + 1 log 5 3 log 5 3, 2, 2, , Level II, 1. log xy = log|x| + log| y|, Now, as x < 0,|x| = − x and y < 0,| y| = − y, ∴, log xy = log (− x) + log (− y), 2. From the given relation, we have, a = y1 − log a x = z1 − log a y, ∴, log a a = (1 − log a x) log a y, and, log a a = (1 − log a y) log a z, ⇒ log a y(1 − log a x) = 1 and log a z (1 − log a y) = 1, 1, ⇒, log a y =, 1 − log a x, 1, and, log a z =, 1 − log a y, 1, 1 − log a x, 1, ∴log a z =, =, =, 1, 1 − log a y 1 −, − log a x, 1 − log a x, 1, 1, Now,, =, = log a x, 1 − log a z 1 + 1 − log a x, log a x, 1, , ⇒ log a x =, , 1, ⇒ x = a 1 − log a z, 1 − log a z, , 3. Given that,, log10 2, log10 (2x − 1), log10 (2x + 3) are in AP., Then, 2 log10 (2x − 1) = log10 2 + log10 (2x + 3), ⇒, log10 (2x − 1)2 = log10 (2x + 1 + 6), ⇒, (2x − 1)2 = (2x + 1 + 6), 2x, ⇒, 2 + 1 − 2 ⋅ 2x = 2 ⋅ 2x + 6 ⇒ 22x − 4 ⋅ 2x − 5 = 0, ⇒, (2x )2 − 4(2x ) − 5 = 0, Let, …(i), y = 2x, 2, ⇒, y − 4 y − 5 = 0 ⇒ y3 − 5 y + y − 5 = 0, ⇒, ( y − 5)( y + 1) = 0, ⇒, y = − 1, 5, [from Eq. (i)], ⇒, 2x = − 1, 5, ⇒, x = log 2 (−1), log 2 (5), [but x ≠ log 2 (−1)], ⇒, x = log 2 (5), , , 1, , 4. log (a + a 2 + 1 ) + log , a + a2 + 1, , , = log (a +, , a 2 + 1 ) + log (a +, , a 2 + 1 )− 1, , = log (a +, , a 2 + 1 ) − log (a +, , a2 + 1) = 0
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95, , Logarithms, 3, 1− , 4, 3 3, 5. 1 + + + .... + n terms =, 3, 4 4, 1−, 4, , n, , 9. We have,, log r 6 = m and log r 3 = n, r, ∴ log r = log r r − log r 2, 2, , 2, , 2, , and, , 1+, , 3 3, 1, =4, + + .... + ∞ terms =, 3, , , 4, 4, 1−, 4, n, , 3, 1− , 4, According to the question,, = 4 − 10−6, 1, 4, n, n, 3, 10−6, 3, 1, , , ⇒ 1 − = 1 − (10−6 ) ⇒ =, 4, 4, 4, 4, 3, n log10 = log10 10−6 − log10 4, 4, , ⇒, ⇒, , n (0.47712 − 2 × 0.30103) = − 6 − 2 × (0.30103), 6.60206, ⇒, n=, = 53, 0.12494, 1, 1, 6. (a), +, = log π 3 + log π 4 = log π 12 > 2, log3 π log 4 π, , = 1 − (m − log r 3), [Q log r 6 = m ⇒ m = log r (3 × 2) ⇒ m = log r 3 + log r 2], = 1 − (m − n ), =1 −m + n, 1 2, 10. We have, log 8 m + log 8 =, 6 3, m 2, log 8 =, ⇒, 6 3, m, ⇒, = 823, 6, ⇒, m = 6 × 823, = 6 × 4 = 24, a x = by = cz, x log a = y log b = z log c, log a y, =, log b x, y, logb a =, x, log b z, =, log c y, z, log c b =, y, , 11. If, Then,, ∴, , 12 > π, (b) log3 5 is an irrational number., 10, (c) log 8 x =, ⇒ x = ( 8 )10/3 = 25, 3, ⇒ x = 32, (d) log x (a 2 + 1) < 0, a ≠ 0, ⇒, a2 + 1 > 1, So, log is negative., Hence, base is (0, 1)., 2, , 7. Now, log 2 (x2 − 1) =, , ⇒, and, ⇒, , Then, according to given condition, logb a = log c b, y z, =, ∴, x y, , log e (x2 − 1), log e 2, , ⇒, log, , = log (x − 1), log (x − 1), log1/ 2 (x − 1) =, log 1 / 2, 2, , =, , 12., , αβ, αβγ, , H, H, , =, , = 22 ⋅ 2− log 2 5 = 22 2log 2 (5 ), , −1, , αβ, , , 1 , , Q log a b =, logb a , , , 1, , log a, Q log am b =, , m, , 13. Given, log5 k log k x = 3, ⇒, log5 x = 3, ⇒, ⇒, ⇒, , [Q a log a x = x], , αβγ, , logH, , = log αβ αβγ, , x = 53, x = 125, , 14. log10 {999 +, , 8. We have, 22 − log 2 5, , logH, , = log αβ αβγ, 1, = log αβ (αβγ ), 2, 1 /2, =, log αβ (αβγ ), 1 /2, , log (x − 1), log 1 − log 2, , = − log (x − 1), ⇒ log 2 (x2 − 1) − log1/ 2 (x − 1) = real number, ⇒ log (x2 − 1) + log (x − 1) = real number, ⇒, log [(x2 − 1) (x − 1)] = real number, or, (x2 − 1) (x − 1) = real number, or, x = 1, − 1 real number, But x = 1, − 1 does not satisfy the equation., ∴It has no solution., , = 22 (5)−1, 22 4, =, =, 5 5, , log, , y2 = zx, , ⇒, ⇒, ⇒, ⇒, , 999 +, , x2 − 3 x + 3 } = 3, x2 − 3x + 3 = 1000, , x2 − 3x + 3 = 1, x2 − 3 x + 3 = 1, x2 − 3 x + 2 = 0, x2 − 2 x − x + 2 = 0, , , b,
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96, , NDA/NA Mathematics, ⇒, ⇒, ⇒, , x(x − 2) − 1(x − 2) = 0, (x − 1)(x − 2) = 0, x = 1, 2, , 21., , 15. We have, log( x + 3 ) (x2 + 6x + 9) + log(5 x + 2) (6x2 − 6x), = log( 2x − 1) (8x3 − 12x2 + 6x − 1), ⇒ log(5 x + 2) (6x2 − 6x), = log( 2x − 1) (2x − 1)3 − log( x + 3 ) (x + 3)2, ⇒ log(5 x + 2) (6x2 − 6x) = 3 log( 2x − 1) (2x − 1), − 2 log( x + 3 ) (x + 3), 2, ⇒ log(5 x + 2) (6x − 6x) = 3 − 2, ⇒, (5x + 2) = 6x2 − 6x, 2, ⇒, 6x − 11x − 2 = 0, ⇒, (x − 2) (6x + 1) = 0, ⇒, x = 2, − 1 /6, Thus, the positive value of equation is 2., 16. Let y = 298, Taking log on both sides, we get, log10 y = 98 log10 2, = 98 × 0.30103 = 29.50094, ∴The number of digits in 298 = 30, 17. (log y x log z x − log x x) + (log x y log z y − log y y), + (log x z log y z − log z z ) = 0, log x log x log y log y, , ⋅, − 1 + , ⋅, − 1, ⇒, log y log z log x log z, , log z log z, , +, ⋅, − 1 = 0, log, x, log, y, , , (log x)2, (log y)2, (log z )2, +, +, −3 =0, log y log z log x log z log x log y, , ⇒, , ⇒ (log x)3 + (log y)3 + (log z )3 − 3 log x log y log z = 0, Q If a3 + b3 + c3 − 3abc = 0,, , , then a + b + c = 0, , , ⇒ log x + log y + log z = 0, ⇒, log xyz = 0, ⇒, xyz = e0 = 1, , log x log 36 log 64, =, =, log 5, log 6, log y, log x 2 log 6, Now,, =, log 5, log 6, ⇒, ⇒, and, ⇒, , 2 log y = 2 log 8 ⇒ y = 8, 10log10 | x | = 2 ⇒ log10 | x| = log10 2, , 22. Q, ⇒, , | x| = 2 ⇒ x = 2 or –2, , 23. Let x = 817 ⇒ x = 251, Taking log on both sides of above equation, we get, log x = 51 log 2 = 51 × 0.3010 = 15.381, Number of terms in 817 = 15 + 1 = 16, ∴, (log3 x)2 + (log3 x) < 2, ⇒, (log3 x)2 + (log3 x) − 2 < 0, ⇒, (log3 x + 2) (log3 x − 1) < 0, ⇒, −2 < log3 x < 1, 1, ⇒, < x<3, 9, log x log y log z, 25. Let, =, =, =λ, y−z z−x x− y, 24., , ⇒ log x = λ ( y − z ), log y = λ (z − x), log z = λ (x − y), ⇒, x log x + y log y + z log z, = λx ( y − z ) + λy (z − x) + λz (x − y) = 0, ⇒, log xx + log yy + log z z = 0, ⇒, log (xx yy z z ) = 0, ⇒, xx yy z z = 1, log 18 log (32 × 2), We have a = log12 18 =, =, log 12 log (22 × 3), 2 log 3 + log 2, =, 2 log 2 + log 3, and b = log 24 54 =, , 2 × 15 , 18. Now, log30 8 = log30 23 = 3 log30 , 15 , , Let, , = 3 [1 − {log30 5 + log30 3}] = 3 (1 − a − b), 1, 19. 2 log10 x − log x (0.01) = 2 log10 x − log x, 100, 2 log10 10, 2, = 2 log10 x +, = 2 log10 x +, log10 x, log10 x, , ∴, , x + 2y , ab + 5 (a − b) = , , 2x + y , , x + 3 y, , , 3x + y, , x2 + 5xy + 6 y2, (2x + y) (3x + y), , (3x2 + 7xy + 2 y2) − (2x2 + 7xy + 3 y2) , +5, , (2x + y) (3x + y), , , x2 + 5 yx + 6 y2 + 5 (x2 − y2) 6x2 + 5xy + y2, =1, =, = 2, (2x + y) (3x + y), 6x + 5xy + y2, , a log c1/ 3 b log a1/ 4 c, log a, log b, log c, ⋅, ⋅, 1, 1, 1, log b, log c, log a, 2, 3, 4, = 2 × 3 × 4 = 24, , log 2 = x and log 3 = y, then, 2y+ x, 3y + x, and b =, a=, 2x+ y, 3x + y, , =, , ∴ The least integral value is 4., b, , log 54 log (33 × 2) 3 log 3 + log 2, =, =, log 24 log (23 × 3) 3 log 2 + log 3, , x + 2y x + 3y , + 5, −, , 2x + y 3x + y , , (log10 x)2 + 1 , =2 , ≥4, log10 x , 20. log, , log x = log 25, x = 25, 2 log 6, log 64, =, log 6, log y, , =, , ∴, , A and R are both correct but R is not the correct, explanation of A.
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97, , Logarithms, 26. Since, log a x, logb x and log c x are in AP. Therefore,, 2 logb x = log a x + log c x, 2, 1, 1, ⇒, =, +, log x b log x a log x c, 2, log x a + log x c, =, ⇒, log x b, log x a log x c, ⇒ 2 log x a ⋅ log x c = log x a log x b + log x c ⋅ log x b, ⇒ 2 log x c ⋅ log x a = (log x a + log x c) log x b, ⇒ 2 log x c ⋅ log x a = log x (ca ) ⋅ log x b, log x b, ⇒, 2 log x c = log x (ca ) ⋅, log x a, ⇒, ⇒, ⇒, , log x c2 = log x (ca ) ⋅ (log a b), log x c2 = log a b ⋅ log x (ca ), log x c2 = log x {(ca )log a b }, c2 = (ca )log a b, , We have,, 1, 1, 1, 1, +, +, + ....+, log 2 x log3 x log 4 x, log 2002 x, = log x 2 + log x 3 + log x 4 + .....+ log x 2002, , , 1, Q log a = log a b, b, , , = log x (2 ⋅ 3 ⋅ 4...2002), = log x (2002 !) = log x x = 1, ∴A is correct but R is wrong., f (1) = log 1 = 0, ⇒ f (x) = 0 at x = 1 > 0, ∴, f (x) = log x, and, f (x) ≥ 0, ∀ x > 0, Thus, A is false but R is true., 16, 25, 81, 28. 7 log, + 5 log, + 3 log, 15, 24, 80, 7, , 5, , = log[228 − 15 − 12 × 510 − 7 − 3 × 312− 7 − 5 ], = log (21 × 50 × 30 ) = log 2 = 0 . 3010, 70, 22, 7, 29. log, + log, − log, 33, 135, 18, 7, 70 22 , = log , ×, − log, 33 135, 18, 70 22 , ×, , , = log 33 135 , 7, , , , 18 , 70 22 18, 8, = log , ×, × = log , 33 135 7 , 9, = log 8 − log 9 = log 23 − log 32, = 3 log 2 − 2 log 3, = 3(0 .3010) − 2 (0 .4771), = 0 .9030 − 0 . 9542 = − 0 .5120, 30. 32 × 3log 3( 4 × log 36), , 27., , 16, 25, 81, = log + log + log , 15, 24, 80, , 16 7 25 5 81 3 , = log × × , 80 , 24, 15, , 3, 5, 7, 24 , 52 , 34 , = log , × 3, × 4, , 2 × 3, 2 × 5 , 3 × 5, , , 28, 10, 12, 2, , 5, 3, = log 7 7 × 15, × 12 3 , 5, ×, ×, ×, 3, 5, 2, 3, 2, 5, , , , −1, , = 9 × (4 × log 36)− 1 =, , 9, 4 × log (6)2, 9, 9, =, =, 4 × 2 log 6 8 (log 2 + log 3), 9, 9, =, =, = 1 .45, 8 (0 .4771 + 0 .3010) 8 × 0 .7781, =, , 3, , 9, 4 × log 36
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6, , Matrices, Matrix, A set of mn numbers (real or complex) arranged in the, form of a rectangular array having‘ m ’ rows and‘ n ’ columns, is called m × n matrix [to be read as m by n matrix]., It is usually written as, a11, a, A = 21, M, a, m1, , , a22 K a2n , , M, M, , am 2 K amn , , a12 K a1n, , It is represented by A = [aij ]m × n ., The numbers a11 , a21 ,... etc., are called elements of the, matrix. The element aij belongs to ith row and jth column., 2, 7, 3, , e.g., A = 5 − 4, 6 is a 3 × 3 matrix., , , 8 − 12, 4, %, , Matrix is simply arrangement of numbers, it doesn’t have any, value, e.g., [7] ≠ 7., , Types of Matrices, 1. Null matrix or zero matrix The m × n matrix, whose all elements are zero is called a null matrix of, order m × n. It is usually denoted by O., 2. Row matrix Any 1 × n matrix which has only one, row and n columns is called a row matrix., e.g., X = [ 2 7 8 5]1 × 4 is a row matrix., 3. Column matrix Any m × 1 matrix which has, only one column and m rows is called a column, matrix., 1, e.g.,, Y = 3 is a column matrix., , 53 × 1, , 4. Upper triangular matrix A square matrix, A = [aij ]n × n is called an upper triangular matrix, if, aij = 0 whenever i > j., , e.g.,, , a11 a12, 0 a, 22, A= , M, M, 0, 0, , , a13 K a1n , a23 K a2n , , M, M , 0 K ann n × n, , 5. Lower triangular matrix A square matrix, A = [aij ]n × n is called a lower triangular matrix, if, aij = 0 whenever i < j., , e.g.,, , a11, a, A = 21, M, a, n1, , 0, , , , , M, , L ann n × n, , 0, , K 0, , a22 0, , K 0, , M M, an 2 an 3, , 6. Square matrix A matrix of order m × n in which, m = n (i.e., the number of rows is equal to the number, of columns) is called a square matrix. The elements, aij of a square matrix A = [aij ]n × n , for which i = j, i.e.,, the elements a11 , a22 ,... , ann are called the diagonal, 0 1 2 3, 2 3 1 0, is a square, elements. The matrix A = , 5 0 1 1, 0 0 1 2, , , matrix of order 4. The elements 0, 3, 1, 2 are the, diagonal elements of A., 7. Diagonal matrix A square matrix A = [aij ]n × n is, called a diagonal matrix, if aij, 0 0 K, a11, 0 a 0 K, 22, e.g.,, A= , M M, M, 0, 0 0 L, , , = 0 for all i ≠ j., 0 , 0 , , M , ann , , Also, represented by diagonal ( a11 , a22 ,... , ann )., , 8. Scalar matrix A diagonal matrix whose diagonal, elements are equal, is called a scalar matrix., k 0 0, e.g.,, A = 0 k 0, , , 0 0 k, (Identity matrix is a scalar matrix)
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99, , Matrices, , 9. Unit matrix or identity matrix A square, matrix each of whose principal diagonal element is ‘ 1’, and each of whose non-diagonal element is equal to, zero, is called a unit matrix or an identity matrix and, is denoted by I . I n will denote a unit matrix of order n., 1 0 0, 1 0, e.g.,, I3 = 0 1 0 and I 2 = , , , , 0 1, 0 0 1, , 10. Determinant of a matrix Let A be a square, matrix. The determinant formed by the element of A, is said to be determinant of matrix A. This is denoted, by| A|., 3 4 5, 3 4 5 , e.g.,, A = 6 7 8 , then | A|= 6 7 8, , , 2 −3 5, 2 − 3 5, 11. Equality of two matrices Two matrices, A = [aij ] and B = [bij ] are said to be equal, if, (i) they are of the same size., (ii) the elements in the corresponding places of the, two matrices are the same, i.e.,aij = bij for each, pair of subscripts of i and j. If two matrices A and, B are equal, then we write A = B., , Operation in Matrices, Addition of Two Matrices, Let A and B be two matrices of same order m × n. Then, their sum is defined to be the matrix of order m × n, obtained by adding the corresponding elements of A, and B., a , b , a, b, e.g.,If, A = 11 12 and B = 11 12 , a21 a22, b21 b22, Then,, , a + b a + b , A + B = 11 11 12 12 , a21 + b21 a22 + b22, , Properties of Matrix Addition, 1. Matrix addition is commutative If A and B, be two m × n matrices, then A + B = B + A., , 2. Matrix addition is associative If A, B, C be, three matrices each of the order m × n, then, ( A + B) + C = A + ( B + C )., , 3. Existence of additive identity If O be the, m × n matrix each of whose elements is zero, then, A + O = A = O + A for every m × n matrix A, O is called, additive identity., , Subtraction of Two Matrices, If A and B be two m × n matrices, then we define, A − B = A + ( − B)., a12 , a, e.g., If, A = 11, , a21 a22, and, Then,, , Multiplication of a Matrix by a Scalar, Let A be an m × n matrix and k be any scalar. Then the, matrix obtained by multiplying every element of A by k is, called scalar multiple of A and is denoted by kA., , Properties of Scalar Multiplication, If A and B be two matrices and k and m are scalars,, then, (a) k ( A + B) = kA + kB, (b) ( k + m ) A = kA + mA, (c) ( k ⋅ m ) A = k ( mA) = m ( kA), (d) ( − k) A = − ( kA) = k( − A), (e) ( −1) A = − A, , Multiplication of Two Matrices, Let A = [aij ]m × n and B = [bij ]n × p be two matrices such, that the number of columns in A is equal to the number of, rows in B. Then the m × p matrix C = [cij ]m × p such that, n, , cij = Σ aik bkj = ai 1 b1 j + ai 2 b2 j + ... + ain bnj, k=1, , is called the product of the matrices A and B in that order, and we write C = AB., , Properties of Matrix Multiplication, 1. Matrix multiplication is associative. If A, B, C are, m × n , n × p, p × q matrices respectively. Then,, ( AB) C = A ( BC ), 2. Multiplication of matrices is distributive over, addition of matrices A ( B + C ) = AB + AC, 3. The multiplication of matrices is not always, commutative., 4. Whenever AB and BA both exist and are matrices of, the same order, it is not necessary that AB = BA., 0, 0 1, 1, e.g., If, , B= , A= , , , 1 0, 0 − 1, Then,, , 4. Existence, , of additive inverse Let, A = [aij ]m × n . Then the negative of the matrix A is, defined as the matrix [− aij ]m × n and is denoted by − A., , b , b, B = 11 12 , b, b, 21 22, a11 − b11 a12 − b12 , A− B =, , a21 − b21 a22 − b22, , and, Thus,, , 1 0 0 1 0 1 , AB = , , =, , 0 − 1 1 0 −1 0, 0 0 − 1, 0 1 1, BA = , , =, , , 1 0 0 − 1 1 0 , AB ≠ BA
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100, , NDA/NA Mathematics, , 5. The product of two matrices may be a zero matrix, when none of them is a zero matrix., 0 0, 1 0, 0 1, e.g.,If A = , and B = 0 0 , then AB = 0 0 while, 0, 0, , , , , , , neither A nor B is a null matrix., 6. In the case of matrix multiplication if AB = 0, then it, does not necessarily imply that A = 0 or B = 0 or, BA = 0., 7. If AB = BA ⇒ A and B are square matrices., 8. AI n = A (I n is an identity matrix of size n × n and A is, any matrix of size m × n)., 9. If A is matrix of order n × n, then, AI n = A = I n A, , A11, A, Then, adj ( A) = 12, M, A, 1n, , A21 K An 1 , A22 K An 1 , , M, M, , A2n, An 2, , Properties of Adjoint A, 1. If A be n rowed square, ( adj A) A = A ( adj A) =| A|I n, , matrix,, , then, , 2. adj( AB) = adj( B) ⋅ ( adj A), 3. ( adj A)′ = adj A′, 4. adj( adj A) =| A|n − 2 A, where A is a non-singular, matrix., 2, , 1 0 1, Example 1. If A = 0 0 0, then the value of A 2 is, , , 1 0 1, (a) 3A, (b) 2A, (c) 4A, (d) 5A, , Solution (b), , 1, A2 = 0, , 1, 1, = 2 0, , 1, , 0 1 1 0 1 2 0 2 , 0 0 0 0 0 = 0 0 0 , , , , 0 1 1 0 1 2 0 2 , 0 1, 0 0 = 2A, , 0 1, , Transpose of a Matrix, Let A = [aij ]m × n . Then n × m matrix obtained from A, by changing its rows into columns and columns into rows is, called the transpose of A and is denoted by A′ or by AT , i.e.,, AT = [a ji ]n × m ., e.g.,, , 1 2 3, 1 4 7, T, , , If A = 4 5 6 , then A = 2 5 8 , , , , , 7 8 9 , 3 6 9 , , Properties of Transpose of a Matrix, , 5. | adj( adj A)|=| A|( n − 1) , where A is a non-singular, matrix., 6. Adjoint of a diagonal matrix is a diagonal matrix., , Singular and Non-singular, Matrix, A square matrix A is said to be non-singular matrix, according as | A|≠ 0. Thus, the necessary and sufficient, condition for a matrix to be invertible is that it is, non-singular. A square matrix A is said to be singular, matrix, if it is not non-singular matrix., , Inverse of a Square Matrix, Let A be any n-rowed non-singular matrix. Then a, matrix B, such that AB = BA = I n , is called inverse of A., Existence of Inverse The necessary and, sufficient condition for a square matrix A to possess the, inverse is that| A|≠ 0. Inverse of the matrix A is denoted by, adj( A), A−1 and A−1 =, ., | A|, , Properties of Invertible Matrices, , 1. ( A′ )′ = A, , 2. ( A + B)′ = A′ + B′, , 1. ( AB)−1 = B−1 A−1, , 3. ( kA)′ = kA′, , 4. ( AB)′ = B′ A′ (Reversal law), , 2. ( ABC )−1 = C −1B−1 A−1, , Adjoint of a Square Matrix, Let A = [aij ]n × n be any matrix. The transpose B′ of the, matrix B = [ Aij ]n × n ′ , where Aij denoted by the cofactor of, the element aij in the determinant| A|, is called the adjoint, of the matrix A and is denoted by the symbol adj (A)., Thus, the adjoint of a matrix A is the transpose of the, matrix formed by the cofactors of A., a11 a12 K a1n , a, a K a 2n , , e.g., If, A = 21 22, M , M M, a, ann , n 1 an 2, , 3.|adj A|=| A|n − 1, 1 2 3, Example 2. The inverse of 2 3 4 is, , , 3 4 6, 1, 1 0 0, −2 0, (b) 0 1 0, (a) 0, 3 −2, , , , , 0 0 1, 1 −2 1 , 4 0 1, (d) None of these, (c) 1 2 3, , , 1, 6, 7, ,
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102, , NDA/NA Mathematics, , 2. Skew-symmetric matrix, A square matrix A = [aij ] is called skew-symmetric, matrix, if aij = a ji for all i, j or AT = − A., h g, 0, – h, e.g.,, 0, f, , , – g – f 0, All principal diagonal elements of a skew-symmetric, matrix are always zero because for any diagonal, element., aij = − aij ⇒ aij = 0, , Properties of Symmetric and, Skew-symmetric Matrices, (i) If A is a square matrix, then A + A T , AA T , A T A are, symmetric matrices, while A − AT is skew-symmetric, matrix., (ii) If A, B are two symmetric matrices, then, (a) A ± B, AB + BA are also symmetric matrices,, (b) AB − BA is a skew-symmetric matrix,, (c) AB is a symmetric matrix, when AB = BA., (iii) If A, B are two skew-symmetric matrices, then, (a) A ± B, AB − BA are skew-symmetric matrices,, (b) AB + BA is a symmetric matrix., (iv) Every square matrix A can uniquely be expressed as, sum of a symmetric and skew-symmetric matrix, , 1, 1, i.e.,, A = ( A + AT ) + ( A − AT )., , 2, 2, , Test of Consistency by, Using Matrices, System AX = B be a system of n linear equations in n, unknowns., 1. If | A|≠ 0, then the system is consistent and has a, unique solution and solution is X = ( A−1B)., 2. If | A|= 0 and ( adj A) B = 0, then the system is, consistent and has infinitely many solutions., 3. If | A|= 0 and ( adj A) B ≠ 0, then the system is, inconsistent., , Example 5. If x + y + z = 6, x − y + z = 2, 2 x + y − z = 1, then, the value of x, y and z is, respectively, (a) 1, 2, 3, (b) 1, 2, −1 (c) 2, 1, 3, 1, , 1 x, , 6 , , , 2, , 1, , , −1 z , , , −1, , Solution (a) 1 −1 1 y = 2 ⇒ A X = B, 1, Here,, A = 1, , 2, 1, ∴, | A| = 1, , 2, 0, , and adj (A) = 3, , 3, , Example 4. If A and B be symmetric matrices of the same, order, then AB − BA will be, (a) symmetric matrix, (c) null matrix, , 1, , (d) 2, 2, 3, , ∴, , (b) skew-symmetric matrix, (d) None of these, Now,, , Solution (b) Since, A, B are symmetric matrices., ∴, ∴, , A = A′ and B = B′, ( AB − BA)′ = ( AB)′ − (BA)′ = B′ A′ − A′ B′, = − ( A′ B′ − B′ A′ ) = − ( AB − BA), Hence, ( AB − BA) is a skew-symmetric matrix., , ⇒, , 1, −1 1 , , 1 −1, 1, 1 1 1 1, , , −1 1 = 0 −2 0 = 6 ≠ 0, , , 1 −1 0 −1 −3, 2, 2, −3 0 , , 1 −2, 2, 0 2, adj( A) 1 , A− 1 =, = 3 −3 0 , , | A|, 6, 3 1 −2, 2 6 x 1, 0 2, 1, X = A− 1B; X = 3 − 3 0 2 ⇒ y = 2, , 6, 3 1 − 2 1 z 3, x = 1, y = 2, z = 3, 1, , Comprehensive Approach, n, n, n, n, n, , n, n, , ( AB) −1 = B−1 A −1, A + A ′ is always a symmetric matrix., A − A ′ is always a skew-symmetric matrix., AA′ is always a symmetric matrix., The matrix B′ AB is symmetric or skew-symmetric matrix, according as A is symmetric or skew-symmetric., All positive integral powers of a symmetric matrix are symmetric., Positive odd integral powers of skew-symmetric matrix are, skew-symmetric and positive even integral powers of, skew-symmetric matrix are symmetric., , n, n, n, , n, n, n, n, , The inverse of a symmetric matrix is symmetric., The inverse of a diagonal matrix is diagonal., If A and B are symmetric matrices of order n, then ABA is a, symmetric matrix., Every orthogonal matrix is invertible., Every invertible matrix is not necessary orthogonal., If A is an orthogonal matrix, then A −1 is also orthogonal., Determinant of a skew-symmetric matrix of odd order is zero and, of even order is a non-zero perfect square.
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Exercise, Level I, 2 6 1, 1. The matrix , is of order, 1 9 3, (a) 3 × 2, (b) 2 × 3, (c) 6 × 9, (d) 2 × 1, 3 5 , 1 2, 2. If A = , , B = x −4 and A + B =, 5, 6, , , , , value of x must be, (a) 6, (b) 1, (c) 0, , 4 7, 6 2, then the, , , (d) –1, , 2 4, 3. The matrix , does not have an inverse, if x is, −8 x, equal to, (a) 16, (b) –16, (c) 8, (d) – 6, 1 2, 2, 4. If A = , and A − kA − I 2 = 0, then the value of k, 2 3, is equal to, (a) 4, (b) – 4, (c) 8, (d) – 8, i 0, n, 5. If A = , ; i = −1 ,then A is equal to, 0 i , (a) A for n = 4, (b) − A for n = 6, (c) −I for n = 5, (d) I for n = 8, 0 2, 6. If A = , 2 1, the matrix, −14 6, (a) , 4 5, −14 −6, (c) , 4 −5, , 3, and B =, 4, −2, 2 , 3, 2, , 5 10, 7. The matrix −2 −4, , −1 −2, (a) –3, (b) 3, (c) 0, (d) for any value of, , 7 6 3, 1 4 5, then 3 A − 2B is, , , 14, (b) , −4, 14, (d) , 4, , −6 −3, −5 −2, 6 −3, 5 −2, , 3, 6 is a singular matrix equal to, , b, , b, , α β , 8. If the matrix A = , , β α, 2, is such that A = I , then which one of the following is, correct?, (NDA 2011 I), (a) α = 0, β = 1 or α = 1, β = 0, (b) α = 0, β ≠ 1 or α ≠ 1, β = 1, (c) α = 1, β ≠ 0 or α ≠ 1, β = 1, (d) α ≠ 0, β ≠ 0, , −1 2 , 3, 9. If A = , and B = , AX = B,then X is equal to, , 2 −1, 1, 5, 1 5, 1, (c), (a) [0 7], (b), [5 7] (d) , 3 7, 3, 7, 1 2 , −1, 10. If A = , , then A is equal to, 3, 5, −, , , 2 , 5, 11 11 , −5 −2, (b) , (a) , , 3, 1, −3 1 , − , , 11, 11, 2, 5, −, −, , 5 2 , 11, (d) , (c) 11, , , 3, 1, 3 −1, − , −, 11, 11, 3 −3 4, 11. The adjoint matrix of 2 −3 4 is, , , 0 −1 1, 4 8 3, 1 −1 0 , , , (a) 2 1 6, (b) −2 3 −4, , , , , 0 2 1, −2 3 −3, 11 9 3, 1 −2 1 , , , (c) 1 2 8, (d) −1 3 3 , , , , , 6 9 1, −2 3 −3, 12. The values of x , y , z in order, if the system of, equations, 3x + y + 2z = 3,, 2x − 3 y − z = − 3,, x + 2 y + z = 4, are, (a) 2, 1, 5 (b) 1, 1, 1 (c) 1, –2, –1 (d) 1, 2, –1, 1 2 −1, 13. If A = 3 0 2 , B =, , , 4 5 0 , 5 1 −3, (a) 3 2 6 , , , 14 5 0 , 1 2 4, (c) 5 6 7, , , 2 3 5, , 1 0 0, 2 1 0 , then AB is, , , 0 1 3, 1 0 0, (b) 0 1 0, , , 0 0 1, (d) None of these, , 1 0, α 0, and B = , 14. If A = , , , 2 1, 1 1, 2, such that A = B, then what is the value of α?, (a) −1, (b) 1, (NDA 2011 I), (c) 2, (d) 4
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104, , NDA/NA Mathematics, , 1, 15. If 2, , 4, , −3, −8, 2, , 2, 5 is not an invertible matrix, then what, , λ , , is the value of λ?, (a) −1, (b) 0, , (NDA 2010 II), , (c) 1, , (d) 2, , 16. If a matrix A is such that 3 A3 + 2 A2 + 5 A + I = 0,, then what is A−1 equal to?, (NDA 2010 II), (a) − ( 3 A2 + 2 A + 5), (b) 3 A2 + 2 A + 5I, (c) 3 A2 − 2 A − 5I, (d) − ( 3 A2 + 2 A + 5I ), 1 2 3 8, 17. What must be the matrix X, if 2X + , ?, =, 3 4 7 2, 1 −3, 1 3 , (b) , (a) , , , 2, −, 1, 2 −1, , , 2 −6, 2 6 , (d) , (c) , , , 4 −2, 4 −2, 1 1 1 x 0, x, , , , , , , 18. If 1 −2 −2 y = 3 , then y is equal to, , , , 1 3 1 z 4, z , 0, 1, 5, 1, (a) 1, (b) 2 , (c) −2, (d) −2, , , , , 1, −3, 1 , 3 , 1 1 0, 19. For the matrix A = 1 2 1 , which of the following, , , 2 1 0, is correct?, (a) A3 + 3 A2 − I = 0, (b) A3 − 3 A2 − I = 0, 3, 2, (c) A + 2 A − I = 0, (d) A3 − A2 + I = 0, a 2, 20. The value of a for which the matrix A = , is, 2 4, singular, if, (a) a ≠ 1, (b) a = 1, (c) a = 0, (d) a = − 1, 21. A and B are two matrices such that AB = A and, BA = B, then what is the value of B2?, (NDA 2012 I), (d) −1, , (a) B, (b) A, (c) 1, where, I is the identity matrix., 0 − i, − i, i, 22. If P = 0 − i i and Q = 0, , , , i 0 , i, − i, equal to, 2, −2 2 , , , (b) −1, (a) 1 −1, , , , −1, 1 −1, 1, 2 −2, (d) 0, (c) , , , −1 1 , 0, , i, 0 , then PQ is, , − i, −2, 1, , 1 , 0 0, 1 0, , 0 1, , a, 23. What is the order of the product [x y z ]h, , g, (a) 3 × 1, (c) 1 × 3, , (b) 1 × 1, (d) 3 × 3, , h, b, f, , g x, f y, , c z , , (NDA 2012 I ), , 0, i, 0 − i, 0 1, 24. If A = , , B= , , C=, , , then, , 0 − i, − i 0 , −1 0, which one of the following is not correct? (NDA 2010 II), (a) A2 = B2 (b) B2 = C 2 (c) AB = C (d) AB = BA, 25. If A is a square matrix, then what is, adj ( AT ) − ( adj A)T equal to?, (a) 2| A|, (c) Null matrix, , (NDA 2010 II), , (b) 2| A| I, (d) Unit matrix, , 5, 26. Let A = , 2, , 6 1, . Let there exist a matrix B such, −1 5, 35 49, that AB = , .What is B equal to? (NDA 2010 I), 29 13, 2 6 3, 5 1 4, (b) , (a) , , , 5 1 4, 2 6 3, 2 5, 5 2, (d) 6 1, (c) 1 6, , , , , 3 4, 4 3, , cos θ sin θ, 27. The adjoint of , is equal to, sin θ cos θ, cos θ − sin θ, cos θ sin θ, (a) , (b) , , , −, sin, θ, cos, θ, , sin θ cos θ, , cos θ − sin θ, cos θ sin θ, (d) , (c) , , , −, sin, θ, cos, θ, sin θ cos θ , , , 0 1, 4, 28. If a matrix A = , , then A is equal to, 1, 0, , , 1 0, (a) , , 0 1, , 1 1, (b) , , 0 0, , 0 1, (d) , , 1 0, , 0 0, (c) , , 1 1, , cos α − sin α, 29. The inverse of the matrix , , is, sin α cos α , cos α sin α, cos α sin α, (b) , (a) , , , − sin α cos α, sin α cos α, − cos α, (c) , sin α, , sin α, cos α, , cos α, (d) , sin α, , sin α , − cos α, , 3 2, 30. If A = , , then A⋅ adj (A) is equal to, 1 4, 0 10, 10 0 , (b) , (a) , , , 10 0 , 0 10, 0 0, 10 10, (d) , (c) , , , 10 10, 0 0
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105, , Matrices, 1 −2, 5 4, 31. If , X = 1 3 , then X is equal to, 1, 1, , , , , 3 14 , 3 14, (b) , (a) , , , −4 −17, −4 17, −3 14, −3 −14, (d) , (c) , , , 4 17, 4 17 , 32. The matrix X satisfies the following equation, 1 1 , 1 3, 0 1 X = 0 −1., , , , , Which one of the following represents X ?, 1 −4, 1 4, (b) , (a) , , , −, 1, 0, , 1 0 , , 1 4 , (c) , , 0 −1, , 1 −2, (d) , , 0 −1, , 0 1 2, 33. The matrix M = 1 2 3 and its inverse N = [n ij ]., , , 3 1 1, What is the element n 23 of the matrix N ?, (a) 2, (b) –2, (c) 1, (d) –1, 34. If X and Y are the matrices of order 2 × 2 each and, 0, 9 13, −7, and 3X + 2Y = , 2X − 3Y = , , , then, 4 13, 7 −13, what is Y equal to?, (NDA 2009 II), 1, (a) , −2, 3, (c) , −1, , 3, 1, 2, 5, , 1, (b) , 2, 3, (d) , 1, , 3, 1, 2, −5, , 1 2, 35. Let A = , = [aij ], where i, j = 1, 2. If its inverse, 3 4, matrix is [bij ], what is b22?, (NDA 2010 I), 3, 1, (d) −, (a) − 2, (b) 1, (c), 2, 2, ω 0, 36. If A = , , where ω is cube root of unity, then, 0 ω, what is A100 equal to?, (NDA 2009 II), (a) A, (b) − A, (c) Null matrix, (d) Identity matrix, 1 −2 −3, 37. If A = 2, 1 −2 , then which one of the following, , , 2, 1, 3, is correct?, (NDA 2009 II), (a) A is symmetric matrix., (b) A is anti-symmetric matrix., (c) A is singular matrix., (d) A is non-singular matrix., , cos θ sin θ, 38. What is the type of the matrix , ?, − sin θ cos θ, (a) Skew-symmetric, (b) Symmetric, (c) Orthogonal, (d) Singular, 39. If a matrix A has inverses B and C, then which one of, the following is correct?, (NDA 2012 I), (a) B may not be equal to C, (b) B should be equal to C, (c) B and C should be unit matrices, (d) None of the above, 19 24, 3 2, and AC = , 40. If matrices A = , , then the, , 37 46, 4 5, matrix C is equal to, , 19, 3 12 , 5 4, (b) , (a) , , 37 46 , 2 6, , , 4, 5, 3 5, 3 4, (d) , (c) , , , 4 6, 5 6, 1 2 3, 2 0 0, , , 41. If A = 0 2 0 , B = 0 1 2 , then what is the value, , , , , 0 0 1, 0 0 2, of det (AB) ?, (a) 4, (b) 8, (c) 16, (d) 0, 3 6 , 5 2, 42. If A + B = , and A − B = 0 −1, then what is, , , 0 9, the value of AB?, 4 12, 4 −4, 0 4, 4 4, (d) , (c) , (b) , (a) , , , , , 0 20, 0 20, 4 4, 0 4, 1 −2, and I is a 2 × 2 identity matrix, then, 43. If X = , 3, 0, X 2 − 2X + 3I equals to which one of the following?, (NDA 2008 I), , (a) −I, , (b) −2X, , (c) 2X, , (d) 4X, , 3 1, −1, 2, 44. If A = , and A + 7I 2 = 5 A, then what is A ?, −1 2, 1 2 −1, 1 2 1, (a) , (b) , , 1, 3, 7, 7 −1 3, , (c), , 1 2 1, 7 −1 3, , (d), , 1 −2 1 , 7 −1 −3, , 45. If A = [aij ]m × n and B = [bij ]n × p are the two matrices,, then their product AB will be of order, (a) m × n, (b) m × p (c) n × p, (d) n × n, 2, 46. If A is a square matrix such that A = 1, where, I is, the identity matrix, then what is the value of A−1?, (NDA 2012 I), , (a) A + 1, (c) A, , (b) Null matrix, (d) Transpose of A
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106, , NDA/NA Mathematics, , 1 0, 2x 0, −1, 47. If A = , and A = −1 2 , then what is the, x, x, , , , , value of x?, (NDA 2007 I), 1, 1, (c) 1, (d) 2, (b), (a) −, 2, 2, 3 2, 48. If A = , , then what is A ( adj A) equal to?, 1 4, (NDA 2009 I), 0, 10, 0 10, (b) , (a) , , 0, 0 10, 10, 1, 10, 1 10, (d) , (c) , , , 1, 1 10, 10, 2, −1, A= , 3, 6, , 1, 49. If A is any 2 × 2 matrix such that , 0, then what is A equal to?, , 0, ,, 3, , (NDA 2007 I), , −5, (a) , −2, , 1, 2, , −5, (b) , 1, , −2, 2, , −5, (c) , 2, , −2, 1, , 5, (d) , −2, , 2, −1, , 50. If a matrix A is symmetric as well as anti-symmetric,, then which one of the following is correct?, , (a), (b), (c), (d), , A is a diagonal matrix., A is a null matrix., A is a unit matrix., A is a triangular matrix., , (NDA 2009 II), , 1 2 3 2 , 51. If 2X − , , then X is equal to, =, 7 4 0 −2, 1 2, 2 2, (b) , (a) , , , 7/ 2 2, 7 4, 2 2, (c) , , 7/ 2 1, , (d) None of these, , 52. If A is a square matrix, then A + AT is, (a) non-singular matrix, (b) symmetric matrix, (c) skew-symmetric matrix, (d) unit matrix, 3 4, 53. If A = , , then A ⋅ ( adj A) is equal to, 5 7, (a) A, (b) | A|, (c) | A|I, (d) None of these, 54. For two invertible matrices A and B of suitable, orders, the value of ( AB)−1 is, (a) ( BA)−1 (b) B−1 A−1 (c) A−1B−1 (d) ( AB ′ )−1, , Level II, x + y 2x + z 4 7 , 1. If , , then the values of, =, x − y 2z + w 0 10, x , y , z , w are, (a) 2, 2, 3, 4, (b) 2, 3, 1, 2, (c) 3, 3, 0, 1, (d) None of these, 0 2 , 2. If A = , and kA =, 3 −4, k, a , b are respectively, (a) −6, − 12, − 18, (c) −6, − 4, − 9, , 0 3a, 2b 24 , then the values of, , , (b) −6, 4, 9, (d) −6, 12, 18, , 3. Assuming that the sum and product given below are, defined, which of the following is not true for, matrices?, (a) A + B = B + A, (b) AB = AC does not imply B = C, (c) AB = O implies A = O or B = O, (d) ( AB) ′ = B ′ A ′, 1 1, 100, 4. If A = , , then A is equal to, 1, 1, , , (a) 2100 A, (c) 100 A, , (b) 299 A, (d) 299 A, , cos 2 θ − sin 2 θ, 5. Inverse of the matrix , is, sin 2 θ cos 2 θ , , cos 2 θ − sin 2 θ, (a) , , sin 2 θ cos 2 θ , cos 2 θ sin 2 θ, (c) , , sin 2 θ cos 2 θ, , cos 2 θ sin 2 θ , (b) , , sin 2 θ − cos 2 θ, cos 2 θ sin 2 θ, (d) , , − sin 2 θ cos 2 θ, , 1, 1, 2 − x, , 6. If the matrix A =, 1, 3− x, 0 is singular,, , , −3, − x, −1, then what is the solution set S?, (NDA 2011 I), (a) S = { 0, 2, 3}, (b) S = { −1, 2, 3}, (c) S = { 1, 2, 3}, (d) S = { 2, 3}, 7. Consider the following statements in respect of, symmetric matrices A and B, I. A + B is symmetric., II. AB is symmetric., III. AB + BA is symmetric., IV. AB − BA is symmetric., Which of the statements given above are correct?, (a) I and II, (b) I and III, (c) II and III, (d) II and IV, 8. If A be a real skew-symmetric matrix of order n such, that A2 + I = 0, I being the identity matrix of the, same order as that of A, then what is the order of A?, (a) 3, (b) Odd, (c) Prime number, (d) None of these
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107, , Matrices, 9. A matrix X has ( a + b) rows and ( a + 2) columns and a, matrix Y has ( b + 1) rows and ( a + 3) columns. If both, XY and YX exist, then what are the values of a, b, respectively?, (NDA 2009 II), (a) 3, 2, (b) 2, 3, (c) 2, 4, (d) 4, 3, a 0, 1 2, and B = , 10. Let A = , , where a and b are, , 0 b, 3 4, positive integers, then which one of the following is, correct?, (NDA 2009 I), (a) There exists more than one but finite number of, BS’s such that AB = BA, (b) There exists exactly one B such that AB = BA, (c) There exist infinitely many B’s such that, AB = BA, (d) There cannot exist any B such that AB = BA, cos θ − sin θ, 11. If A = , , then which of the following, sin θ cos θ , statements is not correct?, (a) A is orthogonal matrix, (b) A ′ is orthogonal matrix, (c) determinant ( A) = 1, (d) A is not invertible, 0 −1, 1 2, and B = , 12. A = , , then what is the value of, , 1 2 , 1 1, (NDA 2012 I), B−1 A−1?, −1 3 , 1 −3, (b) , (a) , , , 1 −2, −1 2 , −1 −3, −1 3 , (d) , (c) , , , 1 −2, −1 −2, 2 1, 13. The multiplicative inverse of matrix , is, 7 4, −4 −1, 4 −1, (b) , (a) , , , 7 −2, −7 −2, 4 −1, 4 −7, (d) , (c) , , , −7 2 , 7 2 , 14. A( α ), A(β ), , cos α, =, sin α, cos β, =, sin β, , − sin α, cos α , − sin β , ., cos β, , Which one of the following is correct?, (a) A( − α ) A( −β ) = A( α + β ), (b) A( − α ) A(β ) = A(β − α ), (c) A( α ) + A( −β ) = A{ − (β − α )}, (d) A( α ) + A(β ) = A( α + β ), 15. A matrix obtained from a unit matrix by a single, elementary operations is known as, (a) unit matrix, (b) identity matrix, (c) elementary matrix, (d) None of these, , 3 − 2 i 3 + 5 i, , then adj (A) is equal to, 16. If A = , 3 − 2 i, 2, −2 , 2 , 3 − 2i, 3 − 2 i, (b) , (a) , , , −3 − 5 i 3 − 2 i, 3 + 5 i 3 − 2 i, 3 − 2 i −3 − 5 i, (c) , 3 − 2 i , −2, , 3 − 2 i 3 + 5 i , (d) , −3 + 2 i, −2, , 17. If A and B are two matrices such that AB = A and, BA = B, then which one of the following is correct?, (NDA 2008 II), , (a) ( A ) = A, (c) ( AT )2 = ( A−1 )−1, T 2, , T, , a, 18. If adj ( A) = , −1, −1, of| A |?, (a) 1, , (b) ( A ) = B, (d) None of these, T 2, , T, , 0, and ab ≠ 0,then what is the value, b, , (b) ab, , (NDA 2008 II), , (c), , 1, ab, , (d), , 1, ab, , 19. If l + m + n = 0, then the system of equations, −2x + y + z = l; x − 2 y + z = m; x + y − 2z = n has, (a) a trivial solution, (NDA 2008 II), (b) no solution, (c) a unique solution, (d) infinitely many solutions, 20. Which one of the following is correct in respect of the, matrix, 0 −1, 0, (NDA 2009 I), A = 0 −1, 0 ?, , , 0, 0, −1, (a) A−1 does not exist, (c) A is a unit matrix, , (b) A = ( −1)I, (d) A2 = I, , 21. If A is a square matrix of order n × n and k is a scalar,, then adj ( kA) is equal to which one of the following?, (a) k adj(A), (b) k2adj (A), (c) kn − 1 adj(A), , (d) kn adj(A), , 22. If A is a non-singular matrix of order n × n, then, which one of the following is equal to | adj (A) |?, (b) | A|n, (a) | A|n + 1, (d) | A|, (c) | A|n − 1, x, 23. If M = y and G = [3 4 5] are two matrices, then, , z , which one of the following is correct?, (a) In the product matrix MG = [hij ], the element h32, is 5y, (b) In the product matrix MG = [hij ], the element h32, is 3x + 4 y + 5z, (c) In the product matrix MG = [hij ], the element h32, is 4z, (d) In the product matrix MG = [hij ], the element h32, does not exist
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108, , NDA/NA Mathematics, , 0 1, 24. A = , , then which one of the following is equal to, 0 0, ( aI + bA)n ?, (b) a n + na n − 1 bA, (d) a n I + bn An, , (a) a n I + na n − 1 bA, (c) a n I + a n − 1 bA, , 25. A is a non-singular matrix such that A−1 = AT , then, which one of the following represents A? Here, AT, denotes the transpose of A, 1 2 2, 1 2 2, 1, 1, , (b), (a), 2 1 −1, 2 1 −2, , , 3, 3, −2 2 −1, −2 2 −1, 1 2 2, 1 2 2 , 1, 1, , (d), (c), 2 1 1, 2 1 −2, , , 2, 2, −2 2 −1, 2 2 −1, 1 2, 26. The matrix A = , satisfies which one of the, 2 2, following polynomial equations?, (NDA 2007 II), 2, 2, (a) A + 3 A + 2I = O, (b) A + 3 A − 2I = O, (c) A2 − 3 A − 2I = O, (d) A2 − 3 A + 2I = O, 1, 27. If , 0, 1, (a) , 0, , 3, A=, 1, , 1, 0, , , −1, , then what is the matrix A?, 1, (NDA 2008 II), , −3, 2, (b) , 1, 0, , 2, 2, , −4, (c) , 1, , −1, 1, (d) , 0, 0, , −4, 1, , 28. Under what condition does A( BC ) = ( AB) C hold,, where A, B, C are three matrices?, (NDA 2008 I), (a) AB and BC both must exist, (b) Only AB must exist, (c) Only BC must exist, (d) Always true, −1, , 5 0 x −1, 29. If , = , then which one of the, 0 7 − y 2, following is correct?, (NDA 2008 I), (a) x = 5, y = 14, (b) x = − 5, y = 14, (c) x = − 5, y = − 14, (d) x = 5, y = − 14, 30. Which one of the following matrix equations is not, correct?, (a) ( AB)′ = A′ B′, (b) ( ABC )′ = C ′ B′ A′, (c) ( A + B)′ = A′ + B′, (d) ( λA)′ = λA′, 31. If A and B are symmetric matrices of the same order,, then AB is, (a) always symmetric, (b) never symmetric, (c) symmetric only when AB = BA, (d) symmetric only when A = B, 0 7 4, 32. The matrix −7 0 −5 is, , , −4 5 0 , , (a) symmetric, (c) non-singular, 0, 33. What is the inverse of 0, , 1, 1 0 0, (a) 0 1 0, , , 0 0 1, 1, 0, −1, , (c), 0, 0 −1, , , 0 −1, 0, , (b), (d), 0, 1, , skew-symmetric, orthogonal, 1, (NDA 2009 I), 0 ?, , 0 0, 0 0 1, (b) 0 1 0, , , 1 0 0, , 0 0 −1, (d) 0 −1, 0, , , 0, −1 0, 0, 1, 2 −1, 2, and B = , 34. If A = , , then ( A + B) is not, , 1, 1, 0, 1, −, −, , , , , equal to, (a) A2 + AB + BA + B2, (b) A2 + 2 AB + B2, 2, 2, (c) A + AB + BA + B I, (d) A2I + AB + BA + B2, 0 1, 2, 35. If A = , = (α I + βA) , then the values of α and β, −1 0, are given by, i, i, 1, 1, (a) α =, (b) α =, ,β =, ,β =, 3, 2, 2, 2, 1, 1, (c) α = β = ±, (d) α = − β = ±, 2, 2, , 36. If the least number of zeroes in a lower triangular, matrix is 10, then what is the order of the matrix?, (NDA 2007 II), , (a) 3 × 3, (c) 5 × 5, 1, 37. If the inverse of 0, , 0, , (b) 4 × 4, (d) 10 × 10, p q, 1 − p, 1, x 0 is 0, , , 0, 0 1, 0, , what is the value of x?, (a) 1, , (b) Zero, , − q, 0 , then, , 1, (NDA 2007 II), , (c) −1, , (d), , 1 1, +, p q, , 3 2, 4 11, and A = , 38. If AB = , , then what is the, , 5, 1 2, 4, value of the determinant of the matrix B?, (NDA 2007 II), , 1, (a) 4, (b) − 6, (c) −, (d) − 28, 4, −2 3, 2 −3, and B = , 39. A = , are two 2 × 2 matrices., , −1 2, 1 −2, For the integral values of k, which one of the, following is correct?, (a) A2k + B2k = O, (b) A2k + 1 − B2k + 1 = O, 2k, 2k, (c) A − B = O, (d) A3 k − B3 k = O, 40. How many matrices of different order can be formed, out of 36 elements (using all the elements at a time)?, (a) 4, (b) 5, (c) 8, (d) 9
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109, , Matrices, 3 2 0, 41. If B = 2 4 0 which one of the following is correct, , , 1 1 0, in respect of the adjoint of B?, 0 0 −2, (a) The adjoint of B has a unique value 0 0 −1, , , 0 0 8 , 0 0 0, (b) The adjoint of B has a unique value 0 0 0, , , −2 −1 8, (c) The adjoint of B can have many possible values,, since|B|= 0, (d) The adjoint of B does not exist, since| B|= 0, 42. If 1, ω , ω 2 are the cube roots of unity, for what value of, 1 ω m, m is the matrix ω m 1 singular?, , , m 1 ω , (a) 0, , (b) 1, , (c) ω, , (d) ω 2, , 43. The sum and product of matrices A and B exist., Which of the following imlications are necessarily, true?, (NDA 2012 I), I. A and B are square matrices of same order., II. A and B are non-singular matrices., Select the correct answer using the code given below, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, x x2 1 + x2, , , 44. If A = y y 2 1 + y 2, where x, y, z are distinct., z z 2 1 + z 2, , , What is| A|?, (NDA 2007 I), 2, 2, (a) 0, (b) x y − y x + xyz, (c) ( x − y )( y − z )( z − x ) (d) xyz, 45. Under which of the following condition(s), will the, 0 0 q , matrix A = 2 5 1 be singular?, , , 8 p p, I. q = 0, II. p = 0, III. p = 20, Select the correct answer using the code given below, (NDA 2007 I), , (a) I and II, (c) I and III, , (b) Only III, (d) I or III, , 46. Consider the following statements, I. Every zero matrix is a squar matrix., II. A matrix has a numerical value., III. A unit matrix is a diagonal matrix., Which of the above statements is/are correct?, (NDA 2007 I), , (a) Only II, (c) Both II and III, , (b) Only III, (d) Both I and III, , 47. Let A and B be two matrices of order n × n. Let A be, non-singular and B be singular. Consider the, following, I. AB is singular., II. AB is non-singular., IV. A−1B is non-singular., III. A−1B is singular., Which of the above is/are correct?, (NDA 2007 I), (a) I and III, (c) I only, , (b) II and IV, (d) III only, , 48. Consider the following statements, I. The inverse of a square matrix, if it exists, is, unique., II. If A and B are singular matrices of order n, then, AB is also a singular matrix of order n., Which of the statements given above is/are correct?, (a) I only, , (b) II only, , (c) Both I and II, , (d) Neither I nor II, , (NDA 2011 I), , 49. Consider the following statements in respect of a, square matrix A and its transpose AT, I. A + AT is always symmetric., II. A − AT is always anti-symmetric., Which of the statements given above is/are correct?, (a) I only, , (b) II only, , (c) Both I and II, , (d) Neither I nor II, , (NDA 2010 II), , 50. Consider the following statements, I. If A′ = A, then A is a singular matrix, where A′ is, the transpose of A., II. If A is a square matrix such that A3 = I , then A is, non-singular., Which of the statements given above is/are correct?, (a) I only, (b) II only, (NDA 2010 I), (c) Both I and II, (d) Neither I nor II, 51. Let A = [aij ]m × m be a matrix and C = [cij ]m × m be, another matrix where cij is the cofactor of aij . Then,, what is the value of| AC|?, (NDA 2007 II), (a) | A|m − 1 (b) | A|m, , (c) | A|m + 1 (d) Zero, , 52. Consider the following statements in respect of, symmetric matrices A and B, (NDA 2009 I), I. AB is symmetric., II. A2 + B2 is symmetric., Which of the above statement(s) is/are correct?, (a) I only, (b) II only, (c) Both I and II, (d) Neither I nor II, 3 4 0, 53. Consider a matrix M = 2 1 0 and the following, , , 3 1 k, statements, Statement (A) Inverse of M exists., Statement (B) k ≠ 0, Which one of the following in respect of the above, matrix and statement is correct?, (NDA 2009 I), (a) A implies B, but B does not imply A, (b) B implies A, but A does not imply B
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110, , NDA/NA Mathematics, Reason (R) The product of two matrices can never, be equal to an identity matrix., (NDA 2007 I), , (c) Neither A implies B nor B implies A, (d) A implies B as well as B implies A, 54. Consider the following statements in respect of the, square matrices A and B of same order, I. A and B are non-zero and AB = 0, ⇒ Either| A| = 0 or| B| = 0, II., AB = 0 ⇒ A = 0 or B = 0, Which of the above statements is/are correct?, (a) I only, (b) II only ( NDA 2011 II), (c) Both I and II, (d) Neither I nor II, 2 3 x 5 , 55. In respect of the equation , , =, 4 6 y c − 5, correctly match List I with List II and select the, correct answer using the code given below the lists, List I, List II, (Value of c), (Nature of the equation), A. 5, 1. The equation has no solution, B. 10, 2. The equation has a unique solution, C. 15, , 3. The equation has an infinite set of, solutions, , 4. The equation has two infinite sets of, independent solutions, , Codes, A B, (a) 4 2, (c) 2 2, , C, 3, 4, , A, (b) 1, (d) 4, , B, 1, 1, , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, , (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 5 10, is invertible., 56. Assertion (A) M = , 8, 4, Reason (R) M is singular., (NDA 2009 I), 3, , B=, 4, , 1, 0, , , 0, , then, 1, (NDA 2007 I), , Reason (R) In the above, AB = BA., cos α, 58. Assertion (A) If A = , cos α, cos α, B= , sin α, , Directions (Q. Nos. 60-62), 1 2 3, A = 3 4 5 ,, , , 6 7 9, , Consider the matrix, 1 1 −1, B = 2 − 3 4 , , , 3 − 2 3 , , 60. The inverse of the matrix A is, − 1/ 2 − 3/ 2 1 , 1 0 0, (b) 0 1 0, (a) − 3/ 2 9/ 2 − 2, , , , , 3/ 2 − 5/ 2 1 , 0 0 1, 4 5 7 , (c) 8 9 10, (d) None of these, , , 0, 2, 3, , , 61. Find the value of| A − B|., (a) − 50, (c) − 49, , C, 3, 3, , Directions (Q. Nos. 56-59), , 2, 57. Assertion (A) If A = , 1, ( A + B)2 = A2 + B2 + 2 AB., , 1 2, 59. Assertion (A) A = , is neither symmetric nor, 5 9, anti-symmetric., Reason (R) The matrix A cannot be expressed as a, sum of symmetric and anti-symmetric matrices., , sin α, and, sin α, cos α, , then AB ≠ I ., sin α, , 62. Find the value of AB., 14 − 11 16, (a) 26 − 19 28, , , 47 − 33 49, 14 − 11 16, (c) − 26 19 28, , , 47 − 33 49, , (b) − 51, (d) None of these, − 14 − 11 − 16, (b) 26 − 19 28 , , , 47 − 33 49 , (d) None of these, , Directions (Q. Nos. 63-65), , Consider the, simultaneous equations, x + y + z = 6; x + 2 y + 3z = 10;, x + 2 y + λz = µ, Then, for what values of λ , µ the simultaneous, equations posses, , 63. A unique solution, if, (a) λ ≠ 3, (c) λ = 2, , (b) λ = 3, (d) λ ≠ 4, , 64. An infinite number of solutions, if, (a) µ= 10, (b) µ = 5, (c) µ = 15, (d) µ = 20, 65. No solution, if, (a) µ ≠ 10, (b) µ = 10, (c) µ = 20, (d) µ = 15
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Answers, Level I, 1., 11., 21., 31., 41., 51., , (b), (b), (a), (c), (b), (c), , 2., 12., 22., 32., 42., 52., , (b), (d), (b), (c), (d), (b), , 3., 13., 23., 33., 43., 53., , (b), (a), (b), (d), (c), (c), , 4., 14., 24., 34., 44., 54., , (a), (b), (c), (c), (a), (b), , 5., 15., 25., 35., 45., , (d), (c), (c), (d), (b), , 6., 16., 26., 36., 46., , (c), (d), (c), (a), (c), , 7., 17., 27., 37., 47., , (d), (a), (a), (d), (b), , 8., 18., 28., 38., 48., , (a), (b), (a), (c), (b), , 9., 19., 29., 39., 49., , (b), (b), (b), (b), (c), , 10., 20., 30., 40., 50., , (b), (b), (a), (c), (b), , 2., 12., 22., 32., 42., 52., 62., , (c), (b), (c), (b), (d), (b), (a), , 3., 13., 23., 33., 43., 53., 63., , (c), (d), (c), (b), (a), (d), (a), , 4., 14., 24., 34., 44., 54., 64., , (b), (a), (a), (b), (a), (d), (a), , 5., 15., 25., 35., 45., 55., 65., , (d), (a), (b), (c), (c), (b), (a), , 6., 16., 26., 36., 46., 56., , (a), (c), (c), (c), (b), (d), , 7., 17., 27., 37., 47., 57., , (b), (d), (d), (a), (b), (a), , 8., 18., 28., 38., 48., 58., , (d), (a), (a), (b), (a), (c), , 9., 19., 29., 39., 49., 59., , (b), (d), (c), (c), (c), (c), , 10., 20., 30., 40., 50., 60., , (c), (d), (a), (d), (b), (a), , Level II, 1., 11., 21., 31., 41., 51., 61., , (a), (d), (c), (c), (b), (b), (b), , Hints & Solutions, Level I, 2 6 1, 1. Let A = , , 1 9 3, ∴ The given matrix has 2 rows and 3 columns. Then, the given matrix is of order (2 × 3)., 1 2, 3 5 , 2. We have, A = , , B = x −4 , 5, 6, , , , , Now,, But, , 7, 4, A+ B=, , +, x, 5, 2, , , 4 7, A+ B=, , 6 2 , , …(i), (given)…(ii), , On comparing Eqs. (i) and (ii), we get 5 + x = 6, , ⇒, , x=1, , ⇒, , 2x + 32 = 0 ⇒ 2x = − 32 ⇒ x = − 16, 1 2, 4. We have, A = , , 2 3, 1 2 1 2 5 8 , ⇒, A2 = , , =, , 2 3 2 3 8 13, ⇒, , A 2 − kA − I 2 = 0, 5 8 , 1 2 1 0, 8 13 − k 2 3 − 0 1 = 0, , , , , , , i3, i 0 i 0 i 2 0 , 2, ⇒, A, A, =, A2 = , ⋅, =, , , , , , 2, 0 i 0 i 0 i , 0, n, , , i, 0, ∴ An = , n, 0 i , 1 0, For n = 8, we get I = , , 0 1, , ∴, , 0 2 3, 7, 6. We have, A = , , B = 1, 2, 1, 4, , , , 0 6 9 , ∴, 3A = , , 6 3 12, 14 12 6 , and, 2B = , , 2 8 10, 0 6 9 14, ∴ 3 A − 2B = , −, 6 3 12 2, −14 −6 3, =, −5 2, 4, , 2 4, 3. We have, A = , , −8 x, Since, the matrix does not have an inverse., ⇒ A is singular. ⇒ | A | = 0, 2 4, ∴, =0, −8 x, , ∴, , 5 − k 8 − 2k 1 0, 8 − 2k 13 − 3k = 0 1, , , , ⇒, 5 − k =1⇒ k =4, i 0, 5. We have, A = , , i = −1, 0 i , ⇒, , (given), , 0, , i3 , , 6 3, 4 5, , 12, 8, , 6, 10, , 5 10, 5 10 3 , , , 7. The matrix −2 −4 6 is singular, if −2 −4, , , −1 −2, −1 −2 b, ⇒ −1 (60 + 12) + 2 (30 + 6) + b (−20 + 20) = 0, ⇒, −72 + 72 + 0b = 0, ⇒ The given matrix is singular for any value of, , 3, 6 =0, b, , b.
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112, , NDA/NA Mathematics, , 8. Q, , α, A=, β, , β, α , , α, ∴ A2 = , β, , =3, β α 2 + β 2, =, α 2αβ, , β α, α β, , ⇒, , = 3 (−3 + 2) − 1 (2 + 1) + 2 (4 + 3), , 2, , = − 3 − 3 + 14 = 8, 5, 7 T −1 3, −1 −3, , 7, 1, adj ( A ) = 3, − 5 = −3 1, , , , , 7 −11, 7 −5 −11, 5, , A =I, 2, , Now,, ⇒, , 2αβ , , α + β2 , , 2αβ 1, =, α + β 2 0, , α 2 + β 2, , 2αβ, , 2, , 0, 1, , α 2 + β 2 = 1, αβ = 0, , A −1 =, , ⇒, , α = 0, β = 1 or β = 0, α = 1, −1 2 , 3, 9. A = , and B = 1, −, 2, 1, , , , 1 −1 −2 1 1 2, −1, A =− , =, 3 −2 −1 3 2 1, AX = B ⇒ X = A −1B =, , Q, , =, 10. Q, , 1, 3, , =, , ⇒, , −5 −2 , ∴| A | =, = − 5 − 6 = − 11 and adj ( A ) = , , 3 −5, −3 1 , 2, , Hence, A −1 =, , 1, 1, adj ( A ) = −, | A|, 11, , 1, =, 11, , 5, 5 2 11, 3 −1 = 3, , , 11, , −5 −2 , −3 1 , , , 2 , 11 , 1, − , 11 , , T, 1 −2 −2 , 3 −3 4, 11. Q A = 2 −3 4, adj ( A ) = −1 3, 3, , , , , 0 −4 −3, 0 −1 1, , 12. Given equations are, 3x + y + 2z = 3, 2x − 3 y − z = − 3, , Let, , x + 2y + z = 4, 2, 3 1, 3, x, , , , , A = 2 −3 −1 , B = −3 , X = y, , , , , 1 , 1 2, 4 , z , 3 1, 2, | A | = 2 −3 −1, 1 2, 1, , …(i), …(ii), …(iii), , − 3 − 9 + 20 , 1, 1, −9 − 3 + 28 =, 8, 8, 21 + 15 − 44, , 3, −3 , , 4 , , 8, 16 , , −8, , x 1 , y = 2 ⇒ x = 1, y = 2, z = − 1, , z −1, −1 , 1 0, 2 and B = 2 1, , , 0 , 0 1, 2 −1 1 0 0, 0 2 2 1 0, , , 5 0 0 1 3, 1 + 4 + 0 0 + 2 − 1, = 3 + 0 + 0 0 + 0 + 2, , 4 + 10 + 0 0 + 5 + 0, , 1 2, 13. Q A = 3 0, , 4 5, 1, AB = 3, , 4, , 0, 0, , 3, , 0 + 0 − 3 5 1 −3 , 0 + 0 + 6 = 3 2 6 , , , 0 + 0 + 0 14 5 0 , , α, Q A=, 1, , 0, α 0 α 0, ⇒ A2 = , , , 1, 1 1 1 1, α2, 0, ⇒, A2 = , 1, +, α, 1, , α2, 0 1 0, 2, But, A2 = B ⇒ , =, ⇒α = 1, α + 1 1 2 1, and, α + 1 =2 ⇒ α =1, 1 −3 2 , 15. Since, the matrix 2 −8 5 is not an invertible, , , 2 λ , 4, matrix, i.e., it should be a singular matrix., 1 −3 2, 2 −8 5 = 0, ∴, , 14., , 1 −1 0 , = −2 3 −4 , , , −2 3 −3, , and, , 5 , −1 3, 1, X = A B=, −3 1, 7 , , 8, 7 −5 −11, , 1 1 2 3, 3 2 1 1, , 3 + 2 1 5 , 6 + 1 = 3 7, , , , , 5, −1 3, 1, 1, adj ( A ) = −3 1, 7, , | A|, 8, 7 −5 −11, −1, , Now,, , 1 2 , A=, , 3 −5, 1, , 2 −3, 2 −1, −3 −1, −1, +2, 1 2, 1 1, 2, 1, , 4, ⇒, ⇒, ⇒, , 2, , λ, , 1 (−8λ − 10) + 3 (2λ − 20) + 2 (4 + 32) = 0, −8λ − 10 + 6λ − 60 + 72 = 0, −2 λ + 2 = 0 ⇒ λ = 1
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113, , Matrices, , ⇒, , a 2, 20. For matrix A = , to be singular,, 2 4, a 2, =0, 2 4, , 3 A3 + 2 A 2 + 5 A + I = 0, , 16. Q, 3, , 3A A, , −1, , + 2A A, 2, , −1, , + 5 AA, , −1, , −1, , =0, , −1, , =0, , + IA, , ⇒, , 3 A I + 2 AI + 5I + A, , ⇒, , A −1 = − (3 A 2 + 2 A + 5I ), , 2, , 1 2 3 8, 17. Given that, 2X + , =, , 3 4 7 2, It can be rewritten as,, 3 8 1 2, 2X = , −, , 7 2 3 4, 2 6 , 1 3 , ⇒, 2X = , = 2 2 −1, 4, −, 2, , , , , 1 3 , ⇒, X =, , 2 −1, 1 x 0, 1 1, , 18. Given that, 1 −2 −2 y = 3, , , 1 z 4, 1 3, We have, to use option, to satisfy the given condition., x 1 , Now, let y = 2 satisfy it, , z −3, 1 1 1 + 2 − 3 0, 1 1, 1 −2 −2 2 = 1 − 4 + 6 = 3, , , , 1 −3 1 + 6 − 3 4, 1 3, where option (b) satisfy the conditions., Alternative Method, 1 x 0, 1 1, 1 −2 −2 y = 3, , , 1 z 4, 1 3, x + y + z 0, x − 2 y − 2z = 3, ⇒, , , x + 3 y + z 4, On comparing both sides, we get, x+ y+ z =0, x − 2 y − 2z = 3, and, x + 3y + z = 4, On solving equations., we get, x = 1, y = 2 and z = − 3, ∴, , 19. Q, , ∴, , x 1 , y = 2 , , z −3, 1 1 0, A = 1 2 1, , , 2 1 0, 7 9 3, 2 3 1, 2, 3, , , A = 5 6 2 and A = 15 19 6, , , , , 9 12 4, 3 4 1, , Hence,, , A3 − 3 A 2 − I = 0, , …(i), …(ii), …(iii), , ⇒, , 4a − 4 = 0, , ⇒, , a =1, , 21. Given that, AB = A, and, BA = B, Now,, B2 = B ⋅ B, = (BA ) ⋅ B, = B ⋅ ( AB), = B⋅ A, =B, 0 − i, i, 22. P = 0 − i, i, , , 0 , i, − i, i, − i, and, Q= 0, 0, , , i − i , 0 − i − i, i, PQ = 0 − i i 0, , , 0 i, − i i, , ... (i), ...(ii), [from Eq. (ii)], [from Eq. (i)], [from Eq. (ii)], , i , 0, , − i , , − i 2 − i 2 i 2 + i 2 , , , = i2, − i2 , 2, 2, i, −i , , , 1 + 1 − 1 − 1 2 −2, = −1, 1 = −1 1 , , , , 1 −1 1 , −1, a h g , x, y, 23. Here, [x y z ]1 × 3 h b f , , , , g f c 3 × 3 z 3 × 1, Order of matrix = 1 × 3 : 3 × 3 : 3 × 1 = 1 × 3 : 3 × 1 = 1 × 1, 0, i, 0 1, 24. Q A = , ,B = , , , 0 − i , −1 0 , −i , 0, and, C=, i, −, 0 , , 0 −1 i 0 0 − i , ⇒, AB = , =, =C, 0 0 − i − i 0 , −1, AB = C, , Hence,, , (adj AT ) = (adj A )T, , 25. Q, , ⇒ (adj AT ) − (adj A )T = Null matrix, 26., , 5, A=, 2, , 1, −1 5 2 × 3, 5 2, and let B = 1 6, , , 4 3 3 × 2, 6
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114, , NDA/NA Mathematics, , ∴, , 5 2, 1 , 1 6, , 5 , 4 3, 10 + 36 + 3, 25 + 6 + 4, 35, =, =, , 10 − 1 + 20 4 − 6 + 15 2 × 2 29, , 5, AB = , 2, , Now,, , 6, −1, , 49, 13, , cos θ sin θ , A=, , sin θ cos θ , C11 = cos θ ,, C12 = − sin θ, Here,, C 21 = − sin θ, C 22 = cos θ, cos θ − sin θ , adj ( A ) = , ∴, , − sin θ cos θ , 0 1, 28. We have, A = , , 1 0, Since, A is an identity matrix ., 1 0, So,, A4 = , , 0 1, , X = A −1B, 1 −4 1 −2 1 − 4 −2 − 12, =, , =, , −1 5 1 3 −1 + 5 2 + 15 , −3 −14, =, 17 , 4, , x y1 , X = 1, , then, x2 y2 , 1 3 x1 y1 1 1 , 0 1 x y = 0 −1, , , 2, 2 , x1 + 3x2 y1 + 3 y2 1 1 , ⇒ , =, y2 0 −1, x2, ⇒ x1 + 3x2 = 1, y1 + 3 y2 = 1, x2 = 0, y2 = − 1, ⇒, x1 = 1, y1 = 4, 1 4 , Thus,, X =, , 0 −1, , 27. Let, , 32. Let, , 29. Let, , 0 1 2, 33. Given that, the matrix M = 1 2 3, , , 3 1 1, and its inverse, = N = [nij ], ⇒, N = M −1, , ⇒, , ∴, 30. Here,, ∴, ∴, , 31. Here,, Let, ∴, , cos α − sin α , A=, , sin α cos α , 2, 2, | A |= cos α + sin α = 1, cos α sin α , adj (A) = , , − sin α cos α , cos α sin α , 1, adj ( A ) = , A −1 =, , | A|, − sin α cos α , 3 2, A=, , 1 4, 4 −2 , adj ( A ) = , , −1 3 , 3 2 4 −2, A adj ( A ) = , , , 1 4 −1 3 , 12 − 2 − 6 + 6 , =, , 4 − 4 −2 + 12, 10 0 , =, , 0 10, 5 4, 1 −2, 1 1X = 1 3 , , , , , 5 4, 1 −2, A=, , B = 1 3 , 1, 1, , , , , | A| = 5 − 4 = 1 ≠ 0, , So, A is non-singular and therefore invertible., , The given system is, ⇒, ⇒, Now,, ∴, ∴, , A, , −1, , AX = B, ( AX ) = A −1B ⇒ ( A −1 A )X = A −1B, , I X = A −1 B ⇒ X = A −1B, 1, adj (A), A −1 =, | A|, 1 −4 , adj ( A ) = , , −1 5 , 1 1 −4 1 −4 , =, A −1 = , 1 −1 5 −1 5 , , Now,, , M −1 =, , adj (M −1 ), |M |, , C11 = (2 − 3) = − 1, C12 = − (1 − 9) = 8 ,, C13 = (1 − 6) = − 5, C 21 = − (1 − 2) = 1, C 22 = (0 − 6) = − 6,, C 23 = − (0 − 3) = 3, C31 = (3 − 4) = − 1, C32 = − (0 − 2) = 2,, C33 = (0 − 1) = − 1, and | M | = 0 (2 − 3) − 1(1 − 9) + 2 (1 − 6), = 8 − 10 = − 2, −1 8 −5 , ∴Cofactor matrix = 1 −6 3 , , , −1 2 −1, T, −1 8 −5 , , , Transpose of cofactor matrix = adj (M ) = 1 −6 3, , , −1 2 −1, −1 1 −1 , ⇒ adj (M ) = 8 −6 2 , , , −5 3 −1, adj (M ), M −1 =, ∴, |M |, 1, 1, 1, −, 2, 2, 2, ⇒, M −1 = − 4, 3 −1 , 5, 3, 1, −, , , 2, 2, 2, 1 1, 1, −, n11 n12 n13 , 2, 2 2, n, = −4, n, n, 3, −1 , ⇒, 22, 23 , 21, 5, 3, 1, n, n, n, , 31, −, 32, 33 , , 2, 2 2, On comparing, we get n23 = − 1
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115, , Matrices, 0, −7, 34. Given, 2X − 3Y = , …(i), , −, 7, 13, , , 9 13, and 3X + 2Y = , …(ii), , 4 13, On multiplying Eq. (i) by 3 and Eq. (ii) by 2 and, subtracting Eq. (i) from Eq. (ii), we get, 0, 9 13, −7, 3Y = 2 , − 3 7 −13, 4, 13, , , , , 26, 39, ⇒, 13Y = , 65, −13, 3 2, ⇒, Y =, , −1 5 , T, , 2, 4 −3 , 4 −2 , , adj ( A ) = , = −3, 4, −, 2, 1, 1, , , , | A| = 4 − 6 = − 2, 1 4 −2 , adj ( A ), −1, ∴, A −1 = − , Q A =, , , −, 3, 1, 2, | A| , , 1 4 −2 , [bij ] = − , ⇒, 1, 2 −3, 1, b22 = −, ⇒, 2, ω 0 , 36. Given, A = , , 0 ω , 0, ω 0 ω 0 ω 2, Now,, =, A2 = , , , , , ω2 , 0 ω 0 ω 0, 0 ω 0 ω3, 0, ω 2, A3 = , =, , , 2 0, ω 0, ω , ω3 , 0, Similarly,, 0 (ω3 )33 ⋅ ω, 0, , ω100, =, A100 = , , 100 , 3 33, ω , (ω ) ⋅ ω , 0, 0, 0 ω 0 , 1 ⋅ ω, =, =, (Q ω3 = 1), ⋅ 1 0 ω , 0, ω, , =A, 1, 35. Q A = , 3, and, , 39. We know that , every matrix possesses a unique inverse., ∴ B and C should be equal., 3 2, 40. Given that, A = , , 4 5, 5 −2 , 1, (15 − 8) −4 3 , 19 24, 1 5 −2 , and AC = , = , , 7 −4 3 , 37 46, 19 24, ⇒, A −1 AC = A −1 , , 37 46, 120 − 92 , 1 5 −2 19 24 1 95 − 74, ⇒ C= , = , , , , 7 −4 3 37 46 7 −76 + 111 −96 + 138, 1 21 28 3 4, = , =, 7 35 42 5 6, , ∴, , 2 0 0, 1 2 3, 41. Given that, A = 0 2 0 and B = 0 1 2, , , , , 0 0 2, 0 0 1, 2 0 0 1 2 3 2 4 6, AB = 0 2 0 0 1 2 = 0 2 4, , , , , 0 0 2 0 0 1 0 0 2, 2 4 6, 2 4, = 2 (4) = 8, ∴ adj ( AB) = 0 2 4 = 2, 0 2, 0 0 2, 5 2, 3 6 , 42. Given that, A + B = , and A − B = 0 −1, 0 9, , , ∴, , , cos 2 θ + sin 2 θ, ∴ AA′ = , − cos θ sin θ + cos θ sin θ, − sin θ cos θ + sin θ cos θ , , sin 2 θ + cos 2 θ, , 1 0, =, =I, 0 1, So, this type of the matrix is orthogonal., , 1, 2, , 5 2 3 6 , , + , , 0 9 0 −1, , =, , 1, 2, , 8 8 4 4, 0 8 = 0 4, , , , , 1 5 2 3 6 , −, , , 2 0 9 0 −1, , 1 2 −4 1 −2, =, 2 0 10 0 5 , 4 4 1 −2, AB = , , , 0 4 0 5 , 4 −8 + 20, =, 20 , 0, =, , | A | = 1 (1 + 4) + 2 (2 + 6) − 3 (4 − 3), , = 5 + 16 − 3 = 18 ≠ 0, Hence, it is non-singular matrix., cos θ sin θ , 38. Let A = , , − sin θ cos θ , cos θ − sin θ , ⇒ A′ = , , sin θ cos θ , , A=, , B=, , and, , 37. Here, we see that its diagonal elements are not zero, so, it is not anti-symmetric matrix., Now,, , A −1 =, , ∴, , 43. Q, ∴, , 1, X =, 0, 1, X2 = , 0, 1, =, 0, , 4 12, =, , 0 20, −2 , 3, −2 1 −2, 3 0, 3, −2 − 6 1 −8, =, 9 0, 9
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116, , NDA/NA Mathematics, 1 −8, 1 −2, 1 0, ∴ X 2 − 2X + 3I = , −2 , +3, , , , 3, 9, 0, 0, 0 1, 0, 1 −8 −2 +4 3, =, + 0 −6 + 0, 0, 9, 3, , , , −8 + 4 , 1 − 2 + 3, =, − 6 + 3, 0, 9, , 2 −4, 1 −2, = 2X, =, = 2 0, 0, 6, 3, , , , 3 1, A=, , −1 2 , , 44. We have,, , 2x, x, , , 0 1 0 1, =, x −1 2 0, 2x 0 1, ⇒, 0 2x = 0, , , On comparing, we get 2x = 1, 1, ⇒, x=, 2, 3 2, 48. Q A = , , 1 4, ∴A (adj A ) = I 2 | A |, 1 0 3 2, 1, =, 1 4 = 0, 0, 1, , , , 1 0, 10, =, × 10 = 0, 0 1, , , A 2 + 7I 2 = 5 A, , ∴, , A 2 ⋅ A −1 + 7I 2 A −1 = 5 AA −1, , ⇒, , A + 7 A −1 = 5 I 2, , ⇒, , 7 A −1 = 5 I 2 − A, , ⇒, , A −1, , ⇒, , 5 0 3 1, =, −, , 0 5 −1 2, 1 0 3 1, =5, −, , 0 1 −1 2, 1 2 −1, = , 7 1 3 , , 1, 49. Let , 0, , 45. We know that, if A = [a ij ]m × n and B = [bij ]n × p are the, two matrices, then the product matrix AB is of order, m × p., , B −1 =, , 1 3, 3 0, , T, , 0, 3, =, , 1, 0, −2 , 1, , 2, −1, A=, 3, 6, , 1, Now, , 0, , A −1 ⋅ A 2 = A −1 ⋅ 1, , ⇒, ⇒, , A, , −1, , ( A ⋅ A) = A, , −1, , ( A A) ⋅ A = A, , ⇒, , −1, , 1 ⋅ A = A −1, , ⇒, , A −1 = A, 2x, A=, x, , (Q A ⋅ 1 = A), , −1, , ⇒, , 47. Given,, , ⇒, , (Q A −1 A = 1), , 0, 1, and A −1 = , , x, −1, , 0, 2, , , , −1 0 , , Q BA = , , 6, 3, , , , , , , 1, −, 0, , , , , A = B −1 , , , 6 3, , 0, 3, , ⇒, 51., , 2, , T, , x − x, x, adj ( A ) = , = −x, 0, 2, x, , , , adj ( A ), −1, A =, | A|, , 0, 2x, , 1 x 0, 2x2 − x 2x, 1, , 0, 2x, , =, , 1, − 1, , x, 2x, , 1, 0, , , 1, 1, But, ∴ =1 ⇒ x=, A −1 = , , 2, 2x, −1 2 , Alternative Method, We know that,, AA −1 = I, A −1 =, , −2 −1, 3 6, −2 , 1, , 0 1 −3 − 12, =, 3 3 6, , −6 , 3, , 50. Since, A′ = A and A′ = − A, , | A | = 2x − 0 = 2x, 2, , 1 3, 3 0, −5, =, 2, , A=, , −2 , 1, , , 1, −1, Q B = | B| adj (B), , , , Given, condition is A 2 = 1, , 46. Q, , 0, (12 − 2), 1, 0, 10, , 2, = B, | B| = 3, 3, , 3, and adj (B) = , −2, ∴, , 0, 1, 0, 1, , ⇒, ⇒, , A = − A ⇒ A =0, 1 2 3 2 , 2X − , =, , 7 4 0 −2, 3 2 1 2, 2X = , + , , 0 −2 7 4, 2 2, 4 4, ⇒X =, 2X = , , , 7 / 2 1, 7 2, , 52. A + AT is a square matrix., Now,, , ( A + AT )T = AT + ( AT )T = AT + A, , Hence, A + AT is symmetric matrix., 53. Q | A | = 1, 3 4 7 −4 1 0, 1 0, ∴ A adj ( A ) = , −5 3 = 0 1 = 1 0 1 =| A | I, 5, 7, , , , , , , 54. Given that, square matrices A and B are of same order., We know that, if A and B are non-singular matrices of, the same orders, then ( AB)−1 = B−1 A −1.
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117, , Matrices, , Level II, x + y 2x + z 4 7 , 1. Q , =, , x − y 2z + w 0 10, ⇒, x+ y=4, , …(i), , x− y=0, , …(ii), , 2x + z = 7, , …(iii), , 2z + w = 10, , …(iv), , On solving these equations, we get, x = 2 , y = 2 , z = 3, w = 4, 2. Q, ⇒, , 0, kA = , 2b, 0 2k 0, 3k −4k = 2b, , , , ⇒, , 3a , 0 2 0 3a , ⇒k, =, , 24 , 3 −4 2b 24 , 3a , 24 , , 2k = 3a , 3k = 2b, − 4k = 24, 2k, 3k, a=, ,b=, ,k = −6, 3, 2, , ⇒, ∴, , 3. If AB = O, then A and B may be equal to O individually., It is not necessary any condition., 1 1, 4., A=, , 1 1, 1 1 1 1, 1 1, A2 = A ⋅ A = , =2 , , , , , 1 1 1 1, 1 1, , ∴, , A4 = 2 A ⋅ 2 A = 4 A2 = 4 × 2 A = 8 A, , = 2A, (Q A 2 = 2 A ), , A 4 = 23 A, Similarly,, , A100 = 299 A, , cos 2 θ − sin 2 θ , 5. Given that, A = , , sin 2 θ cos 2 θ , 2, 2, ∴, | A | = |cos 2θ + sin 2θ | = 1, cos 2 θ sin 2 θ , and adj ( A ) = , , − sin 2 θ cos 2 θ , 1 cos 2 θ sin 2 θ cos 2 θ sin 2 θ , ∴, A −1 = , =, 1 − sin 2 θ cos 2 θ − sin 2 θ cos 2 θ , 6. Since, the given matrix is, 1, 1, 2 − x, 0, A= 1, 3−x, , , −3, − x, −1, This matrix is singular., ∴, | A| = 0, 1, 1, 2−x, 0 =0, 3−x, ⇒, 1, −1, ⇒, , 2−x, 0, , −3, , −x, 1, −x, , 1, −x = 0, , −1, −3 − x, ⇒ (2 − x)(x2 − 3x) + 1(x) + 1(− x) = 0, , 7. If A and B are two symmetric matrices, then A + B and, AB + BA are also symmetric matrices., ∴ Option (b) is correct., 8. Let the real symmetric matrix of order 2 is,, 0 a, A=, , −a 0, 0, 0 a 0 a −a 2, A2 = , =, , , , 2, −a 0 −a 0 0 −a , 0 1 0 1 − a 2, 0 , −a 2, +, =, ⇒ A2 + I = , , , 2, 0, 1, 1 − a2, 0, 0 −a , Here, when a = ± 1, then A 2 + I = 0, , a = − 4, b = − 9, k = − 6, , ∴, , ⇒, (2 − x)(x)(x − 3) = 0, ⇒, x = 2, 0, 3, Hence, solution set S = {0, 2, 3}, , So, the given relation A 2 + I = 0 is not always true for, even order., Now, let the real symmetric matrix of order 3 is,, a, b, 0, , c, 0, A = −a, , , − b − c 0 , a b 0 a b, 0, 2 , A = − a 0 c −a 0 c , , , , − b − c 0 − b − c 0 , − bc, −a 2 − b2, ac , , , 2, 2, = − bc, −a − c, − ab , ac, − ab, − b2 − c2 , , − bc, , −a 2 − b2 + 1, − ac, , , 2, 2, − ab, A +I=, − bc, − a − c2 + 1, ≠0, 2, 2, , ac, − ab, − b − c + 1, , 2, So, the given relation A + I = 0 also is not always true, for an odd order., So, none of the option is true., 9. The order of a given matrices are, [X ]( a + b) × ( a + 2) and [Y ]( b + 1) × ( a + 3 ), As [XY ] and [YX ] exist., ∴, a + 2 = b + 1 and a + 3 = a + b ⇒ a = 2, b = 3, 1 2, a 0 , 10. Let, A=, and B = 0 b, 3, 4, , , , , 2b, 1 2 a 0 a, ∴, AB = , 0 b = 3a 4b, 3, 4, , , , , a 0 1 2 a 2a , and BA = , , =, , 0 b 3 4 3b 4b , If, AB = BA, 2b a 2a , a, ⇒, 3a 4b = 3b 4b ⇒ a = b, , , , From the above it is clear that there exist infinitely, many B’s such that AB = BA.
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118, , NDA/NA Mathematics, , cos θ − sin θ , 11. Since, A = , , sin θ cos θ , cos θ − sin θ, = cos 2 θ + sin 2 θ = 1 ⇒| A | ≠ 0, | A| =, sin θ cos θ, ∴ A is invertible and A is orthogonal matrix as well as A′, is orthogonal matrix. So, option (d) is not correct., 1 2, 0 −1, 12. Given, A = , and B = , , , 1 1, 1 2 , 1 2, Here,| A|=, = 1 − 2 = −1, 1 1, 1 −1 1 −2 , adj ( A ) = , =, , −2 1 −1 1 , 1 −2 , adj A, A −1 =, = − 1, , |A|, −1 1 , 0 −1, Here,|B| =, = 0 − (−1) = 1, 1 2, , = (3 − 2 i ) A12 = (−1)1 + 2 (2) = − 2, A21 = (−1)2 + 1 (3 + 5i ) = − (3 + 5i ), ∴, ⇒, , 17. Given that,, AB = A and BA = B, (a), AT = ( AB)T = BT AT, ⇒ ( AT )2 = (BT AT )(BT AT ), = ( AB)T ( AB)T = AT ⋅ AT = ( AT )2, (b) ( AT )2 = (BT AT )(BT AT ) = ( AT BT )( AT BT ), = (BA )T (BA )T = BT BT = (BT )2, (c) A −1 = ( AB)−1 = B−1 A −1, ⇒, , ', , 2 −1 2 1, adj (B) = , =, , 1 0 −1 0, 2 1, adj (B), B −1 =, =1⋅, , |B|, −1 0 , 2 1, B −1 = , , −1 0 , 2 1 −1 2 −2 + 1 4 − 1 , ∴B−1 A −1 = , , , = , −1 0 1 −1 1 + 0 −2 + 0 , −1 3 , B −1 A −1 = , , 1 −2 , 2 1 4 −1, 13. Inverse of A = , , =, 7 4 −7 2 , − sin α , − sin β , cos α, cos β, 14. Q A( α ) = , and A(β ) = , , cos α , cos β , sin α, sin β, cos α sin α , ∴, A( − α ) = , cos α , − sin α, cos β sin β , and A( −β ) = , cos β , − sin β, sin α cos β sin β , cos α, A( − α ) A( − β ) = , sin, α, cos, α − sin β, cos β , −, , cos α sin β + sin α cos β , cos α cos β − sin α sin β, =, , − sin α cos β − cos α sin β − sin α sin β + cos α cos β , sin (α + β ), cos (α + β ), =, −, +, sin, (, α, β, ), cos (α + β ), , = A( α + β ), 15. We know that a matrix obtained from a unit matrix by a, single elementary operation is known as unit matrix., 3 − 2 i 3 + 5 i , T, 16. We have, A = , and let adj (A ) = B ,, 2, 3, −, 2, i, , , where B is the cofactor matrix of A., ∴ We have to find, A11 , A12 , A21 , A22, ⇒, , A11 = (−1)1 + 1 (3 − 2 i ), , A22 = (−1)2 + 2 (3 − 2 i ) = (3 − 2 i ), −2 , 3 −2i, B=, , −3 − 5 i 3 − 2 i , 3 − 2 i −3 − 5 i , adj ( A ) = BT = , 3 − 2 i , −2, , [Q Aij = (−1)i + j a ij ], , 18., , ( A −1 )−1 = (B−1 A −1 )−1 = AB = A, , For 2 × 2 matrix,, | A | = |adj ( A )|= (ab − 0) = ab =, ∴, , A −1 =, , adj ( A ) 1 a, =, ⋅, ab −1, | A|, , 1, (ab) = 1, ab, 1, 1, −2, , 19. Here, A = 1 −2, 1 ,, , , 1 −2, 1, l, x, B = m and X = y, , , n , z , 1, 1, −2, −1, | A| = − 2, ∴, 1, 1 −2, , a, , 0, , −1, , b, , 0, , b, , | A −1 | =, , 1, 1, +1, −2, −2, , 1, 1, , = − 2(4 − 1) − 1 (−2 − 1) + 1 (1 + 2), = −6 + 3 + 3 = 0, 3 3 3, Now, adj ( A ) = 3 3 3, , , 3 3 3, 3 3 3 l , l + m + n , 0, , , , , , , Q (adj A ) B = 3 3 3 m = 3 l + m + n = 3 0, , , , , , 3 3 3 n , l + m + n , 0, ∴ (adj A ) ⋅ B = 0, (Q l + m + n = 0), So, the given system of equations has an infinitely, many solutions., 0 −1 , 0, 20. Q, 0, A = 0 −1, , , 0, 0, −1, 0 −1, 0, ∴, 0 = − 1(−1) = 1 ≠ 0, | A| = 0, −1, −1, , 0, , 0
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119, , Matrices, ∴ A −1 exists., Now,, , ⇒, , 0 −1 0, 0, 2, , 0 0, A = 0 −1, , , 0 −1, 0, −1, 1 0 0, = 0 1 0, , , 0 0 1, A2 = I, , 21. Let A = [a ij ]n, ⇒, , 0, −1, 0, , a 2 2ab a b a3 3a 2b, =, =, , , a 2 0 a 0, a3 , 0, 1 0, 2 0 1 , 3, 2, = a3 , + 3a b 0 0 = a I + 3 ⋅ a bA, 0, 1, , , , , Continue in this process, we get, (aI + bA )n = a n I + na n − 1 bA, , −1 , 0, , 0, , 25. Given that, A is non-singular matrix such that, A −1 = AT, , ×n, , kA = [ka ij ]n, , ×n, , , where k is scalar., , We know that, A adj ( A ) =| A |I n, , ⇒, , AA −1 = AAT, , ⇒, , AAT = I, , …(i), , (kA ) adj (kA ) = | kA |I n, kA adj (kA) = kn | A |I n, (since| kA | = kn | A |), ⇒, , A adj (kA ) = kn, , −1, , | A|I n, , Now,, , …(ii), , From Eqs. (i) and (ii), we get, adj (kA ) = kn, , −1, , adj ( A ), , 22. A square matrix A is said to be non-singular according, as,, A adj ( A ) = | A | I n, ⇒, | A||adj ( A )|=| A|n (Q order of matrix is n × n), |adj (A)| = | A |n − 1, ⇒, x , 23. Given that, M = y, and G = [3 4 5]1 × 3, , z 3 × 1, x, M = y, [3, , z 3 × 1, 3x 4x, 3 y 4 y, ⇒, , 3z 4z, , 3x 4x 5x , 4 5]1 × 3 = 3 y 4 y 5 y = [hij ]3 × 3 (say), , , 3z 4z 5z , 5x h11 h12 h13 , 5 y = h21 h22 h23 , , , 5z h31 h32 h33 , ⇒, h32 = 4z, 0 1, 24. Given that, A = , , 0 0, 1 0, 0 1 a b , aI + bA = a , + b, , =, , 0 1, 0 0 0 a , 2, Now, (aI + bA ), = (aI + bA ) ⋅ (aI + bA ), a b a b , =, , , 0 a 0 a , a 2 2ab a 2 0 0 2ab, =, =, +, a 2 0 a 2 0 0 , 0, 1 0, 0 1, 2, = a2 , + 2ab , , = a I + 2abA, 0 1, 0 0, Similarly, (aI + bA ) = (aI + bA ) ⋅ (aI + bA ), 3, , 2, , 1 2 2, 1, 1, 1, 2 1 −2 AT = 2, , 3, 3, −2 2 −1, 2, 1, 2, 2, 1, 2, , , 1, 1, AAT = 2 1 −2 2 1, , , 3, 3, −2 2 −1 2 −2, 1 + 4 + 4 2 + 2 − 4, 1, =, 2 + 2 −4, 4+1+4, 9, −2 + 4 − 2 −4 + 2 + 2, 9 0 0 1 0 0, 1, = 0 9 0 = 0 1 0 = I, , , 9, 0 0 9 0 0 1, A=, , Let, , Replacing A by kA, we get, , (Q AA −1 = I), 2 −2 , 1, 2, , −2 −1, −2 , 2, , −1, −2 + 4 − 2 , −4 + 2 + 2 , , 4 + 4 + 1 , , ⇒, , AAT = I, , 1, 26. Given that, A = , 2, 1, , ∴, A2 = , 2, 5, =, 6, Now,, , ⇒, , 2, 2, 2 1 2 1 + 4 2 + 4, =, 2 2 2 2 + 4 4 + 4, 6, 8, 5 6 3 6 2 0, A 2 − 3 A − 2I = , −, −, , 6 8 6 6 0 2, 5 − 3 − 2 6 − 6 − 0 0, =, =, 6 − 6 − 0 8 − 6 − 2 0, A 2 − 3 A − 2I = O, , 1, 27. Let B = , 0, Now,, , 3, 1 −1, and C = , ,, , 1, 1, 0, | B| = 1 − 0 = 1, T, 1 0, 1 −3, =, adj (B) = , , 1, −3 1 , 0, adj (B), B −1 =, | B|, 1, B −1 = , 0, , Then,, , BA = C, , Now,, ⇒, , −3 , 1, , −1, , B BA = B−1C ⇒ A = B−1C, 1, =, 0, , −3 1, 1 0, , −1 1, =, 1 0, , −4 , 1, , 0, 0
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120, , NDA/NA Mathematics, , 28. To hold the condition A (BC ) = ( AB) C, AB and BC both must exist., 29. We know that,, a, c, , , b, d , , −1, , 5, 0, , , 0, 7, , −1, , ∴, , =, , 1, ad − bc, , d, −c, , , =, , 1 7, 35 0, , 0, 5, , −b, a , , −1, , 5 0 x −1, Hence, , = , 0 7 − y 2, 1 7 0 x −1, ⇒, =, 35 0 5 − y 2, ⇒, , 1, 35, , (given), , 7x −1, −5 y = 2, , , , On comparing, 5y, 7x, =2, = − 1 and −, ⇒, 35, 35, ⇒, x = − 5 and y = − 14, , ⇒, , 30. Among the given matrix equations, ( AB)′ = A′ B′ is wrong, [Q ( AB)′ = B′ A′ (by transposed matrices)], 31. We know that, if A and B are symmetric matrices of the, same order, then AB is symmetric only when AB = BA., [Q A matrix A is said to be symmetric, if A′ = A, i.e., a ij = a ji ], 0, 7, 4, −, −, 0, 7, 4, , , , , 32. Let A = −7 0 −5 ⇒ A′ = 7 0, 5 =− A, , , , , −4 5 0 , 4 −5 0 , So, it is skew-symmetric matrix., , 0, 33. Let A = 0, , 1, , 1, 0, , 0, , 0, 1, 0, , ∴, | A| = − 1, Cofactors of A, 1 0, a11 =, = 0, a12 = −, 0 0, 0 1, 0, a 21 = −, = 0, a 22 =, 0 0, 1, 0 1, 0, a31 =, = − 1, a32 = −, 1 0, 0, 0, ⇒ adj ( A ) = 0, , −1, Hence, A −1 =, , 0, −1, 0, , A 2 + 2 AB + B2 ≠ ( A + B)2, , 1 0, 0 1, 35. Now, α I + βA = α , +β, , , 0 1, −1 0 , α 0 0 β α β , =, + , =, , 0 α −β 0 −β α , α β α β α 2 − β 2, 2αβ , ∴ (α I + βA )2 = , −β α = −2αβ α 2 − β 2 , −, β, α, , , , , Since, it is given, 0 1, 2, A=, = (α I + βA ), −1 0 , 0 1 α 2 − β 2, 2αβ , ∴, −1 0 = −2αβ α 2 − β 2 , , , , ⇒ α 2 − β 2 = 0 ⇒ α 2 = β 2 ⇒ α = β and 2αβ = 1, 1, 1, ⇒, 2α 2 = 1 ⇒ α 2 =, ∴α =β = ±, 2, 2, 36. If the least number of zeroes in a lower triangular, matrix is 10, then order of the matrix is 5 × 5., , 0, , 0, , 1, , 0, 1, 0, , 1, 0, , = 0, a13 =, , = − 1, a 23 = −, = 0, a33 =, , 0 0, 0 1, , 0, , 1, , 1 0, 0 0, 1, , 0, , =0, , T, , −1 , 0 −1 , 0, , , 0 and adj ( A ) = 0 −1 0 , , , , 0, 0 , −1 0, , 0, −1, 0, , −1 0, 0 = 0, , 0 1, , 0, 1, 0, , a, b, , A = d, g, , k, , = −1, e.g.,, , =0, , 1, adj ( A ), | A|, , 0, 1, =− 0, 1, −1, , 0, 2 −1, 1, 34. We have, A = , and B = , , , 0 1 , −1 −1 , 3 −1 , ∴, A+ B=, , −1 0 , 3 −1 3 −1 10 −3, ⇒ ( A + B )2 = , , =, , −1 0 −1 0 −3 1 , Now, A 2 + 2 AB + B2, 0, 2 −1 2 −1, 2 −1 1, =, 0 1 + 2 0 1 −1 −1, 0, 1, , , , , , , 0 1, 0, 1, +, , , , −1 −1 −1 −1 , 2 1 0, 4 −3 6, =, + −2 −2 + 0 1, 0, 1, , , , , 11 −1, =, , −2 0 , , 1, 0, , 0, , 0, c, e, h, l, , o, 0, , 0, , f, i, m, , 0, , 0, j, n, , 0, 0, , 0, 0, , o, , Q Number of zero’s in lower triangular matrix, n (n − 1) 5 × 4, =, =, = 10, 2, 2, 1, 37. Let A = 0, , 0, , p, x, 0, , q, 1, 0 and B = 0, , , 1 , 0, , −p, 1, 0, , | B| = 1(1) + p(0) − q(0) = 1, C11 = 1, C12 = 0, C13 = 0, C 21 = p, C 22 = 1, C 23 = 0, C31 = q, C32 = 0, C33 = 1, , −q, 0, , 1
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121, , Matrices, , ∴, , 1, adj (B) = p, , q, , 0, 1, 0, , 1 p, = 0 1, , 0 0, adj, (B), B −1 =, | B|, 1 p, −1, B = 0 1, , 0 0, , 39. Here, A = − B ⇒ ( A )2k = (− B)2k, ⇒ A 2k = B 2k ⇒ A 2k − B 2k = O, , T, , 0, 0, , 1, , 40. Total possible order of matrices using 36 elements are, 1 × 36, 2 × 18, 3 × 12, 4 × 9,6 × 6, 9 × 4, 12 × 3, 18 × 2 ,, 36 × 1., , q, 0, , 1 , , So, total number of matrices are 9., 0 0, 0, 3 2 0, 41. B = 2 4 0 adj (B) = 0, 0 0, , , , , −2 −1 8, 1 1 0, , q, Thus,, 0, , 1 , But B is inverse of A, therefore A = B−1, (according to question), q 1 p q, ⇒, x 0 = 0 1 0 , , , 0 1 0 0 1 , ⇒, x=1, Alternative Method, Given, B is the inverse of A, then, BA = I, 1 − p − q 1 p q 1 0 0, 0, 1, 0 0 x 0 = 0 1 0, , , , , 0, 1 0 0 1 0 0 1, 0, 1 p − px 0 1 0 0, 0, ⇒, 1, 0 = 0 1 0, , , , 0, 1 0 0 1, 0, 1, 0, , 0, , p, , On comparing, we get x = 1, 4 11, 3, 38. Given that, AB = , and A = 1, 4, 5, , , , T, , ⇒, , m = − (1 + ω ) = − (−ω 2 ) = ω 2; m = ω 2, , 43. The sum and product of matrices A and B exist, if A and, B are square matrices of same order. It is not, necessarily that A and B are non-singular matrices for, addition and product of two matrices., x x2 1 + x2 , , , 44., A = y y2 1 + y2 , , , 2, 1 + z2 , z z, , x x2 1 + x2, | A| = y, , y2, , 1 + y2, , z, , z2, , 1 + z2, , 2, 2, , 2 −1 , 2 −2 , adj ( A ) = , = −1, −, 2, 3, 3, , , , 1, 1, , , −, , 1 2 −2 2, adj ( A ), 2, −1, −1, =, ∴A =, , Q A =, , , 3 1, 3 , 6 − 2 −1, | A| , −, 4, 4, 4 11, Now,, AB = , 5, 4, 4 11, ⇒, B = A −1 , 5, 4, 1, 1, −, 2, 2 4 11, =, , 3 4, 5, − 1, 4, 4, 11 5 , , −, 2 −2, , 2 2 0 3, =, =, , −1 + 3 − 11 + 15 2 1, , 4, 4, 0 3, Hence,, | B| =, = 0 − 6 = −6, 2 1, , | A|= 6 − 2 = 4,, , 42. The given matrix is singular, if, 1 ω m, 1+ω+m ω m, ω m 1 =0 ⇒ 1 + ω + m m 1 =0, m 1 ω, 1+ω+m 1 ω, 1 ω m, ⇒ (1 + ω + m) 1 m 1 = 0 ⇒ 1 + ω + m = 0, 1 1 ω, , x− y, , (x − y)(x + y), , (x − y)(x + y), , = y−z, , ( y − z )( y + z ), , ( y − z )( y + z ), , z, , 1 + z2, (use operations R1 → R1 − R2 , R2 → R2 − R3 ), 1 x+ y x+ y, y+ z, = (x − y)( y − z ) 1 y + z, z, , z, =0, , 2, , z2, , 1 + z2, , (Q two rows are identical), , 0 0 q , 45. A = 2 5 1 , , , 8 p p, I. For q = 0, 0 0 0 , . A = 2 5 1 ⇒ | A | = 0, , , 8 p p, ∴ A is singular., II. For p = 0, 0 0 q, A = 2 5 5 ⇒ | A | = − 40q, , , 8 0 0 , ∴ A is not singular.
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122, , NDA/NA Mathematics, III. For p = 20, 0, A = 2, , 8, , 0, 5, 20, , 5, = 40 − 40 = 0, 8 20, ∴ A is singular., Thus, codes I and III are correct., | A| = q, , 46., , 2, , I. Every zero matrix is not necessarily square matrix., II. A matrix does not have a numerical value while, every determinant have a numerical value., III. A unit matrix is a diagonal matrix and scalar, matrix also., , 47. If A is non-singular and B is singular, then AB and A −1B, are non-singular because the inverse of A will exist, when A is non-singular., 48. The inverse of a square matrix, if it exists, is unique but, if A and B are singular matrices of order n, then AB is, not a singular matrix of order n., Hence, only statement I is correct., 49. We know that, A + AT is always symmetric and A − AT, is always anti-symmetric., (by property of transpose), 50. I. It is not necessary that if A is a symmetric matrix,, then it is singular., II. A3 = I ⇒| A3 | = | I |, ⇒, , | A |3 = 1 ⇒ | A | = 1, , ∴ A is a non-singular matrix., 51. Let A = [a ij ]m × m be a matrix and C = [cij ]m × m be, another matrix where cij is the cofactor of a ij . i.e.,, X = adj ( A ), Q A is a square matrix., ∴ adj ( A ) = (adj A ) both have the same order., ⇒, |adj A | = | A |m − 1, ⇒ | A ||adj A | = | A |m − 1|, ⋅ A|, ⇒, |adj A | = | A |m ⇒ | AC | = | A |m, 52. Given that, A = A′ , B = B′, Now, we have AB = A′ B′ = (BA )′, Therefore, AB is not symmetric., and, A 2 + B2 = ( A′ )2 + (B′ )2 = ( A 2 + B2 )′, So,, A 2 + B2 is symmetric., 53., , Now,, , 2 3 x 5 , 4 6 y = c − 5 ⇒ 2x + 3 y = 5, , , , 4x + 6 y = c − 5, Put c = 5, then 2x + 3 y = 5,4x + 6 y = 0, ∴ Equation has no solution., Put c = 10, then 2x + 3 y = 5, 4x + 6 y = 5, ∴ Equation has no solution., Put c = 15, then 2x + 3 y = 5,4x + 6 y = 10, ∴ Equation has an infinite set of solution., 5 10, 56., M =, 8, 4, 5 10, |M |=, = 40 − 40 = 0, 4, 8, , 55. Given,, , q, 1, , 20, , 3, M = 2, , 3, 3, , 4, 1, 1, 4, , 0, 0, , k, 0, , |M |= 2, , 1, , 3, , 1, , 0 = k(3 − 8) = − 5k, k, , So that M is not invertible because M is a singular, matrix., (R) M is singular matrix., Therefore, A is false but R is true., 2 3, 1 0, 3 3, 57. Q, A=, ,B = , , A+ B=, , , , 1 4, 0 1, 1 5, 12 24, ⇒, ( A + B )2 = , , 8 28, 2 3 1 0 2 3, ∴, AB = , , =, ,, 1 4 0 1 1 4, 2 3 2 3 7 18, A2 = A ⋅ A = , , =, , 1 4 1 4 6 19, 1 0 2 3 2 3, and, BA = , , =, , 0 1 1 4 1 4, 1 0 1 0 1 0, B2 = B ⋅ B = , , =, , 0 1 0 1 0 1, 7 18 1 0 4 6, A 2 + B2 + 2 AB = , + , + , , 7 19 0 1 2 8, 12 24, =, , 8 28, ∴, AB = BA, ∴ ( A + B)2 = A 2 + B2 + 2 AB, Hence, both A and R are individually true and R is the, correct explanation of A., cos α sin α , cos α cos α , 58. Q, A=, , B = sin α sin α , cos, α, sin, α, , , , , cos, α, sin, α, cos, α, cos, α, , , , ∴, AB = , , , cos α sin α sin α sin α , cos 2 α + sin 2 α cos 2 α + sin 2 α , =, , 2, 2, 2, 2, cos α + sin α cos α + sin α , 1 1, =, (Q sin 2 α + cos 2 α = 1), , 1 1, , If k ≠ 0, then inverse of M exists. Thus, statement A, implies B as well as B implies A., 54. If AB = 0, then we conclude that A = 0 or B = 0. But, remember that it is not a necessary condition that, AB = 0 holds, if A = 0 or B = 0, because that condition is, also possible when A ≠ 0 and B ≠ 0. If A and B are, non-zero matrix and AB = 0 ⇒| A | = 0 and| B| = 0., , ⇒, AB ≠ I, Sometimes the product of two matrices can be equal to, an identity matrix., e.g.,, , AA −1 = I, , Hence, A is true but R is false.
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123, , Matrices, 1, 59. (A) We have, A = , 5, 1, ∴, A′ = , 2, , 2, 9, 5, 9, , Here,, A ≠ A′ and A ≠ − A′, ∴ It is neither symmetric nor anti-symmetric matrices., ∴ It is true., (R) We know that the matrix A cannot be expressed as a, sum of symmetric and anti-symmetric matrices., It is false., Hence, A is true but R is false., , 1 − 1 2 − 1 3 + 1, = 3 − 2 4 + 3 5 − 4 =, , , 6 − 3 7 + 2 9 − 3 , , ∴ | A − B|= 0 − 1 (6 − 3) + 4 (9 − 21) = − 3 − 48 = − 51, 1 2 3, 1 1 − 1, , , 2 − 3 4 , 62. ∴ AB = 3 4 5, , , , , 6 7 9, 3 − 2 3 , 1 −1 −6 −1 + 8 + 9 , 1+4+9, = 3 + 8 + 15 3 − 12 − 10 − 3 + 16 + 15 , , , 6 + 14 + 27 6 − 21 − 18 − 6 + 28 + 27 , 14 − 11 16, = 26 − 19 28, , , 47 − 33 49, , Solutions (Q. Nos. 60- 62), 1 2 3, | A|= 3 4 5 = 1 (36 − 35) − 2 (27 − 30) + 3(21 − 24), 6 7 9, = 1 + 6 − 9 = − 2 (≠ 0), Since,| A|≠ 0, ∴A is non-singular and hence A −1 exist., Now, the cofactors of the elements of the first row of, | A|are respectively, 4 5, 3 5 3 4, ,−, ,, 7 9, 6 9 6 7, i.e.,, 1, 3, −3, Again, the cofactors of the elements of the second row, of| A|are respectively, 2 3 1 3 1 2, ,, ,, −, 7 9 6 9 6 7, i.e.,, 3, − 9, −5, Again, the cofactors of the elements of the third row of, | A|are respectively, 2 3, 1 3 1 2, ,−, ,, 4 5, 3 5 3 4, i.e.,, −2 , 4, −2, Hence, the matrix B, whose elements are the cofactors, of the corresponding elements of| A|are, 3 − 3, 1, , C (let) = 3 − 9 5 , , , − 2 4 −2 , 3 − 2, 1, , Hence, adj (A) = transpose of B = 3 − 9 4 , , , − 3 5 −2 , 60. ∴, , 61., , A −1 =, , 1, | A|, , −1 / 2 −3 / 2 1 , 1, adj (A) =, adj ( A ) = −3 / 2 9 / 2 −2, , , −2, 3 / 2 −5 / 2 1 , 1 2 3 1 1 −1, A − B = 3 4 5 − 2 −3 4 , , , , 6 7 9 3 −2 3 , , 0 1 4, 1 7 1, , , 3 9 6, , Solutions (Q. Nos. 63-65), The matrix form of the given system is, 1 1 1 x 6 , 1 2 3 y = 10, , , 1 2 λ z µ , The system admits of a unique solution if and only if the, coefficient matrix is non-singular i.e., the corresponding, determinant, 1 1 1 , 1 2 3 ≠0, , , 1 2 λ , The value of the determinant when simplified is λ − 3, Thus, the given system has a unique solution, if λ ≠ 3, and µ may have any value, if λ = 3, we can write the, system as, 1 1 1 x 6 , 1 2 3 y = 10, , , 1 2 3 z µ , Which is equivalent to, 1 1 1 x 6 , 0 1 2 y = 4 R2 → R2 − R1, R → R −R, , , 3, 1, 0 1 2 z µ − 6 3, 1 1 1 x 6 , 0 1 2 y = 4 , , , , 0 0 0 z µ − 10, This is equivalent to the system of equations, , i.e.,, , x + y + z =6, y + 2z = 4, 0 = µ − 10, Hence, the given system will possess an infinite number, of solutions, if µ= 10 and the solutions are provided by, the first two equations. The system will not possess any, solution, if µ ≠ 10 for in that case the third equation, becomes false.
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129, , Determinant, , Comprehensive Approach, n, , n, , n, , n, , n, , n, , n, , n, , n, n, n, , n, , det (B) = k det ( A), if B is obtained from A multiplying one row, ( or column ) of A by k., det (kA) = k n det( A), where A is a matrix of order n × n. Hence,, det (kA) = k3 det ( A), where A is a matrix of order 3 × 3., If A, B ,C are square matrices of the same order that ith column, ( or row) of A is the sum of ith columns (or rows) of B and C and all, other columns (or rows) of A, B and C are identical, then, det ( A) = det (B) + det (C )., If A is a non-singular matrix, then, 1, det ( A − 1) =, det ( A), 1, i.e.,, |A − 1| =, |A|, Determinant of a skew-symmetric matrix of odd order is zero and, of even order is a non-zero perfect square., If A = B + C , then it is not necessary that, det( A) = det(B) + det(C )., Determinant of a diagonal matrix = product of its diagonal, elements., Determinant of a triangular matrix = product of its diagonal, elements., Determinant of an unit matrix is 1., Determinant of a null matrix is 0., Symmetric determinant If equidistant elements to a diagonal of a, determinant are of same magnitude and sign, then the determinant, is known as symmetric determinant, e.g.,, a h g, h b f = abc + 2 fgh − af 2 − bg 2 − ch2, , , g f c, Skew-symmetric determinant If equidistant elements to a, diagonal of a determinant are of same magnitude and opposite in, sign, then the determinant is known as skew-symmetric, determinant. e.g.,, , 0 b c, − b, 0 a = 0, , , − c − a 0, a, 1.b, , c, , n, , c, a = − ( a3 + b3 + c3 − 3abc), , b, 1 1 1, 2.a b c = ( a − b) ( b − c) ( c − a), , 2, 2, 2, a b c , 1 1 1 , 3. a b c = ( a − b) ( b − c) ( c − a) ( a + b + c), , 3, 3, 3, a b c , 1 1 1 , 4.a 2 b 2 c 2 = ( a − b) ( b − c) ( c − a) ( ab + bc + ca), , , 3, 3, 3, a b c , f (r) g(r) h( r), If ∆ r = a, b, c, , , c1 , b1, a1, b, c, a, , where r ∈ N and a, b , c , a1 , b1 , c1 are constants, then, n, n, , n, Σ f (r) Σ g(r) Σ h( r), , r = 1, r =1, r =1, n, 1. Σ ∆ r = a, b, c , r =1, , , c1 , b1, a1, , , n, n, , n, Π f (r) Π g(r) Π h(r), , r = 1, r =1, r =1, n, , 2. Π ∆ r =, b, a, c , r =1, , , c1 , b1, a1, ,
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Exercise, Level I, a − b b − c c − a, 1. x − y y − z z − x is equal to, , , p − q q − r r − p, (a) a ( x + y + z ) + b ( p + q + r ) + c, (b) 0, (c) abc + xyz + pqr, (d) None of the above, 1, , 2. 1, 1, , (a), , 2, a a − bc, , b b2 − ac is equal to, c c2 − ab, , , 0, , (c) 3abc, , 9. If a + b + c = 0, then the solution of the equation, c, b , a − x, c, b− x, a = 0, is, , , a, c − x, b, 3, (a) 0, (b) ± ( a 2 + b2 + c2 ), 2, 3 2, 2, 2, (c) 0, ±, ( a + b + c ) (d) 0, ± a 2 + b2 + c2, 2, , (b) a3 + b3 + c3 − 3abc, (d) ( a + b + c)3, , 3. The system of equations 3x − 2 y + z = 0,, λx − 14 y + 15z = 0, x + 2 y − 3z = 0 has a solution, other than x = y = z = 0 for λ equal to, (a) 1, (b) 2, (c) 3, (d) 5, 4 20 , 1, 4. The roots of the equation1 − 2 5 = 0 are, , , 2, 1 2x 5x , (a) − 1, − 2, (b) − 1, 2, (c) 1, − 2, (d) 1, 2, 5. The solution of the simultaneous linear equations, 2x + y = 6 and 3 y = 8 + 4x will also be satisfied by which, one of the following linear equations?, (NDA 2012 I), (a) x + y = 5, (b) 2x + y = 9, (c) 2x − 3 y = 10, (d) 2x + 3 y = 6, 6. The simultaneous equations 3x + 5 y = 7 and, (NDA 2011 II), 6x + 10 y = 18 have, (a) no solution, (b) infinitely many solutions, (c) unique solution, (d) any finite number of solutions, x, 7. The roots of the equation β, , β, independent of, (a) α, (c) γ, , α, x, γ, , 1, 1 = 0 are, , 1, (NDA 2011 II), , (b) β, (d) α, β and γ, , 8. If| A| = 8, where A is square matrix of order 3, then, what is| adj A| equal to?, (NDA 2010 II), (a) 16, (b) 24, (c) 64, (d) 512, , 10. If, the, system, of, equations, 2x + 3 y + 5 = 0, x + ky + 5 = 0, kx − 12 y − 14 = 0 be, consistent, then value of k is, 12, 1, (a) − 2,, (b) − 1,, 5, 5, 17, 12, (d) 6, −, (c) − 6,, 5, 5, a, b, , a + b + 2c, is equal to, 11. , 2a + b + c, b, c, , , a + 2b + c, c, a, , (a) ( a + b + c)2, (b) 2 ( a + b + c)2, 3, (d) 2 ( a + b + c)3, (c) ( a + b + c), i , 1 + i 1 − i, , 12. 1 − i, i, 1 + i is equal to, , , 1 + i 1 − i, i, (a) − 4 − 7i, (b) 8 + 11i, (c) 3 + 7i, (d) 7 + 4i, a, a, b + c, 13. The value of determinant b, c+a, b, , c, a+, c, (a) abc, (c) 3abc, , , is, , b, , (b) 2abc, (d) 4abc, , 1 −2, 0, 14. In the determinant −1, 0 3, the value of the, , , 2 − 3 0, cofactor to its minor of the element − 3, is, (a) − 1, (b) 0, (c) 1, (d) 2, b + c a + b a, 15. The value of determinant c + a b + c bis equal to, , , a + b c + a c, (a) a3 + b3 + c3 − 3abc, (b) 2abc ( a + b + c), (c) 0, (d) None of the above
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131, , Determinant, 16. x + ky − z = 0, 3x − ky − z = 0 and x − 3 y + z = 0 has, non-zero solution for k is equal to, (a) −1, (b) 0, (c) 1, (d) 2, b2 − ab b − c, , 17. Determinant ab − a 2 a − b, bc − ac, c−a, , (a) abc ( a + b + c), (b), (c) 0, (d), , bc − ac , , b2 − ab is equal to, , ab − a 2, 3a 2b2c2, None of these, , 18. If a −1 + b−1 + c−1 = 0 such that, 1, 1 , 1 + a, 1, 1+ b, 1 = λ,, , , 1, 1 + c, 1, then what is λ equal to?, (NDA 2011 II), (a) − abc, (b) abc, (c) 0, (d) 1, 19. What is the value of the determinant, x+ 4, x+1 x+ 2, x + 3 x + 5, x + 8 ?, , , x + 7 x + 10 x + 14, (c) 2, (a) x + 2, (b) x 2 + 2, , a, 21. If l, , p, , b, m, q, , (a) 10, (c) 40, , 3b, m, q, , 15c, 5n ?, , 5r , , x −6 −1 , 26. One root of the equation 2 −3x x − 3 = 0, is, , , −3 2x x + 2, (a) 0, (b) 1, (c) −1, (d) 3, 1− i, 27. What is the value of, , ω +i, , ω2, ω, , 1 − 2i − ω 2, , ω2 − ω, , 2, , where ω is the cube root of unity?, (a) − 1, (b) 1, (c) 2, , ω, −i ,, i−ω, , (NDA 2009 II), , (d) 0, , 28. If n ≠ 3k and 1, ω , ω 2 are the cube roots of unity, then, (d) −2, 4, x, 8, , 5, 7 = 0,, , x, , (NDA 2011 I), , c, n = 2 , then what is the value of the, , r, , 6a, determinant 2l, , 2p, , 25. If ω be a complex cube root of unity, then, 1 ω −ω 2 / 2, 1 1, 1 is equal to, , , 0 , 1 −1, (a) 0, (b) 1, (c) ω, (d) ω 2, , (NDA 2011 I), , x, 20. If 5 and 7 are the roots of the equation 7, , 5, then what is the third root?, (a) − 12, (b) 9, (c) 13, (d) 14, , x + 1 x + 2 x + 3, 24. If x + 2 x + 3 x + 4 = 0, then a , b, c are in, , , x + a x + b x + c, (a) AP, (b) GP, (c) HP, (d) None of these, , 2n, 1, ωn ω , , , ω n has the value, 1, ∆ = ω 2 n, n, 1 , ω 2n, , ω, , (b) ω, , (a) 0, , (c) ω 2, , (d) 1, , x2 + x, x +1 x − 2, 2, , 29. If 2x + 3x − 1, 3x, 3x − 3 = Ax − 12, then the, 2, 2x − 1 2x − 1, x + 2x + 3, , value of A is, (a) 12, (b) 24, , (c) −12, , (d) −24, , 30. The system of equations, λx + y + z = 0, − x + λy + z = 0,, (NDA 2010 I), , (b) 20, (d) 60, , 22. If two rows of a determinant are identical, then what, is the value of the determinant?, (NDA 2012 I), (a) 0, (b) 1, (c) − 1, (d) Can be any real value, ω, x + 1, ω2 , 2, , 23. If ω is a cube root of unity, then ω, 1 , x+ω, , , 2, 1, x + ω, ω, is equal to, (b) x3 + ω (c) x3 + ω 2 (d) x3, (a) x3 + 1, , − x − y + λz = 0, will have a non-zero solution, if real, values of λ are given by, (a) 0, (b) 1, (c) 3, (d) 3, 2, , 4, , 31. If 0, , 5, , 0, , 0, , (a) 0, (c) 2, , 0, 16 = 20, then what is the value of p?, 1+ p, (NDA 2008 II), (b) 1, (d) 5, , 32. If the system of equations 2x + 3 y = 7 and, 2ax + ( a + b) y = 28 has infinitely many solutions, then, which one of the following is correct?, (NDA 2010 I), (a) a = 2b, (b) b = 2a, (c) a = − 2b, (d) b = − 2a
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132, , NDA/NA Mathematics, , 1, 6i − 3i, 33. Let 4, 3i − 1 = x + iy , then, , , i, 20 3, (a) x = 3, y = 1, (b) x = 0, y = 0, (c) x = 0, y = 3, (d) x = 1, y = 3, 34. If a , b, c are positive integers, then the determinant, ab, a 2 + x, ac , 2, , ∆ = ab, b +x, bc is divisible by, , , bc, c2 + x, ac, (a) x3, (c) ( a 2 + b2 + c2 ), , (b) x 2, (d) None of these, , 35. If A and B are square matrices of order 3 such that, |A| = − 1,|B| = 3, then|3AB| is equal to, (a) − 9, (b) − 81, (c) − 27, (d) 81, 36. The value of the determinant, a1b2 + a2b1, 2a1b1, 2a2b2, ∆ = a1b2 + a2b1, , a1b3 + a3 b1 a3 b2 + a2b3, (a) 1, (b), (c) 0, (d), , a1b3 + a3 b1 , a2b3 + a3 b2, is, , 2a3 b3 , 2a1a2a3 b1b2b3, a1a2a3 b1b2b3, , a2, c2 , b2, , , 37. The value of( a + 1)2 ( b + 1)2 ( c + 1)2 is, , 2, 2, 2, ( a − 1) ( b − 1) ( c − 1) , a 2 b2 c2, (a) 4 a b c , , , 1 1 1, a 2 b2 c2, (c) 2 a b c , , , 1 1 1, , a 2 b2 c2, (b) 3 a b c , , , 1 1 1, (d) None of these, , 38. If x + y + z = 1, then what is the value of, 1, z, −y, −z, 1, x ?, (NDA 2008 I), 2, , 2, , −x, , y, , 2, , (b) 1, , 39. What is the determinant, , 1, (a) 1, 1, , a, , a2, , b, , 2, , c, , a, , a3, , (c) 1, , b, , b3, , 1, , c, , c3, , b, c, , 2, , (d) 2 − 2xyz, , (c) 2, 2, , bc, ca, , a, b, , a, , ab, , c, , c2, , b2 equal to?, (NDA 2008 I), , 1, , a2, , a3, , (b) 1, , 2, , 3, , b, , 2, , c3, , 1, , b, c, , (d), , a, , a2, , a3, , b, , b2, , b3, , c, , c2, , c3, , 40. The equations 2x + 3 y + 4 = 0; 3x + 4 y + 6 = 0 and, 4x + 5 y + 8 = 0 are, (a) consistent (with unique solution), (b) inconsistent, (c) consistent (with infinitely many solutions), (d) None of the above, 41. For what value of k, the system of linear equations, x + y + z = 2, 2x + y − z = 3, 3x + 2 y + kz = 4 has a, unique solution?, (a), (b), (c), (d), , k is any real number other than zero, k is any real number, k is any integer, k= 0, , 42. Which one of the following is correct? The system of, equations, and, x + y + z = 6,, 2x + y + z = 3, 3x + 2 y + 2z = 9, (a), (b), (c), (d), , has no solution, has infinite number of solutions, has only one solution, has only three solutions, , 1 0, 1 2, 43. If A = , , then what is determinant of, and , 1 0, 2 3, (NDA 2012 I), AB?, (a) 0, , (b) 1, , (c) 10, , (d) 20, , 8 −5 1, 44. If 5 x 1 = 2, then what is the value of x?, 6 3 1, (NDA 2012 I), (a) 4, , (b) 5, , (c) 6, , (d) 8, , 1 x y +, 45. The value of the determinant1 y z +, , 1 z x +, (a) x, , (b) y, , 1 + x, , 46. If 1 + y3, , 3, 1 + z, , (c) z, , z, x, is, , y, (d) 0, , x, , y y = 0 and x , y , z are all different, then, , z 2 z, the value of xyz, is, 3, , 1, , (a) 0, , 1, , x, , 2, , 2, , (a) ( x − y ) ( y − z ) ( z − x ), (c) 1, , (b) 0, (d) − 1, , 47. The value of z satisfying the equations, 3x + 5 y − 7z = 13, 4x + y − 12z = 6, 2x + 9 y − 3z = 20,, is, (a) 1, , (b) 2, , (c) −1, , (d) 0
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133, , Determinant, x −3i 1, 48. If y, 1 i = 6 + 11i ,then, , , 0 2 i − i, (a) x = − 3, y = 4, (b) x = 3, y = 4, (c) x = 3, y = − 4, (d) x = − 3, y = − 4, y, x, 2, , 49. The expansion of the determinant x, 5 y3, , 3, 5, x 10 y, contains which one of the following as a factor?, (a) x − 3, (b) x − y, (c) y − 3, (d) ( x − 3) ( y − 3), , (a) 0, (c) –1, , 3, 9, , 27, , 50. If ω is the cube root of unity, then what is one root of, the equation, (NDA 2007 II), , (a) 1, , x2, , −2x, , −2 ω 2, , 2, 0, , ω, ω, , −ω, 1, , (b) –2, , = 0?, , (c) 2, , (d) ω, , 51. If a, b, c are in GP, then what is the value of, a, b, a+b, b, c, b+ c ?, (NDA 2008 I), a+b, , b+ c, , 0, , (b) 1, (d) None of these, , 1 1 1, 52. What are the factors of the determinantx y 1?, 2, , 2, x y 1, (a) x − 1, y − 1 and y − x, (b) x , y and x − y, (c) x − 1, y − 1 and y + x, (d) x − 1, y + 1 and y + x, − a 2 ab ac , 53. If ab − b2 bc = ka 2b2c2 , then k is equal to, , , bc − c2, ac, (a) − 4, (b) 2, (c) 4, (d) 8, 54. If a , b, c are in AP,, x + 2 x + 3 x + a, x + 4 x + 5 x + b is, , , x + 6 x + 7 x + c, (a) x − ( a + b + c), (b) 9x 2 + a + b + c, (c) a + b + c, (d) 0, , then, , the, , value, , of, , Level II, 2r − 1 2 ⋅ 3r − 1 4 ⋅ 5r − 1, 1. If Dr = x, y, z , then the value of, , , n, n, n, 2 − 1 3 − 1 5 − 1 , n, , Σ Dr is equal to, , r =1, , (a) 1, (c) 0, , (b) − 1, (d) None of these, , 2. If the system of equations, ax + y + z = 0, x + by + z = 0 and x + y + cz = 0, where, a , b, c ≠ 1, has a non-trivial solution, then the value of, 1, 1, 1, is, +, +, 1− a 1− b 1− c, (a) −1, (b) 0, (c) 1, (d) None of these, x b b, x b, are the given, 3. If ∆1 = a x b and ∆ 2 =, , , a x, a a x, determinants, then, d, (b), (a) ∆1 = 3 ( ∆ 2 )2, ( ∆1 ) = 3∆ 2, dx, d, (d) ∆1 = 3∆32/ 2, (c), ( ∆1 ) = 2 ( ∆ 2 )2, dx, , 4. What, a − b, b− c, , c − a, (a), (b), (c), (d), , is the value, b + c a, c + a b ?, , a + b c, , of, , the, , determinant, (NDA 2011 II), , a3 + b3 + c3, 3bc, a3 + b3 + c3 − 3abc, 0, , p, 5. If 0, , q, , −q, p, 0, , 0, q = 0, then which one of the following, , p, , is correct?, (a) p is one of the cube roots of unity, (b) q is one of the cube roots of unity, p, is one of the cube roots of unity, (c), q, (d) None of the above, , (NDA 2011 II), , 6. Let A be an n × n matrix. If det ( λA) = λ s det (A), what, is the value of s?, (NDA 2010 I), (a) 0, (b) 1, (c) − 1, (d) n
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134, , NDA/NA Mathematics, , 7. If ai2 + bi2 + ci2 = 1 ( i = 1, 2, 3) and, ai a j + bi bj + ci c j = 0 ( i ≠ j, i , j = 1, 2, 3), then the, 2, a1 a2 a3, value ofb1 b2 b3 is, , , c1 c2 c3, (a) 0, (b) 1/2, (c) 1, (d) 2, 8. In a third order determinant, each element of the, first column consists of sum of two terms, each, element of the second column consists of sum of three, terms and each element of the third column consists, of sum of four terms. Then, it can be decomposed into, n determinants, where n has the value, (a) 1, (b) 9, (c) 16, (d) 24, a b aα − b, 1, 9. If b c bα − c = 0 and α ≠ , then, , , 2, 0 , 2 1, (a) a , b, c are in AP, (b) a , b, c are in GP, (c) a , b, c are in HP, (d) None of these, 10 ! 11 ! 12 !, 10. The value of the determinant11 ! 12 ! 13 ! is, , 12 ! 13 ! 14 !, (a) 2 (10 ! 11 !), (b) 2 (10 ! 13 !), (c) 2 (10 ! 11 ! 12 !), (d) 2 (11 ! 12 ! 13 !), 11. If a , b, c be positive and not all equal, then the value, a b c, of the determinant b c a is, , , c a b, (a) negative, (b) positive, (c) depends on a , b, c, (d) None of these, 2a, 12. A = 4b, −2c, , 3r, 6s, −3t, , x, , a, , r, , x, , 2y = λ b, c, −z, , s, t, , y , then what is, z, , the value of λ?, (a) 12, (b) –12, , (NDA 2009 II), , (c) 7, , (d) –7, , x, , y+z, , 13. If z, , y, , x + y = 0, then which one of the following, , x, , z, , z+x, , y, , is correct?, (a) Either x + y = z or x = y, (b) Either x + y = − z or x = z, (c) Either x + z = y or z = y, (d) Either z + y = x or x = y, 14. What is the value of, (a) 0, , (b) 1, , (NDA 2009 I), , sin 10° − cos 10°, , ?, cos 80° (NDA 2008 II), (c) –1, (d) 1/2, , sin 80°, , 15. If x = cy + bz , y = az + cx , z = bx + ay (where x , y , z, are not all zero) have a solution other than, x = 0, y = 0, z = 0, then a , b and c are connected by the, relation, , (a), (b), (c), (d), , a 2 + b2 + c2 + 3abc = 0, a 2 + b2 + c2 + 2abc = 0, a 2 + b2 + c2 + 2abc = 1, a 2 + b2 + c2 − bc − ca − ab = 1, , x, y, , 1, 16. Determinant 2 sin x + 2x sin y + 2 y is equal to, , , 3 cos x + 3x cos y + 3 y, (a) sin ( x − y ), (b) cos ( x − y ), (c) cos ( x + y ), (d) xy sin ( x − y ), N, , 17. The value of Σ U n ,ifU n, n =1, , (a) 0, (c) − 1, , n, 1, 5 , 2, , = n 2N + 1 2N + 1is, , , 3, 2, N, 3, n, N, 3, , , (b) 1, (d) None of these, , a+ b+ c , a+b, a, 18. ∆ = 3a 4a + 3b 5a + 4b + 3c where, , , 6a 9a + 6b 11a + 9b + 6c, , a = i , b = ω , c = ω 2 , then ∆ is equal to, (a) i, (b) − ω 2, (c) ω, (d) − i, , 19. What is the value of k, if, k, k, k, , (NDA 2009 I), , b2 + c2, , b+ c, c+a, , c2 + a 2 = ( a − b)( b − c)( c − a )?, , a+b, , a 2 + b2, , (a) 1, (c) 2, , (b) –1, (d) 0, , 20. If ( a1 / x ) + ( b1 / y ) = c1 , ( a2 / x ) + ( b2 / y ) = c2, ∆1 =, ∆3 =, , c1, , a1, , b1, , a2, , b2, , , ∆2 =, , b1, , c1, , b2, , c2, , ,, , a1, , , then ( x , y ) is equal to which one of the, c2 a2, following?, (NDA 2008 II), (a) ( ∆ 2 /∆1 , ∆3 /∆1 ), (b) ( ∆3 /∆1 , ∆ 2 /∆1 ), (c) ( ∆1 /∆ 2 , ∆1 /∆3 ), (d) ( − ∆1 /∆ 2 , − ∆1 /∆3 ), 21. For positive numbers x , y and z, the numerical value, logx y logx z, 1, , of the determinant log y x, 1, log y z, is, , , 1 , logz x logz y, (a) 0, (b) 1, (c) loge xyz, (d) None of these, 22. The value of the determinant, 4x + 2 y 4x + 3 y 4x + 4 y, 4x + 3 y 4x + 4 y 4x + 5 y is equal to, , , 4x + 6 y 4x + 7 y 4x + 8 y, (a) x + y, (b) x − y, (c) 1, (d) 0
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135, , Determinant, a h g f, 0 b c e, 23. The value of the determinant, , is, 0 0 d k, , , 0 0 0 l, (a) gfkl, (b) abhg, (c) abdl, (d) ablc, a1 b1 c1 , 24. If ∆ = a2 b2 c2, ∆1, , , a3 b3 c3 , (a) ∆ = ∆1, (c) ∆ = − ∆1, , a1 + 2b1 − 3c1 b1 c1 , = a2 + 2b2 − 3c2 b2 c2, then, , , a3 + 2b3 − 3c3 b3 c3 , (b) ∆ = − 6∆1, (d) − 6∆ = ∆1, , a b 0, 25. If the determinant 0 a b = 0, then, , , b 0 a, (a) a/b is one of the cube roots of unity, (b) a is one of the cube roots of unity, (c) b is one of the cube roots of unity, (d) a/b is one of the cube roots of − 1, 26. If| An × n | = 3 and | adj A| = 243, what is the value of, n?, (NDA 2008 I), (a) 4, (b) 5, (c) 6, (d) 7, 27. If A is matrix of order 3 × 2 and B is matrix of order, 2 × 3, then what is | kAB| equal to (where k is any, scalar quantity)?, (NDA 2008 I), (a) k| AB|, (b) k2| AB|, (c) k3 | AB|, , (d) | AB|, , 28. Which statement is correct in the following?, Linear system of equations 2x + 3 y = 4, and 4x + 6 y = 7, has, (NDA 2008 I), (a) no solution, (b) unique solution, (c) exactly 3 solutions, (d) infinite many solutions, 29. If in obtaining the solution of the system of equations, x + y + z = 7; x + 2 y + 3z = 16 and x + 3 y + 4z = 22.By, ∆′, Cramer’s rule, the value of y is given by , where, ∆, 1 1 1, ∆ =1 2 3, then the determinant ∆′ is given by, , , 1 3 4, 1 1 −7 , 1 1 1, , , (b) 1 2 −16, (a) 1 2 3, , , , , 1 3 −22, 1 3 4, 1 1 −7 , 1 1 −7 , (c) 2 3 −16, (d) 1 3 −16, , , , , 1 4 −22, 3 4 −22, 30. The system of equations 2x + 3 y = 5 and, 10x + 5 y = 50, (a) has a unique solution, (b) has infinitely many solutions, (c) is inconsistent, (d) is consistent and has exactly two solutions, , 31. The value of the determinant of the inverse of the, − 4 − 5, matrix , , is, 2, 2, 1, (a), (b) 2, 2, (c) 3, (d) 4, − a2, 32. What is the value of ab, ac, (a) 4 abc, (c) 4 a 2b2c2, , ab, −b, bc, , 2, , ac, bc ? (NDA 2012 I), − c2, , (b) 4a 2bc, (d) − 4a 2b2c2, , 33. How many values for k the system of equations, ( k + 1)x + 8 y = 4k and kx + (k + 3) y = 3k − 1 have an, infinitely many solutions?, (NDA 2007 II), (a) 1, (b) 2, (c) 3, (d) None of these, 34. For what value of p, is the system of equations, p3 x + ( p + 1)3 y = ( p + 2)3, (NDA 2007 II), px + ( p + 1) y = p + 2; x + y = 1 consistent?, (a) p = 0, (b) p = 1, (c) p = − 1, (d) For all p > 1, 35. The determinant, , a+ b+ c, 4a + 3b + 2c, , a+b, 3a + 2b, , a, 2a, , is, , 10a + 6b + 3c 6a + 3b 3a, independent of which one of the following? (NDA 2007 II), (a) a and b, (b) b and c, (c) a and c, (d) All of these, 36. Let A be a square matrix of order n × n, where n ≥ 2., Let B be a matrix obtained from A with first and, second rows interchanged. Then, which one of the, following is correct?, (NDA 2007 I), (a) det ( A) = det ( B), (b) det ( A) = − det ( B), (c) A = B, (d) A = − B, 37. What should be the value of k so that the system of, linear equations x − y + 2z = 0, kx − y + z = 0,, 3x + y − 3z = 0 does not possess a unique solution?, (a) 0, (b) 3, (NDA 2007 I), (c) 4, (d) 5, , Directions (Q. Nos. 38-40), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a),(b),(c) and, (d) given below, Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true.
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136, , NDA/NA Mathematics, , ab, b2 + c2, ac , 2, 2, , 38. Assertion (A) If, ab, c +a, bc = ka 2b2c2, , , bc, a 2 + b2, ac, then the value of k is 4., Reason (R) Q a 2 + b2 + c2 = 0, , a b c, 44. The value of c a b is, b c a, , 39. Assertion (A) If A is an n × n matrix, then, det ( mA) = m n det ( A), where m is any scalar., Reason (R) If U is a matrix obtained from V by, multiplying any row or column by a scalar m,, then det (U ) = m det (V)., , 45. The value of, 1 + 2ω100 + ω 200, , 40. Assertion (A) If two triangles with vertices ( x1 , y1 ),, ( x2 , y2 ), ( x3 , y3 ) and ( a1 , b1 ), ( a2 , b2 ), ( a3 , b3 ) satisfy the, relation, x1 y1 1, a1 b1 1, x2 y2 1 = a2 b2 1 , then the triangles are, x3, , y3, , 1, , a3, , b3, , 1, , congruent., Reason (R) For the given triangles satisfying the, above relation implies that the triangles have equal, area., (NDA 2007 II), 41. Consider the following statements, (NDA 2007 II), I. If det ( A) = 0, then det ( adj A) = 0., II. If A is non-singular, then det ( A−1 ) = (det A)−1., Which of the above statements is/are correct?, (a) I only, (b) II only, (NDA 2007 I), (c) Both I and II, (d) Neither I nor II, 42. Consider the following statements, I. If any two rows or columns of a determinant are, identical, then the value of the determinant is, zero., II. If the corresponding rows and columns of a, determinant are interchanged, then the value of, determinant does not change., III. If any two rows (or columns) of a determinant, are interchanged, then the value of the, determinant changes in sign., Which of these are correct?, (a) I and II, (b) I and III, (c) II and III, (d) I , II and III, , Directions (Q. Nos. 43-45), , Consider ω is a cube, , root of unity, then, ω ω2, 43. The value of ω ω 2 1 is, ω2 1 ω, 1, , (a) 0, , (b) 2, , (c) 3, , (d) 4, , (a) 1, (c) 3, , (b) 2, (d) None of these, , 1+ω, , 1, ω, (a), (b), (c), (d), , ω2, + 2ω, ω2, , 100, , 1, 200, , ω, 2 + ω100 + 2ω 200, , 1, ω, ω2, None of the above, , Directions (Q., , Nos., , 46-47) Consider, , the, , following system of equations, kx+ y+ z = k−1, x + ky + z = k − 1, x + y + kz = k − 1, 46. What value of k the system is inconsistent?, (a) k = 1 or k = − 2, (b) k = 2 or k = 3, (c) k = 3 or k = 4, (d) None of the above, 47. What value of k the system is consistent?, (a) k ≠ 1 or k ≠ − 2, (b) k = 1 or k = 2, (c) k = 3 or k = 4, (d) None of the above, 48. Match List I and List II and select the correct answer, using the codes given below the lists., List I, x a a, , A., , B., , C., , a x a, a a x, b+ c, a, a, b, c+a, b, c, c, a+ b, 1 a bc, 1 b ca, 1 c ab, , Codes, A B, (a) 1 3, (c) 1 2, , C, 2, 3, , List II, 1., , ( x + 2 a)( x − a) 2, , 2., , ( a − b) ( b − c) ( c − a), , 3., , 4abc, , A, (b) 2, (d) 3, , B, 3, 1, , C, 1, 2
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Answers, Level I, 1., 11., 21., 31., 41., 51., , (b), (d), (d), (b), (a), (a), , 2., 12., 22., 32., 42., 52., , (a), (b), (a), (b), (b), (a), , 3., 13., 23., 33., 43., 53., , (d), (d), (d), (b), (a), (c), , 4., 14., 24., 34., 44., 54., , (b), (a), (a), (b), (d), (d), , 5., 15., 25., 35., 45., , (a), (a), (a), (b), (d), , 6., 16., 26., 36., 46., , (a), (c), (b), (c), (d), , 7., 17., 27., 37., 47., , (a), (c), (d), (a), (d), , 8., 18., 28., 38., 48., , (c), (b), (a), (c), (a), , 9., 19., 29., 39., 49., , (c), (d), (b), (b), (a), , 10., 20., 30., 40., 50., , (c), (a), (a), (c), (c), , 2., 12., 22., 32., 42., , (c), (b), (d), (c), (d), , 3., 13., 23., 33., 43., , (b), (b), (c), (b), (a), , 4., 14., 24., 34., 44., , (c), (b), (a), (c), (d), , 5., 15., 25., 35., 45., , (c), (c), (d), (b), (d), , 6., 16., 26., 36., 46., , (d), (a), (c), (b), (a), , 7., 17., 27., 37., 47., , (c), (a), (c), (d), (a), , 8., 18., 28., 38., 48., , (d), (a), (a), (a), (a), , 9., 19., 29., 39., , (b), (a), (d), (b), , 10., 20., 30., 40., , (c), (d), (c), (b), , Level II, 1., 11., 21., 31., 41., , (c), (a), (a), (a), (c), , Hints & Solutions, Level I, The system of equations has, (non-trivial) solutions, if ∆ = 0., , a − b b − c c − a , 1. x − y y − z z − x, , , p − q q − r r − p, 0 b − c c − a , = 0 y − z z − x = 0, , , 0 q − r r − p, (applying C1 → C1 + C 2 + C3 ), 1 a, , 2. 1 b, 1 c, , , a 2 − bc, , b2 − ac, , c2 − ab, , 0 a − b (a − b) (a + b + c), = 0 b − c (b − c) (a + b + c), , , c, c2 − ab, 1, , , 0, , (Q rows R1 and R2 are identical), , 3. Given system of equations are, 3x − 2 y + z = 0,, λx − 14 y + 15z = 0, and, x + 2 y − 3z = 0, , many, , −2, − 14, , ⇒, , 3, ∆ = λ, , 1, , ⇒, ⇒, , 3( 42 − 30) − λ (6 − 2) + 1 (− 30 + 14) = 0, 36 − 4λ − 16 = 0 ⇒ λ = 5, , 1, 15 = 0, , 2 − 3, , 1, 4 20, 4. We have, 1 − 2, 5 = 0, , , 2, 1 2x 5x , ⇒, , 15 , 0, 6, 0 − 2 − 2x 5 (1 − x2) = 0, , , 2x, 5 x2 , 1, applying , R1 → R1 − R2, , , and R2 → R2 − R3, , , (applying R1 → R1 − R2, R2 → R2 − R3 ), 0 1 a + b + c, = (a − b) (b − c)0 1 a + b + c = 0, , , 2, 1 c c − ab , , infinitely, , ⇒, , ⇒, ⇒, ⇒, ⇒, , 1 , 0, 1, , 3 ⋅ 2 ⋅ 5 0 − (1 + x) 1 − x2 = 0, , , x, x2 , 1, 0, 1, 1 , (1 + x)0 − 1 1 − x = 0, , , x, x2 , 1, (1 + x) (2 − x) = 0, x + 1 = 0 or x − 2 = 0, x = − 1, 2
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138, , NDA/NA Mathematics, , 5. The given simultaneous linear equations, 2x + y = 6, 4x − 3 y = − 8, , …(i), …(ii), , ⇒, , Multiplying Eq. (i) by 3 and then adding in Eq. (ii), we, get, , ⇒, , 6x + 3 y = 18, 4x − 3 y = − 8, 10x = 10 ⇒ x = 1, From Eq. (i), y = 4, This intersection point (1, 4) satisfies the following, equation, x + y = 1 + 4 = 5, 6. The simultaneous equations a1x + b1 y + c1 = 0 and, a 2x + b2y + c2 = 0 have infinitely many solutions, if, a1 b1 c1, =, = ., a 2 b2 c2, For simultaneous equations 3x + 5 y = 7 and, 6x + 10 y = 18, 3 5, 7, ⇒, =, ≠, 6 10 18, 1 1, 7, ⇒, = ≠, 2 2 18, ∴ It has no solution., x α 1, 7. Given, β x 1 = 0, , , β γ 1, , ⇒, , ⇒, , (− x) {(b − x − c) (c − x − b) − (a − c) (a − b)} = 0, 3, , , x x2 − (a 2 + b2 + c2) = 0, , , 2, 3 2, x = 0, x = ±, (a + b2 + c2), 2, , 10. The homogeneous linear system of equations is, consistent i. e, possesses non-trivial solutions (one or, many). If, 3, 5, 2, ∆ ≡ 1, k, 5 = 0, , , k − 12 −14, ⇒, , 2 (− 14k + 60) − 3 (− 14 − 5k) + 5 (− 12 − k2) = 0, , ⇒, , 5k2 + 13k − 102 = 0, , ⇒, , (5k − 17) (k + 6) = 0, , 17, ⇒ k = − 6,, 5, b, a, a + b + 2c, 11. Let ∆ = , 2a + b + c, b, c, , a, a + 2b +, c, , , , , , c, , Applying C1 → C1 + C 2 + C3, , (use operations R2 → R2 − R1 ; R3 → R3 − R1 ), α, 1, x, β − x x − α 0 = 0, , , β − x γ − α 0, , Expand with respect to C3, ⇒, , (β − x)(γ − α ) − (x − α )(β − x) = 0, , ⇒, , (β − x) {(− α + γ ) − (x − α )} = 0, , ⇒, , (β − x) { −α + γ − x + α } = 0, , ⇒, , (β − x)(γ − x) = 0, , ⇒, , ⇒, , b, c, 1, , , a − b = 0, (− x) 0 b − x − c, , , c − x − b, a−c, 0, , x = β, γ, , ∴ Roots of the given equation are independent of α., 8. Q | A | = 8 and A is a square matrix of order 3., (Q |adj ( A )| = | A |n − 1 when A have order n), ∴ |adj ( A )| = 83 − 1 = 82 = 64, b , c, a − x, 9. Given that a + b + c = 0 and c, a = 0, b−x, , , c − x, a, b, Applying C1 → C1 + C 2 + C3, ⇒, , b , c, a + b + c − x, a + b + c − x b − x, a = 0, , , c − x, a, a + b + c − x, , ⇒, , b , c, 1, a = 0, (− x)1 b − x, , , c − x, a, 1, , b, a, 2 (a + b + c), , b, = 2 (a + b + c) 2a + b + c, , a + 2b +, a, 2 (a + b + c), , , , , c, , b, a, 1, , , b, ∆ = 2 (a + b + c) 1 2a + b + c, , , a, +, b, +, c, a, 1, 2, , , 0, 0 − (a + b + c), , = 2 (a + b + c)0 (a + b + c) − (a + b + c) , , , a, a + 2b + c , 1, (R1 → R1 − R2, R2 → R2 − R3 ), 0, 0 − 1, 1, = 2 (a + b + c)3 0, −1, , a a + 2b +, 1, , , = 2 (a + b + c)3, , c, , i , 1 + i 1 − i, 12. Let ∆ = 1 − i, i, 1 + i, , , 1 + i 1 − i, i, Applying C1 → C1 + C 2, C 2 → C 2 + C3, i , 1, 2, 1 + 2 i 1 + i, = 1, , , 2, 1 − i, 1 + 2 i, = 2 (1 − i + 2 i + 2) − 1 [1 − i (1 + 3i − 2)], + i [2 − (1 + 4 i − 4)], = 2 (3 + i ) − 1 (−4i + 2) + i (5 − 4i ), = 8 + 11i
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139, , Determinant, a, b + c, , 13., c+ a, b, , c, c, , , − 2b, b , , a + b, [R1 → R1 − (R2 + R3 )], = 2c ⋅ b (a + b − c) − 2b ⋅ c (b − c − a ), = 2abc + 2b2c − 2bc2 − 2b2c + 2bc2 + 2abc = 4abc, a, b, a+, , 0 − 2c, = b c + a, , c, b c, , 14. Ratio of cofactor to its minor of the element − 3, which is, in the 3rd row and 2nd column = (− 1)3 + 2 = − 1, b + c a + b a , 15. Let ∆ = c + a b + c b , , , a + b c + a c , 2, 1, 2, , (R1 → R1 + R2 + R2), ∆ = (a + b + c) c + a b + c b, , , a, b, c, a, c, +, +, , , 0, 1, 0, C1 → C1 − C 2 , = (a + b + c)a − b c − b b, , , , , C 2 → C 2 − 2C3 , −, −, b, c, a, c, c, , , = (a + b + c) {a 2 + b2 + c2 − ab − bc − ca}, = a3 + b3 + c3 − 3abc, 16. It has a non-zero solution, if, , ⇒, ⇒, , k −1 , 1, 3 − k −1 = 0, , , 1, 1 −3, 1(− k − 3) − k (3 + 1) − 1 (− 9 + k) = 0, − 6k + 6 = 0 ⇒ k = 1, , b2 − ab b − c bc − ac, , , 17. Let ∆ ≡ ab − a 2 a − b b2 − ab= 0, , bc − ac c − a, ab − a 2, , b b − c c, ⇒, ∆ = (b − a ) (b − a ) a a − b b, , , c c − a a, b b c , = (a − b)2a a b, (C 2 → C 2 + C3 ), , , c, c, a, , , =0, (Q two columns are same), 18. Given, a −1 + b−1 + c−1 = 0, …(i), +, 1, a, 1, 1, , , 1, 1+ b, 1 = λ, , , 1, 1 + c, 1, ⇒ Expand with respect to R1, ⇒ (1 + a ) {(1 + b)(1 + c) − 1} − 1 {1 + c − 1}, + 1 {1 − 1 − b} = λ, ⇒, (1 + a ) { b + c + bc} − c − b = λ, ⇒ b + c + bc + ab + ac + abc − c − b = λ, ⇒, bc + ab + ac + abc = λ, 1 1 1, ⇒, abc + + + abc = λ, a b c, −1, ⇒, abc {(a + b−1 + c−1 ) + 1} = λ, ⇒, abc (0 + 1) = λ[from Eq. (i)], ⇒, λ = abc, , x+2, x+5, x + 10, , x + 1, 19. x + 3, , x + 7, , x+4 , x+8 , , x + 14 , , (use operations, C 2 → C 2 − C1 ;C3 → C3 − C1 ), x + 1 1 3, = x + 3 2 5 , , , x + 7 3 7, x + 1 1 3, 1 2, = 2, , , 1 2, 4, (use operations, R3 → R3 − R1; R2 → R2 − R1), = (x + 1)(0) − 1(4 − 8) + 3(2 − 4) = 4 − 6 = − 2, x 4 5, 20. 7 x 7 = 0, , , 5 8 x, Expand with respect to R1, ⇒, x(x2 − 56) − 4(7x − 35) + 5(56 − 5x) = 0, ⇒, x3 − 56x − 28x + 140 + 280 − 25x = 0, ⇒, x3 − 109x + 420 = 0, ⇒, (x − 5)(x − 7)(x + 12) = 0, ⇒, x = − 12, c, a b, 21. Q l m n = 2, , , p q r, c, 6a 3b 15c , a b, , , , ∴ 2l m 5n = 30 l m n = 30 × 2 = 60, , , , , q, 5r , 2 p, p q r, 22. By property of determinant, if two row/ column of a, determinant are identical to each other, then the value of, determinant should be zero., a b c, a a x, e. g. , a, x, , b c = b, , b, , y =0, , y z, , c, , z, , c, , ω, x + 1, ω2 , 23. Let ∆ = ω, x + ω2, 1 , , , 2, 1, x + ω, ω, x + 1 + ω + ω2, ω, ω2 , , , 2, 2, = x + 1 + ω + ω x + ω, 1 , , 2, x + ω, 1, , x + 1 + ω + ω, (C1 → C1 + C 2 + C3 ), ω, 1, ω2 , 1 , (Q1 + ω + ω 2 = 0), = x1 x + ω 2, , , 1, x + ω, 1, 2, = x [1 {(x + ω ) (x + ω ) − 1} + ω {1 − (x + ω )}, + ω 2 {1 − (x + ω 2)}], 2, 2, 3, = x (x + ωx + ω x + ω − 1 + ω − ωx, − ω 2 + ω 2 − ω 2x − ω 4 ), 3, =x, (Qω3 = 1)
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142, , NDA/NA Mathematics, 2, 1 0, , and B = , 3, 1 0, 2 1 0 1 + 2 0 3 0, , =, =, , 3 1 0 2 + 3 0 5 0, 3 0, Then,|AB | =, =3 ×0 −5 ×0 =0, 5 0, , 1, 43. Given, A = , 2, 1, Now, AB = , 2, , 8 −5 1, 44. Given that, 5 x 1 = 2, 6 3 1, Using operations; R2 → R2 − R1 , R3 , → R3 − R1, 8, −5 1, −3 x+ 5 0 =2, −2, , 8, , 3 5 13, and ∆z = 4 1 6 , , , 2 9 20, , ∴, , 1, x −3i, 48. We have, y, 1, i = 6 + 11i, 2i −i, 0, ⇒, , 0, , 1 (− 24 + 2x + 10) = 2, 2x − 14 = 2, 2x = 16, x=8, 1 x y + z , 45. Let ∆ = 1 y z + x, , , 1 z x + y, On applying C3 → C3 + C 2, 1 x 1 , 1 x x + y + z , ⇒, ∆ = 1 y x + y + z = (x + y + z )1 y 1, , , , , 1 z 1 , 1 z x + y + z , = (x + y + z ) × 0 (Q two columns are identical), =0, , ⇒, ⇒, ⇒, ⇒, , 1 + x3, , 46. 1 + y3, , 3, 1 + z, ⇒, , ⇒, ⇒, ⇒, , x2 x , , y2 y = 0, , z2 z , 1 x2 x x3, 3, , 2, 1 y y + y, 1 2 z 3, z, z, , x2 x , , y2 y = 0, , z2 z , , x2 x, , y2 + y, , z 2 z, , x2 x3, , y2 y3 = 0, , z 2 z3, , 1 x, , 1 y, 1 z, , , (x − y) ( y − z ) (z − x) (1 + xyz ) = 0, xyz + 1 ⇒ xyz = – 1., , 47. We have, 3x + 5 y − 7z = 13, 4x + y − 12z = 6, and, 2x + 9 y − 3z = 20, ∆x, ∆y, ∆z, ∴, x=, , y=, ,z =, ∆, ∆, ∆, where, , 3 5 − 7, ∆ = 4 1 − 12, , , −3 , 2 9, = 3 (− 3 + 108) − 5 (− 12 + 24) − 7 (36 − 2), = 315 − 60 − 238 = 17, , = 3 (20 − 54) − 5 (80 − 12) + 13 (36 − 2), = − 102 − 340 + 442 = 0, ∆z 0, z=, =, =0, ∆ 17, , x (− i − 2 i 2) + 3 i (− yi ) + 1 (2 yi ) = 6 + 11i, , ⇒, , x (− i + 2) + 3 y + 2 yi = 6 + 11i, , ⇒, , (2x + 3 y) + i (− x + 2 y) = 6 + 11i, , On comparing real and imaginary parts, we get, and, , 2x + 3 y = 6, , ...(i), , − x + 2 y = 11, , ...(ii), , On solving Eqs. (i) and (ii), we get, x = − 3 and y = 4, x, y, 1, 1, , 1, 3, 2, 3, 2, , , , 49. x, 5y 3, 5y, 9 = 3xy x, , , , 3, 5 27, 2, 4, x 10 y, , x 10 y 9, , , , , , , (taking common x from C1 and y from C 2 and 3 from C3 ), 1, 1, 0, 2, , = 3xy x − 3 5 y, (C1 → C1 − C3 ), 3, 2, , 4, −, x, y, 9, 10, 9, , , 1, , 1, 0, , x−3, = 3xy, 5 y2 3, , , 4, (x + 3) (x − 3) 10 y 9, 1, 1, 0, 2, , = 3xy (x − 3) 1, 5y, 3, , , 4, x + 3 10 y 9, Hence, x − 3 is a factor of given determinant., 50., , x2, , −2 x, , 2, 0, , ω, ω, , ⇒, , −2ω 2, −ω = 0, 1, x2, 2, , −2x − 2ω 2, 0, , 0, , 1+ω, , x, 2, , −2x − 2ω, 0, , 0, , −ω 2, , 2, , ⇒, , (use operation, C 2 → C 2 + C3 ), −2ω 2, −ω = 0, 1, 2, , −2ω 2, −ω = 0, 1, (Q 1 + ω + ω 2 = 0 and ω3 = 1), , Expand with respect to R3, x2 −2ω 2, x2, ⇒ ω2, +1, 2, −ω, 2, ⇒, , −2x − 2ω 2, =0, 0, , ω 2 (− ωx2 + 4ω 2) + ( 4x + 4ω 2) = 0
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143, , Determinant, ⇒, ⇒, ⇒, ⇒, ⇒, , − x2 + 4ω + 4x + 4ω 2 = 0, − x2 + 4ω + 4x − 4 − 4ω = 0, − x2 + 4x − 4 = 0 ⇒ x2 − 4x + 4 = 0, (x − 2)2 = 0, x=2, , 51. Q a, b, c are in GP., ∴, b2 = ac, b, a, Now,, c, b, a+b, , b+ c, , − a 2 ab ac , 53. Given that, ab − b2 bc = ka 2b2c2, , , bc − c2, ac, ⇒, , a+b, b+ c, 0, , ⇒, , Expanding along R1,, c, b+ c, b, =a, −b, b+ c, 0, a+b, , b+ c, 0, b, + (a + b), a+b, , c, b+ c, , = − a (b + c)2 + b (a + b)(b + c), + (a + b)(b2 + bc − ac − bc), 2, 2, = − a (b + c + 2bc) + b (ab + ac + b2 + bc), (Q b2 = ac), 2, 2, 2, 2, = − ab − ac − 2abc + ab + 2abc + b c, (Q b2 = ac), 2, 2, 2, 2, = − ac + b c = − ac + ac = 0, (Q b2 = ac), 1 1 1 1, 52. Let ∆ = x y 1= x, , , 2, 2, 2, x y 1 x, , 0, y−x, y2 − x 2, , 0 , 1 − x, , 1 − x2, , 1, , 0, 0, , , y−x, = x, 1−x, , , 2, x ( y − x) ( y + x) (1 − x) (1 + x), 1, = ( y − x) (1 − x)x, , 2, x, , 0 , 1 , , y + x 1 + x, 0, 1, , − a 2 ab ac , ab − b2 bc = ka 2b2c2, , , 2, ac bc − c , −a, abc a, , a, , b, c, c = ka 2b2c2, −b, , b − c, , 1, 1, − 1, , 1 = ka 2b2c2, ⇒ (abc) (abc) 1 − 1, , , 1 − 1, 1, ⇒ a 2b2c2 [(− 1) (1 − 1) − 1 (− 1 − 1) + 1 (1 + 1)] = ka 2b2c2, ⇒ 4a 2b2c2 = ka 2b2c2 ⇒ k = 4, x + 2 x + 3 x + a , 54. Let A = x + 4 x + 5 x + b , , , x + 6 x + 7 x + c , Applying C 2 → C 2 − C1 , we get, x + 2 1 x +, A = x + 4 1 x +, , x + 6 1 x +, Applying R2 → R2 − R1, x + 2 1, 0, ⇒, A = 2, , 0, 4, , , and R3 → R3 − R1, x + a, b − a = − 1 (2c − 2a − 4b + 4a ), , c−a, , = ( y − x) (1 − x) (1 + x − y − x), , ⇒, A = 2 (2b − c − a ), Since, a , b, c are in AP., a+c, i.e.,, b=, 2, , = ( y − x) (1 − x) (1 − y) = (x − 1) ( y − 1) ( y − x), , ∴, , Thus, factor of determinant are x − 1, y − 1 and y − x., , a, b, , c, , A = 2 (a + c − c − a ) = 0, , ...(i)
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144, , NDA/NA Mathematics, , Level II, 2r−1 2 ⋅ 3r − 1 4 ⋅ 5r − 1, 1. We have, Dr = x, y, z , , , n, n, n, 2 − 1 3 − 1 5 − 1, , ⇒, , ⇒, , n r −1, Σ 2, r = 1, n, Σ Dr = x, r =1, , n, 2 −1, , , n, , Σ 2 ⋅ 3r − 1, , r =1, , y, 3 −1, n, , n, , Σ 4 ⋅ 5r − 1, , r =1, , z, , n, 5 −1 , , , 2n − 1 3n − 1 5n − 1, n, y, z ⇒ Σ D r = 0, Σ Dr = x, r =1, , r =1, n, n, n, 2 − 1 3 − 1 5 − 1, (Q two rows are same), n, , 2. Since, the system of equations ax + y + z = 0,, x + by + z = 0 and x + y + cz = 0, has a non-trivial solution, a 1 1, 1 b 1 = 0, ⇒, , , 1 1 c , 1 , 1, a, R2 → R2 − R1 , 1 − a b − 1, 0 = 0, ⇒, , , , , R3 → R3 − R1, 0, c − 1, 1 − a, ⇒ a (b − 1) (c − 1) − 1(1 − a ) (c − 1) − 1 (1 − a ) (b − 1) = 0, On dividing by (1 − a ) (1 − b) (1 − c), we get, a, 1, 1, +, +, =0, 1−a 1−b 1−c, 1, 1, 1, −1 +, +, =0, ⇒, 1−a, 1−b 1−c, 1, 1, 1, ⇒, +, +, =1, 1−a 1−b 1−c, x b b, 3. ∆1 = a x b = x3 − 3abx, , , a a x, d, x b, = x2 − ab, ⇒, ∆1 = 3 (x2 − ab) and ∆ 2 = , dx, a x, d, ∴, (∆1 ) = 3 (x2 − ab) = 3∆ 2, dx, a − b b + c a , b − c c + a b, 4., , , c − a a + b c, a − b b + c a + b + c, b − c c + a a + b + c, ⇒, , , c − a a + b a + b + c, (use C3 → C3 + C 2), 1, +, −, b, c, a, b, , , ⇒ (a + b + c) b − c c + a 1, , , c − a a + b 1, (use R2 → R2 − R1 and R3 → R3 − R1 ), , ⇒, , a − b, (a + b + c) 2b − a − c, , b + c − 2a, , b+ c, a−b, a−c, , 1, 0, , 0, , Expand with respect to C3, ⇒ (a + b + c) {(a − c)(2b − a − c) − (a − b)(b + c − 2a )}, ⇒ (a + b + c) {2ab − a 2 − ac − 2bc + ac + c2, − ab − ac + 2a 2 + b2 + bc − 2ab}, 2, 2, ⇒ (a + b + c) (a + b + c2 − ab − bc − ca ), ⇒ (a3 + b3 + c3 − 3abc), p − q 0, 5. 0, p, q = 0, , , 0, p, q, , Expand with respect to R1, ⇒ p( p2 − 0) + q(0 − q2) + 0 = 0 ⇒ p3 − q3 = 0, ⇒, ( p − q)( p2 + q2 + pq) = 0, ⇒, p − q = 0 and p2 + q2 + pq = 0, ⇒, , p=q, , and, , p2, pq, + 1 + 2 =0, q2, q, 2, , ⇒, , p, p, p, = 1 and + + 1 = 0, q, q, q, , We conclude that, p, is one of the cube roots of unity., q, 6. If A is an n × n matrix, then, det (λA ) = λn det ( A ), But, , det (λA ) = λs det ( A ), On comparing ⇒ s = n, , 7. We have, ai2 + bi2 + ci2 = 1, and, , aia j + bib j + cicj = 0 for (i = 1, 2, 3), 2, , ∴, , a1 a 2 a3 a1 b1 c1a1 b1 c1, b1 b2 b3 = a 2 b2 c2a 2 b2 c2, , , , , c1 c2 c3 a3 b3 c3a3 b3 c3, [Q| A| = | A′|], 1 0 0, = 0 1 0 = 1, , , 0 0 1, , 8. Since, the first column consists of sum of two terms,, second column consists of sum of three terms and third, column consists of sum of four terms., ∴, n = 2 × 3 × 4 = 24, a b aα − b, 9. Given that, b c bα − c = 0, , , 0 , 2 1, ⇒ a [− (bα − c)] − b [− 2(bα − c)]+ (aα − b) (b − 2c) = 0, ⇒ − abα + ac + 2b2α − 2bc + abα − 2acα − b2 + 2bc = 0, ⇒, , ac + 2b2α − 2acα − b2 = 0
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145, , Determinant, ⇒, , (ac − b2) − 2α (ac − b2) = 0, , ⇒, ⇒, Q, , b2 = ac, i. e. , a , b, c are in GP., 1 11 11 × 12, 10 ! 11 ! 12 !, , , 10. Let ∆ = 11 ! 12 ! 13 ! = 10 ! 11 ! 12 !1 12 12 × 13, , , , , 1 13 13 × 14, 12 ! 13 ! 14 !, , x, , 0, −z + y, z−x, , ⇒, , 14., , (x + y + z )(z − x)(− z + y − x − y + 2z ) = 0, , ⇒, x + y = − z or, sin 10° − cos 10°, sin 80°, , z=x, , cos 80°, = sin (10° + 80° ), , = − (a3 + b3 + c3 − 3abc), = − (a + b + c) (a 2 + b2 + c2 − ab − bc − ca ), 1, = − (a + b + c) [(a − b)2 + (b − c)2 + (c − a )2], 2, Since, a , b, c are positive, therefore the above, determinant is negative., 3r, 2a, x, a r x, 12. Given, A = 4b, 6s 2 y = λ b s y, −2 c −3 t − z, c t z, , = sin 90°, =1, [Q sin ( A + B) = sin A ⋅ cos B + cos A ⋅ sin B], 15. The system of homogeneous equations are, x − cy − bz = 0, cx − y + az = 0, and bx + ay − z = 0,, has non-trivial solution (since x, y, z are not all zero)., 1 − c − b, If,, ∆ ≡ c −1 a = 0, , , b a −1 , ⇒, , 1 (1 − a 2) + c (− c − ab) − b (ca + b) = 0, , ⇒, , a 2 + b2 + c2 + 2abc = 1, y, x, 1, , , 16. Let ∆ = 2 sin x + 2x sin y + 2 y , , , 3 cos x + 3x cos y + 3 y , y , x, 1, = 0 sin x sin y , , , 0 cos x cos y, , Taking 2 common from C1 and 3 from C 2 from LHS, a r x, x, r, a, ∴, 2 × 3 2b 2s 2 y = λ b s y, c t z, −c −t −z, , = sin x cos y − cos x sin y, , Taking 2 common from R2 and –1 from R3 in LHS, , = sin (x − y), , a, , r, , x, , a, , r, , x, , −12 b, , s, , y =λ b, , s, , y, , c, , t, , z, , t, , z, , c, , On comparing ⇒ λ = − 12, y x y+ z, 13., z y x+ y =0, x z z+x, ⇒, , ⇒, , x+ y+ z, z, , x+ y+ z, y, , x, , z, , 2(x + y + z ), =0, x+ y, z+x, , 1, (x + y + z ) z, x, , (Q R1 → R1 + R2 + R3 ), 1, 2, y x+ y =0, z, , z−x, , = sin 10° cos 80° + sin 80° cos 10°, , = (10 ! 11 ! 12 !) (50 − 48) = 2 ⋅ (10 ! 11 ! 12 !), a b c, 11. Let ∆ = b c a, , , c a b, = a (bc − a 2) − b (b2 − ac) + c (ab − c2), , ∴, , 0, x + y − 2z = 0, , (Q C 2 → C 2 − C1 , C3 → C3 − 2C1), Expand with respect to R1, −z + y x + y − 2z, ⇒, (x + y + z ), =0, z−x, z−x, , So,, , Applying R2 → R2 − R1 and R3 → R3 − R1, 1 11 11 × 12, 24 , = 10 ! 11 ! 12 ! 0 1, , , 50 , 0 2, , 1, (x + y + z ) z, , ⇒, , ac − b = 0 or 1 − 2α = 0, 1, b2 = ac or α =, 2, 1, (as given in question), α≠, 2, 2, , z+x, , R2 → R2 − 2R1 , , , R3 → R3 − 3R1, , N, 1, 5 , Σ n, n = 1, , N, , N, 17. Σ U n = Σ n 2 2N + 1 2N + 1, n =1, n = 1, , N 3, , , Σ n, 3N , 3N 2, n = 1, , N (N + 1), 1, 5, 2, N (N + 1) (2N + 1), 2N + 1 2N + 1, =, 6, 2, N (N + 1), 3N 2, 3N, , , , , 2, =, , , 5 , 1, 6, N (N + 1) , 2N + 1 2N + 1, 4N + 2, , , 12, 3N , 3N 2, 3N (N + 1)
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147, , Determinant, , ⇒, , ⇒, Take,, , 1, (a + b)1, , 1, 1, Either a + b = 0 or 1, , 1, , b 0, a b = 0, , 0 a, b 0, a b = 0, , 0 a, a, a + b = 0 ⇒ −1 =, b, , ∴, , ac, − a 2 ab, 32. Let ∆ = ab − b2 bc, ac, bc − c2, Taking common a , b and c from rows R1 , R2, R3 ,, −a b, c, respectively. ∆ = abc a − b c, a, b −c, , 26. Q | An × n | = 3 and|adj A | = 243, We know that,, |adj ( A )| = | An × n |n − 1, ⇒, 243 = 3n − 1 ⇒ 35 = 3n − 1, On comparing, ⇒, n −1 =5 ⇒ n =6, 27. Q Order of A and B are 3 × 2 and 2 × 3, respectively., ∴, | kAB| = k3 | AB| (Q order of AB is 3 × 3), 28. If the two lines a1x + b1 y = c1 and a 2x + b2y = c2 have no, a, b, c, solution, then 1 = 1 ≠ 1, a 2 b2 c2, 2 3 4, For the lines 2x + 3 y = 4 and 4x + 6 y = 7 ⇒ = ≠, 4 6 7, So, the given system of equation have no solution., 29. Given equations are x + y + z = 7, x + 2 y + 3z = 16, and, x + 3 y + 4z = 22, 1 1, ∴, ∆ = 1 2, , 1 3, ∆′, Value of y is given by, ∆, 1, 7, 1, , 1 − 7, , where ∆′ = 1 16 3 = − 1 − 16, , , 1 22 4 1 − 22, , ...(i), ...(ii), …(iii), 1, 3, , 4, , 30. We have, 2x + 3 y = 5 and 10x + 15 y = 50, 5 3, 2 3 , , = − 75, = 0, D1 = , ∴, D =, 50 15, 10 15, , 1, , 1, , −1, , ∆ = a b c { − 1 (1 − 1) − 1 (− 1 − 1) + 1 (1 + 1)}, 2 2 2, , ∆ = a 2b2c2 (0 + 2 + 2) = 4a 2b2c2, 33. Given, equation of system,, (k + 1)x + 8 y = 4k, and, kx + (k + 3) y = 3k − 1, Here,, a1 = k + 1, b1 = 8, c1 = 4k,, a 2 = k1 , b2 = k + 3, c2 = 3k − 1, The system of equations have infinitely many, solutions, if, a1 b1 c1, =, =, a 2 b2 c2, 8, 4k, k+1, i.e.,, =, =, k, k + 3 3k − 1, , 34. The given system of equations are, p3 x + ( p + 1)3 y = ( p + 2)3, px + ( p + 1) y = p + 2, x+ y=1, This system is consistent, if, , 2 5 , = 50, D2 = , 10 50, Since, D = 0 and D1 ≠ 0, D2 ≠ 0, ∴The system is inconsistent., −5, − 4, 31. We have, A = , 2, 2, , ⇒, , − 4 − 5, = − 8 + 10 = 2, | A| = , 2, 2, , We know that, AA − 1 = I, ∴, | AA − 1| = |I| = 1, ⇒, | A|| A − 1| = 1, , Again, taking common a , b and c fdrom column, C1 , C 2, C3 , respectively., 1, −1 1, ∆ = a 2b2c2 1 − 1 1, , …(i), …(ii), , ⇒, 8(3k − 1) = 4k(k + 3), ⇒, 24k − 8 = 4k2 + 12k, 2, ⇒, 4k − 12k + 8 = 0, ⇒, k2 − 3k + 2 = 0, ⇒, (k − 1)(k − 2) = 0, ⇒, k = 1, 2, ∴ The number of values of k = 2., , 1, 3, , 4, , −7 , 1 1, ∆′ = 1 3 −16, , , 1 4 −22, , ⇒, , 1, | A|, 1, | A − 1| =, 2, , | A − 1| =, , ⇒, , p3, , ( p + 1)3, , ( p + 2)3, , p, 1, , ( p + 1), , ( p + 2) = 0, 1, , p3, p, , ( p + 1)3 − p3, 1, , 1, , 0, , 1, , …(i), …(ii), …(iii), , ( p + 2)3 − p3, 2, =0, 0, (Q C 2 → C 2 − C1 , C3 → C3 − C1 ), , [Q |I| = 1], , Expanding along R3, ⇒, 2( p + 1)3 − 2 p3 − ( p + 2)3 + p3 = 0
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8, , Binomial Theorem, Binomial Expression, An algebraic expression consisting of two terms with, positive or negative sign between them is called a binomial, expression., q 1, 4, p, e.g., ( a + b), ( 2x − 3 y ), 2 − 4 , + 3 etc., x, x x y , , Binomial Theorem for Positive, Integral Index, If n is a positive integer and x , a ∈ R, then, ( x + a )n = n C0x n + n C1x n − 1a + n C2x n − 2a 2 + ... + n Cn a n, , and ( x − a )n = nC0x n − n C1x n − 1a + n C2x n − 2a 2 − ..., , Independent Term or Constant, Term, Independent term or constant term of a binomial, expansion is the term in which exponent of the variable is, zero., , Properties of Binomial Expansion, 1. Number of terms in the expansion of ( x + a )n is, ( n + 1)., 2. The sum of indices of x and a in each term is n ., 3. The coefficients of the terms equidistant from the, beginning and the end in the binomial expansion, are, equal., 4. (i) ( x + a )n + ( x − a )n = 2 [n C0 x n + n C2 x n − 2 a 2 + ... ], (ii) ( x + a )n − ( x + a )n = 2 [n C1 x n − 1 a, , + n C3 x n − 3 a3 +... ], , + ( − 1)n Cn a n ., Coefficients, , n, , C0 , n C1 , n C2 ,... , n Cn, , are known as, , binomial coefficients., and, , n, , Cr =, , n!, r !( n − r ) !, , General Term, Let (r + 1) th term be the general term in the expansion, of ( x + a )n ., ∴, %, , Tr + 1 = n Cr x n − r a r, , +1, , = nC r x n − r a r (− 1) r ., , %, , General term in the expansion of (1 + x ) n is T r, , %, , General term in the expansion of (1 − x ) n is, Tr, , %, , +1, , +1, , = nC r x r ., , = (− 1) r nC r x r ., , In the binomial expansion of (x + a) n , the pth term from the end is, (n − p + 2)th term from beginning., , 1. The 9th, 12, x 3a , − 2 is, a, x , (a) 12C 4 x12a 4 38, (c) 12C 4 x −12a 4 3−8, , term, , in, , the, , x, a, , 12 − 8, , of, , (b) 12C 4 x −12a −4 38, (d) 12C 4 x −12a 4 38, , Solution (d) 9th term in the expansion of −, x, = 12C 8 , a, , expansion, , 8, , 3a, , x2 , , 3a, 12, 8, − 2 = C4 ⋅ 3, x , , 12, , 4, , x a , 2, a x , , 8, , = 12C 4 x− 12a43 8, , General term in the expansion of (x − a) n is, Tr, , Example, , Example 2. The value of term independent of x in the, 10, , 1 , , expansion of 3x 2 − 3 is, , 2x , 76545, (a), 8, 76845, (c), 6, , (b), , 76545, 6, , (d) None of these
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151, , Binomial Theorem, , Solution (a) Let the value of (r + 1)th term is independent of x., r, , 1, Tr + 1 = 10C r (3x2)10 − r − 3 , 2x , 10, 10 − r 1, ( − 1) r x20 − 5r, = Cr ⋅ 3, 2r, This term is independent of x, ∴, 20 − 5r = 0 ⇒ r = 4, 4, 1, ∴, T5 = 10C 4 (3) 6 − , 2, ∴, , =, , n + 1, 2. If n is an odd number, then , th term and, 2 , n + 3, , th term will be middle term in the expansion of, 2 , ( x + a )n ., ∴, , T n + 3 = nCn + 1 x, , and, , 10 ! 3, 10 × 9 × 8 × 7 729 76545, ×, =, . 4 =, 4 !6 ! 2, 4 × 3 × 2 ×1, 16, 8, 6, , %, , ( x + x 2 − 1 )6 + ( x − x 2 − 1)6 is, (a) 32 x6 − 48 x 4 + 18 x 2 − 1, , 1⋅ 3 ⋅ 5...(2n − 1) n n, ⋅2 ⋅ x, n!, 2 ⋅ 4 ⋅ 6 ...(2n ) n 2n, (c), ⋅2 ⋅x, 2n !, , Solution (b) We know that,, , ( x + a) n + ( x − a) n = 2 [ nC 0 xn + nC 2 xn − 2 a2 + ... ], , = Tn + 1 =, , x2 − 1) 6 + ( x − x2 − 1) 6, , = 2 {x + C 2x ( x − 1) + C 4 x ( x − 1) + C 6 ( x − 1) }, 6, , 4, , 2, , 6, , 2, , 2, , 2, , 6, , 2, , 3, , = 2 (32x6 − 48x4 + 18x2 − 1), , 1 , is, 2 x, , (b) 469296 x3, (d) None of these, 1 , , , 2 x, , Solution (c) T7 = T6 + 1 = 13C6 ( 4x)13 − 6 −, = 13C 6 4 7 x7, , 6, , 1, 2 x, , 6 3, , = 13C 6 2 8 x4, , Middle Term in a Binomial, Expansion, n, , 1. If n is an even number, then + 1 th will be middle, 2, , term in the expansion of ( x + a )n ., ∴, , Tn, 2, , +1, , n +1, 2, , = Cn / 2 x, , n/ 2, , (b), , 1⋅ 3 ⋅ 5 ...(2n − 1) n, ⋅2, n!, , (d) None of these, , C n (1) 2n − n xn, , 2n, , C n ⋅ xn =, , 2n !, xn, (2n − n) ! n !, , 2n, , =, , 1⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ... (2n − 1) (2n) n, x, n !n !, , =, , [1⋅ 3 ⋅ 5 ... (2n − 1)][2 ⋅ 4 ⋅ 6 ... (2n)] n, x, n !n !, , =, , 1⋅ 3 ⋅ 5 ... (2n − 1) n ! 2 n ⋅ xn, n !n !, , =, , 1⋅ 3 ⋅ 5 ... (2n − 1) ⋅ 2 n ⋅ xn, n!, 12, , Example 6. The middle term in the expansion of + bx is, a, x, , , , 3 3, , = 439296 x4, , n, , ⋅a, , =, , 13, , , , , Example 4. The 7th term in the expansion of 4x −, (a) 469296 x 4, (c) 439296 x 4, , n −1, 2, , Solution (a) Middle term in the expansion of (1 + x) 2n, , n = 6, a = x2 − 1,, 6, , 2, , When there are two middle terms in the expansion, then the, binomial coefficients are equal., Binomial coefficient of middle term is the greatest binomial, coefficient., , (a), , (c) 32 x6 + 18 x 2 − 1, (d) None of the above, , (x +, , ⋅a, , Example 5. The middle term in the expansion of (1 + x) 2n is, , (b) 2(32 x6 − 48 x 4 + 18 x 2 − 1), , Let, , n−1, , 2, , 2, %, , n +1, x 2, , 2, , 2, , Example 3. The value of, , then, , T n + 1 = Cn − 1, n, , a, , n/ 2, , (a) 924a b, (b) 924a3b6, (c) 924a6 b3, (d) 924a6 b6, , Solution (d) The number of terms in the expansion of, a, , + bx, x, , , 12, , 12 , is 13 (odd). Its middle term is , + 1 th term., 2, , , a, ∴ Required term = T7 = T6 + 1 = 12C 6 , x, = 12C 6, , 12 − 6, , ( bx) 6, , a6 6 6 12, b x = C 6 a6 b 6 = 924 a6 b 6, x6
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152, , NDA/NA Mathematics, 3, = 2 9 ⋅9 C 6 , 2, , Greatest Term in the, Expansion of ( x + a ) n, , = 29 ⋅, , Let Tr and Tr + 1 be the rth and (r + 1) th term in the, expansion of ( x + a )n , then, n, n − r + 1 a , Tr + 1 , Cr x n − r a r, , =, =, , n, n − r + 1 r − 1, x , r, a , Tr Cr − 1 x, , Let, , 6, , 9 ⋅ 8 ⋅ 7 312, ⋅, 1⋅ 2 ⋅ 3 212, , Important Results, 1. Coefficient of( r + 1)th term in the expansion of(1 + x )n is n C r ., 2. Coefficient of ( r + 1)th term in the expansion of (1 − x )n is, , n − r + 1 a , ≥1, Tr, r, x , (n + 1)| a |, ⇒ (n − r + 1)| a | ≥ r |x| ⇒ r ≤, | x| + | a |, Tr + 1, , Q, , 6, , , 3 3, x = 2 9 ⋅9 C3 × , , 2 2, , n, , ( −1)r C r ., , ≥1 ⇒, , Properties of Binomial Coefficients, In the binomial expansion of (1 + x )n,, , ( n + 1)|a|, =I+f, |x| +|a|, , (1 + x )n = n C0 + n C1x + n C2x 2, , ( where, I is an integer and 0 ≤ f < 1), If f = 0, then Tr and Tr + 1 are same and greatest. If, 0 < f < 1, then Tr + 1 will be the greatest term., , Greatest Coefficient, (i) If n is even, then the greatest coefficient is, (ii) If n is odd, then the greatest coefficient are, , n, , Cn / 2 ., , n, , Cn + 1, 2, , and n C n + 3, , + ... + n Cr x r + ... + n Cn x n, Where, n C0 , n C1 , n C2 ... n Cn are the coefficients of, various powers of x are called binomial coefficients and, they are written as C0 , C1 , C2 , ... , Cn ., Hence,, (1 + x )n = C0 + C1x + C2x 2 + ... + Cr x r + ... + Cn x n ...(i), 1., , n, , C0 + n C1 + ... + n Cn = 2n, , 2., , n, , C0 + n C2 + n C4 + ... = n C1 + n C3 + ... = 2n − 1, , 3., , n, , C0 − n C1 + n C2 − n C3 + ... + ( − 1)n n Cn = 0, n, , 2, , 4., , Example 7. Numerically the greatest term in the, expansion of (2 + 3x) , when x = 3/ 2 is, 9 ⋅8 ⋅ 7, 9 ⋅ 8 ⋅ 7 312, (a) 29 ⋅ 312 ⋅, (b) 29 ⋅, ⋅, 1⋅ 2 ⋅ 3, 1⋅ 2 ⋅ 3 210, 9 ⋅ 8 ⋅ 7 312, (d) None of these, (c) 29 ⋅, ⋅, 1⋅ 2 ⋅ 3 212, 9, , 3, Solution (c) Since, (2 + 3x) 9 = 29 1 + x, , 9, , 2 , , , , 9, , 3 , , Now, in the expansion of 1 + x , we have, , 2 , Tr + 1 (9 − r + 1) 3 (10 − r) 3 3 , =, x =, (Q x = 3 / 2), ×, Tr, r, r 2 2 , 2 , 10 − r 9 90 − 9r, =, =, r 4, 4r, Tr + 1, 90 − 9r, ∴, ≥1⇒, ≥ 1 ⇒ 90 ≥ 13r, Tr, 4r, 90, 12, ⇒, r ≤, =6, 13, 13, 12, r ≤6, ∴, 13, ∴ Maximum value of r is 6., So, greatest term = 2 9 ⋅ T6 + 1, , 5., 6., , Cr, , =, , n −r +1, r, , n, , Cr − 1, , n, , Cr + n Cr − 1 =, , n +1, , Cr + 1 =, , n +1, , Cr, , n +1 n, ⋅ Cr, r +1, , 7., , n, , C1 + 2 n C2 + 3 n C3 + ... + n n Cn = n ⋅ 2n − 1, , 8., , n, , C1 − 2 n C2 + 3 n C3 − ... = 0, , 9., , n, , C0 + 2 n C1 + 3 n C2 + ... + ( n + 1) n Cn, , = ( n + 2) 2n − 1, ( 2n ) !, 10. C0 Cr + C1 Cr + 1 + ... + Cn − r Cn =, ( n − r ) !( n + r ) !, ( 2n )!, 11. C02 + C12 + C22 + ... + Cn2 =, ( n !)2, 12. C02 − C12 + C22 − C32 + ..., ,, 0, =, n/ 2 n, Cn / 2 ,, ( −1), , if n is an odd, if n is aneven, , Example 8. The sum of the coefficients of all the integral, powers of x in the expansion of (1 + 2 x ) 40 is, 1, (b) ⋅ 340, (a) 340 + 1, 2, 1, (c) 340 − 1, (d) (340 + 1), 2
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153, , Binomial Theorem, , Solution (d) The coefficients of the integral powers of x are, 40, , C0 ,, , C2 ⋅ 2 ,, , 40, , C 4 ⋅ 2 , ... ,, , 2 40, , 4, , Now, (1 + 2) 40 =, (1 − 2), , 40, , =, , C0 +, , 40, , C0 −, , 40, , C 40 ⋅ 2 ., , 40, , 40, , C1 ⋅ 2 +, , C1 ⋅ 2 +, , 40, , 5, C 2 ⋅ 2 2 + ..., + 40C 40 ⋅ 2 40, , 40, , C 2 ⋅ 2 − ... +, , 40, , ⇒ 40C 0 +, , C 2 ⋅ 22 +, , 40, , C 2 ⋅ 22 +, , 40, , −, , 1, 2 (1 +, , 40, , 2, , C 40 ⋅ 2, , 40, , C 4 ⋅ 2 4 + ... +, , 40, , C 4 ⋅ 2 4 + ... +, , 40, , ...(ii), , C 40 ⋅ 2 40), , 40, , C 40 ⋅ 2 40 =, , 40, , 1 40, (3 + 1), 2, , Number of Terms in the, Expansion of Trinomial and, Multinomial, ( a + b + c)n can be expanded as, , Important Results, 1. (1 + x )− n = 1 − nx +, , ( − 1)r n ( n + 1) ( n + 2) . . . ( n + r − 1) r, x + ..., r!, n( n + 1) 2 n( n + 1) ( n + 2) 3, 2. (1 − x )− n = 1 + nx +, x +, x + ..., 2!, 3!, n( n + 1) ( n + 2) . . . ( n + r − 1) r, +, x − ..., r!, n( n − 1) 2, 3. (1 − x )n = 1 − nx +, x −. . ., r!, n( n − 1) ( n − 2) . . . ( n − r + 1) r, . . . + ( − 1)r, x + ..., r!, + ... +, , 5. (1 − x )− 1 = 1 + x + x 2 + x 3 + . . . + x r + . . ., , = ( a + b)n + n C1( a + b)n − 1( c)1 + n C2( a + b)n − 2( c)2, + .... + n Cn cn, = ( n + 1) term + n term + ( n − 1) term + ... +1 term, ∴ Total number of terms, ( n + 1)( n + 2), = ( n + 1) + ( n ) + ( n − 1) +...+ 1 =, 2, Similarly, number of terms in the expansion of, ( n + 1)( n + 2)( n + 3), ( a + b + c + d )n =, 6, , 6. (1 + x )− 2 = 1 − 2 x + 3 x 2 − 4 x 3 + . . . + ( −1)r ( r + 1) x r + . . ., 7. (1 − x )− 2 = 1 + 2 x + 3 x 2 + 4 x 3 + . . . + ( r + 1) x r + . . ., , Example 10. The coefficient of xn in the expansion of, , (1 − 2 x + 3x 2 − 4x3 + ... ∞) − nis, (2n !), (2n) !, (b), (a), n!, (n !) 2, , n( n − 1) 2, x, 2!, n( n − 1) ( n − 2) 3, +, x, 3!, n( n − 1)... ( n − r + 1) r, x + ... terms upto ∞, + ... +, r!, , Suppose (1 + x )n = 1 + nx +, , when n is a negative integer or a fraction, where, −1< x < 1, otherwise expansion will not be possible., If first term is not 1, then make first term unity in the, n, y, y, , following way, ( x + y )n − x n 1 + , if, < 1., x, x, , n( n − 1)( n − 2)... ( n − r + 1) r, General Term Tr +1 =, x, r!, 1, , can be expanded by binomial, , theorem, if, (a) x<1, , (b) x < 1, , (c) x <, , 5, 4, , (d) x <, , 4, 5, , (c), , (n !) 2, (2n) !, , (d), , (n !), (2n) !, , Solution (b) We have, (1 − 2 x + 3x 2 − 4x3 + ... ∞) − n, , Binomial Theorem for any Index, , 5 + 4x, , n( n + 1) 2, n( n + 1) ( n + 2) 3, x −, x, 3!, 2!, , 4. (1 + x )− 1 = 1 − x + x 2 − x 3 + . . . + ( − 1)r x r + . . ., , ( a + b + c)n = {( a + b) + c} n, , Example 9., , 1, , 4 −2, 4, 5, x) and it is valid only when,, x < 1⇒ x <, 5, 5, 4, , ...(i), 40, , On adding Eqs. (i) and (ii), we get, 3 40 + 1 = 2 ( 40C 0 +, , Solution (c) The given expression can be written as, , ∴, , = [ (1 + x) − 2]− n = (1 + x) 2n, (2n) !, Coefficient of xn = 2nC n =, (n !) 2, , Example 11. If x is so small that its square and higher, powers may be neglected, then the value of, (8 + 3x) 2/3, is, (2 + 3x) ( 4 − 5 x)1/ 2, 5, 5, (a) 1 + x, (b) 1− x 2, 8, 8, 5, (d) None of these, (c) 1− x, 8, Solution (c), , (8 + 3x) 2/3, (2 + 3x) ( 4 − 5x)1/ 2, =, , 3 , , 8 2/3 1 + x, , 8 , , 2/3, , 3 5 , , 2 1 + x 2 1 − x, , 2 , 4 , , 3 , , = 1 + x, , 8 , , 2/3, , 3 , , 1 + x, , 2 , , 1/ 2, , −1, , 5 , 1 − x, , 4 , , − 1/ 2
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154, , NDA/NA Mathematics, 2 3, 1 5, , , 3, , = 1 + . x + ... 1 − x + ... 1 + ⋅ x + ..., , , , , , , 3 8, 2, 2 4, , Solution (a) (1 + 2 x)6 (1 − x)7 = [1 + 6C1 (2 x) + 6C 2 (2 x) 2, + 6C3 (2x)3 + 6C 4 (2x) 4 + 6C5 (2x)5 + 6C 6 (2x) 6 ], , 1 3 5, =1+ − + x, 4 2 8, 5, =1− x, 8, , × [1 − 7C1 x + 7C 2 x2 − 7C 3 x3 + 7C 4 x4 − ... ], = (1 + 12x + 60 x2 + 160 x3 + 240 x4 + 192x5 + ...), (1 − 7x + 21x2 − 35x3 + 35x4 − 21 x5 + ...), , Example 12. The value of, , ∴Coefficient of x5 in the expansion of (1 + 2x ) 6 (1 − x ) 7, , (1 + x + x + x + ... ) (1 − x + x + x – ... ) is, (a) 1+ x + x 2 + x3 + ... ∞, (b) 1+ x 2 + x 4 + x6 + ... ∞, (d) 1− x + x 2 − x3 + ... ∞, (c) 1− x 2 + x 4 − x6 + ... ∞, 2, , 3, , 2, , 3, , Solution (b) (1 + x + x 2 + x3 + ... ) (1 − x + x 2 − x3 + ... ), , = (1 − x) − 1 (1 + x) − 1 = [ (1 − x) (1 + x) ]− 1, = (1 − x2) − 1 = 1 + x2 + x4 + x6 + ... ∞, , = 1 × ( − 21) + 12 × 35 + 60 × ( − 35), + 160 × 21 + 240 × ( − 7) + 192, = − 21 + 420 − 2100 + 3360 − 1680 + 192 = 171, , Example 15. The approximate value of (1. 0002)3000 is, (a) 1.6, , = (1 + 0.0002), (3000)(2999), = 1 + (3000)(0.0002) +, (0.0002) 2 + ..., 1. 2, , 2, , We want to get answer correct only one decimal places and, as such, we have left further expansion., = 1 + (3000)(0 . 0002) = 1. 6, , (b) 14, (d) 16, , Example 16. If the coefficients of 2nd, 3rd, and 4th terms in, , 1 − x, (1 − x ), (d) , = (1 − x ) 2 (1 + x ) − 2, =, 1 + x, (1 + x ) 2, = (1 − 2x + x2) (1 − 2x + 3x2 − 4x3 + 5x4 − ... ), , the binomial expansion of (1+ x) n are in AP, then n2 − 9n is, equal to, (a) −7, (b) 7, (c) 14, (d) −14, , 2, , Solution, , 2, , 1 − x, ∴Coefficient of x4 in the expansion of , , 1 + x, , 2, , Solution (c) Coefficients of 2nd, 3rd, and 4th terms are, respectively nC1, nC 2 and nC3 , are in AP., , = 1 × 5 − 2 × ( − 4) + 1 × 3 = 5 + 8 + 3 = 16, ⇒, , Example 14. The coefficient of x5 in the expansion of, (1 + 2 x) (1 − x), (a) 171, (c) 191, 6, , (d) 1.2, , 3000, , Solution (a) (10002, ., ), , Example 13. The coefficient of x in the expansion of, , (c) 15, , (c) 1.8, , 3000, , 4, , 1 − x , , is, 1 + x , (a) 12, , (b) 1.4, , 7, , is, , On solving,, , (b) 181, (d) 201, , ⇒, , 2 ⋅ nC 2 = nC1 + nC3, 2n !, n!, n!, =, +, 2 !(n − 2) ! (n − 1) ! 3 !(n − 3) !, n 2 − 9n + 14 = 0, n 2 − 9n = 14, , Comprehensive Approach, n, , n, , n, n, n, , n, , n, , In any term of expansion of ( x + y) n , the sum of the exponents of, x and y is always constant =n., The binomial coefficients n C 0 , nC1 , nC 2 , .... equidistant from, beginning and end are equal i.e., n C r = nC n − r ., ( x + y) n = Sum of odd terms + Sum of even terms., The number of terms in the expansion of( x + y + z) n is n + 2C 2., Sum of coefficients of the expansion ( a + bx + cx2) n is, ( a + b + c) n . Coefficient of xm in the expansion of (1 + xr ) n , if m is, not a multiple of r. e.g., Coefficient of x1000 in the expansion of, (1 + x3 ) 2000 is 0 because 1000 is not a multiple of 3., If the value of x is so small that on neglecting higher powers and, square of x, we get (1 + x) n = 1 + nx., Coefficient of xn − 1 in the expansion of ( x − 1) ( x − 2)... ( x − n) is, n(n + 1), ., −, 2, , n, , n, , n, , n, , n, , Coefficient of xn − 1 in the expansion of ( x + 1) ( x + 2) ... ( x + n) is, n(n + 1), ., 2, If n is odd, then ( x + y) n + ( x − y) n and ( x + y) n − ( x − y) n both have, n + 1, the same number of terms equal to , ., 2 , n, If n is even, then ( x + y) n + ( x − y) n has + 1 terms and, 2 , n, ( x + y) n − ( x − y) n has terms., 2, If the coefficient of the rth, (r +1) th and (r + 2) th terms in the, expansion of (1+ x) n are in HP, then n + (n − 2r) 2 = 0 ., If the coefficient of r th, (r +1) th and (r + 2) th terms and in the, expansion of (1+ x) n are in AP , then n2 − n( 4r + 1) + 4r 2 − 2 = 0 .
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Exercise, Level I, 1. What is the sum of the coefficients of all the terms in, the expansion of ( 45x − 49)4?, (NDA 2010 I), (a) − 256, (c) 100, , (b) − 100, (d) 256, , 2. What is the approximate value of (1 .02)8?, (a) 1.171, (b) 1.175, (NDA 2008 I), (c) 1.177, (d) 1.179, 3. What is the coefficient of x y in ( 2x + 3 y ) ?, 3, , 4, , 2 5, , (NDA 2008 I), , (a) 240, , (b) 360, , (c) 720, , (d) 1080, , 4. What is the middle term in the expansion of, 12, x y, 3 , −, , ?, (NDA 2007 I), y x, 3, (a) C(12, 7) x3 y −3, (c) C(12, 7) x −3 y3, , (b) C(12, 6) x −3 y3, (d) C(12, 6) x3 y −3, , 5. What is the value of, 8, C0 − 8C1 + 8C2 − 8C3 + 8C4 − 8C5 + 8C6 − 8C7 + 8C8?, (NDA 2008 II), , (a) 0, (c) 2, , (b) 1, (d) 28, , 6. The expansion of (1 + 2x )− 1/ 2, by binomial theorem,, valid when, (a) x > 1 / 2, (b) x < 1 / 2, (c) − 1 / 2 < x < 1 / 2, (d) − 2 < x < 2, , 7. The middle term of 2x −, , 4, 2, (a) 10C4 4, 3, 24, (c) − 10C4 5, 3, , 10, , 1, is, 3x , (b) −, , 10, , C5, , (d) 10C5, , 25, 35, , 25, 35, , 8. The coefficient of the middle term in the expansion of, ( 2 + 3x )4 is, (a) 6, (b) 5 !, (c) 8 !, (d) 216, 9. The coefficients of x m and x n , where m and n are, positive integers, in the expansion of (1 + x )m + n are, (a) equal, (b) equal in magnitude but opposite in sign, (c) reciprocal to each other, (d) in the ratio m : n, 10. What are the values of k, if the term independent of x, 10, k, , in the expansion of x + 2 is 405?, , x , (a) ±3, (b) ±6, (c) ±5, (d) ±4, , 11. What is the sum of the coefficients in the expansion, of ( 5x − 4 y )100?, (a) 1, (b) −1, (d) −2100, (c) 5100, 4, , 12. What is the value of Σ, , i=0, , 4, , (a) Σ, , (57 − k), , 10, , (57 + k), , k= 0, , (c) Σ, , k= 6, , ( 46 + i ), , 4, , C5 + Σ, , (50 − j ), , j=0, , 10, , C5, , (b) Σ, , (57 − k), , 4, , C5, , (d) Σ, , (57 + k), , k= 6, , k= 0, , C4?, , C5, C5, , 13. How many terms are there in the expansion of, ( 4x + 7 y )10 + ( 4x − 7 y )10?, (a) 5, (b) 6, (c) 11, (d) 22, 14. The binomial coefficients which are in decreasing, order are, (b) 15 C10 , 15C9 , 15C8, (a) 15 C5 , 15C6 , 15C7, (c) 15 C6 , 15C7 , 15C8, , (d) 15 C7 , 15C6 , 15C5, , 15. Which one of the following statements is correct? The, natural number 610 − 51 is, (a) a prime number, (b) an even number, (c) divisible by 5, (d) a power of 3, 16. What is the coefficient of x n in ( x 2 + 2x )n − 1?, (b) ( n − 1) 2( n − 1), (a) ( n − 1) 2( n − 2), (d) n 2( n − 1), , (c) ( n − 1) 2n, , 17. What is the coefficient of x3 in, (a) −272, , (b) − 540, , (3 − 2 x), , ?, (1 + 3x )3, (c) − 870, (d) − 918, 10, , 1 , , 18. 6th term in expansion of 2x 2 −, is, , 3x 2 , 4580, 896, (b) −, (a), 17, 27, 5580, (c), (d) None of these, 17, 19. If n is even, then the middle term in the expansion of, n, 2 1, 6, x + is 924x , then n is equal to, , x, (a) 10, (b) 12, (c) 14, (d) None of these, 3, 2, 20. If the 4th term in expansion of x −, , 3, 2x , independent of x, then n is equal to, (a) 5, (b) 6, (c) 9, (d) None of these, , n, , is
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156, , NDA/NA Mathematics, , 21. If the coefficient of 7th and 13th term in the, expansion of (1 + x )n are equal, then n is equal to, (a) 10, (b) 15, (c) 18, (d) 20, 15, , 22. The value of ( 0.99), (a) 0.8432, (c) 0.8502, , is, (b) 0.8601, (d) None of these, , 23. If in the expansion of (1 + x ) , the coefficient of rth, and (r + 2) th term be equal, then r is equal to, 2n + 1, n, 2n − 1, (a) 2n, (b), (d), (c), 2, 2, 2, n, , 24. If the ratio of the coefficient of third and fourth term, n, 1, , in the expansion of x −, is 1 : 2, then the value of, , 2x , n will be, (a) 18, (b) 16, (c) 12, (d) –10, 12, , a, , 25. In the expansion of + bx , the coefficient of x − 10, x, , will be, (b) 12b11 a, (a) 12 a11, 11, (c) 12 a b, (d) 12 a11 b11, 8, , 1, , 26. In the expansion of x3 + 2 , then the term, , x , containing x 4 is, (a) 70x 4, (b) 60 x 4, 4, (c) 56 x, (d) None of these, 27. The total number of terms in the expansion of, ( x + a )100 + ( x − a )100 after simplification will be, (a) 202, (b) 51, (c) 50, (d) None of these, , 28. The expansion of, , 1, 3, , 6 − 3x, , is equal to, , , , x 2x 2, (a) 6 1/ 3 1 + + 2 + ..., 6, 6, , , , , , x 2x 2, (b) 6 −1/ 3 1 + + 2 + ..., 6, 6, , , , , , x 2x 2, (c) 6 1/ 3 1 − + 2 − ..., 6, 6, , , , , , x 2x 2, (d) 6 −1/ 3 1 − + 2 − ..., 6, 6, , , , 29. If p and q be positive , then the coefficients of x p and, x q in the expansion of (1 + x ) p + q will be, (a) equal, (b) equal in magnitude but opposite in sign, (c) reciprocal to each other, (d) None of the above, 30. The term independent of x in the expansion of, 10, x, 3 , will be, +, , , 3 2x 2 , (a) 3/2, (b) 5/4, (c) 5/2, (d) None of these, 31. The sum of the coefficients in the expansion of, (1 + x − 3x 2 )2163 will be, (a) 0, (b) 1, (c) –1, (d) 22163, 32. The coefficient of x − 7, in the expansion of, 11, 1 , , ax −, will be, , bx 2 , 462 a 6, 462 a5, (a), (b), 6, b, b6, 3, 462 a, (d) None of these, (c), b7, , Level II, 1. What is the term independent of x in the expansion of, 9, −2, 1, 3 3x, (NDA 2009 I), (1 + x + 2x ) , −, ?, 3x , 2, (a), (b), (c), (d), , 1/ 3, 19 / 54, 1/ 4, No such term exists in the expansion, , 2. For all n ∈ N , 24n − 15n − 1 is divisible by ), (NDA 2011 I), , (a) 125, (c) 450, , (b) 225, (d) None of these, , 3. What is the number of terms in the expansion of, ( a + b + c)n , n ∈ N ?, (NDA 2010 II), (a) n + 1, (b) n + 2, ( n + 1)( n + 2), (c) n( n + 1), (d), 2, , 4. What is the coefficient of x 4 in the expansion of, 2, 1 − x, (NDA 2010 II), , ?, 1 + x, (a) − 16, (c) 8, , (b) 16, (d) − 8, , 5. If x is so small that its square and higher powers may, 1/ 2, 1 − x, be neglected, then , is approximately equal to, 1 + x, (a) 1 − x, (b) 1 + x, (c) 1, (d) None of these, 6. If the coefficients of ( 2r + 4) th term and (r − 2) th, term in the expansion of (1 + x )18 are equal, then r is, equal to, (NDA 2009 I), (a) 6, (b) 5, (c) 4, (d) 2
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157, , Binomial Theorem, 7. The number of integral terms in the expansion of, ( 51/ 2 + 71/ 8 )1024 is, (a) 128, (b) 129, (c) 256, (d) 512, n, 2, , 8. What is the value of Σ, , r=0, , n, , C2r , if n is even?, , (b) 2n, (a) 2( n + 1), (c) 2( n − 1), (d) 2( n − 2), 1, 9. If|x|< , what is the value of, 2, x n( n + 1), 1+ n , +, , 1 − x 2 ! , 1− x , (a) , , 1 − 2x , , n, , 1 − 2x , (c) , , 1− x , , n, , Σ, , (a) 219 +, (c), , 20, , 1, ⋅, 2, , k= 0, , 2, , (b) (1 − x )n, n, , 10. If the coefficients of pth, ( p + 1) th and ( p + 2) th, terms in the expansion of (1 + x )n are in AP, then, (a) n 2 − 2np + 4 p2 = 0, (b) n 2 − n ( 4 p + 1) + 4 p2 − 2 = 0, (c) n 2 − n ( 4 p + 1) + 4 p2 = 0, (d) None of the above, 11. C0Cr + C1Cr + 1 + C2Cr + 2 + ... + Cn − r Cn is equal to, ( 2n ) !, n!, (a), (b), ( n − r ) !( n + r ) !, ( − r ) !( n + r ) !, n!, (c), (d) None of these, (n − r ) !, , 20, , Ck is, , 20, , (b) 219, , C10, , (d) None of these, , C10, , 16. The expansion of, , x , 1 − x + K ∞ ?, , , , 1 , (d) , , 1 − x, , 10, , 15. The value of, , 1, ( 4 − 3x )1/ 2, , be valid, if, (a) x < 1, 2, 2, (c) −, < x<, 3, 3, 17. The value of, (a) 0, (c) 15, , 15, , C02 −, , by binomial theorem will, , (b)| x|< 1, (d) None of these, 15, , 1, , C12 + 15C22 − ... −, (b) − 15, (d) 51, , 15, , 2, is, C15, , 1, , a 2 a 2, 18. The value of , +, , a + x, a − x, 3x 2, (a) 2 +, (b) 1 +, + ..., 4a 2, x 3x 2, (d) 2 −, (c) 2 + +, + ..., a 4a 2, , is, 3x 2, , + ..., 8a 2, x 3x 2, +, + ..., a 4a 2, , 19. (r + 1)th term in the expansion of (1 − x )− 4 will be, xr, (r + 1) (r + 2) (r + 3) r, (a), (b), x, r!, 6, (r + 2) (r + 3) r, (c), (d) None of these, x, 2, 20. If the coefficients of 5th , 6th and 7th terms in the, expansion of (1 + x )n be in AP, then the value of n is, (a)7 only, (b) 14 only, (c) 7 or 14, (d) None of these, , 12. The approximate value of ( 7 . 995)1/ 3 correct to four, decimal places is, (a) 1.9995, (b) 1.9996, (c) 1.9990, (d) 1.9991, , 21. In the expansion of (1 + x )n , what is the sum of even, binomial coefficients?, (NDA 2012 I), n, n −1, (a) 2, (b) 2, (c) 2n + 1, (d) None of these, , 13. rth term in the expansion of ( a + 2x )n is, n ( n + 1) ... ( n − r + 1) n − r + 1, (a), a, ( 2x )r, r!, n ( n − 1) ... ( n − r + 2) n − r + 1, (b), a, ( 2x )r − 1, (r − 1) !, n ( n + 1) ... ( n − r ) n − r, (c), a, ( x )r, (r + 1), , 22. If (1 + ax )n = 1 + 8x + 24x 2 + ... , then the value of a, and n, is, (a) 2, 4, (b) 2, 3, (c) 3, 6, (d) 1, 2, , (d) None of the above, 14. Sum of odd terms is A and sum of even terms is B in, the expansion ( x + a )n , then, 1, (a) AB = ( x − a )2n − ( x + a )2n, 4, (b) 2 AB = ( x + a )2n − ( x − a )2n, (c) 4 AB = ( x + a )2n − ( x − a )2n, (d) None of the above, , 23. The coefficient of the term independent of x in the, 9, 1, 3, expansion of (1 + x + 2x3 ) x 2 −, is, 2, 3x , 1, 19, (b), (a), 3, 54, 17, 1, (d), (c), 54, 4, 24. If for positive integers r > 1, n > 2 the coefficient of the, ( 3r )th and (r + 2)th powers of x in the expansion of, (1 + x )2n are equal, then, (a) n = 2r, (b) n = 3r, (c) n = 2r + 1, (d) None of these
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158, , NDA/NA Mathematics, , 25. If the coefficient of the second, third and fourth terms, in the expansion of (1 + x )n are in AP, then n is, equal to, (a) 7, (b) 2, (c) 6, (d) None of these, 26. The greatest integer when divides the number, 101100 − 1 is, (a) 10000, (b) 100000 (c) 1000, (d) 100, n, , 27. If (1 + x )n = Σ Cr x r , then, r=0, , , C1 , C2 , Cn , is equal to, ... 1 +, 1 +, 1 +, C0 , C1 , Cn − 1 , , n −1, , ( n + 1), ( n − 1) !, ( n + 1)n + 1, (d), n!, , n, !, ( n − 1), ( n + 1)n, (c), n!, , (b), , 28. If (1 − x + x 2 )n = a0 + a1 x + a2 x 2 + ... + a2n x 2n , then, a0 + a2 + a4 + ... + a2n is equal to, 3n + 1, 3n − 1, (a), (b), 2, 2, 1 − 3n, 1, (d) 3n +, (c), 2, 2, 29. The coefficient of, (1 + x + x 2 + x3 )n is, (a) n C4, (c) n C4 + n C3 + n C2, , x4, , in, , the, , expansion, , of, , (b) n C4 + n C2, (d) n C4 + n C2 + n C1 ⋅n C2, , 30. If the value of x is so small that x 2 and greater powers, can be neglected, then, 5, x, 6, 2, (c) 1 + x, 3, , (a) 1 +, , 1 + x + 3 (1 − x )2, 1+ x+ 1+ x, 5, (b) 1 − x, 6, 2, (d) 1 − x, 3, , is equal to, , 31. The value of the term independent of x in the, 9, 1, , expansion of x 2 − is, , x, (NDA 2012 I), (a) 9, (b) 18, (c) 48, (d) 84, 32. The coefficient of x53 in the following expansion, 100, , Σ, , (a), , (c) −, , C47, , 100, , C53, , Statement II Coefficient of (r + 1)th term in the, expansion of (1 − x )n is ( −1)n ( n − 1) Cr ., Which of the above statements is/are correct ?, (a) I only, (b) II only, (c) Both I and II, (d) Neither I nor II, 36. Consider the following statements, Statement I Total number of terms in the expansion, n+2, of ( x + a )n + ( x − a )n is, , if n is an odd number., 2, Statement II Total number of terms in the, n +1, expansion of ( x + a )n + ( x − a )n is, , if n is an, 2, even number., Which of the above statements is/are correct?, (a) I only, (b) II only, (c) Both I and II, (d) Neither I nor II, , Directions (Q. Nos. 37-41) Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b),(c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 5, , ∑ C( 52 − r , 3), , r =1, , (b) 100C53, (d) −, , 8, , 1, ., x, Statement II. The coefficient of the middle term in, the expansion of (1 + x )8 is less than the coefficient of, the fifth term in the expansion of (1 + x )7 ., Which of the above statements is/are correct?, (a) I only, (b) II only, (NDA 2009 II), (c) Both I and II, (d) Neither I nor II, , 37. Assertion (A) The value of C( 47, 4) +, , Cm ( x − 3)100 − m ⋅ 2m is, , 100, , m=0, 100, , , of x +, , , 35. Consider the following statements, Statement I Coefficient of (r + 1)th term in the, expansion of (1 + x )n is n Cr ., , n −1, , (a), , 34. Consider the following statements, Statement I. The coefficient of the middle term in, the expansion of (1 + x )8 is equal to the middle term, , 100, , C100, , 33. The value of x in the expression [x + x log10 ( x ) ]5 , if the, third term in the expansion is 1000000, is, (a) 10, (b) 11, (c) 12, (d) None of these, , is 52C4., Reason (R), , n, , Cr + n Cr + 1 =, , n +1, , Cr + 1., , 38. Assertion (A) The coefficient of x n in the expansion, of (1 + x )(1 − x )n is ( −1)n (1 − n )., Reason (R), n, C0 + n C2 + n C4 + ... = n C 1 + n C 3 + ... = 2n −1.
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159, , Binomial Theorem, 39. Assertion (A) The coefficient of x n in the binomial, expansion of (1 − x )−2 is ( n + 1)., Reason (R) The coefficient of x r in (1 − x )− n , when, n ∈ N is n + r − 1Cr ., , 44. If the magnitude of the coefficient of x7 in the, 8, 1, , expansion of ax 2 + is equal to the magnitude of, , bx , 8, , 1 , , the coefficient of x −7 in the expansion of ax − 2 ,, , bx , then a and b are connected by the relation, (a) ab =1, (b) ab = 2, (c) a 2b = 1, (d) ab2 = 2, , 40. Assertion (A) The term independent of x in the, ( 4m )!, 1, expansion of ( x + + 2)m is, ., ( 2m )!, x, Reason (R) The coefficient of x 6 in the expansion of, (1 + x )n is n C6., 41. Assertion (A) 9950 + 10050 < 10150., Reason (R) m n + ( m + 1)n < ( m + 2)n for all m, n ∈ N ., , 45. The total number of terms in the expansion, 8, 8, 2 1, 2 1, ax + + ax − after simplification is, , , bx , bx , (a) 0, (b) 5, (c) 10, (d) 15, , of, , 46. The correct matching of List I from List II is, List I, , Directions (Q. Nos. 42-45), , Let us consider the, 8, 1 , 1, , , binomial expansion ax 2 + and ax − 2 , where, , , bx , bx , a and b are positive numbers, then, , List II, , A. (1 − x) n, , 8, , x, 1., x+1, , n( n + 1) 2, ⋅ x − ...,, 2!, if x < 1, n( n + 1) 2, C. If x > 1, then, 3. 1 + nx +, ⋅ x + ...,, 1, 1, 2!, 1 + + 2 + ... is, if x < 1, x x, x, D. If| x| > 1, then, 4., 2, 3, 4, x, −, 1, 1 − 2 + 4 − 6 + ... is, x, x, x, x4, 5. 2, ( x + 1) 2, x4, 6. 2, ( x − 1) 2, , B. (1 + x) − n, , 8, , 1, , 42. The coefficient of x7 in the expansion of ax 2 + is, , bx , a3, a5, (a) 8C3 5, (b) 8C3 3, b, b, 5, a, (c) 8C4 3, (d) None of these, b, 43. The coefficient of x −7 in the expansion of, 8, 1 , , ax − 2 is, , bx , 3, a, a5, (b) ( −1)5 8C5 3, (a) ( −1)5 8C5 5, b, b, 5, a, (c) 8C3 3, (d) None of these, b, , Codes, A B, (a) 1 3, (c) 3 2, , 2. 1 − nx +, , C, 4, 4, , D, 5, 5, , A, (b) 2, (d) 2, , B, 3, 3, , C, 4, 1, , D, 1, 5, , Answers, Level I, 1., 11., 21., 31., , (d), (a), (c), (c), , 2., 12., 22., 32., , (a), (b), (b), (b), , 3. (c), 13. (b), 23. (c), , 4. (d), 14. (d), 24. (d), , 5. (a), 15. (c), 25. (c), , 6. (c), 16. (a), 26. (a), , 7. (b), 17. (d), 27. (b), , 8. (d), 18. (b), 28. (b), , 9. (a), 19. (b), 29. (a), , 10. (a), 20. (b), 30. (b), , 2., 12., 22., 32., 42., , (b), (b), (a), (c), (b), , 3., 13., 23., 33., 43., , 4., 14., 24., 34., 44., , 5., 15., 25., 35., 45., , 6., 16., 26., 36., 46., , 7., 17., 27., 37., , 8., 18., 28., 38., , 9., 19., 29., 39., , 10., 20., 30., 40., , Level II, 1., 11., 21., 31., 41., , (d), (a), (b), (d), (a), , (d), (b), (c), (a), (a), , (b), (c), (c), (a), (a), , (a), (a), (a), (a), (b), , (a), (d), (a), (d), (c), , (a), (a), (c), (a), , (c), (a), (a), (b), , (a), (b), (d), (a), , (b), (c), (b), (d)
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161, , Binomial Theorem, C n − 2 2n − 2 = n − 1C1 2n − 2, (n − 1) ! n − 2, =, 2, = (n − 1) 2n − 2, 1 ! (n − 2)!, , Coefficient of xn =, , n −1, , 17. Given, (3 − 2x)(1 + 3x)−3, = (3 − 2x)(1 − 9x + 54x2 − 270x3 + ... ), = coefficient of x3 = − 810 − 108 = − 918, 18. Applying Tr + 1 = nC r xn − r a r for (x + a )n., 10, 1 , , ∴ 6th term in the expansion of 2x2 − 2 is, , 3x , , But according to the given condition,, T3, n (n − 1) × 3 × 2 × 1 × 8, 1, =−, =, T4, n (n − 1) (n − 2) × 2 × 1 × 4 2, ⇒, , n − 2 = − 12 ⇒ n = − 10, 12, , , a, 25. Given expansion is + bx ., , x, ∴General term,, a, Tr + 1 = 12C r , x, , 8, , r, , 1, Tr + 1 = 8C r (x3 )8 − r 2 = 8C r x24 − 5 r, x , , xn/2 = x6 ⇒ n = 12, n, , 3, 2, 20. Given expansion is x − ., 3, 2x, ∴ General term, Tr + 1 = nC r xn − r a r, 2 , ⇒ T4 = nC3 x, 3 , , n −3, , (bx)r, , 1, , 26. The general term in the expansion of x3 + 2 is, , x , , 2, , ⇒, , 12 − r, , = 12C r (a )12 − r ⋅ br x− 12 + 2r, Since, we have to find the coefficient of x− 10., ∴, − 12 + 2r = − 10, ⇒, r =1, Then,the coefficient of x− 10 is 12C1 (a )11 (b)1 = 12a11b., , 5, , 10 !, 1, 896, 1 , T6 = 10C5 (2x2)5 − 2 = −, × 32 ×, =−, 3x , 5 !5 !, 243, 27, n, , 19. Since, n is even, therefore + 1 th term is the middle, 2, , term., n/ 2, 1, ∴, T n = nC n/ 2 (x2)n/ 2 = 924x6 (given), x, +1, , (given), , 3, , 2, 3, n, − = C3 , 3, 2x, , The term containing x4., ∴, n−6, , ⇒ r 2 + r = n 2 − nr + n − nr + r 2 − r, ⇒, n 2 − 2nr − 2r + n = 0, n, ⇒ (n − 2r ) (n + 1) = 0 ⇒ r =, 2, n, , 1, , 24. Given expansion is, x − ., , 2x, , ∴, , xn − 6 (− 1)3, , Since, it is independent of x., ∴, n −6 =0⇒n =6, 21. Given that, nC 6 = nC12 ⇒ nC n − 6 = nC12, ⇒, n − 6 = 12 ⇒ n = 18, 22. (0.99)15 = (1 − 0.01)15, We want to answer correct upto 4 decimal places and as, such, we have left further expansion., (0.99)15 = 1 − 15C1 (0.01) + 15C 2 (0.01)2 − 15C3 (0.01)3 + ..., 15 ⋅ 14, = 1 − 15 (0.01) +, (0.0001), 1 ⋅2, 15 ⋅ 14 ⋅ 13, −, (0.000001) + ..., 1 ⋅2 ⋅3, = 1 − 0.15 + 0.0105 − 0.000455 + ..., = 1 .0105 − 0.150455 = 0.8601, 23. Given that, Tr = Tr + 2 ⇒ nC r − 1 = nC r + 1, n!, ⇒, (n − r + 1) (n − r ) (n − r − 1) ! (r − 1) !, n!, =, (n − r − 1) ! (r + 1) (r ) (r − 1) !, , 1, T3 = nC 2 (x)n − 2 − , 2x, , 2, , 1, T4 = nC3 (x)n − 3 − , 2x, , 3, , and, , T5 = 8C 4x4 =, , 8! 4, x = 70x4, 4 !4 !, , 27. (x + a )100 + (x − a )100, = {100C 0x100 +, , C1x99a +, , 100, , C 2x98a 2 + ... + a100}, , 100, , + {100C 0x100 −100C1x99a + 100C 2x98a 2 – ... + a 100}, = 2 {100C 0x100 +, , C 2x98a 2 + ... + a 99}, , 100, , ∴ Total number of terms is 51., 28., , 1, = (6 − 3x)−1/3 = 6 − 1/3, (6 − 3x)1/3, =6, , − 1/3, , = 6 − 1/3, , , , 1 +, , , , 1 +, , , 1, − , 3, , x, , 1 − 2 , , − 1/3, , , 1 4, − − , 2, , x 3 3 x, − + ..., − +, 2, 2, 2 ⋅1, , , , , x 2x2, + 2 + ..., 6, 6, , , 29. Coefficient of x p is, , p+ q, , C p and coefficient of xq is ( p + q)C q., , ⇒ They are equal., x, 3 , 30. The general term in the expansion of , +, , 3 2 x2 , Tr + 1 =, , x, Cr , 3, , 10, , 10 − r, 2, , r, , 3 , 10, 2 = C r 3, 2x , , − 10 + 3 r, 2, , 10, , is, , 10 − 5 r, 2, , ⋅ 2− r ⋅ x, , 10 − 5r, =0, 2, ⇒, r =2, So, the term independent of x, 4, 2, 10 × 9, 1, 5, 1 3, ×, =, = 10C 2 × =, 3 2, 2 × 1 3 ×3 ×2 ×2 4, , ∴, , ∴, , 24 − 5r = 4 ⇒ r = 4
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162, , NDA/NA Mathematics, ∴, ⇒, ⇒, , 31. Putting x = 1 in the expansion (1 + x − 3x2)2163 , we get, the sum of the coefficients as, (1 + 1 − 3)2163 = (− 1)2163 = − 1, 1 , , 32. The general term in the expansion of ax − 2, , bx , , 1 , Tr + 1 = 11C r (ax)11 − r −, , bx2, , 11, , (11 − r ) (1) + r (− 2) = − 7, 11 − r − 2r = − 7, r =6, , , Thus, the coefficient of x−7 is 11C 6 (a )5 −, , , is, , 6, , 1, 462 5, = 6 a ., , b, b, , r, , Level II, 9, , 3 x− 2 1 , 1. Let Tr + 1 be the term independent of x in , − ., 3x, 2, ∴, , 3x−2, Tr + 1 = 9C r , , 2 , , 9−r, , 3, = (−1)r 9C r , 2, , 1, − , 3x, , 9−r, , 2, , = (1 − 2x + x2)(1 − 2x + 3x2 − 4x3 + 5x4 − K ), 2, 1 − x, ∴ Coefficient of x4 in , = 5 + 8 + 3 = 16, 1 + x, 1/ 2, , = (1 − x)1/ 2 (1 + x)− 1/ 2, 1 − x, , , 1 + x, , 1/ 2, , 1, 1, , , , = 1 − x ... 1 − x K, , , , 2, 2, 1, 1 2, = 1 −2⋅ x + x = 1 − x, 2, 4, (Neglecting higher degree terms.), , 6. We have, T2r + 4 = 18C 2r + 3 ⋅ x18 − ( 2r+ 3 ), and, Tr − 2 = 18C r − 3 ⋅ x18 − ( r − 3 ), According to the given condition,, 18, C 2r + 3 = 18C r − 3, ⇒, 2r + 3 + r − 3 = 18, ⇒, 3r = 18 ⇒ r = 6, , r, , (7) 8, , For integral terms in the expansion, r must be, multiple of 8., i.e.,, 0, 8, 16, 24, 32, ... , 1024, Now,, l = a + (n − 1) d, ⇒, 1024 = 8 + (n − 1) 8, ⇒, n = 128, , ⋅, , 1 − x, 2, −2, 2, −2, 4. , = (1 − x) (1 + x) = (1 − 2x + x )(1 + x), 1 + x, , ⇒, , 1024 − r, 2, , = 1024C r (5), , r, , 1 −18 + 2r − r, x, 3r, To be term independent from x, −18 + r = 0 ⇒ r = 18, (Q In binomial theorem should be n > r), which is not possible., Thus, no such term exists in the expansion of given, expression., 2. Given, 24n − 15n − 1, (24 )n = (16)n = (15 + 1)n = (1 + 15)n, n, = C 0 (15)0 (1)n + nC1 (1)n − 1 (15)2 + nC 2(1)n − 2(15)2 + K, = 1 + n ⋅ 15 + nC 2(15)2 + nC3 (15)3 + K, (24 )n − (15n ) − 1 = (15)2 { nC 2 + nC3 ⋅ 15 + K }, (24 )n − (15n ) − 1 = 225 ⋅ d , ∀ d ∈ N, Hence, (24 )n − (15n ) − 1 is divisible by 225., 3. Required number of terms in (a + b + c )n., (n + 2) !, = n + 2C 2 =, 2!n!, (n + 2)(n + 1)n ! (n + 1)(n + 2), =, =, 2⋅ n !, 2, , 1 − x, 5. , , 1 + x, , 7. Tr + 1 = 1024C r (51/ 2)1024 − r ⋅ (71/ 8 )r, , 8., , n, 2, , Σ, , r=0, , C 2r = nC 0 + nC 2 + ... + nC n = 2n − 1, , n, , 2, , x n (n + 1) x , 9. Given, 1 + n , +, , + ... ∞, 1 − x 2 ! 1 − x, ⇒, , , x , 1 − 1 − x , , , , −n, , 1 − 2x , =, , 1−x , , −n, , 1−x, = , , 1 − 2x, , n, , 10. Coefficient of pth, ( p + 1)th and ( p + 2)th terms in, expansion (1 + x)n are nC p − 1 , nC p and nC p + 1., Since, these are in AP., ∴, 2 nC p = nC p − 1 + nC p + 1, n!, n!, 2, =, (n − p) ! p ! (n − p + 1) ! ( p − 1) !, n!, +, (n − p − 1) ! ( p + 1) !, 2, p, ⇒, =, (n − p) ! p ! (n − p + 1) (n − p) ! p !, n−p, +, (n − p) ! ( p + 1) p !, ⇒, n 2 − n (4 p + 1) + 4 p2 − 2 = 0, 11. We know that, (1 + x)n = C 0 + C1x + C 2 x2 + ... + C r xr + .., ...(i), n, 1, 1, 1, 1, , and 1 + = C 0 + C1 + C r r + C 2 2 + ..., , x, x, x, x, 1, 1, 1, ...(ii), + C r + 1 r + 1 + C r + 2 r + 2 + ... + C n n, x, x, x, Multiplying Eqs. (i) and (ii) and equating the coefficient, 1, of xr in n (1 + x)2n or the coefficient of xn + r in (1 + x)2n ,, x, we get the value of required expression, (2n ) !, = 2n C n + r =, (n − r ) ! (n + r ) !
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163, , Binomial Theorem, 1/3, , a + x, , , a , , 1 1, , , , − 1, 2, , , 0.005, 1 0.005 3 3, = 2 1 − ×, +, + ..., , 8 , 3, 8, 2 ⋅1, , , , , 1, 1, , , ×, , , 0.005 3 3 (0.005)2, = 2 1 −, −, ×, + ..., 24, 1, 64, , , , , = 2 (1 − 0.000208) = 2 × 0.999792 = 1.9996, 13. rth term of (a + 2x)n is nC r − 1 (a )n − r + 1 (2x)r − 1, n!, =, a n − r + 1 (2x)r − 1, (n − r + 1) ! (r − 1) !, =, , =2 +, , (x + a )n = nC 0xn + nC1xn − 1 a, + nC 2 xn − 2 a 2 + nC3 xn − 3 a3 + ..., , From the given condition,, A = nC 0xn + nC 2xn − 2a 2 + nC 4xn − 4a 4 + ..., B = nC1xn − 1a + nC3 xn − 3 a3 + ..., 1, Hence, AB = {(x + a )2n − (x − a )2n}, 4, , and, , ⇒, , 4 AB = (x + a )2n − (x − a )2n, 10, , 15. ∴ ∑, , Ck =, , 20, , C0 +, , 20, , C1 +, , 20, , ⇒, , =, , 20, , 2, , 20, , C0 +, , 20, , C 2 + ... +, , 20, , C1 +, , 20, , C0 +, , 20, , C1 +, , 20, , +, , 20, , C10, , C 2 + ... +, , 20, , C 2 + ... +, , 20, , C 20, C10, , C 9 + ... + 20C 0 (Q nC r = nC n − r ), 20, 20, 20, ⇒ 2 + C10 = 2 [ C 0 + 20C1 + ... + 20C10 ], 1, ⇒ 20C 0 + 20C1 + ... + 20C10 = 219 + ⋅ 20C10, 2, 1, can be rewritten as, 16. The given expression, (4 − 3x)1/ 2, 20, , − 1/ 2, , 3, 3 , , and it is valid only, when x < 1, 4− 1/ 2 1 − x, , 4 , 4 , 4, 4, ⇒, − <x<, 3, 3, 2, 17. ∴15 C 02 − 15C12 + 15C 22 − ... − 15C15, =, =0, , 15, , 18. Given, , C 02, , −, , 15, , C12, , +, , 15, , expression, , rewritten as, , − ... −, , 15, , a , , , a + x, , 1/ 2, , C 22, , − 1/ 2, , 3 x2, + ..., 4a 2, , 4 ⋅5 2, x + ..., 2, 1 ⋅ 2 ⋅ 3 0 2 ⋅ 3 ⋅ 4 1 3 ⋅ 4 ⋅ 5 2 4 ⋅ 5 ⋅ 6 3, x +, x +, x +, x, =, 6, 6, 6, 6, (r + 1) (r + 2) (r + 3) r, , x + ..., + ... +, 6, , (r + 1) (r + 2) (r + 3) r, Therefore, Tr + 1 =, x, 6, 20. Coefficient of T5 = nC 4 , T6 = nC5 and T7 = nC 6, According to the given condition,, 2 nC5 = nC 4 + nC 6, , , , n!, n!, n!, ⇒, 2, = (n − 4) ! 4 ! + (n − 6) ! 6 ! , (, n, −, 5, ), !, 5, !, , , , , , , 1, 1, 1 , 2, = (n − 4) (n − 5) + 6 ⋅ 5 , (, n, −, 5, ), 5, , , , After solving, we get n = 7 or 14, 21. We know that,, (1 + x)n = nC 0 + nC1x + nC 2x2 + nC3 x3 + nC 4x4 + ... + nC nxn, , Q (1 + x)n = nC 0 + nC1x1 + nC 2x2 + ... + nC nxn, Put, x = 1,, 2n = nC 0 + nC1 + nC 2 + ... + nC n, Put, n = 20, 20, , − 1/ 2, , − 1/ 2, , ⇒, , k=0, , 220 =, , a − x, +, , a , , 19. (1 − x)− 4 = 1 ⋅ x0 + 4x1 +, , n (n − 1) ... (n − r + 2) n − r + 1, a, (2x)r − 1, (r − 1) !, , 14. We know that,, , − 1/ 2, , x, x, , , + 1 + , = 1 + , , , a, a, , , 1 3, − − , 2, , , 1 x 2 2 x , = 1 + − +, + ..., , , , , , , a, 2, 1, a, 2, ⋅, , , , , , , 1 3, − − , 2, , , 1 x 2 2 x , + 1 + − − +, − + ..., a, 2 a , 2 ⋅1, , , , , , 0.005 , , 12. (7.995)1/3 = (8 − 0 ⋅ 005)1/3 = (8)1/3 1 −, 8 , , , C 22, , +, , 15, , −, (Q nC r = C n − r ), , C12, , a , +, , a − x, , 15, , C 02, n, , 1/ 2, , can, , be, , (1 + x)n = nC 0 − nC1x + nC 2x2 − nC3 x3, + nC 4x4 − ... + (− 1)nC nxn, (1 + x)n + (1 − x)n = 2 ⋅ nC 0 + 2 ⋅ nC 2x2 + 2 ⋅ nC 4x4 + ..., Put x = 1;, (1 + 1)n + (1 − 1)n = 2 { nC 0 + nC 2 + nC 4 + ... }, 1, ⇒ nC 0 + nC 2 + nC 4 + ... = {2n + 0}, 2, ⇒ nC 0 + nC 2 + nC 4 + ... = 2n − 1, 22. Given that, (1 + ax)n = 1 + 8x + 24x2 + ..., n, n (n − 1) 2 2, a x + ... = 1 + 8x + 24x2 + ..., ⇒ 1 + ax +, 1, 1 ⋅2, On comparing the coefficients of x and x2, we get, n (n − 1) 2, na = 8,, a = 24, ⇒, 1 ⋅2, ⇒, na (n − 1) a = 48, ⇒, 8 (8 − a ) = 48 ⇒ 8 − a = 6 ⇒ a = 2 ⇒ n = 4, 9, , 1, 3, 23. The general term in the expansion of x2 − is, 2, 3x, 3 , Tr + 1 = 9C r x2, 2 , 3, = 9C r , 2, , 9−r, , 9−r, , 1, − , 3x, , r, , r, , 1 18 − 3 r, − x, 3, , ...(i)
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164, , NDA/NA Mathematics, Now, The coefficient of the terms x0 , x−1 and x−3 in, 9, 3 2 1 , x − is, 2, 3x, 0, For x , 18 − 3r = 0 ⇒ r = 6, For x−1, there exists no integer value of r., For x−3 , 18 − 3r = − 3 ⇒ r = 7, Now, the coefficient of the term independent of x in the, expansion of, 9, 1, 3, (1 + x + 2x3 ) x2 − , 2, 3x, 3, = 1 ⋅9 C 6 (− 1)6 , 2, , 9−6, , 6, , 1, 9, 7, + 0 + 2 ⋅ C7 (− 1), 3, 3, × , 2, , 9 −7, , 1, , 3, , 7, , 9 ⋅8, 3 1, 9 ⋅8 ⋅ 7 3 1, (− 1) 2 ⋅ 7, ⋅ ⋅, + 2⋅, 1 ⋅2, 1 ⋅ 2 ⋅ 3 23 36, 2 3, 7, 2 17, =, −, =, 18 27 54, 24. In the expansion of (1 + x)2n , the general term, = 2nC k xk , 0 ≤ k ≤ 2n, As given for r > 1, n > 2 , 2nC3 r = 2nC r + 2, ⇒ Either 3r = r + 2 or 3r = 2n − (r + 2) (Q nC r = nC n − r ), ⇒, r = 1 or n = 2r + 1, We take the relation only, n = 2r + 1, (Q r > 1), 25. In the expansion of (1 + x)n , it is given that, n, C1 , nC 2 and nC3 are in AP., ⇒, 2 ⋅n C 2 = nC1 + nC3, n (n − 1) n n (n − 1) (n − 2), ⇒, 2⋅, = +, 1 ⋅2, 1, 1 ⋅2 ⋅3, ⇒, 6 (n − 1) = 6 + (n − 2) (n − 1), ⇒, 6n − 6 = 6 + n 2 − 3n + 2, ⇒, n 2 − 9n + 14 = 0, ⇒, (n − 2) (n − 7) = 0 ⇒ n = 2 , 7, But n = 2 is not acceptable because, when n = 2, then we, get only three terms in the expansion of (1 + x)2., ∴ We take only n = 7, 26. (101)100 − 1 = (1 + 100)100 − 1, 100 × 99, , , (100)2 + ... − 1, = 1 + 100 ⋅ 100 +, 2, , , 100 ⋅ 99, , 2, = (100) 1 +, + ..., , , 2!, =, , 3, , 30. Given expression, , 1+ x+ 1+ x, , =1 −, , can be rewritten as, , 5, 5, x + ... = 1 − x, 6, 6, (neglecting the highest power of x), , 1, , 31. Given, x2 − , , x, , 9, , General term,, Tr + 1 = nC r (1)n − r xr , in (1 + x)n, r, 1, Similarly, Tr + 1 = 9C r (x2)9 − r ⋅ − , in, x, = 9C r x18 − 2r (− 1)r x− r, = 9C rx(18 − 3 r) (− 1)r, For independent term,, Put, 18 − 3r = 0 ⇒ 3r = 18 ⇒ r = 6, ∴, T( 6 + 1) = 9C 6x(18 − 18) ⋅ (− 1)6, 9⋅8⋅ 7, T7 = 9C 6 ⋅ 1 =, = 84, 3⋅2⋅1, 100, , 32. The given sigma expansion, , Σ, , m= 0, , 2 1, x − , , x, , 9, , …(i), , Cm (x − 3)100 − m ⋅ 2m, , 100, , can be rewritten as, , n , n (n − 1), 1, , , = 1 + 1 +, ... 1 + , , , 1 , 2!n , n, , [(x − 3) + 2]100 = (x − 1)100 = (1 − x)100, ∴ x will occur in T54, T54 = 100C53 (− x)53, ∴Coefficient is − 100C53 ., 33. Given expression is (x + xlog10 x )5 ., (given), ∴, T3 = 5C 2 x3 (xlog10 x)2 = 106, Put, x = 10, then, 10 ⋅ 103 ⋅ 102 = 106 ⇒ 106 = 106, Hence, x = 10 is the required value., 53, , (1 + n ) (1 + n ) (1 + n ) (1 + n ) (n + 1)n, ., ., ..., =, n!, 1, 2, 3, n, , 28. (1 − x + x2)n = a 0 + a1x + a 2x2 + ... + a 2nx2n, Putting x = 1, we get, (1 − 1 + 1)n = a 0 + a1 + a 2 + K + a 2n, ⇒, 1 = a 0 + a1 + a 2 + ... + a 2n, , 1 + x + 3 (1 − x)2, , (1 + x)1/ 2 + (1 − x)2 / 3, 1 + x + (1 + x)1/ 2, 1, 1 2, 2, 1 2, , , , 1 + 2 x − 8 x + ... + 1 − 3 x − 9 x − ..., =, 1, 1, , , 1 + x + 1 + x − x2 + ..., 2, 8, , , 1, 1 2, , , 1, 17 2, x−, x + ..., 2− x−, x + ... 1 −, 12, 144, , , 6, 72, =, =, 3, 1 2, 3, 1 2, , , 2 + x − x + ..., 1 + 4 x − 16 x + ..., 2, 8, −1, 1, 1 2, 3, 1 2, , , , x−, x + ... × 1 + x −, = 1 −, x + ..., 12, 144, 4, 16, , , , , 2, , Hence, given number is divided by (100)2., , , C , C , Cn , , 27. We have, 1 + 1 1 + 2 ... 1 +, C0 , C1 , C n −1 , , , , =, , Putting x = − 1, we get, ...(ii), 3n = a 0 − a1 + a 2 − ... + a 2n, On adding Eqs. (i) and (ii), we get, 3n + 1, = a 0 + a 2 + a 4 + ... + a 2n, 2, 29. (1 + x + x2 + x3 )n = {(1 + x)n (1 + x2)n}, = (1 + nC1x + nC 2x2 + ... + nC n xn ), n, 2, n, 4, (1 + C1x + C 2x + ... + nC nx2n ), Therefore, the coefficient of x4, = nC 2 + nC 2nC1 + nC 4 = nC 4 + nC 2 + nC1nC 2, , ...(i)
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165, , Binomial Theorem, 34. I. The coefficient of middle term in the expansion of, (1 + x)8 = 8C 4, , x +, , , and, , 8, , 1, 8, = C4, x, , Hence, it is equal., II. The coefficient of middle term in the expansion of, (1 + x)8 = 8C 4, The coefficient of fifth term in the expansion of, (1 + x)7 = 7C 4 or 7C3, 8, ∴, C 4 > 7C 4 or 7C3, Hence, only statement I is true and statement II is false., 35. Statement I is true but statement II is false, because, coefficient of (r + 1)th term in the expansion of (1 − x)n th, is (− 1)r nC r., 36. Both statement are not true, because, total number of, n+2, , if n is, terms in the expansion of (x + a )n+(x − a )n is, 2, n+1, an even number and is, , if n is an odd number., 2, 5, , 37. C (47, 4) + Σ C (52 − r , 3), r =1, , =, =, , C4 +, 48, C4 +, , C3 + 50C3 + 49C3 + 48C3 + 47C3, 48, C3 + 50C3 + 51C3 = 52C 4, (QnC r + nC r + 1 = n + 1C r + 1 ), A and R are correct and R is the correct explanation of A., 38. The coefficient of xn in expansion of (1 + x)(1 − x)n, = coefficient of xn + coefficient of xn−1, n!, n!, = (− 1), + (−1)n − 1, 1 ! (n − 1), n !0 !, 47, , 50 ⋅ 49 ⋅ 48, ⋅ 10047 + ... > 10050, 1 ⋅2 ⋅3, ⇒, 10150 > 9950 + 10050, A and R are individually true and R is the correct, explanation of A., 42. Let the term containing x7 in the expansion of, 8, 2 1, is Tr + 1., ax +, , bx, = 10050 + 2 ⋅, , a 8−r, 1, Tr + 1 = 8C r (ax2)8− r = 8C r r x16 − 3 r, bx, b, r, , ∴, , Since, this term contains x7., ∴, 16 − 3r = 7, ⇒, , r =3, , 1, , ∴Coefficient of x7 in the expansion of ax2 +, , , bx, = 8C3 ⋅ a5 /b3, , 43. Also, the term containing x−7 in the expansion of, 8, 1 , , ax − 2 is TR + 1., , bx , , 51, , 1 , TR + 1 = 8CR (ax)8 − R − 2, bx , = (−1)R 8CR, , 1, , , 40. x + + 2, , x , , m, , x2 + 2 x + 1 , =, , x, , , , m, , =, , (1 + x)2m, xm, , Term independent of x is coefficient of xm in the, (2m)!, expansion of (1 + x)2m = 2mCm =, ., (m !)2, Hence, statement I is false and statement II is true., 41. We have, 10150 = (100 + 1)50, 50 ⋅ 49, = 10050 + 50 ⋅ 10049 +, ⋅ 10048 + ... …(i), 1 ⋅2, and 9950 = (100 − 1)50, 50 ⋅ 49, = 10050 − 50 ⋅ 10049 +, ⋅ 10048 − ... …(ii), 1 ⋅2, Subtracting Eq. (i) from Eq. (ii), we get, 50 ⋅ 49 ⋅ 48, , , 10150 − 9950 = 2 50 ⋅ 10049 +, × 10047 + ..., 1 ⋅2 ⋅3, , , , R, , a 8 − R 8 − 3R, ⋅x, bR, , Since, this term contains x−7, ∴ 8 − 3R = − 7 ⇒ R = 5, ∴Coefficient of x−7 in the expansion of (ax −, , n!, , n!, n, = (− 1)n , −, = (−1) (1 − n ), −, !, 0, !, 1, !(, 1, )!, n, n, , , and nC 0 + nC 2 + nC 4 + ... = nC1 + nC3 + ... = 2n−1, A and R are individually true but R is not a correct, explanation A., 39. Since, coefficient of xr in (1 − x)− n = n + r−1C r, ∴Coefficicnt of xn in (1 − x)−2 = 2 + n − 1C n, = n + 1C n = (n + 1), Hence, option (a) is correct., , 8, , = (−1)5 8C5, , 1 8, ), bx2, , a3, b5, , 44. According to the given condition,, Magnitude of coefficient of x7 = Magnitude of coefficient of, x− 7, a3, a5, 8, ⇒, C3 3 = 8C5 5, b, b, ⇒, , a 2b2 = 1 ⇒ ab = 1, 8, , 8, , 1, 2 1, , 45. Let f (x) = ax2 +, + ax − , , , bx, bx, Here, n = 8, which is even., n + 2 8 + 2 10, =, =, =5, ∴Total number of terms =, 2, 2, 2, n (n + 1) 2, 46. A. (1 − x)− n = 1 + nx +, ⋅ x + ... , if|x|< 1, 2!, n (n + 1), B. (1 + x)− n = 1 − nx +, . ⋅ x2−... , if|x|< 1, 2!, x, 1 1, 1, C. 1 + + 2 + ... =, =, 1, x x, x −1, 1−, x, D. 1 −, , 2, 3, 4, 1, , +, −, + ... = 1 + 2, , x2 x4 x6, x , , −2, , =, , x4, (x + 1)2, 2
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Permutations, and Combinations, Solution (b) Exponent of 7 in 100!, , Factorial, Product of first n natural numbers is denoted by n !, and read as Factorial n., Thus, n ! = n ( n − 1) ( n − 2) ... 3 ⋅ 2 ⋅ 1, , %, %, %, %, , 9, , 100 100 , = 14 + 2 = 16, +, =, 2, 7 7 , , Fundamental Principles of Counting, , e.g.,, , 5 ! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120, , Fundamental Principle of Multiplication, , and, , 4 ! = 4 ⋅ 3 ! = 4 ⋅ 3 ⋅ 2 ! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24, , If there are two jobs such that one of them can be, completed in m ways, and when it has been completed in, anyone of these m ways, second job can be completed in n, ways; then the two jobs in succession can be completed in, m × n ways., e.g., There are 12 doors in a hall. A person can come in, the hall by one of 12 ways and go out from the hall in 11, other ways., ∴ Total number of ways = 12 × 11 = 132, , 0! = 1, Factorials of negative integers and fractions are not defined., n!, = n (n − 1) (n − 2) ... (r + 1), r!, n!, = n (n − 1) (n − 2) ... (n − r + 1), (n − r) !, , Exponent of Prime p in n !, , Fundamental Principle of Addition, , Let n is a positive integer and p is a prime number., Then,the last integer amongst 1, 2, 3, ... , ( n − 1), n which is, n, n, divisible by p is p, where denotes the greatest, p, p, n, integer less than or equal to ., p, , If there are two jobs such that they can be performed, independently in m and n ways respectively, then either of, the two jobs can be performed in ( m + n ) ways., e.g., There are 25 students in a class out of which 15 are, boys and 10 are girls, then for monitor a boy can be selected in, 15 ways and a girl can be selected in 10 ways., Hence, total number of ways to select a monitor, , Let E p ( n !) denote the exponent of the prime p in the, positive integer n. Then,, n n , E p ( n !) = + 2 + ... +, p p , , n, a, p , , where, a is a greatest positive integer such that, pa ≤ n < pa + 1., , Example 1. The exponent of 7 in 100! is, (a) 15, (c) 17, , (b) 16, (d) 4, , = 15 + 10 = 25, , Permutation, Each of the different arrangements which can be made, by taking some or all of a number of distinct objects is, called a permutation., e.g., Out of 3 objects ( a , b, c) taking 2 at a, time, total arrangements are ab, bc, ca , cb, ac, ba , then the, total number of arrangements is 6 each of these is called a, permutation.
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167, , Permutations and Combinations, , Important Results, 1. The number of all permutation of n distinct items or objects, taken r at a time, is denoted by n Pr ., n!, n, Pr =, ,∀ 0≤ r ≤ n, (n − r ) !, = n ( n − 1) ( n − 2) . . . . ( n − r + 1),, ∀ n ∈ N and r ∈ W , 0 ≤ r ≤ n, , 4. The number of all permutations of n different objects taken r at, a time, when a particular object is to be always included in each, arrangement is r ⋅n − 1 Pr − 1 ., , The number of all permutations of n distinct objects taken, all at a time = n ! ., n, P0 = 1, n P1 = n and n Pn − 1 = n !, , 5. The number of permutations of n distinct objects taken r at a, time, when a particular object is never taken in each, arrangement is n − 1 Pr ., , %, %, %, %, %, , n, , n, , n ⋅ n − 1 . Pr − 1, , Pr =, , n −1, , Pr = (n − r), , n −1, , Pr = r ⋅ n − 1 Pr − 1 +, , %, , Pr − 1, n −1, , Pr, , 2. The number of mutually distinguishable permutations of n, things, taken all at a time of which p are alike of one kind, q are, alike of second kind, r are alike of third kind and others are of, different kind, n!, =, p ! q! r !, , Circular Permutations, If the objects are arranged along a closed curve, then, the permutation is known as circular permutations., , Important Results, 1. The number of circular permutations of n distinct objects is, ( n − 1)! ., 2. If anti-clockwise and clockwise order of arrangements are not, distinct, then the number of circular permutations of n, 1, distinct objects is {( n − 1)! }., 2, , Example 2. If, r, , 56, , 3. The number of permutations of n different things, taken r at a, time, when each may repeated any number of times in each, arrangement is n r ., , The number of all permutations of n different objects taken r, at a time, when p particular objects are to be always included, in each arrangement is p ! {r − (p − 1)} n − p Pr − p ., , 6. The number of all permutations of n different objects taken all, but m objects always taken together, is m! × (n − m + 1)! ., 7. The number of all permutations of n different objects taken, all but m objects never taken together, is, n ! − m !( n − m + 1)!, , Example 3. In how many different ways of five boys and, five girls form a circle such that the boys and girls alternate, are, (a) 2800, (b) 2880, (c) 290, (d) 140, Solution (b) After fixing up one boy around the circle, remaining can be arranged in 4! ways but boys and girls are, to alternate. There will be 5 places, one place each between, two boys these five places can be filled by 5 girls in 5! ways., B1, B2, , B5, , Pr + 6 : 54Pr + 3 = 30800 : 1, then the value of, B3, , P2 is, (a) 1520, (c) 1600, , (b) 1640, (d) None of these, 56, , Solution (b) We have,, , 54, , Pr + 6, Pr + 3, , =, , 30800, 1, , ⇒, , 56 !, (51 − r) !, ×, = 30800, 54 !, (50 − r) !, , ⇒, , 56 × 55 × (51 − r) = 30800, , ⇒, ∴, , r = 41, r, , P2 =, , P2 = 1640, , 41, , B4, , Hence, by the principle of multiplication, the required, number of ways = 4 ! × 5 ! = 2880, , Combination, Each of the different selections made by taking some or, all of a number of distinct objects or items, irrespective of, their arrangements or order in which they are placed, is, called a combination., e.g., Out of 3 objects ( a , b, c) taking 2 at a time, the total, groups are ab, bc, ca ,then the number of groups is 3 each of, which is known as combination.
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168, , NDA/NA Mathematics, , Important Results, 1. The number of combinations of n different things taken r, n, at a time is denoted by n C r or C ( n, r ) or , r, n, (, n, −, ), ., ., ., (, n − r + 1), 1, n, !, nC =, =, ,, r, r ! ( n − r )! r ( r − 1) ( r − 2) . . . 2 ⋅ 1, ∀ n ∈ N and r ∈ W and 0 ≤ r ≤ n, %, , n, , C r is a natural number., , %, , n, , C 0 = nC n = 1, nC1 = n, , n, , C r + nC r − 1 =, , %, , n, , C x = nC y ⇔ x = y or x + y = n, , Cr, , n, , %, , n⋅, , C r − 1 = (n − r + 1) nC r − 1, , %, , If n is even, then the greatest value of n C r is n C n / 2 ., , %, , If n is odd, then the greatest value of, , n, , n, , C r is n C n, , +1, 2, , n, , or C n − 1, 2, n, , n, C r = ⋅n − 1 C r − 1, r, , Cr, , n−r +1, r, , %, , n, , %, , n, , C 0 + nC1 + nC 2 + ... + nC n = 2n, , %, , n, , C 0 + nC 2 + nC 4 + ... = nC1 + nC3 + nC 5 + ... = 2n − 1, , %, %, , 2n + 1, n, , C0 +, , Cn +, , + ... +, , n +1, , %, , 2n + 1, , C1 +, , Cn +, , n +2, , 2n − 1, , Cn =, , 2n, , n, , Cr − 1, , 2n + 1, , Cn +, , Cn, , = (α 1 + 1) (α 2 + 1) (α 3 + 1) . . . (α k + 1) − 1, (d) The sum of these divisors = ( p10 + p11 + p12 + . . . + p1α 1 ), , (use sum of GP in each bracket), (e) The number of ways in which N can be resolved as a, product of two factors is, , %, , n −1, , = (α 1 + 1) (α 2 + 1) (α 3 + 1) . . . (α k + 1) − 2, (c) The total number of divisors of N excluding 1 or N., , ( p20 + p12 + p22 + . . . + p2α 2 ) . . . ( pk0 + p1k + pk2 + . . . + pkα k ), , n, NOTE, %, C r = nC n − r, n +1, , (b) The total number of divisors of N excluding 1 and N, , =, , C 2 + ... +, , n +3, , 2n + 1, , C n = 22n, , Cn, , +1, , 2. The number of combinations of n different things taken r, at a time, when k particular objects occur is n − k C r − k . If k, particular objects never occur is, , n−k, , Cr ., , 3. The number of ways (or combinations) of n different things, selecting, atleast, one, of, them, is, n, C1 + nC 2 + nC 3 + . . . + nC n = 2n − 1. This can also be stated as, the total number of combinations of n different things., 4. The number of combinations of n identical things taking, r ( r ≤ n ) at a time is 1., 5. The number of ways of selecting r things out of n alike things is, ( n + 1), where r = 0, 1, 2, . . . , n., 6. If out of ( p + q + r + t ) things, p are alike one kind, q are alike, of second kind, r are alike of third kind and t are different, then, the total number of combinations is, ( p + 1) ( q + 1) ( r + 1) 2t − 1, , (f) The number of ways in which a composite number N can, be resolved into two factors which are relatively prime (or, coprime) to each other is equal to 2n − 1 , where n is the, number different factors in N., 8. Division into groups, (a) The number of ways in which ( m + n ) different things can, be divided into two groups which contain m and n things,, respectively is, (m + n) !, m+ n, C m ⋅ nC n =, ,m≠n, m! n !, If m = n, then the groups are equal in size. Divisions of these, groups can be given by two types., % If order of groups is not important The number of ways, in which 2 n different things can be divided equally into two, (2n) !, groups is, ., 2 ! (n !) 2, %, , If order of group is important The number of ways in, which 2n different things can be divided equally into two, 2n !, (2n) !, distinct groups is, ., × 2 !=, 2 ! (n !) 2, (n !) 2, , (b) The number of ways in which( m + n + p ) different things, can be divided into three groups which contain m, n and p, things, respectively is, (m + n + p ) !, m+ n+ p, C m ⋅ n + p C n ⋅ pC p =, m! n ! p !, m≠n≠ p, , %, , 7. Let N = p1α 1 ⋅ p2α 2 . p3α 3 . . . pkα k , where p1 , p2 , p3 , . . . , pk are, different primes and α 1 , α 2 , α 3 , . . . , α k are natural numbers,, then, (a) The total number of divisors of N including 1 and N, = (α 1 + 1) (α 2 + 1) (α 3 + 1) . . . (α k + 1), , 1, (α1 + 1) (α 2 + 1) ... (α k + 1). If N is not a perfect, 2, square., 1, [(α1 + 1) (α 2 + 1) ... (α k + 1) + 1]. If N is a perfect, 2, square., , %, , If m = n = p, then the groups are equal size. Division of, these groups can be given by two types, If order of group is not important The number of, ways in which 3p different things can be divided equally, 3p !, into three groups is, ., 3 !(p !)3, If order of group is important The number of ways in, which 3p different things can be divided equally into, 3p!, (3 p)!, three distinct groups is, ., ×3!=, 3, 3 !(p !), (p !)3
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169, , Permutations and Combinations, , Example 4. If, , C r + 1 : nC r : n − 1C r − 1 = 11: 6 : 3, then find, , the values of n and r, respectively., (a) 10, 5, (b) 4, 6, (c) 2, 5, n+1, , Cr + 1, , Solution (a) Here,, , n, , Cr, , ⇒, , n + 1 C r 11, =, ., r + 1 nC r, 6, , ⇒, , n + 1 11, =, r +1 6, , ⇒, , n, n, Q C r =, , r, , , C r − 1, , , n −1, , 6n + 6 = 11r + 11, ...(i), , Solution (b), , n, , ⇒, , Cr, 6, =, n −1, Cr − 1 3, n, ., r, , n −1, , Cr − 1, , n −1, , Cr − 1, , =, , 6, 3, , are alike of second kind and 3 bananas are alike of third kind., (i) The required number of combinations (when atleast one, fruit) = (5 + 1) ( 4 + 1) (3 + 1) 2 0 − 1 = 120 − 1 = 119, (ii) The required number of combinations (when one fruit of, each kind) = 5C1 × 4C1 × 3C1 = 5 × 4 × 3 = 60, , Example 6. In how many ways the number 18900 can be, split in two factors which are relative prime (or coprime)?, (a) 5, (b) 8, (c) 4, (d) 2, , 6n − 11r = 5, , and, , (b) 110, 20, (d) None of these, , Solution (a) Here, 5 oranges are alike of one kind, 4 mangoes, , (d) 4, 3, , 11, =, 6, , n, , ⇒, , (a) 119, 60, (c) 100, 10, , n+1, , n, n, Q C r =, , r, , n, = 2 ⇒ n = 2r, r, On solving Eqs. (i) and (ii), we get n = 10 , r = 5., ⇒, , Coprime means consecutive primes. Here, coprime are 2, 3,, 5, 7., (numbers of different factor in N ), ∴, n=4, Hence, number of ways in which a composite number N can, be resolved into two factors which are relative prime (or, coprime) = 2 4 − 1 = 23 = 8, , , C r − 1, , , n −1, , ...(ii), , Example 5. Find the number of combinations that can be, formed with 5 oranges, 4 mangoes and 3 bananas when it is, essential to take atleast one fruit and one fruit of each kind, are, respectively., , Here, N = 18900 = 2 2 ⋅ 33 ⋅ 5 2 ⋅ 71, , Derangement, If n things are arranged in a row, the number of ways in, which they can be deranged so that no one of them occupies, 1, 1, 1, 1, , its original place is n ! 1 −, +, −, + ... + ( − 1)n, ., , 1! 2! 3!, n !, , Comprehensive Approach, n, , n, , n, , n, , n, , n, , n, , n, , n, , Out of n non-concurrent and non-parallel straight lines, points of, intersection are n C 2., Out of n points the number of straight lines are (when no three, points are collinear) n C 2., If out of n points m are collinear, then number of straight lines, = nC 2 − mC 2 + 1, In a polygon total number of diagonals out of n points (no three, n (n − 3), points are collinear) =, 2, Number of triangle formed from n points = nC3 . (when no three, points are collinear), Number of triangles out of n points in which m are collinear, = nC3 − mC3, Number of triangles that can be formed out of n points (when none, of the side is common to the sides of polygon), = nC3 − nC1 − nC1 ⋅n − 4 C1, Number of parallelogram in two system of parallel lines (when 1st, set contains m parallel lines and 2nd set contains n parallel lines), = nC 2 × mC 2, Number of squares in two system of perpendicular parallel lines, (when 1st set contains m parallel lines and 2nd set contains, n parallel lines)., m −1, , = Σ ( m − r) (n − r); ( m < n), r =1, , n, , n, , n, , n, , n, n, , n, , n straight lines are drawn in the plane such that no two, lines are parallel and no three lines are concurrent. Then, the, number of parts into which these lines divide the plane is equal to, 1 + Σn., The sum of the digits in the unit place of all numbers formed with, the help of a1 , a2 , ..., an taken all at a time is, (n − 1) ! ( a1 + a2 + ... + an ) (repetition of digits is not allowed)., The sum of all digits numbers that can be formed using the digits, a1 , a2 , ..., an is, (10n − 1), (n − 1) ( a1 + a2 + ... + an ), 9, (repetition of digits is not allowed), Let the equation is, ...(i), x1 + x2 + .... + xr = n, where x1 , x2 , ..., xr and n are non-negative integers, then the, number of solutions of Eq. (i) is n + r − 1C r ., n! + 1is not divisible by any number which lies between 2 and n., Number of terms in ( a1 + a2 + ... + an ) m = n + k − 1C k − 1 , where k is, number of variables., If in a party all guests shake hands with each other, then total, number of handshakes is n C 2.
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Exercise, Level I, 1. The number of diagonals of an octagon, is, (a) 48, (b) 40, (c) 28, (d) 20, , 12. The value of, , 2. If a man and his wife enter a bus, in which five seats, are vacant, then the number of different ways in, which they can be seated, is, (a) 2, (b) 5, (c) 20, (d) 40, 3. If S = {2, 3, 4, 5, 7, 9}, then the number of different, three digit numbers (with all distinct digits) less, than 400 that can be formed from S , is, (a) 20, (b) 40, (c) 80, (d) 120, 4. The number of triangles that can be formed by, choosing the vertices from a set of 12 points, seven of, which lie on the same straight line, is, (a) 185, (b) 175, (c) 115, (d) 105, 5. How many words of 4 consonants and 3 vowels can be, made from 12 consonants and 4 vowels, if all the, letters are different?, (a) 251820, (b) 258120, (c) 281520, (d) 285120, 6. In how many ways 6 girls can be seated in two chairs?, (NDA 2011 I), , (a) 10, (c) 24, 7. What is the value, P (16, n − 2) = 3 : 4?, (a) 10, (b) 12, , (b) 15, (d) 30, of, , n,, , if, , P(15,, , n − 1), , :, , (NDA 2011 I), , (c) 14, , (d) 15, , 8. Using the digits 1, 2, 3, 4 and 5 only once, how many, numbers greater than 41000 can be formed?, (NDA 2011 I), , (a) 41, (c) 50, , (b) 48, (d) 55, , 9. The number of words which can be formed from the, letters of the word MAXIMUM, if two consonants, cannot occur together, is, (a) 4!, (b) 3 ! × 4 !, (c) 7!, (d) None of these, 10. How many numbers lying between 10 and 1000 can, be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, (repetition is allowed)?, (a) 1024, (b) 810, (c) 2346, (d) None of these, 11. If n Cr = n Cr −1 and n Pr = n Pr + 1 ,then the value of n is, (a) 3, (c) 2, , (b) 4, (d) 5, , (a), (c), , 47, 52, , C6, C4, , 47, , C4 +, , 5, , Σ, , r =1, , 52 − r, , C3 is equal to, , (b) 52C5, (d) None of these, , 13. The number of five digits numbers that can be, formed without any restriction is, (a) 990000, (b) 100000, (c) 90000, (d) None of these, 14. What is the number of ways of arranging the letters, of the word ‘BANANA’ so that no two N's appear, together?, (NDA 2010 I), (a) 40, (b) 60, (c) 80, (d) 100, 15. A team of 8 players is to be chosen from a group of, 12 players. Out of the eight players one is to be, elected as captain and another a vice-captain. In how, many ways can this be done?, (NDA 2010 I), (a) 27720, (b) 13860, (c) 6930, (d) 495, 16. If C ( n , 12) = C ( n , 8), then what is the value of, (NDA 2009 II), C( 22 , n )?, (a) 131, (b) 231, (c) 256, (d) 292, 17. There are four balls of different colours and four, boxes of colours same as those of the balls. The, number of ways in which the balls, one in each box,, could be placed such that a ball does not go to box of, its own colour, is, (a) 8, (b) 7, (c) 9, (d) None of these, 18. In how many ways can 5 boys and 5 girls sit in a circle, so that no two boys sit together?, (a) 5 ! × 5 !, (b) 4 ! × 5 !, 5!× 5!, (c), (d) None of these, 2, 19. In how many ways can 15 members of a council sit, along a circular table, when the Secretary is to sit on, one side of the Chairman and the Deputy Secretary, on the other side?, (a) 2 × 12 !, (b) 24, (c) 2 × 15 !, (d) None of these, 20. 4 buses runs between Bhopal and Gwalior. If a man, goes from Gwalior to Bhopal by a bus and comes back, to Gwalior by another bus, then the total possible, ways are, (a) 12, (b) 16, (c) 4, (d) 8
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171, , Permutations and Combinations, 21. The number of numbers of 4 digits which are not, divisible by 5, are, (a) 7200, (b) 3600, (c) 14400, (d) 1800, 22. What is the smallest natural number n such that n! is, divisible by 990?, (NDA 2009 I), (a) 9, (b) 11, (c) 33, (d) 99, 23. What is the value of r, if P ( 5, r ) = P ( 6, r − 1)?, (NDA 2009 I), , (a) 9, (c) 4, , (b) 5, (d) 2, , 24. What is the number of words formed from the letters, of the word ‘JOKE’ so that the vowels and consonants, alternate?, (NDA 2009 I), (a) 4, (b) 8, (c) 12, (d) None of these, 25. The number of 2 digit even numbers that can be, formed from the digits 1, 2, 3, 4 and 5, repetition, being not allowed, is, (a) 25, (b) 5!, (c) 16, (d) 8, 26. The number of ways in which 6 people can be seated, at a round table, is, (a) 6, (b) 60, (c) 120, (d) 720, 27. The number of lines obtained on joining 7 points in a, plane is, (a) 14, (b) 21, (c) 28, (d) 35, 28. In how many ways can a committee of 5 members be, formed out of 7 men and 3 women, so as to include, exactly 3 men?, (a) 36, (b) 41, (c) 105, (d) 210, , 33. If n P3 + n Cn − 2 = 14 n , then n is equal to, (a) 5, (c) 8, , (b) 6, (d) 10, , 34. How many even numbers of 3 different digits can be, formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 (repetition, is not allowed)?, (a) 224, (b) 280, (c) 324, (d) None of these, 35. In how many ways 3 letters can be posted in 4 letter, boxes, if all the letters are not posted in the same, letter box?, (a) 63, (b) 60, (c) 77, (d) 81, 36. What is the number of ways that 4 boys and 3 girls, can be seated so that boys and girls alternate?, (NDA 2012 I), , (a) 12, (c) 120, , (b) 72, (d) 144, , 37. In how many ways can ` 16 be divided into 4 person, when none of them get less than ` 3?, (a) 70, (b) 35, (c) 64, (d) 192, 38. If 7 points out of 12 are in the same straight line, then, what is the number of triangles formed? (NDA 2008 I), (a) 84, (b) 175, (c) 185, (d) 201, ( n + 2)! + ( n + 1)( n − 1)!, 39. What is, equal to?, ( n + 1)( n − 1)!, (NDA 2007 II), (a), (b), (c), (d), , 1, Always an odd integer, A perfect square, None of the above, , 29. The number of ways in which nine different toys can, be distributed among four children so that the, youngest child gets 3 toys and each of the others gets, 2 toys, is, (a) 2520, (b) 5120, (c) 7560, (d) 9072, , 40. Eighteen football teams take part in the national, championship and every team meets the same, opponent twice. How many matches are played, during the championship?, (a) 306, (b) 300, (c) 72, (d) 153, , 30. In how many ways can 10 lions and 6 tigers be, arranged in a row so that no two tigers are together?, (a) 10 ! × 11P6, (b) 10 ! × 10P6, (c) 6 ! × 10P7, (d) 6 ! × 10P6, , 41. How many arrangements can be made out of the, letters of the word ‘MOTHER’ taken four at a time so, that each arrangement contains the letter ‘M’?, (a) 240, (b) 120, (c) 60, (d) 360, , 31. If a polygon has 20 diagonals, then what is the, number of sides?, (NDA 2008 II), (a) 6, (b) 10, (c) 12, (d) 8, , 42. A polygon has 54 diagonals. How many sides does it, have?, (a) 10, (b) 11, (c) 12, (d) 27, , 32. What is the number of five digit numbers formed, with 0, 1, 2, 3, 4 without any repetition of digits?, , 43. A team consists of three men, three women and three, children. They have to stand in a line such that the, men stand together, women stand together and, children stand together. In how many different ways, can this be done?, , (NDA 2008 I), , (a) 24, (c) 96, , (b) 48, (d) 120
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172, , NDA/NA Mathematics, (a) 1296, (c) 1926, , (b) 1692, (d) 1270, , 44. A telephone dial has 10 holes. On rotating the dial,, how many different telephone numbers, each, consisting of 7 digits, can be dialed, if repetitions are, permitted?, (a) 70, (b) 710, (c) 107, (d) 7!, , 47. What is the number of signals that can be sent by, 6 flags of different colours taking one or more at a, time?, (NDA 2010 II), (a) 21, (b) 63, (c) 720, (d) 1956, 48. In how many ways can a committee consisting of, 3 men and 2 women be formed from 7 men and, 5 women?, (NDA 2010 II), (a) 45, (b) 350, (c) 700, (d) 4200, , 45. In how many ways can be bowler take four wickets in, a single 6 balls over?, (a) 6, (b) 15, (c) 20, (d) 30, , 49. The number of ways in which 5 boys and 5 girls can, sit in a ring, are, (a) 10!, (b) 9!, (c) 5!, (d) 6!, , 46. What is the maximum number of straight lines that, can be drawn with any four points in a plane such, that each line contains atleast two of these points?, (a) 2, (b) 4, (NDA 2010 II), (c) 6, (d) 12, , 50. The number of words that can be formed from the, letters of the word ‘INDRAPRASTHA’ when the, vowels are never separated, is, (a) 727560, (b) 725760, (c) 752760, (d) 757260, , Level II, 1. How many words can be formed from the letters of, the word ‘ARTICLE’, if vowels always comes at the, odd places?, (a) 60, (b) 576, 7!, (c), (d) 120, 3!, 2. In how many ways can 5 books be selected out of, 10 books, if two specific books are never selected?, (a) 56, (b) 65, (c) 58, (d) None of these, 3. In an election there are 8 candidates, out of which 5, are to be chosen. If a voter may vote for any number, of candidates but not greater than the number to be, chosen, then in how many ways can a voter vote?, (a) 216, (b) 114, (c) 218, (d) None of these, 4. A man has 7 friends. In how many ways he can invite, one or more of them for a tea party?, (a) 128, (b) 256, (c) 127, (d) 130, 5. There are 12 volleyball players in all in a college, out, of which a team of 9 players is to be formed. If the, captain always remains the same, then in how many, ways can the team be formed?, (a) 36, (b) 108, (c) 99, (d) 165, 6. There are 6 letters and 3 post boxes. The number of, ways in which these letters can be posted, is, (a) 63, (b) 36, 6, (d) 6 P3, (c) C3, 7. There are 4 candidates for the post of a lecturer in, Mathematics and one is to be selected by votes of, 5 men. What is the number of ways in which the, votes can be given?, (NDA 2011 II), , (a) 1048, (c) 1024, , (b) 1072, (d) 625, , 8. How many diagonals will be there in an n-sided, regular polygon?, (NDA 2011 II), n( n − 1), n( n − 3), (a), (b), 2, 2, n( n + 1), 2, (c) n − n, (d), 2, 9. What is the total number of combinations of n, different things taken 1, 2, 3, …, n at a time?, (b) 22n + 1, (a) 2n + 1, (NDA 2011 I), n −1, n, (c) 2, (d) 2 − 1, 10. What is the number of ways in which an examiner, can assign 10 marks to 4 question giving not less, than 2 marks to any question? (All questions carry, marks equal to integral value), (a) 4, (b) 6, (c) 10, (d) 16, 11. What is the sum of digits in the unit place of all, numbers formed using 1, 2, 3, 4, 5, 6 taken all at a, time without repeating any of them?, (a) 1260, (b) 2520, (c) 3780, (d) 5040, 12. A five-digit number divisible by 3 is to be formed, using the digits 0, 1, 2, 3, 4 and 5 without repetition., What is the total number of different ways in which, this can be done?, (a) 216, (b) 212, (c) 240, (d) 600, 13. If eleven members of a committee sit at a round table, so that the President and Secretary always sit, together, then the number of arrangements is, (a) 10 ! × 2, (b) 10 !, (c) 9 ! × 2, (d) None of these
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173, , Permutations and Combinations, 14. If the best and the worst paper never appear, together, then six examination papers can be, arranged in how many ways?, (a) 120, (b) 480, (c) 240, (d) None of these, , 25. What is the number of words that can be formed, from the letters of the word ‘UNIVERSAL’, the, vowels remaining always together?, (NDA 2010 II), (a) 720, (b) 1440, (c) 17280, (d) 21540, , 15. How many numbers divisible by 5 and lying between, 3000 and 4000 can be formed from the digits 1, 2, 3, 4,, 5, 6 (repetition is not allowed)?, (a) 6 P2, (b) 5 P2, 4, (d) 6 P3, (c) P2, , 26. In a football championship, a total of 153 matches, was played. Every two teams played one match with, each other. What is the total number of teams which, took part in the championship?, (a) 17, (b) 18, (c) 19, (d) 21, , n, , 16. What is the value of, , ∑, , r =1, , (a) 2n − 1, (c) 2n − 1, , P( n , r ), ?, r!, , (NDA 2011 II), , (b) 2n, (d) 2n + 1, , 17. 5 books are to be chosen from a lot of 10 books. If m is, the number of ways of choice when one specified book, is always included and n is the number of ways of, choice when a specified book is always excluded,, then which one of the following is correct?, (a) m > n, (b) m = n, (NDA 2011 I), (c) m = n − 1, (d) m = n − 2, 18. A question paper is divided into two parts A and B, and each part contains 5 questions. The number of, ways in which a candidate can answer 6 questions,, selecting atleast two questions from each part is, (a) 80, (b) 100, (c) 200, (d) None of these, 19. The number of triangles that can be formed by, 5 points in a line and 3 points on a parallel line is, (a) 8C3, (b) 8C3 − 5C3, 8, 5, 3, (d) None of these, (c) C3 − C3 − C3, 20. In how many ways can 5 keys be put in a ring?, 1, 1, (a) ⋅ 4 !, (b) ⋅ 5 !, 2, 2, (c) 4!, (d) 5!, 21. If 15 C3 r =, (a) 3, , 15, , Cr + 3 , then the value of r is, (b) 4, , (c) 5, , (d) 8, , 22. In an election the number of candidates is 1 greater, than the persons to be elected. If a voter can vote in, 256 ways, then the number of candidates is, (a) 7, (b) 10, (c) 8, (d) 6, 23. If x + y ≤ 4, then there are how many non-zero, positive integer ordered pair ( x , y )?, (NDA 2011 I), (a) 4, (b) 5, (c) 6, (d) 8, 24. A, B, C, D and E are coplanar points and three of, them lie in a straight line. What is the maximum, number of triangles that can be drawn with these, points as their vertices?, (a) 5, (b) 9, (NDA 20 11 I), (c) 10, (d) 12, , 27. How many 3 digit numbers, each less than 600, can be, formed from {1, 2, 3, 4, 7, 9}, if repetition of digits is, allowed?, (a) 216, (b) 180, (c) 144, (d) 120, 28. There are four chairs with two chairs in each row. In, how many ways can four persons be seated on the, chairs, so that no chair remains unoccupied?, (a) 6, (b) 12, (c) 24, (d) 48, 29. In how many ways can the letters of the word, CORPORATION be arranged so that vowels always, occupy even places?, (a) 120, (b) 2700, (c) 720, (d) 7200, 30. If all permutations of the letters of the word ‘LAGAN’, are arranged as in dictionary, then what is the rank of, ‘NAAGL’?, (a) 48th word, (b) 49th word, (c) 50th word, (d) 51st word, 31. Let P = { p1 , p2 , p3 , p4 },, Q = { q1 , q2 , q3 , q4 } and, R = { r1 , r2 , r3 , r4 }., If S10 = {( pi , q j , rk ) : i + j + k = 10},, elements does S10 have?, (a) 2, (b) 4, (c) 6, (d) 8, , how, , many, , 32. If a secretary and a joint secretary are to be selected, from a committee of 11 members, then in how many, ways can they be selected?, (a) 110, (b) 55, (c) 22, (d) 11, 33. What is the number of three-digit odd numbers, formed by using the digits 1, 2, 3, 4, 5, 6, if repetition, of digits is allowed?, (NDA 2010 I), (a) 60, (b) 108, (c) 120, (d) 216, 34. In a football championship 153 matches were played., Every team played one match with each other team., How many teams participated in the championship?, (a) 21, (b) 18, (NDA 2009 II), (c) 17, (d) 15
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174, , NDA/NA Mathematics, , 35. A coin is tossed 10 times. The number of heads minus, the number of tails in 10 tosses is considered as the, outcome of the experiment. What is the number of, points in the sample space?, (NDA 2009 II), (a) 10, (b) 11, (c) 21, (d) 99, 36. The number of diagonals in a octagon will be, (a) 28, (b) 20, (c) 10, (d) 16, 37. How many words comprising of any three letters of, the word ‘UNIVERSAL’ can be formed?, (a) 504, (b) 405, (c) 540, (d) 450, 38. In how many ways can a committee consisting of one, or more members be formed out of 12 members of the, Muncipal Corporation?, (a) 4095, (b) 5095, (c) 4905, (d) 4090, 39. Ten different letters of an alphabet are given. Words, with five letters are formed from these given letters., Then, the number of words which have atleast one, letter repeated is, (a) 69760, (b) 30240, (c) 99748, (d) None of these, 40. How many words can be made from the letters of the, word DELHI, if L comes in the middle in every word?, (a) 12, (b) 24, (c) 60, (d) 6, 41. Eight chairs are numbered 1 to 8. Two women and, three men wish to occupy one chair each. First the, women choose the chairs from amongst the chairs, marked 1 to 4 and then men select the chairs from, amongst the remaining. The number of possible, arrangements is, (a) 6C3 × 4C2, (b) 4C2 × 4P3, (d) None of these, (c) 4 P2 × 4P3, 42. Two numbers are successively drawn from the set {1, 2,, 3, 4, 5, 6, 7} without replacement and the outcomes, recorded in that order. What is the number of, elementary events in the random experiment?, (NDA 2009 II), , (a) 49, , (b) 42, , (c) 21, , (d) 14, , 43. How many times does the digit 3 appear while, writing the integers from 1 to 1000?, (NDA 2009 II), (a) 269, (b) 271, (c) 300, (d) None of these, 44. How many words, with or without meaning can be, formed by using all the letters of the word, ‘MACHINE’, so that the vowels occurs only the odd, positions?, (NDA 2008 II), (a) 1440, (b) 720, (c) 640, (d) 576, 45. From 7 men and 4 women a committee of 6 is to be, formed such that the committee contains atleast two, women. What is the number of ways to do this?, (a) 210, (b) 371, (NDA 2008 II), (c) 462, (d) 5544, , 46. In how many ways can 3 books on Hindi and 3 books, on English be arranged in a row on a shelf, so that not, all the Hindi books are together?, (NDA 2008 II), (a) 144, (b) 360, (c) 576, (d) 720, 47. A group consists of 5 men and 5 women. If the, number of different five person committees, containing k men and ( 5 − k) women is 100, what is, the value of k?, (NDA 2008 I), (a) 2 only, (b) 3 only, (c) 2 or 3, (d) 4, 48. A five digit number divisible by 3 is to be formed, using the digits, 0, 1, 2, 3, 4 and 5 without repetition., What is the total number of ways in which this can be, done?, (NDA 2007 II), (a) 216, (b) 240, (c) 600, (d) 3125, 49. A meeting is to be addressed by 5 speakers A, B, C, D, and E. In how many ways can the speakers be, ordered, if B must not precede A (immediately or, otherwise)?, (NDA 2007 II), (a) 120, (b) 24, (c) 60, (d) 54 × 4, 50. On a railway route there are 20 stations. What is the, number of different tickets required in order that it, may be possible to travel from every station to every, other station?, (NDA 2007 II), (a) 40, (b) 380, (c) 400, (d) 420, 51. In how many ways can be letters of the word ‘CABLE’, be arranged, so that the vowels should always occupy, odd positions?, (NDA 2007 I), (a) 12, (b) 18, (c) 24, (d) 36, , Directions (Q. Nos. 52-55), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 52. Assertion (A) The number of triangles that can be, formed by 12 points in a plane out of which 6 are, collinear, is 200., Reason (R) The number of lines that can be drawn, by joining n points, is n C2., 53. Assertion (A) The exponent of 7 in 100! is 16., Reason (R) The exponent of 12 in 100! is 49.
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175, , Permutations and Combinations, 54. Assertion (A) 51 × 52 × 53 × 54 × 55 × 56 × 57 × 58, is divisible by 40320., Reason (R) The product of r consecutive natural, numbers is always divisible by r!., 55. Assertion (A), The number of selections of, 20 distinct things taken 8 at a time is same as that, taken 12 at a time., Reason (R) C( n , r ) = C( n , s), if n = r + s., , Directions (Q. Nos. 56-58), , Consider the letters, , of the word 'Krishna’., 56. How many words can be formed the vowels are not, separated?, (a) 1250, (b) 550, (c) 1440, (d) None of these, , (a) 100, (c) 700, , (b) 720, (d) 4, , 58. How many words can be formed begin with s and end, in k ?, (a) 150, (b) 70, (c) 200, (d) 120, , Directions, , (Q. Nos. 59-60) Consider a, committee of 5 is to be formed out of 6 men and 4 ladies., , 59. How many ways can this be done when atleast, 2 ladies are included?, (a) 214, (b) 150, , (c) 200, , (d) 186, , 60. How many ways can this be done when atmost, 2 ladies are included?, (a) 175, (c) 180, , 57. How many words can be formed the vowels may, occupy only odd places?, , (b) 170, (d) 160, , Answers, Level I, 1., 11., 21., 31., 41., , (d), (a), (a), (d), (a), , 2., 12., 22., 32., 42., , (c), (c), (b), (c), (c), , 3., 13., 23., 33., 43., , (b), (c), (c), (a), (a), , 4., 14., 24., 34., 44., , (a), (a), (b), (a), (c), , 5., 15., 25., 35., 45., , (d), (a), (d), (b), (b), , 6., 16., 26., 36., 46., , (d), (b), (c), (d), (c), , 7., 17., 27., 37., 47., , (c), (c), (b), (b), (b), , 8., 18., 28., 38., 48., , (b), (b), (c), (c), (b), , 9., 19., 29., 39., 49., , (a), (a), (c), (c), (b), , 10., 20., 30., 40., 50., , (b), (a), (a), (a), (b), , 2., 12., 22., 32., 42., 52., , (a), (a), (c), (b), (b), (b), , 3., 13., 23., 33., 43., 53., , (c), (c), (c), (b), (b), (c), , 4., 14., 24., 34., 44., 54., , (c), (c), (b), (b), (d), (a), , 5., 15., 25., 35., 45., 55., , (d), (c), (c), (b), (b), (a), , 6., 16., 26., 36., 46., 56., , (b), (a), (b), (b), (c), (c), , 7., 17., 27., 37., 47., 57., , (c), (b), (c), (a), (c), (b), , 8., 18., 28., 38., 48., 58., , (b), (c), (c), (a), (a), (d), , 9., 19., 29., 39., 49., 59., , (d), (c), (d), (a), (b), (d), , 10., 20., 30., 40., 50., 60., , (c), (a), (b), (b), (b), (c), , Level II, 1., 11., 21., 31., 41., 51., , (b), (b), (a), (c), (d), (b)
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Hints & Solutions, Level I, 1. We know that by joining the two vertices of a octagon,, we get the diagonal of the octagon or side of the octagon., (Q An octagon contain eight vertices), ∴Number of line segments by joining the vertices of an, octagon taken two at a time, 8 × 7 ×6! 8 × 7, = 8C 2 =, =, = 28, 2, 2 ×1 ×6!, ∴From 28 lines, there are eight sides of the octagon., ∴Number of diagonals = 28 − 8 = 20 ., 2. In a bus there are five seats are vacant, if one of them, seated in 5 ways, then another can be seated in 4 ways., Thus, total number of ways in which both of them can be, seated = 5 × 4 = 20., 3. S = {2, 3, 4, 5, 7, 9}, Total number of digits = 6, The numbers to be formed are less than 400., Therefore, the hundreds place is to filled by 2 or 3., Next, we have to fill two places with 5 digits which, can be done in 5 C 2 ways., Thus, number of required 3-digit numbers, = 2 × 5 P2 = 2 × 20 = 40, 4. The number of triangles formed by joining the points, from a set of 12 points taking 3 at a time = 12C3 = 220., ∴ Number of triangles formed by 7 points taken 3 at a, time = 7C3 = 35, But 7 points lie on the straight line, therefore they don't, form a triangle., So, required number of triangles = 220 − 35 = 185, 5. 4 consonants out of 12 consonants can be selected in, 12, C 4 ways and 3 vowels out of 4 vowels can be selected in, 4, C3 ways. Thus, total ways of selecting 4 consonants, and 3 vowels, = 12C 4 × 4C3, Now, in the resulting words,, 4 consonants can be arranged in 4! ways, 3 vowels can be arranged in 3! ways., Thus, total ways of arranging 4 consonants and 3 vowels, =4!×3!, ∴Total number of words with all letters different, = 12C 4 × 4C3 × 4 ! × 3 !, = 285120, 6. Required number of ways = 6 × 5 = 30., 15, , n, n! , Q Pr = (n − r ) ! , 16, Pn − 2, , , 15 !, (16 − n + 2) ! 3, (18 − n ) !, 3, ×, = ⇒, =, ⇒, (15 − n + 1) !, 16 !, 4 16(16 − n ) ! 4, , 7. Q, , ⇒, ⇒, , Pn − 1, , =, , 3, 4, , (given), , (18 − n )(17 − n ) = 12, 306 − 17n − 18n + n 2 = 12, , ⇒, ⇒, ⇒, , n 2 − 35n + 294 = 0, (n − 14)(n − 21) = 0, n = 14, , (Q n ≠ 21), , 8. Required number of ways = 2 × 4 × 3 × 2 × 1 = 48., 9. In a word MAXIMUM, there are 4 consonants (i.e., M,, A, M, M,) and three vowels (i.e., A, I, U)., The pointed place to be filled by M, X, M, M., 3!×4!, ∴ Required number of ways =, =4!, 3!, 10. The total number between 10 and 1000 are 989 but we, have to form the numbers by using numbers 1, 2, ..., 9., Between 10 and 1000, the numbers are of 2 digits and, 3 digits., Since, repetition is allowed, so each digit can be filled, in 9 ways., Therefore, number of 2-digits numbers = 9 × 9 = 81, and number of 3-digits numbers = 9 × 9 × 9 = 729, Hence, total ways = 81 + 729 = 810, 11. We have, nC r = nC r − 1, n, ⇒, C n − r = nC r − 1 ⇒ n − r = r − 1 ⇒ 2r − n = 1 ...(i), n!, n!, ...(ii), and, =, ⇒ n − r =1, (n − r ) ! (n − r − 1) !, On solving Eqs. (i) and (ii), we get r = 2, and n = 3, 12. We have,, 47, , 5, , C4 + Σ, , r =1, , 52 − r, , C3 = 51C3 +, , C3 +, , 50, , 49, , C3, , + 48C3 + 47C3 + 47C 4, = C3 + C3 + C3 + C3 + 48C 4, (Q nC r + nC r + 1 = n + 1C r + 1 ), 51, 50, 49, = C3 + C3 + C3 + 49C 4, = 51C3 + 50C3 + 50C 4 = 51C3 + 51C 4 = 52C 4, 51, , 50, , 49, , 48, , 13. Since, total number of numbers 1 to 5 digits are 99999, and total number of numbers 1 to 4 digits are 9999., Hence, the total number of numbers of exact 5 digits, = 99999 − 9999 = 90000., 14. Number of ways that can be formed by using the words, 6!, ‘BANANA’ =, = 60, 3 !2 !, Number of ways in which two ‘N’ comes together, 5!, =, = 20, 3!, ∴ Required number of ways = 60 − 20 = 40, 15. Number of ways to choose 8 players from 12 players, 12 !, = 12C 8 =, = 495, 8 !4 !, and number of ways to choose a captain and a, vice-captain = 8C1 × 7C1 = 8 × 7 = 56, Hence, required number of ways, = 495 × 56 = 27720
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177, , Permutations and Combinations, 16. Given that, C (n , 12) = C (n , 8), n, ⇒, C12 = nC 8, ⇒, n = 12 + 8 = 20 (Q nC x = nC y ⇒ x + y = n ), 22, So,, C n = C (22, n ) = 22C 20, 22 !, =, = 231, 2 ! 20 !, , 25. To get an even number, unit’s place should be hold only, by 2 and 4. So, unit place can be filled in 2 ways. Now,, ten's place is filled by 4 left letters., ∴Total number of two digit even numbers, = 4C1 × 2P1 = 4 × 2, =8, , 17. Four different balls can be placed in four different boxes, = 4C1 + 3C1 + 2C1 + 1C1 = 4 + 3 + 2 + 1 = 10 ways, Only one way in which the same ball have a same box., ∴ Required number ways = 10 − 1 = 9, , 26. Since, table is round, so one seat is fixed., ∴Total number of ways in which 6 people can be seated, = (6 − 1) ! = 5 ! = 120, , 18. First we fix the alternate position of the boys. 5 boys can, be seated around the circle in (5 − 1) ! = 4 ! ., B, B, , B, , B, , B, , 5 girls can be seated in five vacant place by 5 ! ., ∴ Required number of ways = 4 ! × 5 !, 19. Since, total members are 15, but three special member, constitute a members. Therefore, required number of, arrangements are 12 ! × 2, because Chairman remains, between the two specified persons and the person can, sit in two ways., 20. Since, the man can go in 4 ways and he can back in 3, ways., ∴Total number of ways = 4 × 3 = 12, 21. The total number of 4 digits are 9999 − 999 = 9000, The numbers of 4 digits number divisible by 5 are (i.e.,, either 0 or 5 at the unit place and zero is not at the, thousand place) 9 × 10 × 10 × 2 = 1800., Hence, required number of ways= 9000 − 1800, = 7200, 22. Since, 9 ! = 362880, Which is not divisible by 990., Now, 11 ! = 39916800, Which is divisible by 990., Thus, the required smallest natural number 11., 23. Q, ⇒, ⇒, , Pr = 6Pr − 1, 5!, 6!, =, (5 − r ) ! (6 − r + 1)!, 5!, 6 ⋅5 !, =, (5 − r ) ! (5 − r ) ! (7 − r )(6 − r ), 5, , ⇒, (7 − r )(6 − r ) = 6, ⇒, 42 − 13r + r 2 = 6, ⇒, r 2 − 13r + 36 = 0, 2, ⇒ r − 9r − 4r + 36 = 0, ⇒, (r − 9)(r − 4) = 0, ⇒, r =4, (Q r ≠ 9), Here, r = 9 cannot be taken because in binomial, theorem or C (n , r ) notation n ≥ r ≥ 0., 24. Possibilities of words formed from the letters of word, ‘JOKE’ are JOKE, KOJE, KEJO, JEKO, EJOK, EKOJ,, OKEJ, OJEK, Thus, required number of words = 8., , 27. The number of lines obtained by joining 7 points in a, plane, 7!, = 7C 2 =, = 21, 2 !5 !, 28. We have to choose 5 members in which 3 men is must. It, means 2 members will be women., ∴3 men out of 7 and 2 women out of 3 have to selected., ∴Number of ways = 7C3 × 3C 2, 7!, 3!, =, ×, 3!×4! 2!×1!, 7 ×6 ×5 ×4! 1 ×2 ×3, =, ×, 1 ×2 ×3 ×4! 1 ×2 ×1, = 7 × 5 × 3 = 105, 29. Number of ways in which youngest child get 3 toys, = 9C3 and also each of other get 2 toys = 6C 2 × 4C 2 × 2C 2, ∴Number of required ways, = 9C3 × 6C 2 × 4C 2 × 2C 2, 9 ×8 × 7 6 ×5 4 ×3, =, ×, ×, ×1, 3 ×2 ×1 2 ×1 2 ×1, =9 ×8 × 7 ×5 ×3, = 7560, 30. Let L denote the lion and T denote the tiger., They can be arranged, i.e., LTLTLTLTLTLTLLLL, ∴ According to the question, lions can be arranged in, 10 ! ways and tigers are arranged in 11 P6 ways., ∴ Number of ways = 10 ! × 11P6., n (n − 3), 31. Number of diagonals =, 2, where, n is the number of sides of the polygon., n (n − 3), 20 =, ∴, 2, ⇒, 40 = n 2 − 3n, ⇒, n 2 − 3n − 40 = 0, 2, ⇒, n − 8n + 5n − 40 = 0, ⇒, (n − 8)(n + 5) = 0, ⇒, n −8 =0, (Q n + 5 ≠ 0), ⇒, n =8, 32. Required number of digits = 5 ! − 4 !, = 120 − 24 = 96, 33. We have, n P3 + nC n − 2 = 14n, n!, n!, ⇒, +, = 14n, (n − 3) ! 2 ! (n − 2) !, , n! , 1, ⇒, 1+, = 14n, (n − 3) ! , 2 (n − 2)
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178, , NDA/NA Mathematics, 2n − 3 , n (n − 1) (n − 2) , = 14n, 2(n − 2) , ⇒ 2n 2 − 5n − 25 = 0 ⇒ (2n + 5) (n − 5) = 0, Since, n is an integer., ∴, n −5 =0 ⇒ n =5, ⇒, , 34. The number will be even, if last digit is 2, 4, 6, 8 i.e., the, last digit can be filled in 4 ways and remaining two, digits can be filled in 8 P2 ways. Hence, required number, of numbers of three different digits = 8P2 × 4 = 224, 35. Three letters can be posted in 4 letter boxes in 43 = 64, ways but it consists the 4 ways that all letters may be, posted in same box. Hence required number of ways, = 64 − 4 = 60, 36. The required number of ways that 4 boys and 3 girls, can be seated, so that boys and girls alternate = 4 ! × 3 !, = 24 × 6 = 144, 37. Required number of ways, = coefficient of x16 in (x3 + x4 + x5 + ... + x7 )4, = coefficient of x16 in x12 (1 + x + x2 + ... + x4 )4, = coefficient of x16 in x12 (1 − x5 )4 (1 − x)− 4, = coefficient of x4 in (1 − x5 )4 (1 − x)− 4, = coefficient of x4 in (1 − 4x5 + ... ) (1 − x)−4, = coefficient of x4 in (1 − x)−4 = 7C 4 = 35, 38. Required number of triangles formed, = 12C3 − 7C3, 12 !, 7!, 12 ⋅ 11 ⋅ 10 7 ⋅ 6 ⋅ 5, =, −, =, −, 3 !9 ! 3 ! 4 !, 3 ⋅2 ⋅1, 3 ⋅2 ⋅1, = 220 − 35 = 185, (n + 2) ! + (n + 1)(n − 1) !, 39., (n + 1)(n − 1) !, =, , (n + 2)(n + 1)n (n − 1) ! + (n + 1)(n − 1) !, (n + 1)(n − 1) !, , = (n + 2)n + 1 = n 2 + 2n + 1 = (n + 1)2, = A perfect square, 40. Given that, eighteen football teams takes part in the, national championship and every team meets the same, opponent twice., ∴ During the championship total matches are played, 2 × 18 !, = 2 × 18C 2 =, = 2 × 153 = 306, 2 ! 16 !, , 41. There are six letters in MOTHER., The number of arrangements that can be made out of, the letters of the word ‘MOTHER’ taken four at a time, 5!, = 4 × 5 P3 = 4 ×, (5 − 3) !, 4 ×5 ×4 ×3 ×2×1, =, = 240, 2 ×1, 42. Let the sides of a polygon be n., Given that, number of diagonals = 54, n, ∴, C 2 − n = 54, n!, ⇒, − n = 54, 2 ! (n − 2) !, n (n − 1), − n = 54, ⇒, 2, 2, ⇒, n − 3n − 108 = 0, ⇒ (n + 9) (n − 12) = 0, Now,, n = −9, ∴, n − 12 = 0, n = 12, , (not possible), , 43. The number of ways = 3 ! × 3 ! × 3 ! × 3 ! = 1296, 44. A telephone dial has 10 holes, on rotating the dial, the, number of different telephone numbers, each consisting, of 7 digits can be dialled, if repetitions are permitted is, 107., 45. Required number of ways = 6C 4, 6!, 6 ×5, =, =, = 15, 4 !2 !, 2, 46. Required number of lines = 4C 2 =, , 4!, =6, 2 !2 !, , 47. Required number of ways = 26 − 1 = 64 − 1 = 63, 48. Required number of ways, = 7C3 × 5C 2 = 35 × 10 = 350, 49. Total number of members to sit in a ring = 10, ∴ Number of ways of sitting = (10 − 1) ! = 9 !, 50. The given word is ‘INDRAPRASTHA’., So, total number of letters = 12, Consonants are N, D, P, S, T, H, R, R, ∴Total number of consonants = 8, and number of vowels = 4, Hence, required number of ways, 9! 4!, =, ×, = 725760, 2! 3!, , Level II, 1. In a word ARTICLE, there are 7 letters. Out of 7 places,, 4 places are odd and 3 even. Therefore, 3 vowels can be, arranged in 4 odd places in 4 P3 ways remaining, 4 consonants can be arranged in 4 P4 ways., Hence, required number of ways, = 4P3 × 4P4, = 576, 2. Required number of ways = 8C5, 8 × 7 ×6, =, = 56, 3 ×2 ×1, , 3. Total number of candidates are 8 and we have to select, 5 candidates., The total number of ways a voter can vote, = 8C1 + 8C 2 + 8C3 + 8C 4 + 8C5, = 8 + 28 + 56 + 70 + 56 = 218, 4. A man has a two option for every friend either they invite it, or not., ∴ Required number of ways = 27 − 1 = 127., (Since, we have to subtract those cases in which they, do not invite any friend i.e., nC 0 = 1)
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179, , Permutations and Combinations, 5. Total number of volleyball players = 12, In a team captain will always be their., 11 ! 11 ⋅ 10 ⋅ 9, Required number of ways = 11C 8 =, =, 3 !8 !, 3 ⋅2 ⋅1, = 165, 6. Since, Ist letter may be post in 3 ways. Similarly, all the, letters individually may be posted in 3 ways., Thus, required number of ways, = 3 × 3 × 3 × 3 × 3 × 3 = 36, 7. A man can give votes for 4 candidates = 4 ways, ⇒ 5 men can give votes for 4 candidates, = 4 × 4 × 4 × 4 × 4 ways, ∴ Required number of ways = (4)5 = 1024 ways, 8. Total number of lines in an n-sided regular polygon = nC 2, and total number of sides in n-sided regular polygon = n, ∴ Number of diagonals in n-sided regular polygon, = nC 2 − n, n (n − 1), n − 1, , =, −n =n, − 1, 2, 2, , , n (n − 3), =, 2, 9. Q Combinations formed after taking 1, 2, 3, …, n things, at a time are nC1 , nC 2, ... , nC n., ∴ Total number of combinations, = nC1 + nC 2 + K + nC n, = 1 + nC1 + nC 2 + K + nC n − 1, = 2n − 1, [Q 2n = nC 0 + nC1 + nC 2 + K + nC n ], 10. Case I If examiner allot 2 marks (to each) to three, questions, then rest marks for fourth question, = − 3 × 2 + 10 = 4, If xi is marks given to ith question, then, x1 + x2 + x3 + x4 = 10, and, 2 ≤ xi ≤ 4, ∀ i = 1, 2, 3, 4., Case II Number of required ways, = coefficient of α 10 in (α 2 + α 3 + α 4 )4, = Coefficient of α 2 in (1 + α + α 2)4, 4, 1 − α3 , = Coefficient of α 2 in , , 1 −α , = Coefficient of α 2 in, (1 − 4α 3 + K ) (1 + 4α +, =, , 4⋅5, = 10, 2!, , 4⋅5 2, α + K), 2!, , 11. Suppose, we fix a 1 at unit place and the other 5 digits, can be arranged in 5 ! i.e., 120 ways., Similarly, if we fix the other numbers in unit place the, rest five digit can be arranged in 120 ways., ∴ Required sum of the unit place, = 120 (1 + 2 + 3 + 4 + 5 + 6) = 120 × 21 = 2520, 12. If a number is divisible by 3, the sum of the digits in it, must be a multiple of 3. The sum of the given six, numerals is 0 + 1 + 2 + 3 + 4 + 5 = 15. So, to make a five, digit number divisible by 3 we can either exclude 0 or 3., If 0 is left out, then 5 ! = 120 number of ways are, possible. If 3 is left out, then the number of ways of, , making a five digit number is 4 × 4 ! = 96, because 0, cannot be placed in the first place from left, as it will, give a number of four digit., Thus, the required number of ways = 120 + 96 = 216., 13. Since, out of eleven members two members sit together ,, then the number of arrangement, =9! ×2, (Q two members can be sit in two ways), 14. If the best and the worst paper always together, then the, number of ways = 5 ! × 2 ! = 240., 15. The number which is divisible by 5 and lying between, 3000 and 4000, 3 must be at thousand place and 5 must, be at unit place. Therefore, rest of the digits (1, 2, 4, 6), fill in two places., The number of ways = 4P2, n P (n , r ), n 1, n, n!, n! , 16. Σ, = Σ, ⋅, , Q Pr =, r =1, r = 1 r ! (n − r ) !, (, n, − r ) !, r!, , n, , = Σ, , r =1, , n, , Cr, , , n, n!, , Q C r =, r ! (n − r ) !, , , = (nC1 + nC 2 + nC3 + K + nC n ), = {1 + nC1 + nC 2 + nC3 + K + nC n } − 1, = { nC 0 + nC1 + nC 2 + K + nC n } − 1, = (1 + 1)n − 1 = 2n − 1, 17. Number of ways when one specified book is included, = 9C 4 = m, ⇒, m = 126,, and number of ways when one specified book is, excluded, = 9C5 = n, ⇒, n = 126, ⇒, m=n, 18. The number of ways that the candidate may select 2, questions, from, A, and, 4, questions, from, B = 5C 2 × 5C 4 = 50 ,3 questions from A and 3 questions, from B = 5C3 × 5C3 = 100; 4 questions from A and, 2 questions from B = 5C 4 × 5C 2 = 50 ⋅, Hence, total number of ways = 50 + 100 + 50 = 200, 19. Since, triangles can be formed by taking three points at, a time. The required number of ways = 8C3 − 5C3 − 3C3 ., (Since, total points are 8 but 5 are collinear and other, three are also collinear)., 1, 4!, 20. Required number of ways = (5 − 1) ! = ., 2, 2, (Since, clockwise and anti-clockwise are same in case of, ring), 21. We have, 15 C3 r = 15C r + 3, 15, ⇒, C15 − 3 r = 15C r + 3, ⇒, 15 − 3r = r + 3 ⇒ r = 3, 22. Let the number of candidates be n and the number of, persons to be elected is (n − 1). Therefore, a voter can, vote, = nC1 + nC 2 + nC3 + ...+ nC n − 1, ⇒ 256 = (nC 0 + nC1 + n C 2 + ...+ nC n − 1 + nC n ), − nC 0 − nC n, n, n, = 2 − 2 ⇒ 258 = 2 ⇒ n = 8
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180, , NDA/NA Mathematics, , 23. The possible non-zero positive integer ordered pair (x, y), satisfy the inequality x + y ≤ 4 is (1, 1), (1, 2), (1, 3), (2, 1),, (2, 2), (3, 1)., ∴ The number of required ordered pairs = 6., 24. Number of triangles using 5 points out of three are on a, straight line, 5!, = 5C3 − 3C3 =, 3 !2 !, 5 ×4, =, − 1 = 10 − 1 = 9, 2, 25. Since, the vowels in the word ‘UNIVERSAL’ are U, I, E,, A., Let us consider these as a single letter., Then, number of ways to arrange them = 6 ! = 720, But, vowels can also arranged in 4 ! or 24 ways., Hence, total number of ways = 720 × 24 = 17280, 26. Let their are n teams in a football championship. The, total matches they played with each other = nC 2, ⇒, 153 = nC 2, n!, ⇒, 153 =, (n − 2)! 2 !, n (n − 1), ⇒, 153 =, 2, ⇒, 306 = n (n − 1), ⇒, 18 (18 − 1) = n (n − 1) ⇒ n = 18, 27. Number of ways = 4 × 6 × 6, = 144, 28. Total number of ways = 4 × 3 × 2 × 1 = 24, 29. Total number of ways = 5C3 × 6C5 = 7200, 30. Starting with the letter A and arranging the other four, letters, there are 24 words. There are the first 24 words., Then, starting with G and arranging A, A, L, N in, 4!, different ways, there are, = 12 words. Next the 37th, 2!, word starts with L, there are 12 words starting with L., This accounts up to the 48th word. The 49th word is, ‘NAAGL’., P = { p1 , p2, p3 , p4 }, Q = { q1 , q2, q3 , q4 }, and R = { r1 , r2, r3 , r4 }, S10 = {( p2, q4 , r4 ), ( p3 , q3 , r4 ), ( p4 , q2, r4 ),, ( p4 , q3 , r3 ), ( p4 , q4 , r2), ( p3 , q4 , r3 )}, ∴ Total number of elements in S10 are 6., , 31. Q, , 32. Total number of ways = 11C 2, = 55, 33. Extreme left place can be filled up in 6 ways, the middle, place can be filled up in 6 ways and extreme right place, in only 3 ways., ∴ Required number of numbers = 6 × 6 × 3, = 108, 34. Let total number of teams that participated in a, championship is n., n (n − 1), n, ∴, C 2 = 153 ⇒, = 153, 2, ⇒, n (n − 1) = 306, ⇒, n = 18, , 35., , Head, , Tail, , 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, , 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, , Hence, total number of points in the sample space is 11., 36. In an octagon there are eight sides and eight points. The, diagonal will be formed by joining any two points except, the sides., ∴Required number of ways = 8C 2 − 8 = 28 − 8 = 20, 37. The word ‘UNIVERSAL’ have nine-different letters., ∴ Required number of words, 9!, = 9P3 =, = 504, 6!, 38. The committee consisting of atleast one member, = 12C1 + 12C 2 + 12C3 + ...+ 12C12, (Q nC 0 + nC1 + nC 2 + ... + nC n = 2n ), = 212 − 1 = 4096 − 1 = 4095, 39. Number of words in which all the 5 letters are repeated, = 105 = 100000 and the number of words in which no, 10 !, letter is repeated are 10 P5 =, = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 = 30240, 5!, Hence, the required number of ways, = 100000 − 30240 = 69760, 40. The word ‘DELHI’ have five different letters. If L is, fixed in the middle position, then first two letters can be, arranged in 4 P2 ways and rest two letters can be, arrange in 2P2 ways., ∴Required number of words = 4P2 ⋅2 P2, 4!, = ⋅ 2 ! = 4 ! = 24 ., 2!, 41. 2 women choose the chairs amongst the chairs marked, 1 to 4 is 4 P2 and 3 men can select the chairs out of 6, different numbered chairs in 6 P3 ways., ∴Required possible arrangement = 4P2 ⋅6 P3, 42. Total number of elementary events = 7C1 × 6C1 = 42, 43. Number of times 3 occurs = (when 3 occurs exactly at, one place) + (when 3 occurs exactly at two places) +, (when 3 occurs exactly at three places), = 3C1 × 9 × 9 + 3C 2 × 9 + 3C3, = 243 + 27 + 1 = 271, 44. There are three vowels and they have four odd places to, arrange. Other letters are four and have four places to, arrange., 4!, ∴ The number of words = 4P3 × 4 ! =, × 4 ! = 576, (4 − 3) !
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181, , Permutations and Combinations, 54. Assertion (A) Let N = 51 × 52 × 53 × 54 × 55 × 56, , 45. The number of ways, = 11C 6 − (7C 6 × 4C 0 + 7C5 × 4C1 ), 11 × 10 × 9 × 8 × 7 , 7 ×6, , =, − 7 +, × 4, , , 5 ×4 ×3 ×2, 2, = 462 − (7 + 84) = 371, 46. Total number of arrangements = 6 ! = 720, Total number of arrangements while all the, Hindi books are together = 4 ! × 3 !, = 24 × 6 = 144, The number of ways, in which books are arranged,, ∴, while all the Hindi books are not together, = 720 − 144 = 576, 47., , C k × 5C5 − k = 100, 5!, 5!, ×, = 100, ⇒, k ! (5 − k) ! (5 − k) ! 5 !, 5, , 2, , ⇒, ⇒, , , , 5!, = 100, , k ! (5 − k) !, 5!, = 10, k ! (5 − k) !, , Which is true for k = 2 or 3., 48. If a number is divisible by 3, the sum of the digits in it, must be a multiple of 3. The sum of the given six, numerals is 0 + 1 + 2 + 3 + 4 + 5 = 15. So, to make a five, digit number divisible by 3, we can either exclude 0 or 3., If 0 is left out, then 5 ! = 120 number of ways are, possible. If 3 is left out, then 4 × 4 ! = 96 number of ways, are possible., ∴ Total number of ways = 120 + 96 = 216., 49. If B must not precede A (immediately or otherwise),, then A must follow B, i.e., B should addressed at first, place., ∴ Required number of ways = 4 !, = 24, 50. Required number of tickets = 20 × 19, = 380, 51. Required number of ways = 3 P2 × 3 !, = 3 × 6 = 18, 52. The number of triangles formed by joining 12 points in a, plane out of which 6 are collinear = 12C3 − 6C3 = 200., Both A and R are true but R is not the correct, explanation of A., 100 , 100 , 53. (A) E7 (100 !) = , + 2 = 14 + 2 = 16, , 7, 7 , , , 100 100 100 100 , (R) E 2 (100 !) = , +, +, +, 2, 3, 4, 2 2 2 2 , 100, , , + 5 +, 2 , = 50 + 25 + 12 + 6 + 3 + 1 = 97, 100 , 100 100 100 , and E3 (100 !) = , + 2 + 3 + 4 , 3 , 3 3 3 , = 33 + 11 + 3 + 1 = 48, ∴, E12(100 !) = 48, A is true but R is false., , 100 , 26 , , × 57 × 58, and M = 40320 = 8 !, N 51 × 52 × 53 × 54 × 55 × 56 × 57 × 58 50 !, =, ∴, ×, M, 8!, 50 !, 58 !, =, 50 ! 8 !, This is always divisible., Reason (R) It is true that the product of r consecutive, natural numbers is always divisible by r !., 55. The total number of selections of 20 distinct thing taken, 20 !, 8 at a time = 20C 8 =, 8 ! 12 !, The total number of selections of 20 distinct thing taken, 20 !, 12 at a time = 20C12 =, 12 ! 8 !, ∴Assertion A is correct ., We know that, nC r = nC s, if n = r + s, ∴Both A and R are individually true and R is the correct, explanation of A., 56. There are 7 letters in the word 'Krishna' in which 2 are, vowels (i, a) and 5 are consonants (k, r, s, h, n)., By placing all the vowels in one bracket, we treat them, as one letter., Now, we need to arrange 6 letters k, r, s ,h, n, (i, a), The number of permutation formed of 6 letters, = 6P6 = 6 !, But the letter (i, a) can itself be arranged in 2P2 = 2 !ways, Hence, the required number of permutations, = 6 ! × 2 ! = 1440, 57. Evidently the odd places are 4 which have been, bracketed below (1)2(3)4(5)6(7), The number of vowels is 2. Hence, they can be arranged, in 4 P2 = 12 ways. If the odd places can be filled in anyone, of the above ways, then it remains to fill the remaining 3, even places by 5 consonants and this can be done in, 5, P3 = 5 × 4 × 3 = 60 ways., Since, corresponding to each permutation of the vowels,, the consonants can be arranged in 60 ways, therefore, the required number of permutations = 12 × 60 = 720, 58. We have to find out the number of those words which, begin with s and end in k., Omitting these two, there are only 5 letters left in the, middle., Hence, the required number of permutations, = 5 P5 = 5 ! = 120, 59. The number of ways of these selections are, Case I 4C 2 × 6C3 = 6 × 20 = 120, Case II 4C3 × 6C 2 = 4 × 15 = 60, Case III 4C 4 × 6C1 = 1 × 6 = 6, ∴The required number of ways = 120 + 60 + 6 = 186, 60. We have to make a selection of, (i) (1 lady out of 4) and (4 men out of 6) or, (ii) (2 ladies out of 4) and (3 men out of 6), Case I 4C1 × 6C 4 = 4 × 15 = 60, Case II 4C 2 × 6C3 = 6 × 20 = 120, ∴ Required number of ways = 60 + 120 = 180
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10, , Probability, Random Experiment, An experiment in which total outcomes are known in, advance but occurrence of specific outcome can be told only, after completion of the experiment., e. g. ,in tossing a coin, we know the total outcomes. It is, either a head or a tail., , Sample Space, Set of total possible outcomes in a random experiment, is sample space. In tossing of two coins sample space is, { HH , HT , TH , TT }., , Sample Point, Each outcome from sample space is a sample point, an, element or a case. In the example of tossing two coins there, are 4 sample points (4 elements or 4 cases)., , An Event, Any subset of sample space is an event. In tossing of, two coins, A : {HH, TT} → Same denominations appear on, both the coins., B : {HT, TH} → Different denominations appear, on the coins., C : {HT , TH , T T } → Atleast one tail appears., D : φ → Atleast three head appears (impossible, events)., E : {T T } → Exactly two tail appear., F : { HH , HT , TH , T T } → Arrow two head appear, (sure event)., A, B, C , D , E and F are different subsets of sample, space. Each subset is an event and can be defined as, written above., , Equally Likely Events, Two or more events (or sample points) are equally, likely, if none of them is biased over the other. Suppose a, number is picked from numbers 1 to 20, then events, defined as, , A : Picked number is even and B : Picked number is, odd, are equally likely as in given numbers there are 10 odd, numbers and 10 even numbers., , Mutually Exclusive Events, In a random experiment two or more events are said to, be mutually exclusive events (or disjoint events), if, occurrence of one event rules out the possibility of, occurrence of the others. In above example of tossing of two, coins, considered above events A and B are mutually, exclusive events (as A ∩ B = φ)., B and E are also mutually exclusive events. But A, B, and E are not mutually exclusive events, (as, A ∩ B = φ , B ∩ E = φ but A ∩ E ≠ φ)., , Exhaustive Events, In a random experiment two or more events are said to, exhaustive events, if one of them is sure to occure on, performing the experiment (i.e., Union of all the events, should be the sample space). In the same above example, tossing of two coins, events A and B are exhaustive events, (as A ∪ B is sample space). Also, A, B and C are exhaustive, events., , Favourable Events, The number of cases favourable to an event in a trial is, the number of outcomes which entail the happenting of the, event., e. g. ,In drawing a card from a pack of cards the number, of cases favourable to drawing of an ace is 4, for drawing a, spade is 13 and for drawing a red card is 26., , Complement of an Event, The complement of an event A denoted by A , A′ or Ac is, the set of all sample points of the space other than the, sample points in A., e.g., Let S = {1, 2, 3, 4, 5, 6}., If, A = { 1, 3, 5, 6},, then, Ac = { 2, 4}
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183, , Probability, , Independent Events, Two events A and B are independent events, if the, happening (or non-happening) of any does not affect the, happening (or non-happening) of other. For example an, urn contains 4 red and 5 green balls. A is an Ist event that, one green ball is drawn. B is 2nd event that a red ball is, drawn., Now, there are two cases, Case I When 2nd ball is drawn 1st ball is replaced in, the urn, then B does not depend on A., Case II Ist ball is not replaced, then B depends on, whether in the 1st draw, drawn ball is green or red (means, A has taken place or A′ has taken place), If E1 and E 2 are independent events, then, % E1 and E 2 are independent events., %, , E1 and E 2 are independent events., , %, , E1 and E 2 are independent events., , Probability, The probability of an event E to occur is the ratio of the, number of cases in its favour to the total number of cases., Number of cases favourable to event E, ∴ P( E ) =, Total number of cases, n( E ), =, n(S ), Probability of non-occurrence of event E is, P ( E ) = 1 − P( E ), and 0 ≤ P ( E ) ≤ 1. If P ( E ) = 1, then event E is known as, certain event and if P ( E ) = 0,then E is known as impossible, event., , Let A be the event that the number appearing on dice is greater, than 3, then A = {4, 5, 6}, ∴, n( A) = 3, n( A) 3 1, Hence, required probability =, = =, n( S) 6 2, , Example 2. A bag contains 3 white, 3 black and 2 red, balls, respectively. One by one, three balls are drawn without, replacing them. The probability that the third ball red is, 1, 1, 1, 1, (a), (b), (c), (d), 4, 2, 3, 7, Solution (a) Let W denotes the drawing of white ball,, B denote for black ball and R denote for red ball. Therefore,, required probability, = P(WWR) + P(WBR) + P(BBR) + P(WRR) + P(BWR), + P(BRR) + P(RWR) + P(RBR), 3 ⋅ 2 ⋅ 2 3 ⋅ 3 ⋅ 2 3 ⋅ 2 ⋅ 2 3 ⋅ 2 ⋅1 3 ⋅ 3 ⋅ 2, =, +, +, +, +, 8 ⋅7 ⋅6 8 ⋅7 ⋅6 8 ⋅7 ⋅6 8 ⋅7 ⋅6 8 ⋅7 ⋅6, 3 ⋅ 2 ⋅ 1 2 ⋅ 3 ⋅1 2 ⋅ 3 ⋅ 1, +, +, +, 8 ⋅7 ⋅6 8 ⋅7 ⋅6 8 ⋅7 ⋅6, 12, 18, 12, 6, 18, 6, 6, 6, =, +, +, +, +, +, +, +, 336 336 336 336 336 336 336 336, 84 1, =, =, 336 4, , Venn Diagram, Case I We have only two events A and B, 1. P ( A ∪ B) = P ( A) + P ( B) − P ( AB), (Addition theorem for two events), A, , S, , B, , A∪B, , Odds in Favour and Odds Against, an Event, In a random experiment, if out of ( m + n + p) equally, likely, mutually exclusive and exhaustive sample points,, m sample points are in favour of an event A, then m : n + p, is called odds in favour of A and ( n + p) : m is called odds in, against of A. Also, probability of A, Number of sample points in favour of A, =, Total number of sample points, m, P ( A) =, m+n+ p, , Example 1. An ordinary dice is thrown. The probability that, the number appearing on the dice greater than 3 is, 1, 1, (a), (b), 5, 3, 1, 1, (d), (c), 2, 4, Solution (c) Let S be the sample space, then, ∴, , S = {1, 2, 3, 4, 5, 6}, n( S) = 6, , (Shaded portion), , S = Sample space, A = Event A, B = Event B, P ( A ∪ B) is probability of occurrence of atleast one, out of the events A and B., 2. P ( AcBc ) = 1 − P ( A ∪ B), 3. P ( AcB) = P ( B) − P ( AB), (Probability of occurrence of exactly B event), 4. P ( ABc ) = P ( A) − P ( AB), (Probability of occurrence of exactly A event), Case II When we have three events A, B and C, A, , S, , B, , A ∪ B ∪ C (shaded portion), C
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184, , NDA/NA Mathematics, , 1. P ( A ∪ B ∪ C ) = P ( A) + P ( B) + P (C ), − P ( AB) − P ( BC ) − P (CA) + P ( ABC ), (Addition theorem for three events), 2. If A, B and C are mutually exclusive events, then, P ( A ∪ B ∪ C ) = P ( A) + P ( B) + P (C ), 3. If A, B and C are mutually exclusive and exhaustive, events, then P ( A) + P ( B) + P (C ) = 1., , Example 3. A dice is thrown. The probability that a, number greater than 4 may appear at the dice is, 1, 1, (b), (a), 3, 4, 1, 1, (c), (d), 2, 5, , %, , If A1 and A 2 are events, such that A2 ≠ 0, then, A , A , P 1 + P 1 = 1, A, 2, A2 , , Example 5. A fair coin is tossed repeatedly. The, probability of getting a result in the fifth toss different from, those obtained in the first four tosses is, 1, 1, 1, (a), (b), (c), (d) None of these, 15, 16, 17, Solution (b) Let E be the event of getting head on a coin., Required probability, = P(EEEEE ) + P(E E E E E), = {P(E)} 4 ⋅ P(E) + {P(E )} 4 ⋅ P(E), 1 1, 1 1 1, 1, = 4⋅ + 4⋅ = 4 =, 16, 2 2 2 2 2, , Solution (a) Let S be the sample space. Thus,, S = {1, 2, 3, 4, 5, 6}, ⇒, n( S) = 6, Let A be the event of appearing 5 on the dice, then, A = {5 } ⇒ n ( A) = 1, and, let B be the event of appearing 6 on the dice, then, B = {6 } ⇒ n(B) = 1, 1, 1, P( A) = , P(B) = ., ∴, 6, 6, Now, P (appearance of 5 or 6), 1 1 2 1, = P( A ∪ B ) = P( A) + P(B ) = + = =, 6 6 6 3, (Q A and B are mutually exclusives), , Baye’s Theorem, Let A1 , A2 , A3 , ... , An be certain events which are, mutually exclusive in pairs and which are exhaustive. Let, A be an event which occurs with A1, also with A2 , also with, A3 ,…, also with An . Then, Baye’s theorem states that, A, P ( Ak ) P , Ak , A , P k =, A, A, A, P ( A1 ) P + P ( A2 ) P , A1 , A2 , + .. . + P ( An ) P, , Example 4. Let A and B be the two possible outcomes of, an experiment and P( A) = 0.4, P(B) = x and P( A ∪ B) = 0.7., What is value of x, the events A and B are mutually exclusive?, (a) 0.3, (b) 0.2, (c) 0.5, (d) 0.7, , Solution (a) Since, A and B are mutually exclusive, we have, P( A ∪ B) = P( A) + P(B), 0.7 = 0 . 4 + x, x = 0.7 − 0 . 4 = 0.3, , ⇒, ⇒, , , , , , , , A , , An , , Example 6. The chances of defective screws in three boxes, , 1 1, 1, , and , respectively. A box is selected at, 5 6, 7, random and a screw drawn from it at random is found to be, defective. Find the probability that it came from box A is, 42, 41, 42, (b), (c), (d) None of these, (a), 107, 141, 243, , A, B and C are, , Solution (a) Let E1, E2 and E3 denote the events of selecting, , Conditional Probability, A and B are two events associated with a random, A, experiment, then P = Probability of occurrence of A,, B, when B has occurred., B, Similarly, P = Probability of occurrence of B, when, A, A has occurred., B, P ( A ∩ B) = P ( A)P , A, , (multiplication theorem), , B, If A and B are independent events, then P = P ( B)., A, So,, , P ( A ∩ B) = P ( A) P ( B), , box A, B, C, respectively and A be the event that a screw, selected at random is defective. Then,, P(E1) = P(E 2) = P(E3) =, , 1, 3, , A 1, A 1, A 1, P = ,P = ,P =, E 2 6, E3 7, E1 5, By Baye’s theorem,, , E , Required probability = P 1, A, P(E1) P( A / E1), (, ), (, /, P, E, P, A, E1) + P(E 2)P( A / E 2) , 1, , + P (E3) P ( A / E3) , , 1 1, ×, 42, 3, 5, =, =, 1 1 1 1 1 1 107, × + × + ×, 3 5 3 6 3 7, =
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185, , Probability, , Random Variable, , Solution (a) Let n and p be the parameters of the binomial, , Let S be the sample space in a random experiment., Then, a real valued function X which assigns to each, outcome r ∈S to a unique real number X (r) is called a, random variable., In tossing of two coins S = { HH , HT , TH , TT }, let X, denotes number of heads in tossing of two coins, then there, can be no head, one head or both heads (total sample space, covered). Here, X is a random variable such that, X ( HH ) = 2, X (TH ) = 1, X (TT ) = 0. Clearly, domain of X is, sample space and range of X is 0, 1 and 2., , X (r ) = Number of heads (same above example), , P( X ), %, , 0, 1, 4, , 1, 1, 2, , ∴, , 1, 1, , p = and n = 4, 2, 2, P ( X > 1) = 1 − P ( X ≤ 1), = 1 − P ( X = 0) − P ( X = 1), 4, 4, 1, 1 11, = 1 − 4C 0 − 4C1 =, 2, 2 16, , q=, , Binomial Distribution, , 1, 4, , P ( X = 1) = Probability of getting one head, 1, = P ( HT or TH ) =, 2, P ( X = 2) = Probability of getting two heads, 1, = P ( HH ) =, 4, Probability distribution of X is given by, X, , ⇒, , If probability of success of an event A in a random, experiment is p and probability of its failure is q ( p + q = 1),, experiment is being repeated independently n times, then, probability of exactly r success out of n trials is n Cr pr q n − r ., , Probability Distribution, P ( X = 0 )= Probability of getting no head = P (TT ) =, , distribution, respectively. Then, np = 2 and npq = 1, , 2, 1, 2, , Probability of atleast one success, = 1 − P (no success), = 1 − n C0 p0q n = 1 − q n ., , Example 8. From a box containing 20 tickets of value 1 to, 20, four tickets are drawn one by one. After each draw, the, ticket is replaced. The probability that the largest value of, tickets drawn 15 is, 27, 25, 33, (a), (b), (c), (d) None of these, 320, 743, 211, , Solution (a) The probability of drawing a number less than, or equal to 15 in a drawn =, , 15 3, =, 20 4, , The probability of drawing the ticket of value 15 in a draw, 1, =, 20, 3, 1 3, 4, ∴ Required probability = C1 ⋅, , 20 4, 27, =, 320, , The sum of the probability density function is 1., , Example 7. If the mean and variance of a binomial variate, X are 2 and 1 respectively, then the probability that X takes a, value greater than 1 is, 11, 11, 12, (a), (b), (c), (d) None of these, 16, 243, 121, , Comprehensive Approach, n, n, n, n, n, , n, , n, , P( A ∩ B ) = 1 − P( A ∪ B), P( A ∪ B ) = 1 − P( A ∩ B), P( A) = P ( A ∩ B) + P( A ∩ B ), P(B) = P(B ∩ A) + P(B ∩ A), If A and B are two events such that B ≠ φ, then, P( A /B) + P( A /B) = 1., 1, Probability of inserting allnletters in right addressed envelopes is ., n!, Probability of keeping atleast one letter in wrong envelope, 1, =1−, n!, , n, , n, , n, , Probability of keeping all the n letters in wrong envelopes, ( −1) n, 1 1, =, − + …+, 2 ! 3!, n!, Probability of keeping atleast one letter in right addressed envelope, =1− p, Playing cards A Pack (deck) of cads is 52. There are four suits, viz., spade, heart diamond and club each having 13 cards. There are, two colour red (heart and diamond) and black (spade and club), each having 26 cards. In 13 cards of each suit, there are 3 face cards, viz. king , queen and jack so there are in all 12 face cards in a pack, of playing cards. Also, there are 16 honour cards, 4 of each suit, viz., ace, king, queen and Jack.
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Exercise, Level I, 1. If A and B are two mutually exclusive events, what is, P ( AB)?, (a) 0, (b) P(A) + P ( B), (c) P ( A) P ( B), (d) P ( A) + P ( B / A), , 10. An urn contains 13 balls numbering 1 to 13. The, probability that a ball selected at random is a ball, with a number that is a multiple of 3 or 4 is equal to, (a) 6/13, (b) 5/13, (c) 4/13, (d) 2/13, , 2. A coin is tossed 4 times. The probability that atleast, one head turns up, is, (a) 1/16, (b) 2/16, (c) 14/16, (d) 15/16, , 11. A dice is thrown twice and the sum of the numbers, appearing is observed to be 6. The conditional, probability that the number 4 has appeared atleast, once, is, (a) 1/5, (b) 4/5, (c) 3/5, (d) 2/5, , 3. The mean and variance of a binomial distribution are, 6 and 4, respectively. The parameter n is, (a) 18, (b) 12, (c) 10, (d) 9, 4. A pair of a dice thrown, if 5 appears on atleast one of, the dice, then the probability that the sum is 10 or, greater, is, 11, 2, 3, 1, (b), (c), (d), (a), 36, 9, 11, 12, 5. Five coins whose faces are marked 2, 3 are tossed., The chance of obtaining a total of 12 is, 1, 1, 3, 5, (a), (b), (c), (d), 32, 16, 16, 16, 6. A husband and wife appear in an interview for two, vacancies in the same post. The probability of, 1, husband’s selection is and that of wife’s selection is, 5, 1, . What is the probability that only one of them will, 3, be selected?, (NDA 2011 II), 1, 2, (a), (b), 5, 5, 3, 4, (d), (c), 5, 5, 7. Two letters are drawn at random from the word, ‘HOME’. What is the probability that both the letters, are vowels?, (NDA 2011 II), 1, 5, 1, 1, (a), (b), (c), (d), 6, 6, 2, 3, 8. The probability of solving a problem by three, students is 1/2, 1/3 and 1/4, respectively. The, probability that the problem is not to be solved, is, equal to, (a) 1/3, (b) 1/2, (c) 1/4, (d) 3/4, 9. If A and B are two events such that P ( A ∪ B) = 3 / 4,, P ( A ∩ B) = 1/ 4 and P ( A ) = 2/ 3, then P ( B) is equal to, (a) 1/3, (b) 2/3, (c) 1/9, (d) 2/9, , 12. A box contains 6 distinct dolls. From the box, 3 dolls, are randomly selected one by one with replacement., What is the probability of selecting 3 distinct dolls?, (NDA 2011 I), , 5, (a), 54, 1, (c), 20, , 12, (b), 25, 5, (d), 9, , 13. There are 4 letters and 4 directed envelopes. These, 4 letters are randomly inserted into the 4 envelopes., What is the probability that the letters are inserted, into the corresponding envelopes?, (NDA 2011 I), 11, 23, (a), (b), 12, 24, 1, (c), (d) None of these, 24, 14. If A and B are events, P ( B ) = 0.8 and P ( A / B) =, to?, (a) 0.08, (c) 0.8, , such that P ( A ∪ B) = 0.5,, 0.4, what is P ( A ∩ B) equal, (NDA 2011 I), , (b) 0.02, (d) 0.2, , 15. If a coin be tossed n times, then probability that the, head comes odd times is, (a) 1/ 2, (b) 1/ 2n, n −1, (c) 1/ 2, (d) None of these, 16. The probability that the same number appear on, throwing three dice simultaneously is, (a) 1/6, (b) 1/36, (c) 5/36, (d) None of these, 17. Three dice are thrown. What is the probability that, the same number will appear on each of them?, (NDA 2012 I), , 1, (a), 6, 1, (c), 24, , 1, (b), 18, 1, (d), 36
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187, , Probability, 18. In a binomial distribution the probability of getting a, success is 1/4 and standard deviation is 3, then its, means is respectively, (a) 6, (b) 8, (c) 12, (d) 10, 19. A bag contains 5 black balls, 4 white balls and 3 red, balls. If a ball is selected random wise the probability, that it is a black or red ball, is, (a) 1/3, (b) 1/4, (c) 5/12, (d) 2/3, 20. What is the probability of having 53 Sundays or, 53 Mondays in a leap year?, (NDA 2010 II), 2, 3, 4, 5, (a), (b), (c), (d), 7, 7, 7, 7, 21. Three-digit numbers are formed using the digits 0, 2,, 4, 6 and 8, respectively. A number is chosen at, random out of these numbers. What is the, probability that the number has the same digits?, (NDA 2010 II), , (a), , 1, 16, , (b), , 1, 25, , (c), , 16, 25, , (d), , 1, 645, , 22. What is the probability that a leap year selected at, random contains 53 Mondays?, 1, 2, 7, 26, (b), (c), (d), (a), 7, 7, 366, 183, 23. In tossing a coin twice, let E and F denote occurrence, of head on first toss and second toss, respectively., Then, P ( E ∪ F ) is equal to, 1, 1, 3, 1, (b), (c), (d), (a), 4, 2, 4, 3, 24. The probability of having a king and a queen, when, the two cards are drawn at random from a pack of 52, cards is, 16, 8, (a), (b), 663, 663, 4, 2, (d), (c), 663, 663, 25. A card is drawn from an ordinary pack and a gambler, bets that it is either a spade or an ace. The odds, against his winning are, (a) 9 : 4, (b) 9 : 5, (c) 9 : 6, (d) 9 : 8, 26. In tossing three coins at a time, what is the, probability of getting atmost one head? (NDA 2010 I), 3, 7, (a), (b), 8, 8, 1, 1, (d), (c), 2, 8, 27. Two balls are selected from a box containing 2 blue, and 7 red balls, respectively. What is the probability, that atleast one ball is blue?, (NDA 2010 I), 2, 7, 5, 7, (a), (b), (c), (d), 9, 9, 12, 12, , 28. A bag contains 3 black and 4 red balls, respectively., Two balls are drawn at random one at a time without, replacement. The probability that the first ball, drawn is black, if the second ball is known to be red,, is, 1, 1, 1, 1, (a), (b), (c), (d), 2, 4, 6, 8, 29. If X is a binomial variate with n = 2 and p = 0 . 6, then, X, variance of the random variable will be, 2, (a) 0.12, (b) 0.24, (c) 0.36, (d) 0.48, 30. If A and B are two events, such that P [ A or B] = P [ A] ,, then, (a) events A and B are mutually exclusive, (b) events A and B are statistically independent, (c) event B is a subset of event A, (d) event A is a subset of event B, 31. An electric device consists of two bulbs A and B. From, previous testing procedure, the following results are, known, P ( A is fused ) = 0.20, P(B is fused alone) = 015, ., P(A and B are fused) = 015, ., What is the probability that bulb A is fused alone?, (a) 0.15, (b) 0.20, (c) 0.05, (d) 0.25, 32. If four dice are thrown together, then what is the, probability that the sum of the numbers appearing, on them is 25?, (NDA 2012 I), 1, 1, (a) 0, (b), (c) 1, (d), 2, 1296, x, 33. The probability of guessing a correct answer is . If, 12, the probability of not guessing the correct answer is, 2, , then x is equal to, (NDA 2010 I), 3, (a) 2, (b) 3, (c) 4, (d) 6, 1, 3, 11, 34. Given that, P ( A) = , P ( B) = and P ( A ∪ B) = ,, 3, 4, 12, B, what is the value of P ?, (NDA 2009 II), A, (a), , 1, 6, , (b), , 4, 9, , (c), , 1, 2, , (d), , 1, 3, , 35. A bag contains 5 white and 3 black balls and 4 balls, are successively drawn out and not replaced. The, probability that they are alternately of different, colours, is, (a) 1/196, (b) 1/7, (c) 13/56, (d) 3/7, 36. The probability that in a family of 5 members, exact, 2 members have birthday on sunday, is, 12 × 53, 10 × 62, 10 × 63, (b), (c), 1, (d), (a), 75, 75, 75
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188, , NDA/NA Mathematics, , 37. A and B toss a coin alternately till one of them tosses, heads and wins the game, their respective, probabilities of winning are, 1, 3, 1, 1, (a) and, (b) and, 4, 4, 2, 2, 1, 2, 1, 4, (d) and, (c) and, 3, 3, 5, 5, 38. If birth to a male child and birth to a female child are, equal probable, then what is the probability that, atleast one of the three children born to a couple is, male?, 4, 7, 8, 1, (b), (c), (d), (a), 5, 8, 7, 2, 1, 39. The chance of winning the race of the horse A is and, 5, 1, that of horse B is . What is the probability that the, 6, race will be won by A or B?, (NDA 2008 II), 1, 1, (b), (a), 30, 3, 11, 1, (c), (d), 30, 15, 40. What is the probability of two persons being born on, the same day (ignoring date)?, (NDA 2008 II), 1, 1, (a), (b), 49, 365, 1, 2, (d), (c), 7, 7, 41. If P ( A) = 0.25, P ( B) = 0.50 and P ( A ∩ B) = 014,, then, ., P ( A ∩ B ) is equal to, (a) 0.61, (b) 0.39, (c) 0.48, (d) None of these, 42. One card is drawn randomly from a pack of 52 cards,, then the probability that it is a king or spade, is, (a) 1/26, (b) 3/26, (c) 4/13, (d) 3/13, 43. From a pack of 52 cards two cards are drawn in, succession one by one without replacement. The, probability that both are aces is, (a) 2/13, (b) 1/51, (c) 1/221, (d) 2/21, , 44. Three coins are tossed together, then the probability, of getting atleast one head is, 1, 3, (a), (b), 2, 4, 1, 7, (d), (c), 8, 8, 45. A problem in Mathematics is given to three students, A, B and C and their respective probability of solving, the problem is 1/2, 1/3 and 1/4, respectively., Probability that the problem is solved, is, (a) 3/4, (b) 1/2, (c) 2/3, (d) 1/3, 46. If two dice are thrown simultaneously, then, probability that 1 comes on first dice is, (a) 1/36, (b) 5/36, (c) 1/6, (d) None of these, 47. What is the probability that in a family of 4 children, there will be atleast one boy?, (NDA 2008 I), 15, 3, 1, 7, (a), (b), (c), (d), 16, 8, 16, 8, 48. Each of A and B tosses two coins. What is the, probability that they get equal number of heads?, (NDA 2007 II), , 3, (a), 16, , 5, (b), 16, , 4, (c), 16, , 5, 49. Given P ( A ∪ B) = , P ( A ∩, 6, What is the value of P ( A )?, 1, (a), (b), 6, 2, (d), (c), 3, , B) =, , (d), , 6, 16, , 1, 1, and P ( B) = ., 3, 2, (NDA 2008 I), , 1, 3, 1, 2, , 3, 50. If A and B are any two events such that P ( A ∪ B) = ,, 4, 1, 2, P ( A ∩ B) = and P ( A ) = , where A stands for the, 4, 3, complementary event of A, what is the value of P ( B)?, (NDA 2007 I), , 1, (a), 3, , 2, (b), 3, , 1, (c), 9, , (d), , 2, 9, , Level II, 1. For two events A and B, if, A 1, B 1, P ( A) = P = and P = , then, B 4, A 2, (a) A and B are independent, A′ 3, (b) P =, B 4, B′ 1, (c) P =, A′ 2, (d) All of the above, , 2. If M and N are any two events. The probability that, exactly one of them occurs is, (a) P ( M ) + P ( N ) − P ( M ∩ N ), (b) P ( M ) + P ( N ) + P ( M ∩ N ), (c) P ( M ) + P ( N ), (d) P ( M ) + P ( N ) − 2P ( M ∩ N ), 3. If four dice are thrown together. Probability that the, sum of the number appearing on them is 13, is, 35, 5, 11, 11, (a), (b), (c), (d), 324, 216, 216, 432
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189, , Probability, 4. A coin is tossed 10 times. The probability of getting, exactly six heads is, (a) 512/513, (b) 105/512, (c) 100/153, (d) 10C6, 5. In a random arrangement of the letters of the word, ‘UNIVERSITY’, what is the probability that two I’s, do not come together?, (NDA 2011 II), 4, 1, (b), (a), 5, 5, 1, 9, (c), (d), 10, 10, 6. In a class of 125 students, 70 passed in Mathematics,, 55 passed in Statistics and 30 passed in both, subjects. What is the probability that a student, selected at random from the class has passed in only, one subject?, (NDA 2011 II), 13, 3, (a), (b), 25, 25, 17, 8, (d), (c), 25, 25, 7. An experiment consists of flipping a coin and then, flipping it a second time, if head occurs. If a tail, occurs on the first flip, then a six faced die is tossed, once. Assuming that the outcomes are equally likely,, what is the probability of getting one head and one, tail?, (NDA 2011 I), 1, 1, (a), (b), 4, 36, 1, 1, (d), (c), 6, 8, 8. Two persons A and B throw a die alternately till one, of them gets a 3 and wins the game, the respective, probabilities of winning, if A begins, are, 7 4, 6 5, (a), (b), ,, ,, 11 11, 11 11, 5 1, 4 3, (d) ,, (c) ,, 6 6, 7 7, 9. If three natural numbers from 1 to 100 are selected, randomly, then probability that all are divisible by, both 2 and 3, is, 4, 4, (b), (a), 105, 33, 4, 4, (c), (d), 35, 1155, 10. 5 boys and 5 girls are sitting in a row randomly. The, probability that boys and girls sit alternatively, is, 5, 1, (a), (b), 126, 42, 4, 6, (d), (c), 126, 126, 11. Let A and B are two independent events. The, probability that both A and B occur together is 1/6, and the probability that neither of them occurs is 1/3., The probability of occurrence of A is, , (a) 0 or 1, (c) 1/ 2 or 1 / 4, , (b) 1/ 2 or 1/ 3, (d) 1/ 3 or 1 / 4, , 12. A coin is tossed n times. The probability of getting, head atleast once is greater than 0.8, then the least, value of n is, (a) 2, (b) 3, (c) 4, (d) 5, 13. In an examination, there are 3 multiple choice, questions and each question has 4 choices. If a, student randomly selects answer for all the, 3 questions, what is the probability that the student, will not answer all the 3 questions correctly?, (NDA 2011 I), , 1, (a), 64, , 63, (b), 64, , 1, (c), 12, , (d), , 11, 12, , 14. If a machine is correctly setup, it produces 90%, acceptable items. If it is incorrectly setup, it produces, only 40% acceptable items. Past experience shows, that 80% of the setups are correctly done. If after a, certain setup, the machine produces 2 acceptable, items, find the probability that the machine is, correctly setup., (a) 0.97, (b) 0.95, (c) 0.99, (d) 0.90, 15. In a college, 25% of the boys and 10% of the girls offer, Mathematics. The girls constitute 60% of the total, number of students. If a student is selected at, random and is found to be studying Mathematics., The probability that the student is a girl, is, 1, 3, (b), (a), 6, 8, 5, 5, (d), (c), 8, 6, 16. A, B and C are three mutually exclusive and, exhaustive events associated with a random, 3, 1, experiment. If P ( B) = P ( A) and P (C ) = P ( B), then, 2, 2, P ( A) is equal to, 3, 4, 2, 1, (a), (b), (c), (d), 4, 13, 3, 2, 17. A committee of 3 is to be chosen from a group, consisting of 4 men and 5 women. If the selection is, made at random, then the probability that two of the, members of the committee are men, is given by, (a) 3/14, (b) 5/14, (c) 1/21, (d) 8/21, 18. If X has binomial distribution with mean np and, P ( X = k), variance npq, then, , is equal to, P ( X = k − 1), n−k p, n − k+1 p, (b), (a), ⋅, ⋅, k−1 q, k, q, n +1 q, n −1 q, (c), (d), ⋅, ⋅, k, p, k+1 p
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190, , NDA/NA Mathematics, , 19. If A and B are any two events, then P ( A ∩ B) is, equal to, (a) P ( A ) P ( B ), (b) 1 − P ( A) − P ( B), (c) P ( A) + P ( B) − P ( A ∩ B) (d) P ( B) − P ( A ∩ B), 20. There is a point inside a circle. What is the, probability that this point is close to the, circumference than to the centre?, (NDA 2011 II), 3, 1, 1, 1, (b), (c), (d), (a), 4, 2, 4, 3, 21. In the following Venn diagram circles A and B, represent two events, A, , B, , The probability of the union of shaded region will be, (a) P ( A) + P ( B) − 2 P ( A ∩ B), (b) P ( A) + P ( B) − P ( A ∩ B), (c) P ( A) + P ( B), (d) 2P ( A) + 2P ( B) − P ( A ∩ B), 22. A box contains 10 identical electronic components of, which 4 are defective. If 3 components are selected at, random from the box in succession, without replacing, the units already drawn, what is the probability that, two components of the selected components are, defective?, (NDA 2007 I), 1, 5, (a), (b), 5, 24, 3, 1, (c), (d), 10, 40, 23. The probability distribution of random variable X, with two missing probabilities p1 and p2 is given, below, (NDA 2010 I), X, 1, , II. The probability that there are 5 Mondays in the, month of March is thrice the probability that, there are 5 Mondays in the month of April., Which of the statements given above is/are correct?, (a) I only, (b) II only, (c) Both I and II, (d) Neither I nor II, 25. Let E1 , E2 and E3 be three arbitrary events of a sample, space S. Consider the following statements, I. P (only one of them occurs), = P ( E1E2E3 + E1E2E3 + E1E2E3 ), II. P (none of them occurs) = P ( E1 + E2 + E3 ), III. P (atleast one of them occurs) = P ( E1 + E2 + E3 ), IV. P (all the three occurs) = P ( E1 + E2 + E3 ), where P(Ei ) denotes the probability of Ei and Ei, denotes complement of Ei ., Which of the above statements are correct?, (a) I and II, (b) II and III, (c) I and III, (d) II and IV, 26. Consider the following statements related to a, variable X having a binomial distribution bX ( n , p), 1, I. If p = , then the distribution is symmetrical., 2, II. p remaining constant, P ( X = r ) increases as n, increases., Which of the statements given above is/are correct?, (a) I only, (b) II only, (NDA 2010 I), (c) Both I and II, (d) Neither I nor II, 27. Consider the following statements related to the, nature of Baye’s theorem., I. Baye’s theorem is a formula for computation of a, conditional probability., II. Baye’s theorem modifies an assumed probability, of an event in the light of a related event which is, observed., Which of the statements given above is/are correct?, (NDA 2008 II), , P( X ), k, , 2, , p1, , 3, , 4k, , 4, , p2, , 5, , 2k, , It is further given that, P ( X ≤ 2) = 0.25 and, P ( X ≥ 4) = 0.35. Consider the following statements, I. p1 = p2, II. p1 + p2 = P ( X = 3), Which of the statements given above is/are correct?, (a) I only, (b) II only, (c) Both I and II, (d) Neither I nor II, 24. Consider the following statments, (NDA 2010 I), I. The probability that there are 53 Sundays in a, leap year is twice the probability that there are, 53 Sundays in a non-leap year., , (a) I only, (c) Both I and II, , (b) II only, (d) Neither I nor II, , 28. Match List I (Equality/Inequality) with List II, (Inference) and select the correct answer using the, codes given below the list, List I, (Equality/Inequality), , List II, (Inference), , A. P( E1 ) + P( E 2 ) = 1, , 1. E1 , E 2 are mutually, exclusive events, , B. P ( E1 ) + P ( E 2 ) = 0, , 2. E1 , E 2 are mutually, exhaustive events, , C. P( E1 ) + P( E 2 ) ≤ 0, , 3. E1 , E 2 are sure events, , D. P( E1 ) P( E 2 ) = 1, , 4. E1 , E 2 are impossible, events, 5. E1 , E 2 are not equally, likely events
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191, , Probability, Codes, A B, (a) 2 1, (c) 3 1, , C, 4, 5, , D, 3, 2, , A, (b) 2, (d) 1, , B, 4, 3, , C, 1, 2, , D, 3, 5, , 29. Three letters are randomly selected from the 26, capital letters of the english alphabet. What is the, probability that the letter A will not be included in, the choice?, (NDA 2009 II), 1, 23, (b), (a), 2, 26, 12, 25, (d), (c), 13, 26, 30. The probabilities of two events A and B are given as, P ( A) = 0.8 and P ( B) = 0.7. What is the minimum, value of P ( A ∩ B)?, (NDA 2009 II), (a) 0, (b) 0.1, (c) 0.5, (d) 1, 31. Two numbers X and Y are simultaneously drawn, from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. What is the, conditional probability of exactly one of the two, numbers X and Y being even, given ( X + Y ) = 15?, (NDA 2009 II), , 3, (b), 4, 1, (d), 4, , (a) 1, (c), , 1, 2, , 32. A lot of 4 white and 4 red balls is randomly divided, into two halves. What is the probability that there, will be 2 red and 2 white balls in each half?, (NDA 2010 II), , 18, (a), 35, 1, (c), 2, , 3, (b), 35, (d) None of these, , 33. The mean and variance of a binomial distribution are, 8 and 4 respectively. Then, P ( X = 1) is equal to, (NDA 2010 I), , (a), (c), , 1, , (b), , 212, 1, , (d), , 26, , 1, 28, 1, 24, , 34. For a binomial distribution B ( n , p), np = 4 and variance, npq = 4 / 3. What is value of the probability P ( x ≥ 5) ?, (NDA 2009 I), , 2, (a) , 3, , 6, , 1, (c) , 3, , 6, , (b), (d), , 25, 36, 28, 36, , 35. If P ( A) = 0.8, P ( B) = 0.9, P ( AB) = p, which one of the, following is correct?, (NDA 2008 II), (a) 0.72 ≤ p ≤ 0.8, (b) 0.7 ≤ p ≤ 0.8, (c) 0.72 < p < 0.8, (d) 0.7 < p < 0.8, , 36. The outcomes of an experiment classified as success, A or failure A will follow a binomial distribution, if, (NDA 2008 II), , (a), (b), (c), (d), , 1, P ( A) =, 2, P ( A) = 0, P ( A) = 1, P ( A) remains constant in all the trials, , 37. A bag X contains 2 white and 3 black balls and, another bag Y contains 4 white and 2 black balls. One, bag is selected at random and a ball is drawn from it., Then, the probability for the ball chosen be white, is, 2, 7, (a), (b), 25, 15, 8, 14, (c), (d), 15, 15, 38. A coin is tossed 2n times. The chance that the, number of times one gets head is not equal to the, number of times one gets tail, is, 2n, ( 2n !), ( 2n !) 1 , (a), (b) 1 −, , ( n !)2, ( n !)2 2, ( 2n !) 1, (d) None of these, (c) 1 −, ., ( n !)2 4n, 39. The probability that atleast one of the events A and B, occurs is 3/5. If A and B occur simultaneously with, probability 1/5, then P ( A′ ) + P ( B′ ) is, 2, 4, (a), (b), 5, 5, 6, 7, (c), (d), 5, 5, 1, 5, 1, 40. If P ( A ∩ B) = , P ( A ∪ B) = , and P ( A) = , then, 3, 6, 2, which one of the following is correct?, (a) A and B are independent events, (b) A and B are mutually exclusive events, (c) P ( A) = P ( B), (d) None of the above, 41. In x = 33n , n is a positive integral value, then what is, the probability that x will have 3 at its unit place?, (a) 1/ 3, (b) 1/ 4, (c) 1/ 5, (d) 1/ 2, 42. A die is tossed thrice. If event of getting an even, number is a success, then the probability of getting, atleast two successes is, (a) 7/8, (b) 1/4, (c) 2/3, (d) 1/2, 43. If a committee of 3 is to be chosen from a group of 38, people of which you are a member. What is the, probability that you will be on the committee?, 38, 37, (a) , (b) , 3, 2, 37 38, (c) / , (d) 666/8436, 2 3
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192, , NDA/NA Mathematics, , 44. A coin is tossed. If head is observed, a number is, randomly selected from the set {1, 2, 3} and if tail is, observed, a number is randomly selected from the set, {2, 3, 4, 5}. If the selected number be denoted by X,, what is the probability that X = 3?, (NDA 2008 II), 2, 1, (a), (b), 7, 5, 1, 7, (c), (d), 6, 24, 45. The outcomes of 5 tosses of a coin are recorded in a, single sequence as H (head) and T (tail) for each toss., What is the number of elementary events in the, sample space?, (NDA 2008 I), (a) 5, (b) 10, (c) 25, (d) 32, 46. Which of the following numbers is nearest to the, probability that three randomly selected persons are, born on three different days of the week? (NDA 2008 I), (a) 0.7, (b) 0.6, (c) 0.5, (d) 0.4, 47. One bag contains 5 white and 3 black balls and a, second bag contains 2 white and 4 black balls. One, ball is drawn from the first bag and placed unseen in, the second bag. What is the probability that a ball, now drawn from the second bag is black?(NDA 2008 I), 15, 35, (b), (a), 56, 56, 37, 25, (c), (d), 56, 48, 48. A and B are two events and A denotes the, complements of A., Consider the following statements, I. P ( A ∪ B) ≤ P ( B) + P ( A), II. P ( A) + P ( A ∪ B) ≤ 1 + P ( B), Which of the above statements is/are correct?, (NDA 2007 II), , (a) I only, (c) Both I and II, , (b) II only, (d) Neither I nor II, , 49. Six text books numbered 1, 2, 3, 4, 5 and 6 are, arranged at random. What is the probability that the, text books 2 and 3 will occupy consecutive places?, (NDA 2007 II), , 1, (a), 2, , 1, (b), 3, , 1, (c), 4, , (d), , 1, 6, , 50. In a lottery, 16 tickets are sold and 4 prizes are, awarded. If a person buys 4 tickets, what is the, probability of his winning a prize?, (NDA 2007 I), 4, 175, 1, 81, (c), (d), (b), (a), 4, 256, 4, 256, 16, 51. If two events A and, P ( Ac ) = 0.3, P ( B) = 0.4 and, P ( B / ( A ∪ Bc )) is equal to, , B are such, P ( ABc ) = 0.5,, , that, then, , 1, 2, 1, (c), 4, , (a), , (b), , 1, 3, , (d) None of these, , 52. What is the probability of a well balanced coin when, it flips 12 times and come 5 heads and 7 tails?, (NDA 2007 I), , (a), (c), , C (12, 5), ( 2)5, C (12, 5), ( 2)12, , (b), (d), , C (12, 5), ( 2)7, C (12, 7), ( 2)6, , 53. A can hit a target 4 times in 5 shots, B can hit a target, 3 times in 4 shots and C can hit a target 2 times in 3, shots. All the three fire a shot each. What is the, probability that two shots are atleast hit?(NDA 2007 I), 1, 3, (a), (b), 6, 5, 5, 1, (d), (c), 6, 3, 54. A card is drawn from a pack of 52 cards and a, gambler bets that it is a spade or an ace. Which one of, the following is the odds against his winning this bet?, (NDA 2007 I), , (a) 13 to 4, (c) 9 to 4, , (b) 4 to 13, (d) 4 to 9, , Directions (Q. Nos. 55-58), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 55. In a population of men (two eyed), the probability, that a man’s right eye is brown is p and the, probability that a man’s left eye is brown is also p., Assertion (A) The probability that a man has, atleast one brown eye is not equal to 2p., Reason (R) Probability (left eye brown or right eye, brown) = Probability (left eye brown) + Probability, (right eye brown) = 2p, 56. Assertion (A) If A is any event and P ( B) = 1 , then A, and B are independent., Reason (R) P ( A ∩ B) = P ( A) ⋅ P ( B), if A and B are, independent., 57. Assertion (A) If P ( A) = P ( B) = P (C ) = 1/ 4,, P ( A ∩ B) = P ( B ∩ C ) = 0,, then, P ( A ∩ C ) = 1/ 8
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193, , Probability, Reason (R) P ( A ∪ B ∪ C ) = P ( A) + P ( B) + P (C ), − P ( A ∩ B) − P ( B ∩ C ) − P ( A ∩ C ) + P ( A ∩ B ∩ C )., 58. Assertion (A) The probability of drawing either an, ace or a king from a deck of card in a single draw is, 2, ., 13, Reason (R) For two events E1and E2 which are not, mutually exclusive, the probability is given by, P ( E1 + E2 ) = P ( E1 ) + P ( E2 ) − P ( E1 ∩ E2 ), , Directions (Q. Nos. 59-61), , Consider a die has, one face with number ‘1’ two faces each with number ‘2’, and three faces each with number ‘3’. If a dice is rolled, once, then, 59. P(1) is, 1, (a), 4, 1, (c), 2, , (b), , (b), , (b), , 1, 2, , (d) None of these, , Directions (Q. Nos. 62-64), , Consider A and B are, two non-mutually exclusive events., 1, 2, 1, If P ( A) = , P ( B) = and P ( A ∪ B) = ,, 4, 5, 2, , 62. The value of P ( A ∩ B) is, 4, 3, (b), (a), 13, 20, 3, (c), (d) None of these, 43, 63. The value of P ( A ∩ B′ ) is, 1, 2, (b), (a), 10, 13, 1, (d) None of these, (c), 5, , 1, 6, , (d) None of these, , 60. P( 2 or 3) is, 1, (a), 2, 5, (c), 6, , 61. P( not 3) is, 1, (a), 5, 7, (c), 13, , 64. The value of P ( A′∩ B′ ) is, 1, 1, (a), (b), 3, 2, 1, (c), (d) None of these, 5, , 2, 3, , (d) None of these, , Answers, Level I, 1., 11., 21., 31., 41., , (a), (d), (b), (b), (d), , 2., 12., 22., 32., 42., , (d), (d), (a), (a), (c), , 3., 13., 23., 33., 43., , (a), (c), (c), (c), (c), , 4., 14., 24., 34., 44., , (d), (a), (b), (c), (d), , 5., 15., 25., 35., 45., , (d), (a), (a), (b), (a), , 6., 16., 26., 36., 46., , (b), (b), (c), (d), (c), , 7., 17., 27., 37., 47., , (a), (d), (a), (c), (a), , 8., 18., 28., 38., 48., , (c), (c), (a), (b), (b), , 9., 19., 29., 39., 49., , (b), (d), (a), (c), (b), , 10., 20., 30., 40., 50., , (a), (b), (c), (c), (b), , 2., 12., 22., 32., 42., 52., 62., , (d), (b), (c), (a), (d), (c), (b), , 3., 13., 23., 33., 43., 53., 63., , (a), (b), (d), (a), (c), (c), (a), , 4., 14., 24., 34., 44., 54., 64., , (b), (b), (a), (d), (d), (d), (b), , 5., 15., 25., 35., 45., 55., , (a), (b), (c), (b), (b), (d), , 6., 16., 26., 36., 46., 56., , (a), (b), (c), (d), (b), (d), , 7., 17., 27., 37., 47., 57., , (d), (b), (c), (c), (b), (d), , 8., 18., 28., 38., 48., 58., , (b), (b), (b), (c), (c), (b), , 9., 19., 29., 39., 49., 59., , (d), (d), (b), (c), (b), (b), , 10., 20., 30., 40., 50., 60., , (b), (a), (c), (a), (b), (c), , Level II, 1., 11., 21., 31., 41., 51., 61., , (d), (b), (b), (a), (b), (c), (b)
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Hints & Solutions, Level I, 1. If A and B are mutually exclusive, then, P ( AB) = P ( A ∩ B) = 0, 2. The probability of getting head and tail in one toss is, , 1, 2, , P(atleast one H) = 1 − P(no head in four toss), 4, 1 15, 1, = 1 − P(four tails) = 1 − = 1 −, =, 2, 16 16, 3. Given mean, np = 6, variance, npq = 4, npq 4, 2, 1, = ⇒ q = and p =, ∴, np 6, 3, 3, 1, Now,, np = 6 ⇒ n × = 6 ⇒ n = 18, 3, 4. Favourable cases of getting 10 or greater than 10, if, atleast 5 appears one of the dice, = {(5, 6), (6, 5), (5, 5)}, Number of favourable cases = 3, Total number of cases = 36, 3, 1, Required probability =, =, ∴, 36 12, 5. The probability of getting a number either 2 or 3 in one, 1, toss is ., 2, Condition for getting the sum of 12 in five tossed is, (2, 2, 2, 3, 3), 3, 2, 5, 5 × 4 1, 1 1, ∴ Required probability = 5C3 =, , 2 2, 2 × 1 2, 1, 5, = 10 ⋅ 5 =, 16, 2, 1, 6. Probability of selection of husband P (H ) =, 5, 1 4, ∴, P (H ) = 1 − =, 5 5, 1, and probability of selection of wife P (W ) =, 3, 1 2, P (W ) = 1 − =, ∴, 3 3, ∴ Probability that only one of them is selected, = P (H ) P (W ) + P (H ) P (W ), 1 2 4 1, = + , 5 3 5 3, 2, 4, 6 2, =, +, =, =, 15 15 15 5, 4 ×3, 7. Total number of events n (E ) = 4C 2 =, =6, 2, Favourable number of events n (S ) = 2C 2 = 1, n (S ) 1, =, ∴ Required probability =, n (E ) 6, , 8. The probability that the problem is not being solved by, any of three students, 1 , 1 , 1, , = 1 − 1 − 1 − , , 2 , 3 , 4, 1, =, 4, 9. If A and B are any two events, then, …(i), P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B), And given that,, P ( A ∪ B) = 3 /4, P ( A ∩ B) = 1 /4, and, P ( A ) = 2 /3, Now,, P ( A) = 1 − P ( A ), [Q P ( A ) + P ( A ) = 1], 2 1, ⇒, P ( A) = 1 − =, 3 3, From Eq. (i),, 3 1, 1, = + P (B) −, 4 3, 4, 3 1 1, P (B) = − +, 4 3 4, 9 −4 + 3 8 2, P (B) =, =, =, 12, 12 3, 10. n (S ) = 13, E = Event of multiple of 3 or 4, = {3, 4, 6, 8, 9, 12}, ∴, n (E ) = 6, n (E ) 6, Hence, Required probability =, =, n (S ) 13, 11. The possible set of 6 is {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}, ∴ Total number of cases = 5, Number of cases at which 4 appeared once = 2., 2, ∴ Required probability =, 5, 12. Required probability =, 13. Required probability =, , C1 × 5C1 × 4C1 6 ⋅ 5 ⋅ 4 5, =, =, 6, C1 × 6C1 × 6C1 6 ⋅ 6 ⋅ 6 9, 6, , 1, 1, 1, =, =, 4 ! 4 × 3 × 2 24, , A, P ( A ∪ B) = 0.5, P (B ) = 0.8, P = 0.4, B, A P ( A ∩ B), Now,, P =, B, P (B), A, ⇒ P (B) × P = P ( A ∩ B), [Q PB = 1 − P (B )], B, , 14. Q, , ⇒, , P ( A ∩ B) = 0.4 × (1 − 0.8), = 0.4 × 0.2 = 0.08, , 15. A coin is tossed n times., ∴Total number of ways = 2n, If head comes odd times, then favourable ways = 2n − 1
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195, , Probability, ∴ Required probability of getting odd times head, 2n − 1 1, = n =, 2, 2, 16. Total number of events (sample) n (S ) = 6 × 6 × 6 = 216, and total number of favourable events n (E ) = 6, which, is, (1, 1, 1) (2, 2, 2) (3, 3, 3) (4, 4, 4) (5, 5, 5) and (6, 6, 6), n (E ), 6, 1, =, =, ∴ Required probability =, n (S ) 216 36, 17. Total number of events (sample) n (S ) = 6 × 6 × 6 = 216, and total number of favourable events n (E ) = 6, which is, (1, 1, 1) (2, 2, 2) (3, 3, 3) (4, 4, 4) (5, 5, 5) and (6, 6,6)., n (E ), 6, 1, =, =, ∴ Required probability =, n (S ) 216 36, 18. Given that, probability of success, p =, , 1, 4, , 3, 4, Mean = np, Standard deviation = Variance, 3 = Variance ⇒ Variance = 9 ⇒ npq = 9, 1 3, 9 ⋅4 ⋅4, n. . = 9 ⇒ n =, ⇒ n = 48, 4 4, 3, 1, Mean = np = 48 × = 12, 4, Probability of unsuccess, q =, , ∴, ⇒, ⇒, ∴, , 5, , C1, C1, 3, C, P(selecting a red ball) = 12 1, C1, 5, C1 + 3C1 2, P (black ball or red ball) =, =, 12, 3, C1, , 19. P(selecting a black ball) =, , 12, , 20. A leap year has 366 days, in which 2 days may be any, one of the following pairs, (Sunday, Monday), (Monday, Tuesday), (Tuesday,, Wednesday), (Wednesday, Thursday), (Thursday,, Friday), (Friday, Saturday), (Saturday, Sunday)., 2 2 1 3, ∴ Required probability = + − =, 7 7 7 7, 21. The total number of three-digit numbers using the, digits 0, 2, 4, 6 and 8 = 5 × 5 × 4 = 100, Q Favourable events = {222, 444, 666, 888}, Now, the total number of numbers in which all the three, digits are the same = 4, 4, 1, ∴ Required probability =, =, 100 25, 22. We know that, a leap year contains 366 days in which, 52 weeks and rest 2 days. Now, here 2 days are, arranging like that, (Sunday, Monday), (Monday, Tuesday), (Tuesday,, Wednesday), (Wednesday, Thursday), (Thursday,, Friday), (Friday, Saturday), (Saturday, Sunday), ∴ Total sample events, n (S ) = 7, And numbers of favourable events n (E ) = 2, n (E ) 2, =, ∴Required probability =, n (S ) 7, , 23. Let P (E ) be the probability of getting head on Ist and, P (F ) be the probability of getting head on IInd, 1, 1, P (E ) = and P (F ) =, ∴, 2, 2, 1, and, P (E ∩ F ) =, 4, ∴, P (E ∪ F ) = P (E ) + P (F ) − P (E ∩ F ), 1 1 1, = + −, 2 2 4, 1 3, =1 − =, 4 4, C1 × 4C1, 52, C2, 4 ×4 ×2, =, 52 × 51, 8, =, 663, , 24. Required probability =, , 4, , 25. Probability of getting either a spade or an ace, 16 4, =, =, 52 13, Odds against = (13 − 4) : 4, ∴, = 9 :4, 26. Possible sample is as follow, { HHH , HTH , HHT , THH , TTH , THT , HTT , TTT }, ∴ Total number of samples = 23 = 8, Required probability = Probability of getting one head, + Probability of getting no head, 3 1 4 1, = + = =, 8 8 8 2, 27. Required probability, = P (one ball is blue) + P (both the balls are blue), 2 7 2 1, = × + ×, 9 8 9 8, 14, 2, 16 2, =, +, =, =, 72 72 72 9, 28. The probability that the first ball drawn is black, if the, 3 1, second ball is known to be red = =, 6 2, 29. Since, n = 2 and p = 0.6, ∴, q = 1 − p = 0.4, Variance (X) = npq = 2 × (0.6) × (0.4) = 0.48, X 0.48, Variance =, = 0.12, 2, 4, 30. Since, P ( A ∪ B) = P ( A ), and P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B), ⇒, P (B) = P ( A ∩ B), which is possible only, if B is subset of A., 31. Since, bulb A is fused and bulb B is fused, both the, events are independent, therefore, P (bulb A is fused alone), = P(A is fused) = 0.20, 32. Total sample space occurring to throwing four dice,, n (S ) = 6 × 6 × 6 × 6 = 1296, Total favourable events, n (E ) = 0
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196, , NDA/NA Mathematics, , Because when we throwing four dice, there are no, possibility to get the sum of number is 25. While the, maximum sum occurring on four dice is, = 6 + 6 + 6 + 6 = 24, n (E ), 0, =, =0, ∴ Required probability =, n (S ) 1296, x, 33. Q Probability of guessing a correct answer =, 12, 2, and probability of guessing a wrong answer =, 3, x, 2, + =1, ∴, 12 3, x+8, =1, ⇒, 12, ⇒, x = 12 − 8 = 4, 34. Q P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B), 1 3 11, ⇒ P ( A ∩ B) = + −, 3 4 12, 1, =, 6, B P ( A ∩ B) 1 /6 1, ∴, P =, =, =, A, P ( A), 1 /3 2, , 1, 1, 2, 1, +, =, =, 14 14 14 7, 2, , 3, , Probability of getting king and a spade =, , 1, 52, , ∴ P(king or spade) = P (king ) + P (spade), − P(king and spade), 1, 1 1 4 + 13 − 1 16 4, =, + −, =, =, =, 13 4 52, 52, 52 13, 4, 43. Probability of getting first card is ace =, 52, Probability of getting second card without replacement, 3, is an ace =, 51, 4 3, 1, Required probability =, =, ., 52 51 221, = 1 − P (all three tails) = 1 −, , 5 C × 3C × 4C × 2C , + 8 1 7 1 6 1 5 1, C1 × C1 × C1 × C1 , , 10 × 6, 1 6, 36. Required probability = C 2 =, 7 7, 75, 5, , 4, 1, =, 52 13, 13 1, Probability of getting a spade =, =, 52 4, , 42. Probability of getting a king =, , 44. P(atleast one head) = 1 − P (zero head ), , 35. Required probability, = P (WBWB) + P (BWBW ), 5 C × 3C × 4C × 2C , = 8 1 7 1 6 1 5 1, C1 × C1 × C1 × C1 , , =, , 41. P ( A ∩ B ) = P ( A ) − P ( A ∩ B), = 0.25 − 0.14 = 0.11, , 3, , 37. A and B toss a coin alternately till one of them tosses, heads and win the game, their respectively probabilities, 1, 2, of winning are and , respectively., 3, 3, 38. S = { BBB, BBG , BGB, GBB, GGB, GBG , BGG , GGG }, and E = { BBB, BBG , BGB, GBB, GGB, GBG , BGG }, n (E ) = 7 and n (S ) = 8, n (E ) 7, ∴, P (E ) =, =, n (S ) 8, 39. Probability that the race will be won by A, 1, P ( A) =, 5, Probability that the race will be won by B, 1, P (B) =, 6, ∴ Probability that the race will be won by A or B, = P ( A ) + P (B), 1 1 11, = + =, 5 6 30, 40. Probability that two persons born on the same day, 7, C1, 1, =, =, 7×7 7, , 1 7, =, 8 8, , 45. Probability of solving a problem by students A, B and C, is 1/2, 1/3 and 1/4, respectively., And probability of not solving a problem by students A,, B and C is 1/2, 2/3 and 3/4, respectively., ∴ Probability of solving the problem, = 1 − P(not solved by any student), 1 2 3, 1 3, ∴, P =1 − ⋅ ⋅ =1 − =, 2 3 4, 4 4, 46. Total number of ways, n (S ) = 6 × 6 = 36, Favourable number of ways, n (F ) = 6, n (F ) 6 1, =, =, ∴ Required probability =, n (S ) 36 6, 47. ∴Required probability, = 1 − P (all four children are girls), 1 15, =1 −, =, 16 16, 48. Probability of equal number of one head occurs, 2 2 1 1, = × = ×, 4 4 2 2, Probability of equal number of two heads occur =, ∴ Required probability =, , =, 49. Q P ( A ∪ B) =, and, , 1 1 1 1, × + ×, 4 4 2 2, , 1, 1, 5, + =, 16 4 16, , 5, 1, , P ( A ∩ B) =, 6, 3, 1, P (B ) =, 2, , P (B) = 1 − P (B ) = 1 −, =, , 1, 2, , 1, 2, , 1 1, ×, 4 4
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197, , Probability, We know that,, P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B), 5, 1 1, ⇒, = P ( A) + −, 6, 2 3, 5, 1 1, = 1 − P(A) + −, ⇒, 6, 2 3, 1 1 5, P(A) = 1 + − −, ⇒, 2 3 6, 6 + 3 −2 −5 2 1, =, = =, 6, 6 3, 3, 1, 50. Q P ( A ∪ B) = , P ( A ∩ B) =, 4, 4, 2, and, P(A) =, 3, , 2, 3, 1, P ( A) =, ⇒, 3, We know that,, P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B), 3 1, 1, ⇒, = + P (B) −, 4 3, 4, 1, P (B) = 1 −, ⇒, 3, 2, P (B) =, ⇒, 3, , ⇒, , 1 − P ( A) =, , Level II, P (B ∩ A ) 1, B 1, 1. Given that, P = ⇒, =, A 2, P ( A), 2, 1 1 1, ⇒, P (B ∩ A ) = × =, 2 4 8, P ( A ∩ B) 1, A 1, and, = ,, P = ⇒, B 4, P (B), 4, ⇒, , P (B) = 4P ( A ∩ B) ⇒ P (B) =, , 1, 2, , 1 1 1, = . = P ( A ) ⋅ P (B), 8 2 4, Q Events A and B are independent., A′ P ( A′ ∩ B) P ( A′ ) P (B) 3, Now,, P =, =, =, B, P (B), P (B), 4, B, ′, P, B, A, ′, ′, P, (, B, ), P, (, A, ), 1, (, ′, ∩, ′, ), , and, P =, =, =, A′ , P ( A′ ), P ( A′ ), 2, ∴, , P ( A ∩ B) =, , 2. Given that, M and N are any two events, then the, probability that exactly one of them occurs, = P (M ∩ N ) + P (M ∩ N ), = P (M ) − P (M ∩ N ) + P (N ) − P (M ∩ N ), = P (M ) + P (N ) − 2P (M ∩ N ), 3. ∴ n (S ) = 64 = 1296, and permutation that the sum of the number appearing, on them is 13., 4!, Total permutation of (1, 1, 5, 6) =, = 12, 2!, ∴ Total permutation of ( 1, 2, 4, 6) = 4 ! = 24, 4 ! 24, Total permutation of (1, 3, 3, 6) =, =, = 12, 2! 2, Similarly, for (1, 2, 5, 5,) = 12, (1, 3, 5, 4) = 24, (2, 2, 6, 3) = 12, (2, 2, 5, 4) = 12, (3, 3, 2, 5) = 12, (3, 3, 3, 4) = 4,, (4, 4, 4, 1) = 4 and (4, 4, 3, 2) = 12, n( A), ∴ Required probability =, n (S ), 12 + 24 + 12 + 12 + 24 + 12 + 12 + 12 + 4 + 4 + 12, =, 1296, 140, 35, =, =, 1296 324, , 1, 4. Let p = P (getting a head) = ,, 2, 1, q = P (getting no head) =, 2, By using binomial distribution,, ∴ Required probability = P(six head), 6, 4, 10 ! 1, 1 1, ⋅, = 10C 6 =, 2 2, 6 ! 4 ! 210, 105, 10 × 9 × 8 × 7, 1, =, × 10 =, 512, 4 ×3 ×2, 2, 5. Total number of ways of the arrangement of the letters, 10 !, of the word ‘UNIVERSITY’ =, 2!, Number of ways that both I’s come together = 9 !, ∴ Number of ways that both I’s do not come together, 10 !, =, −9!, 2, , 10 !, 2 − 9 ! 4, =, ∴ Required probability = , 10 ! 5, , , , , 2, 6. The number of students that exactly pass in, Mathematics, = 70 − 30 = 40, and the number of students that exactly pass in, Statistics, = 55 − 30 = 25, 40 + 25 65 13, ∴ Required probability =, =, =, 125, 125 25, 7. The events when flipping a coin and head occurs, = { HT , HH }, The events when flipping a coin and tail occurs, = {T1 , T2, T3 , T4 , T5 , T6 }, Total events = { HT , HH , T1 , T2, T3 , T4 , T5 , T6 }, Favourable events getting one head and one tail, = { HT }, 1, ∴ Required probability =, 8
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198, 8., , NDA/NA Mathematics, 1, 5, 1, 5, P ( A ) = , P ( A ) = and P (B) = , P (B ) =, 6, 6, 6, 6, Hence, probability of winning of A, = P ( A) + P ( A ∩ B ∩ A), + P ( A ∩ B ∩ A ∩ B ∩ A ) + ……, 1, 2, 4, 1 5 1 5 1, 6, 6, =, = + + +… =, 2, 6 6 6 6 6, 11, 5, , 1− , 6, 6, 5, Also, probability of winning B = 1 −, =, 11 11, , 9. As three distinct numbers are to be selected from first, 100 natural numbers, ∴, n (S ) = 100C3, E(Favourable events) = All three of them are divisible by both 2, and 3., ⇒ Divisible by 6 i.e., {6, 12, 18,..., 96}, Thus, out of 16, we have to select 3., ∴, n (E ) = 16C3, 16, C, 4, ∴ Required probability = 100 3 =, C3 1155, 10. Total number of ways = 10!, Total number of ways in which 5 boys and 5 girls are, sitting in a row alternatively = 5 ! × 6 !, 5! × 6! 1, =, ∴ Required probability =, 10 !, 42, 1, 1, 11. It is given, P ( A ∩ B) = and P ( A ′ ∩ B ′ ) =, 6, 3, 1, Now, P ( A ∪ B) ′ = P ( A ′ ∩ B ′ ) =, 3, 1, 2, ⇒, 1 − P ( A ∪ B) = ⇒ P ( A ∪ B) =, 3, 3, But, P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B), ⇒, P ( A ) + P (B) = P ( A ∪ B) + P ( A ∩ B), 5, ⇒, P ( A ) + P (B) =, … (i), 6, Q A and B are independent events., 1, ∴, P ( A ∩ B) = P ( A )P (B) ⇒ P ( A )P (B) =, 6, Now, [P ( A ) − P (B)]2 = [P ( A ) + P (B)]2 − 4P ( A )P (B), 2, 1 25 4 1, 5, = −4 =, − =, 6 36 6 36, 6, 1, P ( A ) − P (B) = ±, … (ii), ⇒, 6, 1, 1, On solving Eqs. (i) and (ii), we get P ( A ) = or, 3, 2, 12. Let X be the number of heads getting in n tossed. X, 1, follows binomial distribution with parameters n , p = ,, 2, 1, q= ., 2, Given that, P (X ≥ 1) ≥ 0.8, ⇒, 1 − P (X = 0) ≥ 0.8 ⇒ P (X = 0) ≤ 0.2, n, 0, 1 1, 1 1, n, C 0 ≤ 0.2 ⇒ n ≤, ⇒, 2 2, 5, 2, , ⇒, 2n ≥ 5, ∴The least value of n is 3., 13. Q Probability of answering all the three questions, correctly, 1 1 1 1, = × × =, 4 4 4 64, ∴ Probability of not answering all the three questions, correctly, 1 63, =1 −, =, 64 64, 14. Let A be the event that the machine produces 2, acceptable items., Also, let B1 represents the event of correct setup and B2, represent the event of incorrect setup., Now, P (B1 ) = 0.8 P (B2) = 0.2, A, A, P = 0.9 × 0.9 and P = 0.4 × 0.4, B1 , B2, P (B1 ) P ( A / B1 ), B1 , ∴ P =, A P (B1 ) P ( A / B1 ) + P (B2) P ( A / B2), 0.8 × 0.9 × 0.9, =, 0.8 × 0.9 × 0.9 + 0.2 × 0.4 × 0.4, 648, =, = 0.95, 680, 15. Let total number of students be 100 in which 60% girls, and 40% boys., Number of boys = 40, Number of girls = 60, 25, 25% of boys offer Mathematics =, × 40 = 10 boys, 100, 10, 10% of girls offer Mathematics =, × 60 = 6 girls, 100, It means, 16 students offer Mathematics., 6 3, =, ∴ Required probability =, 16 8, 3, 16. Given that, P (B) = P ( A ), …(i), 2, 1, and, …(ii), P (C ) = P (B), 2, where, A , B and C are the three mutually exclusive and, exhaustive events., ∴ Let P ( A ) = a, 3, 3, From Eq. (i), we get P (B) = P ( A ) = a, 2, 2, ∴ From, Eq. (ii), we get, 1, 1 3, 3, P (C ) = P (B) = × a = a, 2, 2 2, 4, ∴, A ∪ B ∪C = X, ⇒, P ( A ∪ B ∪ C ) = P (X ), ⇒, P(A ∪ B ∪ C ) = 1, [Q P (X ) = 1], (Q By addition theorem), ⇒ P ( A ) + P (B) + P (C ) = 1, 3, 3, ⇒, a + a + a =1, 2, 4, 4, 4, a=, ⇒ P ( A) =, ⇒, 13, 13, 17. Given that, from the group of 4 men and 5 women a, committee of three persons are chosen., ∴ Total number of persons = 9
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199, , Probability, Out of 9 total number of ways of selecting 3 persons, 9 × 8 × 7 × 6!, = 9C3 =, = 84, 3!×6!, Again, number of ways of selecting two men out of four, men and one woman out of five women, = 4C 2 × 5C1, 4 × 3 × 2!, =, ×5, 2! × 2!, 4 ×3 ×5, =, = 30, 1 ×2, 30 5, Therefore, required probability =, =, 84 14, 18. We know that, if X has binomial distribution with mean, np and variance npq, then, n, P (X = k), C k ( p)k (q)n− k, =n, P (X = k − 1), C k − 1 ( p )k − 1 q n − k + 1, n, , C, p, = n k ⋅ =, Ck − 1 q, , 19., , n!, p, k !(n − k)!, ⋅, n!, q, (k − 1)! (n − k + 1)!, , =, , (k − 1)! (n − k + 1)! p, ⋅, k ! (n − k)!, q, , =, , n−k+1 p, ⋅, k, q, , 22. Required probability =, , 6, , C1 × 4C 2, 10, C3, , 10, 6, , 4, 3, , 1, , =, , 2, , 6 ×6 ×6, 3, =, 10 × 9 × 8 10, , 23. ∴ P (X ≤ 2) = 0.25, ⇒ P (X = 1) + P (X = 2) = 0.25, ⇒, k + p1 = 0.25, ⇒, p1 = 0.25 − k, and, P (X ≥ 4) = 0.35, ⇒ P (X = 4) + P (X = 5) = 0.35, ⇒, p2 + 2k = 0.35, ⇒, p2 = 0.35 − 2k, ⇒, p1 ≠ p2, and, p1 + p2 = 0.25 − k + 0.35 − 2k, = 0.6 − 3k, ≠ P (X = 3), Hence, neither I nor II is correct., 24. I. Q Probability of 53 Sundays in a leap year =, , P ( A ∩ B) = P (B) − P ( A ∩ B), , 2, 7, , and probability of 53 Sundays in a non-leap year =, , A, , B, , 1, 7, , ∴ Statment I is true., , 3, 7, 2, and probability of 5 Mondays in April =, 7, ∴ Statement II is wrong., II. Probability of 5 Mondays in March =, , _, P( A ∩ B), , 20. Let radius of given circle be r. Now, make a concentric, r, circle with radius ., 2, The given point is close to the circumference than to the, centre, if it lies in the shaded region., π { r 2 − (r /2)2}, ∴ Required probability =, πr 2, , 25. I. It is true that P (only one of them occurs), = P (E1E 2E3 + E1E 2E3 + E1E 2E3 ), III. If E1 , E 2 and E3 be three arbitrary events of a, sample space, then, P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B) ≤ P ( A ) + P (B), (Q P ( A ∩ B) ≥ 0), 1, 1 1, 26. p =, ⇒ q =1 − b =1 − =, 2, 2 2, So, distribution is symmetric., ∴ Both statements I and II are correct., 27. Both the statements I and II are correct., , r/2, r, , =, , r, , 3 /4 r 2 3, =, 4, r2, , 21. The probability of the union of the shaded region is, P ( A ) + P (B) − P ( A ∩ B), , 28. We have,, (A) P (E1 ) + P (E 2) = 1, Q E1 and E 2 are mutually exhaustive events., (B) P (E1 ) + P (E 2) = 0, Q E1 and E 2 are impossible events., (C) P (E1 ) + P (E 2) ≤ 0, Q E1 and E 2 are mutually exclusive events., (D) P (E1 ) P (E 2) = 1, E1 and E 2 are sure events.
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200, , NDA/NA Mathematics, , 29. ∴ Required probability =, , 25, 26, , C3 25 × 24 × 23 23, =, =, C3 26 × 25 × 24 26, , 30. Since, P ( A ∪ B) ≤ 1, ⇒, P ( A ) + P (B) − P ( A ∩ B) ≤ 1, ⇒, 0.8 – 0.7 − P ( A ∩ B) ≤ 1, ⇒, P ( A ∩ B) ≥ 1.5 − 1, ⇒, P ( A ∩ B) ≥ 0.5, 31. Given, X + Y = 15, The total number of ordered pairs, = (5, 10), (6, 9), (7, 8), (8, 7), (9, 6), (10, 5), ∴, n (E ) = 6, In each above pairs exactly one is even number., ∴, n (E ) = 6, n (E ) 6, = =1, ∴ Required probability=, n (S ) 6, 4C × 4C 2 2C 2 × 2C 2, 32. Required probability = 28, ×, , 4, C4 , C4 , , 4 ×3 ×4 ×3 ×3 ×2, 18, =, ×1 =, 8 × 7 ×6 ×5, 35, 33. Q np = 8 and npq = 4, npq 4 1, q=, = =, ⇒, np 8 2, and, ⇒, , p+ q =1 ⇒, , p=1 −, , 1 1, =, 2 2, , 1, n = 8 ⇒ n = 16, 2, 16 − 1, , 1, , 1, 1, P (X = 1) = 16C1 , , 2, 2, 16, 1, = 15, =, 2 ⋅ 2 212, 4, 34. Q np = 4 and npq =, (given), 3, 4, 1, ∴ 4q =, ⇒ q=, 3, 3, 1 2, (Q p + q = 1), p=1 − =, ∴, 3 3, 4 ×3, ⇒, n=, =6, 2, Now,, P (X ≥ 5) = 6C5 p5 q1 + 6C 6 p6q0, 5, 6, 2, 2 1, = 6C5 + 6C 6 , 3, 3 3, 6 × 32 64 256 28, =, + 6 = 6 = 6, 36, 3, 3, 3, Now,, , 35. We know that,, P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B), 0.8 + 0.9 − p ≤ 1, ⇒, [Q P ( A ∪ B) ≤ 1], 1.7 − p ≤ 1, ⇒, 0.7 ≤ p, ⇒, Since,, P ( A ) < P (B), ∴, P ( A ∩ B) ≤ P ( A ) ⇒ P ≤ 0.8, Hence, 0.7 ≤ p ≤ 0.8, 36. The outcomes of an experiment classified as success A, or failure A will follow a binomial distribution, if P ( A ), remains constant in all the trials., , 37. Let A be the event of selecting bag X, B be the event of, selecting bag Y and E be the event of drawing a white, ball, then, Probability of selecting a bag, P ( A ) = 1 /2, P (B) = 1 /2,, P (E / A ) = 2 /5, P (E /B) = 4 /6 = 2 /3, Required probability, P (E ) = P ( A ) P (E / A ) + P (B) P (E /B), 1 2 1 2 8, = ⋅ + ⋅ =, 2 5 2 3 15, 38. The required probability, = 1 – probability of equal number of heads and tails., Out of 2n tossed n times head and n times tails., 2n − n, 2n, n, (2n )! 1, (2n )! 1, 1 1, ⋅, = 1 − 2nC n , =1 −, =1−, 2 2, n ! n ! 2, (n !)2 4n, 3, 1, and P ( A ∩ B) =, 5, 5, We know that, P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B), 3, 1, = 1 − P ( A ) + 1 − P (B ) −, 5, 5, 9 3, P ( A ) + P (B ) = −, ⇒, 5 5, 6, ∴, P ( A ) + P (B ) =, 5, 1, 5, 40. Given that, P ( A ∩ B) = , P ( A ∪ B) =, 3, 6, 1, and, P ( A) =, 2, ∴, P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B), 5 1, 1, = + P (B) −, ⇒, 6 2, 3, 5 1 1 4 2, ⇒, P (B) = + − = =, 6 3 2 6 3, 1 2 1, ∴, P ( A ) × P (B) = × = = P ( A ∩ B), 2 3 3, This shows that A and B are independent events., , 39. It is given P ( A ∪ B) =, , 41. Given that, x = 33n, where, n is a positive integral value., Here, only four digits may be at the unit place i. e. , 1, 3,, 7, 9., ∴, n (S ) = 4, Let E be the event of getting 3 at its units place,, n (E ) = 1, n (E ) 1, ∴, P (E ) =, =, n (S ) 4, 42. Let, ∴, , E = {2, 4, 6}, n (E ) = 3 and n (S ) = 6, 3 1, P (E ) = =, 6 2, , 1 1, =, 2 2, ∴ Probability of atleast two success, 3, 2, 1 1, 1, = P (X = 2) + P (X = 3) = 3C 2 + 3C3 , 2 2, 2, 1 1 1, =3 × + =, 8 8 2, Now, probability of failure = 1 −
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201, , Probability, 43. From the given condition, it is clear that a particular, person is always in a committee of 3 persons. It means, we have to select 2 persons out of 37 persons., 37, C, Required probability = 38 2, C3, 44. Probability that, X = 3 =, , 1 1 1 1, 7, × + × =, 2 3 2 4 24, , 45. Required number of elements in the sample space, = 10, and sample space = { H , H , H , H , H , T , T , T , T , T }, 46. 0.6 is nearest to the probability that 3 randomly, selected persons are born on three different days of a, week., Alternate Method, 7 6 5 30, ≈ 0.6, ∴ Required probability = × × =, 7 7 7 49, 47. Case I Let a white ball is drawn from the first bag and, placed unseen in the second bag,, 4, 5, C, C, 20, then probability = 8 1 × 7 1 =, C1, C1 56, Case II Let a black ball is drawn from the first bag, and placed unseen in the second bag, then probability, 4, 3, C, C, 15, = 8 1 × 7 1 =, C1, C1 56, 20 15 35, ∴ Required probability =, +, =, 56 56 56, 48. We know that,, P ( A ∪ B) ≤ P ( A ) + P (B), , (by property), , and P ( A ) + P ( A ∪ B) ≤ 1 + P (B), ∴ Both the given statements I and II are correct., 49. Number of ways of arranging the books 1, 2, 3, 4, 5 and 6, =6!, Now, number of ways of arranging the books 1, (2, 3), 4,, 5 and 6 = 5 !, Number of ways of arranging the books 2 and 3 = 2 !, 5!×2! 2 1, ∴ Required probability =, = =, 6!, 6 3, 50. Probability of the person getting reward, p =, , 4 1, =, 16 4, , Then, the probability not getting the reward,, 1 3, q =1 − =, 4 4, 1, ∴ Required probability = 4C1 , 4, , 3, , 2, , 1 3, 3, 4, + C2 , 4 4, 4, 3, , 1, 1 3, + 4C3 + 4C 4 , 4, 4 4, 4, , 1, = { 4C1 ⋅ 33 + 4C 2 ⋅ 32 + 4C3 ⋅ 31 + 4C 4 }, 4, =, , 1, 175, (108 + 54 + 12 + 1) =, 256, 256, , 2, , 4, , B, P (B ∩ (A ∪ Bc ), , , 51. P , =, c , (A ∪ B ) , P(A ∪ Bc ), P ( A ∩ B), =, P ( A ) + P (Bc ) − P ( A ∩ Bc ), P ( A ) − P ( A ∩ Bc ), =, P ( A ) + P (Bc ) − P ( A ∩ Bc ), 0.7 − 0.5 1, =, =, 0.8, 4, 52. Q The probability, when flipping a coin one time and, 1, coming head = and the probability that it becomes, 2, 1, tail =, 2, 5, , 7, , 1 1, ∴ Required probability = 12C5 , 2 2, (by binomial distribution), 12, C (12, 7), C (12, 5), 1, 12, or, = C5 =, 2, (2)12, (2)12, 1, 53. Probability that no one hits the target =, 60, 9, Probability that one hits the target =, 60, 10 1, ∴Probability of maximum one hit =, =, 60 6, 1 5, Required probability = 1 − =, 6 6, 13, 54. Probability of a spade =, 52, 4, Probability of an ace =, 52, 1, and probability of a spade ace =, 52, 13, 4, 1, Required probability =, +, −, ∴, 52 52 52, 16 4, =, =, 52 13, 4, 9, 1−, 9, 13, 13, Odds against his winning =, =, = = 4 to 9, 4, 4, 4, 13, 13, 55. (A) Let a man left eye is brown P (L ) = P, and a man right eye is brown P (R) = p, ∴ P (L ∪ R) = P (L ) + P (R) − P (L ∩ R), = p + p − 0 = 2p, ∴ A is false., (R) It is true., 56. (A) This statement is false., (R) P ( A ∩ B) = P ( A ) ⋅ P (B), always true, if A and B are independent events., 57. P ( A ∪ B ∪ C ) = P ( A ) + P (B) + P (C ), − P ( A ∩ B) − P (B ∩ C ) − P ( A ∩ C ) + P ( A ∩ B ∩ C ), 1 1 1, P(A ∪ B ∪ C ) = + + − 0 − 0, ⇒, 4 4 4, − P(A ∩ C ) + P(A ∩ B ∩ C )
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202, , NDA/NA Mathematics, , ⇒, , P(A ∩ C ) = P(A ∩ B ∩ C ) − P(A ∪ B ∪ C ) +, , 3, 4, , 1, 8, ∴ A is false, but R is true., 4, 4, 1, 1, 2, 58. Required probability =, +, =, +, =, 52 52 13 13 13, and when two events are not mutually exclusive, then, P (E1 + E 2) = P (E1 ) + P (E 2) − P (E1 ∩ E 2), Both A and R are true but R is not the correct, explanation of A., ⇒, , P(A ∩ C ) ≠, , Solutions (Q. Nos. 59-61), Here, random experiment is throwing the given die., Let S = The sample space, A = The event of getting a face with number 1, B = The event of getting a face with number 2, and C = The event of getting a face with number 3, Now, n (S ) = 6, n ( A ) = 1, n (B) = 2, n (C ) = 3, 1, 2 1, P ( A ) = , P (B) = =, ∴, 6, 6 3, 3 1, and, P (C ) = =, 6 2, 1, We have, P (1) = P ( A ) =, 6, , P (2 or 3) = P (B ∪ C ) = P (B) + P (C ), 1 1 5, = + =, 3 2 6, (Q B and C are mutually exclusive), 1 1, P (not 3) = P (C′ ) = 1 − P (C ) = 1 − =, 2 2, , Solutions (Q. Nos. 62-64), 1, 2, 1, , P (B) = and P ( A ∪ B) =, 4, 5, 2, From addition theorem, we have, P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B), 1 1 2, = + − P ( A ∩ B), 2 4 5, 1 2 1 3, P ( A ∩ B) = + − =, 4 5 2 20, , We have, P ( A ) =, , Now,, , Again,, , P ( A ∩ B′ ) = P ( A ) − P ( A ∩ B), 1 3, 1, = −, =, 4 20 10, P ( A′ ∩ B′ ) = P{( A ∪ B)′ }, = 1 − P ( A ∪ B), 1 1, =1− =, 2 2
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11, , Binary Numbers, Binary Number, A number which contains only 0 and 1 is called binary, number., , Binary System, In the binary system, only two symbols 0 and 1 are, used. Its base is 2. The values of various places in this, system,when expressed in decimal system, are, ... , 24 , 23 , 22 , 21 , 20 , 2− 1 , 2− 2 ,..., Thus , 1111 = 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20, = 8 + 4 + 2 + 1 = 15, and 10010 = 1 × 24 + 0 × 23 + 0 × 22 + 1 × 21 + 0 × 20, , 1. Divide the decimal number by 2 (base of binary, number)., 2. Note the remainder separately as the right most digit, of binary equivalent., 3. Divide the quotient again by 2., 4. Note the remainder as next left digit of binary, number., 5. Repeat the steps 3 and 4 until quotient becomes 0., , Example 1. Convert 175 into a binary number., , Solution (a), , = 16 + 2 = 18, Arithmetically, we can write (10010)2 = (18)10, , Decimal System, In the decimal system we use 10 digits which are 0, 1,, 2, 3, 4, 5, 6, 7, 8 and 9. The value of a digit in this system, depends on its place in the number. The values of various, places in this system, are, ... , 104 , 103 , 102 , 101 , 100 , 10− 1 , 10− 2 , 10− 3 , ..., Thus, 235.35 = 2 × 102 + 3 × 101 + 5 × 100, + 3 × 10− 1 + 5 × 10−2, Since, 10 basic symbols are used in this system, so its, base is 10 and this system is called base ten system., , Conversion of Decimal to, Binary, In Coversion of decimal to binary, we convert the base, 10 into the base 2 in the following ways., , 1. Conversion of a Number, To convert a decimal digit into its binary equivalent,, following steps are performed, , (b) (11111010) 2, (d) (11111111) 2, , (a) (10101111) 2, (c) (10101011) 2, , ∴, , 2, 2, 2, 2, 2, 2, 2, (175)10, , 175, 87, 1, 43, 1, 21, 1, 10, 1, 5, 0, 2, 1, 1, 0, = (10101111) 2, , 2. Conversion of a Fraction, Multiply the given fraction by 2 and note down its, integral part. Multiply the fractional part of this, product by 2 and note down its integral part.Go on, doing it till the complete integer comes (without, fraction) or the same number repeats. (Number from, which we started to multiply)., , Example 2. Convert 0.638 to binary form., (a) (01010101111, ., )2, (c) (01010001101, ., )2, , Solution (c), 0.638 × 2 = 1276, ., 0.276 × 2 = 0.552, 0.552 × 2 = 1104, ., 0.104 × 2 = 0.208, 0.208 × 2 = 0.416, , (b) (0.0000001111) 2, (d) None of these, Binary, 1, 0, 1, 0, 0
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204, , NDA/NA Mathematics, 0.416 × 2 = 0.832, 0.832 × 2 = 1664, ., 0.664 × 2 = 1328, ., 0.328 × 2 = 0.656, 0.656 × 2 = 1312, ., , ∴, , 0, 1, 1, 0, 1, , 1, 1, 1, 1, + 1× 2 + 0 × 3 + 1× 4, 2, 2, 2, 2, 1 1 0, 1, = + + +, 2 4 8 16, , Solution (b) (0.1101) 2 = 1 ×, , = 0.5 + 0.25 + 0 + 0.0625, = 0.8125, , ( 0.638)10 = ( 0.1010001101) 2, , ∴, , Example 3. Convert (0. 35)10 to binary form., (b) (0.010101) 2, (d) (0.010110) 2, , (a) (0101010, ., )2, (c) (0.000111) 2, , Solution (d), 0.70 × 2 = 1.40, , 1, , 0.4 × 2 = 0 .8, , 0, , 0.8 × 2 = 1.6, , 1, , 0.6 × 2 = 1.2, , 1, , 0.2 × 2 = 0 .4, , 0, , Addition, , Now, in next step repeatition will be start., ∴, , (0.35) = (0.010110) 2, , Conversion of Binary to Decimal, In conversion of binary to decimal, we convert the base, 2 into the base 10 in the following ways., , 1. Conversion of a Number, , Example 4. Convert (10111) 2 to decimal form., (b) 22, , (c) 24, , (d) 23, , Solution (d) (10111) 2 = 1 × 2 + 0 × 2 + 1 × 2 + 1 × 21 + 1 × 20, 4, , 3, , Binary addition is similar to that of decimal numbers., A simple table for binary addition is as follows, 0+ 0= 0, 0+1=1, 1+ 0=1, 1+1= 0, with 0 carry of 1., , Example 6. Add two numbers (1011) 2 and (1001) 2., (a) (10111) 2, (c) (10100 ) 2, , Solution (c), , To convert a binary number into its decimal, equivalent,following steps are performed., 1. Each digit of binary number is multiplied by 2 having, powers ( 0, 1, 2, 3, ... )., 2. All the products of multiplication are summed to get, the decimal equivalent of the number., , (a) 21, , Arithmetic Operation on Binary, Like the decimal system, all the arithmetic operations, can be performed using binary system,like addition and, subtraction is as follows, , Binary, 0, , 0.35 × 2 = 0 .70, , (0.1101) 2 = (0.8125)10, , 2, , (b) (10101) 2, (d) (01101) 2, , 1 0 1 1, + 1 0 0 1, 1 0 1 0 0, , ∴ (1011) 2 + (1001) 2 = (10100) 2, , Subtraction, Binary subtraction is similar to that of decimal, numbers. A simple table for binary subtraction is as, follows, 0− 0= 0, , = 16 + 0 + 4 + 2 + 1 = 23, , 1− 0=1, 1−1= 0, , 2. Conversion of a Fraction, To convert a fraction in binary into its decimal, equivalent, following steps are performed., 1. Each digit of binary number is multiplied by 2 having, negative power ( − 1, − 2, − 3, ... ), 2. All the products of multiplication are summed to get, the decimal equivalent of the number., , Example 5. Find the decimal equivalent to (01101, ., ) 2., (a) (0.8124)10, (c) (0.8625)10, , (b) (0.8125)10, (d) (0.8655)10, , 0−1=1, with a borrow of 1 from next higher column., , Example 7. Subtract (10001) 2 from (10011) 2., (b) (10) 2, (d) None of these, , (a) (11) 2, (c) (01) 2, , Solution (b), , 100 11, –10 0 01, 00010, , ∴, , (10011) 2 − (10001) 2 = (10) 2
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205, , Binary Numbers, , Multiplication, Binary multiplication is actually much simpler than, decimal multiplication. In binary multiplication, we only, need to remember the following table, 0× 0= 0, 0×1= 0, 1× 0= 0, 1×1=1, , Example 10. Convert 13.375 into the binary system, number., (a) (1100111, . )2, (b) (1101011, . )2, (c) (0111011, . )2, (d) (0011111, . )2, Solution (b), , 2, 2, 2, , Example 8. Multiplication two numbers (101) 2 and (11) 2., (b) (0111) 2, (d) (1010) 2, , (a) (1111) 2, (c) (1100) 2, , Solution (a), , ∴, , ∴, ∴, , 1111, , (a) ( 43 .375)10, (c) ( 4. 3375)10, , Division, Binary division is almost as easy and involves our, knowledge of binary multiplication., , (b) (10) 2, (d) (11) 2, 11 1011 11, , − 11↓, 101, − 11, 10 Remainder, , Complement of Binary, Number, , (b) ( 433 .75)10, (d) ( 4337. 5)10, , Solution (a) (101011011, . )2, , Example 9. Divide the number (1011) 2 by (11) 2, , Solution (d), , 0, 1, 1, , Example 11. Find the decimal equivalent of (101011011, . ) 2., , Here, 0 is the place holder., , (a) (01) 2, (c) (101) 2, , 1, 0, 1, , (13)10 = (1101) 2 Binary, 0.375 × 2 = 0.750, 0.750 × 2 = 1500, ., 0.500 × 2 = 1000, ., (0.375)10 = (0.011) 2, (13.375)10 = (1101011, . )2, , 101, × 11, 101, +1010, , 13, 6, 3, 1, , ∴, , = 1 × 25 + 0 × 2 4 + 1 × 23 + 0 × 2 2 + 1 × 21 + 1 × 2 0, 1, 1, 1, + 0 × + 1× 2 + 1× 3, 2, 2, 2, = 32 + 0 + 8 + 0 + 2 + 1 + 0 + 0.25 + 0.125 = 43. 375, (101011.011) 2 = ( 43. 375)10, , Example 12. Obtain 1’s complement of (101001010) 2 ., (a) (000011111) 2, (c) (010110101) 2, , (b) (111100000) 2, (d) None of these, , Solution (c) Let x =101001010, 1’s complement of x = 010110101, , 2’s Complement, To get a 2’s complement of a binary number, first we will get, a 1’s complement and after that add 1., , In the binary system, only two digits 0 and 1 are used., So, for finding complement of binary number, we take 1 is, complement of 0 and 0 is complement of 1., , Example 13. Obtain 2's complement of (101001010) 2., , 1’s Complement, , Solution (d) Let, , To get a 1’s complement of a binary number, replace 1 by 0, and 0 by 1 in a binary number ., , (a) (000011111) 2, (c) (010110101) 2, , (b) (111100000) 2, (d) None of these, , x = 101001010, 1’s complement of, x = 010110101, then 2’s complement of, x = 010110101 + 1 = (010110110) 2
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Exercise, Level I, 1. For conversion of a fraction in binary to decimal, form, the binary number is multiplied by, (a) 2 having negative powers, (b) 2 having positive powers, (c) 2 only, (d) None of the above, 2. To convert a decimal digit into binary number, we, divide it by 2 and note the remainder. Then, we write, the digits of binary number, (a) from top to bottom., (b) from bottom to top, (c) in any manner, (d) None of the above, 3. The complement of (1011100)2 is, (a) (1010100)2, (b) ( 0111100)2, (c) ( 0100011)2, (d) (1000011)2, 4. 2’s complement of (1011001100)2 is, (a) ( 0100110100)2, (b) ( 0100110010)2, (c) ( 0100110101)2, (d) ( 0100110111)2, 5. In a binary number system, assume that a = 00111, b, and b = 01110, then in a decimal system , which is, a, equal to, (NDA 2011 I), (a) 1, (b) 2, (c) 4, (d) 5, 6. The base of the binary number system is, (a) 10, (b) 2, (c) 8, (d) 16, 7. The decimal equivalent of (101011)2 is, (a) ( 43)10, (b) ( 59)10, (c) ( 47)10, (d) None of these, , 10. The binary number 10110100001 in decimal system, is, (a) 441, (b) 1441, (c) 1241, (d) 241, 11. The decimal number 0.77 in binary system is, (a) ( 01100101, (b) ( 011000101, ., )2, ., )2, (c) ( 010100101, (d) None of these, ., )2, 12. Binary fraction 0.1011 in decimal system is, (a) 0.6875, (b) 0.8675, (c) 0.7685, (d) None of these, 13. The sum of ( 0110)2 and ( 0101)2 is, (a) (1011)2, (b) ( 01100)2, (c) (10100)2, (d) None of these, 14. The number 292 in decimal system is expressed in, binary system by, (NDA 2012 I), (a) 100001010, (b) 100010001, (c) 100100100, (d) 101010000, 15. The number 1753, in binary notation is written as, (b) (11011111001)2, (a) (11011011101)2, (c) (11011011001)2, (d) (11111010001)2, 16. The binary number 1101101+1011011 is written in, decimal system as, (a) 198, (b) 199, (c) 200, (d) 201, 17. 100101 + 101 + 1101 + 100 is equal to, (a) 1101001, (b) 100000, (c) 111011, (d) None of these, 18. 1111 + 111 + 1111 is equal to, (a) 111011, (b) 1101001, (c) 10001, (d) None of these, , 8. The base of the decimal number system is, (a) 2, (b) 5, (c) 10, (d) e, , 19. 101011 – 10010 is equal to, (a) (1001)2, (b) (1101)2, (c) ( 011001)2, (d) None of these, , 9. The binary number corresponding to the decimal, fraction ( 0125, ., )10 is, (a) ( 0.001)2, (b) ( 0120, ., )2, (c) ( 0100, (d) None of these, ., )2, , 20. What is the decimal number representation of the, (NDA 2012 I), binary number (11101.001)2?, (a) 30.125, (b) 29.025, (c) 29.125, (d) 28.025, , Level II, 1. What is (1111)2 + (1001)2 − (1010)2 equal to?, (a) (111)2, (b) (1100)2 (NDA 2010 II), (d) (1010)2, (c) (1110)2, 2. What is the binary number equivalent of the decimal, number 32.25?, (NDA 2010 II), (a) 100010.10, (b) 100000.10, (c) 100010.01, (d) 100000.01, , 3. What is the value of, ., ( 0101, )(211) 2 + ( 0.011)2(11) 2, ( 0101, ., )2(10) 2, , – ( 010, . 1)(201) 2 ( 0.011)(201) 2 + ( 0.011)2(10) 2, , ?, , (NDA 2010 I), , (a) (0.001)2, (c) (0.1)2, , (b) (0.01)2, (d) (1)2
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207, , Binary Numbers, 4. If x = (1101)2 and y = (110)2 , then what is the value of, x 2 − y 2?, (NDA 2009 II), (a) (1000101)2, (c) (10001101)2, , (b) (10000101)2, (d) (10010101)2, , 5. If (10x 010)2 − (11 y1)2 = (10z11)2 , then what are the, possible values of the binary digits x, y and z,, respectively?, (NDA 2009 II), (a) 0, 0, 1 (b) 0, 1, 0 (c) 1, 1, 0, (d) 0, 0, 0, 6. What is the decimal equivalent of (101101, ., )2?, (a) ( 5.225)10, (b) ( 5.525)10 (NDA 2008 II), (d) ( 5.65)10, (c) ( 5.625)10, 7. The binary number 0.111111 … (where the digit 1 is, recurring) is equivalent in decimal system to which, one of the following?, (NDA 2008 I), 1, 11, (a), (b), 10, 10, 10, (c) 1, (d), 11, 8. What is the product of the binary numbers 1001.01, and 11.1?, (NDA 2007 I), (a) (101110.011)2, (b) (100000.011)2, (c) (101110101, (d) (100000101, ., )2, ., )2, 9. The decimal number corresponding to the binary, number (111000.0101)2 is, (a) ( 5.6312)10, (b) ( 56.3125)10, (c) ( 563125, (d) ( 5631.2)10, ., )10, 10. The binary number corresponding to (13.0625)10 is, (a) (1011 . 0010)2, (b) (1110. 0101)2, (c) (1101 . 0001)2, (d) None of these, 11. The sum of 1011 .01 + 1001 .11 is, (a) 111011, (b) 10001, (c) 10000, (d) 10101, 12. If the sum of the binary numbers (11011)2 ,, (10110110)2 and (10011 x 0 y )2 is the binary number, and y, (101101101)2 , then the values of x, respectively, are, (a) 1 and 1, (b) 1 and 0, (c) 0 and 1, (d) 0 and 0, 13. In the binary addition, where x,y and z are binary, digits, 1 x 1 0 1, + 1 0 y 1, 10 0 z 0 0, the possible values of x,y and z respectively are, (a) 0, 1 and 0, (b) 1, 1and 0, (c) 0, 0 and 1, (d) 1, 0 and 1, 14. The difference of two numbers (1100110011)2 and, (1101001011)2 in binary system is, (b) (101010)2, (a) (100000)2, (c) (11000)2, (d) (10111)2, 15. Suppose A represents the symbol 1, B represents the, symbol 0, C represents the symbol 1, D represents, , the symbol 0 and so on. If we divide INDIA by AGRA,, then which one of the following is the remainder in, binary representation?, (a) (1101)2, (b) (101)2, (c) (11)2, (d) (110)2, 16. What is the 2's complement of the binary number, 1100110011?, (a) 0011001101, (b) 0011001100, (c) 1100110011, (d) 1100110010, 17. Two numbers in binary system are 110010011 and, 101010101, respectively. What is their difference in, decimal system?, (a) 66, (b) 56, (c) 65, (d) 62, 18. What is the value of, (1)2 + (11)2 + (111)2 + (1111)2 + (11111)2?, (a) (111111)2, (b) (111011)2, (c) (111001)2, (d) (100111)2, 19. What is the value of X, if (1010)2 × (111)2 = ( X )10?, (a) 60, (b) 70, (c) 75, (d) 80, 20. If x = (1101)2 , y = (110)2 , then what is x 2 + y 2 equal, to?, (a) (11101011)2, (b) (11001101)2, (c) (111000110)2, (d) (11100101)2, 21. What is the value of, (11), (11), (1001)2 2 − (101)2 2, (10), , (1001)2 2 + (1001)2, (a) (1001)2, (c) (110)2, , ( 01) 2, , ( 01), , (101)2 2 + (101)2, (b) (101)2, (d) (100)2, , (10) 2, , ?, , 22. The maximum three digit integer in the decimal, system will be represented in the binary system by, which one of the following?, (a) (1111110001)2, (b) (111111110)2, (c) (1111100111)2, (d) (1111000111)2, 23. What is the difference between the smallest five digit, binary integer and the largest four digit binary, integer?, (a) The smallest four digit binary integer, (b) The smallest one digit binary integer, (c) The greatest one digit binary integer, (d) The greatest three digit binary integer, 24. Consider the following statements, Statement I Binary equivalent of 225 is, (10110110)2., Statement II To convert a decimal digit C its binary, equivalent. Divide the decimal number by 2., Which of the given statements is/are correct?, (a) I only, (b) II only, (c) Both I and II, (d) Neither I nor II, 25. Consider the following statements, Statement I To obtain the 2's complement of a, binary number. First we find the 1's complement of, that number and then add 1.
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208, , NDA/NA Mathematics, Statement II 2's complement of binary number, (1100110011)2 is ( 0011001100)2, which of the given statement is/ are correct?, (a) I only, (b) II only, (c) Both I and II, (d) Neither I nor II, , Directions (Q. Nos 26-29) Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 26. Assertion (A) The decimal form of (10111)2 is 23., Reason (R) To convert a binary number into its, decimal equivalent, each digit of binary number is, multiplied by 2 having powers (0,1,2,3, ...)., 27. Assertion (A) The base of the binary number, system is 2., Reason (R) In the binary system, only two symbols 0, and 1 are used., , 28. Assertion (A) In the decimal number system we use, 9 digits which are 1-9., Reason (R) The base of the decimal number system, is 10., 29. Assertion (A) In binary addition 1 + 1 = 0 with a, carry of 1., Reason (R) In binary subtraction 1 − 1 = 0 with a, borrow of 1 from next higher column., , Directions (Q. Nos. 30-33), , In a binary number, system, assume x = 00111 and y = 01110 . Then,, 30. The value of ( x ⋅ y )10 is, 1, (a), 2, (c) 84, , (d) 28, , 31. The value of ( x + y )10 is, (a) 01010, (c) 10101, , (b) 00111, (d) None of these, , 32. The value of ( y − x )10 is, (a) 00001, (c) 00110, , (b) 00101, (d) 00111, , (b) 14, , 33. The addition of 2’s complement of x and y is, (a) 111011, (b) 101011, (c) 111000, (d) None of these, , Answers, Level I, 1. (a), 11. (b), , 2. (b), 12. (a), , 3. (c), 13. (a), , 4. (a), 14. (c), , 5. (b), 15. (c), , 6. (b), 16. (c), , 7. (a), 17. (c), , 8. (c), 18. (d), , 9. (a), 19. (c), , 10. (b), 20. (c), , 2., 12., 22., 32., , 3., 13., 23., 33., , 4. (b), 14. (c), 24. (b), , 5. (b), 15. (d), 25. (b), , 6. (c), 16. (a), 26. (a), , 7. (c), 17. (d), 27. (b), , 8. (b), 18. (c), 28. (d), , 9. (b), 19. (b), 29. (c), , 10. (c), 20. (b), 30. (b), , Level II, 1., 11., 21., 31., , (c), (d), (d), (c), , (d), (b), (c), (d), , (d), (a), (d), (b)
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Hints & Solutions, Level I, 1. In this case the binary number is multiplied by 2 having, negative powers., 2. To convert a decimal digit into binary number, we, divide it by 2 and note the remainder. Then, we write, the digits of binary number from bottom to top., 3. Complement of (1011100)2 is (0100011)2, 4. Let x = 1011001100 its 1’s complement is 0100110011, Then, 2’s complement is, 0100110011 + 1 = (0100110100)2, 5. a = 00111 = 24 × 0 + 23 × 0 + 22 × 1 + 21 × 1 + 20 × 1, =4 + 2 + 1 = 7, b = 01110 = 24 × 0 + 23 × 1 + 22 × 1 + 21 × 1 + 20 × 0, = 8 + 4 + 2 = 14, b 14, =, =2, ∴, a, 7, 6. The base of the binary number system is 2., 7. (101011)2 = 1 × 25 + 0 × 24 + 1 × 23, + 0 × 22 + 1 × 21 + 1 × 20, = 32 + 8 + 2 + 1 = 43, ∴ (101011)2 = (43)10, 8. The base of the decimal number system is 10., Binary, 0, 0.125 × 2 = 0.25, 0, 0.25 × 2 = 0.5, 1, 0.5 × 2 = 1.0, ∴, (0.125)10 = (0.001)2, 10. (10110100001)2 = 210 × 1 + 29 × 0 + 28 × 1 + 27 × 1, + 26 × 0 + 25 × 1 + 24 × 0 + 23 × 0, + 22 × 0 + 21 × 0 + 20 × 1, = 1024 + 256 + 128 + 32 + 1 = 1441, , 14., , Binary, 1, 0.77 × 2 = 1 .54, 1, 0.54 × 2 = 1 .08, 0, 0.08 × 2 = 0 .16, 0, 0.16 × 2 = 0.32, 0, 0.32 × 2 = 0.64, 1, 0.64 × 2 = 1 .28, 0, 0.28 × 2 = 0 .56, 1, 0.56 × 2 = 1 .12, ∴, ( 0.77)10 = (0.11000101)2, 1, 1, 1, 1, 12. (0.1011)2 = 1 × + 0 × 2 + 1 × 3 + 1 × 4, 2, 2, 2, 2, = 0.5 + 0.125 + 0.0625 = 0.6875, 13., , ∴ (0.1011)2 = (0.6875)10, 0110, + 0101, 1011, ∴, , (0110)2 + (0101)2 = (1011)2, , 292, , 0, , 2, , 146, , 0, , 2, , 73, , 1, , 2, , 36, , 0, , 2, , 18, , 0, , 2, , 9, , 1, , 2, , 4, , 0, , 2, , 2, , 0, , 1, 15. Given that, the number 1753 and it can be written as, , 9. Now,, , 11. Now,, , 2, , ⇒, , 2, , 1753, , 2, , 876, , 1, , 2, , 438, , 0, , 2, , 219, , 0, , 2, , 109, , 1, , 2, , 54, , 1, , 2, , 27, , 0, , 2, , 13, , 1, , 2, , 6, , 1, , 2, , 3, , 0, , 1, , 1, , 1753 = (11011011001)2, , 16. We have,, (1101101)2 = 26 × 1 + 25 × 1 + 24 × 0 + 23 × 1, + 22 × 1 + 21 × 0 + 20 × 1, = 64 + 32 + 0 + 8 + 4 + 0 + 1 = 109, (1011011)2 = 26 × 1 + 25 × 0 + 24 × 1, + 23 × 1 + 22 × 0 + 21 × 1 + 20 × 1, = 64 + 0 + 16 + 8 + 0 + 2 + 1 = 91, ∴ (1101101)2 + (1011011)2, = (109) + (91) = 200, 17. 100101 + 101 + 1101 + 100 = 111011, 18. 1111 + 111 + 1111 = 100101, 19. (101011)2 − (10010)2 = (011001)2, 20. 1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20, + 0 × 2 −1 + 0 × 2 −2 + 1 × 2 −3, 1, = 16 + 8 + 4 + 0 + 1 + 0 + 0 +, 8, 1, = 29 +, 8, 233, =, = (29.125)10, 8
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210, , NDA/NA Mathematics, , Level II, 1. Q (1111)2 = 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20, = 8 + 4 + 2 + 1 = 15, (1001)2 = 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20, =8 + 1 =9, and (1010)2 = 1 × 23 + 0 × 22 + 1 × 21 + 0 × 20, = 8 + 2 = 10, ∴ (1111)2 + (1001)2 − (1010)2 = 15 + 9 − 10 = 14, Now, converting (14)2 into binary, 14, =7+0, 2, 7, =3 + 1, 2, 3, =1 + 1, 2, 1, =0 + 1, 2, ∴, (14)10 = (1110)2, 2., , 2, 2, 2, 2, 2, , 32, 16, 8, 4, 2, 1, , 0, 0, 0, 0, 0, 1, , ∴, , 0.25, ×2, 0.50 0, ×2, 1 1, , 32.25 = 100000.01, , 3. (0.101)2 = 2−1 × 1 + 2−2 × 0 + 2−3 × 1, 1, 1 5, = +0+ =, 2, 8 8, and (0.011)2 = 0 × 2−1 + 1 × 2−2 + 1 × 2−3, 1 1 3, =0 + + =, 4 8 8, Also,, (11)2 = 1 × 21 + 1 × 20 = 3, (10)2 = 1 × 21 + 0 × 20 = 2, and, (01)2 = 0 × 21 + 1 × 20 = 1, 2 + (0.011)(11)2, (0.101)(11), 2, 2, ∴, 2 – (0.101)(01)2 (0.011)(01)2 + (0.011)(10)2, (0.101)(10), 2, 2, 2, 2, 3, , =, , 3, 5, + , 8, 8, 2, , 3, , 5 3 3, 5, − + , 8 7 8, 8, 5 3 8, = + = = 1 = (1)2, 8 8 8, , Now,, , ∴, , 1, 0, 1, 0, 0, 0, 0, , 133 = (10000101)2, , 6. (101.101)2 = 1 × 22 + 0 × 21 + 1 × 20 + 1 × 2−1, + 0 × 2 −2 + 1 × 2 −3, 1 1 40 + 4 + 1 45, =4 + 1 + + =, =, 2 8, 8, 8, = (5.625)10, 7. 0.111111K, , = 2 −1 + 2 −2 + 2 −3 + 2 −4 + 2 −5 + 2 −6 + K, 1 1 1, 1, 1, 1, = + + +, +, +, +K, 2 4 8 16 32 64, 32 + 16 + 8 + 4 + 2 + 1, =, +K, 64, 63, =, + K = 0.9843 + K ≈ 1, 64, 1, 8. 1001.01 = 1 × 23 + 1 × 20 +, 4, 1 37, =8 + 1 + =, = 9.25, 4 4, 1, and, 11.1 = 1 × 21 + 1 × 20 +, 2, 1 7, = 2 + 1 + = = 3.5, 2 2, ∴ 1001.01 × 11.1 = 9.25 × 3.5, = 32.375, , 2, , 4. Given,, , ∴, , 133, 66, 33, 16, 8, 4, 2, 1, , 5. (10x 010)2 − (11 y1)2 = (10z11)2, ⇒ (25 × 1 + 0 × 24 + x × 23 + 0 × 22 + 1 × 21 + 0 × 20 ), − (23 × 1 + 22 × 1 + y × 21 + 1 × 20 ), 4, = 2 × 1 + 0 × 23 + 22 × z + 21 × 1 + 20 × 1, ⇒, (34 + 8x) − (13 + 2 y) = 19 + 4z, ⇒, 2 = − 8x + 2 y + 4z, ⇒, x = 0, y = 1 and z = 0, , [Q a3 + b3 = (a + b)(a 2 − ab + b2)], , and, , 2, 2, 2, 2, 2, 2, 2, , x = (1101)2 = 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20, = 8 + 4 + 1 = 13, y = (110)2 = 1 × 22 + 1 × 21 + 0 × 20, =4 + 2 =6, x2 − y2 = (13)2 − (6)2, = 169 − 36 = 133, , Now,, , and, ∴, , 2, , 32, , 2, , 16, , 0, , 2, , 8, , 0, , 2, , 4, , 0, , 2, , 2, , 0, , 1, , 0, , (32)10 = (100000)2, 0, 0.375 × 2 = 0.75, 1, 0.75 × 2 = 1.5, 1, 0.5 × 2 = 1.0, (0.375)10 = (0.011)2, (32.375)10 = (100000.011)2
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211, , Binary Numbers, 9. (111000.0101)2, = 1 × 25 + 1 × 24 + 1 × 23 + 0 × 22 + 0 × 21 + 0 × 20, 1, 1, 1, 1, +0× +1× 2+0× 3 +1× 4, 2, 2, 2, 2, 1, 1, = 32 + 16 + 8 + +, = 56.3125, 4 16, (111000.0101)2 = (56.3125)10, 10., , 2, 2, 2, , 13, 6 1, 3 0, 1 1, (13)10 = (1101)2, 0.0625 × 2 = 0.125, 0.125 × 2 = 0.25, 0.25 × 2 = 0.5, 0.5 × 2 = 1.0, (0.0625)10 = (.0001)2, (13.0625)10 = (1101.0001)2., , ∴, Now,, , ∴, , ∴, , 12. We have the sum of, the binary numbers, (11011)2, (10110110)2 and (10011x 0 y)2 is = (101101101)2, (binary number), (11011) 2 =, (10110110) 2 =, (10011 x0 y) 2 =, , 1, 1 0 1 1, 1 0 0 1, 1 0 1 1 0, , 1, 0, 1, 1, , 0, 1, x, 1, , 2, 2, 2, 2, , Binary, 0, 0, 0, 1, , 10 11.01, 10 01 .11, 10 101 .00, , 11., , 14. (1100110011)2, = 1 × 29 + 1 × 28 + 0 × 27 + 0 × 26 + 1 × 25 + 1 × 24, + 0 × 23 + 0 × 22 + 1 × 21 + 1 × 20, 9, 8, 5, = 2 + 2 + 2 + 24 + 21 + 1, = 512 + 256 + 32 + 16 + 2 + 1 = 819, (1101001011)2, = 1 × 29 + 1 × 28 + 0 × 27 + 1 × 26 + 0 × 25 + 0 × 24, + 1 × 23 + 0 × 22 + 1 × 21 + 1 × 20, 9, 8, 6, 3, = 2 + 2 + 2 + 2 + 21 + 1, = 512 + 256 + 64 + 8 + 2 + 1 = 843, ∴, 843 − 819 = 24, , 1, 1, 0, 0, , 1, 0, y, 1, , ∴ From first column, we have 1 + 0 + y = 1, ⇒, y=0, From second column, we have, 1 + 1 + 0 = 0 with carry 1, ∴ From third column, we have, 1+ 0+1+ x=1, ∴ Obviously, 1 + 0 + 1 + 1 = 1 with carry 1, ∴, x=1, , 24, 12, 6, 3, 1, , 0, 0, 0, 1, , 24 = (11000)2, , INDIA, (10011)2 24 × 1 + 0 + 0 + 21 × 1 + 20 × 1, 15. ∴, =, =, AGRA, (1101)2, 23 × 1 + 22 × 1 + 0 + 20 × 1, 16 + 2 + 1 19, =, =, 8 + 4 + 1 13, ⇒ Remainder is 6., i.e.,, 2, , 6, , 0, , 2, , 3, , 1, , 1, ∴ Binary representation is (110)2., 16. 2’s complement of 1100110011 = 1's complement of, 1100110011 + 1, 1’s complement of 1100110011 = 0011001100, 2’s complement of 1100110011 = 0011001100 + 1, = 0011001101, 17. (110010011)2, = 1 × 28 + 1 × 27 + 0 + 0 + 1 × 24 + 0 + 0 + 1 × 21 + 1 × 20, = 256 + 128 + 16 + 2 + 1 = 403, and (101010101)2, , 13. We have,, 1x101, + 10 y 1, , = 1 × 28 + 0 + 1 × 26 + 0 + 1 × 24 + 0 + 1 × 22 + 0 + 2, , 100z00, , = 341, ∴Required difference = 403 − 341 = 62, , ∴In the addition of binary number, we know that, 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 0 with carry of 1, Now, from first column, 1 + 1 = 0, (with carry 1), second column, 1 + 0 + y = 0, (with carry 1), ⇒, y=1, third column, (with carry 1), 1 + 1 + 0 = z =0, fourth column, 1 + x+ 1 =0, (with carry 1), ⇒, x=0, fifth column,, 1 + 1 = 10, Hence, the possible values of x, y and z are respectively, 0,1 and 0., , = 256 + 64 + 16 + 4 + 1, , 18. Since,, , (1)2 = 20 × 1 = 1, (11)2 = 21 × 1 + 20 × 1 = 2 + 1 = 3, (111)2 = 22 × 1 + 21 × 1 + 20 × 1, =4 + 2 + 1 = 7, (1111)2 = 23 × 1 + 22 × 1 + 21 × 1 + 20 × 1, = 8 + 4 + 2 + 1 = 15, , and, (11111)2 = 24 × 1 + 23 × 1 + 22 × 1 + 21 × 1 + 20 × 1, = 16 + 8 + 4 + 2 + 1 = 31, ∴, , (1)2 + (11)2 + (111)2 + (1111)2 + (11111)2, = 1 + 3 + 7 + 15 + 31 = 57, , 0, , ×1
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212, , NDA/NA Mathematics, , 2, 2, 2, 2, 2, , 57, 28, 14, 7, 3, 1, , 23. The smallest five digit binary number is 10000., The greatest four digit binary number is 1001., Now, the difference between them, = (10000)2 − (1001)2 = (111)2, which is the greatest three digit binary integer ., , 1, 0, 0, 1, 1, , 24., , 2, 2, 2, 2, 2, 2, 2, , (57)10 = (111001)2, 19. (1010)2 = 1 × 2 + 0 × 2 + 1 × 2 + 0 × 2, = 8 + 0 + 2 + 0 = 10, (111)2 = 1 × 22 + 1 × 21 + 1 × 20 = 4 + 2 + 1 = 7, ∴ (1010)2 × (111)2 = 10 × 7 = (70)10, ∴ The value of x is 70., 3, , 2, , 1, , 0, , 20. We have, x = (1101)2 and y = (110)2, ⇒, , x = 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20, , = 8 + 4 + 1 = 13 and y = 1 × 2 + 1 × 2 + 0 = 4 + 2 = 6, 2, , ∴, , 1, , x2 + y2 = (13)2 + (6)2 = 169 + 36 = 205, 2, 205, 2, 102, 1, 2, 51, 0, 2, 25, 1, 2, 12, 1, 2, 6, 0, 2, 3, 0, 1, 1, , ∴ (205)10 = (11001101)2, 21. Q, , ∴, , (1001)2 = 23 + 20 = 8 + 1 = 9, (11)2 = 21 + 20 = 2 + 1 = 3, (101)2 = 22 + 20 = 4 + 1 = 5, (10)2 = 21 = 2 and (01)2 = 1, +, , (1001)(201)2, , =, , 9 −5, (9 + 9 × 5 + 52), 3, , (101)(201)2, , +, , (101)(210)2, , ∴, (225)10 = (10000111)2, So, only statement II is correct., 25. Let x = 1100110011, The 1’s complement of x = 0011001100, and then 2’s complement of x = 0011001100 + 1, = 0011001101, but statement I is wrong., So, only statement II is correct., 26. (10111)2 = 1 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 1 × 20, = 16 + 0 + 4 + 2 + 1 = (23)10, So, both A and R are individually true and R is the, correct explanation of A., 27. In the binary system, only two symbols 0 and 1 are uscd, with base 2., but ‘R ’ is not the correct explanation of A., , 29. 1 + 1 = 0 is true, but 0 − 1 = 1 with a borrow of 1 from, next higher column., So, A is true but R is false., , 3, , 2, , (9 − 5) (92 + 9 × 5 + 52), =, (92 + 9 × 5 + 52), = 9 − 5 = 4 = (100)2, 22. The maximum three digit number in decimal system is, 999., Now,, 2, 999, 2, 499, 1, 2, 249, 1, 2, 124, 1, 2, 62, 0, 2, 31, 0, 2, 15, 1, 2, 7, 1, 2, 3, 1, 1, 1, ∴, , 1, 0, 0, 0, 0, 1, 1, , 28. In decimal number system we use 10-digits which are, 0 − 9 and the base of the decimal number system is 10., So, A is false but R is true., , (1001)(211)2 − (101)(211)2, (1001)(210)2, , 225, 112, 56, 28, 14, 7, 3, 1, , (999)10 = (1111100111)2, , Solutions (Q. Nos. 30-33), 30. Given, x = 00111 = 0 × 24 + 0 × 23 + 1 × 22, + 1 × 2 1 + 1 × 20, =0 + 0 + 4 + 2 + 1 = 7, and, y = 01110 = 0 × 24 + 1 × 22 + 1 × 21 + 0 × 20, = 0 + 8 + 4 + 2 + 0 = 14, Now, x ⋅ y = 7.14 = (84)10, 31. (x + y)10 i.e.,, , 32. ( y − x)10 i. e. ,, , 00111, + 01110, 10101, 01110, − 00111, 00111, , 33. Q x = 00111 and y = 01110, Now, 2’s complement of x = 11000 + 1 = 11001 = x′, and 2’s complement of y = 10001 + 1, 10010 = y′, ∴, x′ + y′ i. e. , 11001, +10010, 101011
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12, , Trigonometric, Ratios and Equations, In the figure,, OA = OC = Arc AC = r, , Measurement of Angles, There are three mainly system of measurement of, angles., 1. Sexagesimal system 2. Centesimal system, 3. Circular system, , 1. Sexagesimal System, A right angle is divided into 90 equal parts and each of, part is known as 1 degree. Thus, each right angle is equal, to 90 degree. 1 degree is denoted by 1°., Each degree is divided into 60 equal parts which is, known as one minute and 1 min is denoted by 1 ′. Each, minute is divided into 60 equal parts, each of which is, known as 1 s. One second is denoted by 1′′., Hence,, 1 right angle = 90°, (90 degree), (60 min), 1° = 60 ′, (60 s), 1 ′ = 60 ′′, , Then measure of ∠AOC is, 1 rad and is denoted by 1c., , (100 min), , 1 ′ = 100′′, , (100 s), , 1c, , B, , O, , r, , A, , If r and l be the radius and, length of an arc of a circle, then, angle subtended by the arc is, given by, C, l, r, θ, , B, , O, , θ=, , 1 g = 100 ′, , l, , r, , (radius of circle), , 2. Centesimal System, A right angle is divided into 100 equal parts and each, part is known as one grade. Hence, a right angle is equal to, 100 grade. One grade is denoted by 1 g ., Each grade is divided into 100 equal parts and each, part is known as 1 min and one minute is denoted by 1 ′., Each minute is also divided into 100 equal parts and, each part is known as 1 s. One second is denoted by 1′′., (100 grade), 1 right angle = 100 g, , C, , r, , A, , l, r, , Relation Among Degree, Radian, and Grade, 180° = π c = 100 g, and, %, , 1 rad = 57°16 ′ 22′′, , If the measure of an angle is given in degree. To convert it into, π, radian, the angle should be multiplied by, and to convert an, 180°, angle from radian to degree put 180° at the place of π., , Example 1. Find in degree the angle through which a, , 3. Circular System, Angle subtended by an arc whose length is equal to the, radius of circle, at the centre of circle is known as one, radian., , pendulum swings, if its length is 50 cm and the tip describes, an arc of length 10 cm., (a) 11°27′ 16.36 ′′, (b) 11°27 ′ 17 ′′, (c) 11°27′ 16.50 ′′, (d) None of these
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215, , Trigonometric Ratios and Equations, , Example 3. If 5 tan θ = 4, then, (a) 0, 1, (c), 6, , (a) a 2 + b 2, (c) a + b, , 5 sin θ − 3 cos θ, is equal to, 5 sin θ + 2 cos θ, (b) 1, , (b) a 2 − b 2, (d) None of these, , Solution (a) Since, x = a cos θ + b sin θ, , (d) 6, , and, Now,, , 4, 5, 5 sin θ − 3 cos θ 5 tan θ − 3, =, 5 sin θ + 2 cos θ 5 tan θ + 2, , y = a sin θ − b cos θ, x2 + y 2 = ( a cos θ + b sin θ) 2 + ( a sin θ − b cos θ) 2, , Solution (c) Given, tan θ =, ∴, , = a2 cos2 θ + b 2 sin 2 θ + 2ab sin θ cos θ, + a2 sin 2 θ + b 2 cos2 θ − 2ab sin θ cos θ, = a (sin 2 θ + cos2 θ) + b 2(sin 2 θ + cos2 θ), 2, , (Divide numerator and denominator by cos θ), 4, 5 × −3, 1, 5, =, =, 4, 6, 5× +2, 5, , Example 4. The value of, (a) 3, , = a2 + b 2, , Domain and Range of, Trigonometric Functions, Functions, , Domain, , Range, , sin x, cos x, tan x, , R, R, { x : x ∈ R and, 1, , x ≠ n + π,, , 2, n ∈ I}, { x : x ∈ R and x ≠ nπ,, n ∈ I}, { x : x ∈ R and x ≠ nπ,, n ∈ I}, { x : x ∈ R and, 1, , x ≠ n + π, n ∈ I}, , 2, , [ −1, 1 ], [ −1, 1 ], R, , (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is, (b) 2, (c) − 2, (d) − 3, , Solution (b) (1 + cot θ − cosec θ ) (1 + tan θ + sec θ ), , cos θ, 1 , = 1 +, −, , , sin θ sin θ , , , sin θ, 1 , +, , 1 +, , cos θ cos θ , , sin θ + cos θ − 1 cos θ + sin θ + 1, =, , , , , , sin θ, cos θ, , cot x, , (sin θ + cos θ) 2 − 1 sin 2 θ + cos2 θ + 2 sin θ cos θ − 1, =, sin θ cos θ, sin θ cos θ, 2 sin θ cos θ, =, =2, sin θ cos θ, , cosec x, , =, , sec x, , R, ( −∞, − 1) ∪ (1, ∞), ( −∞, − 1) ∪ (1, ∞), , Example 5. If a cos θ + b sin θ = x and, a sin θ − b cos θ = y, then x 2 + y 2 is equal to, , Trigonometric Ratios in Different Quadrants, y, , y, , In, II Quadrant, , I Quadrant, , sin, cosec are, positive and the, rest are negative, , All positive, , x´, , sine, cosine, tangent, cotangent, secant, cosecant, , decreases, from, decreases, from, increases from, decreases, , x x´, , O, , In, , III Quadrant, , IV Quadrant, , tan, cot are, positive and the, rest are negative, , cos, sec are, positive and the, rest are negative, , y´, , Quadrant, , II, , III, , sine, cosine, tangent, cotangent, secant, cosecant, , In, 1 to 0, 0 to –1, – ∞ to 0, 0 to – ∞, – ∞ to –1, 1 to ∞, , O, , Quadrant, , decreases, from, increases from, increases from, decreases, from, , 0 to 1, –1 to 0, 0 to ∞, ∞ to 0, –1 to – ∞, – ∞ to –1, , I, , sine, cosine, tangent, cotangent, secant, cosecant, , In, , 0 to 1, 1 to 0, 0 to ∞, ∞ to 0, 1 to ∞, ∞ to 1, , x, , IV Quadrant, , sine, cosine, tangent, cotangent, secant, cosecant, , y´, , Quadrant, increases from, decreases, from, increases from, decreases, from, , increases from, increases from, increases from, decreases, from, decreases, , –1 to 0, 0 to 1, – ∞ to 0, 0 to – ∞, ∞ to 1, –1 to – ∞
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217, , Trigonometric Ratios and Equations, , Trigonometric Ratios of, Combined Angles, , Solution (b) We have, tan (α + β) =, , 1. Sum and Difference of Two Angles, (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), , sin ( A +, sin ( A −, cos ( A +, cos ( A −, , B) =, B) =, B) =, B) =, , sin A cos B + cos A sin B, sin A cos B − cos A sin B, cos A cos B − sin A sin B, cos A cos B + sin A sin B, tan A + tan B, tan ( A + B) =, 1 − tan A tan B, tan A − tan B, tan ( A − B) =, 1 + tan A tan B, cot A cot B − 1, cot ( A + B) =, cot A + cot B, cot A cot B − 1, cot ( A − B) =, cot B − cot A, , (ix) sin ( A + B) sin ( A − B), = sin2 A − sin2 B = cos2 B − cos2 A, , Example 8. The value of, , 4, , 2, sin θ + sin π + θ + sin π + θ is, , 3, , 3, (b) 1, , (c) 2, , (d) 3, , Solution (a) sin θ + sin π + θ + sin π + θ, , , , , 2, 3, , 4, 3, 1 , 1, , , , , = sin θ + sin π − π + θ + sin π + π + θ , , 3 , 3, , , , , , , , 1, 1, = sin θ + sin π − π − θ + sinπ + π + θ , , , , , 3, 3, , , , , 1, , 1, = sin θ + sin π − θ − sin π + θ, , 3, , 3, 1, 1, , , = sin θ + sin π cos θ − cos π sin θ, , , 3, 3, 1, 1, , , − sin π cos θ + cos π sin θ, , , 3, 3, 1, = sin θ − 2 cos π sin θ = sin θ − 2 cos 60 ° sin θ, 3, 1, = sin θ − 2 ⋅ ⋅ sin θ = sin θ − sin θ = 0, 2, , Example 9. If tan α =, α + β is equal to, π, (a), 2, π, (c), 3, , m, 1, and tan β =, , then, m +1, 2m + 1, (b), , 1, m, +, 2m2 + m + m + 1, m + 1 2m + 1, =, =, 1, m, 2m2 + 3m + 1 − m, 1−, ×, m + 1 2m + 1, π, 2m2 + 2m + 1, = 1 = tan, 2, 4, 2m + 2m + 1, π, ∴ α+β=, 4, =, , 2. Transformation of Product into Sum or, Difference, (i) 2 sin A cos B = sin ( A + B) + sin ( A − B) , (ii) 2 cos A sin B = sin ( A + B) − sin ( A − B), A> B, (iii) 2 cos A cos B = cos ( A + B) + cos ( A − B) , (iv) 2 sin A sin B = cos ( A − B) − cos ( A + B) , , 3. Transformation of Sum or Difference into, Product, , (x) cos ( A + B) cos ( A − B), = cos2 A − sin2 B = cos2 B − sin2 A, , (a) 0, , tan α + tan β, 1 − tan α tan β, , π, 4, , (d) None of these, , C − D, C + D, (i) sin C + sin D = 2 sin , , ⋅ cos , 2 , 2 , C − D, C + D, (ii) sin C − sin D = 2 cos , , ⋅ sin , 2 , 2 , C + D, C − D, (iii) cos C + cos D = 2 cos , ⋅ cos , , 2 , 2 , D − C, C + D, (iv) cos C − cos D = 2 sin , , ⋅ sin , 2 , 2 , , Example 10. The simplified form of, cos A cos (60 ° − A) cos (60 ° + A) is, 1, 1, (a) sin3A, (b) cos3A, (c), cos 3A (d) cos 3A, 2, 4, , Solution (d) cos A ⋅ cos (60 ° − A) cos (60 ° + A), = cos A ⋅ (cos2 60 ° − sin 2 A), [Q cos ( A + B ) cos ( A − B ) = cos2 A − sin 2 B], , 1, = cos A − sin 2 A, , 4, , 1, = cos A − (1 − cos2 A), , 4, 3, , = cos A − + cos2 A, 4, , 1, = cos A ( −3 + 4 cos2 A), 4, 1, 1, = ( 4 cos3 A − 3 cos A) = cos 3A, 4, 4, , Example 11. The value of, (sin 3A + sin A) sin A + (cos 3A − cos A) cos A is, (a) 1, (b) 0, (c) 2, (d) 3
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Exercise, Level I, 1. The angle subtended at the centre of a circle of radius, 3 m by an arc of length 1 m is equal to, (a) 20°, (b) 60°, 1, (c) rad, (d) 3 rad, 3, 2. The circular wire of diameter 10 cm is cut and placed, along the circumference of a circle of diameter 1 m., The angle subtended by the wire at the centre of the, circle is equal to, π, π, π, π, (a), rad, (b), rad, (c), rad, (d), rad, 4, 3, 5, 10, 3. The maximum value of 3 cos θ + 4 sin θ is, (a) 3, (b) 4, (c) 5, (d) None of these, 4. If sin θ + cos θ = m and sec θ + cosec θ = n, then, n( m + 1)( m − 1) is equal to, (a) m, (b) n, (c) 2m, (d) 2n, 5. What is the value of, sin 420°⋅ cos 390° + cos( −300° ) ⋅ sin( −330° ) ?, (NDA 2012 I), , (a) 0, , (b) 1, , (c) 2, , (d) −1, , 6. The value of cot( 45° + θ ) cot ( 45° − θ ) is, (a) –1, (b) 0, (c) 1, (d) ∞, 7. The value of sin A sin ( 60° − A) sin ( 60° + A) is equal, to, sin 3 A, (a) sin 3A, (b), 2, sin 3 A, (d) None of these, (c), 4, 8. The value of cos 15° is, 1 + cos 30°, (a), 2, 1 + cos 30°, (c) ±, 2, , 1 − cos 30°, 2, 1 − cos 30°, (d) ±, 2, (b), , 9. What is the value of, sin A cos A tan A + cos A sin A cot A ? (NDA 2012 I), (a) sin A, (b) cos A, (c) tan A, (d) 1, 10. What is the maximum value of 3 cos x + 4 sin x + 5?, (NDA 2011 II), , (a) 5, , (b) 7, , (c) 10, (d) 12, sin x, 1 + cos x, 11. What is the value of, ?, +, 1 + cos x, sin x, (NDA 2011 II), , (a) 2 tan x, (c) 2 cos x, , (b) 2 cosec x, (d) 2 sin x, , 12. If sin 3 A = 1, then how many distinct values can sin, A assume?, (NDA 2011 II), (a) 1, (b) 2, (c) 3, (d) 4, 13. If (1 + tan θ )(1 + tan φ ) = 2, then what is (θ + φ ) equal, to?, (NDA 2011 I), (a) 30°, (b) 45°, (c) 60°, (d) 90°, 14. Which one of the following is positive in the third, quadrant?, (NDA 2012 I), (a) sinθ, (b) cosθ, (c) tanθ, (d) sec θ, 15. What is the value of sin1920°?, 1, 1, 3, (a), (b), (c), 2, 2, 2, 16. What is the value of, (a) 1, (c) 1/3, , (NDA 2012 I ), , (d), , sin θ, cos θ, ?, +, cosec θ sec θ, , 1, 3, , (NDA 2012 I), , (b) 1/2, (d) 2, , 17. What is the maximum value of, (NDA 2012 I), sin 3θ cos 2θ + cos 3θ sin 2θ?, (a) 1, (b) 2, (c) 4, (d) 10, cos 12° − sin 12° sin 147°, is equal to, 18. The value of, +, cos 12° + sin 12° cos 147°, (a) 1, (b) –1, (c) 0, (d) None of these, 19. The value of 3 cosec 20° − sec 20° is equal to, 2 sin 20°, (a) 2, (b), sin 40°, 4 sin 20°, (c) 4, (d), sin 40°, 20. The expression tan2 α + cot2 α is equal to, (a) ≥ 2, (b) ≤ 2, (c) ≥ − 2, (d) None of these, 21. One of the angles of a triangle is 1/2 rad and the other, is 99°. What is the third angle in radian measure?, (NDA 2010 II), , 9π − 10, (a), π, 90π − 10, (c), π, , 90π − 100, (b), 7π, (d) None of these, , 22. If y = sec2 θ + cos2 θ , where 0 < θ <, , π, , then which one, 2, , of the following is correct?, (NDA 2010 II), (a) y = 0, (b) 0 ≤ y ≤ 2, (c) y ≥ 2, (d) None of these
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222, , NDA/NA Mathematics, , 3, 12, and tan B = −, , then how many values, 4, 5, can cot ( A − B) have depending on the actual values, of A and B?, (NDA 2010 II), (a) 1, (b) 2, (c) 3, (d) 4, sin θ + cos θ − tan θ, 24. What is the value of, , when, sec θ + cosec θ − cot θ, 3π, (NDA 2010 II), ?, θ=, 4, (a) 0, (b) 1, (c) –1, (d) None of these, 23. If tan A =, , 25. Which one of the following is correct? (NDA 2010 II), (a) sin 1° > sin 1, (b) sin 1° < sin 1, π, (c) sin 1° = sin 1, (d) sin 1° =, sin 1, 180, 26. What is the value of, (NDA 2010 I), cos 15° sin 15°, cos 45° cos 15°, ?, ×, cos 45° sin 45°, sin 45° sin 15°, 1, 3, (b), (a), 2, 4, 1, 3, (d) −, (c) −, 4, 4, 27. The angle A lies in the third quadrant and it satisfies, the equation 4 (sin2 x + cos x ) = 1. What is the, measure of the ∠A?, (NDA 2010 I), (a) 225°, (b) 240°, (c) 210°, (d) None of these, 1, 1, 1 2 1, x + , then x + 2 is equal to, , , 2, x, 2, x , (a) sin 2θ, (b) cos 2θ, (c) tan 2θ, (d) sec 2θ, sin ( B + A) + cos ( B − A), 29. The value of, is equal to, sin ( B − A) + cos ( B + A), cos B + sin B, cos A + sin A, (b), (a), cos B − sin B, cos A − sin A, cos A − sin A, (d) None of these, (c), cos A + sin A, , 28. If cos θ =, , 4, 12, and cos B = −, , where A and B lie in, 5, 13, first and third quadrant respectively, then, cos ( A + B) is equal to, 56, 16, 56, 16, (c), (b) −, (d) −, (a), 65, 65, 65, 65, , 30. If sin A =, , 31. The value of x for the maximum value of, 3 cos x + sin x, is, (a) 30°, (b) 45°, (c) 60°, (d) 90°, 32. If A, B, C, D are the angles of a cyclic quadrilateral,, then cos A + cos B + cos C + cos D is equal to, (a) 2 (cos A + cos C ), (b) 2(cos A + cos B), (c) 2(cos A + cos D ), (d) 0, , 3π, < α < π , then cosec2α + 2 cot α is equal to, 4, (a) 1 + cot α, (b) 1 − cot α, (c) − 1 − cot α, (d) − 1 + cot α, , 33. If, , 34. If (sin x + cosec x )2 + (cos x + sec x )2, = k + tan2 x + cot2 x ,, then what is the value of k?, (a) 8, (b) 7, (c) 4, (d) 3, 41π, 35. If A =, , then what is, 12, 1 − 3 tan2 A, ?, 3 tan A − tan3 A, (a) –1, 1, (c), 3, , (NDA 2009 II), , the, , value, , of, , (NDA 2009 II), , (b) 1, (d) 3, , π, 36. If cot θ = 2 cos θ , where < θ < π , then what is the, 2, value of θ?, (NDA 2009 II), 5π, 2π, (b), (a), 6, 3, 3π, 11π, (d), (c), 4, 12, 5, and θ lies in the third quadrant, then, 37. If cot θ =, 12, what is the value of ( 2 sin θ + 3 cos θ )? (NDA 2009 II), (a) – 4, (b) − p2 for some odd prime p, q, (c) − where p is an odd prime and q is a, p, q, positive integer with not an integer, p, (d) − p for some odd prime p, 38. What is the value of, (NDA 2009 II), π, π, 5, π, 7, π, , , , , cos + cos + cos + cos ?, 9, 9, 3, 9, (a) 1, (b) –1, 1, 1, (c) −, (d), 2, 2, 39. What is the length of arc of a circle of radius 5 cm, subtending a central angle measuring is? (NDA 2009 II), 5π, 7π, cm, (b), cm, (a), 12, 12, π, π, cm, (d), cm, (c), 12, 5, 40. If P = sin ( 989° ) cos ( 991° ), then which is the correct, statement in following?, (NDA 2009 II), (a) P is finite and positive, (b) P is finite and negative, (c) P = 0, (d) P is not defined
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223, , Trigonometric Ratios and Equations, 41. What is the value of 1 − sin 10° sin 50° sin 70° ?, (NDA 2009 I), , (a), , 1, 8, , (b), , 3, 8, , (c), , 5, 8, , (d), , 7, 8, , 42. The sines of two angles of a triangle are equal to, and, , 5, 13, , 99, . What is the cosine of the third angle?, 101, (NDA 2009 I), , 255, (a), 1313, 275, (c), 1313, , 265, (b), 1313, 770, (d), 1313, , 43. One radian is approximately equal to which one of, the following?, (NDA 2009 I), (a) 90°, (b) 180°, (c) 57°, (d) 47°, 44. x = sin θ cos θ and y = sin θ + cos θ are satisfied by, which one of the following equations?, (NDA 2009 I), 2, 2, (a) y − 2x = 1, (b) y + 2x = 1, (c) y 2 − 2x = − 1, (d) y 2 + 2x = − 1, 45. If sin x − cos x = p, then which one of the following, is correct?, (NDA 2009 I), (a) p = 1, (b) p = 0, (c) | p|> 1 (d) | p|≤ 1, 1, 1, 46. If sin A =, and sin B =, , where A and B are, 10, 5, positive acute angles, then A + B is equal to, π, π, π, (c), (d), (a) π, (b), 2, 3, 4, 4, , 4, , 47. If tan A − tan B = x and cot B − cot A = y , then, cot( A − B) is equal to, 1, 1, 1 1, 1 1, (a), (b), (d), (c), +y, −, +, x, xy, x y, x y, 3π, then, the, value, of, 48. If, A+ B+C =, ,, 2, cos 2 A + cos 2B + cos 2C is equal to, (a) 1 − 4 cos A cos B cos C (b) 4 sin A sin B sin C, (c) 1 + 2 cos A cos B cos C (d) 1 − 4 sin A sin B sin C, 2, , 49. The value of, , sin 3A, 2, , sin A, , (a) cos 2A, 1, (c), 8 cos 2 A, 50. The value of, , sin 1°, sin 1c, , (a) greater than 1, (c) equal to 1, , −, , 2, , cos 3A, 2, , is equal to, , cos A, (b) 8 cos 2 A, cos 2 A, (d), 8, , , where 1c represents 1 rad, is, (b) less than 1, (d) equal to π /180, , 51. The value of sin 10° + sin 20° + sin 30° + …+ sin 360°, is equal to, (a) 0, (b) 1/2, (c) 1, (d) 2, , 52. The value of, sin θ + sin (θ + 120° ) + sin (θ + 240° ) is equal to, (a) 0, (b) 1, (c) 3, (d) 2, 53. What is 2 + 2 + 2 + 2 cos A equal to?, (NDA 2008 II), , (a) cos A, A, (c) 2 cos , 2, , (b) cos ( 2A), (d), , 2 cos A, , 54. What is the measure of the angle 114° 35′ 30′ ′ in, radian?, (NDA 2008 I), (a) 1 rad, (b) 2 rad, (c) 3 rad, (d) 4 rad, 55. Which one of the following is correct?, 1° , 1° , , 1 + cos 67 1 + cos 112 is, , 2, 2 , (a), (b), (c), (d), , (NDA 2008 I), , an irrational number and is greater than 1, a rational number but not an integer, an integer, an irrational number and is less than 1, , 56. For what value of x does the equation, 4 sin x + 3 sin 2x − 2 sin 3x + sin 4x = 2 3 hold?, (NDA 2008 I), , π, (a), 6, , π, (b), 4, , π, (c), 3, , (d), , π, 2, , 57. If X = sin( A + B) sin ( A − B) andY = cos ( A + B), cos ( A − B), then which one of the following is not, correct ?, (a) X + Y > 0, if 0° < B < 45° for any A, (b) X + Y = 0, if B = 45° for any A, (c) X + Y is a rational number for any A and B, (d) X + Y < 0, if 45° < B ≤ 90° for any A, 58. If tan θ + sec θ = 4, then what is the value of sinθ?, 15, 8, 15, 3, (b), (c), (d), (a), 28, 15, 17, 5, π, 59. What is the value of tan ?, 12, (a) 2 − 3, (c), , 2− 3, , (NDA 2012 I), , (b) 2 + 3, (d), , 3− 2, , 60. For what values of x is the equation 2 sin θ = x +, valid?, (a) x = ± 1, (c) −1 < x < 1, tan x = b/ a, then, a+b, a−b, ?, +, a−b, a+b, 2 sin x, (a), sin 2x, 2 cos x, (c), sin 2x, , 61. If, , 1, x, , (b) All real values of x, (d) x > 1 and x < − 1, what, , (b), (d), , is, , the, , 2 cos x, cos 2x, 2 sin x, cos 2x, , value, , of
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224, , NDA/NA Mathematics, , 62. Which one of the following pairs is not correctly, matched?, (NDA 2008 I), (a) sin 2π, : sin ( −2π ), (b) tan 45°, : tan ( −315° ), (c) cot (tan−1 0.5) : tan (cot−1 0.5), (d) tan 420°, : tan ( −60° ), 1, 63. If cos 3 A = , then how many values can sin A, 2, assume? ( 0 < A < 360° ), (NDA 2007 II), (a) 3, (b) 4, (c) 5, (d) 6, x = a sec θ cos φ , y = b sec θ sin φ , z = c tan θ, then, x2 y2 z 2, what is the value of 2 + 2 − 2 ?, (NDA 2007 II), a, b, c, (a) 1, (b) 0, (c) −1, (d) a 2 + b2 − c2, , 64. If, , 65. If sin A = sin B and cos A = cos B, then which one of, the following is correct?, (NDA 2007 II), (a) B = nπ + A, (b) A = 2nπ − B, (c) A = 2nπ + B, (d) B = nπ − A (n is an integer), π, 66. If A + B = , what are the greatest and the least, 2, values of cos A cos B, respectively?, (a) 1/2 and 0, (b) 0 and –1/2, (c) 1/2 and –1/2, (d) 0 and –1, 67. The difference of two angles is 1°; the circular, measure of their sum is 1. What is the smaller angle, in circular measure?, π , , , 180, (b) 1 −, (a) , − 1, 180, , , π, π , 1, 1 180, , (c), (d), 1−, − 1, 2 , 180, 2 π, , 68. A positive acute angle is divided into two parts whose, 1, 7, tangents are and . What is the value of this angle?, 8, 9, π, π, π, π, (b), (c), (d), (a), 3, 4, 6, 12, , 71. Given that p = tan α + tan β and q = cot α + cot β ;, 1 1, then what is the value of − ?, p q, (a) cot(α − β ), (c) tan(α + β ), , (b) tan(α − β ), (d) cot(α + β ), , 72. What is the value of, cosec ( π + θ ) cot {( 9π/ 2) − θ } cosec2 ( 2π − θ ), , ?, cot ( 2π − θ ) sec2 ( π − θ ) sec {( 3π/ 2) + θ }, (a) 0, (b) 1, (NDA 2007 I), (c) –1, (d) ∞, , 73. What is the value of, ( sec θ − cos θ )( cosec θ − sin θ )(cot θ + tan θ )?, (NDA 2007 I), , (a) 1, (c) sin θ, , (b) 2, (d) cos θ, , π, and β + γ = α , then which one of the, 2, following is correct?, (NDA 2007 I), (a) 2 tan β + tan γ = tan α (b) tan β + 2 tan γ = tan α, (c) tan β + tan γ = tan α (d) 2 (tan β + tan γ ) = tan α, , 74. If α + β =, , 75. If sin θ + cosec θ = 1, then the general value of θ is, π π, (a) 2n π, (b) n π = ( −1) −, 4 4, π, (c) 2n π +, (d) None of these, 2, 76. If sin θ = 3 cos θ , − π < θ < 0, then θ is equal to, (a), , −5π, 6, , (b), , −4π, 6, , (c), , 4π, 6, , 5π, 6, , (d), , 1, , then the most general value of θ is, 4, π, nπ, π, (a) 2nπ ± ( −1)n, (b), ± ( −1)n, 6, 2, 6, π, π, (d) 2nπ ±, (c) nπ ±, 6, 6, , 77. If sin2 θ =, , 78. The equation sin x cos x = 2 has, (a) one solution, (b) two solutions, (c) infinite solutions, (d) no solution, , 69. If an angle B is complement of an angle A, what are, the greatest and the least values of cos A cos B,, respectively?, 1, 1, (a) 0, −, (b) , − 1, 2, 2, 1, 1, (c) 1, 0, (d) , −, 2, 2, , 79. If tan 2θ tan θ = 1, then the general value of θ is, 1 π, 1, , , (a) n + , (b) n + π, , , 2 3, 2, , 70. In a ∆ ABC, if cos A = cos B cos C , what is the value of, tan A − tan B − tan C?, (a) 0, (b) −1, (c) 1 + tan A tan B tan C, (d) tan A tan B tan C − 1, , 80. If cosθ = −, , 1 π, , (c) 2n ± , , 2 3, , θ is, (a) 2nπ +, (c) nπ +, , π, 4, , 1, 2, π, 4, , (d) None of these, , and tanθ = 1, then the general value of, , (b) ( 2n + 1) π +, (d) nπ ±, , π, 4, , π, 4
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225, , Trigonometric Ratios and Equations, 81. If tan θ + sin θ = m and tan θ − sin θ = n , then, (b) m 2 + n 2 = 16mn, (a) m 2 − n 2 = 16mn, 2, 2 2, (c) ( m − n ) = 16mn, (d) ( m 2 + n 2 )2 = 16mn, , 82. If n is an integer which one of the following is correct?, (a) tan ( nπ + α ) = − tan α, (b) tan ( nπ + α ) = tan α, (c) tan ( nπ + α ) = ± tan α, (d) tan( nπ + α ) = ± cot α, , Level II, 1. The value of sin 20° sin 40° sin 60° sin 80° is equal to, (a) –3/16, (b) 5/16, (c) 3/16, (d) –5/16, , , , 3π, 2. The value of 3 sin4 , − α + sin4( 3π + α ), , , 2, , , , 6 π, , −2 sin + α + sin6( 5π − α ), , , 2, , , is equal to, (a) 0, (b) 1, (c) 3, (d) sin 4α + sin 6 α, cos( A + C ), 3. If cos 2B =, , then tan A, tan B, tanC are in, cos( A − C ), (a) AP, (c) HP, , 10. If tan A − tan B = x and cot B − cot A = y , then what, is the value of cot ( A − B)?, (NDA 2011 II), 1 1, 1 1, 1 1, 1 1, (a), (b), (c), (d) − −, −, −, +, y x, x y, x y, x y, 11. If x = sin θ + cos θ and y = sin θ ⋅ cos θ , then what is, the value of x 4 − 4x 2 y − 2x 2 + 4 y 2 + 4 y + 1 ?, (NDA 2011 I), , (a) 0, (c) 2, , (b) 1, (d) None of these, , 12. If an angle α is divided into two parts A and B such that, A − B = x and tan A : tan B = 2 : 1, then what is the, value of sin x ?, (NDA 2011 I), ( 2 sin α ), (a) 3 sin α, (b), 3, (sin α ), (c), (d) 2 sin α, 3, , (b) GP, (d) None of these, n −1, A+ B, is equal to, 4. If sin A = n sin B, then, tan, n +1, 2, A− B, A− B, (a) sin, (b) tan, 2, 2, A− B, (c) cot, (d) None of these, 2, , 13. If x lies in II quadrant, then, , 5. What is the value of tan 15°⋅ tan 195°? (NDA 2011 II), (b) 7 + 4 3, (a) 7 − 4 3, , 14. If tan x =, , (c) 7 + 2 3, , (d) 7 + 6 3, , equal to, x, (a) sin, 2, , (a), , 6. If tan θ = m, where m is a non-square natural, number m, then sec 2θ is, (NDA 2011 II), (a) a negative number, (b) a transcendental number, (c) an irrational number, (d) a rational number, 7. What is the value of, tan 9° − tan 27° − tan 63° + tan 81°?, (a) 1, (b) 2, (c) 3, (d) 4, , nd, , (NDA 2011 II), , 4π , 2π , 8. If x = y cos = z cos , then what is the value, 3, 3, of xy + yz + zx?, (NDA 2011 II), (a) −1, (b) 0, (c) 1, (d) 2, 9. The equation tan4 x − 2 sec2x + a 2 = 0 will have, atleast one real solution, if, (NDA 2011 II), (a) | a|≤ 4, (b) | a|≤ 2, (d) None of these, (c) | a|≤ 3, , (c), , (b) tan, , b, , then, a, 2 sin x, , sin 2x, 2 cos x, sin 2x, , x, 2, , 1 + sin x + 1 − sin x, 1 + sin x − 1 − sin x, , (c) sec, , x, 2, , (d) cosec, , is, , x, 2, , a−b, is equal to, a+b, 2 cos x, (b), cos 2x, 2 sin x, (d), cos 2x, , a+b, +, a−b, , 15. The value of sin 36° sin 72° sin 108° sin 144° is equal, to, (a) 1/4, (b) 1/16, (c) 3/4, (d) 5/16, 4, 5, 16. If cos(α + β ) = , sin (α − β ) =, and α , β lies between, 5, 13, π, 0 and, , then tan 2α is equal to, 4, 16, 56, (b), (a), 63, 33, 28, (c), (d) None of these, 33, 17. The expression, cos2( A − B) + cos2 B − 2 cos( A − B) cos A cos B is, (a), (b), (c), (d), , dependent on B, dependent on A and B, dependent on A, independent of A and B
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226, , NDA/NA Mathematics, , 18. If A = sin2 θ + cos4 θ, then for all real values of θ, 3, (a) 1 ≤ A ≤ 2, (b) ≤ A ≤ 1, 4, 13, 3, 13, (d) ≤ A ≤, (c), ≤ A≤ 1, 16, 4, 16, 19. If cos A + cos B = m and sin A + sin B = n , where m,, n ≠ 0, then what is the value of sin ( A + B)?, (NDA 2010 II), , mn, , (a), , m +n, m2 + n 2, (c), 2mn, 2, , 2, , 2mn, , (b), , m +n, mn, (d), m+n, 2, , 26. If, 2, , 1, 2+ 3, 3, 1, (c), 2+ 2, 2, , 1, (NDA 2010 II), °?, 2, 1, (b) −, 2− 3, 3, 1, (d) −, 2+ 2, 2, , 21. If angles A, B, C are in AP, then what is the value of, (NDA 2010 I), sin A + 2 sin B + sin C?, 2 A − C, (a) 4 sin B cos , , 2 , A − C, (b) 4 sin B cos2 , , 4 , A − C, (c) 4 sin ( 2B) cos2 , , 2 , A − C, (d) 4 sin ( 2B) cos2 , , 4 , 1 °, 22. What is the value of tan 7 ?, 2, (a), (c), , 6+ 3− 2+2, 6− 3+ 2−2, , 23. What is the value of, 1, 4, 1, (c), 3, , (a), , (b), (d), , (NDA 2010 I), , 6+ 3+ 2+2, 6+ 3+ 2−2, , cos 15° + cos 45°, cos3 15° + cos3 45°, 1, (b), 2, , sin( x + y ) a + b, tan x, is, =, ; then the value of, sin ( x − y ) a − b, tan y, , (a) a/ b, (c) ab, , 20. What is the value of sin 292, (a), , 25. Consider the following statements, I. If sin A = sin B, then we have sin 2 A = sin 2B, always., π, 4π, 5π 1, is ., II. The value of cos cos, cos, 7, 7, 7, 4, Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, , ? (NDA 2010 I), , (d) None of these, , 24. Consider the following statements, I. If A = 30° and area of the ∆ ABC is, , 3 2, a , the, 4, , triangle is right angled., 1 − cos B, , then tan 2A = tan B., II. If tan A =, sin B, tan 3x, 1, never lies between and 3., III., tan x, 3, Which of the statements given above are correct?, (a) I and II, (b) II and III, (c) III and I, (d) All I, II and III, , (b) b/ a, (d) ( a − b)/( a + b), , 1, π, 27. If for A, B ∈ 0, ; sin( A + B) = 1 and sin( A − B) = ,, 2, 2, then the value of tan( A + 2B) tan( 2 A + B) is equal to, (a) –1, (b) 0, (c) 1, (d) 2, 28. If, , 2 + 2 + 2 + 2 +K ∞ = cosec θ, then the value, , of sin θ is equal to, 1, (a) 1, (b), 4, , (c), , 1, 2, , (d), , 1, 2, , 29. Which one of the following statements is correct?, (a) The squares of the tangents of the angles 30°,, 45°, 60° are in GP., (b) The squares of the sines of the angles 30°, 45°,, 60° are in GP., (c) The squares of the secants of the angles 30°,, 45°, 60° are in AP., (d) The squares of the tangents of the angles 30°,, 45°, 60° are in AP., 30. If x + y = z , then 1 + cos x + cos y + cos z is equal to, x, y, z, x, y, z, (b) 4 cos cos cos, (a) 4 cos cos sin, 2, 2, 2, 2, 2, 2, x, y, z, x, y, z, (c) 4 cos sin cos, (d) 4 sin cos cos, 2, 2, 2, 2, 2, 2, sin 3 θ cos 3 θ, 31. If 0° < θ < 90° and x =, and, −, sin θ, cos θ, cos 3 θ sin 3 θ, , then which one of the following, y=, +, cos θ, sin θ, is correct?, (a) x = y implies θ = 15°, (b) x = − y implies θ = 60°, (c) 2x = y implies θ = 30°, (d), , 3 x = y implies θ = ( 221 / 2)°, , 32. Consider the following statements, I. If θ = 1200° , then ( sec θ + tan θ )−1 is positive., II. If θ = 1200° , then ( cosec θ − cot θ ) is negative., Which of the statements given above is/are correct?, (NDA 2009 II), , (a) Only I, (c) Both I and II, , (b) Only II, (d) Neither I nor II
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227, , Trigonometric Ratios and Equations, 33. Match List I with List II and select the correct, answer using the code given below the lists, List I, A., B., C., , tan 15°, tan 75°, tan 105°, , List II, −2 − 3, 2+ 3, −2 + 3, 2− 3, , 1., 2., 3., 4., , (NDA 2009 II), , Codes, A B, (a) 4 1, (c) 3 2, , C, 2, 1, , A, (b) 4, (d) 2, , B, 2, 1, , C, 1, 4, , 34. For which acute angle θ, cosec2 θ = 3 3 cot θ − 5 ?, (NDA 2009 II), , 5π, (a), 12, , π, (b), 3, , π, (c), 6, , (d), , π, 4, , 35. The equation tan2 φ + tan6 φ = tan3 φ ⋅ sec2 φ is, (a) identity for only one value of φ, (NDA 2008 II), (b) not an identity, (c) identity for all values of φ, (d) None of the above, 36. If sec A + tan A = p,then what is the value of sin A ?, (NDA 2008 II), , (a), , p2 − 1, , (b), , p2 + 1, (c) 1, , p2 + 1, , p2 − 1, (d) None of these, , 37. If cot A ⋅ cot B = 2, then what is the value of, (NDA 2012 I), cos ( A + B) ⋅ sec ( A − b)?, (a) 1/3, (b) 2/3, (c) 1, (d) –1, 38. Consider the following, 1, = sec θ cosec θ, x, 1, 1, II. If x + = sin θ , then x 2 + 2 = sin2 θ − 2, x, x, III. If, and, x = p sec θ, y=q, tan θ ,, x 2q 2 − y 2 p2 = p2q 2, I. If cot θ = x , then x +, , (NDA 2008 I), , Assertion (A) The number of elements in X is 2., Reason (R) sin θ and cos θ are both negative both, in second are fourth quadrants., (a) Both A and R are correct and R is the correct, explanation of A., (b) Both A and R are correct but R is not the correct, explanation of A., (c) A is correct but R is wrong., (d) A is wrong but R is correct., 45. What is the correct sequence of the following values?, π, π, I. sin , II. cos , 12, 12, π, III. cot , 12, , (a) III > II > I, (c) I > III > II, then, , 39. If θ = 18°, then what is the value of 4 sin θ + 2 sin θ?, (NDA 2012 I), , (c) 0, , 44. Let X = { θ ∈ [0, 2π ] : sin θ = cos θ }, , (NDA 2008 I), , 2, , (b) 1, , 43. If A ∈ ( − π , π ). For what value of A, is the identity, 2 sin A, 1 − cos A + sin A, true?, =, 1 + cos A + sin A, 1 + sin A, (a) All values of A, π, (b) All values of A, except A = −, 2, (c) All values of A, except A = 0, π, (d) All values of A, except A =, 2, , Select the correct answer using the code given below, , IV. The maximum value of cos θ − 3 sin θ is 3., Which of these are correct?, (a) I and II, (b) II and III, (c) III and IV, (d) I, II and III, , (a) −1, , π, (where n is an integer) and x + y, 4, π, is not an odd multiple of , then what is the value of, 2, sin 2x − sin 2 y, the following expression, ?, cos 2x + cos 2 y, 1, (d) 1, (a) –1, (b) 0, (c), 2, , 42. If x − y = ( 4n + 1), , (d) 2, , 46. Let ABCD be a square and let P be a point on AB such, that AP : PB = 1 : 2. If ∠ APD = θ, then what is the, value of cos θ ?, (NDA 2007 II), 1, 1, 2, 2, (a), (b), (c), (d), 10, 5, 10, 5, 47. Tn = sinn x + cosn x , n ≥ 4. Which one of the following, ratios is independent of n?, Tn − 2 + Tn, Tn − 2 − Tn, (b), (a), Tn − 4, Tn − 4, , 2, , 40. What is the maximum value of sin x? (NDA 2012 I), (a) −1, (b) 0, (c) 1, (d) Infinity, 41. If sin A = p and sin B = q , where| p| and|q| are both, less than 1, then what are the total number of, possible (and distinct) values of sin ( A + B)?, (a) 1, (b) 2, (c) 3, (d) 4, , (b) I > II > III, (d) III > I > II, , (c), , Tn − 4 − Tn − 2, Tn, , (d), , Tn − 4 + Tn − 2, Tn, , 48. If 1 + sin2 A = 3 sin A cos A, then what are the, possible values of tan A ?, (a) 1/4, 2, (b) 1/6, 3, (c) 1/2, 1, (d) 1/8, 4
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228, , NDA/NA Mathematics, 57. Assertion (A) If x = 1 + cot α − sec (α + π / 2), and, y = 1 + cot α + sec (α + π / 2),, then xy > 0, if either 0° < α < 90° or 180° < α < 270°., Reason (R) xy = 2 tanα, , 49. Consider the following, 1 + sin θ, I., = ( sec θ + tan θ )2, 1 − sin θ, II. sec2θ + cosec2θ = tan θ + cot θ, Which of the above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, , 58. Assertion (A) cos 1 < sin 1, Reason (R) In the first quadrant cosine decrease but, sin increases., , 50. Consider the following statements, (NDA 2012 I), I. 1° in radian measure is less than 0.02 radians., II. I radian in degree measure is greater than 45°., Which of the above statements is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 51. Three expressions are given below, Q1 = sin ( A + B) + sin ( B + C ) + sin (C + A), Q2 = cos ( A − B) + cos ( B − C ) + cos (C − A), Q3 = sin A (cos B + cos C ) + sin B (cos C + cos A), + sin C (cos A + cos B), Which one of the following is correct?, (a) Q1 = Q2, (b) Q2 = Q3, (c) Q1 = Q3, (d) All the expressions are different, 52. Let A and B be obtuse angles such that sin A = 4 / 5, and cos B = − 12 / 13. What is the value of sin( A + B)?, (a) –63/65 (b) –33/65 (c) 33/65, (d) 63/65, 53. The angles A, B, C of a triangle are in the ratio 2 : 5 : 5., What is the value of tan B tan C ?, (a) 4 + 3, (b) 4 + 2 3 (c) 7 + 4 3 (d) 3 + 3 3, 54. If sin θ + cos θ = 0, then what is the value of θ?, (a) −π / 4, (b) 0, (c) π / 4, (d) π / 3, 3, , Directions (Q. Nos. 55-59) Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., maximum, , Reason (R), The maximum, sin x + b cos x + c is c + a 2 + b2 ., , value, , value, , of, , of, a, , sin 5 A − sin 3 A, = tan A, cos 3 A + cos 5 A, Reason (R) 2 cos A sin B = sin ( A + B) − sin ( A − B), and 2 cos A cos B = cos ( A + B) + cos ( A − B)., , 56. Assertion (A), , 61. If 2 cos2 x + 3 sin x − 3 = 0, 0 ≤ x ≤180° the value of x is, (a) 30° , 90° , 150°, (b) 60° , 120° , 180°, (c) 0° , 30° , 150°, (d) 45° , 90° , 135°, 62. If 2 sec θ + tan θ = 1, then the general value of θ is, 3π, π, (b) 2nπ +, (a) nπ +, 4, 4, π, π, (c) 2nπ −, (d) 2nπ ±, 4, 4, 63. The number of solutions of the given equation, a sin x + b cos x = c, where| c|> a 2 + b2 is, (a) 1, (b) 2, (c) infinite, (d) None of these, , 3, , 55. Assertion (A), The, 3 sin x + 4 cos x + 7 is 12., , 59. Assertion (A) If y = 1 − sin x , then, x, x, π , for x ∈ , π , y = sin − cos, 2 , 2, 2, Reason (R) Above function is true for every value of x., 1, 60. If cosec θ − cot θ =, , where θ ≠ 0, then what is the, 3, value of cosθ?, (NDA 2012 I), 3, (a) 0, (b), 2, 1, 1, (d), (c), 2, 2, , 64. The number of solutions of the given equation, tanθ + sec θ = 3, where 0 < θ < 2π is, (a) 0, (b) 1, (c) 2, (d) 3, 1 − cos 2θ, 65. If, = 3, then the general value of θ is, 1 + cos 2θ, π, π, (a) 2nπ ±, (b) nπ ±, 6, 6, π, π, (c) 2nπ ±, (d) nπ ±, 3, 3, 66. If tan2 θ − (1 + 3 ) tan θ + 3 = 0, then the general, value of θ is, π, π, π, π, (a) nπ + , nπ +, (b) nπ − , nπ +, 4, 3, 4, 3, π, π, π, π, (d) nπ − , nπ −, (c) nπ + , nπ −, 4, 3, 4, 3, 67. The equation 3 sin2 x + 10 cos x − 6 = 0 is satisfied, if, (b) x = 2nπ ± cos−1(1/ 3), (a) x = nπ ± cos−1(1/ 3), −1, (c) x = nπ ± cos (1/ 6), (d) x = 2nπ ± cos−1(1/ 6)
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229, , Trigonometric Ratios and Equations, 68. If sin θ = sin α, then, θ+α, π, θ−α, (a) 2 is any odd multiple of 2 and 2 is any, multiple of π., θ+α, π, θ−α, (b) 2 is any even multiple of 2 and 2 is any, odd multiple of π, θ+α, π, θ−α, is any multiple of, and, is any odd, (c), 2, 2, 2, multiple of π, θ+α, π, θ−α, (d) 2 is any multiple of 2 and 2 is any even, multiple of π, 69. For what values of a does the equation, cos 2x + a sin x = 2a − 7, possess a real solution?, (a) a < 2, (b) 2 ≤ a ≤ 6, (c) a > 6, (d) a is any integer < − 2, , Directions (Q. Nos. 70-72) Let the angles of a ∆, ABC be in the ratio 2 : 3 : 5, , (a), (b), , 10 + 2 5 + 2 10 + 22 2, , Directions (Q. Nos. 73-75), , Let us consider, A + B + C = 180° Solve the following questions on the, basis of above information, , 73. The value of Σ tan, (a) 0, (c) 2, , A, B, tan is, 2, 2, (b)1, (d) 3, , 74. Find the value of sin 2 A + sin 2B + sin 2 C is, (a) 4 sin A sin B sin C, (b) 4 cos A cos B cos C, (c) 2 cos A cos B cos C, (d) None of these, 75. If A, B and C are acute angle and cot A cot B cos C = k,, then, 1, 1, (a) k ≤, (b) k ≥, 3 3, 3 3, 1, 1, (d) k >, (c) k <, 9, 3, , Directions (Q. Nos. 76-78) Let, , 70. Find the least and the greatest angle of a triangle, (a) 54 ° , 36°, (b) 36°, 90°, (c) 36°, 54°, (d) None of these, 71. The value of cos2 A + sin 3C + cos, , 72. The greatest side of a ∆ ABC is, (a) AB, (b) BC, (c) CA, (d) None of these, , sin ( A + B) = 1 and, , 1, π, sin ( A − B) = , where A, B ∈ 0, ., 2, 2, , C, × (sin 2B)2, 2, , 76. What is the value of A?, π, (a), 6, π, (c), 4, , 16 2, 10 − 2 5 + 2 10 + 22 2, , (NDA 2012 I), , π, 3, π, (d), 8, (b), , 77. What is the value of tan( A + 2B) tan( 2 A + B)?, (a) −1, (b) 0, (c) 1, (d) 2, , 16 2, , 10 + 2 5 + 2 10, (c), 16, (d) None of the above, , 78. What is the value of sin2 A − sin2 B ?, (a) 0, (b) 1/2, (c) 1, , (d) 2, , Answers, Level I, 1., 11., 21., 31., 41., 51., 61., 71., 81., , (c), (b), (d), (a), (d), (a), (b), (d), (c), , 2., 12., 22., 32., 42., 52., 62., 72., 82., , (c), (b), (c), (d), (a), (a), (a), (b), (b), , 3., 13., 23., 33., 43., 53., 63., 73., , (c), (b), (a), (c), (c), (c), (d), (a), , 4., 14., 24., 34., 44., 54., 64., 74., , (c), (c), (b), (b), (a), (b), (a), (b), , 5., 15., 25., 35., 45., 55., 65., 75., , (b), (c), (b), (b), (d), (d), (c), (b), , 6., 16., 26., 36., 46., 56., 66., 76., , (c), (a), (c), (a), (d), (a), (c), (b), , 7., 17., 27., 37., 47., 57., 67., 77., , (c), (a), (c), (d), (d), (c), (c), (c), , 8., 18., 28., 38., 48., 58., 68., 78., , (a), (c), (b), (d), (d), (c), (b), (d), , 9., 19., 29., 39., 49., 59., 69., 79., , (d), (c), (b), (a), (b), (a), (d), (a), , 10., 20., 30., 40., 50., 60., 70., 80., , (c), (a), (d), (b), (b), (a), (a), (b), , 2., 12., 22., 32., 42., 52., 62., 72., , (b), (c), (c), (d), (d), (a), (c), (a), , 3., 13., 23., 33., 43., 53., 63., 73., , (b), (b), (d), (b), (b), (c), (d), (b), , 4., 14., 24., 34., 44., 54., 64., 74., , (b), (b), (b), (c), (c), (a), (c), (a), , 5., 15., 25., 35., 45., 55., 65., 75., , (a), (d), (d), (d), (a), (a), (d), (a), , 6., 16., 26., 36., 46., 56., 66., 76., , (d), (b), (a), (a), (a), (b), (a), (b), , 7., 17., 27., 37., 47., 57., 67., 77., , (d), (c), (c), (a), (b), (c), (b), (c), , 8., 18., 28., 38., 48., 58., 68., 78., , (b), (b), (d), (d), (c), (b), (a), (b), , 9., 19., 29., 39., 49., 59., 69., , (c), (b), (a), (b), (c), (c), (b), , 10., 20., 30., 40., 50., 60., 70., , (c), (d), (b), (c), (a), (b), (b), , Level II, 1., 11., 21., 31., 41., 51., 61., 71., , (c), (a), (b), (b), (d), (c), (c), (a)
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Hints & Solutions, Level I, 1. Given that, radius r = 3 m and arc d = 1m, arc, 1, We know that, angle =, = rad, radius 3, 2. Given that diameter of circular wire = 10 cm, ∴Length of wire = 10π, arc, 10π π, Hence, required angle =, =, = rad, radius, 50, 5, 3. Maximum value of 3 cos θ + 4 sin θ = 32 + 42 = 52 = 5, 4. Given that, sin θ + cos θ = m, …(i), and, …(ii), sec θ + cosec θ = n, Now, n (m + 1) (m − 1) = n (m2 − 1), [from Eqs. (i) and (ii)], = (sec θ + cosec θ ) 2 sin θ cos θ, (Q m2 = 1 + 2 sin θ cos θ ), sin θ + cos θ, [from Eq. (i)], =, ⋅ 2 sin θ cos θ = 2m, sin θ cos θ, 5. We have,, sin 420°⋅ cos 390° + cos (− 300° ) ⋅ sin(− 330° ), = sin (360° + 60° ) ⋅ cos (360° + 30° ), + cos 300° [− sin 330°], = sin 60°⋅ cos 30° − cos (360° − 60° ) ⋅ sin(360° − 30° ), [Q cos (− θ ) = cos θ], = sin 60°⋅ cos 30° − cos 60°⋅( − sin 30° ), = sin 60°⋅ cos 30° + cos 60°⋅ sin 30°, = sin (60° + 30° ) = sin 90° = 1, 6. Now, cot (45° + θ ) cot (45° − θ ), = tan (90° − 45° − θ ) cot (45° − θ ), = tan (45° − θ ) cot (45° − θ ) = 1, 7. Now, sin A sin (60° − A ) sin (60° + A ), = sin A (sin 2 60° − sin 2 A ), , 3, = sin A − sin 2 A, , 4, 3 sin A − 4 sin3 A sin 3 A, =, 4, 4, 1 + cos 30°, 1 + cos 30°, 8. Now, cos 15° =, =+, 2, 2, (taking positive sign because cos 15° > 0), 9. f ( A ) = sin A ⋅ cos A ⋅ tan A + cos A ⋅ sin A ⋅ cot A, , ⇒, 0 ≤ f (x) ≤ 10, ∴ The maximum value of f (x) is 10., sin x, 1 + cos x, 11., +, 1 + cos x, sin x, x, x, x, 2 sin cos, 1 + 2 cos 2 − 1, 2, 2, 2, =, +, x, x, x, 1 + 2 cos 2 − 1, 2 sin ⋅ cos, 2, 2, 2, x, x, x, x, sin, cos, sin 2 + cos 2, 2, 2, 2, 2, 2, =, +, =, ×, x, x, x, x, 2, cos, sin, sin ⋅ cos, 2, 2, 2, 2, 1, =2⋅, = 2 cosec x, sin x, 12. Given,, sin 3 A = 1, ⇒, 3 sin A − 4 sin3 A = 1, ⇒ 4 sin3 A − 3 sin A + 1 = 0 ⇒ sin A = − 1, 1 /2, 1 /2, This is a cubic equation and have three roots in which, two roots are same and one different., So, sin A have 2 distinct values., 13. (1 + tan θ )(1 + tan φ ) = 2, ⇒, 1 + tan θ + tan φ + tan θ tan φ = 2, ⇒, tan θ + tan φ = 1 − tan θ tan φ, tan θ + tan φ, ⇒, =1, 1 − tan θ tan φ, ⇒, tan (θ + φ ) = tan 45°, ⇒, θ + φ = 45°, 14. The conditions of trigonometric function in different, quadrant as follows, y, , =, , x′, , cos A, sin A, = sin A ⋅ cos A ⋅, + cos A ⋅ sin A ⋅, sin A, cos A, 2, 2, 2, = sin A + cos A = 1, (Q sin θ + cos2 θ = 1), `, , − a 2 + b2 ≤ a cos x + b sin x ≤ a 2 + b2, ⇒, ⇒, ⇒, ⇒, , − 3 + 4 ≤ 3 cos x + 4 sin x ≤ 3 + 4, 2, , 2, , 2, , − 25 ≤ 3 cos x + 4 sin x ≤ 25, −5 ≤ 3 cos x + 4 sin x ≤ 5, −5 + 5 ≤ (3 cos x + 4 sin x + 5) ≤ 5 + 5, , x, , O, IVth quadrant, (cosθ, secθ are, positive and rest, trigonometric, functions are, negative), , IIIrd quadrant, (tan θ, cot θ are, positive and rest, trigonometric, functions are, negative), , 10. Let f (x) = (3 cos x + 4 sin x) + 5;, We know that,, 2, , Ist quadrant, (All trigonometric, functions are, positive ), , IInd quadrant, (sinθ, cosecθ are, positive and rest, trigonometric, functions are, negative), , y′, , ∴ tan θ lies in third quadrant., 15. sin (1920° ), = sin(360° × 5° + 120° ), = sin 120°, , [Q sin (360° + θ ) = sin θ]
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231, , Trigonometric Ratios and Equations, [Q sin (90° + θ ) = cos θ], = sin(90° + 30° ), 3, = cos 30° =, 2, sin θ, cos θ, 16. Given that,, +, cosec θ sec θ, sin θ, cos θ, =, +, ⇒ sin 2 θ + cos 2 θ = 1, (1 /sin θ ) (1 / cos θ ), 17. Let f (θ ) = sin 3θ ⋅ cos 2θ + cos 3θ ⋅ sin 2θ, = sin (3θ + 2θ ), = sin 5θ, [Q sin( A + B) = sin A ⋅ cos B + cos A ⋅ sin B], We know that,, − 1 ≤ sin 5θ ≤ 1 ⇒ − 1 ≤ f (θ ) ≤ 1, , 23. Q tan A =, ∴, , 19., , 3 cosec 20° − sec 20°, 3, 1, −, sin 20° cos 20°, 3 cos 20° − sin 20°, =, sin 20° cos 20°, , 3, 1, 2, cos 20° − sin 20°, 2, 2, , = , 2, sin 20° cos 20°, 2, 2 [cos 60° cos 20° − sin 60° sin 20° ], =, 1, sin 40°, 2, 4 cos (20° + 30° ) 4 cos 50°, =, =, sin 40°, sin 40°, 4 cos (90° − 40° ) 4 sin 40°, =, =, =4, sin 40°, sin 40°, =, , 20. Q (tan α − cot α )2 ≥ 0, ⇒ tan 2 α + cot2 α − 2 ≥ 0 ⇒ tan 2 α + cot2 α ≥ 2, 1, 99° × π 11π, 21. Let ∠A = rad, ∠B = 99° =, =, 2, 180°, 20, We know that, ∠A + ∠B + ∠C = π, 1 11π, ⇒, +, + ∠C = π, 2, 20, 11π 1 9π − 10, ∠C = π −, − =, ⇒, 20, 2, 20, π, 22. We know that, sec2 θ + cos 2 θ ≥ 2, ∀ 0 < θ <, 2, Q, AM ≥ GM, 1/ 2, 2, 2, 1 , 1 , sec, ≥, ⋅, 2, θ, , , , sec θ +, , , sec2 θ , sec2 θ , ⇒, ∴, , (sec2 θ + cos 2 θ ) ≥ 2, y≥2, , and tan B = −, , cot ( A − B) =, , 12, 5, , 1, 1 + tan A tan B, =, tan ( A − B), tan A − tan B, , Which shows that cot ( A − B) has only one value of A, and B., 1, 1, 3π, 3π, 3π, −, +1, sin, + cos, − tan, 2, 2, 4, 4, 4, 24., =1, =, 3π, 3π, 3π − 2 + 2 + 1, sec, + cosec, − cot, 4, 4, 4, 25. We know that, 1° < 1 rad, ⇒, sin 1° < sin 1, cos 15° sin 15°, cos 45°, 26., ×, cos 45° sin 45°, sin 45°, , cos 15°, sin 15°, , = (sin 45° cos 15° − cos 45° sin 15° ), × (cos 45° sin 15° − sin 45° cos 15° ), , So, the maximum value of f (θ ) is 1., cos 12° − sin 12° sin 147°, 18., +, cos 12° + sin 12° cos 147°, 1 − tan 12°, =, + tan 147°, 1 + tan 12°, = tan (45° − 12° ) + tan (180° − 33° ), = tan 33° − tan 33° = 0, , 3, 4, , = − sin (45° − 15° ) × sin (45° − 15° ), = − sin 30° × sin 30°, 1 1, 1, =− × =−, 2 2, 4, 27. Q 4 sin 2 x + 4 cos x − 1 = 0, ⇒, , 4 − 4 cos 2 x + 4 cos x − 1 = 0, , ⇒, , −4 cos 2 x + 4 cos x + 3 = 0, , ⇒, , 4 cos 2 x − 4 cos x − 3 = 0, , ⇒ 4 cos 2 x − 6 cos x + 2 cos x − 3 = 0, ⇒, ⇒, , (2 cos x − 3)(2 cos x + 1) = 0, 3, cos x =, 2, (which is not possible), 1, cos x = −, 2, , and, , 1, = cos 210° (Q A lies in IIIrd quadrant), 2, ⇒, A = 210°, 1, 1, 28. Given that, cos θ = x + , , x, 2, 1, …(i), x + = 2 cos θ, ⇒, x, ⇒, , cos A = −, , 2, , We know that, x2 +, , 1 , 1, = x + − 2, x, x2 , = (2 cos θ )2 − 2 = 4 cos 2 θ − 2, , [from Eq. (i)], = 2 cos 2 θ, 1 2 1 1, ∴, x + 2 = × 2 cos 2 θ = cos 2 θ, 2, x 2, sin (B + A ) + cos (B − A ), 29., sin (B − A ) + cos (B + A ), sin (B + A ) + sin (90° − B − A ), sin (B − A ) + sin (90° − A + B), 2 sin ( A + 45° ) cos (45° − B), =, 2 sin (45° − A ) cos(45° − B), =
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232, , NDA/NA Mathematics, 1, sin A +, sin ( A + 45° ), =, = 2, 1, sin (45° − A ), cos A −, 2, cos A + sin A, =, cos A − sin A, , 1, cos A, 2, 1, sin A, 2, , 1, 1, =, π, π, , tan 10π + tan, , 4, 4, =1, , =, , 4, 12, and cos B = −, 5, 13, cos ( A + B) = cos A cos B − sin A sin B, 16 12 4, 144, = 1−, 1−, − −, , , 25, 13, 5, 169, 3 12 4 5 , 36 20, =− ×, − − = −, +, 5 13 5 13, 65 65, 16, =−, 65, , 30. We have, sin A =, Now,, , 36. Given,, cot θ = 2 cos θ, ⇒, cos θ (1 − 2 sin θ ) = 0, π, For, < θ < π , cos θ ≠ 0, 2, ∴, 1 − 2 sin θ = 0, 1, sin θ =, ⇒, 2, 5π, θ=, ⇒, 6, 5, 37. Given, cot θ =, (Q θ lies in IIIrd quadrant), 12, C, , 31. Let f (x) = 3 cos x + sin x, 3, , π, 1, , f (x) = 2 , cos x + sin x = 2 sin x + , ⇒, , 3, 2, 2, , , , 13, , π, , Since, −1 ≤ sin x + ≤ 1, , 3, , A, , π π, Hence, f (x) is maximum, if x + =, 3 2, π, x = = 30°, ⇒, 6, , …(i), …(ii), , = 1 + cot2 α + 2 cot α = |1 + cot α|, 3π, <α < π, 4, ⇒, cot α < − 1 ⇒ 1 + cot α < 0, Hence, |1 + cot α| = − (1 + cot α ), But, , 34. Given, (sin x + cosec x)2 + (cos x + sec x)2, = k + tan 2 x + cot2 x, 2, ⇒, sin x + cosec2 x + 2 + cos 2 x + sec2 x + 2, = k + tan 2 x + cot2 x, 2, ⇒ 1 + (cosec x − cot2 x) + (sec2 x − tan 2 x) + 4 = k, Q cosec2 θ = 1 + cot2 θ , , , , sec2 θ = 1 + tan 2 θ, , 35., , 1+1+1+4=k ⇒ k=7, , 1 − 3 tan 2 A, 1, =, 3 tan A − tan3 A tan 3 A, 1, =, 41π, tan, 4, , B, , sin θ = −, , −24 − 15, 13, −39, =, = −3, 13, π, π, 5π , 7π , 38. cos + cos + cos + cos , 9, 3, 9, 9, =, , 33. We have, cosec2α + 2 cot α, , ⇒, , 5, , 12, 5, , cos θ = −, 13, 13, 12, 5, ∴ 2 sin θ + 3 cos θ = 2 − + 3 − , 13, 13, ⇒, , 32. Given that, ABCD is a cyclic quadrilateral., So,, A + C = 180° ⇒ A = 180° − C, ⇒, cos A = cos(180° − C ) = − cos C, ⇒, cos A + cos C = 0, Similarly,, cos B + cos D = 0, On adding Eqs. (i) and (ii), we get, cos A + cos B + cos C + cos D = 0, , θ, , 12, , 41π , , Q A =, , , 12 , , = cos (20° ) + cos (60° ) + {cos (100° ) + cos (140° )}, 1, = cos 20° + + 2 cos 120° cos 20°, 2, 1, = cos 20° + − 2 sin 30° cos 20°, 2, 1, 1, = cos 20° + − cos 20° =, 2, 2, θ, 39. Arc length of circle = 2πr ⋅, 360°, 15°, = 2π × 5 ×, (Q θ = 15° ), 360°, 5π, cm, =, 12, 40. Given, P = sin (989° ) ⋅ cos (991° ), = sin (1080° − 91° ) ⋅ cos (1080° − 89° ), = − sin 91°⋅ cos 89°, = − sin (90° + 1) ⋅ cos 89°, = − cos 1° ⋅ cos 89°, Here, cos 1° and cos 89° are positive., So, P is rational and negative.
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233, , Trigonometric Ratios and Equations, 41. 1 − sin 10° sin 50° sin 70°, 1, = 1 − (2 sin 70° sin 10° ) sin 50°, 2, 1, = 1 − [(cos 60° − cos 80° ) sin 50° ], 2, 1 1, , =1 −, sin 50° − cos 80° sin 50°, , 2 2, 1, = 1 − [sin 50° − 2 cos 80°⋅ sin 50° ], 4, 1, = 1 − [(sin 50° − sin 130° ) + sin 30° ], 4, 1, = 1 − [−2 cos 90° ⋅ sin 40° + sin 30° ], 4, 1, 1, = 1 − 0 + , 4, 2, 1 7, =1 − =, 8 8, 5, 99, and sin φ =, 42. Q sin θ =, 13, 101, ∴ cos { π − (θ + φ )}, , 5, 1, 1, (2 + 3) =, =, 50, 2, 50, π, π, sin ( A + B) = sin, ⇒ A+ B=, 4, 4, =, , ⇒, , …(i), tan A − tan B = x, …(ii), cot B − cot A = y, 1, Now, cot ( A − B) =, tan ( A − B), 1 + tan A tan B, =, tan A − tan B, 1, tan A tan B, =, +, tan A − tan B tan A − tan B, 1 1, [from Eqs. (i) and (ii)], = +, x y, 3π, 48. Given that, A + B + C =, 2, ∴ cos 2 A + cos 2B + cos 2C, = (cos 2 A + cos 2B) + cos 2C, = 2 cos ( A + B) cos ( A − B) + cos 2 C, = − 2 sin C cos ( A − B) + 1 − 2 sin 2 C, = 1 − 2 sin C [cos ( A − B) + sin C ], = 1 − 2 sin C [cos ( A − B) − cos ( A + B)], = 1 − 2 sin C2 sin A sin B, = 1 − 4 sin A sin B sin C, 47. Given that,, and, , (given), , = − cos (θ + φ ), = − {cos θ cos φ − sin θ sin φ }, 2, , 25, 5, 99 , , 99 , 1−, ×, =− 1−, −, , 101, 169, 13 101 , , , 5, 99 , 12 20, =− ×, −, ×, , 13 101 13 101 , 495 255, 240, =−, −, =, 1313 1313 1313, 43. We know that,, 1 rad = 57° 16′ 22′ ′, Thus, approximate value of 1 rad = 57°., 44. ∴ x = sin θ cos θ and y = sin θ + cos θ, ∴, y2 − 2x = (sin θ + cos θ )2 − 2 sin θ cos θ, = sin 2 θ + cos 2 θ + 2 sin θ cos θ − 2 sin θ cos θ, = sin 2 θ + cos 2 θ = 1, 45. sin 4 x − cos 4 x = p, (given), ⇒ (sin 2 x − cos 2 x)(sin 2 x + cos 2 x) = p, ⇒, sin 2 x − cos 2 x = p, ⇒, − cos 2x = p, ⇒, cos 2x = − p, [Q − 1 ≤ cos 2x ≤ 1 ⇒ |cos 2x| ≤ 1], ∴, | p| ≤ 1, 1, 1, 46. Given that sin A =, and sin B =, 10, 5, We know that,, sin ( A + B) = sin A cos B + cos A sin B, 1, 1, 1 1, =, 1− + 1−, 5, 10 5, 10, 1, 4, 9 1, =, +, 10 5, 10 5, , sin 2 3 A cos 2 3 A, −, sin 2 A, cos 2 A, 2, 2, sin 3 A , cos 3 A , =, −, , , , sin A , cos A , , 49. We have,, , 2, , 4 cos3 A − 3 cos A , 3 sin A − 4 sin3 A , =, , −, sin A, cos A, , , , , , 2, , = (3 − 4 sin 2 A )2 − (4 cos 2 A − 3)2, = 16 (sin 2 A − cos 2 A ) − 24 (sin 2 A − cos 2 A ), = (sin 2 A − cos 2 A )(16 − 24) = 8 cos 2 A, 50. We know that, 1c = 57.27°, sin 1°, sin 1°, =, <1, ∴, c, sin (57.27)°, sin 1, , [Q sin 1° < sin (57.27)° ], , 51. We have, sin 10° + sin 20° + sin 30° + K + sin 360°, = (sin 10° + sin 350° ) + (sin 20° + sin 340° ), + (sin 30° + sin 330° ) + ... + (sin 170° + sin 190° ), + sin 180° + sin 360°, = (sin 10° − sin 10° ) + (sin 20° − sin 20° ), + (sin 30° − sin 30° ) + K + 0 + 0 + 0, =0, 52. The given expression is,, sin θ + sin (θ + 120° ) + sin (θ + 240° ), 2π , 4π , , , , = sin θ + sin θ +, , + sin θ +, , , 3, 3 , , = sin θ + 2 sin (θ + π ) cos, = sin θ − sin θ = 0, , π, = sin θ + sin (π + θ ), 3
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234, 53., , NDA/NA Mathematics, (a) for 0° < B < 45°, obviously X + Y > 0, (b) for B = 45° , we see that X + Y = 0, (d) for 45° < B ≤ 90°, clearly X + Y < 0, ∴ Only option (c) is incorrect for any A and B, the, expression doesn’t given any rational value for X + Y ., , 2 + 2 + 2 + 2 cos 4 A, = 2 + 2 + 2 (1 + cos 4 A ) (Q cos 4 A + 1 = 2 cos 2 2 A ), = 2 + 2 + 4 cos 2 2 A = 2 + 2 + 2 cos 2 A, = 2 + 2(1 + cos 2 A ), , (Q cos 2 A + 1 = 2 cos 2 A ), , 58. tan θ + sec θ = 4, , = 2 + 4 cos 2 A = 2 + 2 cos A, = 2(1 + cos A ), A, A, = 4 cos 2, = 2 cos, 2, 2, , ⇒, , (Q cos A + 1 = 2 cos 2 A /2), , ⇒, , 54. We know that,, , 1′, 60′ ′ = 1′ ⇒ 30′ ′ =, 2, 1 ′ 71, , ∴, 35′ 30 = 35 + = ′, 2, , 2, and, 60′ = 1°, 1 °, 1′ = , ∴, 60, 71, ′ 71 1 ° 71 °, ⇒, × =, , =, 120, 2, 2, 60, 71 ° 13751 °, , 114° 35′ 30′ ′ = 114 +, ∴, , =, 120 , , 120, We know that, 2π rad = 360°, 13751, 13751 ° 2π, ⇒, ×, rad, , =, 120 , 360°, 120, 2 × 22 × 13751, =, rad, 7 × 360 × 120, , = 2.0008069 rad, ⇒, 114° 35′ 30′ ′ = 2 rad (approx), 1 ° , 1 °, , 55. 1 + cos 67 1 + cos 112 , , 2 , 2, 1 ° , 1 ° , , , = 1 + cos 67 1 + cos 180° − 112 , , , 2 , 2, , 1 ° , 1 °, , = 1 + cos 67 1 − cos 67 , , 2 , 2, 1, °, °, 1, = 1 − cos 2 67, = sin 2 67, 2, 2, −1 , 1 − cos 135°, 2+1, , =, =, Q cos 135° =, , , 2, 2, 2 2, Which is an irrational number and is less than 1., , Hence,, , x=, , π, 6, , sin θ + 1 + 2 sin θ = 16 cos 2 θ, , ⇒, , sin 2 θ + 2 sin θ + 1 = 16 − 16 sin 2 θ, , 2, , ⇒, , 17 sin 2 θ + 2 sin θ − 15 = 0, , ⇒, , (sin θ + 1)(17 sin θ − 15) = 0, 15, sin θ =, 17, , ⇒, , (Qsin θ ≠ − 1), , 180°, π, 59. tan = tan , = tan 15°, 12 , 12, = tan (45° − 30° ) =, , tan 45° − tan 30°, 1 + tan 45°⋅ tan 30°, , 1, 3 = 3 −1 × 3 −1, =, 1, 3+1, 3 −1, 1+, 3, 1−, , =, , ( 3 − 1 )2 3 + 1 − 2 3, =, =2 − 3, 3 −1, 2, , 60. Since, sin θ ≤ 1, ⇒, 2 sin θ ≤ 2, 1, Also,, x+ ≥2, x, ∴From Eqs. (i) and (ii), we get, 1, x+ =2, x, ⇒, x=1, b, 61. Given that, tan x =, a, a+b, a−b, Now,, +, a−b, a+b, , …(i), …(ii), , b, b, 1−, 1 + tan x, 1 − tan x, a, a, =, +, =, +, 1 − tan x, 1 + tan x, b, b, 1−, 1+, a, a, 1 + tan x + 1 − tan x, 2 cos x, =, =, 2, 1 − tan x, cos 2 x − sin 2 x, 1+, , 1 3 3, 3, +, −2 +, =2 3, 2, 2, 2, , 57. X + Y = sin ( A + B) sin ( A − B) + cos ( A + B) cos ( A − B), = cos {( A + B) − ( A − B)}, ⇒, X + Y = cos 2B, , sin θ + 1 = 4 cos θ, , ⇒, , 56. Q 4 sin x + 3 sin 2x − 2 sin 3x + sin 4x = 2 3, π, Let x =, 6, 2π, π, π, π, ∴, 4 sin + 3 sin − 2 sin + sin, 6, 3, 2, 3, = 4⋅, , sin θ, 1, +, =4, cos θ cos θ, sin θ + 1, =4, cos θ, , =, , 2 cos x, cos 2x, sin 2π ≠ sin (−2π ), , 62., Q, , sin (−2π ) = − sin 2π
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237, , Trigonometric Ratios and Equations, , 4. We have, sin A = n sin B ⇒, , n sin A, =, 1 sin B, , =, , On applying componendo and dividendo, n − 1 sin A − sin B, ⇒, =, n + 1 sin A + sin B, A+B, A−B, sin, 2 cos, 2, 2, =, A+B, A−B, cos, 2 sin, 2, 2, A−B, A+B, = tan, cot, 2, 2, n −1, A−B, A + B, = tan, ⇒, tan , 2 , n+1, 2, , , 5 −1, 5 + 1, , cos 36° =, Q sin 18° =, , 4, 4 , , , 5. tan 15°⋅ tan 195°, = tan 15°⋅ tan (180° + 15° ), = tan 15°⋅ tan 15°, …(i), = tan 2 (15° ), 2 tan 15°, 1, =, tan 2 (15° ) = tan 30° =, 3, 1 − tan 2 15°, , 2 tan θ , Q tan 2 θ =, , , 1 − tan 2 θ , ⇒, ⇒, ⇒, , 1 − tan 2 15° = 2 3 tan 15°, tan 15° + 2 3 tan 15° − 1 = 0, 2, , tan 15° =, , −2 3 ± 12 + 4 −2 3 ± 4, =, = − 3 ±2, 2 (1), 2, , −k2, k2, −1 + 2 − 1 , + k2 −, = k2 , , 2, 2, 2, , , (2 − 2) 2, =, ⋅k =0, 2, , …(i), , 2 tan θ , sec 2 θ = 1 + tan 2θ = 1 + , , 1 − tan 2 θ , , 2, , 2, , 2 m , = 1+ , , 1 − m , =, =, , [from Eq. (i)], , (1 − m)2 + 4m, 12 + m2 − 2m + 4m, =, |(1 − m)|, |(1 − m)|, (1 + m)2, 1+m, =, |(1 − m)| |1 − m|, , 7. tan 9° − tan 27° − tan 63° + tan 81°, = tan (90° − 81° ) − tan (90° − 63° ) − tan 63° + tan 81°, = cot 81° − cot 63° − tan 63° + tan 81°, = (cot 81° + tan 81° ) − (cot 63° + tan 63° ), 2 cos 2 81° + sin 2 81° cos 2 63° + sin 2 63° 2, = , −, ×, 3 sin 81°⋅ cos 81° sin 63°⋅ cos 63° 2, 2, 2, 2, 2, −, =, −, sin 162° sin 126° sin 18° cos 36°, , 9. Given equation, tan 4 x − 2 sec2 x + a 2 = 0, ⇒, tan 4 x − 2 − 2 tan 2 x + a 2 = 0, ⇒, tan 4 x − 2 tan 2 x + (a 2 − 2) = 0, ⇒, , tan 2 x =, , 2 ± 4 − 4(a 2 − 2), 2, , 2 ± 2 1 − a2 + 2, = 1 ± 3 − a2, 2, For real values of tan 2 x,, 3 − a2 ≥ 0 ⇒ a2 − 3 ≤ 0, ⇒, a 2 ≤ 3 ⇒ |a |≤ 3, 10. Given,, tan A − tan B = x, and, cot B − cot A = y, sin A ⋅ cos B − sin B ⋅ cos A, sin ( A − B), =, ⇒ x=, cos A ⋅ cos B, cos A ⋅ cos B, 1 cos A ⋅ cos B, ⇒, =, x sin ( A − B), cos B ⋅ sin A − cos A ⋅ sin B sin ( A − B), and y =, =, sin A ⋅ sin B, sin A ⋅ sin B, 1 sin A ⋅ sin B, =, ⇒, y sin ( A − B), 1 1 cos A ⋅ cos B + sin A ⋅ sin B, ∴, + =, x y, sin ( A − B), ⇒, , 2, , = A rational number for every non-square, natural number m., , =, , 5 + 1 − 5 + 1, 1, 1 , =8 , −, , =8 , 5 −1, 5 + 1, , , 5 −1, 2, = 8⋅ = 4, 4, 4π , 2π , 8. x = y cos = z cos , 3, 3, π, , π π, ⇒, x = y cos + = z cos π + , , 2 6, 3, π, π, , , x = y − sin = z − cos , ⇒, 6, 3, , , 1, 1, x= − y× = −z ×, ⇒, 2, 2, ⇒, 2x = − y = − z, x, y, z, ⇒, =, =, =k, (1 /2) (−1) (−1), k, ⇒, x = , y = − k, z = − k, 2, k, k, Now, xy + yz + zx = (− k) + (− k)(− k) + (− k) , 2, 2, =, , tan 15° = (− 3 + 2), ⇒, From Eq. (i), we get, tan 15°⋅ tan 195° = (2 − 3 )2, =4 + 3 −4 3, = (7 − 4 3 ), 6. Given, tan θ = m, , 2, 2, −, 5 − 1 5 + 1, , , , 4 4 , , tan 2 x =
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238, , NDA/NA Mathematics, 1 1 cos ( A − B), + =, x y sin ( A − B), 1 1, cot ( A − B) = +, x y, , ⇒, ⇒, , =, =, , 11. Q x = sin θ + cos θ and y = sin θ ⋅ cos θ, Now, x4 − 4x2y − 2x2 + 4 y2 + 4 y + 1, = (sin θ + cos θ )4 − 4(sin θ + cos θ )2 y − 2(sin θ + cos θ )2, + 4 y2 + 4 y + 1, = (sin 2 θ + cos 2 θ + 2 sin θ cos θ )2, − 4(sin 2 θ + cos 2 θ + 2 sin θ cos θ ) y, −2(sin 2 θ + cos 2 θ + 2 sin θ cos θ ), + 4 y2 + 4 y + 1, 2, 2, = (1 + 2 y) − 4(1 + 2 y) y − 2(1 + 2 y) + 4 y + 4 y + 1, = 1 + 4 y2 + 4 y − 4 y − 8 y 2 − 2 − 4 y + 4 y 2 + 4 y + 1 = 0, α=A+B, , 12. Q, , Now,, , ⇒, , ⇒, , ⇒, , α+x, α−x, , B=, 2, 2, tan A 2, =, tan B 1, α − x, cos , , 2 2, =, ⋅, α − x 1, sin , , 2 , , x= A −B⇒ A =, , and, , α + x, sin , , 2 , α + x, cos , , 2 , , 13. We have,, , 1 + sin x − 1 − sin x, , =, , 2, , x, x, , sin − cos , , 2, 2, , 2, , 2, , x, x, , − sin − cos , , 2, 2, , 2, , x, x, , sin + cos +, , 2, 2, x, x, , sin + cos , , 2, 2, , x, 2, , b, a, a+b, a−b, +, a−b, a+b, , Now,, , 1 − tan 2 x, =, , 2, , sin x, cos 2 x, , [from Eq. (i)], , 2 cos x, cos 2x, , 5 + 1 5 + 1 4 , , −, − , 4 4 16 , 1 1 5, −, =, 2 4 16, 4, 5, 16. We have, cos (α + β ) = and sin (α − β ) =, 5, 13, 3, 12, ⇒ sin(α + β ) = and cos (α − β ) =, 5, 13, 3, 5, ⇒, α + β = sin −1 and α − β = sin −1, 5, 13, −1 3, −1 5, ⇒, 2 α = sin, + sin, 5, 13, , 3, 25, 5, 9 , = sin −1 , +, 1−, 1−, 169 13, 25 , 5, 36 20, ⇒, = sin −1 , +, , 65 65, 56, ⇒, 2α = sin −1 , 65, 56, sin 2α =, ⇒, 65, and, , x, x, x, x, + sin, + sin − cos, 2, 2, 2, 2, =, x, x, x, x, cos + sin − sin + cos, 2, 2, 2, 2, x π π , x, x, , π , Q x ∈ 2 , π ⇒ 2 ∈ 4 , 2 ∴ sin 2 ≥ cos 2 , , , , 14. Given that, tan x =, , 2, , =, , 15. sin 36° sin 72° sin 108° sin 144°, 1, = sin 2 36° sin 2 72° = [(2 sin 2 36° ) (2 sin 2 72° )], 4, 1, = [(1 − cos 72° )(1 − cos 144° )], 4, 1, = [(1 − sin 18° )(1 + cos 36° )], 4, 1 , 5 − 1 , 5 + 1 , = 1 −, 1 +, , 4 , 4 , 4 , , …(i), , 1, 4, , 56, cos 2α = 1 − , 65, , 2, , 1089 33, =, (65)2 65, sin 2α 56 /65 56, tan 2α =, =, =, cos 2α 33 /65 33, =, , cos, , = tan, , b2, 1− 2, a, 2, , , 1 +, , 1, =, 1+, 4 , , sin α + sin x = 2 sin α − 2 sin x, 3 sin x = sin α, sin α, sin x =, 3, 1 + sin x + 1 − sin x, , ⇒, , 2, , 1−, , =, , α − x, α + x, 2 sin , , cos , 2 , 2 , =2, α − x, α + x, 2 cos , sin , , 2 , 2 , sin α + sin x, =2, sin α − sin x, , ⇒, ⇒, , =, , 1 + b /a, 1 − b /a, +, 1 − b /a, 1 + b /a, , ∴, , 17. We have,, cos 2( A − B) + cos 2 B − 2 cos ( A − B) cos A cos B, = cos 2( A − B) + cos 2 B − cos ( A − B), [cos ( A − B) + cos ( A + B)], = cos 2 B − cos ( A − B) cos ( A + B), = cos 2 B − (cos 2 A − sin 2 B) = 1 − cos 2 A, Hence, it depends on A.
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239, , Trigonometric Ratios and Equations, 18. We have, A = sin 2 θ + cos 4 θ, = sin 2 θ + cos 2 θ cos 2 θ, ≤ sin 2 θ + cos 2 θ, (since, cos 2 θ ≤ 1), 2, 4, ⇒, sin θ + cos θ ≤ 1 ⇒ A ≤ 1, Again, sin 2 θ + cos 4 θ = 1 − cos 2 θ + cos 4 θ, , 1 °, 2 sin 2 7 , 2, 1, 1 °, °, , 2 sin 7 cos 7 , 2, 2, θ, θ, , , 2θ, = 1 − cos θ and 2 sin ⋅ cos = sin θ, Q 2 sin, , , 2, 2, 2, 1 − cos 15° 1 − cos (45° − 30° ), =, =, sin 15°, sin (45° − 30° ), , 1, 22. tan 7 ° =, 2, , 2, , 3 3, 1, , = cos 4 θ − cos 2 θ + 1 = cos 2 θ − + ≥, , 2, 4 4, Hence,, , 3, ≤ A ≤1, 4, , =, , 19. Q, …(i), cos A + cos B = m, and, …(ii), sin A + sin B = n, From Eqs. (i) and (ii), we get, 2mn, 2 (cos A + cos B)(sin A + sin B), =, 2, 2, (cos A + cos B)2 + (sin A + sin B)2, m +n, sin 2 A + sin 2B + 2 sin ( A + B), =, 1 + 1 + 2 cos ( A − B), sin ( A + B)[2 + 2 cos ( A − B)], =, = sin ( A + B), 2 + 2 cos ( A − B), 2mn, Hence, sin ( A + B) = 2, m + n2, 1, 585, , 20. sin 292 ° = sin, , 2, 2, 1 − cos 585°, =−, 2, 1 − cos 225°, =−, 2, =−, =−, , =, , =, , 3 + 1, 1−, , 2 2 , 3 −1, 2 2, , =, , (2 2 − 3 − 1), 3+1, ×, 3 −1, 3+1, , 2 6 −3 − 3 + 2 2 − 3 −1, 3 −1, , = 6 − 3 + 2 −2, cos 15° + cos 45°, 23., cos3 15° + cos3 45°, cos 15° + cos 45°, =, (cos 15° + cos 45° )(cos 2 45° + cos 2 15°, − cos 45° cos 15° ), [Q a3 + b3 = (a + b)(a 2 − ab + b2)], 1, =, 2, 2, (cos 45° + cos 15° − cos 45° cos 15° ), , 1 + cos 45°, 1 + 1/ 2, =−, 2, 2, , =, , 1, 2, , 1, 1, 2, ⋅ cos 15°, + cos (45° − 15° ) −, 2, 2, 1, =, 1, cos 15°, + (cos 45° cos 30° + sin 45° sin 30° )2 −, 2, 2, 1, =, 2, 1 3, 1 , 1 3 + 1, +, +, −, , , 2 2 2 2 2, 2 2 2 , , 2+1, 2 2, , 2+1, 2, 1, ×, =−, 2+ 2, 2, 2 2, 2, 1, , [Where negative sign shows that sin 292 ° lies in the, , 2, third], =−, , 21. Q A, B and C are in AP., …(i), ∴, 2B = A + C, But, A + B + C = 180°, ⇒, 3B = 180°, ⇒, B = 60°, sin A + 2 sin B + sin C = (sin A + sin C ) + 2 sin B, A+C, A −C, = 2 sin, cos, + 2 sin B, 2, 2, (2B), A − C, [from Eq. (i)], = 2 sin, cos , + 2 sin 2B, 2 , 2, A −C, , , = 2 sin B cos, +1, , , 2, , A − C , 2 A, = 2 sin B 2 cos 2 , Q 1 + cos A = 2 cos, , 4 , 2, , A − C, = 4 sin B cos 2, , 4 , , 1 − (cos 45° ⋅ cos 30° + sin 45° ⋅ sin 30° ), sin 45°⋅ cos 30° − cos 45°⋅ sin 30°, , 1, 1 3+1+2 3, 3+1, +, −, 2, 8, 4, 1, 8 4, =, = =, 4 + 4 + 2 3 −2 3 −2 6 3, 8, =, , 24., , 3 2, a, 4, ⇒ ∆ ABC is an equilateral., , I. Area of ∆ ABC =, , II. tan A =, , 1 − cos B, 2 sin 2 B /2, =, = tan B /2, sin B, 2 sin B /2 cos B /2, 2 tan A, 2 tan B /2, =, 2, 1 − tan A 1 − tan 2 B /2, , Now, , tan 2 A =, , ⇒, , tan 2 A = tan B
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240, , NDA/NA Mathematics, , III. Let y =, , Let, , ∴ tan ( A + 2B) tan (2 A + B), π 2π , 2π π , = tan +, + , tan , 3, 3, 6, 6, 2, π, 5, π, , , = tan tan , 6, 3, π, π, π, , , , = tan + tan π − , , 2 6, 6, π, π, = − cot − tan = 1, 6, 6, , tan 3x, tan x, , =, , 3 tan x − tan3 x, 1 − 3 tan 2 x, , =, , 3 − tan 2 x, 1 − 3 tan 2 x, tan x = t, 3 − t2, 1 − 3t 2, , ∴, , y=, , ⇒, , y − 3 yt 2 = 3 − t 2, , 28. We have, 2 + 2 + 2 + ... ∞ = cosec θ, , t=, , ∴, x= 2+ x, On squaring both sides, we get x2 = 2 + x, ⇒, x2 − x − 2 = 0, ⇒, (x + 1)(x − 2) = 0, ⇒, x = 2 , x = −1, ∴ x = 2 is correct., (Q x is positive), 1, ∴, 2 = cosec θ ⇒ sin θ =, 2, , sin A = sin B, ⇒, A = B or A = π − B, Now, sin 2 A = sin 2B is satisfied by A = B but it is, not satisfied by A = π − B., π, 4π, 5π, II. cos cos, cos, 7, 7, 7, π, 4π, 2π , , cos π −, = cos cos, , , 7, 7, 7, π, 2π, 4π, = − cos cos, cos, 7, 7, 7, 3 π, 8π, sin 2 ⋅ , sin, , 7, 7, =−, =−, π, π, 3, 2 sin, 8 sin, 7, 7, 1, =, 8, sin (x + y) a + b, …(i), 26. We have,, =, sin (x − y) a − b, sin (x + y) + sin (x − y) a + b + a − b, =, ⇒, sin (x + y) − sin (x − y) a + b − a + b, , 25. I., , [by using componendo and dividendo rule in Eq. (i)], sin x cos y + cos x sin y + sin x cos y − cos x sin y, ⇒, sin x cos y + cos x sin y − sin x cos y + cos x sin y, 2a, =, 2b, 2 sin x cos y 2a, tan x a, =, ⇒, =, ⇒, 2 cos x sin y 2b, tan y b, π, 27. We have, sin ( A + B) = 1 ⇒ A + B =, 2, 1, π, and, sin ( A − B) = ⇒ A − B =, 2, 6, On solving Eqs. (i) and (ii), we get, π, π, A = ,B=, 3, 6, , x= 2+ 2+ 2+K∞, , Let, , y −3 1, ×, 1 3, y−, 3, y−3, 1, For existence of t,, ≥ 0, y ≠, y − 1 /3, 3, 1, , y ∈ − ∞ , ∪ (3, ∞), ⇒, , 3, tan 3x, 1, never lies between and 3., tan x, 3, ⇒, , 29. The squares of the tangents of the angles 30°, 45° and, 60° are in GP., ⇒, tan 2 30° , tan 2 45° , tan 2 60° are in GP., 2, 1, 2, ⇒, , 1, ( 3 ) are in GP., 3, 1, , 1, 3 are in GP., ⇒, 3, (Q Condition for GP, b2 = ac is satisfied), 1, i.e.,, 12 = × 3 ⇒ 1 = 1, 3, 30. We have,, …(i), x+ y=z, Now, 1 + cos x + cos y + cos z, = (1 + cos z ) + (cos x + cos y), z, x+ y, x− y, = 2 cos 2 + 2 cos, cos, 2, 2, 2, z, x− y, 2z, [from Eq. (i)], = 2 cos, + 2 cos cos, 2, 2, 2, z, z, x − y, cos + cos, = 2 cos, 2 , 2, 2 , z, 2, z, = 2 cos, 2, x, = 4 cos, 2, = 2 cos, , 31. (a), , …(i), , ⇒, , …(ii), , ⇒, ⇒, Thus,, , z+ x− y, z − x + y, , 2 cos, cos, , , 4, 4, 2x, 2 y, , [from Eq. (i)], 2 cos, cos, 4, 4 , , y, z, cos cos, 2, 2, , x= y, sin 3 θ cos 3 θ cos 3 θ sin 3 θ, −, =, +, sin θ, cos θ, cos θ, sin θ, 2 cos 3 θ, =0, cos θ, cos 3 θ = 0 ⇒ θ = 30°, θ = 15° is incorrect.
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241, , Trigonometric Ratios and Equations, (b) x = − y, sin 3 θ cos 3 θ, cos 3 θ sin 3 θ, −, ⇒, =−, −, sin θ, cos θ, cos θ, sin θ, 2 sin 3 θ, ⇒, = 0 ⇒ sin 3 θ = 0, sin θ, ⇒, 3 θ = 180° or 0°, ⇒, θ = 60° or 0°, Since, (b) comes out to be correct, there is no need for, further checking., 32. I. (sec θ + tan θ )−1 = (sec 1200° + tan 1200° )−1, = [sec (6π + 120° ) + tan (6π + 120° )]−1, = (sec 120° + tan 120° )−1, = (− cosec 30° − cot 30° )−1, 1, = (−2 − 3 )−1 = −, , negative, 2+ 3, II. cosec θ − cot θ = cosec (6π + 120° ) − cot (6π + 120° ), = cosec 120° − cot 120°, = sec 30° + tan 30°, 2, =, + 3 , positive, 3, Hence, both the statements are incorrect., tan 45° − tan 30°, 33. A. tan 15° = tan (45° − 30° ) =, 1 + tan 45° ⋅ tan 30°, 1, 1−, 3 = 3 −1 × 3 −1, =, 1, 3+1, 3 −1, 1+, 3, 3 + 1 −2 3, =, =2 − 3, 2, B. tan 75° = tan (45° + 30° ), tan 45° + tan 30°, =, 1 − tan 45° ⋅ tan 30°, 1, 1+, 3 = 3 + 1 × 3 + 1 =2 + 3, =, 1, 3 −1, 3+1, 1−, 3, tan 60° + tan 45°, C. tan (105° ) = tan (60° + 45° ) =, 1 − tan 60° ⋅ tan 45°, =, , 3+1 1+ 3, ×, 1− 3 1+ 3, , =, , 4+2 3, = −2 − 3, −2, , 35. Given that,, tan 2 φ + tan 6 φ = tan3 φ ⋅ sec2 φ, tan 2 φ {1 + tan 4 φ − tan φ ⋅ sec2 φ } = 0, tan 2 φ {1 + tan 4 φ − tan φ − tan3 φ } = 0, tan 2 φ {(1 − tan φ ) + tan3 φ (tan φ − 1)} = 0, tan 2 φ {(1 − tan φ )(1 − tan3 φ )} = 0, 2, tan φ (1 − tan φ )2(1 + tan 2 φ + tan φ ) = 0, ∴ It is notation identity for all value of φ because at, φ = (2n + 1) π /2 the given identify does not exist., sec A + tan A = p, 1, sin A, +, =p, cos A cos A, 1 + sin A, =p, cos A, , 36., ⇒, ⇒, ⇒, , (1 + sin A )2, = p2, cos 2 A, , ⇒, , (1 + sin A )2, = p2, 1 − sin 2 A, (1 + sin A )2, = p2, (1 + sin A )(1 − sin A ), 1 + sin A, = p2, 1 − sin A, , ⇒, ⇒, , (1 + sin A ) + (1 − sin A ) p2 + 1, =, (1 + sin A ) − (1 − sin A ) p2 − 1, , ⇒, , (using componendo and dividendo), p2 + 1, 2, = 2, 2 sin A p − 1, , ⇒, ⇒, , sin A =, , 37. Given that, cot A ⋅ cot B = 2, cos A ⋅ cos B 2, ⇒, =, sin A ⋅ sin B 1, cos A ⋅ cos B − sin A ⋅ sin B 2 − 1, =, ⇒, cos A ⋅ cos B + sin A ⋅ sin B 2 + 1, , ⇒, ⇒, , cot2 θ − 3 3 cot θ + 6 = 0, 3 3 ± 27 − 24 3 3 ± 3, cot θ =, =, = 2 3, 3, 2, 2, π, cot θ ≠ 2 3 , cot θ = 3 = cot, 6, π, θ=, 6, , ...(i), , (by componendo and dividendo rule), cos ( A + B) 1, =, ⇒, cos ( A − B) 3, 1, ⇒, cos ( A + B) ⋅ sec ( A − B) =, 3, 1, 1, 38. I. If cot θ = x, then x + = cot θ +, x, cot θ, cos θ sin θ cos 2 θ + sin 2 θ, +, =, sin θ cos θ, sin θ cos θ, 1, =, = cosec θ sec θ, sin θ cos θ, =, , 34. Q cosec2 θ = 3 3 cot θ − 5, ⇒ 1 + cot2 θ − 3 3 cot θ + 5 = 0 (Q cosec2 θ = 1 + cot2 θ ), ⇒, , p2 − 1, p2 + 1, , ∴ Statement I is correct., 1, = sin θ, then, x, 2, 1, , 2, x + = sin θ, , x, , II. If x +
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242, , NDA/NA Mathematics, 1, + 2 = sin 2 θ, x2, 1, x2 + 2 = sin 2 θ − 2, ⇒, x, ∴ Statement II is correct., III. If x = p sec θ and y = q tan θ, then, x2q2 − y2p2 = p2q2 sec2 θ − p2q2 tan 2 θ, = p2q2(sec2 θ − tan 2 θ ), = p2q2, ∴Statement III is correct., 1, , 3, IV. Now, cos θ − 3 sin θ = 2 cos θ −, sin θ, 2, 2, , , ⇒, , 2x − 2 y, 2x + 2 y, 2 cos , , sin , 2 , 2 , =, 2x − 2 y, 2x + 2 y, 2 cos , , cos , 2 , 2 , , x2 +, , = 2 [cos 60° cos θ − sin 60° sin θ ], = 2 cos (60° + θ ), ∴ Maximum value = 2 × 1 = 2, ∴ Only I, II and III are correct., 39. Given, θ = 18°, Now, we have 4 sin 2 θ + 2 sin θ = 4 sin 2(18° ) + 2 sin (18° ), , 5 − 1, Q sin 18° =, , 4 , , 2, , 5 − 1, 5 − 1, =4 , , +2, 4, 4 , , , (5 + 1 − 2 5 ) ( 5 − 1) 3 − 5, =4⋅, +, =, +, 16, 2, 2, , 5 −1, =1, 2, , 40. Let f (x) = sin 2 x, 1, 1, f (x) = (2 sin 2 x) = (1 − cos 2x), 2, 2, 1 1, f (x) = − cos 2x, 2 2, 1 1, 1, Q − 1 ≤ cos 2x ≤ 1 ⇒ − ≤ cos 2x ≤, 2 2, 2, 1, 1, −1, ≥ − cos 2x ≥, ⇒, 2, 2, 2, 1 1 1 1, 1 1, ⇒ + ≥ − cos 2x ≥ − ⇒ 0 ≤ f (x) ≤ 1, 2 2 2 2, 2 2, ∴ Maximum value of sin 2 x is 1., 41. Given that, sin A = p and sin B = q, and, | p| < 1,|q| < 1, i.e.,, 1 − p2 = positive and 1 − q2 = positive, Now, sin ( A + B) = sin A cos B + cos A sin B, = ± sin A 1 − sin 2 B ± 1 − sin 2 A sin B, = ± p 1 − q 2 ± 1 − p2 q, Thus, there exists 4 number of solutions., π, 42. Given, x − y = (4n + 1), 4, π, and, x + y ≠ (2n + 1), 2, sin 2x − sin 2 y, Now,, cos 2x + cos 2 y, , =, , cos (x + y) sin (x − y), cos (x + y) cos (x − y), , = tan (x − y), = tan (4n + 1), 43. Given that,, , π, =1, 4, , 1 + sin A − cos A, 2 sin A, =, 1 + sin A, 1 + sin A + cos A, , ⇒, , 2 sin A (1 + sin A ) = (1 + sin A )2 − cos 2 A, , ⇒, , 2 sin A + 2 sin 2 A = 1 + sin 2 A + 2 sin A − cos 2 A, , ⇒, , sin 2 A = 1 − cos 2 A, , ⇒, , sin A + cos 2 A = 1, 2, , π, , given, 2, identity become indeterminate. So, the given identity is, π, true for all values of A except A = − ., 2, , Which is true for all values of A but for A = −, , 44. Q X = { θ ∈ [0, 2π ] : sin θ = cos θ }, Thus, the number of elements in X is 2. Since,, sin θ = cos θ is possible at θ = 45° and 225°., While sin θ positive, negative in IInd, IVth quadrant,, respectively and cos θ positive, negative in IVth, IInd, quadrant, respectively., Hence, A is true but R is false., π, 3 −1, 45. I. sin, = sin 15° =, 12, 2 2, Q, sin 15° = sin (45° − 30° ), = sin 45°⋅ cos 30° − cos 45°⋅ sin 30°, =, II., , 3, 1, 3 −1, −, =, 2 2 2 2, 2 2, , π, 3+1, = cos 15° =, 12, 2 2, cos 15° = cos (45° − 30° ), cos, , Q, , = cos 45°⋅ cos 30° + sin 45°⋅ sin 30°, =, III., , 3, 1, 3+1, +, =, 2 2 2 2, 2 2, , π, = cos 15° = 2 + 3, 12, cot 15° = tan 75° = tan (45° + 30° ), cos, , Q, , 1, 1+, tan 45° + tan 30°, 3, =, =, 1 − tan 45°⋅ tan 30° 1 − 1, 3, 3+1, =2 + 3, =, 3 −1, ∴ The correct sequence is III > II > I.
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243, , Trigonometric Ratios and Equations, 46. Let the length of side of square is a., , sec2 θ + cosec2θ = tan 2 θ + cot2 θ + 2, , II. Q, , = (tan θ + cot θ ), , C, , D, , ∴, , A, , ∴, , θ, k P, , B, , 2k, , AD = a and AB = k + 2k = a, a, a, k=, ⇒ AP =, 3, 3, , ⇒, Now,, , In ∆ APD;, , PD 2 = AP 2 + AD 2, 10a 2, a2, 10a, ⇒ PD =, =, + a2 =, 9, 9, 3, AP, cos θ =, PD, a /3, 1, =, =, 10a /3, 10, , 47. Given that, Tn = sin n x + cos n x, ∴, Tn − 2 = sin n − 2 x + cos n − 2 x, and, Tn − 4 = sin n − 4 x + cos n − 2 x, Now,, Tn − 2 − Tn = sin n − 2 x + cos n− 2 x − sin n x − cos n x, = sin n − 2 x [1 − sin 2 x] + cos n − 2 x [1 − cos 2 x], = sin n − 2 x ⋅ cos 2 x + cos n − 2 x ⋅ sin 2 x, ⇒, Tn − 2 − Tn = sin 2 x cos 2 x {sin n − 4 x + cos n − 4 x}, ⇒, Tn − 2 − Tn = sin 2 x cos 2 x ⋅ Tn − 4, Tn − 2 − Tn, = sin 2 x cos 2 x, ⇒, Tn − 4, ∴, , Tn − 2 − Tn, Tn − 4, , is independent of n., , 48. 1 + sin 2 A = 3 sin A cos A, On dividing both sides of above equation by cos 2 A,, we get,, sec 2A + tan 2 A = 3 tan A, ⇒, 1 + tan 2 A + tan 2 A = 3 tan A, ⇒, 2 tan 2 A − 3 tan A + 1 = 0, 2, ⇒, 2 tan A − 2 tan A − tan A + 1 = 0, ⇒, 2 tan A (tan A − 1) − 1(tan A − 1) = 0, ⇒, (2 tan A − 1) (tan A − 1) = 0, 1, ⇒, tan A = 1,, 2, 1 + sin θ , 49. I. Q (sec θ + tan θ )2 = , , cos θ , , 2, , (1 + sin θ )2, (1 + sin θ )2, =, 2, (1 − sin θ ) (1 + sin θ ), cos θ, 1 + sin θ, =, 1 − sin θ, 1 + sin θ, 2, ∴ (sec θ + tan θ ) =, 1 − sin θ, =, , sec θ + cosec θ = tan θ + cot θ, 2, , 2, , Hence, both of the given statements are correct., π, radian, 50. Ist, 1° =, 180, 3.14, =, = 0 . 017 = 0. 02 (approx), 180, which is equal to 0 . 02, 180, 180, degree =, IInd, 1 radian =, π, 3.14, = 57 . 32 degree, which is greater than 45°., 51. Q Q3 = sin A (cos B + cos C ) + sin B (cos C + cos A ), + sin C (cos A + cos B), = sin A cos B + sin A cos C + sin B cos C, + sin B cos A + sin C cos A + sin C cos B, = sin ( A + B) + sin (B + C ) + sin (C + A ) = Q1, ⇒ Q3 = Q1, 52. sin A =, , 16, 3, 4, =−, ⇒ cos A = − 1 −, 25, 5, 5, , cos B = −, , 12, 144, 5, =, ⇒ sin B = + 1 −, 169 13, 13, , ∴ sin ( A + B) = sin A cos B + cos A sin B, 4 12 3 5, = × − − ×, 5 13 5 13, − 48 − 15, 63, =, =−, 65, 65, 53. Let the angles of a triangle be, Q, , A = 2x, B = 5x and C = 5x, A + B + C = 180°, , ⇒, , 2x + 5x + 5x = 180°, , ⇒, , 12x = 180°, , ⇒, , x = 15°, B = 75° , C = 75°, , ∴, , tan B tan C = tan 75° tan 75° = [tan(45°+30° )]2, =, , tan 45° + tan 30° , 1 − tan 45° tan 30° , , 2, , 2, , 1 , , 2, 1+, , 3 = 3 + 1, =, , , , 1, 3 −1, 1 − 1 ×, , 3, , 2, , ( 3 + 1 )2 , (4 + 2 3 )2, =, =, , 2, (3 − 1)2, ( 3) −1 , = (2 + 3 )2 = 4 + 3 + 4 3, =7+4 3
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244, , NDA/NA Mathematics, , 54. Given, sin3 θ + cos3 θ = 0, ⇒, tan3 θ = − 1, ⇒, tan θ = − 1, π, ⇒, θ=−, 4, 55. The maximum value of, 3 sin x + 4 cos x + 7 = 7 + 32 + 42, =7+5, = 12, 56. We have,, , =, , sin 5 A − sin 3 A, cos 3 A + cos 5 A, 5A + 3A, 5A − 3A, 2 cos, sin, 2, 2, , = tan A, , 5A + 3A, 5A − 3A, cos, 2, 2, ∴From second statement, we have,, 2 cos A sin B = sin ( A + B) − sin ( A − B), and, 2 cos A cos B = cos ( A + B) + cos ( A − B), ∴ Both A and R are individually true but R is not correct, explanation of A., π, , ... (i), 57. We have, x = 1 + cot α − sec α + , , 2, π, , ... (ii), and, y = 1 + cot α + sec α + , , 2, 2 cos, , From Eq. (i), x = 1 + cot α + cosec α, From Eq. (ii), y = 1 + cot α − cosec α, Now,, xy = (1 + cot α + cosec α ) (1 + cot α − cosec α ), = (1 + cot α )2 − cosec2α, = 1 + 2 cot α + cot2 α − cosec2α, ⇒, xy = 2 cot α, Obviously xy > 0, if either 0° < α < 90°, or, 180° < α < 270°, since cot α > 0 either 0° < α < 90°, or, 180° < α < 270°, so A is true., As shown above, xy ≠ 2 tan α, so R is false., 58. (A) Since, π rad = 180°, 180° 180°, ⇒, 1 rad =, =, ×7, π, 22, 630°, 1 rad =, = 57° (approx), 11, We know, when θ ∈ [ 45° , 90° ], cos θ < sin θ, Here,, 1 ∈ [ 45° , 90° ], ∴, cos 1 < sin 1, hence, it is true., (R) It is true that cosine is decreases but sin is increases, in the first quadrant., x, x, x, x, 59. (A) y = 1 − sin x = sin 2 + cos 2 − 2 sin cos, 2, 2, 2, 2, 2, , Since,, , x, x, > cos, 2, 2, (R) Above function is not true for every value of x., x π, When,, ∈ 0, , 2 4, x, x, y = cos − sin, 2, 2, 1, , where θ ≠ 0, 60. Given, cosec θ − cot θ =, 3, cos θ, 1, 1, ⇒, −, =, sin θ sin θ, 3, (1 − cos θ ), 1, =, ⇒, sin θ, 3, then, , x, x, x, x, , = sin − cos = sin − cos, , 2, 2, 2, 2, x, x, = sin − cos, 2, 2, x π π, π , x ∈ , π ⇒ ∈ , , 2 , 2 4 2, , ⇒, , ⇒, , ⇒, ⇒, ⇒, ∴, , sin, , , 2 θ , 1 − 1 − 2 sin 2 1, =, , sin θ, 3, , , , , 2 sin 2, , θ, 2, , θ, θ, 2 ⋅ sin ⋅ cos, 2, 2, θ, tan = tan 30°, 2, θ, = 30°, 2, , =, , 1, 3, , θ = 60°, 1, cos 60° =, 2, , 61. We have, 2 cos 2 x + 3 sin x − 3 = 0, ⇒, 2 − 2 sin 2 x + 3 sin x − 3 = 0, ⇒, (2 sin x − 1) (sin x − 1) = 0, 1, sin x = or sin x = 1, ⇒, 2, π 5π, π, or, ⇒, x= ,, ⇒ 30°, 150°, 90°, 6 6, 2, 2 sec θ + tan θ = 1, sin θ, 2, ⇒, +, =1, cos θ cos θ, sin θ − cos θ = − 2, ⇒, 1, 1, ⇒, sin θ −, cos θ = − 1, 2, 2, 1, 1, cos θ −, sin θ = 1, ⇒, 2, 2, π, π, , , ⇒ cos cos θ − sin sin θ = 1, , , 4, 4, π, , ⇒, cos θ + = cos (0° ), , 4, π, π, θ + = 2 nπ ± 0 ° ⇒ θ = 2 nπ −, ⇒, 4, 4, c, a, b, 63., cos x =, sin x +, a 2 + b2, a 2 + b2, a 2 + b2, c, ⇒, sin (x + α ) =, >1, a 2 + b2, 62. We have,, , (as given)
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245, , Trigonometric Ratios and Equations, , α=, , where, , a, a +b, 2, , a ± a 2 − 16 a + 64, = a ± (a − 8)2, 4, a ± (a − 8) 2a − 8, =, =, ,2, 4, 4, a −4, (Qsin x ≥/ 1), sin x =, 2, − 1 ≤ sin x ≤ 1, a −4, −1 ≤, ≤1, 2, −2 ≤ a − 4 ≤ 2, 2 ≤ a ≤6, =, , 2, , Hence, there is no solution exist, because sin cannot be, greater than 1., 64. sec θ + tan θ = 3, Also, we have, sec2 θ − tan 2 θ = 1, ⇒, (sec θ + tan θ ) (sec θ − tan θ ) = 1, 1, sec θ − tan θ =, ⇒, 3, From Eqs. (i) and (iii), we get, 1, 1, 1, π, = tan , tan θ = 3 −, =, , , 6, 2, 3, 3, π, θ = nπ +, ⇒, 6, π, 7π, ., ∴ Solutions for 0 < θ < 2π are and, 6, 6, Hence, there are two solutions., 1 − cos 2θ, 65. Given that,, =3, 1 + cos 2θ, 2 sin 2 θ, ⇒, = 3 ⇒ tan 2 θ = 3, 2 cos 2 θ, π, tan 2 θ = ( 3 )2 ⇒ tan 2 θ = tan 2, ⇒, 3, π, θ = nπ ±, ⇒, 3, , ...(i), ...(ii), , ∴, Q, , ...(iii), , ∴, ⇒, ⇒, , Solutions (Q. Nos. 70-72), Let angles of a triangle be 2x, 3x and 5x., ∴, 2x + 3x + 5x = 180° ⇒ x = 18°, ∴, A = 36° , B = 54° ,C = 90°, 70. Here, the least angle is 36° and the greatest angle is 90°., C, 71. ∴cos 2 A + sin 3C + cos × (sin 2B)2, 2, = (cos 36° )2 + sin 3 × 90° + cos (45° ) sin (2 × 54° )2, 2, 5 + 1, =, + 1 + cos 45° × [sin (90° + 18° )]2, 4 , =, , 6+2 5, 1 10 + 2 5 , =, +1+, , 4, 16, 2 , , , 66. Given that,, ⇒, ⇒, ⇒, , tan 2 θ − tan θ − 3 tan θ + 3 = 0, tan θ (tan θ − 1) − 3 (tan θ − 1) = 0, (tan θ − 3 ) (tan θ − 1) = 0, π, π, θ = nπ + , nπ +, 3, 4, , 67. We have, 3 sin 2 x + 10 cos x − 6 = 0, ⇒ 3 (1 − cos 2 x) + 10 cos x − 6 = 0, ⇒, (cos x − 3) (3 cos x − 1) = 0, 1, ⇒, cos x ≠ 3 or cos x =, 3, −1 1 , ⇒ x = 2nπ ± cos , 3, 68., , sin θ = sin α, ⇒, sin θ − sin α = 0, θ + α, θ − α, ⇒, 2 sin , =0, cos , 2 , 2 , θ −α, θ + α, ⇒, sin, = 0 or cos , =0, 2 , 2, θ −α, θ+α, π, = nπ or, = (2n + 1) , n ∈ I, ⇒, 2, 2, 2, θ+α, π, θ −α, Thus,, is any odd multiple of, and, is, 2, 2, 2, any multiple of π., , 69. Given, cos 2x + a sin x = 2a − 7, ∴ 1 − 2 sin 2 x + a sin x = 2a − 7, ⇒ 2 sin 2 x − a sin x + (2a − 8) = 0, a ± a 2 − 8 (2a − 8), sin x =, ∴, 2 (2), , 5+1+2 5, 1, +1+, × [cos 18° ]2, 16, 2, 2, , 22 + 2 5 10 + 2 5, +, 16, 16 2, 22 2 + 2 10 + 10 + 2 5, =, 16 2, =, , 72. We know corresponding to the greatest angle have the, greatest side., Here, C is the greatest., Hence, AB is the greatest side., , Solutions (Q. Nos. 73-75), A + B = 180° − C, A B, C, + = 90°−, ⇒, 2 2, 2, C, , A B, tan + = tan 90°− , ⇒, , 2 2, 2, A, B, tan + tan, 2, 2 = cot C, ⇒, A, B, 2, 1 − tan tan, 2, 2, A, B, C, A, B, , ⇒ tan + tan tan = 1 − tan tan, , 2, 2, 2, 2, 2, A, B, B, C, A, C, ⇒ tan tan + tan tan + tan tan = 1, 2, 2, 2, 2, 2, 2, , 73. Given,, , 74. Q, ⇒, ∴, , B+C=π − A, sin (B + C ) = sin (π − A ) = sin A, sin 2 A + sin 2B + sin 2C
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246, , NDA/NA Mathematics, = 2 sin A cos A + 2 sin (B + C ) cos (B − C ), = 2 sin A [cos A + cos (B − C )], = 2 sin A [cos (B − C ) − cos (B + C )], = 2 sin A [2 sin B sin C ], , 75. Since, A + B + C = π, ⇒ tan A + tan B + tan C = tan A tan B tan C, Using, AM ≥ GM, tan A + tan B + tan C, ≥ (tan A tan B tan C )1/3, ⇒, 3, tan A + tan B + tan C ≥ 3 (tan A tan B tan C )1/3, , ⇒, ⇒, , (tan A tan B tan C )2/3 ≥ 3, 1, , k, , 2/3, , ≥3, , 1, 1, ≥ 33 / 2 = k ≤, k, 3 3, , Solutions (Q. Nos. 76-78), , sin ( A − B) = sin, , ⇒, , A−B=, , π, 3, , π, 3, , 76. On adding Eqs. (i) and (ii), we get, , = 4 sin A sin B sin C, , ⇒, , ⇒, , π, Given that, sin ( A + B) = 1, where A , B, C ∈ 0,, 2 , π, ⇒, sin ( A + B) = sin, 2, π, …(i), ⇒, A+ B=, 2, π, and, sin( A − B) =, 2, , 2π, 3, π, π, and B =, A=, 3, 6, , 2A=, ⇒, , 77. Now, tan( A + 2B) ⋅ tan (2 A + B), 2π π , π π, + , = tan + ⋅ tan , 3, 3 3, 6, 2π , 5π , = tan ⋅ tan , 3, 6, π π, π π, = tan + ⋅ tan + , 2 6, 2 3, = [− cot, , π, π, 1, ][− cot ] = (− 3 ) ⋅ −, =1, , 6, 3, 3, , 78. Now, sin 2 A − sin 2 B, = sin 2(π / 3) − sin 2(π / 6), 2, , 2, 3, 1, = − , 2, 2, , =, , 3 1 2 1, − = =, 4 4 4 2, , …(ii)
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13, , Properties of Triangles, ⇒, , Relation between Sides and, Angles of a Triangle, In a ∆ABC the length of sides opposite to the angles, A, B and C are denoted by a , b and c. Area of a triangle and, perimeter of a triangle are denoted by ∆ and 2s,, respectively and, a+ b+ c, s=, 2, , ⇒, , Cosine Rule, In ∆ABC,, , b2 + c2 − a 2, 2bc, a 2 + c2 − b2, 2. cos B =, 2ca, a 2 + b2 − c2, 3. cosC =, 2ab, , In ∆ABC,, , b, , c, , B, , a, , C, , In ∆ ABC,, 2, 2, 2, % If ∠ A = 60 ° , then b + c − a = bc, , A, b, , c, , A, , 1. cos A =, , Sine Rule, 1, sin A sin B sin C, =, =, =, 2R, a, b, c, , 8, 12, 2, x=, 3, x=, , %, , If ∠ B = 60 °, then a 2 + c 2 − b 2 = ac, , %, , If ∠ C = 60 °, then a 2 + b 2 − c 2 = ab, , b +c c + a a+ b, , then, =, =, 11, 12, 13, cos A cos B cos C, (a), =, =, 25, 19, 7, cos A cos B cos C, (b), =, =, 19, 25, 7, cos A cos B cos C, (c), =, =, 7, 19, 25, (d) None of the above, , Example 2. If in a ∆ABC ,, a, , B, , C, , where, R be the radius of circumcircle of the ∆ABC., , Example 1. In a ∆ABC, A = 30 °, b = 8, a = 6, then, , B = sin−1 x, where x is equal to, 1, (a), 2, 2, (c), 3, , (b), , 1, 3, , (d) 1, −1, , Solution (c) Given, A = 30 °, B = sin x ⇒ sin B = x and b = 8,, a = 6,, By sine rule,, ⇒, ⇒, , sin A sin B, =, a, b, sin 30 ° x, =, 6, 8, x 1/ 2, =, 8 6, , b+ c c+ a a+ b, =, =, =k, 11, 12, 13, ⇒, 2 ( a + b + c) = 36k, and, b + c = 11k, c + a = 12k, a + b = 13k, From Eqs. (i) and (ii), we get, a = 7k, b = 6k, c = 5k, , Solution (c) Let, , Hence,, , cos A =, , b 2 + c 2 − a2, 2bc, , ...(i), ...(ii)
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248, , NDA/NA Mathematics, , 36k2 + 25k2 − 49k2, 1 7, = =, 2, 5, 35, 60k, 2, 2, 2, a +c −b, cos B =, 2ac, 19, 49k2 + 25k2 − 36k2 38, =, =, =, 70, 35, 70k2, 2, 2, 2, 2, 2, 49k + 36k − 25k2 60 25, a +b −c, and cos C =, =, =, =, 2ab, 84 35, 84k2, cos A cos B cos C, ⇒, =, =, 7, 19, 25, =, , Example 3. If in a ∆ ABC, a, b, 2 cos A cos B 2 cos C, , then, =, +, +, +, bc ac, a, b, c, (a) ∠A = 90°, (b) ∠B = 90°, (c) ∠C = 90°, (d) None of these, , ⇒, , 2 cos A cos B 2 cos C, a, b, +, +, =, +, a, b, c, bc ac, 2bc cos A + ac cos B + 2ab cos C = a2 + b 2, , ⇒, , b 2 + c 2 − a2 +, , Solution (a) Q, , a2 + c 2 − b 2, + a2 + b 2 − c 2 = a2 + b 2, 2, c 2 + a2 − b 2 = 2 a2 − 2 b 2 ⇒ b 2 + c 2 = a2, , ⇒, , ∴ ∆ ABC is right angled at A., i.e.,, ∠ A = 90 °, , Example 4 . In any ∆ ABC, then the value of, cos A cos B cos C, +, +, a, b, c, and, c ( b cos A − a cos B) is, 2, a + b2 + c 2, a2 + b 2 + c 2, (a), , a 2 − b 2 (b), , b 2 − a2, abc, 2 abc, a2 + b 2 + c 2, (c), , a 2 + b 2 (d) None of these, 2, , Solution (b) Using the cosines formula, we get, b 2 + c 2 − a2 1 a2 + c 2 − b 2 , 1 a2 + b 2 − c 2 , + , + , , , 2bc, 2ac, c, 2ab, , b, , , 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, =, (b + c − a + a + c − b + a + b − c ), 2abc, a2 + b 2 + c 2, =, 2abc, Using cosine formula, we get, a2 + c 2 − b 2 , b 2 + c 2 − a2 , LHS = bc , , − ac , 2bc, 2ac, , , , , =, , 1, a, , 1 2, [ b + c 2 − a2 − a2 − c 2 + b 2 ], 2, 1, = (2b 2 − 2a2), 2, = b 2 − a2, =, , Projection Formulae, In ∆ABC,, A, , c, , B, , b, , a, , C, , 1. a = c cos B + b cos C, 3. c = a cos B + b cos A, , 2. b = a cos C + c cos A, , Napier’s Analogy, In ∆ABC,, C− A c−a, B, A− B a− b, C, 2. tan, 1. tan, =, cot, =, cot, 2, c+a, 2, 2, a+b, 2, B−C b− c, A, 3. tan, =, cot, 2, b+ c, 2, , Example 5. If k be the perimeter of the ∆ ABC, then the, value of b cos2, (a) 2 k, , C, B, + c cos2 is, 2, 2, (b) k/2, , (c) 3 k/2, , (d) k, , C, B, + c cos2, 2, 2, b, c, = (1 + cos C) + (1 + cos B ), 2, 2, b c 1, a+ b+ c k, = + + ( b cos C + c cos B ) =, =, 2 2 2, 2, 2, , Solution (b) b cos2, , Example 6. In any ∆ABC, the value of, ( b + c ) cos A + (c + a) cos B + ( a + b) cos C = a + b + c, and ( b − a cos C ) tan A = a sin C is, (a) abc , b sin A, (b) a + b − c , sin B, (c) a − b − c , sinC, (d) a + b + c , a sin C, , Solution (d) b cos A + c cos A + c cos B + a cos B, + a cos C + b cos C, ( b cos A + a cos B) + ( c cos A + a cos C), + ( b cos C + c cos B), Using the projection formula, c + b + a, Using b = c cos A + a cos C from the projection formula, we, get, = ( c cos A + a cos C − a cos C) tan A, sin A , = c cos A , = c sin A, cos A, Using the sine formula, we get, sin A = ka and sin C = kc, k ≠ 0, Therefore, we have, sin A, = ( ck ) , = a sin C, k
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249, , Properties of Triangles, π, 3, , Example 7. In ∆ABC , a = 2 b and | A − B| = , then the, value of ∠ C is, (a) π /6, (c) π / 4, , (b) π /3, (d) π /2, , Example 8. If the sides of a ∆ABC satisfy 3a = b + c , then, the value of cot (B / 2) cot (C / 2) is, (a) 1, (b) 2, (c) 3, , (d) 4, , s ( s − c), B, C, s ( s − b), cot =, 2, 2, ( s − a) ( s − c) ( s − a) ( s − b), a+ b+, s, a+ b+ c, , =2 ⇒, =2, since, s =, , 2, s−a, b+ c−a, a + b + c = 2b + 2c − 2a ⇒ 3a = b + c, , Solution (b) cot, , ..(i), a = 2b, π, ⇒, ∠ A > ∠ B and| A − B| =, 3, π, ...(ii), A−B=, ⇒, 3, From Napier's Analogy,, C, A − B a − b, (Q ∠ A > ∠ B), tan , cot, =, 2 a+ b, 2, π 2b − b, C, ⇒, tan =, cot, 6 2b + b, 2, C, 1, 1, C, = cot ⇒ cot = 3, ⇒, 2, 2, 3 3, π, ∠C π, =, ⇒∠C =, ⇒, 3, 2, 6, , Solution (b) Q, , ⇒, ⇒, , =2, c, , , , Circumcircle, The circle passing through the vertices of the ∆ABC is, called the circumcircle. Its radius R is called the, circumradius. In the ∆ABC,, A, , O, , Trigonometric Ratios of, Half-angles of any Triangle, A, =, 2, B, 2. sin =, 2, C, 3. sin =, 2, A, 4. cos =, 2, B, 5. cos =, 2, C, 6. cos =, 2, 1. sin, , ( s − b) ( s − c), bc, ( s − c) ( s − a ), ca, ( s − a ) ( s − b), ab, s (s − a), bc, s ( s − b), ca, s ( s − c), ab, , 7. tan, , A, ( s − b) ( s − c), =, 2, s (s − a), , 8. tan, , B, ( s − a ) ( s − c), =, 2, s ( s − b), , 9. tan, , C, ( s − a ) ( s − b), =, 2, s ( s − c), , 1, ab sin C, 2, 1, 3. ∆ = ca sin B, 2, 5. ∆ =, , a 2 sin B sin C, 2 sin A, , B, , a, 2 sin A, c, 3. R =, 2 sin C, , 1. R =, , C, , b, 2 sin B, abc, 4. R =, 4∆, , 2. R =, , Incircle, The circle touching the three sides of the triangle, internally is called the inscribed or the incircle of the, triangle. Its radius r is called the inradius of the circle. In, the ∆ABC,, A, , O, r, , ∆, 1. r =, s, , B, , 1, bc sin A, 2, c2 sin A sin B, 4. ∆ =, 2 sin C, , 2. ∆ =, , 6. ∆ =, , b2 sin C sin A, 2 sin B, , C, , A, B, C, = ( s − b) tan = ( s − c) tan, 2, 2, 2, B, A, B, C, A, C, c sin sin, a sin sin, b sin sin, 2, 2, 2, 2, 2, 2, 3. r =, =, =, C, A, B, cos, cos, cos, 2, 2, 2, A, B, C, 4. r = 4R sin sin sin, 2, 2, 2, r, 5. cos A + cos B + cos C = 1 +, R, 2. r = ( s − a ) tan, , Area of a Triangle, 1. ∆ =, , R
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250, , NDA/NA Mathematics, , Given that, b = 2, c = 3, ∠A = 30 ° , then the, radius of incircle of ∆ ABC is, 1− 3, (a), (b) 3, 2, 3 −1, (c), (d) 3 / 2, 2, , Example 9., , Solution (c) Q b = 2, C = 3 , ∠A = 30 °, ∴, , a = b 2 + c2 − 2bc cos A, , 3, =1, 2, A b+ c−a, A, We know that, r = ( s − a) tan =, tan, 2, 2, 2, 3 +1, =, tan 15°, 2, 3 +1, 3 −1, 3 −1, =, =, ×, 2, 2, 3 +1, = 4 + 3 − 2 ⋅2 ⋅ 3 ⋅, , Escribed Circle, , r, ( a + b + c), 2, r, ∆, = (2s) = rs = s = ∆, s, 2, =, , Solution of a Right Angled, Triangle, (i) When two sides are given Let the triangle be, right angled at C. Then, we can determine the remaining, elements as given in the following table., Given, a, b, c, a, , Required, a, a, , B = 90 ° − A , c =, b, sin A, a, sin A = , b = c cos A, B = 90 ° − A, c, tan A =, , (ii) When a side and an acute angle are, given In this case, we can determine the remaining, elements as given in the following table., , The circle touching BC and the two sides AB and AC, produced of ∆ ABC externally is called the escribed circle, opposite A. Its radius is denoted by r1., Similarly, r2 and r3 denoted the radii of the escribed, circles opposite angles Band C, respectively and r1 , r2 , r3 are, called the exradii of ∆ ABC., , Given, , a, A, c, A, , Required, , a, sin A, B = 90° − A, a = c sin A, b = c cos A, , B = 90° − A, b = a cot A, c =, , Solution of a Triangle in General, B, r1, A, , O1, , C, , B, C, ∆, A, A, 1. r1 =, = s tan = 4R sin cos cos, 2, 2, 2, 2, s−a, 2. r2 =, , B, B, C, A, ∆, = s tan = 4R sin cos cos, s−b, 2, 2, 2, 2, , 3. r3 =, , ∆, C, C, A, B, = s tan = 4R sin cos cos, 2, 2, 2, 2, s−c, , 4. r1 + r2 + r3 = 4R + r, 5. r1r2 + r2r3 + r3r1 = s2 =, , r1r2r3, r, , Example 10. In any ∆ABC, the value of, (a) ∆, (c) 2∆, , Rr(sin A + sin B + sin C ) is, ∆, (b), 2, (d) 4∆, , Solution (a) Rr (sin A + sin B + sin C), b, c, a, = Rr , +, + , 2R 2R 2R , , (i) When three sides a, b and c are given In, this case, the remaining elements are determined by, using, the, following, formulae,, ∆ = s( s − a )( s − b)( s − c) , where, 2s = a + b + c =, perimeter of triangle, 2∆, 2∆, 2∆, , sin B =, , sinC =, ,, sin A =, bc, ac, ab, A, ∆, B, ∆, , tan =, ,, tan =, 2 s( s − a ), 2 s( s − b), C, ∆, tan =, 2 s( s − c), (ii) When two sides a, b and the included ∠C, are given In this case, we use the following, formulae,, 1, ∆ = ab sin C, 2, A− B a− b, C A+ B, C, tan, =, cot ;, = 90° −, 2, a+b, 2, 2, 2, a sin C, and, c=, sin A, (iii) When one sides a and two angles A and B, are given In this case, we use the following, formulae to determine the remaining elements, A + B + C = 180°
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251, , Properties of Triangles, ⇒, , C = 180° − A − B, 1, a sin B, a sin C, and c =, ⇒ ∆ = ca sin B, b=, 2, sin A, sin A, , (iv) When two sides a, b and the ∠ A opposite, to one side is given In this case, we use the, following formulae,, b, ...(i), sin B = sin A, a, a sin C, C = 180° − ( A + B), c =, sin A, , Special Cases, Case I When A is an acute angle, (a) If a < b sin A, there is no triangle. When a < b sin A,, then Eq. (i), sin B > 1; which is impossible., C, , a, b, , B, , b sin A, , (ii) a > b in the following figure Let AC = b,∠CAX = A, and a > b, also a > b sin A., C, a, , b sin A, , A, , A, , B', , a, , b, , N, , X, , B, , Now, taking C as centre , if we draw an arc of radius, a, it will intersect AX at one point B and hence only, one ∆ABC is constructed. Also, this arc will intersect, XA produced at B′ and ∆AB′ C is also formed but this, ∆ is inadmissible (because ∠CAB′, is an obtuse angle, in this triangle ), Hence, if a > b and a > b sin A, then only one triangle, is possible., (iii) b > a ( i.e., b > a > b sin A), In figure let AC = b, ∠CAX = A. Now taking C as, centre if we draw an arc of radius a, then it will, intersect AX at two points B1 and B2, Hence, if b > a > sin A,then there are two triangles., C, , A, A, , N, , C2, , X, , (b) If a = b sin A, then only one triangle is possible which, is right angled at B. When a = b sin A, then from sine, rule, sin B = 1, ∴ ∠B = 90° from figure. It is clear that, CB = a = b sin A, C, , b, , C1, , b sinA, , A, A, , a, , a, , N, , B2, , X, , B1, , Case II When A is an obtuse angle, In this case, there is only one triangle, if a > b, C, , b, a = b sin A, A, , A, , B, , N, , X, , Thus, in this case, only one triangle is possible which, is right angled at B., (c) If a > b sin A, then three possibilities will arise, (i) a = b in this case, from sine rule, sin B = sin A, ∴, B = A or B = 180° − A, But B = 180° − A ⇒ A + B = 180°, which is not, possible in a triangle., C, , A, A, , a, , 90°, , A, , a, , b, , b, , b sinA, , A, , X, , B, , Case III When b > a and B = 90°, Again, the circle with A as centre and, b as radius will cut the line only in one, point. So, only one triangle is possible, , B, , X, , ∴ In this case, we get ∠A = ∠B., Hence, if b = a > b sin A, then only one isosceles ∆ABC, is possible in which ∠A = ∠B., , Bc, b, , A, , Case IV When b ≤ c and B = 90°, The circle with A as centre and b as radius will not, cut the line in any point. So, no triangle is possible., This is sometimes called an ambiguous case., , b sinA, N, , C, , B, , c, b, , A
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252, , NDA/NA Mathematics, , Example 11. If A = 30 °, a = 7, b = 8 in ∆ABC, then B has, (a) one solution, (c) no solution, , (b) two solution, (d) None of these, , Cyclic Quadrilateral, A cyclic quadrilateral is a quadrilateral which can be, circumscribed by a circle., , Solution (b) Here, b sin A = 8 sin 30 ° = 4, a = 7 thus, we have,, b > a > b sin B. Hence, ∠B has two solutions., , π, 3, , %, , Sum of the opposite angles of a cyclic quadrilateral is 180°., , %, , In a cyclic quadrilateral sum of the products of the opposite sides, is equal to the product of the diagonals. This is known as, Ptolemy's theorem., , %, , If sum of the opposite sides of a quadrilateral is equal, then and, only then a circle can be inscribed in the quadrilateral., , Example 12. If b = 3, c = 4 and B = , then the number of, triangle that can be constructed is, (a) infinite, (b) two, (c) one, (d) nil, , Solution (d) Hence, c sin B = 4 sin π / 3 = 2 3 > b ( = 3), Thus, we have b< sinB ⋅ C, Hence, no triangle is possible i.e., the number of triangles, that can be constructed is nil., , Example 13. In a ∆ ABC, if a sin A = b sin B, then the, nature of the triangle is, (a) a > b, (b) a < b, , (c) a = b, , (d) a + b = c, , Solution (c) Given, a sin A = b sin B, ⇒, , a ( ak) = b ( bk), a2 = b 2, , ⇒, , a=b, , (by sine rule), , Example 14. We are given b, c and sinB such that B is, acute and b < c sin B, then, (a) no triangle is possible, (b) one triangle is possible, (c) two triangle are possible, (d) a right angled triangle is possible, , Regular Polygon, A regular polygon is a polygon which has all its sides as, well as its angles equal. If the polygon has ‘n’ sides, sum of, ( n − 2) π, its internal angle is ( n − 2) π and each angle is, ., n, %, , %, , Sum of the exterior angles of a polygon taken in one direction, (clockwise or anti-clockwise) remains constants and it is equal to, 360°., In the regular polygon the centroid, the circumcentre and the, incentre are same., , Example 16. The area of a circle is A1 and the area of a, regular pentagon inscribed in the circle is A2. Then, the value, of A1 : A2 is, π, π, 2π, π, (a) sec, (b), sec, 5, 10, 5, 10, 2π, π, (c), (d) None of these, sec, 5, 5, , Solution (b) In the ∆ OAB, OA = OB = r, D, , Solution (a) Q By sine rule ;, sin B sin C, =, b, c, c, sin C = sin B > 1, ⇒, b, (Q b < c sin B), which is impossible, Hence, no triangle is possible., , Example 15. If in a ∆ABC a, b, c and ∠A is given and, c sin A < a < c , then, (a) b1 + b 2 = 2c cos A, (b) b1 + b 2 = c cos A, (d) b1 + b 2 = 4 c sin A, (c) b1 + b 2 = 3 c cos A, Solution (a) From cosine formula, cos A =, b 2 − (2c cos A) b + (c 2 − a 2) = 0,, equation in b,, , b 2 + c 2 − a2, or, 2 bc, , which, , is, , quadratic, , Q c sin A < a < c, ∴ Two triangle will be obtained , but this is possible when two, values of third side are also obtained. Clearly, two values of, side b will be b1 and b2. Let these are roots of above equation., ∴ Sum of roots = b1 + b2 = 2c cos A., , E, , C, O, r, , A, , B, , ∴, , 360 °, = 72°, 5, 1, ar ( ∆AOB) = ⋅ rr ⋅ sin 72°, 2, 1, = r 2 cos18°, 2, , ⇒, , A2 = ar (pentagon ABCDEA) =, , and, , ∠AOB =, , 5 2, r cos18°, 2, A1 = πr 2, , and, ∴, , 2 πr 2, 5r cos18°, 2π, π, =, sec, 5, 10, , A1 : A2 =, , 2
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253, , Properties of Triangles, , Example 17. If in a ∆ABC, the value of cot A, cot B and, , Example 18. If in ∆ABC, , cot C are in AP, then a 2, b 2 and c 2 are in, (a) GP, (b) HP, (c) AP, (d) None of these, , sin3 A + sin3 B + sin3 C = 3 sin A sin B sin C ,, a b c, then the value of b c a is, , Solution (c) Qcot A, cot B and cotC are in AP., ∴, , 2 cot B = cot A + cot C, , ⇒, , 2 ( a2 + c2 − b 2) b 2 + c2 − a2 a2 + b 2 − c2, =, +, 2ac ⋅ kb, 2bc ⋅ ka, 2ab ⋅ kc, , ⇒, , 2 ( a2 + c2 − b 2) = b 2 + c2 − a2 + a2 + b 2 − c2, , ⇒, , 2 ( a2 + c2 − b 2) = 2b 2, , ⇒, , a +c −b =b, , ⇒, , 2, , 2, , 2, , c, (a) −1, , a, , b, , (b) 1, , (c), , 1, 2, , (d) 0, , 1 1 1, , a b c, , Solution (d) b c a = ( a + b + c) b c a, c, , a b, , c a b, , = ( a + b + c) ( bc + ca + ab − a2 − b 2 − c2), , 2, , = − ( a3 + b3 + c3 − 3abc), , a2 + c 2 = 2 b 2, , = − 8R3(sin3 A + sin3 B + sin3 C − 3 sin A sin B sin C), , Then, a2, b 2 and c2 are in AP., , = − 8R3(3 sin A sin B sin C −3 sin A sin B sin C) = 0, , Comprehensive Approach, A, n, , n, , n, n, n, n, n, n, n, , n, , n, , n, , n, , n, n, n, , n, , n, , n, , 3, , then the triangle is equilateral., 2, 3 3, If sin A + sinB + sinC =, , then the triangle is equilateral., 2, 2, 2, 2, If cos A + cos B + cos C = 1, then the triangle is right angled., If tan A + tanB + tanC =3 3, then the triangle is equilateral., If cot A + cot B + cot C = 3, then the triangle is equilateral., sin A + sinB + sinC is maximum, when A = B = C, cos A + cosB + cosC is maximum, when A = B = C, tan A + tanB + tanC is minimum, when A = B = C, cot A + cot B + cot C is minimum, when A = B = C, A, B s−c, tan tan =, 2, 2, s, A, B, s, and cot cot =, 2, 2 s−c, A, B c, C, c, tan + tan = cot = ( s − c), 2, 2 s, 2 ∆, A, B a−b, tan − tan =, ( s − c), 2, 2, ∆, A, B, c, C, cot + cot =, cot, 2, 2 ( s − c), 2, If cos A + cosB + cosC =, , In an equilateral triangle, Circumcentre, Centroid and orthocentre are collinear., In any right angled triangle, the orthocentre coincides with the, vertex containing the right angled., The mid-point of the hypotenuse of a right angled triangle is, equidistant from the three vertices of the triangle., The mid-point of the hypotenuse of a right angled triangle is the, circumcentre of the triangle., The lenth of the medians AD,BE,CF of ∆ABC are given by, 1, 2 b 2 + 2 c 2 − a2, AD =, 2, 1 2, =, b + c 2 + 2 bc cos A, 2, , E, , F, G, B, , D, , C, , 1, 1 2, 2 c 2 + 2 a2 − b 2 =, c + a 2 + 2 ca ⋅ cosB, BE =, 2, 2, 1, 1 2, 2 a2 + 2 b 2 − c 2 =, a + b 2 + 2 ab ⋅ cosC, CF =, 2, 2, n, , The area of the quadrilateral is ( s − a) ( s − b) ( s − c) ( s − d ) ., c, , D, , C, , d, , b, , A, n, , n, , B, , a, , The length of the median AD, BE, CF of ∆ ABC are, 1, 2 b 2 + 2 c 2 − a2 ,, AD =, 2, 1, 1, 2 c 2 + 2 a 2 − b 2 and CF =, 2 a2 + 2 b 2 − c 2 ., BE =, 2, 2, Circumradius of a cyclic quadrilateral, R=, , 1 ( ac + bd ) ( ad + bc) ( ab + cd ), 4 ( s − a) ( s − b) ( s − c) ( s − d ), D, , c, C, , d, b, , A, , a, , B
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Exercise, Level I, 1. If in a ∆ABC , ( s − a ) ( s − b) = s ( s − c), then ∠C is, equal to, (a) 90°, (b) 45°, (c) 30°, (d) 60°, 2. In a ∆ABC, if 2s = a + b + c and ( s − b) ( s − c), A, = x sin2 , then the value of x is, 2, (a) bc, (b) ca, (c) ab, (d) abc, 3. If the angles of a ∆ABC be in AP, then, (a) c2 = a 2 + b2 − ab, (b) b2 = a 2 + c2 − ac, , 12. If a ∆ABC , ( b − c) sin A + ( c − a ) sin B + ( a − b) sin C is, , (c) a 2 = b2 + c2 − ac, (d) b2 = a 2 + c2, 4. In ∆ABC ,, , b+ c, is equal to, a, , 1, (B − C), 2, (a), 1, sin A, 2, 1, cos ( B + C ), 2, (c), 1, sin A, 2, cos, , 10. In a ∆ABC, if sin A : sin C = sin ( A − B) : sin ( B − C ),, then, (a) a , b, c are in AP, (b) a 2 , b2 , c2 are in AP, (c) a 2 , b2 , c2 are in GP, (d) None of the above, A, C 1, 11. In a ∆ABC,if tan tan = , then a , b and c are in, 2, 2 2, (a) AP, (b) GP, (c) HP, (d) None of these, , 1, (B − C), 2, (b), 1, cos A, 2, 1, cos ( B + C ), 2, (d), 1, cos A, 2, sin, , 5. In a ∆ABC,, ( b + c) cos A + ( c + a ) cos B + ( a + b) cos C is equal to, (a) 0, (b) 1, (c) a + b + c, (d) 2 ( a + b + c), cos A cos B cos C, , then the, 6. In a ∆ABC, if, =, =, a, b, c, triangle is, (a) right angled, (b) obtuse angled, (c) equilateral, (d) isosceles, C, A, , 7. In a ∆ABC, 2 a sin2, + c sin2 is equal to, , 2, 2, (a) a + b − c, (b) c + a − b, (c) b + c − a, (d) a + b + c, 8. In a ∆ABC, if a = 2x, b = 2 y and ∠C = 120°, then the, area of the triangle is, (a) xy, (b) xy 3, (c) 3xy, (d) 2xy, 1, 1, 9. If a ∆ABC , ∠ C = 60°, then, is equal to, +, a+ c b+ c, 1, 2, (a), (b), a+ b+ c, a+ b+ c, 3, (d) None of these, (c), a+ b+ c, , (a) ab + bc + ca, (b) a 2 + b2 + c2, (c) 0, (d) None of these, b − c cos A, 13. In a ∆ABC ,, is equal to, c − b cos A, sin B, cos C, (a), (b), sin C, cos B, cos B, (d) None of these, (c), cos C, sin B, , then the triangle is, 14. In a ∆ABC , cos A =, 2 sin C, (a) equilateral, (b) isosceles, (c) right angled, (d) None of these, 15. If ABCD is a cyclic quadrilateral, then what is the, value of sin A + sin B − sin C − sin D?, (NDA 2012 I), (a) 0, (b) 1, (c) 2, (d) 2 (sin A + sin B), 16. In a ∆ABC, if (sin A + sin B + sin C ), (sin A + sin B − sin C ) = 3 sin A sin B, then the ∠C is, equal to, π, π, (a), (b), 2, 3, π, π, (d), (c), 4, 6, A, C, 17. If in a ∆ABC , a + c = 2b, then the value of cot ⋅ cot, 2, 2, is equal to, (a) 4.5, (b) 3, (c) 1.5, (d) 1, 18. The median AD of a ∆ABC is bisected at F and BF is, produced to meet the side AC in P. If AP = λ AC,, then what is the value of λ?, 1, 1, (a), (b), 4, 2, 2, 1, (d), (c), 3, 3
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255, , Properties of Triangles, , Level II, 1. If the sides of a triangle are in the ratio 2 : 6 : 1 + 3,, then what is the smallest angle of the triangle?, (a) 75°, (b) 60°, (NDA 2011 II), (c) 45°, (d) 30°, 2. In a ∆ABC, a = 8, b = 10 and c = 12. What is ∠C equal, to?, (NDA 2011 II), (a) A / 2, (b) 2A, (c) 3 A, (d) 3 A / 2, 3. The sides a, b, c of a ∆ABC are in arithmetic, progression and ‘a’ is the smallest side. What is cos A, equal to?, (NDA 2011 II), 3c − 4b, 3c − 4b, (a), (b), 2c, 2b, 4c − 3b, 3b − 4c, (d), (c), 2c, 2c, 4. ABC is a triangle in which BC = 10 cm, CA = 6 cm and, AB = 8 cm. Which one of the following is correct?, (a) ABC is an acute angled triangle., (NDA 2010 II), (b) ABC is an obtuse angled triangle., (c) ABC is a right angled triangle., (d) None of the above, 5. If angles of a triangle are in the ratio 1 : 2 : 3, then, what is the ratio of its corresponding sides?, (a) 3 : 2 : 1, (b) 1 : 2 : 3 (NDA 2009 II), (c) 1 : 3 : 2, , (d) 2 : 3 : 4, , 6. If in a ∆ABC, cos B = (sin A) / ( 2 sin C ), then the, triangle is, (a) isosceles triangle, (b) equilateral triangle, (c) right angled triangle (d) scalene triangle, 7. For finding the area of a ∆ ABC, which of the, following entities are required?, (NDA 2009 I), (a) Angles A, B and side a., (b) Angles A, B and side b., (c) Angles A, B and side c., (d) Either (a) or (b) or (c)., 8. ABC is a triangle in which AB = 6 cm, BC = 8 cm and, CA = 10 cm. What is the value of cot ( A / 4)?, (a) 5 − 2, (b) 5 + 2, (NDA 2008 II), (d) 3 + 1, (c) 3 − 1, 9. If median of the ∆ABC through A is perpendicular to, BC, then which one of the following is correct?, (NDA 2007 II), , (a) tan A + tan B = 0, (c) tan C + 2 tan A = 0, , (b) tan B − tan C = 0, (d) tan B + tan C = 0, , 10. In a ∆ABC,if a = 2b and A = 3B,then which one of the, following is correct?, (NDA 2007 I), (a) The triangle is obtuse angled., (b) The triangle is acute angled but not right, angled., (c) The triangle is right angled., (d) The triangle is isosceles, but not obtuse angled., , 3, 11. In a ∆ABC , a = 5, b = 7 and sin A = , then the, 4, number of possible triangles is, (a) 1, (b) 0, (c) 2, (d) infinite, 12. In a ∆ABC, if a = 2 , b = 4, and ∠ C = 60°, then ∠ A, and ∠ B are equal to, (a) 90° , 30°, (b) 60° , 60°, (c) 30° , 90°, (d) 60° , 45°, 13. In a ∆ABC, if b2 + c2 = 3a 2, then cot B + cot C − cot A, is equal to, ab, (a) 1, (b), 4∆, ac, (c) 0, (d), 4∆, A, 14. In a ∆ABC , ( b + c − a ) tan is equal to, 2, 2∆, ∆, (a), (b), s, s, ∆s, s, (d), (c), R, bc, a, 15. If cos2 A + cos2 C = sin2 B, then ∆ABC is, (a) equilateral, (b) right angled, (c) isosceles, (d) None of these, 16. If the angles of a triangle be in the ratio 1 : 2 : 7, then, the ratio of its greatest side to the least side is, (a) 1 : 2, (b) 2 : 1, (c) ( 5 + 1) : ( 5 − 1), (d) ( 5 − 1) : ( 5 + 1), 17. In a triangle, the lengths of the two larger sides are, 10 cm and 9 cm, respectively. If the angles of the, triangle are in AP, then the length of the third side in, cm can be, (a) 5 − 6 only, (b) 5 + 6 only, (c) 5 − 6 or 5 + 6, (d) Neither 5 − 6 nor 5 + 6, 18. In a ∆ABC, if ∠ A = 45° , ∠ B = 75°, then a + c 2 is, equal to, (a) 0, (b) 1, (c) b, (d) 2b, 19. In a ∆ABC,if 8R 2 = a 2 + b2 + c2, then the triangle is, (a) right angled, (b) equilateral, (c) acute angled, (d) obtuse angled, 20. If the sides of a triangle be ( x 2 + x + 1), ( 2x + 1) and, ( x 2 − 1), then the greatest angle is, (a) 105°, (b) 120°, (c) 135°, (d) None of these
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256, , NDA/NA Mathematics, , 21. In a ∆ABC,, a3 cos ( B − C ) + b3 cos (C − A) + c3 cos ( A − B), is equal to, (a) abc, (b) 3 abc, (c) a + b + c, (d) None of these, 22. In a ∆ABC, a 2(cos2 B − cos2 C ) + b2(cos2 C − cos2 A), + c2(cos2 A − cos2 B) is equal to, (a) 0, (b) 1, (d) 2 ( a 2 + b2 + c2 ), (c) a 2 + b2 + c2, 23. In a ∆ABC ,, (a), , a−b, a−c, , 1 + cos ( A − B) cos C, is equal to, 1 + cos ( A − C ) cos B, a+b, a 2 − b2, a 2 + b2, (b), (c) 2, (d), a+c, a − c2, a 2 + c2, sin 3B, is equal to, sin B, c2 − a 2, (b), ca, , 24. If in a ∆ABC , 2b2 = a 2 + c2 , then, (a), , c2 − a 2, 2ca, , c2 − a 2 , , (c) , ca , , 2, , c2 − a 2 , , (d) , 2ca , , 2, , 25. If in a ∆PQR, sin P , sin Q and sin R are in AP, then, (a) the altitudes are in AP, (b) the altitudes are in HP, (c) the medians are in GP, (d) the medians are in AP, 26. If x , y and z are perpendicular drawn from the, vertices of triangle having sides a , b and c, then the, bx cy az, value of, will be, +, +, c, a, b, a 2 + b2 + c2, a 2 + b2 + c2, (a), (b), R, 2R, a 2 + b2 + c2, 2 ( a 2 + b2 + c2 ), (c), (d), 4R, R, 27. If for a ∆ABC, 1 + cos 2 A + cos 2 B + cos 2 C = 0,, then the triangle must be, (a) equilateral, (b) isosceles, (c) right angled, (d) obtuse angled, 28. ABC is a triangle which, BC cos A = AB cos C , then, (a)AB2 + AC 2 = BC 2, , is, , not, , isosceles,, , (b)AC 2 + BC 2 = AB2, (c) Area of ∆ABC =, , 1, 2 2, , ( AB) ( BC ), , (d)AB2 + BC 2 = AC 2, 29. In a ∆ABC, if the angles, A, B and C are in AP, then, which one of the following in correct? (NDA 2012 I), (a) c = a + b, (b) c2 = a 2 + b2 − ab, (c) a 2 = b2 + c2 − bc, (d) b2 = a 2 + c2 − ac, , 30. In a ∆ABC , b2 = c2 + a 2, then what is the value of, tan A + tan C ?, (a) tan B, (b) tan A ⋅ tan C, b, b2, (c), (d), ac, ac, 31. ABCD is rhombus and K is the middle point of the side, AB, CK is perpendicular to AB. What is the value of, ∠A?, (a) 120°, (b) 105°, (c) 90°, (d) 60°, 32. Consider the following statements, I. In a ∆ABC , a = 3 + 1, ∠ B = 30°, ∠ C = 45°, then, c is equal to 2., II. In a triangle, if a 2 + b2 + c2 = 8R 2, then the, triangle is right angled., 7, III. In a ∆ABC , a = 2, b = 3, c = 4, then cos A = ., 8, Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Only III, (d) All of these, 33. Consider the following statements, I. If the length of the median AD , BE and, CF of ∆ ABC, then, 3, AD 2 + BE 2 + CF 2 = ( a 2 + b2 + c2 ), 4, II. The area of the quadrilateral is, s(s − a ) (s − b)(s − c)(s − d ), , Which of the given statements given above is/are, correct?, (a) Only I, (b) Only II, (c) I and II, (d) Neither I nor II, 34. Considor the following statements, I. If r1 , r2 and r3 are the exradii of ∆ABC, then, 1 1 1, + + = r., r1 r2 r3, , II. If r1 = r2 + r3 + r, then the ∆ABC is an equilateral, triangle., Which of the given statements given above is/are, correct?, (a) Only I, (b) Only II, (c) I and II, (d) Neither I nor II, 35. Consider the following statements, I. In an equilateral triangle, the inradius,, circumradius and one of the exradii are in the, ratio 1 : 2 : 3, II. In ∆ABC, the cosine law define as, cos A cos B cos C, =, =, a, b, c, Which of the given statements given above is/are, correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II
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257, , Properties of Triangles, , Directions (Q. Nos. 36-39), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c), and (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not the, correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 36. Assertion (A) If the sides of the triangle are 13,14, 65, and15, then the radius of the circumcircle is ., 8, Reason (R) If a , b and c are the sides of the triangle,, abc, then the radius of circumcircle is given by, ., 4∆, C, A + B, 37. Assertion (A) In a ∆ABC, tan , = cot ., 2 , 2, Reason (R) A + B + C = 180°, and, , tan ( 90° − θ ) = cot θ., , 38. Assertion (A) The sides of a ∆ABC are a, b and, a 2 + ab + b2 , then the greatest angle is 120°., Reason (R) If one angle of a triangle is greater than, 90°, then the triangle is an obtuse triangle., , cos B cos C, , then the, =, b, c, triangle is an equilateral triangle., , 39. Assertion (A) If in a ∆ABC,, , Reason (R) In an equilateral ∆ABC, the length of all, sides are equal as well as each angle is 60°., , Directions (Q. Nos. 40-44), , If A, B, C are in AP, , and b : c = 3 : 2, then, 40. Measure of ∠C, (a) 30°, (c) 60°, , (b) 45°, (d) 90°, , 41. Measure of ∠A, (a) 75°, (c) 35°, , (b) 105°, (d) 180°, , 42. Ratio between the sides i.e., a : b : c, (a) 1 : 3 : 2, (b) (1 + 3) : 3 : 2, 1+ 3, (c) 1 : 3 : 2, (d), : 3: 2, 2, 43. ∆ABC is known as, (a) a right angle triangle, (b) an isosceles triangle, (c) a scalene triangle, (d) None of the above, 44. The value of cos2 A + cos2 B + cos2 C is, (a) 0, (b) 1, 5− 3, 3− 5, (c), (d), 4, 4, , Answers, Level I, 1. (a), 11. (d), , 2. (a), 12. (c), , 3. (b), 13. (b), , 4. (a), 14. (b), , 5. (c), 15. (a), , 6. (c), 16. (b), , 7. (b), 17. (b), , 8. (b), 18. (d), , 2., 12., 22., 32., 42., , 3., 13., 23., 33., 43., , 4., 14., 24., 34., 44., , 5., 15., 25., 35., , 6., 16., 26., 36., , 7., 17., 27., 37., , 8., 18., 28., 38., , 9. (c), , 10. (b), , Level II, 1., 11., 21., 31., 41., , (c), (b), (b), (a), (a), , (b), (c), (a), (d), (d), , (c), (c), (d), (a), (c), , (c), (a), (d), (d), (c), , (c), (b), (b), (a), , (a), (c), (a), (a), , (c), (c), (c), (a), , (b), (d), (d), (a), , 9., 19., 29., 39., , (b), (a), (d), (d), , 10., 20., 30., 40., , (c), (b), (d), (b)
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Hints & Solutions, Level I, 1. We have, tan, , C, =, 2, , (s − a ) (s − b), s (s − c), , Q (s − a ) (s − b) = s (s − c), C, s (s − c), tan =, ∴, 2, s (s − c), C, π, ⇒, tan = tan ⇒ C = 90°, 2, 4, A, (s − b) (s − c), 2. We have, sin, =, 2, bc, 2 A, = (s − b) (s − c), bc sin, ⇒, 2, A, On comparing with x sin 2 = (s − b) (s − c), 2, Hence, x = bc, , (given), , 3. Since, angles A , B and C are in AP., ∴, 2B = A + C, Q, A + B + C = 180° ⇒ B = 60°, a 2 + c2 − b2, ∴, cos B =, 2ac, 1 a 2 + c2 − b2, ⇒, =, 2, 2ac, 2, 2, 2, ⇒, a + c − b = ac, sin A sin B sin C, 4. Q, =, =, a, b, c, B − C, B + C, sin , , cos , 2 , 2 , b + c sin B + sin C, ∴, =, =, A, A, sin A, a, cos, sin, 2, 2, B −C, cos, 2, =, A, sin, 2, [Q A + B + C = π ⇒ A = π − (B + C )], 5. (b + c) cos A + (c + a ) cos B + (a + b) cos C, = b cos A + c cos A + c cos B + a cos B, + a cos C + b cos C, = (b cos A + a cos B) + (c cos A + a cos C ), + (c cos B + b cos C ), =c+, cos A, 6., =, cos B, ⇒, ⇒, ⇒, , b+ a, a sin A, =, b sin B, cos A sin B = sin A cos B, sin ( A − B) = 0, A=B, , Similarly, A = B = C. Hence, it is an equilateral triangle, , C, A, , 7. 2 a sin 2 + c sin 2 , , 2, 2, (s − b) (s − c), (s − a ) (s − b), = 2 a, +c, , , , ab, bc, , (s − b), (s − a + s − c), =2 , , b, 2, = (s − b) b = 2 (s − b) = a − b + c, b, 8. We have, a = 2x, b = 2 y and ∠ C = 120°, Area of triangle, 1, 1, ∆ = ab sin C = × 2x × 2 y × sin (90° + 30° ), 2, 2, = xy 3, a 2 + b2 − c2, 2ab, a 2 + b2 − c2, ⇒ a 2 + b2 − c2 = ab, ⇒ cos 60° =, 2ab, ⇒, b2 + bc + a 2 + ac = ab + ac + bc + c2, ⇒, b (b + c) + a (a + c) = (a + c) (b + c), On dividing by (a + c) (b + c) and add 2 on both sides, we, get, b, a, 1+, +1+, =3, a+c, b+ c, 1, 1, 3, ⇒, +, =, a+c b+ c a + b+ c, , 9. Since, cos C =, , 10. We have,, ⇒, ⇒, ⇒, ⇒, , sin A sin ( A − B), =, sin C sin (B − C ), , sin (B + C ) sin (B − C ) = sin ( A + B) sin ( A − B), sin 2 B − sin 2 C = sin 2 A − sin 2 B, 2, 2 sin B = sin 2 A + sin 2 C, sin A sin B sin C , =, =, 2b2 = a 2 + c2, Q, , , a, b, c , , Hence, a 2, b2 and c2 are in AP., A, C 1, 11. We have, tan, tan =, 2, 2 2, (s − b) (s − c) (s − a ) (s − b) 1, ⇒, =, s (s − a ), s (s − c), 2, s−b 1, =, ⇒, s, 2, ⇒, 2 s − 2b − s = 0, ⇒, a + c − 3b = 0, 12. (b − c) sin A + (c − a ) sin B + (a − b) sin C, = (b − c) ak + (c − a ) bk + (a − b) kc, = k [ab − ac + bc − ab + ac − bc] = 0
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259, , Properties of Triangles, b2 + c2 − a 2, b2 + a 2 − c2 c, 2b, =, 2, 2, 2, b2 + c2 − a 2, c + a − b b, c−, 2c, b2 + a 2 − c2, 2ac, cos C, =, ⋅ 2, =, 2ab, c + a 2 − b2 cos B, , b − c cos A, 13., =, c − b cos A, , 14., , b−, , sin B, 2 sin C, b2 + c2 − a 2 b, ⇒, =, 2bc, 2c, ⇒, b2 + c2 − a 2 − b2 = 0 ⇒ c2 = a 2 ⇒ c = a, ∴The triangle is an isosceles., cos A =, , 15. Since, ABCDs is a cyclic quadrilateral., Then, A + C = 180°, and B + D = 180°, (by property), Then, sin A + sin B − sin C − sin D, = sin (180° − C ) + sin (180° − D ) − sin C − sin D, = sin C + sin D − sin C − sin D, =0, a, b, c, 16. Q, (say), =, =, =k, sin A sin B sin C, ⇒, (a + b + c) (a + b − c) = 3ab, ∴, (sin A + sin B + sin C ) (sin A + sin B − sin C ), , = 3 sin A sin B, ⇒ a 2 + b2 + 2ab − c2 = 3ab, π, a 2 + b2 − c2 1, ⇒, = ⇒ cos C = cos, 3, 2ab, 2, π, ∠C =, ⇒, 3, 17. Given that, in a ∆ ABC , a + c = 2b, then, A, C, s (s − a ), s (s − c), cot ⋅ cot =, ., (s − b) (s − c), 2, 2, (s − a ) (s − b), 2s, s2, s, =, =, (s − b)2 s − b 2s − 2b, a+ b+ c, =, a + b + c − 2b, A, C 3b, =3, (Q a + c = 2 b), ⇒ cot cot =, 2, 2, b, =, , 18. We know that, if the median AD of a ∆ ABC, is bisected, at F and BF is produced to meet the side AC in P, then, AP : PC = 1 : 2, 1, ∴, AP = AC, 3, But, AP = λAC, 1, ∴, λ=, 3, , Level II, 1. Let a, b and c be the sides of ∆ABC , respectively., Given,, …(i), a : b : c = 2 : 6 : (1 + 3 ), We know that,, By sine law,, a, b, c, (say), =, =, =k, sin A sin B sin C, , ⇒, ⇒, Q, , Now, from Eq. (i), we get, k sin A : k sin B : k sin C = 2 : 6 : (1 + 3 ), ⇒, , sin A : sin B : sin C =, , 2, 1+ 3, :1 :, 3, 6, , Q, , ⇒, , sin A : sin B : sin C =, , 1, 3 1+ 3, 3, :, :, ×, 2, 2 2, 6, , ⇒, , =, , 1, 3 1+ 3, :, :, 2 2, 2 2, , = sin 45° : sin 60° : sin 75°, ⇒, A : B : C = 45° : 60° : 75°, So, the required smallest angle of ∆ ABC is 45°., 2. Given, in a ∆ABC , a = 8, b = 10 and c = 12, Then, by cosine law,, 2, 2, 2, a 2 + b2 − c2 (8) + (10) − (12), cos C =, =, 2ab, 2 ⋅ 8 ⋅ 10, ⇒, ⇒, , 64 + 100 − 144, 20, 1, =, =, 2 ⋅ 8 ⋅ 10, 2 ⋅ 8 ⋅ 10 8, 1, cos C =, 8, cos C =, , b2 + c2 − a 2 100 + 144 − 64, =, 2bc, 2 ⋅ 10 ⋅ 12, 180, 3, cos A =, =, 2 ⋅ 10 ⋅ 12 4, 3, cos A =, 4, cos 2 A = − 1 + 2 cos 2 A, 2, 9 1, 3, = −1 + 2 = −1 + =, 4, 8 8, 1 + cos A 1 + 3 /4 7, 2 A, cos, =, =, =, 2, 2, 2, 8, A, 7, =, cos, 2 2 2, A, C≠, 2, cos 3 A = 4 cos3 A − 3 cos A, , Now,, , ∴, Q, , cos A =, , …(i), , …(ii), , …(iii), , 3, , 3, 3, =4 −3 , 4, 4, 27 9 27 − 36 −9, =, − =, =, 16 4, 16, 16, ∴, C ≠ 3A, 3A, Q, cos 3 A = 1 − 2 cos 2, 2, 9 25, 9, 2 3A, ⇒ 2 cos, = 1 − − = 1 +, =, 16, 2, 16 16, 3 A 25, =, ⇒ cos 2, 2, 32, , …(iv)
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260, , NDA/NA Mathematics, 3A, 5, =, 2, 4 2, 3A, C≠, ∴, 2, So, the required relation is, cos C = cos 2 A, ⇒, C = 2A, ⇒, , 8. Here, a = 8, b = 10 and c = 6, , cos, , A, , …(v), b, , c, , [From Eqs. (i) and (ii)], B, , 3. Given, sides a, b and c of a ∆ABC are in AP. Then,, …(i), 2b = a + c, b2 + c2 − a 2, Q, cos A =, 2bc, b2 + c2 − (2b − c)2, cos A =, 2bc, [Q from Eq. (i), a = 2b − c ⇒ a 2 = (2b − c)2], b2 + c2 − 4b2 − c2 + 4bc, ⇒, cos A =, 2bc, 4bc − 3b2 4c − 3b, cos A =, =, ⇒, 2bc, 2c, 4c − 3b, ⇒, cos A =, 2c, , ∴, , ⇒, , Now,, , 4. Q CA 2 + AB2 = 36 + 64 = 100 = BC 2, ∴ ∆ABC is a right angled triangle., 5. Let the angles of a triangle be x, 2x and 3x, respectively, ∴, x + 2x + 3x = 180°, ⇒, x = 30°, ∴ Angles are 30°, 60°, 90°., By using sine rule,, a, b, c, =, =, sin 30° sin 60° sin 90°, a, b, c, ⇒, =, =, 1, 3 1, 2, 2, a, b, c, =, =, ⇒, 1, 3 2, 6. Using sine rule,, sin A sin B sin C, =, =, =k, a, b, c, sin A, Given,, cos B =, 2 sin C, , a 2 + c2 − b2, a 2 + c2 − b2 a, =, ⇒, Q cos B =, , 2ac, 2ac, 2c, , , ⇒ a +c −b =a, ⇒, b2 = c2, ⇒, b=c, Hence, it is an isosceles triangle., 2, , 2, , 2, , Let, , tan, , (12 − 10)(12 − 6), 1 1, =, =, 12 (12 − 8), 4 2, , A, =, 2, , , A, =, Q tan, 2, , A, cot, =2, 2, , where,, A + B + C = 180°, So, finding the area of ∆ABC , angles A, B and side c are, required., , (s − b)(s − c), 1, , and s = (a + b + c), , s(s − a ), 2, , A, −1, cot2, A A, 4, cot + =, A, 4, 4, 2 cot, 4, , cot A ⋅ cot B − 1 , Q cot ( A + B) = cot A + cot B , , , A, cot2, −1, A, 4, cot =, A, 2, 2 cot, 4, A, cot = x, 4, x2 − 1, ⇒ x2 − 4 x − 1 = 0, 2x, 4 ± 16 + 4, x=, 2, 4±2 5, x=, =2 ± 5, 2, A, cot = 2 + 5, 4, 2− 5, , ∴, , 2=, , ⇒, ⇒, So,, or, , A, , 9., , B, a/2, , 2, , 7. We know that, area of ∆ ABC whose sides are a, b and c, are, c2 sin A ⋅ sin B a 2 sin B ⋅ sin C b2 sin C ⋅ sin A, ∆=, =, =, 2 sin C, 2 sin A, 2 sin B, , C, , a, , D, , C, a/2, , a, , Let, , BC = a, , ∴, , BD = CD =, , a, 2, , In ∆ABD,, tan B =, , AD AD, 2 AD, =, ⇒ tan B =, BD a /2, a, , In ∆ACD,, tan C =, , AD AD, =, CD a /2, , …(i)
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261, , Properties of Triangles, , ⇒, , tan C =, , 2 AD, a, , …(ii), , From Eqs. (i) and (ii), we get, ⇒, , tan B = tan C, tan B − tan C = 0, , 10. We know that,, a, b, =, sin A sin B, 2b, b, ⇒, =, sin 3B sin B, , (by sine rule), (Q a = 2b and A = 3B), , ⇒, 2 sin B = sin 3B, ⇒, 2 sin B = 3 sin B − 4 sin3 B, ⇒, 0 = sin B − 4 sin3 B, ⇒, 0 = sin B (1 − 4 sin 2 B), ⇒, sin B = 0, or 1 − 4 sin 2 B = 0, when, sin B = 0, then B = 0°, and, A = 3B = 0°, which is not possible., and when, 1 − 4 sin 2 B = 0, 1, sin B = ±, ⇒, 2, ⇒, B = 30° and A = 90°, (Q A = 3B), ⇒ Given triangle is right angled triangle., 3, 11. Given that, a = 5, b = 7, sin A =, 4, sin A sin B, As, we know that, =, a, b, 21, 3, sin B, =, ⇒ sin B =, ⇒, 20, 4 ×5, 7, Which is not possible because its value is greater than, one., ∠C = 60° , a = 2 , b = 4, a 2 + b2 − c2, ⇒, cos C =, 2ab, ⇒, 2ab cos 60° = a 2 + b2 − c2, ⇒, ab = a 2 + b2 − c2, ⇒, 8 = 4 + 16 − c2, 2, ⇒, c = 12 ⇒ c = 12 = 2 3, 3, 2⋅, a sin C, 2 = 1 ⇒ A = 30°, We have, sin A =, =, c, 2, 2 3, 3, 4⋅, b sin C, 2 = 1 ⇒ B = 90°, and, sin B =, =, c, 2 3, cos B cos C, 13. Now, cot B + cot C − cot A =, +, − cot A, sin B sin C, sin C cos B + cos C sin B, =, − cot A, sin B sin C, sin (B + C ) cos A, =, −, sin B sin C sin A, sin 2 A − sin B sin C cos A, (Q A + B + C = π ), =, sin A sin B sin C, a 2 − bc cos A, =, k (abc), 12. Given,, , sin A sin B sin C, , , =, =, = k (say ), since,, a, b, c, , , b2 + c2 − a 2, , , , and cos A =, , , 2bc, 2, 2, 2, +, −, b, c, a, (, ), a 2 − bc, 2a 2 − (3a 2 − a 2), 2bc, =, =, (abc) k, 2 (abc) k, [Q b2 + c2 = 3a 2 (given )], (a 2 − a 2), =, =0, abc k, A, A, = (2s − 2a ) tan, 2, 2, (s − b) (s − c), = 2 (s − a ), s (s − a ), (s − a ) (s − b) (s − c) 2 ∆, =2, =, s, s, , 14. Now, (b + c − a ) tan, , 15. Given that, cos 2 A + cos 2 C = sin 2 B, Obviously, it is not an equilateral triangle because, A = B = C = 60° does not satisfy the given condition. But, we take, B = 90°, then sin 2 B = 1 and, , π, cos 2 A + cos 2 C = cos 2 A + cos 2 − A, , 2, = cos 2 A + sin 2 A = 1, Hence, this satisfies the condition, so it is a right angled, triangle but not necessarily isosceles triangle., 16. Let angles of a triangle are x, 2x and 7x, respectively, ∴, x + 2x + 7x = 180° ⇒ x = 18°, Hence, the angles are 18° , 36° , 126°, Greatest side ∝ sin (126° ), Smallest side ∝ sin (18° ), sin (126° ) sin (90° + 36° ), Ratio =, ∴, =, sin (18° ), sin (18° ), cos 36°, 5 +1, =, =, sin 18°, 5 −1, 17. We know that, in triangle larger side have an larger, angle. Since, angles ∠ A , ∠ B and ∠ C are in AP., ⇒, 2B = A + C Q A + B + C = π ⇒ B = 60°, a 2 + c2 − b2, ∴, cos B =, 2ac, 1 100 + a 2 − 81, ⇒, cos 60° = =, 2, 20a, ⇒, a 2 + 19 = 10a ⇒ a 2 − 10a + 19 = 0, 10 ± 100 − 76, ∴, a=, =5 ± 6, 2, 18. Given that, ∠ A = 45° , ∠ B = 75°, ⇒, ∠ C = 180° − 45° − 75° = 60°, a + c 2 = k (sin A + 2 sin C ), ∴, = k (sin 45° + 2 sin 60° ), 1, 1 + 3, 3, =k, + 2⋅ = k , , 2, 2 , 2, , 2 2b, b, b, and, k=, =, =, sin B sin 75°, 3+1, On putting the value of k in Eq. (i), we get, a + c 2 = 2b, , …(i)
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262, , NDA/NA Mathematics, , 19. 8R2 = a 2 + b2 + c2 = 4R2 (sin 2 A + sin 2 B + sin 2 C ), ⇒, , sin A + sin B + sin C = 2, , ⇒, , (cos 2 A − sin 2 C ) + cos 2 B = 0, , ⇒, , cos ( A − C ) cos ( A + C ) + cos 2 B = 0, , ⇒, , 2 cos B cos A cos C = 0, , ⇒, , cos A = 0 or cos B = 0 or C = 0, π, π, π, or B = or C =, A=, 2, 2, 2, , 2, , ⇒, , 2, , 2, , 24., , sin 3B 3 sin B − 4 sin3 B, = 3 − 4 sin 2 B, =, sin B, sin B, a 2 + c2 − b2, = 3 − 4 + 4 cos 2 B = − 1 + 4 , , 2 ac, , , a 2 + c2, , , 2 , = −1 +, (ac)2, , 20. Sides of a triangle are (x2 + x + 1), (2x + 1) and (x2 − 1)., The greatest side subtends the greatest angle. Hence,, x2 + x + 1 is the greatest side which makes an angle θ., (2x + 1)2 + (x2 − 1)2 − (x2 + x + 1)2, ∴, cos θ =, 2 (2x + 1) (x2 − 1), , , 4x2 + 1 + 4x + x4 + 1 − 2x2 − x4 − x2, , − 1 − 2x3 − 2x − 2x2 , cos θ = , 2 (2x + 1) (x2 − 1), , ⇒, ⇒, , = −1 +, , + sin 2 C {2 sin ( A + B) cos ( A − B)}, = k [sin A sin B (sin A cos B + cos A sin B), 3, , + sin B sin C (sin B cos C + cos B sin C ), + sin C sin A (sin C cos A + cos C sin A )], = k3 [sin A sin B sin C + sin B sin C sin A, + sin C sin A sin B], = 3 k3 sin A sin B sin C = 3abc, 22. a 2(cos 2 B − cos 2 C ) + b2(cos 2 C − cos 2 A ), , + c2 (sin 2 B − sin 2 A ), = k2a 2 (c2 − b2) + k2b2(a 2 − c2) + k2c2(b2 − c2) = 0, 1 + cos C cos ( A − B) 1 − cos ( A + B) cos ( A − B), ., 23., =, 1 + cos ( A − C ) cos B 1 − cos ( A − C ) cos ( A + C ), (Q A + B + C = π ), 1 − cos 2 A + sin 2 B sin 2 A + sin 2 B a 2 + b2, ⇒, =, =, 1 − cos 2 A + sin 2 C, sin 2 A + sin 2 C a 2 + c2, , 2, , 25. sin P , sin Q , sin R are in AP., ⇒, a , b, c are in AP., sin P sin Q sin R, =, =, =λ, Q, a, b, c, , (say), , P, , c, , b, p1, , Q, , a, , R, , Let p1 , p2, p3 be altitudes from P , Q , R., p1 = c sin Q = λbc,, p2 = a sin R = λac,, p3 = b sin P = λab,, Since, a , b, c in AP., 1 1 1, Hence, , , are in HP., a b c, abc abc abc, are in HP ⇒ bc, ac, ab are in HP., ⇒, ,, ,, a, b, c, ⇒ λbc, λac, λab are in HP ⇒ p1 , p2, p3 are in HP., 26. Let area of triangle be ∆, then according to question,, 1, 1, 1, ∆ = ax = by = cz, 2, 2, 2, bx cy az b 2∆ , c 2∆ a 2∆ , ∴, +, +, =, + + , c, a, b, c a a b b c, =, , = a 2(1 − sin 2 B − 1 + sin 2 C ) + b2(1 − sin 2 C − 1 + sin 2 A ), = a 2(sin 2 C − sin 2 B) + b2 (sin 2 A − sin 2 C ), , [Q2b2 = a 2 + c2 (given)], , (a 2 + c2)2 − 4a 2c2 c2 − a 2, =, =, , 4 (ac)2, 2ac , , + c2(cos 2 A − cos 2 B), + c2(1 − sin 2 A − 1 + sin 2 B), , 2, , (a 2 + c2)2, 4 (ac)2, , −2x3 − x2 + 2x + 1 − (x2 − 1) (2x + 1), =, =, 2 (2x + 1) (x2 − 1) 2 (2x + 1) (x2 − 1), 1, cos θ = − = cos 120°, 2, θ = 120°, , 21. a3 cos (B − C ) + b3 cos (C − A ) + c3 cos ( A − B), = k3 sin3 A cos (B − C ) + k3 sin3 B cos (C − A ), + k3 sin3 C cos ( A − B), a, b, c, , , Q, =, =, = k (say ), sin A sin B sin C, , 1, = k3 {sin 2 A (2 sin (B + C ) cos (B − C )}, 2, + sin 2 B {2 sin (C + A ) cos (C − A }, , 2, , =, , 2∆ (b2 + c2 + a 2) 2 (a 2 + b2 + c2) abc, =, ⋅, 4R, abc, abc, abc, , , Q ∆ =, , 4R , a 2 + b2 + c2, 2R, , 27. Since, 1 + cos 2 A + cos 2B + cos 2C = 0, ⇒ 1 + 2 cos 2 A − 1 + 2 cos 2 B − 1 + 2 cos 2 C − 1 = 0, ⇒, 2 cos 2 A + 2 cos 2 B + 2 cos 2 C − 2 = 0, ⇒, cos 2 A + cos 2 B + cos 2 C = 1, If ∆ABC is right angled triangle in which ∠A = 90°,, then, cos 2 90° + cos 2 B + cos 2 (90° − B), = 0 + cos 2 B + sin 2 B = 1, ∴The given triangle is right angled triangle.
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263, , Properties of Triangles, 28. In ∆ ABC ,, , D, , C, , A, , a, b, , c, , A, B, , a, , We have, BC cos A = AB cos C, ⇒, a cos A = c cos C, a 2 + b2 − c2, b2 + c2 − a 2, ⇒, a, , =c, 2bc, 2ab, , , , , ⇒, ⇒, ⇒, ⇒, ⇒, ⇒, ⇒, , a 2(b2 + c2 − a 2) = c2(a 2 + b2 − c2), a 2b2 + a 2c2 − a 4 = a 2c2 + b2c2 − c4, a 2b2 − a 4 = b2c2 − c4, c4 − a 4 = b2c2 − a 2b2, (c2 − a 2) (c2 + a 2) = b2(c2 − a 2), c2 + a 2 = b2, 2, AB + BC 2 = AC 2, , 29. Given that, angles A, B and C are in AP., i.e.,, , 2B = A + C, , ...(i), , Also, in ∆ ABC, A + B + C = 180°, ⇒, , 2B + B = 180°, , [from Eq. (i)], , B = 60°, From cosine law,, a 2 + c2 − b2, 1 a 2 + c2 − b2, cos B =, = cos 60° ⇒ =, 2ac, 2, 2ac, ⇒, , ac = a 2 + c2 − b2 ⇒ b2 = a 2 + c2 − ac, , 30. Given that, b = c + a 2, ∴, ∠ B is a right angled., ⇒, A + C = 90°, sin A sin C, +, tan A + tan C =, cos A cos C, sin A cos C + cos A sin C sin ( A + C ), ∴, =, =, cos A cos C, cos A cos C, 1, 1, =, =, cos A cos C b2 + c2 − a 2 a 2 + b2 − c2, ., 2bc, 2ab, 2, 4ab c, = 2, (b + c2 − a 2) (a 2 + b2 − c2), 4ab2c, = 2, 2, 2, (a + c + c − a 2) (a 2 + c2 + a 2 − c2), 2, , 2, , b2, 4ab2c, = 2, =, 2c × 2a 2 ac, , (Q b2 = c2 + a 2), , 31. Let a be the side of rhombus ABCD and K be the, a, mid-point of AB. Therefore, BK =, 2, In ∆ BCK ,, a2, a 2 = CK 2 +, 4, a 2 3a 2, 3a, 2, 2, ⇒, =, ⇒ CK =, CK = a −, 4, 4, 2, , B, , K, a/2, , C, , a/2, , Now, in ∆ BCK ,, CK, 3 a /2, 3, sin B =, =, =, BC, a, 2, ⇒, ∠ B = 60°, ∴ In rhombus ABCD,, ∠ A + ∠ B = 180°, (Q AD || BC ), ⇒, ∠ A = 180° − 60° ⇒ ∠ A = 120°, 32. I. Q A + B + C = 180°, ⇒, A = 180° − (45° + 30° ) = 105°, By sine law,, sin C, sin 45°, sin A sin C, ⋅a=, × ( 3 + 1), =, ⇒ c=, sin A, sin 105°, a, c, ( 3 + 1) 2 2, =, ×, =2, ( 3 + 1), 2, II. Q a 2 + b2 + c2 = 8R2, ⇒ (2R sin A )2 + (2R sin B)2 + (2R sin C )2 = 8R2,, (by sine rule), ⇒, 2R2{2 sin 2 A + 2 sin 2 B + 2 sin 2 C } = 8R2, ⇒ (1 − cos 2 A ) + (1 − cos 2B) + (1 − cos 2C ) = 4, ⇒, , cos 2 A + cos 2B + cos 2C = − 1, , ⇒, , 2 cos( A + B) ⋅ cos ( A − B) + cos 2 C = − 1, , ⇒, , cos (π − C ) ⋅ cos ( A − B) + cos 2 C = − 1, , (Q A + B + C = π ), ⇒, 2 cos C ⋅ cos ( A − B) + 2 cos 2 C = 0, ⇒, 2 cos C{cos ( A − B) + cos C } = 0, ⇒ 2 cos C{cos ( A − B) + cos ( A + B)} = 0, ⇒, 2 cos C ⋅ cos B ⋅ cos A = 0, If cos A = 0 ⇒ ∠A = 90°, if cos B = 0 ⇒ ∠B = 90° and if, cos C = 0 ⇒ ∠C = 90°, thus ∆ABC is right angled., III. By cosine law;, 7, b2 + c2 − a 2 9 + 16 − 4 21, =, ⇒ cos A =, =, cos A =, 8, 2bc, 2 ⋅3 ⋅4, 24, So, all statements are correct., 33. We know that, the length of the median AD , BE and CF, of ∆ABC are, 1, 1, AD =, 2b2 + 2c2 − a 2, BE =, 2c2 + 2a 2 − b2 and, 2, 2, 1, CF =, 2a 2 + 2b2 − c2, 2, Now, AD 2 + BE 2 + CF 2, 1, = {2b2 + 2c2 − a 2 + 2c2 + 2a 2 − b2 + 2a 2 + 2b2 − c2}, 4, 3 2, = (a + b2 + c2), 4, and the area of quadrilateral is, ∆ = (s − a )(s − b)(s − c)(s − d )
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264, , NDA/NA Mathematics, ∆, ∆, ∆, ∆, , r2 =, , r3 =, and r =, s− a, s− b, s− c, s, 1 1 1 s− a s− b s− c, Now,, + + =, +, +, r1 r2 r3, ∆, ∆, ∆, , 34. I. Q r1 =, , (Q 2s = a + b + c), 1, 1, s 1, {(3s) − (a + b + c)} = (3s − 2s) = =, ∆, ∆, ∆ r, II. Given, r1 = r2 + r3 + r, ⇒, r1 − r = r2 + r3, ∆, ∆, ∆, ∆, ⇒, − =, +, s− a s s− b s− c, s− s+ a s− c+ s− b, =, ⇒, s(s − a ) (s − b)(s − c), a, 2s − (b + c) 2s − (2s − a ), ⇒, =, =, s(s − a ) (s − b)(s − c) (s − b)(s − c), a, a, =, ⇒ s2 − sa = s2 − sb − sc + bc, ⇒, s(s − a ) (s − b)(s − c), ⇒, s(− a + b + c) = bc ⇒ (a + b + c)(b + c − a ) = 2bc, ⇒, (b + c)2 − a 2 = 2bc ⇒ b2 + c2 = a 2, ∴∆ABC is a right angled triangle., abc, 35. I. Circumradius, R =, 4∆, ∆, Inradius, r =, s, ∆, and exradii r1 =, s− a, and in equilateral triangle, a = b = c, 3, 3a, and, (a )2, s =, ∆=, 4, 2, abc ∆, ∆, : :, ∴ R : r : r1 =, 4∆ s s − a, a3, 4, 3a 2, 2, 3a 2 2, =, ×, :, ×, :, ×, 4, 4, 3a, 4, a, 3a 2, a, a a 3, 1, 1, 3, =, =, = 2 :1 :3, :, :, :, :, 2, 3 2 3, 3 2 3 2, ⇒, r : R : r1 = 1 : 2 : 3, II. We know that, cosine law is, b2 + c2 − a 2, a 2 + c2 − b2, cos A =, , cos B =, 2bc, 2ac, a 2 + b2 − c2, and cos C =, 2ab, cos A b2 + c2 − a 2 cos B a 2 + c2 − b2, ⇒, ,, ,, =, =, 2abc, 2abc, a, b, cos C a 2 + b2 − c2, =, 2abc, c, cos A cos B cos C, ≠, ≠, ∴, a, b, c, abc, 36. Circumradius, R =, …(i), 4∆, Here,, 2s = a + b + c = 13 + 14 + 15 = 42, ⇒, s = 21, ∆ 2 = s(s − a ) (s − b) (s − c) = 21 ⋅ 8 ⋅ 7 ⋅ 6, ∆ = 84, 13 ⋅ 14 ⋅ 15 65, R=, =, ∴, 4 ⋅ 84, 8, =, , 37. Q A + B + C = π, , A + B π C , A + B = π −C ⇒, = − , 2 2 2, C, π C, A + B, tan , = tan − = cot, 2 2, 2 , 2, , 38. cos C =, =, , a 2 + b2 − ( a 2 + ab + b2), 2ab, , 2, , 1, a 2 + b2 − (a 2 + ab + b2) − ab, = − = cos 120°, =, 2, 2ab, 2ab, , ⇒ ∠C = 120°, cos B cos C, 39. Q, =, b, c, cos B, cos C, =, ⇒, k sin B k sin C, , (greatest angle), A, , E, , F, G, , (by sine rule), B, ⇒ cot B = cot C ⇒ B = C ⇒ b = c, So, the triangle is an isosceles triangle., , D, , C, , Solutions (Q. Nos. 40-44), 40. Given, A,B and C are in AP., ⇒, 2B = A + C, In ∆ ABC , A + B + C = 180°, From Eqs. (i) and (ii), we get, 3B = 180° ⇒ B = 60°, Now, by sine rule;, sin B sin C, =, b, c, c, ⋅ sin B =, b, 2 3, sin C =, ⋅, =, 3 2, , ⇒, , sin C =, , ⇒, , …(i), …(ii), , b, 3, Q =, , 2, c, , 2, sin 60°, 3, 1, = sin 45° ⇒ ∠ C = 45°, 2, , 41. From Eq. (ii), we get, …(ii), A + B + C = 180°, A = 180°− (B + C ), A = 180° − (60°+ 45° ), ∠A = 180°−105° = 75°, sin A sin B sin C, 42. By sine rule,, =, =, a, b, c, a : b : c = sin A : sin B : sin C, sin 75° : sin 60° : sin 45°, 3 +1 3 1, = ( 3 + 1) : 6 : 2, :, :, =, 2, 2 2, 2, (1 + 3), : 3: 2, 2, 43. Since, all sides as well as all angles of ∆ ABC is different, or unequal., ∴∆ ABC is a scalene triangle., 44. cos 2 A + cos 2 B + cos 2 C = cos 2 75° + cos 2 60°+ cos 2 45°, 2, , 2, , 2, 3 − 1, (3 + 1 − 2 3 ) 1 1, 1, 1, =, + +, + + =, 2, 2, 4 ⋅2, 4 2, 2 2 , , =, , (2 − 3 ) 1 1 2 − 3 + 1 + 2 5 − 3, + + =, =, 4, 4 2, 4, 4
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14, , Height and, Distance, Angle of Elevation, , m-n Theorem, , If ‘ O ’ be the observer’s eye and OX be the horizontal, line through O. If the object P is at a higher level than eye,, then angle POX is called the angle of elevation., , If D divides BC in the ratio m : n, then, A, α β, , P, Line of sight, , B, , θ, , O, , Horizontal line, , m, , D, , C, m:n, , n, , C, , 1. ( m + n ) cot θ = n cot B − m cot C, 2. ( m + n ) cot θ = m cot α − n cot β, , Angle of Depression, If the object P is at a lower level than O, then angle, POX is called the angle of depression., Horizontal line, O, , θ, , B, , X, , Apollonius Theorem, A, , X, , θ, , Line of sight, , C, , P, , B, D, , If in ∆ ABC, AD is median, then, AB2 + AC 2 = 2 ( AD 2 + BD 2 ), , %, , a = h (cot α − cot β), A, , %, , α + β, If AB = CD, then x = y tan , , 2 , A, y, , h, D, , B, , β, , α, a, , D, , x, , C, , C, , α, x, , β, B, , E
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266, , %, , NDA/NA Mathematics, , h=, , %, , H sin (β − α), h cot α, and H =, cot α − cot β, cos α sin β, , H = x cot α tan (α + β), A, , A, , D, , α, , E, , B, , B, , %, , D, , β, , x, , β, , h, , hcot β, H =, cot α, , %, , C, , α, , C, , a sin (α + β), H =, sin (β − α), A, , D, , α, β, , B, , A, , h, , H, , E, h, , h, , α, C, , B, , C, , A, , Example 1. The angle of elevation of the Sun when the, length of the shadow of a pole is 3 times the height of the, pole is, (a) 30°, (b) 40°, (c) 45°, (d) None of these, , Solution (a) Let the length of pole = h m, , C, , x, , B, , x + 20 = h 3, ⇒, From Eqs. (i) and (ii),, x + 20 = 3x ⇒ x = 10 m, And from Eq. (i), h = 3 × 10 = 10 3 m, , In ∆ ABC,, C, , h, , ...(ii), , Example 3. The upper part of a tree broken over by the, , θ, , wind makes an angle of 30° with the ground and the distance, from the foot to where the top of the tree touches the ground is, 10 m. The height of the tree is, (a) 14 m, (b) 15 3 m, (c) 20 m, (d) None of these, , B, 3h, , tan θ =, , h, , D, , ∴ Length of the shadow of the pole = 3 h, , A, , H, , H, , BC, h, 1, =, =, = tan 30 °, AB, 3h, 3, θ = 30 °, , Example 2. A person standing on the bank of a river, , Solution (b) Let AB be the tree and BC be the broken part of the, tree, then BC takes the position CD. Let CD = BC = x m., , observes the angle subtended by a tree on the opposite bank, is 60° when he retreats 20 m away from the bank, he finds, the angle of tree be 30°. The height of the tree, (a) 14 3 m, , (b) 20 m, , (c) 10 3 m, , (d) None of these, , B, x, C, , x, , Solution (c) Let the height of the tree = h m, and breadth of the river = x m, h, In ∆ ABC,, tan 60 ° =, x, h= 3x, ⇒, h, Now in ∆ ABD, tan 30 ° =, x + 20, , h, , 30°, D, 15 m, , ...(i), , In ∆ ACD, we have, cos 30 ° =, , 15, x, , A
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267, , Height and Distance, , ⇒, and, ⇒, , 15, 30, =, = 10 3 m, cos 30 °, 3, h, tan 30 ° =, 15, h = 15 tan 30 ° = 5 3 m, x=, , BP = 100 cot 60 ° =, Hence,, , Hence, height of the tree = x + h = 10 3 + 5 3 = 15 3 m, , Example 4. A man on the top of a cliff 100 m high, observes the angle of depression of two points on the, opposite sides of the cliff as 30° and 60° respectively. The, distance between the two points is, 200, 400, m, (b), m, (a), 3, 3, (c) 100 m, (d) None of these, Solution (b) Let PQ be the cliff and A and B be the points under, observation., In ∆ OAP ,, AP, ,, cot 30 ° =, 100, , 1 400, , m, AB = AP + BP = 100 3 +, =, , 3, 3, , Example 5. The shadow of a tower standing on a level, plane is found to be 60 m longer when the Sun’s altitude is, 30° than when it is 45°. The height of the tower is, (a) 40( 3 + 1) m, (b) 30( 3 + 1) m, (c) 200 m, (d) None of these, Solution (b) Let OP be the tower. Let OA and OB be its, shadows when the altitude of the Sun is 45° and 30°,, respectively. Then, AB = 60 m; ∠OAP = 45° and ∠OBP = 30 °., Let, OP = h, OA, Then,, = cot 45° = 1⇒ OA = OP = h, OP, 45° 30°, , O, 30°, , A, , 60°, , Also,, 100, , ⇒, 30°, A, , 60°, P, , 100, 3, , ⇒, , B, , AP = 100 cot 30 ° = 100 3, BP, In ∆ BOP , cot 60 ° =, 100, , ⇒, , 45°, , 30°, B, , P, , O, , OB, = cot 30 °, OP, h + 60, = 3 ⇒ h + 60 = 3h, h, ( 3 − 1) h = 60, 60, 3 + 1, h=, ×, = 30 ( 3 + 1) m, 3 + 1, 3 −1, , Hence, the height of the tower is 30 ( 3 + 1) m., , Comprehensive Approach, n, n, , n, , Angle of depression and angle of elevation are always acute angle., A line perpendicular to the plane is perpendicular to each point, that lies on the plane., In an isosceles triangle median is perpendicular to base., , n, n, , n, , In similar triangle corresponding sides are proportional ., Exterior angle of a triangle is equal to the sum of interior opposite, angle., Angle of same segment are equal.
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Exercise, Level I, 1. A flagstaff 6 m high is placed on the top of a tower., The flagstaff casts a shadow, which is 2 3 m long, when measured along the ground. The angle in, degrees that the Sun-rays make with the ground is, (a) 60°, (b) 45°, (c) 30°, (d) 15°, 2. The angle of elevation of the tip of a flagstaff from a, point 10 m due South of its base is 60°. What is the, height of the flagstaff correct to the nearest metre?, (NDA 2012 I), , (a) 15 m, , (b) 16 m, , (c) 17 m, , (d) 18 m, , 3. A tower of height 15 m stands vertically on the, ground. From a point on the ground the angle of, elevation of the top of the tower is found to be 30°., What is the distance of the point from the foot of the, tower?, (NDA 2011 II), (a) 15 3 m (b) 10 3 m (c) 5 3 m (d) 30 m, 4. What is the angle subtended by 1 m pole at a distance, 1 km on the ground in sexagesimal measure?, (NDA 2012 I), , 9, degree, (a), 50 π, (c) 3.4 min, , 9, (b), degree, 5π, (d) 3.5 min, , 5. A person at the top of a hill observes that the angles, of depression of two consecutive kilometre stones on, a road leading to the foot of the hill are 30° and 60°., The height of the hill is, 3, 5, 6, 7, km (b), km (c), km (d), km, (a), 2, 2, 2, 2, 6. The angle of elevation of top of a tree on the bank of a, river from its other bank is 60° and from a point 20 m, further away from this is 30°. The width of the river, is, (d) 20 3 m, (a) 10 m, (b) 10 3 m (c) 20 m, 7. The angles of elevation of the top of a tower as, observed from the bottom and top of a building of, height 60 m are 60° and 45° respectively. The, distance of the base of the tower from the base of the, building is, (a) 30 ( 3 − 1) m, (b) 30 ( 3 + 3 ) m, (d) 30 ( 3 + 1) m, (c) 30 ( 3 − 3 ) m, 8. Two poles are 10 m and 20 m high. The line joining, their tops makes an angle of 15° with the horizontal., What is the approximate distance between the poles?, (NDA 2010 II, 2012 I), , (a) 35.3 m, , (b) 37.3 m, , (c) 41 m, , (d) 44 m, , 9. The angle of elevation of a tower at a level ground is, 30°. The angle of elevation becomes θ when moved, 10 m towards the tower. If the height of tower is 5 3, m, then what is the value of θ?, (NDA 2012 I), (a) 45°, (b) 60°, (c) 75°, (d) None of these, 10. The angle of elevation of the top of a tower at a point, on the ground is 30°. If on walking 20 m toward the, tower, the angle of elevation becomes 60°, then the, height of the tower is, 10, m, (a) 10 m, (b), 3, (c) 10 3 m, (d) None of these, 11. When the elevation of Sun changes from 45° to 30° the, shadow of a tower increases by 60 m. The height of the, tower is, (a) 30 3 m, (b) 30 ( 2 + 1) m, (d) 30 ( 3 + 1) m, (c) 30 ( 3 − 1) m, 12. A person standing on the bank of a river observes, that the angle subtended by a tree on the opposite, bank is 60°. When he retire 40 m from the bank, he, finds the angle to be 30°. The breadth of the river is, (a) 20 m, (b) 40 m, (c) 30 m, (d) 60 m, 13. From of 60 m high tower angles of depression of the, top and bottom of a house are α and β respectively. If, 60 sin (β − α ), the height of the house is, , then x is, x, equal to, (a) sin α sin β, (b) cos α cos β, (c) sin α cos β, (d) cos α sin β, 14. From the top of a building of height h metre, the, angle of depression of an object on the ground is θ., What is the distance (in metre) of the object from the, foot of the building?, (NDA 2012 I), (a) h cotθ, (b) h tanθ, (c) h cosθ, (d) h sinθ, 15. The angle of elevation of the top of a flag post from a, point 5 m away from its base is 75°. What is the, approximate height of the flag post?, (NDA 2010 I), (a) 15 m, (b) 17 m, (c) 19 m, (d) 21 m, 16. Looking from the top of a 20 m high building, the, angle of elevation of the top of a tower is 60° and the, angle of depression of its bottom is 30°. What is the, height of the tower?, (NDA 2009 II), (a) 50 m, (b) 60 m, (c) 70 m, (d) 80 m
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269, , Height and Distance, 17. The horizontal distance between two towers is 60 m, and the angular depression of the top of the first tower, as seen from the top of the second is 30°. If the height, of the second tower be 150 m, then the height of the, first tower is, (b) 90 m, (a) (150 − 60 3) m, (c) (150 − 20 3 ) m, (d) None of these, , angle α at a point P on the ground. If AP = n AB,, then the correct relation is, (a) n = ( n 2 + 1) tan α, (b) n = ( 2n 2 − 1) tan α, (c) n 2 = ( 2n 2 + 1) tanα, (d) n = ( 2n 2 + 1) tan α, , 18. From the top of a light house 60 m high with its base, at the sea level the angle of depression of a boat is, 15°. The distance of the boat from the foot of light, house is, 3 − 1, 3 + 1, (b) , (a) , 60 m, 60 m, 3 + 1, 3 − 1, 3 + 1, (d) None of these, (c) , m, 3 − 1, , 22. The angle of elevation from a point on the bank of a, river of the top of a temple on the other bank is 45°., Retreating 50 m, the observer finds the new angle of, elevation as 30°. What is the width of the river?, (a) 50 m, (b) 50 3 m (NDA 2009 I), (d) 100 m, (c) 50 / ( 3 − 1) m, , 19. A tower subtends an angle α at a point A in the plane, of its base and angle of depression of the foot of the, tower at a point l m just above A is β. The height of, the tower is, (a) l tan β cot α, (b) l tan α cot β, (c) l tan α tan β, (d) l cot α cot β, 20. Two poles of equal height stand on either side of a, 100 m wide road. At a point between the poles, the, angles of elevation of the topes of the poles are 30°, and 60°. The height of each pole is, (a) 25 m, (b) 25 3 m, 100, m, (d) None of these, (c), 3, 21. AB is vertical tower. The point A is on the ground and, C is the middle point of AB. The part CB subtend an, , 23. The tower 24 m portion of a 50 m tall tower is painted, green and the remaining portion red. What is the, distance of a point on the ground from the base of the, tower where the two different portions of the tower, subtend equal angles?, (NDA 2007 I), (a) 60 m, (b) 72 m, (c) 90 m, (d) 120 m, 24. An observer on the top of tree, finds the angle of, depression of a car moving towards the tree to be 30°., After 3 min this angle becomes 60°. After how much, more time, the car will reach the tree?, (a) 4 min, (b) 4.5 min, (c) 1.5 min, (d) 2 min, 25. A house of height 100 m subtends a right angle at the, window of an opposite house. If the height of the, window be 64 m, then the distance between the two, houses is, (a) 48 m, (b) 36 m, (c) 54 m, (d) 72 m, , Level II, 1. A ladder rests against a wall making an angle α with, the horizontal. The foot of the ladder is pulled away, from the wall through a distance x, so that it slides a, distance y down the wall making an angle β with the, horizontal. The correct relation is, α +β, α +β, (a) x = y tan, (b) y = x tan, 2, 2, (c) x = y tan (α + β ), (d) y = x tan (α + β ), 2. The angle of elevation of the top of the tower observed, from each of the three points A, B and C on the, ground forming a triangle is the same angle α. If R is, the circumradius of the ∆ABC, then the height of the, tower is, (a) R sinα, (b) R cosα, (c) R cotα, (d) R tanα, 3. The angle of elevation of the top of a tower from a, point A due South of the tower is α and from a point B, due East of the tower is β. If AB = d, then the height, of the tower is, , (a), (c), , d, tan α − tan β, d, 2, , 2, , cot2 α + cot2 β, , (b), (d), , d, tan α + tan2 β, d, 2, , cot2 α − cot2 β, , 4. From the bottom of a pole of height h the angle of, elevation of the top of a tower is α and the pole, subtends an angle β at the top of the tower. The height, of the tower is, h tan (α − β ), h cot (α − β ), (a), (b), tan (α − β ) − tan α, cot (α − β ) − cot α, cot (α − β ), (c), (d) None of these, cot (α − β ) − cot α, 5. An aeroplane flying horizontally 1 km above the, ground is observed at an elevation of 60° and after, 10 s the elevation is observed to be 30°. The uniform, speed of the aeroplane (in km/h) is, (a) 240, (b) 240 3, (c) 60 3, (d) None of these
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270, 6. From a point a m above a lake the angle of elevation, of a cloud is α and the angle of depression of its, reflection is β. The height of the cloud is, a sin (α + β ), a sin (α + β ), (a), m, (b), m, sin (α − β ), sin (β − α ), a sin (β − α ), m, (d) None of these, (c), sin (α + β ), 7. A vertical tower stands on a horizontal plane and is, surmounted by a vertical flagstaff of height h. At a, point P on the plane, the angle of elevation of the, bottom of the flagstaff is β and that of the top is α., What is the height of the tower?, (NDA 2011 II), h tan β, h tan β, (a), (b), tan α − tan β, tan α + tan β, h cos β, h, (d), (c), cos α − cos β, cos (α − β ), 8. An aeroplane flying at a height of 300 m above the, ground passes vertically above another plane at an, instant when the angles of elevation of two planes, from the same point on the ground are 60° and 45°,, respectively. What is the height of the lower plane, from the ground?, (NDA 2011 II), 100, (a) 50 m, (b), m, 3, (c) 100 3 m, (d) 150 ( 3 + 1) m, 9. An observer measures angles of elevation of two, towers of equal heights from a point between the, towers. If the angles of elevation are 60° and 30° and, distance of the nearer tower is 100 m, then the height, of each tower and the distance between the towers,, respectively are, 100, 100, (a), m and 400 m, (b), m and 300 m, 3, 3, (c) 100 3 m and 400 m (d) 100 3 m and 300 m, 10. The upper part of a tree broken over by the wind makes, an angle of 60° with the ground and the distance from, the roots to the point where the top of the tree meets, the ground is 20 m. The length of the broken part of the, tree is, (a) 20 m, (b) 40 m, (c) 20 2 m (d) 40 3 m, 11. From the top of a tower 60 m high, the angles of, depression of two objects which are on the horizontal, plane and in a line with the foot of the tower are α and, β with β > α. What is the distance between the two, objects in metres?, (a) 60 sin (β − α ) cosec α cosec β, (b) 60 cos (β − α ) sec α sec β, (c) 60 (cot α + cot β ), (d) 60 (tan β − tan α ), 12. A man on the top of a rock rising on a sea-shore, observes a boat coming towards it. If it takes 10 min, for the angle of depression to change from 30° to 60°,, how soon will the boat reach the shore?, (a) 20 min (b) 15 min (c) 10 min (d) 5 min, , NDA/NA Mathematics, 13. A flag of height h stands on the top of a hemispherical, dome of radius 10 m. From a point on the ground,, when the flag is just visible the angle of elevation is, 45°. What is the height of the flag?, (a) 5 m, (b) 10 m, (c) 10 ( 2 − 1) m, (d) 10 ( 2 + 1) m, 14. A man standing on the bank of a river observes that, the angle of elevation of the top of a tree just on the, opposite bank is 60°. The angle of elevation is 30°, from a point at a distance y m from the bank. What is, the height of the tree?, (NDA 2011 I), 3y, y, (a) y m, (b) 2y m, (c), m, m (d), 2, 2, 15. A man observes the elevation of a balloon to be 30°., He, then walks 1 km towards the balloon and finds, that the elevation is 60°. What is the height of the, balloon?, (NDA 2009 I), (a) 1/ 2 km, (b) 3/ 2 km, (c) 1/ 3 km, (d) 1 km, 16. The foot of a tower of height h m is in a direct line, between two observers A and B. If the angles of, elevation of the top of the tower as seen from A and B, are α and β respectively and if AB = d m, then what is, the value of h / d ?, (NDA 2008 II), tan (α + β ), cot (α + β ), (a), (b), (cot α cot β − 1), (cot α cot β − 1), tan (α + β ), cot (α + β ), (c), (d), (cot α cot β + 1), (cot α cot β + 1), 17. The length of the shadow of a tree is 10 3 m, when, the angle of elevation of the Sun is 60°. What is the, length of the shadow of the tree when the angle of, elevation of the Sun is 30°?, (a) 30 3 m, (b) 10 3 m, (d) 4 3 m, (c) 5 3 m, 18. The shadow of a pole standing on a horizontal plane, is d m longer when the Sun’s altitude is α, then when, it is β. What is the height of the pole?, cos α cos β, sin α cos β, (a) d, (b) d, cos (α − β ), sin (α − β ), sin α sin β, sin β cos α, (d), (c) d, sin (β − α ), cos (α + β ), 19. A vertical pole with height more than 100 m consists, of two parts, the lower being one-third of the whole., At a point on a horizontal plane through the foot and, 40 m from it, the upper part subtends an angle whose, 1, tangent is . What is the height of the pole?, 2, (a) 110 m (b) 200 m (c) 120 m, (d) 150 m, 20. Assertion (A) A flagstaff of length 100 m stands on a, tower of height h. If a point on the ground the angle of, elevation of the tower and top of the flagstaff be 30°, and 45°, respectively, then h = 50( 3 + 1) m
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271, , Height and Distance, Reason (R) A flagstaff of length d stand on tower of, height h. If at a point on the ground the angle of, elevation of the tower and top of the flagstaff be, d cot β, ., α and β , then h =, cot α − cot β, (a) A and R are true and R is the correct explanation, of A., (b) A and R are true but R is not the correct, explanation of A., (c) A is true but R is false., (d) A is false but R is true., , Directions (Q. Nos. 21-22) Let 20 ft long ladder, reaches 20 ft below the flag. The angle of elevation of the, top of the flag at the foot of the ladder is 60°?, 21. Find the height of the flag., (a) 20 ft, (b) 30 ft, (c) 40 ft, (d) 20 2 ft, 22. Find the distance between the foot of the ladder and, bottom of the building., (b) 10 3 ft, (a) 5 3 ft, (c) 7 3 ft, (d) None of these, , Answers, Level I, 1. (a), 11. (d), 21. (d), , 2. (c), 12. (b), 22. (c), , 3. (a), 13. (d), 23. (d), , 4. (a), 14. (a), 24. (c), , 5. (a), 15. (c), 25. (a), , 6. (a), 16. (d), , 7. (d), 17. (d), , 8. (b), 18. (b), , 9. (b), 19. (b), , 10. (c), 20. (b), , 2. (d), 12. (d), 22. (b), , 3. (c), 13. (c), , 4. (b), 14. (c), , 5. (b), 15. (b), , 6. (b), 16. (b), , 7. (a), 17. (a), , 8. (c), 18. (c), , 9. (c), 19. (c), , 10. (b), 20. (a), , Level II, 1. (a), 11. (a), 21. (b), , Hints & Solutions, Level I, 1. Given that the height of the flagstaff = 6 m, and the length of shadow on the ground = 2 3 m, , O, , θ, 2√3 m, , tan 60° =, , h, 10, , R, , ⇒, , h = 10 tan 60° = 10 ⋅ 3, , 6m, , ⇒, , h = 10 ⋅ (1 .732), , Q, , ⇒, , h = 17 .32 m, , ⇒, , h = 17 m, , (approx.), , 3. Let AB = x m, P, , C, , 6, tan θ =, 2 3, tan θ = 3, tan θ = tan 60° ⇒ θ = 60°, , In ∆ OQR,, ⇒, ⇒, 2. In ∆ABC,, , C (flagstaff), , h, , A, , 60°, 10 m, , 15 m, A, , x, , Then in ∆ ABC ,, BC 15, =, AB, x, 1, 15, =, x, 3, , tan 30° =, ⇒, , (tip), B, , 30°, , ⇒, , x = 15 3 m, , B
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272, , NDA/NA Mathematics, , 4. Let ∠CAB = θ, , D, C, , h, 1 m (Pole), , 1 km, = 1000 m, , tan θ =, , B, , BC, 1, radian, =, AB 1000, , tan θ = 0.001 radian, , ⇒, , θ = tan −1 (0.001) radian, , ⇒, , θ = 9.999 × 10−4 radian, 180, degree, θ = 9.999 × 10−4 ×, π, , ⇒, ⇒, , θ = 0.057324 degree, 9, degree, θ=, 50π, , h, E, , A, , 0°, , ⇒, ⇒, , 60°, , h = 3h − 3, , ⇒, ⇒, ⇒, , ( 3 − 1) x = 60 ⇒ x =, , ⇒, , 8. Q, ⇒, , B, , x, , 3x = 60 + h, [from Eq. (i)], 60, 3+1, ×, 3 −1, 3+1, , 60 ( 3 + 1), 2, x = 30 ( 3 + 1) m, ⇒, ∴ Distance between the bases = 30 ( 3 + 1) m, , B, x, , − 2h = − 3, , x=, , AC = 20 m, ED = 10 m, A, , 3, km, 2, CD, 6. In ∆BCD, tan 60° =, BC, ⇒, , C, , 60°, C, , From Eqs. (i) and (ii), we get, h, = 3h − 1, 3, ⇒, , 45°, , BD, tan 60° =, AB, 60 + h, 3=, ⇒, x, 3x = 60 + x, 3x − x = 60, , ⇒, h, , 1 km, , A, , In ∆ABD,, , 30°, , 30°, , D, , ... (ii), , 6, , h=, , 3=, x=, , E, , h, x, , h, 3, , ...(ii), , ...(i), , ...(i), , 1 + x = 3h, x= 3 h −1, , D, , tan 30° =, , 7. Let CD = h m and AB = x m, CD , In, ∆CDE , tan 45° = , , CE , h, 1= ⇒ x=h, ⇒, x, , 5. Let CB = x km and AB = h km, AB, h, In ∆ABC ,, tan 60° =, ⇒ 3=, BC, x, h, ⇒, x=, 3, AB, In ∆ ABD,, tan 30° =, BD, 1, h, =, ⇒, 3 1+ x, ⇒, ⇒, , C, , x, , CD, AC, h, 1, =, ⇒ h 3 = 20 + x, ⇒, 3 20 + x, ⇒, x = h 3 − 20, From Eqs. (i) and (ii),, h, = h 3 − 20 ⇒ h = 3h − 20 3, 3, 2h = 20 3 ⇒, h = 10 3 m, ⇒, Now, putting h = 10 3 in Eq. (i), we get, x = 10 m, Thus, the width of the river is 10 m., In ∆ACD,, , ⇒, , ⇒, , B, , 20 m, , θ, , A, , 60°, , 30°, , A, , 15°, , B 20 m, , 10 m, , ... (i), , D, , C, , ∴ AB = AC − BC = AC − ED = 10 m
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273, , Height and Distance, Now in ∆ABE,, , ⇒, , AB, BE, 10, tan (45° − 30° ) =, BE, tan 15° =, , ⇒, , ⇒, , 1, 1−, tan 45° − tan 30°, 10, 3 ⇒ 3 − 1 = 10, =, =, ⇒, 1, 1 + tan 45° tan 30° BE 1 +, 3 + 1 BE, 3, 3 + 1 10 × ( 3 + 1)2, ⇒ BE = 10 , =, 2, 3 − 1, , h=, , 20, , 1, 3−, 3, h = 10 3 m, , 11. Let the height of the tower be h., h, In ∆ BCD,, tan 45° =, BC, ⇒, BC = h, h, h, In ∆ ACD, tan 30° =, ⇒ AC =, tan 30°, AC, , h, , Again in ∆ADC ,, ⇒, ⇒, , A, , 30°, 10 m, , 30°, 60 m, , A, , 53m, , 45°, C, , B, , AB + BC = 3h ⇒ 60 + h = 3h, 60, 60 ( 3 + 1), [from Eq. (i)], h=, =, 2, 3 −1, , ⇒, , h = 30 ( 3 + 1) m, , 12. Let the height of the tree be h m and breadth of the river, be b m., , θ, B, , ...(i), , D, , = 5 (3 + 1 + 2 3 ) = 5 (4 + 2 × 1.73) (Q 3 = 1.732), = 5 ( 4 + 3.46 ), ⇒ CD = BE = 5 × 7.46 = 37.3 m, 5 3, 9. In ∆BCD,, tan θ =, x, D, , 20 3, 2, , ⇒h =, , xm, , D, , C, , 5 3, 1, =, 10 + x, 3, 10 + x = 15, x=5m, , tan 30° =, ⇒, , h, , From Eq. (i), , 30°, , ⇒, , 60°, , A, , 5 3, tan θ =, = 3 = tan 60°, 5, θ = 60°, , B, , 40 m, , h, ⇒ h=b 3, b, h, In ∆ DAC , tan 30° =, 40 + b, 40 + b, ⇒, h=, 3, From Eqs. (i) and (ii), we get, 40 + b, b 3=, 3, ⇒, 2b = 40 ⇒ b = 20 m, 60, 13. In ∆ ABD, tan β =, d, ⇒, d = 60 cot β, In ∆ DEC ,, tan 60° =, , P, , βα, , h, , 60°, , B, , ⇒, ⇒, , 20 m, , AB + AO = 3h, h, 20 +, = 3h, 3, , A, , C, , In ∆ DBC ,, , 10. Let the height of the tower be h m., h, In, ∆PAO, tan 60° =, OA, ⇒, OA = h cot 60°, h, =, 3, h, In ∆ PBO,, tan 30° =, OB, h, ⇒, OB =, 1, 3, , 30°, , b, , α, , E, , O, , d, , D, , C 60 m, , h, A, , β, d, , B, , ...(i), , ...(ii), , ...(i)
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274, , NDA/NA Mathematics, DC, EC, DC = d tan α, 60 − h = d tan α, [Q BC = EA = h ], [from Eq. (i)], 60 − h = 60 cot β tan α, cos β sin α , , h = 60 1 −, ⋅, , , sin β cos α , 60 sin (β − α ), h=, cos α sin β, 60 sin ( β − α ) 60 sin (β − α ), (given), =, x, cos α sin β, tan α =, , ⇒, ⇒, ⇒, ⇒, ⇒, ⇒, ⇒, , x = cos α sin β, , 14. In ∆ABC,, , F, , hm, 60º, 30º, , D, , C, , 20 m, , 20 m, 30º, , A, , ⇒, h = 3 ⋅ 20 3 (Q AB = DC = 20 3 ), [from Eq. (i)], ⇒, h = 60 m, ∴ Height of the tower, BF = 60 + 20 = 80 m, (Q BF = h + 20), BC, 17. In ∆ ABC , tan 30° =, AC, , hm, (object), , h, , 30º, , A, , (foot), , θ, , B, , 30º, , C, , A, , B, , C, , 150 m, , B, , xm, , O, , D, , 60 m, , 1, h − 150, ⇒, =, 60, 3, 60, ⇒ h = (150 + 20 3 ) m, h − 150 =, ⇒, 3, 60, 18. In ∆ ABC , tan 15° =, d, , h, tan θ =, x, x = h cot θ, 15. Let h be the height of the flag-post., In ∆ABC ,, AB h, tan 75° =, =, BC 5, , C, 15º, , A, , 60 m, 15º, d, , A, , hm, , 3 + 1, d = 60 cot 15° = 60 , m, 3 − 1, , ⇒, 75º, C, , ⇒, ⇒, , 5m, , B, , 19. In ∆ AOT , tan α =, , tan 45° + tan 30°, h, =, 1 − tan 45° tan 30°, 5, , ⇒, , h=, , T, , α, , ⇒, , 3 + 1 + 2 3, h =5 , , 3 −1, , , , In, , 16. In ∆ABD, tan 30° =, , In ∆DCF ,, , β, , A, , = 5 (2 + 3 ), (Q 3 = 1.732), (approx), = 5 × 3.732 = 18.660 = 19 m, , ⇒, , h, , l, , ( 3 + 1 )2, ×5, ( 3 )2 − (1)2, , AD 20, =, AB AB, AB = 20 3 m, h, tan 60° =, = 3, DC, , h, OA, , P, , 1+ 3 h, =, 3 −1 5, , ⇒, , B, , ...(i), , ⇒, ⇒, , O, , OA = h cot α, l, ∆ AOP, tan β =, AO, l, tan β =, h cot α, , ...(i), , [from Eq. (i)], , h = l tan α cot β, , 20. Let the height of pole AD = BC = h m, h, In ∆ OBC ,, tan 60° =, x, h, ⇒, x=, 3, , ...(i)
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275, , Height and Distance, , tan 30° =, , In ∆ AOD,, , h, 100 − x, C, , D, , ⇒, ⇒, , x + 50 = h 3, h + 50 = h 3, 50, h=, 3 −1, , So, width of the river is x = h =, , 30º, 60º, 100 m, x, , A, , h, 1, =, 3 x + 50, , ⇒, , h, , h, , ⇒, , B, , 1, ⇒, h = (100 − x), 3, h 1, , = 100 −, , , 3 3, 3h = 100 3 − h, ⇒, 100 3, ⇒, h=, 4, h = 25 3 m, ⇒, AB, 21. In ∆ PAB, tan β =, AP, , [from Eq. (i)], , [from Eq. (i)], , 50, m., 3 −1, , 23. Let AC be the tower whose length is 50 m., Let, AB = 24 m and BC = 26 m, AB, In ∆ABD,, tan θ =, AD, 24, ⇒, tan θ =, x, C, 26 m, , B, , B, =, , P, , β, , D, , =, , tan 2θ =, , AC, AP, , In ∆ PAC ,, , tan θ =, , ∴, , tan α = tan (β − θ ) =, , tan β − tan θ, 1 + tan β tan θ, , AB AC, −, ...(i), = AP AP, AB AC, ⋅, 1+, AP AP, Q, AP = n ( AB), = n (2 AC ) (QC is the mid-point of BA), 1, 1, −, n, From Eq. (i) tan α = n 2n =, 2, 1, n, +1, 2, 1+, 2n 2, ⇒, n = (2n 2 + 1) tan α, tan 45° =, , A, , x, , and in ∆ACD,, , A, , 22. In ∆ABC ,, , 24 m, , θ, θ, , C, α, θ, , AC, 2 tan θ, 50, ⇒, =, AD, x, 1 − tan 2 θ, , ⇒, , 24, x = 50, 576, x, 1− 2, x, , ⇒, , 24x2, = 25, x2 − 576, , 2⋅, , ⇒, , [from Eq. (i)], , 24x2 = 25x2 − 576 × 25, , ⇒, ⇒, , x2 = 576 × 25, x = 24 × 5 = 120 m, h, 24. In ∆ ABC , tan 60° =, BC, BC = h cot 60°, , A, , h, , 60º, B, , 45º, C, , D, , ⇒, , x, , 1=, , tan 30° =, , B, , h, ⇒h = x, x, , Now in ∆ABD,, AB, BD, , ....(i), , A, , AB, BC, , 30º, , …(i), , …(i), , 30º, C, , d, , D, , h, In ∆ABD,, tan 30° =, BD, ⇒, BD = h cot 30°, ⇒, BC + CD = h cot 30°, ⇒, CD = h cot 30° − BC, [from Eq. (i)], ⇒, d = h cot 30° − h cot 60°
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278, , NDA/NA Mathematics, , 9. Given that the angle of elevation of two towers of equal, heights from a point between the towers is 60° and 30°,, respectively. Let the height of each tower be h m., B, , D, , 12. Let CD = h m, AB = x m and BC = y m, h, In ∆ BCD, tan 60° =, y, h=y 3, ⇒, , ...(i), , D (man), , h, , h, , 60º, A, 100 m, , h, , 30º, P, , C, , x, , 30º, , h, or h = 100 3 m, 100, h, 1, tan 30° = =, x, 3, , In ∆ APB,, , tan 60° =, , In ∆ CPD,, , A, , x = 100 3 × 3, = 300 m, ∴ The total distance between the towers, = AP + PC = 100 + x, = 100 + 300 = 400 m, 10. Let the length of the tree be x m. Suppose AQ be the, broken part of the tree OP whose upper, P, part touches the ground A such that, ∠ QAO = 60°, Given that,, OA = 20 m, Q, OQ, 20, OQ = 20 3 m, , tan 60° =, , ⇒, Now in, , 60º, A 20 m O, , ∆ OAQ, AQ = (OA )2 + (OQ )2, , x+ y, 11. In ∆ ADC , cot α =, 60, , y, , B, , C, , h, and in ∆ ACD, tan 30° =, x+ y, y 3, 1, ⇒, =, 3 x+ y, , ⇒, , In ∆ AOQ,, , 60º, , x, , [using Eq. (i)], , ⇒, , 3y = x + y, , ⇒, , 2y = x ⇒ y =, , x, 2, ∴ Time taken to cover a distance from A to B = 10, 10, Time taken to cover a unit distance =, x, Time taken to cover a distance of y m, 10, 10 x, =, × y=, × = 5 min, x, x 2, 13. A flag BC of height h stands on the top of a, hemispherical dome of radius 10 m ( AB). From a point E, on the ground when the flag is just visible, the angle of, elevation is 45°., ∴, ∠ DCA = 45°, , = 400 + 1200 = 40 m, , C, , ...(i), , h, , 45º, , B, , D, , D, α β, , 10, 10, 45º, E, , 60 m, , A, , In ∆DCA,, , A, , In ∆ BCD,, , α, x, , cot β =, , β, B, , y, , ⇒, C, , y, 60, , ...(ii), , From Eqs. (i) and (ii), we get, x + 60 cot β, cot α =, 60, ⇒, 60 cot α = x + 60 cot β, ⇒, x = 60 (− cot β + cot α ), cos β cos α , x = 60 −, +, , sin β sin α , ⇒ x = 60 (− sin α ⋅ cos β + cos α ⋅ sin β ) cosec α cosec β, x = 60 sin (β − α ) cosec α ⋅ cosec β, , N, , DA, sin 45° =, AC, 1, 10, =, ⇒ 10 + h = 10 2, 10, +h, 2, , ⇒, ⇒, , h = 10 2 − 10, h = 10 ( 2 − 1) m, , 14. In ∆ACD,, D, h, A, , 60º, , 30º, y, , tan 30° =, , CD, AC, , B, , C, x
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280, , NDA/NA Mathematics, (A) Put d = 100 m, α = 30, β = 45°, 3+1, 100 cot 45°, 100 × 1, ∴, h=, =, ×, cot 30° − cot 45°, ( 3 + 1), 3−1, 100( 3 + 1), =, = 50( 3 + 1)m, 3−1, Hence, both the statements are true and R is the, correct explanation of A., , 1, 2, Let h be the height of pole, then, in ∆ ABC,, BC, tan φ =, AB, h/3, h, =, =, 40 120, , 19. Q, , tan θ =, , 21. Let BD is a flag and BD = (h + 20) ft, , D, , D, 20 ft, , h, , C, θ, A, , φ, , 20, , 40 m, , and in ∆ ABD,, tan (θ + φ ) =, ⇒, ⇒, , C, , h/3, , 60º, , BD, AB, , A, , 80 (60 + h ) = 240 h − h 2, 2, h − 160 h + 4800 = 0, (h − 120) (h − 40) = 0, h = 120 m, , ⇒, ⇒, , (Q h > 100), , d, C, h, B, , In ∆ ABC,, h, AB, AB = h cot α, , tan α =, ⇒, In ∆ ABD ,, , h+ d, tan β =, AB, ⇒, AB = cot β (h + d ), From Eqs. (i) and (ii),, h cot α = cot β (h + d ), d cot β, ⇒, h =, cot α − cot β, , BD, AB, h + 20, 3=, AB, (h + 20), AB =, 3, 3, , tan 60° =, , D, , A, , B, , In ∆ABD,, , 20. (R) Let BC = h be the height of the tower and CD = d be, the height of the flagstaff., , α β, , h, , B, , 1, h, +, tan θ + tan φ, h, h, 2, 120, ⇒, =, =, h, 40, 1 − tan θ tan φ 40, 1−, 240, 2 (60 + h ) h, =, (240 − h ) 40, , ⇒, ⇒, ⇒, ⇒, , ft, , …(i), , Now, in ∆ABC ,, AC 2 = AB2 + BC 2, 3 (h + 20)2, [from Eq. (i)], ⇒, 202 =, + h 2,, 9, (h + 20)2 + 3h 2, ⇒, 400 =, 3, ⇒, 1200 = h 2 + 40h + 400 + 3h 2, ⇒, 4h 2 + 40h − 800 = 0, ⇒, h 2 + 10h − 200 = 0, 2, ⇒, h + 20h − 10h − 200 = 0, ⇒, h (h + 20) − 10(h + 20) = 0, ⇒, h = 10, (Q h ≠ − 20), ∴ Height of flag = BD = BC + CD, = 10 + 20, = 30 ft, 22. In ∆ ABC,, BD, AB, 30, 3=, AB, 30, 3, AB =, ×, 3, 3, ft, = 10 3, , tan 60° =, …(ii), , …(i), , ⇒, ⇒, ⇒
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15, Inverse Trigonometric, Functions, Properties of Inverse Trigonometric Function, , Inverse Functions, If a function f is one-one and onto from A to B, then, function of which associates each element y Î B to one and, only one element x Î A, such that y = f ( x ) , then g is called, the inverse of function of y., \, g = f -1 Þ x = f -1( y ), %, , If y = f (x ) and x = g( y) are two functions such that f { g( y)} = y, and g { f ( y)} = x , then f and g are said to be inverse function of, each other., , Domain and Range of Inverse Functions, S.No., 1., 2., , Function, sin -1 x, cos, , -1, , [-p / 2 , p / 2], , x, , [-1, 1], , [0, p ], , x, , R, , (-p / 2 , p / 2), , (-¥ , - 1] È [1, ¥ ), , [- p / 2 , p / 2] - {0}, , (-¥ , - 1] È [1, ¥ ), , [0, p ] - { p / 2}, , R, , (0, p ), , -1, , tan, , 4., , cosec -1 x, , 6., , -1, , sec x, cot, , -1, , Range, , [-1, 1], , 3., , 5., , Domain, , x, , Principal Value of Inverse Functions, S.No. Function, , Principal value branch, , 1., , sin -1 x, , - p /2 £ y £ p /2 , where y = sin -1 x, , 2., , cos -1 x, , 0 £ y £ p , where y = cos -1 x, , 3., , tan -1 x, , - p /2 < y < p /2 , where y = tan -1 x, , 4., , cosec-1x, , - p /2 £ y £ p /2 , where y = cosec-1x, y ¹ 0, , 5., , sec-1 x, , 0 £ y £ p, where y = sec-1 x, y ¹ p /2, , 6., , cot-1 x, , 0 < y < p , where y = cot-1 x, , p p, 1. (i) sin -1 (sin q ) = q ; " q Î é - , ù, êë 2 2 úû, (ii) cos -1 (cos q ) = q ; " q Î [0, p ], p p, (iii) tan -1 (tan q ) = q ; " q Î æç - , ö÷, è 2 2ø, p p, (iv) cosec -1 ( cosec q ) = q ; " q Î é - , ù , q ¹ 0, êë 2 2 úû, p, (v) sec -1 ( sec q ) = q ; " q Î [0, p ], q ¹, 2, (vi) cot -1 (cot q ) = q ;" q Î ( 0, p ), 2. (i) sin(sin -1 x) = x, " x Î [-1, 1], (ii) cos(cos -1 x) = x, " x Î [-1, 1], (iii) tan(tan -1 x) = x," x Î R, (iv) cosec( cosec -1 x ) = x," x Î ( -¥ , - 1] È [1, ¥ ), (v) sec( sec -1 x) = x," x Î ( -¥ , - 1] È [1, ¥ ), (vi) cot(cot -1 x) = x," x Î R, 3. (i) sin -1 ( - x) = - sin -1 ( x)," x Î [-1, 1], (ii) cos -1 ( - x) = p - cos -1 ( x)," x Î [-1, 1], (iii) tan -1 ( - x) = - tan -1 x, " x Î R, (iv) cosec -1 ( - x) = - cosec -1 x,, " x Î ( -¥ , - 1] È [1, ¥ ), (v) sec -1 ( - x) = p - sec -1 x," x Î ( -¥ , - 1] È [1, ¥ ), (vi) cot -1 ( - x) = p - cot -1 x," x Î R, 1, 4. (i) sin -1 æç ö÷ = cosec -1 x," x Î ( -¥ , - 1] È [1, ¥ ), è xø, 1, (ii) cos -1 æç ö÷ = sec -1 x," x Î ( -¥ , - 1] È [1, ¥ ), è xø, ì, cot -1 x, " x > 0, 1, (iii) tan -1 æç ö÷ = í, è x ø î - p + cot -1 x, " x < 0
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Exercise, Level I, æ 5ö, 1. What is the value of sec2 tan-1 ç ÷ ?, è 11 ø, , (NDA 2012 I), , (a) 121 / 96, (c) 146 / 121, , (b) 217 / 921, (d) 267 / 121, 1, 1, 2. sin-1 x + sin-1 + cos-1 x + cos-1 is equal to, x, x, p, (a) p, (b), 2, 3p, (d) None of these, (c), 2, 3. The value of sin (cot-1 x ) is, (a) 1 + x 2, (c) (1 + x 2 )-3/ 2, -1, , 4. If tan x - tan, (a) x - y, x- y, (c), 1 + xy, , (b) x, (d) (1 + x 2 )-1/ 2, -1, , y = tan, , -1, , A, then A is equal to, (b) x + y, x+ y, (d), 1 - xy, , é 1 - sin x + 1 + sin x ù, 5. cot-1 ê, ú is equal to, ë 1 - sin x - 1 + sin x û, (a) p - x, (b) 2p - x, x, x, (c), (d) p 2, 2, -1, 6. The value of sin ( 2 sin 0.8) is, (a) 0.96, (b) 0.48, (c) 0.64, (d) None of these, -1 3, -1 5, is, 7. The value of cot, + sin, 4, 13, 63, 12, (b) sin-1, (a) sin-1, 65, 13, 65, 5, (c) sin-1, (d) sin-1, 68, 12, 8. tan (cos-1 x ) is equal to, (a), , 1- x, x, , (NDA 2008 I), , 2, , (b), , x, , 2, , 1 + x2, , 2, , 1+ x, (d) 1 - x 2, x, æ 1 ö, 9. If cos-1 ç, ÷ = q, then what is the value of, è 5ø, cosec-1 ( 5) ?, (NDA 2007 II), (c), , æpö, (a) ç ÷ + q, è 2ø, p, (c), 2, , æpö, (b) ç ÷ - q, è 2ø, (d) - q, , 10. If sin-1 x + cot-1 (1 / 2) = p / 2, then what is the value, of x?, (NDA 2009 II), 1, (a) 0, (b), 5, 2, 3, (c), (d), 2, 5, 11. sin-1, , 1, , p, 6, p, (c), 3, , 5, , + cot-1 3 is equal to, p, 4, p, (d), 2, , (a), , (b), , 12. The value of sin cot-1 tan cos-1 x is equal to, p, (a) x, (b), 2, (c) 1, (d) None of these, 1, 1, 1, 13. tan-1 + tan-1 + tan-1 is equal to, 2, 3, 4, 5, -1 3, (b) tan-1, (a) tan, 5, 3, 1, 7, (c) tan-1, (d) tan-1, 5, 3, 4, 2ù, é, 14. The value of tan ê cos-1 + tan-1 ú is, 5, 3û, ë, 16, 7, (a), (b), 17, 16, 16, 17, (d), (c), 7, 6, 1, 1, 15. If tan-1 - tan-1 = tan-1 x, then x is equal to, 3, 4, 1, 1, (a), (b), 11, 12, 1, 3, (d), (c), 13, 4, 1, ö, æ, 16. If sin ç sin-1 + cos-1 x ÷ = 1, then x is equal to, ø, è, 5, (NDA 2011 I), (a) 0, (c) 4 / 5, , (b) 1, (d) 1 / 5, , 17. If sin-1 x - cos-1 x =, (a) x = (c) x =, , 1, 2, , 1, 2, , p, , then what is the value of x?, 6, (NDA 2008 I), (b) x = 1, (d) x =, , 3, 2
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285, , Inverse Trigonometric Functions, ép, æ, 3 öù, 18. The value of sin ê - sin-1 ç ÷ ú is, è 2 øû, ë2, 3, 3, (a), (b) 2, 2, 1, 1, (d) (c), 2, 2, , 1, = 2, what is the principal value of sin-1 x?, x, (a) 30°, (b) 45°, , 28. If x +, , (c) 60°, , 30. What is the value of sin-1cos sin-1 x + cos-1sin cos-1 x ?, (a) 2x, (b) p / 2, (c) - p / 2, , 4, 12ü, ì, 20. What is the value of cos í cos-1 + cos-1 ý?, 5, 13þ, î, (NDA 2012 I), , (b) 33/65, (d) 11/65, , 21. If tan-1 2, tan-1 3 are two angles of a triangle, then, what is the third angle?, (NDA 2012 I), (b) tan-1 4, (a) tan-1 2, (c) p/4, (d) p/3, 22. What is the value of sin-1, p, (a), 3, 23. If sin-1, (a) 1, 24. If sin-1, (a) 1, (c) 0, , 1, 4, + 2 tan-1 ?, 3 (NDA 2010 II), 5, p, p, (c), (d), 4, 6, , p, (b), 2, 5, 12, p, + sin-1, = , then what is the value of x?, x, x, 2, (NDA 2010 I), (b) 7, (c) 13, (d) 17, 1, 1, x = tan-1 y, what is the value of 2 - 2 ?, x, y, (b) -1, (NDA 2007 II), (d) 2, , 25. If cos-1 x + cos-1 y + cos-1 z = p, then, (a) x 2 + y 2 + z 2 + xyz = 0, (b) x 2 + y 2 + z 2 + 2xyz = 0, (c) x 2 + y 2 + z 2 + xyz = 1, (d) x 2 + y 2 + z 2 + 2xyz = 1, x-1, x+1 p, = , then x is equal to, + tan-1, x+2, x+2 4, 1, 1, (b) (a), 2, 2, 1, 5, (c) ±, (d) ±, 2, 2, , 26. If tan-1, , 27. If sin-1 x + sin-1 y + sin-1 z =, value of x + y + z?, (a) -3, 1, (c) 3, , 3p, , then what is the, 2, , (b) 3, 1, (d), 3, , -1, , 29. If, cos x + cos y = p, what is the value of, sin-1 x + sin-1 y?, (a) 0, (b) p / 2, (c) p, (d) 2 p, , é, æ1ö p ù, 19. tan ê 2 tan-1 ç ÷ - ú is equal to, è 5ø 4 û, ë, 17, 17, (b) (a), 7, 7, 7, 7, (d) (c), 17, 17, , (a) 63/65, (c) 22/65, , (d) 90°, -1, , (d) 2x + p, , p, 31. If sin-1 x = for some x Î ( -1, 1) , then the value of, 5, cos-1 x is, 3p, 5p, 7p, 9p, (a), (b), (c), (d), 10, 10, 10, 10, 32. If x 2 + y 2 + z 2 = r 2, then, æ xz ö, æ xy ö, æ yz ö, tan -1 ç ÷ + tan -1 ç ÷ + tan -1 ç ÷ is equal to, è zr ø, è xr ø, è yr ø, , p, 2, (d) None of these, , (a) p, , (b), , (c) 0, , 2p ö, æ, 33. What is the value of sin-1 ç sin, ÷?, è, 3ø, (a) -, , p, 3, , (b), , 2p, 3, , (c) -, , 2p, 3, , (d), , p, 3, , 34. In a D ABC, if A = tan-1 2 and B = tan-1 3, then C is, equal to, (NDA 2011 II), p, p, p, p, (a), (b), (c), (d), 3, 4, 6, 2, 35. What is the value of tan-1 x + cot-1 x ?, p, -p, for x > 0 and, for x < 0, (a), 2, 2, p, (b), for all x, 2, -p, for all x, (c), 2, p, for integral x, (d), 2, 36. What is the value of, tan (tan-1 x + tan-1 y + tan-1 z ) cot (cot, (a) 0, 3p, (c), 2, , -1, , x + cot-1 y + cot-1 z ) ?, , (b) 2 ( x + y + z ), 3p, +x+ y+ z, 2, , (d), , 37. What is the value of x that satisfies the equation, cos-1 x = 2 sin-1 x ?, 1, 1, (a), (b) -1, (c) 1, (d) 2, 2
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287, , Inverse Trigonometric Functions, 14. Which one of the following is not correct?, (NDA 2008 II), , (a), (b), (c), (d), , sin-1 {sin ( 5p / 4)} = - p / 4, sec-1 { sec ( 5p / 4)} = 3p / 4, tan-1 {tan ( 5p / 4)} = p / 4, cosec-1 { cosec ( 7p / 4)} = p / 4, , 15. If sin2 x + sin2 y = 1, then what is the value of, cot ( x + y )?, (NDA 2009 I), , (a) 1, , (b), , 3, , (c) 0, , (d) 1 /, , 16. Let - 1 £ x £ 1. If cos (sin-1 x ) =, value does tan (cos-1 x ) assume?, (a) One, (c) Four, , 3, , 1, , then how many, 2, (NDA 2008 I), , (b) Two, (d) Infinite, , 17. The equation sin-1 ( 3x - 4x3 ) = 3 sin-1 ( x ) is true for, all values of x lying in which one of the following, intervals?, (NDA 2008 I), , é 1 1ù, (a) ê - , ú, ë 2 2û, 1ù, é, (c) ê - 1 , - ú, 2û, ë, , é1 ù, (b) ê , 1ú, ë2 û, , (c), , 2, , 1, 2, , (d), , 2 2, , D., , p, 2, p, (d), 3, , 20. What is the value of sin [cot-1 {cos (tan-1 x )}], where, x > 0,, (a), (c), , ( x 2 + 1), ( x 2 + 2), ( x 2 + 1), ( x 2 + 1), , 21. Consider the following, p, æ 2 ö, I. cosec-1 ç ÷=è, ø, 3, 3, , B., C., , (b), , p, 4, , A., , 1, , (NDA 2007 II), , (c), , 23. Match List I (function ) with List II ( their values), and select the correct answer using the codes given, below the lists., , (b) 0, , æm - nö, æmö, 19. What is the value of tan-1 ç ÷ - tan-1 ç, ÷?, ènø, èm + nø, (a) p, , Which of the following statements given above are, correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, , List I, (Function), , (NDA 2007 II), , 1, , (b), (d), , ( x 2 + 2), ( x 2 + 1), ( x 2 + 2), ( x 2 + 1), , (NDA 2011 I), , 22. Consider the following statements, æbö p, æaö, I. If tan -1 ç ÷ + tan -1 ç ÷ = , then x = ab ., è xø 2, è xø, II. If tan -1 ( x - 1) + tan -1 ( x ) + tan -1 ( x + 1) = tan -1 ( 3x ),, 1, then x = 0, ± ., 2, , (d) [- 1, 1], , é, ì, æ 15p öü ù, 18. What is the value of cos ê tan-1í tanç, ÷ý ?, è 4 øþ úû, î, ë, (a) -, , æ 2 ö p, II. sec-1 ç, ÷=, è 3ø 6, Which of the above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, , List II, (Their values), , æ, æ3öö, cos ç tan -1 ç ÷ ÷, è4øø, è, , 1., , 1, 1ü, ì, sinísin -1 + cos -1 ý, 2, 2þ, î, 1, 4, æ, ö, sin ç cos -1 ÷, è2, 5ø, 4, 2ü, ì, tan í cos -1 + tan -1 ý, 5, 3þ, î, , 1, 17, 6, 4, 5, 1, 10, , 2., 3., 4., , Codes, , A, (a) 3, (c) 3, , B, 1, 1, , C, 4, 2, , D, 2, 4, , A, (b) 1, (d) 1, , B, 3, 3, , C, 4, 2, , D, 2, 4, , Directions (Q. Nos. 24-26) Each of these questions, contain two statements, one is Assertion (A) and other is, Reason (R). Each of these questions also has four, alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true.
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288, , NDA/NA Mathematics, , 24. Assertion (A) sin-1 x = tan-1, , x, , Directions (Q. Nos. 27-29), , 1 - x2, , Consider, , the, , statement, , Reason (R) -1 < x < 0, æ1ö, æ1ö, 25. Assertion (A) The value of 2 tan-1 ç ÷ + tan-1 ç ÷ is, è 7ø, è 3ø, p, equal to ., 2, 2x, Reason (R), 2 tan-1( x ) =, 1 - x2, æx+ yö, and, tan-1 x + tan-1 y = tan-1 ç, ÷, è 1 - xy ø, 26. Assertion, (A) If, - 1 £ x < 0,, cos(sin-1 x ) = - 1 - x 2, Reason (R) If - 1 £ x < 0, then, , then, , If sin-1 x + sin-1 y + sin-1 z =, when, , 3p, 2, , - 1 £ x , y , z £ 1, then, , 27. The value of x 2000 + y 2001 + 32002 is, (a) 1, (b) 2, (c) 3, (d) None of these, 28. The value of x 4 + y 4 + z 4 - 4x 2 y 2z 2 + 4xyz is, (a) 1 + 2x 2z 2 + 2 y 2z 2 - 2x 2 y 2, (b) x 2z 2 + y 2x 2, (c) x3 y3, (d) None of the above, 29. The value of cos(sin-1 x + sin-1 y ) is, (a) xyz, (b) - z, (c) x 2 + y 2, (d) None of these, , sin(cos-1 x ) = 1 - x 2, , Answers, Level I, 1., 11., 21., 31., , (c), (b), (c), (a), , 2., 12., 22., 32., , (a), (a), (c), (b), , 3., 13., 23., 33., , (d), (b), (c), (d), , 4., 14., 24., 34., , (c), (d), (a), (b), , 5., 15., 25., 35., , (d), (c), (d), (b), , 6., 16., 26., 36., , (a), (d), (c), (a), , 7., 17., 27., 37., , (a), (d), (b), (a), , 8., 18., 28., 38., , (a), (c), (d), (d), , 9., 19., 29., 39., , (b), (d), (a), (d), , 10., 20., 30., 40., , (b), (b), (b), (a), , Level II, 1. (c), 11. (c), 21. (c), , 2. (c), 12. (b), 22. (c), , 3. (d), 13. (d), 23. (a), , 4. (c), 14. (d), 24. (a), , 5. (d), 15. (c), 25. (d), , 6. (c), 16. (b), 26. (d), , 7. (d), 17. (a), 27. (c), , 8. (c), 18. (c), 28. (a), , 9. (a), 19. (c), 29. (b), , 10. (b), 20. (a)
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16, , Functions, Constant, A quantity which retains the same value throughout a, set of mathematical operation is called a constant., , Variable, A quantity which can take a number of values is called, a variable., Variables are of two types, (i) Independent variable A variable which can, take an arbitrary value from a given set is called an, independent variable., (ii) Dependent variable A variable whose value, depends upon the value of independent variable is, called a dependent variable., , Function, Let X and Y be two non-empty sets. A function f of X, into Y (or from X to Y ) written as f : X → Y is a rule or a, correspondence which connects every member, say x of X, to exactly one member say y of Y., , Domain and Range of a Function, The set X is domain (or definition or inputs) of f and Y is, called codomain of f. The set of f ( x ), {f ( x ), x ∈ X } consisting, of all images f (x), is called the range of f i.e., range is, always a subset of the codomain. Range is also set of, outputs., %, %, , y = f (x ) is not a function ( y as a function of x), if there is an input, for which we get more than one outputs, y = f (x ) is not a function, if there is an element in the set of X for, which we do not have it’s image (corresponding element in set Y)., , Algebraic Operation on Functions, (Combination of Functions), If f and g are two functions with domains D1 and D2, respectively, then, , 1. ( f + g) x = f ( x ) + g( x ) with domain D1 ∩ D2, 2. ( f − g) x = f ( x ) − g( x ) with domain D1 ∩ D2, 3. ( fg) x = f ( x ) ⋅ g( x ) with domain D1 ∩ D2, f(x), f, 4. ( x ) =, , provided g( x ) ≠ 0 with domain D1 ∩ D2, g( x ), g, , Example 1. The domain of the function f ( x) = x2 − 4 is, (a) (−∞, +∞), (c) (−∞, −2 ] ∪ [2, ∞), , (b) (−∞, −2], (d) None of these, , Solution (c) Since, f ( x) is real number for which x2 − 4 ≥ 0, ⇒, , x2 ≥ 4 ⇒ x ≥ ± 2, , ⇒, x ∈ ( −∞ , − 2] ∪ [2, ∞), ∴ Required domain of f is ( −∞ , − 2] ∪ [2, ∞)., , Example 2. The domain and range of the function, f ( x) =, , 1, are, respectively, 2 − cos 3x, 1 , 1 , (b) R − {}, (a) R, ,1, 1 , ,2, 3 , 3 , (c) R − {2}, [2, 3], (d) None of these, , Since, −1 ≤ cos 3x ≤ 1 for all x ∈ R ,, therefore −1 ≤ − cos 3x ≤ 1 ⇒1 ≤ 2 − cos 3x ≤ 3 for all x ∈ R, ⇒ f ( x) is defined for all x ∈ R., ∴ Domain of f is a set of real number R., 1, Range Let, f ( x) = y ⇒, =y, 2 − cos 3x, 1, 1, 2 − cos 3x =, ⇒ cos 3x = 2 −, ⇒, y, y, 1, ⇒, −1 ≤ 2 − ≤ 1, [Q − 1 ≤ cos 3x ≤ 1], y, 1, ⇒, −3 ≤ − ≤ −1, y, 1, ⇒, 3 ≥ ≥1, y, 1, 1 , ≤ y ≤ 1 ⇒ y ∈ , 1, ⇒, 3, 3 , 1 , ∴ Range of f is , 1., 3 , , Solution (a) Domain
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298, , NDA/NA Mathematics, , Intervals, , Onto and Into Functions, , Closed interval If a real number x is such that, a ≤ x ≤ b,then x lies in closed interval a and b and is written, as x ∈ [a , b]. e.g., if x ∈[2, 3] ⇒ 2 ≤ x ≤ 3 (here end points are, included)., Open interval If a real number x is such that, a < x < b,then x lies in the open interval a and b and written, as x ∈( a , b). e.g., x ∈ ( 2,3) ⇒ 2 < x < 3. (here, end points are, not included). Interval of real number R is written as, ( −∞ , ∞ )., Half open, half closed interval If x is such that, a < x ≤ b. It is written as ( a , b] or x ∈ ] a , b ] e.g.,, x ∈ ( 2, 3] ⇒ 2 < x ≤ 3 (here, left end point is not included, but right end point is included)., Half closed, half open interval If x is such that, a ≤ x < b. It is written as [a , b) or [a , b[ e.g., x ∈ [2, 3) or [2, 3[, ⇒ 2 ≤ x < 3. (here, left end point is included but right end, point is not included.)., One-one and many-one functions If, f ( x1 ) = f ( x2 ) ⇒ x1 = x2 , then function f ( x ) is one-one. i.e., If, distinct elements in domain of a function f have distinct, images in codomain, then f is said to be one-one and if,, there exist atleast two distinct elements in domain having, same image,then it is many-one function. One-one, functions are also called injective functions., , If each element in codomain have atleast one, pre-image in the set of domain, then the function is onto, and if there is even single element in the set of codomain, which does not have it’s pre-image, then it is into function, or if range is same as codomain, then function is onto and if, range is proper subset of codomain, then the function is, into. Onto functions are also called ‘surjective functions’., e.g.,, π π, 1. If a function f : − , → R is defined as y = sin x., 2 2, Here, codomain of this function is R while range is, [−1, 1]. Hence, the function is into function., 2. If a function f: [− 1, 1] → [0, 1] is defined as y = x 2., Here, codomain of this function is [0, 1] and range is, also [0, 1], hence, the function is onto function., , y-axis, , y-axis, , %, , Functions which are one-one functions as well as onto functions, are called ‘bijective functions’., , Example 3. If the function f : R → A given by f ( x) =, a surjection, then the set A is, (a) [0, 1), (c) (2,3), , x-axis, One-one function, , O, , x-axis, Many-one function, , Methods to Find one-one and, Many-one Functions, 1. If x1 ≠ x2 ⇒ f ( x1 ) ≠ f ( x2 ), then f ( x ) is one-one., 2. If f ( x1 ) = f ( x2 ) ⇒ x1 = x2 , then f ( x ) is one-one., 3. Any function, which increases monotonically or, decreases monotonically is one-one function i.e.,, f ′( x ) > 0 for all x in domain or f ′( x ) < 0 for all x in, domain., 4. Any continuous function f(x) which has atleast one, local maxima or local minima, is many-one., 5. If there is any line parallel to x-axis intersecting the, graph of y = f ( x ) in more than one point, then, function is many-one., 6. We put f ( x1 ) = f ( x2 ). Since, x1 = x2 always satisfies, f ( x1 ) = f ( x2 ), so ( x1 − x2 ) will be a factor of f ( x1 ) − f ( x2 )., Hence,, we, can, write, as, f ( x1 ) − f ( x2 ) = ( x1 − x2 ) g( x1 , x2 ), where g( x1 , x2 ) is, some function of x1 and x2. Now, if g( x1 , x2 ) = 0, and x1 ≠ x2, hence f ( x ) is many-one., , is, , (d) None of these, x2, x2 + 1, , y ( x2 + 1) = x2 ⇒ x2 (1 − y) = y, , y, y, ⇒x=, 1− y, 1− y, y, Since, x is real, therefore, ≥ 0 ⇒0 ≤ y <1, 1− y, ⇒, , O, , x2 + 1, , (b) (1,2), , Solution (a) We have, y = f ( x ) =, ⇒, , x2, , x2 =, , Therefore, the set A = [0 , 1), , Some Special Functions, Constant function The function f : R → R is said, to be a constant function, if f ( x ) = c, ∀ x ∈ R, Domain of f = R and Range of f = {c}., Identity function The function f : R → R is said, to be an identity function, iff, f(x) = x, ∀ x ∈ R, Domain of f = R and Range of f = R, Polynomial function A function f : R → R, defined by f ( x ) = a0 + a1 x + a2 x 2 + ... + an x n , where, a0 , a1 , a2 , ... , an are real constants and n is a non-negative, integer, is called a polynomial function., Rational function Let, P ( x ) = a0x n + a1x n − 1 + K + an, and Q( x ) = b0x m + b1x m − 1 + K + bm, Then, f ( x ) =, , P( x), is called a rational function., Q( x )
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299, , Functions, Exponential function The function defined by, f ( x ) = a x , x ∈ R is called an exponential function., Domain of f = R and Range of f = ( 0, ∞ ), Logarithmic function The function defined by, f ( x ) = loga x is called a logarithmic function., Domain of logarithmic function = ( 0, ∞ ), Range of, logarithmic function = R, Trigonometric, function The, functions, involving circular functions of variable angles are known, as trigonometric function., Function, , Domain, , Range, , sin x, , R, , [ − 1, 1 ], , cos x, , R, , [ − 1, 1 ], , tan x, cot x, sec x, cosec x, , R − {( 2 n + 1), , π, : n ∈ I}, 2, , R − {nπ : n ∈ I}, π, R − {(2 n + 1) : n ∈ I}, 2, , ( − ∞, − 1 ] ∪ [1, ∞), , R − {nπ : n ∈ I}, , ( − ∞, − 1 ] ∪ [1, ∞), , sin x, cos −1 x, tan −1 x, , [ − 1, 1 ], [ − 1, 1 ], R, , R, R, , cot−1 x, sec −1 x, , R, ( − ∞, − 1 ] ∪ [1, ∞), , cosec −1 x, , ( − ∞ , − 1 ] ∪ [1, ∞), , π, : n ∈ I}, 2, R − {nπ : n ∈ I}, π, R − {( 2 n + 1) : n ∈ I}, 2, R − {n π : n ∈ I}, R − {( 2 n + 1), , Absolute Value Function, (Modulus Function), x, x ≥ 0, y =|x| = , − x, x < 0, x 2 =|x|, y-axis, , y=–x, , 1, if x > 0, |x|, ,x ≠ 0 , , = −1, if x < 0, f(x) = x, 0 , x = 0 0, if x = 0, , y-axis, , 1, , x-axis, , R, , Range, , Also,, , Signum Function, , O, , Domain, , −1, , 1.|x|≤ a ⇔ − a ≤ x ≤ a 2.|x|≥ a ⇒ x ≤ − a or x ≥ a, 3.|x ± y|≤|x| +| y|, 4.|x ± y|≥||x| −| y||, , R, , Inverse, trigonometric, function The, functions involving inverse trigonometric ratios are known, as inverse trigonometric functions., Function, , Properties of Modulus Functions, , –1, , It is written as sign x. Domain ∈ R, Range ∈ {−1, 0, 1} ., , Even and Odd Functions, A function is said to be an even function, if f ( − x ) = f ( x ),, e.g.,, f ( x ) = cos x , f ( x ) = x 2 etc., On the other hand a function is said to be an odd, function, if, f(− x) = − f(x), e.g., f ( x ) = sin x , f ( x ) = x3 etc., , Properties of Even and Odd, Functions, 1. Derivative of an odd function is an even function and, derivative of an even function is an odd function., 2. Each function can be expressed as the sum of an even, and an odd function., 3. Graph of an even function is symmetrical about y-axis, and the graph of an odd function is symmetric about, origin., 4. f ( x ) = 0 is both an even and an odd function., 5. f ( x ) = constant, is an even function., , Even Extension and Odd, Extension of a Function, , y=x, , Let y = f ( x ) is defined in [a , b], then y = f ( − x ) is it’s, even extension in [− b, − a ] and y = − f ( − x ) is it’s odd, extension in [− b, − a ]., 45°, , 45°, O, , x-axis, , %, , Here, a and b both are either non-negative or non-positive.
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300, , NDA/NA Mathematics, , Example 4. If f is an even function defined on the interval, [ −5, 5] , then the number of real values of x satisfying the, x + 1, equation f ( x) = f , is, x + 2, (a) 1, (b) 2, (c) 3, (d) 4, Solution (d) Since, f ( x ) is an even function defined on [ −5, 5]., Therefore, f ( − x ) = f ( x ) for all x ∈ [ −5, 5], x + 1, f ( x) = f , Q, , x + 2, ⇒, , x=, , x + 1, x+1, or x = − , , x+2, x + 2, , ⇒, or, , x2 + x − 1 = 0, 2, x + 3x + 1 = 0, −1 ± 5, ⇒, x=, 2, −3 ± 5, or, x=, 2, −1 + 5 −1 − 5 −3 + 5, Hence, the values of x are, ,, ,, 2, 2, 2, −3 − 5, and, ., 2, , Periodic Function, A function f : X → Y is said to be a periodic function, if, there exist a positive real number T such that, f ( x + T ) = f ( x ), ∀ x ∈ X . The least value of T is called the, fundamental period of a function. In general, the, fundamental period (principal period) is called the period, of a function. Graphically, graph gets repeated after each, period of the function., , Properties of a Periodic Function, 1. If f ( x ) is a periodic function with fundamental period, 1, will also be a periodic function with same, T, then, f(x), fundamental period T., 2. If f ( x ) is a periodic function with period T, then, f ( ax + b) is also a periodic function with fundamental, T, period, ., |a|, 3. If f ( x ) is a periodic function with period T, then, af ( x ) + b is also a periodic function with same, fundamental period T., 4. If f ( x ) and g( x ) are two functions with fundamental, periods T1 and T2 respectively, then f ( x ) + g( x ) is a, periodic function with fundamental period LCM of, T1 and T2 , provided f ( x ) and g( x ) cannot be, interchanged by adding a positive number in x, which is less than LCM of T1 and T2, in that case, period becomes that number and also LCM of T1, and T2 should exist otherwise this is not a periodic, function., , 5. If f ( x ) is a periodic function with period T and g (x) is, a monotonic function, then g (f (x)) is also a periodic, function with same period T as that of f (x)., x, 2x, e.g., f ( x ) = sin + cos, is a periodic function with, 3, 3, LCM of 6π and 3π, so period of f ( x ) is 6π., , Example 5. Let f ( x) be a function and k be a positive real, number such that f ( x + k) + f ( x) = 0, ∀ x ∈ R, then f ( x) is a, periodic function with period, (a) k, (b) 2k, (c) 3k, (d) 4k, , Solution (b) We have,, f ( x + k ) + f ( x ) = 0 , ∀ x ∈R, ⇒, f ( x + k ) = − f ( x ), ∀ x ∈ R, ⇒, f ( x + 2k) = − f ( x + k), ∀ x ∈ R, ⇒, f ( x + 2k) = f ( x ), ∴ f ( x) is a periodic function with period 2k., , Identical Functions or Equal, Functions, Two functions y = f ( x )and y = g( x )are equal functions,, if f ( x ) = g( x ) and they have same domain and range., x2, x3, e.g., f ( x ) = x and g( x ) =, and g( x ) = 2 are equal, x, x, functions (identical functions)., , Explicit Functions, y is said to be an explicit function of the independent, variable x, if we can write y = f ( x ) i.e., if y can be expressed, directly in terms of the independent variable x., e.g.,, y = x 2 , y = 2x3 + 3x 2 + x + 9, , Implicit Functions, y is said to be an implicit functions of the independent, variable x, if it cannot be expressed directly in terms of x., e.g., x3 y 2 + 3xy 2 + x3 + y3 + 3 y + 2 x + 5 = 0, , Invertible Functions, A function y = f ( x )is said to be invertible if and only if f, is one-one onto., All one-one and onto functions (bijective functions) are, invertible functions. Their graphs are symmetrical about, the line y = x., , Method of Finding Inverse of a, Function, Suppose, y is written as a function of x., l First find x in terms of y, −1, l and write it as f, ( y), , (step 1), (step 2)
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301, , Functions, , l, , Now, replace y by x., (step 3), π π, e.g., y = sin x , D ∈ − , and codomain ∈ [−1,1], 2 2, x = sin, , −1, , sin, %, , −1, , y =f, , −1, , x = f (x), , If f −1 (x ) is inverse function of y = f (x ), then D f = R f−1 and, D f−1 = R f, , Example 6. The inverse of the function f ( x) =, y , (b) log , , 2 − y, , y, 2 −y, 1, (c), 1− y, , ⇒, , ( e x − e − x), ex + e− x, , +1, , 1/ 2, , ex − e− x, +1, ex + e− x, , 5, and g = 1, then the value of gof ( x) is, 4, (a) 2, , y , x = log , , 2 − y, , =, , Let f : X → Y and g : Y → Z be two functions, we define, a function h : X → Z by putting h ( x ) = g ( f ( x ))., , (d) 4, , 1, 2π , π, , , 1 − cos 2x + 1 − cos 2x +, + cos 2x + , , , 3, 2 , 3, + cos, , π, 3 , , =, , 1, 2π , π 1, , , , 2 − cos 2x + cos 2x +, + cos 2x + + , , , , 2 , 3, 3 2, , , , =, , 1, 2, , π, π , 5, , , π, 2 − 2 cos 3 cos 2x + 3 + cos 2x + 3 , , , , 1 5, π, π , , , − cos 2x + + cos 2x + , , , 2 2, 3, 3 , 5, = , ∀ x ∈R, 4, 5, gof ( x) = g ( f ( x)) = g , 4, =, , 1/ 2, , Composite Functions, , (c) 3, , π, π, , , f ( x) = sin 2 x + sin 2 x + + cos x cos x + , , , 3, 3, , y , 1, log , , 2, 2 − y, , ⇒, , (b) 1, , Solution (b) Given that,, , y , y, = e2x ⇒ 2 x = log , , 2 −y, 2 − y, x=, , 3, π, , + cos x cos x + , , 3, , e2x − 1, y − 1 = 2x, e +1, , ⇒, , g (x), , π, Example 7. If f ( x) = sin2 x + sin2 x + , , Applying componendo and dividendo rule,, 2e2x, y − 1 + 1 e2x − 1 + e2x + 1, y, ⇒, = 2x, =, −2, y − 1 − 1 e − 1 − e2x − 1 y − 2, ⇒, , f (x), , , , (d) None of these, , Solution (b) Let y = f ( x) =, , x, , (step 3), , is, , g, , f, , (let) (step 2), , ( y), , −1, , (a), , h = gof, , ∴, , =1, , , , 5, given g 4 = 1 , ,
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302, , NDA/NA Mathematics, , Comprehensive Approach, n, , n, n, n, n, , n, , n, n, , n, , n, n, , n, , If a polynomial satisfies the property, 1, 1, f ( x) + f = f ( x) f , then f ( x ) = 1 ± xn ., x, x, Domain of polynomial functions is ( − ∞ , ∞)., Range of odd degree polynomial functions is ( − ∞ , ∞)., Even functions are many-one functions., Any function can be written as sum of an even function and an, odd function., f ( x) + f ( − x) f ( x) − f ( − x), i.e.,, f ( x) =, +, 2, 2, Every constant function is an even function and f ( x) = 0is an even, function as well as an odd function., f ( x) = x function is an identity function. (Df ∈ R and Rf ∈ R), Constant functions are periodic functions with no fundamental, period., If f ( x) is inverse function of g( x) , then f ( g( x)) = x or g( f ( x)) = x, i.e.,, same as f −1( f ( x)) = x or f ( f −1( x)) = x., f ( x) = 0 is the only function which is both even and odd., If f ( x) is an odd function (or even function), f '( x) is an even function, (or odd function). f ( x) is differentiable of R., If A and B are finite having m and n elements respectively, then, n P , if n ≥ m, (a) Number of one-one function from A to B = m, 0 , if n < m, , a, , r =1, , n, , n, , (c) Number of bijection = m!, Number of disjoint elements in A and B are m and n respectively ,, then number of mappings from A to B = mn ., If A = B, then number of mappings from A to B = nn ., , = f ( x), , = 2n log |x| , where n ∈ N., 1, 1, ≥ 2 (If x ∈ R+ ) and x + ≤ − 2 (If x ∈ R− ), x, x, Two functions y = f ( x) and y = g( x) can be combined (added,, subtracted, multiplied or divided) only in common domains., Range of f ( x) = a sin x + b cos x + c is, 2n, , log x, , x+, , n, , n, , [c −, n, , n, , n, n, n, n, , n, , (b) Number of onto functions = Σ ( −1) n − r nC r r m, , log a f ( n ), , n, , n, , a2 + b 2 , c +, , a2 + b 2 ], , Range of f ( x) = ax + bx + c is, D, D, , , − 4 a , ∞ , if a > 0 and − ∞ , − 4 a , if a < 0, , , (where D = b 2 − 4 ac), a, c LCM of a and c, LCM of and =, b, d HCF of b and d, LCM of a rational number and an irrational number does not exist., The graph of an odd function is symmetric about origin., The graph of an even function is symmetric about y-axis., If A and B are two different sets, the number of elements in each of, these is n, then number of one-one onto functions= n!., (a) If gof is one-one ⇒ f is one-one, (b) If gof is onto ⇒ f is onto, (c) If f and g are one-one onto, then gof will also be one-one, onto., If f ( x) is any function, then, (a) f ( x) + f ( − x) is an even function., (b) f ( x) − f ( − x) is an odd function., ax + b, If y = f ( x) =, , then ( fof ) ( x) = x., x−a, 2
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Exercise, Level I, 1. Suppose that g ( x ) = 1 + x and, f ( g (x)) = 3 + 2 x + x, then f (x) is, , (a) 1 + 2 x 2, (c) 1 + x, , (b) 2 + x 2, (d) 2 + x, , 2. The domain of the function log( x 2 − 6 x + 6) is, (a) ( − ∞ , ∞ ), (b) ( − ∞ , 3 − 3 ) ∪ ( 3 + 3 , ∞ ), (c) ( − ∞ , 1] ∪ [ 5, ∞ ), (d) [0, ∞ ), 3. If n is an integer, then domain of the function sin 2x, is, π, π, , , , (a) nπ − , nπ, (b) nπ , nπ + , 2, 2, , , , (c) [( 2n − 1) π , 2nπ ], (d) [2nπ , ( 2n + 1) π ], 4. The domain of the function f ( x ) = exp ( 5x − 3 − 2 x 2 ), is, (a) [ 3/ 2, ∞ ), (b) [1, 3/ 2], (c) ( − ∞ , 1], (d) (1, 3/ 2), 5. If f : R → R, g : R → R and g( x ) = x + 3 and, ( fog) ( x ) = ( x + 3)2, then what is the value of f( − 3)?, (NDA 2010 I), , (a) − 9, , (b) 0, , (c) 9, (d) 3, 1, 2, 6. Given that f ( x ) = x + , then f ( x ) is equal to, x, (NDA 2007 II), , x +1, x, + 2, x, x +1, 1, (c) x 4 + 4, x, 2, , (a), , 1, , (b) x + , , x, 1, (d) x 2 + 2, x, , 2, , 7. The inverse f −1( x ) of the function f : [R{ 1}] → [R/{ 1}], x+1, is, defined by f ( x ) =, x−1, x+1, x−1, (a), (b), x−1, x+1, 1, (d) x, (c), x, 8. If f ( x + 1) = x 2 − 3x + 2 , then f ( x ) is equal to, (a) x 2 − 5x − 6, (b) x 2 + 5x − 6, 2, (d) x 2 − 5x + 6, (c) x + 5x + 6, 9. If y = f ( x ) =, (a) 2x, (c) −x, , ax + b, , then f ( y ) is equal to, cx − a, (b) x, (d) −2 x, , 10. Let g( x ) = x3 − 4x + 6. If f ′ ( x ) = g′ ( x ) and f(1) = 2,, then what is the value of f ( x )?, (NDA 2009 II), (b) x3 − 4x + 6, (a) x3 − 4x + 3, (d) x3 − 4x + 5, (c) x3 − 4x + 1, 3x + 2, , then the value of f −1( x ) is, 11. If f ( x ) =, 5x − 3, 3x + 2, (a) 4x − 2, (b), 5x − 3, x−1, (c), (d) None of these, x+1, 12. The period of function f ( x ) = sin4 x + cos4 x is, π, (a) π, (b), (c) 2π, (d) 3π, 2, 13. The, domain, of, the, function, f ( x ) = log ( x − 4 + 6 − x ) is, (a) [4, ∞ ), (c) [4, 6], , (b) ( − ∞ , 6], (d) None of these, , 14. The range of f ( x ) = cos ( x/ 3) is, (a) ( − 1 / 3, 1 / 3), (b) [− 1, 1], (c) (1 / 3, − 1 / 3), (d) ( − 3, 3), 15. The domain of the function f ( x ) = sin−1 [log2 ( x/ 2)] is, (a) [ 1, 4], (b) [− 4, 1], (c) [− 1, 4], (d) None of these, x+2, 16. The range of the function f ( x ) =, is, |x + 2|, (a) { 0, 1} (b) { − 1, 1}, (c) R, (d) R − { − 2}, 2, 3, 17. If f ( x ) = x + , x ∈ R, then what is the value of, 3, 2, f −1( x )?, (NDA 2010 II), 3x 2, 3x 9, (a), (b), +, −, 2, 3, 2, 4, 2x 4, 2x 2, (d), (c), −, −, 3, 9, 3, 3, 18. If f ( x ) = x and g( x ) =| x|, then what is the value of, (NDA 2008 I), ( f + g) ( x )?, (a) 0, ∀x ∈ R, (b) 2 x,∀x ∈ R, 0, for x ≥ 0, 2x , for x ≥ 0, (d) , (c) , 2x , for x < 0, 0, for x < 0, 19. The inverse of the function y =, (a) log10 ( 2 − x ), (c), , 1, 2x, log10, 4, 2− x, , 10 x − 10− x, , is equal to, 10 x + 10− x, 1, (b), log10 ( 2x − 1), 2, , (d), , 1, 1+ x, log10, 2, 1− x
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304, , NDA/NA Mathematics, , 2x , 1 + x, 20. If f ( x ) = log , is equal to, , then f , 1 − x, 1 + x2 , 1 −, (a) f , 1 +, , x, , x, , (c) 1, , (b) f ( x )2, (d) 2f ( x ), , 1 − x, 21. If f ( x ) = log , . What is the value of f ( a ) + f ( b) ?, 1 + x, (a) f ( a + b), , a − b, (b) f , , 1 + ab, , a + b, (c) f , , 1 + ab, , (d) f ( a − b), , 22. Let fog ( x ) = F ( x ). If F ( x ) = a 2 − x 2 and g( x ) = − x 2 ,, what is the value of f ( x ) ?, (a), , a2 − x, , (b), , a2 + x, , (c), , a2 + x2, , (d), , a2 − x2, , 1, π, 23. If g( x ) = sin x, x ∈ R and f ( x ) =, , x ∈ 0, , what, 2, sin x, is the value of ( gof ) ( x )?, (NDA 2008 I), 1, (a) 1, (b), sin (sin x ), 1 , 1, (c), (d) sin , , sin x , sin2( x ), 1, 24. If f ( x ) = 2 + , then which one of the following is, x, correct?, (a) As x increases, f ( x ) also increases., (b) As x becomes larger and larger, f ( x ) assumes, values nearer to 2., (c) As x increases, f ( x ) takes values nearer and, nearer to 3 for sufficiently large values of x., (d) As x becomes larger and larger, it is not possible, to find the value of f ( x )., 1 − x, 1− x, , then f , is equal to, 1 + x, 1+ x, 1− x, (a) x, (b), 1+ x, 1+ x, 1, (d), (c), 1− x, x, , 25. If f ( x ) =, , 26. Let y 2 = 4ax , a ≠ 0. Now, consider the following, statements, I. y = 2 ax expresses y as a function of x., II. y = − 2 ax expresses y as a function of x., III. y = ± 2 ax expresses y as a function of x., Which of these is/are correct?, (a) I and II, (b) I and III, (c) II and III, (d) Only III, , 2x − 1, , ( x ≠ 5), then f −1( x ) is equal to, x+5, 1, x+5, 5x + 1, (b), (a), ,x ≠, ,x ≠2, 2x − 1, 2, 2− x, 1, x−5, 5x − 1, (d), (c), ,x ≠ −, ,x ≠2, 2x + 1, 2, 2− x, , 27. If f ( x ) =, , 2x + 1, , then ( fof ) (2) is equal to, 3x − 2, (a) 1, (b) 3, (c) 4, (d) 2, 1+ x, , then f ( x ) is, 29. If f ( x ) = log, 1− x, (a) even function, (b) f ( x1 ) f ( x2 ) = f ( x1 + x2 ), f ( x1 ), (c), = f ( x1 − x2 ), f ( x2 ), (d) odd function, 28. If f ( x ) =, , 30. What, is, the, period, of, f ( x ) =|sin x| +|cos x| ?, (a) π/ 2, (b) π, (c) 2π, (d) π/ 4, , the, , function, , 31. What is the inverse of f ( x ) = (1 − x3 )1/ 5 + 2 ?, (a) [1 − ( x − 2)5 ]1/ 3, (b) [1 + ( x − 2)5 ]1/ 3, 1/ 3, (d) [1 + ( x + 2)5 ]1/ 3, (c) ( x + 2), 32. A, , function, satisfies, the, condition, f(x), 1, 1, , 2, f x + = x + 2 , x ≠ 0. What is the value of f ( x ) ?, , x, x, 2, (a) x − 2 for all x ≠ 0, (b) x 2 − 2 for all x satisfying|x|> 2, (c) x 2 − 2 for all x satisfying|x|< 2, (d) x 2 + 2 for all x ≠ 0, , 3x + 2 , 3, x ≠ , then which one of the, 5x − 3 , 5, following is correct?, (a) f −1( x ) = f ( x ), (b) f −1( x ) = − f ( x ), 1, (c) ( fof ) ( x ) = − x, (d) f −1x = −, f(x), 19, x( x − p) x ( x − q ), 34. f ( x ) =, +, , p ≠ q. What is the value of, q− p, p− q, f ( p) + f ( q ) ?, (a) f ( p − q ), (b) f ( p + q ), (c) f ( p ( p + q )), (d) f ( q ( p − q )), 33. If f ( x ) =, , 1 , 35. If 2 f ( x + 1) + f , = 2 x and x ≠ − 1, then f ( 2) is, x + 1, equal to, (a) −1, (b) 2, (c) 5/3, (d) 5/2, 36. The function f : R → R defined by f ( x ) = ex is, (a) onto, (b) many-one, (c) one-one and into, (d) many-one and onto
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305, , Functions, , f ( x ) = ( 3 − cos 2 x )−1 ?, 1 1, , , 4 2, 1 1, 4 , 2, , 37. If R denotes the set of all real number, then the, function f : R → R defined f ( x ) =| x| is, (a) one-one only, (b) onto only, (c) Both one-one and onto, (d) Neither one-one nor onto, , 44. What is the range of function, 1 , (b), (a) , 1, 4 , 1, 1, (d), (c) − , − , 4, 2, , 38. Let f : R → R be defined by f ( x ) = 2 x +| x|, then, f ( 2x ) + f ( − x ) − f ( x ) is equal to, (a) 2 x, (b) 2|x|, (c) − 2x, (d) − 2|x|, , 45. If X = { 1, 2, 3} and Y = { 0, 1} and f : X → Y defined by, f = {(1, 1), ( 2, 1), ( 3, 0)}, then f is, (a) one-to-one but not onto, (b) onto but not one-to-one, (c) one-to-one and onto, (d) neither one-to-one nor onto, , 39. If f : R → R and g : R → R defined by f ( x ) = 2 x + 3 and, g( x ) = x 2 + 7, then the value of x for which, f ( g ( x )) = 25, are, (a) ± 1, (b) ± 2, (c) ± 3, (d) ± 4, 40. If f ( x ) =, , cos2 x + sin4 x, sin2 x + cos4 x, , for x ∈ R, then f( 2002) equals, , 47. If f ( x ) = cos (loge x ) , then, , to, (a) 1, (c) 3, 41. If f ( x ) =, , (b) 2, (d) 4, 1, x + 2 2x − 4, , then f(11) is equal to, (a)7/6, (c) 6/7, , 46. Domain of the function f ( x ) = sin−1 (log2 x ) in the set, of real numbers is, (a) { x : 1 ≤ x ≤ 2}, (b) { x : 1 ≤ x ≤ 3}, , 1, (c) { x : − 1 ≤ x ≤ 2 }, (d) x : ≤ x ≤ 2, , 2, , +, , f(x) f( y) −, 1, , x − 2 2x − 4, , , for x > 2,, , (b) 5/6, (d) 5/7, , 42. Let f : R → R : f ( x ) = x 2 and g : R → R: g( x ) = x + 5,, then gof is, (a) ( x + 5), (b) ( x + 52 ), 2, (c) ( x + 5), (d) ( x + 5)2, 43. The function f satisfies the functional equation, x + 59, 3 f(x) + 2 f , = 10x + 30 for all real x ≠ 1. The, x−1 , value of f( 7) is, (a) 8, (b) 4, (c) −8, (d) 11, , (a) 1, (c) –2, , 1 y, , f + f ( xy ) has the value, , 2 x, , (b) 1/2, (d) 0, , 48. The range of the function sin( sin−1 x + cos−1 x ),| x|≤ 1, is, (a) [− 1, 1], (b) [1, − 1], (c) { 0}, (d) { 1}, 49. The domain of sin−1 (log3 x ) is, (a) [−1, 1], (b) [0, 1], 1 , (c) [0, ∞ ], (d) , 3, 3 , αx, , x ≠ − 1 for what value of α is f ( f ( x )) = x?, x+1, (a) 2, (b) − 2, (c) −1, (d) 2, , 50. If f ( x ) =, , Level II, 1. The largest possible set of real numbers which can be, 1, the domain of f ( x ) = 1 − is, x, (a) ( 0, 1) ∪ ( 0, ∞ ), (b) ( −1, 0) ∪ (1, ∞ ), (c) ( −∞ , − 1) ∪ ( 0 , ∞ ), (d) ( −∞ , 0) ∪ (1, ∞ ), 1, is, 2. The domain of the function f ( x ) = log, |sin x|, (a) R − { 2nπ, n ∈ I }, (b) R − { nπ, n ∈ I }, (c) R − { − π, π }, (d) ( − ∞ , ∞ ), 10 x − 10− x, 3. The inverse of the function, is, 10 x + 10− x, 1 + x, 1 − x, 1, 1, (a), (b), log10 , log10 , , , 1 − x, 1 + x, 2, 2, 2x , 1, (d) None of these, (c) log10 , , 2 − x, 4, , 4. Domain of f ( x ) = log|log x| is, (a) ( 0, ∞ ), (b) (1, ∞ ), (c) ( 0, 1) ∪ (1, ∞ ), (d) ( −∞ , 1), 5. Which one of the following is a bijective function on, the set of real numbers?, (a) 2x − 5, (b) | x|, (c) x 2, (d) x 2 + 1, 6. If p, q and r are positive integers, ω is the cube root of, unity and f ( x ) = x3 p + x3 q + 1 + x3 r + 2, then what is the, value of f(w)?, (NDA 2011 II), 2, (a) ω, (b) − ω, (c) − ω, (d) 0, 7. The function y = f ( x ) = mx + c has, (NDA 2011 II), (a) maximum point but no minimum point, (b) minimum point but no maximum point, (c) both maximum and minimum points, (d) neither maximum point nor minimum point
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306, , NDA/NA Mathematics, , 8. Let X and Y be subsets of R, the set of all real, numbers. The function F : X → Y defined by f ( x ) = x 2, for x ∈ X is one-one but not onto, (here R + is the set of all positive real numbers)., (b) X = R , Y = R +, (a) X = Y = R +, (c) X = R + , Y = R, (d) X = Y = R, 9. The, function, is, defined, by, f :R→ R, f ( x ) = cos2 x + sin4 x for x ∈ R, then f ( R ) equals to, 3 , 3 , 3 , 3 , (b) , 1, (c) , 1, (d) , 1, (a) ,1, 4 , 4 , , 4, 4, , , , 10 + x, , x ∈( − 10, 10) and, 10 − x, 200x , f ( x ) = kf , , then k is equal to, 100 + x 2 , (a) 0.5, (b) 0.6, (c) 0.7, (d) 0.8, , 10. If e f ( x ) =, , 12. Suppose f : [2, 2] → R is defined by, , for − 2 ≤ x ≤ 0, −1, ,, f (x) = , x − 1 for 0 ≤ x ≤ 2, then { x ∈{ −2, 2}: x ≤ 0 and f (| x|) = x} is equal to, , (b) { 0}, (d) φ, , 13. If f ( xy ) = f ( x ) f ( y ), then f(t) may be of the form, (NDA 2012 I), , (a) t + k, (b) ct + k, where R is constant., , (c) t k + c, , 3 − 1, (a) 0 < x < , , 2 , −1 − 5 , (c) , < x< 0, 2 , , , 1, x, , x + 1 is, , 5 − 1, (b) 0 < x ≤ , , 2 , 5 − 1, (d) 0 ≤ x < , , 2 , , 18. If f ( x ) = 2 x + 7 and g( x ) = x 2 + 7 , x ∈ R, then which, value of x will satisfy fog( x ) = 25?, (NDA 2010 II), (a) − 1 and 1, (b) − 2 and 2, (c) − 2 and 2, (d) None of these, 19. Let f : R → R be defined as f ( x ) = x|x|., , 11. If f : R → R and g : R → R are defined by f ( x ) = 2x + 3, and g( x ) = x 2 + 7, then the values of x such that, g( f ( x )) = 8, are, (a) 1, 2, (b) − 1, 2, (c) − 1, − 2, (d) 1, − 2, , (a) { − 1}, (c) { − 1 / 2}, , 17. The domain of the function, , (d) t k, , 14. If the roots of the equation x 2 − 4x − log3 N = 0 are, real, then what is the minimum value of N?, (NDA 2011 II), , 1, (a), 256, , 1, 1, 1, (b), (c), (d), 27, 64, 81, a, 15. If sin θ = x + , ∀ x ∈ R − { 0}, then which one of the, x, following is correct?, (NDA 2011 II), 1, (a) a ≥ 4, (b) a ≥, 2, 1, 1, (c) a ≤, (d) a ≤, 4, 2, 16. If a real valued function of a real variable is defined, as f ( x ) = cos−1( 3x − 1), then the domain of the, function f, is given by, 1, , (a) x : 0 ≤ x ≤ , (b) { x : 1 ≤ x ≤ 2}, 3, , 2, 3, , , (c) x : 0 ≤ x ≤ , (d) x : 0 ≤ x ≤ , 3, 2, , , , Which one of the following is correct?, (a) f is only into., (b) f is only one-one., (c) f is neither onto nor one-one., (d) f is one-one and onto., 20. Let f ( x ) be a differentiable even function. Consider, the following statements, I. f ′( x ) is an even function., II. f ′( x ) is an odd function., III. f ′( x ) may be even or odd., Which of the above statements is/are correct?, (a) Only I, (b) Only II, (c) I and III, (d) II and III, 21. Let g( x ) be the inverse of an invertible function f ( x ), which is differentiable at x = c. Which one of the, following is equal to g ′ [ f ( c)]?, 1, (a) f ′( c), (b), f ′( c), 1, (c) f ( c), (d), f ( c), 22. Let f : R → R be defined by f ( x ) =| x|/ x, x ≠ 0, f( 0) = 2., What is the range of f ?, (NDA 2009 II), (a) { 1, 2}, (b) { 1, − 1} (c) { −1, 1, 2} (d) { 1}, 23. A function f : R → R is defined by f ( x ) = ( x − 1) ( x − 2)., Which one of the following is correct in respect of, function?, (a) It is one-one but not onto., (b) It is onto but not one-one., (c) It is both one-one and onto., (d) It is neither one-one nor onto., 24. What is period of the function, f ( x ) =|sin x + cos x| +|sin x − cos x| ?, (a) π/ 6, (b) π/ 4, (c) π/ 2, , (d) π, , 25. Which one of the following is correct for the graph of, y =|x|?, (a) It lies only in the first quadrant of the xy-plane., (b) It lies only in the first and third quadrants of the, xy-plane., (c) It lies only in the first and second quadrants of, the xy-plane., (d) It lies only in the third and fourth quadrants of, the xy-plane.
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307, , Functions, 26. If f ( x ) = l x 2 + m x + n , x ∈ R, then what is the value of, f ( x + 3) − f ( x ), ?, f ( x + 2) − f ( x + 1), (a) 0, (b) 3, 3l − m + n, l+m+n, (c), (d), l − 2m + n, l+m−n, 27. If f ( x ) = ax + b and g( x ) = cx + d such that, fog( x ) = gof ( x ), then which one of the following is, correct?, (a) f ( a ) = g( c), (b) f ( b) = g( a ), (c) f ( c) = g( d ), (d) f ( d ) = g( b), 28. A rectangular box with a cover is to have a square, base. The volume is to be 10 cu cm. The surface area, of the box in terms of the side x is given by which one, of the following functions?, (NDA 2008 II), 2, (a) f ( x ) = ( 40 / x ) + 2 x, (b) f ( x ) = ( 40 / x ) + x 2, (c) f ( x ) = ( 40 / x ) + x, (d) f ( x ) = ( 60 / x ) + 2 x, 29. Which one of the following real valued function is, never zero?, (NDA 2008 I), (a) Polynomial function, (b) Trigonometric function, (c) Logarithmic function, (d) Exponential function, 30. Let f : [− 100 π , 1000 π ] → [− 1, 1] be defined by, f(θ ) = sin θ. Then, what is the number of values of, θ ∈ [− 100 π , 1000 π ] for which f(θ ) = 0? (NDA 2007 I), (a) 1000, (b) 1101, (c) 1100, (d) 1110, 31. Let f : R → R be defined as f ( x ) = ax 2 + bx + c, a , b, and c being fixed non-zero real numbers. Which one, of the following statements is correct in general?, (NDA 2007 I), , (a), (b), (c), (d), , If b2 − 4ac > 0, then f −1( 0) does not contain 0, If b2 − 4ac < 0, then f −1( 0) must contain 0, If b2 − 4ac > 0, then f −1( 0) may contain 0, If b2 − 4ac < 0, then f −1( 0) may contain 0, , Directions (Q. Nos. 32-36), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a),(b),(c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., , 32. Assertion (A) The function f ( x ) =, , sec4x + cosec4x, x3 + x 4 cot x, , is, , an odd function., Reason (R) f ( x ) is an odd function, since,, f ( x ) = − f ( − x )., 33. Assertion (A) If f ( x ) is a periodic function, then, T, ., f ( ax + b) is a periodic function with period, |a|, T, ., Reason (R) Since, f ( ax − b) has a period, |a|, 34. Assertion (A) The modulus function f ( x ) =|x| is not, one-one., Reason (R) The negative real numbers are not the, images of any real numbers., If f ( x ) = x and F ( x ) =, , 35. Assertion (A), , F ( x ) = f ( x ) always., Reason (R) At x = 0, F ( x ) is not defined., , x2, , then, x, , 36. Assertion (A) The function f ( x ) =|x|/ x does not, pass through the origin., Reason (R) f ( x ) is undefined for x ≤ 0., 37. Match List I with List II and select the correct, answer using the code given below the lists, List I, A., , List II, , f ( x) = cos x, , 1. The graph cuts y-axis in, infinite number of points, B., 2. The graph cuts x-axis in, f ( x) = ln x, two points, C. f ( x) = x2 − 5 x + 4 3. The graph cuts y-axis in, only one point, D., 4. The graph cuts x-axis in, f ( x) = e x, only one point, 5. The graph cuts x-axis in, infinite number of points, (NDA 2007 I), , Codes, A B, (a) 1 4, (c) 5 4, , C, 5, 2, , D, 3, 3, , A, (b) 1, (d) 5, , 38. Function ‘f ’ defined by f ( x ) = x +, , B, 3, 3, , C, 5, 2, , D, 4, 4, , 1, , then consider the, x, , following statements, (NDA 2008 II), 2, 2, I. { f ( x )} = f ( x ) + 2, II. { f ( x )}3 = f ( x3 ) + 3 f ( x ), Which of the correct in above statements?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 39. Consider the following statements, I. Every function has a primitive., II. A primitive of a function is unique.
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308, , NDA/NA Mathematics, Which of the statements given above is/are correct?, (NDA 2010 I), , (a) Only I, (c) Both I and II, , (b) Only II, (d) Neither I nor II, , 40. Consider the following statements, 1− x, is inverse of itself., I. f ( x ) =, 1+ x, II. f ( x ) = 5log x is inverse of itself., Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 41. Consider the following statements, I. The function log ( x + ( x 2 + 1)) is an odd, function., II. The function cos (log ( x + x 2 + 1 )) is an even, function., III. If f ( x ) is an even function, then curve y = f ( x ) is, symmetric about x-axis., Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Only III, (d) All I, II and III, , Directions (Q. Nos. 42-44) Consider, x 2 − 4x + 3, x < 3, ;, f(x) = , x − 4 , x ≥ 3, x − 3 ; x < 4, g( x ) = 2, x + 2x + 2, x ≥ 4 , then, , 42. The value of f + g, when x < 3 is, (b) x − 2, (a) x 2 − 3x, (c) x − 1, (d) None of these, f, 43. The value of when x ≥ 4 is, g, x−4, (a), x+2, x−4, (b) 2, x + 2x + 2, x+1, (c), x−1, (d) None of these, 44. The value of f ( g( x )) when x < 4 is, (a) ( x − 6)( x − 4), (b) ( x − 6)( x − 2), (c) ( x − 3), (d) None of these, , Directions (Q. Nos. 45-46) Consider the function, f(x) = x − x2, 45. The domain of f ( x ) is, (a) [0, 2], (c) [0, 3], , (b) [0, 1], (d) None of these, , 46. The range of f ( x ) is, 1, (b) 0, , 2, (d) None of these, , (a) [0, 3], (c) [0, 1], , Answers, Level I, 1., 11., 21., 31., 41., , (b), (b), (c), (a), (c), , 2., 12., 22., 32., 42., , (c), (b), (b), (a), (c), , 3., 13., 23., 33., 43., , (b), (c), (d), (a), (b), , 4., 14., 24., 34., 44., , (b), (b), (b), (b), (d), , 5., 15., 25., 35., 45., , (c), (a), (a), (c), (b), , 6., 16., 26., 36., 46., , (b), (b), (d), (c), (d), , 7., 17., 27., 37., 47., , (a), (b), (b), (d), (d), , 8., 18., 28., 38., 48., , (d), (c), (d), (b), (d), , 9., 19., 29., 39., 49., , (b), (d), (d), (b), (d), , 10., 20., 30., 40., 50., , (d), (d), (a), (a), (c), , 2., 12., 22., 32., 42., , (b), (c), (c), (a), (a), , 3., 13., 23., 33., 43., , (a), (d), (b), (b), (b), , 4., 14., 24., 34., 44., , (c), (d), (c), (a), (a), , 5., 15., 25., 35., 45., , (a), (c), (c), (d), (b), , 6., 16., 26., 36., 46., , (d), (c), (b), (c), (b), , 7., 17., 27., 37., , (d), (b), (d), (c), , 8., 18., 28., 38., , (d), (c), (a), (c), , 9., 19., 29., 39., , (c), (a), (d), (b), , 10., 20., 30., 40., , (a), (b), (c), (a), , Level II, 1., 11., 21., 31., 41., , (d), (c), (b), (a), (d)
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Hints & Solutions, Level I, 1. Given that, g (x) = 1 + x and f ( g (x)) = 3 + 2 x + x ... (i), ⇒ f (1 + x ) = 3 + 2 x + x, Put 1 + x = y ⇒ x = y − 1 ⇒ x = ( y − 1)2, Then, f ( y) = 3 + 2( y − 1) + ( y − 1)2, = 3 + 2 y − 2 + y2 + 1 − 2 y = 2 + y 2, Replacing y on x, we get f (x) = 2 + x2, 2. Let the function be f (x) = log (x2 − 6x + 6 ) is defined,, when log (x2 − 6x + 6) ≥ 0, ⇒, x2 − 6x + 6 ≥ 1 ⇒ x2 − 6x + 5 ≥ 0, ⇒, (x − 5) (x − 1) ≥ 0, This inequality hold, if x ≤ 1 or x ≥ 5. Hence, the domain, of the function will be (− ∞ , 1] ∪ [5, ∞ )., 3. Let f (x) = sin 2x, As sin 2x can’t be negative., We know sin x is positive, if 0 ≤ x ≤ π, π, ⇒, sin 2x is positive, if 0 ≤ x ≤, 2, ∴ Domain of function sin 2x is [nπ , nπ + π /2]., 4. Given that, f (x) = exp ( 5x − 3 − 2x2 ), or, , f (x) = e, , 5 x − 3 − 2x 2, , For domain, 5x − 3 − 2x2 > 0, ⇒, ⇒, ⇒, ∴, , 2 x2 − 5 x + 3 ≤ 0 ⇒ 2 x2 − 2 x − 3 x + 3 ≤ 0, 2x(x − 1) − 3 (x − 1) ≤ 0 ⇒ (2x − 3) (x − 1) ≤ 0, 3, 3, and x ≥ 1 ⇒ 1 ≤ x ≤, x≤, 2, 2, 3, Domain of f (x) = 1, , 2, , 5. Q fog (x) = (x + 3)2 = f ( g (x)), and, g (x) = x + 3, ⇒, f (x + 3) = (x + 3)2, ⇒, f (x) = x2, ⇒, f (−3) = (−3)2 = 9, 1, 1, , , 6. Given that, f (x) = x + ⇒ f 2(x) = x + , , , x, x, x+1, x+1, 7. Let y = f (x) =, . Now, y =, x−1, x−1, ⇒, ⇒, , 2, , xy − y = x + 1 ⇒ x ( y − 1) = y + 1, y+1, y+1, x=, ⇒ f −1 ( y) =, y−1, y−1, x+1, (replacing y by x), f −1 (x) =, x−1, , 8. We have f (x + 1) = x2 − 3x + 2, Put x = x − 1, , ⇒, , f (x) = (x − 1)2 − 3(x − 1) + 2, f (x) = x2 − 2x + 1 − 3x + 3 + 2, f (x) = x2 − 5 x + 6, ax + b, 9. y = f (x) =, ⇒ f ( y) = f { f (x)}, cx − a, ax + b, a, +b, cx − a , a 2x + ab + bcx − ab, =, =, ax + b, acx + bc − acx + a 2, c, −a, cx − a , x (a 2 + bc), ⇒ f ( y) =, =x, (a 2 + bc), 10. Given, f ′ (x) = g′ (x), On integrating both sides, we get, f (x) = g (x) + C, …(i), ⇒, f (x) = x3 − 4x + 6 + C, (given), Q, f (1) = 2, ∴, 2 = 1 − 4 + 6 + C ⇒ C = −1, [from Eq. (i)], ∴, f (x) = x3 − 4x + 5, 3x + 2, 11., y=, 5x − 3, ⇒ 5xy − 3 y = 3x + 2 ⇒ (5 y − 3) x = 3 y + 2, 3y + 2, x=, ⇒, 5y −3, 3x + 2, 3x + 2, −1, ⇒ f ( y) =, ⇒ f −1 (x) =, 5y −3, 5x − 3, 12. Qsin 4 x + cos 4 x = (sin 2 + cos 2 x)2 − 2 sin 2 x cos 2 x, 1 1 − cos 4 x 3 1, 1, = 1 − (sin 2x )2 = 1 − , = + cos 4x, 4 4, 2, 2, 2, Since, cos x is periodic with period 2π, then cos 4x is, 2π π, periodic with period, = ., 4 2, π, So, the period of function f (x) = sin 4 x + cos 4 x is ., 2, 13. Given that, f (x) = log ( x − 4 + 6 − x ), ⇒, x − 4 ≥ 0 and 6 − x ≥ 0 ⇒ x ≥ 4 and x ≤ 6, ∴ Domain of f (x) = [4, 6 ], 14. Given that, f (x) = cos (x/3), We know that − 1 ≤ cos (x/3) ≤ 1, Range of f (x) = [−1, 1], ∴, 15. Given that, f (x) = sin −1 [log 2 (x/2)], Domain of sin −1 x is x ∈ [− 1, 1], x, x, ⇒, −1 ≤ log 2 ≤ 1 ⇒ 2−1 ≤ ≤ 21, 2, 2, 1 x, ⇒, ≤ ≤ 2 ⇒ 1 ≤ x ≤4, 2 2, ∴, x ∈[1, 4]
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310, , NDA/NA Mathematics, , 16. Given that, f (x) =, , x+2, | x + 2|, , (1 + ab) − (a + b) , = log , , (1 + ab) + (a + b), , Now, we can define the function, − 1 , x < − 2, f (x) = , 1, x > − 2, ∴ Range of f (x) is { − 1, 1}, 2, 3, 17., f (x) = x + = y, 3, 2, ⇒, 4x + 9 = 6 y, 6y − 9, ⇒, x=, = f −1 ( y), 4, 6x − 9 3x 9, ⇒, f −1(x) =, =, −, 4, 2 4, , (say), , 18. Q f (x) = x and g (x) = | x|, ∴ ( f + g ) (x) = f (x) + g (x), = x + | x|, x + x, x ≥ 0, (by redefining the function), =, x − x, x < 0, 2x, x ≥ 0, =, 0, x < 0, 19. We have, y =, , 10 x − 10− x, 10 x + 10− x, , Using componendo and dividendo,, 1+ y, 1 + y 2 ⋅ 10 x, = 102x, ⇒, ⇒, =, −x, 1− y, 1 − y 2 ⋅ 10, , x=, , ⇒, , 1, log10, 2, , 1 +, , 1 −, , y, , y, , ⇒, , f, , ( y) =, , 1 + y, 1, log10 , , 2, 1 − y, , ∴, , f −1 (x) =, , 1 + x, 1, log10 , , 2, 1 − x, , −1, , 1 + x, 20. We have, f (x) = log , , 1 − x, 2x , , , 1 +, 1 + x 2 + 2 x, 2x , 1, x 2, +, , log, f , =, log, =, , , , 2, 2, 1 − 2x , 1 + x , 1 + x − 2 x, , 2, 1+ x , , 2, , 1 + x, 1 + x, = log , = 2 f (x), = 2 log , 1 − x, 1 − x, 1 − x, 1 − a, 21. We have, f (x) = log , , f (a ) = log , , 1 + x, 1 + a, 1 − b, f (b) = log , , 1 + b, ∴, , 22. Given that, F (x) = a 2 − x2 and g (x) = − x2, , 1 − a, 1 − b, f (a ) + f (b) = log , + log , , 1 + a, 1 + b, , f (x) = a 2 + x, , Let, then, , ( fog )(x) = f [ g (x)], , ⇒, , a 2 − x2 = F (x), , 23. Q g (x) = sin x and, , f (x) =, , 1, sin x, , 1 , ( gof )(x) = g [ f (x)] = sin f (x) = sin , , sin x, 1, x, Then, as x becomes larger and larger, f (x) assumes, values nearer to 2., 1−x, 25. We have, f (x) =, 1+ x, , 24. If f (x) = 2 +, , 1 − x, 1−, , 1 + x, 1 − x, 1+ , , 1 + x, 1 + x − 1 + x 2x, =, =, 1+ x+1−x 2, , 1 − x, Then, f , =, 1 + x, , 1 + y, 2x = log10 , , 1 − y, , ⇒, , ∴, , [Q log a + log b = log ab], a+b, , −, 1, , , a + b, 1 + ab , = log , =f , , +, a, b, 1 +, , 1 + ab, , , 1 + ab , , , So,, , 1 − x, f , =x, 1 + x, , 26. We have, y2 = 4ax, a ≠ 0. So, clearly y = ± 2 ax, 2x − 1, 27. We have, f (x) =, ...(i), x+5, Let, ...(ii), y = f (x) ⇒ x = f −1 ( y), 2x−1, where,, y=, x+5, ⇒, yx + 5 y − 2x + 1 = 0, 5y + 1, We get,, x=, 2− y, 5y + 1, −1, [using Eq. (ii)], f ( y) =, ∴, 2− y, 5x + 1, or, , when x ≠ 2, f −1 (x) =, 2−x, 4+1 5, 2x+ 1, 28. Given that, f (x) =, ⇒ f (2) =, =, 6 −2 4, 3x − 2, 5, 2 × +1, 5, 4, ∴, ( fof ) (2) = f ( f (2)) = f =, 5, 4, 3 × −2, 4, 10, +1, 14, = 4, =, =2, 15, 7, −2, 4
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311, , Functions, 1 + x, 29. Here, f (x) = log , and replacing x by − x, we get, 1 − x, 1 − x, 1 + x, f (− x) = log , = log , , 1 + x, 1 − x, 1 + x, = − log , = − f (x), 1 − x, , −1, , ⇒ f (x) is an odd function., 30. |sin x| and |cos x| has period π. Here, f (x) is an even, function and sin x, cos x are complementary., 1, π, Thus, period of f (x) = |LCM of π and π |=, 2, 2, 31. We have, f (x) = (1 − x3 )1/5 + 2, Let, y = (1 − x3 )1/5 + 2 y − 2 = (1 − x3 )1/5, 3, 1 − x = ( y − 2)5 , x3 = 1 − ( y − 2)5, x = [1 − ( y − 2)5 ]1/3, −1, f ( y) = [1 − ( y − 2)5 ]1/3, On interchanging y by x, we get, f −1 (x) = [1 − (x − 2)5 ]1/3, 2, , 1, 1, 1 , 1, , , f x + = x2 + 2 , f x + = x + − 2, , , x, x , x, x, ∴, f (x) = x2 − 2, Hence, f (x) = x2 − 2 , ∀ x ≠ 0, 3x + 2, 33. Given that, f (x) =, 5x − 3, 3x + 2, Let, =y, f (x) = y , then, 5x − 3, 3y + 2, 3y + 2, ⇒, x=, ⇒ f −1 ( y) =, 5y − 3, 5y − 3, x, +, 3, 2, ∴, f −1 (x) =, = f (x) for all x, 5x − 3, 32., , f −1 (x) = f (x), x (x − p) x (x − q), 34., f (x) =, +, q− p, p−q, p ( p − p) p ( p − q ), ⇒, f ( p) =, +, =p, (q − p), ( p − q), q (q − p) q (q − q), and, f (q) =, +, =q, (q − p), ( p − q), ( p + q ) ( p + q − p) ( p + q ) ( p + q − q ), Now, f ( p + q) =, +, (q − p), ( p − q), q ( p + q ) ( p + q ) ( p), =, +, q− p, ( p − q), 2, 2, pq + q − p − pq, =, (q − p)2, 2, q − p2 (q − p) (q + p), =, =, q− p, (q − p), Hence,, , = q + p = f (q) + f ( p), ∴ f ( p) + f (q) = f ( p + q), 1 , 35. We have, 2 f (x + 1) + f , = 2x and x ≠ − 1, x + 1, Put, , x = 1, we get, , 1 , 2 f (1 + 1) + f , =2 ×1, 1 + 1, , 1, 2 f (2) + f = 2, 2, Putting x = − 1 /2 , we get, 1, 2 f + f (2) = − 1, 2, , ⇒, , ...(i), , ...(ii), , Multiplying Eq. (i) by 2 and subtracting Eq. (ii) from, Eq. (i), we get, 1, 4 f (2) + 2 f = 4, 2, 1, f (2) + 2 f = − 1, 2, –, –, +, 3 f (2) = 5, ⇒, f (2) = 5 /3, 36. Function f : R → R is defined by f (x) = ex ., Let x1 , x2 ∈ R and f (x1 ) = f (x2) ⇒ ex1 = ex2 ⇒ x1 = x2., Therefore, f is one-one, let f (x) = ex = y. Taking log on, both sides, we get x = log y, We know that log is defined only of non-zero positive, number, we conclude that there is some element left in, codomain which does not have pre-image., Therefore, function f is into., x, x ≥ 0, 37. The given function is f (x) =|x|= , − x, x < 0, and, f :R→ R, Then, it is clear that function is neither one-one nor, onto., 38. Given that, f (x) = 2x + | x|, ∴, f (2x) = 2 (2x) + |2x| = 4x + 2| x|, f (− x) = − 2x + | − x| = − 2x + | x|, ∴, f (2x) + f (− x) − f (x), = 4x + 2| x| + | x| − 2x − 2x − | x| = 2| x|, 39. Given that, f ( g (x)) = 25 ⇒ f (x2 + 7) = 25, ⇒, 2 (x2 + 7) + 3 = 25 ⇒ 2x2 = 8 ⇒ x2 = 4, ⇒, x=±2, cos 2 x + sin 4 x, sin 2 x + cos 4 x, cos 2 x + sin 2 x (1 − cos 2 x), f (x) =, sin 2 x + cos 2 x (1 − sin 2 x), sin 2 x + cos 2 x − sin 2 x cos 2 x, f (x) =, sin 2 x + cos 2 x − sin 2 x cos 2 x, , 40. Given that, f (x) =, ⇒, ⇒, , ∴, f (x) = 1, This is a constant function ⇒ f (2002) = 1, 1, 1, 41. f (x) =, +, x + 2 2 x −4, x −2 2 x −4, 1, 1, +, f (11) =, 11 + 2 22 − 4, 11 − 2 22 − 4, 1, 1, +, f (11) =, 11 + 2 18, 11 − 2 18, 1, 1, =, +, 3 2 + ( 2 )2 + 2 9 × 2, 3 2 + ( 2 )2 − 2 9 × 2, =, , 1, 1, 3− 2 3+ 2 6, +, =, +, =, 7, 7, 7, 3+ 2 3− 2
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312, , NDA/NA Mathematics, , 42. gof = g { f (x)} = g (x2) = x2 + 5, x + 59, 43. Given that, 3 f (x) + 2 f , = 10x + 30, x −1 , Or it can be written as, x + 59, 40x + 560, 3f , + 2 f (x) =, x −1, x −1 , , …(i), , …(ii), , On solving Eqs. (i) and (ii), we get, 6x2 − 4x − 242, f (x) =, x−1, 6 × 49 − 28 − 242 294 − 270 24, ∴, f (7) =, =, =4, =, 6, 6, 6, 1, 44. We have, f (x) = (3 − cos 2 x)−1 =, 3 − cos 2 x, The function f (x) is defined for all x ∈ R., ∴ Domain of f (x) is R. Let f (x) = y,, 1, then, = y and y > 0, (3 − cos 2 x), 1, 3y − 1, ⇒, 3 − cos 2x =, ⇒ cos 2x =, y, y, 1, −1 3 y − 1 , ⇒, x = cos , , y , 2, Now, x ∈ R, if, , 3y − 1, ≤1, y, 1, ⇒, −1 ≤ 3 − ≤ 1, y, −1, 1, ⇒, −4 ≤, ≤ −2 ⇒ 2 ≤ ≤ 4, y, y, 1, 1, < y≤, ⇒, 4, 2, 1 1 , ∴ Range of function y ∈ , ., 4 2 , −1 ≤, , 46. Q, ⇒, ⇒, , , 1, ∴ Domain of the function = x : ≤ x ≤ 2, , 2, 47. Q f (x) = cos (log e x), , 1 y, , f + f (xy), , , , 2 x, , = cos (log e x) cos (log e y), 1, , y, ⇒, − cos log e + cos log e(xy), x, 2, , 2, = cos (log e x) cos log e y − cos (log e x) cos (log e y), 2, =0, , ∴, , f (x) f ( y) −, , 48. sin (sin −1 x + cos −1 x) = sin π / 2 = 1, ∴ Range of sin (sin −1 x + cos −1 x) = 1, 49. For domain of sin −1 (log3 x), − 1 ≤ log3 x ≤ 1, ⇒, 3 –1 ≤ x ≤ 3, 1 , ∴ Domain of sin −1 (log3 x) = , 3, 3 , αx, 50., f (x) =, x+1, αx, α, , αx, x + 1, α 2x, =, ∴, f ( f (x)) = f , =, α x+ x+1, x + 1 α x , +1, , x + 1, But, ∴, , 45. We have, X = {1, 2, 3} and Y = {0, 1}, and f : X → Y is defined by f = {(1, 1), (2, 1), (3, 0)}, Here, f shows the property of onto not one-to-one., , −1 ≤ log 2 x ≤ 1, 2−1 ≤ x ≤ 21, 1, ≤ x ≤2, 2, , ⇒, ⇒, ⇒, ⇒, ⇒, , f ( f (x)) = x, α 2x, =x, αx+ x+1, α 2 = α x + x + 1 ⇒ α 2 − 1 = (α + 1) x, (α − 1) (α + 1) − (α + 1) x = 0, (α + 1) (α − 1 − x) = 0, α + 1 =0, (Qα − 1 − x ≠ 0), α = −1, , Level II, 1−, , 1., ⇒, , 1, >0, x, , Taking log, we get x =, , x −1, >0⇒x<0, x>1, x, , ⇒, 2. f (x) = log, , (−∞ , 0) ∪ (1, ∞ ), 1, ⇒ sin x ≠ 0, |sin x|, , ⇒ x ≠ n π + (−1)n0 ⇒ x ≠ n π ., All real values of x except {n π}, i.e., Domain of f (x) = R − { nπ, n ∈ I }, −x, , 10 − 10, 10 − 1, ⇒ y = 2x, −x, x, 10 + 10, 10 + 1, 1+ y, 2x, 2x, ⇒ 10 (1 − y) = (1 + y) ⇒ 10 =, 1− y, , 3. Let f (x) = y =, , x, , 2x, , ⇒, , f, , ⇒, , f, , 1 + y, 1, log10 , , 2, 1 − y, , −1, , ( y) =, , 1 + y, 1, log10 , , 2, 1 − y, , −1, , (x) =, , 1 + x, 1, log10 , , 2, 1 − x, , 4. Given that, f (x) = log|log x|, f (x) is defined, if|log x| > 0, and x > 0 i. e., if x > 0 and x ≠ 1, (Q|log x| > 0, if x ≠ 1), ⇒, x ∈ (0, 1) ∪ (1, ∞ ), 5. |x|is not one-one, x2 is not one-one, x2 + 1 is not one-one., But 2 x − 5 is one-one because, f (x) = f ( y) ⇒ 2x − 5 = 2 y − 5 ⇒ x = y, 5+ y, Now,, y=2 x−5 ⇒ x=, 2
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313, , Functions, ⇒ For every value of y, we get pre-image., ∴ It is onto., ∴, f (x) = 2x − 5 is bijective., 6. Given, p, q and r ∈ z + ,, and ω is the cube root of unity., Then,, , f (x) = x3 p + x3 q + 1 + x3 r + 2, , ⇒, , f (ω ) = ω3 p + ω3 q + 1 + ω3 r + 2, = (ω ) + (ω ) ⋅ ω + (ω ) ⋅ ω, 3 p, , 3 q, , 3 r, , 2, , = 1 + ω + ω2 = 0, 7. Given function, f (x) = y = mx + c., Which is the general equation of straight line and a, straight has neither maximum point nor minimum, point, dy, = m = constant, Q, dx, So, the function has not any point to define its, maximum or minimum position., 8. Now, f (x1 ) = f (x2) ⇒ x12 = x22 ⇒ x1 = x2, [if X = R+ ], Since, f (x) is onto. It mean that we take all real number, in codomain either it has pre-image or not, ∴Y = R, y = f (x) = cos 2 x + sin 4 x, ⇒, ⇒, ⇒, , y = f (x) = cos 2 x + sin 2 x (1 − cos 2 x), y = cos 2 x + sin 2 x − sin 2 x cos 2 x, y = 1 − sin 2 x cos 2 x, 1, 1, ⇒, y = 1 − (2 sin x cos x)2 ⇒ y = 1 − sin 2 2 x, 4, 4, 2, 1, 2x, sin, We know, 0 ≤ sin 2 2 x ≤ 1 ⇒ − ≤ −, ≤0, 4, 4, 3, 3, sin 2 2x, ⇒, ≤1 −, ≤1 ⇒, ≤ f (x) ≤ 1, 4, 4, 4, 3 , ⇒, f (x) ∈ , 1, 4 , 10 + x, 10 + x, 10. ef ( x ) =, , x ∈ ( − 10, 10) ⇒ f (x) = log , , 10 − x, 10 − x , ⇒, , ⇒, , 200x , , 10 + 100 + x2 , 200x , f, , = log , 2, 100 + x , 10 − 200x , , 100 + x2 , 10 (102 + x2 + 2 × 10x) , 200x , f, = log , , 2, 2, 2, 100 + x , 10 (10 + x − 2 × 10x) , 2, , ∴, , Hence,, f (| x|) = x, This, option satisfies the condition., 13. Given that, f (xy) = f (x ) f ( y), , Q ω3 = 1 and , , = (1) p + (1)q ⋅ ω + (1)r ⋅ ω 2 , , 2, 1 + ω + ω = 0, , 9., , 12. Now, taking option (c), 1, 1, 1 1, By verification f − = f = − 1 = −, , , 2, 2, 2, 2, , 10 + x, (10 + x) , = log , = 2 log 10 − x = 2 f (x), (, x, ), 10, −, , , , , 1 200x , 1, f (x) = f , ⇒ k = = 0.5, 2 100 + x2, 2, , 11. Given that, g [ f (x)] = 8 ⇒ g (2x + 3) = 8, ⇒, (2 x + 3)2 + 7 = 8, (Q f (x) = x2 + 7 given), ⇒ 2x + 3 = ± 1 ⇒ 2x = ± 1 − 3 ⇒ x = − 1, − 2, , From option, we take f (t ) = t k, then,, f (xy) = (xy)k = (x k ) ( y) y = f (x ) ⋅ f ( y), 14. Given equation is, x2 − 4x − log3 N = 0, Then, ∆ ≥ 0 for real roots, ⇒, B2 − 4 AC ≥ 0, ⇒, (−4)2 − 4(1)(− log3 N ) ≥ 0, ⇒, 16 + 4 log3 N ≥ 0, ⇒, log3 (3)16 + log3 (N )4 ≥ 0, ⇒, log3 {(3)16 ⋅ (N )4 } ≥ 0, ⇒, 316 ⋅ N 4 ≥ 30 = 1, 1, 1, 1, 4, ⇒, N ≥ 16 ⇒ N ≥ 4 =, 81, 3, 3, 1, ∴ Minimum value of N is ., 81, a, 15. Given , sin θ = x + ,∀ x ∈ R − {0}, x, We know that,, AM ≥ GM, 1/ 2, a, a, , ⇒, x + ≥ 2 x. , x, , x, ⇒, sin θ ≥ 2 a, 2 a ≤ sin θ, ⇒, We know that −1 ≤ sin θ ≤ 1, , …(i), , [from Eq.(i)], …(ii), ...(iii), , From Eqs. (ii) and (iii), we get, ⇒, , 2 a ≤ sin θ ≤ 1 ⇒ 2 a ≤ 1 ⇒, , a≤, , 1, 1, ⇒a≤, 2, 4, , 16. Given that, f (x) = cos −1 (3x − 1), Then, the domain of cos −1 (3x − 1) exist between the, interval [–1, 1]., ⇒, −1 + 1 ≤ 3 x ≤ 1 + 1, [Q − 1 ≤ 3x − 1 ≤ 1], 2, 2, ⇒, 0 ≤ 3x ≤ 2 ⇒ 0 ≤ x ≤ ⇒ x ∈ 0, , 3, 3, 2, , −1, ∴ Domain of cos (3x − 1) is x : 0 ≤ x ≤ ., 3, , 1, +, x, x>0, , 17. The given function f (x) =, , 1, For domain, +, x, ⇒, ⇒, ⇒, , x + 1 is defined for, , x+ 1 ≥0, x2 + x ≥ −1 ⇒ x2 + x − 1 ≥ 0, , , −1 − 5 , −1 + 5 , x − , ≥0, x − , 2, 2, , , , , , , , −1 − 5 , −1 + 5 , ≥ 0 or x − , ≥0, x − , 2, 2, , , , , , , , …(i)
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314, , NDA/NA Mathematics, −1 − 5, −1 + 5, or x ≥, …(ii), 2, 2, , , −1 − 5 , −1 + 5 , and, ≤ 0 or x − , ≤0, x − , 2 , 2 , , , , , −1 − 5, 5 −1, or x ≤, ⇒, x≤, 2, 2, 5 −1, From all these conditions, we get 0 < x ≤, ., 2, ⇒, , x≥, , 18. f (x) = 2 x + 7 and g (x) = x2 + 7, ∴, , fog (x) = f { g (x)} = f (x2 + 7), = 2(x2 + 7 ) + 7 = 2 x2 + 14 + 7 = 2x2 + 21, , But, , fog (x) = 25, , ⇒, , 2 x2 + 21 = 25⇒ x2 = 2, , ⇒, , x=± 2, , (given), , 24. Given that f (x) = |sin x + cos x| + |sin x − cos x|, π, π, , π, , Now, f + x = sin + x + cos + x , 2, 2, , 2, , π, , π, , + sin + x − cos + x , , 2, , 2, = |(cos x − sin x)| + |(cos x + sin x)|, = |sin x + cos x| + |sin x − cos x|, π, , f + x = f (x), ⇒, 2, , π, This shows that period of given function is ., 2, 25. The graph of y =|x| lies only in the first and second, quadrants of the xy-plane., y, y, , 19. We have, f : R → R is defined as f (x) = x|x|, when x < 0 ⇒ f (x) = − x2, ∀ x ∈ R, and when x > 0 ⇒ f (x) = x2, ∀ x ∈ R, which is only into., , =, , |, |x, , O, , x, , 20. Since, f (x) is an even function, therefore, f (− x) = f (x) for all x , − f ′ (− x) = f ′ (x) for all x, f ′ (− x) = − f ′ (x) for all x, ∴ f ′ (x) is an function., 21. Since, g (x) is the inverse of function f (x), therefore, gof (x) = I (x) for all x, Now, gof (x) = I (x), gof (x) = x, ( gof )′ (x) = 1,, (using chain rule), g′ ( f (x)) f ′ (x) = 1,, 1, g ′ ( f (x)) =, f ′ (x), 1, (by putting x = c), ⇒, g ′ ( f (c)) =, f ′ (c), 1, ⇒, g ′ ( f (c)) =, f ′ (c), 22., , | x|, ,, , Given, f (x) = x, 2,, , x≠0, x=0, , 1,, , Now, redefine the given function = 2,, − 1 ,, , ∴ Range of f (x) is { −1, 1, 2}., 23. Q f (x) = (x − 1) (x − 2), For one-one, Let, f (x1 ) = f (x2), ⇒, , (x1 − 1) (x1 − 2) = (x2 − 1)(x2 − 2), , ⇒, , x12 − 3x1 = x22 − 3x2, , ⇒ (x1 − x2) (x1 + x2 − 3) = 0, ⇒, , x1 = x2 or x1 + x2 = 3, , Since, at x1 = 1 and x2 = 2 ⇒ 1 + 2 = 3, ∴ Function is not one-one., For onto, Range of f (x) is R., ∴ Function is onto., , x>0, x=0, x<0, , 26. We have, f (x) = l x2 + mx + n, f (x + 1) = l (x + 1)2 + m (x + 1) + n, = l (x2 + 2 x + 1) + m (x + 1) + n, = l x2 + (2l + m) x + l + m + n, f (x + 2) = l (x + 2)2 + m (x + 2) + n, = l (x2 + 4x + 4) + m (x + 2) + n, = l x2 + (4l + m) x + 4l + 2m + n, and, , f (x + 3) = l (x + 3)2 + m (x + 3) + n, = l (x2 + 6x + 9) + m (x + 3) + n, , = l x2 + (m + 6 l) x + 9 l + 3m + n, f (x + 3) − f (x), Now,, f (x + 2) − f (x + 1), lx2 + mx + 6lx + 9l + 3m + lx2 − mx − n, = 2, lx + 4 xl + mx + 4 l + 2m + n − lx2 − 2l x − mx − l − m − n, =, , 6 lx + 9 l + 3m 3 (2 lx + 3 l + m), =, =3, 2 lx + 3 l + m, 2 lx + 3 l + m, f (x) = ax + b and g (x) = cx + d, fog (x) = f [ f (x)] = a (cx + d ) + b, = acx + ad + b, gof (x) = g [ f (x)] = c(ax + b) + d, = acx + bc + d, fog (x) = gof (x), acx + ad + b = acx + bc + d, ad + b = bc + d, f (d ) = g (b), , 27. We have,, , and, Also,, ⇒, ⇒, ⇒, , 28. Let the height of rectangular box be y cm., ∴ Volume of rectangular box = x × x × y = 10, , (given)
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315, , Functions, 10, (Q volume = l × b × h) …(i), x2, Now, surface area of box, = 2(x2 + xy + yx), [Q surface area = 2(lb + bh + hl)], 2 20, 2, [from Eq. (i)], = 2(x + 2xy) = 2 x +, , , x, 40, = 2 x2 +, x, ⇒, , y=, , Q f −1 (0) does not contain 0 for non-zero real numbers, a, b and c., So, option (a) is correct., 32. Q, , f (x) =, , ∴, , f (− x) =, , sec 4 (− x) + cosec 4 (− x), (− x)3 + (− x)4 cot (− x), sec 4x + cosec 4x, = − f (− x), =− 3, x + x4 cot x, , 29. We know that the exponential function is never zero., Because its graph never passes through the origin., y, , x′, , ⇒ f (x) = − f (− x), Thus, both A and R are true and R is the correct, explanation of A., , y = ex, , (0,1), , 33. If f (x) is a periodic function with period T, then f (ax + b), T, is a periodic function with period, . But also f (ax − b), |a |, T, ., is a periodic function with period, |a |, Thus, both A and R are true but R is not the correct, explanation of A., , x, , O, , y′, , 30. Given that,, f : [−100π , 1000π ] → [−1, 1], And, Now,, , f (θ ) = sin θ, sin θ = 0 = sin 0, , ⇒, , θ = nπ , where n ∈ Z, , Then,, , n = − 100 to n = 1000, , …(i), , Now, the number of values of θ which satisfy the Eq.(i), = 1000 + 100 = 1100, Q f (θ ) is defined for every integer value of n., 31. Given, f (x) = ax2 + bx + c, b, c, , f (x) = a x2 + x + , ⇒, a, a, , 2 bx, b2, b2, c, = a x +, +, −, + , 2, 2, a, a, 4a, 4a, , , 2, 2, , , b, b − 4ac, , = a x +, , −a, 2, , 2a , 4a , 2, , Let, ⇒, , (b2 − 4ac), b, , y = f (x) = a x +, −, , 2a , 4a, 2, b2 − 4ac, b, , a x +, , = y+ , , 2a , 4a , , ⇒, , x+, , ⇒, , f, , −1, , b, =, 2a, , y b2 − 4ac, +, , a 4a 2 , , ( y) = −, , b, +, 2a, , y b2 − 4ac, +, , a 4a 2 , , b, +, 2a, , b2 − 4ac, 4a 2, , f, , ⇒, , 34. The assertion and reason both are true and R is the, correct explanation of A. The correct reason may stated, as follows., For y = f (x) = |x|to be one-one every element must have, a distinct image but in the present case it is not so,, since, distinct elements have same images. For, example, the correct image of x and (−x) comes out to, be x. Thus, the function cannot be one-to-one., x2, 35. Assertion (A) f (x) = x and F (x) =, x, At x = 0, F (x) ≠ f (x), ∴ It is false statement., Reason (R) It is true that F (x) is not defined at x = 0., ∴ Option (d) is correct., 36. Assertion (A) At x = 0, f (x) =, , −1, , (0) = −, , 37. (A) The graph of f (x) = cos x cuts x-axis in infinite, number of points., (B) The graph of f (x) = ln x cuts x-axis in only one point., (C) The graph of f (x) = x2 − 5x + 4 cuts x-axis in two points., (D) The graph of f (x) = ex cuts y-axis in only one point., y, , x′, , b, +, 2a, , b2 − 4ac, 2a, , Graph of cos x, , 5π/2, , π/2, O, , b2 − 4ac > 0, f −1 (0) = −, , |x|, does not defined, x, , ∴ It is true., Reason (R) Since, f (x) is defined for every value of x, except x = 0., ∴ It is false., , y = 0,, , At, , Here,, , sec 4x + cosec 4x, x3 + x4 cot x, , y′, , 3π/2, , x
Page 443 : 316, , NDA/NA Mathematics, , Graph of loga x, , y, , (1,0), x′, , O, , x, , x′, , y, , O, , y′, 0<a<1, , (1,0), a>1, , y′, , B (4,0), , A, O (1,0), , x, , x, , O, , 2, , 1, 1, , + 2 = x + = { f (x)}2, , x, x2, 1, 1, , 3, 3, f (x ) + 3 f (x) = x + 3 + 3 x + , , x, x, 3, 1, , , = x + = { f (x)}3, , x, , 38. f (x2) + 2 = x2 +, , , 1, x2 + 1 ) = log , x2 + 1 +, , , ⇒ f (− x) = − f (x), ∴ f (x) is an odd function., , x2 + 1 )), , = cos (log (x +, , x2 + 1 )) = g (x), , ⇒, g (− x) = g (x), ∴ g (x) is an even function., III. If f (x) is an even function, then the curve y = f (x), is symmetric about x-axis., , = x2 − 6x + 9 − 4x + 12 + 3 = x2 − 10x + 24, , x2 + 1 ), , x2 + 1 ) = − f (x), , = cos (− log (x +, , 44. ∴ f ( g (x) ) = (x − 3)2 − 4 (x − 3) + 3, , 39. Every function doesn’t have a primitive, but a primitive, of a function is unique, by definition of function., 1−x, 1− y, 40. I. Let y =, ⇒ y(1 + x) = (1 − x) ⇒ x =, 1+ x, 1+ y, 1−x 1−x, is, ., ∴ Inverse of, 1+ x 1+ x, Thus, f (x) is inverse of itself., II. Let y = 5 log x, ⇒, log y = log x log 5, log y, log x =, ⇒ log x = log5 y ⇒ x = elog 5 y, ⇒, log 5, i.e., inverse of 5log x is elog 5 y, Thus, f (x) is not inverse of itself., Hence, it is clear that the statement I is correct., , = − log(x +, , x2 + 1 )), , , , x − 1, when x < 3, , x− 4, =, , when 3 < x < 4, x− 3, , x− 4, , when x ≥ 4, x 2 + 2x + 2, For x < 4 , g (x) = x − 3, , y′, , ∴ f (− x) = log (− x +, , g (− x) = cos (log (− x +, , 42. ( f + g ) x = f (x) + g (x), = (x2 − 4x + 3 ) + (x − 3), when x < 3, = x2 − 3x , when x < 3, x2 − 4 x + 3, x − 3 , when x < 3, , f (x) , x −4, 43., =, , when 3 < x < 4, g (x) , x −3, x −4, , , when x ≥ 4, x2 + 2 x + 2, , , (0, 1), , 41. I. Let f (x) = log (x +, , ∴, , x −3 ,, x2 − 4 x + 3 ,, x<3, , , 3 ≤ x < 4, g (x) = x − 3 ,, f (x) = x − 4 ,, x2 + 2 x + 2 ,, x −4 ,, x ≥4, , , , Graph of e x, y, , and, , x + 1 )), , Solutions (Q. Nos. 42-44), , y′, , x′, , g (x) = cos (log (x +, , x, , Graph of f(x) = x 2 – 5x + 4, f(x) = (x – 1) (x – 4), y, , x′, , II. Let, , , , x, , = (x − 6) (x − 4), , Solutions (Q. Nos. 45-46), 45. The given function is defined only if, x − x2 ≥ 0 ⇒ x(1 − x) ≥ 0, ⇒, x(x − 1) ≤ 0 ⇒ 0 ≤ x ≤ 1, Hence, domain of f (x) is [0,1]., 46. To find out its range, Let, y = x − x2,, so that, ⇒, , y2 = x − x2, x2 − x + y2 = 0 ⇒ x =, , 1 ± 1 − 4 y2, 2, , Since, x is real., ∴, 1 − 4 y2 ≥ 0, ⇒, 4 y2 − 1 ≤ 0 ⇒ (2 y + 1)(2 y − 1) ≤ 0, 1, 1, ⇒, − ≤ y≤, 2, 2, But x ≥ 0, ∴y cannot be negative., 1, ∴, 0 ≤ y≤ ⋅, 2, 1, i.e., The range is 0, ., 2, , x<3, 3 ≤ x <4, x ≥4
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17, Limits, Continuity, and Differentiability, Limit of a function exists, if, left hand limit = right hand limit, , Limits, Limit of a function y = f ( x )at a point x = a exists, if as x, approaches a from left or from right, y = f ( x ) approaches to, same value., y-axis, , Definition of Limit, A real number l is called the limit of the function f, if, for every ε > 0, however small, there exists δ > 0 such that, | f ( x ) − l|< ε , whenever 0 <| x − a|< δ, or we write lim f ( x ) = l, , y = x + 2, x ≠ 2, , x→a, , 2, , Indeterminate Forms, –2, , e. g. ,, , f(x) =, , O, , x-axis, , 2, , If a function f ( x ) takes any of the following forms at, 0 ∞, x = a , , , ∞ − ∞ , 0 × ∞ , 00 , ∞ 0 , 1∞ , then f ( x ) is said to be, 0 ∞, indeterminate at x = a., , x2 − 4, x−2, , (we have to check limit at x = 2), f(x) = x + 2, ( if x ≠ 2), %, , Function is not defined at x = 2., Now,, if x = 1.9, 1.999, 1.99999, and, x = 2.1, 2.001, 2.0001, , 1. lim ( mx + c) = ma + c, x→a, , 2. lim, , x→a, , f (x ) = 3.9, 3.999, 3.99999, f (x ) = 4.1, 4.001, 4.0001, , n, , x=, , n, , a, , 3. lim ( cf ( x )) = c lim f ( x ), x→a, , x→a, , 4. lim ( f ( x ) + g( x )) = lim f ( x ) + lim g( x ), x→a, , x→a, , x→a, , 5. lim ( f ( x ) − g( x )) = lim f ( x ) − lim g( x ), x→a, , So, as x approaches 2 either from left or from right, f ( x ), gets closer and closer to 4. So, limit of f ( x )at x = 2is 4. Limit, at x = a of a function y = f ( x ) is written as lim f ( x ), left, x→a, , hand limit at x = a is written as lim f ( x ) or f ( a − h ) or, x → a−, , −, , f ( a ), similarly right hand limit is written as lim f ( x ) or, x→a, , f ( a + h ) or f ( a + ) ., , Some Important Properties of Limits, , +, , x→a, , x→a, , 6. lim ( f ( x ) g( x )) = lim f ( x ) lim g( x ), , x → a, x → a, x→a, 7. lim ( f ( x ))n = lim f ( x ), , x → a, x→a, lim f ( x ), f ( x ) x → a, 8. lim , =, x → a g( x ), lim g( x ), x→a, , n, , Provided lim g( x ) ≠ 0, , , , , x→a
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318, , NDA/NA Mathematics, , 9. lim f ( g( x )) = f lim g( x ) (Only, if f ( x ) does not have, , x → a, x→a, non-removable discontinuity at x = g( a )., (i) lim log f ( x ) = log lim f ( x ), x→a, x → a, , , Important Results, 1., 2., , lim f ( x ), , (ii), , lim e f ( x ) = ex → a, , 3., , x→a, , lim g ( x ), , x→ a, 10. lim [ f ( x )] g ( x ) = lim f ( x ), , x → a, x→a, , 4., , 11. If f ( x ) ≤ g( x ), then lim f ( x ) ≤ lim g( x ), , sin f ( x ), =1, f ( x), , Provided lim f(x) = 0, , , x→ a, , lim cos f ( x ) = 1, , Provided lim f(x) = 0, , , x→ a, , lim, , x→a, , x→a, , lim, , x→a, , Provided lim f(x) = 0, , , x→ a, , tan f ( x ), =1, f ( x), , Provided lim f(x) = 0, , , x→ a, , lim (1 + f ( x ))1 / f (x ) = e, , x→a, , f (x ), , 5., , , 1 , lim 1 +, , x→a , f ( x ), , 1, 13. If lim f ( x ) = + ∞ or − ∞, then lim, =0, x→a, x → a f(x), , 6., , lim, , e f (x ) − 1, =1, f ( x), , x − 7x + 15 x − 9, , 7., , lim, , b f (x ) − 1, = log b ( b > 0), f ( x), , lim, , xm − am m m − n, = a, n, xn − an, , x→a, , x→a, , 12. lim | f ( x )| = lim f ( x ), x→a, , x→a, , 3, , Example 1. The value of lim, , x→ 3, , (a) 4, (c), , 2, 3, , 2, , x − 5 x + 27x − 27, 2, (b), 9, 4, , 3, , is, , (d) None of these, , 8., , x→a, , x→a, , x→a, , Provided lim f(x) = ∞, , , x→ a, , =e, , Provided lim f(x) = 0, , , x→ a, , Example 3. The value of lim, , x→ 0, , Solution (b) We have, lim, x→3, , x3 − 7x2 + 15x − 9, x4 − 5x3 + 27x − 27, , ( x − 3) ( x2 − 4x + 3), x → 3 ( x − 3) ( x3 − 2 x2 − 6 x + 9), , = lim, , ( x − 3) ( x − 1), x2 − 4x + 3, = lim, = lim 3, x→ 3 ( x − 3) ( x2 + x − 3), x → 3 x − 2 x2 − 6 x + 9, , 4, 3, 2, (c), 3, , (a), , x→ 0, , x→ 0, , = lim, , ( x + x2 + x3 + K + xn) − n, is, x→ 1, x −1, , (b), , n(n + 1), 3, , (d) None of these, , Example 4. The value of lim, , π, x→, 4, , ( x − 1), ( x3 − 1), x2 − 1, + lim, + lim, x → 1 ( x − 1), x → 1 ( x − 1), x → 1 ( x − 1), , = lim, , (x n − 1), + K + lim, x → 1 (x − 1), , (a) 4, , (b) 2, , Solution (c) lim, x→, , π, 4, , cos x − sin x, is, , π, − x (cos x + sin x), , 4, (c) 1, (d) 3, , cos x − sin x, π, , − x (cos x + sin x), 4, , , = 1 + 2 (1) 2 − 1 + 3 (1)3 − 1 + K + n (1) n − 1, n (n + 1), 2, , x, x, − log 5 − log 1 − , 5, , 5, x, , x, x, , , log 1 + , log 1 − , , , , 5 1 1 2, 5, = lim, = + =, − lim, x→ 0, x, 0, →, x, x, 5 5 5, , ( −5) − , 5 , 5, 5, , 2, , ( x + x2 + x3 + K + xn) − n, Solution (a) lim, x→ 1, x −1, , =1+ 2 + 3 + K + n =, , x , x , , − log 5 1 − , , , 5 , 5 , , x, , , log 5 + log 1 +, , , x→ 0, , lim, , n(n + 1), 2, n2, (c), 4, , log (5 + x) − log (5 − x), x, , , log 5 1 +, , = lim, , Example 2. The value of, , (a), , (d) None of these, , Solution (b) lim, , 2, 3 −1, x −1, =, = lim 2, =, x→ 3 x + x − 3, 9 + 3 −3 9, , log (5 + x) − log (5 − x), is, x, 2, (b), 5, , = lim, x→, , π, 4, , 1, 1, , 2, cos x −, sin x, 2, , 2, π, , − x (cos x + sin x), 4,
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319, , Limits, Continuity and Differentiability, , = lim, , π, x→, 4, , =, , , π, 2 sin − x, , 4, 1, ⋅, π, (cos x + sin x), , , − x, , 4, , Example 6. The value of lim, , x→ 0, , (a) 4, , 2, 2, 2, =, =, =1, 1, 1, π, π, , 2, +, cos + sin , , 2, 2, 4, 4, , (b), , x2, x3, +, +…, 2! 3!, x2, x3, a x = 1 + x log a +, (log a )2 +, (log a )3 + …, 2!, 3!, x2, x3 x4, log (1 + x ) = x −, +, −, + …, | x | < 1, 2, 3, 4, n( n − 1) 2, (1 + x )n = 1 + nx +, x +…, 2!, x3, x5, sin x = x −, +, −…, 3! 5!, x2, x4, cos x = 1 −, +, −…, 2! 4 !, 3, 5, 2x, x, tan x = x +, +, +…, 3, 15, 1 2 3 1 2 ⋅ 3 2 5 12 ⋅ 3 2 ⋅ 5 2 7, x +, x +, sin −1 x = x +, x +…, 3!, 5!, 7!, x3, cos−1 x = x −, +…, 6, 3, x, x5, tan −1 x = x −, +, −…, 3, 5, xn − an, = xn − 1 + xn − 2 a + xn − 3a2 + … + an − 1, x−a, , 4., 5., 6., 7., 8., 9., 10., 11., , Example 5. The value of lim, , ex + e− x − 2, x2, (b) 1, (d) 3, , x→ 0, , (a) 4, (c) 2, , is, , = lim, , ex + e− x − 2, lim, x→ 0, x2, , , x2 x3 x4, = lim 1 + x +, +, +, + K, x→ 0 , 2! 3! 4!, , , , x2 x3 x4, + 1 − x +, −, +, − K − 2, 2! 3! 4!, , , x2, , 1, x, x, 2x2 +, +, + K, !, !, !, 2, 4, 6, , , 2, , = lim, , x→ 0, , 2, , 4, , x, 1, , = 2 + 0 + 0 + K = 1, 2!, , , 2, 3, , (d) None of these, , x cos x − log (1 + x), x2, 2, , , , x, x4 x6, +, −, + K, x 1 −, , , 2! 4! 6!, , , 2, 3, 4, , , x, x, x, , − x −, +, −, + K , 2, 3, 4, , , , , x2, 1 x x, 1, = lim −, − + K =, 2, x→ 0 2, 2! 3, x →0, , Standard Results, 1., 2., 3., 4., 5., 6., , lim c = c, , x→a, , lim sin x = 0, , x→0, , lim cos x = 1, , x→0, , lim tan x = 0, , x→0, , sin x, tan x, = lim, =1, x, →, 0, x, x, lim (1 + x )1 / x = e, lim, , x→0, x→0, , x, , 7., 8., , 1, , lim 1 + = e, x→∞ , x, lim (1 + λx )1 / x = e λ, , x→0, , x, , 9., 10., 11., 12., , Solution (b), , (c), , is, , x→ 0, , 1. e x = 1 + x +, , 3., , 1, 2, , x2, , Solution (b) lim, , Important Results, , 2., , x cos x − log (1 + x), , 13., 14., 15., , λ, , lim 1 + = e λ, x→∞ , x, x, a −1, = loge a, lim, x→0, x, ex − 1, lim, =1, x→0, x, xn − an, lim, = na n − 1, x→0 x − a, loga (1 + x ), = loga e, lim, x→0, x, sin −1 x, tan −1 x, lim, = lim, =1, x→0, x→0, x, x, If m and n are positive integers and a0 , b0 are non-zero real, numbers, then, a x m + a1 x m − 1 + … + am, lim 0 n, x → ∞ b x + b xn − 1 + … + b, 0, 1, n, a0, , if m = n, b, 0, = 0, if m < n, ∞, if m > n, ,
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320, , NDA/NA Mathematics, , , , , Example 7. The value of lim tan , x→ 0, , (a) e, , (b), , 1, e, , (c) e 2, , π, , , Solution (c) lim tan + x , , 4, x→ 0 , , 2 tan x , = lim 1 +, , x→ 0 , 1 − tan x, , 1/ x, , 1/ x, , lim, , Example 9. If f ( a) = 2, f ′ ( a) = 1, g ( a) = − 1, g ′ ( a) = 2. Then,, , 1/ x, , π, , + x is, , 4, , the value of, lim, , (d) e3, , x→a, , 1 + tan x, = lim , , x → 0 1 − tan x, , 2 tan x, , ⋅, , = ex → 0 1 − tan x, , 1, x, , 2, , lim, , (a) 5, , 1/ x, , = ex → 0 1 − tan x, , Solution (a) We have, lim, , x →a, , ⋅, , tan x, x, , = e2, , 1/ x, , = 2 (2) − ( −1) (1) = 4 + 1 = 5, , 1/3, , (a) 4abc, (c) 2a, , (b) ( abc ), (d) None of these, , x →0, , x 1/ x, , a + b + c, , 3, , , , Solution (b) lim , , x, , x, , , ax + b x + c x − 3 , = lim 1 +, , x→ 0 , 3, , , 1/ x, , ( ax − 1) + ( b x − 1) + ( cx − 1) , = lim 1 +, , x →0 , 3, , a x − 1 b x − 1 c x −1, lim , +, +, 3x, 3x, 3x, , x → 0 , , 1, 3, , =, , g ( x) f ( a) − g ( a) f ( x), x−a, , = g ′ ( a) f ( a) − g ( a) f ′ ( a), , ax + b x + c x , is, x→0 , 3, , , =e, , (d) 1, , Using L’ Hospital’s rule,, g ′ ( x) f ( a) − g ( a) f ′ ( x), = lim, x→ a, 1, , Example 8. The value of lim , , =e, , g ( x) f ( a) − g ( a) f ( x), is, x−a, (b) 4, (c) 3, , 1/ x, , , , , , , Continuous Function, The word ‘continuous’ means without any break or, gap. If the graph of a function has no break or gap or jump,, then it is said to be contiuous., A function which is not continuous is called a, discountiuous function. While studying graphs of, functions, we see that graphs of functions sin x, x cos x, ex, 1, etc., are continuous but reciprocal function has break at, x, x = 0, so it is not contiuous. Similarly, tan x, cot x, sec x etc.,, are also discontinous function., y, , x, x, x, , , lim a − 1 + lim b − 1 + lim c − 1, x→ 0 x, x, x , x, →, 0, x, →, 0, , , 1, (log a + log b + log c ), e3, , (0,1), , =e, , = ( abc), , 1/3, , –2π, , –π, , Let f ( x )and g( x )be two functions such that f ( a ) = 0and, g( a ) = 0, then, f(x), f ′ (x), lim, = lim, x → a g( x ), x → a g′ ( x ), , Sandwitch Theorem or, Squeeze Principle, If f , g and h are functions such that f ( x ) ≤ g( x ) ≤ h( x ),, then, lim f ( x ) ≤ lim g( x ) ≤ lim h( x ), x→a, , and let lim f ( x ) = lim h( x ) = l , then lim g( x ) = l, x→a, , x→a, , x, , Continuous Function, y, f (x) = x, , For other indeterminate forms we have to convert, 0, ∞, them into either or form and then we may apply, 0, ∞, L’ Hospital’s rule., , x→a, , 2π, , Oπ/2, (0,–1) y = sinx, , L’ Hospital’s Rule, , x→a, , π, , –π/2, log( abc )1/ 3, , x→a, , x′, , O, , x, , y′, , (1) Continuity at a point, A function y = f ( x ) is said to be continuous at x = a, if, lim f ( x ) = f ( a ) i. e. ,, lim f ( x ) = lim f ( x ) = f ( a ), x → a−, , x→a, , x → a+, , (2) Continuity in an interval, (a) A function y = f ( x ) is said to be continuous in a, closed interval [a , b], if, (i) lim f ( x ) = f ( a ), x → a+
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321, , Limits, Continuity and Differentiability, , π, cos − h, , 2, sin h, =1, = lim, = lim, h→ 0 π, h, →, 0, π, h, , , − − h, , 2 2, , (ii) lim f ( x ) = f ( c) (Here, c is every point in ( a , b)), x→c, , (iii) lim f ( x ) = f ( b), x → b−, , (b) Continuity in an open interval ( a , b) is, lim f ( x ) = f ( c) [Here, c is every point in ( a , b)]., , , π, RHL = lim f (t ) = lim f + h, , h→ 0 2, π+, , x→c, , %, , %, , x→, , If y = f (x ) and y = g(x ) are continuous functions at x = a, then, +, functions f (x ), g(x ) are also continuous at x = a, only in case of, ×, ÷, f (x ) ÷ g(x ), g(a) ≠ 0, If y = f (x ) and y = g(x ) are discontinuous functions at x = a, then, +, f (x ), g(x ) may be continuous function at x = a, ×, ÷, , Discontinuous Function, A function f which is not continuous at a point x = a in, its domain is said to be discontinuous. The point a is called, a point of discontinuity of the function., y, , x', , O, , x, , f (x) = 1/x, , π, , cos + h, 2, , sin h, = lim =, =1, = lim −, h→ 0, π π, −h, h→ 0, − + h, , 2 2, LHL = RHL = f(p / 2) = 1, π, ∴ Function is continuous at t =, 2, Q, , Example 11. The function, , e1/x − 1, , f(x) = e1/x + 1 , when x ≠ 0 is discontinuous at, 0, , when x = 0, , (a) x = 1, (c) x = 3, , (b) x = 0, (d) x = 4, , Solution (b) At x = 0, LHL = lim f ( x ) = lim f (0 − h) = lim f ( −h), x → 0−, , Discontinuous Function, The discontinuity may arise due to any of the following, situations., (i) lim f ( x ) or lim f ( x ) or both may not exist, x→ a, , +, , x→ a, , (iii) lim f ( x ) as well as lim f ( x ) both may exist, but, x→ a +, , x→ a −, , either of the two or both may not be equal tof ( a )., , Example 10. The function, cos t, , t ≠ π/ 2, , is continuous at, f ( t ) = π/ 2 − t, , t = π/ 2, 1, π, (a) t = 4π, (b) t =, 2, (c) t = π, (d) None of these, π, 2, , π, , LHL = lim f (t ) = lim f − h, −, 2, , h, 0, →, π, x→, , 2, , h→ 0, , e, −1, e−1/ h + 1, 0 −1, =, = −1, 0 +1, h→ 0, , 1, , , Q lim 1/ h = 0, , h→ 0 e, , RHL = lim f ( x ) = lim f (0 + h), x→ 0+, , h→ 0, , −, , (ii) lim f ( x ) as well as lim f ( x ) may exist, but are, x→ a +, x→ a −, unequal., , Solution (b) At t =, , h→ 0, , −1/ h, , = lim, y', , 2, , = lim f (h) = lim, h→ 0, , h→ 0, , e1/ h − 1, e1/ h + 1, , 1, 1/ h, 1− 0, e, = lim, =, =1, 1, h→ 0, 1, +0, 1 + 1/ h, e, 1−, , ∴LHL ≠ RHL, So, f ( x) is discontinuous at x = 0., , Differentiability of a Function, at a Point, The function, f ( x ) is differentiable at point P, iff there, exists a unique tangent at a point P. In other words, f ( x ) is, differentiable at a point P iff the curve does not have P as a, corner point. i.e., ‘‘the function is not differntiable at a, those points on which function has jumps (or holes) and, sharp edges.’’
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322, , NDA/NA Mathematics, , Let us consider the function f ( x ) =|x − 1|, which can be, graphically shown,, y, , f (x)= – x + 1, f ′ (x)= –1, , O, , Differentiability in an Interval, , f (x)= x – 1, f ′ (x) = 1, 1, , 2, , 3, , and the common limit is called the derivative of f ( x ), f(x) − f(a), at x = a, denoted by f ′ ( a ). Clearly, f ′ ( a ) = lim, x→ a, x−a, {x → a from the left as well as from the right}., , x, , Which shows f ( x ) is not differentiable at x = 1. Since,, f ( x ) gas sharp edge at x = 1., (i) Right hand derivative Right hand derivative of, f ( x ) at x = a, denoted by f ′ ( a + 0) of f ′ ( a + ), is the, f(a + h ) − f(a), ., lim, h→0, h, (ii) Left hand derivative Left hand derivative of f ( x ), at x = a, denoted by f ′ ( a − 0) or f ′ ( a − ), is the, f(a − h ) − f(a), ., lim, h→0, −h, (iii) A function f ( x ) is said to be differentiable (finitely) at, x = a if f ′ ( a + 0) = f ′ ( a − 0) = finite, f(a + h ) − f(a), f(a − h ) − f(a), i.e., lim, = finite, = lim, h→0, h, →, 0, h, −h, , A function y = f ( x ) defined on an open interval ( a , b) is, said to be differentiable in open interval ( a , b), if it is, differentiable at each point of ( a , b)., A function y = f ( x ) defined on a closed interval [a , b] is, said to be differentiable in closed interval [a , b], if it is, f(x) − f(a), differentiable at each point of ( a , b) and lim, (x − a), x → a+, f ( x ) − f ( b), and lim, both exist., ( x − b), x → b–, , Continuity for Composite, Functions, If the function u = f ( x ) is continuous at the point x = a, and the function y = g( u ) is continuous at the point, u = f ( a ),then the composite function y = gof ( x ) = g ( f ( x ))is, continuous at the point x = a., , Continuity and Differentiability Table, Types of Function, , Curve, , Domain and Range, , Continuity and Differentiability, , Continuous and differentiable in their, domain, Continuous and differentiable in (0, ∞), , Identity function, Exponential, function, , f ( x) = x, f ( x) = ax , a > 0, a ≠ 1, , Logarithmic, function, , f ( x) = log a x; x, a > 0, and a ≠ 1, , Domain = R, Range = ] −∞ , ∞[ , Domain = R, Range = ] 0, ∞ [ , , , , Domian = ( 0, ∞), Range = R, , , Root function, , f ( x) = x, , Domain = [0, ∞], Range = (0, ∞), , Sine function, , y = sin x, , Domain = R, Range = [−1, 1], , Cosine function, , y = cos x, , Domain = R, Range = [−1, 1], , Tangent function, , y = tan x, , π, , Domain = R − ( 2 n + 1) , 2, , Range = R, , Cosecant function, , y = cosec x, , Secant function, , y = sec x, , Domain = R − nπ,, Range = ( −∞, − 1 ] ∪ [1, ∞), π, , Domain = R − ( 2 n + 1) , 2, , Range = ( −∞, − 1 ] ∪ [1, ∞), , , , , , , , , , , , , , , , , , Continuous and differentiable in their, domain
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323, , Limits, Continuity and Differentiability, , Continuity and Differentiability Table, Cotangent function, Arc sine function, , y = cot x, y = sin −1 x, , Domain = R − {nπ }, Range = R, , Arc cosine function, , y = cos −1 x, , Arc tangent function, , y = tan −1 x, , Domain = [−1, 1], Range = [ 0, π ], −π π , Domain = R, Range = , , , 2 2, , Arc cosecant, function, , y = cosec −1 x, , Domain = ( −∞, 1 ] ∪ [1, ∞),, π π, Range = − , = {0}, 2 2, , Arc secant function, , y = sec − 1 x, , Domain = ( − ∞, − 1 ] ∪ [1, ∞), π , Range = [ 0, π ] − , 2, , Arc cotangent, function, , y = cot− 1 x, , Domain = R, Range = ( 0, π ), , π π, Domain [1, 1 ], Range = , , , − 2 2, , Continuous and differentiable in their, domain, , Comprehensive Approach, n, , limcos−1 x, lim x does not exist because one side of limit cannot, x→1, , x→ 0, , −1 +, , be evaluated as cos (1 ) is not defined and, , n, , −, , 0 is also not, , defined., n, , 1, lim sin does not exist because we are not sure what value it, x, takes in [ −1, 1,] through it’s a finite value., 1, if x is rational, , does not exist because, lim f ( x), where f(x) = , x→1, −1, if x is irrational, x→ 0, , n, , n, , it is not known what is the number just adjacent to any number, so, we do not know just adjacent to a rational number or irrational, number, number is rational or irrational., lim e1/ x does not exist, as we have to find LHL and RHL separately, , n, , All polynomials, logarithmic functions, exponential functions,, trigonometric functions, modulus function are continuous in their, domains. The greatest integer function is discontinuous at, integers., If lim f ( x) = l and lim g( x) = m, then lim [ f ( x)] g( x )= l m, x→ a, , n, , lim log f ( x) = log (lim f ( x)), , x→ a, n, , x→ a, , lim fog( x) = f (lim g( x)), , x→ a, n, , x→ a, , If lim f ( x) = 1 and lim g( x) = ∞ , then, x→ a, , x→ a, , 1, changes it’s characteristics at 0 (discontinuous graph,, x, rectangular hyperbola). Here, LHL = 0 and RHL = ∞., If a function is continuous at x = a,then lim f ( x) exist, but converse, x→ a, may not be true., , n, , lim g( x ) [ f ( x ) − 1], , lim [ f ( x)] g( x ) = e x→ a, , x→ 0, , as, , x→ a, , x→ a, , x→ a, n, , If lim f ( x) = lim g( x) = 0, then, x→ a, , x→ a, , lim, , f ( x), , lim [1 + f ( x)]1/g( x ) = e x→ a g( x ), , x→ a
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Exercise, Level I, 3x − a − x + a, is equal to, x−a, (b) 1 / 2a, (a) 2a, (c) 2a, (d) 1 / 2 a, , 10. Which of the following is not true?, (a) A polynomial function is always continuous, (b) A continuous function is always differentiable, (c) A differentiable function is always continuous, (d) ex is continuous for all x, , 1. lim, , x→ a, , 2. The value of lim, , 3x / 2 − 3, 3 −9, x, , x→2, , is, , 11. lim, , ex − 1, a, (a) log , b, (c) log ( ab), , x→0, , (a) 0, (b) 1/3, (c) 1/6, (d) ln 3, 1 + log x − x, is equal to, 3. lim, x → 1 1 − 2x + x 2, (a) 1, 4. lim, , (b) –1, , asin x − 1, , bsin x − 1, a, (a), b, log a, (c), log b, , x→0, , (c) 0, , x→0, , b, a, log b, (d), log a, , 15., , 16., , (NDA 2008 II), , x→0, , (b) ∞, (d) Does not exist, , (a) 0, (c) e, , x → 2+, , (b) lim f ( x ) ≠ 0, x → 2−, , xa − aa, , x+ 2, 17. What is the value of lim 3, ?, x → − 2 x + 8, 1, 4, 1, (c), 12, , −, , = − 1, then a is equal to, (b) 0, (d) 2, , (NDA 2012 I), , 1, 4, 1, (d) −, 12, (b) −, , x −1, , 2x 2 − 7x + 5, (a) 1 / 3, (c) −1 / 3, x →1, , (d) f ( x ) is continuous at x = 2, ax − xa, , b, (b) log , a, (d) log ( a + b), , x2 − 9, , if x ≠ 3, , , is continuous at x = 3,, If f ( x ) = x − 3, 2x + k, otherwise, , then k is equal to, (a) 3, (b) 0, (c) – 6, (d) 1/6, sin 2x, If lim, is equal to, x→0, x, (a) 0, (b) 1, (c) 1 / 2, (d) 2, tan x − sin x, is equal to, lim, x→0, x3, (a) 0, (b) 1, (c) 1/2, (d) –1/2, log ( x − a ), is equal to, lim, x → a log ( e x − e a ), (a) 0, (b) 1, (c) a, (d) Does not exist, sin x, What is the value of lim, ?, (NDA 2008 I), x→∞, x, (a) 1, (b) 0, (c) ∞, (d) − 1, , 18. lim, , (c) lim f ( x ) ≠ lim f ( x ), x→ 2, , is equal to, , (a), , 8. If f ( x ) =|x − 2|, then, (a) lim f ( x ) ≠ 0, , (a) –1, (c) 1, , (d) 0, (NDA 2010 II), , 7. lim e−1/ x is equal to, , x→0, , 14., , (b), , a, (c) log , b, , 9. If lim, , 13., , log a, log b, b, (d) log , a, , (a) log ( ab), , +, , 12., , (b), , a x − bx, is equal to, x, , x→ 2, , 1, (d) −, 2, , is equal to, , ex − 1, 5. The value of lim , is, x→0, x , 1, (a), (b) ∞, (c) 1, 2, 6. lim, , a x − bx, , is equal to, (b) 1 / 11, (d) None of these, , , 2 , 19. What is the value of lim x sin ?, x→∞ , x , (a) 2, (b) 1, (c) 1/2, (d) ∞
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325, , Limits, Continuity and Differentiability, , 20. Let f ( x ) =, , 1, 18 − x 2, , ., , What is the value of lim, x →3, , (a) 0, , (b) −, , 1, 9, , f ( x ) − f ( 3), ?, x−3, 1, (c), 3, , (d), , 1, 9, , sin2 α − sin2 β , 21. lim , is equal to, α→β, α2 − β2, , , (a) 0, (b) 1, sin β, sin 2β, (c), (d), β, 2β, , x( x − 2), , x ≠ ± 2 is continuous at, x2 − 4, (NDA 2010 II), x = 2, then what is the value of f( 2)?, 1, (a) 0, (b), (c) 1, (d) 2, 2, x +4, , ?, , (NDA 2010 I), , (b) e2, (d) e5, , a + x − a − x, 27. The value of lim , is, x→0, x, , , (a) 1, (b) 0, 1, (d), (c) a, a, sin πx, , x≠, , 28. Let f ( x ) = 5x, k,, x=, x = 0, then k is equal to, π, (a), 5, (c) 1, , 0, , . If f ( x ) is continuous at, , 0, 5, π, (d) 0, (b), , x< 2, , when, , x≥ 2, , , then f ′ ( 2) is equal, , (b) 1, (d) Does not exist, , 30. What is the value of lim, , x→0, , 25. If the function f ( x ) =, , (a) e, (c) e4, , when, , constants)?, (a) 0, a2, (c), b, , 2− x+ 4, 22. If f ( x ) =, , ( x ≠ 0), is continuous function at, sin 2x, x = 0, then f ( 0) is equal to, 1, 1, 1, 1, (c), (b) −, (d) −, (a), 4, 8, 4, 8, loge (1 + x ), 23. lim, is equal to, x→0, 3x − 1, (a) loge 3, (b) 0, (c) 1, (d) log3 e, cos( ax ) − cos( bx ), ?, 24. What is the value of lim, x→0, (NDA 2010 II), x2, (a) a − b, (b) a + b, b2 − a 2, b2 + a 2, (d), (c), 2, 2, , x + 6, 26. What is the value of lim , , x → ∞ x + 1, , x + 1,, 29. Let f ( x ) = , 2x − 1,, to, (a) 0, (c) 2, , sin2 ax, ( a and b are, bx, (NDA 2009 I), , (b) a, (d) Does not exist, , 3x − 4, 0 ≤ x ≤ 2, 31. If f ( x ) = , 2x + λ , 2 < x ≤ 3, is continuous at x = 2, then what is the value of λ?, (NDA 2009 I), , (a) 1, , (b) − 1, , x , 32. What is the value of lim , , x → ∞ 3 + x, (a) e, 33. lim, , (a) 0, , 3x, , ?, , (NDA 2009 II), , (c) e−9, , (b) e3, , sin−1 x − tan−1 x, , x→0, , (d) − 2, , (c) 2, , x3, (b) 1, , (d) e9, , is equal to, (c) –1, 2x − sin, , (d) 1/2, −1, , x, , , ( x ≠ 0) is, 2x + tan x, continuous at each point of its domain, then the value, of f( 0) is, (a) 2, (b) 1/3, (c) 2/3, (d) –1/3, , 34. If, , the, , function, , f(x) =, , −1, , 5x − 4,, if 0 < x ≤ 1, 35. If the function f ( x ) = 2, is, if 1 < x < 2, x, bx, ,, +, 4, 3, , continuous at every point of its domain, then the, value of b is, (a) –1, (b) 0, (c) 1, (d) None of these, 2x + 1,, , 36. If f ( x ) = k,, 5x − 2,, , , x>1, x = 1 , is continuous at x = 1, then, x<1, , the value of k is, (a) 1, (b) 2, 1 − sin x, , x≠, , 37. If f ( x ) = π − 2x, λ, , x=, , then the value of λ is, (a) –1, (b) 1, , (c) 3, , (d) 4, , π, 2 , be continuous at x = π/ 2,, π, 2, (c) 0, , (d) 2, , x +1, , x + 3, 38. lim , is equal to, , x → ∞ x + 1, (a) e2, (b) e3, (c) e, , (d) e−1
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326, , NDA/NA Mathematics, x +b, , x + a, is equal to, 39. lim , , x → ∞ x + b, (a) 1, (b) eb − a, (c) ea − b, 1, 40. What is the value of lim x 2 sin ?, x→∞, x, (a) 0, 1, (c), 2, , (d) eb, (NDA 2012 I), , (b) 1, (d) Limit does not exist, 3x −1, , , 4 , is equal to, 41. lim 1 −, , x→∞ , x − 1, (b) e− 12, (c) e4, (a) e12, 42., , (d) e3, , a cot x − a cos x, is equal to, x → π / 2 cot x − cos x, (a) log a, (b) log 2, (c) a, (d) log x, lim, , 43. lim, x→, , π, 6, , 2 sin2 x + sin x − 1, 2 sin2 x − 3 sin x + 1, , (a) 3, , (b) –3, , 1 + tan x, 46. lim , , x → 0 1 + sin x , 1, (a), e, (c) e, , cosec x, , is equal to, (b) 1, (d) e2, , 4 × 5x for x < 0, 47. If the function f ( x ) = , is continuous,, 8a + x for x ≥ 0, then the value of a is, (a) 1/2, (b) 2, (c) 3, (d) 4, f ( a ) = 2, f ′ ( a ) = 1; g( a ) = − 1, g′ ( a ) = 2 ,, g( x ) ⋅ f ( a ) − g( a ) ⋅ f ( x ), , is equal to, lim, x→a, x−a, (a) –5, (b) 0, (c) 1/5, (d) 5, , 48. If, , is equal to, (c) 6, , 2x − 2− x, , x ≠ 0, is continuous at x = 0, is, x, (a) 0, (b) log 4, (c) 4, (d) e4, , f(x) =, , (d) 0, , 2 −1, , −1 ≤ x < ∞ , x ≠ 0, , 44. If f ( x ) = 1 + x − 1, is, k, , x=0, , continuous everywhere, then k is equal to, 1, (a), (b) log 4, log 2, 2, (c) log 8, (d) log 2, x, , 45. The value of f at x = 0 so that function, , then, , mx + 1 , x ≤ π / 2, 49. If f ( x ) = , sin x + n , x > π / 2, is continuous at x = π / 2, then which one of the, following is correct ?, nπ, (a) m = 1, n = 0, (b) m =, +1, 2, (c) n = m( π / 2), (d) m = n = π / 2, x sin 5x, 50. What is the value of lim, ?, x → 0 sin2 4x, (a) 0, (b) 5/4, (c) 5/16, (d) 25/4, , Level II, x3 + 1, , 1. If lim 2, − ( ax + b) = 2, then, x→∞ x +1, , , (a) a = 1 and b = 1, (b) a = 1 and b = − 1, (c) a = 1 and b = − 2, (d) a = 1 and b = 2, 1, for x ≠ 0. Then,, 2. Given, f( 0) = 0 and f ( x ) =, (1 − e−1/ x ), only one of the following statements on f ( x ) is true., That is f ( x ), is, (a) continuous at x = 0, (b) not continuous at x = 0, (c) both continuous and differentiable at x = 0, (d) not defined at x = 0, sin 2x + sin 6x, is equal to, 3. lim, x → 0 sin 5x − sin 3x, (a) 1 / 2, (b) 1 / 4, (c) 2, (d) 4, 4. lim ( 3 + 4 ), n, , n→∞, , (a) 3, , n 1/ n, , is equal to, , (b) 4, , (c) ∞, , (d) e, , |x|, is, x, (a) continuous at the origin, (b) discontinuous at the origin because| x| is, discontinuous there, |x|, (c) discontinuous at the origin because, is, x, discontinuous there, (d) discontinuous at the origin because both|x| and, |x|, are discontinuous there, x, , 5. The function f ( x ) =|x| +, , , 3x + 4 tan x, , 6. If f ( x ) = , ; x ≠ 0 is continuous at x = 0,, x, , k; x = 0, then the value of k is, (a) 7, (c) − 5, , (b) 6, (d) − 1
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327, , Limits, Continuity and Differentiability, 7. f ( x ) = cos(| x|) is a continuous function because, (a) composition of continuous functions is a, continuous function, (b) product of continuous functions is a continuous, function, (c) cosine is an even function, (d) sum of continuous functions is continuous, 1 − cos θ, is equal to, θ→ 0, θ, (a) 2, (b) 2 2, 1, (c), (d) Does not exist, 2, , 8. lim, , | x − 3|, x−3, (a) is equal to 1, (c) is equal to –1, , 9. lim, , x→3, , (b) is equal to 0, (d) Does not exist, , 10. If f ( x ) = x 2 + 4x − 5, then the domain of the, real-valued function f ( x ) is, (a) set of all real numbers, (b) set of all integers, (c) [–5, 1], (d) ( −∞ , − 5] ∪ [1, ∞ ), 11. If f ( x ) = a sin| x| + be| x | is differentiable at x = 0 and, f( 0) = 0 ,then, (a) a = 0, (b) b = 0, (c) a − b = 0 (d) a + b = 0, 12. If f ( x ) =|x − 3| +|x − 4|, then in the interval [0, 5],, the function f ( x )is, (a) differentiable at x = 3, (b) differentiable at x = 4, (c) not differentiable at x = 3 and x = 4, (d) not continuous in the interval [0, 5], , 1 − cos 4x, , , if x < 0, x2, , 13. If f ( x ) = , a, , if x = 0, , x, , if x > 0, , 16 + x − 4, is continuous at x = 0, then the value of a is, (a) 4, (b) 6, (c) 8, (d) 16, 14. Two functions f and g are continuous at x = a ; and, f + g, f − g, f / g, [g( a ) ≠ 0] and f × g are also, continuous at x = a. If a function F is defined as, ex + e− x − 2, π π, on − , , then which one of the, F( x) =, x sin x, 2 2, following is correct?, (a) F ( x ) is continuous on [−π/ 2, π/ 2], (b) F ( x ) is not continuous on [−π / 2, π / 2]/{ 0}, (c) F ( x ) is continuous on [−π / 2, π / 2]/{ 0}, (d) F ( x ) is continuous on ( −π/ 2, π/ 2), 15. The value of lim, , log ( x − a ), , is, log ( ex − ea ), (b) ea, (c) e− a, , x→ a, , (a) 1, , (d) –1, , 16. If f : R → R is differentiable function and f(1) = 4, f (x ), 2t, then the value of lim ∫, dt is, x→1 4, ( x − 1), (a) 8 f ′ (1), (b) 4 f ′ (1), (c) 2 f ′ (1), (d) f ′ (1), 17. The, , function, is, given, f(x), x 2 , if x is rational, then, it is, f(x) = 2, − x , if x is irrational, (a) continuous at x = 0, 1, (b) continuous at x =, 2, (c) discontinuous at x = 0, (d) None of the above, , by, , sin x , x ≠ nπ , n ∈ Z, 18. If f ( x ) = , otherwise, 0,, 2, x + 1, x ≠ 0, 2, , and g( x ) = 4,, x = 0 , then lim g {f ( x )}is equal to, x→0, 5,, x, =2, , (a) 1, 1, (c), 2, , (b) 0, 1, (d), 4, , 1 − cos x, , x≠0, , 19. If f ( x ) = , is continuous at x = 0,, x, k, , x=0, then the value of k is, 1, (a) 0, (b), 2, 1, 1, (c), (d) −, 4, 2, 20. What is lim ( a 2x 2 + ax + 1 − a 2x 2 + 1 ) equal to?, x→∞, , 1, (a), 2, (c) 2, , (NDA 2011 II), , (b) 1, (d) 0, , 21. What is the value of k for which the following, function f ( x ) is continuous for all x?, (NDA 2011 I), x3 − 3x + 2, , for x ≠ 1, , f(x) = ( x − 1)2, , k,, for x = 1, , (a) 3, (b) 2, (c) 1, (d) − 1, 22. Which one of the following is correct in respect of the, function f ( x ) =| x| + x 2 ?, (NDA 2011 I), (a), (b), (c), (d), , f ( x ) is not continuous at x = 0, f ( x ) is differentiable at x = 0, f ( x ) is continuous but not differentiable at x = 0, None of the above
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328, , NDA/NA Mathematics, , , − 2 sin x , if, , , 23. Let f ( x ) = A sin x + B, if, , cos x ,, if, , , x≤, , −π, 2, , −π, π, < x<, 2, 2, π, x≥, 2, , The values of A and B, so that f ( x ) is continuous, everywhere are, (a) A = 0, B = 1, (b) A = 1, B = 1, (c) A = − 1, B = 1, (d) A = − 1, B = 0, 24. Consider the following statements, I. f ( x ) =| x − 3| is continuous at x = 0., II. f ( x ) =| x − 3| is differentiable at x = 0., Which of the statements given above is/are correct?, (NDA 2010 II), , (a), (b), (c), (d), , Only I, Only II, Both I and II, Neither I nor II, , ( x − 1)2, 25. What is the value of lim, ?, x → 1 | x − 1|, (a), (b), (c), (d), , (NDA 2010 I), , 0, 1, −1, The limit does not exist, , 26. Match List I with List II and select the correct, answer using the codes given below the lists., List I, A., B., C., D., , lim x3 / 2 ( x3 + 1 −, , x→ ∞, , lim ( 3 x +, , x→ ∞, , C, 2, 4, , x+1, 3, , D, 4, 2, , 29. Consider the following function f : R → R such that, f ( x ) = x , if x ≥ 0 and f ( x ) = − x 2, if x < 0. Then, which, one of the following is correct?, (NDA 2009 II), (a) f ( x ) is continuous at every x ∈ R, (b) f ( x ) is continuous at x = 0 only, (c) f ( x ) is discontinuous at x = 0 only, (d) f ( x ) is discontinuous at every x ∈ R, 30. What is the set of all points, where the function, x, is differentiable?, f(x) =, (NDA 2009 I), 1 +| x|, (a) Only ( − ∞ , ∞ ), (b) Only ( 0, ∞ ), (c) Only ( − ∞ , 0) ∪ ( 0, ∞ ) (d) Only ( − ∞ , 0), 1, 31. Let f ( x ) =, . Then, what is the, 1 −|1 − x|, value of, , lim f ( x ) ?, , (NDA 2008 II), , x→0, , (b) ∞, , (a) 0, , 32. What is the value of lim, , x→α, , (a), , 4. e −2 / 3, , B, 3, 4, , C, 2, 1, , α + 2x − 3x, 3α + x − 2 x, , ?, , (NDA 2008 II), , 3. 1, , A, (b) 1, (d) 3, , (d) − 1, , (c) 1, , 2. e −1, , 9 x2 − x), , 3x − 4, lim , , x → ∞ 3x + 2, log x − 1, lim, x→e, x−e, , Codes, A B, (a) 3 1, (c) 3 1, , x3 − 1 ), , List II, 1, 1., 6, , 28. Consider the following statements, (NDA 2008 II), x2, I. lim, exists., x→0 x, x2 , II. is not continuous at x = 0., x, | x|, doesn’t exist., III. lim, x→0 x, Which of the statements given above are correct?, (a) I, II and III, (b) Only I and II, (c) Only II and III, (d) Only I and III, , D, 4, 2, , 27. It the function defined by, π, 1 − sin x, when x ≠, ( π − 2x )2, 2 is continuous at, f(x) = , π, , when x =, k,, , 2, π, x = , then the value of k is, 2, 1, 1, (b), (a), 4, 8, 1, (c), (d) None of these, 2, , 2, 3, , (b), , 1, (3 3), , (c), , 2, (3 3), , (d), , 1, 3, , 33. Let f : R → R be defined as, f ( x ) = sin(| x|), Which one of the following is correct?, (NDA 2008 I), (a) f is not differentiable only at 0, (b) f is differentiable at 0 only, (c) f is differentiable everywhere, (d) f is non-differentiable at many points, 34. A function f is defined as follows, 1, f ( x ) = x p cos , x ≠ 0, f( 0) = 0., x, What conditions should be imposed on p, so that f, may be continuous at x = 0?, (NDA 2007 II), (a) p = 0, (b) p > 0, (c) p < 0, (d) No value of p, 35. If f ( x ) = ( x + 1)cot x is continuous at x = 0, then what is, the value of f( 0) ?, (NDA 2007 I), 1, (a) 1, (b) e, (c), (d) e2, e
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329, , Limits, Continuity and Differentiability, 36. If f ( x ) is differentiable everywhere, then which one of, the following is correct?, (NDA 2007 I), (a)| f| is differentiable everywhere, (b)| f|2 is differentiable everywhere, (c) f| f|is not differentiable at same points, (d) None of the above, , Reason (R) Both h( x ) = x and, 1, , x sin , x ≠ 0, are continuous at x = 0., g( x ) = , x, 0, x = 0, , 40. Assertion (A) | x| is not differentiable at x = 0., | x|/ x ; x ≠ 0, , then lim f ( x ), Reason (R) If f ( x ) = , x→0, x=0, 0;, does not exist., , Directions (Q. Nos. 37-40), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 1, f ( x ) = x sin is differentiable at, x, , 37. Assertion (A), x = 0., , Reason (R) f ( x ) is continuous at x = 0., 1, 1, 38. Assertion (A) lim, =, x → 0 x cot x, 2, x→a, , { lim ( f ( x ) ⋅ g( x )} exists., x→a, , 39. Assertion, , (A), , continuous at x = 0., , lim, , 4 + sin 2x + a sin x + b cos x, x2, , x→ 0, , exists. Then,, , 41. The value of a is, (a) −2, (c) 5, , (b) 4, (d) None of these, , 42. The value of b is, (a) −2, (c) 3, , (b) −4, (d) None of these, , 43. The value of limit is, (a) 4, (c) 2, , (b) 5, (d) None of these, , Directions (Q. Nos. 44-45) Consider the function, , Reason (R) If lim f ( x ) and lim g( x ) exist, then, x→a, , Directions (Q. Nos. 41-43) Consider, , 1, 2, x sin , x ≠ 0, f(x) = , x, 0, , x=0, , is, , ax 2 + b,, x < −1, is continuous everywhere,, f(x) = 2, x ≥ −1, bx, ax, +, +, ,, 4, , then, , 44. The value of a is, (a) 1, (c) 4, , (b) 2, (d) None of these, , 45. The value of b is, (a) 2, (c) 3, , (b) 4, (d) None of these, , Answers, Level I, 1., 11., 21., 31., 41., , (b), (a), (d), (d), (b), , 2., 12., 22., 32., 42., , (c), (b), (d), (c), (a), , 3., 13., 23., 33., 43., , (d), (d), (d), (d), (b), , 4., 14., 24., 34., 44., , (c), (c), (c), (b), (b), , 5., 15., 25., 35., 45., , (c), (b), (b), (a), (b), , 6., 16., 26., 36., 46., , (c), (b), (d), (c), (b), , 7., 17., 27., 37., 47., , (d), (c), (d), (c), (a), , 8., 18., 28., 38., 48., , (d), (c), (a), (a), (d), , 9., 19., 29., 39., 49., , (c), (a), (d), (c), (c), , 10., 20., 30., 40., 50., , (b), (d), (a), (a), (c), , 2., 12., 22., 32., 42., , (b), (c), (c), (c), (b), , 3., 13., 23., 33., 43., , (d), (c), (c), (d), (c), , 4., 14., 24., 34., 44., , (b), (c), (c), (b), (b), , 5., 15., 25., 35., 45., , (c), (a), (a), (b), (c), , 6., 16., 26., 36., , (a), (a), (c), (c), , 7., 17., 27., 37., , (a), (a), (b), (d), , 8., 18., 28., 38., , (d), (a), (d), (d), , 9., 19., 29., 39., , (d), (a), (a), (a), , 10., 20., 30., 40., , (d), (a), (a), (b), , Level II, 1., 11., 21., 31., 41., , (c), (b), (a), (c), (a)
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Hints & Solutions, Level I, 1. lim, x→ a, , 3x − a − x + a, 3x − a + x + a, ×, x−a, 3x − a + x + a, , −, , LHL = lim e, , 7., , h→ 0, ∞, , (3x − a ) − (x + a ), = lim, x→ a (x − a ) ( 3 x − a + x + a ), , RHL = lim, , 2 (x − a ), x→ a (x − a ) ( 3 x − a + x + a ), , h→ 0, , LHL ≠ RHL, −, , So, lim e, , 3x/ 2 − 3, 3x/ 2 − 3 , 2. lim x, , = lim x / 2 2, x → 2 (3, x→ 2 3 − 9 , ) − 32, , h→ 0, , 8. Given,, ∴, , 1, 1, 1, =, =, + 3 31 + 3 6, , x→ 2−, , h→ 0, , x→ 2 +, , h→ 0, , 9. We have,, , 11. lim, , x→ 0, , , , x x2, +, + ... − 1, 1 +, 1! 2!, , , e −1, 5. lim, = lim, x→ 0, x→ 0, x, x, , , x x2, = lim 1 +, +, + ... ∞, x→ 0 , 2! 3!, , x, , x→3, , , 0, form, , 0, , a log a − b log b, (by L’Hospital’s rule), 1, a log a − b0 log b, =, 1, a, = log a − log b = log, b, x→ 0, 0, , h→ 0, , a x − xa, = −1, xa − a a, 1 −0, 1, = −1 ⇒, = −1, 0 − aa, −a a, a, a =1⇒a =1, , a x − bx, a x − bx, x, = lim, ⋅ x, x, x→ 0, x, e −1, e −1, a x − 1 bx − 1 x, = lim , −, , x→ 0, x ex − 1, x, , a x − 1, b x − 1 , 1, = lim , − lim , lim x, x → 0 x x → 0 x x → 0 e − 1, x, 1, a, = (log e a − log e b) = log e , b, 1, , 12. lim f (x) = lim, , =1, , = lim, , x→ 0, , ⇒, , Using L’ Hospital’s rule,, log e a asin x log e a, = lim, =, x → 0 log b bsin x, log e b, e, , x, , h→ 0, , 10. A continuous function may or may not be differentiable., So, option (b) is not true., 0, , form, 0, , , x, , lim, , ⇒, , Again, applying L’ Hospital’s rule,, 1, 1, = lim −, =−, x → 1 2x, 2, , x→ 0, , f (x) =|x − 2|, f (2) =|2 − 2| = 0, , Hence, f (x) is continuous at x = 2., , (On applying L’ Hospital’s rule), 1−x, = lim, x → 1 2 x (x − 1 ), , a x − bx, x, , doesn’t exist., , lim f (x) = lim f (2 + h ) = lim|2 + h − 2| = 0, , 1, −1, 1 + log x − x, x, 3. lim, =, lim, x → 1 1 − 2 x + x2, x → 1 − 2 + 2x, , 6. lim, , 1, x, , lim f (x) = lim f (2 − h ) = lim|2 − h − 2| = 0, , x → 2 3x / 2, , x→ 0, , e, , 1, (0 + h), , h→ 0, , Q, , asin x − 1, bsin x − 1, , −, , = lim e−1/ h = e−∞ = 0, , 2, 1, =, =, 2a + 2a, 2a, , 4. lim, , −1, , 1, , = lim eh, h→ 0, , =e =∞, , = lim, , = lim, , 1, (0 − h), , x→3, , x2 − 9, = lim (x + 3) = 6, x − 3 x→3, f (3) = 2 (3) + k = 6 + k, , and, , Q f is continuous at x = 3,, ∴ f (3) = lim f (x), x→3, , ⇒6 + k = 6 ⇒ k = 0, sin 2x 2, sin 2x, 13. lim, × = 2 lim, = 2 (1), x→ 0, x→ 0 2 x, x, 2, =2, , sin x , , = 1, Q lim, x→ 0 x,
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331, , Limits, Continuity and Differentiability, , 14. lim, , x→ 0, , tan x − sin x, x3, , 1, ∴, , tan x (1 − cos x), x→ 0, x3, tan x 2 sin 2(x / 2) 1, = lim, ⋅, =, x→ 0, x, 2, 4 ⋅ (x / 2)2, , = lim, , log (x − a ), log (ex − ea ), 1, x−a, (by L’ Hospital’s rule), = lim, 1, x→ a, ex, x, a, e −e, ex − ea, = lim x, x → a e ⋅ (x − a ), ex, (again, by L’ Hospital’s rule), = lim x, x → a e (x − a ) + ex, ea, = a =1, e, sin x, 16. lim, x→ ∞, x, 1, Put, and as x → ∞, h → 0, x=, h, sin x, 1, ∴, lim, = lim h sin = 0, x→ ∞, h→ 0, x, h, 1, , , lies between −1 and 1, Q value of sin, , , h, , lim, x→3, , 15. lim, , x→ a, , x+2, 17. lim 3, , x → − 2 x + 8, (x + 2), (x + 2) (x2 + 4 − 2x), 1, 1, 1, = lim, =, =, x → − 2 (x2 − 2 x + 4 ), (4 + 4 + 4) 12, = lim, , = lim, x→1, , sin 2 α − sin 2 β, , applying L’ Hospital’s rule, α→ β, α 2 − β2, = lim, , 2 sin α cos α − 0, 2α − 0, , = lim, , sin 2α, sin 2 β, =, 2α, 2β, , α→ β, , α→ β, , 22. If f (x) is continuous at x = 0, then, f (0) = lim f (x) = lim, x→ 0, , =, , 19. lim x sin, x→ ∞ , , 2 sin 2 /x, 2 , = lim , , x x → ∞ 2 /x , h→ 0, , =2, 20. Q, , f (x) =, , 1, , 18 − x2, 1, ∴ f (3) =, 18 − 9, 1, =, 3, , sin h, h, , 2− x+ 4, sin 2x, , 23. lim, , x→ 0, , log e (1 + x), 3x − 1, , , 0, form, , 0, , Using L’ Hospital’s rule,, 1, 1, 1+ x, 1+0, = lim x, = 0, x → 0 3 log 3, 3 log e 3, e, =, , 1, = log3 e., log e 3, , , log e e, = log3 e, Q, , log e 3, , cos ax − cos bx, x2, Using L’Hospital’s rule,, − a sin ax + b sin bx, = lim, x→ 0, 2x, Using L’Hospital’s rule,, − a 2 cos ax + b2 cos bx, = lim, x→ 0, 2, , − a 2 cos 0° + b2 cos 0°, 2, b2 − a 2, =, 2, x(x − 2), x, 25. Q f (x) = 2, =, 2, +, x, x −4, =, , (where h = 2/x), , 0, , form, 0, , , Using L’ Hospital’s rule,, , 1 , 1, −, −, 2 0 + 4 −1 / 4, 2 x + 4, 1, ⇒ f (0) = lim, =, =, =−, x→ 0, 2 cos 2x, 2 cos 0, 2, 8, , x→ 0, , 1, −3, , = 2 lim, , x→ 0, , 24. lim, , 1, (2x − 5), , 1, 3, , 21. lim, , x→ − 2, , x −1, (x − 1), 18. lim 2, = lim, x→1 2 x − 7 x + 5, x→1 (x − 1 ) (2 x − 5 ), , −, , 18 − x, f (x) − f (3), = lim, x→3, x−3, x−3, 1, = lim − (18 − x2)−3/ 2 (−2x), x→3, 2, 1, = − (9)−3/ 2 (−2 × 3), 2, 1, 1, =, ×3 =, 27, 9, 2, , and f (x) is continuous at x = 2., x, lim, = f (2), x→ 2 x + 2, 2 1, ⇒, f (2) = =, 4 2, , , 0, form, , 0, , 0, , form, 0,
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332, , NDA/NA Mathematics, , x + 6, 26. lim , , x → ∞ x + 1, , x+ 4, , , 5 , = lim 1 +, , x→ ∞ , x + 1, , , , 5 , = lim 1 +, , x → ∞ , x + 1, , 4, x, 1, x→ ∞, 1+, x, , x+ 4, 5, ⋅, ⋅ ( x + 1), 5, x+1, , x+ 4, x + 1 5 ⋅ x + 1, 5 , , , , , x, 32. lim , x → ∞ 3 +, , =e, , [(1) form], , , , Q lim 1 +, x → ∞ , , x, , 1, = e, , x, , , a + x − a − x, 27. lim , , x→ 0, x, , , ( a + x − a − x) ( a + x + a − x), = lim , , x→ 0, x ( a + x + a − x), , , (a + x) − (a − x), = lim, x → 0 x ( a + x + a − x), , , 2x, 2, 1, =, = lim , =, x → 0 x ( a + x + a − x), a, a, a, +, , , 28. Since, f (x) is continuous at x = 0, therefore, lim f (x) = f (0), , x→ 0, , sin πx, ⇒, lim, =k, x→ 0, 5x, sin πx π, =k, lim , ⇒, , x → 0 πx 5, π, π, sin x, , , ⇒, =k ⇒ k=, (1), = 1, Q lim, , , x, →, 0, x, 5, 5, f (2 + h ) − f (2), 29. Rf ′ (2) = lim, h→ 0, h, 2 (2 + h ) − 1 − (4 − 1), 4 + 2h − 1 − 3, = lim, = lim, h→ 0, h, →, 0, h, h, 2h, = lim, =2, h→ 0 h, f (2 − h ) − f (2), and Lf ′ (2) = lim, h→ 0, −h, 2 − h + 1 −3, = lim, =1, h→ 0, −h, ∴, Lf ′ (2) ≠ Rf ′ (2), Thus, f′ (2) does not exist., 30. lim, , sin θ, , , = 1, Q lim, , θ→ 0 θ, , Also, f (x) is continuous at x = 2., ∴, lim f (x) = f (2), x→ 2, , ⇒, ⇒, , lim (2x + λ ) = 6 − 4, , x→ 2, , lim 2(2 − h ) + λ = 2, 4 + λ = 2 ⇒ λ = −2, , h→ 0, , −3 x, , , = lim 1 +, x→ ∞ , , 3, , x, , , = lim 1 +, x→ ∞ , , 3 3, , x, , x, , ( −9 ), , [(1)∞ form], , x, , , 1, , Q lim 1 + = e, , x→ ∞ , x, , , , = e−9, , = e5, , sin 2 ax a 2 x, ⋅ 2⋅, x→ 0, bx, a x, 2, 2, sin ax a, x=0, = lim , ⋅, x → 0 ax , b, 3x − 4, 0 ≤ x ≤ 2, 31. Q f (x) = , 2x + λ , 2 < x ≤ 3, , 3x, , ∞, , 1+, , 5 lim, , , , x, , sin −1 x − tan −1 x, x→ 0, x3, Applying L’ Hospital’s rule,, 1, 1, −, 2, 2, 1−x 1+ x, = lim, x→ 0, 3 x2, , , 0, form, , 0, , 33. lim, , 0, , form, 0, , , Again, applying L’ Hospital’s rule,, 2x, −1, −2 x, +, ×, 2, (1 − x2)3/ 2 (1 + x2)2, = lim, x→ 0, 6x, = lim, , x→ 0, , 1, 6, , 1, , 1, 2, 1, +, = (3) =, , 2 3/ 2, 2 2, 6, 2, 1, 1, (, x, ), (, x, ), −, +, , , , 34. Since, f (x) is continuous., 2x − sin −1 x , f (x) = lim , ∴, = f (0), x → 0 2 x + tan −1 x, Applying L’ Hospital’s rule,, , 1 , 2 −, , 2−, , 1 − x2 , , f (0) = lim, =, x→ 0 , , 1 , 2 +, 2 +, , , 1 + x2 , , 2 −1 1, =, =, 2+1 3, , , 0, form, , 0, , 1, , 1, 1, , 1, , 35. f (x) is continuous at every point., ∴, lim f (x) = lim f (x) = f (1), x → 1−, x →1+, ⇒, 5 ×1 −4 =4 ×1 + 3 × b ×1, ⇒, 1 = 4 + 3b ⇒ 3b = − 3, ⇒, b = −1, 2x + 1 , when x > 1, , 36. Given, that,, ,, f (x) = k, , when x = 1, 5x − 2 ,, x, <, when, 1, , is continuous at x = 1, then, lim f (x) = lim f (x) = f (1), x → 1−, x →1+, ⇒, lim 5 (1 − h ) − 2 = lim 2 (1 + h ) + 1 = k, h→ 0, , ⇒, , 1+2=k, , ⇒, , k =3, , h→ 0, , 37. Since, f (x) is continuous at x = π /2 ., π, ∴, f = lim f (x), 2 x → π/ 2, ⇒, , λ = lim, , x → π/ 2, , 1 − sin x, π − 2x, , , 0, form, , 0
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333, , Limits, Continuity and Differentiability, , 2x − 1, , x≠0, , 44. Q f (x) = 1 + x − 1, k, x=0, ,, , x, 2 −1, 2x log e 2, (by L’ Hospital’s rule), ∴ lim, = lim, 1, x→ 0 1 + x − 1, x→ 0, 2 1+ x, , Applying L’ Hospital’s rule,, − cos x, ⇒, λ = lim, ⇒λ =0, x → π/ 2, −2, x + 3, 38. lim , , x → ∞ x + 1, , x+1, , , 2 , = lim 1 +, , x→ ∞ , x + 1, , x+1, 2, 2, , 2, , x +1, , 2 2 , , 2, = lim 1 +, , =e, x→ ∞ , +, x, 1, , , , , , , x + a, 39. lim , , x → ∞ x + b, , x+b, , , a − b, = lim 1 +, , x→ ∞ , x + b, , , a − b, , = lim 1 +, , x→ ∞ , x + b, , , = ea − b, 1, 40. lim x2 sin , x, x→0, , x, , , 1, , Q lim 1 + = e, x, x → ∞ , , , = 2 log e 2 = log e 4, Q f (x) is continuous at k = 0., ∴, lim f (x) = f (0), , x+b, , ⇒, 45. lim, , , , Q lim 1 +, x → ∞ , , , , , , 2 −2, x, x, , x→ 0, , x + b a − b, a −b , , x→ 0, , , 1, = e, , x, , x, , , 0, form, , 0, , −x, , log e 4 = k, , = lim 2x log 2 + 2− x log 2, x→ 0, , (by L’ Hospital’s rule), = log 2 + log 2 = log 4, Q Function is continuous at x = 0., 2x − 2−x, ∴, f (0) = lim, x→ 0, x, ⇒, f (0) = log 4, 1/cos x, , Put x = 1 / y, , 1, ⋅ sin ( y), y2, y, lim sin 2, y→∞, y, , ⇒, , lim, , y→ ∞, , ⇒, , 1 + tan x , 46. lim , , x → 0 1 + sin x, , , , cosec x, , lim 1/cos x, , (Q − 1 ≤ sin θ ≤ 1, θ ∈ R), , sin ∞, = 0 × (finite value) = 0, ∞, , ⇒, , , 4 , 41. Given that, lim 1 −, , x→ ∞ , x − 1, , , 4 , = lim 1 −, , x → ∞ , x − 1, , , 3x −1, , 3 x − 1, − ( x − 1 ) −4 x − 1 , 4, , , 1 , 1, , lim 3 − / 1 − , x , x, −4 x → ∞ , , =e, , , , , = e−4 × 3 = e−12, , a cot x − cos x − 1, = a cos ( π / 2) lim , = 1 log a = log a, x → π / 2 cot x − cos x , , , ax − 1, = log a , Q xlim, →0, x, , , x→, , π, 6, , 2 sin 2 x + sin x − 1, 2 sin 2 x − 3 sin x + 1, , = lim, , π, x→, 6, , 4 sin x cos x + cos x, 4 sin x cos x − 3 cos x, , =, , e x→ 0, e, , =, , e, =1, e, , 4 × 5x for x < 0, 47. We have, f (x) = , 8a + x for x ≥ 0, LHL = lim f (0 − h ) = lim 4 × 5− h = 4, h→ 0, , and, , h→ 0, , RHL = lim f (0 + h ) = lim (8a + h ) = 8a, h→ 0, , h→ 0, , Since, if f (x) continuous function., LHL = RHL ⇒ 8a = 4 ⇒ a =, , cot x − cos x, a cot x − a cos x , − 1, cos x a, 42. lim , , , = lim a, x → π / 2 cot x − cos x , x → π/ 2, cot x − cos x , , 43. lim, , cos x , , 1 + sin x sin x , , , , cos x, , , = lim, x→ 0, (1 + sin x)1/sin x, , (by L’ Hospital’s rule), , π, 4 sin + 1, cos x (4 sin x + 1), 2+1, 6, = lim, =, =, = −3, π cos x (4 sin x − 3 ), π, x→, 4 sin − 3 2 − 3, 6, 6, , 4 1, =, 8 2, , 48. Given that, f (a ) = 2, f ′ (a ) = 1, g (a ) = − 1, g ′ (a ) = 2, g (x) f (a ) − g (a ) f (x), lim, ∴, x→ a, x−a, On putting x = a , above expression become, 0, indeterminate form of ., 0, f (a ) g ′ (x) − g (a ) f ′ (x), = lim, x→ a, 1 −0, (on applying L’ Hospital’s rule), = f (a ) g ′ (a ) − g (a ) f ′ (a ), = 2 × 2 − (−1) × 1 = 4 + 1 = 5, mx, + 1, x ≤ π / 2, , 49. f (x) = , sin x + n , x > π / 2, π, Since, f (x) is continuous at x =, 2, π, , π, , π, f + 0 = f − 0 = f , ∴, 2, , 2, , 2
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334, , NDA/NA Mathematics, , π, lim f + h = lim f, h→ 0, 2, , ⇒, , π, , π, − h = f , 2, , 2, , h→ 0, , 50. lim, x→0, , mπ, π, , π, , ⇒ lim sin + h + n = lim m − h + 1 =, +1, 2, , 2, , h→ 0, h→ 0, 2, ⇒, ⇒, ⇒, , m π, mπ, − mh + 1 =, lim( cos h + n) = lim , +1, , h→ 0 2, 2, mπ, mπ, 1+ n=, +1 =, +1, 2, 2, π, n = m , 2, , h→ 0, , 5x, x sin 5x, sin 5x, = lim x , ⋅, 2, 2, , , →, x, 0, 5x, sin 4x, sin 4x, 2, × 16x, , 4x , sin 5x, 5, 1, lim, =, ⋅, 2, 16 x→ 0 5x , sin 4x, , lim, x→ 0 4x , 5, 1, =, ×1 ×, 16, 1, 5, =, 16, , Level II, x3 + 1, , lim 2, − (ax + b) = 2, x→ ∞ x + 1, , , 3, x (1 − a ) − bx2 − ax + (1 − b) , lim , ⇒, =2, x→ ∞, x2 + 1, , , a (1 − b) , , x (1 − a ) − b − x + x2 , ⇒, lim , =2, 1, x→ ∞, , , 1+ 2, x, , , This limit will exist, if, 1 − a = 0 and b = − 2 ⇒ a = 1 and b = − 2, 1, 1, 1, 2. lim, ⇒, =, = 1 and f (0) = 0, x → 0 1 − e− 1/ x, 1 − e−1/ 0 1 − 1, ∞, e, 1, lim, ≠ f (0), ⇒, x → 0 1 − e−1/ x, 1. Given that,, , ∴ Function is not continuous at x = 0., sin 2x + sin 6x, sin 4x cos 2x, 3. lim, = lim, x→ 0 sin 5 x − sin 3 x, x→ 0 2 sin x cos 4x, sin 4x x cos 2x, = lim 4, , , , x→ 0 4 x sin x cos 4 x, 1, =4 ×1 ×1 × =4, 1, 1/ n, , 3n, , 4. lim (3n + 4n )1/ n = lim 4 n + 1, n→ ∞, n→ ∞, 4, , , , 1 +, = lim 4 , n→ ∞, , , , 0, 1, , = 4 1 + , ∞, , = 4 × (1)0 = 4, , 1/ n, , , 1 , n, 4 , , 3 , , , 5. Since, |x| is discontinuous at x = 0 and, discontinuous at x = 0., |x|, is discontinuous at x = 0., ∴ f (x) = |x| +, x, , |x|, is, x, , 3x + 4 tan x, 4 tan x , , = lim3 +, , x→ 0 , x, x , tan x, = 3 + 4 lim, =3 + 4 = 7, x→ 0, x, If the given function f (x) is continuous at x = 0, then, lim f (x) = f (0) ⇒ k = 7, , 6. lim f (x) = lim, x→ 0, , x→ 0, , x→ 0, , 7. f (x) = cos (|x|), Let, φ (x) = |x|and g (x) = cos x,, then, ( go φ ) (x) = g (φ (x)) = g (|x|), ⇒, ( go φ ) (x) = cos (|x|), Since, φ and g are continuous functions, so ( go φ ) is also, a continuous function., 1 − cos θ, 8. Let y = lim, θ→ 0, θ, θ, 2 sin 2, , 2, 2 θ, ⇒ y = lim, Q 1 − cos θ = 2 sin 2 , θ→ 0, θ, θ, 2 sin, 2, y = lim, θ→0, θ, θ, sin, 2, −1 −1, LHL = 2 lim, = 2 × =, 2, 2, θ → 0− θ, ×2, 2, θ, sin, 1, 1, 2, RHL = 2 lim, = 2× =, 2, 2, x→ 0 + θ, ×2, 2, Q RHL ≠ LHL, ∴ The limit does not exist., − (x − 3); x < 3, 9. We know that,|x − 3| = , (x − 3); x > 3, |x − 3|, Now, LHL = lim f (x) = lim, −, −, x−3, x→ 3, x→3, − (x − 3), = lim, = −1, x→3, x−3, |x − 3|, RHL = lim f (x) = lim, x→ 3 +, x→3 + x − 3
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335, , Limits, Continuity and Differentiability, , = lim, , x→3, , Since,, , (x − 3), =1, x−3, , Now,, lim f (x) = lim f (0 − h ), , x → 0−, , lim f (x) ≠ lim f (x), , x→ 3, , −, , x→3, , +, , h→ 0, , 1 − cos 4 (0 − h ), 1 − cos 4 h, 2 sin 2 2 h, lim, =, =, lim, h→ 0, h→ 0, h→ 0, (0 − h )2, h2, h2, , = lim, , Thus, the limit does not exist., 10. We have, f (x) = x2 + 4x − 5, ∴, , 2, , sin 2h , =2 ×4 =8, = 2 × 4 lim , h → 0 2h , , , f (x) = x2 + 5x − x − 5 ⇒ f (x) = (x − 1) (x + 5), , Now,, f (0) = a, From Eq. (i), lim f (x) = f (0), x → 0−, ⇒, a =8, , For f (x) to be real-valued function., (x − 1) (x + 5) ≥ 0 ⇒ x ≥ 1 and x ≤ −5, ⇒ x ∈ (−∞ , − 5] ∪ [1, ∞ ), which is required domain., 11. f (x) = a sin|x| + be |x |, a sin (− x) + be( − x ), , f (x) = , 0, a sin x + bex, , , ex + e− x − 2, x sin x, The function ex and e− x both are continuous on, [−π / 2, π / 2] ., , 14. We have, F (x) =, , , x<0, , x=0, , x>0, , So, that ex + e−x − 2 is also continuous on [−π / 2, π / 2]., , LHL = lim f (x) = lim [a sin (− x) + be( − x ) ], x→ 0, , −, , x→ 0, , Also, x and sin x both are continuous on [−π / 2, π / 2]., , = lim [− a sin x + be− x ], x→ 0, , ex + e− x − 2, is continuous on [−π / 2, π / 2] ., x sin x, , = − a sin (0) + be0 = b, RHL = lim f (x) = lim [a sin x + bex ], , Then, F (x) =, , = a sin(0) + be0 = b, Also,, f (0) = 0, Since, the given function is differentiable at x = 0,, therefore it is also continuous at x = 0., LHL = RHL = f (0) ⇒ b = 0, , Provided x sin x ≠ 0 at every point of [−π / 2, π / 2]., , x→ 0 +, , x→ 0, , But x sin x = 0 at x = 0, ex + e− x − 2, is continuous except at x = 0 on, x sin x, the interval [−π / 2, π / 2]., log (x − a ), 15. lim, x → a log (ex − ea ), So, F (x) =, , 12. At x = 3,, , f (3 − h ) − f (3), −h, |3 − h − 3| + |3 − h − 4| − (0 + |3 − 4|), = lim, h→ 0, −h, 2h + 1 − 1, = lim, = −2, h→ 0, −h, f (3 + h ) − f (3), RHD = lim, h→ 0, h, |3 + h − 3| + |3 + h − 4| − 1, = lim, h→ 0, h, 2h − 2, = lim, =∞, h→ 0, h, Q, LHD ≠ RHD, ∴ f (x) is not differentiable at x = 3., Now, at x = 4,, |4 − h − 3| + |4 − h − 4| − 1, LHD = lim, h→ 0, 4 − h −4, 1 − h + h −1, = lim, =0, h→ 0, −h, |4 + h − 3| + |4 + h − 4| − 1, RHD = lim, h→ 0, 4 + h −4, 1 + h + h −1, = lim, =2, h→ 0, h, Q, LHD ≠ RHD, ∴ f (x) is not differentiable at x = 4., LHD = lim, , h→ 0, , 13. Since, f (x) is continuous at x = 0., ⇒, lim f (x) = f (0), x → 0−, , Using L’Hospital’s rule, we get, 1, ex − ea, x−a, = lim, = lim, x, x → a (x − a ) ex, x→ a, e, x, a, e −e, Again, by L’ Hospital’s rule, we get, ex, ea, = lim, = a =1, x, x, x → a (x − a ) e + e, e, f ( x) 2 t, 16. Here, lim ∫, dt, x→1 4, (x − 1), On integrating wrt t, we have, f ( x), , 2 t2 , lim , , x → 1 2 (x − 1 ), 4, , , = lim, , x→1, , 1, f ( x), [t 2]4, x−1, , 1, [( f (x))2 − 16], x−1, [ f (x)]2 − 16, = lim, x→1, x−1, = lim, , x→1, , Using L’ Hospital’s rule,, 2 f (x) ⋅ f ′ (x) − 0, lim, = lim 2 f (x) f ′ (x) = 2 f (1) f ′ (1), x→1, x→1, 1 −0, = 2 (4) f ′ (1), ...(i), , = 8 f ′ (1), , [given f (1) = 4]
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336, , NDA/NA Mathematics, , 17. At the rational point x = 0, we have f (0) = 0, lim, x2 = 0, , Also,, lim f (x) = x→ 0, x→ 0, (− x2) = 0, lim, x→ 0, ∴ f (x) is continuous at x = 0., , ⇒, LHL = RHL = f (0), ⇒ f (x) is continuous at x = 0., , 18. lim g {f (x)} = lim {f (x)}2 + 1, , Now,, , x→ 0, , RHL = lim f (x) = lim f (0 + h ), , and, , x→ 0, , h→ 0, , x→ 0, , h→ 0, , = lim sin 2 x + 1 = 1, , 1 − cos x, ,, , 19. Given that, f (x) = , x, k, ,, , = lim, , h→ 0, , x≠0, x=0, , 1 − cos x, x, 2 sin 2 x/ 2, x=0, = lim, x → 0 4 (x / 2 )2, , 2, , k =0, , −π , −π , f = −2 sin = + 2, 2, 2, π, lim f (x) = A sin − + B = B − A, π, 2, x→ − + 0, , 20. lim [ a 2x2 + ax + 1 − a 2x2 + 1 ], x→ ∞, , After rationalization,, a 2x2 + ax + 1 − a 2x2 − 1 , , = lim , 2 2, 2 2, x→ ∞ , a x + ax + 1 + a x + 1 , a, a, 1, a2 + + 2 +, x x, a, a, 1, =, =, a 2 + a 2 2a 2, , = lim, , x→ ∞, , x3 − 3x + 2, ,, , 21. Q f (x) = (x − 1)2, , k, ,, , , 2, , If f (x) is continuous at x = −, B − A =2, , 1, a2 + 2, x, , Again,, , lim f (x) = cos, , π, x→ + 0, 2, , By L’Hospital’s rule,, , π, , then we must have, 2, ...(ii), A + B =0, On solving Eqs. (i) and (ii), we get A = − 1 , B = 1, x − 3 , x ≥ 3, 24. Q f (x) = | x − 3| = , 3 − x, x < 3, , , 0, Q form, , 0, , ∴ LHL = lim f (x) = lim f (0 − h ) = lim (3 + h ) = 3, x → 0−, , By L’Hospital’s rule,, = lim, , x→1, , 6x, 2, , and, , LHL = lim f (x) = lim f (0 − h ), h→ 0, , = lim (− h )2 + h = 0, h→ 0, , h→ 0, , h→ 0, , RHL = lim f (x) = lim f (0 + h ), x → 0+, , h→ 0, , = lim (3 − h ) = 3, , =3, f (x) = | x| + x2, x2 + x, x ≥ 0, f (x) = , 2, x − x, x < 0, x → 0−, , π, =0, 2, , ∴ If f (x) is continuous at x = +, , , 0, Q form, , 0, , 3 x2 − 3, x → 1 2 (x − 1 ), , ...(i), , π, lim f (x) = A sin + B = A + B, π, 2, x→ − 0, π, π, f = cos = 0, 2, 2, , ∀x=1, , k = lim, , π, , then we must have, 2, , 2, , ∀x≠1, , and f (x) is continuous., x3 − 3x + 2, ∴, lim, =k, x→1, (x − 1)2, ⇒, , f (0 + h ) − f (0), h, , h2 + h, = lim h + 1 = 1, h→ 0, h→ 0, h, ⇒, LHD ≠ RHD, ⇒ f (x) is not differentiable at x = 0., π, 23. Here, lim f (x) = −2 sin − = + 2, −π, 2, x→, , x→ 0, , ⇒, , h→ 0, , [Q f (0) = 0], , = lim, , and, f (0) = k, Q Function is continuous at x = 0,, ∴, lim f (x) = f (0), , =, , Rf ′ (0) = RHD = lim, , and, , x→ 0, , ⇒, , h2 + h, = − lim h + 1, h→ 0, −h, , = −1, , lim f (x) = lim, , x→ 0, , f (0 − h ) − f (0), −h, , Lf ′ (0) = LHD = lim, , x→ 0, , 22. Q, , h→ 0, , = lim (+ h 2) + h = 0, , h→ 0, , ⇒, LHL = RHL, ∴ f (x) is continuous at x = 0., Now,, , f (0) − f (0 − h ), h, 3 − (3 + h ), = lim, = −1, h→ 0, h, , LHD = f ′ (0− ) = lim, , h→ 0
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337, , Limits, Continuity and Differentiability, , and, , RHD = f ′ (0+ ) = lim, , h→ 0, , f (0 + h ) − f (0), h, , 3 − h −3, = lim, = −1, h→ 0, h, ⇒, LHD = RHD, ∴ f (x) is differentiable at x = 1., Hence, both statements I and II are correct., (x − 1)2, | x − 1|, (x − 1),, =, − (x − 1),, , f (x) =, , 25. Let, , Now,, , x≥1, x<1, , h→ 0, , h→ 0, , h→ 0, , RHL = lim f (1 + h ), h→ 0, , = lim (1 + h − 1) = lim h = 0, h→ 0, , h→ 0, , (x − 1)2, =0, LHL = RHL ⇒ lim, x → 1 | x − 1|, , Q, , 26. (A) lim x3/ 2 ( x3 + 1 − x3 − 1 ), x→∞, , 2x3/ 2, , = lim, , x→ ∞, , x3 + 1 + x3 − 1, 2, 2, = lim, =, =1, x→∞, 1+ 1, 1, 1, 1+ 3 + 1− 3, x, x, , (B) lim (3x + 9x2 − x ), x→ −∞, , y→∞, , lim, , y, 3y + 9y + y, , 3x − 4 , (C) lim , , x→ ∞ 3 x + 2, , 2, , x+1, 3, , =, , 1, 1, =, 3+3 6, , , −6 , = lim 1 + , , x→ ∞, 3x + 2 , , lim, , =, , 2, , 1, 1 − sin x, 1 sin(h / 2) , = lim ⋅ , =, x → π / 2 (π − 2 x)2, h→ 0 8 , 8, h /2 , π, Since, the function f (x) is continuous at x = ., 2, 1, π, ∴, lim f (x) = f ⇒ k =, π, 2, 8, x→, Hence, lim, , x2, 28. I. lim, = lim (x) = 0, x→ 0 x, x→ 0, So, it is exist., x2, II., = x,, x, Since, a polynomial is continuous everywhere, so it, is continuous at x = 0., |0 − h |, h, III. LHL = lim, = lim, = −1, h → 0 (0 − h ), h → 0 −h, |0 + h |, h, RHL = lim, = lim = 1, h → 0 (0 + h ), h→ 0 h, Q LHL ≠ RHL, So, it does not exist., x,, x≥0, 29. Given, f (x) = 2, − x , x < 0, LHL = lim f (x) = − lim x2 = 0, x→ 0, x → 0−, RHL = lim f (x) = lim x = 0, x → 0+, , Let − x = y, lim (−3 y + 9 y2 + y ), ∴, y→ ∞, , 2, , 2, , LHL = lim f (1 − h ), = lim [− (1 − h − 1)] = lim h = 0, , and, , , π, 1 − sin + h, , 2, , 1 − cos h, 4h, 4h 2, 2, 1 2 sin (h / 2) 1 sin 2(h / 2), = ⋅, = ⋅, 4 (h / 2)2 ⋅ 4, 8 (h / 2)2, , 1 − sin x, =, ∴, (π − 2x)2, , −6, , x→ ∞ 3x + 2, , ⋅, , x+1, 3, , x+1, 3, , =e, = e−2/ 3, log x − 1, log x − log e, (D) lim, = lim, x→ e, x→ e, x−e, x−e, log (e + h ) − log e, = lim, h→ 0, e+ h −e, h, , log 1 + , , 1, e 1, = lim, = = e−1, h, e h→ 0, e, e, 1 − sin x, π, as x → ., 27. We first of all, find out the limit of, 2, (π − 2x)2, π, π, For this, let x − = h, so that when x → , h → 0, 2, 2, , x→ 0, , and, f (0) = 0, ∴, LHL = RHL = f (0), ∴ It is continuous at x = 0., Also, f (x) is continuous in the given interval., Hence, f (x) is continuous in every x ∈ R., x, 30. Q f (x) =, 1 + | x|, x, , x<0, 1 − x, =, x , x≥0, 1 + x, f (0 − h ) − f (0), ∴ LHD = f ′ (0− ) = lim, h→ 0, −h, −h, −0, 1+ h, = lim, h→ 0, −h, 1, = lim, =1, h→ 0 1 + h, and RHD = f ′ (0+ ), f (0 + h ) − f (0), = lim, h→ 0, h
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338, , NDA/NA Mathematics, 1, lim f (x) = lim x p cos , x, x→ 0, x→ 0, , h, −0, 1, 1+ h, = lim, = lim, =1, h, →, 0, h→ 0, 1+ h, h, , ∴, 32. lim, , x→ α, , = lim, , x→ α, , = lim, , x→ α, , α + 2x + 3x, , ×, , α + 2x + 3x, , ×, , =, , 3α + x + 2 x, 1, lim, 3 x → α α + 2x + 3x, , =, , 1 4α + 2 α, ⋅, 3 3α + 3α, , =, , 1 4 α, 2, ⋅, =, 3 2 3 α 3 3, , lim x cot x, , lim, , x, , x → 0 tan x, , =e, , = e1 = e, , , , x, = 1, (1 + x)1/ x = e and lim, Q xlim, x → 0 tan x, →0, , , Since, f (x) is continuous at x = 0., ∴, f (0) = lim f (x) = e, x→ 0, , 36. If f (x) is differentiable everywhere, then f | f| is not, differentiable at same points., Since, the product of two functions in which one is, differentiable and other is not differentiable, then the, resultant is not differentiable., 1, 37. Q f (x) = x sin , x, 3α + x + 2 x, , ( 3α + x )2 − (2 x )2, , 3α + 2x + 2 x, α + 2x − 3x, ×, 3α + x − 4x, α + 2x + 3x, 3α + x + 2 x, , = lim [(1 + x)1/ x ]x cot x, =e, , α + 2x − 3x, , x→ α, , x→ 0, , x→ 0, , 3α + x − 2 x, = lim, , [(1)∞ form], , lim f (x) = lim (x + 1)cot x, , x→ 0, , x→ 0, , x→1, , 2, , f (x) is continuous at x = 0, 1, lim x p cos = f (0) = 0, x, x→ 0, , Which is possible only, if p > 0 ., 35., , lim f (x) = 1, , ( α + 2x) − ( 3x )2, , Q, ∴, , Q, LHD = RHD, ∴ f (x) is differentiable at x = 0., Hence, f (x) is differentiable in (−∞ , ∞ )., 1, 31., LHL = lim, h → 0 1 − |1 − (1 − h )|, 1, = lim, h → 0 1 −| h |, 1, = lim, =1, h→ 0 1 − h, 1, RHL = lim, h → 0 1 − |1 − (1 + h )|, 1, = lim, h → 0 1 −| − h |, 1, = lim, =1, h→ 0 1 − h, , (α − x), 3(α − x), , 33. f (x) = sin| x|, x≥0, sin x,, =, sin (− x), x < 0, sin (x), x ≥ 0, =, − sin x, x < 0, cos x, x ≥ 0, ⇒ f ′ (x) = , − cos x, x < 0, From above it is clear that f (x) is not differentiable, at x = 0 and f (x) is a periodic function. Thus, f (x) is, non-differentiable at many points., P, 1, x ≠0, x cos ,, (, ), =, f, x, x, , 34. Q, 0,, x =0, , , For differentiability at x = 0, f (0 − h ) − f (0), LHD = lim, h→ 0, −h, 1, (− h ) sin − , h, = lim, h→ 0, −h, 1, h sin , h, = lim, h→ 0, h, 1, = lim sin = a value lies between –1 and 1., h, h→ 0, = does not exist, f (0 + h ) − f (0), h, 1, h sin , h, = lim, h→ 0, h, 1, = lim sin , h, h→ 0, , RHD = lim, , h→ 0, , = a value lies between –1 and 1 = does not exist., LHD ≠ RHD, Hence, f (x) is not differentiable at x = 0. For continuity, at x = 0, LHL = lim f (0 − h ), x → 0−, 1, = lim − h sin − , h, h→ 0, = lim h sin, h→0, , 1, = 0 × (finite value lies between −1 to +1), h, , =0, RHL = lim f (0 + h ), x→0 +
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339, , Limits, Continuity and Differentiability, , lim h sin, , h→ 0, , 1, = 0 × (finite value lies between −1 to +1) = 0, h, , f (0) = 0, ⇒, , LHL = RHL = f (0), , Hence, f (x) is continuous at x = 0, Thus, A is false but R is true., tan x, 1, 38. We have, lim, = lim, x→ 0, x → 0 x cot x, x, Using L’ Hospital’s rule,, 1, sec 2x, =1 ≠, = lim, x→ 0, 2, 1, If lim f (x) exist and lim g (x) exist., ∴, , x→ a, , x→ a, , lim { f (x) ⋅ g (x)} exists., , x→ a, , (by property of limit), , ∴ A is false, but R is true., 1, 2, x sin , x ≠ 0, 39. Assertion (A) f (x) = , x, , x=0, 0, 1, LHL = lim f (x) = lim (− x)2 sin − = 0, −, x, x, 0, →, x→ 0, 1, RHL = lim f (x) = lim x2 sin = 0, x, x→ 0, x → 0+, ∴, , f (0) = lim f (x) = lim f (x) = 0, x→ 0 −, , x → 0+, , ∴ It is continuous at x = 0., Reason (R) We have,, h (x) = x, it is continuous for every values of x., 1, , x sin , x ≠ 0, Now,, g (x) = , x, , x=0, 0, 1, LHL = lim g (x) = lim x sin = 0, − x, x→ 0, x→ 0 −, 1, RHL = lim g (x) = lim x sin = 0, x, x→ 0, x→ 0 +, ∴, , f (0) = lim f (x) = lim f (x) = 0, x → 0−, , x→ 0, , ∴ Both A and R are true and R is the correct explanation, of A., 40. Assertion (A) Let f (x) = |x|, f (0 − h ) − f (0) h − 0, LHD = lim, =, = −1, h→ 0, −h, −h, f (0 + h ) − f (0) h, RHD = lim, = =1, h→ 0, h, h, ⇒, LHD ≠ RHD, ∴ f (x) is not differentiable at x = 0 ., |x|, , x≠0, Reason (R) We have, f (x) = x, 0 , x = 0, , x, LHL = lim f (x) = lim − = − 1, −, x, →, x, 0, x→ 0, x, RHL = lim f (x) = lim = 1, +, x, →, 0, x, x→ 0, Limit does not exist., ∴Both A and R are true, but R is not the correct, explanation of A ., , Solutions (Q. Nos. 41-43), Since, the given limit exists and denominator, approaches 0 as x → 0., The numerator must approach zero as x → 0, which is, ∴, possible, if 4 + b = 0, which is obtained by putting x = 0, in the numerator., ∴, b = −4, Hence, the given function becomes, 4 + sin 2x + a sin x − 4 cos x, x2, Now, the numerator is (on expansion), , , , x3, , 8x3, x2 x 4, 4 + 2x −, −..., +, + ... − 41 −, + ... + a x −, 3!, 2! 4!, , , , 3!, , a, 2, 3 −8, = (4 − 4) + (2 + a ) x + 2x + x − + ..., 6 6, 1, 2, = 0 + (2 + a ) x + 2x − (8 + a )x3 + K, 6, 1, (2 + a ) + 2x2 − (8 + a )3 + K, 6, The expression =, x2, If the limit exists, then we must have 2 + a = 0, i.e., a = − 2 and in that case, the limit = 2, Hence, a = − 2, b = − 4 and the limit = 2, , Solutions (Q. Nos. 44-45), ax2 + b,, x< −1, Given that, f (x) = 2, +, +, ≥ −1, ,, bx, ax, 4, x, , Since, f ′ (x) is continuous everywhere, sof (x) is also, continuous everywhere., So,, f (− 1) = lim f (x) = lim a (−1 − h )2 + b, ⇒, ⇒, , x→ 1 −, , b−a+4=a+ b, a =2, 2ax, Now,, f ′ (x) = , 2bx + a, , h→ 0, , , x < −1, ,, , x≥ −1, , Now, given that f ′ (x) is everywhere continuous., ∴, f (−1) = lim f (x) = lim (2a ) (− 1 − h ) = − 2a, x→ − 1 −, , h→ 0, , ⇒ 2b(− 1) + a = − 2a, ⇒, 3a − 2b = 0 ; put a = 2, we get b = 3
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18, Differentiation, Differentiation is a method to compute the rate at, which a dependent variable y changes with respect to the, change in the independent variable x. This rate of change is, called the derivative of y with respect to x., , Geometrical Meaning of Derivative, Let us consider a function y = f ( x ) and x and y are real, numbers. If the graph of y is plotted against x. The, derivative measures the slope of this graph at each point., Let a point Q( x + h , f ( x + h )) is very near to point, P ( x , f ( x )) on y = f ( x ). The value g / h is an approximation to, the slope of the tangent which we require., y, , It gives the instantaneous rate of change of y with, respect to x., , Differentiation of Some, Important Functions, 1., 2., 3., 4., , Q, g, , y = f (x), , 5., , P, h, , 6., , O, , Also, it can be written as, , x, , change in y, ∆y, or m =, change in x, ∆x, , If we move Q closer and closer to P, the line PQ will get, closer and closer to the tangent at P and so the slope of PQ, gets closer to the slope that we require. If we let Q go all the, way to touch P ( i.e . , h = 0), then we would have the exact, slope of the tangent., g f(x + h ) − f(x), Now,, =, h, h, So, also the slope PQ will be given by, f(x + h ) − f(x), m=, h, But we require the slope at P, so let h → 0, then in, effect, Q will approach P and g / h will approach the, required slope. Putting this together, the slope of the, tangent at P is, dy, f(x + h ) − f(x), = lim, dx h → 0, h, , 7., 8., 9., 10., , d, ( c) = 0, c is independent of x, dx, d, ( x n ) = nx n − 1, dx, d, (sin x ) = cos x, dx, d, (cos x ) = − sin x, dx, π, d, , (tan x ) = sec2 x , x ≠ nπ + , 2, dx, , d, 2, (cot x ) = − cosec x , { x ≠ nπ}, dx, d, , (sec x ) = sec x tan x , x ≠ nπ +, dx, , d, ( cosec x ) = − cosec x cot x , { x ≠, dx, d, 1, (sin−1 x ) =, ,|x|< 1, dx, 1 − x2, , π, , 2, nπ }, , d, 1, (cos−1 x ) = −, ,|x|< 1, dx, 1 − x2, , d, 1, (tan−1 x ) =, dx, 1 + x2, d, 1, 12., (cot−1 x ) = −, dx, 1 + x2, , 11., , 13., , d, 1, (sec−1 x ) =, ,| x|> 1, dx, | x| x 2 − 1, , 14., , d, 1, ( cosec−1x ) = −, ,|x|> 1, dx, |x| x 2 − 1, , 15., , d, ( a x ) = a x loge a, dx
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341, , Differentiation, , 16., 17., 18., 19., 20., 21., , d x, ( e ) = ex, dx, 1, d, (log x ) = , ( x > 0), dx, x, d, 1, (loga x ) =, dx, x loge a, x, d, |x|, or, |x| =, , { x ≠ 0}, |x|, dx, x, 1, d, x=, dx, 2 x, d x, x = x x (1 + log x ), dx, , xx − x− x , , 2 , , Solution (a) f ( x) = cot −1 , , On differentiating, we get, f ′( x) = −, , =−, , ∴, , d, F ( x ) = f ( x ),, dx, d, then, F ( ax + b) = af ( ax + b), dx, d, d, ( kf ( x )) = k, f ( x ), where k is a constant., dx, dx, d, d, d, [( f ( x ) ± g( x ))] =, g( x ), f(x) ±, dx, dx, dx, d, d, d, f(x), g( x ) + g( x ), [ f ( x ) g( x )] = f ( x ), dx, dx, dx, d f ( x ) g( x ) f ′ ( x ) − f ( x ) g′ ( x ), , g( x ) ≠ 0, , =, dx g( x ), [g( x )]2, d, d, ( f ( x )) g ( x ) = ( f ( x )) g ( x ), ( g( x ) log f ( x )), dx, dx, 1, d 1, d, =−, ⋅, f ( x ), f ( x ) ≠ 0, 2, dx f ( x ), dx, ( f ( x )), , 1. If, , 3., 4., 5., 6., 7., , ⇒, , xx − x, , 1+ , 2 , , 2, 4 + ( x − x− x) 2, x, , ⋅, , d xx − x− x , , , dx 2 , , d −x , d x, dx ( x ) − dx ( x ) , , , , 2, [ xx (1 + log x) + x− x(1 + log x)], ( xx + x− x) 2, 2, =− x, ( xx + x− x) (1 + log x), ( x + x− x) 2, 2, =− x, (1 + log x), x + x−x, 2, f ′ (1) = −, (1 + 0), 1+ 1, 2, = − = −1, 2, f ′( x ) = −, , Example 3. The derivative of the function, f ( x) =, , x2, 1 − x2, , 4x, 1− x, 1, (c), 1− x, , Example 1. If f ( x) = | x − 2 | and g ( x) = f ( f ( x)), then the, (b) 1, (d) 3, , is, (b), , (a), , 8. Chain rule If y is a function of z and z is a function, dy dy dz, of x, then, =, ⋅, dx dz dx, value of g ′ ( x) for x > 20 is, (a) 4, (c) 2, , −x 2, , , d x, x, , Q dx ( x ) = x (1 + log x), , , and d ( x− x) = − x− x (1 + log x ) , , , dx, , Algebra of Differentiable, Functions, , 2., , 1, , 2x, (1 − x 2) 2, , (d) None of these, , Solution (b) We have, f ( x) =, , x2, 1 − x2, , On differentiating both sides wrt x, we get, (1 − x2) (2x ) − x2 ( −2x ), f ′ ( x) =, (1 − x2) 2, = 2x ⋅, , (1 − x2 + x2), 2x, =, (1 − x2) 2, (1 − x2) 2, , Solution (b) We have, f ( x) = | x − 2|, for, , ∴, , x > 20 ,, f ( x) = x − 2, g ( x) = f ( f ( x)), = f ( x − 2) = x − 2 − 2 = x − 4, g ′ ( x) = 1, , (given), , xx − x− x , , then the value of f ′(1), 2 , , Example 2. If f ( x) = cot −1 , is, , (a) −1, , (b) 7, , (c) −2, , (d) 3, , Differentiation of Functions in, Parametric Form, Sometimes x and y are given as functions of a single, variable, for example, x = φ( t ), y = ψ( t ) are two functions, and t is a variable. In such a case x and y are called, parametric functions or parametric equations and t is, dy, called the parameter. To find, in case of parametric, dx, functions, we first obtain the relationship between x and y, by eliminating the parameter t and then we differentiable x
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343, , Differentiation, , Important Trigonometrical, Formulae, The following formulae will be helpful to solve the, problems involving inverse trigonometric function etc., 2 tan x, 1. sin 2x = 2 sin x cos x =, 1 + tan2 x, 2. sin 3x = 3 sin x − 4 sin3 x, 3. cos 2x = 2 cos2 x − 1, = 1 − 2 sin x = cos x − sin x =, 2, , 2, , 2, , 1 − tan2 x, 1 + tan2 x, , 4. cos 3x = 4 cos3 x − 3 cos x, 5. tan 2x =, 6. tan 3x =, , 2 tan x, , The following table will be useful to reduce the, function into simple form, Expressions, 1., 2., 3., 4., , 3 tan x − tan3 x, 1 − 3 tan x, , a+ x, a− x, or, a− x, a+ x, 2x, 2x, 6., or, 2, 1+ x, 1 − x2, , x = a cos θ, x = tan θ, , 1+ x + 1−, 2, , −1, , π, 2, π, (ii) tan−1 x + cot−1 x =, 2, π, (iii) sec−1x + cosec−1x =, 2, , 2 cos θ /2 + 2 sin θ /2, , = sin −1, 2, , , 1, θ, θ, 1, = sin −1 , cos +, sin , 2, 2, 2, 2, π, θ, π, θ, , = sin −1 sin cos + cos sin , , 4, 2, 4, 2, , π θ π θ, = sin −1 sin + = +, 4 2 4 2, , ⇒, , 12. sin−1 x ± sin−1 y = sin−1 ( x 1 − y 2 ± y 1 − x 2 ), , 1− x 1, = cos−1 x, 1+ x 2, , (d) None of these, , Putting x = cos θ , we get, 1 + cos θ + 1 − cos θ , , y = sin −1 , 2, , , , (i) sin−1 x + cos−1 x =, , 1 − x2 , = 2 tan−1 x, 15. cos−1 , 2, 1 + x , , 2 1− x 2, , (1 + x) + (1 − x) , , 2, , , , , −1 x + y , ∀ x , y > 0 and xy < 1, tan 1 xy ,, − , , =, π + tan−1 x + y , ∀ x , y > 0 and xy > 1, , 1 − xy , , 2x , 14. sin−1 , = 2 tan−1 x, 1 + x2 , , (b), , x, wrt x is, , , Solution (b) Let y = sin−1 , , 10. tan−1 x + tan−1 y, , 13. cos−1 x ± cos−1 y = cos−1 ( xy m, , 1, , 1− x 2, 1, (c), 3x, , 1 ± tan x, π, , = tan ± x, , , 1 m tan x, 4, , 16. tan−1, , x = a tan θ, x = a sin θ or x = a cos θ, x = asec θ or x = a cosec θ, x = a tan θ, , 2, , 5., , (a), , 2, , mx, 2, 2 mx, 8. 1 − cos mx = 2 sin, 2, , 11., , Substitutions, , a + x, a 2 − x2, x2 − a 2, a+ x, a− x, or, a− x, a+ x, 2, , Example 7. The derivative of sin−1 , , 1 − tan2 x, , 7. 1 + cos mx = 2 cos2, , 9., , Differentiation by Substitution, , 1 − x2, , 1 − y2 ), , y=, , π 1, + cos−1 x, 4 2, , On differentiating, we get, dy, 1, 1, =0 −, =−, 2, dx, 2 1− x, 2 1 − x2, , 1 + x 2 − 1, wrt, , , x, , , , Example 8. The derivative of tan−1 , 2 x 1 − x2 , at x = 0, is, tan−1 , 1 − 2 x2 , , , 2, 1, (a), (b), 3, 4, , (c) 2, , (d), , 4, 3
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344, , NDA/NA Mathematics, 1 + x2 − 1, , , , x, , , , Solution (b) Let y = tan−1 , and, , …(i), , Example 9. If f ( x) = θ sec x, , 2x 1 − x2 , , z = tan −1 , 1 − 2x2 , , , , …(ii), , ∴, ⇒, , Solution (b) We have, f ( x) = θ sec x, 1, , 1, , sec θ, ∴ f ′(θ) = 0 + θ sec θ tan θ, 1, , sec θ tan 2 θ 1, sec θ 1 + θ sec θ tan θ θ, 0, 0, 0, sec2θ 0, , tan 2 θ 0, 2, , = 1( −1) + 0 = − 1, , Successive Differentiation, , To differentiate a determinant, we differentiate one, row (or column) at a time, keeping others unchanged., f ( x ) g( x ) h( x ), If, y = p( x ) q( x ) r( x ), u( x ) v( x ) w( x ), q( x ), v( x ), , tan x − tan θ 0, , = 1 (tan 2 θ − sec2θ) − sec2θ (θ sec θ − θ sec θ), , 1− 0, 1, dy , =, =, , dz ( x = 0) 4 (1 + 0) 4, , dy, = p( x ), dx, u( x ), , f(x), , g( x ), , dy, is called first, dx, derivative of a function and derivative of first derivative,, d2y, is called second derivative of original function and so, dx 2, on., The process of differentiating a function more than, once is called successive differentiation., Let y = f ( x ) be a function of x, then, , %, , h( x ), , dn, dx, , r( x ) + p′( x ) q ′( x ) r ′( x ), w( x ), u( x ) v( x ) w( x ), , %, , %, , n, , dn, dx, , f (x) g (x) h (x), + p(x) q(x) r (x), u′(x) v′(x) w′(x), , n, , dn, dx n, , (x n ) = n !, , %, , e mx = m n e mx, , %, , dn, dx n, dn, dx n, , (log x ) = (−1) n − 1 (n − 1)! x − n, , Comprehensive Approach, n, , n, , n, , 1, , tan x, x, tan x − tan θ 0, , 1, tan 2 θ, sec θ, sec θ tan 2 θ 1, 2, 1 + θ sec x tan x x, sec x, + θ sec x tan x, 1, tan x − tan θ 0, sec2x 0, 0, , 1 − x2, dy dy /dx, 1, =, =, ⋅, 2, dz dz /dx 2 (1 + x ), 2, , Differentiation of a Function Given, in the Form of Determinant, , ∴, , tan 2 θ, , On differentiating both sides, wrt x, we get, 0, 0, 0, f ′( x) = θ sec x, tan x, x, , 2 sin θ cos θ , −1, = tan −1 , = tan (tan 2 θ) = 2θ, cos 2θ , dz, 2, z = 2 sin −1 x ⇒, =, dx, 1 − x2, , f ′( x ) g′( x ) h ′( x ), , (b) −1, (d) 3, sec θ, , 1 − cos θ , θ, , = tan −1 , = tan −1 tan , , 2, sin θ , θ 1, dy, 1, y = = tan −1 x ⇒, =, 2 2, dx 2 (1 + x2), , 1, , tan x, x , then the, tan x − tan θ 0, , value of f ′ (θ) is, (a) 4, (c) 2, , Again, putting x = sin θ in Eq. (ii), we get, 2 sin θ 1 − sin 2 θ , , z = tan −1 , 1 − 2 sin 2 θ , , , , ⇒, , tan2 θ, , 1, , Putting x = tan θ in Eq. (i), we get, 1 + tan 2 θ − 1, = tan −1 sec θ − 1, y = tan −1 , , , tan θ , θ, tan, , , , ⇒, , sec θ, , f ′( x), d, , , [ f ( x)] g( x ) = [ f ( x)] g( x ) g( x), + g ′( x)log f ( x) , dx, f ( x), , , d, f ( x), f ( x), [ f ( x)] = [ f ( x)] [1 + log f ( x)] f ′( x), dx, y 2f ′( x), dy, If y = [ f ( x)]y , then, =, dx f ( x) (1 − log y), , n, , If y =, , n, , If xmy n = ( x + y) m +, , f ′( x), dy, =, dx 2y − 1, dy y, n, =, , then, dx x, , f ( x) + y , then, , , nπ , (sin x ) = sin x +, , , 2 , , nπ , (cos x ) = cos x +, , , 2
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Exercise, Level I, 1. If f ( x ) = sin (cos x ), then f ′ ( x ) is, (a) cos (cos x ), (b) sin ( − sin x ), (c) − sin (cos x ), (d) − sin x cos (cos x ), 2x , 2x , 2. The derivative of sin−1 , wrt tan−1 , is, 2, 1 + x , 1 − x2 , (a) −2, (c) 1, , (b) −1, (d) 2, , 3. If y = cosec−1, , x+1, x−1, , + cos−1, , x−1, x+1, , 2, , ( x − 1), 1, (c), x−1, , 2, , (b), , dy, is equal, dx, , , then, 2, , ( x + 1)2, , (d) 0, , 4. If y = sin x + sin x + sin x + .. ,then, (a), , sin x, 2y − 1, , (b), , cos x, 2y − 1, , (d), , 2y − 1, cos x, , 2, , (c), , y, cos x − x, , 2, , , tan−1 1 + x − 1 is equal to, , , x, , , 1, x2, (a), (b), 1 + x2, 2 1 + x 2 ( 1 + x 2 − 1), 2, 1, (d), (c), 2, 1+ x, 2 (1 + x 2 ), , d, dx, , 10. If f ( x ) = sin2 x 2, then what is the value of f ′ ( x )?, , to, (a), , 9., , dy, is equal to, dx, , (NDA 2009 I), 2, , (a) 4x sin ( x ) cos ( x ), (c) 4 sin ( x 2 ) sin2 x, , 2, , (b) 2 sin ( x ) cos ( x 2 ), (d) 2x cos2 ( x 2 ), , 11. If x = cos 2t and y = sin2 t, then what is the value of, d2y, (NDA 2010 II), ?, dx 2, (a) 0, (b) sin ( 2t ), 1, (c) − cos ( 2t ), (d) −, 2, 4x , dy, ?, 12. If y = sin−1 , , then what is the value of, 2, dx, 1 + 4x , (NDA 2010 I), , dy, π, at x = ?, dx, 4, , (a), , (NDA 2011 II), , (c), , 5. If y = log tan x , then what is the value of, (b) − 1, , (a) 0, 1, (c), 2, , 2, , (d) 1, , 1, , (b) −, , 1 + 4x 2, 4, , (d), , 1 + 4x 2, , 1, 1 + 4x 2, 4x, , 1 + 4x 2, , 13. If x = t 2, y = t3 , then what is the value of, , 6. What is the derivative of sin2x wrt cos2 x ?, (NDA 2010 II), , (a) tan2 x, , (b) cot2 x, , (c) − 1, , (d) 1, , 7. If x +, , y = 2 , then what is the value of, , (NDA 2010 II), , (a) 5, (c) 2, , (b) 4, (d) − 1, , 8. If x m y n = 2 ( x + y )m + n , then the value of, (a) x + y, x, (b), y, y, (c), x, (d) x − y, , dy, is, dx, , dx 2, , ?, , (NDA 2010 I), , 3, (b), 2t, 3, (d), 2, , (a) 1, (c), , dy, at y = 1?, dx, , d2y, , 3, 4t, , x +1, dy, , then what is the value of, ?, x −1, dx (NDA 2012 I), −2, −2, (b), (a), x −1, ( x − 1)2, 2, 2, (d), (c), 2, ( x − 1), ( x − 1), , 14. If y =, , 3, , 15. The differential of ex wrt log x is, (a), (b), (c), (d), , 3, , ex, 3, 3x 2 2ex, 3, 3x3 ex, 3, 3x 2ex + 3x 2
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346, , NDA/NA Mathematics, , 16. If x + y = t −, , 1 2, 1, , x + y 2 = t 2 + 2 , what is the value of, t, t, , dy, ?, dx, 1, (a), x, 1, (c) 2, x, , (a), (b) −, (d) −, , 1, x, 1, , equal to, 1, (a), + x log a, x, 1, (c), + x log a, x log a, 1, , 18. If y = e 2, , log (1 + tan 2 x ), , (c) sec x tan x, 19. If y = 2x ⋅ 32x − 1, then, (a) (log 2) (log 3), (c) (log 182 ) y 2, , 1, 3a, 4a, (c), 3, , dy, is, dx, , 27. If x y = ex − y , then, , log a, x, +, x, log a, 1, log a, (d), −, x log a x (log x )2, , 1, sec2x, 2, , following?, (x − y), (a), (1 + log x )2, (x + y), (c), (1 + log x ), , dy, is equal to, dx, (b) sec2x, (d), , dy, is equal to, dx, (b) (log 18), (d) (log 18) y, dy, ?, dx, , (c), , 1 − x2, 1 − y2, , (b), , y, x, , d2y, dx 2, (b) −, , (d) 1, , at t = 2 is, 1, 16a, , (d) None of these, dy, is equal to which one of the, dx, (NDA 2009 I), , y, (b), (1 + log x ), (log x ), (d), (1 + log x )2, , 29. If f ( x ) = x + x + x + K ∞ , then what is the value, of f ′ ( x )?, 1, (a), 1 − 2 f(x), 1, (c), 1 + 2 f(x), , (NDA 2009 II), , (1 − x 2 )(1 − y 2 ), , (c), , dy, ?, dx, , 28. If x = sin t − t cos t and y = t sin t + cos t ,then what is, π, dy, at point t = ?, (NDA 2008 I), 2, dx, π, (a) 0, (b), 2, π, (d) 1, (c) −, 2, , 1, log (1 + tan 2 x ), e2, , 20. If 1 − x 2 + 1 − y 2 = a, then what is the value of, , (a), , (b) xy, , (a) −, , x2, , (b), , , then, , x, y, , 26. If x = at 2 and y = 2at, then, , 17. If y = loga x + logx a + logx x + loga a , then, , (a), , 25. If ( x + y )m + n = x m y n , then what is the value of, , 1 − y2, 1 − x2, , (d) None of these, , 21. If x = log t and y = t 2 − 1, then what is the value of, d2y, (NDA 2009 II), at t = 1 ?, dx 2, (a) 2, (b) 3, (c) − 4, (d) 4, 22. If f ( x ) = tan x + e−2x − 7x3 , then what is the value of, (NDA 2009 I), f ′ ( 0)?, (a) − 2, (b) − 1, (c) 0, (d) 3, dy, 3, is, 23. If x = 3 cos θ − 2 cos θ , y = 3 sin θ − 2 sin3 θ , then, dx, (a) cotθ, (b) tanθ, (c) sec θ, (d) cosec θ, dy, 24. If y = cos x + cos x + cos x .... ∞ , then, is equal, dx, to, sin x, sin x, sin x, sin x, (d) −, (b), (c) −, (a), 2y + 1, 2y − 1, 2y − 1, 2y + 1, , (NDA 2007 II), , 1, (b), 2 f(x) − 1, 1, (d), 2 + f(x), , 30. If 2x + 2 y = 2x + y , then the value of, (b) −1, (d) 2, , (a) 0, (c) 1, , dy, at x = y = 1 is, dx, , 2, 2x , −1 1 − x , , is, , wrt, 31. The derivative of sin−1 , cos, , 2, 1 + x2 , 1 + x , (a) −1, (b) 1, (c) 2, (d) 4, , 1 − t2, , dy, 2t, is equal to, and y =, , then, dx, 1 + t2, 1 + t2, y, x, y, x, (c) −, (a) −, (b), (d), x, y, x, y, , 32. If x =, , 33., , d , −1 a − x , is equal to, tan , dx , 1 + ax , 1, 1, 1, (b), (a) −, −, 2, 2, 1+ x, 1+ a, 1 + x2, 1, −1, (d), (c), 2, 2, a−x, a−x, 1+ , , 1− , , 1 + ax , 1 + ax
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347, , Differentiation, 34. If x = k (θ + sin θ ) and y = k (1 + cos θ ), then what is, the derivative of y wrt x at θ = π / 2?, (NDA 2010 II), (a) − 1, (b) 0, (c) 1, (d) 2, 35. If f ( x ) = 2x , then what is the value of f ′ ′ ( x )?, (NDA 2011 I), x +1, , (c) 2, , 2, , x, , (d) 2 (log10 2), , (log 2), , 37. If u = sin−1 ( x − y ), x = 3 t , y = 4 t3 , then what is the, derivative of u wrt t?, −, , (b) 3 (1 − t 2 ), , 1, 2, , 39. If y = x + e , then what is the value of, (a) e, , (b) −, , (c) −, , e, , x, , (d) −, , (1 + e x ), , dy, ?, dx, , d 2x, dy 2, , ?, , (1 + e x )3, , π, 2, , 1 1, , ?, 4 4, 1, (a), 2, , (b) 1, , dy, ?, dx, , 1, 1, +, 1 − x 2 (b), cos x cos 1 − x 2, , dy, at, dx, , (d) 2, , 1, 1, 42. If x + y = t + , x 4 + y 4 = t 2 + 2 , then what is the, t, t, 3 dy, value of −x y ?, dx, 2, , 2, , (d), , 2xyz, cos( x 2 ), t, xyz t, cos ( x 2 ), , sec x + tan x, ?, sec x − tan x, , 48. What is the differential coefficient of f (log x ), where, f ( x ) = log x ?, x, (b) ( x log x )−1, (a), (log x ), (log x ), 1, (c), (d) f , x, x, d 2x, cos x, is, , then, 2, dy 2, 2 cos x, 4 cos x, (a) −, (b) −, ( 6 + sin x )2, ( 6 + sin x )2, 4 cos x, 4 sin x, (d) −, (c) −, ( 6 + sin x )3, ( 6 + sin x )3, , 49. If y = 3x −, , y = 1, what is the value of, , (c) –1, , (b), , (d) sec x ( sec x + tan x )2, , (1 + e x )2, , (d) 0, , 41. For the curve x +, , xyz, t, xyz cos ( x 2 ), (c), t, , (c) 2 sec x ( sec x + tan x )2, , x, , (NDA 2008 I), , (c), , dt, is equal, dx, , (a) 2 sec x ( sec x + tan x ), (b) 2 sec2x ( sec x + tan x )2, , 40. If y = sin−1 x + sin−1 1 − x 2 , what is the value of, (a) cos−1 x + cos−1, , xx − x− x , , then f ′ (1) is equal to, 45. If f ( x ) = cot−1 , 2, , , (a) −1, (b) 1, (c) log 2, (d) − log 2, , 47. What is the derivative of, , ex, e, , sin ( a + y ), sin a, , a cos x − b sin x , dy, is equal to, 44. If y = tan−1 , , then, dx, b cos x + a sin x , a, b, (a) 2, (b) − 1, (c), (d), b, a, , (d) − x / y, , x, , x, , (d), , (a), , (NDA 2012 I), , (c) − y / x, , sin2( a + y ), sin a, , to, , (d) 5 (1 − t 2 ), , (b) x/y, , dy, is, dx, (b) sin2 ( a + y ), , 2, , 38. If y = cos t and x = sin t, then what is the value of, (a) xy, , (d) 1, , 46. If y = sin ( x 2 ), z = e y , t = z , the value of, , 1, , (c) 5 (1 − t 2 ) 2, , 1, 3, , 43. If sin y = x sin ( a + y ), then, , (c), , , t , wrt, 36. What is the derivative of sin−1 , 1 + t2 , , , 1 , ?, cos−1 , 1 + t2 , , , (a) 1, (b) –1, (c) 2, (d) –2, , (a) 3 (1 − t 2 ), , (b), , (a) sin ( a + y ), , (b) x ( x − 1) 2x − 2, , (a) 2x (log 2)2, , 1, 4, 1, (c), 2, (a), , 50. If y = a cos(log x ) − b sin(log x ), then the value of, dy, d2y, x2, +x, + y is, dx, dx 2, (a) 4, (b) 0, (c) 2, (d) 3
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348, , NDA/NA Mathematics, , Level II, 1+ x − 1− x, ,, 1. The differential coefficient of tan−1 , 1+ x + 1− x, is, 1, (b), (a) 1 − x 2, 1 − x2, 1, (c), (d) x, 2 1 − x2, , x3, 9. f ( x ) = 1, , p, is, , (a) proportional to x 2, (c) proportional to x3, , x + 2 , − 1 ≤ x < 3, , 5. If f ( x ) = 5, ,, x = 3 ,then at x = 5, f ′ ( x ) is, 8 − x ,, x> 3, , equal to, (a) 1, (c) 0, , (b) −1, (d) Does not exist, , 6. If y = (1 + x1/ 4 ) (1 + x1/ 2 ) (1 − x1/ 4 ), then what is the, dy, (NDA 2011 II), ?, value of, dx, (a) 1, (b) − 1, (c) x, (d) x1/ 2, x, 7. What is the derivative of x a 2 − x 2 + a 2 sin−1 ?, a, (NDA 2010 II), , (a), , a −x, , (c), , x −a, , 2, , 2, , 2, , (b) 2 a − x, , 2, , (d) 2 x 2 − a 2, , 2, , , d 2, −1 1 − x , 8., sin cot , equals, dx , 1 + x , (a) −1, 1, (b), 2, 1, (c) −, 2, (d) 1, , 2, , (b) proportional to x, (d) a constant, , 10. What is the differentiation of logx x wrt log x ?, , d, 2. The value of, (|x − 1|+|x − 5|) at x = 3 is, dx, (a) − 2, (b) 0, (c) 2, (d) 4, 1, 3. Let 3 f ( x ) − 2 f = x, then f ′ ( 2) is equal to, x, 2, 1, (b), (a), 7, 2, 7, (c) 2, (d), 2, dy, is equal to, 4. If x = y 1 − y 2 , then, dx, (a) 0, (b) x, 1 − y2, 1 − y2, (d), (c), 1 − 2 y2, 1 + 2 y2, , x 2 3x 2, d3 f ( x ), −6 4 , here p is a constant, then, , dx3, p2 p3 , , (NDA 2010 I), , (a) 0, 1, (c), x, , (b) 1, (d) x, , 11. If e y + xy = e, then what is the value of, , d2y, dx 2, , at x = 0?, , (NDA 2009 II), , (a) e−1, (c) e, , (b) e−2, (d) 1, , , y − x2 , dy, , then, 12. If x = exp tan−1 , equals, 2, dx, , x , (a) 2x [ 1 + tan (log x )] + x sec2 (log x ), (b) x [1 + tan (log x )] + sec2 (log x ), (c) 2x [1 + tan (log x )] + x 2sec2 (log x ), (d) 2x [1 + tan (log x )] + sec2 (log x ), 13., , d, [sin−1( x 1 − x − x 1 − x 2 )] is equal to, dx, 1, 1, (a), −, 2 x (1 − x ), 1 − x2, 1, (b), 1 − { x 1 − x − x (1 − x 2 )} 2, 1, 1, (c), −, 2, 2 x (1 − x ), 1− x, 1, (d), x (1 − x )(1 − x )2, , 14. The derivative of y = a x log a sin x is equal to, (a) log sin x + x tan x, (b) log sin x + x cot x, (c) y log (sin x ex cot x ), (d) y log (sin x ex tan x ), 15. If y =, , x sin−1 x, 1 − x2, , to, (a) 0, 1, (c), 2, , d 2 y, is equal, + loge 1 − x 2 , then 2 , dx x = 0, (b) 1, (d) 2, , 16. What is the derivative of logx 5 wrt log5 x?, (NDA 2009 II), −2, , (a) − (log5 x ), (c) − (logx 5)2, , −2, , (b) (log5 x ), (d) (logx 5)−2
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349, , Differentiation, 17. If y = (1 + x1/ 4 ) (1 + x1/ 2 ) (1 − x1/ 4 ), then what is the, dy, (NDA 2009 II), ?, value of, dx, (a) 1, (b) − 1, (c) 0, (d) − 2x, , 26. Which one of the following functions is differentiable, for all real values of x?, (NDA 2012 I), x, 1, 1, (b) x|x|, (c), (d), (a), |x|, |x|, x, , 18. What is the derivative of|x − 1| +|x − 4| at x = 3?, (a) – 3, (b) 3, (c) 0, (d) 2, , 27. If f ( x ) = cos x , g( x ) = log x and y = ( gof ) ( x ), then what, dy, is the value of, at x = 0?, (NDA 2009 I), dx, (a) 0, (b) 1, (c) − 1, (d) 2, , 19. y = A sin (log x ) + B cos (log x )., d dy , What is the value of x, x ?, dx dx , (a) 0, (b) 1, (c) y, , 28. If f ( x ) = log| x|, x = 0, then what is the value of f ′ ( x )?, (NDA 2008 II), , (d) –y, , 20. Consider the following in respect of the function, f ( x ) =|x − 3|, I. f ( x ) is continuous at x = 3., II. f ( x ) is differentiable at x = 0., Which of the above statements is/are correct?, (NDA 2012 I), , (a) Only I, (c) Both I and II, , (b) Only II, (d) Neither I nor II, , 21. The second derivative of the function, f ( x ) = ( |x| − ex )2 − e2x + 2ex |x| − cos 2x + 2 cos2 x, is identically zero. What is the value of f( 2)?, (a) 0, (b) 1, (c) 2, (d) 3, , 1, (a), | x|, −1, (c), x, , 23. If f ( x ) = ex and g( x ) = log x, then what is the value of, (NDA 2009 II), ( gof )′ ( x )?, (a) 0, (b) 1, (c) e, (d) None of these, dy, 24. If 3x + 3 y = 3x + y , then what is the value of, ?, dx, (NDA 2009 I), , (a), (c), , 3x +, , y, , − 3x, , 3y, x, 3 + 3y, 3x − 3 y, , (b), , 1 − 3x, 3 + 3y, x, , (d), , 25. If x = f ( t ), y = g( t ) such that, following is correct?, dx d 2 y dy d 2x, (a), ⋅, =, ⋅, dt dt 2, dt dt 2, d 2x d 2 y, (c), +, =0, dt 2, dt 2, , 3x − y ( 3 y − 1), , 1 + 3x +, , d2y, dx 2, , (b), , y, , = 0. Which one of the, , d 2x, 2, , =, , d2y, , dt, dt 2, 2, dx d y dy d 2x, (d), ⋅, +, ⋅, =0, dt dt 2, dt dt 2, , (d) None of these, , 29. If y = sin ( m sin−1 x ), then what is the value of, x = 0?, (a) m, (c) m 2 + 2, , value of, (a) −, (c), , d 2x, dy 2, , d2y, dx 2, , at, , (NDA 2008 II), 2, , (b) m, (d) None of these, , 30. If y = f ( x ), p =, , 22. The derivative of the function, π, , f ( x ) = 2 sin + x cos x − sin x ( 3 cos x − sin x ), 6, , π, is identically zero. What is the value of f ?, 12, 5−1, 1, (a), (b) 0, (c) 1, (d), 2, 2, , 1, (b), x, , dy, d2y, and q =, , then what is the, dx, dx 2, , ?, (NDA 2008 I), , q, , (b) −, , p2, , 1, q, , (d), , q, p3, , q, p2, , Directions (Q. Nos. 31-33), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a),(b),(c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 31. Assertion (A) If x =, , 1 − t2, 1 + t2, , dy a ( t 2 − 1), =, dx, 2t, 1 + t2, Reason (R) If x = f ( t ) and y = g( t ), then, dy g ′ ( t ) dy dt, =, ⋅, =, dx f ′ ( t ) dt dx, and y =, , 2at, , , then
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350, , NDA/NA Mathematics, , 32. Assertion (A) If f ( x ) is an even function, then f ′ ( x ), is an odd function., Reason (R) Derivative of an even function is always, an odd function., 33. Assertion (A) If xy = c2, where c is a constant and if, du, du, u is any function of x, then x, +y, =0, dy, dx, Reason (R) If x = a cos3 t and y = a sin3 t, then, d2y, 4 2, 2, =, 3a, dx at t = π, 4, , 34. Consider the following statements, dy, cos x, I. If y = sin x + y , then, =, dx 2 y − 1, d n y ( −1)n −1( n − 1)!, II. If y = log x, then, =, dx n, xn, Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 35. Consider the following statements, I. If y = aex + be− x + c, where a,b and c are, parameters, then y ′′′ is equal to y ′., II. If f ( x + y ) = f ( x ) f ( y ) and f ( x ) = 1 + sin( 3x ) ⋅ g( x ) ,, where g( x ) is continuous, then f ′ ( x ) is 3 f ( x )g( 0)., Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 36. Match List I with List II and select the correct, answer using the codes given below the lists., List I, A., B., C., D., , List II, , d, {log (log x)}, dx, 2, d , 1 , x+, , dx , x, d, 2 x, ( x e sin x), dx, d, ( xx ), dx, , Codes, A B, (a) 2 4, (c) 1 3, , 1. 1 − 1, x2, 2. ( x log x) −1, , 39. If y =, , 1, dy, ?, , then what is the value of, log10 x, dx, (NDA 2008 I), , (a) x, (c) −, , (logx 10)2 (log10 e), x, , Consider x =, , Directions (Q. Nos. 40-42), y=, , D, 1, 2, , (b) e, , 1 + t3, , ,, , . On the basis of this parametric equation solve, 1 + t3, the following questions., 40. The value of, (a), (c), , dx, is, dt, , 5a, , (b), , 1+ t, 3a(1 − 2t3 ), 2, , (a), , dy, is, dt, , 5t, , (b), , 1+ t, 7at, (c), 1+ t, , 3, , 42. The value of, , (1 + t3 )2, , (d) None of these, , (1 + t3 )2, , 41. The value of, , 2a, , 3at( 2 − t3 ), (1 + t3 )2, , (d) None of these, 1, dy, at t = is, 2, dx, , 5, 4, 1, (c), 5, , (b), , (a), , 2, 3, , (d) None of these, , (NDA 2007 II), , 3. xx log ex, 4. xe x ( 2 sin x + x sin x + x cos x), , 1, (b), 2, , 1, (a) −, 4, 3, (c), 2, , (d) 1, , 44. If x 1 + y + y 1 + x = 0, what is the value of, C, 3, 4, , 3at, , 3at 2, , A, (b) 2, (d) 4, , B, 1, 2, , C, 4, 3, , D, 3, 1, , (c) 1, , (d) 0, , 38. If f ( x ) = esin (log cos x ) and g( x ) = log cos x , then what is, the derivative of f ( x ) wrt g( x )?, (NDA 2008 I), (a) f ( x ) cos [g( x )], (b) f ( x ) sin [g( x )], (c) g( x ) cos [ f ( x )], (d) g( x ) sin [ f ( x )], , dy, ?, dx, , (NDA 2007 II), , (a) −, , 1, 1+ x, , (b) −, , (NDA 2008 I), , (a) e, , (d) x log10 e, , x − x, at x = 1?, 43. What is the derivative of tan−1 , 3/ 2, 1 + x , , 37. If f ( x ) = loge [loge x ], then what is the value of f ′ ( e)?, −1, , (b) x loge 10, , 45. If sin x cos y =, π π, , ?, 4 4, (a) – 4, (c) – 6, , 1, (1 + x )2, , (c), , 1, (1 + x )2, , (d), , x, 1+ x, , d2y, 1, at, , then what is the value of, 2, dx 2, (NDA 2007 I), , (b) – 2, (d) 0
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Answers, Level I, 1., 11., 21., 31., 41., , (d), (a), (d), (b), (c), , 2., 12., 22., 32., 42., , (c), (c), (b), (c), (d), , 3., 13., 23., 33., 43., , (d), (c), (a), (a), (c), , 4., 14., 24., 34., 44., , (b), (b), (d), (a), (b), , 5., 15., 25., 35., 45., , (d), (c), (c), (a), (a), , 6., 16., 26., 36., 46., , (c), (c), (b), (a), (b), , 7., 17., 27., 37., 47., , (d), (d), (d), (b), (c), , 8., 18., 28., 38., 48., , (c), (c), (a), (d), (b), , 9., 19., 29., 39., 49., , (d), (d), (b), (b), (c), , 10., 20., 30., 40., 50., , (a), (d), (b), (d), (b), , 2., 12., 22., 32., 42., , (b), (a), (c), (a), (a), , 3., 13., 23., 33., 43., , (b), (c), (b), (b), (a), , 4., 14., 24., 34., 44., , (c), (c), (b), (c), (b), , 5., 15., 25., 35., 45., , (b), (b), (a), (c), (a), , 6., 16., 26., 36., , (b), (c), (b), (b), , 7., 17., 27., 37., , (b), (b), (a), (a), , 8., 18., 28., 38., , (b), (c), (b), (a), , 9., 19., 29., 39., , (d), (d), (d), (c), , 10., 20., 30., 40., , (a), (c), (b), (c), , Level II, 1., 11., 21., 31., 41., , (c), (b), (d), (a), (b), , Hints & Solutions, Level I, 1. We have, f (x) = sin (cos x), On differentiating wrt x, we get, d, f ′ (x) = cos (cos x), (cos x), dx, ∴, f ′ (x) = cos (cos x) ⋅ (− sin x), = − sin x ⋅ cos (cos x), , 2x , 2x , and tan −1 , , 2. We have, sin −1 , 2, 1 + x , 1 − x2, 2x , , Suppose,, p = sin −1 , 1 + x2 , 2x , , and, q = tan −1 , 1 − x2 , , It can be rewritten as, x−1, x−1, y = sin −1, + cos −1, x+1, x+ 1, π, π, , −1, −1, y=, ⇒, Q sin x + cos x = 2 , 2, , , dy, =0, ⇒, dx, 4. We have, y = sin x + sin x + sin x + K, …(i), …(ii), , Now, putting x = tan θ in Eqs. (i) and (ii), From Eq. (i),, 2 tan θ , , p = sin −1 , 1 + tan 2 θ , = sin −1 (sin 2θ ) = 2θ = 2 tan −1 x, p = 2 tan −1 x, dp, 2, =, dx 1 + x2, , ∴, ⇒, Similarly,, ⇒, ∴, , q = 2 tan −1 x, dq, 2, =, dx 1 + x2, dp dp dq, =, =1 ⇒, dq dx dx, , ⇒, y = sin x + y, On squaring both sides, we get, y2 = sin x + y, On differentiating the above equation wrt x, we get, dy, dy, 2y, = cos x +, dx, dx, dy, dy dy, (2 y − 1) = cos x, ⇒, 2y, −, = cos x ⇒, dx, dx dx, dy, cos x, ⇒, =, dx (2 y − 1), 5. Given,, ⇒, ⇒, , dp, =1, dq, , ⇒, , y = log tan x, 1, y = log tan x, 2, dy 1, 1, = ⋅, ⋅ sec2 x, dx 2 tan x, , dy, , dx, , π, , at x = , , 4, , 1, = ⋅, 2, , 3. We have,, −1, , y = cosec, , x+1, x−1, , + cos, , −1, , x−1, x+1, , =, , π, sec2 , 4, π, tan , 4, , 1 ( 2 )2 1 2, ⋅, = ⋅ =1, 2, 1, 2 1
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356, , NDA/NA Mathematics, Again, differentiating wrt y, we get, 2, 3d 2x cos x dx, sin x d 2x, +, 0=, ⋅ 2, +, 2, 2 dy, 2, dy, dy, 2, 1, sin x d x cos x, , = 3 +, +, ⋅, , 2, , 2 dy2, 2, sin x, , 3 +, , , 2 , , ⇒, , =−, , 4 cos x, (6 + sin x)3, , y = a cos (log x) − b sin (log x), dy a (− sin (log x)) b cos (log x), ∴, =, −, dx, x, x, (a sin (log x) + b cos (log x)), =−, x, dy, ⇒, x, = − (a sin (log x) + b cos (log x)), dx, d 2y dy, y, a cos (log x) b sin (log x), ⇒ x 2+, =−, −, =−, , , dx, x, x, x, dx, dy, d 2y, + y =0, ⇒ x2 2 + x, dx, dx, , 50. Q, , [from Eq. (i)], sin x d 2x, 2 cos x, , = 3 +, +, , , 2 dy2 (6 + sin x)2, sin x d 2x, 2 cos x, , =−, 3 +, , , 2 dy2, (6 + sin x)2, 2 cos x, 1, d 2x, =−, ⋅, sin x, (6 + sin x)2 , dy2, 3 +, , , 2 , , Level II, 1. Let, Put, ∴, , 1 + x − 1 − x, , y = tan −1 , 1 + x + 1 − x, 1, x = cos 2θ ⇒ θ = cos −1 x, 2, , +, cos, 1, 2θ − 1 − cos 2θ , , y = tan −1 , 1 + cos 2θ + 1 − cos 2θ , , ⇒, , On differentiating wrt x, we get, 1, 1, 2, 2 1, f ′ (x) = 3 − 2 ⇒ f ′ (2) = 3 − =, 5, 5, 4 2, x , , cos θ − sin θ , 1 − tan θ , y = tan −1 , = tan −1 , , cos θ + sin θ , 1 + tan θ , , π, , y = tan −1 tan − θ , 4, , , π, y= −θ, ⇒, 4, π 1, y = − cos −1 x, 4 2, On differentiating wrt x, we get, dy, 1 −1 , 1, =− , =, dx, 2 1 − x2 2 1 − x2, , 4. Given that, x = y 1 − y2, On differentiating wrt x, we get, dy, 1, dy, 1=, 1 − y2 + y ⋅, (− 2 y), 2, dx, dx, 2 1− y, ⇒, , 1=, , dy 1 − y2 − y2 , dy 1 − 2 y2 , , , ⇒1 =, , dx 1 − y2 , dx 1 − y2 , , , , , , 1 − y2, dy, =, dx 1 − 2 y2, x + 2 , − 1 ≤ x < 3, , 5. Given, f (x) = 5, ,, x=3, 8 − x ,, x>3, , , 2. Given that, f (x) = | x − 1| +|x − 5|, − (x − 1) − (x − 5), x < 1, , f (x) = (x − 1) − (x − 5), 1 ≤ x ≤ 5, x − 1 + x − 5, x > 5, , For x = 5, we take the function, f (x) = 8 − x, On differentiating wrt x, we get, ⇒, f ′ (x) = − 1, , [Q x = 3 ∈ (1, 5)], , 6. Given, y = (1 + x1/ 4 )(1 + x1/ 2)(1 − x1/ 4 ), ⇒, y = {(1 + x1/ 4 )(1 − x1/ 4 )} (1 + x1/ 2), ⇒, , ∴ For x = 3, f (x) = 4 ⇒ f ′ (x) = 0, 3. Given that, 3 f (x) − 2 f (1 /x) = x, 1, Let, = y, then 3 f (1 / y) − 2 f ( y) = 1 / y, x, ⇒, −2 f ( y) + 3 f (1 / y) = 1 / y, , …(ii), , Multiply Eq. (i) by 3 and Eq. (ii) by 2 and adding, equations, we get, 2, 1, 2, 5 f (x) = 3x +, ⇒ f (x) = 3x + , x, x, 5, , 2 cos 2 θ − 2 sin 2 θ , , y = tan −1 , 2 cos 2 θ + 2 sin 2 θ , , , , 6 − 2x, x < 1, , f (x) = 4, 1 ≤ x ≤ 5, 2x − 6, x > 5, , −2 f (x) + 3 f (1 /x) = 1 /x, , …(i), , y = (1 − x1/ 2) (1 + x1/ 2), , ⇒, y=1 − x, On differentiating, we get, dy, ⇒, = −1, dx
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358, , NDA/NA Mathematics, , 14. y = a x log a sin x, ⇒, , y=a, , ⇒, y = (sin x)x, Taking log on both sides, we get, log y = x log sin x, On differentiating wrt x, we get, 1 dy, x, ⋅, = log (sin x) +, (cos x), y dx, sin x, = log (sin x) + x cot x, dy, = y [log sin x + log ex cot x ], dx, dy, = y log (sin x ⋅ ex cot x ), dx, 15. y =, , x sin −1 x, 1 − x2, , log 5, =−, , log x, , 17. Given, y = (1 + x1/ 4 )(1 + x1/ 2)(1 − x1/ 4 ), = (1 + x1/ 2)[1 − (x1/ 4 )2], = (1 + x1/ 2)(1 − x1/ 2), ⇒, y=1 − x, On differentiating wrt x, we get, dy, = −1, dx, , + log e 1 − x2, , x , du, x sin −1 x 1, 1, 1, +, …(i), =, ⋅, +, 2 1/ 2, −1, dx (1 − x ) x sin x 1 − x2 1 − x2 , , , v = log 1 − x2, 1, log (1 − x2), 2, dv 1, 1, x, = ×, × (−2x) = −, 2, dx 2 1 − x, 1 − x2, v=, , Q, ⇒, ∴, , …(ii), , y=u+ v, dy du dv, =, +, dx dx dx, dy sin −1 x, x, x2 sin −1 x, x, =, +, +, −, 2, 2 3/ 2, 2, dx, (1 − x ) (1 − x ), 1 − x2, 1−x, [from Eqs. (i) and (ii)], sin −1 x, sin −1 x, [1 − x2 + x2] =, =, 2 3/ 2, (1 − x2)3/ 2, (1 − x ), (1 − x ) − sin, 2, , ∴, , d 2y, =, dx2, , −1, , 3, x (1 − x2 )1/ 2 × (−2x), 2, (1 − x2)3, , d 2y (1 − x2) + sin −1 x ⋅ 3 x(1 − x2)1/ 2, =, (1 − x2)3, dx2, ∴, , d 2y, 2, =1, dx x = 0, , 16. Let u = log x 5 and v = log5 x, du − log 5 1, Then,, =, ⋅, dx (log x)2 x, , 2, , = − (log x 5)2, , y=u+ v, x sin −1 x, and v = log e 1 − x2, where, u=, 1 − x2, 1, log u = log x + log sin −1 x − log (1 − x2), ∴, 2, 1 du 1, 1, 1, 1, = +, ⋅, −, (−2x), ⇒ ⋅, u dx x sin −1 x 1 − x2 2 (1 − x2), , Now,, , − log 5 1, ×, du du /dx (log x)2 x, =, =, dv dv/dx 1 1, , ×, log 5 x, , (Q a log a = x), , Let, , ∴, , dv, 1, =, dx x log 5, , and, , log a (sin x )x, , 18. Let f (x) =|x − 1| + |x − 4|, x<1, 1−x+4−x ,, , = x − 1 − (x − 4) , 1 ≤ x < 4, x − 1 + x −4 ,, x≥4, , x≥4, 2x − 5 ,, , , 1 ≤ x<4, = 3, 5 − 2 x ,, x<1, , For x = 3, we take the function, f (x) = 3, On differentiating wrt x, we get, f ′ (x) = 0, ⇒, f ′ (3) = 0, 19. y = A sin (log x) + B cos (log x), …(i), On differentiating Eq. (i), we get, dy, 1, 1, = A cos (log x) ⋅ − B sin (log x) ⋅, dx, x, x, dy, …(ii), x, = A cos (log x) − B sin (log x), dx, On differentiating Eq. (ii), we get, d dy, 1, 1, x = − A sin (log x) − B cos (log x), , , dx, dx, x, x, d, dx, d, ⇒ x, dx, ⇒ x, , dy, x = − ( A sin (log x) + B cos (log x)), dx, dy, x = − y, dx, , 20. Given function, f (x) =|x − 3|, I. LHL at x = 3, f (3 − 0) = lim f (3), x→3 −, = lim (3 − h ), h→ 0, , = lim|3 − h − 3|= lim|− h|= lim h = 0, h→ 0, , h→ 0, , h→ 0, , RHL a ⋅ x = 3, f (3 + 0) = lim f (3) = lim f (3 + h ), x→3 +, , = lim|3 + h − 3|= lim| h| = 0, h→ 3, , h→ 0, , Here, LHL = RHL = f (3), , h→ 0
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359, , Differentiation, So, f (x) is continuous at x = 3., f (0 − h ) − f (0), |− h − 3|−|− 3|, II. Lf ′ (0) = lim, = lim, x→0, h→ 0, −h, −h, h + 3 −3, h, = lim, = lim, = −1, h→ 0, h→ 0 − h, −h, f (0 + h ) − f (0), |h − 3|−|− 3|, Rf ′ (0) = lim, = lim, h→ 0, h→ 0, h, h, − h + 3 −3, − h, = lim, = lim, h→ 0, h→ 0 h, −h, , dy g′ (t ), =, dx f ′ (t ), , ⇒, , On differentiating Eq. (i) wrt t, we get, d 2x, d 2y, = f ′ ′ (t ), 2 = g′ ′ (t ), 2, dt, dt, d 2y f ′ (t ) g′ ′ (t ) − g′ (t ) f ′ ′ (t ), =0, =, ( f ′ (t ))2, dx2, , =1, ∴ Lf ′ (0) = Rf ′ (0), ⇒ f (x) is differentiable at x = 0., , ⇒, , f ′ (t ) g′ ′ (t ) − g′ (t ) f ′ ′ (t ) = 0, , ⇒, , dx d 2y dy d 2x, ⋅, −, ⋅, =0, dt dt 2 dt dt 2, , = ( |x|)2 + e2x − 2ex |x| − e2x, + 2ex |x| − 2 cos 2 x + 1 + 2 cos 2 x, f (x) = |x| + 1, f (2) = |2| + 1 = 2 + 1 = 3, π, , 22. f (x) = 2 sin + x cos x − sin x ( 3 cos x − sin x), 6, , π, π, , , ⇒ f (x) = 2 sin cos x + cos sin x cos x, , , 6, 6, , 26. Let us take the function f (x) = x|x|, Redefine this function,, x2,, if x ≥ 0, f (x) = , 2, − x , if x < 0, f (0 − h ) − f (0), Lf ′ (0) = lim, h→ 0, −h, − (− h )2 − 0, h→ 0, −h, , = lim, , − sin x ( 3 cos x − sin x), ⇒ f (x) = (cos x + 3 sin x) cos x − sin x ( 3 cos x − sin x), ⇒ f (x) = cos 2 x + 3 sin x cos x − 3 sin x cos x + sin 2 x, ⇒ f (x) = cos 2 x + sin 2 x = 1, ⇒ f (x) = 1, ⇒ f (x) is a constant function, π, ∴, f =1, 12, , y, , x', , 23. ( gof )x = g ( f (x)), = g (ex ) = log ex = x, On differentiating wrt x, we get, ( gof )′ (x) = 1, , O, , x, , f(x) = x |x|, , y', , 24. 3x + 3y = 3x + y, On differentiating wrt x, we get, dy, dy, , 3x log 3 + 3y log 3, = 3( x + y ) log 3 1 +, , , dx, dx, dy, dy, 3x + 3y, = 3x + y + 3( x + y ), ⇒, dx, dx, dy, ⇒, (−3x + y + 3y ) = 3x + y − 3x, dx, , − h2, = lim(+ h ) = 0, h→ 0 − h, h→ 0, , = lim, , f (0 + h ) − f (0), h, h2 − 0, = lim, h→ 0, h, lim h = 0, , Rf ′ (0) = lim, , h→ 0, , h→ 0, , dy 3x (3y − 1) 3x − y (3y − 1), =, =, dx 3y (1 − 3x ), (1 − 3x ), , ∴ L f ′ (0) = R f ′(0), ∴ f (x) is differentiable for all real values of x., f (x) = cos x, g (x) = log x, y = gof (x), = g{ f (x)} = g{cos x}, = log (cos x), dy, 1, ∴, =, (− sin x) = − tan x, dx cos x, dy, = − tan 0 = 0, ⇒, , dx at x = 0, , 27. Q, ∴, , 25. We have,, x = f (t ) and y = g (t ), On differentiating wrt t, we get, dx, dy, = f ′(t ),, = g′(t ), dt, dt, dy dy dt, ∴, =, ⋅, dx dt dx, , (given), , dx d 2y dy d 2x, ⋅, =, ⋅, dt dt 2 dt dt 2, , ⇒, , 21. The given function is, f (x) = ( |x| − ex )2 − e2x + 2ex |x| − cos 2x + 2 cos 2 x, , ⇒, , …(ii), , …(i)
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361, , Differentiation, , sin 3 h , , = 3 f (x) lim, lim g (h ), , h→ 0 3 h h→ 0, 36. (A), , (B), , (C), , (D), , = 3 f (x) g (0), 1 1, 1, d, ⋅ =, {log (log x)} =, log x x x log x, dx, d, ∴, {log (log x)} = (x log x)−1, dx, 2, d , 1, d , 1, , x + + 2, x+, =, , dx , dx , x, x, 1, =1 − 2, x, d 2x, (x e sin x), dx, = x2 [ex cos x + sin x ex ] + 2x ex sin x, = x ex [2 sin x + x cos x + x sin x], d x, (x ) = xx (1 + log x), dx, = xx (log e + log x), = xx log ex, , 37. Q f (x) = log e [log e x], On differentiating wrt x, we get, 1, 1, f ′ (x) =, ⋅, log e x x, 1, 1 1, f ′ (e) =, ⋅ = = e−1, ⇒, log e e e e, 38. Q f (x) = esin (log cos x ), ∴ f ′ (x) = esin (log cos x ) ⋅ cos (log cos x) ⋅, , 1, (− sin x), cos x, , = − esin (log cos x ) ⋅ cos (log cos x) ⋅ tan x, and g (x) = log cos x, 1, ∴ g′ (x) =, (− sin x) = − tan x, cos x, Hence,, f ′ (x) − esin (log cos x ) ⋅ cos (log cos x) ⋅ tan x, =, g′ (x), − tan x, = esin (log cos x ) ⋅ cos (log cos x), = f (x) ⋅ cos [ g (x)], 39. Q, ∴, , 1, y=, log10 x, dy, 1, 1, =−, ⋅ log10 e, dx, (log10 x)2 x, , , 1 , Q logb a =, , log a b, , , (log x 10)2 ⋅ log10 e, =−, x, , Solutions (Q. Nos. 40-42), Given, x =, , 3at, 1 + t3, , y=, , 3at 2, 1 + t3, , Taking log of x and y, we have, log x = log (3at ) − log (1 + t3 ), = log (3a ) + log t − log (1 + t3 ), , and, log y = log 3a + 2 log t − log (1 + t3 ), Now, differentiating them wrt t, we get, 1 dx, 1, 3t 2, ⋅, =0 + −, x dt, t 1 + t3, =, , (1 + t3 ) − 3t3, 1 − 2t3, =, t (1 + t3 ), t (1 + t3 ), , dx x(1 − 2t3 ), (1 − 2t3 ) 3a (1 − 2t3 ), 3at, =, =, =, ⋅, 3, 3, dt t (1 + t ) (1 + t ) t (1 + t3 ), (1 + t3 )2, and, ⇒, , 2 + 2t3 − 3t3, 2 − t3, 1 dy 2, 3t 2, =, =, ⋅, = −, 3, 3, y dt t 1 + t, t (1 + t ), t (1 + t3 ), 3at (2 − t3 ), dy 3at 2 2 − t3, , =, =, dt 1 + t3 t (1 + t3 ), (1 + t3 )2, dy dy, =, dx dt, , dx y(2 − t3 ), =, dt t (1 + t3 ), , x(1 − 2t3 ), t (1 + t3 ), , =, , y(2 − t3 ) 3at 2(2 − t3 ), 1 + t3, ×, =, 3, 3, x(1 − 2t ), (1 + t ), 3at (1 − 2t3 ), , =, , t (2 − t3 ), 1 − 2t3, , 1, 1, 2 − , 1 dy 2 , 8 15 4 5, ∴ At t = ,, =, =, × =, 2, 16 3 4, 2 dx, 1−, 8, , , x−x , −, x, x, = tan −1 , 43. Let y = tan −1 , , 3/ 2, 1 + x , 1 + x ⋅ x , = tan −1 x − tan −1 x, On differentiating wrt x, we get, dy, 1, 1, 1, =, ⋅, −, dx 1 + x 2 x 1 + x2, 1, 1, 1, 1 1, 1, dy, Now, , =, ⋅ −, = − =−, dx x = 1 1 + 1 2 1 + 1 4 2, 4, 44. Given that, x 1 + y + y 1 + x = 0, x 1+ y=− y 1+ x, ⇒, On squaring both sides, we get, x2(1 + y) = y2(1 + x), ⇒, ⇒, , x2 + x2y = y2 + y2x, x2 − y2 = y2x − x2y, , ⇒ (x − y)(x + y) = − xy(x − y), ⇒, ⇒, , x + y = − xy, y(1 + x) = − x, −x, y=, 1+ x, , On differentiating wrt x, we get, −1, dy (1 + x)(−1) + x(1), =, =, dx, (1 + x)2, (1 + x)2
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19, , Application of, Derivative, Velocity =, , Motion in a Straight Line, If x and v denotes the displacement and velocity of a, particle at any instant t, then velocity and acceleration is, given by, dx, v=, dt, dv, dv d 2x, and, a=, =v, =, dt, dx dt 2, Where, a is acceleration of particle. If the sign of, acceleration is opposite to that of velocity, then the, acceleration is called retardation which means decrease in, magnitude of the velocity., , Example 1. A particle moves in a straight line in such a way, , that its velocity at any point is given by v 2 = 2 − 3x, where x, is measured from a fixed point. The acceleration is, 3, (a) − 4, (b) −, 2, (c) 3, (d) None of these, , Solution (b) We have, v 2 = 2 − 3x, On differentiating wrt t, we get, dv, dx, 2v, =0 −3, dt, dt, dv, 3, dv, =−, 2v, = − 3v ⇒, ⇒, dt, 2, dt, 3, ∴Required acceleration = −, 2, , ⇒, , ds, = 15, , dt ( t = 0), , and, , ds, = 15 − 12 = 3, , dt ( t = 3), , ∴ Average velocity =, , s = 15 t − 2 t 2, , On differentiating wrt t, we get, , 15 + 3, =9, 2, , Increasing Functions, A function y = f ( x ) is a increasing function, if f ( x ), increases as x increases i.e., x1 > x2, dy, >0, ⇒ f ( x1 ) > f ( x2 ), ( x1 , x2 ) ∈ Domain of f ( x ). Here,, dx, y, , y = loga x, (a > 1), x, , O, , Decreasing Functions, A function y = f ( x ) is a decreasing function, if f ( x ), decreases as x increases i.e., x1 > x2, dy, ⇒, f ( x1 ) < f ( x2 ). Here,, <0, dx, y, , Example 2. A point moves in a straight line during the time, t = 0 to t = 3 according to the laws s = 15 t − 2 t 2. The average, velocity of the point is, (a) 4, (b) 9, (c) 3, (d) 2, Solution (b) We have, , ds, = 15 − 4 t, dt, , y = loga x, (0 < a < 1), O, , x
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364, , NDA/NA Mathematics, , Non-decreasing Functions, If x1 > x2 ⇒ f ( x1 ) ≥ f ( x2 ), then f ( x ) is non-decreasing, dy, dy, function. Here,, ≥ 0 (but, = 0 for atleast one set of, dx, dx, interval of values of x e.g.,, ( x − 1 )2 ,, x≥1, , f(x) = 0 ,, 0≤ x < 1, − x2 ,, x< 0, , y, , f ( x ) = − 2x + 3 are monotonic functions. f ( x ) = x 2 is, monotonic in ( −∞ , 0) or ( 0, ∞ ) but is not monotonic in R., , Conditions for Monotonicity of, Functions, Let f be a differentiable real function defined on an, open interval ( a , b)., (a) If f ′ ( x ) > 0 for all x ∈( a , b), then f ( x ) is increasing, on ( a , b)., (b) If f ′ ( x ) < 0 for all x ∈( a , b), then f ( x ) is, decreasing on ( a , b)., , Properties of Monotonic Functions, , x, , O, , Many-one function, , x≥1, 2( x − 1),, , 0,, 0≤ x < 1, ⇒, f ′ (x) = , − 2x ,, x, <0, , dy, dy, Here,, = 0for x ∈ [0, 1] and, > 0for rest values of x., dx, dx, So, f ( x ) is non-decreasing., and when, f ( x ) = x3, ⇒, , f ′( x ) = 3x 2, f ′ ( x ) = 0 for x = 0, y, , 1. If f ( x ) is continuous on [a , b] such that, f ′ ( c) ≤ 0 [ f ′ ( c) < 0] for each c ∈( a , b), then f ( x ) is, monotonically (or strictly) decreasing function on, [a , b]., 2. If f ( x ) and g( x ) are monotonically (or strictly), increasing (or decreasing) functions on [a , b],, then gof(x) is a monotonically (or strictly), increasing function on [a , b]., 3. If f ( x ) is strictly (or monotonically) increasing, and g( x ) is strictly (or monotonically) decreasing,, then gof (x) is strictly (or monotonically), decreasing on [a , b]., 4. If f ( x ) is strictly increasing function on an, interval [a , b], then f −1 exists and it is also strictly, increasing function., 5. If f ( x ) is strictly increasing function on an, interval [a , b] such that it is continuous,then f −1, is continuous on [ f ( a ), f ( b)]., , Example, x, O, One-one function, , and f ′ ( x ) > 0 for all x except 0. Here, f ( x ) is increasing, dy, function as, = 0 only at a point not in any interval., dx, , Non-increasing Function, If x1 > x2 ⇒ f ( x1 ) ≤ f ( x2 ), then f ( x ) is non-increasing, function., Y, , O, , X, , Monotonic Function, or, , Monotonic function means either increasing function, decreasing function. i.e., f ( x ) = log x , f ( x ) = 2x ,, , 3. The intervals in which, f ( x) = 2 x 2 − log| x |, x ≠ 0 is increasing in, (a) ( − ∞, + ∞), 1 1 , (b) − , 0 ∪ , ∞, 2 2 , 1 , (c) − , 1, 2 , (d) None of the above, , the, , Solution (b) We have,, 2x2 − log x, x > 0, f ( x) = 2x2 − log| x| = 2, 2x − log ( − x ), x < 0, 1, for all x ≠ 0, f ′ ( x) = 4x − ,, ⇒, x, For f ( x) to be increasing,, 1, f ′ ( x) > 0 ⇒ 4x − > 0, x, 4x2 − 1, >0, ⇒, x, (2x − 1) (2x + 1), ⇒, >0, x, ⇒, x (2x − 1) (2x + 1) > 0, , function
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365, , Application of Derivative, −, , +, , −, , +, , ←•→, , −∞, , −1/2, , 0, , Example 5. The equation of the tangent to the curve, y = (2 x − 1) e 2 ( 1 − x ) at the point of its maximum is, (a) y = 2 x + 3, (b) y =1, (c) y = 3x, (d) None of these, , ∞, , 1/2, , From figure,, x ∈ ( −1/2 , 0) ∪ (1/2 , ∞), , Example, , 4. The interval in which, f ( x) = ( x + 2) e − x is decreasing in, (a) (− 1, ∞), (b) (−11, , ), (c) ( −1, 2), (d) (1, 2), , Solution (a) We have,, , the, , function, , ⇒, , f ( x) = ( x + 2) e− x, , On differentiating wrt x, we get, f ′ ( x) = e− x − ( x + 2) e− x, = − e− x ( x + 1), For decreasing function, f ′ ( x) < 0, ⇒, − e− x ( x + 1) < 0 ⇒ e− x ( x + 1) > 0 ⇒ x > −1, Hence, f ( x) is decreasing in ( −1, ∞)., , Tangents and Normals, dy , Slope of tangent at P = , dx ( x, , Solution (b) We have,, , = tan ψ = m, , 1 , y1 ), , y, , y = (2x − 1) e2 (1 − x), dy, = 2e2 (1 − x) − 2 (2x − 1) e2 (1 − x), dx, = 2 e2 (1 − x) (2 − 2x ), , = 4 e2 (1 − x) (1 − x ), dy, For maximum, put, =0, dx, ⇒, x =1, d 2y, = − 4 e2 (1 − x) − 8 (1 − x ) e2 (1 − x), dx2, d 2y , = − 4 <0, ⇒, 2, dx x = 1, So, y is maximum at x = 1, when x = 1, y = 1 the point of, maximum is (1, 1)., ∴ The equation of the tangent at (1, 1) is, y − 1 = 0 ( x − 1), ⇒, y =1, , Example 6. The point on the curve 3y = 6 x − 5 x3 in which, y = f (x), P, , (x1, y1), , ψ, , x, , O, , dy , ( y − y1 ) = , dx ( x, , ⇒, , ( x − x1 ), , 1 , y1 ), , and equation of normal at ( x1 , y1 ) is, dx , ( y − y1 ) = − , ( x − x1 ), dy ( x , y, 1, , 1), , 1. If tangent is parallel to x-axis, then, dy , =0, , dx ( x , y ), 1, , 2. If tangent is parallel to y-axis or perpendicular to, x-axis, then, dx , , dy ( x, , Solution (a) Let the required point be ( x1, y1)., We have,, , Equation of tangent at ( x1 , y1 ) is, , 1, , the normal is passing through the origin is, 1, (b) (2, 3), (a) 1, , 3, (c) (1, 2), (d) ( − 3, 3), , =0, , 1 , y1 ), , 3. If the tangent is equally inclined to the axes, then, dy, = tan 45° or tan 135°, dx, = ±1, , ⇒, ⇒, , 3y = 6x − 5x3, dy, 3, = 6 − 15x2, dx, dy, = 2 − 5x2, dx, dy , , dx ( x, , = 2 − 5x12, , 1 , y1 ), , The equation of the normal at ( x1, y1) is, 1, y − y1 = −, ( x − x1), 2 − 5x12, If it passes through the origin, then, 1, (0 − x1), 0 − y1 = −, 2 − 5x12, x1, ⇒, y1 = −, 2 − 5x12, Since, ( x1, y1) lies on the given curve,, ∴, 3y1 = 6x1 − 5x13, , …(ii), , On solving Eqs. (i) and (ii), we get, x1 = 1 and y1 =, 1, ∴ Required point is 1, ., 3, , …(i), , 1, 3
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366, , NDA/NA Mathematics, , Common Tangents, Slope of tangents should be equal at point of tangency., , Angle of Intersection, Angle of intersection is angle between tangents at, intersection points of two curves. Let they intersect at, P ( x1 , y1 ),, then, df ( x ), m1 = , dx ( x1 , y1 ), y, y = f(x), , θ, , P (x1, y1), , y = g(x), , x, , O, , and, ∴, , dg( x ), m2 = , dx ( x1 , y1 ), , Solution (b) Given curves are, y = x2 and 6y = 7 − x3, , Example 8. If at any point on a curve the subtangent and, subnormal are equal, then the length of the tangent is, (a) 2 x, (b) 2y, (c) x, (d) None of these, subtangent = subnormal, y, dy, dy, = ±1, =y, ⇒, dy, dx, dx, dx, dy , y 1+ , dx , ∴ Length of the tangent =, dy, dx, , 2, , = 2y, , Maxima and Minima, , These two curves intersect at a point (1, 1., ), Now,, y = x2, dy, dy , = 2x ⇒ m1 = , =2, dx (1, 1), dx, x2, 1, dy , and 6y = 7 − x3 ⇒ m2 = , =−, =−, dx (1, 1), 2, 2, 1, m1m2 = 2 − = − 1, 2, , ∴ The angle of intersection of two curves is, , 2, , dy , y 1+ , dx , , 1. Length of tangent =, dy, , , , , dx, , , 2, , dy , 2. Length of normal = y 1 + , dx , , , , , y , , 3. Length of subtangent =, dy , dx , dy , , 4. Length of subnormal = y , dx , , Solution (b) We have,, , m − m2 , , tan θ = 1, 1 + m1m2 , , Example 7. The angle of intersection of the curves, y = x 2, 6y = 7 − x3 at point (1, 1) is, π, (a) π, (b), (c) 2 π, (d) 4 π, 2, , ∴, , Lengths of Tangent, Normal,, Subtangent and Subnormal, , π, ., 2, , Let f be a function defined on an interval I and let c be, an interior point of I (i.e., not an end point of I)., (a) f is said to have a local (or relative) maximum value, (or simply a local maximum) at x = c, if the ordinate at x = c, is the highest as compared to the surrounding ordinates., i.e., f ( c) > f ( x ) for a small deleted neighbourhood of, f ( x ) < f ( c) for c − δ < x < c + δ , x ≠ c, δ > 0, f ( x ) < f ( c) for 0<|x − c|< δ, y, , Orthogonal and Touching Curves, Two curves are said to be orthogonal curves, if the, angle of intersection of two curves is right angle i.e., if, m1m2 = − 1., Two curves touch each other, if m1 = m2., , c–d, O, , c+d, , x
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367, , Application of Derivative, (b) f is said to have a local (or relative) minimum value, (or simply a local minimum) at x = c, if the ordinate at x = c, is the lowest as compared to the surrounding ordinates., i.e.,, f ( c) < f ( x ) for small deleted neighbourhood of c, f ( x ) > f ( c) for c − δ < x < c + δ , x ≠ c, δ > 0, f ( x ) > f ( c) for 0<|x − c|< δ, , i.e., for x slightly < c and a decreasing function in the right, neighbourhood of x = c i.e., for x slightly > c. Also, for x = c,, the graph has a horizontal tangent., Thus, f ′ ( x ) changes continuously from positive to, negative as x increases through c., y, , L, , c)<, f '(, , 0, , f' (c) = 0, , Rf, , ' (c, , )>0, , x, , O, , O, , c–d, , c+d, , f is said to have a local extreme value (or simply a local, extremum) at x = c, if it has either a local maximum or a, local minimum at x = c., %, %, %, , The pluerals of maximum and minimum are maxima and, minima, respectively., The end points of the interval cannot be the points of local, extremum., A function may have more than one local maximum and/or more, than one local minimum., , Case II Let f have a local minimum value at x = c, then, f is a decreasing function in the left neighbourhood of x = c, i.e., for x slightly < c and an increasing function in the right, neighbourhood of x = c i.e., for x slightly > c. Also, for x = c,, the graph has a horizontal tangent., Thus, f ′ ( x ) changes continuously from negative to, positive as x increases through c., y, , Lf, , ' (c, )>, , '(c, Rf, , 0, , )>, , 0, , f' (c) = 0, , C, A, B, , D, O, , C, , x, , Second Derivative Test for Finding, Local Extremum, %, , %, , %, , Also, a local minimum may be greater than a local maximum. In, the above figure, the points A and C correspond to local, maximum while the points B and D correspond to local, maximum. Moreover, the local minimum at D is greater than the, local maximum at A., Local maxima and local minima occur alternately. Between two, local maximum values, there is a local minimum value and, between two local minimum values, there is a local maximum, value., A local maximum (minimum) value may not be the absolute, maximum (minimum) or the greatest (smallest) value., , A necessary condition for the existence of extreme, values at c is that f ' ( c) = 0., , First Derivative Test for Finding, Local Extremum, Let y = f ( x ) be a function defined on an interval I and, let f be derivable at an interior point c of I. Let f have an, extreme value at x = c, then f ′ ( c) = 0., Case I Let f have a local maximum value at x = c, then, f is an increasing function in the left neighbourhood of x = c, , Case I Let f have a local maximum value at x = c, then, f ′ ( c) = 0. Also, f ′ ( x ) > 0 for x slightly < c and f ′ ( x ) < 0 for x, slightly > c., Thus, f ′ ( x ) changes continuously from positive to, negative as x increases through c., ⇒ f ′ ( x ) is a decreasing functions at x = c., d, ⇒, [ f ′ ( x )] < 0 at x = c, dx, ⇒, f ′′ ( x ) < 0 at x = c ⇒ f ′′ ( c) < 0, Case II Let f have a local minimum value at x = c, then, f ′ ( c) = 0. Also, f ′ ( x ) < 0 for x slightly < c and f ′ ( x ) > 0 for x, slightly > c., Thus, f ′ ( x ) changes continuously from negative to, positive as x increases through c., ⇒ f ′ ( x ) is an increasing function at x = c, d, ⇒, [ f ′ ( x )] > 0 at x = c, dx, ⇒, f ′′ ( x ) > 0 at x = c, ⇒, f ′′ ( c) > 0
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368, , NDA/NA Mathematics, , Working Rules, To find a local maximum or a local minimum value of, f ( x ) defined on an interval I (open or closed)., (a) Find f ′ ( x )., (b) Solve f ′ ( x ) = 0 and find the solutions which are, interior points of I., These are the critical points or turning points., Let c be one of the critical points, Working Rule (i) (Second derivative test) Find f ′′( x ), and hence f ′′( c)., If f ′′( c) < 0, then f has a local maximum value at x = c., If f ′′( c) > 0, then f has a local minimum value at x = c., %, , This test should be preferred, if it is not tedius to find f ′′(x ) or, f ′′(c ) ≠ 0. But, if it is failed to find f ′′(x ) is harder [or even, if you, don't feel like finding f ′′(x ) or f ′′(c ) = 0], then we should apply, Working Rule (ii) (First derivative test)., , Working Rule (ii) Find the signs of f ′( x ), when x is, slightly <c and when x is slightly >c., If f ′( x ) changes sign from positive to negative as x, increases through c, then f has a local maximum value at, x = c., If f ′( x ) changes sign from negative to positive as x, increases through c, then f has a local minimum value at, x = c., If f ′( x ) does not change sign as x increases through c,, then f has a point of inflection at x = c., , Critical Points of a Function, Points, where a function f ( x ) is not differentiable and, points, where is derivative (differentiable coefficient) is, zero are called the critical points of the function f ( x )., Maximum and minimum value of a function f ( x ) can, occur only at critical points. However, this does not mean, that the function will have maximum or minimum values, at all critical points. Thus, the points, where maximum or, minimum value occurs are necessarily critical points but a, function may or may not have maximum or minimum, value a critical point., , Point of Inflection, Consider a function f ( x ) = x3 . At x = 0, f ′( x ) = 0. Also,, f ′′ ( x ) = 0 at x = 0 such point is called point of inflection,, where second derivative is zero. Consider another function, f ( x ) = sin x, f ′′ ( x ) = − sin x. Now, f ′′ ( x ) = 0 when x = n π,, then this points are called points of inflection., , For maximum or minimum, put f ′ ( x) = 0, ⇒, 6x2 − 24 = 0 ⇒ x2 = 4 ⇒ x = ± 2, But, x = − 2 ∉ [1, 3], ∴ x = 2 is the stationary point., Now,, f(1) = 2 − 24 + 107 = 85, f(2) = 2 (2)3 − 24 (2) + 107 = 75, f(3) = 2 (3)3 − 24 × 3 + 107 = 89, Hence, the maximum value of f ( x ) is 89 at x = 3., , Example 10. The minimum value of px + q y, when xy = r 2 is, (b) pq, (d) 2p, , (a) 2r pq, (c) 4p, , Solution (a) Let z = px + q y, , r 2, z = px + q , x, , (Q xy = r 2, given), , dz, qr 2, =p− 2, dx, x, For maximum or minimum, put, ⇒, At, , x=±, , qr 2, p, d 2z 2qr 2, = 3 >0, dx2, x, , qr 2, ,, p, , x=, , dz, =0, dx, , qr 2, p, , ∴ z is minimum at x =, The minimum value is, z=p, , qr 2, qr 2, +, p, qr 2, p, , = pqr 2 +, = 2r pq, , pqr 2, , Example 11. An isosceles triangle of vertical angle 2θ is, inscribed in a circle of radius a. The angle of triangle in which, area is maximum is, π, π, (b), (a), 2, 6, (c) 2π, (d) 3π, , Solution (b) We have,, AD = a + a cos 2θ and BC = 2BD = 2a sin 2θ, A, , 2θ, , Example 9. The maximum value of f ( x) = 2 x3 − 24x + 107 in, , a, , the interval [1, 3] is, (a) 89, (c) 75, , (b) 85, (d) 40, , Solution (a) f ( x) = 2x3 − 24x + 107, f ′ ( x) = 6x2 − 24, , O, a, B, , 2θ 2θ, D, , a, C
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369, , Application of Derivative, Area of ∆ ABC,, 1, ∆ = BC ⋅ AD, 2, 2, = a2 (sin 2θ + sin 2θ cos 2θ), 2, 1, = a2 sin 2θ + a2 sin 4θ, 2, d∆, ∴, = 2a2 cos 2θ + 2a2 cos 4θ, dθ, d∆, For maximum or minimaum,, =0, dθ, , π, ⇒, cos 2θ = − cos 4θ ⇒ 2θ = π − 4θ ⇒ θ =, 6, π, d 2∆, 2, 2, = − 4a sin 2θ − 8a sin 4θ < 0 at θ = ., 6, dθ 2, π, ∴ Triangle is maximum for θ = ., 6, , Example 12. If sum of two number is 3. The maximum, value of the product of first and the square of second is, (a) 4, (b) 5, (c) 3, (d) 2, Solution (a) Let the first number be 3 − x, then the second, number will be x., According to given condition,, let, f ( x) = (3 − x) x2, = 3x2 − x3 ⇒ f ′( x) = 6x − 3x2, For maximum or minimum, put f ′ ( x) = 0, ⇒, x = 0,2, Also,, f ′ ′ ( x) = 6 − 6x, At, x = 2,, f ′ ′ (2) = − 6 < 0, ∴, f(2) = (3 − 2) (2) 2 = 4, , Example, , 13. It is given that for the function, f ( x) = x3 − 6 x 2 + ax + b on [1, 3] , Rolle’s theorem holds with, 1, . The values of a and b, if f (1) = f (3) = 0 are, c =2 +, 3, respectively, is, (a) 11, − 6, (b) 2, 4, (c) 3, 2, (d) None of these, Solution (a) Given that, f (1) = f (3) = 0, ⇒, (1)3 − 6 (1) 2 + a + b = 33 − 6 (3) 2 + 3a + b = 0, ⇒, , a + b = 5 and 3a + b = 27, a = 11 and b = − 6, We have,, f ( x) = x3 − 6x2 + ax + b, , On differentiating wrt x, we get, f ′ ( x) = 3x2 − 12x + a = 3x2 − 12x + 11, At x = c,, f ′ ( c) = 3c2 − 12c + 11, , (Q a = 11), , 2, , ∴, , 1, 1, , , = 3 2 +, − 12 2 +, + 11, , , 3, 3, 12, 12, = 12 +, + 1 − 24 −, + 11 = 0, 3, 3, a = 11 and b = − 6, , Lagrange’s Mean Value, Theorem, Statement If a function f ( x ) is, 1. continuous in the closed interval [a , b]., y, f (c), , Rolle’s Theorem, , O, , Statement If a function f ( x ) is, 1. continuous in the closed interval [a , b]., 2. differentiable in an open interval ( a , b) i.e.,, differentiable at each point in the open interval, ( a , b)., 3. f ( a ) = f ( b), y, , a, , b, , c, , x, , 2. differentiable in an open interval ( a , b). Then,, there will be atleast one point c, where a < c < b, such that, f ( b) − f ( a ), f ′ ( c) =, b− a, , Example 14. Let f be differentiable for all x. If f (1) = − 2 and, , f ' (c) = 0, , f ′ ( x) ≥ 2 for all x ∈[1, 6] , then the minimum value of f (6) is, (a) 4, (b) 2, (c) 8, (d) None of these, , O, , a, , b, , c, , x, , Then, there will be atleast one point c in the interval, ( a , b) such that f ′ ( c) = 0., , Solution (c) By Lagrange’s mean value theorem there exists, c ∈(1, 6), such that, f (6) − f (1), f (6) + 2, f ′ ( c) =, ⇒, = f ′ ( c), 6 −1, 5, (Q f ′ ( x) ≥ 2 for all x ∈ [1, 6]), ⇒, ⇒, , f(6) + 2 ≥ 10, f(6) ≥ 8
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370, , NDA/NA Mathematics, , Comprehensive Approach, n, , n, , n, , n, , n, , n, , n, , If f ( x) is increasing, then f −1( x) is also increasing., If f ( x) is decreasing, then f −1( x) is also decreasing., , n, , If f ( x) and g( x) are monotonic on [ a, b ], then g( f ( x)) is also, monotonic of same nature., , n, , If y = f ( x) is continuous in [ a, b ] and f ( a) f ( b) < 0, then y = f ( x), intersect x-axis atleast once., , n, , If y = f ( x) is continuous function and it’s the least value is m and, the greatest value is M, then m ≤ f ( x) ≤ M (Range of y = f ( x))., , n, , Rectangle of the largest area inscribed in a given circle is a square, whose length of diagonal is diameter of circle., c, If f ( x) =, (where, c is positive and independent of x), now to, g( x), find points of extreme for f ( x) first find points of extreme for g( x), and points of maxima for g(x) are points of minima for f (x) and, similarly points of minima for g( x) are points of maxima for f ( x)., But g( a) should not be zero such point is not in the domain of f ( x)., If f ( x) = g( x) ( g( x) ≥ 0) here the points where g( x) is maximum f ( x), is also maximum and where g( x) is minimum, f ( x) is also, minimum., , n, n, , n, , n, , n, , If y = f ( x) is an increasing and continuous function in [ a, b ], (Domain ∈[ a, b ]), then Range ∈[ f ( a), f ( b)]., Similarly, if y = f ( x) is a decreasing and continuous function in, [ a, b ], then Range ∈[ f ( b), f ( a)]., If y = f ( x) is an increasing and continuous function in ( a, b), (Domain ∈( a, b)), then Range ∈( f ( a), f ( b))., Geometrically, the tangent to the curve y = f ( x) at a point, where, the ordinate is maximum or minimum is parallel to the x-axis., Maxima and minima occur alternately., If f ( x) → ∞ as x → a or b and f ′( x) = 0 only for one value of x, between a and b, then f (x) is necessarily the minimum and the, least value., If y is maximum or minimum, then log y = z is also maxima or, minimum provided y > 0., 1, is, y = f ( x) is maximum or minimum according as z =, f ( x), minimum or maximum., If the sum of two positive numbers is constant, then their product, is the greatest when the numbers are equal.
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Exercise, Level I, 1. Maximum slope of the curve y = − x3 + 3x 2 + 9x − 27, is, (a) 0, (b) 12, (c) 16, (d) 32, 2. If the function f ( x ) = x3 − 6x 2 + ax + b satisfies, Rolle’s theorem in the interval [1, 3] and, 2 3 + 1, f′, = 0, then, 3 , , (a) a = − 11, (c) a = 6, , (b) a = − 6, (d) a = 11, , x, , 4. The function x is increasing, when, 1, 1, (b) x <, (a) x >, e, e, (c) x < 0, (d) for all real x, 5. What is the value of p for which the function, sin 3x, π, has an extremum at x = ?, f ( x ) = p sin x +, 3, 3, (NDA 2010 II), , (b) 1, , (c) −1, , (d) 2, , 6. What is the minimum value of cos θ + cos 2θ ?, (NDA 2007 I), , (a) –2, (c) 0, , 9, (b) −, 8, 9, (d) −, 16, , 11. The angle of intersection of the curves y = x 2 and, x = y 2 at (1, 1) is, 4, (b) tan−1 (1), (a) tan−1 , 3, 3, (c) 90°, (d) tan−1 , 4, , (NDA 2010 I), , (a) (1, e), (c) ( e, 1), , (b) (1, e−1 ), (d) ( e−1 , 1), , 13. The function y = tan−1 x − x, (a), (b), (c), (d), , (NDA 2009 II), , is always decreasing, is always increasing, first increases and then decreases, first decreases and then increases, , 14. The function f ( x ) = 2x3 − 15x 2 + 36x + 4 is maximum, at, (a) x = 2, (b) x = 4, (c) x = 0, (d) x = 3, 15. The length of subtangent to the curve x 2 y 2 = a 4 at the, point ( − a , a ) is, (a) 3a, (b) 2a, (c) a, (d) 4a, 16. The abscissa of the points, where the tangent to the, curve y = x3 − 3x 2 − 9x + 5 is parallel to x-axis are, (a) x = 0 and 0, (b) x = 1 and − 1, (c) x = − 1 and 3, (d) None of these, , 7. If x + y = 12, what is the maximum value of xy?, (NDA 2007 I), , (a) 25, (c) 49, , (b) (1, e), (d) None of these, , 12. What is the maximum point on the curve x = ex y?, , 3. The speed v of a particle moving along a straight line, is given by a + bv 2 = x 2 (where x is its distance from, the origin). The acceleration of the particle is, (a) bx, (b) x/ a, (c) x/ b, (d) x/ ab, , (a) 0, , (a) (− ∞, e), (c) ( 2, 3), , (b) 36, (d) 64, , 8. How many tangents are parallel to x-axis for the, curve? y = x 2 − 4x + 3, (NDA 2012 I), (a) 1, (b) 2, (c) 3, (d) No tangent is parallel to x-axis, 9. At which point the tangent to the curve x 2 + y 2 = 25 is, parallel to the line 3x − 4 y = 7 is, (a) (3, − 4), (b) (1, 2), (d) (1, 3), (d) None of these, 10. The interval in which the function f ( x ) = x1/ x is, increasing is, , 17. The equation of tangent at ( − 4, − 4) on the curve, x 2 = − 4 y is, (a) 2x + y + 4 = 0, (b) 2x − y − 12 = 0, (c) 2x + y − 4 = 0, (d) 2x − y + 4 = 0, 18. The point at which the tangent to the curve, y = 2x 2 − x + 1 is parallel to y = 3x + 9 will be, (a) (2, 1), (b) (1, 2), (c) (3, 9), (d) ( − 2 , 1), 19. What is the least value of f ( x ) = 2x3 − 3x 2 − 12x + 1, on [−2, 2.5]?, (NDA 2010 I), (a) –3, (b) 8, (c) –19, (d) –16.5, 20. What is the maximum value of the function log x − x ?, (NDA 2008 II), , (a) –1, (c) 1, , (b) 0, (d) ∞
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372, , NDA/NA Mathematics, , 21. What is the maximum value of x ⋅ y subject to the, condition x + y = 8?, (NDA 2008 I), (a) 8, (b) 16, (c) 24, (d) 32, , 32. The point on the curve x + y = a at which the, normal is parallel to the x-axis, is, (a) (0, 0), (b) ( 0, a ), (c) ( a , 0), (d) ( a , a ), , 22. If a and b are non-zero roots of x 2 + ax + b = 0, then, the least value of x 2 + ax + b is, 2, 9, (b) −, (a), 3, 4, 9, (c), (d) 1, 4, , 33. The function f defined by f ( x ) = 4x 4 − 2x + 1 is, increasing for, (a) x < 1, (b) x > 0, (c) x < 1/ 2, (d) x > 1/ 2, , 23. If x is real, the minimum value of x 2 − 8x + 17 is, (a) –1, (b) 0, (c) 1, (d) 2, 24. The rate of change of the surface area of a sphere of, radius r, when the radius is increasing at the rate of, 2 cm/s is proportional to, 1, 1, (a), (b) 2, r, r, (c) r, (d) r 2, 25. If the distance ‘s’ metres travelled by a particle in t, seconds is given by s = t3 − 3 t 2 , then the velocity of, the particle when the acceleration is zero (in m/s) is, (a) 3, (b) − 2, (c) − 3, (d) 2, 26. What is the rate of change of x 2 + 16 with respect to, x 2 at x = 3 ?, (NDA 2012 I), (a) 1/5, (c) 1/20, , (b) 1/10, (d) None of these, , 27. Find the minimum value of 2x 2 − 3x + 5 ? (NDA 2008 II), 3, (a) 0, (b), 4, 31, 31, (d), (c), 4, 8, 28. What is the maximum slope of the curve, y = − x3 + 3x 2 + 2x − 27?, (NDA 2007 II), (a) 1, (c) 5, , (b) 2, (d) –23, , 29. What is the product of two parts of 20 such that the, product of one part and the cube of the other is, maximum?, (NDA 2007 II), (a) 75, (b) 91, (c) 84, (d) 96, 30. The value of x for which the polynomial, 2x3 − 9x 2 + 12x + 4 is a decreasing function of x, is, (a) − 1 < x < 1, (b) 0 < x < 2, (c) x > 3, (d) 1 < x < 2, 31. The function y = a (1 − cos x ) is maximum when x is, equal to, 3π, (a) π, (b), 2, π, π, (c), (d) –, 2, 6, , 34. If f ( x ) = 3x 2 + 6x − 9, then, (a), (b), (c), (d), , (NDA 2009 I), , f ( x ) is increasing in ( −1, 3), f ( x ) is decreasing in ( 3, ∞ ), f ( x ) is increasing in ( −∞ , − 1), f ( x ) is decreasing in ( −∞ , − 1), , 35. The function f ( x ) = x 2 − 2x increases for all (NDA 2009 I), (a) x > − 1 only, (c) x > 1 only, , (b) x < − 1 only, (d) x < 1 only, 1, 36. The function f ( x ) = k sin x + sin 3x has maximum, 3, π, value at x = . What is the value of k? (NDA 2011 I), 3, 1, 1, (a) 3, (b), (c) 2, (d), 3, 2, 37. If the function f : R → R be defined, f ( x ) = tan x − x , then f ( x ), (a) increases, (b) decreases, (c) remains constant, (d) becomes zero, , by, , 38. The equation of tangent to the curve y = be− x / a at, the point, where it crosses y-axis, is, (a) ax + by = 1, (b) ax − by = 1, x y, x y, (d), (c), − =1, + =1, a b, a b, 39. What is the slope of the tangent to the curve, x = t 2 + 3t − 8, y = 2t 2 − 2t − 5 at = 2 ?, (NDA 2012 I), (a) 7/6, (b) 6/7, (c) 1, (d) 5/6, 40. The function f ( x ) = 1 − x3 − x5 is decreasing for, (a) 1 ≤ x ≤ 5, (b) x ≤ 1, (c) x ≥ 1, (d) all values of x, log x, is equal to, 41. The maximum value of, x, 2, 1, (a), (b), e, e, (c) e, (d) 1, 42. In which one of the following intervals, the function, x 2 − 5x + 6 is decreasing?, (a) ( −∞ , 2), (b) [3, ∞ ], (c) ( −∞ , ∞ ), (d) ( 2, 3), 43. The function f ( x ) = x 2 − 2x is strictly increasing in, the interval, (a) ( −2 , − 1), (b) ( −1, 0), (c) ( 0, 1), (d) (1, 2)
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373, , Application of Derivative, 44. If f ( x ) = x log x, then f ( x ) attains the minimum value, at which one of the following points?, (NDA 2011 II), (c) x = e−1 (d) x = 2e−1, (a) x = e−2 (b) x = e, 45. What is the interval over which the function, f ( x ) = 6x − x 2, x > 0 is increasing?, (NDA 2010 II), (a) ( 0, 3), (c) ( 6, 9), , (b) ( 3, 6), (d) None of these, , 46. What are the points on the curve x 2 + y 2 − 2x − 3 = 0,, where the tangents are parallel to x-axis? (NDA 2011 II), (a) (1, 2) and (1, − 2), (b) ( 0, 3 ) and ( 0, − 3 ), (c) ( 3, 0) and ( − 3, 0), (d) ( 2, 1) and ( 2, − 1), x, on [ − 1, 1] is, 47. The maximum value of f ( x ) =, 4 + x + x2, 1, 1, (b) −, (a) −, 4, 3, 1, 1, (d), (c), 6, 5, , 48. The equation of the tangent to the curve, (1 + x 2 ) y = 2 − x , where it crosses the x-axis, is, (a) x + 5 y = 2, (b) x − 5 y = 2, (c) 5x − y = 2, (d) 5x + y − 2 = 0, 2, 3, 49. y = 4x − 5 is a tangent to the curve y = px + q at, ( 2 , 3). Which one of the following is correct?, (a) p = − 2 , q = 7, (b) p = − 2 , q = − 7, (c) p = 2 , q = 7, (d) p = 2 , q = − 7, 50. If the curve y = a x and y = bx intersect at angle α,, then tan α is equal to, a−b, log a − log b, (a), (b), 1 + ab, 1 + log a log b, a+b, log a + log b, (d), (c), 1 − ab, 1 − log a log b, , Level II, 1. Function, , f(x) =, , increasing, if, (a) λ > 1, (c) λ < 4, , λ sin x + 6 cos x, 2 sin x + 3 cos x, , is, , monotonic, , (b) λ < 1, (d) λ > 4, , 2. If x > 0, xy = 1, then what is the minimum value of, x + y?, (a) 2, (b) –2, (c) 1, (d) –1, 3. If g( x ) = min ( x , x 2 ), where x is real number, then, (a), (b), (c), (d), , g ( x ) is an increasing function, g ( x ) is a decreasing function, f ( x ) is a constant function, g ( x ) is a continuous function except at x = 0, , 4. The position of a point in time ‘t’ is given by, x = a + bt − ct 2 , y = at + bt 2. Its acceleration at time, ‘t’ is, (a) b − c, (b) b + c, (c) 2b − 2c, (d) 2 b2 + c2, 5. A function f is defined by f ( x ) = 2 + ( x − 1)2/ 3 in [0, 2]., Which of the following is not correct?, (a) f is not derivable in (0, 2), (b) f is continuous in [0, 2], (c) f ( 0) = f ( 2), (d) Rolle’s theorem is true in [0, 2], 6. The point in the interval ( 0, 2π ), where f ( x ) = ex sin x, has maximum slope, is, (NDA 2011 II), π, π, (b), (a), 4, 2, 3π, (c) π, (d), 2, , 7. The largest value of 2x3 − 3x 2 − 12x + 5 for − 2 ≤ x ≤ 2,, occurs when, (NDA 2011 I), (a) x = − 2, (b) x = − 1, (c) x = 2, (d) x = 0, 8. At an extreme point of a function f ( x ), the tangent to, the curve is, (NDA 2011 II), (a) parallel to the x-axis, (b) perpendicular to the x-axis, (c) inclined at an angle 45° to the x-axis, (d) inclined at an angle 60° to the x-axis, 9. A balloon is pumped at the rate of 4 cm 3 /s. What is, the rate at which its surface area increases when its, radius is 4 cm?, (NDA 2010 II), 2, 2, (a) 1 cm /s, (b) 2 cm /s, (c) 3 cm 2/s, , (d) 4 cm 2/s, , is, monotonically, 10. If, f ( x ) = kx3 − 9x 2 + 9x + 3, increasing in every interval, which one of the, following is correct?, (NDA 2010 I), (a) k < 3, (b) k ≤ 3, (c) k > 3, (d) k ≥ 3, 11. Which one of the following statements is correct?, (NDA 2012 I), x, , (a) e is an increasing function, (b) ex is a decreasing function, (c) ex is neither increasing nor decreasing function, (d) ex is a constant function, 12. What is the value of b for which f ( x ) = sin x − bx + c is, decreasing in the interval ( −∞ , ∞ )?, (a) b < 1, (b) b ≥ 1, (c) b > 1, (d) b ≤ 1
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374, , NDA/NA Mathematics, , 13. A body moves in a straight line according to the law, s = t3 − 4 t 2 − 3 t. What are the values of time and, acceleration (time, acceleration); each time when, velocity of the body is zero?, (a) ( 3, − 10), (b) ( 3, 10), (c) ( 2, − 10), (d) ( 2, 10), 14. Which one of the following is correct?, x, , x ∈ ( −∞ , ∞ ), f(x) =, 1 +|x|, (a) is monotonically increasing at every point, ( −∞ , ∞ ), (b) is monotonically decreasing at every point, ( −∞ , ∞ ), (c) is monotonically increasing on ( −∞ , ∞ ) except, x=0, (d) is monotonically decreasing on ( −∞ , ∞ ) except, x=0, , of, of, at, at, , 15. f ( x ) = cos x is monotonic decreasing under which one, of the following conditions?, (NDA 2008 II), π, π, (a) 0 < x < only, (d), < x < π only, 2, 2, (c) 0 < x < π, (d) 0 < x < 2π, 16. What is the smallest value of m for which, f ( x ) = x 2 + mx + 5 is increasing in the interval, (NDA 2008 I), 1 ≤ x ≤ 2?, (a) m = 0, (b) m = − 1, (c) m = − 2, (d) m = − 3, 17. What is/are the critical point(s) of the function, f ( x ) = x 2/ 3 ( 5 − 2x ) on the interval [−1, 2]? (NDA 2007 I), 1, (a) Only 1, (b) 0 and, 3, 3, 3, (d) 0 and, (c) Only, 2, 2, 18. If the rate of change in volume of spherical soap, bubble is uniform, then the rate of change of surface, area varies as, (NDA 2011 II), (a) square of radius, (b) square root of radius, (c) inversely proportional to radius, (d) cube of the radius, f ( b) − f ( a ), 19. In the mean value theorem, = f ′ ( c), if, b− a, 1, a = 0, b = and f ( x ) = x ( x − 1) ( x − 2), the value of c is, 2, 15, (b) 1 + 15, (a) 1 −, 6, 21, (d) 1 + 21, (c) 1 −, 6, 20. Moving along the x-axis are two points with, x = 10 + 6 t ; x = 3 + t 2. The speed with which they are, reaching from each other at the time of encounter is, (x is in cm and t is in seconds), (a) 16 cm/s (b) 20 cm/s (c) 8 cm/s (d) 12 cm/s, , 21. Under which conditions are the two curves, y = x 2 + ax + b and y = cx − x 2 tangent to each other, at the point (1, 0)?, (a) a = 2 , b = − 3, c = − 1 (b) a = 3, b = − 4, c = − 1, (c) a = − 1, b = 0, c = 1, (d) a = − 3, b = 2, c = 1, 22. If f ( x ) = x 2 − 6x + 8 and there exists a point c in the, interval [2, 4] such that f ′ ( c) = 0, then what is the, value of c?, (a) 2.5, (b) 2.8, (c) 3, (d) 3.5, dy, 23. A particle P moves along the curve x 2 y3 = 27., = 10, dt, at the time, when P is at the point (1, 3). What is the, dx, at that instant?, value of, dt, (a) 5, (b) –10, (c) 10, (d) –5, 24. The velocity of telegraphic communication is given, 1, by v = x 2 log , where x is the displacement. For, x, maximum velocity, x is equal to, (NDA 2009 II), (b) e−1/ 2, (a) e1/ 2, (c) ( 2e)−1, (d) 2e−1/ 2, 25. The function f ( x ) = ( x − 1)ex + 1 is, , (NDA 2007 II), , (a) negative for all x > 0 (b) positive for all x > 0, (c) increasing for all x, (d) decreasing for all x, 26. If a differentiable function f defined for x > 0 satisfies, the relation f ( x 2 ) = x3 , x > 0, then what is the value of, (NDA 2007 II), f ′ ( 4)?, (a) 1, (b) 2, (c) 3, (d) 4, 27. If at any instant t, for a sphere, r denotes the radius,, S denotes the surface area and V denotes the volume,, dV, (NDA 2010 II), then what is the value of, ?, dt, 1 dr, 1 dS, (a) S, (b) r, 2 dt, 2 dt, 1, dS, dS, (c) r, (d) r 2, 2, dt, dt, 28. The second derivative of f ( ex ) with respect to x,, where f is a polynomial, is, (a) f ′′ ( ex ) ex + f ′ ( ex ), (b) f ′′ ( ex ) e2x + f ′′ ( ex ) ex, (c) f ′′ ( ex ), (d) f ′′ ( ex ) e2x + f ′ ( ex ) ex, 29. For what values of p, the function, px 2 + 1, if x ≤ 1, is derivable at x = 1?, f(x) = , x + p, if x > 1, 1, (a), (b) 2, 2, 1, (c) −, (d) –2, 2
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375, , Application of Derivative, 30. Each side of an equilateral triangle expands at the, rate of 2 cm/s. What is the rate of increase of area of, the triangle when each side is 10 cm?, (a) 10 2 cm 2/s, (b) 10 3 cm 2/s, (c) 10 cm 2/s, (d) 5 3 cm 2/s, 31. Let, be, real, numbers, such, that, x, y, Let, and, let, − 4 ≤ x ≤ 4, − 5 ≤ y ≤ 5., θ∈R, A = x cos θ − y sin θ , B = x cos θ + y sin θ. What is the, maximum value of A2 − B2?, (a) 32, (b) 40, (c) 50, (d) 80, , 37. Match List I (Function) with List II (Property) and, select the correct answer using the codes given below, the lists., , A., B., , List I, (Function), tan x, f ( x) =, x, f ( x) = ( x − 1) − log x, , 34. The motion of a particle is described as, s = 2 − 3t + 4t3 . What is the acceleration of the, particle at the point, where its velocity is zero?, (NDA 2007 II), , (a) 0, (c) 8 unit, , (b) 4 unit, (d) 12 unit, , 35. What is the interval in which the function, f ( x ) = 9 − x 2 is increasing? ( f ( x ) > 0), (NDA 2007 I), (a) 0 < x < 3, (c) 0 < x < 9, , (b) −3 < x < 0, (d) −3 < x < 3, , 36. What is the x-coordinate of the point on the curve, f ( x ) = x ( 7x − 6), where the tangent is parallel to, x-axis?, (NDA 2007 I), 2, 1, (b), (a) −, 7, 3, 6, 1, (c), (d), 7, 3, , 2., , f ( x) =, , sin x, x, , 3., , D., , f ( x) =, , log (1 + x), x, , 4., 5., , (NDA 2012 I), , 33. y = a log x + bx 2 + x has its extremum value at x = − 1, and x = 2. What are the values of a and b?, (a) a = 2 , b = − 1, 1, (b) a = 2 , b = −, 2, 1, (c) a = − 2 , b =, 2, 1, 1, (d) a = , b = −, 2, 2, , 1., , C., , 32. Which one of the following statements is correct?, (a) The derivative of a function f ( x ) at a point will, exist, if there is one tangent to the curve y = f ( x ), at that point and the tangent is parallel to y-axis., (b) The derivative of a function f ( x ) at a point will, exist, if there is one tangent to the curve y = f ( x ), at that point and the tangent must be parallel to, x-axis., (c) The derivative of a function f ( x ) at a point will, exist, if there is one and only one tangent to the, curve y = f ( x ) at that point and the tangent is not, parallel to y-axis., (d) None of the above, , List II, (Property), , Codes, A, (a) 2, (b) 5, (c) 5, (d) 2, , B, 4, 3, 1, 4, , C, 3, 1, 4, 1, , Increasing for every, x>1, Decreasing for, every x > 0, Neither increasing, nor decreasing for, x>0, Decreasing for x in, ( 0, π / 2), Increasing for x in, ( 0, π / 2), , D, 5, 2, 2, 5, , 38. Consider the following statements in respect of the, (NDA 2011 II), function f ( x ) = x3 − 1, x ∈ [− 1, 1], I. f ( x ) is increasing in [− 1, 1]., II. f ( x ) has no root in ( − 1, 1)., Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 39. Consider the following statements, I. The function f ( x ) = tan x − 4x is increasing in the, π, interval − < x < 0., 3, II. The set of all x for which log(1 + x ) ≤ x is (0, ∞)., Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 40. What is the area of the largest rectangular field, which can be enclosed with 200 m of fencing?, (NDA 2008 I), 2, , 2, , (a) 1600 m, (c) 2400 m 2, , (b) 2100 m, (d) 2500 m 2, , 41. A stone thrown vertically upward satisfies the, equation s = 64t − 16t 2, where s is in metre and t is in, second. What is the time required to reach the, maximum height?, (NDA 2009 I), (a) 1 s, (b) 2 s, (c) 3 s, (d) 4 s, 42. The profit function, in rupees, of a firm selling, x items ( x ≥ 0) per week is given by, P ( x ) = − 3500 + ( 400 − x ) x. How many items should, the firm sell, so that the firm has maximum profit?, (NDA 2009 I), , (a) 400, , (b) 300, , (c) 200, , (d) 100
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376, , NDA/NA Mathematics, , 43. Given two squares of sides x and y such that, y = x + x 2. What is the rate of change of the area of, the second square with respect to the area of the first, square?, (NDA 2010 I), (b) 1 + 2x + 3x 2, (a) 1 + 3x + 2x 2, (c) 1 − 2x + 3x 2, (d) 1 − 2x − 3x 2, , Directions (Q. Nos. 44-46) Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also, has four alternative choices, only one of which is the, correct answer. You have to select one of the codes (a),, (b), (c) and (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not the, correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 44. Assertion (A) Curve y = xex is minimum at the, point x = − 1., dy, Reason (R), < 0 at x = − 1., dx, 45. Assertion (A) y = x 2e− x increases on the interval, ( 0, 2)., , Reason (R) since,, , dy, > 0 on the interval ( 0, 2)., dx, , 46. Assertion (A) The tangent to the curve, y = x3 − x 2 − x + 2 at (1, 1) is parallel to the x-axis., Reason (R) The slope of the tangent to the curve at, (1, 1) is zero., , Directions (Q. Nos. 47-49) Consider the function, f ( x ) = 2x 2 − log|x| is monotonically for values of x ≠ 0., , 47. The interval in which the function f ( x ) is, monotonically increasing, 1 , (b) (1, 2), (a) − , 0, 2 , (c) ( 2, 3), (d) None of these, 48. The interval in which the function f ( x ) is, monotonically decreasing?, 1, (a) 0, , (b) ( 2, 4), 2, (c) (5, 3), (d) None of these, df ( x ), is, 49. The value of, dx, 1, (a) 4x 2 −, (b) 4x, x, (c) 4x − 2, (d) None of these, , Answers, Level I, 1., 11., 21., 31., 41., , (b), (d), (b), (a), (b), , 2., 12., 22., 32., 42., , (d), (b), (b), (b), (a), , 3., 13., 23., 33., 43., , (c), (a), (c), (d), (d), , 4., 14., 24., 34., 44., , (a), (a), (c), (d), (c), , 5., 15., 25., 35., 45., , (d), (c), (c), (c), (a), , 6., 16., 26., 36., 46., , (b), (c), (b), (c), (a), , 7., 17., 27., 37., 47., , (b), (d), (d), (a), (c), , 8., 18., 28., 38., 48., , (a), (b), (c), (d), (a), , 9., 19., 29., 39., 49., , (a), (c), (a), (b), (d), , 10., 20., 30., 40., 50., , (a), (a), (d), (d), (b), , 2., 12., 22., 32., 42., , (a), (c), (c), (d), (c), , 3., 13., 23., 33., 43., , (a), (b), (d), (b), (a), , 4., 14., 24., 34., 44., , (d), (c), (b), (d), (c), , 5., 15., 25., 35., 45., , (d), (c), (b), (b), (a), , 6., 16., 26., 36., 46., , (b), (c), (c), (b), (a), , 7., 17., 27., 37., 47., , (b), (a), (b), (c), (a), , 8., 18., 28., 38., 48., , (a), (c), (d), (c), (a), , 9., 19., 29., 39., 49., , (b), (c), (a), (b), (a), , 10., 20., 30., 40., , (c), (c), (b), (d), , Level II, 1., 11., 21., 31., 41., , (d), (a), (d), (b), (b)
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Hints & Solutions, Level I, 1. Let y = f (x) = − x3 + 3x2 + 9x − 27, The slope of this curve f ′ (x) = − 3x2 + 6x + 9, Let, g (x) = f ′ (x) = − 3x2 + 6x + 9, On differentiating wrt x, we get, g ′ (x) = − 6x + 6, For maxima or minima, put g ′ (x) = 0 ⇒ x = 1, Now, g ′ ′ (x) = − 6 < 0 and hence at x = 1, g (x) slope will, have maximum value., ∴, [ g (1)]max = − 3 × 1 + 6(1) + 9 = 12, 2. f (x) = x3 − 6x2 + ax + b, On differentiating wrt x, we get, f ′ (x) = 3x2 − 12x + a, By the definition of Rolle’s theorem,, ⇒, f ′ (c) = 0, 1, , f ′ 2 +, ⇒, =0, , 3, 2, , ⇒, ⇒, ⇒, ⇒, , 1, 1, , , 3 2 +, + a =0, − 12 2 +, , , 3, 3, 1, 4, 1, , , 3 4 + +, + a =0, − 12 2 +, , , , 3, 3, 3, 12 + 1 + 4 3 − 24 − 4 3 + a = 0, a = 11, , 3. Given equation is a + bv2 = x2, On differentiating, we get, dv, dx, dx, dv, 0 + b 2v = 2x, ⇒ vb, =x, ⇒, dt , dt, dt, dt, dv, x dx, dv x, dx , ⇒, = v, =, ⋅, ⇒, =, Q, dt, , dt vb dt, dt b, 4. Let y = xx, On differentiating wrt x, we get, dy, = xx (1 + log x), dx, dy, For increasing function,, >0, dx, x, ∴, x (1 + log x) > 0 ⇒ 1 + log x > 0, 1, 1, ⇒, log e x > log e, ⇒ x>, e, e, 1, Function is increasing when x > ., e, sin 3x, 5. f (x) = p sin x +, (given), 3, 3 cos 3x, ⇒ f ′ (x) = p cos x +, = p cos x + cos 3x, 3, π, f (x) has an extremum value at x = ., 3, π, ∴, f′ =0, 3, π, p cos + cos π = 0, ⇒, 3, , 1, 1, p −1 =0 ⇒ p=, =2, 2, 1 /2, , ⇒, , 6. Let A = cos θ + cos 2θ, On differentiating wrt θ, we get, dA, = − sin θ − 2 sin 2θ, dθ, dA, Put, = 0, for maxima or minima, dθ, sin θ = − 2 sin 2θ = − 4 sin θ ⋅ cos θ, ⇒, 4 cos θ = − 1 or sin θ = 0, 1, or, cos θ = −, θ =0, ⇒, 4, d 2A, Now,, = − cos θ − 4 cos 2θ, dθ 2, = − cos θ − 4 (2 cos 2 θ − 1), 2, d A, 1, 1, , , ⇒, = − 4 2 ⋅, − 1 > 0, , , 16, , 4, 1, dθ 2 , cos θ = −, , 4, , 1, ∴ A is minimum at θ = cos −1 − , 4, Now, minimum value of, cos θ + cos 2θ = cos θ + 2 cos 2 θ − 1, 1, 1, = − + 2 − 1, 4, 16, , =−, , 9, 1 1, −2 + 1 − 8, + −1=, =−, 8, 8, 4 8, , 7. Q x + y = 12 ⇒ y = 12 − x, Let, A = xy, = x(12 − x) = 12x − x2, dA, = 12 − 2x, dx, dA, Put, = 0, for maxima or minima, dx, 12 − 2x = 0 ⇒ x = 6, d 2A , d 2A, Now,, =, −, 2, ⇒, = −2 < 0, 2, dx2, dx x = 6, , …(i), , ∴ A is maximum at x = 6., ∴ Maximum value of, A = (12x − x2) = 12 × 6 − 62, = 72 − 36 = 36, 8. Given, curve y = x2 − 4x + 3, Now, differentiating wrt x, we get, dy, ...(i), = 2x − 4 = 2 (x − 2), dx, dy, Here at x = 2,, =0, dx, i.e., for the given curve only one tangent is possible, because slope of tangent parallel to x-axis is zero.
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378, , NDA/NA Mathematics, , 9. We have, x2 + y2 = 25, dy, ⇒, 2x + 2 y, =0, dx, dy − x, =, ⇒, dx, y, , …(i), , 3, 4, Since, tangent is parallel to given line, then, dy 3, x 3, =, ⇒ − =, dx 4, y 4, 4, …(ii), y=− x, ⇒, 3, From Eqs. (i) and (ii), we get, 16 2, x2 +, x = 25 ⇒ x = ± 3, 9, if, [using Eq. (ii)], x = 3, y = − 4, if, [using Eq. (ii)], x = − 3, y = 4, ∴ Points of contact are (3, − 4) and (–3, 4)., Now, slope of the line 3x − 4 y = 7 is m =, , 10. Given that, f (x) = x1/ x, 1, f ′ (x) = 2 (1 − log x) ⋅ x1/ x, ⇒, x, f ′ (x) > 0, if 1 − log x > 0, ⇒, log x < 1, ⇒, x<e, ∴ f (x) is increasing in the interval (−∞ , e)., , ⇒, and, ⇒, , dy, = 2 = m1, , dx (1, 1), , dy, = 2x, dx, , x = y2 ⇒ 1 = 2 y, , (say), dy, dx, , dy 1, 1, dy, =, ⇒ , = = m2, dx (1, 1) 2, dx 2 y, , ∴ Angle of intersection at the point of (1, 1), 1, 2−, m1 − m2, 2 =3, tan θ =, =, 1 + m1m2 1 + 2 × 1 4, 2, −1 3 , θ = tan , ⇒, 4, 12. The given equation of curve can be rewritten as, y = xe− x, On differentiating wrt x, we get, dy, = − xe− x + e− x, dx, dy, On putting, = 0, for maxima or minima, dx, − xe− x + e−x = 0, ⇒, x=1, d 2y, Now,, = − e−x + xe− x − e−x, dx2, d 2y, ⇒, = − e−1 < 0, 2, dx x = 1, , 13. Given, y = tan −1 x − x, On differentiating wrt x, we get, dy, 1, − x2, =, −1 =, 2, dx 1 + x, 1 + x2, dy, < 0, ∀ x ∈ R, ⇒, dx, Hence, function is always decreasing., 14. f (x) = 2x3 − 15x2 + 36x + 4, On differentiating wrt x, we get, f ′ (x) = 6x2 − 30x + 36, Either maxima or minima, f ′ (x) = 0, ∴, 6x2 − 30x + 36 = 0 ⇒ x2 − 5x + 6 = 0, ⇒, (x − 2) (x − 3) = 0 ⇒ x = 2 , 3, Again, differentiating Eq. (i), we get, f ′ ′ (x) = 12x − 30, ⇒, f ′ ′ (2) = 24 − 30 = − 6 < 0, Therefore, f (x) is maximum at x = 2, , …(i), , 15. Equation of the curve x2y2 = a 4, On differentiating the given equation, we get, dy − y, dy, =, x22 y, + y22x = 0 ⇒, dx, x, dx, a , dy, ⇒, =−, , =1, dx ( − a, a), − a, , 11. The slope of the curve,, y = x2 ⇒, , ∴ y is maximum at x = 1, then y = 1 ⋅ e−1 = e−1, Hence, maximum point on the curve x = ex y is (1, e−1 )., , (say), , Therefore, subtangent at the point (−a , a ), y, a, =, = =a, dy 1, , dx, 16. y = x3 − 3x2 − 9x + 5, On differentiating, we get, dy, = 3 x2 − 6 x − 9, dx, Since, tangent is parallel to x-axis., dy, ∴, = 0 ⇒ 3 x2 − 6 x − 9 = 0, dx, ⇒, (x + 1) (x − 3) = 0 ⇒ x = − 1, 3, 17. Given equation is x2 = − 4 y, On differentiating, we get, dy, dy, x, 2x = − 4, ⇒, =–, 2, dx, dx, Slope of tangent at (−4,−4)., 4, dy, ⇒, = =2, , dx ( −4, − 4) 2, We know that, equation of tangent at the point (x1 , y1 ) is, dy, ( y − y1 ) = , (x − x1 ), dx ( x , y ), 1, , 1, , If the point (– 4, – 4), then, y + 4 = 2 (x + 4), ⇒, 2x − y + 4 = 0, 18. Given equation is y = 2x2 − x + 1, On differentiating, we get, dy, = 4x − 1, dx
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379, , Application of Derivative, Since, this is parallel to the given curve y = 3x + 9, ∴ These slopes are equal, ⇒, 4x − 1 = 3 ⇒ x = 1, At x = 1,, y = 2(1)2 − 1 + 1 ⇒ y = 2, Thus, the point is (1, 2)., , 22. Given that, a and b are roots of the equation, x2 + ax + b = 0, Then, sum of roots, a + b = − a, …(i), ⇒, b = − 2a, and product of roots, ab = b, ⇒, ab − b = 0, ⇒, b (a − 1) = 0, Since, a and b are non-zero, then we take b ≠ 0., Then,, a = 1 and b = − 2, ∴ The expression will be, f (x) = x2 + x − 2, f ′ (x) = 2x + 1, For maxima and minima, put f ′ (x) = 0, 1, 2x + 1 = 0, x = −, 2, 1, f ′ ′ (x) = 2 f ′ ′ − = 2 > 0, 2, 1, ⇒ Function is minimum at x =, 2, 9, 1 1 1, ∴ Minimum value of function f − = − − 2 = −, 2 4 2, 4, , 19. Q, f (x) = 2x3 − 3x2 − 12x + 1, ⇒ f ′ (x) = 6x2 − 6x − 12, Put f ′ (x) = 0, for maxima or minima, 6x2 − 6x − 12 = 0, ⇒, x2 − x − 2 = 0, ⇒, (x − 2)(x + 1) = 0, ⇒, x = 2 and –1, Now,, f ′ ′ (x) = 12x − 6, ⇒, f′ ′ (2) = 24 − 6 = 18 > 0, ∴ f (x) is minimum at x = 2, Hence,, f (2) = 2(2)3 − 3(2)2 − 12 × 2 + 1, = 16 − 12 − 24 + 1, = − 19, 20. Let y = log x − x, , dy 1, = −1, ∴, dx x, 2, 1, d y, and, =− 2, dx2, x, For maximum and minimum values of y,, dy 1, = −1 =0, dx x, 1, ⇒, =1, x, ⇒, x=1, For x = 1,, d 2y, = − ve, dx2, Thus, the value of the given function is maximum for, x = 1., So, the maximum value of the function, = log (1) − 1 = − 1, , 21. Q, ⇒, and let, ∴, ⇒, and, , x+ y=8, y=8 − x, P = xy, P = x (8 − x), = 8 x − x2, dP, = 8 − 2x, dx, 2, dP, = −2, dx2, , dP, = 0, for maxima or minima, dx, 8 − 2x = 0, 8, x= =4, ⇒, 2, d 2P , and, = −2 < 0, 2, dx x = 4, Put, , ∴ P is maximum at x = 4., Maximum value of P = 4 ⋅ (8 − 4) = 4 ⋅ 4 = 16, , …(i), , 23. Here,, f (x) = x2 − 8x + 17, ∴, f ′ (x) = 2x − 8, For maximum or minimum, put f (x) = 0, ⇒, 2x − 8 = 0 ⇒ x = 4, f ′ ′ (x) = 2 > 0, hence minima at x = 4, The minimum value is, f (4) = 42 − 8 × 4 + 17 = 16 − 32 + 17 = 1, dr, 24. Surface area of sphere, S = 4πr 2 and, =2, dt, dS, dr, dS, ∴, = 4π × 2r, = 8πr × 2 = 16πr ⇒, ∝r, dt, dt, dt, 25. We have, s = t3 − 3 t 2, On differentiating wrt t, we get, ds, = 3 t2 − 6 t, dt, Again, differentiating Eq. (i), we get, d 2s, =6 t −6 =0 ⇒t=1, dt 2, On putting the value of t = 1 in Eq. (i), we get, ds, = 3 × 1 − 6 × 1 = 3 − 6 = − 3 m/s, dt, 26. Let u = x2 + 16 and v = x2, du, 1, Now,, =, × 2x =, dx 2 x2 + 16, , x, x + 16, 2, , and, , ....(i), , dv, = 2x, dx, , ∴ Now, rate of change of u wrt to v is,, x, 1, du du / dx, =, ×, =, 2, dv dv / dx, x + 16 2x, , du , , dv , , ⇒, ∴, , du, 1, =, 2, dv 2 x + 16, 1, 1, 1, 1, =, =, =, at ( x =3 ) =, 2 9 + 16, 2 25 2 × 5 10, , d (x2 + 16), d (x2), , =, , 1, at x = 3, 10
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380, , NDA/NA Mathematics, Q First part = 15, and, second part = 20 − 15 = 5, Q Required product = 15 × 5 = 75, , 27. Let y = 2x2 − 3x + 5, 3x , , y = 2 x2 − + 5, 2, , 2, , 3, 9, y = 2 x − −, +5, , , 4, 16, , , 2, 3, 9, , , y = 2 x − − + 5, , 4, 8, , 30. Q, f (x) = 2x3 − 9x2 + 12x + 4, ⇒, f ′ (x) = 6x2 − 18x + 12, For function to be decreasing, f ′ (x) < 0, ⇒, 6 (x2 − 3x + 2) < 0 ⇒ (x2 − 2x − x + 2) < 0, ⇒, (x − 2) (x − 1) < 0 ⇒ 1 < x < 2, , 2, , 3, 31, , y = 2 x − +, , 4, 8, , …(i), 2, , , 3, , Q 2 x − ≥ 0, 4, , , , 2, , ⇒, ⇒, , 3, 31 31, , ≥, 2 x − +, , , 4, 8, 8, 31, y≥, 8, , ∴ The minimum value of y =, , [from Eq. (i)], 31, 8, , 28. Given curve,, y = − x3 + 3x2 + 2x − 27, On differentiating wrt x, we get, dy, = − 3 x2 + 6 x + 2, dx, dy, Let, A=, = − 3 x2 + 6 x + 2, dx, dA, Now,, = − 6x + 6, dx, d 2A, and, = −6, dx2, For max or min value of slope of the curve, put, dA, =0, dx, −6 x + 6 = 0 ⇒ x = 1, d 2A , Now,, = −6 < 0, 2, dx x = 1, ∴ At (x = 1), A is maximum., ∴ Maximum slope of curve = Maximum value of A, = −3 + 6 + 2 = 5, 29. Let first part = x, ∴ Second part = 20 − x, Let, P = x3 (20 − x) = 20x3 − x4, dP, Now,, = 60x2 − 4x3, dx, d 2P, and, = 120x − 12x2, dx2, For max or min value,, dP, =0, dx, 2, ⇒, 4x (15 − x) = 0, x = 0 and 15, d 2P , ∴, = 120 × 15 − 12 × (225), 2, dx x = 15, = 1800 − 2700 = − 900 < 0, ∴ At x = 15, P is maximum., , 31. y = a (1 − cos x), On differentiating, we get y ′ = a sin x, Put y′ = 0, for maxima or minima, ⇒, sin x = 0, ⇒, x = 0, π, Again, differentiating, we get, y ′ ′ = a cos x ⇒ y′ ′ (0) = a and y ′ ′ (π ) = − a, Hence, y is maximum when x = π., 32. The equation of given curve is x +, 1, 1 dy, ∴, +, ⋅, =0, 2 x 2 y dx, dy, y, ⇒, =−, dx, x, The normal is parallel to x-axis, if, dx, = 0 ⇒ x1 = 0, , dy ( x , y ), 1, , ∴, ∴, , y= a, , 1, , From equation of curve, y1 = a, Required point is (0, a )., , 33. We have, f (x) = 4x4 − 2x + 1, On differentiating, we get f ′ (x) = 16x3 − 2, Put f ′ (x) = 0 for maxima, ⇒, 16x3 − 2 = 0 ⇒ x = 1 / 2, 1, , 1 , Taking interval − ∞ , and , ∞ ., , 2 , 2, 1 , In interval , ∞ , put x = 1, 2 , f ′ (1) = 16 − 2 = 14, , (increasing), , 1, ∴ Function f is increasing for x > ., 2, 34. Q f (x) = 3x2 + 6x − 9, On differentiating wrt x, we get, f ′ (x) = 6x + 6, For a decreasing function,, f ′ (x) < 0 ⇒ 6x + 6 < 0 ⇒ x < − 1, ∴ f (x) is decreasing in (− ∞ , − 1),, and for increasing function,, f ′ (x) > 0, ⇒, 6x + 6 > 0 ⇒ x > − 1, ∴ f (x) is increasing in (−1, ∞ )., 35. Q f (x) = x2 − 2x, On differentiating wrt x, we get, f ′ (x) = 2x − 2, f (x) is increasing, if f ′ (x) > 0, 2x − 2 > 0, ⇒, x > 1 only
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381, , Application of Derivative, 36. f (x) = k sin x +, ⇒, Put, At, ⇒, , 1, sin 3x, 3, , (given), Now,, , 3, cos 3x, 3, f ′ (x) = 0, for maxima, k cos x + cos 3x = 0, π, π, x = , k cos + cos π = 0, 3, 3, 1, k =1 ⇒ k =2, 2, f ′ (x) = k cos x +, , 1, 1 − cos 2 x, −1 =, 2, cos x, cos 2 x, , Since, 0 ≤ cos2 x ≤ 1 for all values of x., ∴ f ′ (x) > 0 for all values of x. Thus, f (x) always, increases., 38. Given equation of curve is y = be− x/ a, Since, the curve crosses y-axis (i.e., x = 0), ⇒, y = be−0, ⇒, y=b, , On differentiating Eq. (i), we get, dy − b − x/ a, =, e, dx a, − b −0/ a − b, dy, At point (0, b), , =, =, e, dx ( 0, b), a, a, ∴ Required equation of tangent is, −b, y−b=, (x − 0), a, y, x, x y, ⇒, −1 = – ⇒ + =1, b, a, a b, 39. Given that, x = t 2 + 3t − 8, and y = 2t 2 − 2t − 5, dx, dy, Now,, = 2t + 3 and, = 4t − 2, dt, dt, ∴Slope of the tangent to the curve =, =, , dy / dt 4t − 2, =, dx / dt 2t + 3, , dy, , dx, , =, at ( t = 2), , 1, x2 ⋅ − − (1 − log x) ⋅ 2x, x, , x4, d 2y, − (3 − 2 log x), 1, and 2 , =, = − 3 <0, 3, x, e, dx x = e, , ∴Function is maximum at x = e., log e 1, ∴, =, ymax =, e, e, , 37. f (x) = tan x − x, On differentiating, we get, f ′ (x) = sec2 x − 1 =, , d 2y, =, dx2, , dy, dx, , 4(2) − 2 8 − 2 6, =, =, 2(2) + 3 4 + 3 7, , 40. Given that, f (x) = 1 − x3 − x5, On differentiating wrt x, we get, f ′ (x) = − 3x2 − 5x4 ⇒ f ′ (x) = − (3x2 + 5x4 ), ⇒, f ′ (x) < 0, for all values of x., log x, 41. Let y =, x, 1, x ⋅ − log x, dy, dy 1 − log x, x, =, =, ⇒, ∴, 2, dx, dx, x, x2, dy, Put, = 0, for maxima or minima, dx, ∴, x=e, , …(i), , 42. Given that, f (x) = x2 − 5x + 6, On differentiating Eq. (i), wrt x, we get, f ′ (x) = 2x − 5, For f (x) to be decreasing, f ′ (x) < 0, 5, i.e.,, 2x − 5 < 0 ⇒ 2x < 5 ⇒ x <, 2, 5, , ∴ f (x) is decreasing in the interval −∞ , ., , 2, i.e., f (x) is decreasing in the interval ( − ∞ , 2)., 43. f (x) = x2 − 2x, f ′ (x) = 2x − 2 = 2 (x − 1), For f (x) to be strictly increasing, then, f ′ (x) > 0 ⇒ 2 (x − 1) > 0, ⇒, x−1 >0⇒x>1, f (x) = x log x, 1, f ′ (x) = x ⋅ + log x ⋅ 1, x, = 1 + log x, For maximum or minimum value of f (x), f ′ (x) = 0, ⇒, 1 + log e x = 0, ⇒, log e x = − 1, ⇒, x = e−1, 1, Now,, f ′ ′ (x) =, x, 1, −1, At (x = e ), f ′ ′ (x) = −1 = e > 0, e, So, at (x = e−1 ), then f (x) attains minimum value., , 44. Given,, , 45. We have, f (x) = 6x − x2, On differentiating wrt x, we get, f ′ (x) = 6 − 2x, f (x) will be increasing function, if, f ′ (x) > 0 ⇒ 6 − 2x > 0 ⇒ x < 3, Thus, required interval is (0, 3)., 46. Given that, x2 + y2 − 2x − 3 = 0, Differentiating both sides wrt x, we get, dy, 2x + 2 y, −2 =0, dx, dy 1 − x, ∴, =, dx, y, If the tangent is parallel to the x-axis, then, dy, =0, dx, , …(i)
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382, , NDA/NA Mathematics, ∴, , 1−x, =0, y, , dy (1 + x2) (−1) − (2 − x) (2x), =, dx, (1 + x2)2, , ⇒, 1 − x =0, ⇒, x =1, Putting x = 1 in the given equation, we get, 1 + y2 − 2(1) − 3 = 0, ⇒, y2 = 4, ⇒, y=±2, Therefore, the coordinates of the points at which the, tangent is parallel to the x-axis are (1, ± 2)., x, 47. f (x) =, 4 + x + x2, On differentiating, we get, 4 + x + x2 − x (1 + 2x), f ′ (x) =, (4 + x + x2)2, For maximum, f ′ (x) = 0 ⇒, ⇒, , dy x2 − 4x − 1, =, dx, (1 + x2)2, (2)2 − 4 (2) − 1, (1 + 22)2, 4 − 8 − 1 −5 −1, =, =, =, 5, (1 + 4)2 25, , ∴ Slope of tangent at (2 , 0) =, , ∴ Equation of tangent at (2, 0) and slope –1/5 is, 1, y − 0 = − (x − 2) ⇒ 5 y = − x + 2, 5, ⇒, x + 5y = 2, 49. y = 4x − 5 is a tangent to the curve y2 = px3 + q at (2, 3)., Therefore, (2, 3) satisfy the equation of curve, ∴, (3)2 = p (2)3 + q, ⇒, 9 = 8 p + q …(i), From the given options, option (d) satisfied Eq. (i)., Therefore, the option (d) is correct., , 4 − x2, =0, (4 + x + x2)2, , x=2,−2, , Both the values of x are not in a interval [–1, 1]., −1, −1, f (− 1) =, =, ∴, 4 −1 + 1, 4, 1, 1, (maximum), f (1) =, =, 4+1+1 6, 48. The given curve (1 + x2) y = 2 − x, meets x-axis, where y = 0 ⇒ 0 = 2 − x ⇒ x = 2, So, Eq. (i) meets x-axis at the point (2, 0)., 2−x, Also, from Eq. (i), y =, 1 + x2, , ⇒, , …(i), , 50. Given equations of curves are y = a x and y = bx ., Point of intersection of curves is (0, 1)., Now, slope of tangent of the curve, y = a x, dy, dy, m1 =, = a x log a ⇒ , = m1 = log a, dx ( 0, 1), dx, Slope of tangent of the curve, y = bx, dy, dy, m2 =, = bx log b ⇒ m2 = , = log b, dx ( 0, 1), dx, ∴ tan α =, , log a − log b, m1 − m2, =, 1 + m1m2 1 + log a log b, , On differentiating wrt x, we get, , Level II, 1. Given function, f (x) =, , λ sin x + 6 cos x, 2 sin x + 3 cos x, , …(i), , On differentiating, we get, [(2 sin x + 3 cos x) (λ cos x − 6 sin x), − (λ sin x + 6 cos x) (2 cos x − 3 sin x)], f ′ (x) =, (2 sin x + 3 cos x)2, The function is monotonic increasing, if f ′ (x) > 0., ⇒, 3λ (sin 2 x + cos 2 x) − 12 (sin 2 x + cos 2 x) > 0, ⇒, 3λ − 12 > 0, (Q sin 2 x + cos 2 x = 1), ⇒, λ >4, 2. Given that, xy = 1, Let, A=x+ y, 1, A=x+, x, On differentiating wrt x, we get, dA, 1, d 2A 2, =, = 1 − 2 and, dx, x, dx2 x3, , …(i), …(ii), , For maxima or minima, put, ⇒, , dA, 1, = 0 ⇒1 − 2 = 0, dx, x, x2 = 1 ⇒ x = 1 , − 1, , d 2A , 2, = =2 >0, 2, dx x = 1 1, , Now,, , Thus, function is minimum at x = 1, y = 1, So, minimum value of A = 1 + 1 = 2, 3. Q g (x) = min (x, x2), y-axis, y = x2 y = x, , (0,0), , x-axis, , ∴ g (x) is an increasing function.
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384, , NDA/NA Mathematics, , 12. Given that, f (x) = sin x − bx + c, Now,, f ′ (x) = cos x − b, For function to be decreasing,, f ′ (x) < 0, ∴, cos x − b < 0 ⇒ cos x < b, We know that, cos x lies between −1 and 1., ∴, 1 < b ⇒ b >1, 13. We have,, s = t3 − 4 t 2 − 3 t, On differentiating wrt t, we get, ds, = 3t 2 − 8t − 3, dt, ds, Since, velocity is zero i.e.,, =0, dt, ⇒, 3t 2 − 8t − 3 = 0, ⇒, (t + 1) (t − 3) = 0, ⇒, t = 3 (Q t ≠ − 1), Again, differentiating, we get, d 2s, =6t −8, dt 2, 2, d s, = 6 (3) − 8 = 10, ∴ Acceleration = 2, dt ( t = 3 ), 14. We have,, f (x) =, , x, , x ∈ (−∞ , ∞ ), 1 + |x|, , x, , x>0, 1 + x, , f (x) = 0, , x=0, x, 1 − x , x < 0, , When x > 0,, ∴, , x, 1+ x, (1 + x) (1) − x (1), 1, f ′ (x) =, =, >0, (1 + x)2, (1 + x)2, f (x) =, , ∴ It is increasing, when x > 0., When x < 0,, x, ∴, f (x) =, 1−x, (1 − x) (1) − x (−1), f ′ (x) =, (1 − x)2, 1, =, >0, (1 − x)2, ∴ It is increasing, when x < 0., ∴ Function f (x) is monotonic increasing in the interval, (−∞ , ∞ ) except at x = 0., 15. Given, f (x) = cos x, ⇒, f ′ (x) = − sin x, Since, f (x) is monotonic decreasing function., Now,, f ′ (x) < 0, ⇒, − sin x < 0, sin x > 0, ⇒, sin x > sin (0), ⇒, x>0, , Clearly, f ′ (x) < 0, when 0 < x < π., Hence, f (x) is decreasing, when 0 < x < π., Alternative Method, y, , x¢, , p, , 2p, , O, , x, , y¢, Graph of cos x, , So, f (x) is monotonically decreasing in (0, π ) by graph., 16. f (x) = x2 + mx + 5, ∴, f ′ (x) = 2x + m, Since, f (x) is an increasing function., ∴, f ′ (x) ≥ 0, ⇒, 2x + m ≥ 0, m, x≥−, ⇒, 2, Given that, f (x) is increasing in the interval x ∈ [1, 2]., m, ∴, x=−, = 1 ⇒ m = −2, 2, Thus, the smallest value of m is –2 for which f (x) is, increasing in 1 ≤ x ≤ 2, 17. Q f (x) = x2 /3 (5 − 2x) = 5x2 /3 − 2x5 /3, 10 −1 /3 10 2 /3, Now,, f ′ (x) =, x, −, x, 3, 3, For critical points, put f ′ (x) = 0, 10 −1 /3 10 2 /3, x, =, x, ⇒ x=1, ⇒, 3, 3, which is the required critical point lie in the interval, [−1, 2]., 4, 18. Let the volume of spherical soap bubble is V = πr3 ,, 3, where r → radius., dV, dr, ⇒, = 4 πr 2, = C (constant) (given) …(i), dt, dt, And the surface area, S = 4πr 2, dS, dr, ⇒, = 4π ⋅ 2 ⋅ r, dt, dt, dS, C, [from Eq. (i)], = 8 πr ×, ⇒, dt, 4 πr 2, dS 2C, dS 1, =, ⇒, ∝, ⇒, dt, r, dt, r, ∴ Rate of change of surface area varies as inversely, proportional to radius., f (b) − f (a ), 19. From mean value theorem, f ′ (c) =, b−a, 1, 3, Given, a = 0 ⇒ f (a ) = 0 and b =, ⇒ f (b) =, 2, 8, Now, f ′ (x) = (x − 1) (x − 2) + x (x − 2) + x (x − 1), ∴, f ′ (c) = (c − 1) (c − 2) + c (c − 2) + c (c − 1), = c2 − 3c + 2 + c2 − 2c + c2 − c, ⇒, f ′ (c) = 3c2 − 6c + 2
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385, , Application of Derivative, By definition of mean value theorem,, f (b) − f (a ), f ′ (c) =, b−a, 3, 8, 5, (, /, )−0 3, ⇒, 3c2 − 6c + 2 =, =, ⇒ 3c2 − 6c + = 0, 4, (1 / 2) − 0 4, This is a quadratic equation in c,, 6 ± 36 − 15 6 ± 21, c=, =, =1±, 2 ×3, 6, Since, ‘c’ lies between [0, 1 /2],, 21, (neglecting c = 1 +, c=1 −, ∴, 6, , 21, 6, 21, ), 6, , 20. Time of encounter, 10 + 6 t = 3 + t 2 ⇒ t 2 − 6 t − 7 = 0 ⇒ t = 7, At t = 7 s, moving in a first point,, d, v1 =, (10 + 6 t ) = 6 cm/s, dt, At t = 7 sec, moving in a second point,, d, v2 =, (3 + t 2) = 2 t = 2 × 7 = 14 cm/s, dt, ∴ Resultant velocity = v2 − v1 = 14 − 6 = 8 cm/s, 21. Since, both curves has the point of tangents., ∴ It passes through the same point (1, 0), ⇒, 0 = 12 + a + b, …(i), ⇒, a + b + 1 =0, and, …(ii), 0 = c − 12 ⇒ c = 1, dy, Now,, m1 =, = 2x + a ⇒ m1(1 , 0) = 2 + a, dx, dy, m2 =, = c − 2x ⇒ m2(1 , 0) = c − 2, dx, Since, both the slopes are equal., ⇒, m1 = m2, ⇒, 2 + a = c−2, [from Eq. (ii)], ⇒, a =1−4, ⇒, a = −3,b =2,c=1, , ⇒, , − 2 = 10 ⋅, , ⇒, , dx, = −5, dt, , dt, dx, , dy, , , given, = 10, , , dt, , 24. Given, v = − x2 log x, On differentiating wrt x, we get, dv, x2, = − 2x log x −, = − 2x log x − x, dx, x, For maximum or minimum value of velocity,, dv, put, = 0 ⇒ − 2x log x − x = 0, dx, 1, ⇒, log x = −, ⇒ x = e−1/ 2, 2, d 2v, 2x, Now,, =−, − 2 log x − 1, x, dx2, = − 3 − 2 log x, At, x = e−1/ 2, d 2v, 1, = − 3 − 2 − = − 2, maxima, 2, dx2, ∴ At x = e−1/ 2, the velocity is maximum., 25. The given function is, f (x) = (x − 1) ex + 1, ∴, f ′ (x) = (x − 1) ex + ex = (x − 1 + 1) ex = xex, Thus, it is clear that, f (x) increases for all x > 0 and, decreases for all x < 0. So, the function f (x) is positive for, all x > 0., 26. Q f (x2) = x3, Let, ⇒, ∴, ⇒, ⇒, , t = x2 ⇒ x = t, f (t ) = ( t )3 = t3/ 2, f (x) = x3/ 2, 3, f ′ (x) = x1/ 2, 2, 3 1/ 2 3, f′ (4) = ⋅ 4 = ⋅ 2 = 3, 2, 2, , 22. The given function is f (x) = x2 − 6x + 8 and internal [2, 4]., Here, f (x) is continuous in [2, 4], As well as differentiable in ] 2,4 [. Since, f (x) is a, polynomial function, so that every polynomial is, continuous as well as differentiable in its domain., Now, f (2) = (2)2 − 6(2) + 8 = 4 − 12 + 8 = 0, f (4) = (4)2 − 6(4) + 8 = 16 − 24 + 8 = 0, ∴, f (2) = f (4), So, there exist a point c in the interval [2, 4], such that, f ′ (c) = 0 ⇒ 2c − 6 = 0 ⇒ c = 3, which lies in the interval [2, 4]., , 27. Q The surface area of sphere, S = 4πr 2, (given), dS 8πr dr, …(i), =, ⇒, dt, dt, and the volume of the sphere, 4, V = πr3, 3, dr, dV 4, dr, = 4πr 2, ⇒, = π ⋅ 3r 2, dt, dt 3, dt, 4πr 2 dS 1 dS, [from Eq. (i)], =, ⋅, = r, 8πr dt 2 dt, , 23. We have, x2y3 = 27, On differentiating wrt x, we get, dy, 2xy3 + x3 3 y2, =0, dx, dy, 2y, =−, ⇒, dx, 3x, 2 3, dy, = − ⋅ = −2, ⇒, , dx (1, 3 ), 3 1, dy dy dt, Q, =, ⋅, dx dt dx, , 28. We have,, y = f (ex ), On differentiating Eq. (i) wrt x, we get, dy, d (ex ), = f ′ (ex ) ×, = f ′ (ex ) ⋅ ex, dx, dx, Again, differentiating, we get, d 2y, = f ′ ′ (ex ) ex ⋅ ex + f ′ (ex ) ⋅ ex, dx2, d 2y, ⇒, = f ′ ′ (ex )e2x + f ′ (ex ) ex, dx2, , …(i)
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387, , Application of Derivative, 35. Q f (x) = 9 − x2, On differentiating wrt x, we get, 1, x, f ′ (x) =, (−2x) = −, 2, 2 9−x, 9 − x2, f (x) is increasing, if f ′ (x) > 0., x, −, > 0, ⇒, 9 − x2, x, <0, ⇒, 9 − x2, ⇒, , x(9 − x2)3/ 2, <0, (9 − x2)2, , ⇒, ⇒, ⇒, , x(9 − x2)3/ 2 < 0, x < 0 or (9 − x2)3/ 2 < 0, x < 0 or (x2 − 9) > 0, +, –∞, , ⇒, ⇒, , –, –3, , 3, , +, +∞, , x < 0 or −3 < x < 3, −3 < x<0, , 36. Q f (x) = x (7x − 6), = 7x3/ 2 − 6x1/ 2, 3, 21 1/ 2, 1, ∴ f ′ (x) = 7 × x1/ 2 − 6 × x−1/ 2 =, x − 3x−1/ 2, 2, 2, 2, Since, tangent is parallel to x-axis., ∴, f ′ (x) = 0, 21 1/ 2, x = 3x−1/ 2, ⇒, 2, 3 ×2 2, x=, =, ⇒, 21, 7, 2 4, x3, +, x +K, tan x, 3, 15, 37. A. f (x) =, =, x, x, 2 3, x2, f (x) = 1 +, +, x +K, 3 15, π, f (x) is increasing for x in 0, ., 2, B., f (x) = (x − 1) − log x, 1 x−1, f ′ (x) = 1 − =, x, x, For f (x) to be increasing., f ′ (x) > 0 ⇒ x > 1, x3, x5, x−, +, −K, sin x, 3! 5!, C., f (x) =, =, x, x, x 2 x4 x 6, =1 −, +, −, + ..., 3! 5! 7!, π, f (x) is decreasing for x in 0, ., 2, x+, , x2 x3 x4, +, −, +K, 2, 3, 4, x, x x2, =1 − +, −K, 2 3, f (x) is decreasing for x > 0., , log (1 + x), D. f (x) =, =, x, , x−, , 38. Given that, f (x) = x3 − 1 ⇒ f ′ (x) = 3x2, For increasing function, f ′ (x) > 0, ⇒, 3 x2 > 0, ∴ The function will increase, for x ∈ R, i.e., the function, will increase in [−1, 1]., The curve does not have any sign change in the interval, (−1, 1)., ∴ f (x) has no root in (−1, 1)., 39. We have, f (x) = tan x − 4x, ⇒ f ' (x) = sec2 x − 4 = 1 + tan 2 x − 4 = tan 2 x − 3, Now,, f ' (x) > 0, ⇒, tan 2 x − 3 > 0, (tan x − 3) (tan x + 3) > 0, ⇒, tan x < − 3 and tan x > 3, π, π, and x >, ⇒, x<−, 3, 3, π, π, and x > ., Thus, f (x) is increasing for x < −, 3, 3, π, Hence, f (x) is not increasing in the interval − < x < 0., 3, Hence, statement I is false., Consider the statement II,, Let, f (x) = log (1 + x) − x, In order that, log (1 + x) is defined, we must have,, 1 + x ≥ 0 i.e.,, x > −1, Now,, 1, −x, f ' (x) =, −1 =, 1+ x, 1+ x, −x, <0 ⇒ x >0, ∴, f ' (x) < 0, ⇒, 1+ x, −x, and, f ' (x) > 0 ⇒, >0 ⇒ x <0, 1+ x, Therefore, f decreases on (0, ∞) and increases on (− 1, 0)., That is, f (x) ≤ f (0), if x ≥ 0 and f (x) ≥ f (0) if − 1 < x ≤ 0, Thus,, f (x) ≤ f (0) if x ∈ (− 1, ∞ ), Hence,, log (1 + x) − x ≤ 0 for x ∈ (0, ∞ )., So, statement II is true., 40. We know that, the area of the largest rectangular field, to be enclosed with 200 m of fencing is possible, if length, and breadth of the rectangular field are equal., ∴, 2(x + x) = 200, 200, ⇒, x=, = 50 m, 4, ∴ Area of the largest rectangular field, = 50 × 50 = 2500 m2, Alternative Method, Let length and breadth of rectangular field are x and y,, respectively., ∴, 2(x + y) = 200, ⇒, y = 100 − x, and area, A = xy = x (100 − x) = 100x − x2, dA, = 100 − 2x, ∴, dx, dA, Put, = 0, for maxima or minima, dx, 100 − 2x = 0, ⇒, x = 50
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388, , NDA/NA Mathematics, d 2A, = − 2x, dx2, , Now,, , ⇒, , d 2A , = − 100 < 0, 2, dx x = 50, , ⇒, , Hence, at x = − 1, y is minimum., Thus, A is true but R is false., 45. Q y = x2e− x, , ∴ A is maximum at x = 50, Hence, the required area = 50(100 − 50), = 50 × 50, = 2500 m2, , (given), , 43. Q Area of first square,, A1 = x2, and area of second square, A2 = y2, = (x + x2)2, (Q y = x + x2), = x2 + x4 + 2x3, dA1, Now,, = 2x, dx, dA2, and, = 2x + 4x3 + 6x2, dx, dA2 2x + 4x3 + 6x2, Hence,, =, dA1, 2x, = 1 + 2 x2 + 3 x, 44. A. y = x e, , x, , ∴, , dy, = x ex + ex, dx, , dy, = 0, for maxima or minima, dx, xex + ex = 0, ⇒, x= −1, d 2y, Now,, = 2ex + xex, dx2, , Put, , dy, = 2xe− x − x2e− x, dx, Function will be increasing, if, dy, >0, dx, −x, 2 −x, ∴, 2xe − x e > 0, ⇒, e− x (− x2 + 2x) > 0 ⇒ − x2 + 2x > 0, ⇒, x ∈ (0, 2)., Both A and R are individually true and R is the correct, explanation of A., ⇒, , (given), 41. Q s = 64t − 16t 2, On differentiating wrt t, we get, ds, = 64 − 32t, dt, ds, Put, = 0, for maximum height, dt, 64 − 32t = 0, ⇒, 64 = 32t, ⇒, t = 2s, 42. P (x) = − 3500 + (400 − x) x, On differentiating wrt x, we get, P′ (x) = 400 − 2x, Put, P′ (x) = 0, for maxima or minima, 400 − 2x = 0, ⇒, x = 200, Now,, P′ ′ (x) = − 2x, ⇒, P′ ′ (200) = − 2 < 0, ∴ P (x) is maximum at x = 200, Hence, required number of items = 200, , d 2y, = e−1 (2 − 1) > 0, 2, dx x = − 1, , 46. Given that, the tangent to the curve,, y = x3 − x2 − x + 2 at (1, 1) is parallel to the x-axis., dy, ∴, = 3 x2 − 2 x − 1, dx, dy, =3 −2 −1 =0, , dx (1, 1), ∴ The equation of tangent at (1, 1) is given by, dy, y−1 = , (x − 1) ⇒ y − 1 = 0 ⇒ y = 1, dx (1, 1), dy, Which is parallel to x-axis and , =0, dx (1, 1), ∴ Both A and R are true and R is the correct explanation, of A., , Solutions (Q. Nos. 47-49), 1 4 x2 − 1, =, x, x, ∴ f ' (x) ≥ 0, if (i) 4x2 − 1 ≥ 0 and x > 0 or, (ii), 4x2 − 1 ≤ 0 and x < 0, Now,, 4x2 − 1 ≥ 0 ⇒ (2x − 1) (2x + 1) ≥ 0, 1, 1, or x ≥, x≤−, ⇒, 2, 2, 1, ∴ (i) holds, if, x≥, 2, Again,, 4x2 − 1 ≤ 0 ⇒ (2x − 1) (2x + 1) ≤ 0, 1, 1, ⇒, − ≤x≤, 2, 2, 1, ∴ (ii) holds, if − ≤ x < 0, 2, 1, 1, Hence, f increases for x ≥ or for − ≤ x < 0., 2, 2, Similarly, it can be shown that, f decreases for, 1, 1, or for x ≤ −, 0≤x≤, 2, 2, We have,, , f ' (x) = 4x −
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20, Indefinite Integration, Indefinite Integration, If f ′( x ) is derivative of f ( x ), then f ( x ) is primitive or, anti-derivative or integration of f ′ ( x ). So, differentiation, and integration are inverse to each other., d, e. g. ,, (sin x ) = cos x , so integration of cos x is sin x., dx, d, (sin x + c) = cos x, so integration of cos x is sin x + C., dx, d, ( f ( x ) + c) = F ( x ) ⇒ f ( x ) + C is anti-derivative of, dx, F ( x )., Let ∫ F ( x ) dx = f ( x ) + C, , ∫a, , 5., , ∫ sin x dx = − cos x + C, ∫ cos x dx = sin x + C, 2, ∫ sec x dx = tan x + C, , 6., 7., , ‘∫ ’ is integral sign and ∫ F ( x ) dx means integration of, F ( x ) with respect to x and it is called indefinite integration., , 2., 3., , ∫, ∫, , { f ( x ) ± g( x )} dx =, , ∫, , f ( x ) dx ±, , ∫, , 9., , ∫ sec x tan x dx = sec x + C, ∫ cosec x cot x dx = − cosec x + C, ∫ cot x dx = log|sin x| + C, ∫ tan x dx = − log|cos x| + C, ∫ sec x dx = log|sec x + tan x| + C, , 10., 11., , 13., , f(x), +C, n +1, , 4., , ∫ ( f ( x )), , 5., , ∫, , f ( x ) dx = F ( x ) + C, , ⇒, , ∫ f ( ax + b) dx =, , n, , f ( x ) dx =, , n +1, , f ( ax + b), +C, a, , Useful Formulae, 1., , ∫x, , n, , dx =, , n +1, , x, + C,, n +1, , dx, 2. ∫, = loge |x| + C, x, 3. ∫ ex dx = ex + C, , x dx = − cot x + C, , ∫ cosec x dx = log|cosec x − cot x| + C, x, = log tan + C, 2, , kf ( x ) dx = k ∫ f ( x ) dx , where k is a constant., , d, ( f ( x ) dx ) = f ( x ) + C, dx ∫, , 2, , π x, = log tan + + C, 4 2, , 14., , g( x ) dx, , dx =, , ∫ cosec, , Properties of Indefinite Integration, 1., , x, , 8., , 12., , where, C is constant of integration., , ax, +C, log e a, , 4., , x, = sin−1 + C, a, a −x, dx, , 15., , ∫, , 16., , ∫ x 2 + a 2 = a tan, , 17., , ∫x, , 18., , ∫ x 2 − a 2 = 2a logx + a + C , x > a, , 19., , ∫ a2 − x2, , 20., , ∫, , ( n ≠ − 1), , 2, , 2, , 1, , dx, , dx, x −a, 2, , 2, , 1, , dx, , dx, , dx, a + x2, 2, , =, , =, , −1, , x, +C, a, , 1, x, sec−1 + C, a, a, x − a, , 1, a+x, log, + C, a > x, 2a, a−x, , = log|x + a 2 + x 2| + C
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390, , 21., , NDA/NA Mathematics, , ∫, , dx, x − a2, 2, , sin 2 x + cos2 x, dx, sin 2 x cos2 x , , = log|x + x 2 − a 2| + C, , Solution (b) Let I = ∫ , , = ∫ (sec2 x + cosec2x ) dx, , x, = cosh−1 + C, a, , = tan x − cot x + C, , 1, 1, x, x a 2 − x 2 + a 2 sin−1 + C, a, 2, 2, 1, 23. ∫ a 2 + x 2 dx = x x 2 + a 2, 2, 1, + a 2 log|x + a 2 + x 2| + C, 2, 1, 2, 2, 2, 24. ∫ x − a dx = x x − a 2, 2, 1, − a 2 logx + x 2 − a 2 + C, , , 2, 25. ∫ eax + b ( af ( x ) + f ′( x )) dx = eax + b f ( x ) + C, 22., , ∫, , a 2 − x 2 dx =, , 26., , ∫e, , ax, , 27., , ∫e, , ax, , sin bx dx =, cos bx dx =, , eax, , ( a cos bx + b sin bx ) + C, , =∫, , x, , b, , cos bx − tan−1 + C, 2, 2, , , a, a +b, , Example 1. The value of, , dx, , ∫, , 2 x − 1, tan−1, +C, 3 , 3, 2, 2 x − 1, (b) −, tan−1, +C, 3 , 3, , (a), , x2 − x + 1, , =a∫, , dx, , dx, a −x, 2, , 2, , x 1, +, a 2, , −∫, , ∫, , x, a − x2, 2, , −2 x, a2 − x2, , dx, , dx, , 2, 2, x 1 a −x, + ⋅, +C, a 2, 1/ 2, x, = a sin −1 + a2 − x2 + C, a, , Geometrical Interpretation of, Indefinite Integral, , 1, dx, dx, =∫, 2, 2, x − x+1, 3, 1, x− + , 2, 2 , , ∫ f ( x ) dx = F( x ) + C = y, , 2, , , 1, x− , 1, 2 + C, =, tan −1 , 3, 3 , , , 2 , 2, 2, 2x − 1, =, tan −1 , +C, 3 , 3, , (a) tan x + cot x + C, (b) tan x − cot x + C, (c) tan x − sec x + C, (d) None of the above, , a2 − x2, , = a sin −1, , 1, , Example 2. The value of ∫, , a−x, , is, , 2, , Solution (a) Let I = ∫, , a−x, dx, a+ x, , = a sin −1, , 2 x − 1, tan−1, +C, 3 , 3, (d) None of the above, , (c), , (d) None of the above, , 2, , a +b, ea, , x, + a 2 − x2 + C, a, x, (c) sin−1 + a 2 + x 2 + C, a, (b) sin−1, , Solution (a) Let I = ∫, , eax, , =, , x, + a 2 − x2 + C, a, , ( a sin bx − b cos bx ) + C, , a +b, 2, , (a) a sin−1, , 2, , 2, , a−x, dx., a+x, , Example 3. Find the value of ∫, , dx, 2, , sin x cos2 x, , (say), , y, y = x2 + 2, y = x2 + 1, y = x2, y = x2 – 1, y = x2 – 2, x, , x′, , is, y′, , y = F ( x ) + C represents a family of curves. By giving, different values to C, we get different members of family., These members can be obtained by shifting any of the curves, parallel to itself.
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391, , Indefinite Integration, , 1, dt = log|t| = log| f ( x )| + C, t, (ii) Integrals of the form ∫ ( f ( x ))n f ′ ( x ) dx , then put, , Integrals of the Form, ∫ f ( ax + b ) dx, ∫, , f ( x ) dx = φ ( x ) + C , then, 1, f ( ax + b) dx = φ ( ax + b) + C, a, , If, , ∫, e.g.,, , ∫, , ( ax + b)n dx =, , ( ax + b)n + 1, + C, n ≠ − 1, a ( n + 1), , 1, tan ( ax + b) + C, a, 1, ∫ tan ( ax + b) dx = − a log |cos ( ax + b)| + C, , 3., 4., 5., 6., , ∫ sinh x dx = cosh x + C, ∫ cosh x dx = sinh x + C, 2, ∫ sech x dx = tanh x + C, 2, ∫ cosech x dx = − coth x + C, ∫ sech x tanh x dx = − sech x + C, ∫ cosech coth x dx = − cosech x + C, , Example 4. The value of ∫, , ( e x − e − x), ( e x + e − x) log cos h x, , 1, f ( x) + C, 2, 1, (c) f ( x) + C, 2, , ∴, , dx is, , Solution (b) Put f ( x) = t ⇒ f '( x) dx = dt, , put, put, , x = a tan θ or a cotθ, x = a sec θ or a cosec θ, , (iv) For the terms in the form a 2 − x 2 or a 2 − x 2 ,, , ( ex − e− x ), dx, ( e + e− x) log cosh x, 1, log(cosh x) = t ⇒, sinh xdx = dt, cosh x, , ex − e− x, dx = dt, ex + e− x, , ex − e− x, ex + e− x , and cosh x =, Q sinh x =, , 2, 2 , , 1, I = ∫ dt = log t + C, t, = log(log cosh x) + C, , f ′ ( x )dx = dt, , 1, dt = 2 t + C, t, = 2 f ( x) + C, , I=∫, , (a) If m is an odd integer, put cos x = t, (b) If n is an odd integer, put sin x = t, (c) If m + n is negative even integer, then put, tan x = t or cot x = t, (ii) For the terms in the form x 2 + a 2 or x 2 + a 2 ,, , put, x = a sinθ or a cosθ, (v) If a + x and a − x both are present, then, put, , f ′ (x), dx , then put f ( x ) = t, f(x), , x = a cos 2θ, , (vi) For the type ( x − a ) ( b − x ) and, put, (vii) For the type, put, , (x − a), ( x − b), , x = a cos2 θ + b sin2 θ, x−a, or ( x − a ) ( x − b), x−b, x = a sec2θ − b tan2 θ, , (viii) For the type ( x 2 + a 2 ± x )n or ( x ± x 2 − a 2 )n ,, put the expression, , Integration by Substitution, ⇒, , (d) 2f ( x) + C, , (iii) For the terms in the form x 2 − a 2 or x 2 − a 2 ,, , x, , (i) Integrals of the form ∫, , dx is, , (i) Integral of the form ∫ sinm x cosn x dx, , (d) log(log(cosh x)) + C, , ⇒, , f ( x), , (b) 2 f ( x) + C, , (a), , (c) 2 log( e x − e − x) + C, , Put, , f ′( x), , ( n ≠ − 1), , Some Important Standard, Substitutions, , (a) log(tanh x) + C, (b) 2 log( e x + e − x) + C, , Solution (d) I = ∫, , f ( x ) = t ⇒ f ′ ( x ) dx = dt, ( f ( x ))n + 1, tn + 1, t n dt =, +C=, +C, ( n + 1), n +1, , Example 5. The value of ∫, , ∴, , Integration of Hyperbolic, Functions, 2., , ∫, , ∴, , 2, ∫ sec ( ax + b) dx =, , 1., , ∫, , ∴, , (ix), , within the bracket = t, dx, , ∫, , ( where, m + n = 2),, ( ax + b)m ( cx + d )n, ax + b, put, =t, cx + d
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392, , NDA/NA Mathematics, x2 + 1, , (x) Integral of the form ∫, , dx and ∫, , x2 − 1, , dx,, x +1, x4 + 1, 1, 1, put, x − = t and x + = t, respectively., x, x, (xi) Integral of the form, f, ∫ (sin 2x ) (sin x + cos x ) dx, put, and, ∫ f (sin 2x ) (sin x − cos x ) dx,, sin x − cos x = t and sin x + cos x = t, respectively., 4, , 1, 1 dx, (xii) Integral of the form ∫ f n n + 1 , put n = t, x x, x, dx, (xiii) Integral of the form ∫, ,, ( ax + b) lx 2 + mx + n, 1, put, ax + b =, t, (xiv) Integral of the form, dx, , ∫ ( ax 2 + bx + c), , lx + mx + n, 2, , (xv) Integral of the form, dx, , ∫ ( ax 2 + b), , lx + m, 2, , and, , and, , + bx + c, , put a sin x + b cos x = A ( c sin x + d cos x ), + B ( c cos x − d sin x ), Now, compare the coefficients of sin x and cos x and, we get similar form as in the above., , = t2, , , put x =, , dx, , ∫ ax 2, , f ( x ) = cex + de− x, a sin x + b cos x, (xxi) Integral of the form ∫, dx,, c sin x + d cos x, where,, , 1, t, , Example 6. The value of ∫ cos3x e log(sin x ) dx is, cos4 x, +C, 4, (c) sin4 x + C, , lx + m, , dx , ∫, , , put lx + m = t 2, , Solution (a) Let I = ∫ cos x elog(sinx) dx = ∫ cos3x sin x dx, , ax 2 + bx + c, , I = ∫ t 3 dt =, , ∴, , dx,, , Example 7. The value of ∫, , Find A and B by comparing the coefficients of x and, constant term on both the sides of equation. Then,, ax + b, px + q, ∫ ax 2 + bx + c dx = A ∫ ax 2 + bx + c dx, 1, ax + bx + c, 2, , dx, , Now, we can integrates it easily. Similarly, other, two cases., (xviii) Integrals of the form, dx, (a) ∫, a cos x + b sin x + C, x, , put tan = t, 2, 2 x, 2 x, + b cos, +C, a sin, 2, 2, , t4, +C, 4, , cos4 x, +C, 4, , =, , 2, ∫ ( px + q ) ax + bx + c dx, , In such integrals put px + q = A, (differential coefficient of ax 2 + bx + c) + B., , (b) ∫, , (d) None of these, 3, , Put cos x = t ⇒ sin x dx = dt, , px + q, , + B∫, , (b) cos4 x + C, , (a), , ( ax + b) lx + m, , ∫ ( ax 2 + bx + c), , (xvii) Integral of the form, px + q, , Now, find A and B by comparing the coefficients of, ex and e− x ., aex + be− x, f ′ (x), Now, ∫ x, dx = A ∫ dx + B ∫, dx,, f (x), ce + de− x, , dx, , (xvi) Integral of the form ∫, , In all such integration, put tan x = t., aex + be− x, (xx) Integrals of the form ∫ x, dx, ce + de− x, put aex + be− x = A ( cex + de− x ) + B ( cex − de− x ), , lx 2 + mx + n, , ax 2 + bx + c, , put, , (xix) Integrals of the form, dx, (a) ∫, a cos 2x + b sin 2x + C, dx, (b) ∫ 2, a sin2 x + b cos2 x + C, , dx, , (1 + x ) 1 − x 2, 1− x 2 , 1 π, +C′, − cot −1, (a), 2x , 2 2, , , , , 1− x 2 , 1 π, +C′, − cot −1, (b) −, 2x , 2 2, , , , 2, , π, 1− x , +C′, (c) − cot −1, 2x , 2, , , , (d) None of the above, , Solution (b) Let I = ∫, , dx, , Put, , 2, , dx, , (1 + x ) 1 − x2, 1, 1, x = ⇒ dx = − 2 dt, t, t, 2, , is
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393, , Indefinite Integration, 1, dt, − t dt, t2, ∴, =, I=∫, 1, 1 ∫ (t 2 + 1) t 2 − 1, , 1 + 2 1 − 2, , t , t, 2, 2, Again, put t − 1 = u, −, , ⇒, , 2 t d t = 2 u du, I=−∫, , ⇒, , u du, (u + 2) u, 2, , 2, , =−∫, , 1, du, u + ( 2) 2, 2, , 1, u , =−, tan −1 + C′, 2, 2, =−, , 1 − x2 , 1, + C′, tan −1 , 2x , 2, , , , =−, , 1, 2, , π, 1 − x2 , + C′, − cot −1, 2 x , 2, , , , , 1 − x2 , , 1, + C′ − π , cot −1, 2x , , 2, 2, 2, , , 2x , 1, +C, =, tan −1 , 1 − x2 , 2, , =, , π , , where, C = C′ −, , , 2 2, , Example 8. The value of ∫, , x+2, , ( x + 3x + 3) x + 1, , , x, 2, (a), tan−1, +C, 3, 3 ( x + 1) , x , 2, (b), tan−1, +C, 3, x + 1, , , 1, x, (c), tan−1, +C, 3, 3 ( x + 1) , (d) None of the above, , Solution (a) Let I = ∫, , 2, , dx is, , x+2, dx, ( x2 + 3x + 3) x + 1, , I = 2∫, , t2 + 1, , dt, + t2 + 1, 1, 1+ 2, t, = 2∫, dt = ∫, 1, t2 + 1+ 2, t, t, , 4, , 1+, , 1, t2, 2, , 1, t − + 3, t, , dx, , ∫ 5 + 4 cos x = A tan, , −1, , value of A is, (a) 2, 1, (c), 3, , x, , B tan + C , then the, , 2, , (b) 3, 2, (d), 3, , Solution (d) We have,, dx, x, , = A tan −1 B tan + C, , 5 + 4 cos x, 2, dx, 1, Now,, dx, ∫ 5 + 4 cos x = ∫, , 2 x, 1 − tan , 2, 5+ 4, 2, 1 + tan x , , 2, x, sec2, 2, =∫, dx, x, tan 2 + 9, 2, 1, x, 2 x, Put tan = t ⇒, sec dx = dt, 2, 2, 2, 2 dt, I= ∫ 2, ∴, t + 32, 2, t, = tan −1 + C, 3, 3, x, x, , , tan , tan , 2, 2, −1, −, 1, 2 + C = tan , 2 + C, = tan , 3, 3, 3 , 3 , , , , , I=∫, , ⇒ A tan, , ⇒ A=, , −1, , x, , tan , x, 2, , −1, 2, +C, B tan + C = tan , , 2, 3, 3 , , , , 2, 3, , Example 10. The value of, , dx, , ∫ 1 + 2 sin x + cos x is, , x, (a) log(2 + 4 tan ) + C, 2, x, 1, (b) log(2 + 4 tan ) + C, 2, 2, x, 1, (c) log(2 − 4 tan ) + C, 2, 2, (d) None of the above, , Put x + 1 = t 2 ⇒ dx = 2t dt, ∴, , Example 9. If, , dt, , 1, t − , , , 2, x, 2, −1, +C, =, tan −1, tan t + C =, 3, 3, 3 ( x + 1) , 3, , , , Solution (b) Let I = ∫, , dx, 1 + 2 sin x + cos x, , x, 2, cos x =, x, 1 + tan 2, 2, 1 − tan 2, , Put
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394, , NDA/NA Mathematics, , ∴, , Let, ∴, , dx, x, x, 1 − tan 2, 4 tan, 2, 2, +, 1+, x, x, 1 + tan 2, 1 + tan 2, 2, 2, x, sec2 dx, 2, =∫, x, x, x, 1 + tan 2 + 4 tan + 1 − tan 2, 2, 2, 2, x, sec2, 2, =∫, dx, x, 2 + 4 tan, 2, x, 2 x 1, 2 + 4 tan = t ⇒ 4 sec, ⋅ dx = dt, 2, 2 2, 1 dt, I= ∫, 2 t, 1, 1, x, , = log t + C = log 2 + 4 tan + C, , 2, 2, 2, I =∫, , Integration by Parts, If u and v are the differentiable functions of x, then, , d, , ∫ u ⋅ v dx = u ∫ vdx − ∫ dx ( u ) ( ∫ vdx ) dx, i.e., the integral of the product of two functions = (First, Function) × (Integrals of the second function) − Integral of, {(Differential of first function) × (Integral of second, function)}., , Assume that, ⇒, ∴, , x, dx, x+1, 1, x =t ⇒, dx = dt, 2 x, I1 = ∫, , Let, , dx = 2 t dt, t ⋅ 2t, dt, I1 = ∫, 1+ t2, 1 + t 2, 1 , −, =2 ∫ , dt = 2 t − tan −1 t, 2, 1 + t 2, 1 + t, , ∴, , = 2 x − tan −1 x + C, 1, I = x tan −1 x − (2 x − tan −1 x) + C, 2, 1, = x tan −1 x − x + (tan −1 x) + C, 2, , Example 12. The value of, , %, %, %, , If the integral contains a single logarithmic or single inverse, trigonometric function take unity as the second function., If the integrals of both the functions are known, the function which, is easy to integrate is taken as the second function., In certain cases, integration by parts will lead to a simple equation, involving the integral. Solve the equation and determine the, integral., , Example 11. The value of, , 1, (tan−1 x ) + C, 2, 1, (b) x tan−1 x − x + (tan−1 x ) + C, 2, (c) x tan−1 x − x + tan−1 x + C, (d) None of the above, , Solution (b) Let I = ∫ 1⋅ tan−1 x dx, = tan −1 x ⋅ x − ∫, , 1, 1, ⋅, ⋅ x dx, 1+ x 2 x, , e x (1 + sin x), dx is, (1 + cos x), , x, +C, 2, x, (c) 2 e x tan + C, 2, , (b) e x tan x + C, (d) None of these, , ex, ex sin x, dx + ∫, dx, 1 + cos x, 1 + cos x, 1, x, x, = ∫ ex ⋅ sec2 dx + ∫ ex tan dx, 2, 2, 2, x, = ex tan + C, 2, , Solution (a) Let I = ∫, , Example 13. The value of, (a), (c), , − ex, ( x + 1) 2, ex, ( x + 1) 2, , +C, +C, , ∫e, , x, , x −1 , , dx is, ( x + 1)3 , ex, (b), +C, ( x + 1), (d) None of these, , x + 1 − 2 x, e dx, ( x + 1)3 , , Solution (c) I = ∫ , , 1, ex, 2 x, e, dx, =∫ , −, =, +C, , 2, ( x + 1) 2, ( x + 1)3 , ( x + 1), , −1, ∫ tan x dx is, , (a) x tan−1 x + x +, , ∫, , (a) e x tan, , How to Choose Ist and IInd Functions, If two functions are of different types take the function, as Ist which comes first in the word ILATE, where I stands, for inverse circular function, L stands for logarithmic, function, A stands for algebraic functions, T stands for, trigonometric functions and E for exponential functions., , x, 1, dx, 2 ∫ 1+ x, , = x tan −1 x −, , Example 14. The value of ∫ x log x dx is, 1, x2, log x + x 2 + C, 2, 4, x2, 1, (b) log x − x 2 + C, 2, 4, x2, (c) log x + x 2 + C, 2, (d) None of the above, , (a)
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395, , Indefinite Integration, , Solution (b) I = ∫ x log xdx, II, , I, , d, = log x {∫ xdx} − ∫ { (log x ) ⋅ ∫ xdx}dx, dx, 1 x2, 1, x2, x2, = log x . − ∫ . dx = log x − ∫ xdx, 2, 2, 2, x 2, x2, 1, 1 x2, x2, log x − + C = log x − x2 + C, =, 2, 4, 2, 2 2, , Integration by Partial Fraction, P( x), , where P ( x )andQ( x )are, If a function is of the form, Q( x ), polynomials of degree m and n, respectively., If m < n , this fraction is proper fraction and if m ≥ n, this fraction is improper fraction to integrate first convert, it into proper fraction on dividing it by denominator., P1( x ), P (x) , P( x), , is proper fraction, = M (x) + 1, where,, , Q( x ), Q( x ) , Q( x ), Case I If Q( x ) has non-repeating linear factors, then, P1( x ), A1, A2, An, =, +, +K+, Q( x ) a1x + b1 a2x + b2, an x + bn, A1 , A2 , K , An are independent of x, to find A1 put, b, x = − 1 in the LHS except in the factor( a1x + b1 ), we get A1., a1, Similarly, A2 , A3 , ... , An . Now, integration, , ∫, , P1( x ), dx =, Q( x ), , A, , A, A, ∫ a1x +1 b1 + a2x +2 b2 + K + an x +n bn dx, , P (x), P1( x ), Case II 1, =, Q( x ) ( a1x + b1 ) ( a2x + b2 )n, =, , An + 1, A1, A2, +, +K+, a1x + b1 a2x + b2, ( an x + bn )n, , On comparing, we have to find constants, then, integrate., P1( x ), P1( x ), Case III, =, Q( x ) ( a1 x + b )... ( lx 2 + mx + n ), Now, it is not possible to find real linear factor of, (lx 2 + mx + n), then, P1( x ), ( a1 x + b) ( lx 2 + mx + n ), +, , =, , A, a1x + b, , B (differential coefficient of lx 2 + mx + n ) + C, lx 2 + mx + n, A, B ( 2lx + m ) + C, =, +, a1x + b, lx 2 + mx + n, , Find constants A, B and C by comparing them and, then integrate., , Example 15. The value of ∫, (a) log, , ( x + 2) 2, +C, ( x − 1), , (b) log, , (x + 2 )2, +C, ( x + 1), , (c) log, , ( x − 2) 2, +C, ( x + 1), , x, dx is, ( x + 1)( x + 2 ), , (d) None of these, , Solution (b) Let, , x, A, B, =, +, ( x + 1)( x + 2 ) x + 1 x + 2, , ∴, x = A( x + 2 ) + B( x + 1), On equating the coefficient of x and constant terms,, we get, 1= A + B, 0 = 2A + B, On solving, we get A = −1, B = 2, 1, 2, x, ∴ ∫, dx = ∫ −, dx + ∫, dx, ( x + 1)( x + 2 ), ( x + 1), ( x + 2), = − log x + 1 + 2 log x + 2 + C, ( x + 2) 2, = log, +C, ( x + 1), , Example 16. The value of, , 2x, , ∫ (x2 + 1) (x2 + 2) dx is, , (a) log( x 2 + 1) − log( x 2 + 2 ) + C, (b) log( x 2 − 1) + log( x 2 + 2 ) + C, (c) log( x 2 + 1) + log( x 2 + 2 ) + C, (d) None of the above, , Solution (a) We have,, I=∫, Let, ∴, Let, ⇒, , 2x, dx, ( x + 1) ( x2 + 2 ), 2, , x2 = t ⇒ 2x dx = dt, I= ∫, , 1 dt, (t + 1) (t + 2), , 1, A, B, =, +, (t + 1) (t + 2 ) t + 1 t + 2, 1 = A (t + 2) + B (t + 1), , Put t = − 2 and t = − 1, we get, 1 = B ( −1), ⇒, , B = −1, , and, , 1 = A ( +1) = 0, , ⇒, , A =1, , ∴, , I=∫, , 1, 1, dt − ∫, dt, (t + 1), (t + 2 ), , = log (t + 1) − log (t + 2 ) + C, = log ( x2 + 1) − log ( x2 + 2 ) + C
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396, , NDA/NA Mathematics, , Let, , Example 17. The value of, cos θ, , ⇒, , ∫ (2 + sin θ) (3 + 4 sin θ) dθ is, , Put, , 1, 1, log(sin θ + 2) + log(3 + 4 sin θ) + C, 5, 5, 1, 1, (b) log(sin θ + 2) − log(3 + 4 sin θ) + C, 5, 5, 1, (c) log(sin θ + 2) + log(3 + 4 sin θ) + C, 5, (d) None of the above, , (a) −, , Solution (a) Let I = ∫, , 1, A, B, =, +, (2 + t ) (3 + 4 t ) 2 + t 3 + 4 t, 1 = A(3 + 4 t ) + B(2 + t ), 3, t = − , then, 4, 3, 4, 5, , 1 = B 2 − = B ⇒ B =, 4, , 4, 5, , Again, put t = − 2, 1 = A (3 − 8), 1, ⇒, A=−, 5, 1, 4, I=∫−, +, dt, ∴, 5 (2 + t ) ∫ 5 (3 + 4 t ), 1, 4 log (3 + 4 t ), = − log (t + 2) +, +C, 5, 5, 4, 1, 1, = − log (sin θ + 2) + log (3 + 4 sin θ) + C, 5, 5, , cos θ, dθ, (2 + sin θ) (3 + 4 sin θ), , Put sin θ = t ⇒ cos θ dθ = dt, 1, ∴, I=∫, dt, (2 + t ) (3 + 4 t ), , Comprehensive Approach, n, , n, , n, , n, , ∫ [xf ′ ( x) + f ( x)] dx = x f ( x) + C, f ′ ( x), ∫ f ( x) dx = 2 f ( x) + C, ∫ (xe + e ) dx = xe + C, x, x, ∫ e [ f ( x) + f ′ ( x)] dx = e f ( x) + C, x, , x, , n, , n, , x, , n, , ∫ log x dx = xlog x − x + C, If In = ∫ (log x) n dx , then, In = x( log x) n − n ⋅ In − 1, The anti-derivative of every odd function is an even function and, vice-versa.
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Exercise, Level I, 1, , 1. The value of ∫, , ( x − 5), 1, (a), +C, ( x − 5), 2, (c), +C, ( x − 5)3, , 2. The value of ∫, , 2, , dx is, (b) −, , 2 dx, , is, , 1 − 4x 2, , (b) cot−1( 2x ) + C, (d) sin−1( 2x ) + C, , xπ x 2, −, +C, 2, 2, xπ x 2, (c) −, −, +C, 2, 2, 12. ∫ x3 log x dx is equal to, , x2, , (a), , 1, tan−1( x3 ) + C, 3, (d) None of these, , (b), , (c) log(1 + x3 ), , ∫, , cos 2x − 1, dx is equal to, cos 2x + 1, , (a) tan x − x + C, (c) x − tan x + C, 5., , 11. What is the value of ∫ sin−1(cos x ) dx?, , ∫ 1 + x 6 dx is equal to, (a) x3 + C, , 4., , ∫e, , x, , x 4 log x, +C, 4, 1, (c) ( x 4 log x − 4x 2 ) + C, 8, , (a), (b) x + tan x + C, (d) − x − cot x + C, , (1 − cot x + cot2 x ) dx is equal to, , (a) ex cot x + C, , (b) ex cosec x + C, , (c) − ex cot x + C, , (d) − ex cosec x + C, , (a) e2x sin 3x + C, (c) e, , (b) e2x cos 3x + C, , +C, , 7. What is the value of ∫, , (d) e, sin x, , cos x, +C, 2, cos x, (c) −, +C, 2, (a), , 8., , ( x + 1)2, , ∫ x( x 2 + 1), , ∫, , ( 2 sin 3x ) + C, , ∫, , dx?, , 14., , ∫ 13, , 1, ( 4x 4 log x − x 4 ) + C, 16, 1, (d), ( 4x 4 log x + x 4 ) + C, 16, , (b), , dx is equal to, (b) 13x + 1 + C, (d) 14x + 1 + C, , 15. What is the value of ∫ elog x sin x dx?, , (b) 2 cos x + C, , (NDA 2010 I), , (a) e, (sin x − cos x ) + C (b) (sin x − x cos x ) + C, (c) ( x sin x + cos x ) + C, (d) (sin x + x cos x ) − C, log x, , (d) − 2 cos x + C, , 16. If I n = ∫ (log x )n dx , then I n + nI n − 1 is equal to, (b) loge x + 2 tan, , −1, , x+C, , (d) loge { x( x 2 + 1)} + C, , sin−1 x dx is equal to, , (a) cos−1 x + C, 1, (c), +C, 1 − x2, , x, , 13x, +C, log 13, (c) 14x + C, , (NDA 2011 I), , π x2, +, +C, 2, 2, π x2, (d), −, +C, 2, 2, , (b), , 1 + sin, , (a), , dx is equal to, , (a) loge x + C, 1, (c) loge 2, +C, x +1, 9., , x, , 2x, , (NDA 2011 I), , x, dx is equal to, 4, x, x, x, x, , , (a) 8 sin − cos + C (b) sin + cos + C, , , 8, 8, 8, 8, x, x, 1, x, x, , (c) sin − cos + C (d) 8 cos − sin + C, , 8, 8, 8, 8, 8, , 13., , 6. The value of ∫ e2x ( 2 sin 3x + 3 cos 3x ) dx is, 2x, , sin x dx, , ∫ 3 + 4 cos2 x is equal to, (a) log( 3 + 4 cos2 x ) + C, 1, cos x , (b), tan−1 , +C, 3 , 2 3, 1, 2 cos x , (c) −, tan−1 , +C, , 2 3, 3 , 1, 2 cos x , (d), tan−1 , +C, , 2 3, 3 , , 1, +C, ( x − 5), , (d) −2( x − 5)3 + C, , (a) tan−1( 2x ) + C, (c) cos−1( 2x ) + C, 3., , 10., , (b) x sin−1 x + 1 − x 2 + C, (d) x sin−1 x − 1 − x 2 + C, , 17., , (a) x(log x )n, , (b) ( x log x )n, , (c) (log x )n − 1, , (d) n(log x )n, , dx, , ∫ x( x7 + 1) is equal to, x7 , (a) log 7, +C, x + 1, x7 + 1, (c) log 7 + C, x , , x7 , 1, log 7, +C, 7, x + 1, x7 + 1, 1, (d) log 7 + C, 7, x , (b)
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398, 18., , NDA/NA Mathematics, , ∫, , xe, , x, , 2 x − e − 4 xe + C, ( 2x − 4 x + 4) e x + C, ( 2x + 4 x + 4) e x + C, (1 − 4 x ) e x + C, dx, is equal to, 2, x + 4x + 13, x, , (a), (b), (c), (d), 19., , ∫, , 25. The value of the integral ∫, , dx is equal to, x, , (a) log( x 2 + 4x + 13) + C, (c) log( 2x + 4) + C, , (a) log (1 + ex ) + C, ex , (c) − log x, +C, e + 1, , 1, x + 2, tan−1 , +C, 3 , 3, 2x + 4, (d), +C, ( x 2 + 4x + 13)2, , (b), , 5, , 20. What is the value of ∫ ( x 2 + 1) 2 x dx?, , (c) log ( ex + x e ) + C, 28., , x, , 7, , (c) ex (sin x − cos x ) + C, , 1 2, ( x + 1) 2 + C, 7, (d) None of the above, , 29., cos x, , dx?, , a sec x + b tan x + C, a tan x + b sec x + C, a cot x + b cosec x + C, a cosec x + b cot x + C, f ′ (x), 22. ∫, dx is equal to, f ( x ) log [ f ( x )], f(x), (b) f ( x ) ⋅ log f ( x ) + C, (a), +C, log f ( x ), 1, (c) log [log f ( x )] + C, (d), +C, log[log f ( x )], dx, 23. For any positive integer n , ∫ n + 1, is equal to, +x, x, 1, (a), loge ( x n + 1) + C, n, 1 , 1, (b), loge n, +C, n, x + 1, xn , 1, (c), loge n, +C, n, x + 1, xn , 1, (d), loge n, +C, n +1, x + 1, dx, is equal to, x (1 + log x )2, 1, (b) −, +C, 1 + log x, 1, (d), +C, 1 + x2, , 24. The value of the integral ∫, 1, +C, 1+ x, 1, (c), +C, 1 + log x, , (a), , 30., , ex, +C, cos x, ex, (d), (sin x − cos x ) + C, 2, (b), , sin3 x cos x dx is equal to, , cos4 x, +C, 4, sin4 x, (c), +C, 4, , (NDA 2009 II), , (a), (b), (c), (d), , ∫, , (d) log ( x e + ex ) e + C, , sin x dx is equal to, , (a) ex cos x + C, , (c), , (a) −, , ∫e, , 7, , 2, (b) ( x 2 + 1) 2 + C, 7, , 2, , (d) log (1 + ex ) + x + C, , 1, , 7, , a + b sin x, , dx is equal to, 1 + ex, (b) − log (1 + e− x ) + C, , 26. A function f is such that f ′ ( x ) = 6 − 4 sin 2x and, (NDA 2009 II), f( 0) = 3. What is the value of f ( x )?, (a) 6x + 2 cos 2x, (b) 6x − 2 cos 2x, (c) 6x − 2 cos 2x + 1, (d) 6x + 2 cos 2x + 1, e −1, x −1, x, +e, 27. The value of ∫, dx is equal to, x e + ex, (a) x + C, (b) log ( x + e) + C, , (NDA 2012 I), , (a) ( x 2 + 1) 2 + C, , 21. What is the value of ∫, , 1, , ∫e, , log (tan x ), , (b) sin4 x + C, (d), , sin 4x, +C, 4, , dx is equal to, , (a) log tan x + C, (c) tan x + C, , (b) log sec x + C, (d) etan x + C, , 31. What is the value of ∫ sec x ° dx?, (NDA 2009 I), (a) log (sec x ° + tan x ° ) + C, π π, π log tan + , 4 2, (b), +C, 180°, π π, 180° log tan + , 4 2, (c), +C, π, πx , π, 180° log tan +, , 4 360° , (d), +C, π, dx, 32. ∫, is equal to, 1 + e− x, 1, (b), (a) 1 + ex + C, log (1 + ex ) + C, 2, (c) log (1 + ex ) + C, (d) 2 log (1 + ex ) + C, 33., , 3, , ∫ cosh, , x dx is equal to, 1, 1, (a) cosh x + cosh 2 x + C (b) cosh x + sinh3 x + C, 3, 3, 1, 1, 2, (c) sinh x + cosh x + C (d) sinh x + sinh3 x + C, 3, 3
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399, , Indefinite Integration, 34. What is the value of ∫ ( ex + 1)−1 dx?, (a) log( e + 1) + C, (c) − log( e− x + 1) + C, , ∫, , (NDA 2008 II), −x, , (b) log ( e + 1) + C, (d) − ( ex + 1) + C, , x, , sin x + cos x, , dx is equal to, 1 + sin 2x, (a) log (sin x + cos x ) + C (b) x + C, (c) log (1 + sin 2x ) + C, (d) sin x + cos x + C, dx, 36. What is the value of ∫, ?, 2ax − x 2, (a) (1/ a ) sin−1 (( x − a )/ a ) + C, (b) sin−1 [( x − a )/ a ] + C, 35., , (c) log|( x − a ) + 2ax − x 2| + C, , 38. If ∫ f ( x ) dx = g( x ) and also ∫ f ( x ) dx = h( x ), then which, one of the following is correct?, (a) g( x ) = h( x ), (b) g( x ) + h( x ) = constant, (c) g( x ) h( x ) = constant (d) g( x ) − h( x ) = constant, 39. What is the value of the integral, x2 + 1, , , x , , , , ∫e, , dx − ∫, , x2 + 1, , , x , , , , e, , (a) e + C, x, , (c) xe, , (b) e, , x 2 + 1, , , x , , , , x 2 + 1, , , x , , , , +C, , 1, , (d) x + ex + C, , x, , +C, , sin x, , 40. What is the value of ∫, , (d) (1 / a ) log|( x − a ) + 2ax − x 2| + C, , dx ?, , x2, , dx ?, sin x − sin2 α, (a) sin−1( sec α cos x ) + C (b) cos−1( sec α cos x ) + C, (c) sinh −1 ( sec α cos x ) + C (d) cosh −1( sec α cos x ) + C, , 37. If ∫ f ( x ) dx = f ( x )/ 2, then which one of the following is, correct?, (a) f ( x ) = e2x + constant (b) f ( x ) = x + constant, (c) f ( x ) = constant, (d) f ( x ) = e2x, , 2, , Level II, 1., , ∫e, , 3 log x, , ( x 4 + 1)−1 dx is equal to, 1, log( x 4 + 1) + C, 4, (d) None of these, , (a) log( x 4 + 1) + C, , (b), , (c) − log( x 4 + 1) + C, 2., , 3., , 6., , dx, , ∫ sin x − cos x +, , is equal to, 2, 1, 1, x π, x π, (a) −, tan + + C (b), tan + + C, 2 8, 2 8, 2, 2, 1, 1, x π, x π, (d) −, (c), cot + + C, cot + + C, , , 2 8, 2, 8, 2, 2, , ∫ cos1(loge x ) dx is equal to, , x [cos(loge x ) + sin(loge x )], 2, (b) x [cos(loge x ) + sin(loge x )], 1, (c) x [cos(loge x ) − sin(loge x )], 2, (d) x [cos(loge x ) − sin(loge x )], , 4., , ∫ 32 x ( log x ), 3, , (a), (b), (c), (d), , 2, , (a), (b), (c), (d), , dx, sin2 x cos2 x, , x2 − 1, , ∫ x 4 + x 2 + 1 dx is equal to, , (c), , 1, x2 − x + 1, +C, log 2, 2, x + x+1, , 9. The primitive of ∫, , 8x 4 (log x )2 + C, x 4{ 8 (log x )2 − 4 log x + 1} + C, x 4{ 8 (log x )2 − 4 log x } + C, x3 {(log x )2 + 2 log x } + C, , tan x + cot x + C, tan x − cot x + C, (tan x + cot x )2 + C, (tan x − cot x )2 + C, , (a) x f −1( x ) + C, (b) f ( g−1( x )) + C, −1, −1, (c) x f ( x ) − g( f ( x )) + C (d) g−1( x ) + C, , (a) log ( x 4 + x 2 + 1) + C, , dx is equal to, , 5. What is the value of ∫, , 7. If ∫ f ( x ) dx = g( x ) + C, then ∫ f −1( x ) dx is equal to, , 8., , (a), , cos x − 1 x, e dx is equal to, sin x + 1, ex sin x, ex cos x, (b) C −, (a), +C, 1 + sin x, 1 + sin x, ex, ex cos x, (c) C −, (d) C −, 1 + sin x, 1 + sin x, , ∫, , (a), (c), ?, , (NDA 2011 II), , 1, ( x + 4), , 2, , +C, , ex, +C, x+4, , (b) log, (d), , ( x + 3)ex, ( x + 4)2, , x2 − x + 1, x2 + x + 1, , +C, , 1, x2 + x + 1, +C, log 2, 2, x − x+1, , dx is, , (b), (d), , ex, ( x + 4)2, , +C, , ex, +C, x+3, , 1 , , 10. What is the value of ∫ ex x +, dx? (NDA 2011 I), , 2 x, (a) xex + C, , (b) ex ( x ) + C, , (c) 2ex ( x ) + C, , (d) 2 xex + C
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400, 11., , NDA/NA Mathematics, , ∫ x ( ax + b), , −2, , 2, , 2 , x −, a2 , 2 , (b) 2 x −, a , (a), , (c), , dx is equal to, , 18., , b, , log( ax + b) + C, , a, b, x2, , log( ax + b) −, +C, a( ax + b), a, , (a), (b), (c), (d), , x2, b, 2 , , +, +, log(, ax, b, ), +, +C, x, , , a( ax + b), a, a2 , , 12., , ∫, , x, b, 2 , , +C, x + log( ax + b) −, 2 , a( ax + b), a, a , ax/ 2, , dx is equal to, a− x − ax, 1, 1, (a), (b), sin−1( a x ) + C, tan−1( a x ) + C, log a, log a, (d) log ( a x − 1) + C, (c) 2 a − x − a x + C, , 13. What is the value of ∫, , ex (1 + x ), cos2 ( xex ), , (a) xe + C, (c) tan( xex ) + C, , (NDA 2011 II), , (b) cos ( xe) + C, (d) x cosec ( xex ) + C, , x, , 14. What is the value of ∫, , dx?, x, , x4 + 1, x2 + 1, , dx?, , (NDA 2010 I), , x3, x3, − x + 4 tan−1 x + C (b), + x + 4 tan−1 x + C, 3, 3, x3, x3, (c), − x + 2 tan−1 x + C (d), − x − 4 tan−1 x + C, 3, 3, ex, 15. ∫, dx is equal to, ( 2 + ex ) ( ex + 1), ex + 1, ex + 2 , (a) log x, (b) log x, +C, +C, e + 2, e + 1, ex + 1, ex + 2, (d) x, (c) x, +C, +C, e + 2, e + 1, (a), , 16. The anti-derivative of, , ∫, , , 1− x , cos 2 cot−1, dx is, 1+ x , , , equal to, 1, (a) x 2 + C, 2, , 1− x , 1, (b) sin 2 cot−1, +C, 1+ x , 2, , 1, (c) − x 2 + C, 2, 1, (d) x + C, 2, 17. If f ( x ) = cos x − cos2 x + cos3 x − …. ∞, then ∫ f ( x ) dx is, equal to, x, (a) tan + C, 2, 1, x, (c) x − tan + C, 2, 2, , x, (b) x + tan + C, 2, x, (d) x − tan + C, 2, , log (tanh x ) + C, 2 log ( ex + e− x ) + C, log [log (cosh x )] + C, 2 log [log ( ex + e− x )] + C, , x3, x3, log x +, + C , what are the, m, n, values of m and n, respectively?, (NDA 2010 I), 1, 1, (b) 3 and − 9, (a) and −, 3, 9, (c) 3 and 9, (d) 3 and 3, dx, is equal to, 20. ∫, sin ( x − a ) sin ( x − b), sin ( x − a ), 1, (a), +C, log, sin ( x − b), sin ( a − b), , 19. If, , 2, , (d), , ( ex − e− x ) dx, , ∫ ( ex + e− x ) log (cosh x ) is equal to, , ∫x, , (b) −, , 2, , log x dx =, , sin ( x − b), 1, +C, log + C, sin ( x − a ), sin ( a − b), , (c) log sin ( x − a ) sin ( x − b) + C, sin( x − a ), (d), sin( x − b), 21. Consider the following statements., I. The anti-derivative of f ( x ) = ex / 2 whose graph, passes through the point ( 0, 3) is 2 ex/ 2 + 1., II. If a function f is such that f ′ ′ ( x ) = sec4x + 4,, f ′ ( 0) = 0 and f( 0) = 0, then the function is, f (x) =, , 2, 1, log sec x + tan 2 x + 2x2., 3, 6, , Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, log x, 22. What is the value of ∫, dx? (NDA 2009 II), (1 + log x )2, 1, 1, (b), (a), +C, +C, 3, (1 + log x ), (1 + log x )2, x, x, (c), (d), +C, +C, (1 + log x ), (1 + log x )2, 23. If I1 =, , d sin x, esin ( x + h ) − esin x, and, (e, ), I 2 = lim, h→0, dx, h, , I3 = ∫ esin x ⋅ cos x dx, then, , (a) I1 ≠ I 2, (c), 24. If, , d, ( I3 ) = I 2, dx, (d) I 2 = I3, (b), , ∫ I3 dx = I 2, u = ∫ ex cos x dx , v = ∫ ex sin x dx ,, , equal to, du, (a), dx, du dv, (c), +, dx dx, , then ( u + v ) is, , dv, dx, du dv, (d), −, dx dx, (b)
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401, , Indefinite Integration, 25. The value of the integral ∫, to, (a) tan−1( ex ) + C, 1, (c), tan−1( ex ) + C, 2, , dx, is equal, 3 sinh x + 5 cosh x, , 2 + sin 2x x, 27. The value of integral ∫ , e dx is equal to, 1 + cos 2x , (a) ex sin x + C, , (b) ex cos x + C, , (c) ex tan x + C, , (d) ex cot x + C, , 28. The value of the integral, x3 (tan−1 x 4 ), 1 , 2, cosech, x, +, , dx is equal to, 8∫ , 1 + x8 , 1, 1, coth x + tan−1 x 4 + C, 8, 8, 1, 1, (b) coth x + tan−1 x 4 + C, 8, 8, 1, 1, (c) coth x +, (tan−1 x 4 )2 + C, 8, 64, 1, 1, (d) − coth x +, (tan−1 x 4 )2 + C, 8, 64, , (c), , (b), (c), (d), , ( a 2 + b2 ), −1, [{tan ( x/ a )}/ a + {tan−1( x/ b)}/ b], (a + b ), [{tan−1 ( x/ a )}/ a + {tan−1 ( x/ b)}/ b], 2, , 2, , ( b2 − a 2 ), [{tan−1 ( x/ a )}/ a − {tan−1 ( x/ b)}/ b], ( b2 − a 2 ), , 1, log 2, 2, 1, (d), log 2, (b), , 1, 2, , 4/ 3, , 31, , +C, − 1, , , 8 x2, 4/ 3, 1, 1, (c), +C, 1 − 2 , 8, x , , (a), , ( x − x3 )1/ 3, , ?, x4, 4/ 3, 31, , (b) − 2 − 1, +C, , 8 x, ( x − x3 )1/ 3, (d), +C, x4, , 33. What is the integration of e x ?, (a) e x ( x − 1) + C, (b) 2e x ( x − 1) + C, (d) e x + C, (c) 2e x ( x + 1) + C, 34. What is the value of ∫ sin x log(tan x )dx?, , (NDA 2008 II), , (a), (b), (c), (d), , cos x log tan x + log tan ( x/ 2) + C, − cos x log tan x + log tan ( x/ 2) + C, cos x log tan x + log cot ( x/ 2) + C, − cos x log tan x + log cot ( x/ 2) + C, , 35. If l r ( x ) means loge loge ... loge x , log being repeated r, times,, then, what, is, the, value, of, 1, dx, ?, ∫ x l( x ) l 2( x ) l3 ( x ) K l r ( x ), (a) l r + 1( x ) + C, , 29. Consider the following statements, f ′ (x), 1, I. ∫, dx = −, +C, 2, f(x), [ f ( x )], II. ∫ e f ( x ) f ′ ( x ) dx = e f ( x ) + C, 1, III. ∫ log [ f ( x )] dx =, +C, f(x), Which of the statements given above is/are correct?, (a) Only II and III, (b) Only I and II, (c) Only I and III, (d) I, II and III, dx, 30. What is the value of ∫ 2, ? (NDA 2007 I), ( x + a 2 )( x 2 + b2 ), (a), , dx = k sin−1( 2x ) + C, then k is equal to, , 32. What is the integration of, , (a) −, , [{tan−1 ( x/ a )}/ a − {tan−1( x/ b)}/ b], , 1− 4, , x, , (a) log 2, , (b) tan−1( 2ex ) + C, 1, (d) tan−1( 2ex ) + C, 2, , 26. The value of k for which the integral of, 3x3 + 7x 2 − 2 3kx + 1, +, , ( x ≠ 0) may be a rational, x, x2, function, is, 3, 2, 3, 2, (b), (c) −, (d) −, (a), 2, 3, 2, 3, , 2x, , 31. If ∫, , (c) l r ( x ) + C, , dx, = log { f ( x )} 2 + C , what is the value of f ( x )?, f(x), (NDA 2008 I), x, (a) 2x + α, (b) x + α, (c), (d) x 2 + α, +α, 2, , 36. If ∫, , 37. If f ( x ) = log ( x − 1 + x 2 ), then what is the value of, ?, ∫ f ′ ′ ( x ) dx, 1, (a), +C, (x − 1 + x2 ), 1, (b) −, +C, 1 + x2, (c) − 1 + x 2 + C, , +C, +C, +C, +C, , l r + 1( x ), +C, r +1, (d) l r − 1( x ) + C, (b), , (d) log ( x − 1 + x 2 ) + C, 38., , ∫, , cos x − sin x, , dx is equal to, sin 2x, (a) cos h−1 (sin x + cos x ) + C, (b) sinh−1 (sin x + cos x ) + C, (c) − cos h−1 (sin x + cos x ) + C, (d) − sinh −1 (sin x + cos x ) + C
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402, , NDA/NA Mathematics, , 39. Consider the following statements, I. ∫ log10 dx = x + C, , 43. Assertion (A) Let us define f ′ ( x ) =, , II. ∫ 10x dx = 10x + C, , where C is the constant of integration. Which of the, above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, , x ex, , Reason (R), , ∫e, , 41. Assertion (A), Reason (R), , x, , ex, , ( f ( x ) + f ′ ( x )) dx = f ( x ) ex + C, , ∫e, , x log a, , x, ∫ a dx =, , ⋅ ex dx =, , ex, x, ∫ x (1 + x log x ) dx = e log x + C, x, x, ∫ e [ f ( x ) + f ′ ( x )]dx = e f ( x ) + C, , 44. Assertion (A), Reason (R), , ∫ (1 + x )2 dx = x + 1 + C, , 40. Assertion (A), , and, , − x + x2 + 1, , 1 + 2, f( 0) = − , , then f(1) is equal to log 1 + 2 ., 2 , Reason (R) f ( x ) is not defined for every value of x., , Directions (Q. Nos. 40-44), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., , 1, , (NDA 2009 I), , Directions (Q. Nos. 45-47) An integral is defined, as below, I=∫, , 1, x + x5, , dx = f ( x ) + C, , 45. The value of f ( x ) is, x4 , 1, (a) log, , 4, 1 + x4 , x4 , (b) log , 4, 1 + x , 1 + x 4 , 1, (c) log 4 , 4, x , (d) None of the above, 46. The value of ∫, , x4, , dx is, x + x5, (a) log x + f ( x ) + C, (b) log x − f ( x ) + C, (c) 2 log x + f ( x ) + C, (d) None of these, cos x, 47. The value of ∫, dx is, sin x + sin 5x, (a) f (sin x ) + C, (b) f (cos x ) + C, (d) None of these, (c) f (cos2 x ), , x, , ( ae), +C, log ( ae), , ax, +C, loge a, , ex, x, ∫ x (1 + x log x ) dx = e log x + C, Reason (R) ∫ ex [ f ( x )) + f ′ ( x )] dx = ex f ′ ( x ) + C, , 42. Assertion (A), , Answers, Level I, 1., 11., 21., 31., , (b), (a), (b), (d), , 2., 12., 22., 32., , (d), (b), (c), (c), , 3., 13., 23., 33., , (b), (a), (c), (d), , 4., 14., 24., 34., , (c), (a), (b), (c), , 5., 15., 25., 35., , (c), (b), (b), (b), , 6., 16., 26., 36., , (a), (a), (d), (b), , 7., 17., 27., 37., , (d), (b), (d), (d), , 8., 18., 28., 38., , (b), (b), (d), (a), , 9., 19., 29., 39., , (b), (b), (c), (b), , 10., 20., 30., 40., , (c), (c), (b), (b), , 2., 12., 22., 32., 42., , (d), (a), (c), (b), (c), , 3., 13., 23., 33., 43., , (a), (c), (b), (b), (c), , 4., 14., 24., 34., 44., , (b), (c), (b), (b), (a), , 5., 15., 25., 35., 45., , (b), (a), (d), (a), (a), , 6., 16., 26., 36., 46., , (a), (c), (b), (c), (b), , 7., 17., 27., 37., 47., , (c), (d), (c), (b), (a), , 8., 18., 28., 38., , (c), (c), (c), (a), , 9., 19., 29., 39., , (c), (b), (b), (a), , 10., 20., 30., 40., , (b), (b), (d), (a), , Level II, 1., 11., 21., 31., 41., , (b), (b), (c), (d), (a)
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404, , NDA/NA Mathematics, , 13. Let I = ∫ 1 + sin, =∫, , x, dx, 4, , I = ∫ t5/ 2 ⋅, , x , x, x, 2x, + cos 2 + 2 sin cos dx, sin, , 8, 8 , 8, 8, , x, , sin, , 8, x, − cos, 8 +, =, 1, , 8, , =∫, , 15., , ∫, , = − x cos x +, , 21., , , ax , Q ∫ a x dx =, , log e a , , , elog x sin x dx = ∫ x sin x dx, , ∫, , (Q elog a = a ), 1 ⋅ cos x dx, , = sin x − x cos x + C, 16. I n = ∫ (log x) dx, , …(i), , n, , ∴, , I n − 1 = ∫ (log x), , n −1, , dx, , 22. Let I = ∫, 23. Let, , 1, = (log x)n x − n ∫ (log x)n − 1 ⋅ ⋅ x dx, x, = x(log x)n − n ∫ (log x)n − 1 dx, , I n = x (log x)n − nI n − 1, I n + nI n − 1 = x(log x)n, dx, 17. Given that, I = ∫, x(x7 + 1), Put, , x7 = t, , ⇒, , dx =, , ∴, , ⇒ 7x6dx = dt, , 1, dt, 7 x6, dt, dt, 1, I=∫ 7⋅, =∫, 7t (t + 1), 7x (t + 1), 1 1, 1 , = ∫ −, dt, 7 t t + 1 , 1, = [log t − log(t + 1)] + C, 7, t, x7, 1, 1, = log, + C = log 7, +C, t+1, 7, 7, x +1, , 18. Given that, I = ∫, Put, , x=t, , ∴, , I =2 ∫, , x e x dx, ⇒, , = (2t 2 − 4t + 4) et + C, = (2x − 4 x + 4)e x + C, , 19. Let I = ∫, , ∫, , dx, dx, =∫ 2, x + 4x + 13, x + 4x + 4 + 9, dx, 1, x + 2, = tan − 1 , +C, 3 , (x + 2)2 + 9 3, 2, , 20. Let I = ∫ (x2 + 1)5/ 2 x dx, , t = x2 + 1 ⇒ dt = 2x dx, , I=∫, , dx, xn − 1, ⇒ I=∫ n n, dx, n, x (x + 1), x (x + 1), , xn = t ⇒ nxn − 1dx = dt, 1, ⇒ xn − 1dx = dt, n, 1 1, 1 , ∴, I= ∫ −, dt, n t t + 1, =, , xn , t 1, 1, log e , +C, = log e n, t + 1 n, n, x + 1, , dx, x (1 + log x)2, 1, Put (1 + log x) = t ⇒, dx = dt, x, 1, ∴, I = ∫ 2 dt, t, 1, 1, t −2 + 1, ⇒, I=, +C=− +C=−, +C, 1 + log x, −2 + 1, t, , 24. We have, I = ∫, , 25. Given that, I = ∫, , 1, dx, 1 + ex, , It can be rewritten as, I = ∫, , (1 + ex − ex ), dx, (1 + ex ), , ⇒, , 1 + ex , I=∫ , dx +, 1 + ex , , ⇒, , I = ∫ dx − ∫, , ⇒, , I = x − log (1 + ex ) + C, ex , I = log , +C, 1 + ex , , ⇒, , 1, , dx = dt, 2 x, t 2et dt = 2 [t 2et − (2 t )et + 2et ] + C, , = a tan x + b sec x + C, f ′ (x), dx = log [log f (x)] + C, f (x) log [ f (x)], , Put, , I n = ∫ (log x) dx, , ⇒, ∴, , Let, , … (ii), , ∫, , n, , Now,, , 1 2, (x + 1)7/ 2 + C, 7, a + b sin x, dx = ∫ (a sec2 x + b tan x sec x) dx, cos 2 x, I=, , x, x, x, , + cos dx = ∫ sin + cos dx, , , 8, 8, 8, x, sin, 8 + C = 8 sin x − cos x + C, , 1, 8, 8, , 8, , 13x, 14. Let I = ∫ 13x dx =, +C, log 13, , 1 t7/ 2 , +C, 2 7 / 2 , , I=∫, , 2, , dt, 2, , ⇒, , − ex, , ∫ 1 + ex, , dx, , ex, dx, 1 + ex, [Q log e ex = x], , 1 + ex , I = − log x + C, e , = − log (1 + e− x ) + C, , 26. Given, f ′ (x) = 6 − 4 sin 2x, On integrating both the sides, we get, 4 cos 2x, f (x) = 6x +, +C, 2, As, f (0) = 3, ∴, f (0) = 3 = 0 + 2(1) + C, ⇒, C =1, ∴, f (x) = 6x + 2 cos 2x + 1, , (given)
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405, , Indefinite Integration, , 27. Given that,, Put,, ∴, , xe −1 + ex − 1, dx, xe + ex, , ∫, , =∫, , xe + ex = t ⇒ (exe − 1 + ex ) dx = dt, 1 dt 1, I=∫ ⋅, = log t + C, t e, e, 1, e, = log (x + ex ) + C = log [(xe + ex )]1/ e + C, e, , ⇒, , I = ex sin x − ∫ cos x ex dx, , ⇒, , I = e sin x − [e cos x − ∫ (− sin x) e dx] + C, , ⇒, , I = ex sin x − [ex cos x +, , ⇒, , I = e sin x − e cos x − I + C, , ⇒, ⇒, , x, , x, , x, , 35. We have,, , ∴, , =∫, , (Q e, , I = ∫ sec x° dx = ∫ sec, πx, =t, 180°, 180°, dx =, dt, π, , ∴, , 1 + ex = t ⇒ ex dx = dt, 1, I = ∫ dt = log t + C = log (1 + ex ) + C, t, , ⇒, , ∫ f (x) dx = ∫ e, , ⇒, , ∫ f (x) dx =, , 2x, , = ∫ cosh x ⋅ cosh x dx, 2, , t3, I = ∫ (1 + t ) dt = t +, +C, 3, sinh3 x, = sinh x +, +C, 3, 2, , e2x, 2, , 38. If ∫ f (x) dx = g (x) and ∫ f (x) dx = h (x), then g (x) = h (x), 39. Let I = ∫, , x 2 +1, e x, , e, , dx − ∫, , x2 + 1, x, , x2, , dx, , 2, , Put, , 40. We have,, , x2+ 1, , , x , , , , I = ∫ et dt = et + C = e, I=∫, =∫, , Put, ∴, ⇒, , x +1, x, , 1, , 1 − 2 dx, , x , 2, 1, x +1, , = t ⇒ 1 − 2 dx = dt, , x, x , =∫e, , = ∫ (1 + sinh 2 x) cosh x dx, sinh x = t, cosh x dx = dt, , dx =, , f (x), 2, f (x) = e2x, , ∴, , 3, , ∴, , f (x) = e2x, , 37. Let, , dx, ex, =∫, dx, −x, 1+ e, 1 + ex, , 33. Let I = ∫ cosh x dx, , Put, ⇒, , − (x − 2ax + a 2) + a 2, dx, , Thus,, , 180°, 180°, π t, =, log tan + + C, 4 2, π, π, 180°, πx , π, =, log tan +, +C, 4 360°, π, I=∫, , dx, , a 2 − (x − a )2, x − a, = sin −1 , +C, a , , = x), , π , , Q 1° =, , , 180, , I = ∫ sec t dt ⋅, , 32. We have,, Put, , πx, dx, 180°, , sin x + cos x, dx = ∫ 1 dx, sin x + cos x, , 2, , =∫, , log x, , dx, , 2ax − x2, , =∫, , = log (sec x) + C, , ∴, , sin x + cos 2 x + 2 sin x cos x, , =x+C, dx, , 36. Let I = ∫, , I = ∫ sin3 x cos x dx, , = ∫ tan x dx, , ⇒, , sin x + cos x, 2, , x, , 30. We have, I = ∫ elog (tan x )dx, , Put, , sin x + cos x, dx, 1 + sin 2x, , =∫, , sin x = t ⇒ cos x dx = dt, 1, t4, I = ∫ t3 dt =, + C = sin 4 x + C, 4, 4, , Put, , I=∫, , sin x dx] + C, , 2I = ex (sin x − cos x) + C, ex, I=, (sin x − cos x) + C, 2, , 29. We have,, , 31. Let, , 1, dt = − log t + C, t, = − log (1 + e− x ) + C, , I=−∫, , ∴, , x, , ∫e, , 1, e− x, dx, =, ∫ 1 + e−x dx, ex + 1, , 1 + e− x = t, − e− x dx = dt, , Let, ⇒, , 28. We have, I = ∫ ex sin x dx, x, , I = ∫ (ex + 1)−1 dx, , 34. Let, , sin x, sin x − sin 2 α, sin x, 2, , cos 2 α − cos 2 x, , +C, , dx, dx, , cos x = t ⇒ − sin x dx = dt, t , dt, I=−∫, = cos −1 , +C, 2, 2, cos α , cos α − t, I = cos −1 (cos x secα ) + C
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406, , NDA/NA Mathematics, , Level II, I = ∫ e3 log x (x4 + 1)−1 dx, , 1. Let, , = 8x4 (log x)2 − 16 {log x ⋅, , 3, , = ∫ elog x (x4 + 1)−1 dx, , = 8x4 (log x)2 − 4x4 log x + 4∫ x3 dx, , −1, , = ∫ x (x + 1) dx, 3, , =, , 4, , = 8x4 (log x)2 − 4x4 log x + x4 + C, , 3, , 1, 4x, dx, 4 ∫ (x4 + 1), , Let x4 + 1 = t, ⇒ 4x3 dx = dt, 1, 1 1, I = ∫ dt = log t + C, ∴, 4, 4 t, 1, 4, ⇒, I = log(x + 1) + C, 4, dx, 2. Let I = ∫, sin x − cos x + 2, dx, =∫, 2, 2, − cos x ⋅, + 2, sin x ⋅, 2, 2, dx, =∫, π, π, 2 (sin x ⋅ sin − cos x ⋅ cos + 1), 4, 4, 1, 1, dx, dx, =, =, π, x π, , 2 ∫, 2 ∫, 1 − cos x + , 1 − cos 2 + , 2 8, , 4, 1, dx, =, x π, 2 ∫, 2 sin 2 + , 2 8, =, , 1, 2 2, , 2 x, , = x4 [8 (log x)2 − 4 log x + 1] + C, dx, 5. Given, ∫, sin 2 x ⋅ cos 2 x, , =4 ∫, , = −4, , =−, , = (tan x − cot x) + C, cos x − 1 x, 6. Let I = ∫, e dx, sin x + 1, =∫, , ⇒, , −∫, , − sin(log e x) , x dx, x, , , sin(log e x) dx, , cos (log e x), x dx, x, , = x cos (log e x) + x sin (log e x) − I, 2 I = x [cos (log e x) + sin (log e x)], x, I = [cos (log e x) + sin (log e x)], 2, , =, , ex cos x, +, 1 + sin x, , =, , ex cos x, +C, 1 + sin x, , 7. Let I = ∫ f, and, , x4, 1 x4, dx}, − ∫ 2 log x ⋅ ⋅, x 4, 4, = 8x4 (log x)2 − 16∫ x3 log x dx, , −1, , 1, , x, , ex dx, , … (i), , (x) dx, , ∫ f (x) dx = g(x) + C, , ⇒, , ex, dx + C, sin x + 1, , ∫ 1 + sin x e dx − ∫ 1 + sin x + C, , ∴ From Eq. (i), let, , f, , −1, , (given) … (ii), , (x) = u, , x = f (u ), , ⇒, , dx = f ′ (u )du, I = ∫ uf ′ (u ) du, , ∴, , By using integration by parts, we get, I = u f (u ) − ∫ f (u ) du, , 4. Let I = ∫ 32 x3 (log x)2dx, = 32 {(log x)2 ⋅, , cos x, 1, ex dx − ∫, ex dx, 1 + sin x, 1 + sin x, , (cos x) ex, – (1 + sin x) sin x − cos 2 x x, −∫, e dx, 1 + sin x, (1 + sin x)2, , =, , ∫, = x cos (log e x) + ∫ sin(log e x) ⋅ 1 dx, , ⇒, , 2 cos 2x − 2 (cos 2 x − sin 2 x), =, +C, 2 sin x ⋅ cos x, sin 2x, , = − {cot x − tan x } + C, , π, , = x cos (log e x) + sin (log e x)x − ∫, , cot 2x, +C, 2, , = − 2 cot 2x + C, , I = ∫ cos (log e x)dx = ∫ cos(log e x) ⋅ 1 dx, , = x cos (log e x) +, , dx, (sin 2x)2, , = 4 ∫ cosec2 2x dx, , ∫ cosec 2 + 8 dx, , I = cos (log e x)x − ∫, , dx, (2 sin x ⋅ cos x)2, , ⇒4∫, , x π, − cot + , 2 8, 1, +C, =, ⋅, 1 /2, 2 2, 1, x π, cot + + C, =−, 2 8, 2, 3. Let, , 1 x4, x4, −∫ ⋅, dx}, 4, x 4, , = u f (u ) − g (u ) + C, −1, , ∴ Put, , u=f, , We get,, , I =xf, , (x), f (u ) = x, , −1, , (x) − g ( f, , −1, , (x)) + C, , [from Eq. (ii)]
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407, , Indefinite Integration, x2 − 1, dx, x + x2 + 1, 1, 1− 2, x, I=∫, dx, 1, x2 + 2 + 1, x, 1, 1− 2, x, dx, =∫, 2, 1, , x + − 1, , x, 1, x+ =t, x, 1, , 1 − 2 dx = dt, , x , dt, t −1, 1, I=∫ 2, =, +C, log, t+1, t − 12 2 × 1, , 8. We have, I = ∫, ⇒, , Put, ⇒, ∴, , 4, , 9. Let, , I=∫, =∫, , 1, ex ⋅ 1, dx, dx − ∫, x+4, (x + 4)2, 1, ex, ex, =, + ∫ ex, −, dx, ∫ (x + 4)2 dx + C, x+4, (x + 4)2, ex, I=, +C, x+4, , ∴, 10., , ∫e, , x, , ∴, , 1, dx +, 2 x, , x, , ⋅, , ∫e, , x, , ⋅, , Put, ∴, , x2, dx, (ax + b)2, 1, t − b, Put ax + b = t ⇒ dx = dt and x = , , a , a, 2, 1, (t − b), I= 3 ∫, dt, ∴, a, t2, , 1, b 2 2b, = 3 ∫ 1 + 2 − dt, t, a, t, , =, , , 1 , b2, − 2b log t + C, t −, 3, t, a , , , =, , , 1 , b2, − 2b log (ax + b) + C, ax + b −, 3, ax + b, a , , , =, , 2 , b, x2, , +C, x − log(ax + b) −, 2 , a (ax + b), a, a , , ⇒ a x dx =, , 1, log a, , ∫, , 1, dt, log a, , dt, 1 − t2, , 14., , ∫, , xex ⇒ ex (1 + x) dx = dt, I = ∫ sec2 t dt = tan t + C, , = tan (xex ) + C, x4 − 1, 2 , x4 + 1, +, dx = ∫ 2, dx, 2, x +1, x + 1 x2 + 1 , , 2 , = ∫ x2 − 1 + 2, dx, , x + 1, x3, =, − x + 2 tan −1 x + C, 3, , 15. Let, , I=∫, , ex, dx, (2 + e ) (ex + 1), x, , Put, ex = t, x, ⇒ e dx = dt, , 1, dt, (2 + t ) (1 + t ), 1 1 , = ∫ , −, dt, 1 + t 2 + t , = log (1 + t ) − log (2 + t ) + C, 1 + ex , = log , +C, 2 + ex , , I=∫, , 1, dx, 2 x, , I=∫, , dx, , ex (1 + x), dx, cos 2 (xex ), , 13. Let I = ∫, , = ex ⋅ x + C, 11. Let, , dx, , 1, sin −1 (t ) + C, log a, 1, sin −1 (a x ) + C, =, log a, , 1, dx, 2 x, , = ex ⋅ x − ∫ ex ⋅, , I, , 1 − a 2x, , I=, , ∴, , ∫e, , II, , − ax, , =, , 1 , , dx, x+, , 2 x, = ∫ ex ⋅ x dx +, , a, , ax = t, , Put, , I = ∫ ex, , ⇒, , a x/2, −x, , ax, , =∫, , x2 + 1 − x, 1, +C, = log 2, 2, x +1+ x, (x + 3)ex, (x + 4 − 1)ex dx, dx = ∫, 2, (x + 4), (x + 4)2, x+4, 1 x, −, , e dx, (x + 4)2 (x + 4)2, , I=∫, , 12. Let, , , 1−x , 16. Let I = ∫ cos 2 cot−1, dx, 1+ x, , = ∫ cos [cos −1 (− x)] dx = ∫ ( − x) dx, =−, 17., , x2, +C, 2, , f (x) = cos x − cos 2 x + cos3 x − … ∞, cos x, =, 1 + cos x, 1 + cos x, 1, ∴ ∫ f (x) dx = ∫, dx − ∫, dx, 1 + cos x, 1 + cos x, 1, x, = x − tan ⋅ 2 + C, 2, 2, x, = x − tan + C, 2
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408, , NDA/NA Mathematics, , I=∫, , 18. Let, Put, ⇒, ⇒, ∴, , ex − e− x, dx, (e + e ) log (cosh x), x, , −x, , log (cosh x) = t, 1, ⋅ sinh x dx = dt, cosh x, ex − e− x, dx = dt, ex + e− x, 1, I = ∫ dt = log t + C, t, = log (log cosh x) + C, , 1 x3, x3, −, ⋅, dx, ∫, 3 ∫ x 3, 3, 2, x, x, =, log x − ∫, dx, 3, 3, 3, 3, 1 x, x, log x − ⋅, =, +C, 3, 3 3, 3, 3, x, x, log x −, =, +C, 3, 9, 3, x, x3, (on comparing), But ∫ x2 log x dx =, log x +, +C, m, n, ∴, m = 3 and n = − 9, dx, 20. Let I = ∫, sin (x − a ) sin (x − b), 1, sin (b − a ), dx, =, sin (b − a ) ∫ sin (x − a ) sin (x − b), 1, sin [(x − a ) − (x − b)], =, dx, sin (b − a ) ∫ sin (x − a ) sin (x − b), 1, =, sin (b − a ), {sin (x − a ) cos (x − b) − cos (x − a ) sin (x − b)}, dx, ∫, sin (x − a ) sin (x − b), 1, =, {cot (x − b) − cot (x − a )}dx, sin (b − a ) ∫, 1, =, [log sin (x − b) − log sin (x − a )] + C, sin (b − a ), , 19., , x2 log x dx = log x ⋅, , , sin (x − b) , −1, =, log , + C, , sin (a − b) , sin (x − a ) , 21. I. Now,, , ∫ f (x) dx = ∫ e, , x/ 2, , dx, , x/ 2, , e, + C = 2ex/ 2 + C, 1 /2, Since, it passes through (0, 3)., ⇒, 3 = 2e0 + C ⇒ C = 1, x/ 2, ∴, ∫ f (x)dx = 2e + 1, y=, , II. Given, f ′ ′ (x) = sec4 x + 4, ⇒ ∫ f ′ ′ (x) dx = ∫ sec2 x (1 + tan 2 x)dx + ∫ 4dx, , = ∫ (sec2 x + sec2 x tan 2 x)dx + 4∫ dx, , tan3 x, f ′ (x) = tan x +, + 4x + C1, ⇒, 3, Here, f ′ (0) = 0, ⇒, 0 = 0 + 0 + 0 + C1 ⇒ C1 = 0, , ⇒, , f ′ (x) = tan x +, , Also,, , ∫ f ′ (x) dx = ∫, , tan3 x, + 4x, 3, , , , tan3 x, + 4x dx, tan x +, 3, , , , 1, x2, tan x(−1 + sec2 x)dx + 4 ⋅, ∫, 3, 2, 1, tan 2 x 4x2, + C2, f (x) = log sec x + − log sec x +, +, 3, 2 , 2, = log sec x +, , ⇒, , Here, f (0) = 0, , 1, (0 + 0) + 0 + C 2 ⇒ C 2 = 0, 3, 1, tan 2 x, f (x) = log sec x − log sec x +, + 2 x2, ∴, 3, 6, 2, tan 2 x, = log sec x +, + 2 x2, 3, 6, ∴ Both the statements I and II are correct., log x, 22. Let, dx, I=∫, (1 + log x)2, 1, Put, log x = t ⇒, dx = dt and x = et, x, 1, 1 , et ⋅ t, ∴, −, dt = ∫ et , dt, I=∫, 2, 2, (1 + t ), (1 + t ) (1 + t ) , ⇒, , 0 = 0+, , et, et, dt, dt − ∫, 1+ t, (1 + t )2, 1, et, et, dt, dt − ∫, =, − ∫ − et ⋅, 2, 1+ t, (1 + t ), (1 + t )2, x, =, +C, (1 + log x), =∫, , d sin x, (e, ), dx, esin ( x + h) − esin x, I 2 = lim, h→ 0, h, I3 = ∫ esin x ⋅ cos x dx, , 23. Given that I1 =, , and, , …(i), …(ii), …(iii), , ∴ From Eq. (i), we have, I1 = esin x ⋅ cos x, d (esin x ) sin x, From Eq. (ii), I 2 =, =e, ⋅ cos x, dx, and from Eq. (iii), on putting, sin x = p, ⇒, cos x dx = dp, we get, I3 = ∫ e pdp = e p = esin x, ∴, , d (I3 ) d (esin x ) sin x, =, =e, ⋅ cos x = I 2, dx, dx, , 24. We have, u = ∫ ex cos x dx, and, , v = ∫ ex sin x dx, , Now,, , (u + v) =, , ⇒, , ∫ e cos x dx + ∫ e sin x dx, (u + v) = ∫ ex cos x dx + sin x ∫ ex dx, x, , x, , −∫, , , d, sin x ∫ ex dx dx, , dx
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410, , NDA/NA Mathematics, 4/3, , 1, , − 1, , 1 x2, =− ⋅, ⋅3 + C, 4, 2, 4/3, 31, , +C, = − 2 − 1, , 8 x, 33. We have, I = e, , x=t ⇒, , Put, , I =2 ∫, , ∴, , =∫, Put sin x + cos x = t, , x, , 1, , 2 x, et ⋅ t dt, , dx = dt, , ∫, , II. ∫ 10x dx =, , = − cos x log (tan x) +, = − cos x log (tan x) +, , ∫, ∫, , 40. A., , ∫e, , x, , , x , , dx, (x + 1)2, x + 1 − 1, = ∫ ex , dx, (x + 1)2 , , 1, ⋅ sec2 x dx, tan x, , 1, 1 , = ∫ ex , −, dx, 1 + x (x + 1)2, , 1, dx, sin x, , =, , cosec x dx, , x, , = − cos x log (tan x) + log tan + C, , 2, 1, 35. We have, ∫, dx, 2, xl(x) l (x) K lr (x), , 41. A. Let I = ∫ ex log a ⋅ ex dx, x, , = ∫ elog a ⋅ ex dx, , dx, = dt, xl(x) ⋅ l (x) K lr − 1 (x), 1, = ∫ dt = log t + C, t, = log lr (x) + C = lr + 1 (x) + C, x, f (x) = + α, 2, dx, 2 dx, =, ∫x, ∫ (x + 2α ) = 2 log (x + 2α ) + C1, +α, 2, = log (x + 2α )2 + C1, 2, x, , = log + α + log 22 + C1, 2, , , ⇒, , 2, , 36. Let, ∴, , 2, , x, , = log + α + C, 2, , where,, , = ∫ (ae)x dx =, , ∫, , =, , ∴ Both A and R are individually true and R is the, correct explanation of A., 42. A. ∫, , R. ∫, , ex, , 1, (1 + x log x) dx = ∫ ex + log x dx, , x, x, x 1, = ∫ e ⋅ dx + ∫ ex log x dx, x, = ex log x + C, ex [ f (x) + f ′ (x)] dx, , ∴, , f ′′ (x) dx = f ′ (x) + C, , , 2x, +C, 1 −, , 2, 2, (x − 1 + x ) , 2 1+ x , , ( 1 + x2 ) (x − 1 + x2 ), 1, =−, +C, 1 + x2, , = ∫ ex f (x) dx +, , ∫e, , x, , f ′ (x) dx, , f (x) = ∫ (x + x2 + 1 ) dx, x2 x 2, +, x + 1 + log x + x2 + 1 + C, 2 2, x = 0, we get, 1 + 2, f (0) = C = − , , 2 , =, , 1, , − (x − 1 + x2 ), , ax, +C, log e a, , = e f (x) + C, Here, A is true but R is false., 1, 43. A. Given, f ′ (x) =, − x + x2 + 1, , C = log 22 + C1, , =, , R. ∫ a x dx =, , (ae)x, +C, log e ae, , x, , f (x) = log (x − 1 + x2 ), , 37. Q, , ex, +C, 1+x, , Reason (R) is always true., , lr (x) = t, , Let, , (cos x − sin x)dx = dt, , 10x, +C, log e 10, , I, , II, , ⇒, dt, , dx, , I. ∫ log 10dx = ∫ 1 ⋅ dx = x + C, , sin x log (tan x) dx, = − cos x log (tan x) − ∫ (− cos x) ⋅, , ∴, , 39., , cos x − sin x, (sin x + cos x)2 − 1, , = cosh −1 t + C, t2 − 1, = cos h −1 (sin x + cos x) + C, , I=∫, , ∴, , , , d, = 2 t ∫ et dt − ∫ t ∫ et dt dt , , dt, , = 2 [t et − et ] + C = 2et [t − 1] + C, = 2e x [ x − 1] + C, , 34., , cos x − sin x, dx, sin 2x, , 38. I = ∫, , Put, , +C, ∴, , f (1) =, , 1 + 2, 1 1, +, 2 + log 1 + 2 − , , 2 2, 2
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21, , Definite Integration, Let f ( x ) be a continuous function defined on a closed, interval [a , b] and ∫ f ( x ) dx = F ( x ) + C , then, b, , ∫ a f ( x) dx = [F( x)]a, b, , b, , ∫a, , or, , to, , a, , a, , Put −, , +, +, x=b, , +, , O, , a, , ∫0, , π, , π, , ∫ 0 sin x dx = [− cos x]0, , ⇒, , = [1 + 1] = 2, , a, 0, a, 0, , =3 ∫, , 3., , f ( x) dx +, , x, f − dx, 2, , 0, , ∫ −2a, , x, = t ⇒ − dx = 2 dt , in the second integration, 2, , =∫, , x, , ∫ sin x dx = − cos x + C, , e. g. ,, , (d) None of these, , 0, , =∫, , –, , 0, , (c) −3∫ f ( x)dx, , y, , x=a, , (b) 2 ∫ f ( x)dx, , 0, , Solution (a), , f ( x ) dx +, , 0, , ∫a, , f(x) + 2 ∫, a, , a, 0, , − 2f (t ) dt, f ( x) dx, , f ( x ) dx, , 0, , b, , c, , b, , ∫ a f ( x ) dx = ∫ a f ( x ) dx + ∫ c f ( x ) dx, ( a < c < b), , Example 2. The value of ∫, , 1, , x dx is, , −1, , y, , x, f − dx is equal, 2, , a, , (a) 3∫ f ( x)dx, , f ( x) dx = F( b) − F( a), , Geometrically it represents an algebraic sum of the, areas of regions bounded by graph of the function y = f ( x ),, the x-axis and the straight lines x = a and x = b., , +, , 0, , a, , ∫ 0 f (x) dx + ∫ − 2a, , Example 1. Integral, , (b) −1, (d) None of these, , (a) 1, (c) 0, 0, , 1, , −1, , 0, , Solution (a) I = ∫ −xdx + ∫ xdx, O, , π, —, 2, , π, , 2π, , b, , 4. 1∫ f ( x ) dx =, a, , b, , b, , b, , ∫ a f ( x ) dx = ∫ a f ( t) dt = ∫ a, , f ( u ) du, , Here, x is a dummy variable, it can be replaced by, any other variable t , u ,K, e. g. , ∫, 2., , b, , p/ 2, , 0, , sin x dx =, a, , p2, , ∫0, , ∫ a f ( x ) dx = − ∫ b f ( x ) dx, , sin t dt =, , p/ 2, , ∫0, , sin u du, , π /2, , 1 1, , , 0 − 2 + 2 − 0 =1, , b, , ∫ a f ( a + b − x ) dx, , In 1particular, ∫, , Properties of Definite Integration, 1., , 0, , x2 0, x2 , = − −1 + 1 = −, 2 , 2 , , x, , a, 0, , f ( x ) dx =, π /2, , , π, sin − x dx, , 2, , e. g. ,, , ∫0, , or, , ∫ π / 6 tan x dx = ∫ π / 6 tan 6 + 3 − x dx, , sin x dx =, , ∫0, , a, , ∫ 0 f ( a − x ) dx, , π /3, , π /3, , =, , π /3, , π, , ∫ π / 6 cot x dx, , π, ,
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413, , Definite Integration, , Special Case, , π, , Solution (b) Let I = ∫ 2, , f(x), dx, I=∫, a f(x) + f(a + b − x), b, f(a + b − x), dx, =∫, a f(x) + f(a + b − x), b f(x) + f(a + b − x), dx, 2I = ∫, a f(x) + f(a + b − x), b, , ⇒, ⇒, ∴, , 2I =, I=, , 0, , =∫, , b, , ∫ a dx = ( b − a ), b, , ∫a, , is an even integer., (a) 2, (c) 1, , ∫0, , 2I = ∫, , sin kx, dx, where k, sin x, , 0, , Example 4. The value of ∫, , π /2, 0, , (a) 0, (c) −1, , sin x − cos x, Solution (a) Let I = ∫, dx, 0, 1 + sin x cos x, , π, , π, sin − x − cos − x, π /2, , 2, , 2, ⇒, I=∫, dx, 0, , π, , π, 1 + sin − x cos − x, , 2, , 2, π / 2 cos x − sin x, I=∫, dx, ⇒, 0, 1 + cos x sin x, On adding Eqs. (i) and (ii), we get, π / 2 sin x − cos x, cos x − sin x , 2I = ∫, +, , dx, 0, 1, x, x, 1, +, + cos x sin x, sin, cos, , π /2, 0, , Example 5. The value of ∫, π, 2, (c) 0, , π, , f ( x ) dx +, , a, , ∫ 0 f ( 2a − x ) dx, , 2π, 0, , cos5x dx is, (b) 0, (d) 2, , 2π, , cos5 x dx, , 0, , f ( x ) = cos5 x, , Let, , f (2π − x ) = cos5(2π − x) = cos5 x, π, , I =2 ∫, , ∴, , 0, , cos5 x dx, , f ( π − x ) = cos5( π − x ), , Now,, , = − cos5 x, …(i), , ∫ 0 cos x dx = 0, , ∴, , Hence,∫, , …(ii), , = − f ( x), , π, , 2π, 0, , 5, , cos5 x dx = 2 ∫, , π, 0, , Example 7. The value of, , cos5 x dx = 0, 2π, , ∫0, , (a) −1, (c) 1, , cos99 x dx is, (b) 0, (d) 2, , Solution (b) I =, =2, , 0 dx = 0 ⇒ I = 0, π, 2, 0, , ...(ii), , (In general), 2 a f ( x ) dx , if f ( 2a − x ) = f ( x ), , = ∫0, , if f ( 2a − x ) = − f ( x ), 0,, , Solution (b) Let I = ∫, , sin x − cos x, dx is, 1 + sin x cos x, (b) 1, (d) −2, , π /2, , a, 0, , (a) 1, (c) −1, , (Q k is an even integer), , (a), , f ( x ) dx = ∫, , Example 6. The value of ∫, , 2I = 0 ⇒ I = 0, , =∫, , 2a, , ∫0, , π, , sin kx, dx, sin x, π sin k( π − x), dx, =∫, 0 sin ( π − x), π sin kx, = −∫, dx = −I, 0 sin x, , ⇒, , ⇒, , 5., , π, 2, 0, , (b) −1, (d) 0, , Solution (d) Let I = ∫, , sin x, dx, cos x + sin x, , ...(i), , sin x + cos x, dx = ∫ 2 1 dx, 0, sin x + cos x, π, π, 2I = ⇒ I =, 2, 4, , ...(i), , Eq. (i) it is a special case of 4th property and is used as, standard result., , Example 3. Find the value of integral, , π, 2, 0, , On adding Eqs. (i) and (ii), we get, , b− a, f(x), dx =, f(x) + f(a + b − x), 2, , π, , I=∫, , ⇒, , π, 2, 0, , cos x, dx, sin x + cos x, π , cos − x, 2 , dx, π, π, sin( − x) + cos ( − x), 2, 2, , 2π, , ∫0, , π, , cos99 x dx, , ∫ 0 cos, , 99, , xdx, , [Q cos99 (2π − x ) = cos 99 x], , =0, , cos x, dx is, sin x + cos x, π, (b), 4, (d) None of these, , 2 a f ( x ) dx , if f ( − x ) = f ( x ), , 6. ∫ f ( x ) dx = ∫ 0, −a, , 0,, if f ( − x ) = − f ( x ), a, , e. g. , ∫, , π /2, , − π /2, , sin x dx = 0, , (as sin x is an odd function)
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414, , NDA/NA Mathematics, , and, , π /2, , ∫ − π / 2 cos x dx = 2, , (as cos x is an even function), , , d x3 1, dt ., 2, dx ∫ x log t , 1, 1, (a) −, (b), ( x 2 + x), ( x 2 − x), log x, log x, (d) None of these, (c) log x( x 2 + 2 x), , Example 11. Find, , If we plot the graph of sin x and cos x in the interval, π π, − 2 , 2 , then also we can confirm this., , , , Example 8. Find the value of ∫, (a) −1, (c) 2, , π /4, − π /4, , x3 sin4 x dx., , Solution (b), , (b) 0, (d) 3, , Solution (b) Let I = ∫, , π /4, −π / 4, , =, , x3 sin 4 x dx, , Let, , 3x2, 2x, −, 3 log x 2 log x, 1, ( x2 − x), =, log x, , 4, , f ( − x) = ( − x)3 sin 4 ( − x) = − x3 sin 4( x) = − f ( x), f ( x) is an odd function., π /4, , ∫ −π / 4 x, , ∴, , 3, , sin 4 x dx = 0, , 8., , Example 9. If f ( x), g ( x), x ∈ R are continuous function, then, value of integral, π, 2, π, −, 2, , ∫, , [ f ( x) + f ( − x)] {g ( x) − g ( − x)} dx is, , (a) π, , (b), , Solution (d) Let I = ∫, , π, 2, , π, 2, π, −, 2, , (c) 1, , (d) 0, , [ f ( x) + f ( − x)] {g ( x) − g ( − x)} dx, , h( x) = [ f ( x) + f ( − x)] [ g ( x) − g ( − x)], h( − x) = [ f ( − x) + f ( x)] [ g ( − x) − g ( x)], = − h( x), Hence, h( x) is an odd function, ∴, I= 0, Let, , Example 10. The value of, (a) −2, (c) 0, , π, 2, π, −, 2, , ∫, , 1, (1 − cos 2 x dx is, 2, , ∫, , π, 2, π, −, 2, , sin x dx = 2∫, , π, 2, 0, , f ( x ) dx = n ∫, , T, , f ( x ) dx ,, , 0, , if f ( x + T ) = f ( x ) and n ∈ N, In particular,, (a) ∫, , a +T, , a, , f ( x ) dx =, , T, , ∫0, , f ( x ) dx, , where f ( x ) is a periodic function with period T, (b) ∫, , nT, , T, , f ( x ) dx = n ∫ f ( x ) dx (if f ( x + T ) = f ( x )), , 0, , 0, , Example 12. The value of ∫, , 30π, , sin x dx is, , 0, , (a) − 60, (c) 60, , (b) 0, (d) None of these, 30π, , Solution (c) I = ∫ |sin x| dx, 0, , Since,|sin x| is periodic with period π, π, , I = 30 ∫ sin xdx, , ∴, , 0, , π, , = 30[ − cos x] 0 = 30(1 + 1) = 60, b, , 9. If f ( x ) ≥ g( x ), then ∫ f ( x ) dx ≥, , b, , ∫ a g( x ) dx, (where b > a ), , sin x dx, , π, , = 2 [ − cos x] 2, 0, , π, , , = 2 − cos − cos 0 = 2, 2, , , , 7., , a + nT, , ∫a, , a, , (b) 2, (d) 2, , Solution (b) I =, , d 3 1 d 2 1 , , −, (x ) , (x ) , dx, log x2, log x3 dx, , =, , f ( x) = x sin x, 3, , , d x3 1, dt , , dx ∫ x2 log t , , d h (x ), f ( t ) dt = h ′( x ) f ( h( x )) − g′ ( x ) f ( g( x )), dx ∫ g ( x ), (Leibnitz’ s Rule), d h (x ), In particular, f ( t ) dt = h ′ ( x ) f ( h( x )),, dx ∫ a, {a is any constant independent of x }, d x, or, f ( t ) dt = f ( x ), dx ∫ a, , Example 13. Let I1 = ∫, , 2, 1, , 1, 1+ x, , (a) I1 ≤ I2, (c) I1 > I2, , 2, , dx and I2 =, (b) I1 < I2, (d) I1 ≥ I2, , Solution (b) Since, 1 + x2 > x, ∀ x ∈ [1, 2], ⇒, 2, , ∫1, , ⇒, , 1, 1, < , ∀ x ∈ [1, 2], 1 + x2 x, 2 1, 1, dx < ∫, dx, 2, 1 x, 1+ x, , ⇒, , I1 < I2, b, , 10. |∫ f ( x ) dx|≤, a, , b, , ∫ a | f ( x )| dx, , 2, , ∫1, , 1, dx. Then,, x
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415, , Definite Integration, , Example 14. Find the absolute value of ∫, , 19, , cos x, , 1+ x, (b) 9 × 10 −4, (d) None of these, , (a) 9 × 10 −8, (c) 10 −4, , 10, , 8, , Solution (a) For x ≥ 10 , we have|cos x| < 1and 1 + x ≥ 10, 8, , 11. If m and M are global minima and global maxima of, f ( x ) in [a , b], then, , dx., , y, M, 8, , m, , 1, ≤ 10 −8, 1 + x8, , ⇒, , x=a, , 19 |cos x|, 19, 19 cos x dx , ≤ ∫, dx ≤ ∫ 10 −8dx, ∴∫, 10 1 + x8, 10, 10 1 + x8, , , , x=b, , O, , m( b − a ) ≤, , x, , b, , ∫ a f ( x ) dx ≤ M ( b − a ), , = 9 × 10 −8, , Comprehensive Approach, π /2, , n, , ∫0, , n, , ∫0, , log(sin x) dx =, , π /2, , log (tan x) dx =, , n, , n, , n, , n, , log (cot x) dx = 0, , n, , ∫0, , tan x dx +, , If In =, , n, , π /2, , ∫0, , ∫0, , n−2, , tan, , cosn x dx, then In =, a, , n −1, In − 2, n, , a, , ∫0, , sin n x dx =, , π /2, , ∫0, , cosn x dx =, , (n − 1) (n − 3 ) K 1 π , =, , n (n − 2) K2 2 , π /2, , n, , π /4, , π /2, , ∫0, , ∫ 0 f ( x) dx = ∫ 0 f ( x) dx + ∫ 0 f (2 a − x) dx, a, a, = ∫ f ( x) dx + ∫ f ( a + x) dx, 0, 0, π /2, , π, , dx, , If In =, 2a, , n, , n, , ∫ 0 x2 + a2 = 2 a, π /4, , n, , π /2, , ∫0, , π, log 2, 2, , ∫0, , ∞, , n, , log(cos x)dx = −, , sin x, π, dx =, 4, sin x + cos x, a, dx, π, ∫ 0 1 + esin x = 2, a, f ( x), a, ∫ 0 f ( x) + f ( a − x) dx = 2, a, π, dx, ∫ 0 a2 − x2 = 2, π /2, , n, , π /2, , ∫0, , 1, dx =, n −1, , n −1, sinn x dx, then In =, In − 2, n, , ∫0, =, , (n − 1) (n − 3) K 2, n (n − 2) K 3 ⋅1, (if n is odd positive integer), , (if n is even positive integer), sinm x cosn xdx, , {( m − 1) ( m − 3) K (2 or 1)} {(n − 1) (n − 3) K (2 or 1)}, ( m + n) ( m + n − 2 ) K (2 or 1), , (if m and n both are not simultaneously even positive integers), {( m − 1) ( m − 3) K (1)} {(n − 1) (n − 3) K (1)} π , =, , 2, ( m + n ) ( m + n − 2 ) K (2 ), (if m and n are both even positive integers)
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Exercise, Level I, 1., , dx, , 0, , ∫ −1, , x 2 + 2x + 2, (a) 0, (c) π / 2, , 2. The value of ∫, , is equal to, (b) π/ 4, (d) − π/ 4, , 1, , dx, , 0, , x + 1 − x2, , π, 3, 1, (c), 2, x dx, , x + a+ b− x, , is, , 1, (b − a), 2, (d) b − a, , ∫ − 2 |x| dx is equal to, , (a), , (b) 1, (d) 4, , 5. Suppose f is such that f ( − x ) = − f ( x ) for every real x, 1, 0, and ∫ f ( x ) dx = 5, then ∫, f ( t ) dt is equal to, −1, , 0, , (a) 10, (c) 0, , (b) 5, (d) –5, , 6. The value of ∫, , 3, , x+1, , 2, , x ( x − 1), 2, , 16 1, +, 9, 6, 1, (c) 2 log 2 −, 6, (a) log, , dx is, 16 1, −, 9, 6, 4 1, (d) log −, 3 6, , (b) log, , 7. If f ( x ) is an even function, what is ∫, , π, 0, , to?, (a) 0, (c) 2 ∫, 8., , 1, , ∫ −1, , (b), π/ 2, 0, , f (cos x ) dx, , ∫0, , f (cos x ) dx, , (d) 1, , an odd function, an even function, Neither even nor odd, zero, , x, , ∫0, , π/ 4, − π/ 4, , (b), , x3 sin4 x dx is equal to, , π, 2, , π, 8, , (c), , (d) 0, , π/ 2, , cos x, dx equals to, 1 + sin x, (a) log 2, (b) 2 log 2, 1, (d), (c) (log 2)2, log 2, 2, π /3, dx, is equal to, 15. ∫, π / 6 1 + tan x, π, π, (b), (a), 12, 2, π, π, (c), (d), 6, 4, 14., , ∫0, , 16., , ∫− π, , sin4 x, , π, , π, 4, (c) π, , sin4 x + cos4 x, , dx is equal to, , (a), , (c), f ( t ) dt is, , (NDA 2010 II), , (b) log 3, (d) 4 log 3, , π, 2, (d) 2π, (b), , sin3 x, , π /2, 0, , sin3 x + cos3 x, , (a) π, , (b) 0, (d) 4, , 9. If f ( t ) is an odd function, then, , π, 4, , 17. What is ∫, , |1 − x| dx is equal to, , (a) –2, (c) 2, , (a), (b), (c), (d), , f (cos x ) dx equal, (NDA 2011 I), , π /2, , dθ is equal to, (b), , 13. The value of ∫, , 2, , (a) 0, (c) 2, , 4 − sin2 θ, , (a) 2 log 3, (c) 2 log 3, , (b), , (c) π / 2, , cos θ, , π, 6, π, (d), 5, 2a, f ( x ) dx, is, 11. The value of ∫, 0, f ( x ) + f ( 2a − x ), (a) f ( a ), (b) f ( 2a ), (c) f( 0), (d) a, π /4, dx, 12. What is the value of ∫, ?, π / 6 sin x cos x, , is, , (a) π, , 4., , π, 2, π, (c), 3, , (b), , b, , ∫a, , π/ 2, , ∫0, , (a), , π, 2, π, (d), 4, , (a), , 3., , 10., , π, 4, , (NDA 2010 I), , π, 2, , (b), , (d) 0, , 18. The value of the integral, equal to, (a) 2 ( 2 − 1), (c) ( 2 2 + 1), , dx ?, , π/ 2, , ∫0, , |cos x − sin x| dx , is, , (b) ( 2 2 − 1), (d) 2 ( 2 + 1)
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417, , Definite Integration, 19. Let f : R → R and g : R → R be two continuous, functions,, then, the, value, a, is, equal, to, f, x, g, −, x, −, f, −, x, g, x, dx, [, (, ), (, ), (, ), (, )], ∫, −a, , (a) eπ, , (b) 2 ∫, , (c) 1, , (d) 0, , 20. What is ∫, (a), (c), 21., , π/ 4, −π / 4, , (b), , 1, 2, , ∫π /4, , 1, 3, , 28. To, 2, , ∫ −2, , cosθ cosec2θ dθ is equal to, , 2−1, 1, (c) 1 +, 2, a, , ∫ − a (x, , 3, , 2+1, 1, (d) 1 −, 2, , (b) 3 + sin a + cos a, (d) 4 − sin a, π/ 2, , ∫− π / 2, , (a) 2, (c) π, 1, 0, , 1, n( n + 1), (c) 1, , ∫0, , sin x, , e, , (c), , π, 0, , x 4 + ( π − x )4, , (b) q only, (d) p, q, s, , (b), , 1, m−n, , π /2, , ∫0, , 1, m+n, , (d) mn, , sin x cos8x dx is equal to, , 35π, 256, 16, (c), 35, , (NDA 2009 I), , (NDA 2007 II), , 70, 256, 8π, (d), 35, (b), , (a), , 2, , 31. The value of ∫ [x ] dx is, 0, , cos x dx is equal to, , 26. What is the value of ∫, , find, out, the, numerical, value, of, 2, ( px + qx + s) dx, it is necessary to know the, , (a) 0, , 30., , 1, ( n + 1)( n + 2), (d) 0, , (b) e − 1, (d) e, cos x, , (NDA 2008 II), , 0, , (b), , (a) e + 1, (c) e + 2, , ( ax3 + bx + c) dx depends on which of, , 29. If m and n are integers, then what is the value of, π, ∫ sin mx sin nx dx, if m ≠ n?, , (NDA 2012 I), , x(1 − x )n dx is equal to?, , (a), , π /2, , sin x dx ?, (b) 1, (d) 0, , 24. What is ∫, , −2, , (a) p only, (c) p and s, , + 5x + sin x ) dx is equal to, , 23. What is the value of, , 2, , value/values of which of the following?, , (b), , (a) 4 + sin a, (c) 0, , 25., , 27. The value of ∫, , the following?, (a) Values of x only, (b) Values of each of a, b and c, (c) Value of c only, (d) Value of b only, , (d) 0, , (a), , 22., , 0, , f ( x ) g( x ) dx, , tan3 x dx equal to?, , 3, , π /2, , a, , (b) π, π, (d), 2, , (a) 0, π, (c), 4, , (where [ ] is the greatest integer less than or equal to, x), (a) 0, (b) 1, (c) 2, (d) 3, , dx ?, , Level II, 1., , 1, , ∫ −1, , log ( x + x 2 + 1 ) dx is equal to, , (a) 0, , (b) log 2, , 1, (c) log, 2, 2. If F ( x ) =, , (d) None of these, x3, , ∫x, , 2, , log t dt , ( x > 0), then F ′ ( x ) is equal to, , (a) ( 9x − 4x ) log x, (c) ( 9x 2 + 4x ) log x, 2, , 3., , π, , ∫0, , esin, , (a) –1, (c) 1, , 2, , x, , 4., , (b) ( 4x − 9x ) log x, (d) None of these, 2, , cos3 x dx is equal to, (b) 0, (d) π, , π/ 2, , ∫0, , log sin x dx is equal to, , π, (a) − log 2, 2, 1, (c) − π log, 2, , 5. If, , 2, , ∫1{ k, , 2, , (b) π log, (d), , 1, 2, , π, log 2, 2, , + ( 4 − 4k)x + 4x3 } dx ≤ 12, which one of the, , following is correct?, (a) k = 3, (b) 0 ≤ k < 3, (c) k ≤ 4, (d) k = 0, , (NDA 2011 II)
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418, , NDA/NA Mathematics, , 6. The value of, , , x , −1 , + tan−1, tan 2, x + 1, , , 3, , ∫ −1, , is, (a) 2π, π, (c), 2, 7., , 3, , ∫1, , ∫ −1/ 2, , 9., , 10., , π, , x dx, , ∫0, , a cos x + b sin x, π, (a), ab, π2, (c), ab, 2, , 2π, , ∫0, , 2, , 2, , 2, , (a), 12., , 1, n, , π/ 8, , ∫0, , π/ 2, , ∫0, , π, (a), 4, , ∫0, , 4, , f ( x − 1) dx, , f ( x ) dx =, , 6, , f ( x − 1) dx, , ∫ −3, , π, 2ab, π2, (d), 2ab, , (d), , (b), , (c), , 5, , ∫ −3, , A., , ∫0, , B., , ∫1, , (NDA 2011 I), , C., , ∫ −1, , 1, ( n − 2), , D., , ∫1, , (d), , 0, , π, (b), 8, x dx, , [x + 1 − x 2 ] 1 − x 2, (a) 0, π, (c), 4, , tan−1 x, 1+ x, , 2, , dx ?, , π2, (c), 8, is equal to, (b) 1, π2, (d), 2, , 2, , ∫0, , [ f ( x ) − f ( − x )] dx, 10, , ∫ −6, , ∫ −2, , f ( x − 1) dx, , 1, , 2, , List II, e −1, e, 2. − log 2, , πx, log sin dx, 2, 1, 1, , , e x − 2 dx, x x , , 1., , 3. e 2, −e, 2, 4., 0, , x| x| dx, 1 −1 / x, e, dx, x2, , Codes, A B, (a) 3 2, (c) 2 3, , C, 4, 4, , D, 1, 1, , A, (b) 1, (d) 2, , B, 2, 3, , C, 3, 1, , D, 4, 4, , 21. Consider the following statements, I. The value of the integral, 2a, f ( x ) dx, ∫ 0 f ( x ) + f ( 2a − x ) is equal to a., 4, , II. The value of ∫ (|x − 1| +|x − 3|) dx is 10, 0, , (b) π / 2, (d) 2π, 1, , 3, , dx is, , List I, , 2, , n, ( n − 1), , 2π, , 20. Match List I with List II and select the correct, answer using the codes given below the lists, , sin 2x log tan x dx is equal to, , 15. What is the value of ∫, , 16., , ∫ −4, , (c), , cos3 4θ dθ is equal to, , (a) π, (c) 0, , 1, , f ( x ) dx =, , is equal to, , (a) 5 / 3, (b) 5/4, (c) 1/3, (d) 1/6, π/ 2, cos x − sin x, 13. ∫, dx is equal to, 0, 1 + cos x sin x, (a) 0, (b) π/ 2, (c) π/ 4, (d) π/ 6, 14., , 5, , ∫ −3, , 1, ( n − 1), , + 2 cos x, (b) π, , 2 f ( x ) dx =, , (b), , tann x dx , what is I n + I n − 2 equal to?, (b), , 2, , 3, , 5, , (b) 1, (d) 2e1/ 2, , (b) 4, (d) 1, , ∫, , sin x, , (d), , sin x, , f ( x ) dx =, , 1, , 11. If I n =, , 2, , 0, , (NDA 2011 I), , π, , (c), , 2, , ∫ −2, , (sin x +|sin x| ) dx is equal to, , π, 4, 0, , π/ 2, , equal to?, , (d) 2π, , (a), , (a) 0, (c) 8, , 1 + 2 sin2 x, π, (b), 3, , 19. If f is continuous function, then, , x , dx is equal to, x , , (a) 0, (c) e1/ 2, , dx, , (a) 2, π, (c), 4, , (b) 2, (d) 0, , 1 −, (cos x ) log , 1 +, , , π, 0, , 18. The value of ∫, , ( x − 1) ( x − 2) ( x − 3) dx is equal to, , 1/ 2, , 17. What is ∫, (a) π, , (b) π, π, (d), 4, , (a) 3, (c) 1, 8., , x 2 + 1 , , dx, x , , III. If f ( x ) is a periodic function with period T, then, b + nT, , ∫ a + nT, , (NDA 2012 I), , π2, (d), 32, , f ( x ) dx =, , b, , ∫a, , f ( x ) dx., , Which of the statements given above are correct?, (a) II and III, (b) I and II, (c) I and III, (d) All I, II and III, 1, , 22. If f ( x ) = a + bx + cx 2 , what is ∫ f ( x ) dx equal to?, 0, , (a), (b), (c), (d), , [ f ( 0) + 4 f (1/ 2) + f (1)]/ 6, [ f ( 0) + 4 f (1/ 2) + f (1)]/ 3, [ f ( 0) + 4 f (1/ 2) + f (1)], [ f ( 0) + 2 f (1/ 2) + f (1)]/ 6, , (NDA 2009 II)
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419, , Definite Integration, , 23. If I1 =, , e2, , dx, and I 2 =, log x, , ∫e, , (a) I1 = I 2, (c) I 2 + I1 = 0, 24., , 2, , ∫ −2, , ex, dx ,then, x, (b) 2I1 = I 2, (d) I1 = 2I 2, , 33. What is the value of ∫, , 2, , ∫1, , (a) 1, , |1 − x 2| dx is equal to, , (a) 2, , (b) 4, , (c) –2, , 25. What is the value of ∫, (a) 0, , π/ 2, 0, , (b) 1, , (d) 0, , log (tan x ) dx ? (NDA 2009 I), (c) –1, , π, (d), 4, , II. ∫, , 1, and 2, 2, 1, (d) − and –2, 2, 2, , 28. What is the value of ∫ xex dx ?, 0, , ( e − 1), 2, (c) 2( e − 1), , (NDA 2008 I), , (b) e2 − 1, , 29. What is the value of ∫, , −π/ 2, , (a) 2, (c) 1, , 1, , m, , 32. If ∫, , 1, , 0, , x, log 2, , (b) 1, ( ex − 1)−1, , (c) –1, (d) 2, 3, dx = log , what is the value of x ?, 2, (NDA 2007 I), , (a) e2, , (b), , 1, e, , f ( x ) dx = −, , (d) −1/ e, 5, , what is the, 6, (NDA 2007 I), , (c) log 4, , 16, 9, , (b), , (c), , 3, 2, , (d) −, , 3, 2, , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., π/ 4, , ∫0, , π/ 2, , ∫0, , b, , (1 − x )n dx = k ∫ x n (1 − x )m dx , what is the value, , of k ?, (a) 0, , 19, 6, , 38. Assertion (A), , log x, dx equal to?, (NDA 2007 II), x, log( b), b, 1, (b), (a) log( ab) log , a, 2, log( a ), b, 1, ( a + b), (c) log , (d) log , , a, 2, ab , , ∫0 x, , ∫−3, , 1/ e, , cos5 x dx = 0., , a, , a, , 31., , (c), 9, , Reason (R) If f ( x ) is an odd function, then, b, ∫ f ( x ) dx = 0., , sin|x| dx ?, (b) – 2, (d) 0, , 30. What is ∫, , 1, , 37. Assertion (A), , (d) e − 1, π /2, , (d) 1, , Directions (Q. Nos. 37-41), , (b) −, , (a), , and, , 35. What is the value of ∫ ( x − 1) e− x dx ?, , (a) −, , 0, , 1, , (d) –2, 1, f′ = 2, 2, , 2, , 27. What are the values of p which satisfy the equation, p, 2, 3, (NDA 2008 II), ∫ ( 3x + 4x − 5) dx = p − 2 ?, 1, and 2, 2, 1, (c) and –2, 2, , dx ?, , π x, 34. If f ( x ) = A sin , + B and, 2 , 1, 2A, ∫0 f ( x ) dx = π , then, what is the value of B?, 2, 4, (b), (c) 0, (a), π, π, , (b) e, 7, 36. If ∫ f ( x ) dx = and, −3, 3, 9, value of ∫ f ( x ) dx ?, , Which of these are correct ?, (a) I and II, (b) II and IV, (c) III and I, (d) II and III, , (a), , (b) 0, , 2, , ∫− π / 2sin|x| dx = 2, π /2, IV. ∫, sin x dx = − 2, −π/ 2, III., , x2, (c) π, , (a) 0, , cos x dx = 2, , −π/ 2, π/ 2, , sin x, , 0, , 26. Consider the following equations, ∞, 1, I. ∫, dx = ∞, 0 1 + x2, π/ 2, , π/ 2, −π/ 2, , (d) 1, , Reason (R), , b, , ∫a, , 39. Assertion (A), , cos x − sin x, dx = 0., 1 + sin x cos x, , f ( x ) dx =, e, , ∫1, , Reason (R) I n =, , b, , ∫a, , f ( a + b − x ) dx., , log2 x dx = e − 2., e, , ∫1, , 40. Assertion (A) 16 <, , logn x dx = e − n I n − 1, 6, , ∫ 4 2x dx < 24, , Reason (R) If m is the smallest and M is the, greatest value of a function f ( x ) in an interval ( a , b),, b, then the value of the integral ∫ f ( x ) dx is such that, a, , for a < b, we have, m (b − a) ≤, , b, , ∫ a f ( x ) dx ≤ M ( b − a ).
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420, , NDA/NA Mathematics, , 41. Assertion (A), , π, , ∫0, , sin7 x dx = 2 ∫, , π /2, , 0, , sin7 x dx, (NDA 2007 I), , 7, , Reason (R) sin x is an odd function., , Directions (Q. Nos. 45-47) Let us define an, integral function, a, a, ∫ f ( x ) dx = ∫ f ( a − x ) dx, 0, , 45. The value of, , Directions (Q. Nos. 42-44), defined as, π /2, , ∫0, , log sin x dx = −, , If an integral is, , π, log 2, then solve the, 2, , (c), , following question., , π /2, 0, , π, log 2, 2, (d) None of these, , (b) −, , (c) 0, 44. Find the value of ∫, , π /2, 0, , (b) −, , π, log 2, 2, , (c), , 47. The value of ∫, , log tan x dx., , (a) 0, , π, 4, , sin x, sin x + cos x, π, (b), 4, π, (d), 3, , dx, , π /2, , log cos x dx, , (a) − log 2, , π, 2, , π /2, , ∫0, , , π, φ − x, , 2, 46. The value of the integral ∫, 0, π, φ( x ) + φ −, 2, π, π, (a), (b) −, 4, 4, π, (c), (d) 0, 2, , 1, πx , 42. Find the value of ∫ log sin dx., 0, 2, (a) 3 log 2, (b) − log 2, (c) 2 log 3, (d) log 2, , 43. Find the value of ∫, , (a) −, , 0, , π, log 2, 2, , (d) None of these, , 1, , x, , 0, , x + 1− x, , , x, , , dx is, , dx, , (a) 2, , (b) −, , (c) 0, , (d), , 1, 2, , 1, 2, , Answers, Level I, 1., 11., 21., 31., , (b), (d), (a), (b), , 2. (d), 12. (b), 22. (c), , 3. (b), 13. (d), 23. (a), , 4. (d), 14. (a), 24. (b), , 5. (d), 15. (a), 25. (b), , 6. (b), 16. (c), 26. (a), , 7. (c), 17. (c), 27. (c), , 8. (c), 18. (a), 28. (c), , 9. (b), 19. (d), 29. (a), , 10. (b), 20. (d), 30. (a), , 2., 12., 22., 32., 42., , 3., 13., 23., 33., 43., , 4., 14., 24., 34., 44., , 5., 15., 25., 35., 45., , 6., 16., 26., 36., 46., , 7., 17., 27., 37., 47., , 8., 18., 28., 38., , 9., 19., 29., 39., , 10., 20., 30., 40., , Level II, 1., 11., 21., 31., 41., , (a), (b), (d), (b), (b), , (a), (d), (a), (c), (b), , (b), (a), (a), (b), (b), , (a), (c), (b), (c), (a), , (a), (d), (a), (d), (b), , (a), (c), (d), (a), (a), , (d), (c), (a), (d), (d), , (a), (c), (a), (a), , (d), (b), (a), (a), , (b), (c), (a), (a)
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Hints & Solutions, Level I, 1. Let I = ∫, , 0, dx, dx, =, x2 + 2x + 2 ∫ −1 (x + 1)2 + 1, , 0, −1, , = [tan −1 (x + 1)]−01, = [tan, 2. Let I = ∫, , −1, , 1 − tan, dx, , 0, , x + 1 − x2, , 2I = ∫, 3. We have, I = ∫, , π/ 2, , 0, , 3, , x dx, x+ a+ b−x, , b, a, , , x − 1 1 , = 2 log , +, x x 2, , , 2, 1 1 1 , = 2 log − log + − , , 3, 2 3 2 , , 4 1, = 2 log −, 3 6, 16 1, = log, −, 9 6, 7. Q f (x) is an even function., , ...(i), , ∫a, , i.e.,, , f (x) dx = ∫, , I=∫, , ⇒, , b, , 8. We have, I = ∫, , b, , a+ b−x, , a, , a+ b−x+ x, , 2I = ∫, 4. Let I = ∫, , 2, −2, , |x| dx = − ∫, , 0, , −2, 2 2, , 0, , x dx +, , I=∫, , 2, , ∫0, , b−a, 2, , x dx, , ⇒, , 0, , ∫ −1, , f (x) dx = 0 = ∫, f (x) dx +, , 0, −a, , π/ 2, , ∫0, , f (x) dx +, , ∫0, , ⇒, , ∫ −1, , f (x) dx = − 5, , ⇒, , 0, , f (t ) dt = − 5, , ∫ −1, I=∫, , 3, 2, , f (x) dx, , x+1, dx, x2(x − 1), , 1, , −1, , (1 − x) dx, , [Q ∫, , 1, 0, , x, 0, , f (t ) dt is an even, , π/ 2, , , sin θ , dθ = sin −1 , , 2, 2 0, , 4 − sin θ, 1 π, = sin −1 =, 2 6, 2a, f (x) dx, I=∫, 0 f (x) + f (2 a − x), 2a, f (2a − x), I=∫, dx, 0 f (x) + f (2 a − x), cos θ, , ...(i), …(ii), , On adding Eqs. (i) and (ii), we get, , f (x) dx = 0, , 0, , 6. Given that, , a, , ∫0, , 11. Let, , [Q f (x) = − f (x)], , 1, , |1 − x| dx, , function., 10., , 5. Given f (− x) = − f (x), ∀ values of real x., We know that,, a, , −1, , 9. Since, f (t ) is an odd function, then ∫, , x, x2 , = − + = − (0 − 2) + (2 − 0) = 4, 2 −2 2 0, , ∫ −a, , 1, , , 1, 1, x2 , = x − = 1 − + 1 + = 2, 2, 2, 2, , −1, , ...(ii), , dx, , 1dx = [x]ba = b − a ⇒ I =, , a, , f (cos x) dx, , 1, , On adding Eqs. (i) and (ii), we get, b, , π /2, , 0, , − 1 ≤ x ≤ 1 ⇒1 − x ≥ 0, , Here,, ∴, , f (a + b − x) dx, , a, , f (cos x) dx = 2 ∫, , a, , , , Q 2a f (x) = 2 ∫ 0 f (x) dx, if f (2a − x) = f (x), (even ) , , ∫0, 0,, if f (2a − x) = − f (x), (odd ), , , Now, using property of definite integral, b, , π, , ∫0, , ∴, , π, 4, , 1 dθ ⇒ I =, , −2 1, 2 , − 2+, , dx, x − 1, x, x, 3, , Put x = sin θ ⇒ dx = cos θ dθ, π/ 2, cos θ dθ, ...(i), I=∫, 0, sin θ + cos θ, , π, cos − θ dθ, π/ 2, , 2, ⇒, I=∫, 0, , π, , π, sin − θ + cos − θ, , 2, , 2, π/ 2, sin θ, ...(ii), dθ, =∫, 0, cos θ + sin θ, On adding Eqs. (i) and (ii), we get, , 3, 2, , 1, , , = − 2 log x + + 2 log (x − 1), x, 2, , , π, π, 0] = − 0 =, 4, 4, , −1, , 1, , =∫, , 2I=∫, , f (x) dx = 5], 12., , π/ 4, , 2a, 0, , 2a, f (x) + f (2a − x), dx = ∫, 1 dx = 2a ⇒ I = a, 0, f (x) + f (2a − x), π /4, , dx, , ∫ π/ 6 sin x cos x = 2 ∫ π/ 6, =2 ∫, , π/ 4, π /6, , dx, sin 2x, cosec 2x dx, , = 2 [log tan x] ππ //64 ⋅, , 1, 2
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423, , Definite Integration, =∫, , 1, 0, , (t n − t n + 1 ) dt, 1, , tn + 1, tn + 2 , =, −, , n + 1 n + 2 0, 1, 1, 1, =, −, =, n + 1 n + 2 (n + 1)(n + 2), 25. Let I = ∫, , π / 2 sin x, , e, , 29. Since, sin mx, sin nx is an odd function, if m ≠ n, then, π, , ∫0, 30. I = ∫, , π/ 2, 0, , m + 1 n + 1 , !, !, , 2 2 , m, n, x, x, dx, sin, ⋅, cos, =, ∫0, m + n + 2 , 2, ! , , , 2, 1 9, 0 + 1 8 + 1, ! ! !, !, , 2 2, 2 2 , =, I=, 0 + 8 + 2, 2 (5) !, 2, !, , , 2, 7 5 3 1, ⋅ ⋅ ⋅, π, 2, 2 2 2, = π⋅, 2 ⋅4 ⋅3 ⋅2 ⋅1, 1, , 7 ⋅5, 35π, =, π=, Q ! = π , , , , , 2, 16 ⋅ 16, 256, , , , Q, , , , , 0, , 26. Let, , …(i), , …(ii), , On adding Eqs. (i) and (ii), we get, π (cos x − cos x), π, dx = ∫ 0 dx, 2I = ∫, 0 x4 + (π − x)4, 0, ∴, 27., , I =0, 2, , ∫ −2 (ax, , 3, , + bx + c) dx, , sin mx ⋅ sin nx dx = 0, , sin x ⋅ cos 8 x dx, , 1, , cos x, dx, I=∫ 4, 0 x + (π − x)4, π cos (π − x), dx, I=∫, 0 (π − x)4 + (x)4, π, − cos x, dx, I=∫, 0 (π − x)4 + (x)4, , ( px2 + s) dx, , 0, , For finding a numerical value of I, it is necessary to, know the values of p and s only., , cos x dx, , π, , 2, , I =2 ∫, , I = ∫ et dt = [et ] 10 = e1 − e0 = e − 1, , ∴, , (px2 + qx + s) dx, , ∴, , sin x = t ⇒ cos x dx = dt, x = 0, t = 0, π, x= ,t =1, 2, , and when, , −2, , Q qx is an odd function, therefore its integral value is, zero., , 0, , Put, When, , 2, , 28. Let I = ∫, , 1, , 2, , ax4 bx2, , =, +, + cx, 4, 2, , −2, , a (16) b(4), a (16) b(4), +, − 2c = 4c, =, +, + 2c − , 2, 2, , 4, 4, So, the value of given integral depends on only c., , π/ 2, , 31. I = ∫ [x]dx +, 0, , 2, , ∫ 1 [x]dx, 0 < x < 1, [x] = 0, 1 < x < 2, [x] = 1, , For, , 1, , I = ∫ 0 dx +, , ∴, , 0, , 2, , ∫ 1 1dx, , = 0 + [x]12 = 1, , Level II, = 3 log x ⋅ 3x2 − 2 log x ⋅ 2x, = (9x2 − 4x) log x, , 1. Let f (x) = log (x + 1 + x2 ) and replacing x by – x, we get, f (− x) = log ( 1 + x2 − x), , ( 1 + x2 + x) , , = log ( 1 + x2 − x), , ( 1 + x2 + x) , , [(1 + x2) − x2], = log 1 − log ( 1 + x2 + x), = log, ( 1 + x2 + x), = − log ( 1 + x2 + x), ⇒, f (− x) = − f (x), Hence, f (x) is an odd function., ∴, 2. F (x) = ∫, , 1, , ∫ −1, x, , log (x + 1 + x2 )dx = 0, , 3, , x2, , log t dt, , Applying Leibnitz’s theorem, we get, d 3, d 2, F ′ (x) = log x3, x − log x2, x, dx, dx, , π, , I=∫, , 0, , ⇒, , I=∫, , 0, , ⇒, , I=−∫, , 3. Let, , π, , e sin, esin, π, 0, , 2, , cos3 x dx, , x, , 2, , ( π − x), , e sin, , 2, , x, , ...(i), , cos3 (π − x) dx, , cos3 x dx, , ...(ii), , On adding Eqs. (i) and (ii), we get, ⇒, 2I = 0 ⇒ I = 0, π/ 2, , 4. Let I = ∫, , 0, , I=∫, , 0, , π/ 2, , ...(i), , log sin x dx, , π/ 2, , π, log sin − x dx = ∫, log cos x dx ...(ii), 0, , 2, , On adding Eqs. (i) and (ii), we get, 2I = ∫, =∫, , π/ 2, , 0, π/ 2, 0, , log sin x cos x dx, log sin 2x dx − ∫, , π/ 2, 0, , log 2 dx
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426, , NDA/NA Mathematics, , 19. Since, f (x) is continuous function., Let us consider f (x) = x, 5, , ∫ −3, , ∴, , 10, , ∫ −6, , and, , f (x − 1) dx = ∫, 5, , ∫ −3, , ∴, , 5, , 2 f (x) dx = ∫, , −3, 10, −6, 10, , 2 f (x) dx = ∫, , −6, , I=∫, , C., , =∫, , −1, 0, , −1, , 4, , 3, , 1 2, , 2a, , f (2a − x), dx, f (x) + f (2a − x), , On adding Eqs. (i) and (ii), we get, 2a f (x) + f (2 a − x), 2I = ∫, dx, 0 f (x) + f (2 a − x), ⇒, , 1, , 2I = ∫, , III. I = ∫, , 2a, 0, , 4, , b + nT, , a + nT, , f (x) dx, , Put x = nT + θ, ⇒ dx = dθ, b, , b, , a, , a, , ⇒, , I = ∫ f (nT + θ )dθ = ∫ f (θ )dθ, , or, , I = ∫ f (x)dx, , [Q f (x + T ) = f (x) , where T = period of f (x)], b, , a, , ∴ IIIrd statement is correct., 22. Given, f (x) = a + bx + cx2, ∴, , 0, , 0, , 4, , 3, , , 2. x2, , 2x2, − 4x, = −, + 4x + 2[x]31 + , 3, 2, 0, 2, = − 1 + 4 + 2 [3 − 1] + [(16 − 16) − (9 − 12)], = 3 + 4 + 3 = 10, ∴ IInd statement is correct, , 1, , 0, , 1, , Using property,, I=∫, , 3, , 1, , 1, , ∫0, , f (x) dx = ∫, , 1, , 0, , …(i), , (a + bx + cx ) dx, 2, , 1, , x3 , x3 , = − ∫ x dx + ∫ x dx = − + , −1, 0, 3 −1 3 0, 1 1, = − + =0, 3 3, 2 e−1/ x, D., dx, I=∫, 1 x2, 1, 1, Put − = t ⇒ 2 dx = dt, x, x, −1/ 2 t, 1 1, ⇒, I=∫, e dt = [et ]−− 11/ 2 = e−1/ 2 − e−1 =, −, −1, e e, e −1, =, e, ∴ A → 2, B → 3, C → 4, D → 1, 2a, f (x)dx, …(i), 21. I. I = ∫, 0 f (x) + f (2 a − x), or, , 1, , 0, , = ∫ (−2x + 4)dx + ∫ 2dx + ∫ (2x − 4)dx, , f (x − 1) dx, , x(− x) dx + ∫ x (x) dx, 2, , 1, , + ∫ (x − 1 + x − 3)dx, , x|x|dx, , 0, , 3, , 0, , (x − 1) dx = 16, , 1, , 1, , 0, 1, , = ∫ { − (x − 1) − (x − 3)}dx + ∫ { x − 1 − (x − 3)}dx, , 2x dx = 16, , πx, log sin dx, 0, 2, πx, π, Put, = t ⇒ dx = dt, 2, 2, 2 π /2, log sin tdt, I= ∫, π 0, π /2, 2, π, π, , , = (− log 2) Q ∫, log sin x dx = − log 2, π, 2, 2, , 0, = − log 2, 2, 1 1, B., I = ∫ ex ( − 2 )dx, 1, x x, 1, 1, Let f (x) = ⇒ f ′ (x) = − 2, x, x, 2, 1, [Q ∫ ex [ f (x) + f ′ (x) ] dx = ex f (x)], I = ex , ⇒, x 1, e2, = −e, 2, , 20. A. I = ∫, , 4, , I = ∫ (|x − 1|+|x − 3|)dx, , II., , dx = [x]20a = 2a, , ∴, I=a, Ist statement is correct., , …(ii), , Here,, and, Now,, , , bx2 cx3 , = ax +, +, , 2, 3 0, , b c, …(ii), =a+ +, 2 3, b c, 1, f (0) = a , f = a + +, 2, 2 4, [from Eq. (i)], f (1) = a + b + c, 1, f (0) + 4 f + f (1), 2, 6, b c, , a + 4 a + + + a + b + c, , 2 4, =, 6, b c, …(iii), =a+ +, 2 3, , From Eqs. (ii) and (iii), we get, 1, f (0) + 4 f + f (1), 1, 2, ∫ 0 f (x) dx =, 6, 23. We have, I1 = ∫, Now,, Put, ⇒, ∴, , I1 = ∫, , e2, e, e2, e, , 2 ex, dx, dx, and I 2 = ∫, 1 x, log x, , dx, log x, , log x = t ⇒ x = et, dx = et dt, 2 et, 2 ex, I1 = ∫, dt, I1 = ∫, dx = I 2, 1 t, 1 x, , …(i)
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428, , NDA/NA Mathematics, , ex − 1 = t and dx =, , Put, , = [− (1 − 1)e−1 − { − (0 − 1)e− 0 }] + [− e−x ]10, , dt, 1+ t, , e dx = dt, x, 1, dt = ∫, log, 2, (1 + t ), , = − 1 + (− e−1 + e−0 ) = − 1 − e−1 + 1, , x, , ∴, , 1, 1 , I=∫, dt, −, log 2, t 1 + t, = [log t − log (1 + t )] xlog 2, x, = [log (ex − 1) − log ex ] log, 2, x, , , ex − 1, ex − 1 , 2 − 1 , = log x − log , = log x , , 2 , e , e log 2 , , , ex − 1, 1, = log x − log, 2, e , x, , But, , I, ex − 1, 2 x , e , , ∴, ⇒, ⇒, , ex − 1, = log 2 x , e , 3, = log, 2, 3, =, 2, , 4e − 4 = 3e ⇒ e = 4, x = log 4, x, , 33. Let I = ∫, , π / 2 sin, −π/ 2, , x2, , x, , x, , x, , sin x, x2, sin (− x), sin x, f (− x) =, =− 2, Q, (− x)2, x, ⇒, f (− x) = − f (x), ⇒ f (x) is an odd function., ∴, I =0, π x, 34., f (x) = A sin , +B, 2 , π x π, ∴ f ′ (x) = A cos , ⋅, 2 2, π π, , 2 = A cos , , 4 2, ( 2 × 2) × 2 4, ⇒, A=, =, π, π, 1, 2A, Now,, ∫ 0 f (x) dx = π, 1 , , 2 ×4, π x, ⇒, ∫ 0 A sin 2 + B dx = π 2, 1, , ⇒, ⇒, ⇒, 35., , 1, , πx 2, , , − A cos 2 ⋅ π + Bx =, 0, π, 4 2, 4 2, − ⋅ cos + B + ⋅ cos 0 =, π π, π π, 2, 8, 8, B + 2 = 2 ⇒B =0, π, π, −x, , ∫ 0 (x −I 1) eII, , 36. We know that,, 9, , 2, , dx, , 8, π2, 8, π2, , But, , 9, , ∫ −3, , 1, , = [− (x − 1)e−x ]10 − ∫ − e−x dx, 0, , ...(i), , (by property), Q b f (x) dx = c f (x) dx + a f (x) dx, ∫a, , ∫a, ∫c, , , whare a ≤ c ≤ b, 2, 7, −5, f (x) dx =, and ∫, f (x) dx =, −3, 6, 3, , From Eq. (i), 9, 5 7, = + ∫ f (x) dx, 2, 6 3, 9, −5 7 −5 − 14 − 19, ∫2 f (x) dx = 6 − 3 = 6 = 6, f (x) = cos5 x, , −, , 37. Let, ∴, , f (− x) = cos5 (− x) = cos5 x = f (x), , ∴f (x) is an even function, π/ 4, , ∫0, , ∴, , cos5 x dx ≠ 0, , Thus, A is false but R is true., π/ 2, cos x − sin x, 38. Let I = ∫, dx, 0, 1 + sin x cos x, , π, , π, cos − x − sin − x, π/ 2, , 2, , 2, dx, ∴ I=∫, 0, , π, , π, 1 + sin − x cos − x, , 2, , 2, π/ 2, sin x − cos x, I=∫, dx, 0, 1 + sin x cos x, , …(i), , …(ii), , On adding Eqs. (i) and (ii), we get, π/ 2, , 2I = ∫, , 0, , 0 dx = 0 ⇒ I = 0, , Thus, both A and R are individually true and R is the, correct explanation of A., 39. (A) I = ∫, , e, 1, , log 2 x dx, , Integrating by parts, we get, e, , e, , I = ∫ 1. (log x)2 = [x(log x)2] 1e − ∫ x ×, 1 II, , 1, , I, , 2 log x, dx, x, , e, , = e(log e)2 − log 1 − 2∫ 1 . log x dx, 1, , Again, integrating by parts, we get, e, 1 , , = e(1)2 − 0 − 2[x log x] 1e − ∫ x × dx, 1, x , , e, , = e − 2(e log e − log 1 − ∫ dx), 1, , = e − 2(e − 0 − [x]1e ) = e − 2e + 2(e − 1) = e − 2, (R), , Integrating by parts, we get, , 9, , ∫−3 f (x) dx = ∫−3 f (x) dx + ∫2 f (x) dx, , dx, , f (x) =, , Again let, , = − e−1 = − 1/e, , e, , I n = ∫ 1 ⋅ log nx dx, 1, , Integrating by parts, we get, e, , = [x(log x)n ] 1e − ∫ x × n, 1, , (log x)n − 1, dx, x
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22, , Area Bounded, by Region, Curve Tracing, For the evaluation of area of bounded regions it is very, essential to know the rough sketch of the curves. The, following points are very useful to draw a rough sketch of a, curve., , Symmetry, (i) Symmetry about x-axis If the equation of the, curve remains unaltered when y is replaced by − y,, then the curve is symmetrical about x-axis., (ii) Symmetry about y-axis If the equation of the, curve remains unaltered when x is replaced by −x,, then the curve is symmetrical about y-axis., (iii) Symmetry about y = x If the equation of the, curve remains unaltered, if x and y are interchanged,, then the curve is symmetrical about y = x., (iv) Symmetry about y = − x If x and y are replaced by, − y and −x and the equation of the curve is unaltered,, then the curve is symmetrical about the line y = − x., (v) Symmetry in opposite quadrants If the, equation of the curve is unaltered, when x and y are, replaced by −x and − y, then it is symmetrical in, opposite quadrants., , Intersection with Origin, If the constant term in the equation of curve is zero,, then curve passes through the origin., , Intersection with Coordinate Axes, (i) For finding intersection points of the curve with the, x-axis put y = 0 and solve equation for x. Roots of, equation gives points of intersection with x-axis., , (ii) Similarly for finding intersection points with the, y-axis put x = 0 and solve the equation for y., Roots of equation give points of intersection with, y-axis., , Area of Curves Under Different, Conditions, 1. (a) Area of shaded portion A =, , b, , ∫ a f ( x ) dx, , y, y = f(x), , O, , x=a, , x, , x=b, , (b) Area of shaded portion, A=, , b, , ∫a f ( y )dy, , y, y=b, x = f (y ), y=a, O, , x, , Example 1. Find the area of the region bounded by the, curve y = 2 x − x 2 and x-axis., 1, 4, (b) sq units, (a) sq unit, 3, 3, 2, (d) None of these, (c) sq unit, 3
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431, , Area Bounded by Region, , Solution (b) We have,, , (b), , y = 2x − x2, 2, − ( x − 2x + 1) = y − 1 ⇒, , ⇒, , A=, , b, , ∫0, , f ( y ) dy +, , 0, , ∫−a, , f ( y ) dy, , y, , ( x − 1) 2 = − (y − 1), y=b, , This is the equation of the parabola, whose meet the x-axis at, x = 0 , 2., 2, , ∴ Required area = ∫ ydx, 0, , x′, , x, , O, , y, y = 2 x –x2, , y=a, y′, , x’, , (0,0), , x, , (2,0), , Example 3. Find the area bounded by the curve y = sin x, between x = 0 and x = 2π ., (a) 1 sq unit (b) 2 sq units (c) 3 sq units (d) 4 sq units, , y’, , =∫, , 2, 0, , π, , 2, , , 8 4, x3 , (2x − x2) dx = x2 − = 4 − = sq units, 3 0, 3 3, , , Solution (d) Required area = ∫ y dx +, 0, π, , = ∫ sin x dx − ∫, 0, , 2π, π, , sin x dx, , y = sin x, , 8, sq units, 3, (d) 5 sq units, (b), , B(0,3), , y, , x2 = y –1, y =3, , x, , = [ − cos x] π0 − [ − cos x] 2ππ, , Example 4. The area of the region bounded by, y = − 1, y = 2 , x = y3 and x = 0 is, 17, 15, (a), sq units, (b), sq units, 4, 4, 13, sq units, (d) None of these, (c), 4, , A (0,1), x, , y′, 3, 3, = 2 ∫ xdy = 2∫ y − 1dy, , 1, 1, 3, − 1) 2 , , 2π, , = − cos π + cos 0 + cos 2π − cos π, = 1 + 1 + 1 + 1 = 4 sq units, , C (√2, 3), , x′, , π, , O, , Solution (a) ∴Required area = 2 area of curve ABC, , , (y, =2, 3, , 2, , ( − y) dx, , y, , Example 2. The area of region bounded by the curve., x 2 = y − 1 and y = 3 is, 8 2, (a), sq units, 3, (c) 8 sq units, , 2π, , ∫π, , Solution (a) Required area, , 3, , 3, 8 2, 4 2, sq units, =, −0 =, (, ), 2, , , 3, 3 , , , 1, , = Shaded area of the curve (OAB + ODC), y, (0,2) A, , B, , x = y3, y=2, , 2. (a) Area of the shaded portion, y, , x′, y = –1, , C, , D(0,–1), y′, , x=a, x, , O, , 2, , = ∫ x dy +, 0, , 2, , x=c, , x=b, , A = ∫ f ( x ) dx+ ∫ f ( x ) dx, c, a, c, , x, , O(0,0), , b, , 0, , ∫ −1, , 2, , 0, , 0, , −1, , − xdy = ∫ y3dy − ∫ y3dy, 0, , y 4 , y 4 , = − , 4 0 4 −1, 16 1, 16 1 17, sq units, =, − [ 0 − 1] =, + =, 4, 4, 4, 4, 4
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432, , NDA/NA Mathematics, , 3. (a) Area of shaded portion, A=, , b, , ∫a, , Solution (b), , { f ( x ) − g( x )} dx, , y, , y, , y 2 = 12 x, , x 2 + y 2 = 64, , y = f (x), , O, O, , x′, O, , (b), , A=, , y = g (x), x=a, x=b, , x, , x=0, , b, , ∫a { f ( y ) − g( y )} dy, , y ′x = 4 x = 8, , Required area = Area of circle, 4, −2 ∫ 2 3 x dx +, 0, , y, , x, , O, , = 64π −, , y′, , Example 5. Find the area of the region bounded by the, curves y = x 2 + 2 , y = x, x = 0 and x = 3 ., 15, 21, sq units, (b), sq units, (a), 2, 2, 15, sq units, (d) None of these, (c), 3, y = x2 + 2 , y = x, x = 0 and x = 3, Required area, , x′, , 3, 0, , =, , 64, 32π, − 32π + 16 3 +, 3, 3, , 128π 16 3 16, −, =, (8π − 3) sq units, 3, 3, 3, , Example 7. The area bounded by the parabolas, y = 4x 2, y =, 5 2, sq units, 3, 15 2, (c), sq units, 3, , (a), , Solution (b) The equation of given curves are, , y, , (y 2 − y1) dx, , x2, and the line y = 2 is, 9, 10 2, (b), sq units, 3, 20 2, (d), sq units, 3, , Solution (d) ∴ Required area, 1, , 2, , 2, , 0, , = 2∫ ( x2 − x1) = 2∫ (3 y −, , y = x 2+ 2, y=x, C, B, , y, )dy, 2, y = 4x2, , (0, 2) D, , y, , O, , (3,0), , 64 − x2 dx, , , 8, x , 64, x, +, sin −1 , 64 − x2 +, 2, 2, 8 4 , , y= a, , = Area of curve OBCD = ∫, , 8, , ∫4, , 4, , 2, = 64π − 2 2 3 x3 / 2 × , 3 0, , , x = f(y), , x = g (y ), , y=b, , x′, , x, , x, , x2 = 9 y, y=2, , y′, , =∫, , 3, 0, , 9 21, sq units, =9 + 6 − =, 2 2, , Example 6. Find the area of the portion of the circle, x 2 + y 2 = 64 which is exterior to the parabola y 2 = 12 x., 10 2, 16, sq units, (b), (a), (8 π − 3) sq units, 3, 3, (c) 16(8 π − 3) sq units, (d) None of these, , x, , x′, , 3, , x3, x2 , [ x2 + 2 − x] dx = + 2x − , 2 0, 3, , y′, , 2 5 y , = ∫ , dy , 0 2 , 3, y 2 , 10, = 5 2 =, 3, 3, , 2 0, =, , 2, , 3, 2 2 , 0, , 10, 20 2, sq units, [ 8 −0]=, 3, 3
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433, , Area Bounded by Region, =∫, , 4. (a) Area of shaded portion, y, , π /4, 0, , sin xdx + ∫, , π /2, π /4, , cos xdx, , = − [cos x] π0 / 4 + [sin x] ππ // 24, 1, , = 2 1 −, = (2 − 2) sq unit, , 2, , y = g(x), , y =f (x), , Example 9. Using integration, find the area of the region, x=c x=b, , O x=a, , (b), , c, , b, , A=, , ∫a, , f ( x ) dx +, , ∫c, , A=, , ∫a, , c, , f ( y ) dy +, , ∫c, , b, , x, , g( x ) dx, , bounded by the curves y = |1 + x | + 1, x = − 3, x = 3, y = 0., (a) 12 sq units, (b) 15 sq units, (c) 3 sq units, (d) None of the above, 1 + x + 1, if x ≥ − 1, 1 − ( x + 1), if x < −1, , g( y ) dy, , Solution (a) y = f ( x) = , , y, , 2 + x, if x ≥ − 1, =, − x, if x < − 1, , y=b, x = g(y), , y, , y, , D, , =, , y=c, , –, x, , A, , x = f(y), , x, y=, , + 2F, , y=a, x′, , x′, , x, , O, , C, B, (–3,0) (–1,0), , E, (3,0), , x, , y′, y′, , Example 8. Find the area of one curvilinear triangles, , Required area, = Area of curve CBEFAD, = Area of curve CBAD + Area of curve BEFA, , formed by the curves y = sin x, y = cos x and x-axis., (b) (2 − 2 ) sq units, (a) (2 + 2 ) sq units, (d) (2 + 2 2 ) sq units, , (c) 2 2 sq units, , =∫, , 3, , 1 9 9, 1, , = − − + + 6 − − 2 , , , 2, 2, 2, 2, , , 3, 9, =4+ +6+ , 2, 2, 12, = 10 +, = 16 sq units, 2, , y = cos x, , y′, , 3, , ∫ −1( x + 2) dx, , x2 , x2, , = − + + 2x, 2, 2, −3 , −1, , y, , O, , − xdx +, −1, , Solution (b) Required area, , x’, , −1, , −3, , y = sin x, x, , x = π/4 x = π/2, , Comprehensive Approach, n, , n, , The area of the region bounded by y 2 = 4 ax and x2 = 4 by is, 16 ab, sq units, 3, Area of the region bounded by y 2 = 4 ax and y = mx is, 8 a2, sq units, 3m3, , n, , n, , n, , Area of the region bounded by y 2 = 4 ax and its latusrectum is, 8 a2, sq units, 3, Area of the region bounded by one arch of sin ax or cos ax and, 2, x-axis is sq units, a, x2 y 2, Area of ellipse 2 + 2 = 1 is πab sq units, a, b
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Exercise, Level I, 1. The area bounded by the x-axis, the curve y = f ( x ), and the lines x = 1, x = b is equal to b2 + 1 − 2 for, all b > 1, then f ( x ) is, (a) ( x − 1 ), , (b) ( x + 1 ), , (c) ( x − 1 ), , (d) ( x / 1 + x 2 ), , 2, , 2. The area formed by triangular shaped region, bounded by the curves y = sin x , y = cos x and x = 0 is, (a) ( 2 − 1) sq unit, (b) 1 sq unit, (c) 2 sq unit, (d) (1 + 2 ) sq units, 3. The area bounded by y = log x , x-axis and ordinates, x = 1, x = 2 is, 2, 1, (b) log sq units, (a) (log 2)2 sq units, e, 2, 4, (c) log sq units, (d) log 4 sq units, e, 4. The area enclosed between the curves y = x3 and, y = x is [in sq unit], (a) 5/3, (b) 5/4, (c) 5/12, (d) 12/5, 5. What is the area bounded by the curve y = cos 3x ,, π, (NDA 2011 II), 0 ≤ x ≤ and the coordinate axes?, 6, 1, (a) 1 sq unit, (b) sq unit, 2, 1, 1, (c) sq unit, (d) sq unit, 3, 4, 6. The area bounded by the curve y = f ( x ), x-axis and, ordinates x = 1 and x = b is ( b − 1) sin ( 3b + 4), then f ( x ), is, (a) 3 ( x − 1) cos ( 3x + 4) + sin ( 3x + 4), (b) ( b − 1) sin ( 3x + 4) + 3 cos ( 3x + 4), (c) ( b − 1) cos ( 3x + 4) + 3 sin ( 3x + 4), (d) None of the above, 7. The area between the curve y = 4 + 3x − x 2 and, x-axis is, (a) 125/6 sq units, (b) 125/3 sq units, (c) 125/2 sq units, (d) None of the above, 8. The area bounded by the curve y = x , x-axis and, ordinates x = − 1 to x = 2, is, (a) zero, (b) 1/2 sq unit, (c) 3/2 sq units, (d) 5/2 sq units, , 9. What is the area of the region bounded by the line, 3x − 5 y = 15, x = 1, x = 3 and x-axis in sq unit?, 36, 18, (NDA 2008 I), (a), (b), 5, 5, 9, 3, (d), (c), 5, 5, 10. Area of the region bounded by the curve y = tan x , at, π, x = and the x-axis is, 4, 1, 1, , (a) sq unit, (b) log 2 + sq unit, , 4, 4, 1, , (c) log 2 − sq unit (d) None of these, , 4, 11. If the area above the x-axis, bounded by the curves, 3, y = 2kx and x = 0 and x = 2 is, ,then the value of k, log 2, is, (a) 1/2, (b) 1, (c) – 1, (d) 2, 12. What is the area bounded by the curves y = ex ,, (NDA 2011 I), y = e− x and the straight line x = 1 ?, 1, , (a) e + sq unit, , e, 1, , , (c) e + − 2 sq unit, , , e, , , (b) e −, , , (d) e −, , , 1, sq unit, e, 1, , − 2 sq unit, , e, , 13. The area bounded by the straight lines x = 0, x = 2, and the curves y = 2x , y = 2x − x 2 is, 4, 3, 1 , 4, (a) −, + sq units, sq units (b) , 3 log 2, log 2 3, , 4, (c) , − 1 sq units, , log 2, , 3, 4, (d) , − sq units, log 2 3, , 14. The area enclosed within the curve|x| +| y| = 1, is, (a) 2 sq unit, (b) 1 sq unit, (c) 3 sq unit, (d) 2 sq units, 15. The area bounded by the curve x = f ( y ), the y-axis, and the two lines y = a and y = b is equal to, (a), (c), , b, , ∫a, b, ∫a, , b, , ∫a, , (NDA 2012 I), , y 2 dx, , y dx, , (b), , x dy, , (d) None of these, , 16. The area between the curve y = sin x and the x-axis, from x = 0 to x = 2π is equal to, (a) 2 sq units, (b) 4 sq units, (c) 1/2 sq unit, (d) 1/4 sq unit
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435, , Area Bounded by Region, 17. What is the area under the curve f ( x ) = xex above the, x-axis and between the lines x = 0 and x = 1?, (NDA 2010 II), , 1, (a) sq unit, 2, 3, (c) sq units, 2, , (b) 1 sq unit, (d) 2 sq units, , 20. What is the value of K, if the area bounded by the, curve y = sin Kx, y = 0, x = π/ K , x = π/( 3K ) is, 3 sq units?, (NDA 2009 II), 1, (a), (b) 1, 2, 3, (d) 2, (c), 2, , 18. What is the value of k ( k > 0), if the area of the triangle, bounded by the lines y = 0, x = k and y = kx is, 4 sq units?, 1, (a) 1, (b) 2, (c) 4, (d), 2, , 21. The area bounded by the curve y = x3 , x-axis and two, ordinates x = 1 to x = 2 is equal to, 15, 15, sq units, (b), sq units, (a), 2, 4, 17, 17, (c), sq units, (d), sq units, 2, 4, , 19. The area of the region bounded by the parabola, y 2 = 4ax and its latusrectum is, 8a 2, 4a 2, (a), sq units, (b), sq units, 3, 3, 2, 2, 2a, a, sq unit, sq unit, (d), (c), 3, 3, , 22. The area between the parabola y = x 2 and the line, y = x is, 1, 1, (b) sq unit, (a) sq unit, 6, 3, 1, (c) sq unit, (d) None of these, 2, , Level II, 1. Area enclosed between the curve y 2( 2a − x ) = x3 and, line x = 2a above x-axis is, 3πa 2, (b), (a) π a 2 sq units, sq units, 2, 2, 2, (d) 3πa sq units, (c) 2πa sq units, 2. If a curve y = a x + bx passes through the point (1, 2), and the area bounded by the curve, line x = 4 and, x-axis is 8 sq units, then, (a) a = 3, b = − 1, (b) a = 3, b = 1, (c) a = − 3, b = 1, (d) a = − 3, b = − 1, 3. The area of region {( x , y ) : x 2 + y 2 ≤ 1 ≤ x + y } is, π2, π2, sq units, (b), sq units, (a), 5, 2, π2, π 1, sq units, (d) − sq units, (c), 4 2, 4, 4. If the area bounded by parabola y = 2 − x 2, the line, 2x + y = 2 is A sq unit, then A equals, (a) 1/2 sq unit, (b) 1/3 sq unit, (c) 2/9 sq unit, (d) 2 / 3 sq unit, 5. The area between the curves y = xex and y = xe− x , the, line x = 1 in sq unit, is, 1, , (b) 0 sq unit, (a) 2 e + sq units, , e, 2, (c) 2e sq units, (d) sq unit, e, 6. What is the area bounded by the curve, x + y = a ( x , y ≥ 0), and the coordinate axes?, , (NDA 2011 II), , 5a 2, 6, a2, (c), 2, , (a), , a2, 3, a2, (d), 6, (b), , 7. The area of the smaller segment cut-off from the, circle x 2 + y 2 = 9 by x = 1 is, 1, (a) ( 9 sec−1 3 − 8 ) sq units, 2, (b) ( 9 sec−1( 3) − 8 ) sq units, (c) ( 8 − 9 sec−1 3) sq units, (d) None of the above, 8. The area bounded by y = − x 2 + 2x + 3 and y = 0 is, (a) 32 sq units, (b) 32/3 sq units, (c) 1/32 sq unit, (d) 1/3 sq unit, 9. The area of the region bounded by the curve, y = x| x|, x- axis and the ordinates x = 1, x = − 1, is, given by, 1, (a) 0 sq unit, (b) sq unit, 3, 2, (c) sq unit, (d) 1 sq unit, 3, 10. If A is the area between the curve y = sin x and x-axis, π, in the interval 0, , then in the same interval, area, 4, between the curve y = cos x and x-axis, is equal to, π, (a) A′, (b), −A, 2, (c) 1 − A, (d) 2A
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436, , NDA/NA Mathematics, , 11. The area bounded by the curves y = x 2 and y = 2| x|is, equal to, 4, 8, (a) sq units, (b) sq units, 3, 3, 2, 1, (d) sq unit, (c) sq unit, 3, 3, 12. What is the area bounded by the curve y = 4x − x 2 − 3, and the x-axis?, (NDA 2009 II), 2, 4, (b) sq units, (a) sq unit, 3, 3, 5, 4, (c) sq units, (d) sq unit, 3, 5, 13. The area in the first quadrant enclosed by the x-axis, 1, the line y =, x and the circle x 2 + y 2 = 4,is given by, 3, (a), (b), , 3, , ∫ 0 ( x + 4 − x ) dx, 2, 2, ∫ 0 ( x + 4 − x ) dx, 2, , x, , + 4 − x 2 dx, , 3, , 3 x, 2, (d) ∫, dx + ∫, 4 − x 2 dx, 0, 3, 3, , (c), , 2, , ∫0, , 14. The area bounded by the coordinate axes and the, curve x + y = 1 is equal to, 1, (a) 1 sq unit, (b) sq unit, 2, 1, 1, (c) sq unit, (d) sq unit, 3, 6, 15. What is the area enclosed by the curve 2x 2 + y 2 = 1?, (NDA 2007 II), , (a) 2π, π, (c), 2, , (b) π, π, (d), 2, , 16. Which one of the following definite integrals, represents the area included between the parabola, 4 y = 3x 2 and the straight line 2 y = 3x + 12 ?, 2, 4 3x, (a) ∫, dx, −2, 4, 4 3x + 12, 3 , (b) ∫ , − x 2 dx, 0 , 2, 4 , 4 3x + 12, 3 2, (c) ∫ , − x dx, −2 , 2, 4 , 2 3x + 12, 3 , (d) ∫, − x 2 dx, , −2 , 2, 4 , 17. What is the area of the region bounded by y =| x − 1|, and y = 1 (in sq unit)?, (a) 2, (b) 1, 1, 1, (c), (d), 2, 4, , 18. What is the area under the curve y =|x| +|x − 1|, between x = 0 and x = 1?, 1, (b) 1, (a), 2, 3, (d) 2, (c), 2, 19. f ( x ) = 1 −, , x2, , x ∈ [− 2, 2], find the area covered by, 4, , x-axis., 8, (a) sq units, 3, 2, (c) sq unit, 3, , (NDA 2010 II), , 4, (b) sq units, 3, 1, (d) sq unit, 3, , 20. The ratio of the areas bounded by the curves y = cos x, π, and y = cos 2x between x = 0, x = and x-axis is, 3, (a) 1 : 1, (b) 2 : 1, (c) 2 : 1, (d) 1 : 2, 21. The area in the first quadrant between x 2 + y 2 = π 2, and y = sin x is, π3, π3 − 16, sq unit, sq unit, (b), (a), 4, 4, π3 − 8, (π3 − 8), (c), sq unit, (d), sq unit, 2, 4, , Directions (Q. Nos. 22-23), , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer.You have to select one of the codes (a),(b),(c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not the, correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 22. Assertion (A) The area of the region bounded by the, parabola y = x 2 + 1 and the straight line x + y = 3 is, 9, given by ., 2, Reason (R) When we rotate the above figure the, area of the bounded region is change., 23. Assertion (A) The area between the curve y = 1 −|x|, 1, and the positive x-axis is ., 2, Reason (R) The area between the curve and the, x-axis is half of the area between the curve and, positive x-axis.
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437, , Area Bounded by Region, , Directions (Q. Nos 24-26), , Let us consider the, parabola y = 2x whose area is bounded by the lines y = 4, and x = 8., 2, , 24. Find the area A1 between the line y = 4, curve y = 2x, and y-axis., (a) 32, (b) 32/3, (c) 35/3, (d) 16/3, 2, , 25. Find the area A2 between the line x = 8, curve y 2 = 2x, and below x-axis., , 62, 3, 64, (c), 3, (a), , (b), , 61, 3, , (d) None of these, , 26. The ratio between A2 and A1 is, (a) 2 : 1, (b) 1 : 2, (c) 1: 1, (d) None of the above, , Answers, Level I, 1. (d), 11. (b), 21. (b), , 2. (a), 12. (c), 22. (a), , 3. (c), 13. (d), , 4. (c), 14. (d), , 5. (c), 15. (c), , 6. (a), 16. (b), , 7. (a), 17. (b), , 8. (d), 18. (b), , 9. (b), 19. (a), , 10. (d), 20. (a), , 2. (a), 12. (b), 22. (c), , 3. (d), 13. (d), 23. (a), , 4. (d), 14. (d), 24. (b), , 5. (d), 15. (d), 25. (c), , 6. (d), 16. (c), 26. (a), , 7. (b), 17. (b), , 8. (b), 18. (b), , 9. (c), 19. (a), , 10. (c), 20. (c), , Level II, 1. (b), 11. (b), 21. (d), , Hints & Solutions, Level I, ∫ 1 f (x) dx =, , ∴, , f (x) =, , π, π, , , = sin + cos − cos 0, , , 4, 4, , b, , b 2 + 1 − 2 = x2 + 1 , , 1, , b, , 1. Q, , d, dx, , x2 + 1 =, , 2x, 2 x +1, 2, , =, , x, , 1, 2, 1, − 1 = ( 2 − 1) sq unit, =, +, − 1 =, 2, 2, , 2, , x +1, 2, , 3. We have, y = log x, x = 1, x = 2, , 2. The given equation of curves are, y = sin x, and, y = cos x, From Eqs. (i) and (ii), we get, π, sin x = cos x ⇒ x =, 4, , …(i), …(ii), , ∴, ⇒, , Required area = ∫, , 2, , 1, , y dx, , 2, , A = ∫ log x dx = [x log x − x] 12, 1, , = 2 log 2 − 2 − log 1 + 1, = 2 log 2 − 1 = 2 log 2 − log e, = log 4 − log e, , y, , y, , y = cos x, , x=1, , x, , y = log e, , y = sin x, , O, , x, , π, 4, , ∴ Required area = ∫, , π /4, 0, , (cos x − sin x) dx, , = [sin x + cos x]π0/ 4, , x′, , O, , y′, , 1, , 2, , x, , x=2, , m, 4, , = log sq units Q log m − log n = log , e, , n
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438, , NDA/NA Mathematics, y, , 4. Solving y = x or y2 = x, ( y ≥ 0) and, y = x3, We get points of intersection (0, 0), and (1, 1), ∴ Required area = ∫, , 1, 0, , ( x − x3 ) dx, , 4, , , 3x2 x3 , = 4x +, − , 2, 3 −1, , 64, 3 1, , = 16 + 24 −, +4− − , 3, 2 3, , 65 3, = 44 −, −, 3 2, 264 − 130 − 9 125, sq units, =, =, 6, 6, , 2, , y =x, 3, , y, , A(1, 1), , =x, , x, , O, , 1, , x3/ 2 x4 , =, − , 3 /2 4 0, 5, sq unit, =, 12, 5. Given curve, y = cos 3x, 0 ≤ x ≤, ∴, , Required area = ∫, =∫, , π /6, x=0, , 0, 2, 8. Required area = ∫ x dx + ∫ x dx, , 0, −1, x2 0 x2 2 , =, , 2 , + , 2 , −1 0 , , π, 6, y dx, π/ 6, , 0, , y, , sin 3x , cos 3x dx = , 3 0, , π/ 6, , y=x, , y, y = cos 3x, O, , x′, , π/6, , π/2, , x=–1, , x, , O, , x=2, , x, , y′, , 1, =, 3, 1, =, 3, , π, , , sin − sin 0, , , 2, 1, × 1 = sq unit, 3, , 6. Area bounded by the curve y = f (x), x-axis and the, b, ordinates x = 1 and x = b is ∫ f (x) dx, 1, , ∴, , b, , ∫ 1 f (x) dx = (b − 1) sin (3b + 4), , On differentiating wrt b, we get, f (b) ⋅ 1 = 3 (b − 1) cos (3b + 4) + sin (3b + 4), Hence,, f (x) = 3(x − 1) cos (3x + 4) + sin (3x + 4), 7. Equation of curve are y = 0, and, y = 4 + 3 x − x2, On solving Eqs. (i) and (ii), we get, x = − 1, 4, y, , + |2| = 2 + 1, − 1 , =, 2, 2, 5, = sq units, 2, 9. The given equation of line can be rewritten as, x y, − =1, 5 3, 3x − 15, and, y=, 5, y, , …(i), …(ii), , x=1 x=3, , x, O, , x′, , (1, 0) (3, 0), B, C, D, (5, 0), , x, , (0, –3) A, , y = 4+3x–x2, , y′, 3, , ∴Required area = ∫ y dx, , –1, , 4, , O, , x, , =∫, , 1, , ∴ Required area = ∫, , 4, , −1, , (4 + 3x − x2) dx, , 1, 3x − 15, dx =, , 5 , 5, 3, , 3, , ∫1, , (3x − 15) dx, , , 1 3 x2, 1 27, 3, , − 45 − + 15, − 15x = , , 5 2, 5, 2, 2, , , 1, 1 24, 1, = , − 30 = [12 − 30], 52, 5, 18, 18, sq units, =−, =, 5, 5, (neglecting negative sign), , =, ∴ Curve does not intersect x-axis between x = − 1, and x = 4, , 1, 3
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439, , Area Bounded by Region, 10. Required area = ∫, , π /4, 0, , 4, 8, 1, −4 + −, log 2, 3 log 2, 3, 4, =, − sq units, log 2 3, =, , tan x dx, , π, = [log sec x]π0 / 4 = log sec − log sec 0, 4, = log 2 − log 1 = log 2 sq unit, y, , (–π /2, 0), , x, , x = π/4, , (π /2, 0), , 14. The given lines are ± x ± y = 1 i.e., x + y = 1, x − y = 1,, x + y = − 1, x − y = − 1., These lines form a quadrilateral where vertices are, A (−1, 0), B(0, − 1), C (1, 0) and D(1, 0) obviously ABCD is, a square., Length of each sides of this square is 12 + 12 = 2, Hence, area of square = 2 × 2 = 2 sq units, , 11. Area bounded by the curves y = 2, given by, , kx, , y, , and x = 0 and x = 2 is, , 2, , =, y, –x, , x′, , 1, , 22k − 1 , 2kx , =, , =, k log 2 0 k log 2 , 3, A=, log 2, 22k − 1, 3, =, ⇒ 22k − 1 = 3k, k log 2 log 2, , =, , +, , y, , 2, , x, , 1, , x, , =, , –, , y, , y, , –, , =, , 1, , –x, , ∴, , +, , 0, , But, , x, , 1, , A = ∫ 2kx dx, , y′, , This, relation is satisfied by only option (b)., 12. The equation of curves are y = ex and y = e− x ., 1, ∴, ex = x, e, ⇒, e2x = e0 ⇒ x = 0, , 15. ∴ Required area = ∫, , y =b, y=a, , x dy, , y, y=b, x = f (y), , 1, , ∴ Required area = ∫ (ex − e−x ) dx, , y=a, , 0, , = [ex + e− x ]10 = e + e−1 − e0 − e0, y, , x′, , B, A, , D (1,1), C, x, , O, , x, , O, y′, , 16. The required area is given by, y, , 1, , , = e + − 2 sq unit, , , e, 13. Required area = ∫, , 2, 0, , y = sin x, , [2 − (2x − x )] dx, , 2π, , π, , O, , 2, , x, , x, , 2, , 2x, x3 , =, − x2 +, , 3 0, log 2, , A=∫, , y, , sin x dx +, , 2π, , ∫π, , − sin x dx, , = [− cos x] π0 + [cos x] 2ππ, =, , 2, , x, , = [1 + 1] + [1 + 1], , y, , =2 + 2, ⇒, , x′, , π, 0, , x, , O, , 1, , 17. Required area = ∫ xex dx = [xex − ∫ ex dx]10, 0, , y = 2 x – x2, x=2, y′, , A = 4 sq units, , = [xex − ex ]10 = (e − e) − (0 − 1), = 1 sq unit
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440, , NDA/NA Mathematics, , 18. The given lines are y = 0 and x = k and y = kx., , 2, , 21. Required area = ∫ x 3 dx, 1, , y, y = kx, , x′, , y, , x, , x=k, , O, , x′, , y = x3, , x, , O, , y′, , ∴ Required area = ∫, , k, , y′, , y dx, , 0, , 4=∫, , ⇒, , k2 , 4 = k ⇒ k3 = 23 ⇒ k = 2, 2, , 0, , 22. Equation of parabola is,, y = x2, Equation of the straight line is, y=x, , 19. We have equation of parabola y2 = 4ax, y, L, s (a, 0), , x′ O, (0, 0), , = 2∫, , a, 0, , a, 0, , P (1,1), , y dx, , O (0,0) (1,0), , 2 a, , ⇒, ⇒, , A=∫, , π /K, π /3 K, , sin Kx dx, π /K, , cos Kx , 3=−, K π /3K, 3=−, , ⇒, , 3=−, , ⇒, , 3=, , 1, K, , 1, K, , 3, 2K, , π, , cos π − cos 3 , 1, , − 1 − 2 , ⇒ K =, , 1, 2, , (given), , x, , y=x, , a, , , 2, x dx = 4 a x 3/ 2, 0, 3, , 8 a 3/ 2 8 2, =, [a ] = a sq units, 3, 3, 20. Area,, , ...(ii), , y, , y = x2, , x, , ...(i), , L′, , y′, , ∴ Required area = 2 ∫, , x=2, , x4 , 1, = = (16 − 1), 4, 1 4, 15, =, sq units, 4, , k, , x2 , kx dx ⇒ 4 = k , 2 0, , k, , ⇒, , x=1, 2, , From Eqs. (i) and (ii), we get, x2 − x = 0 ⇒ x (x − 1) = 0, ∴, x = 0 or x = 1, y = 0 or y = 1, Hence, the coordinates of their points of intersection are, O (0,0) and P (1,1)., ∴Required area between parabola and straight line is, 1, , 1, , 0, , 0, 1, , = ∫ x dx − ∫ x2 dx, x2 x3 , = − , 2 3 0, 1 1, , = − − 0, 2 3, , 1, = sq unit, 6
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441, , Area Bounded by Region, , Level II, 1. Given, equation of curve is y2(2a − x) = x3 ., Which is symmetrical about x-axis and passes through, origin., , y, (0, 1) B, , y, , x 2 + y 2= 1, A(1, 0), , x = 2a, , x, , x, +, y, =, 1, , x, , O, , y, , 4., , (0, 2), , y=, , 2–, , x, , 2, , x3, < 0 for x > 2a or x < 0, 2a − x, , Also,, , So, curve does not lie in x > 2a and x < 0, therefore curve, lies wholly on 0 ≤ x ≤ 2a., 2a, x 3/ 2, ∴ Required area = ∫, dx, 0, 2a − x, , ( 2, 0), x, , x′, , 2x, , +y, , x = 2a sin θ ⇒ dx = 2a ⋅ 2 sin θ cos θ dθ, 2, , Put, , ∴ Required area = ∫, , π /2, , 0, , 2. Given, equation of curve is y = a x + bx. This curve, passes through (1, 2)., …(i), ∴, 2=a+ b, and area bounded by the curve and line x = 4 and, 4, x-axis is 8 sq units, then ∫ (a x + bx) dx = 8, 0, , 2a 3/ 2 4 b 2 4, [x ]0 + [x ]0 = 8, 3, 2, , 2a, ⋅ 8 + 8b = 8 ⇒ 2a + 3b = 3, 3, On solving Eqs. (i) and (ii), we get, a = 3 and b = − 1, ⇒, , …(i), …(ii), , x2 + (1 − x)2 = 1, , =∫, , ⇒ 2x − 2x = 0 ⇒ 2x (x − 1) = 0, 2, , ⇒, x = 0, x = 1 ⇒ y = 1, y = 0, ∴ Point of intersection of circle and line are, A (1, 0) , B(0, 1), 1, , ∴ Required area = ∫ [ 1 − x2 − (1 − x)] dx, 0, , 1, , x 1 − x2 1, x2 , =, + sin −1 x − x + , 2, 2, 2, , 0, 1 π, 1 π 1, ⋅ − 1 + = − sq units, 2 2, 2 4 2, , 2, , [(2 − x2) − (2 − 2 x)] dx, , 0, 2, 0, , (− x2 + 2x) dx, 2, , x3, 2 2 2 2, 2x2 , +, = −, +, = −, , 2 0 3, 2 , 3, 2 2, 2, sq unit, =, =, 6, 3, 1, , 5. Required area = ∫ (xex − xe− x ) dx, 0, , = [xex − ex + xe− x + e− x ] 10, 1 1 2, = e − e + + = sq units, e e e, 6. The given equation of curve,, x+ y= a, (x, y ≥ 0), y= a − x, ⇒, ⇒, ( y )2 = ( a − x )2, y = ( a − x )2, ⇒, At x = 0,, y = a ⇒ y=a, At y = 0,, x = a ⇒ x=a, So, curve cuts the axes at (a , 0) and (0, a ), respectively., ∴, , x + 1 + x2 − 2 x = 1, 2, , =, , ∴ Required area = ∫, , …(ii), , 3. Given equations of circle and line are, x2 + y2 = 1, and, x+ y=1, From Eqs. (i) and (ii), we get, ⇒, , 2, , y′, , 8 a 2 sin 4 θ dθ, , 3 1 π , = 8 a 2 ⋅ ⋅ (using gamma function), 4 2 2 , 2, 3πa, sq units, =, 2, , ⇒, , =, , Required area = ∫, , x=a, , x=0, a, , y dx, , = ∫ ( a − x )2 dx, 0, a, , = ∫ (a + x − 2 a x ) dx, 0, , , x2 4, = ax +, −, 2 3, , , a, , , ax3/ 2, 0, , a2 4, −, a ⋅ a3 / 2, 2 3, 3a 2 4 2 (9 − 8) 2 a 2, =, − a =, a =, 2, 3, 6, 6, = a2 +
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442, , NDA/NA Mathematics, , 7. The equation of circle is x2 + y2 = 9, ∴Area of the smaller segment cut off from the circle, y, , 9. Given that, y = x| x|, x2 : x ≥ 0 , or, y= 2, , −x : x < 0, y, , A (1, 2√2), , x, (3, 2), , x′, , (–1,0), , (1,0), O, , B (1, –2√2), x=1, , y′, , y′, , A =2∫, = 2⋅, , 1, , 1, 2, , 0, (− x2) dx +, = ∫, −1, , , 9 − x dx, 2, , 3, , , −1 x , 2, x 9 − x + 9 sin 3 , 1, , =, , π, , 1 , = 9 − sin −1 − 8 , , , 3 , 2, , , , 1, = 9 cos −1 − 8 , 3, , , , 1, , ∫0, , y dx, 0, , 1, , x 3 , −x 3 , (x ) dx = , + , , 3 0, 3 −1, 2, , 1 1 2, + = sq unit, 3 3 3, , A=∫, , π /4, , 0, , …(i), …(ii), , sin x dx, 1, 2, , =1 −, , = [9 sec−1 (3) − 8 ] sq units, 8. Given that, y = − x2 + 2x + 3, and, y=0, At, y = 0, − x2 + 2x + 3 = 0, ⇒, x2 − 2 x − 3 = 0, ⇒, (x + 1) (x − 3) = 0, ⇒, x = − 1, 3, At x = − 1, y = 0, ∴, and, at x = 3 , y = 0, , 1, , 10. Given that, the area between the curve y = sin x and, π, x-axis in the interval 0, is A, then in the same, 4, interval, area between the curve, y = cos x and x-axis is, , 9π, 1 , =, − 8 − 9 sin −1 , 3 , 2, , y, , 0, , ∫ −1 y dx + ∫ 0, , ∴ Area of the curve =, , x2 + y2 = 9 by x = 1, is given by, 3, , x, , π/ 4, , 1, 2, If the area along y = sin x = A, then area along, y = cos x = 1 − A, A1 = ∫, , and, , 0, , cos x dx =, , y1 = 2|x|, 2x, x > 0, y1 = , − 2x, x < 0, y2 = x2, , 11. We have,, or, and, , y, , …(ii), , y, , A, , y, , =, , y = – x2 + 2x + 3, , …(i), , 2x, , ), ,0, (0, , O, , y =x2, , x′, , y = x |x |, , =, –2, , (2,4), , x, , x′, , x′, , –1, , O, , O, , x, , x, , 3, , y′, , y′, , ∴ Required area = ∫, , B, x=1, , 3, −1, , (−x2 + 2x + 3) dx, 3, , , − x3, =, + x2 + 3x, 3, −1, , , , 1, = − 9 + 9 + 9 − + 1 − 3 , , 3, , , 5 , = 9 − − , 3 , , 32, sq units, =, 3, , On solving Eqs. (i) and (ii), the point of intersection of, Eqs. (i) and (ii) is (2, 4). The required area is given by the, shaded portion., ∴ Required area = 2 × area (OBA ), = 2∫, , 2, , 0, , = 2∫, , 2, 0, , ( y1 − y2) dx, 2, , , x3 , (2x − x2) dx = 2 x2 − , 3 0, , , 8, 4 8, , = 2 4 − = 2 × = sq units, 3, 3 3,
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443, , Area Bounded by Region, 12. Given, y = 4x − x2 − 3, ⇒, − (x2 − 4x) = y + 3, 2, ⇒, − (x − 4x + 4) = y + 3 − 4, ⇒, (x − 2)2 = − ( y − 1), This is a equation of parabola., , Now, when y = 0, then x = 1 and when x = 0, then y = 1, ∴Required area = ∫, =∫, =∫, , y, , 1, , 0, 1, 0, 1, 0, , y dx, (1 − x )2 dx, (1 + x − 2 x ) dx, 1, , , 2x3/ 2 , x2, = x +, −2⋅, , 2, 3 0, , 1 4, , , = 1 + − − (0), , , 2, 3, , , 9 −8 1, =, = sq unit, 6, 6, , y = 4x – x2 – 3, O, , x′, , x, (3,0), , (1,0), , y′, 3, , ∴ Required area = ∫ y dx, 1, , 3, , 3, , , x3, − 3x, = ∫ (4x − x2 − 3) dx = 2x2 −, 1, 3, , 1, , 1, 4, , = 18 − 9 − 9 − 2 − − 3 = sq units, 3, , 3, x, 3, x2 + y2 = 4, y2 = 4 − x2, , 13. We have, y =, and, ⇒, , 15. Given curve is 2x2 + y2 = 1, This curve can be written as, x2, y2, +, = 1, which is an ellipse., 1 /2, 1, 1, 1, where a 2 = , a =, and b2 = 1, b = 1, 2, 2, y, B (0, 1), , …(i), x′, , …(ii), , O (0, 0), , A, x, (1/√2, 0), , y, , y′, , ∴ Area of an ellipse = 4 × Area of circle ABO, , ( 3, 1), x´, O, , ( 3, 0), , (2, 0), , =4 ∫, , x, , =4 ∫, Let, , y´, , From Eqs. (i) and (ii), we get, x2, = 4 − x2 ⇒ x2 = 12 − 3x2, 3, ⇒, x2 = 3 ⇒ x = ± 3, From Eq. (i), y = 1, Radius of circle = 2, 3, 2, x, (4 − x2) dx, dx + ∫, ∴ Required area, A = ∫, 0, 3, 3, 14. Here, we have, x + y =1, y, √x + √y = 1, x′, , O, , y′, , x, , 1/ 2, 0, 1/ 2, , 0, , y dx = 4 ∫, , 1/ 2, 0, , 1 − 2x2 dx, , 1 − ( 2x)2dx, , t = 2x, dt = 2dx, 4 1, =, 1 − t 2 dt , , 2 ∫0, =, , 4 t, 1, 1 − t 2 + sin −1, , 2, 2 2, , 1, , , t, 0, , 4 1, 1, , ⋅ 0 + ⋅ sin −1 (1) − 0 − 0, 2, 2 2, , π, 4 1 π, =, ⋅ ⋅ =, 2, 2 2 2, Alternate Method, We know that the area of an ellipse = πab, π, 1, = π (1) =, 2, 2, =, , 16. Given, parabola is 4 y = 3x2, and the straight line 2 y = 3x + 12, From Eqs. (i) and (ii), we get, 3x2, = 3x + 12, 2, ⇒, 3x2 − 6x − 24 = 0, , …(i), …(ii)
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444, , NDA/NA Mathematics, y, , (–2, 3), , y, (0,1), , B(4, 12), , C, A (–4, 0)O (0, 0), , ⇒, , x2 − 2 x − 8 = 0, , ⇒, , x2 − 4 x + 2 x − 8 = 0, , ⇒, , x (x − 4) + 2 (x − 4) = 0, , ⇒, , (x − 4) (x + 2) = 0, , ⇒, , (–2,0), , x, , x′, , , x2 , [Q f (x) = 1 − is an even function ⇒ f (− x) = f (x)], 4, , 2 8, , = 2 2 − = sq units, , 3 3, , From Eq. (i), we get, x =4, y=, , 20. Given curves are y = cos x, y = cos 2x, x = 0, x =, , 3, ⋅ (4)2 = 12, 4, , 3, ⋅ (−2)2 = 3, 4, Then, the points of intersection are (4, 12) and (–2, 3)., The required area, 4 3 x + 12 3 2, =∫, − x dx, , −2 , 2, 4 , x = − 2, y =, , and at, , 2, , 2, , (x − 1) dx − ∫, , 1, , 1, , 0, , = sin, A2 = ∫, , =, , 1, , 2, , , x2, , x2, + x, = 2 − − x − −, 2, 2, 0, 1 , , 1, , 1, , = 2 − 2 − 2 − + 1 − − + 1, , 2, , 2, D (0,1) y = x – 1 C (2,1), , π, 3, −0 =, 3, 2, π, , π, 3, 0, , sin 2x 3, cos 2x dx = , 2 0, , 1, 2π, 1, sin, − 0 = cos 30°, , 2, 3, 2, 3, 4, , 3, A1, 2, 2, =, =, A2, 3 1, 4, , ∴, , y, y = –x + 1, , =, , (− x + 1) dx, , 0, , π, , π, , = Area (OBCD) – Area (∆ABC ) – Area (∆OAD ), 0, , π, 3, , A1 = ∫ 3 cos x dx = [sin x]03, , Let, , 17. Required area = Area of (∆ACD ), = ∫ 1 dx − ∫, , x, , (2,0), , y′, , x = 4, − 2, , At, , O, , 21. Given equation of circle is x2 + y2 = π 2 and curve is, y = sin x, y, , x′, (0,0) O, , B (2,0), , A, (1,0), , x, , π, , x′, , O, , y′, , =2−, , x, , 1 1, + − 1 = 2 − 1 = 1 sq unit, 2 2, , 18. Required area = ∫, , 1, 0, , x dx − ∫, , 1, 0, , (x − 1) dx, π, , 1, , 1, , ∴ Required area = ∫ ( y1 − y2) dx, , , x2, x2 , = − − x, 0, 2 0 2, =, , 0, , π, , 1 1, , − − 1 = 1 sq unit, , 2 2, , 19. Required area = ∫, , 2, −2, , , x2 , 1 − dx, 4, , 3 2, , y′, , , , x, 2 , = 2 x −, , = 2 2 −, 12 0, 12 , , , 3, , = ∫ ( π 2 − x2 − sin x)dx, 0, , π, , , x, π2, x, π 2 − x2 +, sin −1 + cos x, =, 2, 2, π, 0, , , , π2 π, = 0 +, − 1 − (0 + 0 + 1), , , 2, 2, , , =, , π3, π3 − 8, sq unit, −2 =, 4, 4
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445, , Area Bounded by Region, y, , 22. (A) The intersection point of, y = x + 1 and x + y = 3, 2, , B (0,1), , y = x 2 +1, , y, (–2,5) C, , x′, A (1,2), B (0,1), x+y=3, x, O, , x′, , y′, , (A) Area of bounded region = Area of curve OAB, 1, , 1, , 0, , 0, , = ∫ y dx = ∫ (1 − x) dx, 1, , y′, , , 1 1, x2 , , = x − = 1 − =, 2, 2 2, , 0, , , 3−x=x +1, 2, , is, , A (1,0), (–1,0), x, O (0,0), C, y=1+x, y=1–x, , ⇒, , x + x−2 =0, , ⇒, , (x + 2)(x − 1) = 0, , 2, , (R) It is clear from the above figure that curve is, symmetrical about y-axis., , ⇒, , x = − 2, 1, , ∴ It is true that total area of curve ABC is, , ⇒, , y = 5, 2, , area of curve OAB., , ∴ Area of bounded region, = Area of shaded region ABC, =∫, =∫, =∫, , 1, −2, 1, −2, 1, −2, , ( y2 − y1 ) dx, , Solutions (Q. Nos. 24-26), The intersection point of given curve y2 = 2x and lines, x = 8 and y = 4 is A(8, 4)., y, , [(3 − x) − (x2 + 1)] dx, , A (8, 4), (0, 4) B A1, , (2 − x − x2) dx, 3 1, , , x2 x, = 2x − − , 2 3 − 2, , 8 , 1 1 , = 2 − − − −4 − 2 + , 3 , 2 3 , 8, 3 1, = − +6− , 3, 2 3, =, , 1, of the, 2, , 9 − 2 + 36 − 16 27 9, =, =, 6, 6 2, , (R) When we rotate the figure the area of the bounded, region is not change., Hence, A is true but R is false., , x′, , O, , C (8, 0), x, A2, D, , y′, , 24. Required area, A1 = Area of curve OAB, 4, 4, 4 y2, y3 , = ∫ x dy = ∫, dy = , 0, 0 2, 6 0, 64 32, sq units, =, =, 6, 3, 25. Required area, A2 = Area of curve OCD, 8, , 8, , 0, , 0, , = ∫ y dx = ∫, , 1 9, , 23. Given curves is, y = 1 − |x|, 1 + x, x < 0, =, 1 − x, x > 0, , +, 2 2 3/ 2 2, 26 64, [8 ] = × 2 2 2 =, =, 3, 3, 3, 3, A2 64 / 3 2, 26. ∴ Required ratio =, =, =, A1 32 / 3 1, , =, , 8, , x3/ 2 , 2 x dx = 2 , , 3 / 2 0
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23, , Differential Equations, An equation involving the independent variable and, dependent variable and also the derivatives of the, dependent variable with respect to independent variable is, known as a differential equation., d2y, dy, e.g., x, + xy, + 4 = 0 is a differential equation., dx, dx 2, , Order of a Differential Equation, The order of a differential equation is the order of the, highest derivative (differential coefficient) involved in its, expression., d3 y, dy, d2y, e.g., differential equation, +x, + xy, + 4= 0, 3, dx, dx, dx 2, is of order 3. The differential equation, of order 2., , Degree of Differential, Equation, The highest exponent of the highest derivative is, called degree of a differential equation provided exponent, of each derivative and the unknown variable appearing in, the differential equation is a non-negative integer., d2y, dy, e.g., (i), +, +x=0, 2, dx, dx, , dy d 2 y, = 2 + x, dx dx, , , ⇒, 2, , d 2 y, d2y, dy , 2 − + 2x, + x2 = 0, 2, , , dx, dx, dx, , , , ⇒, , Here, degree is 2., (ii), , 2, , d3 y, 3, dx, , , , , , 2/ 3, , +x+ y=0, , d3 y , 3 = ( − x − y )3, dx , , So, degree is 2., , Example 1., , The order and degree of the differential, 3/ 2, dy 2, d 2y , equation 1 + = k 2 are respectively, dx , dx , (a) 2, 2, (b) 2, 3, (c) 3, 4, (d) 1, 5, , Solution (a) The given equation is, dy 2, 1 + , dx , , , 3, , dy , + = 0 is, 2, dx , dx, , d2y, , 2, , ⇒, , 3/ 2, , d 2y , = k 2, dx , , 3, , 2, , 2, dy 2, 2 d y , 1 + = k 2 , dx , dx , , This shows that the degree and order of the given differential, equation are 2 and 2., , ⇒, , Formation of a Differential, Equation, The differential equation of a family of curves of one, parameter is a differential equation of the first order. The, differential equation of a family of curves of two, parameters is a differential equation of the order two and, the differential equation of a family of curves of n, parameters is a differential equation of n order., If the family of curves have one parameter, then we, differentiate it once and eliminate parameter using, equation of family of curves and equation, we get after, differentiation. e.g., x 2 + y 2 = a 2 ( a is a parameter),, represents family of concentric circles., …(i), x2 + y2 = a2, Differentiate Eq. (i), we get, dy, =0, 2x + 2 y ⋅, dx, , …(ii)
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447, , Differential Equations, ⇒ x dx + y dy = 0 is a differential equation., Family of curves of two parameters will be, differentiated twice to get a relation independent of any, parameter., Similarly, for family of curves of n parameters will be, differentiated n times and then eliminate all the, parameters., e. g. ,, y = a sin µx + b cos µx, (where a and b are parameters), dy, ⇒, = aµ cos µx − bµ sin µx, dx, d2y, = − aµ 2 sin µx − bµ 2 cos µx, ⇒, dx 2, d2y, ⇒, = − µ 2 ( a sin µx + b cos µx ), dx 2, d2y, + µ 2 y = 0 is a differential equation, ⇒, dx 2, , Example 2. The differential equation corresponding to, y 2 = m ( a 2 − x 2) is, d 2y dy 2, dy, dy, (b) 2x, (a) x y 2 + = y, =y, dx , dx, dx, dx, , dy, (d) None of these, (c) x 2, =1, dx, Solution (a) Given differential equation is, y 2 = m ( a2 − x2), On differentiating wrt x, we get, dy, 2y, = m ( −2x), ⇒, dx, dy, y, = − mx, ⇒, dx, Again, differentiating wrt x, we get, 2, d 2y dy , y 2 + = −m, ⇒, dx , dx, , …(i), …(ii), , …(iii), , From Eqs. (ii) and (iii), we get, d 2y dy 2, dy, x y 2 + = y, , , dx , dx, dx, %, , Here, parameters are also called arbitrary constants., , Solution of a Differential Equation, Any relation connecting the variables of an equation, and not involving the differential coefficients such that, this relation and the derivatives obtained from it satisfy, the given differential equation, is called a solution of the, differential equation., , General Solution, The solution which contains a number of arbitrary, independent constants equal to the order of the differential, , equation is called the general solution or the complete, primitive of the equation., , Particular Solution, The solution obtained from the general solution by, assigning particular values to one or more of the arbitrary, constants are called particular solutions., , Different Forms of First Order and, First Degree Differential Equations, Variable Separable Differential Equations, f ( x ) dx = g( y ) dy, Method Integrate it on both sides, we get, ∫ f ( x ) dx = ∫ g( y ) dy + C, , Example 3. The solution of, (a) x = e y + C, (c) y = e x + C, , dy, = e x + y is, dx, (b) − e − y = e x + C, (d) None of these, , dy, = ex + y, dx, dy, ⇒, = ex ⋅ ey, dx, Separating the variables, we get, ⇒, e− ydy = ex dx, , Solution (b) Q, , On integrating both sides, we get, ⇒, − e− y = ex + C, , Reducible to Variable Separable Differential, Equation, Some times differential equation does not take directly, form of the type f ( x ) dx = g ( y )dy but after some, substitution, we get this form., dy, e.g.,, =x+ y, dx, dy dt, Put, x+ y= t ⇒ 1+, =, dx dx, dy dt, =, −1, ⇒, dx dx, dt, dt, So,, −1= t ⇒, = dx, dx, t+1, Now, this reduces to variable separable differential, equation., , Example 4. The solution of, (a) sin( x + y) = C, ( x + y), (c) tan, = x+C, 2, , Solution (c) Given,, Let x + y = t, , dy, = cos( x + y) is, dx, (b) cos( x 2 + y 2) = 2C, (d) None of these, , dy, = cos ( x + y), dx
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450, , NDA/NA Mathematics, , P=, ∴, , IF = e∫, , y − constant, , P dx, 2x, , =e, , dy = C, , ∫ 1 + x2, , dx, , y ⋅ (1 + x2) = ∫, , provided, , 2, , = elog(1 + x ) = 1 + x2, , ∴ The general solution is, y ⋅ (IF) = ∫ Q ⋅ (IF) dx + C, ⇒, , The solution of Mdx + Ndy = 0 is, Mdx + ∫, (Terms of N not tonating x), ∫, , 2x, 1, and Q =, 2, 1+ x, 1 + x2, , 1, ⋅ (1 + x2) dx + C, (1 + x2), , ⇒, , y (1 + x2) = ∫ 1dx + C, , ⇒, , y (1 + x2) = x + C, , Since, y = 0 wh en x = 0, ∴, C =0, Thus, the required solution is y (1 + x2) = x, , Equation Reducible to the Exact Form, Sometimes a differential equation of the form, Mdx + Ndy = 0 which is not exact can be reduced to an, exact form by multiplying by a suitable function f ( x , y ), which is not identically zero. This function f ( x , y ) which, then multiplied to a non-exact differential equation makes, it exact is known as integrating factor., One can find integrating factors by inspection but for, that some experience and practice is required., , Example 9. The solution of, , ( x 2 − ay) dx + (y 2 − ax) dy = 0 is, (a) x − 3axy + y3 = a, (b) x3 + y3 = a, 2, 2, (c) x + y = 2 a, (d) None of these, , Exact Differential Equation, A, differential, equation, of, the, form, M ( x , y )dx + N ( x , y )dy = 0is said to be exact (or total), if its, left hand expression is the exact differential of some, function u( x , y )., i.e.,, du = Mdx + Ndy, Hence, its solution is u( x , y ) = C (where, C is an, arbitrary constant). But then, there is a question that how, do we confirm whether the above mentioned equation is, exact. The answer to this question is the following, theorem:, Theorem The necessary and sufficient condition for, the differential equation of Mdx + Ndy = 0 to be exact is, ∂M ∂N, ., =, ∂y, ∂x, , ∂M ∂N, =, ∂x, ∂y, , 3, , Solution (a) Here, we have M = x2 − ay and N = y 2 − ax, ∴, , ∂M, ∂N, =−a=, ∂y, ∂x, , Thus, the equation is exact., ∴ The solution is, 2, ∫ y −constant ( x − ay) dx +, ⇒, ⇒, , ∫ y dy = C, , x3, y3, − axy +, =C, 3, 3, x3 − 3axy + y3 = a, where a = 3 C, , Comprehensive Approach, n, n, , n, , n, , n, , n, , d ( x + y) = dx + dy, d ( xy) = y dx + xdy, x y dx − xdy, d =, y , y2, xdy − y dx, y, d =, x, x2, y dx + xdy, d [log( xy)] =, xy, xdy − y dx dy dx, y, d log =, = −, , x , xy, y, x, , n, , n, , n, , n, , n, , 1, x + y xdy − y dx, d log, = 2, 2, x−y, , x − y2, xdy − y dx, y, d tan−1 = 2, , x , x + y2, xdx + y dy, d ( x2 + y 2 ) =, x2 + y 2, d ( x2 + y 2) = dx2 + dy 2 = 2 xdx + 2y dy, 1 dx dy, 1 1, 1, d − = d − d = 2 − 2, x x y, y x, y , , 2
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Exercise, Level I, 1. The, , degree, , of, , differential, , equation, , (a) 1, , (b) 3, , y=x, (c) 2, , (d) 5, , 2. The solution of the equation ( 2 y − 1) dx − ( 2x + 3), dy = 0 is, 2x − 1, 2y + 1, (b), (a), =C, =C, 2y + 3, 2x − 3, 2x + 3, 2x − 1, (c), (d), =C, =C, 2y − 1, 2y − 1, 3. The solution of, (a) y = Ce∫ Pdx, (c) y = Ce− ∫ P dx, , dy, + P ( x ) y = 0 is, dx, (b) x = Ce− ∫ Pdy, (d) x = Ce∫ P dy, , dy, 4. The solution of, + y = e− x , y ( 0) = 0 is, dx, (b) y = xex, (a) y = e− x ( x − 1), −x, (d) y = xe− x, (c) y = xe + 1, , (c) 4, , (NDA 2010 II), , (c) 8, , 7. What is the solution of the differential equation, dy, = ex − y ( e y − x − e y )?, (NDA 2009 I), dx, (a) y = x − ex + C, (b) y = x + ex + C, x − y, y, (c) y = e, (d) None of these, − e +C, 8. What is the solution of the differential equation, dy, = xy + x + y + 1?, (NDA 2008 I), dx, 2, x, x2, (b) log ( y + 1) =, (a) y =, + x+C, + x+C, 2, 2, 2, 2, (c) y = x + x + C, (d) log ( y + 1) = x + x + C, 9. What is the general solution of, (1 + ex ) y dy = ex dx?, , (a) 1, (c) − 1, , (NDA 2012 I), , (b) 2, (d) Degree does not exist, , 11. The order and degree of the differential equation, 3, 4, d 2 y, dy , 2 + − xy = 0 are respectively, dx , dx , (a) 2 and 4, (b) 3 and 2, (c) 4 and 5, (d) 2 and 3, 12. The, differential, equation, sin−1 x + sin−1 y = c, is given by, , for, , which, , (a) 1 − x 2 dx + 1 − y 2 dy = 0, (b) 1 − x 2 dy + 1 − y 2 dx = 0, 1 − y 2 dy = 0, dx dy, +, = 0 is, x, y, (b) x + y = C, (d) x 2 + y 2 = C, , 13. Solution of the differential equation, , 2, , (b) 2, , −1, , (d) 1 − x 2 dx −, , 6. What is the degree of the differential equation, 4, d 2 y, dy , , 1, +, = 2 ?, , , dx , dx , , dy dy , + , dx dx , , (c) 1 − x 2 dy − 1 − y 2 dx = 0, , 5. The, solution, of, differential, equation, ( x + y ) ( dx − dy ) = dx + dy is, (b) x + y = Kex + y, (a) x − y = Kex − y, x − y, (c) x + y = Ke, (d) ( x − y ) = Kex + y, , (a) 1, , 10. What is the degree of the differential equation?, , 3, , dy , + + 6 y = 0 is, 2, dx , dx, , d2y, , (NDA 2010 II), , (a) y 2 = log [C 2( ex + 1)2 ] (b) dy = log [C( ex + 1)], (d) None of these, (c) y 2 = log [C( ex + 1)], (where C is a constant of integration), , (a) xy = C, (c) log x log y = C, , 14. What is the solution of the differential equation, 1 − y2, dy, = 0?, +, (NDA 2011 I), dx, 1 − x2, (a) sin−1 y + sin−1 x = C (b) sin−1 y − sin−1 x = C, (c) 2 sin−1 y + sin−1 x = C (d) 2 sin−1 y − sin−1 x = C, (where, C is a constant of integration.), 15. What is the degree of the differential equation, 3, , dy , − 1 + = 0?, dx , dx 2, , d2y, (a) 1, (c) 3, , (NDA 2010 I), , (b) 2, (d) 6, , 16. What is the solution of the differential equation, (NDA 2010 I), 3ex tan y dx + (1 + ex ) sec2 y dy = 0 ?, (b) (1 + ex )3 tan y = C, (a) (1 + ex ) tan y = C, (c) (1 + ex )2 tan y = C, (d) (1 + ex ) sec2 y = C, (where, C is a constant of integration.), dy, 17. The solution of, + 1 = cosec ( x + y ) is, dx, (a) cos ( x + y ) + x = C, (b) cos ( x + y ) = C, (c) sin( x + y ) + x = C, (d) sin ( x + y ) + sin ( x + y ) = C
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452, , NDA/NA Mathematics, , 18. The order and degree of the differential equation, 2, d 2 y dy , + y 2 = 0 are respectively, x, +, , , 2, , , dx, dx, (a) 2 and 2, (c) 2 and 1, , (b) 1 and 1, (d) 1 and 2, , 19. What is the degree of the differential equation, 2, , d3 y 3, d2y, dy, 3 − 3 2 + 5, + 4 = 0?, dx, dx, dx, , , 2, (a) 3, (b), 3, (c) 2, (d) 6, 20. y = A cos ωt + B sin ωt is a solution of the differential, equation, d2y, d2y, (a), (b), − ω2y = 0, − ωy = 0, 2, dt, dt 2, d2y, d2y, (d), (c), +, ω, =, 0, + ω2y = 0, y, dt 2, dt 2, 21. y = aex + be− x satisfies the differential equation, d2y, d2y, (a), (b), − y=0, + y=0, 2, dx, dx 2, d 2 y dy, d 2 y dy, (d), (c), y, +, +, =, 0, +, − y=0, dx 2 dx, dx 2 dx, 22. What is the solution of the differential equation, (NDA 2007 II), ( x + y )( dx − dy ) = dx + dy ?, (a) 2 log ( x + y ) = C( y − x ), (b) ( y − x ) + log ( x + y ) = C, y , y , (c) + log = C, x, x , (d) None of the above, 23. What is the solution of the differential equation, x dy − y dx = xy 2 dx ?, (NDA 2008 II), (a), (b), (c), (d), , yx 2 + 2x = 2 Cy, y 2x + 2 y = 2 Cx, y 2x 2 + 2x = 2 Cy, None of the above, , (1 + x ), , (a) e 2 − 1, (c) log (1 + x ) − 1, , (a) x3 y3 = x 2 + C, (c) x3 y3 = x 2 + x + C, , (1 − x ), , 2, , (b) e 2, (d) log (1 − x ), , (b) 2x3 y3 = 3x 2 + C, (d) x3 y3 = 3x 2 + C, , 27. What is the solution of the differential equation, dy, y, ?, =, dx ( x + 2 y3 ), (a) y (1 − xy ) = Cx, (c) x (1 − xy ) = Cy, , (b) y3 − x = Cy, (d) x (1 + xy ) = Cy, , 28. y = aemx + be− mx satisfies which of the following, differential equation?, dy, dy, (a), (b), − my = 0, + my = 0, dx, dx, 2, 2, d y, d y, (d), (c), + m2 y = 0, − m2 y = 0, 2, dx, dx 2, 29. The differential equation of system of concentric, circles with centre (1, 2) is, dy, dy, (a) ( x − 2) + ( y − 1), = 0 (b) ( x − 1) + ( y − 2), =0, dx, dx, dy, dy, (c) ( x − 1), + ( y − 2) = 0 (d) ( x + 2), + ( y − 1) = 0, dx, dx, 30. The solution of the differential equation, 1, dy, 2 yx, is, =, +, 2, dx 1 + x, (1 + x 2 )2, (a) y (1 + x 2 ) = C + tan−1 x, y, (b), = C + tan−1 x, 2, 1+ x, (c) y log (1 + x 2 ) = C + tan−1 x, (d) y (1 + x 2 ) = C + sin−1 x, 31. What is the degree of the following differential, equation?, d3 y , 3 , dx , , 2/ 3, , + 4− 3, , (a) 1, (c) 3, , dy, 24. What does the differential equation y, +x= A, dx, represent?, (NDA 2011 I), (a) A set of circles having centre on the y-axis, (b) A set of circles having centre on the x-axis, (c) A set of ellipses, (d) A pair of straight lines, (where A is a constant.), dy, 25. If, = 1 + x + y + xy and y( −1) = 0, what is the value, dx, of y( x )?, (NDA 2009 I), 2, , 26. What is the solution of x 2 y 2dy = (1 − xy3 ) dx ?, , d2y, dx, , 2, , +5, , dy, =0, dx, , (NDA 2011 I), , (b) 2, (d) 4, , 32. What is the equation of the curve whose slope at any, point is equal to 2x and which passes through the, origin?, (NDA 2008 II), 2, 2, 2, (a) y(1 − x ) = x, (b) y (1 + x ) = x 4, 2, 2, (c) y = ( x + 1), (d) y = x 2, 33. The solution of 2 ( y + 3) − xy, where x = 1, is, (a) y + 3 = x 2, (c) x 4 ( y + 3) = 1, , dy, = 0 with y = − 2,, dx, , (b) x 2 ( y + 3) = 1, (d) x 2 ( y + 3)3 = e y + 2, , dy, + y tan x = sec x is, dx, (a) y sec x = tan x + C, (b) y tan x = sec x + C, (c) tan x = y tan x + C, (d) x sec x = tan y + C, , 34. The solution of
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453, , Differential Equations, 35. An integrating factor of the differential equation, 1, − log x, dy, x, + y log x = xex x 2, ,( x , 0) is, dx, log x, (a) x, (b) ( x )log x, (c) ( e )( log x ), , 2, , (d) ex, , 2, , 36. Under which one of the following conditions does the, dy ax + b, represent a parabola?, solution of, =, dx cy + d, (NDA 2007 I), (a) a = 0 and c = 0, (b) a = 1, b = 2 and c ≠ 0, (c) a = 0, c ≠ 0 and b ≠ 0, (d) a = 1 and c = 1, 37. What does the equation x dy = y dx represent?, (NDA 2009 II), , (a), (b), (c), (d), , (a), , (b) 0, (d) 2, , 39. If y = a cos 2x + b sin 2x , then, d2y, (b), (a), + y=0, dx 2, d2y, (d), (c), + 3y = 0, dx 2, , d2y, dx 2, d2y, dx 2, , + 2y = 0, + 4y = 0, , 40. What is the solution of the differential equation?, dy y, (NDA 2012 I), + =0, dx x, (a) xy = C, (c) y = Cx, where C is a constant., , (b) x = Cy, (d) None of these, , 41. A general solution of the differential equation, is given by, (a) y = Ae x, (c) y = ex, 42. The solution of, y dx + x dy = 0 is, (a) xy = C, x, (c) = C, y, , dy, =y, dx, , (b) y = eAx, (d) y = e− Ax, the, , 45. Which one of the following differential equations is, 2, , 38. What is the order of the differential equation, 1, dy, + y=, ?, dy , dx, (NDA 2008 II), , dx , , differential, (b) x + y = C, y, (d) = C, x, , equation, , x2 + y2, , , y(1) = 2 has, x2 − y2, the slope at the point (1, 0) of the curve, equals to, (a) − 5/ 3, (b) − 1, (c) 1, (d) 5/ 3, , 44. Integral curve satisfying y ′ =, , not linear?, , A family of circles, A family of parabolas, A family of hyperbolas, A family of straight lines, , (a) –1, (c) 1, , dy, + y = sin x is, dx, 1, 1, (a) y = Ce−2x + sin x − cos x, 4, 2, 1, 1, −x, (b) y = Ce + sin x − cos x, 2, 2, (c) y = Ce−3 x + sin x, (d) y = Ce− x, , 43. The general solution of, , d y, dx, , 2, , (NDA 2012 I), , + 4y = 0, , (c) ( x − y )2, , dy, =9, dx, , dy, + y = x3, dx, dy, (d) cos2 x, + y = tan x, dx, (b) x, , 46. An integrating factor of the differential equation, dy, (1 − x 2 ), − xy = 1 is, dx, x, (a) − x, (b) −, (1 − x 2 ), 1, (c) (1 − x 2 ), (d) log (1 − x 2 ), 2, 47. The second order differential equation of a parabola, with its principal axis along the x-axis is, (a) y ′′ + 2 ( y ′ )2 = 0, (b) 3 yy ′′ + 2 ( y ′ )3 = 0, 2, (c) yy ′′ + ( y ′ ) = 0, (d) y ′′ + 2( y ′ )4 = 0, dy, = 2 y − x is, dx, (a) 2x + 2 y = C, (b) 2x − 2 y = C, 1, 1, 1, 1, (d) x + y = C, (c) x − y = C, 2, 2, 2, 2, , 48. The solution of, , 49. The integrating factor of the differential equation, dy , cos x + y sin x = 1 is, dx , (a) sec x, (b) tan x, (c) sin x, (d) cos x, d2y, , + sin x = 0, then the solution of differential, dx 2, equation is, (a) y = sin x + Cx + D, (b) y = cos x + Cx 2 + D, (c) y = tan x + C, (d) y = log sin x + Cx, , 50. If
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454, , NDA/NA Mathematics, , Level II, 1. The order and degree of the differential equation, 2, , dy , 1 + , dx , , ρ=, 2, d y / dx2, , 3/ 2, , are respectively, , (a) 2 and 2, (c) 2 and 1, , (b) 2 and 3, (d) None of these, , 2. What is the degree of the differential equation, 3/ 2, 3, d2y , dy , K, = 1 + ?, (NDA 2007 I), dx , dx 2 , , (a) 1, , (b) 2, , (c) 3, , (d) 4, , (where K is a constant.), , 3. What are the order and degree, respectively of the, differential equation, {( d 4 y/ dx 4 )3 } 2/ 3 − 7x( d3 y/ dx3 )2 = 8 ? (NDA 2008 II), (a) 3 and 2, (c) 4 and 2, , (b) 4 and 3, (d) 3 and 3, , 4. The differential equation of the family of curves, for, which the length of the normal is equal to a constant, K is given by, 2, dy, dy , 2, 2, (a) y 2, (b) y, = K 2 − y2, =K −y, dx , dx, 2, 2, dy , dy , (c) y = K 2 + y 2, (d) y = K 2 + y 2, dx , dx , 5. The degree of the differential equation, 2, d 2y, d 2y, dy, + 3 = x2 log 2 is, 2, dx, dx, dx , , (a) 1, (c) 3, , (b) 2, (d) None of these, , 6. The slope of the tangent at ( x , y ) to a curve passing, x2 + y2, , then the equation of, through a point (2, 1) is, 2xy, the curve is, (a) 2 ( x 2 − y 2 ) = 3x, (c) x ( x 2 − y 2 ) = 6, , (b) 2 ( x 2 − y 2 ) = 6 y, (d) x ( x 2 + y 2 ) = 10, , 7. The solution of the differential equation, x dy − y dx = x 2 + y 2 dx is, (a) x + x 2 + y 2 = Cx 2, , (b) y − x 2 + y 2 = Cx, , (c) x − x 2 + y 2 = Cx, , (d) y + x 2 + y 2 = Cx 2, , 8. The solution of the differential equation, dy, = y tan x − 2 sin x is, dx, (a) y sin x = C + sin 2x, (c) y cos x = C − sin 2x, , 1, sin 2x, 2, 1, (d) y cos x = C + cos 2x, 2, , (b) y cos x = C +, , 9. Which one of the following represents the differential, equation of all parabolas having the axes of, symmetry coincident with the axis of x?, 2, 2, d 2 y dy , d3 y d 2 y , , , (b), (a) y, =, +, 0, y, +, = 0, , dx 2 dx , dx3 dx 2 , 2, , (c) y, , dy d 2 y , =0, +, dx dx 2 , , 2, , (d) y, , dy d3 y , = 0, +, dx dx3 , , 10. What are the order and degree, respectively of the, differential equation representing the family of, curves y 2 = 2c ( x + c), where c is a positive, parameter?, (a) 1, 2, (b) 1, 1, (c) 1, 3, (d) 1, 4, 11. Which one of the following differential equations, represents the system of circles touching the y-axis at, the origin?, dy , dy , (a) x 2 + y 2 − 2xy = 0 (b) x 2 + y 2 + 2xy = 0, dx , dx , dy, , , dy , (c) x 2 − y 2 + 2xy = 0 (d) x 2 − y 2 − 2xy = 0, dx , dx , 12. What is the integrating factor of ay dx + bx dy = 0?, (a) x a − 1 y b − 1, (c) x a + 1 y b + 1, , (b) x a y b, xb, (d) a, y, , 13. What is the differential equation of all parabolas, whose axes are parallel to y-axis?, (NDA 2011 I), 3, 2, d y, d x, (a), (b), =0, =C, dx3, dy 2, d3 x, d3 y, (d), (c), =, 1, =C, dy3, dx3, 14. What does the solution of the differential equation, dy, (NDA 2011 II), = y represent?, x, dx, (a) Family of straight lines passing through the, origin, (b) Family of circles with their centres at the origin, (c) Family of parabolas with their vertices at the, origin, (d) Family of straight lines having slope 1 and not, passing through the origin, 15. What is the solution of the differential equation, dy, = sec ( x + y )?, (NDA 2007 I), dx, ( x + y ), (a) y + tan ( x + y ) = C, (b) y − tan , =C, 2 , ( x + y ), ( x − y ), (c) y + tan , = C (d) y + tan , =C, 2, , , 2
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455, , Differential Equations, 16. The differential equation satisfied by the family of, , 1, curves y = ax cos + b , where a , b are parameters,, , x, is, (a) x 2 y2 + y = 0, (b) x 4 y2 + y = 0, (d) x 4 y2 − y = 0, (c) xy2 − y = 0, 17. Solution, of, the, differential, equation, tan y sec2 x dx + tan x sec2 y dy = 0 is, (a) tan x + tan y = K, (b) tan x − tan y = K, tan x, (d) tan x tan y = K, (c), =K, tan y, 18. Which one of the following differential equations is, 2, satisfied by the family of curves y = 2 + c ⋅ e−2x ?, dy, dy, (b), (a), = 8x, + 4xy = 8x, dx, dx, dy, dy, (d), (c), + 4xy = 0, − 4xy = 0, dx, dx, 19. The solution of the differential equation, dy, x4, + x3 y + cosec ( xy ) = 0 is equal to, dx, (a) 2 cos( xy ) + x −2 = C, (b) 2 cos( xy ) + y −2 = C, −2, (c) 2 sin( xy ) + x = C, (d) 2 sin( xy ) + y −2 = C, 20. What is the equation of the curve passing through, π, the point 0, satisfying the differential equation, 3, (NDA 2011 II), sin x cos y dx + cos x sin y dy = 0?, 3, 3, (b) sin x sin y =, (a) cos x cos y =, 2, 2, 1, 1, (c) sin x sin y =, (d) cos x cos y =, 2, 2, 21. If the solution of the differential equation, dy ax + 3, represents a circle, then what is the, =, dx 2 y + f, value of a?, (NDA 2010 II), (a) 2, (b) 1, (c) − 2, (d) − 1, 22. Which one of the following is the differential, equation to family of circles having centre at the, origin?, (NDA 2010 II), dy, dy, (a) ( x 2 − y 2 ), (b) ( x 2 + y 2 ), = 2xy, = 2xy, dx, dx, dy, (c), (d) x dx + y dy = 0, = (x2 + y2 ), dx, dy, 23. The solution of, = 1 − x 2 − y 2 + x 2 y 2 is, dx, (a) sin−1 y = sin−1 x + C, 1, 1, (b) sin−1 y =, 1 − x 2 + sin−1 x + C, 2, 2, 1, 1, −1, 2, (c) sin y = x 1 − x + sin−1 x + C, 2, 2, 1, 1, (d) sin−1 y = x 1 − x 2 + cos−1 x + C, 4, 2, , 24. The differential equation of the system of circles, touching the y-axis at the origin, is given by, dy, dy, (a) x 2 + y 2 − 2xy, = 0 (b) x 2 + y 2 + 2xy, =0, dx, dx, dy, dy, (c) x 2 − y 2 + 2xy, = 0 (d) x 2 − y 2 − 2xy, =0, dx, dx, 25. In, , order to solve the differential equation, dy, + y ( x sin x + cos x ) = 1, the integrating, dx, factor is, (a) x cos x, (b) x sec x, (c) x sin x, (d) x cosec x, , x cos x, , 26. The solution of the differential equation, dy, y, 1, +, = is, dx x log x x, 1, (a) y = log x + C(log x )−1 (b) y = log x + C(log x )−1, 2, 1, 1, C, (c) y = log x +, (d) y = log x − C (log x )−1, 2, 3, 2, (log x ), 27. The, differential, equation, ( x 2 y3 + xy 2 + y ) dx + ( x3 y 2 − x 2 y + x ) dy = 0 is not, exact. The integrating factor in order to convert it, into exact form is, 1, (b) 2 2, (a) x 2 y 2, x y, 1, 2 2, (d) 2 2, (c) 2x y, 2x y, 28. What does the differential equation y, , dy, +x=K, dx, , represent?, (NDA 2010 II), (a) A family of a circle whose points are on y-axis, (b) A family of a circle whose points are on x-axis, (c) Touching the x-axis a family of a circle, (d) None of the above, 29. What is the differential equation to family of, parabolas having their vertices at the origin and foci, on the x-axis?, (NDA 2010 II), (a) y = 2xy ′, (b) x = 2 yy ′, (c) xy = y ′, (d) x = yy ′, 30. What is the solution of the differential equation, dy, , dy, ?, + 2 y = xy, a x, (NDA 2010 I), , dx, dx, (a) x 2 = Kye y/ a, (c) y 2x 2 = Kye y, , (b) yx 2 = Kye y/ a, 2, , /a, , (d) None of these, , 31. If y = p ( x ) is a polynomial of degree 3, then what is, d 3 d 2 y, the value of 2, ?, y, dx dx 2 , 2, , (a) p′ ( x ) p′ ′ ′ ( x ), (b) p′ ′ ( x ) p′ ′ ′ ( x ), (c) p ( x ) p′ ′ ′ ( x ), (d) A constant
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456, , NDA/NA Mathematics, , 32. What is the degree of the equation, d 2 y , 2 = y +, dx , (a) 1, (c) 3, , 1, 2 4, , dy , ?, dx , , (b) 2, (d) 4, , 33. The general solution of the differential equation, dy, x − y, x + y, + sin , is, = sin , 2 , 2 , dx, y, (a) log tan = C − 2 sin x, 2, x, y, (b) log tan = C − 2 sin , 2, 4, y π, (c) log tan + = C − 2 sin x, 2 4, x, y π, (d) log tan + = C − 2 sin , 2, 4 4, 34. When a and b are eliminated from the equation, xy = aex + be− x , the resulting differential equation is, of, (NDA 2009 II), (a) first order and first degree, (b) first order and second degree, (c) second order and first degree, (d) second order and second degree, 35. What is the solution of the differential equation, − cosec2 ( x + y ) dy = dx ?, (NDA 2008 II), (a), (b), (c), (d), , y − C = sin ( x + y ), x − C = sin ( x + y ), y − C = tan ( x + y ), None of the above, , 36. What are the degree and order, respectively of, differential equation of the family of rectangular, hyperbolas whose axes of symmetry are the, coordinate axes?, (NDA 2009 I), (a) 1 and 1, (b) 1 and 2, (c) 2 and 1, (d) 2 and 2, 37. Which one of the following equations represents the, differential equation of circles, with centres on the, x-axis and all passing through the origin?, dy x 2 + y 2, dy x 2 − y 2, (a), (b), =, =, dx, dx, 2xy, 2xy, 2, 2, dy y − x, dy, x, (d), (c), =, =−, dx, 2xy, dx, y, 38. What is the degree of the differential equation, −4, dy , dy, , + x = y − x, ?, , dx , dx, (a) 2, (b) 3, (c) 4, (d) 5, , 39. What does the solution of the differential equation, x dy − y dx = 0 represent?, (a) Rectangular hyperbola, (b) Straight line passing through (0, 0), (c) Parabola with vertex at (0, 0), (d) Circle with centre at (0, 0), 40. Which one of the following differential equations, represents the system of circles touching y-axis at, the origin?, dy, dy, (a), (b) 2 xy, = x2 − y2, = y2 − x2, dx, dx, dy, dy, (d), (c) 2 xy, = y2 − x2, = x2 − y2, dx, dx, 41. What is the only solution of the initial value problem, (NDA 2007 II), y ′ = t(1 + y ), y( 0) = 0 ?, (a) y = − 1 + et, (c) y = − t, , 2, , (b) y = 1 + et, (d) y = t, , /2, , 2, , /2, , 42. What is the differential equation of the curve, y = ax 2 + bx ?, (NDA 2007 II), d2y, , dy, + 2y = 0, dx, dx, 2, d2y, dy , (b) x 2, −, y, + 2= 0, dx , dx 2, 2, 2, d y dy , (c) (1 − x 2 ), y, −, =0, , dx 2 dx , (d) None of the above, , (a) x 2, , 2, , − 2x, , 43. What is the equation of the curve passing through, the origin and satisfying the differential equation, (NDA 2007 I), dy = ( y tan x + sec x ) dx ?, (a) y = x cos x, (b) y cos x = x, (c) xy = cos x, (d) y sin x = x, 44. Match List I (Differential equation) with List II, (Its solution) and select the correct answer using the, codes given below the lists., List I, , List II, , A. yy′ = sec 2 x, , 1. y sec 2 x = sec x + C, , B. y′ = x sec y, , 2. xy = sin x + C, , C. y′ + ( 2 tan x) y = sin x, , 3. y2 = 2 tan x + C, , D. xy′ + y = cos x, , 4. x2 = 2 sin y + C, , Codes, A, (a) 3, (b) 4, (c) 3, (d) 3, , B, 2, 1, 4, 2, , C, 1, 2, 1, 4, , D, 4, 3, 2, 1
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457, , Differential Equations, , Directions (Q. Nos. 45-47) Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., , 47. Assertion (A) The elimination of two arbitrary, constants in y = ( a + b) x results into a differential, dy, equation of the first order x, = y., dx, Reason (R) Elimination of n arbitrary requires in, general, a differential equation of the nth order., , Directions (Q. Nos. 48-50) Consider the, differential equation corresponding to the family of, curves y = C ( x − C )2 , C being an arbitrary constant., 48. The differential equation of the family of curves, y = C (x − C )2 is, 3, , dy, , dy, dy , (a) = 4 y x, − 2 y (b), = 2, , dx, dx , dx, d 2 y 2x dy, (c), (d) None of these, −, =0, dx, dx 2, , 45. Assertion (A), The degree of the differential, 2, 2, d 2 y, d 2 y, dy , equation 2 + = x sin 2 is 2., dx , dx , dx , Reason (R) The degree of the differential equation, which is not a polynomial in differential coefficients,, cannot be defined., 46. Assertion (A) The differential equation of all, parabola whose axis of symmetry is parallel to x-axis, is of order 3., Reason (R) The order of equation depends upon the, number of unknown in equation of the curve., , 49. The order of the differential equation when the, differential equation is formed from the given family,, is, (a) 3, (b) 1, (c) 2, (d) 4, 50. The degree of the differential equation when the, differential equation is formed from the given family,, is, (a) 5, (b) 1, (c) 3, (d) 2, , Answers, Level I, 1., 11., 21., 31., 41., , (a), (d), (a), (b), (a), , 2., 12., 22., 32., 42., , (c), (b), (b), (d), (a), , 3., 13., 23., 33., 43., , (c), (a), (a), (d), (b), , 4., 14., 24., 34., 44., , (d), (a), (b), (a), (c), , 5., 15., 25., 35., 45., , (c), (b), (a), (c), (c), , 6., 16., 26., 36., 46., , (b), (b), (b), (c), (c), , 7., 17., 27., 37., 47., , (a), (a), (b), (d), (c), , 8., 18., 28., 38., 48., , (b), (c), (d), (c), (c), , 9., 19., 29., 39., 49., , (a), (c), (b), (d), (a), , 10., 20., 30., 40., 50., , (b), (d), (a), (a), (a), , 2., 12., 22., 32., 42., , (b), (b), (d), (d), (a), , 3., 13., 23., 33., 43., , (c), (a), (c), (b), (b), , 4., 14., 24., 34., 44., , (b), (a), (c), (c), (c), , 5., 15., 25., 35., 45., , (d), (b), (b), (d), (d), , 6., 16., 26., 36., 46., , (a), (b), (a), (a), (a), , 7., 17., 27., 37., 47., , (d), (d), (d), (c), (b), , 8., 18., 28., 38., 48., , (d), (b), (b), (c), (a), , 9., 19., 29., 39., 49., , (a), (a), (a), (b), (b), , 10., 20., 30., 40., 50., , (c), (d), (d), (b), (c), , Level II, 1., 11., 21., 31., 41., , (a), (c), (c), (c), (a)
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Hints & Solutions, Level I, 1. The degree of given equation is 1., , ⇒, , 2. Given, (2 y − 1) dx − (2x + 3) dy = 0, dy, dx, `, =, 2 y − 1 2x + 3, , ⇒, , On integrating both sides, we get, 1, 1, ∫ 2 y − 1 dy = ∫ 2x + 3 dx, 1, 1, ⇒, log (2 y − 1) = log (2x + 3) + log C1, 2, 2, 2x + 3, 1, log , ⇒, = − log C1, 2, 2 y − 1, 2x + 3, ⇒, =C, 2y − 1, 3. Given equation,, dy, + P (x) y = 0, dx, It is a standard linear differential equation, P dx, IF = e∫, ∴ Required solution is, P dx, −, ye∫, = 0 + C ⇒ y = Ce ∫, , ⇒, , It is a linear differential equation, comparing with the, standard equation, dy, + Py = Q, dx, ⇒, P = 1, Q = e− x, P dx, IF = e∫, = ex, ∴ Required solution is, yex = ∫ e− x ex dx + C = ∫ 1 dx + C, , ⇒, yex = x + C, At, x= y=0, ∴, C =0, Hence, the required solution is, yex = x ⇒ y = xe− x, , 5. ∴ (x + y) dx − (x + y) dy = dx + dy, ⇒, (x + y − 1) dx = (x + y + 1) dy, dy x + y − 1 , =, ⇒, , dx x + y + 1, dy dv, Let x + y = v and, =, −1, dx dx, dv, v−1, ⇒, −1 =, dx, v+1, dv v − 1 + v + 1, =, ⇒, dx, v+1, , x + y = Ke x − y, , 6. The given differential equation is, 2, 4, d 2y, dy, , 1 +, = 2, , dx, dx , From above, it is clear that degree of given differential, equation is 2., (Q Degree of differential equation = (Degree of the, highest order derivative), dy, x−y, y−x, y, −y, 7. Q, =e, (e, − e ) = e ⋅ ey (ex ⋅ e− x − ex ), dx, = 1 ⋅ (1 − ex ), x, ⇒, ∫ 1 dy = ∫ (1 − e ) dx, ⇒, , P dx, , dy, 4. Given equation,, + y = e− x, dx, , On integrating both sides,, , ⇒, , v + 1, , dv = ∫ 1 dx, 2v , 1, 1, v + log v = x + log C, 2, 2, x + y, log 2 = x − y, C , , ∫, , y = x − ex + C, , 8. The given differential equation is, dy, = xy + x + y + 1, dx, dy, ⇒, = (x + 1)( y + 1), dx, 1, ⇒, ∫ (1 + y) dy = ∫ (x + 1)dx, x2, ⇒, log (1 + y) =, + x+C, 2, 9. The given differential equation is, (1 + ex ) y dy = ex dx, ex , ⇒, ∫ y dy = ∫ 1 + ex dx, ⇒, ⇒, ⇒, , y2, = log (1 + ex ) + log C, 2, y2 = 2 log [C (1 + ex )], y2 = log [C 2 (1 + ex )2], , 10. Given, differential equation is, −1, dy dy, dy, 1, +, y=x, + ⇒ y=x, dx dx, dx (dy / dx), 2, , dy, dy, y =x +1, dx, dx, Here, Degree = Power of highest derivative = 2, , ⇒, , 11. From the given differential equation, it is clear that, order = 2, degree = 3, 12. Given equation is, sin −1 x + sin −1 y = c, , …(i)
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461, , Differential Equations, , Now, differentiating Eq. (ii) wrt x, we get, d 2y, = − 4a cos 2x − 4b sin 2x, dx2, = − 4(a cos 2x + b sin 2x) = − 4 y [from Eq. (i)], d 2y, ⇒ 2 + 4y = 0, dx, , 35. Given differential equation is, 1, − log x, dy, x, + y log x = xex x 2, dx, 1, − log x, dy y, + log x = ex x 2, ⇒, dx x, 1, − log x, 1, Here,, P = log x and Q = ex x 2, x, ∴, , ∫, IF = e, , log x, dx, x, , (log x ), 2, , 40. Given differential equation is, dy y, dy, + = 0 comparing with, + Py = Q, dx x, dx, 1, Here, P = and Q = 0, x, dx, Pdx, ∫, Now, IF = e∫, = e x = elog x = x, , 2, , =e, , = ( e ) ( log x ), , 2, , 36. The given equation is, dy ax + b, =, dx cy + d, ⇒, , ∫ (cy + d ) dy = ∫ (ax + b) dx, 2, , y ⋅ (IF) = ∫ Q ⋅ (IF) dx + C, , 2, , cy, ax, + dy =, + bx + K, 2, 2, This equation will represent a parabola, if a = 0, c ≠ 0, and b ≠ 0., cy2, ⇒, + dy = 0 + bx + K, 2, c 2 2dy , ⇒, y +, = bx + K, 2, c , c 2, dy d 2 d 2 , ⇒, + 2 − 2 = bx + K, y + 2, 2, c, c, c , 2, c, d, d2, ⇒, = bx + K, y+ −, 2, c, 2c, ⇒, , c, 2, , ⇒, , 2, , ⇒, , dy, =y, dx, dy, ⇒, = dx, y, On integrating both sides, we get, dy, ∫ y = ∫ dx, log y = x + log A, , 41. We have,, , ⇒, , d, d, , y + = bx + K +, , c, 2c, , d2, K, d, b , ,0 ., + , − and focus, − , 2c , 2, bc, b, c, , , , , y, =x, A, , ⇒ y = Aex, , 43. Here, we have, , Since,, (given), , dy, + y = sin x, dx, dx, IF = e∫ = ex, , ∴ Complete solution is, yex = ∫ ex sin x dx, And also, we have, , 2, , dy, dy, + y =1, dx, dx, , ∫e, , ax, , Thus, the order of the differential equation is 1., 39. We have, y = a cos 2x + b sin 2x, On differentiating Eq. (i) wrt x, we get, dy, = − 2a sin 2x + 2b cos 2x, dx, , y, = ex, A, , ⇒, , log, , y dx + x dy = 0, dx, dy, ⇒, =−, x, y, dx dy, ⇒, +, =0, x, y, On integrating both sides, we get, dx, dy, ∫ x + ∫ y =0, ⇒, log x + log y = log C, ⇒, log xy = log C, ⇒, xy = C, , d, 2 bx d 2 2 K, , + 2 +, y+ =, , c, c, c, c, 2, 2, , 2, d, b, d, K, , , + , x +, y+ =, , , 2 bc b , c, c , , 37. Given, x dy = y dx, dy dx, ⇒, =, y, x, On integrating both sides, we get, log y = log x + log C ⇒ y = Cx, It represents a family of straight lines., dy, 1, 38., + y=, dx, dy, , dx, , ⇒, , 42. We have,, , Which forms a parabola, whose vertices, , ⇒, , log y − log A = x, , 2, , 2, , ⇒, , y ⋅ x = ∫ 0 dx + C ⇒ xy = C, , Solution,, , …(i), , ∴, ⇒, , …(ii), , eax, (a sin bx − b cos bx) + C, a + b2, ex, yex =, (sin x − cos x) + C, 2, 1, 1, y = sin x − cos x + Ce− x, 2, 2, , sin bx dx =, , 2
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468, , NDA/NA Mathematics, y cos x = ∫ 1 dx + C, , ⇒, , ⇒, y cos x = x + C, Since, this curve passes through (0, 0), ∴, 0 =0 + C ⇒ C =0, ∴ Required equation of curve is, y cos x = x, dy, 44. A., y⋅, = sec2 x, dx, y dy = sec2x dx, On integrating both sides, we get, 2, ∫ y dy = ∫ sec x dx, y2, = tan x + C, 2, 2, y = 2 tan x + C, dy, = x sec y, dx, , ⇒, ⇒, B., ⇒, ⇒, , ∫, , x2, cos y dy = ∫ x dx ⇒ sin y + C =, 2, x2 = 2 sin y + C, , dy, + (2 tan x) y = sin x, dx, It is of the form, dy, + Py = Q, dx, 2tan x dx, So,, IF = e∫, C., , 2 tan x dx, =e∫, = e2logsec x, , = sec2x, ∴ Required solution is, y ⋅ sec 2x = ∫ sin x ⋅ sec2 x dx + C, ⇒, , = ∫ sec x tan x dx + C, , y sec2x = sec x + C, dy, dy 1, cos x, D. x, + y = cos x ⇒, + ⋅y=, dx, dx x, x, It is of the form, dy, + Py = Q, dx, 1, ∫ dx, Now, IF = e x = elog x = x, So, required solution is, cos x, y⋅ x = ∫ x⋅, dx + C, x, y ⋅ x = ∫ cos x dx + C, y ⋅ x = sin x + C, , 45. The given differential equation is, 2, 2, d 2y, d 2y, dy, 2 + = x sin 2 , dx, dx , dx , , This equation is not a polynomial in differential, coefficients. So, its degree is not defined., Thus, A is false but R is true., 46. Let the general equation of the given parabolas is, x = ay2 + by + c, where a , b and c are constants., Hence, the order of the required differential equation is, 3., Thus, both A and R are true and R is the correct, explanation of A., 47. A. We have,, ⇒, , y = (a + b) x, dy, x, =y, dx, , It is true., ∴ Both A and R are true and R is not the correct, explanation of A., Solutions (Q. Nos. 48-50), Given,, y = C (x − C )2, Differentiating both sides wrt x, we get, dy, = 2C (x − C ), dx, On division, we get, y, x −C, =, dy / dx, 2, dy, y, x −C, , where y1 =, ⇒, =, dx, 2, y1, y, ⇒, x − C =2 , y1 , 2y, C = x −, ⇒, y1, , ....(i), , On putting the value of C in Eq. (i), we get, 2, , 2 y y , y = x − 2 , y1 y1 , , xy1 − 2 y 4 y2, =, ×, y1, y12, ⇒, , y13 = 4 y (xy1 − 2 y), dy, in Eq. (ii), we get, On putting y1 =, dx, 3, dy, dy, , − 2 y, ⇒, = 4y x, dx, dx, , , ...(ii), , ...(iii), , which is required differential equation., The order of a differential equation is the order of the, highest derivative of Eq. (iii)., Hence, order = 1, The degree of a differential equation is the degree of the, highest derivative of Eq. (iii)., Hence, degree = 3
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24, , Rectangular, Cartesian System, Rectangular Coordinate Axes, , Quadrants, , The lines XOX ′ and YOY ′ are, called the axis of x and axis of y,, respectively and the two lines taken, together are called the coordinate X′, axes or the axes of coordinates., These two lines X ′OX and, Y ′OY are mutually intersect each, other at the point O, which is known, as origin., , Y, The axes X ′ OX and Y ′ OY, divide the whole plane into four, II, I, parts, which, are, called, (–, +), (+, +), quadrants. Here, OX and OX ′, X, are called the positive and X′, O, negative directions, respectively, III, IV, (–, –), (+, –), of X-axis and similarly for, Y -axis, OY and OY ′ are the, positive and negative directions,, Y′, respectively., In Ist quadrant, : x > 0, y > 0, In IInd quadrant : x < 0, y > 0, In IIIrd quadrant : x < 0, y < 0, In IVth quadrant : x > 0, y < 0, The coordinates of any point on X-axis are of the form, ( x , 0) and on Y-axis as (0 , y)., ∴ If the X-coordinate or abscissa of a point is zero, it, would be somewhere on the Y-axis and if its Y-coordinate, or ordinate is zero, then it would be on X-axis., , Y, , O, , X, , Y′, , Cartesian Coordinates of a Point, The position of the point P can be completely, determined with reference to the rectangular axes X ′OX, andY ′OY by means of an ordered pair of real numbers (x, y), called cartesian, Y, coordinates of P. From, figure, x and y are the, P (x , y ), M, distances of the point P from, y and x axes, respectively x, y, with proper sign is called the, X-coordinate or abscissa of, point P and y is called the X ′, X, O, x, L, Y-coordinate or ordinate of, point P. Thus, for a given, Y′, point, the abscissa and, ordinate are distances of the, point from Y-axis and X-axis, respectively., ∴ x = PM = perpendicular length from P to M on Y, axis, and y = PL = perpendicular length from P to L on, X-axis., , Distance Formula, The distance between any two A ( x1 , y1 ) and B ( x2 , y2 ), points in the plane is the length of the line segment joining, them., (x 2 , y 2 ), , (x1, y1), A, , i.e.,, or, , B, , D = ( x2 − x1 ) + ( y2 − y1 )2, 2, , D = (Difference of abscissae)2, + (Difference of ordinates )2, , If O is the origin and P ( x , y )is any point, then from the, distance formula, we have, OP = ( x − 0)2 + ( y − 0)2 = x 2 + y 2
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470, , NDA/NA Mathematics, , Example 1. The points (0, 0), (3, 3) and (3, − 3) are the, vertices of, (a) an isosceles triangle, (c) a right angled triangle, , (b) an equilateral triangle, (d) None of these, , Solution (b) Let A(0 , 0), B (3, 3) and C(3, − 3) are the given, points., Now,, , AB = (3 − 0) 2 + ( 3 − 0) 2 = 9 + 3 = 12, BC = (3 − 3) 2 + ( − 3 − 3) 2 = 12, , and, , Example 2. Let A(6, − 1), B(1, 3) and C ( x, 8) be three points, such that AB = BC . Find the values of x., (a) − 3, 5, (b) 3, 5, (c) 3, − 5, (d) − 3, − 5, Solution (a) We have, AB = BC, (6 − 1) 2 + ( −1 − 3) 2 = (1 − x) 2 + (3 − 8) 2, , On squaring both sides, we get, 5 2 + 4 2 = (1 − x) 2 + 5 2, ⇒, , (1 − x) 2 = 16 or x = 1 ± 4, x = − 3, 5, , i.e.,, , Section Formulae, Let O be the origin and OX , OY are the X-axis and, Y-axis, respectively. Let A( x1 , y1 ) and B ( x2 , y2 ) be the given, points., Y, , y, x,, P(, , ), , B (x2 , y2 ), , O, , L, , the points P(3 , − 4) and Q( −2 , 5) that is twice as far from P as, from Q., (a) ( − 7, 14), (b) ( 7, − 14), (c) ( − 7, − 14), (d) ( 7, 14), , Solution (a) Let A( x, y) be the required point., (given), ∴, PA = 2 ⋅ PQ, Thus, A divides PQ internally or externally in the ratio 2 :1., If A divides PQ internally in the ratio 2 : 1, then coordinates, of A are, 2 × ( −2) + 1 × 3, 2 × 5 + 1 × ( −4), and y =, x=, 2 +1, 2 +1, 1, x = − and y = 2, ⇒, 3, 1 , ∴ Coordinates of A are − , 2 ., 3 , Again, if A divides PQ externally in the ratio 2 : 1,, then coordinates of A are, 2 × ( −2) − 1 × 3, 2 × 5 − 1 × ( −4), and y =, x=, 2 −1, 2 −1, ⇒, x = − 7 and y = 14, ∴ Coordinates of A are (–7, 14)., , Example 5. Find the coordinates of point which divides the, line segment joining the points (5, − 2) and (9, 6) internally, and externally in the ratio 3 : 1., (a) (10, 11), (b) (−1110, , ), (c) (11, 10), (d) (−10, 11 ), , (x1 , y1 )A, , X′, , Given that, A(0 , 0) and B(12, 0) are two points and, m : n = 5 :1, 5 × 12 − 1 × 0 5 × 12, ∴, x=, =, = 15, 5 −1, 4, 5 × 0 −1× 0, and, y=, =0, 5 −1, ∴ Coordinates of C are (15, 0)., , Example 4. Find the coordinates of point on the line joining, , CA = (0 − 3) 2 + (0 + 3) 2 = 9 + 3 = 12, , Since, AB = BC = CA, then the given points are the vertices of, an equilateral triangle., , ⇒, , Solution (c) Let the coordinates of the point C are ( x, y)., , N, , M, , X, , Y′, , P ( x , y ) be the point which divides the line segment, joining two given points A ( x1 , y1 ) and B ( x2 , y2 ) in the ratio, m : n., Case I For external division, mx2 − nx1, my2 − ny1, ,y=, x=, m−n, m−n, Case II For internal division, my2 + ny1, mx2 + nx1, x=, ,y=, m+n, m+n, , Example 3. If A(0, 0) and B(12, 0) are two points, then find, the coordinates of the point C on line joining AB dividing it in, the ratio 5 : 1 (externally)., (a) (0 , 15), (b) ( −15, 0) (c) (15, 0), (d) (0, 0), , Solution (c) Given that, P ≡ (5, − 2) and Q ≡ (9, 6)., Now, let R( x, y) divide PQ in the ratio 3 : 1., If R divide PQ internally, then coordinates of R are, 3 × 9 + 1× 5, x=, 3 +1, and, , y=, , 3 × 6 + 1 × ( −2), 3 +1, , Hence, the coordinates of the required point are (8, 4)., If R divide PQ externally, then coordinates of R are, 3 × 9 −1× 5, x=, 3 −1, and, , y=, , 3 × 6 − 1 × ( −2), 3 −1, , Hence, the coordinates of the required point are (11, 10).
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472, , NDA/NA Mathematics, , Some Points of a Triangle, Centroid of a Triangle, The centroid of a triangle is the point of intersection of, its medians. The centroid divides the medians in the ratio, 2 : 1 (vertex : base)., A (x1, y1), , F, , 1, , E, 2, , 1, G, , (x 2 , y 2 ) B, , vertices are ( −2, − 3), ( −1, 0), ( 7, − 6). Also, find the radius of, the circumcircle., (a) ( −3, −3), 4, (b) (3, −3), 5, (c) ( −3, −3), 5, (d) None of these, , Solution (b) Let the vertices of triangle are P( −2, − 3), Q ( −1, 0), and R(7, − 6). Let A( x, y) be the circumcentre of ∆ PQR., ∴, AP 2 = AQ 2, , 2, , 2, , Example 11. Find the circumcentre of the triangle whose, , P(–2, –3), , 1, D, , C (x3, y3), , If A( x1, y1 ), B( x2 , y2 ) and C( x3 , y3 ) are the vertices of a, triangle. If G be the centroid upon one of the median (say),, AD, then AG : GD = 2 : 1, x + x2 + x3 y1 + y2 + y3 , ⇒ Coordinates of G are 1, ,, , , , 3, 3, , Example 10. If vertices of coordinates of triangle are (3, − 5), and ( −7, 4) and its centroid is (2, − 1). Then, the coordinates of, third vertex is, (a) (10, 2), (b) (10, − 2 ), (c) (−10, 2), , (d) None of these, , Solution (b) Let ( x, y) be the coordinates of third vertex, then, x+ 3 −7, y −5 + 4, = 2 and, = −1, 3, 3, ⇒, , x − 4 = 6 and y − 1 = − 3, , ⇒, x = 10 and y = − 2, Thus, the coordinates of the third vertex are (10 , − 2)., , A(x, y), R(7, –6), , Q, (–1, 0), , ⇒, , ( x + 2) 2 + (y + 3) 2 = ( x + 1) 2 + y 2, , ⇒, ⇒, ⇒, Similarly,, ⇒, , 4x + 6y + 13 = 2x + 1, 2x + 6y = − 12, x + 3y = − 6, AP 2 = AR 2, , …(i), , ( x + 2) 2 + (y + 3) 2 = ( x − 7) 2 + (y + 6) 2, , ⇒, 4x + 6y + 13 = − 14x + 12y + 85, ⇒, 18x − 6y = 72, ⇒, 3x − y = 12, From Eqs. (i) and (ii), we get ( x, y) ≡ (3, − 3), Hence, circumcentre is (3, − 3)., And radius of the circumcircle, , …(ii), , AP = ( x + 2) 2 + (y + 3) 2 = 5 2 + 0 2 = 5, , Circumcentre, The circumcentre of a triangle is the point of, intersection of the perpendicular bisectors of the sides of a, triangle. It is the centre of the circle, which passes through, the vertices of the triangle and so, its distance from the, vertices of the triangle is the same and this distance is, known as the circumradius of the triangle., , Incentre, The incentre of a triangle is the point of intersection of, internal bisector of the angles . Also, it is a centre of a circle, touching all the sides of a triangle., A ( x 1 , y1 ), , A ( x1 , y 1), , cF, F, , Eb, , E, , I, O, (x2, y2) B, , D, , (x2, y2) B, C (x3, y3), , If a triangle is right angle, then its circumcentre is the, mid-point of hypotenuse. If angles of triangle i.e., A, B,C, and vertices of triangle A( x1 , y1 ), B( x2 , y2 ) and C( x3 , y3 ) are, given, then circumcentre of the ∆ ABC is, x1 sin 2 A + x2 sin 2B + x3 sin 2C, ,, , sin 2 A + sin 2B + sin 2C, , y1 sin 2 A + y2 sin 2B + y3 sin 2C , , sin 2 A + sin 2B + sin 2C, , , D, a, , C (x3, y3), , Coordinates of incentre, ax1 + bx2 + cx3 ay1 + by2 + cy3 , ,, , , a + b+ c, a + b+ c , , Where, a , b and c are the sides of a ∆ ABC., , Example 12. The incentre of the triangle formed by (0, 0),, (5, 12) and (16, 12) is, (a) (7, 9), (c) (− 9, 7), , (b) (9, 7), (d) (− 7, 9)
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473, , Rectangular Cartesian System, , Solution (a) Let A = (0 , 0) , B = (5,12), C = (16,12), a = BC = (16 − 5) + (12 − 12) = 11, 2, , 2, , b = AC = (16 − 0) 2 + (12 − 0) 2 = 256 + 144 = 400 = 20, c = AB = (5 − 0) 2 + (12 − 0) 2 = 25 + 144 = 169 = 13, ax + bx2 + cx3 ay1 + by 2 + cy3 , ∴ Incentre = 1, ,, , a+ b+ c , a+ b+ c, 11 × 0 + 20 × 5 + 13 × 16 11 × 0 + 20 × 12 + 13 × 12, =, ,, , 11 + 20 + 13, 11 + 20 + 13, , , 0 + 100 + 208 0 + 240 + 156, =, ,, = (7, 9), , , 44, 44, , Example 13. Find the orthocentre of the triangle formed by, A(0, 0), B(8, 0) and C( 4, 6)., 8, 8, , (a) 4, , (b) − 4, , 3, , 3, 8, , , (d) None of these, (c) − 4, − , , 3, Solution (a) Line perpendicular to AB passing through C is, x = 4., , y, , C (4,6), , Excircle, A circle touches one side outside the triangle and other, two extended sides, then circle is known as excircle. Let, ABC be a triangle, then there are three excircles with three, excentres. Let I1 , I 2 and I3 be the centres of excircles, opposite to vertices A, B and C, respectively. If vertices of, triangle are A( x1 , y1 ), B( x2 , y2 ) and C( x3 , y3 ), then, , A (0,0), , B(8,0), , x, , Line perpendicular to BC through origin is y =, , 2, x, 3, , 2, 8, x is 4, ., 3, 3, Which is the orthocentre of ∆ ABC., Intersection point of x = 4 and y =, , I3, , A, , B, , I2, , Locus, , C, I1, , The curve described by any point, which moves under, given conditions is called locus of the point., − ax1 + bx2 + cx3 − ay1 + by2 + cy3 , I1 ≡ , ,, ,, − a + b+ c , − a + b+ c, ax − bx2 + cx3 ay1 − by2 + cy3 , I2 ≡ 1, ,, ,, a − b+ c, a − b+ c , , ax + bx2 − cx3 ay1 + by2 − cy3 , I3 ≡ 1, ,, , a + b− c, a + b− c , , , Orthocentre, It is the point of intersection of perpendiculars drawn, from vertices on opposite sides (called altitudes) of a, triangle and can be obtained by solving the equation of any, two altitudes., A (x1, y1), , D, , The equation of locus of a point is the relation, which, is satisfied by the coordinates of every point on the locus, of the point., , Algorithm to Find the Locus of a Point, Step I Assume the coordinates of the point say (h , k), whose locus is to be found., Step II Write the given condition in mathematical, form involving h , k., Step III Eliminate the variable (s), if any., Step IV Replace h by x and k by y in the result, obtained in step III. The equation so obtained is the locus of, the point, which moves under some stated condition (s)., , F, O, , (x2, y2) B, , Equation of Locus, , E, , C (x3, y3), , Here, O is the orthocentre. Since, AE ⊥ BC, BF ⊥ AC,, CD ⊥ AB, then OE ⊥ BC, OF ⊥ AC and OD ⊥ AB., The orthocentre of the ∆ ABC is, x1 tan A + x2 tan B + x3 tan C y1 tan A + y2 tan B + y3 tan C , ,, , , tan A + tan B + tan C, tan A + tan B + tan C, , , If a triangle is right angled triangle, then orthocentre, is the point, where right angle is formed., , Example 14. The equation of the locus of a point equidistant, from the points A(1, 3) and B( −2, 1) is, (a) 6 x − 4y = 5, (b) 6 x + 4y = 5, (c) 3x + 2y = 5, (d) 3x − 2y = 5, Solution (b) Let P(h, k) be any point on the locus, then, PA2 = PB2, ⇒, (h − 1) + (k − 3) 2 = (h + 2) 2 + (k − 1) 2, ⇒, 6h + 4k = 5, Hence, locus of (h, k) is, 6x + 4y = 5, 2
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Exercise, Level I, 1. The centroid of a triangle is (2, 7) and two of its, vertices are (4, 8) and ( −2, 6). The third vertex is, (a) (0, 0), (b) (4, 7), (c) (7, 4), (d) (7, 7), 2. If the points ( k, 3), ( 2, k) and ( −k, 3) are collinear, then, the values of k are, (a) 2, 3, (b) 1, 0, (c) 1, 2, (d) 0, 3, 3. If the points ( − 2, − 5),( 2, − 2) and ( 8, a ) are collinear,, then the value of a is, 5, 3, 1, 5, (a) −, (c), (d), (b), 2, 2, 2, 2, 4. If the points (1, 1), ( −1, − 1), ( − 3 , 3 ) are the vertices, of a triangle, then this triangle is, (a) right angled, (b) isosceles, (c) equilateral, (d) None of these, 5. What is the equation of the locus of a point, which, moves such that 4 times its distance from the x-axis,, is the square of its distance from the origin?, (a) x 2 + y 2 − 4 y = 0, (b) x 2 + y 2 − 4| y| = 0, (d) x 2 + y 2 − 4| x| = 0, (c) x 2 + y 2 − 4x = 0, 6. If ( p, q) be the point on the x-axis equidistant from, the points (1, 2) and (2, 3), then which one of the, following is correct?, (NDA 2011 II), (a) p = 0, q = 4, (b) p = 4, q = 0, (c) p = 3 / 2, q = 0, (d) p = 1, q = 0, 7. If the sum of the squares of the distances of the point, (x , y) from the points (a, 0) and (− a, 0) be 2b2, then, which one of the following is correct?, (NDA 2011 I), (a) x 2 + a 2 = b2 + y 2, (b) x 2 + a 2 = 2b2 − y 2, (d) x 2 + a 2 = b2 − y 2, (c) x 2 − a 2 = b2 + y 2, 8. If the area of the triangle with vertices ( x , 0), (1, 1), and (0, 2) is 4 sq units, then the value of x is, (a) −2, (b) −4, (c) −6, (d) 8, 9. Three vertices of a parallelogram taken in order are, ( −1, − 6), ( 2, − 5) and (7, 2). The fourth vertex is, (a) (1, 4), (b) (4, 1), (c) (1, 1), (d) (4, 4), 10. The points (1, 1), ( −5, 5) and (13, λ ) lie on a straight, line, if λ is equal to, (a) 7, (b) −7, (c) ±7, (d) 0, 11. If A( 3, 5), B( −5, − 4), C( 7, 10) are the vertices of a, parallelogram taken in the order, then the, coordinates of the fourth vertex are, (a) (10, 19), (b) (15, 10), (c) (15, 19), (d) (19, 15), , 12. What is the set of points (x , y) satisfying the, equations x 2 + y 2 = 4 and x + y = 2?, (NDA 2011 I), (a) {( 2, 0),( − 2, 0),( 0, 2)}, (b) {( 0, 2),( 0, − 2)}, (c) {( 0, 2),( 2, 0)}, (d) {( 2, 0),( − 2, 0),( 0, 2),( 0, − 2)}, 13. If (a, 0), (0, b) and (1, 1) are collinear, what is the value, (a + b − ab)?, (NDA 2011 I), (a) 2, (b) 1, (c) 0, (d) − 1, 14. If A( x1 , y1 ), B( x2 , y2 ) and C( x3 , y3 ) are the vertices of a, triangle, then the excentre with respect to B is, ax − bx2 + cx3 ay1 − by2 + cy3 , (a) 1, ,, , a − b+ c, a − b+ c, , , ax + bx2 − cx3 ay1 + by2 − cy3 , (b) 1, ,, , a + b− c, a + b− c, , , ax − bx2 − cx3 ay1 − by2 − cy3 , (c) 1, ,, , a − b− c, a − b− c , , (d) None of the above, 15. If the points ( x , y ), (1, 2) and ( −3, 4) are collinear, then, (a) x + 2 y − 5 = 0, (b) x + y − 1 = 0, (c) 2x + y − 4 = 0, (d) 2x − y + 10 = 0, 16. If P, Q and R are three non-collinear points, then, what is the value of PQ ∩ PR?, (NDA 2011 I), (a) Null set, (b) { P }, (c) { P , Q , R}, (d) { Q , R }, 17. If the area of a triangle with vertices ( − 3, 0), ( 3, 0) and, (0, k) is 9 sq units, then what is the value of k?, (a) 3, (b) 6, (NDA 2010 II), (c) 9, (d) 12, 18. After subtending an angle of 1000° from its initial, position, the revolving line will be situated in which, one of the following quadrants?, (NDA 2009 I), (a) First quadrant, (b) Second quadrant, (c) Third quadrant, (d) Fourth quadrant, 19. Points ( at12 , 2at1 ), ( at22 , 2at2 ) and ( a , 0) are collinear., Which one of the following is correct?, (b) t1t2 = 1, (a) t1t2 = − 1, (c) t1 = t2, (d) t1 + t2 = 1, 20. The cartesian form of the polar equation θ = tan− 1 2 is, (a) x = 2 y, (b) y = 2x, (c) x = 4 y, (d) y = 4x, 21. The area of the triangle formed by the coordinate, axes and the line 4x + 5 y = 20 is (in sq unit), (a) 5, (b) 10, (c) 15, (d) 20
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476, 22. If A = ( − 3, 4), B = ( − 1, − 2), C = ( 5, 6) and D = ( x , − 4) are, the vertices of a quadrilateral such that, ∆ABD = 2 ∆ACD, then x is, (a) 6, (b) 9, (c) 69, (d) 96, 23. What does the equation x3 y + xy3 − xy = 0 represent?, (a) A pair of straight lines only, (NDA 2009 I), (b) A pair of straight lines and a circle, (c) A rectangular hyperbola only, (d) A rectangular hyperbola and a circle, 24. If ( a , b), ( c, d ) and ( a − c, b − d ) are collinear, then, which one of the following is correct?, (NDA 2008 I), (a) bc − ad = 0, (b) ab − cd = 0, (c) bc + ad = 0, (d) ab + cd = 0, 25. If A = ( 0, 4), B = ( 0, − 4) and| AP − BP|= 6, then locus of, the point P is, (b) 7x 2 − 9 y 2 = 63, (a) 7x 2 − 9 y 2 + 63 = 0, (d) 9x 2 − 7 y 2 = 63, (c) 9x 2 − 7 y 2 + 63 = 0, 26. If the point (x , y) is equidistant from points, ( a + b, b − a ) and (a − b, a + b), then, (a) ax + by = 0, (b) bx = ay, (c) ax = by, (d) bx + ay = 0, 27. The triangle joining the points P ( 2, 7), Q ( 4, − 1),, R ( − 2, 6) is, (a) scalene triangle, (b) isosceles triangle, (c) right angled triangle, (d) equilateral triangle, 28. If p1 , p2 denotes the lengths of the perpendiculars, from the origin on the lines x sec α + y cosec α = 2a and, respectively,, then, x cos α + y sin α = a cos 2 α, 2, p1 p2 , + is equal to, , p2 p1 , (a) 4 sin2 4 α, (b) 4 cos2 4 α, (d) 4 sec2 4 α, (c) 4 cosec2 4 α, 29. What is the perimeter of the triangle with vertices, (NDA 2012 I), A ( − 4, 2), B ( 0, − 1) and C ( 3, 3)?, (b) 10 + 5 2, (a) 7 + 3 2, (c) 11 + 6 2, (d) 5 + 2, 30. The coordinates of a point are (0, 1) and the ordinate, of another point is − 3. If the distance between the, two points is 5, then the abscissa of another point is, (a) 3, (b) − 3, (c) ± 3, (d) 1, 31. The point whose abscissa is equal to its ordinate and, which is equidistant from the points (1, 0) and (0, 3) is, (a) (1, 1), (b) (2, 2), (c) (3, 3), (d) (4, 4), 32. If the points (0, 0), (2, 2 3) and (a , b) be the vertices of, an equilateral triangle, then (a , b) is, (a) ( 0, − 4), (b) ( 0, 4), (c) (4, 0), (d) (− 4, 0), , NDA/NA Mathematics, 33. The length of altitude through A of the ∆ ABC, where, A ≡ ( − 3, 0), B ≡ ( 4, − 1), C ≡ ( 5, 2) is, 2, 4, (a), (b), 10, 10, 11, 22, (d), (c), 10, 10, 34. If the point dividing internally the line segment, joining the points (a , b) and (5, 7) in the ratio 2 : 1 be, (4, 6), then, (a) a = 1, b = 2, (b) a = 2, b = − 4, (c) a = 2, b = 4, (d) a = − 2, b = 4, 35. The point, which divides externally the line joining, the points (a + b, a − b) and (a − b, a + b) in the ratio, a : b , is, a 2 − 2ab − b2 a 2 + b2 , (a) , ,, , a−b, a−b , , a 2 − 2ab − b2 a 2 − b2 , (b) , ,, , a−b, a−b , , a 2 − 2ab + b2 a 2 + b2 , (c) , ,, , a−b, a−b , , (d) None of the above, 36. The points, which trisect the line segment joining the, points (0, 0) and (9, 12), are, (a) (3, 4), (6, 8), (b) (4, 3), (6, 8), (c) (4, 3), (8, 6), (d) (3, 4), (8, 6), 37. The mid-points of sides of a triangle are (2, 1),, (− 1, − 3) and (4, 5). Then, the coordinates of its, vertices are, (a) (7, 9), (− 3, − 7), (1, 1), (b) (− 3, − 7,), (1, 1), (2, 3), (c) (1, 1), (2, 3), (− 5, 8), (d) None of the above, 38. The quadrilateral formed by the vertices (− 1, 1),, ( 0, − 3),( 5, 2) and (4, 6) will be, (a) square, (b) parallelogram, (c) rectangle, (d) rhombus, 39. The points (3a, 0), (0, 3b) and (a , 2b) are, (a) vertices of an equilateral triangle, (b) vertices of an isosceles triangle, (c) vertices of a right angled isosceles triangle, (d) collinear, 40. The quadrilateral formed by the vertices (− 2, 2),, ( − 2, − 1), (3, − 1) and (3, 2) will be, (a) square, (b) rhombus, (c) rectangle, (d) parallelogram, 41. Two vertices of a triangle are (5, 4) and (− 2, 4). If its, centroid is (5, 6), then the third vertex has the, coordinates, (a) (12, 10), (b) (10, 12), (c) (− 10, 12), (d) (12, − 10)
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477, , Rectangular Cartesian System, , Level II, 1. The medians BE and AD of a triangle with vertices, A ( 0, b), B ( 0, 0) and C ( a , 0) are perpendicular to each, other, if, b, a, (a) a =, (b) b =, (c) ab = 1, (d) a = ± 2b, 2, 2, 2. The length of altitude through A of the ∆ ABC, where, A ≡ ( −3, 0) , B ≡ ( 4, − 1) , C ≡ ( 5, 2), is, 2, 4, 11, 22, (b), (c), (d), (a), 10, 10, 10, 10, 3. If the distance of any point P from the points, A( a + b, a − b) and B( a − b, a + b) are equal, then the, locus of P is, (a) x − y = 0, (b) ax + by = 0, (c) bx − ay = 0, (d) x + y = 0, 4. The locus of a point whose difference of distance from, points (3, 0) and ( −3, 0) is 4, is, x2 y2, x2 y2, (a), (b), −, =1, −, =1, 4, 5, 5, 4, 2, 2, 2, 2, x, y, x, y, (d), (c), −, =1, −, =1, 2, 3, 3, 2, 5. What is the locus of a point which moves equidistant, from the coordinates axes?, (NDA 2011 I), (a) x ± y = 0, (b) x + 2 y = 0, (c) 2x + y = 0, (d) None of these, 6. The x-coordinate of the incentre of the triangle,, where the mid-point of the sides are (0, 1), (1, 1) and, (1, 0), is, (b) 1 + 2, (a) 2 + 2, (c) 2 − 2, (d) 1 − 2, 7. ABC is a triangle with vertices A( −1, 4), B ( 6, − 2) and, C( −2, 4). D , E and F are the points, which divide each, AB, BC and CA, respectively in the ratio 3 : 1, internally. Then, the centroid of the ∆ DEF is, (a) (3, 6), (b) (1, 2), (c) (4, 8), (d) ( −3, 6), 8. If A( at 2 , 2at ), B( a/ t 2 , − 2a/ t ) and C( a , 0), then 2a is, equal to, (a) AM of CA and CB, (b) GM of CA and CB, (c) HM of CA and CB, (d) None of these, 1, 2 , 9. If the points ( x + 1, 2), (1, x + 2), , ,, are, x + 1 x + 1, collinear, then x is, (a) 4, (b) 5, (c) −4, (d) None of these, 10. A point P moves such that the difference of its, distances from two given points ( c, 0) and ( − c, 0) is, constant. What is the locus of the point P?, (NDA 2010 II), , (a) Circle, (c) Hyperbola, , (b) Ellipse, (d) Parabola, , 11. The coordinate axes rotated through an angle 135°. If, the coordinates of a point P in the new system are, known to be ( 4, − 3), then the coordinates of P in the, original system are, 7, 7, 1, 1, (b) , (a) , ,, ,−, , , 2 2, 2, 2, 7, 7, 1, 1, (c) −, (d) −, ,−, ,, , , , , 2, 2, 2, 2, 12. The middle point of the segment of the straight line, joining the points ( p, q) and (q , − p) is (r / 2, s / 2)., What is the length of the segment?, (NDA 2009 II), 2, 2 1/ 2, 2, 2 1/ 2, (a) [( s + r ) ]/ 2, (b) [( s + r ) ]/ 4, (c) ( s2 + r 2 )1/ 2, (d) s + r, 13. What is the locus of a point, which is equidistant from, the point (m + n , n − m) and the point (m − n , n + m)?, (NDA 2009 II), , (a) mx = ny, (c) nx = my, , (b) nx = − my, (d) mx = − ny, , 14. Consider the following statements, I. A triangle is isosceles, if any two of its medians, are equal., II. The circumcentre of a right angled triangle is, the mid-point of the hypotenuse., III. Orthocentre, centroid and circumcentre of a, triangle are collinear., Which of the statements given above are correct?, (a) I and II, (b) II and III, (c) I and III, (d) All I, II and III, 15. Consider the following statements, I. Orthocentre of the right angled ∆ ABC right, angled at A, is A., II. x-axis divides the line segment joining ( x1 , y1 ) and, ( x2 , y2 ) in the ratio − y1 : y2., Which of the statements given above is /are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 16. The points ( a , b),( 0, 0), ( − a , − b), ( ab, b2 ) are, (a) vertices of a parallelogram, (b) vertices of a rectangle, (c) vertices of a square, (d) collinear, 17. What is the area of the triangle formed by the lines, (NDA 2009 I), y − x = 0 , y + x = 0,and x = c?, 2, (a) c / 2, (b) c, (d) c2 / 2, (c) 2c2, 18. The point of intersection of the two lines, 2x + 3 y + 4 = 0 and 4x + 3 y + 2 = 0 is at a distance d, from origin. What is the value of d?, (NDA 2009 I), (b) 3, (a) 2, (c) 5, (d) 7
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478, , NDA/NA Mathematics, , 19. What is the locus of a point which is equidistant from, the points (a + b, a − b) and (b − a , a + b)? (NDA 2009 I), (a) bx − ay = 0, (b) bx + ay = 0, (c) − ax + by = 0, (d) ax + by = 0, , 30. If the points A(1, 2), B( 2, 4) and C( 3, a ) are collinear,, what is the length of BC?, (a) 2 units, (b) 3 units, (d) 5 units, (c) 5 units, , 20. If the points with coordinates ( −a , 0) , ( ap2 , 2ap) and, ( ap12 , 2ap1 ) are collinear, then the value of pp1 is, (a) 3, (b) 2, (c) 1, (d) –1, , 31. Which one of the following points on the line, 2x − 3 y = 5 is equidistant from (1, 2) and (3, 4)?, , 21. If the vertices B and D of a square ABCD be ( 2, 3) and, ( 4, 1) respectively, then the length of its side is, (a) 1 unit, (b) 2 units (c) 3 units (d) 4 units, 22. The equation of the set of all points equidistant from, the point ( 4, 2) and the x-axis is, (a) x 2 − 6x + 4 y + 10 = 0, (b) x 2 − 6x − 4 y − 10 = 0, (c) x 2 − 8x − 4 y + 20 = 0, (d) y = 3, 23. Points (1, 3, 4), (− 1, 6, 10 ), ( − 7, 4, 7) and ( − 5, 1, 1) are, the vertices of a, (NDA 2009 I), (a) rhombus, (b) rectangle, (c) parallelogram, (d) square, 24. The triangle formed by the points(1, 1),( 2, 0) and( 3, 1) is, (a) an equilateral triangle, (b) a right angled triangle, which is not isosceles, (c) a right angled triangle, which is isosceles, (d) neither a right angled triangle nor an isosceles, triangle, 25. An equilateral triangle is inscribed in the circle, x 2 + y 2 = a 2 with one of the vertices at ( a , 0). What is, the equation of the side opposite to this vertex?, (a) 2 x − a = 0, (b) x + a = 0, (c) 2 x + a = 0, (d) 3x − 2 a = 0, 26. The middle point of A(1, 2) and B( x , y ) is C( 2, 4). If BD, is perpendicular to AB such that CD = 3 units, then, what is the length of BD?, (b) 2 units, (a) 2 2 units, (c) 3 units, (d) 3 2 units, 27. If A( 2, 3), B(1, 4), C ( 0, − 2) and D ( x , y ) are the vertices, of a parallelogram, then what is the value of ( x , y )?, (NDA 2008 I), , (a) (1, − 3), , (b) ( 2, 4), , (c) (1, 1), , (d) ( 0, 0), , 28. If O be the origin and A( x1 , y1 ), B( x2 , y2 ) are two, points, then what is the value of (OA) (OB) cos ∠ AOB?, (a) x12 + x22, (b) y12 + y22, (NDA 2008 I), (c) x1x2 + y1 y2, , (d) x1 y1 + x2 y2, , 29. If the points with the coordinates, (a , ma), { b,( m + 1)b}, { c,( m + 2)c} are collinear, then, which one of the following is correct?, (NDA 2007 II), (a) a , b, c are in arithmetic progression for all m., (b) a , b, c are in geometric progression for all m., (c) a , b, c are in harmonic progression for all m., (d) a , b, c are in arithmetic progression only for m = 1., , (NDA 2007 II ), , (a) (7, 3), (c) (1, − 1), , (b) (4, 1), (d) ( − 2, − 3), , 32. The points (2, − 2), ( 8, 4), ( 4, 6) and (− 1, 1) in order are, the vertices of which one of the following, quadrilaterals?, (NDA 2007 I), (a) Square, (b) Rhombus, (c) Rectangle (but nor square), (d) Trapezium, 33. The area (in sq unit) of the triangle formed by the, lines x = 0,y = 0 and 3x + 4 y = 12 is, (a) 3 sq units, (b) 4 sq units, (c) 6 sq units, (d) 12 sq units, 34. The point P is equidistant from A(1, 3), B( − 3, 5) and, C( 5, − 1), then PA is equal to, (a) 5, (b) 5 5, (c) 25, (d) 5 10, 35. If a point ( x , y ) = (tan θ + sin θ, tan θ − sin θ ), then the, locus of (x , y) is, (a) ( x 2 y )2/ 3 + ( xy 2 )2/ 3 = 1, (b) x 2 − y 2 = 4xy, (c) x 2 − y 2 = 12xy, (d) ( x 2 − y 2 )2 = 16xy, 36. A straight rod of length 9 units slides with its ends. A, and B always on the x-axis and y-axis respectively,, then the locus of the centroid of ∆ OAB is, (b) x 2 + y 2 = 9, (a) x 2 + y 2 = 3, 2, 2, (c) x + y = 1, (d) x 2 + y 2 = 81, 37. The vertices of a triangle are (6, 6), (0, 6) and (6, 0), the distance between its circumcentre and centroid is, (a) 2 2, (b) 2, (d) 1, (c) 2, 38. Orthocentre of the triangle formed by the lines, x + y = 1 and xy = 0 is, (a) (0, 0), (b) (0, 1), (c) (1, 0), (d) ( − 1, 1), 39. The incentre of the triangle with vertices (1, 3),, (0, 0) and (2, 0) is, , 3, (a) 1,, , 2, , , 2 3, (c) ,, , 3 2 , , 2 1 , (b) ,, , 3 3, 1 , (d) 1,, , , 3
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479, , Rectangular Cartesian System, , Directions (Q. Nos. 40-42) Each of these question, contain two statements, one is Assertion (A) and other is, Reason (R). Each of these questions also has four, alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 40. Assertion (A) Three points ( p + 1, 1), ( 2 p + 1, 3) and, ( 2 p + 2, 2 p) are collinear, if p is equal to 2., Reason (R) Three given points are collinear, if the, area of triangle formed by these points should be zero., 41. Assertion (A) Three points (0, 0), (2, 2 3) and ( p, q), are the vertices of an equilateral triangle, then, ( p, q ) = ( 4, 0)., , 42. Assertion (A) Orthocentre of the triangle whose, vertices are (0, 0), (3, 0) and (0, 4) is (1, 1)., Reason (R), In a right angled triangle, the, orthocentre is the coordinate in which it is right, angled., , Directions (Q. Nos. 43-45), , Let us consider the, three vertices of a triangle be A(1, 1), B( − 1, − 1) and, C ( − 3 , k). On the basis of above information solve the, following questions., 43. If the triangle is an equilateral, then value of k is, (b) 3, (a) − 3, (c) 2 3, (d) None of these, 44. If the area of triangle is 4, then the value of k is, (b) 4 + 3, (a) − 4 − 3, (d) None of these, (c) 4 − 3, 45. The altitude of an equilateral triangle is, (a) 2, (b) 3, (d) 6, (c) 5, , Reason (R) In an equilateral triangle all sides are, equal., , Answers, Level I, 1., 11., 21., 31., 41., , (b), (c), (b), (b), (a), , 2., 12., 22., 32., , (d), (c), (c), (c), , 3., 13., 23., 33., , (b), (c), (b), (d), , 4., 14., 24., 34., , (c), (a), (a), (c), , 5., 15., 25., 35., , (d), (a), (d), (a), , 6., 16., 26., 36., , (b), (b), (b), (a), , 7., 17., 27., 37., , (d), (a), (c), (a), , 8., 18., 28., 38., , (c), (d), (c), (b), , 9., 19., 29., 39., , (b), (a), (b), (d), , 10., 20., 30., 40., , (b), (b), (c), (c), , 2., 12., 22., 32., 42., , (d), (c), (c), (d), (d), , 3., 13., 23., 33., 43., , (a), (c), (a), (c), (b), , 4., 14., 24., 34., 44., , (a), (d), (c), (d), (c), , 5., 15., 25., 35., 45., , (a), (c), (c), (b), (d), , 6., 16., 26., 36., , (c), (d), (b), (b), , 7., 17., 27., 37., , (b), (b), (a), (c), , 8., 18., 28., 38., , (c), (c), (c), (a), , 9., 19., 29., 39., , (c), (c), (c), (d), , 10., 20., 30., 40., , (c), (c), (c), (a), , Level II, 1., 11., 21., 31., 41., , (d), (d), (b), (b), (a)
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Hints & Solutions, Level I, 1. Let the third vertex of the triangle be ( x , y ), then, x + 4 + (−2), =2 ⇒x=4, 3, y+8+6, and, =7⇒ y=7, 3, ∴ The coordinates of the third vertex are (4, 7)., , 2. Since, points ( k, 3),( 2, k) and ( −k, 3) are collinear, so, k, 2, , 3 1, k 1 =0, , −k 3 1, ⇒, ⇒, ⇒, ⇒, ⇒, , k (k − 3) − 3 (2 + k) + 1 (6 + k2) = 0, k2 − 3k − 6 − 3k + 6 + k2 = 0, 2k2 − 6k = 0, k (k − 3) = 0, k = 0, 3, , 3. Let the three points be A( −2, − 5), B( 2, − 2) and, C( 8, a )., If three points are collinear, then, Slope of AB = Slope of BC, −2 + 5 a + 2, =, ⇒, 2+2, 8 −2, 5, 3 a+2, ⇒, =, ⇒ 9 = 2a + 4 ⇒ a =, 2, 4, 6, , 7. Let the sum of the distances of the point P ( x , y ) from, the points A( a , 0) and B( − a , 0) be 2b2., ∴, , PA 2 + PB2 = 2b2, (x − a ) + ( y − 0) + (x + a )2 + ( y − 0)2 = 2b2, x2 + a 2 − 2ax + y2 + x2 + a 2 + 2ax + y2 = 2b2, x2 + a 2 + y 2 = b 2, x2 + a 2 = b2 − y2, 2, , ⇒, ⇒, ⇒, , 8. Given that ,, , 2, , x1 = x , x2 = 1, x3 = 0, , and, y1 = 0, y2 = 1, y3 = 2, ∴ Area of triangle, 1, = [x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2)], 2, 1, = [x (1 − 2) + 1 (2 − 0) + 0 (0 − 1)], 2, 1, 1, = [− x + 2 + 0] = (2 − x), 2, 2, But area of triangle is 4 sq units, 1, ⇒, (2 − x) = 4, 2, ⇒, 2 − x=8⇒x= −6, , 9. Let the fourth vertex be D( x , y )., D, ( x, y), , C, (7, 2), , 4. Let P (1, 1), Q( −1, − 1) and R( − 3 , 3 ) are the vertices, of ∆ PQR, ∴, , PQ = (1 + 1)2 + (1 + 1)2 = 8 = 2 2, , QR = (− 3 + 1)2 + ( 3 + 1)2 = 8 = 2 2, Similarly,, PR = (− 3 − 1)2 + ( 3 − 1)2 = 8 = 2 2, ⇒, PQ = QR = PR, Which shows, ∆ PQR is an equilateral., , 5. Let ( h , k) be the point., According to question,, 4 (h − h )2 + k2 = h 2 + k2, ⇒, 4| k| = h 2 + k2, Locus of the point is, 4| y| = x2 + y2 ⇒ x2 + y2 − 4| y| = 0, , A, , We know that two diagonals of a parallelogram are, perpendicular bisector to each other., , −1 + 7 2 + x, =, 2, 2, ⇒, x=4, −6 + 2 −5 + y, and, =, 2, 2, ⇒, y=1, ∴ Fourth vertex of D is (4, 1)., ∴, , 10. The given points lie on a straight line, if, 1 1 1, −5 5 1 = 0, 13 λ 1, , 6. Let A( p, q ) be the point on the x-axis, which is, equidistant from the points B(1, 2) and C( 2, 3), then, AB = AC, ⇒, AB2 = AC 2, ⇒, ( p − 1)2 + (q − 2)2 = ( p − 2)2 + (q − 3)2, 2, ⇒ p + 1 − 2 p + q 2 + 4 − 4 q = p2 + 4 − 4 p + q 2 + 9 − 6 q, ⇒, 2 p + 2q = 8, ...(i), ⇒, p + q =4, Since, the value of p = 4 and q = 0 satisfies the Eq. (i), and on the x-axis q must be zero, so ( p, q) = (4, 0)., , B, , ⇒, ⇒, ⇒, , 1 (5 − λ ) − 1 (−5 − 13) + 1 (−5 λ − 65) = 0, 5 − λ + 18 − 5λ − 65 = 0, −6 λ = 42 ⇒ λ = − 7, , 11. Let the coordinates of fourth vertex are ( x , y )., Since, mid-point of a diagonals are same, x−5 7 + 3, =, ⇒ x = 15, ∴, 2, 2
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481, , Rectangular Cartesian System, y − 4 10 + 5, =, ⇒ y = 19, 2, 2, ∴ Coordinates of fourth vertex are (15, 19)., , 20. Given, θ = tan− 1 2, , and, , ⇒, , tan θ = 2, , 12. The given equations are, ...(i), x2 + y2 = 4, and, ...(ii), x+ y =2, These equations are satisfied by only (2, 0) and (0, 2)., Hence, the required set is {(0, 2), (2, 0)}., , 13. Q ( a , 0), ( 0, b) and (1, 1) are collinear., b 1 =0, , θ, 1, , x = r cos θ, y = r sin θ, y, tan θ =, x, y = 2x, , ⇒, , 1 1 1, ⇒, ⇒, ⇒, , ∴, , a (b − 1) + 1(0 − b) = 0, ab − a − b = 0, a + b − ab = 0, , 21. Given line is 4x + 5 y = 20, x y, + =1, 5 4, , or, , 14. We know, if the vertices of a triangle are, A ( x1 , y1 ), B ( x2 , y2 ) and C ( x3 , y3 ) and the respective, sides are a , b and c, then the excentre with respect to, angle B is, , y, , B (0, 4), , ax1 − bx2 + cx3 ay1 − by2 + cy3 , ,, , ., a−b + c, a −b+ c , , , 4x+5y = 20, , 15. If ( x , y ), (1, 2) and (–3, 4) are collinear, then area of, triangle formed by these points should be zero., ∴, ⇒, ⇒, , 1, [x(2 − 4) + 1(4 − y) − 3( y − 2)] = 0, 2, −2 x + 4 − y − 3 y + 6 = 0, −2x − 4 y + 10 = 0 ⇒ x + 2 y − 5 = 0, , x′, , 17. Let the vertices of the ∆ ABC be A( − 3, 0), B( 3, 0) and, C( 0, k)., , x, , 1, × OA × OB, 2, 1, = × 5 × 4 = 10 sq units, 2, , 22. Given,, , x1 = − 3, y1 = 4, x2 = − 1, y2 = − 2, , −3 0 1, 1, 3 0 1, ∴ Area of ∆ ABC =, 2, 0 k 1, , ⇒, , A, (5, 0), , ∴ Area of ∆ OAB =, , PQ ∩ PR = { P }., , ⇒, , O, , y′, , 16. Since, P, Q and R are three non-collinear points, then, , 1, 9 = { − 3 (− k) + 1(3k)}, 2, 18 = 3k + 3k, 18, k=, =3, 6, , 2, , We know that,, , a 0 1, 0, , √5, , x3 = 5, y3 = 6, x4 = x, y4 = − 4, (given), , 18. Q 1000° = 2 × 360° + 280°, ∴ From above it is clear that the revolving line will be, in the fourth quadrant., , 19. We know that, the points A( x1 , y1 ), B( x2 , y2 ) and, C( x3 , y3 ) are collinear, if, x1 ( y2 − y3 ) + x2( y3 − y1 ) + x3 ( y1 − y2) = 0, Here, x1 = at12, y1 = 2at1 , x2 = at22, y2 = 2at2, x3 = a , y3 = 0, ⇒ at12(2at2 − 0) + at22(0 − 2at1 ) + a (2at1 − 2at2) = 0, ⇒, 2a 2t1t2(t1 − t2) + 2a 2(t1 − t2) = 0, ⇒, 2a 2(t1 − t2) (t1t2 + 1) = 0, ⇒, t1t2 + 1 = 0 ⇒ t1t2 = − 1, , According to question,, ∆ABD = 2 ∆ACD, 1, [− 3 ( − 2 − ( − 4)) + ( − 1) ( − 4 − 4) + x ( 4 − ( − 2))], 2, 1, = 2 × [( − 3) ( 6 + 4) + ( 5) ( − 4 − 4) + x ( 4 − 6)], 2, 1, ⇒, ( − 6 + 8 + 6x ) = ( − 30 − 40 − 2x ), 2, 1, ( 2 + 6x ) = − ( 70 + 2x ), 2, Neglecting negative sign, we get, ⇒, , 1, [2 + 6x] = 70 + 2x, 2, , ⇒, , 1 + 3x = 70 + 2x, , ⇒, , x = 69
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482, , NDA/NA Mathematics, , 23. Given, equation, , 28. ∴ p12 + p22 =, , x y + xy − xy = 0, 3, , ⇒, , 3, , ⇒, , xy(x + y − 1) = 0, 2, , ⇒, x2 + y2 = 1, xy = 0, Thus, the equation represents a pair of straight lines, and a circle., , 24. Q ( a , b),( c, d ) and {( a − c),( b − d )} are collinear, then, the area of the triangle should be zero., Q, ∴, ⇒, ⇒, ⇒, , 1, ∆ = { x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2)} = 0, 2, a (d − b + d ) + c(b − d − b) + (a − c) (b − d ) = 0, 2ad − ab − cd + ab − ad − bc + cd = 0, ad − bc = 0, bc − ad = 0, , 25. Let the point be P ( x , y )., Given, points are A = (0, 4), B = (0, − 4), We have,, | AP − BP| = 6, ⇒, ( AP − BP ) = ± 6, 2, 2, ⇒ (x − 0) + ( y − 4)2, = ± 6 +, , (x − 0)2 + ( y + 4)2, , ⇒, , − 16 y − 36 = ± 12 x + y + 8 y + 16, , ⇒, , − 4 y − 9 = ± 3 x + y + 8 y + 16, , 2, , ⇒ [x − (a + b) ] + [ y − (b − a )], = [x − (a − b)]2 + [ y − (a + b)]2, 2, 2, ⇒ x + (a + b) − 2x(a + b) + y2 + (b − a )2 − 2 y (b − a ), = x2 + (a − b)2 − 2x (a − b) + y2 + (a + b)2 − 2 y (a + b), ⇒, −2x (a + b − a + b) = − 2 y (a + b − b + a ), ⇒, 4bx = 4ay, ⇒, bx = ay, 2, , 27. The vertices of the triangle are, P ( 2, 7), Q( 4, − 1), R( − 2, 6), ∴ PQ = (4 − 2)2 + (− 1 − 7)2 = 4 + 64 = 68, QR = (− 2 − 4) + (6 + 1) = 36 + 49 = 85, 2, , and, ∴, ⇒, ⇒, , cos2 α + sin2 α, , p22 = a 4 sin 2 2α cos 2 2α, 1 4, 2, = a sin 4α, 4, 2, , ∴, , p1 p2, ( p12 + p22)2, + =, p2 p1 , p12 p22, 4, =, = 4 cos ec2 4α, sin 2 4α, , 29. Given that, the vertices of the triangle are A ( − 4, 2),, B ( 0, − 1) and C ( 3, 3)., Then, length of AB = ( − 4 − 0)2 + ( 2 + 1)2, = 16 + 9 = 25 = 5, Length of BC = ( 0 − 3) + ( − 1 − 3)2, = 9 + 16 = 25 = 5, , = 49 + 1 = 50 = 5 2, ∴ The perimeter of ∆ABC = AB + BC + CA, = 5+ 5+ 5 2, , 2, , 26. Let the points A ( a + b, b − a ) and B ( a − b, a + b) and, P ( x , y ) be any point, then according to given condition, PA2 = PB2, 2, , a 2 cos2 2α, , = a 2 (sin 2 2α + cos 2 2 α ) = a 2 and, , p12, , 2, , 16 y2 + 81 + 72 y = 9(x2 + y2 + 8 y + 16), 9x2 − 7 y2 + 63 = 0, , ⇒, ⇒, , +, , Length of CA = ( 3 + 4)2 + ( 3 − 2)2, , 2, , ⇒ x2 + y2 + 16 − 8 y, = 36 + x2 + y2 + 16 + 8 y ± 12 x2 + y2 + 8 y + 16, 2, , sec α + cos ec α, 2, , 4 cos 2 α sin 2 α cos 2 2 α , = a2 , +, , 1 , cos 2 α + sin 2 α, , xy(x2 + y2) = xy, 2, , 4a 2, 2, , 2, , = 10 + 5 2, 30 . According to the given condition,, ⇒, , 25 = x2 + 16, x2 = 9 ⇒ x = ± 3, , 31. Let the point be (x , x), so according to the condition, (x − 1)2 + (x − 0)2 = (x − 0)2 + (x − 3)2, ⇒, − 2x + 1 = − 6x + 9, ⇒, x =2, Hence, the point is (2, 2)., , 32. I = 4 + 12 = 4, ⇒, a 2 + b2 = 16, 2, and, (a − 2) + (b − 2 3) 2 = 16, ⇒, a + 3b = 4, Hence,, (a , b) = (4, 0), , 33. In ∆ ABC, A ≡ ( − 3, 0), B ≡ ( 4, − 1) and C ≡ ( 5, 2), We know that, BC = (5 − 4)2 + (2 + 1)2, = 1 + 9 = 10, , RP = (− 2 − 2)2 + (6 − 7)2 = 16 + 1, , A, , = 17, QR2 = RP 2 + PQ 2, 85 = 17 + 68, 85 = 85, , ∆ PQR is a right angled., , B, , L, , C
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483, , Rectangular Cartesian System, and Area of ∆ABC, 1, = [− 3(− 1 − 2) + 4(2 − 0) + 5 (0 + 1)] = 11, 2, 2 ∆ ABC 2 × 11 22, Therefore, altitude AL =, =, =, BC, 10, 10, , 2 × 5 + 1( a ), =4, 2+1, , 34., ⇒, , ⇒, , y=, , AB = 17, CD = 17, BC = 50,, AD = 50, AC = 37 and BD = 97, Obviously, AB = CD and BC = AD. Also, diagonal, AC ≠ BD. Therefore, quadrilateral is parallelogram., l2 = a 2 + b2, l3 = (2a )2 + (2b)2 = 2 a 2 + b2, ⇒ l1 = l2 + l3, Hence, the points are collinear., , b =4, , 35. Here, x =, , ⇒, , 39. l1 = ( 3a )2 + ( 3b)2 = 3 a 2 + b2, , a =2, 2 × 7 + 1(b), =6, 2+1, , and, , 38. Let A( − 1, 1), B( 0, − 3), C( 5, 2) and D( 4, 6)., , a( a − b) − b ( a + b) a 2 − 2ab − b2, =, a−b, a−b, , 40. Let the points A, B, C and D are (− 2, 2), ( − 2, − 1),, ( 3, − 1) and (3, 2), respectively., , a (a + b) − b(a − b) a 2 + b2, =, a−b, a−b, , (–2,2) A, , D (3,2), , (–2,–1)B, , C (3,–1), , 36. Let the point be ( x , y )., A, , B, , C, , D, , (i) Point B (x, y) divides AD in 1 : 2, 0+9, 0 + 12, = 3 and y =, =4, 3, 3, (ii) Now point C (x, y) divides AD in 2 : 1., 0 + 18, 0 + 24, Then,, x=, = 6 and y =, =8, 3, 3, , ∴, , 37., , x=, , x1 + x2, x +x, x + x1, = 2, 2 3 = − 1, 3, =4, 2, 2, 2, x1 = 7, x2 = − 3, x3 = 1, Similarly, y1, y2 and y3 can be found., , Then, AB = 3, BC = 5, CD = 3, DA = 5, Clearly, angle between the diagonals is 90°. So, it is, rectangle., , 41. Let the third vertex be (x , y), then, x + 5 −2, 3, y+4+4, x = 12 and 6 =, 3, y = 10, 5=, , ⇒, ⇒, , Level II, 1. The vertices of a ∆ABC are A( 0, b), B ( 0, 0) and C( a , 0)., a b, a , Mid-point of E and D are , and , 0 ., 2 2, 2 , b, The slope of median BE , m1 = and, a, A (0, b), , 2. In ∆ABC, the vertices are, A (−3, 0), B(4, − 1) and C (5, 2), Distance between B and C,, BC = (5 − 4)2 + (2 + 1)2, = 1 + 9 = 10, and Area of ∆ABC, A (–3, 0), , E, L, (0, 0) B, , D, , C (a, 0), (4,–1) B, , 2b, . Since, the medians are, a, perpendicular to each other, , =, , ∴, , m1m2 = − 1 ⇒, , =, , ⇒, ⇒, , −2 b 2 = − a 2, a = ± 2b, , Slope of AD , m2 = −, , b −2 b , ×, = −1, a a , , L, , C (5, 2), , 1, [x1( y2 − y3 ) + x2( y3 − y1 ) + x3 ( y1 − y2)], 2, , 1, [−3 ( −1 − 2) + 4 ( 2 − 0) + 5 ( 0 + 1)], 2, 1, = [9 + 8 + 5] = 11, 2
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484, , NDA/NA Mathematics, 1, × BC × AL, 2, 1, 11 = × 10 × AL, 2, 2 × 11, 22, AL =, =, 10, 10, , As we know, ∆ ABC =, ⇒, ⇒, , y, , y=–x, , y=x, , x′, , x, , 3. Let the coordinates of point P are ( x , y )., It is given that, PA = PB, ⇒, (PA )2 = (PB)2, ⇒, { x − (a + b)}2 + { y − (a − b)}2, = { x − (a − b)}2 + { y − (a + b)}2, 2, 2, ⇒ x + (a + b) − 2x (a + b) + y2 + (a − b)2 − 2 y (a − b), = x2 + (a − b)2 − 2x (a − b) + y2 + (a + b)2 − 2 y (a + b), ⇒ −2x (a + b) − 2 y (a − b) = − 2x (a − b) − 2 y (a + b), ⇒, 2x [− a − b + a − b] + 2 y [− a + b + a + b] = 0, ⇒, x (−2b) + y (2b) = 0, ⇒, −x + y = 0 ⇒ x − y = 0, ∴ Locus of point P is x − y = 0, , y′, , So, x ± y = 0 is the locus of a point which moves, equidistant from the coordinate axes., , 6. Since, (0, 1), (1, 1) and (1, 0) are mid-point, of sides, AB, BC and CA, respectively., A (0, 0), , c, , (0, 2) B, , 4. Let the point be P ( h , k)., , b, , C (2 , 0 ), , a, , ∴Coordinates of A, B and C are (0, 0), (0, 2) and (2, 0),, respectively., , P (h,k), , AB = 2, BC = 2 2 , CA = 2, 0 + 0 + 2 ⋅2, x-coordinate of incentre =, 2+2 2+2, , Now,, (–3,0) B, , ∴, , A (3, 0), , =, , It is given that difference of the distance from points, A( 3, 0) and B( −3, 0) is 4, i.e.,, , PA − PB = 4, , ⇒, , (h − 3) + k − (h + 3) + k = 4, , ⇒, , (h − 3)2 + k2 = 4 +, , 2, , 2, , 2, , 2, , (h + 3)2 + k2, , On squaring both sides, we get, (h − 3)2 + k2 = 16 + (h + 3)2 + k2 + 8 (h + 3)2 + k2, ⇒ h + 9 − 6h + k = 16 + h + 9 + 6h + k, 2, , 2, , 2, , 7. Let ( x1 , y1 ), ( x2 , y2 ) and ( x3 , y3 ) are coordinates of the, points D , E and F, which divide each AB, BC and CA, respectively, in the ratio 3 : 1 (internally)., 3 × 6 − 1 × 1 17, =, 4, 4, 2, 1, −2 × 3 + 4 × 1, =− =−, y1 =, 4, 4, 2, , ∴, , x1 =, , and, , 2, , A (–1, 4), , + 8 (h + 3)2 + k2, ⇒, ⇒, , − 6h = 16 + 6h + 8 (h + 3)2 + k2, (x1, y1) D, , −8 (h + 3)2 + k2 = 12h + 16, , Again, squaring both sides, we get, 64 [(h + 3)2 + k2] = (12h + 16)2, ⇒, 64(h 2 + 9 + 6h + k2) = 144h 2 + 256 + 2⋅ 16 ⋅ 12h, ⇒, 64(h 2 + 9 + 6h + k2) = 16(9h 2 + 16 + 24h ), ⇒, 4(h 2 + 9 + 6h + k2) = 9h 2 + 16 + 24h, ⇒, 4h 2 + 36 + 24h + 4k2 = 9h 2 + 16 + 24h, h 2 k2, ⇒, 5h 2 − 4k2 = 20 ⇒, −, =1, 4, 5, 2, x, y2, Hence, the locus of point P is, −, =1, 4, 5, , 5. Q The lines y = x and y = − x lie at the same distances, in coordinate axes., ∴, , 2, =2 − 2, 2+ 2, , y = ± x ⇒ x ± y =0, , (6,–2) B, , Similarly,, , F (x3, y3), , E (x2, y2), , x2 = 0, y2 =, , C (–2, 4), , 5, 2, , 5, , y3 = 4, 4, Let (x, y) be the coordinates of centroid of ∆ DEF, 1 17, 5, x= , + 0 − =1, 3 4, 4, and, , x3 = −, , and, , y=, , ∴, , 1 1 5, , − + + 4 = 2, , 3 2 2, , Coordinates of centroid are (1, 2).
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485, , Rectangular Cartesian System, 8. Now, CA = ( at 2 − a )2 + ( 2at )2, = a (t 2 − 1)2 + 4 t 2= a t 4 + 1 + 2 t 2 = a (1 + t 2), 2, , 2, , −2 a , a, , and CB = 2 − a + , =, t , t, , =a, , 2a 2 4a 2, a2, + a2 − 2 + 2, 4, t, t, t, , 10. We know that, if P is such that the difference of its, distances from two given points is constant, then, locus of P is a hyperbola, which is the definition of, hyperbola., 11. We know, if coordinate axes are rotated, then, P = ( x cos θ − y sin θ , x sin θ + y cos θ )., , 1, 2, 1, , 4 + 1 + 2 = a 1 + 2, t, , , t, t , , It is rotated at an angle 135° i.e., θ = 135° and the new, point be, , 1, , 2a 2(1 + t 2) 1 + 2, , t , Now, HM of CA and CB =, 1, , , a 1 + t 2 + 1 + 2 , t , , , 2xy , Q HM of x and y = x + y , , , 1, , 2, 2a 2 + 2 + t , , , t, =, = 2a, 1, , 2, +, +, 2, t, , , , , t2, , P = [(4 cos (90° + 45° ) + 3 sin (90° + 45° ),, 4 sin (90° + 45° ) − 3 cos (90° + 45° )], = (− 4 sin 45° + 3 cos 45° , 4 cos 45° + 3 cos 45° ), 1 1 7, 1, 1, , 1, ,4⋅, + 3⋅, = −4 ⋅ + 3 ⋅, = −, , , , , 2 , 2, 2, 2, 2 2, , , p + q q − p, 12. Mid-point of ( p, q ) and (q , − p) is , ,, , which, 2, 2 , r s, is given by , ., 2 2, , 9. Let the points be, , ∴, , 1, 2 , ,, A = (x + 1, 2), B = (1, x + 2), C = , ., x + 1 x + 1, , If A, B and C are collinear, then area of ∆ ABC must, be zero., x1, x2, , i.e.,, , x3, x+1, ⇒, , 1, 1, x+1, , y1 1, y2 1 = 0, y3 1, 2, , x, 1, 1, x+1, , = s2 + r 2, , 13. Let the locus of the point be ( x1 , y1 )., ∴, , ⇒, , ⇒ x12 + (m + n )2 − 2x1 (m + n ) + y12 + (n − m)2 − 2 y1 (n − m), , 1, , = x12 + (m − n )2 − 2x1 (m − n ) + y12 + (n + m)2 − 2 y1 (n + m), , x+ 2 1 =0, 2, 1, x+1, , ⇒, ⇒, ⇒, , 2x1 (m − n − m − n ) + 2 y1 (n + m − n + m) = 0, − 4x1n + 4 y1m = 0, my1 = nx1, , Hence, locus of the point is, , −x 0, x+ 2 1 =0, 2, 1, x+1, , 0, , nx = my, , 14., , I. It is a correct statement that a triangle is, isosceles, if any two of its medians are equal., II. If we draw a circumcircle of a right angled ∆ ABC,, then AC is a diameter of that circle., A, , 0, , 1, x+ 3 1 =0, 1, 3, 1, 1+ x 1+ x, , ⇒, , , 3 , x x + 3 −, =0, x + 1, , , ⇒, , x (x2 + 3 + 4x − 3), =0, (x + 1), , ⇒, ∴, , [x1 − (m + n )]2 + [ y1 − (n − m)]2, = [x1 − (m − n )]2 + [ y1 − (n + m)]2, , Applying C 2 → C 2 + C1, x, , q− p s, =, 2, 2, , and, , Now, length of segment = ( p − q)2 + (q + p)2, , Applying R1 → R1 − R2, ⇒, , p+ q r, =, 2, 2, , x2 (x + 4) = 0 ⇒, x= −4, , O, , B, , x = 0, − 4, , C, , So, mid-point of a diameter AC is a centre of a, circle., III. We know, centroid of a triangle divides the, orthocentre and circumcentre in 2 : 1 internally., So, orthocentre, centroid and circumcentre are, collinear.
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486, , NDA/NA Mathematics, , 15. I. Altitudes BA and CA meets at A, then A is the, orthocentre of a triangle., B, , = x2 + (b − a )2 − 2x(b − a ) + y2 + (a + b)2 − 2 y(a + b), ⇒ 2x(a + b) + 2 y(a − b) = 2x(b − a ) + 2 y(a + b), ⇒ x {(a + b) − (b − a )} + y{(a − b) − (a + b)} = 0, ⇒, 2ax + (− 2by) = 0, ⇒, ax − by = 0 ⇒ − ax + by = 0, , 20. Since, the points with coordinates ( −a , 0), ( ap2 , 2ap), and ( ap12 , 2ap1 ) are collinear, then, A, , 0, 1, −a, ap2 2ap 1 = 0, ap12 2ap1 1, , C, , II. Let x-axis divides line segment in k:1, then, k x + 1x1 k y2 + 1 y1 , ,, C 2, =0, k+1 , k+1, , ⇒, , y, k y2 + y1, =0⇒k = − 1, y2, k+1, , ∴, , 16. Let the four points A ( a , b), O( 0, 0), B ( − a , − b) and, C ( ab, b2 )., , and, , …(i), …(ii), …(iii), , ∴ From Eqs. (i), (ii) and (iii), we see that the slopes c, m1 , m2 and m3 are equal., m1 = m2 = m3, , i.e.,, , 17. Required area = 2 area (∆ AOD), , A (c,c), , O, x, , x=, 0, , D (c,0), y–, , x=, , y+, 0, , y′, , = c× c= c, , 18. Given equations of two lines as, 2x + 3 y + 4 = 0, 4x + 3 y + 2 = 0, , ...(i), ...(ii), , On solving Eqs. (i) and (ii), the coordinates of the, intersecting point are (1, − 2)., Now,, , (0 − 1) + {0 − (− 2)} = d, 2, , d= 1+4 = 5, , ⇒, , 19. Let the coordinates of the point be ( x , y )., ∴, , 0, , −1, , 1, , p, p1, , p2 = 0, p12, , Now, using R2 → R2 − R1 and R3 → R3 − R1 , we get, 1 0, −1, 0 p p2 + 1 = 0, 0 p1 p12 + 1, ⇒, ⇒, , pp12 + p − p2p1 − p1 = 0, ( p1 − p) ( pp1 − 1) = 0, pp1 = 1, p1 = p, , = { x − (b − a )} + { y − (a + b)}, , A, , D (4,1), , C, , or, (x − 4)2 + ( y − 2)2 = y2, 2, or, x − 8x + 16 + y2 − 4 y + 4 = y2, or, x2 − 8x − 4 y + 20 = 0, This is the required equation of the set of all points., , 23. Let the coordinates of the points A, B, C and D be, (1, 3, 4), ( − 1, 6, 10) (− 7, 4, 7) and ( −5, 1, 1), respectively., AB = (− 1 − 1)2 + (6 − 3)2 + (10 − 4)2, = 4 + 9 + 36 = 7, BC = (− 7 + 1)2 + (4 − 6)2 + (7 − 10)2, = 36 + 4 + 9 = 7, , { x − (a + b)}2 + { y − (a − b)}2, 2, , 1, , (x − 4)2 + ( y − 2)2 = (x − x)2 + ( y − 0)2, , 2, , 2, , p2 = 0, p12, , 22. Let P ( x , y ) be any point. Since, this point is, equidistant from ( 4, 2) and x-axis, therefore, , B (c,–c), , x=c, , and, , 1, , p, p1, , = 4 + 4 = 8 =2 2, (which is the length of diagonal BD), ∴ Side of square ABCD, B (2,3), 1, 1, =, × BD =, ×2 2, 2, 2, = 2 units, , y, , O, , − 2a 1, , ∴ The distance between B and D, = (2 − 4)2 + (3 − 1)2, , 1, = 2 × × OD × AD, 2, , x′, , −1, , 21. Given that, the vertices B and D of a square ABCD be, ( 2, 3) and ( 4, 1) ,respectively., , Therefore, these points are collinear., , (0,0), , 0, , 1, , b, a, b, m2(slope of BO ) =, a, b2 b, m3 (slope of CO ) =, =, ab a, m1 (slope of AO ) =, , ∴, , ⇒, , 1, 2, , 2, , On squaring, we get, ⇒ x2 + (a + b)2 − 2x (a + b) + y2 + (a − b)2 − 2 y (a − b), , DA = (1 + 5)2 + (3 − 1)2 + (4 − 1)2, = 36 + 4 + 9 = 7
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487, , Rectangular Cartesian System, 0 + 2 3 − 2, Mid-point of AC = , ,, , 2, 2 , x + 1 y + 4, Mid-point of BD = , ,, , 2, 2 , , AC = (− 7 − 1)2 + (4 − 3)2 + (7 − 4)2, and, , = 64 + 1 + 9 = 74, CD = (− 5 + 7)2 + (1 − 4)2 + (1 − 7)2, = 4 + 9 + 36 = 7, BD = (− 5 + 1)2 + (1 − 6)2 + (1 − 10)2, , (x, y) D, , C (0,–2), , (2, 3) A, , B (1, 4), , = 16 + 25 + 81 = 122, Q, AB = BC = CD = DA, But, BD ≠ AC, ∴ Points A, B,C and D form a rhombus., , 24. Let the points A(1, 1), B( 2, 0) and C( 3, 1) be the, vertices of ∆ ABC., ∴, , AB = (2 − 1)2 + (0 − 1)2 = 2, BC = (3 − 2)2 + (1 − 0)2 = 2, , and, , 28. OA = ( x1 − 0)2 + ( y1 − 0)2 = x12 + y12, , CA = (3 − 1)2 + (1 − 1)2 = 2, , ∴ We see that the sum of the squares of two sides is, equal to the square of third side which is the property, of right angled triangle and also, two sides are equal, i.e., isosceles right angled triangle., 25. We know that, the other vertices of the triangle are, a − 3 a, a, 3a , , and − ,, − ,, 2 , 2 , 2, 2, ∴Equation of line BC is, y, , C, , O, , (a, 0), , y′, , x=−, , a, 2, , 2x + a = 0, , ⇒ a, b and c are in harmonic progression for all m., , 26. Since, middle point of A(1, 2) and B ( x , y ) is C( 2, 4)., ∴, , By cosine law,, y, OA 2 + OB2 − AB2, cos ∠AOB =, 2 OA ⋅ OB, ⇒ OA ⋅ OB ⋅ cos ∠AOB, B (x2, y2), 1, = (x12 + y12 + x12 + y22) − (x12 + x22 − 2x1x2), 2, A (x1, y1), − ( y12 + y22 − 2 y1 y2), 1, x, = {2x1x2 + 2 y1 y2} = (x1x2 + y1 y2), O, 2, , 1, [x1 ( y2 − y3 ) + x2 ( y2 − y1 ) + x3 ( y1 − y2)], 2, ∴ a{(m + 1)b − (m + 2)c} + b{(m + 2)}c − ma }, + c {ma − (m + 1)b} = 0, ⇒, mab + ab − mac − 2ac + mbc + 2bc − mab, + mac − mbc − bc = 0, ⇒, ab − 2ac + 2bc − bc = 0, ⇒, ab + bc = 2ac, 2ac, b=, ⇒, a+c, , x, , B, , 1+ x, =2, 2, x =3, , AB = (x1 − x2)2 + ( y1 − y2)2, , ∆=, , A, , – a , – √3a, 2, 2, , ⇒, , and, , 29. Since, the points A( a , ma ), B[b,( m + 1)b] and, C [c,( m + 2)c] are collinear area of the triangle should, be zero formed these points., , – a , √3a, 2 2, , x′, , 2+0 1+ x, 3 −2 4 + y, and, =, =, 2, 2, 2, 2, ⇒, x = 1 and y = − 3, ∴ Value of (x, y) = (1, − 3), , ∴, , 2+ y, and, =4, 2, and, y =6, , D, , ⇒, ∴ B(3, 6), BC = (2 − 3)2 + (4 − 6)2 = 1 + 4, = 5, In ∆BCD,, , 1 2 1, 30. For collinear 2 4 1 = 0, 3 a 1, , 3, , A, , C, , B, , BD = (CD )2 − (BC )2, = 3 2 − ( 5 )2 = 9 − 5, = 4 =2, , 27. Given, A( 2, 3), B (1, 4), C( 0, − 2) and D( x , y ) are the, vertices of a parallelogram. We known that the, mid-points of diagonals of a parallelogram are same., , 1(4 − a)− 2(2 − 3) + 1 (2a − 12) = 0, ⇒, ⇒, −6 + a = 0, ⇒, a =6, ∴ Point C is (3, 6), ∴, BC = (3 − 2)2 + (6 − 4)2 = 1 + 4 = 5, , 31. Let point P ( x1 , y1 ) be equidistant from points A(1, 2), and B( 3, 4)., ∴, ⇒, , PA = PB, PA 2 = PB2
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488, , NDA/NA Mathematics, ⇒, (1 − x1 )2 + (2 − y1 )2 = (3 − x1 )2 + (4 − y1 )2, 2, ⇒ 1 + x1 − 2x1 + 4 + y12 − 4 y1 = 9 + x12 − 6x1 + 16 + y12 − 8 y1, ...(i), ⇒, x1 + y1 = 5, Point P (x1 , y1 ) lies on 2x − 3 y = 5, ...(ii), ∴, 2x1 − 3 y1 = 5, On solving Eqs (i) and (ii), we get, x1 = 4 and y1 = 1, ∴ Coordinates of P are (4, 1)., , 32. Let A( 2, − 2), B( 8, 4), C( 4, 6) and D( − 1, 1) are in order., AB = (8 − 2)2 + (4 + 2)2 = 36 + 36 = 72, 2, , y, , , , x − y, , = tan θ sin θ, 2 , , 1 − cos 2 θ , sin 2 θ , ⇒ x2 − y2 = 4 , , =4 , cos θ , cos θ , x2 − y2 = 4 (sec θ − cos θ ), (x2 − y2)2 = 16 (sec2 θ + cos 2 θ − 2), (x2 − y2)2 = 16 (1 + tan 2 θ + 1 − sin 2 θ − 2), (x2 − y2)2 = 16 (tan 2 θ − sin 2 θ ), (x2 − y2)2 = 16xy, , CD = (− 1 − 4)2 + (1 − 6)2 = 25 + 25 = 50, and, , x+, , 2, , (x2 − y2)2 = 16 (tan θ + sin θ ) (tan θ − sin θ ), , BC = (4 − 8) + (6 − 4) = 16 + 4 = 20, 2, , Now,, , 36. Let the straight rod intercept the coordinate axes be, A( x1 , 0) and B(0, y1)., , DA = (2 + 1)2 + (2 − 1)2 = 9 + 1 = 10, , y, , Q, AB ≠ BC ≠ CD ≠ DA, ∴ The quadrilateral ABCD is a trapezium., , B (0, y1), , 33. The intersection point of x = 0,y = 0 and 3x + 4 y = 12 is, A( 4, 0),B( 0, 3) and C( 0, 0)., , 9, , y, , A (x1, 0), , x, , B (0, 3), , Let coordinate of centroid be G (h , k), then h =, x′, , C (0,0), , A (4, 0), , In ∆ AOB, OA 2 + OB2 = AB2, ⇒, x12 + y12 = AB2, ⇒, x12 + y12 = 81, , x, , 2, , 1, 1, × OA × OB = × 4 × 3 = 6 sq units, 2, 2, , 34. Let the coordinates of P are (x , y)., Since, P is equidistant from A, B and C., AP 2 = BP 2, and, BP 2 = CP 2, From Eq. (i),, (x − 1)2 + ( y − 3)2 = (x + 3)2 + ( y − 5)2, ⇒, x2 + 1 − 2 x + y 2 + 9 − 6 y, , ...(i), ...(ii), , = x2 + 9 + 6x + y2 + 25 − 10 y, ⇒, 8x − 4 y + 24 = 0, ...(iii), ⇒, 2x − y + 6 = 0, From Eq (ii),, (x + 3)2 + ( y − 5)2 = (x − 5)2 + ( y + 1)2, 2, ⇒, x + 9 + 6x + y2 + 25 − 10 y, = x2 + 25 − 10x + y2 + 1 + 2 y, ⇒, 16x − 12 y + 8 = 0, … (iv), ⇒, 4x − 3 y + 2 = 0, On solving Eqs. (iii) and (iv), we get, x = − 8, y = − 10, Now, PA 2 = (− 8 − 1)2 + (− 10 − 3)2, = 81 + 169 = 250 ⇒ PA = 250 = 5 10, , 35. Given , tan θ + sin θ = x and tan θ − sin θ = y, ∴, , x+ y, x− y, and sin θ =, tan θ =, 2, 2, , 2, , x1 , y , + 1 =9, 3, 3, ⇒, h 2 + k2 = 9, Hence, locus of a point is x2 + y2 = 9, , ⇒, , y′, , Area of ∆ ABC =, , x1, y, , k= 1, 3, 3, , 6 + 0 + 6 6 + 6 + 0, 37. Centroid of ∆ ABC = , ,, = ( 4, 4), , 3, 3 , y, C (0, 6), , B (6, 6), , 90º, M, x′, , O, , A (6, 0), , x, , y′, , Mid-point of AC is called the circumcentre., ∴ Circumcentre, 0 + 6 6 + 0, ,, M, = M (3, 3), 2, 2 , The distance between circumcentre and centroid, = (4 − 3)2 + (4 − 3)2 = 2, , 38. Given lines are x + y = 1 and xy = 0, When x = 0, then y = 1, When x = 1, theny = 0
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489, , Rectangular Cartesian System, y, , 41. A. Here, A( 0, 0), B( 2, 2 3 ) and C( p, q ), Here, AB = CA = BC, , (0, 1), , 2 2 + (2 3) 2 = p 2 + q 2 = ( p − 2) 2 + (q − 2 3) 2, ⇒, , ∴ (0, 1) and (1, 0) are the vertices of triangle. Clearly,, triangle is right angled isosceles. Orthocentre of right, angled triangle is same as the vertex of right angle., Therefore, point of intersection of x + y = 1 and xy = 0 is, (0, 0)., , 39., , 16 = p2 + q2 = p2 + 4 − 4 p + q2 + 12 − 4 3q, , ⇒, , x, , (1, 0), , (0, 0), , 4 + 12 = p2 + q2 = ( p − 2)2 + (q − 2 3 )2, 0 = 16 − 4 p − 4 3q, , ⇒, , 4 p + 4 3q = 16, , ⇒, ⇒, , p = 4, q = 0, , 42. A. It is clear that the given points are the coordinate, of a right angle triangle at (0, 0)., So, the orthocentre is (0, 0)., Hence, it is not true., But R is true., , A (1, √3), , 2, , Solutions (Q. Nos. 43-45), , √3, , 43. Since, triangle is an equilateral., 60º, (0, 0) B, , (1 , 0 ), 2, , C (2, 0), , ( AB)2 = (BC )2 = (CA )2, , x, , 2, , Clearly, the triangle is equilateral., So, the incentre is the same as the centroid., 1 + 0 + 2 3 + 0 + 0 1 , Incentre = , ∴, ,, = 1, , 3, 3, 3, , , , 40. Since, points ( p + 1, 1), ( 2 p + 1, 3) and ( 2 p + 2, 2 p) are, collinear, then, p+1 1 1, 2p + 1 3 1 = 0, 2p + 2 2p 1, ⇒ ( p + 1), , 3, , 1, , 2p 1, , −1, , 2p + 1 1, 2p + 2 1, , +1, , 2p + 1, , 3, , 2p + 2 2p, , =0, , ⇒ ( p + 1) (3 − 2 p) − (2 p + 1 − 2 p − 2), + (4 p2 + 2 p − 6 p − 6) = 0, ⇒, ⇒, , 2 p2 − 3 p − 2 = 0, p = 2 , − 0.5, , Thus, both A and R are individually true and R is the, correct explanation of A., , (− 1 − 1)2 + (− 1 − 1)2 = (− 3 + 1)2 + (k + 1)2, , , , , 8 = 3 + 1 − 2 3 + k2 + 2k + 1, , ⇒, , 8 = 5 − 2 3 + k2 + 2k, , ⇒, ⇒, ⇒, , k + 2k − 2 3 − 3 = 0, k= 3, 2, , 44. Area of ∆ =, , 1, 1 1, 1, −1 −1 1, 2, − 3 k 1, , 1, 1 (− 1 − k) − 1(− 1 + 3 ) + 1(− k − 3 ), 2, 1, = − 2k − 2 3, 2, , ⇒, , 4=, , ⇒, , 4 =|− k − 3|= k + 3, , ⇒, , k =4 − 3, , 45. Side of a triangle, AB = ( − 1 − 1)2 + ( − 1 − 1)2, = 4 + 4 =2 2, 3, 3, side =, ∴ Altitude of ∆ ABC =, ×2 2 = 6, 2, 2, , 2
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25, , The Straight Line, Any curve is said to be a straight line, if two points are, taken on the curve such that every point on the line, segment joining any two 2points on it lies on the curve., , Slope or Gradient of a Line, , Hence, the equation of a line parallel to x-axis at a, distance b, is y = b . If a line 2is parallel to x-axis at a distance, b and below x-axis, then its equation is y = − b. Equation of, x-axis is y = 0., , Equation of a Line Parallel to y-axis, , If a line AB makes an angle θ with the positive x-axis,, then m or tanθ is called the slope or gradient of the line., i. e. , m = tanθ, y, The angle of inclination of a, line with the positive direction of, B, x-axis in anti-clockwise sense, θ, always lies between 0° and 180° x ′, x, O, A, i.e.,, 0≤ θ < π, If 0° < θ < 90° , then θ is an, acute angle., y′, If θ = 0°, either the line is, x-axis or it is parallel to x-axis., If 90° < θ < 180° , then θ is an obtuse angle., Also, the slope of a line equally inclined with axes is 1, or − 1 as it makes 45° or 135° angle with x-axis., , Let AB be a line parallel to y-axis and at a distance a, from it. Then, the abscissa of every point on AB is a. So, it, can be treated as the locus of a point at a distance a from, y-axis, then equation of line is x = a . Since, if a line parallel, to y-axis at a distance a to the left of y-axis, then equation is, x = − a, equation of y-axis is x = 0., , Equation of Lines Parallel to Axes, , 5 units from y-axis is, (a) x = 5, (c) x + y = 5, , Equation of a Line Parallel to x-axis, Let AB be a straight line parallel to x-axis at a distance, b from it., Thus, if P ( x , y ) is any point on AB, then y = b, y, , A, , y=b, , B, , y′, , a, O, , y′, , x, , x=a, , x′, , B, , Example 1. The equation of line which is at a distance, (b) y = 5, (d) None of these, , Solution (a) Since, line is at a distance 5 units from y-axis, then, it is parallel to y-axis. Then, equation of line is x = 5., , Example 2. The equation of a line which is at equal, (d) x = 1, , Solution (d) Since, the given lines are both parallel to y-axis, x, , O, , A, , distance from the lines x = − 2 and x = 4 is, (a) y = 1, (b) x = − 1, (c) y = − 1, , b, x′, , y, , and the required line is equidistant from these lines, so it is also, parallel to y-axis and its distance from y-axis, 1, = ( −2 + 4) = 1 unit, 2, Hence, its equation is x = 1.
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491, , The Straight Line, , Equation of Straight Lines in, Different Forms, , Solution (a) Qθ = 60 °, ∴, , Cartesian Form, If the equation of the form y = f ( x )or φ ( x , y ) = 0, where, x and y are the cartesian coordinates of the moving point,, then it is said to be cartesian, y, equation of line., P (x, y ), , Slope-intercept Form, The equation of a line with x ′, slope m and making an intercept, c on y-axis, is, , θ, O, , c, L, , x, , y′, , y = mx + c, , Example 3. The equation of a line with slope 1/2 and, making an intercept 5 on y-axis is, (a) 2y = m + 4, (b) 2y = m + 5, (c) y = m + 5, (d) None of these, , Solution (b) Here, m =, , 1, m+5, 2, 2y = m + 5, , ∴ Equation of required line is y =, , (b) 3x − y − 4 = 0, (d) None of these, , Solution (a) We know that,, m = tan θ, m = tan 120 °, m = tan (90 ° + 30 ° ), = − cot 30 ° = − 3, , (Slope), ( given θ = 120 ° ), , and given that, c=–4, ∴ Equation of line is y = mx + c, ⇒, y = − 3x − 4, or, , Two-point Form, Equation of line passing through two points (x1 , y1) and, (x2 , y2) is given by, y − y1 x − x1, =, y2 − y1 x2 − x1, , Example 6. The equation of line joining the points, ( at12 , 2 at1) and ( at 22 , 2 at 2 ) is, (a) y (t1 − t 2) = 2 x + 2 at1 t 2, (c) y (t1 + t 2) = x + 2 at1 t 2, , 3 x+ y + 4= 0, , ⇒, , y − y , y − y1 = 2 1 ( x − x1) ,, x2 − x1 , 2at 2 − 2at1, ( x − at12), at 22 − at12, 2, y − 2at 1 =, ( x − at12), t1 + t 2, , y(t1 + t 2) − 2at12 − 2at1t 2 = 2x − 2at12, , ⇒, , y (t1 + t 2) = 2x + 2at1t 2, , Intercept Form, Equation of the line if their intercept on the axes are a, x y, and b, is given by + = 1., a b, , Example 7. The equation of the line which cut off an, intercept 8 on the positive direction of x-axis and an intercept, 6 on the negative direction of y-axis is, (a) 3x − y = 24, (b) 3x + 4y = 24, (c) 3x − 4y = 24, (d) 2 x − 3y = 24, , Solution (c) Using, , Parametric Form, Equation of line passing through ( x1 , y1 ) and making, an angle θ from the positive direction of x-axis in, anti-clockwise rotation is given by x = x1 + r cosθ and, y = y1 + r sinθ, r is known as parameter., , (b) y (t1 + t 2) = 2 x + 2 at1 t 2, (d) None of these, , (y − 2at1) =, , ⇒, , intercept of 4 units on negative direction of y-axis and makes, an angle of 120° with the positive direction of x-axis is, , ⇒, ⇒, , 3x − y = 3 − 2, , ⇒, , we get, , Example 4. The equation of a straight line which cut off an, (a) 3x + y + 4 = 0, (c) x + y + 4 = 0, , ∴ Equation of line whose slope is 3 and passing through, the point (1, 2) is, y − 2 = 3 ( x − 1) ⇒ y − 2 = 3x − 3, , Solution (b) Using, , 1, and c = 5, 2, , ⇒, , m = tan 60 ° = 3, , or, , x y, + = 1, we get, a b, x, y, +, =1, 8, −6, 6x − 8y = 48 or 3x − 4y = 24, , Normal Form, , The equation of a line which passes through the point, ( x1 , y1 ) and has the slope m is y − y1 = m ( x − x1 )., , Equation of a line, if perpendicular distance of origin, from the line is p and perpendicular drawn from origin to, the line is making an angle α from the positive direction of, x-axis, then equation of line is, x cos α + y sin α = p, , Example 5. The equation of line which passes through, , Example 8. The equation of the perpendicular bisector of, , Point-slope Form, , (1, 2) and making an angle of 60° with x-axis., (a) 3 x − y = 3 − 2, (b) 3 x + y = 3 − 2, (c) 3 x − 2y = 3 + 2, (d) 3 x − y = 2 3 − 4, , the line segment joining the points A (1, 2) and B(3 , – 7) is, (a) 18y − 4x + 53 = 0, (b) 9y − 3x + 50 = 0, (c) 18y + 4x + 53 = 0, (d) None of these
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492, , NDA/NA Mathematics, , Example 10. The angle between the lines 3x + 4y + 6 = 0, and 3x + 4y + 7 = 0 is, (a) 0°, (b) 90°, (c) 45°, (d) None of these, , Solution (a) The slope of AB is given by, −7 − 2 − 9, m=, =, 3 −1, 2, , 1 2, =, m 9, Let P be the mid-point of AB. Then, the coordinates of P are, 5, , 1+ 3 2 − 7, ,, , i.e., 2 , − ., , 2, 2, 2 , 5, , Thus, the required line passes through P 2, − and has, , 2, 2, the slope ., 9, 5 2, , So, its equation is y + = ( x − 2), , 2 9, or, 9 (2y + 5) = 4x − 8, ⇒, 18y − 4x + 53 = 0, , ∴, , Slope of a line perpendicular to AB = −, , Solution (a) The equation of lines are, , 3, 4, 3, and slope of line (ii) is, m2 = −, 4, Since,, m1 = m2, ∴ Lines are parallel to each other i.e., the required angle is 0°., , Example 11. The angle between the line x + y = 3 and the, line joining the points (1, 1) and ( −3 , 4) is, , distance from the origin is 5 and perpendicular drawn from, the origin to the line is making an angle 45° from the positive, direction of x-axis is, (a) x + y = 5 3, (b) x + y = 2, (d) None of these, (c) x + y = 5 2, ∴ Equation of required line is, x cos 45° + y sin 45° = 5, 1, 1, x⋅, + y⋅, =5, ⇒, 2, 2, x+ y =5 2, ⇒, , General linear equation in x and y, say ax + by + c = 0,, represents an equation of line., , Angle between Two Straight, Lines, Angle between two lines, if their slopes are m1 and m2 ,, is given by, y, m1 − m2, tan θ = ±, D, B, 1 + m1m2, a2b1 − a1b2, a1a2 + b1b2, , P, θ, , i.e., a2b1 − a1b2 = 0, a1 b1, or, = ., a2 b2, , θ1, , θ2, , O, y′, , Case II If m1m2 = − 1, then lines are perpendicular, i.e.,, , a1a2 + b1b2 = 0, , Slope of line PQ,m1 =, , ⇒, , 1 − 4 −3, =, 1+ 3, 4, x+ y −3 =0, , ...(i), , −3, +1, 1, m1 − m2, = 4, =, tan θ =, 3, 7, 1 + m1 m2, 1+, 4, 1, θ = tan −1 , 7, , Condition of Concurrency, of Three Lines, Let, a1x + b1 y + c1 = 0,, a2x + b2 y + c2 = 0 and, a3 x + b3 y + c3 = 0 are the three concurrent lines. Then, the, condition of concurrency of three lines are, a1 b1 c1, a 2 b2 c2 = 0., a3 b3 c3, , If m1 = m2,, x′, , (d) None of these, , Solution (a) Let P = (1, 1) and Q = ( −3 , 4), , ∴, , then lines are parallel, , 1, (b) tan− 1 , 4, , Given, line is, or, x+ y =3, −1, Its slope, m2 =, = −1, 1, , General Form, , Case I, , 1, (a) tan− 1 , 7, 3, (c) tan− 1 , 7, , ∴, , Solution (a) Given that p = 5 and α = 45°, , ±, , …(i), …(ii), , m1 = −, , Slope of line (i) is, , Example 9. The equation of line whose perpendicular, , or, , 3x + 4y + 6 = 0, 3x + 4y + 7 = 0, , and, , x, , Example 12. The value of λ,, 8 x − 11y − 33 = 0 and 2 x − 3y + λ, (a) 7, (c) 3, , if the lines 3x − 4y − 13 = 0,, = 0 are concurrent, is, (b) − 7, (d) − 3, , Solution (b) Given that 3x − 4y − 13 = 0, 8x − 11y − 33 = 0 and, 2x − 3y + λ = 0 are concurrent lines.
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493, , The Straight Line, 3 −4 −13, 8 −11 −33 = 0, 2 −3, λ, , ∴, ⇒, ⇒, ∴, , −2 − 2 + λ = 0 ⇒ λ = 4, On putting the value of λ in Eq. (ii),, ∴ The equation of the required line is 2x − y + 4 = 0 ., , 3( − 11λ − 99) + 4(8λ + 66) − 13( − 24 + 22) = 0, − λ −7 =0, λ = −7, , Example 13. If the lines x + 2 ay + a = 0, x + 3by + b = 0, and x + 4 cy + c = 0 are concurrent, then a, b, c are in, (a) AP, (b) GP, (c) HP, (d) None of these, , Solution (c) Since, the lines x + 2ay + a = 0 , x + 3by + b = 0, and x + 4cy + c = 0 are concurrent, then, 1 2a a, 1 3b b = 0, , ⇒, ⇒, ⇒, ⇒, ⇒, , 1, , 4c c, , − 2a, , 1 b, 1 c, , +a, , 1 3b, , Example 16. The equation of a line perpendicular to the, line 3x + 2y = 6 and passes through (2, 3) is, (a) 2 x + 3y + 5 = 0, (b) 2 x − 3y + 5 = 0, (c) x − 3y + 6 = 0, (d) None of these, , 1 4c, , The equation of line perpendicular to given line is, 2x − 3y + λ = 0, This equation passes through (2, 3), then, 4 −9 + λ =0, ⇒, λ =5, ∴ The equation of required line is 2x − 3y + 5 = 0, , =0, , (3bc − 4bc) − 2a( c − b) + a( 4c − 3b) = 0, − bc − 2ac + 2ab + 4ac − 3ab = 0, − ab − bc + 2ac = 0, 2ac, b=, a+ c, , Example 17. The equation of a line which divides the join, of (1, 0) and (3, 0) in the ratio 2 :1 and perpendicular to it is, (a) 3x = 8, (b) x = 6, (c) 2 x = 5, (d) 3x = 7, , a, b, c are in HP., , Solution (d) Let C be the point which divides the join of (1, 0), , Equation of a Line Parallel, to a Given Line, Let ax + by + c = 0 be any straight line, then, ax + by + λ = 0 be the equation of a line parallel to the, given line, where λ is a constant., , Example 14. The equation of a line which is parallel to, x + 4y + 5 = 0 and passes through (1, 2) is, (a) x + 4y + 9 = 0, (b) x + 4y − 9 = 0, (c) x + 2y + 9 = 0, (d) None of these, , Example 15. The equation of the line through ( − 1, 2) and, (b) x + y + 5 = 0, (d) None of these, , Equation of any line parallel to line (i) is, 2x − y + λ = 0, If line (ii) passes through the point ( − 1, 2), then, , Since, AB is along x-axis , therefore a line perpendicular to AB is, 7 , parallel to y-axis . As it passes through C , 0 , therefore its, 3 , 7, ⇒ 3x = 7, 3, Hence, the equation of the required line is 3x = 7., x=, , x + 4y + 5 = 0, ∴ The equation of line parallel to this line is x + 4y + λ = 0, This line passes through (1, 2), then, 1+ 8 + λ = 0 ⇒ λ = − 9, ∴ The equation of required line is x + 4y − 9 = 0., , Solution (a) Given line is 2x − y + 3 = 0, , and (3, 0) in the ratio 2 :1., Then, the coordinates of C are, 2 × 3 +1 × 1 2 × 0 + 1 × 0 7 , ,, = , 0, , 2 +1, 3 , 2+1, , equation is, , Solution (b) Since, the equation of given line is, , parallel to the line y = 2 x + 3 is, (a) 2 x − y + 4 = 0, (c) 2 x + y + 4 = 0, , Let ax + by + c = 0 be any straight line, then, bx − ay + λ = 0 is perpendicular to it, where λ is a, constant., , Solution (b) The equation of given line is 3x + 2y = 6, , 1 4c c, 3b b, , Equation of a Line Perpendicular, to a Given Line, , …(i), …(ii), , Perpendicular Distance of a, Line from a Point, Let the length of the perpendicular from a point ( x1 , y1 ), to a line ax + by + c = 0 is, ax + by + c, 1, 1, , ., 2, a + b2 , , The length of the perpendicular from the origin to the, |c |, ., line ax + by + c = 0 is, 2, a + b2
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495, , The Straight Line, and at point (1, 2), 3 (1) + 4 (2) − 7 = 3 + 8 − 7, = 11 − 7, = 4 >0, Since, both the values are of opposite signs, then both lines, are on the opposite side to the line 3x + 4y − 7 = 0 ., , The Image of a Point wrt the Line, The image of a point A ( x1 , y1 ) wrt the line minor, ax + by + c = 0 be B ( h , k) is given by, A (x1, y1) ax + by + c = 0, , Example 23. The position of points (1, 2) and ( 4, − 6) on the, line 3x − 4y = 8 is, (a) same side, (c) don’t say, , B, , h − x1 k − y1 − 2( a x1 + b y1 + c), =, =, a, b, a 2 + b2, , (b) opposite side, (d) None of these, , Solution (b) Let z = 3x − 4y − 8, z1 = 3 × 1 − 4 × 2 − 8, = 3 − 8 − 8 = − 13, at point ( 4, − 6) z 2 = 3 × 4 − 4 × ( − 6) − 8, = 12 + 24 − 8 = 28, ⇒ z 1 and z 2 are of opposite signs, therefore the two, at point (1, 2), , points are on the opposite sides of the given line., , Family of Lines through the, Intersection of Two Lines, The equation of the family of lines passing through the, intersection of the lines a1x + b1 y + c1 = 0 and, a2x + b2 y + c2 = 0 is( a1x + b1 y + c1 ) + λ ( a2x + b2 y + c2 ) = 0,, where λ is a parameter., , Comprehensive Approach, n, , n, , n, , n, , The equation of a line whose mid-point is ( x1 , y1) in between the, x, y, axes, is +, = 2., x1 y1, The foot of perpendicular on a line from a given point can be found, by solving the equation of the line and the equation of, perpendicular from the given point on the line. The coordinates of, the foot of perpendicular from ( x1 , y1) to the line are, ax + by + c = 0 are, b 2x1 − aby1 − ac a 2y1 − abx1 − bc , ,, , , a2 + b 2, a2 + b 2, , , The foot of the perpendicular (h, k) from ( x1 , y1) to the line, ax + by + c = 0 is given by, h − x1 k − y1 − ( ax1 + by1 + c), =, =, a, b, a2 + b 2, The image or (reflection) of the point ( x1 , y1), ax + by + c = 0 is (h, k), h − x1 k − y1 − 2 ( ax1 + by1 + c), where,, =, =, a, b, a2 + b 2, , in the line, , n, , n, , n, , n, , n, , n, , A straight line is such that the algebraic sum of the perpendicular, drawn upon it from any number of fixed point is zero. Then, the line, always passes through a fixed point., If the vertices of a triangle have integral coordinates, then the, triangle can’t be equilateral., If coordinates of the vertices of a triangle are rational, then triangle, cannot be equilateral., Orthocentre of a triangle will lie inside, if triangle is acute and in, case of obtuse triangle orthocentre will lie outside and in case of, right angle triangle orthocentre is the vertex at which it is right, angled., Three or more straight lines are said to be concurrent lines, if they, meet in a point., The distance between parallel lines, ax + by + c1 = 0 and ax + by + c2 = 0, c −c , 2 ., is 1, , 2, 2, +, a, b, ,
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Exercise, Level I, 1. The distance of the point (–2, 3) from the line, x − y = 5 is, (a) 5 2, (b) 2 5, (c) 3 5, (d) 5 3, 2. The equation of the straight line joining the origin to, the point of intersection of y − x + 7 = 0 and, y + 2x − 2 = 0 is, (a) 3x + 4 y = 0, (b) 3x − 4 y = 0, (c) 4x − 3 y = 0, (d) 4x + 3 y = 0, 3. The equation of the straight line which is, perpendicular to y = x and passes through (3, 2) is, (a) x − y = 5, (b) x + y = 5, (c) x + y = 1, (d) x − y = 1, 4. The angle between the straight lines x − y 3 = 5 and, 3x + y = 7 is, (a) 90°, (b) 60°, (c) 75°, (d) 30°, 5. What is the perpendicular distance of the point (x, y), from x-axis?, (NDA 2012 I), (a) x, (b) y, (c)|x|, (d)| y|, 6. In what ratio is the line y − x + 2 = 0 divides the line, joining the points (3, –1) and (8, 9)?, (a) 1 : 2, (b) 2 : 1, (c) 2 : 3, (d) 3 : 4, 7. For what value of k, are the lines x + 2 y − 9 = 0 and, (NDA 2011 II), kx + 4 y + 5 = 0 parallel?, (a) 2, (b) − 1, (c) 1, (d) 0, 8. Equation of the straight line making equal intercepts, on the axes and passing through the point (2, 4) is, (a) 4x − y − 4 = 0, (b) 2x + y − 8 = 0, (c) x + y − 6 = 0, (d) x + 2 y − 10 = 0, 9. The angle between the lines 2x − y + 3 = 0 and, x + 2 y + 3 = 0 is, (a) 90°, (b) 60°, (c) 45°, (d) 30°, 10. What is the equation of a line through (0, 1) and making, an angle with the y-axis equal to the inclination of the, line x − y = 4 with x-axis?, (NDA 2012 I), (a) y = x + 1, (b) x = y + 1, (c) 2x = y + 2, (d) None of these, 11. The point moves such that the area of the triangle, formed by it with the points (1, 5) and (3, –7) is, 21 sq units. The locus of the point is, (a) 6x + y − 32 = 0, (b) 6x − y + 32 = 0, (c) x + 6 y − 32 = 0, (d) 6x − y − 32 = 0, 12. If p is the length of the perpendicular drawn from the, x y, origin to the line + = 1, then which one of the, a b, following is correct?, (NDA 2011 II), , 1, 1, 1, =, +, p2 a 2 b2, 1 1 1, (c) = +, p a b, , (a), , 1, 1, 1, =, −, p2 a 2 b2, 1 1 1, (d) = −, p a b, (b), , 13. The value of λ for which the lines 3x + 4 y = 5,, 5x + 4 y = 4 and λx + 4 y = 6 meet at a point is, (a) 2, (b) 1, (c) 4, (d) 3, 14. Distance between the lines 5x + 3 y − 7 = 0 and, 15x + 9 y + 14 = 0 is, 35, 1, 35, 35, (a), (b), (c), (d), 34, 3 34, 3 34, 2 34, 15. If (– 4, 5) is one vertex and 7x − y + 8 = 0 is one, diagonal of a square, then the equation of second, diagonal is, (a) x + 3 y = 21, (b) 2x − 3 y = 7, (c) x + 7 y = 31, (d) 2x + 3 y = 21, 16. What is the equation of the line passing through, (2, − 3) and parallel to Y -axis?, (NDA 2011 I), (a) Y = − 3, (b) Y = 2, (c) X = 2, (d) X = − 3, 17. Two straight lines x − 3 y − 2 = 0 and 2x − 6 y − 6 = 0, (NDA 2011I), , (a), (b), (c), (d), , never intersect, intersect at a single point, intersect at infinite number of points, intersect at more than one points (but finite, number of points), , 18. If the lines 3x + 4 y + 1 = 0, 5x + λy + 3 = 0 and, 2x + y − 1 = 0 are concurrent , then λ is equal to, (a) – 8, (b) 8, (c) 4, (d) – 4, 19. Two consecutive sides of a parallelogram are, 4x + 5 y = 0 and 7x + 2 y = 0. One diagonal of the, parallelogram is 11x + 7 y = 9. If the other diagonal is, ax + by + c = 0, then, (a) a = − 1, b = − 1, c = 2 (b) a = 1, b = − 1, c = 0, (c) a = − 1, b = − 1, c = 0, (d) a = 1, b = 1, c = 1, 20. The length of perpendicular from the, upon, the, straight, ( a cos α , a sin α ), y = x tan α + c, c > 0, is, (a) c, (b) c sin2 α, (c) c cosα, (d) c sec2 α, , point, line, , 21. If the straight lines x − 2 y = 0 and kx + y = 1 intersect, 1, at the point 1, , then what is the value of k?, 2, (NDA 2010 II), (a) 1, , (b) 2, , (c) 1 / 2, , (d) − 1 / 2
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497, , The Straight Line, 22. The image of the point (4, –3) with respect to the line, y = x is, (a) (– 4, – 3) (b) (3, 4), (c) (– 4, 3) (d) (– 3, 4), 23. If the mid-point between the points ( a + b, a − b) and, ( −a , b) lies on the line ax + by = k, what is the value, of k?, (NDA 2012 I), (a) a / b, (b) a + b, (c) ab, (d) a − b, 24. Equation of a line passing through (1, − 2) and, perpendicular to the line 3x − 5 y + 7 = 0 is, (a) 5x + 3 y + 1 = 0, (b) 3x + 5 y + 1 = 0, (c) 5x − 3 y − 1 = 0, (d) 3x − 5 y + 1 = 0, 25. What is the distance between the lines 3x + 4 y = 9, and 6x + 8 y = 18?, (NDA 2012 I), (a) 0, (b) 3 units, (c) 9 units, (d) 18 units, 26. If x cos θ + y sin θ = 2 is perpendicular to the line, x − y = 3, then what is one of the values of θ?, (NDA 2009 I), (a) π / 6, (b) π / 4, (c) π / 2, (d) π / 3, 27. The acute angle which the perpendicular from origin, on the line 7x − 3 y = 4 makes with the x-axis is, (NDA 2012 I), , (a) zero, (c) negative, , (b) positive but not π / 4, (d) π / 4, , 28. The equation of the line passing through ( −1, − 2) and, 4, having a slope of is, 7, 4, 10, (a) 7 y + 10 = 4x, (b) y = x +, 7, 7, 4, 10, (d) 4x + 7 y = 10, (c) x = y +, 7, 7, 29. The perpendicular form of the straight, 3x + 2 y = 7 is, 3, 7, x y, (b) + = 1, (a) y = −, x+, 2, 2, 7 7, 3, 2, 3, 2, (d), (c), x+, y= 7, x+, y=1, 7, 7, 7, 7, , line, , 30. The x-intercept and the y-intercept of the line, 5x − 7 = 6 y , respectively are, 7, 7, 7, 7, (a) and, (b) and−, 5, 6, 5, 6, 5, 6, 5, 6, (d) − and, (c) and, 7, 7, 7, 7, 31. The line through the points (4, 3) and (2, 5) cuts off, intercepts of lengths λ and µ on the axes. Which one, of the following is correct?, (NDA 2009 I), (a) λ > µ, (b) λ < µ, (c) λ > − µ, (d) λ = µ, x y, x y, 32. The lines + = 1 and + = 1, where p ≠ − q are, p q, q p, (a) parallel for all values of p and q, (b) perpendicular for all values of p and q, , pq, pq , (c) not parallel and meet at , ,, , ( p + q ) ( p − q ), pq, pq, (d) not parallel and meet at, ,, ( p + q) ( p + q), 33. The lines 4x + 4 y = 1, 8x − 3 y = 2 and y = 0 are, (a) sides of an isosceles triangle, (b) concurrent, (c) parallel, (d) sides of an equilateral triangle, 34. Which one of the following is the nearest point on the, line 3x − 4 y = 25 from the origin?, (a) ( −1, − 7), (b) ( 3, − 4), (c) ( −5, − 8), (d) ( 3, 4), 35. Which one of the following is the reflection of the, point ( 4, 3) on the line x + y = 0?, (a) ( −4, 3), (b) ( −3, − 4), (c) ( −3, 4), (d) ( 4, − 3), 36. Which one of the following is the orthocentre of the, triangle whose sides are x = 0, y = 1 and x + y − 2 = 0 ?, (a) (1, 1), (b) ( 0, 1), (c) ( −1, 0), (d) ( 0, − 1), 37. What is the gradient of the coplanar and, non-intersecting line which lies midway between the, lines 9x + 6 y − 7 = 0 and 3x + 2 y + 6 = 0 ?, 2, 3, 2, 3, (a), (b) −, (d) −, (c), 3, 2, 3, 2, 38. If p be the length of the perpendicular from the origin, on the straight line x + 2by = 2 p, then what is the, value of b?, (NDA 2007 I), 1, 1, 3, (b) p, (c), (d), (a), p, 2, 2, 39. If the mid-point of the section of a straight line, intercepted between the axes is (1, 1), then what is, the equation of this line?, (a) 2x + y = 3, (b) 2x − y = 1, (c) x − y = 0, (d) x + y = 2, 40. What is the equation of the line which passes, through ( 4, − 5) and is parallel to the line, 3x + 4 y + 5 = 0 ?, (a) 3x − 4 y − 32 = 0, (b) 3x + 4 y + 8 = 0, (c) 4x − 3 y − 31 = 0, (d) 3x + 4 y − 8 = 0, 41. The lines ( p + 2 q ) x + ( p − 3 q ) y = p − q for different, values of p and q pass through the fixed point given, by which one of the following ?, 3 5, 2 2, (a) , , (b) , , 2 2, 5 5, 3 3, 2 3, (c) , , (d) , , 5 5, 5 5, 42. What is the angle between the two straight lines, y = ( 2 − 3 ) x + 5 and y = ( 2 + 3 ) x − 7 ?, (a) 60°, (b) 45°, (c) 30°, (d) 15°
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498, , NDA/NA Mathematics, , Level II, 1. The equation of straight line through the, intersection of the lines x − 2 y = 1 and x + 3 y = 2 and, parallel to 3x + 4 y = 0 is, (a) 3x + 4 y + 5 = 0, (b) 3x + 4 y − 10 = 0, (c) 3x + 4 y − 5 = 0, (d) 3x + 4 y + 6 = 0, 2. If we reduce 3x + 3 y + 7 = 0 to the form, x cos α + y sin α = p, then the value of p is, 7, 7, 3 7, 7, (a), (d), (c), (b), 2, 3, 2 3, 3 2, 3. Equation to the straight line cutting off an intercept, 2 from the negative direction of the axis of y and, inclined at 30° to the positive direction of axis of x, is, (a) y + x − 3 = 0, (b) y − x + 2 = 0, (d) 3 y − x + 2 3 = 0, (c) y − 3x − 2 = 0, 4. The three straight lines ax + by = c, bx + cy = a and, cx + ay = b are collinear, if, (a) b + c = a, (b) c + a = b, (c) a + b + c = 0, (d) a + b = c, 5. What are the coordinates of the foot of the, perpendicular from the point (2, 3) on the line, (NDA 2011 II), x + y − 11 = 0?, (a) (2, 9), (b) (5, 6), (c) (− 5, 6), (d) ( 6, 5), 6. What is the equation of the line joining the origin, with the point of intersection of the lines 4x + 3 y = 12, and 3x + 4 y = 12?, (NDA 2011 I), (a) x + y = 1 (b) x − y = 1 (c) 3 y = 4x (d) x = y, 7. A line AB makes zero intercepts on x-axis and y-axis, and it is perpendicular to another line CD,, 3x + 4 y + 6 = 0. The equation of line AB is, (a) y = 4, (b) 4x − 3 y + 8 = 0, (c) 4x − 3 y = 0, (d) 4x − 3 y + 6 = 0, 8. The equation of the line bisecting perpendicularly, the segment joining the points ( −4, 6) and (8, 8) is, (a) 6x + y − 19 = 0, (b) y = 7, (c) 6x + 2 y − 19 = 0, (d) x + 2 y − 7 = 0, 9. If a tangent to the curve y = 6x − x 2 is parallel to the, line 4x − 2 y − 1 = 0, then the point of tangency on the, curve is, (a) (2, 8), (b) (8, 2), (c) (6, 1), (d) (4, 2), x y, 10. The line − = 1 cuts the x-axis at P. The equation of, a b, the line through P perpendicular to the given line is, (a) x + y = ab, (b) x + y = a + b, (c) ax + by = a 2, (d) bx + ay = b2, 11. What is the equation to the straight line joining the, x y, origin to the point of intersection of the lines + = 1, a b, x y, and + = 1?, (NDA 2010 II), b a, (a) x + y = 0, (b) x + y + 1 = 0, (c) x − y = 0, (d) x + y + 2 = 0, , 12. What is the image of the point (1, 2) on the line, (NDA 2010 I), 3x + 4 y − 1 = 0?, 7 1, 7 6, (b) , , (a) − , − , 8 2, 5 5, 7 1, 7 1, (c) , − , (d) − , , 5 2, 8 2, 13. A straight line through P(1, 2) is such that its, intercept between the axes is bisected at P. Its, equation is, (a) x + y = − 1, (b) x + y = 3, (c) x + 2 y = 5, (d) 2x + y = 4, 14. The equation of the sides of a triangle are x − 3 y = 0,, 4x + 3 y = 5 and 3x + y = 0.The line 3x − 4 y = 0 passes, through, (a) the incentre, (b) the centroid, (c) the orthocentre, (d) the circumcentre, 15. Two lines are drawn through (3, 4), each of which, makes angle of 45° with the line x − y = 2, then area, of the triangle formed by these lines is, (a) 9 sq units, (b) 9/2 sq units, (c) 2 sq units, (d) 2/9 sq units, 16. Equation of a line passing through the line of, intersection of lines 2x − 3 y + 4 = 0, 3x + 4 y − 5 = 0, and perpendicular to 6x − 7 y + 3 = 0, then its, equation is, (a) 119x + 102 y + 125 = 0 (b) 119x + 102 y = 125, (c) 119x − 102 y = 125, (d) None of these, 17. If the lines 3 y + 4x = 1, y = x + 5 and 5 y + bx = 3 are, concurrent, then what is the value of b? (NDA 2010 I), (a) 1, (b) 3, (c) 6, (d) 0, 18. If (− 5, 4) divides the line segment between the, coordinate axes in the ratio 1 : 2, then what is its, equation?, (NDA 2010 I), (a) 8x + 5 y + 20 = 0, (b) 5x + 8 y − 7 = 0, (c) 8x − 5 y + 60 = 0, (d) 5x − 8 y + 57 = 0, 19. The triangle formed by the lines x + y = 0,, 3x + y = 4 and x + 3 y = 4 is, (a) isosceles, (b) equilateral, (c) right angled, (d) None of these, 20. Consider the following statements, I. The straight line 5x + 4 y = 0 passes through the, point of intersection of the straight lines, x + 2 y − 10 = 0 and 2x + y + 5 = 0., II. Only one straight line can be drawn through the, origin at equal distances from the points A ( 2, 2), and B( 4, 0)., III. The product of the perpendiculars from the, points ( 5 , 0), ( − 5 , 0) to the straight line, 2x cos θ − 3 y sin θ = 6 is independent of θ., Which of the statements given above are correct?, (a) I and II, (b) II and III, (c) Both III and I, (d) All I, II and III
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499, , The Straight Line, 21. Consider the following statements, I. The points ( 2 , − 5) and ( − 1, 4) are equidistant, from the line 3x + y + 5 = 0., II. If 2 p is the length of the perpendicular from the, x y, origin to the lines + = 1, then a 2 , 8 p2 , b2are, a b, in HP., Which of the statements given above are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) None of these, 22. What is the product of the perpendiculars from the, ± b2 − a 2 , 0, two, points, to, the, line, , , (NDA 2009 II), ax cos φ + by sin φ = ab?, (b) b2, (c) ab, (d) a / b, (a) a 2, 23. What is the foot of the perpendicular from the point, (NDA 2009 I), ( 2, 3) on the line x + y − 11 = 0?, (a) (1, 10), (b) ( 5, 6), (c) ( 6, 5), (d) ( 7, 4), 24. The perpendicular segment from the origin to a line, is of length 4 units and is inclined to the positive, direction of x-axis at an angle of 30°. The equation of, the line is, 3, (a) 3x + y = 8, (b), x+ y=4, 2, 3, (c) x + 3 y = 8, (d) x +, y=4, 2, 25. The angle between the lines x cos α + y sin α = a and, x sin β − y cos β = a is, (a) β − α, (b) π + β − α, π, π, (c) + α + β, (d) − β + α, 2, 2, 26. The perpendicular distance between two parallel, lines 3x + 4 y − 6 = 0 and 6x + 8 y + 7 = 0 is, 1, 13, 19, 1, (b), (a), unit, units (c), units (d) unit, 5, 5, 10, 2, 27. Consider the following statements, I. The equation to a straight line parallel to the axis, of x is, y = d, where d is a constant., II. The equation to the axis of x is x = 0., Which of the statement(s) given above is/are correct?, (NDA 2009 I), , (a) Only I, (c) Both I and II, , (b) Only II, (d) Neither I nor II, , 28. The coordinates of P and Q are (− 3, 4) and ( 2, 1),, respectively. If PQ is extended to R such that, PR = 2QR, then what are the coordinates of R?, (a) (3, 7 ), (b) ( 2, 4), (NDA 2007 II), 1 5, (c) − , , (d) ( 7, − 2), 2 2, 29. If β is the acute angle between the lines, px + qy = p + q and p ( x − y ) + q( x + y ) = 2q , then the, value of sinβ is, , 3, 2, 1, (c), 2, (a), , (b), (d), , 3, 4, 1, 2, , 30. The line 3x + 4 y − 24 = 0 cuts the x-axis at A and, y-axis at B. The incentre of the ∆ OAB, where O is the, origin, is at, (a) (2, 2), (b) (3, 3), (c) (4, 3), (d) (3, 4), 31. The points ( 3, 4) and ( 2, − 6) lie on opposite sides of, which one of the following lines?, (a) 10x − y = 26, (b) 5x + 3 y = − 9, (c) 3x − 4 y = 8, (d) x − y = 10, 32. What is the acute angle between the lines, Ax + By = A + B and A( x − y ) + B ( x + y ) = 2B ?, (NDA 2007 I), , (a) 45°, , , B, , (c) tan− 1 , A2 + B2 , , , , , , A, , (b) tan− 1 , A2 + B2 , , , (d) 60°, , 33. In what ratio does the line y − x + 2 = 0 cut the line, joining (3,− 1) and (8, 9)?, (NDA 2007 I), (a) 2 : 3, (b) 3 : 2, (c) 3 : − 2, (d) 1 : 2, 34. What, is, the, angle, between, the, lines, x cos α + y sin α = p and x cos β + y sin β = p′ , α > β ?, (a) α + β, (b) α − β, α +β, α −β, (c), (d), 2, 2, 35. What are the coordinates of the centroid of the, triangle whose sides are x = 0, y = 0, x + y = 2?, (a) (0, 0), (b) ( 0, 2), 3 3, (c) , , 2 2, , 2 2, (d) , , 3 3, , 36. What is the locus of a moving point equidistant from, the straight lines x + y = 0 and x − y = 0?, (a) xy = 0, (b) xy = constant, (c) x = 0, (d) y = 0, 37. The intercepts of a straight line upon the coordinate, axes are a and b. If the length of the perpendicular on, this line from the origin be 1, then which one of the, following relations is correct?, 1, 1, 1, 1, 1, (b) 2 + 2 =, (a) 2 + 2 = 2, 2, a, b, a, b, 1, 1 1, 1, 1, (d) 2 + 2 = 1, (c) 2 + 2 =, 2, a, b, a, b, 38. If p be the length of the perpendicular from the origin, 3, , then what, on the straight line ax + by = p and b =, 2, is the angle between the perpendicular and the, positive direction of x-axis?, (NDA 2007 I), (a) 30°, (b) 45°, (c) 60°, (d) 90°
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500, , NDA/NA Mathematics, , 39. The straight line ax + by + c = 0 and the coordinate, axes from an isosceles triangle under which one of, the following conditions?, (NDA 2007 I), (a)|a | = |b |, (b)|a|=|c|, (c)|b|=|c|, (d) None of these, , a1, a2, , b1, b2, , c1, c2 = 0, , a3, , b3, , c3, , 43. Assertion (A) Distance between two parallel lines, 1, 3x + 4 y + 6 = 0 and 6x + 8 y + 7 = 0 is ., 2, Reason (R) Distance between two parallel lines, c −c , 2, ax + by + c1 = 0 and ax + by + c2 = 0 is 1, ., 2, 2, a +b , , 40. What is the locus of the point of intersection of the, straight line x cos θ + y sin θ = a and the straight line, x sin θ − y cos θ = b?, (a) A circle, (b) An ellipse, (c) A hyperbola, (d) A parabola, 41. The bisector of the acute angle between the straight, lines 3x − 4 y − 3 = 0 and 12x + 5 y + 6 = 0 passes, through which one of the following points?, (a) (5, 3), (b) ( −3, 6), (c) (2, 7), (d) (–1, 4), , 44. Assertion (A) Two lines 5x + 3 y + 7 = 0 and, 15x + 9 y + 39 = 0 are parallel., Reason (R) Two lines a1x + b1 y + c1 = 0 and, a2x + b2 y + c2 = 0 are parallel, if a1b2 − a2b1 = 0., 45. Assertion (A) The lines 2x + 3 y + 4 = 0 and, 3x + 2 y + 3 = 0 are perpendicular., Reason (R) The lines ax + by + c = 0 and, Ax + By + C = 0 are perpendicular, if aA + bB = 0, , Directions (Q. Nos. 42-45) Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., , Directions (Q. Nos. 46-48) Suppose there are, two lines given by 3x − 4 y + 9 = 0 and 6x − 8 y − 15 = 0., 46. Find the angle between two lines., (a) 0 °, (b) 45°, (c) 60°, , (d) 90°, , 47. The distance between two lines is, (a) 5, (b) 33/8, (c) 3, , (d) 33/10, , 48. Find the equation of a line passing through (2, − 5), and parallel to the line 6x − 8 y − 15 = 0., (a) 2x − 3 y + 4 = 0, (b) 3x − 4 y − 26 = 0, (c) 3x + 4 y − 25 = 0, (d) x + 2 y + 3 = 0, , 42. Assertion (A), The lines ax + by + c = 0,, bx + cy + a = 0 and cx + ay + b = 0 are concurrent, if, a3 + b3 + c3 = 3abc., Reason, (R) The, lines, a1x + b1 y + c1 = 0,, and, are, a2x + b2 y + c2 = 0, a3 x + b3 y + c3 = 0, concurrent, if, , 49. The equation of line which is perpendicular to, 3x − 4 y + 9 = 0 and passing through (0, 3) is, (a) 4x − 3 y − 9 = 0, (b) 3x − 4 y − 8 = 0, (c) 4x + 3 y − 9 = 0, (d) 3x + 4 y + 8 = 0, , Answers, Level I, 1., 11., 21., 31., 41., , (a), (a), (c), (d), (d), , 2., 12., 22., 32., 42., , (d), (a), (d), (d), (a), , 3., 13., 23., 33., , (b), (b), (c), (b), , 4., 14., 24., 34., , (a), (c), (a), (b), , 5., 15., 25., 35., , (d), (c), (a), (b), , 6., 16., 26., 36., , (c), (c), (b), (b), , 7., 17., 27., 37., , (a), (a), (b), (d), , 8., 18., 28., 38., , (c), (b), (a), (d), , 9., 19., 29., 39., , (a), (b), (c), (d), , 10. (a), 20. (c), 30. (b), 40. (b), , 2., 12., 22., 32., 42., , (d), (a), (a), (a), (a), , 3., 13., 23., 33., 43., , (d), (d), (b), (a), (a), , 4. (c), 14. (c), 24. (a), 34. (b), 44. (a), , 5., 15., 25., 35., 45., , (b), (b), (d), (d), (d), , 6., 16., 26., 36., 46., , (d), (b), (c), (a), (a), , 7., 17., 27., 37., 47., , (c), (c), (a), (d), (d), , 8., 18., 28., 38., 48., , (a), (c), (d), (c), (b), , 9., 19., 29., 39., 49., , (a), (a), (d), (a), (c), , 10. (c), 20. (c), 30. (a), 40. (a), , Level II, 1., 11., 21., 31., 41., , (c), (c), (c), (c), (c)
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Hints & Solutions, Level I, 1. The distance of the point (–2, 3) from the line, x− y=5, −2 − 3 − 5 −10 , , p =, =, 2, 2, 2 , , (, ), (, ), 1, 1, +, −, , , 10, =, =5 2, 2, , 2. The intersection point, y + 2x − 2 = 0 is ( 3,− 4)., , of, , y− x+ 7= 0, , 8. Line making equal intercepts therefore, its equation, is, and, , ∴ Equation of straight line joining from origin to the, −4, point (3, – 4) is y − 0 =, ( x − 0), 3, ⇒, , Then, for the lines x + 2 y − 9 = 0 and kx + 4 y + 5 = 0, 1 2, 9, = =−, k 4, 5, Taking the first two parts, ⇒, k =2, , 3 y = − 4x ⇒ 4x + 3 y = 0, , 3. Q Slope of given line y = x is 1., ∴ Slope of required line which is perpendicular to, given line is –1., , Thus, the equation of required line passing through, (3, 2) and slope –1, is, , x± y=a, Since, it passes through (2, 4), ∴, a = – 2, 6, ∴ Equation of the required lines are, x± y=a, i.e.,, x + y = −2, or, x+ y=6 ⇒ x+ y−6 =0, , 9. We know that, angle between two lines is, m − m2, ., tanθ = 1, 1 + m1m2, The slopes of the given lines are, , y − 2 = − 1(x − 3) ⇒ x + y = 5, , 4. Given equation is compared with a1x + b1 y = 0 and, a2x + b2 y = 0., Now,, a1a 2 + b1b2 = (1) ( 3 ) + (− 3 ) (1) = 0, ∴ Lines are perpendicular., ∴, θ = 90°, , 5. The perpendicular distance of the point ( x1 , y1 ) to line, |ax1 + by1 + c|, ax + by + c = 0 p =, a 2 + b2, The equation of x-axis y = 0, | y|, p=, =| y|, (1)2, 6. Any point on the line joining ( 3,−1) and ( 8, 9), if it is, 8λ + 3 9λ − 1, dividing in the ratio λ : 1 is , ,, and if it, λ +1 λ +1, lies on y − x + 2 = 0, ⇒, , 9λ − 1 8λ + 3, −, + 2 =0, λ+1, λ+1, , ⇒, ⇒, , 9λ − 1 − (8λ + 3) + 2(λ + 1) = 0, 3λ − 2 = 0, 2, λ=, ⇒, 3, ∴ The ratio is 2 : 3., , 7. If two lines a1x + b1 y + c1 = 0, , a, b c, and a 2x + b2y + c2 = 0 are parallel, then 1 = 1 = 1 ., a 2 b2 c2, , …(i), , m1 = 2, m2 = −, Now,, , 1, 2, , 1, m1m2 = − (2) = − 1, 2, , So, the lines are perpendicular., ∴, θ = 90°, , 10. Since, the line passes through the point (0, 1) and, making an angle with y-axis which is equivalent to, the slope of the line y = x − 4., i.e., θ = 45° ⇒ tanθ = 1 = m, ∴ Equation of line is, ( y − 1) = m ( x − 0) = 1 ( x ) ⇒ y = x + 1, 11. Let the locus of the point be ( x , y )., ∴ Area of triangle with points ( x , y ), (1, 5) and ( 3, − 7), is 21 sq units., ∴, ⇒, ⇒, ⇒, , x y 1, 1, 1 5 1 = 21, 2, 3 −7 1, 1, [x (5 + 7) − y (1 − 3) + 1 (−7 − 15)] = 21, 2, 1, [12x + 2 y − 22] = 21, 2, 6x + y − 32 = 0, , 12. Given equation of line is, , x y, + =1, a b, , ...(i), , Then, the length of perpendicular drawn from the, origin to the line (i) = p, (given)
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502, , NDA/NA Mathematics, 1, , 1, a × 0 + b × 0 − 1, =p, , ⇒, , 2, , 18. The given lines are concurrent, if, 3 4, 5 λ, 2 1, , 2, , 1, 1, + , a, b, |− 1|, =p, 1, 1, +, a 2 b2, 1, 1, 1, =, + 2, 2, p, a, b, 1, 1, 1, =, +, p2 a 2 b 2, , ⇒, , ⇒, ⇒, , ⇒, ⇒, ⇒, , 13. Given, lines are 3x + 4 y = 5, 5x + 4 y = 4 and, λx + 4 y = 6. These three lines meet at a point, if the, point of intersection of first two lines lies on the third, line., Now, point of intersection of line 3x + 4 y = 5 and, 1 13, 5x + 4 y = 4 is − , ., 2 8, The line λx + 4 y = 6 passes through the point, 1 13, − , ., 2 8, ∴, , 1, 13, λ − + 4 = 6, 2, 8, , ⇒, ⇒, , − λ + 13 = 12, λ =1, , 14. Equation of lines are 5x + 3 y − 7 = 0, and 15x + 9 y + 14 = 0 or 5x + 3 y +, , Q, , 14, =0, 3, , …(i), …(ii), , Lines (i) and (ii) are parallel and c1 and c2 are of, opposite signs, therefore these lines are on, opposite sides of the origin., , So, the distance between them is, , c, c, 7 14, 2, 1, +, = −, , , 2, + , , 2, 2, 2, 2, 2, 2, 2, +, +, a, b, a, b, +, 5, 3, 2, 2, 1, 1, , , , 3 5 + 3, 7, 14, 35, = −, +, =, 34, 3 34 3 34, , 15. Equation of perpendicular line to 7x − y + 8 = 0 is, x + 7 y = λ which passes through ( −4, 5)., ∴, , λ = 31, , So, equation of another diagonal is x + 7 y = 31., 16. The equation of line passing through (2, − 3) and, parallel to Y -axis is (Y + 3) = tan 90° ( X − 2), ⇒, , X −2 =0 ⇒ X =2, , 17. Here, a1 = 1, a2 = 2, b1 = − 3, b2 = − 6, c1 = − 2, c2 = − 6, Q, ⇒, , a1 1 b1 1 c1 1, = , = , =, a 2 2 b2 2 c2 3, a1 b1 c1, = ≠, a 2 b2 c2, , ∴ Both the straight lines never intersect because both, lines are parallel to each other., , 1, 3 =0, −1, , 3 (− λ − 3) − 4 (−5 − 6) + 1 (5 − 2λ ) = 0, −3λ − 9 + 44 + 5 − 2λ = 0, 5 λ = 40 ⇒ λ = 8, , 19. Since, the coordinates of three vertices A, B and C are, 4, 5, 2 7, , − , ( 0, 0) and − , , respectively. Also, the, 3, 3 3, 3, 1 1, mid-point of AC is , ., 2 2, Therefore, the equation of line passing through, 1 1, , and (0, 0) is given by x − y = 0, which is the, 2 2, required equation of another diagonal., So,, , a = 1, b = − 1 and c = 0, , 20. The given line is x tanα − y + c = 0, or x sin α − y cos α + c cos α = 0, ∴ Length of perpendicular from (a cos α , a sin α ), a cos α sin α − a sin α cos α + c cos α, =, sin 2 α + cos 2 α, c cos α, =, = c cos α, 1, , 21. Since, the straight lines x − 2 y = 0 and kx + y = 1, 1, intersect at the point 1, ., 2, 1, ∴ The point 1, will satisfy the equation kx + y = 1, 2, 1, 1, ∴ k ⋅1 + = 1 ⇒ k =, 2, 2, , 22. Let P ( x1 , y1 ) be the image of point Q( 4, − 3)., x + 4 y1 − 3, Mid-point of PQ is 1, ,, ., 2, 2 , This point lies on y = x, y1 − 3 x1 + 4, ∴, =, ⇒ x1 − y1 = − 7, 2, 2, −3 − y1, and slope of y = x is 1., Slope of PQ =, 4 − x1, Q, ∴, , …(i), , PQ is perpendicular to y = x, −3 − y1 , (1) = − 1, , 4 − x1 , , …(ii), ⇒, y1 + x1 = 1, On solving Eqs. (i) and (ii), we get x1 = − 3, y1 = 4, , 23. Given that, mid-point of the point ( a + b, a − b) and, ( −a , b) lies on the line ax + by = k, ( a + b) + ( − a ) ( a − b) + b, ,, ∴ Mid-point = , , 2, 2, , , b a, = , , 2 2
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503, , The Straight Line, , 28. Equation of line passing through the point ( −1, − 2), 4, and having slope is, 7, , Which lies on the line ax + by = k, b, a, a ⋅ + b ⋅ = k ⇒ k = ab, ⇒, 2, 2, 24. Equation of a line perpendicular to 3x − 5 y + 7 = 0 is, 5x + 3 y + λ = 0, Q This equation passes through (1, –2)., ∴, 5 × 1 + 3 × (−2) + λ = 0, ⇒, 5 −6 + λ =0 ⇒ λ =1, , ⇒, ⇒, , ∴ The required equation is 5x + 3 y + 1 = 0, , 29. We have, equation of line is, , 25. Since, both equation of lines are same that means, both lines are coincident to each other, 3x + 4 y = 9, and, , 6x + 8 y = 18, , ⇒, , 3x + 4 y = 9, , 3x + 2 y = 7, The perpendicular form of Eq. (i) is, 3, 2, 7, or, x+, y=, 7, 7, 7, , …(i), …(ii), , 26. Since, slope of line x cos θ + y sin θ = 2 is − cotθ and, slope of line x − y = 3 is 1., Also, these lines are perpendicular to each other, ∴, (− cot θ ) (1) = − 1, π, ⇒, cot θ = 1 = cot θ, 4, π, θ=, ⇒, 4, , 27. The equation of perpendicular line to 7x − 3 y = 4 is, 3x + 7 y = λ which is passing through the origin., 3× 0+ 7× 0= λ, λ=0, 3x + 7 y = 0, , i.e.,, , y, , 7x – 3y = 4, , 5x 6 y, −, =1 ⇒, 7, 7, , −, , ⇒, , x y, + =1, λ µ, Since, it passes through (4, 3) and (2, 5), 4 3, + =1, ∴, λ µ, 2 3, and, + =1, λ µ, On solving Eqs. (i) and (ii), we get, µ =λ =7, , Here,, , 3, x ⇒ θ = tan−1( −3/ 7 ), 7, α = 180° − θ = 180° − tan−1( −3 / 7), α = 180° + tan−1( 3/ 7), , ⇒, , tan−1( 3 / 7) = α − 180°, 3 / 7 = tan(α − 180° ), , ⇒, ⇒, ⇒, , 3 / 7 = − tan(180° − α ), tan α = 3 / 7, α = tan−1( 3 / 7 ), , ⇒ Which is positive but not, , π, ., 4, , …(i), …(ii), , 32. Given, lines are, , Here,, , y=−, , 7, and, 5, , 31. Let the equation of the line be, , or, and, , y', , x, y, =1, +, 7, 7, −, , , , 6, 5, , 7, , respectively., 6, , x, , α, , 3, 2, x+, y= 7, 7, 7, , The x-intercept and y-intercept of the line are, , θ, x', , …(i), , 30. The intercept form of the line 5x − 7 = 6 y is, , ∴ Distance between two coincident line is zero., , ⇒, , 4, (x − (−1)), 7, 4, y + 2 = (x + 1), 7, 7 y + 10 = 4x, , y − (−2) =, , x y, x y, + = 1 and, + =1, p q, q p, qx + py = pq, px + qy = pq, q, p, m1 = − and m2 = −, p, q, , Since, m1 ≠ m2, therefore the given lines are not, parallel. Their point of intersection is obtained by, eliminating x and y from these two equations which, pq, pq , comes out to be , ,, ., p + q′ p + q, 33. The given lines are, and, Now,, , 4x + 4 y − 1 = 0,, y=0, 4 4 −1, , 8x − 3 y − 2 = 0,, −1, , 4, , −1, , 8 −3 −2 = −4 −2 −3 −2 = 0, 0, 1, 0, 0 1, 0, , (Since, two columns are same), ∴ Lines are concurrent.
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504, , NDA/NA Mathematics, , 34. Let the point A( −1, − 7) and B( 3, − 4) are satisfied the, given equation of the line 3x − 4 y = 25., , y, B(0,2), , Now, distance of A(−1,−7) from the origin O., = (0 + 1)2 + (0 + 7)2 = 50 = 5 2, , P(1,1), , Distance of B(3, − 4) from the origin O., = (0 − 3)2 + (0 + 4)2 = 9 + 16 = 25 = 5, , A(2,0), , ∴ The nearest point is (3, − 4)., , 35. We have,, , O, , x+ y=0, , So, equation of straight line is, x y, + =1, 2 2, , x = 4 ⇒ 4 + y = 0 ⇒ y = −4, y=3 ⇒ x+ 3 =0 ⇒ x= −3, , when, and when, , ∴ Reflection of the point ( 4, 3) on the line x + y = 0 is, ( −3, − 4)., 36. Sides of a triangle are x = 0, y = 1 and x + y − 2 = 0, y, A(0,2), C(1,1), , y, x+, =2, , From figure, B is the orthocentre of ∆ ABC which is( 0, 1)., 37. The, given, lines, are, 9x + 6 y − 7 = 0 and, 3x + 2 y + 6 = 0 and these lines are parallel to each, other, therefore their slope is – 3/2., The required line will also be parallel to these lines,, so the gradient of the required line will be – 3/2., 38. The given equation of straight line is, …(i), x + 2by − 2 p = 0, Length of perpendicular from origin to line (i) = p, , ⇒, ⇒, , 1 + 4b2, 2p, 1 + 4b, , 2, , =p, , = p ⇒ 4 = 1 + 4b2, , 4b2 = 3 ⇒ b =, , 40. Equation of line parallel to line 3x + 4 y + 5 = 0 is, 3x + 4 y + λ = 0, , …(i), , This line passes through ( 4, − 5)., ∴, , 3 (4) + 4 (−5) + λ = 0, , ⇒, , 12 − 20 + λ = 0, , ⇒, , λ =8, , 41. ( p + 2q)x + ( p − 3q ) y = p − q, x, , 0 + 0 −2p, , x+ y=2, , ∴ Required equation of line is 3x + 4 y + 8 = 0., , (0,0)O, , ∴, , ⇒, , On putting the value of λ in Eq. (i)., , y =1, , B (0,1), x=0, , x, , ⇒, , The line always pass through the fixed point, the, coefficient of p and q are zero., ⇒, x + y − 1 = 0 and 2x − 3 y + 1 = 0, On solving, we get, 2, 3, x= , y=, ⇒, 5, 5, 2 3, ∴ Fixed point is , ., 5 5, , 42. The given lines are, y = (2 − 3 ) x + 5, and, y = (2 + 3 ) x − 7, Therefore, m1 = 2 − 3 and m2 = 2 + 3, m − m1 2 + 3 − 2 + 3, =, , ∴ tan θ = 2, 1 + m1m2 1 + (4 − 3) , 2 3, π, = = 3 = tan, 2, 3, , , 3, 2, , 39. Since, P is the mid-point of straight line, then end, points of AB are A( 2, 0) and B( 0, 2)., , p(x + y − 1) + q(2x − 3 y + 1) = 0, , ⇒, , θ=, , π, = 60°, 3, , ...(i), ...(ii)
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505, , The Straight Line, , Level II, 1. The intersection point of lines x − 2 y = 1 and, 7 1, x + 3 y = 2 is , ., 5 5, Since, required line is parallel to 3x + 4 y = 0., 3, Therefore, the slope of required line = −, 4, ∴ Equation of required line which passes through, 3, 7 1, , and slope − , is, 5 5, 4, 1 −3 , 7, =, x − , 5, 4 , 5, 3x, 21 1, + y=, +, 4, 20 5, 3x + 4 y 21 + 4, =, 4, 20, 3x + 4 y = 5, 3x + 4 y − 5 = 0, y−, , ⇒, ⇒, ⇒, ⇒, , ⇒, ⇒, , 3, , 3, , x+, , y+, , 7, , =0, , 1, 1, , sin α =, 2, 2, −, 7, , , , p =, 3 2 , 7, , −7 , =, p =, 3 2 3 2, , y = mx + c, , 1, 3, , [QIt is intercepted in negative axes of y with an angle, of 30° ], ∴ The required line is, y=, , 4. We have,, and, , x, −2 ⇒, 3, , ⇒, y1 − 3 = x1 − 2, …(i), ⇒, x1 − y1 + 1 = 0, Since, the point M lies on the line AB, then, …(ii), x1 + y1 − 11 = 0, On solving Eqs. (i) and (ii), we get, 2x1 − 10 = 0, ⇒, x1 = 5 and y1 = 6, ∴ Required foot of the perpendicular M is (5, 6)., , ⇒, , c= −2, , and, , y1 − 3, x1 − 2, , m1m2 = − 1, y1 − 3, −1 , = −1, x1 − 2 , , and, , 3. Let the equation of line is, m = tan 30° =, , m2 =, , 6. The equation of given lines are, , cos α =, , Q, , B, M, x + y –11=0, (x1, y1), , Now,, PM ⊥ AB, Slope of line AB is, Let, m1 = − 1, and slope of line MP is, , ⇒, , This equation is of the form x cos α + y sin α = p, on, comparing these two equations, we get, , ∴, , A, , ⇒, , 32 + 32, 32 + 32, 32 + 32, 3, 3, −7, x+, y=, 3 2, 3 2, 3 2, 1, 1, −7, x+, y=, 2, 2, 3 2, , and, , P (2,3), , Let, , 2. Given equation is 3x + 3 y + 7 = 0, ⇒, , 5. Let the point of foot of the perpendicular be M ( x1 , y1 )., , 3y − x + 2 3 = 0, , x=, , 12, 7, , 4x + 3 y = 12, 3x + 4 y = 12, and, , y=, , …(i), …(ii), 12, 7, , 12 12, ∴ Point of intersection of the given line is , , ., 7 7, Hence, the equation of line passing through (0, 0) and, 12 12, , is, , 7 7, 12, −0, y −0 = 7, (x − 0), 12, −0, 7, y=x, , ax + by = c, , …(i), , 7. Given, line AB makes 0 intercepts on x-axis and, y-axis or ( x1 , y1 ) ≡ ( 0, 0) and the line is perpendicular, to line CD , 3x + 4 y + 6 = 0., , bx + cy = a, cx + ay = b, , …(ii), …(iii), , ∴ Slope of required line which is perpendicular to, 3x + 4 y + 6 = 0 is 4/3., , On adding Eqs. (i), (ii) and (iii), we get, ax + by + bx + cy + cx + ay = a + b + c, ⇒, (a + b + c)x + (a + b + c) y = (a + b + c), On comparing with 0x + 0 y = 0, (for collinearity), we get,, a + b + c = 0., , ∴ Required line which is passing through origin and, slope 4/3, is, 4, (x − 0), 3, 4x − 3 y = 0, y−0 =, , ⇒
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506, , NDA/NA Mathematics, , 8. Equation of the line passing through( −4, 6) and( 8, 8) is, , ⇒, ⇒, , 8 − 6, y−6 = , (x + 4), 8 + 4, 2, y−6 =, (x + 4), 12, 6 y − 36 = x + 4 ⇒ 6 y − x − 40 = 0, , …(i), , Now, equation of any line perpendicular to the Eq. (i),, is, 6x + y + λ = 0, , …(ii), , This line passes through the mid-point of ( −4, 6) and, −4 + 8 6 + 8, ( 8, 8) is , ,, i.e., (2, 7), 2, 2 , ∴, ⇒, , On putting the value of λ in Eq. (i), we get, x y , x y , + − 1 − 1 + − 1 = 0, a b , b a, , x y, x y, + −1 − − + 1 =0, ⇒, a b, b a, 1 1, 1 1, ⇒, x − − y − =0, a b, a b, ⇒, , x − y =0, , 12. Let (α, β) be the image of point (1, 2) with respect to, line 3x + 4 y − 1 = 0., y, , 6 ×2 + 7 + λ =0, 19 + λ = 0 ⇒ λ = − 19, , P (1, 2), M, , Putting λ = − 19 in Eq. (ii), we get the required line, 6x + y − 19 = 0., 1, 9. Since, tangent is parallel to y = 2x −, 2, ∴ Equation of tangent is, , x′, P′, (α, β), , y = 2x + λ, , y′, , The point of tangency will be the point of, intersections of tangent and curve but in the given, options only option (a) satisfied the equation of curve,, then (2, 8) will be the point of tangency., x y, 10. Line perpendicular to the given line − = 1 is, a b, 1, 1, x+ y+ λ =0, b, a, , …(i), , According to the question, line (i) is passing through, the point P ( a , 0), y, x, y, —– —=1, a, b, P(a, 0), O, , O, , α+1 , β+1, 2, 2, x, , α + 1 β + 2, ∴ , ,, will be on the line, 2, 2 , ⇒, , 3x + 4 y − 1 = 0, α + 1, β + 2, 3, +4, −1 =0, 2 , 2 , , ⇒, ⇒, , 3α + 3 + 4β + 8 − 2 = 0, 3α + 4β + 9 = 0, 7 6, Which is satisfied by − , − ., 5 5, 7 6, Thus, the image of point (1, 2) is − , − ., 5 5, , 13. Q Point P(1, 2) is the mid-point of AB. Therefore,, coordinates of A and B are respectively (2, 0) and (0, 4)., y, , x, , Q (0, –b), (0, 4)B, , ∴, , a, a, + 0 + λ =0 ⇒ λ = −, b, b, , O, , x, , y', , x y a, + − =0, b a b, ax + by = a 2, , ∴, , Equation of line AB is, 4, (x − 2), −2, 2x + y = 4, y−0 =, , 11. We know that the equation of the straight line, passing through the intersection point of two lines, x y, x y, + = 1 and + = 1 is, a b, b a, x y , x y , + − 1 + λ + − 1 = 0, , b a, a b , This line passes through the origin., ∴, (0 − 1) + λ (0 − 1) = 0 ⇒ λ = − 1, , A (2, 0), , x', , On putting the value of λ in Eq. (i), we get, ⇒, , P(1, 2), , ...(i), , ⇒, , 14. Two sides x − 3 y = 0 and 3x + y = 0 are perpendicular, to each other. Therefore, its orthocentre is the point, of intersection of x − 3 y = 0 and 3x + y = 0 i.e., (0, 0)., So, the line 3x − 4 y = 0 passes, orthocentre of triangle., , through, , the
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507, , The Straight Line, 15. The equation of lines are, y − y1 =, Since,, ⇒, ⇒, , m1 ± m2, (x − x1 ), 1 m m1m2, , 4=, , ⇒, , b = 12, y, , m1 = 1, m2 = 1, 1±1, y−4 =, (x − 3), 1m1, y=4, , or, , B, , x=3, , The intersection points of these lines are (6, 4), ( 3, 1), and (3, 4)., 1, ∆ = |6(1 − 4) + 3(4 − 4) + 3(4 − 1)|, 2, 1, = |6(−3) + 3(0) + 3(3)|, 2, 1, 9, = |− 18 + 0 + 9| = sq units, 2, 2, , 16. The point of intersection of lines 2x − 3 y + 4 = 0 and, 2 22, , ., 3x + 4 y − 5 = 0 is −, 34 17 , The slope of required line which is perpendicular to, 6x − 7 y + 3 = 0 is –7/6., ∴Equation of required line which is passing through, 7, 2 22, , and slope − , is, −, 34 17 , 6, 22, 7, 2, = − x +, , 17, 6, 34, 6 (17 y − 22), 7(34x + 2), =−, 17, 34, y−, , ⇒, ⇒, , (–5,4), 1, x′, , and, , 5 y + bx = 3, , 19. The vertices of triangle are the intersection points of, these lines, , and, , …(ii), …(iii), , ⇒ 2b = 12, , 18. Let A( a , 0) and B ( 0, b) be two points on respective, coordinate axes and (− 5, 4) divides AB in the ratio, 1 : 2., , ⇒, , 1 ×0 + 2 × a, 3, − 15, a=, 2, , AC = (2 + 2)2 + (−2 − 2)2, , …(i), , b =6, , −5 =, , AB = (1 + 2)2 + (1 − 2)2, , = 12 + (−3)2 = 10, , If three lines are concurrent, then these values must, satisfy the third equation., , ∴, , …(i), …(ii), …(iii), , = 9 + 1 = 10, BC = (2 − 1)2 + (−2 − 1)2, , y =3, , 15 − 2b = 3, , x+ y=0, 3x + y = 4, x + 3y = 4, , On solving Eqs. (i) and (ii), Eqs. (ii) and (iii), Eqs. (iii), and (i), we get, Now,, , x = −2, , ⇒, , x, , the vertices of ∆ are A ( −2, 2), B (1, 1) and C( 2, − 2), , On solving Eqs. (i) and (ii), we get, and, , O, , A (a,0), , − 15 , Hence, equation of line joining , , 0 and (0, 12) is, 2, , 12 − 0 , 15, ⋅ x + , ( y − 0) =, 15 , 2, 0+, 2, 4, y = (2x + 15), ⇒, 5, ⇒, 5 y = (8x + 60), ⇒, 8x − 5 y + 60 = 0, , 17. The equation of given lines are, y=x+5, , M, , y′, , 119x + 102 y = 125, 3 y + 4x = 1, , (0, b), , 2, , Hence, the lines which makes the triangle are, x − y = 2, x = 3 and y = 4., , ∴, , 1 × b + 2 ×0, 3, , and, , = 16 + 16 = 4 2, Q, AB = BC, ∴Triangle is isosceles., , 20. I. Point of intersection of lines x + 2 y − 10 = 0 and, − 20 25, , ., 2x + y + 5 = 0 is , 3, 3, 5x + 4 y = 0, 25, − 20, i.e.,, 5×, +4×, =0, 3, 3, − 100 100, +, =0, ⇒, 3, 3, ⇒, 0 =0, Line 5x + 4 y = 0 passes through the point of, ∴, intersection of given lines., II. Let equation of line through origin is y = mx, 2 −2m, Distance from point A(2, 2) =, 1 + m2, Now,
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508, , NDA/NA Mathematics, , B(4, 0) =, , and from point, , 0 − 4m, 1 + m2, , − 4m , , =± , , 2, 2, 1+m, 1+m , , 2 − 2m, , Now,, , ⇒, 2 − 2m = ± (− 4m), ⇒, 2 − 2m = − 4m, and, 2 − 2m = − (− 4m), ⇒, m = −1, and, m = 1 /3, ∴ Two lines are possible., III. Line is 2x cos θ − 3 y sin θ = 6 perpendicular from, ( 5 , 0),, |2 5 cos θ − 0 − 6|, P1 =, (2 cos θ )2 + (− 3 sin θ )2, =, =, , On squaring both sides, we get, 1, = 4 P2, 1, 1, +, a 2 b2, 1, 1, 1, =, +, ⇒, 4P 2 a 2 b2, 2, 1, 1, or, = 2 + 2 are in HP., 2, 8P, a, b, Hence, both the statements are true., , 22. Given, ax cos φ + by sin φ − ab = 0, At point ( b2 − a 2 , 0), d1 =, , d2 =, , 6 − 2 5 cos θ, 6 − 2 5 cos θ, , ∴, , 4 + 5 sin 2 θ, , ×, , 6 + 2 5 cos θ, 4 + 5 sin 2 θ, , =, , 36 − 20 cos 2 θ 36 − 20 (1 − sin 2 θ ), =, 4 + 5 sin 2 θ, 4 + 5 sin 2 θ, , =, , 16 + 20 sin 2 θ 4 (4 + 5 sin 2 θ ), =4, =, 4 + 5 sin 2 θ, 4 + 5 sin 2 θ, , 21. I. Line is 3x + y + 5 = 0, Distance from point (2, − 5) =, , 3 ×2 −5 + 5, 32 + 12, , 6, =, 10, and distance from point (−1, 4), 3 × (− 1) + 4 + 5, 6, =, =, 2, 2, 10, 3 +1, ∴ Points are equidistant from the given line., x y, II. Line is + = 1, a b, Now, length of perpendicular from origin, =, 1, 1, 1, +, a 2 b2, , 0 + 0 −1, 1, 1, +, a 2 b2, , =±2P, , [a 2 (b2 − a 2) cos 2 φ − a 2 b2], a 2 cos 2 φ + b2 sin 2 φ, , =−, , a 2 (− b2 sin 2 φ − a 2 cos 2 φ ), a 2 cos 2 φ + b2 sin 2 φ, , = a2, , 4 + 5 sin 2 θ, 6 − 2 5 cos θ, , a 2 cos 2 φ + b2 sin 2 φ, , d1 d2 = −, , 4 + 5 sin 2 θ, , ∴ Product of perpendicular is independent of θ., Hence, Ist and IIIrd statements are true., , ⇒, , − a b2 − a 2 cos φ − ab, , 4 cos 2 θ + 9 sin 2 θ, , |− 2 5 cos θ − 0 − 6|, , Now, P1 × P2 =, , a 2 cos 2 φ + b2 sin 2 φ, , At point (− b2 − a 2 , 0), , and perpendicular from (− 5 , 0) ,, P2 =, , a b2 − a 2 cos φ − ab, , 23. The equation of line perpendicular to the given line, …(i), x + y − 11 = 0, is, …(ii), − x + y + λ =0, This equation passes through (2, 3)., − 2 + 3 + λ = 0, λ = − 1, ∴ From Eq. (ii), we get, − x + y −1 =0 ⇒ y = x + 1, ∴ From Eq. (i), we get, x + x + 1 − 11 = 0, ⇒, 2x = 10, ⇒, x = 5 and y = 5 + 1 = 6, Hence, coordinates of foot of perpendicular from (2, 3), to given line is (5, 6)., , 24. Given that perpendicular segment from ( 0, 0) to any, line x cos α + y sin α = A is of length A units which is, inclined to the positive direction of x-axis at an angle, of 30°., ∴ We have,, …(i), x cos α + y sin α = A, and, α = 30° , A = 4, On putting the value of α and A in Eq. (i), we get, 3x y, + =4, 2, 2, 3x + y = 8, ⇒, Which is the required equation of line., , 25. We have, x cos α + y sin α = a and x sin β − y cos β = a, Above equations can be rewritten as,, y = − x cot α + a cosec α, and, y = x tanβ − a sec β, ∴ Angle between the lines (i) and (ii) is, , …(i), …(ii)
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509, , The Straight Line, , ⇒, ⇒, ⇒, ⇒, , − cot α − tan β , θ = tan −1 , , 1 − cot α tan β , , −1 m1 − m2 , , Q θ = tan , 1 + m1m2 , , cot α cot β + 1 , θ = tan −1 −, , cot β − cot α , θ = tan −1 [− cot (α − β )], , π, , θ = tan −1 tan + (α − β ), 2, , , , π, θ = −β + α, 2, , 26. Given, 3x + 4 y − 6 = 0, and, or, , 6x + 8 y + 7 = 0, 7, 3x + 4 y + = 0, 2, , and, , ∴, , ⇒, , …(i), …(ii), , are two parallel lines., , ∴, , 30. Since, the line 3x + 4 y − 24 = 0 cuts x-axis at A and, y-axis at B., y, , ∴ Perpendicular distance between two parallel lines, is given by, |c1 − c2|, , (0,6) B, , …(iii), , +, , a 2 + b2, , 3x, , d=, , p, Coefficient of x, =−, q, Coefficient of y, p+ q, m2 =, p−q, p p+ q, − −, q p−q, tan β =, p p + q , 1 + − , , q p − q, − p2 + pq − pq − q2, =, pq − q2 − p2 − pq, π, tan β = 1 ⇒ β =, 4, π, 1, sin β = sin, =, 4, 2, m1 =, , 4y, –2, 4, , From Eqs. (i) and (ii), we get, , =, , =, , 0, , 7, 2, On putting all these values in Eq. (iii), we get, , , − 6 − 7 , − 19, , 2 2, , d=, =, 25, 32 + 42, a = 3, b = 4, c1 = −6, , and c2 =, , ∴, , 19, 19, units, =, 2 × 5 10, , Q, , 1, , ( x, y), R, , =, y=, =, , 6 × 8 + 8 × 0 + 10 × 0 48, =, =2, 6 + 8 + 10, 24, ay1 + by2 + cy3, a+ b+ c, 6 × 0 + 8 × 6 + 10 × 0 48, =, =2, 6 + 8 + 10, 24, , ∴ Incentre = (2, 2), , 31. (a) Let L = 10x − y − 26, , ∴ Coordinates of R are (7, − 2)., , 29. Given that β is the acute angle between the lines, px + qy = p + q, p(x − y) + q(x + y) = 2q, , ∴ For incentre of ∆ OAB,, ax + bx2 + cx3, x= 1, a+ b+ c, , and, , Let coordinates of R be (x, y)., Also,, PR = 2QR, ⇒, m : n = 2 :1, By external section formula, 2 ×2 + 1 × 3, ∴, x=, =4 + 3 = 7, 2 −1, 2 ×1 − 1 ×4, and, y=, =2 − 4 = −2, 2 −1, , and, , AB = c = 82 + 62 = 10, OA = b = 8 and OB = a = 6, , 28. Since, the coordinates of P and Q are (− 3, 4) and, (2, 1), respectively., 2, , x, , ∴ Coordinates of A and B are (8, 0) and ( 0, 6),, respectively., , 27. We know that the equation of x-axis is y = 0. Thus,, only statement I is correct., , P, , (8,0) A, , (0,0)O, , …(i), …(ii), , ∴ From Eqs. (i) and (ii), slopes of both equations are, , At point (3, 4), L = 30 − 4 − 26 ⇒ L = 0, and at point (2, − 6), L = 20 + 6 − 26 ⇒ L = 0, (b) Let L = 5x + 3 y + 9, At point (3, 4), ⇒, L = 15 + 12 + 9 = 36, At point (2, − 6), ⇒, L = 10 − 18 + 9 = 1
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510, , NDA/NA Mathematics, (c) Let L = 3x − 4 y − 8, At point (3, 4), ⇒, L, and at point (2, − 6), ⇒, L, (d) Let L = x − y − 10, At point (3, 4), ⇒, L, and at point (2, − 6), ⇒, L, , = 9 − 16 − 8 = − 15, , = 3 − 4 − 10 = − 11, = 2 + 6 − 10 = − 2, , ∴ From the above discussion, we see that the points, ( 3, 4) and ( 2 , − 6) are the opposite sides to the, equation 3x − 4 y = 8., , m2 = −, , …(i), , tan θ =, , ⇒, , tan θ =, , θ =α −β, , 35. The equation of sides, x = 0, y = 0, x + y = 2., , of, , the, , triangle, , are, , x=0, , …(ii), , (0, 2) B, , ( A + B), (B − A ), , y=0, A, , (0,0)O, , Let θ be the angle between both the lines, then, m − m2, Q, tan θ = 1, 1 + m1 m2, ∴, , cos α, cos β, and m2 = −, sin α, sin β, Let θ be the angle between these two lines, then, cos α cos β, −, +, m1 − m2, sin α sin β, tan θ =, =, 1 + m1 m2 1 + cos α ⋅ cos β, sin α sin β, sin α cos β − cos α sin β, ⇒, tan θ =, cos α cos β + sin α sin β, sin (α − β ), ⇒, tan θ =, = tan (α − β ), cos (α − β ), , ∴, , 32. The given equations of lines be, , ⇒, , and, , m1 = −, , = 6 + 24 − 8 = 22, , Ax + By = A + B, and, A (x − y) + B(x − y) = 2B, ⇒, ( A + B)x + (B − A ) y = 2B, Let slope of line (i) is m1, A, m1 = −, ⇒, B, and the slope of the line (ii) is m2, , x cos α + y sin α = p, , 34. The given lines are, x cos β + y sin β = p′, , A A + B, − +, , B B − A, , y, , =, , 2, , From figure, the vertices of ∆OAB are O( 0, 0), A( 2, 0), and B( 0, 2)., So, the coordinates of the centroid are, , A A + B, 1+ , , B B − A, , 0 + 2 + 0 0 + 0 + 2, 2 2, ,, , i.e., , ., , 3 3, , 3, 3, , − A (B − A ) + B ( A + B), B (B − A ) + A ( A + B), , 36. Let P ( x1 , y1 ) be a moving point and, Let, , = tan − 1, , AB + B − AB + A, A 2 + AB + B2 − AB, , = tan −1, , A 2 + B2, = tan −1 (1) = 45°, A 2 + B2, , 2, , x+, , 2, , 33. Let C be the point which divides the join of (3, − 1) and, (8 , 9) in the ratio (λ : 1)., Then, the coordinates of C are, λ × 8 + 1 × 3 9 × λ − 1 × 1 8λ + 3 9λ − 1, ,, ,, , =, , λ +1, λ +1 λ +1 λ +1, , (by internal section formula), This point also lies on, y − x + 2 =0, 9λ − 1 8λ + 3, −, + 2 =0, ∴, λ +1, λ +1, 9λ − 1 − 8λ − 3 + 2λ + 2 = 0, 2, 3λ − 2 = 0, ⇒, λ=, 3, ∴ Required ratio = 2 : 3, , d1 =, , x1 + y1, 1 +1, 2, , 2, , and d2 =, , x1 − y1, 12 + 12, , According to given condition,, d12 = d22, ⇒, (x1 + y1 )2 = (x1 − y1 )2, ⇒, 4x1 y1 = 0 ⇒ x1 y1 = 0, ∴ Locus of a point is xy = 0., , 37. Let the equation of intercept form of line is, , x y, + = 1., a b, , The perpendicular distance from origin to the line, =, , ⇒, , ⇒, , ⇒, , ⇒, , 1, , |−1|, , =1, 2, 1, 1, + , b, a, 1, 1, + 2 =1, 2, a, b, 2, , 2, , 1, 1, + , a, b, , 2, , (given)
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511, , The Straight Line, 38. Since, pis the length of perpendicular from the origin, on the straight line ax + by − p = 0., 0+0− p, , ∴, , a 2 + b2, , =p, , 1 = a 2 + b2, 3, 1 = a2 +, 4, 1, 2 1, a = ⇒a=, 4, 2, , ⇒, ⇒, ⇒, , , 3, , Q b =, 2, , , y, , p, , M, 60º, , x, , A, , O, , − 3x + 4 y + 3 12x + 5 y + 6, =, 5, 13, ⇒, − 39x + 52 y + 39 = 60x + 25 y + 30, ⇒, 99x − 27 y − 9 = 0, Now, point (2, 7), 99 × 2 − 27 × 7 − 9 = 0, 0 =0, ∴ Point (2, 7) satisfy the bisector line., , 42. The lines ax + by + c = 0, bx + cy + a = 0, cx + ay + b = 0 are concurrent, if, , y′, , Thus, equation of straight line is, 1, 3, x+, y= p, 2, 2, ⇒, x cos 60° + y sin 60° = p, Hence, required angle is 60°., Which is the equation of line in normal form, where, 60° be the angle between the perpendicular and the, positive direction of x-axis., , 39. The straight line ax + by + c = 0 and the coordinates, axes form an isosceles triangle., y, , ⇒, ⇒, ⇒, ⇒, , a, , c, , a, , a, , b, , −b, , b a, c, , b, , a, b, , b c, c a =0, , c, , a b, b c, , +c, , distance, , =, x, , ax + by = − c, x, y, (intercept form), ⇒, +, =1, (− c / a ) (− c / b), c, c, If, − =−, a, b, i.e., intercept on x-axis = intercept on y-axis, 1 1, =, ⇒ |a|=|b|, ⇒, a b, , 40. Let the point of intersection be ( x1 , y1 ). Then, this, point lies on both the lines., …(i), ∴, x1 cos θ + y1 sin θ = a, and, …(ii), x1 sin θ − y1 cos θ = b, On squaring and adding Eqs. (i) and (ii), we get, x12(cos 2 θ + sin 2 θ ) + y12(sin 2 θ + cos 2 θ ) = a 2 + b2, ⇒, x12 + y12 = a 2 + b2, ∴ The locus of intersecting point is a circle., , =0, , a (bc − a 2) − b(b2 − ac) + c (ab − c2) = 0, 3abc − a3 − b3 − c3 = 0, 3abc = a3 + b3 + c3, , between, , two, , 3x + 4 y + 6 = 0 and, , A, , c a, , and, , Hence, both A and R are individually true and R is, the correct explanation of A., 43. The, , B, , O, , − 3x + 4 y + 3 = 0 and 12x + 5 y + 6 = 0, a1 = − 3, b1 = 4, c1 = 3 and a 2 = 12, b2 = 5, c2 = 6, Here,, a1a 2 + b1b2 = − 3 × 12 + 4 × 5 = − 16 < 0, ⇒ For acute angle we take positive sign., ∴ Equation of acute angle bisector is, − 3x + 4 y + 3 12x + 5 y + 6, =, (− 3)2 + 42 122 + 52 , or, , ⇒, , B, , x′, , 41. Lines are , 3x − 4 y − 3 = 0 and 12x + 5 y + 6 = 0, , parallel, 3x + 4 y +, , lines, 7, =0, 2, , 7, 2 = 5 =1, 2 ×5 2, 32 + 42, , 6−, , Hence, both A and R are individually true and R is, the correct explanation of A., 44. Lines are 5x + 3 y + 7 = 0 and 15x + 9 y + 39 = 0, ⇒, , a1 = 5, b1 = 3, c1 = 7 and a 2 = 15, b2 = 9, c2 = 39, a1 b1, =, a 2 b2, 5 3, =, ⇒, 15 9, 1 1, ⇒, =, 3 3, ∴ The two lines are parallel., Hence, Both A and R are true and R is the correct, explanation of A., For parallel lines,, , 45. Assertion (A) Given, lines are, 2x + 3 y + 4 = 0 and 3x + 2 y + 3 = 0, If these lines are perpendicular to each other, then, a1a 2 + b1b2 = 0, ⇒, 2 ×3 + 3 ×2 =0
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512, , NDA/NA Mathematics, ⇒, ∴ It is false., , 12 ≠ 0, , =, , Reason (R) If the lines ax + by + c = 0 and, Ax + By + C = 0 are perpendicular, then aA + bB = 0, ∴ It is true., ∴ A is false, but R is true., , 46. 3x − 4 y + 9 = 0, ⇒, and, ⇒, , m1 =, , − ( 3) 3, =, −4 4, , 6x − 8 y − 15 = 0, − (6) 3, m2 =, =, −8 4, , ⇒, m1 = m2, ⇒ Lines are parallel., ∴ Angle between lines is 0°., , 48. 6x − 8 y − 15 = 0, , 3, 4, 3, ⇒ Slope of parallel line (m) =, 4, ∴ Equation of line passing through (2, − 5) and, 3, slope, = is, 4, 3, y − (− 5) = (x − 2), 4, 4 y + 20 = 3x − 6, ⇒, 3x − 4 y − 26 = 0, Slope of line =, , 3, 49. Slope of line 3x − 4 y + 9 = 0 is m = ., 4, , 1 −4, =, 3, 3, 4, −4, ∴ Required equation is, y −3 =, (x − 0), 3, ⇒, 3 y − 9 = − 4x, ⇒, 4x + 3 y − 9 = 0, ⇒ Slope of perpendicular line = −, , 47. Lines are, 3x − 4 y + 9 = 0, 6x − 8 y − 15 = 0, 15, or, 3x − 4 y −, =0, 2, Since, two lines are parallel, , …(i), , and, , ∴ Distance between lines =, , 33 / 2 33, =, 5, 10, , − 15, 9−, , 2 , 32 + (− 4)2, , …(ii)
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26, , The Circle, Y, , A circle is defined as the, locus of all such points in a, plane, which remains at, constant distance from a fixed, point,where the fixed point (C ), and the given distance (CP ) is, called the centre and radius of, the circle, respectively as shown, in given figure., , Example 2. The equation of circle whose centre is (0,0) and, area of the circle is 154 sq units is, (a) 2 x 2 + 2y 2 = 48, (b) x 2 + y 2 = 48, (d) None of these, (c) x 2 + y 2 = 49, , N, P, , C, , M, , X, , ⇒, , Let C( h , k) be the centre of circle and CP ( = r ) be the, radius of circle, then equation of circle is, P (x, y), , …(i), , Now, if origin (0,0) be the centre of circle, then Eq.(i), becomes, x2 + y2 = r 2, , …(ii), , Which is the required equation of circle., The area of the circle is given by π r 2 sq units., , Example 1. The equation of a circle whose centre is (2, − 3), and radius 5 is, (a) x 2 + y 2 − 4x + 6y − 12 = 0, (b) x 2 + y 2 + 4x − 6y − 12 = 0, (c) x 2 + y 2 − 4x + 6y + 12 = 0, (d) None of the above, Solution (a) The equation of the required circle is, ( x − 2) 2 + (y + 3) 2 = 5 2, ⇒, , x + y − 4x + 6y − 12 = 0, , πr 2 = 154, , (Q area of circle = π r 2 ), , r2 = 7 × 7, r =7, , Now, equation of circle is given by, ( x − h) 2+ (y − k) 2 = r 2, x2 + y 2 = (7) 2 ⇒ x2 + y 2 = 49, , General Equation of Circle, , C(h , k), , ( x − h )2 + ( y − k)2 = r 2, , ⇒, , ⇒, , r, , 2, , Area of circle = 154 sq units, ⇒, , Equation of Circle, , 2, , Solution (c) Given that,, , We know that, the general equation of second degree, may represents a circle, if the coefficient of x 2 and, coefficient of y 2 is identical and the coefficient of xy, becomes zero., Let the second degree equation be, ax 2 + by 2 + 2hxy + 2gx + 2 fy + c = 0, , …(i), , It represents a circle, if, (a) a = b i. e., coefficient of x 2 = coefficient of y 2, (b) h = 0 , i. e. , coefficient of xy = 0,, then Eq. (i) reduces as, x 2 + y 2 + 2gx + 2 fy + c = 0,, whose centre and radius are ( − g, − f ) and, g2 + f 2 − c, respectively., , Nature of the Circle, Let the equation of circle is x 2 + y 2 + 2gx + 2 fy + c = 0,, whose centre and radius is ( − g, − f ) and, respectively., , g2 + f 2 − c ,
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514, , NDA/NA Mathematics, , l, , l, , l, , If g2 + f 2 − c > 0,then the radius of circle will be real, and real circle is possible., If g2 + f 2 − c = 0, radius of circle = 0, then the circle, becomes a point circle., If g2 + f 2 − c < 0, then radius of circle will be, imaginary, so in this case, no real circle is possible., , Intercepts Made on the, Axes by a Circle, Case I. When the circle passes through the origin, (0, 0) Let the equation of circle be, …(i), ( x − h )2 + ( y − k)2 = r 2, , Example 3. The given curve 2 x2 + 2y 2 + 14x + 10y − 26 = 0, represents, (a) parabola, (c) ellipse, , y, , (b) circle, (d) hyperbola, , Solution (b) The equation of given curve is, , 2x2 + 2y 2 + 14x + 10y − 26 = 0, (i) Here, coefficient of x2 = coefficient of y 2, (ii) Coefficient of xy = 0, ∴ The given curve represents a circle., , C (h , k), r, , the three points (0,1), (2,0) and (5,6) is, (a) x 2 + y 2 − 5 x − 7y + 6 = 0, (b) x 2 + y 2 − 5 x − 7y − 6 = 0, (c) x 2 + y 2 + 5 x − 7y + 6 = 0, (d) None of the above, , 1 + 2f + c =0, −, −, −, 3 + 4g − 2f = 0, On subtracting Eq. (iii) from Eq. (iv), we get, 61 + 10 g + 12f + c = 0, 4 + 4g, + c =0, – –, –, 57 +6g + 12f, = 0, , M, , x, , Q It passes through origin (0,0), ∴, h 2 + k2 = r 2, ∴ Eq. (i) becomes,, ( x − h )2 + ( y − k)2 = h 2 + k2, or, , Solution (a) The general equation of circle be, Since, the given points satisfy this equation, we have, 1 + 2f + c = 0 ,, 4 + 4g + c = 0, and, 61 + 10 g + 12f + c = 0, On subtracting Eq. (ii) from Eq. (iii), we get, 4 + 4g + c = 0, , h, , O(0, 0), , Example 4. The equation of circle, which passes through, , x2 + y 2 + 2gx + 2fy + c = 0, , k, , …(i), …(ii), …(iii), …(iv), , x 2 + h 2 − 2hx + y 2 − 2ky + k2 = h 2 + k2, x 2 + y 2 − 2hx − 2ky = 0, , or, , Example 5. The equation of the circle which passes through, the origin and cuts off intercepts 3 and 4 from the positive, parts of the axes, respectively is, (a) 4x 2 + 12 xy + 4y 2 − 16y = 0, (b) 4x 2 − 12 x + 4y 2 − 16y = 0, (c) 4x 2 − 12 x + 4y 2 + 16y = 0, , …(v), , (d) None of the above, , Solution (b) We have,, OA = 3, OB = 4, 3, OL =, and CL = 2, 2, , ∴, …(vi), , In ∆OCL, we have, , On multiplying Eq. (v) by 6 and adding Eqs. (v) and (vi), we get, 57 + 6g + 12f = 0, 18 + 24g − 12f = 0, 75 + 30 g, = 0, , y, , − 75, 5, g =, =−, ⇒, 30, 2, 7, and, f =–, 2, ∴ From Eq. (iv), we get, −7 , −5 , 61 + 10 + 12 + c = 0, 2, 2, , 4, , ⇒, 61 − 25 − 42 + c = 0 ⇒ c = 6, Put all these values in Eq. (i), we get required equation of circle, which is x2 + y 2 − 5x − 7y + 6 = 0, , C (3/2,2), 2, 3/2, , O, , L, , A, , OC 2 = OL2 + LC 2, 2, , ⇒, ⇒, , 9, 25, 3, OC 2 = + 2 2 = + 4 =, 2, 4, 4, 5, OC =, 2, , x
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515, , The Circle, 3 , The required equation of circle whose centre is , 2 and radius, 2 , 5, is is, 2, 2, 2, 3, 5, , 2, x − + (y − 2) = , 2, , 2, 9, 25, 2, 2, ⇒, x − 3x + + y − 4y + 4 =, 4, 4, ⇒, 4x2 − 12x + 4y 2 − 16y = 0, , Case II. When the circle touches abscissa (x-axis), If the centre of circle be C( h , k)and it touches x-axis and the, radius of circle is CP = k, y, , h, P, , C (h , k), , x, , O, , ∴ Equation of circle is, ( x − h )2+ ( y − k)2 = (CP )2 = h 2, x 2 + y 2 − 2ky − 2hx + k2 = 0, , or, , Example 7. The equation of circle whose centre is ( −3, 5), and touches the y-axis is, (a) x 2 + y 2 + 16 x + 10y + 25 = 0, (b) x 2 + y 2 + 6 x − 10y + 25 = 0, (c) 2 x 2 + 2y 2 + 6 x − 10y + 25 = 0, (d) None of the above, , C(h , k), , O, , y, , x, , P, , Solution (b) Here, radius CM = 3, y, , ∴ Equation of circle is ( x − h )2+ ( y − k)2 = (CP )2 = k2, ⇒, , x 2 + y 2 − 2hx − 2ky + h 2 = 0, , Example 6. The equation of the circle whose centre is ( 4, 2), and touches the x-axis is, (a) x 2 + y 2 − 8 x − 4y + 16 = 0, (b) x 2 + y 2 − 8 x + 4y + 16 = 0, (c) x 2 + y 2 − 8 x − 4y − 16 = 0, (d) None of the above, Solution (a) Since, the circle touches x-axis, therefore its radius, , CM = 2., , C(–3,5), x', , M, , x, , O, y', , ∴ Required equation of circle is, ( x + 3) 2 + (y − 5) 2 = (3) 2, ⇒, , x2 + y 2 + 6x − 10y + 9 + 25 = 9, , ⇒, , x2 + y 2 + 6x − 10y + 25 = 0, , Case IV. When the circle touches both abscissa, and ordinate (x-axis and y-axis) In this case h = k. Then,, the equation of circle is ( x − h )2 + ( y − k)2 = r 2, where, radius h = k = α, (say), , y, , y, , C (4,2), 2, O, , M, , x, , ∴ Required equation of circle is, ( x − 4) 2 + (y − 2) 2 = 2 2, ⇒, , x + y − 8x − 4y + 16 + 4 = 4, , ⇒, , x + y − 8x − 4y + 16 = 0, , 2, , M, , h, , C (h, k), k, , 2, , 2, , O, , 2, , Case III. When the circle touches ordinate, (y-axis) Let the centre of circle be C( h , k) and it touches, y-axis at point P with radius CP = h., , P, , ∴, , ( x − α )2 + ( y − α )2 = α 2, , or, , x 2 + y 2 − 2 αx − 2 αy + α 2 = 0, , x
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516, , NDA/NA Mathematics, y, , Example 8. The equation of a circle of radius 6 units, touches the coordinate axes in the second quadrant is, (a) 2 x 2 + 2y 2 + 12 x − 12y + 36 = 0, (b) x 2 + y 2 + 12 x − 12y − 40 = 0, (c) x 2 + y 2 − 12 x − 12y + 37 = 0, (d) x 2 + y 2 + 12 x − 12y + 36 = 0, , C (0,k), x', , x, , (0, 0) O, , Solution (d) ∴ Radius = 6 units and centre is ( −6, 6), y', , y, , (–6, 6)C, , 6, , 6, x, , x', O, y', , ∴ Required equation of circle is, ( x + 6) 2 + (y − 6) 2 = 6 2, 2, 2, ⇒, x + y + 12x − 12y + 36 + 36 = 36, ⇒, x2 + y 2 + 12x − 12y + 36 = 0, , Case V. When the circle passes through O(0,0) and, centre lies on abscissa (x-axis) In this case k = 0, then, equation of circle is, y, , x, , (0, 0) O, , C (h, 0), , x', , Example 10. The equation of circle passes through origin, and whose radius is 4 units and their centre lies on ordinate is, (a) x 2 + y 2 − 8y = 0, (b) x 2 + y 2 + 8y = 0, (c) x 2 + y 2 − 8y − 8 x = 0, (d) None of the above, Solution (a) ∴Centre is (0 , 4) and radius is 4., ∴The required equation of circle is, ( x − 0) 2 + (y − 4) 2 = 4 2, ⇒, ⇒, , x2 + y 2 − 8y + 16 = 16, x2 + y 2 − 8y = 0, , Equation of Circle in Diameter, Form, Let A ( x1 , y1 ) and B ( x2 , y2 ) be the end points of a, diameter of the given circle and let P ( x , y ) be any point on, the circle., P(, , 90, , y', , whose centre lies on the x-axis of radius 3 units is, (a) x 2 + y 2 + 6 x = 0, (b) x 2 + y 2 − 6 x = 0, 2, 2, (d) None of the above, (c) x + y − 6 x + 6y = 0, , ∴ From figure ∠ APB = 90°, ∴, and, , Solution (b) The centre of circle is (3, 0) and radius is 3., ∴ Equation of circle is, ( x − 3) 2 + (y − 0) 2 = 3 2, ⇒, , x2 + y 2 − 6x + 9 = 9 ⇒ x2 + y 2 − 6x = 0, , Case VI. When the circle passes through O(0,0), and centre lies on ordinate (y-axis), In this case h = 0, then equation of circle is, ( x − 0)2 + ( y − k)2 = k2, ⇒, , x 2 + y 2 − 2ky = 0, , B(x 2 ,y2 ), , (x1, y1) A, , x 2 + y 2 − 2hx = 0, , Example 9. The equation of circle passes through origin,, , °, , C(h , k ), , ( x − h )2 + ( y − 0)2 = h 2, ⇒, , y), x,, , y − y1 , Slope of AP = , , x − x1 , y − y2 , slope of BP = , , x − x2 , , ∴ For perpendicular, m1 m2 = − 1, ⇒, , AP ⋅ BP = − 1, , ⇒, , y − y1 y − y2 , , ⋅, = −1, x − x1 x − x2 , , ⇒ ( x − x1 ) ( x − x2 ) + ( y − y1 ) ( y − y2 ) = 0, Which is the required equation of circle in diameter, form.
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517, , The Circle, , Example 11. The equation of the circle, the coordinates of, the end points of whose diameter are (2, 4) and (3, 5) is, (a) x 2 + y 2 − 5 x − 9y + 26 = 0, (b) x 2 + y 2 − 5 x − 9y − 26 = 0, (c) x 2 + y 2 + 5 x + 9y − 26 = 0, (d) None of the above, , On resolving the component, we get, … (i), x = OM = r cosθ, and, … (ii), y = PM = r sinθ, Here, Eqs. (i) and (ii) are the required parametric form, of the circle x 2 + y 2 = r 2, where ' θ' is a parameter., Case II. Parametric form of equation of circle, if ( h , k), is the centre and r being the radius is, , Solution (a) The equation of the circle described on the line, , y, , segment joining ( x1, y1) and ( x2, y 2) as diameter is, , P, , ( x − x1) ( x − x2) + (y − y1) (y − y 2) = 0, and it is given that x1 = 2 , x2 = 3, y1 = 4, y 2 = 5, So, the equation of required circle is, ( x − 2) ( x − 3) + (y − 4) (y − 5) = 0, ⇒, x2 − 5x + 6 + y 2 − 9y + 20 = 0, ⇒, x2 + y 2 − 5x − 9y + 26 = 0, , r, , (h, k), C, , x', , Example 12. If them y = 2 x is a chord of the circle, x 2 + y 2 − 10 x = 0, then the equation of a circle with this chord, as diameter is, (b) x 2 + y 2 − 2 x − 4y = 0, (a) x 2 + y 2 + 2 x − 4y = 0, (c) x 2 + y 2 + 2 x + 4y = 0, (d) None of these, , Solution (b) Given, equation of chord is y = 2x, and equation of circle is x2 + y 2 − 10 x = 0, , …(i), …(ii), , From Eqs. (i) and (ii), we get, x2 + 4x2 − 10 x = 0, , θ, , N, , x, O, , x = h + r cos θ ; y = k + r sin θ, where, θ being the parameter., , Example 13. The parametric equations of the circle, x 2 + y 2 − 2 x − 4y − 4 = 0 is, , (a) x = 1+ 3 cos θ, y = 2 + 3 sin θ, (b) x = 1− 3 cos θ, y = 2 + 3 sin θ, (c) x = 1+ 3 cos θ, y = 2 − 3 sin θ, , ⇒, 5x ( x − 2) = 0 ⇒ x = 0 , x = 2, at, x = 0,y = 0, and at, x = 2, y = 4, These are the coordinates of end points of the chord. Since,, the chord is a diameter, therefore these points are the end, points of a diameter., ∴ Equation of circle is, ( x − 0) ( x − 2) + (y − 0) (y − 4) = 0, ⇒, x2 + y 2 − 2x − 4y = 0, , (d) None of the above, , Solution (a) The equation of circle is, x2 + y 2 − 2x − 4y − 4 = 0, ⇒, , ( x2 − 2x + 1) + (y 2 − 4y + 4) = 9, , ⇒, , ( x − 1) 2 + (y − 2) 2 = 3 2, , ∴ The parametric equations of the circle are x = 1 + 3 cos θ,, y = 2 + 3 sin θ, where θ is the angle lying between 0 to 2π., , Example 14. The parametric equations of the circles, x 2 + y 2 + mx + my = 0 is, , Equation of Circle in, Parametric Form, , m m, m m, − cos θ, y = − +, cos θ, 2 2, 2, 2, m m, m m, (b) x = − + cos θ, y = − +, sin θ, 2 2, 2, 2, , (a) x = −, , Case I. Let P ( x , y ) be any point on the circle, x 2 + y 2 = r 2, then from figure ∠ MOP = θ, , (c) x =, , y, , m m, m m, +, cos θ, y = − +, sin θ, 2, 2, 2, 2, , (d) None of the above, P(x, y), r, x', , θ, M, , O, , Solution (b) Here,, x, , ⇒, ⇒, , x2 + y 2 + mx + my = 0, ( x + mx) + (y 2 + my) = 0, 2, , 2, m2 2, m2 m2, x + mx +, + y + my +, =, 4 , 4, 2, , 2, , ⇒, y', , 2, , m , m m , x − − 2 + y − − 2 = 2 , , , , , 2
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518, , NDA/NA Mathematics, , So, the parametric equations of the circle are, m m, x=−, +, cos θ, 2, 2, m m, and, y=−, +, sin θ, 2, 2, , Position of a Point with, Respect to a Circle, , Concentric Circles, Two circles having the same centre C(h , k) but different, radii r1 and r2, respectively are called concentric circles., and, Thus,, the, circles, ( x − h )2 + ( y − k)2 = r12, 2, 2, 2, ( x − h ) + ( y − k) = r2 , r1 ≠ r2 are concentric circles., Therefore, the equation of concentric circles differ only in, constant term., , Let C( h , k)be the centre and r be the radius of the circle, and P ( a , b)be any point in the plane of the circle, then three, cases arises, Case I. Let point P lies outside the circle, then, equation of circle is ( a − h )2 + ( b − k)2 > r 2., P(a, b), Q, , r, C (h, k), , Case II. Let point ‘ P ’ lies on the circle, then equation, of circle is ( a − h )2 + ( b − k)2 = r 2., , C (h,k), r, r2 1, , P (a,b), , r, C (h,k), , Example 15. The equation of circle passing through (1, 2), and, which, is, concentric, with, the, circle, x 2 + y 2 + 11x − 5y + 3 = 0 is, (a) x 2 + y 2 + 11x + 5y + 6 = 0 (b) x 2 + y 2 + 11x − 5y − 6 = 0, (c) x 2 + y 2 + 11x − 6y + 7 = 0 (d) None of these, , Case III. Let point P lies inside the circle, then, equation of circle is ( a − h )2 + ( b − k)2 < r 2., r, C(h,k), , Solution (b) We have,, x2 + y 2 + 11x − 5y + 3 = 0, , Since, it passes through (1, 2)., ∴, 1 + 4 + 11 − 10 + λ = 0, ⇒, 6 + λ =0, ⇒, λ = −6, ∴ Required equation of circle is, x2 + y 2 + 11x − 5y − 6 = 0, , Cyclic Quadrilateral, If all four vertices of a quadrilateral lie on a circle is, called a cyclic quadrilateral as shown in figure., S, , P, , Q, , These four vertices are said to be concyclic., Here, ∠S + ∠Q = 180° and ∠P + ∠R = 180°, , b, (a,, , ), , …(i), , The equation of a circle, which is concentric with the circle (i), is, x2 + y 2 + 11x − 5y + λ = 0, , R, , P, , Example 16. The position of a point (3, 4) with respect to, the circle x 2 + y 2 = 36, is, (a) inside, (b) on, (c) outside, (d) None of the above, , Solution (a) Let S ≡ x2 + y 2 − 36 = 0, At point (3, 4), S = 9 + 16 − 36 = − 9 < 0, ∴ Point lies inside the circle., , Maximum and Minimum, Distance of a Point from a, Circle, Let any point P ( x1 , y1 ) inside or outside the circle, x 2 + y 2 + 2gx + 2 fy + c = 0., The centre and radius of the circle are ( − g, − f ) and, g2 + f 2 − c, respectively.
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519, , The Circle, Case I. Roots of Eq. (i) are real and distinct, if, c, ., D = r 2(1 + m 2 ) − c2 > 0, i.e., if r >, 1 + m2, , P (x1,y1), A, P, C (–g –f), , Hence, the line meets the circle at two distinct points., Case II. Roots of Eq. (i) are coincident, if, c, ., D = r 2(1 + m 2 ) − c2 = 0 i.e., if r =, 1 + m2, , B, , The maximum and minimum distance from P ( x1 , y1 ) to, the circle are PB = CB + PC = r + PC, and, PA =|CP − CA|=|PC − r|, , Hence, the line touches the circle., Case III. Roots of Eq. (i) are imaginary, if, , Example 17. The greatest distance of the point P(10, 7) from, , D = r 2(1 + m 2 ) − c2 < 0, i.e., if r <, , the circle x + y − 4x − 2y − 20 = 0 is, (a) 10, (b) 15, (c) 5, (d) None of these, 2, , 2, , c, 1 + m2, , ., , Hence, the line will not intersect the circle at all., %, , Length of perpendicular from (0,0) on y = mx + c is, , Solution (b) At point P(10 , 7), , c, 1 + m2, , ., , S1 = 10 2 + 7 2 − 4 × 10 − 2 × 7 − 20, , Condition of Tangency of a, Line to a Circle, , 100 + 49 − 40 − 14 − 20 = 75 > 0, So, point lies outside the circle., Here, centre of circle C(2,1., ), Radius,, r = 4 + 1 + 20 = 5, CP = (10 − 2) + (7 − 1), 2, , 1. The line y = mx + c touches the circle x 2 + y 2 = r 2,, , 2, , iff c = ± r 1 + m 2 ., 2. The line lx + my + n = 0 will touch the circle, x 2 + y 2 + 2gx + 2 fy + c = 0, iff, , = 64 + 36 = 10, ∴ Greatest distance = r + CP = 5 + 10 = 15, , ( l 2 + m 2 ) ( g2 + f 2 − c) = ( − lg − mf + n )2., , Chord, , %, , Chord if A and B are any two points on a curve, then, the segment AB is called a chord of the curve., , x 2 + y2 = r 2., , Example 18. The range of values of n for which the line, y = mx + 2 cuts the circle x 2 + y 2 = 1 at distinct on coincident, points is, (a) [ − ∞, − 3] ∪ [ 3, ∞], (b) [ − 3, 3], (d) None of these, (c) [ 3, ∞], , B, O, , A, , Solution (a) The length of the perpendicular from the centre, (0 , 0 ) to the line =, , Secant, The chord AB produced on either or both sides is, known as a secant., , Intersection of a Line and a, Circle, , (1 + m 2 ) x 2 + 2mcx + ( c2 − r 2 ) = 0, , Which is quadratic in x, then three cases arises., , …(i), , 2, , 1 + m2, , The radius of the circle =1, For the line to cut the circle at distance or coincident points,, 2, ≤ 1 or 1 + m2 ≥ 4 or m2 ≥ 3, 1 + m2, ⇒, , Let y = mx + c and x 2 + y 2 = r 2 be the equation of a line, and a circle, respectively., ∴ On solving both the equations, we get, x 2 + ( mx + c)2 = r 2, (Q y = mx + c), ⇒, , The line y = mx ± r 1 + m 2 is always a tangent to the circle, , m ≤ − 3 and m ≥ 3, , Length of the Tangent, Suppose P is any point outside the circle, then the, length of tangent drawn from the point ( x1 , y1 ) to a circle, x 2 + y 2 + 2gx + 2 fy + c = 0 is given by, x12 + y12 + 2gx1 + 2 fy1 + c .
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520, , NDA/NA Mathematics, , Solution (a) In a right angle ∆ C1PC 2., , T, , P, a, , P (x1,y1), , C, , a, , C1, (2,3), , C2, (5,6), , (C1C 2) 2 = (PC1) 2 + (PC 2) 2, , Example 19. The length of tangent drawn from the point, , ( −14, , ) to the circle 2 x + 2y = 9 is, (b) 5 / 2, (a) 5 / 2, (d) 5 / 3 2, (c) 2 / 5, 2, , 2, , Solution (a) The given equation of circle can be written as, 9, x + y − =0, 2, ∴ Required length of tangent, 2, , 2, , = x12 + y12 − r 2, = ( −1) 2 + 4 2 −, , 9, 2, , 5, 9, units, = 17 − =, 2, 2, , Angle of Intersection of Two, Circles, The angle of intersection of two intersecting curves is, the angle between their tangents at the point of, intersection., R, , ⇒, , (2 − 5) 2 + (3 − 6) 2 = a2 + a2, , ⇒, , 9 + 9 = 2 a2 ⇒ a2 = 9, a =3, , (Q radius cannot be negative), , Family of Circles, We have general equation of a circle is, x 2 + y 2 + 2gx + 2 fy + c = 0, then, 1. Family of circles passing through the points of, intersection of line P ≡ lx + my + n = 0 and circle, S ≡ x 2 + y 2 + 2gx + 2 fy + c = 0 is, S + λP = 0, 2. Family of circles passing through the points of, intersection of two given circles, S1 ≡ x 2 + y 2 + 2g1 x + 2 f1 y + c1 = 0, and, , S1 ≡ x 2 + y 2 + 2g2 x + 2 f2 y + c = 0, , are, S1 + λ S 2 = 0, 3. Equation of family of circles passing through two, points ( x1 , y1 ) and ( x2 , y2 ) is, x y 1, ( x − x1 ) ( x − x2 ) + ( y − y1 ) ( y − y2 ) + λ x1 y1 1 = 0 ., x2, , P, T, , Since,, ∴, , y2 1, , Q, , θ, , T', , RQ ⊥ RT and RP ⊥ RT ′, ∠ TRT ′ = π −∠ QRP = π − θ, , Orthogonal Circles, When two circles intersect at right angles known as, orthogonal circles., , Condition of Orthogonality, Let S1 = 0 and S 2 = 0 be any two circles, then, 2 ( g1g2 + f1 f2 ) = c1 + c2., , Example 20. If the circles of same radius a and centres at, (2, 3) and (5, 6) cut orthogonally, then a is, (a) 3, (b) 4, (c) 6, (d) 10, , Example 21. The equation of the circle through the, intersection of the circles x 2 + y 2 − 8 x − 2y + 7 = 0 and, x 2 + y 2 − 4x + 10y + 8 = 0 and that passes through the point, ( −1, − 2) is, (a) 9 x 2 + 9y 2 − 40 x + 78y + 71= 0, (b) x 2 + y 2 − 40 x + 78y + 71= 0, (c) 9 x 2 + 9y 2 − 40 x − 78y + 71= 0, (d) None of the above, Solution (a) The equation of any circle is, x2 + y 2 − 8x − 2y + 7 + λ( x2 + y 2 − 4x + 10y + 8) = 0, ⇒ x2(1 + λ ) + y 2 (1 + λ ) − 4x (2 + λ ), −2y (1 − 5λ ) + 7 + 8λ = 0, This circle passes through the point ( −1, − 2). Therefore,, (1 + λ ) + 4 (1 + λ ) + 4(2 + λ ) + 4(1 − 5λ ) + 7 + 8λ = 0, ⇒, , 24 − 3λ = 0, , ⇒ λ =8, , On putting λ = 8 in Eq. (i), we get the required circle as, 9x2 + 9y 2 − 40 x + 78y + 71 = 0, , …(i)
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521, , The Circle, , Condition of Two Circles, Touch Each Other, , Case IV. If the circles touch each other externally at a, point A, then C1 C2 = r1 + r2, Hence, only three common tangents are drawn., A, , Case I. If one circle lies inside the other, then, C1C2 ≥|r1 − r2|, , r1, , r2, , C1, , r1, , C2, , r2, C1 C 2, , Hence, no common tangent can be drawn., Case II. If the two circles touch each other internally, at a point A, then C1 C2 =|r1 − r2|, Hence, only one common tangent is drawn., , (i.e., Distance between their centres C1and C2 is equal, to the sum of their radii). Where, A is the point of contact, (touching)., Case V. If two circles lies outside the other, then, C1C2 >|r1 + r2|, C1, , r1, , C2, , r2, , Hence, there are four common tangents can be drawn., r1 r2, C1 C2, , A, , (i.e., Distance between their centres c1and c2is equal to, the mode of difference of their radii r1 and r2)., Case III. If two circles intersect each other, then, |r1 − r2|< C1C2 < r1 + r2, r1, , r2, , C1, , C2, , Example 22. The circles x2 + y 2 + 2 x + 2y + 1 = 0 and, x 2 + y 2 − 4x − 6y − 3 = 0 touch each other are, (a) internally, (b) externally, (c) no where, (d) None of these, , Solution (b) The equation of the given circles are, and, , x2 + y 2 + 2x + 2y + 1 = 0, , …(i), , x + y − 4x − 6y + 3 = 0, , …(ii), , 2, , 2, , The coordinates of the centres of Eqs. (i) and (ii), are C1( −1, − 1), and C 2(2, 3), respectively., Let r1 and r2 be the radii of Eqs. (i) and (ii), we get, Then, r1 = 1 + 1 − 1 = 1 and r2 = 4 + 9 + 3 = 4, Now,, , C1C 2 = (2 + 1) 2 + (3 + 1) 2 = 5, , ⇒, C1C 2 = r1 + r2 = 5, So, the circle touch each other externally., , Hence, two common tangents can be drawn., , Comprehensive Approach, n, , n, , ax2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 represents a circle, if, a = b , i. e., coefficient of x2 = coefficient of y 2 and coefficient of, xy = 0., The centre of a circle is mid-point of the diameter and radius is half, of the length of diameter., , Intercepts on Axes, , n, , n, , The length of intercept made by circle x + y + 2 gx + 2 fy + c = 0, with x-axis is 2 g 2 − c , ( g 2 > c) and with y-axis is, 2, , n, , n, , n, , 2, , 2 f 2 − c , ( f 2 > c)., Point ( x1 , y1) lies outside, on or inside a circle S = x2 + y 2 − a 2 = 0, according as S1 > = < 0., , n, , n, , n, , Number of Tangents, n, , n, n, , In general two real tangents can be drawn from a point outside a, circle., The point θ on the circle x2 + y 2 = a 2 is ( a cos θ, a sin θ)., A line touches a circle, if the length of the perpendicular from the, centre is equal to the radius of the circle., , n, , n, , A line intersects a given circle at two distinct real points, if the, length of the perpendicular from the centre is less than the radius of, the circle., A line does not intersect a circle, if the length of the perpendicular, from the centre is greater than the radius of the circle., If circle x2 + y 2 + 2 gx + 2 fy + c = 0 touches the x-axis, then, g 2 = c., If circle x2 + y 2 + 2 gx + 2 fy + c = 0 touches the y-axis, then, f 2 = c., If the circle x2 + y 2 + 2 gx + 2 fy + c = 0 touches both the axes,, then g 2 = f 2., A line intersects a given circle at two distinct points, if the length of, the perpendicular from the centre is less than the radius of the, circle., A line touches a circle, if the length of the perpendicular from the, centre is equal to the radius of the circle., The diameter corresponding to a system of parallel chords of a, circle always passes through the centre of the circle and is, perpendicular to the parallel chords.
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Exercise, Level I, 1. The coordinates of the centre and the radius of the, circle x 2 + y 2 + 4x − 6 y − 36 = 0 are respectively,, given by, (a) ( −4, 6) and 6, (b) ( 4, − 6) and 7, (c) ( 2, − 3) and 6, (d) ( −2, 3) and 7, , 10. Which of the following is a point on the common, chord of the circles x 2 + y 2 + 2x − 3 y + 6 = 0 and, x 2 + y 2 + x − 8 y − 13 = 0?, (a) (1, − 2), (b) (1, 4), (c) (1, 2), (d) (1, − 4), , 2. The equation x 2 + y 2 + 2gx + 2 fy + c = 0, represents a, circle of non-zero radius, if, (a) g2 + f 2 > c, (b) g2 + f 2 < c, 2, 2, (c) g > f + c, (d) g2 < f 2 + c, , 11. Radius of circle ( x − 5)( x − 1) + ( y − 7)( y − 4) = 0 is, (a) 3, (b) 4, (c) 5 / 2, (d) 7 / 2, , 3. If g + f = c, then the equation, 2, , 2, , x 2 + y 2 + 2gx + 2 fy + c = 0 will represent, (a) a circle of radius g, (b) a circle of radius f, (c) a circle of diameter c (d) a circle of radius 0, 4. The circle x 2 + y 2 − 8x + 4 y + 4 = 0 touches, (a) x-axis, (b) y-axis, (c) Both axes, (d) Neither x-axis nor y-axis, 5. What is the radius of the circle touching x-axis at, (3, 0) and y-axis at (0, 3)?, (NDA 2011 II), (a) 3 units, (b) 4 units, (c) 5 units, (d) 6 units, 6. For the equation, ax 2 + by 2 + 2hxy + 2gx + 2 fy + c = 0, where a ≠ 0, to, represent a circle, the required condition will be, (a) a = b and c = 0, (NDA 2011 I), (b) f = g and h = 0, (c) a = b and h = 0, (d) f = g and c = 0, 7. The circle x 2 + y 2 − 3x − 4 y + 2 = 0 cuts x-axis at, (a) ( 2, 0), ( −3, 0), (b) ( 3, 0), ( 4, 0), (c) (1, 0), ( −1, 0), (d) (1, 0), ( 2, 0), 8. The limiting points of the system of circles, represented by the equation 2 ( x 2 + y 2 ) + λ x + 9/ 2 = 0, are, 9 , 3 , (b) ( 0, 0) and , 0, (a) ± , 0, 2 , 2 , 9 , (c) ± , 0, 2 , , (d) ( ± 3 , 0), , 9. The value of c for which the line y = 2x + c is a, tangent to the circle x 2 + y 2 = 16 is, (a) −16 5, (c) 16 5, , (b) 4 5, (d) 20, , 12. The square of the length of the tangent from ( 3, − 4) to, the circle x 2 + y 2 − 4x − 6 y + 3 = 0 is, (a) 20, (b) 30, (c) 40, (d) 50, 13. The radius of any circle touching the lines, 3x − 4 y + 5 = 0 and 6x − 8 y − 9 = 0 is, (a) 1.9, (b) 0.95, (c) 2.9, (d) 1.45, 14. If the area of the circle 4x 2 + 4 y 2 − 8x + 16 y + k = 0 is, 9π sq units, then the value of k is, (a) 4, (b) 16, (c) −16, (d) ±16, 15. Let AB be the intercept of the line y = x by the circle, x 2 + y 2 − 2x = 0. Then, the equation of the circle with, AB as its diameter is, (a) x 2 + y 2 − x − y = 0, (b) x 2 + y 2 + x + y = 0, (c) x 2 + y 2 + 2( x − y ) = 0 (d) x 2 + y 2 − 2x + y = 0, 16. If, x-axis, is, tangent, to, the, circle, x 2 + y 2 + 2gx + 2 fy + k = 0, then which one of the, following is correct?, (NDA 2009 I), 2, 2, 2, (a) g = k, (b) g = f, (c) f = k, (d) f 2 = g, 17. The equation of the circle passing through ( −1, 2) and, concentric with x 2 + y 2 − 2x − 4 y − 4 = 0 is given by, (a) x 2 + y 2 − 2x − 4 y + 1 = 0, (b) x 2 + y 2 − 2x + 4 y + 1 = 0, (c) x 2 + y 2 − 2x − 4 y + 4 = 0, (d) x 2 + y 2 − 2x + 4 y + 4 = 0, 18. The lines 5x − 12 y − 5 = 0 and 10x − 24 y + 3 = 0 are, tangents to the same circle. What is the diameter of, the circle?, (a) 1 unit, (b) 5 units, 1, (c) 8 units, (d) unit, 2, 19. Point, (1,, 2), relative, x 2 + y 2 + 4x − 2 y − 4 = 0 is a/an, (a) exterior point, (b) interior point but not centre, (c) boundary point, (d) centre, , to, , the, , circle, , (NDA 2008 I)
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523, , The Circle, 20. If the circle x 2 + y 2 + 2gx + 2 fy + c = 0 (c > 0) touches, the y-axis, then which one of the following is correct?, (NDA 2007 II), , (a) Only g = − c, (c) Only f = c, , (b) g = ± c, (d) f = ± c, , 21. For what value of k, does the equation, 9x 2 + y 2 = k( x 2 − y 2 − 2x ) represents equation of a, circle?, (a) 1, (b) 2, (c) –1, (d) 4, 22. What is the length of the intercept made on the x-axis, by the circle, x 2 + y 2 + 2gx + 2 fy + c = 0?, (a), , ( g2 − c ), 2, , (c) 2 ( g2 − 4c), , (b), , (a) 5x 2 + 5 y 2 − 10x + 30 y + 49 = 0, (b) 5x 2 + 5 y 2 + 10x − 30 y + 49 = 0, (c) 5x 2 + 5 y 2 − 10x + 30 y − 49 = 0, (d) None of the above, 28. Centre of circle ( x − x1 )( x − x2 ) + ( y − y1 )( y − y2 ) = 0 is, x + y1 x2 + y2 , x − y1 x2 − y2 , (a) 1, (b) 1, ,, ,, , , 2, 2, 2 , 2 , x + x2 y1 + y2 , x − x2 y1 − y2 , (c) 1, (d) 1, ,, ,, , , 2, 2, 2 , 2 , 29. The area of the circle whose centre is at (1, 2) and, which passes through the point (4, 6) is, (a) 5π, (b) 10π, (c) 25π, (d) None of these, 30. The number of circle having radius 5 and passing, through the points (− 2, 0) and (4, 0) is, (a) one, (b) two, (c) four, (d) infinite, , ( g2 − 4c), 2, , (d) 2 ( g2 − c), , 23. Under which one of the following conditions does the, circle x 2 + y 2 + 2gx + 2 fy + c = 0 meet the x-axis in two, points on opposite sides of the origin?, (NDA 2007 I), (a) c > 0, (b) c < 0, (c) c = 0, (d) c ≤ 0, 24. If the line x + 2by + 7 = 0 is a diameter of the circle, x 2 + y 2 − 6x + 2 y = 0, then b is equal to, (a) 3, (b) − 5, (c) − 1, (d) 5, 25. The, centres, of, the, circles, x 2 + y 2 = 1,, 2, 2, 2, 2, x + y + 6x − 2 y = 1 and x + y − 12x + 4 y = 1 are, (a) same, (b) collinear, (c) non-collinear, (d) None of these, 26. If a circle passes through the point (0, 0), (a, 0), (0, b),, then its centre is, (a) ( a , b), (b) (b, a), a b, b a, (c) , , (d) , − , 2 2, 2, 2, 27. The equation of the circle whose centre is (1, − 3) and, which touches the line 2x − y − 4 = 0 is, , 31. The equation of the circle, which touches x-axis and, whose centre is (1, 2) is, (a) x 2 + y 2 − 2x + 4 y + 1 = 0, (b) x 2 + y 2 − 2x − 4 y + 1 = 0, (c) x 2 + y 2 + 2x + 4 y + 1 = 0, (d) x 2 + y 2 + 4x + 2 y + 4 = 0, 32. The equation of the circle having centre (1, − 2) and, passing through the point of intersection of lines, 3x + y = 14, 2x + 5 y = 18 is, (a) x 2 + y 2 − 2x + 4 y − 20 = 0, (b) x 2 + y 2 − 2x − 4 y − 20 = 0, (c) x 2 + y 2 + 2x − 4 y − 20 = 0, (d) x 2 + y 2 + 2x + 4 y − 20 = 0, 33. For the line 3x + 2 y = 12 and the circle, x 2 + y 2 − 4x − 6 y + 3 = 0, which of the following, statements is true?, (a) Line is a tangent to the circle, (b) Line is a cord of the circle, (c) Line is a diameter of the circle, (d) None of the above, , Level II, 1. If a1 , b1 , c1 , f1 , g1 and h1 are real numbers such that, g12 + f12 > c1a1 , then the equation, a1x 2 + 2h1xy + b1 y 2 + 2g1x + 2 f1 y + c1 = 0, represents a circle if and only if, (b) a1 = b1 , h1 = 0, (a) a1 = b1, (c) a1 = b1 , a1 ≠ 0, h1 = 0 (d) a1 = b1 , a1 ≠ 0, h1 ≠ 0, 2. The line 3x − 2 y = k meets the circle x + y = 4 r at, only one point, if k2 is, (b) 52 r 2, (a) 20 r 2, 52 2, 20 2, (d), (c), r, r, 9, 9, 2, , 2, , 2, , 3. The radius of the circle passing through the point, (6, 2) and two of whose diameters are x + y = 6 and, x + 2 y = 4 is, (a) 4, (b) 6, (c) 20, (d) 20, 4. Two circles x 2 + y 2 = 6 and x 2 + y 2 − 6x + 8 = 0 are, given. The equation of the circle through their point, of intersection and the point (1, 1) is given by, (b) x 2 + y 2 − 4 y + 2 = 0, (a) x 2 + y 2 − 3x + 1 = 0, (c) x 2 + y 2 − 6x + 4 = 0, , (d) x 2 + y 2 − 3x + 2 = 0
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524, , NDA/NA Mathematics, + 2 y 2 − 3x + 6 y + k = 0 and, 0 cut orthogonally, then the, , 5. If r1 , r2 and r3 are the radii of the circles, and, x 2 + y 2 + 6x − 4 y = 3, x 2 + y 2 − 4x + 6 y = 5,, 2, 2, x + y − 2x + 4 y = 8, respectively, then, (b) r2 > r3 > r1, (a) r1 > r2 > r3, (c) r3 > r1 > r2, (d) r1 > r3 > r2, , 14. If the two circles 2x 2, x 2 + y 2 − 4x + 10 y + 16 =, value of k is, (a) 41, (c) 4, , 6. What is the equation to circle, which touches both the, axes and has centre on the line x + y = 4? (NDA 2010 II), (a) x 2 + y 2 − 4x + 4 y + 4 = 0, (b) x 2 + y 2 − 4x − 4 y + 4 = 0, (c) x 2 + y 2 + 4x − 4 y − 4 = 0, (d) x 2 + y 2 + 4x + 4 y − 4 = 0, , 15. The circle x 2 + y 2 + 4x − 4 y + 4 = 0 touches (NDA 2009 I), , 7. The equation of the circumcircle of the triangle, formed by the lines x = 0, y = 0 and 2x + 3 y = 5, is, (a) 6 ( x 2 + y 2 ) + 5( 3x − 2 y ) = 0, (b) x 2 + y 2 − 2x − 3 y + 5 = 0, (c) x 2 + y 2 + 2x − 3 y − 5 = 0, (d) 6 ( x 2 + y 2 ) − 5( 3x + 2 y ) = 0, 8. The value of k, so that x 2 + y 2 + kx + 4 y + 2 = 0 and, 2( x 2 + y 2 ) − 4x − 3 y + k = 0 cut orthogonally is, (a) 10/ 3, (b) −8/ 3, (c) −10/ 3, (d) 8/ 3, 9. The equation of the tangents to the circle x 2 + y 2 = 4,, which are parallel to x + 2 y + 3 = 0, are, (a) x − 2 y = 2, (b) x + 2 y = ± 2 3, (c) x + 2 y = ± 2 5, (d) x − 2 y = ± 2 5, 10. Consider the following statements in respect of, circles x 2 + y 2 − 2x − 2 y = 0 and x 2 + y 2 = 1 (NDA 2010 I), I. The radius of the first circle is twice that of the, second circle., II. Both the circles pass through origin., Which of the statement(s) given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 11. The value of λ, for which, x 2 + y 2 + 2λx + 6 y + 1 = 0 intersects, x 2 + y 2 + 4x + 2 y = 0 orthogonally, is, (a) 11/ 8, (b) −1, (c) −5/ 4, (d) 5/ 2, , the, the, , 12. The two circles x 2 + y 2 − 2x + 6 y + 6 = 0, x 2 + y 2 − 5x + 6 y + 15 = 0, (a) intersect, (b) are concentric, (c) touch internally, (d) touch externally, , circle, circle, , and, , 13. The centre of a circle is ( 2, − 3) and the circumference, is 10π. Then, the equation of the circle is, (a) x 2 + y 2 + 4x + 6 y + 12 = 0, (b) x 2 + y 2 − 4x + 6 y + 12 = 0, (c) x 2 + y 2 − 4x + 6 y − 12 = 0, (d) x 2 + y 2 − 4x − 6 y − 12 = 0, , (a) only the x-axis, (c) Both the axes, , (b) 14, (d) 0, (b) only the y-axis, (d) Neither of the axes, , 16. If two circles of the same radius r and centres at (2, 3), and (5, 6), respectively cut orthogonally, then the, value of r is, (a) 3, (b) 2, (c) 1, (d) 5, 17. The equations of the tangents to the circle, x 2 + y 2 − 6x + 4 y − 12 = 0, which are parallel to the, line 4x + 3 y + 5 = 0, are, (a) 4x + 3 y + 11 = 0 and 4x + 3 y + 8 = 0, (b) 4x + 3 y − 9 = 0 and 4x + 3 y + 7 = 0, (c) 4x + 3 y + 19 = 0 and 4x + 3 y − 31 = 0, (d) 4x + 3 y − 10 = 0 and 4x + 3 y + 12 = 0, 18. The equation of the circle, which touches the axes at, a, distance, 5, from, the, origin, is, y 2 + x 2 − αx − 2αy + α 2 = 0. What is the value of α?, (NDA 2008 II), , (a) 4, (c) 6, , (b) 5, (d) 7, , 19. The lines 2x − 3 y = 5 and 3x − 4 y = 7 are diameters of, a circle having area as 154 sq units. Then, the, equation of the circle is, (a) x 2 + y 2 + 2x − 2 y = 62 (b) x 2 + y 2 + 2x − 2 y = 47, (c) x 2 + y 2 − 2x + 2 y = 47 (d) x 2 + y 2 − 2x + 2 y = 62, 20. A circle is drawn to cut a chord of length 2a units, along X-axis and to touch the Y-axis. The locus of the, centre of the circle is, (a) x 2 + y 2 = a 2, (b) x 2 − y 2 = a 2, 2, (c) x + y = a, (d) x 2 − y 2 = 4a 2, 21. ABC is an equilateral triangle inscribed in a circle of, centre O and radius 5 cm. Let the diameter through C, meet the circle again at, (NDA 2008 II), Assertion (A) AD ⋅ BD < OB ⋅ OC, Reason (R), ( AD 2 + BD 2 ) = CD 2 = 100 sq cm, (a) A and R are both correct and R is correct, explanation of A., (b) A and R are both correct but R is not correct, explanation of A., (c) A is correct but R is wrong., (d) A is wrong but R is correct., 22. Consider the following statements, I. The area of the triangle formed by the tangent at, ( 3, 4) to the circle x 2 + y 2 = 25 and the coordinate, 625, sq units., axes is, 24
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525, , The Circle, II. The equation of the circle having centre at (3, – 4), and touching the line 5x + 12 y − 12 = 0 is, 45, ( x − 3)2 + ( y + 4)2 = , 13 , , 2, , Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) None of the above, 23. Consider the following statements, I. The shortest distance from the point ( 2, − 7) to the, circle x 2 + y 2 − 14x − 10 y − 151 = 0 is equal to 5., and, II. The, circles, x 2 + y 2 − 2x − 4 y = 0, 2, 2, x + y + 8 y − 4 = 0 touch each other., Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 24. Consider the following statements, I. Two tangents can be drawn from the point ( 0, 1) to, the circle x 2 + y 2 − 2x − 4 y = 0., II. The line x + 3 y = 0 is a diameter of the circle, x 2 + y 2 + 6x + 2 y = 0., Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 25. Which one of the following is correct?, The circles x 2 + y 2 − 2x − 2 y − 7 = 0, and 3( x 2 + y 2 ) − 8x + 29 y = 0, (a) touch externally, (b) touch internally, (c) cut each other orthogonally, (d) do not cut each other, 26. Equation of a circle passing through origin is, x 2 + y 2 − 6x + 2 y = 0. What is the equation of one of its, diameters?, (NDA 2008 I), (a) x + 3 y = 0, (b) x + y = 0, (c) x = y, (d) 3x + y = 0, 27. Which one of the following statements is correct?, The two circles x 2 + y 2 − 2x − 3 = 0, and, 2, 2, x + y − 4x − 6 y − 64 = 0 are such that, (a) they touch each other, (b) they intersect each other, (c) one lies inside the other, (d) each lies outside the other, 28. The centre of circle passing through the points ( 0, 0),, (1, 0) and touching the circle x 2 + y 2 = 9, is, 3 1, 1 1, (a) , , (b) , , 2 2, 2 2, 1 1, , 1, (d) , − 2, (c) −, , , , 2, , 2 2, , 29. Match List I (Equations of circles) with List II (Their, centres) and select the correct answer using the codes, given below the lists, List I, (Equations of circles), , List II, (Their centres), , A. ( x + 2) 2 + ( y + 1) 2 = 1, , 1. (1, − 2), , B. ( x − 1) 2 + ( y + 2) 2 = 1, , 2. ( −1, 2), , C. ( x + 1) + ( y − 2) = 4, , 3. ( 2, 1), , D. ( x − 2) 2 + ( y − 1) 2 = 4, , 4. ( −2, − 1), , 2, , Codes, A, (a) 4, (b) 4, (c) 3, (d) 3, , B, 1, 2, 1, 2, , 2, , C, 2, 1, 2, 1, , D, 3, 3, 4, 4, , 30. ABCD is a square whose side is a. If AB and AD be, the coordinate axes, then the equation of the, circumcircle is, (a) x 2 + y 2 + 2a ( x + y ) = 0, (b) x 2 + y 2 + a ( x + y ) = 0, (c) x 2 + y 2 − a ( x + y ) + a 2 = 0, (d) x 2 + y 2 − a ( x + y ) = 0, 31. What is the equation of circle, which touches the, lines x = 0, y = 0 and x = 2?, (NDA 2007 II), (a) x 2 + y 2 + 2x + 2 y + 1 = 0 (b) x 2 + y 2 − 4x − 4 y + 1 = 0, (c) x 2 + y 2 − 2x − 2 y + 1 = 0 (d) None of these, 32. Which of the following are the equations of circles,, which touch the x-axis at a distance 3 from the origin, and intercept a distance 6 on the y-axis?, (a) x 2 + y 2 − 6x ± 6 2 y + 9 = 0, (b) x 2 + y 2 + 6x ± 6 2 y + 9 = 0, (c) x 2 + y 2 ± 6x + 6 2 y + 9 = 0, (d) x 2 + y 2 ± 6 2 y − 6 y + 9 = 0, 33. A tangent is drawn from the origin to the circle, x 2 + y 2 + 2x + 2 y + 1 = 0. What is the point of contact?, (a) ( −1, − 1), (b) ( 0, − 1), (c) ( 0, 1), (d) (1, 1), 34. What is the number of common tangents to the, circles x 2 + y 2 = 1 and x 2 + y 2 − 4x + 3 = 0?, (a) One, (c) Three, , (b) Two, (d) Four, , 35. Which one of the following statements is correct?, The circles x 2 + y 2 + x − 2 y = 0, and, x 2 + y 2 − 2x + y = 0, (a) touch each other, (b) intersect each other, (c) do not intersect each other, (d) are concentric
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526, , NDA/NA Mathematics, , 36. The centre of the circle ( x − α )2 + ( y − β )2 = 9 lies on, the straight line x = y and the circle touches the circle, x 2 + y 2 = 1 externally. What are the values of α , β ?, (a) α = ± 2 2 , β = ± 2 2 (b) α = ± 2 , β = ± 2, (c) α = 0, β = 0, (d) α = 2, β = 2, 37. What is the radius of the circle passing through the, points (0, 0), ( a , 0) and ( 0, b)?, (a) a − b, 1, (c), a 2 + b2, 2, 2, , 2, , (b) a + b, 2, , 2, , (d) 2 a 2 + b2, , 38. If two circles A and B of equal radii pass through the, centres of each other, then what is the ratio of the, length of the smaller arc to the circumference of the, circle A cut-off by the circle B?, 1, 1, (a), (b), 2, 4, 1, 2, (d), (c), 3, 3, 39. If the extremities of a diameter of a circle are (0, 0), and ( a3 , 1 / a3 ), then the circle passes through, which, one of the following points?, (a) ( a 2 , 1 / a 2 ), (b) ( a , 1 / a ), (c) ( a , − a ), (d) (1 / a , a ), 40. What is the equation of a circle whose centre lies on, the x-axis at a distance h from the origin and the, circle passes through the origin?, (NDA 2007 I), 2, 2, (a) x + y − 2hx = 0, (b) x 2 + y 2 − 2hx + h 2 = 0, (c) x 2 + y 2 + 2hxy = 0, (d) x 2 + y 2 − h 2 = 0, , Directions (Q. Nos. 41-43) Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 41. Assertion (A) The equation, 4x 2 + 4 y 2 + 2x + 8 y + 7 = 0, represents an equation of a circle., Reason (R) The equation, ax 2 + by 2 + 2gx + 2 fy + c = 0, represents an equation of a circle, if a = b., , 42. Assertion (A) Length of tangent drawn from the, point (1, 2) to a circle x 2 + y 2 + 8x + 7 y + 5 = 0 is 3 2., Reason (R) Length of tangent drawn from the, point ( x1 , y1 ) to a circle x 2 + y 2 + 2gx + 2 fy + c = 0 is, x12 + y12 + 2gx1 + 2 fy1 + c ., 43. Assertion (A) The line y = 4x + 153 touches the, circle x 2 + y 2 = 32., Reason (R) The line y = mx + c touches the circle, x 2 + y 2 = r 2, iff c = ± r 1 + m 2 ., , Directions (Q. Nos. 44-47) Let there are two points, A( 2, 3) and B( 4, 5) in xy-plane and one line y − 4x + 3 = 0., , 44. Find the equation of circle passing through A and B, and centre lie on a given line., (a) x 2 + y 2 + 4x − 10 y + 25 = 0, (b) x 2 + y 2 − 4x − 10 y + 25 = 0, (c) x 2 + y 2 − 4x − 10 y + 6 = 0, (d) None of the above, 45. Find the position of a point (4, 5) relative to circle., (a) inside, (b) on, (c) outside, (d) None of these, 46. Find the area of the circle., (a) 12.56 sq units, (b) 12.50 sq units, (c) 12.80 sq units, (d) None of these, 47. If we shift the centre to the point (4, 6), then the, equation of circle having radius 3 is, (a) x 2 + y 2 − 8x − 12 y + 40 = 0, (b) x 2 + y 2 − 8x − 12 y + 49 = 0, (c) x 2 + y 2 + 8x + 12 y + 49 = 0, (d) None of the above, , Directions (Q. Nos. 48-50), , The equation of two, , circles is given by, S1 = x 2 + y 2 − 2x + 6 y + 6 = 0 and, S 2 = x 2 + y 2 − 5x + 6 y + 15 = 0, 48. Two circles S1 and S 2, (a) intersect, (c) touch internally, , (b) are concentric, (d) touch externally, , 49. Find the equation of circle having centre S1 and, radius 5 is, (a) x 2 + y 2 + 2x − 6 y = 15 (b) x 2 + y 2 − 2x + 6 y = 15, (c) x 2 + y 2 + 2x + 6 y = 15 (d) None of these, 50. Find the intersecting equation of circles S1 and S 2, passing through (1, 2) is, (a) 2x 2 + 2 y 2 + 17x + 12 y − 51 = 0, (b) 2x 2 + 2 y 2 + 17x + 12 y + 51 = 0, (c) 2x 2 + 2 y 2 + 17x + y + 51 = 0, (d) None of the above
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Answers, Level I, 1., 11., 21., 31., , (d), (c), (d), (b), , 2., 12., 22., 32., , (a), (c), (d), (a), , 3., 13., 23., 33., , (d), (b), (b), (c), , 4. (b), 14. (c), 24. (d), , 2., 12., 22., 32., 42., , (b), (c), (c), (a), (d), , 3., 13., 23., 33., 43., , (d), (c), (b), (b), (a), , 4., 14., 24., 34., 44., , 5. (a), 15. (a), 25. (b), , 6. (c), 16. (a), 26. (c), , 7. (d), 17. (a), 27. (a), , (a), (d), (c), , 8., 18., 28., , 9. (b), 19. (a), 29. (c), , 10. (d), 20. (d), 30. (b), , Level II, 1., 11., 21., 31., 41., , (c), (c), (d), (c), (a), , (a), (c), (d), (c), (b), , 5., 15., 25., 35., 45., , (a), (a), (c), (b), (b), , 6., 16., 26., 36., 46., , (b), (a), (a), (a), (a), , (d), (c), (c), (c), (b), , 7., 17., 27., 37., 47., , 8., 18., 28., 38., 48., , (c), (b), (d), (c), (c), , 9., 19., 29., 39., 49., , (c), (c), (a), (d), (b), , 10., 20., 30., 40., 50., , (d), (b), (d), (a), (a), , Hints & Solutions, Level I, 1. Given that, the equation of circle is, …(i), x2 + y2 + 4x − 6 y − 36 = 0, ∴ The standard equation of circle is, …(ii), x2 + y2 + 2 gx + 2 fy + c = 0, On comparing the given equation with the standard, equation of circle, we get, g = 2 , f = − 3, c = − 36, ∴ Centre of circle is (− g , − f ) = (−2 , 3), and the radius is r =, , g2 + f 2 − c, , 2. We, know, that,, the, equation, of, circle, x2 + y2 + 2 gx + 2 fy + c = 0 represents a circle of non-zero, radius, when, or, , g + f − c >0, 2, , g2 + f 2 > c, , 3. Given that, x2 + y2 + 2 gx + 2 fy + c = 0 and g 2 + f 2 = c, ∴ Radius of circle = g 2 + f 2 − c, ⇒, , Radius = 0, , [Q g + f = c], 2, , 2, , 4. We have, equation of circle is, x2 + y2 − 8x + 4 y + 4 = 0, On comparing with standard equation of circle, x2 + y2 + 2 gx + 2 fy + c = 0, we get, g = − 4, f = 2 and c = 4, ∴ Coordinates of the centre are (− g , − f ) = (4, − 2), ∴ Radius of the circle = g 2 + f 2 − c, = (− 4)2 + (2)2 − 4 = 16 + 4 − 4, =4, , 5. Radius of the circle = AC = BC, y-axis, , C (3, 3), , B, , = 4 + 9 + 36 = 49 = 7, , 2, , Here, radius of circle is equal to x- coordinate of the, centre., ∴ Circle touches the y-axis., , 3, O, , 3, , ∴ Radius = 3 units, , A, , x-axis, , (Q AC = OB = 3 and BC = OA = 3), , 6. The equation ax2 + by2 + 2hxy + 2 gx + 2 fy + c = 0, represents a circle, if a = b and h = 0., Then, the equation becomes the general equation of a, circle, x2 + y2 + 2 gx + 2 fy + c = 0, 7. Given, equation of circle is x2 + y2 − 3x − 4 y + 2 = 0 and, it cuts the x-axis., ∴, y=0, The equation of the circle becomes, x2 + 0 − 3x − 4(0) + 2 = 0, ⇒, x2 − 3x + 2 = 0 ⇒ x2 − 2x − x + 2 = 0, ⇒, (x − 1)(x − 2) = 0 ⇒ x = 1, 2, Therefore, the points are (1, 0), (2, 0).
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528, , NDA/NA Mathematics, 9, = 0., 2, On comparing with standard equation of circle, we get, λ2 9, λ , centre − , 0 and radius r =, − ., 4 , 16 4, We know that, for limiting point, λ2 9, − =0, 16 4, ⇒, λ2 = 36 ⇒ λ = ± 6, 6 , 3 , ∴ Limiting points are ± , 0 or ± , 0 ., 4 , 2 , , 15. The coordinates of the point of intersection of line y = x, and x2 + y2 − 2x = 0 are A(0,0) and B(1,1)., ∴ Equation of circle with AB as its diameter is, x(x − 1) + y( y − 1) = 0, ⇒, x2 + y2 − x − y = 0, , 9. Given that,, y = 2x + c, … (i), and, x2 + y2 = 16, … (ii), We know that, if y = mx + c is tangent to the circle, , 16. Since, x-axis is a tangent to the given circle, it means, the circle touches the x-axis., 2 g2 − k = 0, ∴, , 8. We have, 2(x2 + y2) + λx +, , x2 + y2 = a 2, then c = ± a 1 + m2, here, m = 2 , a = 4., ∴, , …(i), , S 2 ≡ x2 + y2 + x − 8 y − 13 = 0, , …(ii), , ∴ Equation of common chord is, S1 − S 2 = 0, ⇒ (x2 + y2 + 2x − 3 y + 6) − (x2 + y2 + x − 8 y − 13) = 0, x + 5 y + 19 = 0, , … (iii), , In the given options, only the point (1, – 4) satisfied the, Eq. (iii)., 11. Extremities of diameter are (5, 7) and (1, 4) and radius, 1, 5, is half of the distance between them =, (4)2 + (3)2 = ., 2, 2, 12. Length of tangent from the point (x1 , y1 ) to the circle, x2 + y2 + 2 gx + 2 fy + c = 0 is, x12 + y12 + 2 gx1 + 2 fy1 + c, ∴ Required length of tangent from the point (3,−4), to the circle x2 + y2 − 4x − 6 y + 3 = 0, = 32 + (−4)2 − 4 (3) − 6 (−4) + 3 = 40, ∴ Square of length of tangent = 40, 13. Since, given lines are parallel to each other, so the line, segment joining the points of contact is diameter of the, circle. Distance between the lines 3x − 4 y + 5 = 0 and, 9, 3x − 4 y − = 0 is, 2, 5+ 9 , , 2 = 19 = 1.9, 32 + 42 10, , , Length of diameter of the circle is 1.9., 1.9, ∴ Radius of circle =, = 0.95, 2, 14. The equation of circle is, x2 + y2 − 2x + 4 y +, , k, =0, 4, , k, k, = 5−, 4, 4, , Area of circle = 9π, k, k, , ⇒, π 5 − = 9π ⇒ 5 − 9 =, , , 4, 4, ⇒, k = − 16, , ⇒, , (given), , g2 = k, , 17. Let the equation of the circle concentric with the circle, x2 + y2 − 2x − 4 y − 4 = 0, , c = ± 4 1 + 22 = ± 4 5, , 10. Let the equation of circles are, S1 ≡ x2 + y2 + 2x − 3 y + 6 = 0, , ⇒, , ∴ Radius of circle = 1 + 4 −, , is, , x2 + y2 − 2x − 4 y + λ = 0, , …(i), , Eq. (i) passes through the point (−1, 2)., ∴, (−1)2 + (2)2 − 2 (−1) − 4 (2) + λ = 0, ⇒, , 1 + 4 + 2 −8 + λ =0, , ⇒, , λ =1, , Hence, the required equation is, x2 + y2 − 2x − 4 y + 1 = 0, 18. Two given tangents are parallel to each other therefore,, the distance between them is equal to the diameter of, the circle., Thus, diameter = [distance between 5x − 12 y − 5 = 0, 3, and 5x − 12 y + = 0], 2, −5 − 3 , 2 =, −13 , = 1 unit, =, 25 + 144 2 × 13 2, , , 19. The given equation of circle is, x2 + y 2 + 4 x − 2 y − 4 = 0, At (1, 2), (1)2 + (2)2 + 4(1) − 2(2) − 4 = 1 + 4 + 4 − 4 − 4 = 1 > 0, Thus, the point (1, 2) lies outside the circle, i.e., this, point is an exterior point., 20. The circle x2 + y2 + 2 gx + 2 fy + c = 0 touches y-axis, then, 2 f 2 − c =0, ⇒, ⇒, , f2=c, f =± c, , 21. Given, equation is, 9x2 + y2 = k (x2 − y2 − 2x), This equation represents a circle, if, coefficient of x2 = coefficient of y2, ⇒, 9 − k =1 + k, ⇒, 2k = 8, ⇒, k =4
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529, , The Circle, 22. The length of the intercept made on the x-axis is given, by 2 g 2 − c., 23. The circle x2 + y2 + 2 gx + 2 fy + c = 0 meets the x-axis in, two points on opposite of the origin, if c < 0 (by, property)., 24. Here, the centre of circle (3, − 1 ) must lie on the line, x + 2by + 7 = 0, Therefore,, 3 − 2b + 7 = 0, ⇒, b =5, 25. Centres are (0, 0), (− 3, 1) and (6, − 2) and a line passing, 1, through any two points say (0, 0) and (− 3, 1) is y = − x, 3, and point (6, − 2) lies on it. Hence, points are collinear., 26. Let the equation of circle be x2 + y2 + 2 gx + 2 fy + c = 0., Now, on passing through the points, we get three, equations., …(i), c=0, …(ii), a 2 + 2 ga + c = 0, …(iii), b2 + 2 fb + c = 0, a, b, On solving them, we get g = − , f = −, 2, 2, a b, Hence, the centre is , ., 2 2, 27. Radius of circle is, , 2 + 3 −4, 1, =, 5, 5, , Therefore, equation is (x − 1)2 + ( y + 3)2 =, , 1, 5, , 1, 5, or, 5x2 + 5 y2 − 10x + 30 y + 49 = 0, 28. (x1 , y1 ) and (x2, y2) are extreme points of diameter., x + x2 y1 + y2, Hence, centre is 1, ,, ., 2, 2 , or, , x2 + y 2 − 2 x + 6 y + 1 + 9 =, , 29. Obviously, radius = (1 − 4)2 + (2 − 6)2 = 5, Hence, the area is given by πr 2 = 25π sq units, 30. Two centre of each lying on the perpendicular bisector, of the join of the two points., 31. Centre (1, 2) and since, circle touches x-axis, therefore, radius is equal to 2., Hence, the equation is (x − 1)2 + ( y − 2)2 = 22, ⇒, x2 + y 2 − 2 x − 4 y + 1 = 0, 32. The point of intersection of 3x + y − 14 = 0, 2x + 5 y − 18 = 0 are, x = 4 and y = 2, i.e., point is (4, 2)., Therefore, radius is (9) + (16) = 5 and equation is, x2 + y2 − 2x + 4 y − 20 = 0., 33. Centre of the circle is (2, 3). Obviously, the line, 3x + 2 y = 12 passes through the centre of the circle., Hence, it is a diameter of the circle., , Level II, 1. A general equation of second degree is, ax2 + 2hxy + by2 + 2 gx + 2 fy + c = 0, in x, y represent a circle, if, (i) coefficient of x2 is non-zero, i.e., a ≠ 0, (ii) coefficient of x2 = coefficient of y2, i.e.,, a=b, Coefficient of xy is zero, i.e.,, h =0, Thus, from the given equation, the condition comes out, to be a1 = b1 , a1 ≠ 0, h1 = 0., , 4. The equation of point of intersection of circle x2 + y2 = 6, and, x2 + y2 − 6x + 8 = 0 is given by, ...(i), (x2 + y2 − 6) + λ (x2 + y2 − 6x + 8) = 0, Since it passes through the point (1, 1), ∴ (1 + 1 − 6) + λ (1 + 1 − 6(1) + 8) = 0, ⇒, − 4 + λ (4) = 0 ⇒ λ = 1, Putting λ = 1 in Eq. (i), we get, 2 x2 + 2 y 2 − 6 x + 2 = 0, x2 + y 2 − 3 x + 1 = 0, , 2. Given, equation of line is 3x − 2 y = k, and equation of circle is, x2 + y2 = 4r 2, 3, k, Eq.(i) can be rewritten as y = x −, 2, 2, 3, k, m= ,c=−, ⇒, 2, 2, The line will meet the circle in one point, if, , 5. Given equation of circles are,, x2 + y2 − 4x + 6 y − 5 = 0, … (i), ... (ii), x2 + y2 + 6x − 4 y − 3 = 0, ...(iii), x2 + y2 − 2x + 4 y − 8 = 0, Let C1, C 2, C3 be the centre of equation of circle (i), (ii), and (iii), respectively., Centre C1 = (2, − 3) , C1 = − 5, r1 = (−2)2 + (3)2 − (−5), ∴, , c = a 1 + m2, On squaring, we get, , ⇒, , −, , k, 3, = (2r ) 1 + , 2, 2, , … (i), … (ii), , 2, , 13, k2, = 4r 2 ×, ⇒ k2 = 52r 2, 4, 4, , 3. Centre is the point of intersection of two diameters, i. e. ,, the point is C(8, − 2), therefore, the distance from the, centre to the point P(6, 2) is, r = CP = 4 + 16 = 20, , = 4 + 9 + 5 = 18 = 3 2, Centre C 2 = (− 3, 2), C 2 = − 3, ∴r2 = (3)2 + (− 2)2 − (− 3) =, , 9+4+3 =, , 16 = 4, , Centre C3 = (+ 1, − 2), C3 = − 8, ∴r3 = (− 1)2 + ( 2)2 − (− 8) =, , 1+4+8 =, , 13, , It is obvious from above r1 > r2 > r3 ., 6. We know that, the equation of circle, which touches, both the axes, is
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530, , NDA/NA Mathematics, ...(i), x2 + y2 − 2rx − 2ry + r 2 = 0, The centre (r , r) of this circle lies on the line x + y = 4, ∴, r + r =4, ⇒, r =2, On putting the value of r in Eq. (i), we get, x2 + y2 − 4 x − 4 y + 4 = 0, Which is the required equation of circle., , 7. Given that, x = 0, y = 0, 2x + 3 y = 5, On solving these equations, we get the required points is, 5 5 , (0, 0), 0, , , 0, 3 2 , Let equation of circle is, x2 + y2 + 2 gx + 2 fy + c = 0, …(i), Eq.(i) passes through (0,0), we get, c=0, 5, 5 , Similarly, Eq. (i) passes through 0, and , 0 ,, 3, 2 , 5, 5, we get, 2 f = − and 2 g = −, 3, 2, ∴ Required equation of circle is, 5, 5, x2 + y2 − x − y = 0, 2, 3, ⇒, 6 x2 + 6 y2 − 15x − 10 y = 0, 8. We have, equation of circles are, x2 + y2 + kx + 4 y + 2 = 0, … (i), and, 2(x2 + y2) − 4x − 3 y + k = 0, … (ii), From Eq.(i), we get, ⇒, g1 = k / 2 , f1 = 2 , c1 = 2, and from Eq.(ii), we get, ⇒, g2 = − 1, k, 3, f2 = − , c2 =, 2, 4, Condition for two circles cut orthogonally is, 2 g1 g2 + 2 f1 f2 = c1 + c2, k, k, 3, ⇒ 2 × × (−1) + 2 × 2 × − = 2 +, 4, 2, 2, k, −k − 3 = 2 +, ⇒, 2, 3k, 10, = −5 ⇒ k = −, ⇒, 2, 3, 9. Centre of circle is (0,0)., Equation of tangent, which is parallel to x + 2 y + 3 = 0,, is, x + 2y + λ = 0, … (i), As we know, perpendicular distance from centre (0,0) to, x + 2 y + λ = 0 should be equal to radius., , (0, 0), , x + 2y + λ = 0, , ∴, , 0 + 2 ×0 + λ, 1 +2, 2, , 2, , = ±2, , ⇒ λ = ± 2 5, put the value of λ in Eq.(i)., ∴ Equation of tangents to the circle are x + 2 y = ± 2 5., 10. The equation of first circle is x2 + y2 − 2x − 2 y = 0, Radius of this circle = (− 1)2 + (− 1)2 = 2,, and equation of second circle is x2 + y2 = 1, Radius of this circle = 1., From above it is clear that the radius of first circle is not, twice that of the second circle., Also, the first circle passes through the origin while the, second circle does not pass through the origin., Hence, neither statements I or II is correct., 11. We have, x2 + y2 + 2λx + 6 y + 1 = 0, and, x2 + y2 + 4x + 2 y = 0., Since, the circles cut orthogonally, ∴, 2 gg′ + 2 ff ′ = c + c′, ⇒, 2 λ × 2 + 6 × 1 = 1 + 0 ⇒ 4λ + 6 = 1, 5, ⇒, λ=−, 4, 12. Given, equation of circles are, x2 + y2 − 2x + 6 y + 6 = 0, … (i), 2, and, x + y2 − 5x + 6 y + 15 = 0, … (ii), The standard equation of a circle is, x2 + y2 + 2 gx + 2 fy + c = 0, From Eq. (i), we get, g = − 1, f = 3, c = 6 ⇒ centre of the circle C1 = (1, − 3), and radius of the circle,, r1 =, , g 2 + f 2 − c = (−1)2 + 32 − 6, , = 1 + 9 −6 =2, Similarly, from Eq. (ii), g = −, ⇒, , 5, , f = 3, c = 15, 2, , 5, , Centre C 2 = , − 3 and, 2, , 2, , 5, r2 = + (3)2 − 15, 2, , radius of the circle, , =, , 25, + 9 − 15, 4, , =, , 25 − 24 1, = ., 4, 2, , Therefore, distance between their centres C1 and C 2, 2, , , 5, = − 1 + (−3 + 3)2, , 2, 2, , 3, 3, = =, 2, 2, Also, difference of radii, (r1 − r2) = 2 −, , 1 3, = ., 2 2, , Since, distance between C1 and C 2 is equal to r1 − r2,, therefore, the circles touch each other internally., 13. It is given, centre is (2, –3) and, circumference of circle = 10π ⇒ 2πr = 10π ⇒ r = 5.
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531, , The Circle, The equation of circle, if centre is (2, − 3) and radius is 5,, is, ⇒, ⇒, , (x − 2)2 + ( y + 3)2 = 52, x + y2 − 4x + 6 y + 13 = 25, x2 + y2 − 4x + 6 y − 12 = 0, 2, , …(i), , So, equation of the circle from the figure is, (x − 5)2 + ( y − 5)2 = (5)2, 2, x + 25 − 10x + y2 + 25 − 10 y = 25, ⇒, x2 + y2 − 10x − 10 y + 25 = 0, Comparing with,, x2 + y2 − 2αx − 2αy + α 2 = 0, ⇒, α =5, 19. The equation of diameters are, 2x − 3 y = 5, and, 3x − 4 y = 7, On solving Eqs. (i) and (ii), we get, x = 1 and y = − 1, , y, , ∴, , O, , Since, area of circle is 154 sq units, then radius of circle, is r = 7., ∴ Equation of circle is, (x − 1)2 + ( y + 1)2 = 49, ⇒, x2 + y2 − 2x + 2 y = 47, , x, , y', , ⇒, ⇒, , x + y − 4x − 6 y − r + 13 = 0, 2, , 2, , Centre of circle = (1, − 1), , 20. Let the equation of circle is, , Here, we see that the circle touches both the axes., , x2 + y2 + 2 gx + 2 fy + c = 0, , 16. The equation of circles are, (x − 2)2 + ( y − 3)2 = r 2, 2, , …(i), …(ii), , B, , 2, A, , x, , (5, 0), , y′, , 15. Given, equation is, x2 + 4 x + 4 + y2 − 4 y = 0, ⇒, (x + 2)2 + ( y − 2)2 = 22, , x', , A, , x′, , …(ii), , Q Circles (i) and (ii) cut orthogonally., ∴, 2 g1 g2 + 2 f1 f2 = c1 + c2, k, 3, 3, 2 − (−2) + 2 5 = + 16, ⇒, 4, 2, 2, k, k, ⇒, 3 + 15 = + 16 ⇒ 18 = + 16, 2, 2, ⇒, k =4, , (–2,2)C 2, , O, (5, 5), , (0, 5) B, , 14. Given, equation of circles are, 2x2 + 2 y2 − 3x + 6 y + k = 0, 3, k, or, x2 + y2 − x + 3 y + = 0, 2, 2, and, x2 + y2 − 4x + 10 y + 16 = 0, , and, , y, , Since, the circle cuts the X-axis and touch the Y -axis., …(i), , Y, , (x − 5)2 + ( y − 6)2 = r 2, x2 + y2 − 10x − 12 y − r 2 + 61 = 0, , …(ii), , The given circles cut orthogonally, if, ⇒, 2 g1 g2 + 2 f1 f2 = c1 + c2, ⇒, ⇒, , ⇒, , 2 (−2) (−5) + 2 (−3) (−6) = − r 2 + 13 − r 2 + 61, 20 + 36 − 13 − 61 = − 2r, , O, , 2, , r=3, , 17. The centre and radius of given circle are (3, − 2) and 5,, respectively. The equation of a line parallel to, 4x + 3 y + 5 = 0 is 4x + 3 y + λ = 0, As we know, that perpendicular distance from centre, (3,−2) to the circle = radius of the circle, 4 × 3 + 3 × (−2) + λ , ⇒, , = 5 ⇒ λ = 19, − 31, 42 + 32, , , ∴ Equation of tangents are, 4x + 3 y + 19 = 0 and 4x + 3 y − 31 = 0, 18. Given that, the circle touches the axes at a distance 5, from the origin, then, , A, , Q, , 2, , 2a, , B, , X, , g 2 − c = 2a, , ⇒, g2 − c = a 2, and, f2 = c, On solving Eqs.(i) and (ii), we get, g2 − f 2 = a 2, ∴ Locus of centre of the circle is, x2 − y 2 = a 2, 21. From figure,, ∆ABC is an equilateral triangle,, , … (i), … (ii)
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532, , NDA/NA Mathematics, ∴ Distance between P( 2, 7) andC( 7, 5), = ( 7 − 2)2 + ( 5 − 7)2 = 25 + 4 = 29, , y, A, , Radius,, , D, O, , x′, , = 49 + 25 + 151, = 15, ∴Shortest distance = 15 − 29, II. Centres of given circles are, C1 (1, 2) and C 2(0, − 4), ∴, r1 = 1 + 4 + 0 = 5, r2 = 0 + 16 + 4 = 20 = 2 5, , x, , C, , B, , y′, , Then, ∆ODA is also an equilateral triangle., Similarly, ∆OBD is also an equilateral triangle., ∴, , AD ⋅ BD = 25 and OB ⋅ OC = 25, and 2( AD + BD ) = 2(25 + 25) = 100 = CD, 2, , 22. I. Given, x2 + y2 = 25, ∴, , 2x + 2 y, , ⇒, , 2, , dy, =0, dx, dy − x, =, dx y, , r1 + r2 = 5 + 2 5 = 3 5 ≠ 37 (C1C 2), Hence, they do not touch., , ∴Equation of tangent at (3, 4) is, −3, y −4 =, (x − 3), 4, 4 y − 16 = − 3x + 9 ⇒ 3x + 4 y = 25, y, B (0, 25/4), 3x + 4y = 25, , O, , 24. I. Let S = x2 + y2 − 2x − 4 y = 0, At point (0, 1), S1 ≡ 02 + 12 − 2(0) − 4(1), =1 −4 = −3 < 0, , Point lies inside the circle, hence, no tangents, can be drawn., II. Centre of a circle (− 3, − 1)., If the given line is a diameter of the circle, then, centre lies on the line., , At point (3, 4), −3, dy, =, = m (say), , dx (3 , 4) 4, , x′, , C1C 2 = (0 − 1)2 + (− 4 − 2)2 = 1 + 36 = 37, , Now,, , OB = OA = AD = BD = 5, 2, , A (25/3, 0), x, , ∴ Line is x + 3 y = 0, ∴, − 3 + 3 (− 1) = 0, ⇒, −6 ≠0, Hence, given line is not a diameter of the circle., 25. Given, circles are x2 + y2 − 2x − 2 y − 7 = 0, 8, 29, and, x2 + y 2 − x +, y =0, 3, 3, 4 − 29, Centre of circles are C1 (1, 1) and ,, ., 3 6 , C1 = − 7, C 2 = 0, Condition for orthogonally,, 2( g1 g2 + f1 f2) = C1 + C 2, 4, − 29, , 2 1 × + 1 ×, =−7+0, , 3, 6 , 4 29, 2 − =−7, 3 6 , 21, 2 =−7, 6, , ∴, y′, , ⇒, , ⇒, , x, y, +, =1, 25 / 3 25 / 4, , ⇒, , 1 25 25 625, sq units, ×, =, ∴ Area of ∆OAB = ×, 2 3, 4, 24, II. Radius = perpendicular distance from centre (3, − 4), to the line 5x + 12 y − 12 = 0, |5(3) + 12(− 4) − 12| 45, =, =, 13, 25 + 144, , 2, , 2, , 16 841, 905, 905, 4, 29, +, =, =, r2 = + =, 3, 6, 9, 36, 36, 6, 2, , Now,, , 2, , Hence, both statements are correct., 23. I. Centre of circle is C (7, 5)., , r1 = 12 + 12 + 7 = 3, , Here,, , ∴ Equation of circle is, 45, (x − 3)2 + ( y + 4)2 = , 13, , R = 72 + 52 + 151, , ∴, , 2, , 1 35, 4 , − 29 , C1C 2 = − 1 + , − 1 =, + , 3 , 6, , 9 6, 1 1225, 1229 35.06, =, +, =, =, = 5.84, 9, 36, 6, 6, 905, 905 + 18 30.08 + 18, r1 + r2 =, +3=, =, 6, 6, 6, = 8.01, C1C 2 < r1 + r2, , 2
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533, , The Circle, 26. The given equation of circle is, x2 + y2 − 6 x + 2 y = 0, ⇒, x(x − 6) + y( y + 2) = 0, ⇒, (x − 0)(x − 6) + ( y − 0) ( y + 2) = 0, This is the equation of circle in diameter form, where, end points of the diameter are (0, 0) and (6, − 2)., Now, equation of diameter is a line, which passes, through the points (0, 0) and (6, − 2), which is, −2, ( y − 0) =, (x − 0), 6, ⇒, , Centre of circle (x + 1)2 + ( y − 2)2 = 4 is (−1, 2), Centre of circle (x − 2)2 + ( y − 1)2 = 4 is (2 , 1), 30. Since, AB and AD are coordinate axes., a a, The equation of circumcircle whose centre is , and, 2 2, 2, 2, a, a, a, a2, , , , is x − + y − =, radius is, , , 2, 2, 2, 2, y, a, , x + 3y =0, , D, , 27. Let S1 = x + y − 2x − 3 = 0, 2, , 2, , C1 = (1, 0), r1 = 12 + 02 − (− 3) = 10, , a, a, 2 2, , a, , S 2 = x + y − 4x − 6 y −. 64 = 0, 2, , And, , a, , 2, , a/ 2, , C 2 = (2, 3), r2 = 22 + 32 − (− 64) = 77, Now,, , C, , a/ 2 E a/ 2, , A, , x, , B, , C1C 2 = (2 − 1)2 + (3 − 0)2 = 10, ⇒, , |r1 − r2|=| 10 − 77|> 10, ⇒, C1C 2 <|(r1 − r2)|, ∴ One lies inside the other., , x2 + y2 − a (x + y) = 0, , 31. From the figure, it is clear that coordinates of centre of, circle are (1, 1) and radius of circle is 1., y, , 28. The general equation of circle is, x2 + y2 + 2 gx + 2 fy + c = 0, , (1,1), , Since, it passes through (0, 0), it gives c = 0 and it passes, through (1, 0), then, 1, g=−, 2, , 1, (2,0), O, , 2, , y=0, , x, , 2, , ∴ Radius, , 1, R1 = − + f 2 − 0, 2, , 1, + f2, 4, , 1, and its centre, C1 = , − f , , 2, and this circle touches the circle, x2 + y2 = 9, =, , x=0, , (x − 1)2 + ( y − 1)2 = 1, , ∴Its centre C 2 = (0, 0) and radius R2 = 3., Since, the circle is touching the required circle, then, C1C 2 = R2 − R1, 2, , ⇒, ⇒, ⇒, , 1, 1, , 2, + f2, 0 − + (0 + f ) = 3 −, , 2, 4, 1, 1, + f2 =3 −, + f2, 4, 4, 1, 2, + f2 =3, 4, , , 1, ⇒ 4 + f 2 = 9, , 4, , ⇒, , 1, 9, + f2 = ⇒ f2 =2, 4, 4, , ⇒, , f = 2, , 1, C1 = , − 2 ., , 2, , ∴ Centre, , x = +2, , ∴ Equation of circle is, , 29. Centre of circle (x + 2)2 + ( y + 1)2 = 1 is (−2 , − 1), Centre of circle (x − 1)2 + ( y + 2)2 = 1 is (1, − 2), , ⇒, , x2 − 2x + 1 + y2 − 2 y + 1 = 1, , ⇒, , x2 + y 2 − 2 x − 2 y + 1 = 0, , 32. Let the equation of circle is, x2 + y2 + 2 gx + 2 fy + c = 0, This touches x-axis at (3, 0), ∴, , x2 + 2 gx + c = 0, , Let this equation has roots x1 and x2., ∴, , x1 + x2 = − 2 g and x1x2 = c, , At x-axis,, ∴, , x1 = x2 = 3, , 6 = − 2 g ⇒ g = − 3 and 3 ⋅ 3 = c ⇒ c = 9, , Also, y-intercept = 6, ⇒, ⇒, , 2, , f2 − c =6, f2 −9 =9, , ⇒, f 2 = 18, ⇒, f = ±3 2, ∴ Equation of circle is, x2 + y2 − 6x ± 6 2 y + 9 = 0
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534, , NDA/NA Mathematics, Since, the centre of circle (α , β ) lies on the line x = y, ∴, α =β, And these two circles touch externally, then, C1C 2 = r1 + r2, , 33. The equation of circle is, y, P, , x', , O, , x, , ⇒, , 1, , (α − 0)2 + (β − 0)2 = 3 + 1, , ⇒, , α 2 + β 2 = 16, [ Q α = β], α 2 + α 2 = 16, 2, α = 8 ⇒ α = ± 2 2 and β = ± 2 2, ⇒, ∴ The values of α and β are ± 2 2 and ± 2 2,, respectively., , C, (–1, –1), , y', , x + y + 2x + 2 y + 1 = 0, x2 + 2x + 1 + y2 + 2 y + 1 = 1, (x + 1)2 + ( y + 1)2 = 1, 2, , ⇒, ⇒, , 2, , This shows that the centre of circle is (−1, − 1) and, radius is 1. It is clear that the circle touches both the, axes and two tangents can be drawn from the origin., Therefore, the points of contact are (0, − 1) and (−1, 0)., 34. Equation of given circles are, x2 + y2 = 1, and, x2 + y2 − 4x + 3 = 0, , …(i), …(ii), , Since, centre and radius of given circles are C1 (0, 0), 1, and C 2(2 , 0), 1, respectively, ∴, , C1C 2 = (2 − 0)2 + (0 − 0)2 = 2, , 38. Let r1 = r2 = r, A, , r1 + r2 = 1 + 1 = 2, C1C 2 = r1 + r2, , and, ⇒, , r, , This shows that the two circles touch each other, externally. Therefore, the number of common tangents, is 3., 35. Given, equation of circles are, x2 + y2 + x − 2 y = 0, and, x2 + y2 − 2x + y = 0, , , C1C 2 = 1 +, , , 2, , 1, 1, , + − − 1, , , , 2, 2, , r, , C2, , BRAHAMPAL, , …(i), …(ii), , ⇒, C1C 2 = r, Angle between two circles is, r2 + r2 − r2 1, cos θ =, =, 2r × r, 2, ⇒, θ = 60°, ⇒, ∠C1 AC 2 = 60°, ⇒, ∠AC1B = 60° + 60° = 120°, ∴ Angle of the arc of circle A cut by B = 120°, ∴ Ratio of arc length = ratio of their angles, 120 1, =, =, 360 3, 1, , 39. End points of a diameter are (0, 0) and a3 , 3 , , a , , 2, , 9 9, +, 4 4, 9, 3, =, =, 2, 2, 3 2, =, 2, 5, 5, and, +, = 5, r1 + r2 =, 2, 2, ⇒, C1C 2 < r1 + r2, Thus, both circles intersect each other., =, , 36. Equation of circles are, (x − α )2 + ( y − β )2 = 9, and, x2 + y2 = 1, And equation of line is, x= y, , C1, , r, , B, , ∴Centre and radius of circles are, 1 5, 1 5, , and C 21, − ,, , respectively., C1 − , 1 ,, 2 2, , 2 2, Now,, , 37. Let (α , β ) be the centre of the circle., ∴, α 2 + β 2 = (α − a )2 + β 2 = α 2 + (β − b)2, ⇒, α 2 + β 2 = α 2 + β 2 + a 2 − 2 aα, = α 2 + β 2 + b 2 − 2 bβ, 2, 2, ⇒, α + β = α 2 + β 2 + a 2 − 2 aα, a, ⇒, α=, 2, b, Similarly,, β=, 2, ∴ Radius of circle = α 2 + β 2, 1, =, a 2 + b2, 2, , Then, equation of a circle is, , …(i), …(ii), …(iii), , 1, , (x − 0) (x − a3 ) + ( y − 0) y − 3 = 0, , a , y, ⇒, x2 − a3 x + y2 − 3 = 0, a, y, 2, 2, 3, x + y − a x − 3 =0, ⇒, a, 1 , Clearly, , a satisfy the circle, a , 1, 1 a, i.e.,, + a 2 − a3 × − 3 = 0, a a, a2
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535, , The Circle, 1, 1, + a2 − a2 − 2 =0, 2, a, a, 0 =0, , ⇒, , 1, , Hence, circle passes through the point , a ., a, , 40. Let (h,0) be the centre of circle and radius is h., y, , x′, , (h, 0), , O, h, , C, , x, , On solving Eqs. (i) and (ii), we get, h = 2, k = 5, ∴, C (2, 5), Also, radius, r = AC, = (2 − 2)2 + (5 − 3)2 = 2, ∴ Equation of circle, (x − 2)2 + ( y − 5)2 = x2, 2, ⇒, x + y2 − 4x − 10 y + 25 = 0, 45. Let S ≡ x2 + y2 − 4x − 10 y + 25 = 0, S1 ≡ (4)2 + (5)2 − 4(4) − 10(5) + 25, ≡ 16 + 25 − 16 − 50 + 25 = 0, Hence, the point lies on the circle., , At point (4, 5),, , 46. Area of circle = πr 2, = 3.14 × (2)2, = 12.56 sq units, y′, , ∴ The equation of circle is, ⇒, (x − h )2 + y2 = h 2, 2, ⇒, x − 2hx + h 2 + y2 = h 2, ⇒, x2 + y2 − 2xh = 0, 41. The given equation is, 4x2 + 4 y2 + 2x + 8 y + 7 = 0, Since, coefficient of x2 = coefficient of y2, ∴ The given equation represents a circle., ∴ Both A and R are individually true and R is the, correct explanation of A., 42. The length of tangent drawn from the point (1, 2) to a, circle x2 + y2 + 8x + 7 y + 5 = 0, = (1)2 + (2)2 + 8 (1) + 7 (2) + 5, = 1 + 4 + 8 + 14 + 5 = 32, =4 2, ∴ A is false but R is true., 43. The equation of circle is x2 + y2 = 32 and equation of line, is y = 4x + 153, ⇒, 153 = 153 [Q c = a 2(1 + m2)], ∴ The given line is a tangent to the circle x2 + y2 = 32., ∴ Both A and R are true and R is the correct explanation, of A., , Solutions (Q. Nos. 44-47), 44. Let C (h , k) be the centre of circle., ∴, AC = BC, ⇒, (h − 2)2 + (k − 3)2 = (h − 4)2 + (k − 5)2, ⇒ h 2 − 4h + 4 + k2 − 6k + 9 = h 2 − 8h + 16 + k2 − 10k + 25, …(i), ⇒, 4h + 4k = 28, Also, centre lies on a given line,, …(ii), ∴, k − 4h + 3 = 0, , 47. ∴ Equation of required circle is, (x − 4)2 + ( y − 6)2 = ( 3 )2, 2, 2, ⇒, x + y − 8x − 12 y + 16 + 36 = 3, ⇒, x2 + y2 − 8x − 12 y + 49 = 0, , Solutions (Q. Nos. 48-50), 48. Centres of given circles are, 5, , C1 (1, − 3) and C 2 , − 3, 2, , ∴, , r1 = g 2 + f 2 − c = 1 + 9 − 6 = 2, r2 =, , 25, 1, + 9 − 15 =, 4, 2, 2, , 9 3, 5 , C1C 2 = − 1 + (3 − 3)2 =, =, 2 , 4 2, 1 3, Now, r1 − r2 = 2 − =, 2 2, ∴, C1C 2 = r1 − r2, Hence, the circle touch each other internally., ∴, , 49. ∴ Required equation of circles, is, (x − 1)2 + ( y + 3)2 = 52, 2, 2, ⇒, x + y − 2x + 6 y + 9 + 1 = 25, ⇒, x2 + y2 − 2x + 6 y = 15, 50. Intersection of two circles is, x2 + y2 − 2x + 6 y + 6 + λ (x2 + y2 − 5x + 6 y + 15) = 0, At point (1, 2),, 1 + 4 − 2 + 12 + 6 + λ (1 + 4 − 5 + 12 + 15) = 0, ⇒, 21 + λ (27) = 0, 7, λ =−, ⇒, 9, 7, ∴ x2 + y2 − 2x + 6 y + 6 − (x2 + y2 − 5x + 6 y + 15) = 0, 9, ⇒, 2x2 + 2 y2 + 17x + 12 y − 51 = 0
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27, Conic Sections, The locus of a point P which, moves in a plane such that the, distance from a fixed point is always, in a constant ratio to its, perpendicular distance from a, fixed straight line, is known as conic, section., ∴From figure, section of a right, circular cone by a plane parallel to, a generator of the cone is a, parabola., , Definition of Terms Related to, Parabola, Focus, , The fixed point S is called the focus of the, , conic section., , Directrix The fixed straight line NM is called the, directrix of the conic section., Axis The straight line passing through the focus and, perpendicular to the directrix, is known as axis., Eccentricity The constant ratio is called the, eccentricity and is denoted by e., Vertex The point of intersection of the conic section, and the axis is the vertex A of conic section., Latusrectum The chord passing through the focus, and perpendicular to the axis i.e., DE., Focal chord A chord of a parabola passing through, the focus is called a focal chord RS., , Parabola, Locus of all such points which is equidistant from a given, fixed point and from a given fixed line is known as parabola., The straight line perpendicular from the focus to the, directrix is called the axis of the parabola., , Terms Related to the Parabola, 2. y2 = − 4ax, y, , y, N, , 3. x 2 = 4ay, y, , N, , P (x, y), , 4. x 2 = − 4ay, y, N, , (0, ,, , Figures →, , 1. y2 = 4ax, , a), , Parabola →, , Q, , S, , M, x, , x′, , S ′ (–a,0), , O, , x, , O, , Q, , O, y′, , ( −a, 0), x=a, (0, 0), y=0, , ( 0, a), y=−a, (0, 0), y- axis i. e., x = 0, , ), , ( a, 0), x= −a, (0, 0), x-axis i.e., y = 0, , N, , –a, , Focus, Directrix, Vertex, Axis, , ,, (0, , Q, , 1., 2., 3., 4., , x, , x, S, , Terms related, to parabola, ↓, , Q a O a S (a, 0), , ( 0, − a), y=a, (0, 0), x=0
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537, , Conic Sections, , Parabola →, 5., 6., 7., 8., 9., , 1. y2 = 4ax, , Length of, latusrectum, Equation of, latusrectum, Extremities of, latusrectum, Eccentricity, Focal distance, , 2. y2 = − 4ax, , 3. x 2 = 4ay, , 4a, , 4a, , 4a, , 4a, , x=a, , x+ a=0, , y=a, , y+ a=0, , ( a, 2 a) and ( a, − 2 a), , ( −a, 2 a) and ( − a, − 2 a), , ( 2a, a) and ( −2a, a), , ( 2a, − a) and ( −2a, − a), , e =1, x+ a, , e =1, x+ a, , e =1, y+ a, , e =1, y−a, , ⇒, , General Equation of a, Parabola, , ⇒, , The second degree equation, ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 represents, if, abc + 2 fgh − af − bg − ch ≠ 0, ab − h = 0and e = 1, 2, , 2, , 2, , 2, , Example 1. Find the equation of the parabola whose focus, is the point (0,0) and whose directrix is the straight line, 3x − 4y + 2 = 0., (a) 16 x 2 + 9y 2 + 24xy − 12 x + 16y − 4 = 0, (b) 16 x 2 + 9y 2 − 24xy + 12 x + 16y + 4 = 0, (c) 16 x 2 + 9y 2 + 20 xy − 12 x + 16y = 0, (d) None of the above, , 3x – 4 y + 2 = 0, , Solution (a) By definition of parabola,, M, , 2, , ⇒, ⇒, , 16x + 9y + 24xy − 12x + 16y − 4 = 0, , 2, , 2, , 2, , This is the required equation of the parabola., , Example 2. The focus of the parabola y 2 − x − 2y + 2 = 0 is, (b), , 5, 4, , (c), , 4, 5, , Solution (b) We have, y 2 − 2y = x − 2, (y – 1) = 1( x – 1), 2, , Q, ∴, , (d) ( − 2, ± 4), , Solution (c) If the coordinates of a point on the parabola, y 2 = 4ax are P( x, y) , then its focal distance is SP = x + a., Here, a = 2 and SP = 4, , ∴, ⇒, , 4= x+2 ⇒ x=2, y2 = 8 × 2, , ⇒, y=± 4, ∴ Coordinates of required point are (2, ± 4)., , %, , x2 + y 2 =, , 3, 4, , y 2 = 8 x and whose focal distance is 4., (a) (2, − 4), (b) (2, + 4), (c) (2, ± 4), , Parametric form of the parabola y 2 = 4ax is x = at 2 and, y = 2at, where t being the parameter., , (3x − 4y + 2), 25, 25x2 + 25y 2 = 9x2 + 16y 2 + 4 − 24xy + 12x − 16y, , (a), , Example 3. Find the coordinates of a point on the parabola, , Equation of Parabola in, Parametric Form, , S(0, 0), , (3x − 4y + 2) , , ( x − 0) 2 + (y − 0) 2 = , 9 + 16 , , , ⇒, , 1, , y −1= 0, 4, 5, Focus of parabola is ., 4, x −1=, , P(x, y), , SP = PM, SP 2 = PM 2, ⇒, , 4. x 2 = − 4ay, , 4a =1, Focus is X = a, Y = 0, , (d), , 1, 5, , The parametric relation between the coordinates of the ends of a, focal chord of a parabola is, t1 t 2 = − 1 or t 2 = − 1 / t1, If one extremity of a focal chord is (at 12 ,2at1 ), then the other, a 2a , extremity (at 22 ,2at 2 ) becomes ,− ., t12 t1 , , Example 4. Write the parametric equations of the parabola, (y − 1) 2 = 12 ( x + 1)., (a) x = 3t 2 − 1 and y = 6t + 1, (b) x = 3t 2 + 1 and y = 6t − 1, (c) x = 3t 2 − 1 and y = t + 1, (d) None of the above, Solution (a) The given equation of parabola is, (y − 1) 2 = 12 ( x + 1), , …(i), , This parabola is of the form, (y − k) 2 = 4a ( x − h), , …(ii), , On comparing Eqs. (i) and (ii), we get, a = 3, h = −1
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538, , NDA/NA Mathematics, , and, k =1, ∴ The parametric equations of the given parabola are, x + 1 = 3t 2 and y − 1 = 6 t, , Definition of Terms Related, to an Ellipse, , ⇒, , x = 3t 2 − 1, , and, , y = 6t + 1, , Vertices The point A and A', where the curve meets, the line joining the foci S and S', are called the vertices of, an ellipse., Major and minor axes, , Point P ( h , k) lies, outside the parabola, if, ( k2 − 4ah ) > 0. Point P ( h , k) lies on the parabola, if, ( k2 − 4ah ) = 0 and point P ( h , k) lies inside the parabola, if, ( k2 − 4ah ) < 0., , Example 5. The position of point ( −1, 7) lies relative to the, parabola y 2 = 12 x is, (a) outside, (b) inside, (c) on, (d) None of these, Solution (a) Since, k = −1 and, , h =7, , ∴, 49 − 12 > 0, ∴ The given point lies outside the given parabola., , (1,1) and eccentricity, , The locus of a point in a plane which moves in the plane, in such a way that the ratio of its distance from a fixed, point (focus) in the same plane to its distance from a fixed, straight line (directrix) is always constants, which is, always less than unity (one)., y, , Q, , M', , B, , L, , P(x, y), , S', , Z' A', Directrix, , (b) 7x 2 + 7y 2 + 2 xy − 22 x − 10y + 7 = 0, (c) 7x 2 + 7y 2 + 2 xy + 22 x + 10y + 7 = 0, , Centre Major axis, A, C, S, Foci, , Z, , x, , ∴, , SP = e ⋅ PM, , ⇒, , ( x − 1) 2 + (y − 1) 2 =, , Vertex Directrix, Q', , Vertex, L', Double ordinate, , B', Latusrectum, , PS ′, PM ′, i.e.,, PS ′ = e ⋅ PM ′, On squaring both sides of Eq. (i), we get, PS ′ 2 = e2 ⋅ PM ′ 2, , a, ( x − ae) + ( y − 0) = e − x, , e, 2, , x, , 2, , a, , 2, , +, , 2, , y, , 2, , a (1 − e ), 2, , 2, , = 1 or, , b = a (1 − e ), b < a, 2, , 2, , 2, , …(i), , x, a, , 2, , +, , y, , 1 x−y + 3 , , , 2 (1) 2 + ( − 1) 2 , , , , ⇒, , 8 ( x − 1) 2 + 8 (y − 1) 2 = ( x − y + 3) 2, , ⇒, , 8x2 − 16x + 8 + 8y 2 − 16y + 8, , ⇒, , 7x2 + 7y 2 + 2xy − 22x − 10y + 7 = 0, , Which is the required equation of an ellipse., , Example 7. The foci of the ellipse, 25( x + 1) 2 + 9(y + 2) 2 = 225 is, , 2, , 2, , 2, , (by definition), , = x2 + y 2 + 9 − 2xy + 6x − 6y, , ∴ Eccentricity (e) =, , where,, , (a) 7x 2 + 7y 2 + 2 xy − 22 x + 10y − 7 = 0, , Solution (b) Let P ( x, y) be the point on the ellipse, , M, , x', , 1, and directrix x − y + 3 = 0, 2, , (d) None of the above, , Minor axis, , ⇒, , P (outside), , Example 6. Find the equation of an ellipse with focus at, , Ellipse, , ⇒, , y, , P (on), Major axis is the one which lie along, the line passing through focus and, P (inside), perpendicular to directrix and x ′, minor axis is the one which is, perpendicular to major axis and, passes through the mid-point of the, y′, foci., Centre Since, all chords passing through C are, bisected at C. Hence C ( 0, 0) is the centre of an ellipse., Ordinate and double ordinate Let P be a point, on the ellipse and let PN be perpendicular to the major axis, AA' such that PN produced meets the ellipse at P' Then, PN, is called the ordinate of P and PNP' the double ordiante of, P., Latusrectum The double ordinate passing, through an focus is called latusrectum ., x, , Position of a Point with Respect, to a Parabola, , 2, , b2, , =1, , (a) ( − 1, 2), ( − 1, − 6), (c) ( − 1, 2), (1, 6), , (b) (1, 2), ( − 1, − 6), (d) None of these, , Solution (a) The given equation can be rewritten as, ( x + 1) 2 (y + 2) 2, +, =1, 9, 25
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539, , Conic Sections, Which represents an ellipse of centre ( − 1, − 2) and semimajor, and semiminor axes are 5 and 3, respectively., i.e.,, b = 5 and a = 3, ∴ The eccentricity of the ellipse is given by, a2 = b 2 (1 − e2), ⇒, , 9 = 25 (1 − e2) ⇒ e =, , 4, 5, , Shifting the origin at ( −1,−2) , given equation reduces to, X2 Y2, +, =1, 9, 25, where,, x = X − 1and y = Y − 2., ∴ Coordinates of the foci are ( X = 0 , Y = ± be), i.e.,, ( −1, 2) and ( −1,−6), , …(i), …(ii), , Terms Related to an Ellipse, Ellipse →, , x2, 2, , a, Figure →, , +, , y2, 2, , b, , x2, , = 1, a > b, , 2, , a, , x′, , A′ S ′(– ae, 0) C(0, 0)S(ae, 0) A, (a, 0), (– a, 0), , B ′(0, – b), y′, , y2, b2, , = 1, a < b, , y, , y, B (0, b) P(x, y), , +, , A(0, b), , M, x, , S, (0, be), , x′, , (– a, 0)B′, , C (0, 0) B(a, 0), , Terms related to, an ellipse, , ↓, , S ′ be), ,–, (0, A′(0, – b), , y′, , 1., , Centre, , ( 0, 0), , ( 0, 0), , 2., , Vertices, , ( ± a, 0), , ( 0, ± b), , 3., , Length, axis, , of, , major, , 2a, , 2b, , 4., , Length, axis, , of, , minor, , 2b, , 2a, , 5., , Foci, , ( ± ae, 0), , ( 0, ± be), , 6., , Equations, directrices, , x = ± a/ e, , y = ± b/ e, , 7., , Relation in a, b and e, , b2 = a2 (1 − e 2 ), , a2 = b2 (1 − e 2 ), , of, , 2b, a, , b2 , ± ae, ±, , a, , , 2a2, b, a2, , , ± be, ±, b, , , , Parametric, coordinates, , ( a cos φ, b sin φ) , 0 ≤ φ ≤ 2π, , ( a cos φ, b sin φ), 0 ≤ φ ≤ 2π, , 11., , Focal radii, , SP = a − ex1 and S′ P = a + ex1, , SP = b − ey1 and S′ P = b + ey1, , 12., , SP + S′P, , 2a, , 2b, , 13., , Distance, foci, , 2 ae, , 2 be, , 14., , Distance between, directrices, , 2a / e, , 2 b/ e, , 8., , Length, latusrectum, , of, , 9., , Ends, latusrectum, , of, , 10., , between, , 2, , x
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540, , NDA/NA Mathematics, , Example 8. Find the eccentricity and latusrectum of the, ellipse 25 x 2 + 9y 2 − 150 x − 90y + 225 = 0., (a) 17/5, (b) 18/5, (c) 5/18, (d) None of the above, , Position of a Point with Respect to an, Ellipse, y, P(outside), P(on), , B, , Solution (b) The given equation of an ellipse is, 25x2 + 9y 2 − 150 x − 90y + 225 = 0, ⇒, , 25( x2 − 6x) + 9(y 2 − 10y) + 225 = 0, , ⇒, , 25( x2 − 6x + 9) + 9(y 2 − 10y + 25) − 225 = 0, , 25 ( x − 3) 2 9 (y − 5) 2, +, =1, 225, 225, 2, 2, ( x − 3), (y − 5), ⇒, +, =1, 9, 25, Let x − 3 = X and y − 5 = Y , then the equation of an ellipse, X2 Y2, becomes, +, = 1., 9, 25, Here,, a = 3 and b = 5., , P(inside), , x′, , A′, , A, B′, , ⇒, , 9, 4, a2, = 1−, =, 2, 25, 5, b, 2a2 2 × 9 18, and length of latusrectum =, =, =, 5, 5, b, , ∴ Eccentricity = 1 −, , General Equations of an Ellipse, The second degree equation, ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0, represents an, ellipse, if, abc + 2 fgh − af 2 − bg2 − ch 2 ≠ 0, ab − h 2 > 0, e < 1, , Parametric Equations of, an Ellipse, x = a cos θ and y = b sinθ, since (a cos θ, b sin θ ) is, x2 y2, satisfying the equation 2 + 2 = 1, a, b, ∴x = a cosθ and y = b sinθ is the required parametric form., , Example, , y′, , Point P ( h , k) lies outside the ellipse, if, h 2 k2, +, −1> 0, a 2 b2, h 2 k2, Point P ( h , k) lies on the ellipse, if, +, −1= 0, a 2 b2, h 2 k2, Point P ( h , k) lies inside the ellipse, if 2 + 2 − 1 < 0, a, b, , Example 10. The position of a point (3, 4) relative to the, x2 y 2, +, = 1 is, 9 16, (a) inside, (c) on, , ellipse, , (b) outside, (d) None of these, , (3) 2 ( 4) 2, +, −1, 9, 16, 9 16, = +, −1, 9 16, =1+ 1−1=1> 0, ∴ The point P(3, 4) lies out side the ellipse., , Solution (b) Since, S1 =, , Auxiliary Circle, The circle which is described on the major axis AA′ of, an ellipse as diameter is called the auxiliary circle., y, , 9. The curve with parametric equations, , x = 1 + 4 cosθ and y = 2 + 3 sinθ is a/an, (a) circle, (b) parabola, (c) ellipse, (d) hyperbola, , Q', Q, B, , Solution (c) We have, x = 1 + 4 cos θ and y = 2 + 3 sin θ, 2, , ∴, , A, , P', , C N', , P, N, , A', , 2, , y − 2, x − 1, =1, + , , 3 , 4 , , ( x − 1) 2 (y − 2) 2, +, =1, 16, 9, Which is an ellipse., , x, , ⇒, , ⇒, , PN, BC b, =, =, QN, AC a, , x
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541, , Conic Sections, , %, , %, , P (h, k), , The sum of the focal distances of any point on an ellipse is a, constant and is equal to the length of the major axis of the ellipse, i.e., SP + S ′ P = 2 a., , a, Since, S ′ P = ePN ′ = e (CQ ′ + CS) = e + x = a + ex, , e, Also,, SP = ePN = e ( CQ − CS), , a, = e − x = a − ex, , e, Thus, S' P = a + ex and SP = a − ex, x2, 2, , +, , B, A′, , y2, , = 1 formed by, a, b2, drawing PM perpendicular from P on the major axis and produce, MP to meet the auxiliary circle in Q, ∴The ∠ XCQ = φ is called eccentric angle., Eccentric angle of a point on the ellipse, , 90°, , i.e., Equation of director circle is x 2 + y 2 = a 2 + b2, , Example 12. The locus of the point of intersection of the, perpendicular tangents to ellipse, , y, , A, , C, B′, , (a) x 2 + y 2 = 4, (c) x 2 + y 2 = 5, , Q, , x2 y 2, +, = 1 is, 9, 4, 2, (b) x + y 2 = 9, (d) x 2 + y 2 = 13, , P, x', , Solution (d) The locus of the point of intersection of the, , φ, , C, , M, , x, , x2 y 2, +, = 1 is a director circle, a2 b 2, and whose equation is given by x2 + y 2 = a2 + b 2, , perpendicular tangents to ellipse, , ∴Required equation of director circle of an ellipse, x2 y 2, +, = 1 is x2 + y 2 = 9 + 4 = 13, 9, 4, , y´, , Example 11. The line lx + my + n = 0 will cut the ellipse, π, + 2 = 1 in points whose eccentric angles differ by , if, 2, 2, a, b, (a) l 2a 2 + m 2b 2 = 2n2, (b) l 2a 2 − m 2b 2 = 2n2, (c) l 2a 2 + m 2b 2 = n2, (d) None of the above, x2, , y2, , Hyperbola, A hyperbola is the locus of a point in a plane which, moves in a plane in such a way that the ratio of its distance, from a focus to a fixed line (directrix) is always greater, than unity., y, Directrix, , Directrix, , Solution (a) Let the line lx + my + n = 0 cuts the ellipse at two, , ∴, la cos θ + mb sin θ + n = 0, and, − la sin θ + mb cos θ + n = 0, ⇒, la cos θ + mb sin θ = − n, and, − la sin θ + mb cos θ = − n, Now, (la cos θ + mb sin θ) 2 + ( − la sin θ + mb cos θ) 2 = n 2 + n 2, , La, x', , S'(– ae, 0), um, e ct, u sr, Lat, , A', , l 2a2 + m2b 2 = 2n 2, , Director Circle of an Ellipse, The locus of the point of intersection of the tangent to, x2 y2, an ellipse 2 + 2 = 1, which are perpendicular to each, a, b, other is called director circle., , A, (a, 0), , (– a, 0), , x = –a/e, , Double, ordinate, , x = a/e, , ∴, , SP = e ⋅ PM, , ⇒, , SP 2 = e2 ⋅ PM 2, , ⇒, , a, , ( x − ae)2 + y 2 = e2 x − , , e, , ⇒, , x 2( e2 − 1) − y 2 = a 2( e2 − 1), x2, a, , 2, , −, , y2, a ( e2 − 1), 2, , ctu, , S(ae, 0), , y', , ⇒, , tu, , e, sr, , x, , C, , l 2a2 (sin 2 θ + cos2 θ) + m2l 2 (cos2 θ + sin 2 θ) = 2n 2, ⇒, , P, , M, , M', , Centre, , points P and Q whose coordinates are ( a cos θ , b sin θ) and, , , π, , π, a cos + θ , sin + θ , respectively., , , 2, 2, , =1, , 2, , m
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542, , NDA/NA Mathematics, x2, , ⇒, , a, , where,, , 2, , −, , y2, , Double ordinates If Q be a point on the, hyperbola draw QN perpendicular to the axis of the, hyperbola and produced to meet the curve again at Q’., Then, QQ' is called a double ordinate of Q., Latusrectum The double ordinate passing, through focus is called latusrectum., , = 1 (standard form of hyperbola), b2, b2 = a 2 ( e2 − 1), , Definitions of Terms Related, to Hyperbola, , Parametric Form of Hyperbola, , Vertices The point A and A' where the curve meets, , x = a sec θ and y = b tanθ is the parametric form of, hyperbola., , the line joining the foci S and S', are called the vertices of, the hyperbola., Transverse and conjugate axes Transverse, axis is the one which lie along the line passing through the, foci and perpendicular to the directrices. Conjugate axis is, the one which is perpendicular to the transverse axis and, passes through the mid-point of the foci (i.e., centre)., Centre The mid-point C of AA' bisects every chord, of the hyperbola passing through it and is called the centre, of the hyperbola., , Focal Distance of a Point, The difference of the focal distances of any point on the, hyperbola is constant and is equal to the length of the, transverse axis of the hyperbola., S ′ P – SP = transverse axis., , Terms Related to the Hyperbola, Figure → Hyperbola, , x 2 y2, –, =1, a 2 b2, , Hyperbola, , x 2 y2, –, =–1, a 2 b2, y, , Figure →, , y, , S, (ae, 0), B'(0, b), y = b/e, , Terms related to hyperbola, , X', , O, , S(– ae, 0), A', (– a, 0), , A, (a, 0), , x, S' ( ae, 0), , x, , O, , y = – b/e, B'(0, – b), , x = – a/e x = a/e, , S', (– ae, 0), , 1., 2., 3., 4., 5., 6., , Centre, Foci, Length of transverse axis, Length of conjugate axis, Equations of directrices, Eccentricity, , 7., , Length of latus rectum, , 8., , Parametric coordinates, , 9., 10., 11., 12., , Focal radii, Difference of focal radii, Equation of transverse axis, Equation of conjugate axis, , (0 , 0 ), ( ± ae, 0), 2a, 2b, x=± a / e, , ( 0, 0), ( 0, ± be), 2b, 2a, y = ± b/ e, , a2 + b2, a2, 2, 2b, a, (a sec φ , b tan φ), 0 ≤ φ < 2π, SP = ex1 − a and S′ P = ex1 + a, 2a, y=0, x=0, , a2 + b2, b2, 2, 2a, b, ( a tan φ, b sec φ), 0 ≤ φ < 2π, SP = ey1 − b and S′ P = ey1 + b, 2b, x=0, y=0, , e=, , e=
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543, , Conic Sections, , Example 13. Find the equation of the hyperbola whose, directrix is 2 x + y = 1, focus (1, 2) and eccentricity 3., (a) 7x 2 − 2y 2 + 12 xy − 2 x + 14y − 22 = 0, (b) 7x 2 + 2y 2 + 12 xy − 2 x + 14y + 22 = 0, (c) 7x 2 − 2y 2 + 12 xy + 2 x + 14y = 0, (d) None of the above, , Solution (c) Let the equation of the hyperbola be, x2 y 2, −, =1, a2 b 2, The coordinates of the foci are ( ae, 0) and ( −ae, 0)., ∴, 2ae = 16 ⇒ 2a 2 = 16 ⇒ a = 4 2, b 2 = a2( e2 − 1) = 32 (2 − 1) = 32, , Also,, , (Q a = 4 2 and e = 2), , Solution (a) Let P ( x, y) be a point on the hyperbola, then, SP = e ⋅ PM, , a = 32 and b = 32, x2 y 2, ∴ The required equation is, −, =1, 32 32, Thus,, , (by definition), P (x, y), , M, , 2, , 2, , 2x+y=1, , ⇒, , ⇒, ⇒, ⇒, ⇒, , S (1, 2), , General Equation of a, Hyperbola, ( x − 1) + (y − 2) = 3, 2, , 2, , The second degree equation, ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0, represents a hyperbola, if, abc + 2 fgh − af 2 − bg2 − ch 2 ≠ 0, , 2x + y − 1, 2 2 + 12, , (2x + y − 1) 2, 5, 5 [( x − 1) 2 + (y − 2) 2] = 3 (2x + y − 1) 2, ( x − 1) 2 + (y − 2) 2 = 3, , ab − h 2 < 0 and e > 1., , 5x2 + 5y 2 − 10 x − 20y + 25, = 3 ( 4x2 + y 2 + 1 + 4xy − 4x − 2y),, , ⇒, , x2 − y 2 = 32, , 7x2 − 2y 2 + 12xy − 2x + 14y − 22 = 0, , Position of a Point with Respect, to the Hyperbola, y, , P (on), , This is the required equation of the hyperbola., x′, , Example 14. The eccentricity of the conic represented by, x − y − 4x + 4y + 16 = 0 is, (b) − 2, (a) 2, 2, , x, , 2, , (c) 3, , (d) 5, , y′, , Solution (a) The equation of conic is, , Let the equation of hyperbola be, , x − y − 4x + 4y + 16 = 0, 2, , ⇒, , 2, , ( x2 − 4x) − (y 2 − 4y) = 16, , ⇒, , ( x − 4x + 4) − (y − 4y + 4) = − 16, , ⇒, , ( x − 2) 2 − (y − 2) 2 = − 16, , 2, , 2, , ( x − 2) 2 (y − 2) 2, −, =1, 42, 42, Let, x − 2 = X and y − 2 = Y, X2 Y2, ∴ The equation becomes 2 − 2 = 1, 4, 4, ⇒, , ∴, , e=, , 1+, , 42, a2, = 1+ 2 = 2, 2, 4, b, , Example 15. Find the equation of the hyperbola whose, eccentricity is 2 and the distance between the foci is 16,, taking transverse and conjugate axes of the hyperbola as x, and y-axes, respectively., (b) x 2 + y 2 = 32, (a) x 2 − y 2 = 16, 2, 2, (c) x − y = 32, (d) None of these, , x2, a, , 2, , −, , y2, b2, , = 1 and, , coordinates of P are ( h , k)., Now, P will lie outside, on or inside the hyperbola, h 2 k2, according as 2 − 2 − 1 < , = , > 0., a, b, %, , The line y = mx + c will be a tangent to the hyperbola, x 2 y2, −, = 1, if c 2 = a 2 m 2 − b 2 and point of contact is, a2 b2, a2m, b2 , ,±, ±, , c, c , , , Example 16. The position of the point (2,3) with respect, to the hyperbola, , x2, , 5, , 2, , −, , y2, , 32, , = 1 is, , (a) inside, (c) on, , (b) outside, (d) None of these, , 4 32, 4, −46, −, − 1=, −2 =, <0, 25 3 2, 25, 25, ∴ The point lies inside the hyperbola., , Solution (a) Since,, , S1 =
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544, , NDA/NA Mathematics, , Comprehensive Approach, n, n, n, , n, , n, , n, , n, , n, , n, , n, , n, , n, , n, n, , n, , n, , For PQ to be focal chord t1t 2 = − 1, Length of focal chord having t1 and t 2 as end points is a(t 2 − t1) 2., Angle between tangents at two points P( at12 , 2 at1), Q( at 22 , 2 at 2) on, t −t , the parabola y 2 = 4 ax is given by θ = tan−1 2 1 ., 1 + t1t 2 , x = ly 2 + my + n is a parabola with it’s axis parallel to x-axis and, 4nl − m2 m , it’s vertex is , ,– ., 4l, 2l , , If the distance between directrix and latusrectum is 2a, then, length of latusrectum is 4a., The locus of the points of intersection of perpendicular tangents to, the parabola y 2 = 4 ax is its directrix., An equilateral triangle is inscribed in the parabolay 2 = 4 ax whose, vertices are at the parabola, then the length of its side is equal to, 8 a 3., The angle of intersection of the parabolas y 2 = 4 ax and x2 = 4 by, 3a1/3 b1/3, ., is given by tan−1, 2 ( a 2/3 + b 2/3 ), The length of the intercept made by the line y = mx + c on the, 4, parabola y 2 = 4 ax is 2 a (1 + m2) ( a − mc) ., m, The orthocentre of the triangle formed by three tangents to the, parabola lies on the directrix., y = ax2 + bx + c is a parabola with it’s axis parallel to y-axis and, D, b, 2, it’s vertex is −, ,−, , where D = b − 4 ac., 2a, 4a , For the ends of latusrectum of the parabola y 2 = 4 ax, the values of, the parameter are ± 1., Tangents drawn from any point on the directrix are at right angle., The tangents at the ends of any chord of parabola meet on the, diameter which bisects the chord., The tangent at one extremity of a focal chord of a parabola is, parallel to the normal at the other extremity., Any tangent is the polar of its point of contact., , n, , n, , n, , n, , n, , n, , n, n, , n, n, , n, , n, , n, , The sum of focal distances of any points on the ellipse is equal to, the major axis., Two tangents can be drawn from a point to an ellipse. The two, tangent are real and distinct or coincidental or imaginary, according as the given points lies outside or inside the ellipse., x2 y 2, The equation of director circle of an ellipse 2 + 2 = 1 is, a, b, x2 + y 2 = a 2 + b 2., The product of perpendiculars from the foci on any tangent to the, x2 y 2, ellipse 2 + 2 = 1 is equal to b 2., a, b, Two tangents can be drawn from a point to a hyperbola., x2 y 2, The equation of the director circle of the hyperbola 2 − 2 = 1is, a, b, x2 + y 2 = a 2 − b 2., Difference of the focal distances = 2a (length of transverse axis), Locus of the point of intersection of perpendicular tangents to the, x2 y 2, hyperbola 2 − 2 = 1 is a circle ., a, b, 2, x + y 2 = a 2 − b 2., Centre of the hyperbola is the mid-point of line joining two foci., The locus of the point of intersection of mutually perpendicular, x2 y 2, tangents to the hyperbola 2 − 2 = 1 is a circle., a, b, The locus of the point of intersection of the lines, ax sec θ + by tan θ = a and ax tan θ + by sec θ = b, where θ is the, parameter, is a hyperbola., If the line l x + my + n = 0 is a tangent to the hyperbola, x2 y 2, −, = 1, then a 2l 2 − b 2m2 = n2., a2 b 2, The product of lengths of perpendiculars drawn from foci on any, x2 y 2, tangent to the hyperbola 2 − 2 = 1 is b 2., a, b
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Exercise, Level I, 1. The focus of the parabola ( y − 3)2 = 4x is, (a) (–1, –3), (b) (–1, 3), (c) (1, –3), (d) (1, 3), , 11. What is the focal distance of any point P ( x1 , y1 ) on the, parabola Y 2 = 4ax?, (NDA 2011 II), , 2. For the parabola y 2 − 8 y − x + 19 = 0, the focus and, directrix are, 11, 13 , 19 , (b) , 8 and y = 7, (a) , 4 and x =, 4 , 7 , 4, 7 , (c) , 3 and y = 9, 2 , 3. The equation of, x 2 − 4 y + 8 = 0 is, (a) y = 0, (c) x = 0, , (d) ( 6, 3) and x = 7, the, , axis, , of, , the, , parabola, , (b) y = 2, (d) x = 2, , 4. The equation of the parabola whose focus is (–3, 0), and the directrix x + 5 = 0 is, (a) y 2 = − 4( x + 4), (b) y 2 = 4 ( x + 4), (c) y = 4 ( x − 4), 2, , (d) y = − 4 ( x − 4), 2, , 5. The length of the latusrectum of the parabola, x 2 − 4x − 8 y + 12 = 0 is, (a) 4, (b) 6, (c) 8, (d) 10, 6. The equation of the parabola with its vertex at the, origin, axis on the y-axis and passing through the, point ( 6, − 3), is, (b) x 2 = 12 y, (a) y 2 = 12x + 6, (c) x 2 = − 12 y, , (d) y 2 = − 12x + 6, , 7. The equation of parabola whose focus is (5, 3) and, directrix is 3x − 4 y + 1 = 0, is, (a) ( 4x + 3 y )2 − 256x − 142 y + 849 = 0, (b) ( 4x − 3 y )2 − 256x − 142 y + 849 = 0, (c) ( 3x + 4 y )2 − 142x − 256 y + 849 = 0, (d) ( 3x − 4 y )2 − 256x − 142 y + 849 = 0, 8. The point of the parabola y 2 = 18x for which the, ordinate is three times the abscissa, is, (a) (6, 2), (b) ( − 2, − 6), (c) ( 3, 18), (d) (2, 6), 9. The directrix of the parabola x 2 − 4x − 8 y + 12 = 0 is, (a) x = 1, (b) y = 0, (c) x = − 1, (d) y = − 1, 10. The equation of the parabola with vertex at the origin, and directrix y = 2 is, (b) y 2 = − 8x, (a) y 2 = 8x, 2, (c) y = 8x, (d) x 2 = − 8 y, , (a) x1 + y1, (c) ax1, , (b) x1 y1, (d) a + x1, , 12. If a focal chord of the parabola y 2 = ax is, 2x − y − 9 = 4 = 0, then the equation of the directrix is, (a) x + 4 = 0, (b) x − 4 = 0, (c) y − 4 = 0, (d) y + 4 = 0, 13. If the line x + y − 1 = 0, is a tangent to the parabola, y 2 − y + x = 0, then the point of contact is, (a) ( 0, 1), (b) (1, 0), (c) ( 0, − 1), (d) ( − 1, 0), 14. The equation of directrix ( x − 1)2 = 2 ( y − 2) is, (a) 2 y + 3 = 0, (b) 2x + 1 = 0, (c) 2 y − 3 = 0, (d) 2 y − 1 = 0, 15. The point at which the line y = mx + c touches the, parabola y 2 = 4ax is, 2a , a 2a , a, (a) 2 , , (b) 2 , −, , m m, m, m, 2a , a 2a , a, (c) − 2 , , (d) − 2 , −, , m m, m, m, 16. The focal distance of a point on the parabola y 2 = 8x, is 4. Its ordinates are, (a) ± 1, (b) ± 2, (c) ± 3, (d) ± 4, 17. What is the length of the smallest focal chord of the, parabola y 2 = 4ax?, (a) a, (b) 2a, (c) 4a, (d) 8a, 18. Which of the following is the correct set of the, parametric coordinates of the parabola y 2 = ax ?, (a) ( at 2 , 2at ), , at , , (b) at 2 , , , 2, , at 2, at , (c) , ,− , 2, 4, , at 2, , (d) , , − at, 2, , , 19. What is the slope of the normal at the point ( at 2 , 2 at ), of the parabola y 2 = 4ax ?, (a), , 1, t, , (b) t, , (c) − t, , (d) −, , 1, t, , 20. What is the area of the triangle formed by the lines, joining the vertex of the parabola x 2 = 12 y to the end, of the latusrectum?, (NDA 2011 II), (a) 9 sq units, (b) 12 sq units, (c) 14 sq units, (d) 18 sq units
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546, , NDA/NA Mathematics, , 21. The eccentricity of the ellipse 25 x 2 + 16 y 2 = 400 is, 3, 1, (a), (b), 5, 3, 2, 1, (d), (c), 5, 5, , 30. The, curve, represented, by, the, 4x 2 + 16 y 2 − 24x − 32 y − 12 = 0 is, (a) a parabola, (b) a pair of straight lines, (c) an ellipse with eccentricity 1/2, (d) an ellipse with eccentricity 3/ 2, , 22. The sum of focal distance of any point on the ellipse, with major and minor axes as 2a and 2b respectively,, is equal to, a, (a) 2a, (b) 2, b, b, b2, (d), (c) 2, a, a, , 31. The latusrectum of the ellipse 9x 2 + 16 y 2 = 144 is, 11, (a) 4, (b), 4, 7, 9, (c), (d), 2, 2, , 23. The, eccentricity, of, 4x 2 + 16 y 2 − 24x − 32 y = 1 is, 1, (b) 3, (a), 2, 3, 3, (d), (c), 4, 2, , the, , conic, , 24. The distance between the foci of an ellipse is 16 and, 1, eccentricity is . Length of the major axis of the, 2, ellipse is, (a) 8, (b) 64, (c) 16, (d) 32, 1, 25. Equation of the ellipse with eccentricity and foci at, 2, ( ± 1, 0) is, x2 y2, x2 y2, (a), (b), +, =1, +, =1, 3, 4, 4, 3, x2 y2 4, (d) None of these, (c), +, =, 4, 3 3, 26. If the latusrectum of an ellipse is equal to half its, minor axis, then what is its eccentricity? (NDA 2010 II), 1, (a), (b) 3, 2, 3, 1, (c), (d), 2, 2, 27. In an ellipse 9x 2 + 5 y 2 = 45, the distance between the, foci is, (a) 4 5, (b) 3 5, (c) 3, (d) 4, 28. The locus of a point which moves such that the, difference of its distances from two fixed points is, always a constant is, (a) a straight line, (b) a circle, (c) an ellipse, (d) a hyperbola, 29. What is the eccentricity of an ellipse, if its, latusrectum is equal to one-half of its minor axis?, (NDA 2009 I), , 1, (a), 4, (c), , 3, 4, , 1, (b), 2, (d), , 3, 2, , 32. The, eccentricity, of, 25x 2 + 16 y 2 − 150x − 175 = 0 is, 2, 2, (b), (a), 5, 3, 4, 3, (d), (c), 5, 5, , the, , equation, , ellipse, , 33. If a bar of given length moves with its extremities on, two fixed straight lines at right angles, then the locus, of any point on bar marked on the bar describes a/an, (a) circle, (b) parabola, (c) ellipse, (d) hyperbola, 34. What is the sum of focal radii of any point on an, ellipse equal to, (NDA 2009 I), (a) length of latusrectum, (b) length of major axis, (c) length of minor axis, (d) length of semi-latusrectum, 35. The equation of the ellipse with foci at ( ±5, 0) and, 36, as one directrix, is, x=, 5, 2, x, y2, x2 y2, (b), (a), +, =1, +, =1, 3, 5, 36 11, 2, 2, 2, 2, x, y, x, y, (c), (d), +, =1, +, =1, 36, 9, 11 36, 36. The eccentricity of the ellipse whose major axis is, three times the minor axis, is, 2, 3, 2 2, 2, (a), (b), (c), (d), 3, 2, 3, 3, 37. If the foci are ( ±1, 0) and the major axis is 6, then the, equation of the ellipse in its standard form is given by, x2 y2, x2 y2, (a), (b), +, =1, +, =1, 8, 9, 9, 8, 2, 2, 2, 2, ( x − 2), ( y − 1), x, y, (c), +, = 1 (d), +, =1, 9, 8, 3, 4, 38. What does an equation of the first degree containing, one arbitrary parameter passing through a fixed, point represent?, (NDA 2009 I), (a) Circle, (b) Straight line, (c) Parabola, (d) Ellipse
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547, , Conic Sections, 2, 1 1, = − + cosθ is a/an, r, 2 4, (a) parabola, (b) hyperbola, (c) ellipse, (d) straight line, , 39. The curve, , 48. Locus of the point of intersection of straight lines, x y, x y 1, is, − = m and + =, a b, a b m, (a) an ellipse, (b) a circle, (c) a hyperbola, (d) a parabola, , 40. What is the eccentricity of the ellipse whose length of, minor axis is equal to the distance between the two, foci?, 2, 1, (b), (a), 3, 2, 1, 2, (d), (c), 3, 2, , 49. The length of transverse axis of the hyperbola, 3x 2 − 4 y 2 = 32 is, 8 2, 16 2, (a), (b), 3, 3, 3, 64, (c), (d), 32, 3, , 41. If the latusrectum of an ellipse is equal to one-half its, minor axis, what is the eccentricity of the ellipse?, , 50. Any point on the hyperbola, , (NDA 2012 I), , 1, (a), 2, 3, (c), 4, , 3, (b), 2, 15, (d), 4, , 42. Which one of the following points lies outside the, ellipse ( x 2 / a 2 ) + ( y 2 / b2 ) = 1 ?, (a) ( a , 0), (b) ( 0, b), (c) ( − a , 0), (d) ( a , b), 43. In how many points do the ellipse, circle x 2 + y 2 = 9 intersect?, (a) One, (c) Four, , x2 y2, +, = 1 and the, 4, 8, (NDA 2007 II), , (b) Two, (d) None of these, , 44. The foci of the hyperbola 2x 2 − 3 y 2 = 5 is, , 5, (a) ±, , 0, , 6 , , 5 , (c) ±, , 0, 6, , , , (d) None of these, , 45. The length of latusrectum, 16x 2 − 9 y 2 = 144 is, 16, 3, 8, (c), 3, , (a), , of, , the, , hyperbola, , 32, 3, 4, (d), 3, , (b), , 46. What is the eccentricity of the conic 4x 2 + 9 y 2 = 144 ?, ( NDA 2012 I), , 5, (a), 3, 3, (c), 5, , the form, (a) ( 4 sec θ , 2 tan θ ), (b) ( 4 sec θ + 1, 2 tan θ − 2), (c) ( 4 sec θ − 1, 2 tan θ − 2), (d) ( 4 sec θ − 1, 2 tan θ + 2), 1 1 3, 51. Equation = +, cosθ represents, r 8 8, (a) a circle, (b) a hyperbola, (c) an ellipse, (d) a parabola, 52. What are the points of intersection of the curve, 4x 2 − 9 y 2 = 1 with its conjugate axis?, (NDA 2011 I), (a), (b), (c), (d), , (1 / 2, 0) and ( − 1 / 2, 0), ( 0, 2) and ( 0, − 2), ( 0, 3) and ( 0, − 3), No such point exists, , 53. The, locus, of, the, x cos α − y sin α = a and x, (a) ellipse, (c) parabola, , 5 , (b) ± , 0, 6 , , 5, (b), 4, 2, (d), 3, , 47. The distance between the directrices of a rectangular, hyperbola is 10 units, then distance between its foci, is, (a) 10 2, (b) 5, (d) 20, (c) 5 2, , ( x + 1)2 ( y − 2)2, −, = 1 is of, 16, 4, , intersection point, of, sin α − y cos α = b is a/an, (b) hyperbola, (d) None of these, , 54. Which one of the following is correct?, If the equation ax 2 + 2 hxy + by 2 + 2 g x + 2 fy + c = 0, represents a hyperbola, then, (a) ∆ ≠ 0, h 2 > ab, (b) ∆ ≠ 0, h 2 < ab, a + b = 0, (c) ∆ ≠ 0, h = ab, a + b = 0, (d) None of the above, 55. The equation 2x 2 − 3 y 2 − 6 = 0 represents a/an, (a) circle, (b) parabola, (c) ellipse, (d) hyperbola, 56. The eccentricity of the hyperbola 25x 2 − 9 y 2 = 144 is, 34, 34, (a), (b), 12, 3, 6, 3, (d), (c), 34, 34, 57. What is the difference of the focal distances of any, point on the hyperbola?, (a) Eccentricity, (b) Distance between foci, (c) Length of transverse axis, (d) Length of semi-transverse axis
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548, , NDA/NA Mathematics, , Level II, 1. The equation of latusrectum of a parabola is x + y = 8, and the equation of the tangent at the vertex is, x + y = 12, then length of the latusrectum is, (a) 4 2, (b) 2 2, (c) 8, (d) 8 2, 2. Equation of the parabola with its vertex at (1, 1) and, focus ( 3, 1) is, (a) ( x − 1)2 = 8 ( y − 1), (b) ( y − 1)2 = 8 ( x − 3), (c) ( y − 1)2 = 8 ( x − 1), (d) ( x − 3)2 = 8 ( y − 1), 3. If (0, 6) and (0, 3) are, respectively the vertex and, focus of a parabola, then its equation is, (a) x 2 + 12 y = 72, (b) x 2 − 12 y = 72, 2, (c) y − 12x = 72, (d) y 2 + 12x = 72, 4. The tangents drawn from the ends of latusrectum of, y 2 = 12x meets at, (a) directrix, (b) vertex, (c) focus, (d) None of these, 5. An equilateral triangle is inscribed in a parabola, y 2 = x whose one vertex is the vertex of the parabola., What is the length of side of the triangle?, (b) 2 3 units, (a) 3 units, (d) 1 unit, (c) 3 3 units, 6. Two equal parabolas have the same vertex and their, axes are at right angles. What is the angle between, the tangents drawn to them at their point of, intersection (other than vertex)?, π, π, 3, (a), (b) tan−1 2 (c), (d) tan−1 , 4, 4, 3, 7. Consider the following statements, I. Chords parallel to latusrectum of the parabola, y 2 = 4ax are called double ordinate., II. Equation of the chord joining points P ( at12 ,2at1 ), and Q( at22 , 2at2 ) is ( t1 + t2 ) y = 2x + 2a t1t2., Which of the statements is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 8. Consider the following statements, I. The orthocentre of the triangle formed by three, tangents to the parabola lies on the directrix., II. Tangents drawn at the ends of any focal chord, intersect at directrix of the parabola., III. The locus of the points of intersection of, perpendicular tangents to the parabola is its, directrix., Which of the statements given above are correct ?, (a) I and II, (b) II and III, (c) III and I, (d) All I, II and III, 9. Consider the parabolas S1 ≡ y 2 − 4ax = 0 and, S 2 ≡ y 2 − 4bx = 0, S 2 will contain S1, if, (NDA 2008 II), (a) a > b > 0, (b) b > a > 0, (c) a > 0, b < 0 but| b|> a (d) a < 0, b > 0 but b >| a|, , 10. What is the length of the focal distance from the point, t of the parabola y 2 = 4ax?, (b) a (1 + t 2 ), (a) at 2, , (c) a t +, , , 1, , t, , 2, , (d) at −2, , 11. What are coordinates of the point of intersection of, 1, the line y = mx +, and the curve y 2 = 4x?, m, 2 , 1 2, 1, (a) 2 , , (b) 2 , 2 , m m, m m , 1 2 , 1 2, (c) , , (d) , 2 , m m , m m, 12. What is the length of the chord of the parabola, y 2 = 9x, which is inclined at an angle of 45° to the axis, of the parabola and passes through the point (1, 3)?, (a) 2, (b) 2 2, (c) 3 2, (d) 4 2, 13. P ( 2 , 2) is a point on the parabola y 2 = 2 x and A is its, vertex. Q is another point on the parabola such that, PQ is perpendicular to AP. What is the length of PQ?, (a) 2, (b) 2 2, (d) 6 2, (c) 4 2, 14. What is the equation to the parabola, whose vertex, and focus are on the x-axis at distances a and b from, the origin, respectively? ( b > a > 0), (NDA 2007 I), 2, 2, (a) y = 8( b − a )( x − a ), (b) y = 4( b + a )( x − a ), (c) y 2 = 4( b − a )( x + a ), (d) y 2 = 4( b − a )( x − a ), 15. In an ellipse length of minor axis is 8 and eccentricity, 5, . The major axis is, is, 3, (a) 6, (b) 12, (c) 10, (d) 16, 16. The eccentricity of the curve 2x 2 + y 2 − 8x − 2 y + 1 = 0, is, 1, 1, (a), (b), 2, 2, 2, 3, (d), (c), 3, 4, 17. The locus of the point of intersection of the, x2 y2, perpendicular tangents to the ellipse, +, = 1 is, 9, 4, (b) x 2 + y 2 = 4, (a) x 2 + y 2 = 9, 2, 2, (c) x + y = 13, (d) x 2 + y 2 = 5, 18. If ( 4, 0) and ( − 4, 0) are the foci of an ellipse and the, semi-minor axis is 3, then the ellipse passes through, which one of the following points?, (NDA 2010 I), (a) ( 2, 0), (b) ( 0, 5), (c) ( 0, 0), (d) ( 5, 0)
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549, , Conic Sections, 19. Equation, , of, , the, , circle passing through the, x2 y2, x2 y2, intersection of ellipses 2 + 2 = 1 and 2 + 2 = 1 is, a, b, b, a, (a) x 2 + y 2 = a 2, (b) x 2 + y 2 = b2, 2a 2b2, (c) x 2 + y 2 = 2, (d) x 2 + y 2 = 1, a + b2, , 20. On the ellipse 4x 2 + 9 y 2 = 1, the point at which the, tangent are parallel to 8x = 9 y are, 1, 2 1, 2 1, 2 1, 2, (a) , or − , − (b) − , or , − , 5 5, 5 5, 5 5, 5, 5, 2, (c) − , −, 5, , 1, , 5, , 3 2, 3 2, (d) − , − or , , 5 5, 5 5, , 21. The angle between the pair of tangents drawn from, the point (1, 2) to the ellipse 3x 2 + 2 y 2 = 5 is, (b) tan−1 ( 6 / 5 ), (a) tan−1 (12 / 5), −1, (d) tan−1 ( 6 / 5), (c) tan (12 / 5 ), 22. A circle is drawn with the, x2 y2, +, = 1 at the end of the, a 2 b2, equation of the circle?, (b), (a) x 2 + y 2 = a 2 + b2, (d), (c) x 2 + y 2 = 2( a 2 + b2 ), , two foci of an ellipse, diameter. What is the, (NDA 2010 I), , x 2 + y 2 = a 2 − b2, x 2 + y 2 = 2( a 2 − b2 ), , x2 y2, +, = 1. If, 25 16, P is variable point on the ellipse and if ∆ is area of the, triangle PSS ′, then the maximum value of ∆ is, (a) 8, (b) 12, (c) 16, (d) 20, , 23. Suppose S and S ′ are foci of the ellipse, , 24. If the angle between the lines joining the end points, of minor axis of an ellipse with its foci is π / 2, then the, eccentricity of the ellipse is, (a) 1/ 2, (b) 1 / 2, (c) 3 / 2, (d) 1 / 2 2, 25. If M1 and M 2 are the feet of the perpendiculars from, x2 y2, the foci S1 and S 2 of the ellipse, +, = 1 on the, 9 16, tangent at any point P on the ellipse, then, (S1M1 ) (S 2 M 2 ) is equal to, (a) 16, (b) 9, (c) 4, (d) 3, 26. The length of the axes, 9x 2 + 4 y 2 − 6x + 4 y + 1 = 0 are, 1, 2, 2, (b) 3,, (a) , 9, (c) 1,, 2, 5, 3, , of, , the, , conic, , 28. Consider the following statements, I. Maximum four normals can be drawn from a, point to an ellipse., II. The sum of focal distances of any point on the, ellipse is equal to the major axis., III. The equation of director circle of an ellipse, x2 y2, +, = 1 is x 2 + y 2 = a 2 − b2., a 2 b2, Which of the statements given above are correct?, (a) I and II, (b) III and I, (c) II and III, (d) All I, II and III, 29. Consider the following statements, I. The product of perpendiculars from the foci on any, x2 y2, tangent to the ellipse 2 + 2 = 1 is equal to b2., a, b, II. Locus of mid-points of focal chord of an ellipse, x2 y2, x 2 y 2 ex, is, +, =, 1, +, = ., a 2 b2, a 2 b2 b2, Which of the statements is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 30. Consider the following statements, I. The line lx + my + n = 0 is a normal to the ellipse, x2 y2, a2, b2 ( a 2 − b2 )2, ., if, +, =, 1, ,, −, =, a 2 b2, l2 m2, n2, II. If the line lx + my + n = 0 is a tangent to the, x2 y2, ellipse 2 + 2 = 1, then n 2 = b2m 2 + a 2l 2., a, b, Which of the statements is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 31. What is the area of the ellipse 4x 2 + 9 y 2 = 1 ?, π, (a) 6π, (b), (NDA 2008 II), 36, π, π, (c), (d), 6, 6, 32. Match List I (Equation of ellipse) with List II (Their, centre) and select the correct answer using the codes, given below the lists., List I, (Equation of ellipse), , (d) 3, 2, , x2 y2, 27. Let E be the ellipse, +, = 1 and C be the circle, 9, 4, 2, 2, x + y = 9. Let P = (1, 2) and Q = ( 2, 1). Which one of, the following is correct?, (NDA 2010 I), (a) Q lies inside C but outside E, (b) Q lies outside both C and E, (c) P lies inside both C and E, (d) P lies inside C but outside E, , x2, y2, +, =1, a2 b2, ( x − 2) 2 ( y − 3) 2, +, =1, B., 16, 25, A., , List II, (Their centre), 1. ( 3, 5), 2. ( 0, 0), , 2, 2, C. 25 x + 9 y − 150 x − 90 y + 225 = 0 3. (7, − 8), , D., , ( x − 7) 2 ( y + 8) 2, +, =1, 25, 9, , 4. ( 2, 3)
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550, , NDA/NA Mathematics, Codes, A B, (a) 4 1, (b) 2 3, (c) 2 4, (d) 2 1, , C, 3, 4, 1, 3, , D, 2, 1, 3, 4, , 33. The distance from the major axis of any point on the, x2 y2, ellipse 2 + 2 = 1 and its corresponding point on, a, b, auxiliary circle are in the ratio, a, b, a2, b2, (b), (c) 2, (a), (d) 2, b, a, b, a, x2 y2, +, = 1 from, 9, 4, the centre is 2. The eccentric angle of the point is, π, π, 3π, (b), (c), (d) π, (a), 4, 2, 4, , 34. The distance of a point on the ellipse, , 35. If the angle between the lines joining the end points, π, of minor axis of an ellipse with its foci is , then the, 2, eccentricity of the ellipse is, 1, 1, 3, 1, (b), (c), (d), (a), 2, 2, 2, ( 2 2), 36. Which one of the following is correct? The, x2, y2, eccentricity of the conic 2, + 2, = 1, ( λ ≥ 0), a +λ b +λ, (NDA 2008 I), , (a), (b), (c), (d), , increases with increase in λ, decreases with increase in λ, does not change with λ, None of the above, , 37. The equation 14x 2 − 4xy + 11 y 2 = 60 represents a, certain locus. If the axes be rotated through an angle, tan−1 2, without change of origin, then the equation of, the locus referred to new axes becomes, x2 y2, x2 y2, (b), (a), +, =1, +, =1, 4, 5, 6, 4, x2 y2, x2 y2, (c), (d), −, =1, −, =1, 6, 4, 5, 4, 38. S and T are the foci of an ellipse and B is an end of its, minor axis. If STB is an equilateral triangle. What is, the eccentricity of the ellipse?, 1, 1, 1, 2, (a), (b), (c), (d), 4, 3, 2, 3, 39. The eccentricities of the ellipse, , x, , 2, , α, , 2, , +, , y, , 2, , β2, , = 1, α > β and, , x2 y2, +, = 1 are equal. Which one of the following is, 9 16, correct?, (a) 4α = 3β, (b) α β = 12, (c) 4β = 3α, (d) 9 α = 16 β, , 40. If t is a variable, then what does the curve, a(1 − t 2 ), 2bt, represent?, x=, ,y=, (1 + t 2 ), (1 + t 2 ), (a) An ellipse with centre at ( 0, 0), (b) An ellipse with centre at ( a , b), (c) A hyperbola with centre at ( 0, 0), (d) A hyperbola with centre at ( a , b), 41. If the eccentricity of a conic section is equal to, , x2, , ,, x2 + 1, , x ∈ R/{ 0}, then the conic section is, (a) a circle, (b) a parabola, (c) an ellipse, (d) a hyperbola, 42. What is the locus of the point of intersection of the, straight, lines, and, ( x / a ) + ( y/ b) = m, ( x / a ) − ( y/ b) = 1/ m?, (a) Circle, (b) Parabola, (c) Ellipse, (d) Hyperbola, 43. The distance between the directrices of the hyperbola, x = 8 sec θ , y = 8 tan θ is, (b) 2, (c) 8 2, (d) 4 2, (a) 16 2, 44. If the foci of the ellipse, , x2 y2, +, = 1 and the hyperbola, 16 b2, , 1, x2, y2, coincide, then the value of b2 is, −, =, 144 81 25, (a) 1, (b) 7, (c) 5, (d) 9, 45. The equation of the hyperbola in the standard, form (with transverse axis along the x-axis), having the length of the latusrectum = 9 units and, 5, eccentricity = is, 4, x2 y2, x2 y2, (a), (b), −, =1, −, =1, 16 18, 36 27, 2, 2, 2, 2, x, y, x, y, (d), (c), −, =1, −, =1, 64 36, 36 64, 46. If a point ( x , y ) = (tan θ + sin θ , tan θ − sin θ ), then, locus of ( x , y ) is, (a) ( x 2 y )2/ 3 + ( xy 2 )2/ 3 = 1, (c) ( x 2 − y 2 )2 = 16xy, , (b) x 2 − y 2 = 4xy, (d) x 2 − y 2 = 6xy, , 47. If e and e′ are the eccentricities of the ellipse, 5x 2 + 9 y 2 = 45 and the hyperbola 5x 2 − 4 y 2 = 45, respectively, then ee′ is equal to, (a) 9, (b) 4, (c) 5, (d) 1, 48. Consider the following statements, I. The equation of the director circle of the, x2 y2, hyperbola 2 − 2 = 1 is x 2 + y 2 = a 2 − b2., a, b, x2 y2, II. The angle between the asymptotes of 2 − 2 = 1, a, b, −1 b , is 2 tan ., a
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551, , Conic Sections, III. Asymptotes always passes through the centre of, the hyperbola., Which of the statements given above are correct?, (a) I and II, (b) II and III, (c) III and I, (d) All I, II and III, 49. Equation of the hyperbola with eccentricity 3 / 2 and, foci at ( ± 2, 0) is 5x 2 − 4 y 2 = k2. What is the value of, k?, (NDA 2008 II), (a) 4 / 3, (b) 3 / 4, (c) ( 4 / 3) 5, (d) ( 3 / 4) 5, 50. The standard equation of the hyperbola having the, distance between foci as 32 and eccentricity 2 2 is, (b) x 2 − 7 y 2 = 56, (a) 7x 2 − y 2 = 56, (c) 7x 2 − y 2 = 224, , (d) x 2 − 7 y 2 = 224, , 51. The equation of the hyperbola whose eccentricity is, 5/4 and distance between foci is 10, is given by, x2 y2, x2 y2, (b), (a), −, =1, −, =1, 25 16, 16, 9, x2 y2, x2 y2, (d), (c), −, =1, −, =1, 9 16, 16 25, 1, y2, x2, y2, = 1 and, −, =, 2, 7, 144 81 25, a, were to coincide, then what is the value of a?, (a) 2, (b) 3, (NDA 2007 II), (c) 4, (d) 16, , 52. If the foci of the conics, , x2, , +, , 53. Match List I (Equations of curves) with List II (Their, types) and select the correct answer using the codes, given below the lists, List I, (Equation of curves), , List II, (Their types), , A., , x2 + x + 1 = y, , 1., , Circle, , B., , x + y + 2x + 2y − 6 = 0, , 2., , Parabola, , C., , 2 x + 3y + 4x + 6 y = 0, , 3., , Ellipse, , D., , 3 x2 − 2 y 2 + 6 x − 4 y = 0, , 4., , Hyperbola, , 5., , Straight lines pair, , 2, , 2, , 2, , Codes, A B, (a) 2 3, (c) 2 1, , 2, , C, 4, 3, , D, 4, 4, , A, (b) 4, (d) 4, , B, 3, 1, , C, 5, 5, , x2 y2, +, = 1, where µ < 4 and, 16 µ 2, x2, y2, 1, , coincide., the foci of the hyperbola, −, =, 144 81 25, What is the value of µ 2?, (a) 1, (b) 5, (c) 7, (d) 9, , 55. The foci of the ellipse, , 56. If the eccentricity and length of latusrectum of a, 13, 10, units respectively, then, and, 3, 3, what is the length of the transverse axis? (NDA 2007 I), 7, (a) units, (b) 12 units, 2, 15, 15, units, (d), units, (c), 2, 4, , hyperbola are, , Directions (Q. Nos. 57-64) Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 57. Assertion (A) The length of the intercept made by, 1, the line y = 3 x +, on the parabola y 2 = 16x is, 3, 16, 3 units., 3, Reason (R) The length of the intercept made by, the line y = mx + c on the parabola y 2 = 4ax is, 4, a (1 + m 2 ) ( a − mc)., m2, , D, 5, 5, , 54. If e and e′ are the eccentricities of a hyperbola and its, 1, 1, conjugate respectively, then 2 + 2 is equal to, e, e′, (a) 0, (b) 1, 1, (c) 2, (d), 2, , 58. Assertion (A) The equation 5x 2 + 10xy + 5 y 2 + 7x, + 14 y + 5 = 0 represents a parabola., Reason (R) The equation ax 2 + 2hxy + by 2 + 2gx, represents, a, parabola,, if, + 2 fy + c = 0, abc + 2 fgh − af 2 − bg2 − ch 2 ≠ 0 and ab − h 2 = 0, 59. Assertion (A) The slopes of tangents drawn from a, 1, 9, point ( 4, 10) to the parabola y 2 = 9x are and ., 4, 4, Reason (R) The equation of a tangent of slope m to, 9, the parabola y 2 = 9x is y = mx +, ., 4m
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552, , NDA/NA Mathematics, , 60. Assertion (A) The number of values of c such that, the straight line y = 4x + c touches the curve, x2, + y 2 = 1 is 2., 4, , Reason (R) The equation ax 2 + 2hxy + by 2 + 2 gx, represents, a, hyperbola,, if, + 2 fy + c = 0, 2, 2, 2, 2, and, abc + 2 f gh − af − bg − ch ≠ 0, h > ab., , Reason (R) The line y = mx + c touches the curve, x2 y2, +, = 1, iff c2 = a 2m 2 + b2., a 2 b2, , Directions (Q. Nos. 65-67) A parabola has the, origin as its focus and the line x = 2 as the directrix., , 61. Assertion (A) The point ( 3, 5) lies outside the, ellipse 4x 2 + 3 y 2 = 12., Reason (R) The point ( x1 , y1 ) lies outside the, x2 y2, x2 y2, ellipse 2 + 2 = 1, iff 12 + 12 − 1 > 0., a, b, a, b, 62. Assertion (A) If the point P( 3, 5) lies on the ellipse, x2 y2, +, = 1, then SP + S ′ P = 2., 4, 5, Reason (R) If the point ( x1 , y1 ) lies on the ellipse, x2 y2, +, = 1, then SP + S ′ P = 2a., a 2 b2, 63. Assertion (A) For an ellipse with major axis and, minor axis equal to 9 and 5 units of length, the, distance of a point P on its periphery from directrix is, 9 units, while its distance from the focus has been, measured to be 11 units of length., Reason (R) Eccentricity of an ellipse is less than, unity., , 65. The vertex of the parabola is, (a) (2,0), (b) (0,2), (c) (1,0), (d) None of these, 66. Find the equation of the parabola, (a) y 2 = 4 ( x − 1), (b) y 2 = − 4 ( x − 1), 2, (d) None of these, (c) y = − 4 ( x + 1), 67. The position of a point (1,4) relative to the parabola, is, (a) inside, (b) outside, (c) on, (d) None of these, , Directions (Q. Nos. 68-70), , Let the equation of, , x2 y2, ellipse be, +, =1, 16 9, 68. Find the foci of an ellipse., (b) ( ± 7,0), (a) ( ± 7 ,0), (c) ( ± 5 ,0), (d) None of these, 69. The distance between two foci is, (b) 2 7, (c) 3 7, (a) 7, , (d) 4 7, , 70. The equation of a circle passing through focus having, centre (0,3) is, (a) x 2 + y 2 − 6 y − 7 = 0, (b) x 2 + y 2 − 6 y + 7 = 0, 2, 2, (c) x + y + 6 y + 7 = 0, (d) None of these, , 64. Assertion (A) The equation, 9x 2 − 16 y 2 − 18x + 32y − 151 = 0, represents a hyperbola., , Answers, Level I, 1., 11., 21., 31., 41., 51., , (d), (d), (a), (d), (b), (b), , 2., 12., 22., 32., 42., 52., , (a), (a), (a), (d), (d), (d), , 3., 13., 23., 33., 43., 53., , (c), (a), (c), (c), (d), (d), , 4., 14., 24., 34., 44., 54., , (b), (c), (d), (b), (a), (a), , 5., 15., 25., 35., 45., 55., , (c), (b), (b), (b), (b), (d), , 6., 16., 26., 36., 46., 56., , (c), (d), (c), (c), (a), (b), , 7., 17., 27., 37., 47., 57., , (a), (c), (d), (b), (d), (c), , 8., 18., 28., 38., 48., , (d), (c), (c), (b), (c), , 9., 19., 29., 39., 49., , (d), (c), (d), (c), (a), , 10., 20., 30., 40., 50., , (d), (d), (d), (c), (d), , 2., 12., 22., 32., 42., 52., 62., , (c), (c), (b), (c), (d), (c), (d), , 3., 13., 23., 33., 43., 53., 63., , (a), (d), (b), (b), (c), (c), (d), , 4., 14., 24., 34., 44., 54., 64., , (a), (d), (b), (b), (b), (b), (a), , 5., 15., 25., 35., 45., 55., 65., , (b), (b), (a), (b), (c), (c), (c), , 6., 16., 26., 36., 46., 56., 66., , (d), (b), (c), (b), (c), (c), (b), , 7., 17., 27., 37., 47., 57., 67., , (c), (c), (d), (b), (d), (a), (b), , 8., 18., 28., 38., 48., 58., 68., , (d), (d), (a), (c), (d), (a), (a), , 9., 19., 29., 39., 49., 59., 69., , (b), (c), (a), (b), (c), (a), (b), , 10., 20., 30., 40., 50., 60., 70., , (c), (b), (b), (a), (c), (a), (a), , Level II, 1., 11., 21., 31., 41., 51., 61., , (d), (a), (c), (c), (c), (b), (a)
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Hints & Solutions, Level I, 1. We have, ( y − 3)2 = 4x, and the standard equation of parabola is, Y 2 = 4aX, From Eqs. (i) and (ii), we get, X = x, Y = y − 3 and 4a = 4 i.e., a = 1, The focus of the parabola is given by, X = a ,Y = 0, i.e.,, x = 1 and y − 3 = 0 ⇒ y = 3, ∴ Required focus is (1, 3)., , …(i), …(ii), , From Eq. (i), (x + 3)2 + y2 = x + 5, On squaring both sides, we get, (x + 3)2 + y2 = (x + 5)2, 2, ⇒, x + 6x + 9 + y2 = x2 + 10x + 25, ⇒, y2 − 4x = 16, ⇒, y2 = 4(x + 4), This is the equation of the required parabola., , 2. Given, equation of parabola is y2 − 8 y − x + 19 = 0, ⇒, y2 − 8 y = x − 19 ⇒ ( y − 4)2 = x − 3, This is the form of Y 2 = 4aX, Here, Y = y − 4, X = x − 3, 1, and, 4a = 1 ⇒ a =, 4, ∴ Focus is, X = a ,Y = 0, 1, x − 3 = and y − 4 = 0, ⇒, 4, 13, and y = 4, ∴, x=, 4, 13 , Hence, the focus is , , 4 ., 4 , , 5. The given equation of parabola is, x2 − 4x − 8 y + 12 = 0 ⇒ x2 − 4x = 8 y − 12, ⇒, x2 − 4x + 4 = 8 y − 12 + 4, ⇒, (x − 2)2 = 8( y − 1), ∴The length of latusrectum = 4a = 8, 6. Given that the axis of parabola is y-axis and vertex is, origin., ∴ Equation of parabola, x2 = 4ay, Since, it passes through (6, − 3), ∴, (6)2 = 4a (− 3), ⇒, 36 = − 12a ⇒ a = − 3, ∴ Equation of parabola is x2 = − 12 y., , ∴ Equation of directrix, X = − a, 1, ⇒, x−3 = −, 4, 11, ⇒, x=, 4, , 7. By definition of parabola, PM 2 = PS 2, 2, , 3. Equation of given parabola is, x2 − 4 y + 8 = 0, or, x2 = 4 ( y − 2 ), Which is the form of, X 2 = 4aY, Here, X = x, Y = y − 2 and 4a = 4, The axis of the parabola is y-axis i.e., X = 0, ⇒, x=0, , (given), …(i), …(ii), , 4. Let S(−3, 0) be the focus and ZM be the directrix whose, equation is x + 5 = 0, x+5=0, , y, M, , x', , P(x, y), , S(–3,0) O, , Z, , and PM = Perpendicular distance from P to ZM, x+5, =, =x+5, 1+0, , ⇒, , 9x2 + 16 y2 + 12 − 24xy − 8 y + 6x, = 25 (x2 + 25 − 10x + y2 + 9 − 6 y), 2, 2, ⇒ 16x + 9 y − 256x − 142 y + 24xy + 849 = 0, ⇒, (4x + 3 y)2 − 256x − 142 y + 849 = 0, 8. Let x be any point on the parabola, then y = 3x, putting, this value in the given equation y2 = 18x, we get, (3x)2 = 18x, ∴ Point is (2,6)., , ⇒, , x ⇒ 2 and y = 6, , 9. Equation of parabola is x2 − 4x − 8 y + 12 = 0, ⇒, x2 − 4x + 4 = 8 y − 8 ⇒ (x − 2)2 = 8 ( y − 1), 2, ⇒ X = 8Y , this is a standard form of parabola, On comparing with X 2 = 4aY , we get a = 2., ∴ Directrix is Y = − a ⇒ y − 1 = − 2 ⇒ y = − 1, , x, , y', , Let P (x, y) be any point on the parabola, then, PS = PM, ⇒, PS = (x + 3)2 + y2, , 3x − 4 y + 1 , = (x − 5)2 + ( y − 3)2, , 2, 2, 3 + (− 4) , , …(i), , 10. Directrix of parabola is y = 2, ∴, a = −2, ∴ Required equation of parabola is, x2 = − 4 ⋅ 2 ⋅ y ⇒ x2 = − 8 y, 11. Given equation of parabola is y2 = 4ax, and P (x1 , y1 ) be a point on the parabola.
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554, , NDA/NA Mathematics, , Y, , (–a,0), , X′, , Z, , (0,0) (a,0), , A, , dx, Slope of normal = − , dy ( at 2 , 2at ), , P (x1, y1), , M, , S, , N, , y, 2 at, =−, =−, 2 a, 2a, , X, , = −t, 20., , Equation of parabola is x2 = 12 y, , …(i), , y-axis, , Y′, , We know that,, (by definition of parabola), PS = ePM, ⇒, PS = PM, (Q for parabola, e = 1), ⇒, PS = ZN, ⇒, PS = ZA + AN, ⇒, PS = a + x1, Which is the required focal distance., 12. Since, focal chord of parabola y2 = ax is 2x − y − 8 = 0, a , Q This chord passes through focus i. e. , , 0, 4 , a, 2⋅ −0 −8 =0, ∴, 4, ⇒, a = 16, ∴Directrix is x = − 4 ⇒ x + 4 = 0, 13. Since, x + y − 1 = 0 is a tangent to the parabola, y2 − y + x = 0, then the point of contact is (0, 1)., 14. Equation of directrix (x − 1)2 = 2 ( y − 2) is, 1, ( y − 2) = −, 2, ⇒, 2y − 3 = 0, 15. We know, if the line y = mx + c touches the parabola, 2a , a, y2 = 4ax, then the point of contact is 2 , − ., m, m, 16. Let the point on the parabola is (x1 , y1 ), then focal, distance, = a + x1, ⇒, 2 + x1 = 4, (Q a = 2), ⇒, x1 = 2, On putting this value in y2 = 8x, ⇒, y12 = 8 × 2, ⇒, y1 = ± 4, 17. The smallest length of the focal chord is the length of, the latusrectum. So, the length of required chord = 4a., at 2 − at , 18. Let the point , ,, lies on the parabola y2 = ax., 2 , 4, Q, ⇒, , 2, at 2, at , , − = a , 2, 4 , 2 2, , (0,3), , x', , 19. Equation of parabola is, y2 = 4ax, dy, ∴, 2y, = 4a, dx, dy 2 a, =, ⇒, dx, y, , 12, , (0,0) O, , B (6,3), x2 = 12y, x-axis, , y', , So, area of ∆ ABO is, 0 0, 1, 1, =, 6 3, 1, 2, −6 3, 1, 1, 1, = (18 + 18) = × 36 = 18 sq units, 2, 2, x2, y2, +, =1, 16 25, 2, 2, a = 16, b = 25, 16, a2, 9, 3, e= 1− 2 = 1−, =, =, 25, 25 5, b, , 21. Given equation,, Here,, ∴, , 22. We know that, the sum of focal distance of any point on, the ellipse always equal to the length of major axis i.e.,, it is equal to 2a., 23. We have, 4x2 + 16 y2 − 24x − 32 y = 1, ⇒, 4 (x2 − 6x) + 16 ( y2 − 2 y) = 1, 2, ⇒ 4 (x − 6x + 9) + 16 ( y2 − 2 y + 1) −36 − 16 = 1, ⇒, 4 (x − 3)2 + 16 ( y − 1)2 = 53, (x − 3)2 ( y − 1)2, +, =1, ⇒, 53, 53, 4, 16, x2, y2, On comparing with 2 + 2 = 1, we get, a, b, a 2 = 53 / 4 and b2 = 53 / 16, a 2 − b2, a2, (53 /4) − (53 /16), e=, (53 /4), , ∴ Eccentricity of ellipse is e =, ⇒, , 2 2, , a t, a t, =, 4, 4, , S, , (–6,3) A, , ⇒, , e=, , 3, 2, , 24. Given, the distance between the foci = 2ae = 16 and, 1, eccentricity of ellipse (e) = . We know that length of the, 2, 2ae 16, major axis of the ellipse = 2a =, =, = 32., 1 /2, e
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555, , Conic Sections, 1, and foci (± 1, 0), 2, 1, ⇒, ae = 1, ⇒ a = =2, 1, 2, 1, 3, 2, 2, 2, 2, Now, b = a (1 − e ) = 2 1 − = 4 = 3, 4, , , 4, ∴ The equation of ellipse is, x2, y2, + 2 =1, 2, a, b, x2 y2, ⇒, +, =1, 4, 3, , 25. Given, eccentricity e =, , 26. According to the question,, 2b2 1, = (2b), 2, a, 2b2, ⇒, =b, a, ⇒, 2b = a ⇒ 4b2 = a 2, ⇒, 4a 2(1 − e2) = a 2, [Q a 2 = b2(1 − e2)], 3, 3, ⇒, e2 =, ⇒ e=, 2, 4, x2 y2, +, =1, 5, 9, (here, a < b), 5 2, a2, Eccentricity e = 1 − 2 = 1 − =, 9 3, b, 2, Distance between foci is 2be = 2 × 3 × = 4, 3, , 27. Given equation can be rewritten as, , 28. It is an ellipse., 29. Latusrectum of an ellipse =, , 2b2, a, , and minor axis = 2b, ∴, ⇒, Also,, , 2b2, a, a = 2b, b=, , (according to question), b2, 3, 3, b2, e= 1− 2 = 1− 2 =, =, 4, 2, a, 4b, , 30. The given equation can be rewritten as, 4x2 − 24x + 36 + 16 y2 − 32 y + 16 − 36 − 16 − 12 = 0, ⇒, (2x − 6)2 + (4 y − 4)2 = 64, (x − 3)2 ( y − 1)2, ⇒, +, =1, 16, 4, This represents an ellipse and a 2 = 16, b2 = 4, 4, 3, ∴, e= 1−, =, 16, 2, 31. The given equation of ellipse can be rewritten as, x2, y2, +, =1, 16, 9, Here,, a = 4, b = 3, 2b2 2 × 9 9, ∴ Length of latusrectum =, =, =, 4, 2, a, , 32. The equation of ellipse is, 25x2 + 16 y2 − 150x − 175 = 0, ⇒, 25 (x2 − 6x + 9) − 225 + 16 y2 − 175 = 0, 25 (x − 3)2 16 y2, (x − 3)2 y2, +, =1, +, =1⇒, ⇒, 400, 400, 16, 25, The major axis of ellipse is a line parallel to y-axis,, therefore eccentricity of ellipse is given by, 16, 9, 3, e= 1−, =, =, 25, 25 5, 33. It is an ellipse., 34. We know that the sum of focal radii of any point on an, ellipse is equal to length of major axis. i.e., sum of focal, radii, = (a + x) + (a − x), = 2a = major axis, 36, 35. Given, the foci at (±5, 0) and the directrix is x =, 5, a 36, …(i), ⇒, ae = 5 and =, e, 5, a, 36, ae × = 5 ×, ⇒ a =6, ∴, e, 5, 5, From Eq. (i), ae = 5 ⇒ e =, 6, 25, , 2, 2, 2, 2, b = a (1 − e ) ⇒ b = 36 1 − = 11, ∴, , 36, 2, 2, x, y, ∴ Equation of the ellipse is, +, =1, 36 11, 36. Given that major axis = 3 minor axis, i.e.,, a = 3b, and eccentricity is given by, b2 = a 2(1 − e2) ⇒ b2 = 9b2(1 − e2), 1, 1, ⇒, = (1 − e2) ⇒, = (1 − e2), 9, 9, 1, 8, 2 2, 1 − = e2 ⇒ e2 =, ⇒ e=, ⇒, 9, 9, 3, 37. Given that, foci = (±1, 0), i.e.,, ae = ± 1, and major axis = 6, i.e.,, 2a = 6, ⇒, a =3, Q, ae = ± 1, ∴, 3e = 1, 1, ⇒, e=, 3, ∴ By the relation, b2 = a 2(1 − e2), we get, b2 = 8, and we know that the standard equation of ellipse is, x2, y2, …(i), +, =1, a 2 b2, On putting the values of a 2 and b2 in Eq. (i), we get, the required equation of ellipse which is, x2 y2, +, =1, 9, 8, 38. The required equation represents a straight line., (by definition)
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556, , NDA/NA Mathematics, 2, 1 1, = − + cos θ, r, 2 4, ⇒, 8 = − 2r + r cos θ, ⇒, 8 − r cos θ = − 2r, ⇒, (8 − r cos θ )2 = 4r 2, ⇒, (8 − x)2 = 4 (x2 + y2), 2, where, x = r cos θ , r = x2 + y2, ⇒, 64 + x2 − 16x = 4x2 + 4 y2, 2, ⇒, 3x + 4 y2 + 16x − 64 = 0, Which represents the equation of an ellipse., , 39. Since,, , 40. Given that the length of minor axis is equal to distance, between the two foci., ∴, 2b = 2ae, ⇒, b = ae, b2, We know that, e = 1 − 2, a, 2 2, a, e, ⇒, e2 = 1 − 2, a, ⇒, 2e2 = 1, 1, e=, ⇒, 2, 41. Length of latusrectum of an ellipse is, ∴, ⇒, , 2 b2, ., a, , 2 b2, =b, a, 2b = a, b2, a2, 1, 3, = 1− =, 4, 2, , e= 1−, , We know that,, , 42. Given equation of ellipse is, , x2 y2, +, =1, a 2 b2, , a 2 b2, +, = −1, a 2 b2, Let point (a,b), then, a 2 b2, S = 2 + 2 −1 =1 + 1 −1 =1, a, b, i.e.,, S >0, ⇒ (a,b) lies outside the ellipse., i.e.,, , S≡, , 43. The given equation of circle and ellipse are, respectively, …(i), x2 + y2 = 9, x2 y2, and, …(ii), +, =1, 4, 8, From Eqs. (i) and (ii), we get, x2 9 − x2, +, =1, 4, 8, ⇒, 2x2 + 9 − x2 = 8 ⇒ x2 = − 1, Thus, it is clear that circle and ellipse never intersect, each other., 44. The given equation is 2x2 − 3 y2 = 5 , it can be rewritten, as, , x2 y2, −, =1, 5, 5, 2, 3, , 5 5 2, = (e − 1), 3 2, 2 25, 5, = ⇒e=, 56, 3, , Now,, , b2 = a 2(e2 − 1) ⇒, , ⇒, , e2 =, , 2, 5, , 5 5, + , 3 2, , The foci of hyperbola is (± ae, 0), , 5 5 5, , = ±, ⋅, , 0 = ±, , 0, 2 3 , 6 , , 45. The given equation of hyperbola is 16x2 − 9 y2 = 144, x2 y2, It can be rewritten as, −, =1, 9 16, 2b2 2 × 16 32, ∴ Length of latusrectum =, =, =, 3, 3, a, 46. The given equation of conic is 4x2 + 9 y2 = 144, x2, y2, +, =1, 36 16, which represents an ellipse, here a > b., Q, a 2 = 36 ⇒ a = 6, b2 = 16 ⇒ b = 4, Now,eccentricity b2 = a 2(1 − e2), ⇒, 16 = 36 ( 1 − e2), 4, 5, 4, ⇒, = 1 − e2 ⇒ e = 1 − =, 9, 9, 9, 5, e=, ⇒, 3, 2a, 47. Distance between directrices =, e, and eccentricity of rectangular hyperbola = 2, 2a, ∴ Distance between directrices =, 2, From the given condition,, 2a, = 10, 2, 2a = 10 2, ⇒, Distance between foci = 2ae = 10 2 ⋅ 2 = 20, 48. Given equation of lines are, x y, − =m, a b, x y 1, and, + =, a b m, Multiplying Eqs. (i) and (ii), we get, 1, x y x y, − + = m⋅, a b a b, m, x2 y2, −, =1, a 2 b2, which is the required equation of hyperbola., 49. The given equation may be rewritten as, x2, y2, −, =1, 32 /3 8, x2, y2, ⇒, −, =1, 2, (2 2 )2, 4 2 , , , 3 , , …(i), …(ii)
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557, , Conic Sections, On comparing the given equation with the standard, x2 y 2, equation 2 − 2 = 1, we get, a, b, 2, 4 2, 2, a =, and b2 = (2 2 )2, 3 , ∴ Length of transverse axis of a hyperbola, 4 2 8 2, = 2a = 2 ×, =, 3, 3, (x + 1)2 ( y − 2)2, −, = 1 is of the, 16, 4, form (4 sec θ − 1, 2 tan θ + 2)., 1 1 3, 51. Given equation is = + cos θ, it can be rewritten as, r 8 8, 8, l, = 1 + 3 cos θ which is of the form = 1 + e cos θ, r, r, On comparing, we get, e=3 >1, ∴ Given equation is of hyperbola., …(i), , This is an equation of a hyperbola and the equation of, conjugate axes is y-axis i.e., x = 0., On putting x = 0 in Eq. (i), we get, 1, 1, or y = i.e., imaginary points, y2 = −, 9, 3, Hence, no point of intersection exists., 53. Let (h , k) be the point of intersection of, x cos α − y sin α = a and x sin α − y cos α = b, ∴, h, k, 1, =, =, b sin α − a cos α b cos α − a sin α − (cos 2 α − sin 2 α ), ⇒, , h=, , x4 + y4 − 2x2y2 = (a 2 + b2) (x2 + y2) + 4 abxy, which is, not a locus of any given curves., 54. The, equation, a x2 + 2 hxy + by2 + 2 gx + 2 fy + c = 0, represents a hyperbola, if, ∆ ≠ 0, h 2 > ab, a + b = 0, 2 x2 − 3 y2 − 6 = 0, x 2 y2, or, −, =1, 3, 2, Which represents a hyperbola of the form, x2 y2, −, =1, a 2 b2, , 55. We have,, , 50. Any point on hyperbola, , 52. The equation of curve is, 4 x2 − 9 y2 = 1, x2, y2, ⇒, −, =1, 1 /4 1 /9, , Then, the locus of point (h , k) is, , a cos α − b sin α, a sin α − b cos α, ,k=, cos 2α, cos 2α, , …(i), , 56. Given equation is 25x2 − 9 y2 = 144, x2, y2, ⇒, −, =1, 144 144, 25, 9, x2, y2, −, =1, ⇒, 144 16, 25, 144 2, Here,, a2 =, , b = 16, 25, 2, 2 2, Now,, b = a (e − 1), 144 2, 16 =, (e − 1), ⇒, 25, 16 × 25, ⇒, e2 =, +1, 144, 25, ⇒, e2 =, +1, 9, 34, e2 =, ⇒, 9, 34, ⇒, e=, 3, 57. The difference between the focal distance from any, point P (x1 , y1 ) of the hyperbola is equal to the length of, transverse axis., i.e.,, S′ P − SP = (ex1 + a ) − (ex1 − a ) = 2a, , Level II, 1. Since, the equation of latusrectum and equation of, tangent both are parallel and they lie in the same side of, the origin, −8 + 12 , 4, ∴, a =, =2 2, =, 2, 2, 2, 1 +1, ∴ Length of latusrectum = 4a, = 4 (2 2 ) = 8 2, 2. Given, vertex of parabola (h , k) ≡ (1, 1) and its focus, (a + h , k) ≡ (3, 1) or a + h = 3, ⇒, a =2, Since, y coordinate of vertex and focus are same,, therefore axis of parabola is parallel to x-axis. Thus,, equation of parabola is, ( y − k)2 = 4a (x − h ) ⇒ ( y − 1)2 = 8 (x − 1), , 3. Since, the focus and vertex of the parabola are on y-axis., Therefore, directrix is parallel to x-axis and axis of, parabola is y-axis., Let the equation of the directrix be y = k. The directrix, meets the axis of the parabola at (0, k) . But vertex is the, mid-point of the line segment joining the focus to the, point, where directrix meets axis of the parabola., k+3, =6⇒k =9, ∴, 2, Thus, the equation of directrix is y = 9, ∴Equation of parabola is, 2, | y − 9|, (x − 0)2 + ( y − 3)2 = , [Q PS 2 = PM 2], , 12 , ⇒, x2 + y2 − 6 y + 9 = y2 − 18 y + 81, ⇒, x2 + 12 y − 72 = 0
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559, , Conic Sections, 11. The given equations of line and curve are, 1, y = mx +, m, and, y2 = 4 x, or, y=2 x, On putting the value of y in Eq. (i), we get, 1, 2 x = mx +, m, m2 x + 1, ⇒, 2 x=, m, m2x − 2 m x + 1 = 0, ⇒, (m x − 1)2 = 0, ⇒, 1, ⇒, m x −1 =0⇒ x =, m, 1, ⇒, x= 2, m, From Eq. (ii), we get, 1, y = 2⋅, m2, 2, =, m, 1 2, So, points are 2 , ., m m, , …(i), …(ii), , = 36 + 36 = 72, =6 2, , (a > b), Length of minor axis = 2b = 8 ⇒ b = 4 and, eccentricity,, 5, e=, 3, Now, b2 = a 2 (1 − e2), 5, , 4, ⇒, (4)2 = a 2 1 − ⇒ 16 = a 2 , , 9, 9, , P(x1,y1), , O, , 14. Since, the focus and vertex of the parabola are on x-axis,, therefore, its direction is parallel to x-axis and axis of, parabola is x-axis. Let the equation of the directrix be, x = k. The directrix meets the axis of parabola at (k, 0)., But vertex is mid-point of the line segment joining the, focus to the point, where directrix meets the axis of the, parabola., k+ b, = a ⇒ k = 2a − b, ∴, 2, ∴ Equation of the directrix is x = 2a − b, Let (x, y) be a point on the parabola, then, 2, x − 2a + b, (x − b)2 + y2 =, 1, ⇒ x2 + b2 − 2bx + y2 = x2 + 4a 2 + b2 − 4ax − 4ab + 2bx, ⇒, y2 = 4bx − 4ax − 4ab + 4a 2, ⇒, = 4x(b − a ) − 4a (b − a ), ⇒, y2 = 4(b − a )(x − a ), Which is the required equation of parabola., 15. Let the standard equation of ellipse is, , y, , Q(x2,y2), , ...(ii), , ∴ Required distance PQ = (8 − 2)2 + (−4 − 2)2, , 12. Equation of parabola is y2 = 9x., …(i), The chord of the parabola is inclined at an angle of 45°, to the axis of the parabola and passes through the, point (1, 3)., , 45°, , ⇒, y1 − 2 = − x1 + 2, ⇒, x1 + y1 = 4, From Eqs. (i) and (ii), we get, y1 = − 4 and 2, ∴ Coordinates of point Q are (8, − 4)., , x, , ∴ Equation of chord is ( y − 3) = 1 (x − 1), …(ii), y=x+2, On solving Eqs. (i) and (ii), we get x = 1 and 4., ∴ Point of intersections of parabola and chord are, P(4, 6) and Q (1, 3), So, length of chord = PQ, = (1 − 4)2 + (3 − 6)2 = 32 + 32 = 18 = 3 2, 13. Equation of parabola is y2 = 2 x, ∴ Coordinates of vertex are A (0, 0)., Let (x1 , y1 ) be the coordinates of the point Q, ∴, y12 = 2 x1, y −2, and slope of PQ = 1, x1 − 2, 2 −0, Also, slope of AP =, =1, 2 −0, Since, PQ and AP are perpendicular to each other, y1 − 2, ∴, , = −1, x1 − 2 , , ...(i), , x2, y2, + 2 =1, 2, a, b, , ⇒, a 2 = 36 ⇒ a = 6, Length of major axis = 2a = 12., 16. The equation of curve is, 2 x2 − 8 x + y 2 − 2 y + 1 = 0, 2, ⇒ 2 (x − 4x + 4 − 4) + ( y2 − 2 y + 1 − 1) + 1 = 0, ⇒, 2 [(x − 2)2 − 4] + ( y − 1)2 = 0, ⇒, 2 (x − 2)2 + ( y − 1)2 = 8, (x − 2)2 ( y − 1)2, +, =1, ⇒, 4, 8, 2, 2, x, y, This equation is of the form 2 + 2 = 1, a, b, Here,, a 2 = 4 and b2 = 8, Q, ∴, ⇒, , b2 − a 2, b2, 8 −4, e=, 8, 1, 1, e=, =, 2, 2, e=, , …(i)
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560, , NDA/NA Mathematics, , 17. We know, the locus of point of intersection of two, perpendicular tangents drawn on the ellipse is, x2 + y2 = a 2 + b2, which is called director circle., For the given equation of ellipse, x2 y2, +, = 1 ⇒ a 2 = 9, b2 = 4, 9, 4, ∴ Locus is, x2 + y2 = a 2 + b2, 2, 2, ⇒, x + y = 9 + 4 ⇒ x2 + y2 = 13, 18. Q Foci of an ellipse are (4, 0) and (−4, 0)., ∴, 2ae = 8, ⇒, ae = 4, and, (semi-minor axis), b =3, b2, We know that, e= 1− 2, a, 2, 9, , 4, ⇒, = 1 − 2, , a, a , 16 a − 9, =, a2, a2, 2, ⇒, a = 25 ⇒ a = 5, Thus, the equation of an ellipse is, x2, y2, +, =1, 25, 9, Which is satisfied by (5, 0). Hence, the ellipse passes, through (5, 0)., 2, , ⇒, , 19. The equation of circle passing through the intersection, of ellipses, x2, y2, x2, y2, and, +, =, 1, +, = 1 is, a 2 b2, b2 a 2, 2 2, 2a b, x2 + y2 = 2, a + b2, , ⇒, ⇒, , tan θ =, , θ = tan −1 (12 / 5 ), x2, y2, + 2 = 1 are (ae, 0) and (−ae, 0)., 2, a, b, Equation of circle whose end points of diameter, is, (x − ae)(x + ae) + ( y − 0)( y − 0) = 0, x2 − a 2e2 + y2 = 0, , b2 , x2 + y 2 − a 2 1 − 2 = 0, a , , 2 , b2 , Q e = 1 − 2 , a , , , x2 + y2 − a 2 + b2 = 0, x2 + y2 = a 2 − b2, , 22. Q Foci of an ellipse, ∴, ⇒, ⇒, , ⇒, ⇒, , 23. We have, a 2 = 25 and b2 = 16, 16, b2, ∴, e= 1− 2 = 1−, 25, a, 9, 3, =, =, 25 5, So, the coordinates of foci S and S′ are (3, 0) and (− 3, 0),, respectively. Let P (5 cos θ , 4 sin θ ) be a variable point, on the ellipse., 1, Then, ∆ = area of ∆PSS′ = × 6 × 4 sin θ = 12 sin θ, 2, So, maximum value of area of ∆PSS′ is 12., Since, value of sin θ lies between −1 and 1., 24. Let the equation of ellipse be, , 1 2 1, 8, , b = ,m =, 4, 9, 9, The required points are, , , a 2m, b2, ±, , ,m, , a 2m2 + b2, a 2m2 + b2 , , ⇒, , ⇒, , 1 8, ×, 4 9, ,m, 1 64 1, ×, +, 4 81 9, 1, 2, ± , m , 5, 5, , 1, 9, 1 64 1, ×, +, 4 81 9, , x2, y2, +, =1, a 2 b2, , (a > b), , y, B, , 20. We have, a 2 =, , , , ±, , , , , 2 144 + 36 2 180, = 12 / 5, =, 9 −4, 5, , F′, , , , , , , , , The angle between these lines is tan θ =, , 2 h 2 − ab, a+b, , F, , x, , π, 2, π, ∴, ∠ FBC = and ∠ CFB is also an angle of π / 4, 4, ⇒, BC = CF ⇒ b = ae, ...(i), ⇒, b2 = a 2e2, 2, We know, b = a 2 (1 − e2), [from Eq. (i)], ∴, a 2e2 = a 2 (1 − e2), ⇒, e2 = 1 − e2 ⇒ 2e2 = 1, 1, ⇒, e=, 2, , Q, , 21. The combined equation of the pair of tangents drawn, from (1, 2) to the ellipse 3x2 + 2 y2 = 5 is SS ′ = T 2, ⇒, (3x2 + 2 y2 − 5) [3 (1)2 + 2(2)2 − 5], = [3x (1) + 2 y (2) − 5]2, 2, 2, ⇒ (3x + 2 y − 5) (3 + 8 − 5) = (3x + 4 y − 5)2, ⇒, 9x2 − 24xy − 4 y2 + 40 y + 30x − 55 = 0, This is the equation of pair of straight lines, where,, a = 9, h = − 12, b = − 4, , C, , ∠ FBF ′ =, , 25. We know that the product of perpendiculars from two, x2 y2, foci S1 and S 2 of an ellipse, +, = 1 on the tangent at, 9 16, any point P on the ellipse is equal to the square of the, semi-minor axis., ∴, (S1M1 ) ⋅ (S 2M 2) = 16
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561, , Conic Sections, 26. Given, equation is 9x2 + 4 y2 − 6x + 4 y + 1 = 0, 2, 1, 1, , , ⇒ 9 x2 − x + 2 + 4 y2 + y + + 1 − 1 − 1 = 0, , , , 3, 4, 3, 2, 2, 1, 1, , , y+ , x − , , , 2, 3, [ here, a < b], +, ⇒, =1, 2, 2, 1, 1, , , , , 2, 3, 1, Length of major axis = 2b = 2 = 1, 2, 1 2, Length of minor axis = 2a = 2 =, 3 3, 27. For a point P(1, 2), 4(1)2 + 9(2)2 − 36 = 40 − 36 > 0, and, 12 + 22 − 9 = 5 − 9 < 0, ∴ Point P lies outside of E and inside of C, For a point Q(2, 1)., 4(2)2 + 9(1)2 − 36 = 16 + 9 − 36 < 0, and, (2)2 + (1)2 − 9 = 4 + 1 − 9 < 0, ∴ Point Q lies inside of E and C., 28. I. Let any given point be P(x1 , y1 ) and any point on an, ellipse be φ i.e., (a cos θ , b sin θ ), then, Equation of normal is, ax sec φ − by cosec φ = a 2 − b2, φ, Now putting , tan = t, then, 2, 2t, 1 − t2, and, sin φ =, cos, φ, =, 1 + t2, 1 + t2, 2, 2, 1 + t , 1 + t , ax , − by , = a 2 − b2, 1 − t2, 2t , It passes through the point P (x1 , y1 )., ⇒ by1t 4 + 2t3 (ax1 + a 2e2) + 2t (ax1 − a 2e2) − by1 = 0, Which is a four degree equation in t., ⇒ It has maximum four normal can be drawn from, point P., II. Let the focus be S and S' and any point on an ellipse, be P, then sum of focal distances of a point, = a − ex + a + ex = 2a (major axis), x2, y2, III. 2 + 2 = 1, a, b, Equation of director circle is, x2 + y2 = a 2 + b2, Hence, statements I and II are correct., x2, y2, + 2 =1, 2, a, b, Then, foci are S = (ae, 0) and S′ = (− ae, 0), , 29. I. Let the equation of ellipse be, , Equation of tangent at any point P is, y = mx +, , a 2m2 + b2, , Now, length of perpendicular from foci are, mae + a 2m2 + b2, L1 =, 1 + m2, and, , L2 =, , − mae +, , a 2m2 + b2, , 1 + m2, , ⇒, , a 2m2 + b2 − m2a 2e2, 1 + m2, 2 2, 2, a m (1 − e ) + b2, =, 1 + m2, 2 2, m b + b2, [Q b2 = a 2 (1 − e2)], =, 1 + m2, b2(1 + m2), = b2, =, 1 + m2, , L1 × L 2 =, , II. Let the mid-point of the focal chord of the given, ellipse be (h , k). Then, its equation is, hx ky h 2 k2, +, =, +, a 2 b2 a 2 b2, Since, it passes through the focus i.e., (ae, 0), hae h 2 k2, = 2+ 2, ∴, a2, a, b, he h 2 k2, ⇒, = 2+ 2, a, a, b, ex x2, y2, = 2+ 2, ∴ Focus of mid-point is, a a, b, Hence, only statement It is correct., 30. I. Line lx + my + n = 0, − lx n, ⇒, y=, −, m, m, −l, Here, slope m′ =, , c = − n /m, m, Condition for normal is, m′ (a 2 − b2), c=±, a 2 + b2(m′ )2, l2 (a 2 − b2)2, − n, m2, On squaring, , =, m, l2, a 2 + b2 × 2, m, n 2 l2(a 2 − b2)2, ⇒, =, m2 a 2m2 + b2l2, a 2m2 + b2l2 (a 2 − b2)2, ⇒, =, m2l2, n2, 2, 2, 2, (a − b2)2, a, b, ⇒, + 2=, 2, l, m, n2, II. Condition of tangency is, c = a 2(m′ )2 + b2, 2, , 2, , − n, 2, 2, 2, On squaring, , = a (− l/m) + b, m, n 2 a 2l2 + m2b2, =, m2, m2, 2, 22, ⇒, n = a l + b2m2, Hence, only II statement is correct., 31. Given, equation of ellipse, 4x2 + 9 y2 = 1, 2, x, y2, =1, +, ⇒, 2, 2, 1, 1, , , 3, 2
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564, , NDA/NA Mathematics, , 41. Let e =, , x2, , x ∈ R /{0}, x +1, , 9, 9, =, ⇒ a =8, 2a 16, On putting this value in Eq. (i), we get, 9 ×8, b2 =, = 36 ⇒ b = 6, 2, ∴ Equation of hyperbola is, x2 y2, x2 y2, − 2 =1 ⇒, −, =1, 2, 64 36, 8, 6, 46. Put the value of, (x, y) = (tan θ + sin θ , tan θ − sin θ ) in the given option,, we get the required result., Now, putting the value of x and y in option (c), we get, [(tan θ + sin θ )2 − (tan θ − sin θ )2]2, = 16 (tan θ + sin θ ) × (tan θ − sin θ ), ⇒ [tan 2 θ + sin 2 θ − tan 2 θ − sin 2 θ + 4 tan θ sin θ]2, = 16 (tan 2 θ − sin 2 θ ), ⇒, (4 tan θ sin θ )2 = 16 (tan 2 θ − sin 2 θ ), ⇒, 16 tan 2 θ sin 2 θ = 16 tan 2 θ (1 − cos 2 θ ), ⇒, 16 tan 2 θ sin 2 θ = 16 tan 2 θ sin 2 θ, Hence, the option (c) satisfies., ⇒, , 2, , For every non-zero values of x, we will get the value of, e in the interval (0, 1)., The given eccentricity is of an ellipse., x y, x y 1, 42. Lines are + = m and − =, a b, a b m, On multiplying both equations, we get, 2, 2, 1, x, y, − =m ×, a, b, m, x2 y 2, −, =1, ⇒, a 2 b2, Which is a equations of hyperbola., Hence, locus of the point of intersection of given two, lines is hyperbola., 43. Given, x = 8 sec θ, y = 8 tan θ, x, y, ⇒, secθ = and tan θ =, 8, 8, We know,, sec2 θ − tan 2 θ = 1, x2 y2, −, =1, ⇒, 64 64, , 47. Given, equation of ellipse, , 64, = 2, 64, 2a 2 × 8, ∴ Distance between directrices =, =, =8 2, e, 2, ⇒, , e = 1+, , 44. We have, equation of ellipse is, ∴, , e= 1−, , x2, y2, + 2 =1, 16 b, , ∴, , e= 1−, , …(i), , b2, 16, , ∴, ∴, , and equation of hyperbola is, 1, x2, y2, −, =, 144 81 25, 81 225, b2, =, e2 = 1 + 2 = 1 +, ∴, 144 144, a, 225 15 5, ⇒, e=, =, =, 144 12 4, 144, also,, a2 =, 25, Hence, foci are (± ae, 0)., 12 5 , i.e.,, × , 0 = (± 3, 0), ±, 5 4 , , e′ = 1 +, , and, ⇒, , ee′ =, , x2, y2, −, =1, 32 45 2, , , 4 , , 45 /4, 3, b2, = 1+, =, 9, 2, a2, , 2 3, × =1, 3 2, , 48. Obviously, statement I is correct., II. Combine equation of asymptotes of hyperbola, x2 y2, x2 y2, x y, is, −, =, 1, − 2 = 0 i.e., − = 0, 2, 2, 2, a, b, a, b, a, b, x y, and, + =0, a b, b, −b, and m2 =, m1 =, ⇒, a, a, b/a − (− b /a ), ⇒, tan θ =, 1 + (− b /a ) (b /a ), 2ab, i.e., θ = 2 tan −1 (b /a ), ⇒, tan θ = 2, a − b2, , b2 = 16 − 9, , III. Asymptotes passes through centre of hyperbola., Hence, all the three statements are correct., , 2, , 2b, a, 5, b2 25, e=, ⇒ 1+ 2=, 16, 4, a, 9a 25, 1+, =, 2a 2 16, , 45. Length of latusrectum = 9 =, , 5, b2, 2, ⇒e= 1 − ⇒e=, 2, 9, 3, a, , and equation of hyperbola, , Hence, foci are (± ae, 0) i.e., (± 16 − b2 , 0), , Since, the foci coincide., ⇒, 16 − b2 = 9, ⇒, ⇒, b2 = 7, , x2, y2, +, =1, 2, 2, 3, 5, , …(i), , [using Eq. (i)], , 49., , Given, equation of hyperbola, 5 x2 − 4 y2 = k 2, x2, y2, − 2 =1, ⇒, 2, k, k, 5, 4
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565, , Conic Sections, , ∴, , a=, , k, 5, , and b =, , The eccentricity and foci are, ∴, ⇒, ⇒, , k, 2, , 3, and (± 2, 0), respectively, 2, , 3, and ± ae = 2, 2, k 3, ⋅ =2, 5 2, 4, k=, 5, 3, , e=, , 50. Let the standard equation of hyperbola be, x2 y 2, …(i), −, =1, a 2 b2, Given that distance between foci = 32, and eccentricity, e = 2 2, ∴, 2 ae = 32, a =4 2, (Q e = 2 2 ), ⇒, Also,, b2 = a 2 (e2 − 1), ⇒, b2 = 32 (8 − 1) = 224, ⇒, b2 = 224, 2, On putting a = 32 and b2 = 224 in Eq. (i), we get the, equation of hyperbola which is, x2, y2, −, =1, 32 224, ⇒, 7x2 − y2 = 224, 51. Distance between the foci = 10, ⇒, 2ae = 10 ⇒ ae = 5, 5, and, e=, 4, ∴, ae = 5, 5, a × =5, ⇒, 4, ⇒, a = 4 ⇒ a 2 = 16, Using the relation, b2 = a 2 (e2 − 1), , 25, b2 = 16 , − 1 ⇒ b2 = 9, ⇒, , 16, ∴ Equation of hyperbola is, ⇒, 52., , ∴ Eccentricity, e′ = 1 +, , x2 y2, −, =1, a 2 b2, , x2 y 2, −, =1, 16 9, , The equation of ellipse is, x2, y2, +, =1, 7, a2, 7, a2, Therefore, foci of ellipse are (± ae, 0) i.e.,, , 7 , ± a 1 − 2 , 0, , , a, , Here, eccentricity e = 1 −, , Now, the equation of hyperbola is, 1, x2, y2, −, =, 144 81 25, x2, y2, −, =1, ⇒, 144 81, 25, 25, , =, , 81 /25, 144 + 81, 225, =, =, 144 /25, 144, 144, 2, , Q e′ = 1 + b , , a 2 , , , 15, 12, , ∴ Foci of hyperbola are (± ae, 0) i.e.,, , 12 15, ., , 0 , i.e., ( ± 3 , 0), ±, , 15 12, Given that these foci coincides, ∴, , 3=a 1−, , ⇒, , 9, 7, =1 − 2, a2, a, 16, =1, a2, a =4, , ⇒, ⇒, , 7, a2, , 53. General equation is, …(i), ax2 + 2 hxy + b y2 + 2 gx + 2 fy + c = 0, 2, (A) x + x + 1 − y = 0, On comparing with Eq.(i), we get, 1, 1, a = 1, h = 0, b = 0, g = , f = − , c = 1, 2, 2, ∴, ∆ = abc + 2 fgh − af 2 − bg 2 − ch 2, 2, 1, ⇒, ∆ = 0 + 0 − 1 − − 0 − 0, 2, 1, ∆ = − ≠0, ⇒, 4, and h 2 = 0 = ab, ∴ It is parabola., (B) x2 + y2 + 2x + 2 y − 6 = 0, On comparing with Eq. (i), we get, a = 1, b = 1, h = 0, g = 1, f = 1, c = − 6, ∴, ∆ = abc + 2 fgh − af 2 − bg 2 − ch 2, = 1 ⋅ 1 ⋅ (−6) + 0 − 1 (1)2 − 1 (1)2 − 0, = −6 − 1 −1 = − 8, ⇒ ∆ ≠0, and, a = b, h = 0, ∴ It is circle., (C), 2 x2 + 3 y2 + 4 x + 6 y = 0, On comparing with Eq. (i), we get, a = 2 , b = 3 , g = 2, f = 3 , h = 0 , c = 0, ∆ = 0 + 0 − 2 (3)2 − 3 (2)2 − 0, = − 18 − 12 = − 30 ≠ 0, 2, Now,, h = 0 and ab = 6, ⇒, h 2 < ab, ∴ It is an ellipse., (D) 3x2 − 2 y2 + 6x − 4 y = 0, Here, a = 3, b = − 2, c = 0, h = 0, g = 3, f = − 2, ∆ = 0 + 0 − 3( − 2)2 + 2 (3)2 − 0 = 6, ⇒, ∆ ≠0, Now, ab = − 6 and h 2 = 0, ∴, h 2 > ab, ∴ It is a hyperbola.
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566, , NDA/NA Mathematics, , 54. If e and e′ are the eccentricities of a hyperbola and its, conjugate, respectively, b2, Then,, e= 1+ 2, a, b2, 2, ⇒, e =1 + 2, a, a2, and, e′ = 1 + 2, b, a2, 2, ⇒, e′ = 1 + 2, b, 1, 1, 1, 1, ∴, + 2=, +, 2, 2, e, e′, b, a2, 1+ 2 1+ 2, a, b, a 2 + b2, a2, b2, = 2, + 2, = 2, 2, 2, a + b2, a +b, a +b, 1, 1, +, =1, ⇒, e2 e′ 2, 55. Given equation of ellipse is, x2, y2, + 2 =1, 16 µ, Here, a 2 = 16 and b2 = µ 2, µ2, µ2, ∴, e= 1− 2 = 1−, 16, 4, , (µ < 4), , 16 − µ 2, 16, ∴ Foci of the ellipse are (± ae, 0),, ± (16 − µ 2 , 0, i.e.,, , , Given, equation of hyperbola is, 1, x2, y2, −, =, 144 81 25, x2, y2, =1, −, ⇒, 2, 2, 9, 12, , , 5, 5, 2, , 9, 12, a 2 = , b2 = , 5, 5, , 2, , 81 15 5, b2, = 1+, =, =, 2, 144 12 4, a, ∴ Foci of the hyperbola are (± ae, 0) i.e., (±3, 0)., By the given condition, 16 − µ 2 = 3, , ∴, , e= 1+, , 16 − µ 2 = 9, µ 2 = 16 − 9, ⇒, µ2 = 7, 13, 10, 56. Q e =, and length of latusrectum =, 3, 3, 5a, 2b2 10, 2, =, ⇒ b =, ⇒, 3, a, 3, We know that,, b2 = a 2(e2 − 1), 5a, , 13, = a2 , − 1, ⇒, , 9, 3, ⇒, , 5a 4a, =, 3, 9, , 57. The length of the intercept made by the line, 1, on the parabola y2 = 16x, y= 3 x+, 3, 4, 1, , =, 4 (1 + 3) 4 − 3 ⋅ , , 3, 3, 4, = ⋅2 ⋅2 4 − 1, 3, 16, =, 3 units, 3, Hence, A and R both are true and R is the correct, explanation of A., , 59. The equation of a tangent of slope m to the parabola, y2 = 9x is, , 9, 4m, It passes through (4, 10), then, 9, 10 = 4m +, 4m, ⇒, 16m2 − 40m + 9 = 0, ⇒, (4m − 1) (4m − 9) = 0, 1 9, m= ,, ⇒, 4 4, Hence, both A and R are true and R is the correct, explanation of A., y = mx +, , ⇒, , 2, , 4a 2 − 15a = 0, a (4a − 15) = 0, 15, a=, ⇒, (Q a ≠ 0), 4, 5 × 15, [from Eq. (i)], ∴, b2 =, 3 ×4, 5, ⇒, b=, 2, 2 × 15 15, units, =, ∴Length of transverse axis = 2a =, 4, 2, , 58. Given equation is 5x2 + 10xy + 5 y2 + 7x + 14 y + 5 = 0, On comparing with general equation of second degree, in two variables, i. e. , ax2 + 2hxy + by2 + 2 gx + 2 fy + c = 0,, we get, 7, a = 5, h = 5, b = 5, g = f = 7, c = 5, 2, Now,, 7, abc + 2 fgh − af 2 − bg 2 − ch 2 = 5 × 5 × 5 + 2 × 7 × × 5, 2, − 5 (7)2 − 5 (7 /2)2 − 5 (5)2, 225, = 125 + 225 − 225 −, − 125, 4, − 225, =, ≠0, 4, and, ab − h 2 = 5 × 5 − 52, =0, ∴ It represents a parabola., Hence, both A and R are true and R is the correct, explanation of A., , =, , Here,, , ⇒, ⇒, , …(i), , 60. The equation of given line is, y = 4x + c, On comparing with y = mx + c, we get, m= y
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567, , Conic Sections, and equation of an ellipse is, x2 y2, +, =1, 4, 1, Here,, a 2 = 4, b2 = 1, Now,, c2 = 4 × 16 + 1, [Q c2 = a 2m2 + b2], ⇒, = 65, ⇒, c = ± 65, Thus, the number of values of c such that the straight, x2, line y = 4x + c touches the curve, + y2 = 1 is 2., 4, Hence, both A and R are individually true and R is the, correct explanation of A., 61. Since, 4(3)2 + 3(5)2 − 12 = 4 × 9 + 3 × 25 − 12, = 36 + 75 − 12, = 99 > 0, Thus, the point (3, 5) lies outside the ellipse., Hence, both A and R are individually true and R is the, correct explanation of A., x2 y2, +, =1, 4, 5, Here,, a = 5, b = 2, 4, 1, ⇒, e= 1− =, 5, 5, Now, SP + S′ P = Sum of focal distances from point, P (3, 5), = 2a = 2 5, ⇒ A is false but R is true., , 62. Equation of ellipse is, , 63. Assertion (A) From figure,, y, B, , x′, , A′, , P, S, , O, , M, A, , Z, , 64. Given equation is 9x2 − 16 y2 − 18x + 32 y − 151 = 0, On comparing, with general equation of second degree, in two variables i.e., ax2 + 2hxy + by2 + 2 gx + 2 fy + c = 0,, we get, a = 9, h = 0, b = − 16, g = − 9, f = 16, c = − 151, Now, abc + 2 fgh − af 2 − bg 2 − ch 2 = 9 × − 16 × − 151, + 2 × 16 (− 9) × 0 − 9 (16)2 − (− 16) (− 9)2 − (− 151) (0)2, = 21744 − 0 − 2304 + 1296 − 0, = 20736 ≠ 0, and, h 2 − ab = 02 − 9 (− 16), = 14 > 0, ⇒, h 2 > ab, ∴ It represents a hyperbola., Hence, both A and R are true and R is the correct, explanation of A., , Solutions (Q.Nos. 65-67), 65. The vertex of parabola is the mid-point of foot directrix, and focus., 0 + 2 0 + 0, i.e.,, ,, , = (1, 0), 2, 2 , 66. Using (PS )2 = (PM )2, , (x − 2)2, 1, x2 + y2 = x2 + 4 − 4x, y2 = − 4 (x − 1), , ∴ (x − 0)2 + ( y − 0)2 =, ⇒, ⇒, , 67. Let S1 ≡ y2 + 4 (x − 1), At point (1, 4), S1 = 42 + 4 (1 − 1), = 16 > 0, Hence, point lies outside the parabola., , Solutions (Q. Nos. 68-70), x, , B′, y′, , AA′ = Length of major axis = 9, BB′ = Length of minor axis = 5, ZM = Directrix, S = Focus, SP = 11, PM = 9, SP, For ellipse,, = e < 1, e is the eccentricity., PM, 11, ⇒, <1, 9, ⇒ 11 < 9, which is not possible., Reason (R) It is true that eccentricity of an ellipse, is always less than unity., ∴ A is false but R is true., , x2, y2, +, =1, 16, 9, Here,, a = 4, b = 3, 9, 7, ∴, e= 1−, =, 16, 4, ∴ Coordinator of foci are (± ae, 0) i. e. , (±, , 68. Given,, , 69. ∴Distance between foci is 2ae, 7, =2 ×4 ×, 4, =2 7, 70. Here, radius = (0 ±, , 7 )2 + (3 − 0)2, , = 7 + 9 =4, ∴ Equation of circle is, (x − 0)2 + ( y − 3)2 = 42, ⇒, x2 + y2 − 6 y + 9 = 16, ⇒, x2 + y2 − 6 y − 7 = 0, , 7 , 0)
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28, , Vector Algebra, Scalar and Vector Quantities, Those quantities, which have only magnitude and are, not related to any direction in space are called scalar, quantities e.g., mass, speed, distance etc., Those quantities, which have both magnitude and, direction are called vector quantities. e.g., displacement,, velocity etc., , Representation of Vectors, A vector quantity can be represented by directed line, segment because this line segment has both magnitude, and direction. On joining O and P, we get a line segment OP, whose length is its magnitude and its direction is from O to, P. The direction is represented by an arrow. On this line, segment O is called origin and P is called end point or, terminus point. Vector OP is denoted by a or a, Modulus of vector OP is denoted by|OP|., , Types of Vectors, (i) Free vectors Those vectors, which are not, related to specific points are called free vectors. These, vectors do not alter due to transfer value of free vector, depends only upon its direction and magnitude. It does not, depend upon its position in space., (ii) Localised vector A vector which can be, shifted only along its line of action is called localised vector., (iii) Like vectors The two vectors of same direction, are called like vectors. In the given figure vectorsa and b are, like vectors., a, , (v) Unit vector A vector whose modulus is unity,, is known as unit vector. Unit vector of a is denoted by a$ and, a, ., a$ =, |a|, (vi) Equal vectors If magnitude and direction of, two vectors are same, then those are called equal vectors., (vii) Equality of vectors Two vectorsa and b are, said to be equal, if (i) a = b (ii) They have the same or, parallel support and (iii) The same sense., (viii) Zero vector A vector whose modulus is zero,, is known as zero vector. The direction of zero vector cannot, be determined., (ix) Negative vector A vector having the same, magnitude as that of a given vector and direction opposite, to that given vector is called as negative vector., (x) Collinear or parallel vectors Two or more, vectors are known as collinear vectors, if they are parallel, to a given straight line. The magnitude of collinear vectors, can be different., (xi) Coplanar vectors Those vectors, which are, parallel to the same plane are called coplanar vectors., (xii) Position vector If O is a fixed point in a, space, then vector OP which represents the position of a, point P, is known as position vector of P., P, , (xiii) Displacement vectors If a particle is, displaced from A to B, then displacement AB is a vector and, AB is called the displacement vector of that particle., AB = OB − OA, , b, , A, , (iv) Unlike vectors The vectors having opposite, line of action are called unlike vectors. In the given figure a, and b are unlike vectors., a, b, , O, , B
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569, , Vector Algebra, , Addition of Vectors, , %, , The addition of two vectors a and b is denoted by a + b, and it is known as resultant of a and b ., There are three methods of addition of vectors., (a) Triangle law of vector addition If two, vectors a and b lie along the two sides of a triangle in, consecutive order, , If ABCDEF is a hexagon, then, AB + AC + AD + EA + FA = 4AB, , C, , F, , B, , A, , b, , a+, , D, , E, , b, , %, , Resultant vector of vectors λ OA and µ OB is (λ + µ ) OG, where, G divides line segment AB in the ratio µ : λ., , B, , a, , (as shown), then third side represents the sum, (resultant) a + b., (b) Parallelogram law of vector addition, If two vectors lie along two adjacent sides of a, parallelogram (as shown) the diagonal of the, parallelogram through the common vertex represents, their sum., , a, , O, , G, , µ, , G, µ )O, (λ+, O, %, , b, , λ, , µ OB, , +b, , A, , λ OA, , If position vectors of point A and B are a and b which are wrt, origin, then G, which divides line AB in the ratio p : q, is, qa + pb , ., , p+ q , , Example 1. If a, b, c be the vectors represented by the sides, , a, , (c) Polygon law of vector addition If ( n − 1), sides of a polygon represents vector a1 , a2 , a3 ,... in, consecutive order, then nth side represents their sum., a3, , a1+ a2+ a3, , of a triangle taken in order, then a + b + c is equal to, (a) a, (b) − a, (c) 0, (d) None of the above, , Solution (c) Let ABC be a triangle such that, BC = a , CA = b and AB = c. Then,, A, , a2, , O, a1, , c, , Properties of Addition of Vectors, 1. Addition of vectors is commutative. i.e.,, a+ b= b+ a, 2. Addition of vectors is associative. i.e.,, (a + b ) + c = a + ( b + c), %, , If AD, BE and CF are medians of ∆ ABC, then, AD + BE + CF = 0, , A, , E, , F, , B, , D, , C, , b, , B, , C, a, , a + b + c = BC + CA + AB, = BA + AB, = − AB + AB, a + b + c =0, , (QBC + CA = BA), , Example 2. P is the point of intersection of the diagonals of, the parallelogram ABCD. If, O is any point, then, OA + OB + OC + OD is equal to, (a) OP, (b) 2 OP, (c) 3 OP, (d) 4 OP, Solution (d) We know that, P will be the mid-point of AC and, BD
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570, , NDA/NA Mathematics, , Orthonormal Triad of Unit Vectors, , O, , D, , Let i, j and k represents unit vectors along x, y and z, axes. Let a can be represents in terms of i, j and k as, a = ax i + a y j + az k, , C, , Y, , P, A, , j, , B, , ∴, , OA + OC = 2 OP, , ...(i), , and, , OB + OD = 2 OP, , ...(ii), , k, , On adding Eqs. (i) and (ii), we get,, , i, , X, , Z, , OA + OB + OC+ OD = 4 OP, , Subtraction of Vectors, Ifa and b are two vector, then their subtractiona − b is, defined as a − b = a + ( − b ) where −b is the negative of b, having magnitude equal to that of b and direction opposite, to b . If a = a1i + a2 j + a3 k, b = b1i + b2 j + b3 k., , where ax , a y and az are components of a along x, y and z, axes., Modulus of a = a =|a|= ax2 + a 2y + az2 ., %, , %, , If two vectors a = (a1 , a 2 , a3 ) and b = (b1 , b 2 , b3 ) are equal, then, their resolved parts will also equal., i.e.,, a1 = b1 , a 2 = b 2 and a3 = b3 ., The resolved parts of a resultant vector of addition of two vectors, are equal to the sum of resolved parts of those vectors., , Then, a − b = ( a1 − b1 )i + ( a2 − b2 ) j + ( a3 − b3 )k, , Important Results, , B, a+b, O, , b, A, , a, , –b, a + (–b) = a – b, , 1. If α,β and γ are the angles subtended by vector a on, coordinate axes x, y and z. Then, cos α , cos β and cos γ are, known as direction cosines of vector a., ay, a, a, cos α = x , cos β =, , cos γ = z, | a|, | a|, | a|, cos2 α + cos2 β + cos2 γ = 1., , and, B′, , Properties of Vector Subtraction, (i) a − b ≠ b − a, (ii) (a − b ) − c ≠ a − ( b − c), (iii) Since, any one side of a triangle is less than the sum, and greater than the difference of the other two sides,, so for any two vectors a and b, we have, (b) a + b ≥ a − b, (a)|a + b|≤ a +| b|, (c) a − b ≤ a + b, , (d) a − b ≥ a − b, , 2. Unit vector along vector a is, ay, a, a, a$ = x i +, j + z k., | a|, | a|, | a|, , Example 3. The unit vector along 3i + 4j + 7k is, (a) (3i + 4j + 7k), (b) (3i + 4j + 7k) / 74, (c) (3i − 4j + 7k) / 74, (d) None of the above, , Solution (b) Unit vector along vector 3i + 4j + 7k, , Multiplication of a Vector by, Scalar, Let a is a vector and λ be any scalar, then product of, vector by a scalar will be λ a or a λ., , Properties of Multiplication of a Vector by, Scalar, If λ and µ be two scalars, then, (i) λ (µ a) = λµ a, (ii) ( λ + µ ) a = λ a + µ a, (iii) λ (a + b ) = λ a + λ b, , 1, (3i + 4j + 7k), 9 + 16 + 49, 1, =, (3i + 4j + 7k), 74, =, , Example 4. The unit vector in ZOX plane and making, , angles 45° and 60°, respectively with a = 2i + 2 j − k and, b = 0i + j − k is, 1, 1, 1, 1, (a) −, (b), i−, k, i+, k, 2, 2, 2, 2, 1, 4, 1, (c), i+, j+, k (d) None of these, 3 2, 3 2, 3 2
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571, , Vector Algebra, , Solution (b) Let the required vector be c = xi + zk, Since ,, , |c|=1 ⇒ x + z = 1, 2, , 2, , a and c are inclined at the angle 45°., 2 x− z, 3, ⇒ 2x − z =, cos 45° =, ∴, 4 + 4 +1, 2, b and c are inclined at the angle 60 °., z, 1, ∴, −, = cos 60 ° ⇒ z = −, 2, 2, 1, From Eqs. (ii) and (iii), we get x =, 2, 1, 1, c=, i−, k, ∴, 2, 2, , …(i), …(ii), , If a1v1 + a2v 2 + a3 v3 + ... + an v n = 0,, then, combination of vectors is known as linearly dependent, combination of vectors., , Condition for Three Points to be Collinear, Let the position vectors of three points are a, b , c and, a + λb + µc = 0, where 1 + λ + µ = 0, then three points will, be collinear., , …(iii), , Position Vector, If a point O is fixed as the origin in space (or plane ) and, P is any point, thenOP is called the position vector of P with, respect to O., P, r, , O (Origin), , If we say that P is the point r, then we mean that the, position vector of P is r with respect to some orgin O., (1) AB in terms of the position vectors of point A, and B If a and b are position vectors of points A and B, respectively. Then , AB = a, OB = b, ∴ AB = (Position vector of B) − (Position vector ofA), = OB − OA = b − a, (2) Position vector of a dividing point The, position vectors of the points dividing the line AB in the, mb + na, ration m : n internally or externally are, or, m+n, mb − na, ., m−n, , Example 5. The position vectors of P and Q are, respectively, a and b. If R is a point on PQ such that, PR = 5 PQ, then the position vector of R is, (a) 5 b − 4a, (b) 5 b + 4a, (c) 4b − 5 a, (d) 4b + 5 a, Solution (a) Given, PR = 5PQ, It means R divides PQ externally in the ratio 5 : 4., 5b − 4a, ∴ Position vector of R =, = 5b − 4a, 5−4, , Linear Combination of Vectors, If some vectors v1 , v 2 , v3 , ... , v n can be written as a, where, vector, a1v1 + a2v 2 + a3 v3 + ... + an v n ,, a1 , a2 , a3 ,... , an are scalars, then this vector is known as, linear combination of vectors v1 , v 2 , K , v n ., , Example 6. If a and b are non-collinear vectors, find the, value of x for which vectors, α = ( x − 2) a + b, and, β = (3 + 2 x) a − 2 b are collinear., 1, 1, (b), (a), 4, 3, 1, 1, (c), (d), 2, 5, , Solution (a) Since, the vectors α and β are collinear., ∴ There exist scalar λ such that, α = λβ, ⇒, ( x − 2) a + b = λ {(3 + 2x) a − 2b }, ⇒, , ( x − 2) − λ (3 + 2x) a + (1 + 2λ ) b = 0, , ⇒, , x − 2 − λ (3 + 2x) = 0 and 1 + 2λ = 0, 1, x − 2 − λ (3 + 2x) = 0 and λ = −, 2, 1, x − 2 + (3 + 2x) = 0, 2, , ⇒, ⇒, ⇒, ⇒, , 4x − 1 = 0, 1, x=, 4, , Thus, if a , b are non-collinear vectors, but α, β are collinear,, 1, ., 4, , then x =, , Linear Independent and Dependence of, Vectors, 1. Linearly independent vectors A set of, non-zero vectors a1, a2 ... an is said to be linearly, independent, if x1a1 + x2a2 + .... + xn an = 0, ⇒ x1 = x2 = .... = xn = 0, 2. Linearly dependent vectors A set of, vectors a1 , a2 ,... an is said to be linearly dependent, if there, not all zero such that, exist scalars x1 , x2 ,... , xn, x1a1 + x2a2 + ....+ xn an = 0., Three vectors a = a1i + a2 j + a3 k , b = b1i + b2 j + b3 k, and c = c1i + c2 j + c3 k will be linearly dependent vectors iff, a1, , a2, , a3, , b1, c1, , b2, c2, , b3 = 0, c3
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572, , NDA/NA Mathematics, , Properties of Linearly Independent and, Dependent Vectors, (i) Two non-zero, non-collinear vectors are linearly, independent., (ii) Any two collinear vectors are linearly dependent, (iii) Any three non-coplanar vectors are linearly, independent., (iv) Any three coplanar vectors are linearly, dependent., (v) Any four vectors in 3-dimensional space are, linearly dependent., , 7., If, and, a = i + j + k, b = 4i + 3j + 4k, c = i + α j + β k are linearly dependent vectors and | c| = 3,, then the value of α and β are, (a) ±2, 1, (b) ±11, ,, (c) ±1, 2, (d) None of these, , Example, , Solution (b) Q a , b and c are linearly dependent vectors., ⇒, ⇒, , [a b c ] = 0, 1 1 1, 4 3 4 =0, 1 α β, , ⇒, ⇒, Now ,, ⇒, ⇒, , 1 (3β − 4α ) − 1 ( 4β − 4) + 1( 4α − 3) = 0, −β + 1= 0 ⇒ β =1, | c| = 3, 1 + α 2 + β2 = 3, 1+ 1+ α 2 = 3, , ⇒ α 2 = 1⇒ α = ± 1, , Example 8. The vectors i − 3j + 2 k and 2i − 4j − 4k and, 3i + 2 j − k are, (a) linearly dependent, (b) linearly independent, (c) Both ‘a’ and ‘b’, (d) None of these, Solution (b) Let x, y , and z be scalars such that, x ( i − 3j + 2k) + y(2i − 4j − 4k) + z(3i + 2j − k) = 0, ⇒ ( x + 2y + 3z) i + ( −3x − 4y + 2z) j + (2x − 4y − z) k = 0, ⇒ x + 2y + 3z = 0 , − 3x − 4y + 2z = 0 and 2x − 4y − z = 0, This is a homogeneous system of equation. The determinant, 1, , 2, , b, , b, , θ, , θ, a, , Properties of Scalar Product, (i) The scalar product of two vectors is commutative, i.e., a ⋅ b = b ⋅ a., (ii) If m and n be any two scalars and a and b be any, two vectors, then ( m a) ⋅ ( n b ) = ( n a) ⋅ ( m b )., (iii) a ⋅ ( b + c) = a ⋅ b + a ⋅ c (distributive law), (iv) a ⋅ a =|a|2, (v) If two vectors a and b are perpendicular to each, other, then a ⋅ b = 0., (vi) If a = a1i + a2 j + a3 k and b = b1i + b2 j + b3 k, then, a ⋅ b = a1b1 + a2b2 + a3 b3 ., (vii) i ⋅ i = j ⋅ j = k ⋅ k = 1 and i ⋅ j = j ⋅ k = k ⋅ i = 0., (viii) If a = a1i + a2i + a3 k and b = b1i + b2 j + b3 k , then, , a⋅b , a1b1 + a2b2 + a3 b3, , cos θ = , =, , |a||b| a12 + a22 + a32 b12 + b22 + b32 , (ix) If a1a2 + b1b2 + c1c2 = 0, then both vectors are, a, a, a, perpendicular to each other and if 1 = 2 = 3 ,, b1 b2 b3, then both vectors are parallel to each other., (x) (a + b + c)2 = a2 + b 2 + c2+2(a ⋅ b + b ⋅ c + c ⋅ a), b ⋅a, (xi) Projection of b along a =, |a|, a⋅b, and projection of a along b =, ., |b|, (xii) Let a particle be placed at O and a force F, represented by OB be acting on the particle at O., Due to the application of force F, the particle is, displaced in the direction of OA., Let OA be the displacement d., B, , F, , 3, , of coefficient matrix is −3 − 4 2 = 11 ≠ 0, 2 − 4 −1, So, the system of equations has only the trivial solution given, by x = 0 = y = z., Hence, the set of given vectors is linearly independent., , Scalar Product, If acute angle between two vectors a and b be θ and, their modulus, respectively a and b. Then, their scalar, product is ab cosθ., ∴, a ⋅ b = ab cosθ, , a, , O, , θ, , A, , Then, total work done = F ⋅ d = Fd cosθ., , Example 9. Find the angle between the vectors, 3i + 2j − 6k and 4i − 3j + k., , (a) 90°, (c) 60°, , (b) 45°, (d) 0°, , Solution (a) Let a = 3i + 2j − 6k and b = 4i − 3j + k, Now, a ⋅ b = 12 − 6 − 6 = 0, ∴ Two vectors a and b are perpendicular., i.e., the angle between the vectors a and b = 90 °
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573, , Vector Algebra, , Example 10. Two forces F1 = 3i − 2 j + k and F2 = i + 3j − 5 k, acting on a particle at A move it to B. Find the work done, if, the position vectors of A and B are − 2i + 5 k and 3i − 7j + 2 k., (a) 24, (b) 23, (c) 22, (d) 25, , Solution (d) Let R be the resultant of two forces F1 and F2 and d, be the displacement., Then,, , R = (3i − 2j + k) + ( i + 3j − 5k), , and, , d = (3i − 7j + 2k) − ( − 2i + 5k), , = 4i + j − 4k, = 5i − 7j − 3k, ∴ The total work done = the work done by resultant, = R⋅d, = ( 4i + j − 4k) ⋅ (5i − 7j − 3k), , (iv) If a = a1i + a2 j + a3 k and b = b1i + b2 j + b3 k, then, a × b = ( a2b3 − a3 b2 )i + ( a3 b1 − a1b3 ) j, + ( a1b2 − a2b1 )k, i j k, or, a × b = a1 a2 a3, b1 b2 b3, (v) a × b = 0, If a = 0 or b = 0 or a and b are two collinear vectors., (vi) i × i = j × j = k × k = 0, and, i × j = k, j × k = i, k × i = j, and, j × i = − k, k × j = − i, i × k = − j, (vii) Perpendicular unit vector to a and, a× b, $ =±, b, n, |a × b|, (viii) Area of parallelogram, Area of parallelogram OACB, , = 20 − 7 + 12, = 25 units, , B, , Example 11. Let u, v and w be vectors such that, if u = 3, v = 4 and, u + v + w = 0,, u ⋅ v + v ⋅ w + w ⋅ u is equal to, (a) 25, (b) −25, (c) 30, (d) −30, 2, , 2, , 2, , Solution (b) u + v + w = u + v + w, ⇒, ⇒, , w = 5,, , C, , b, , then, θ, O, , M, , + 2 (u ⋅ v ⋅ + v ⋅ w + w ⋅ u), 0 = 9 + 16 + 25 + 2 ( u ⋅ v + v ⋅ w + w ⋅ v), u ⋅ v + v ⋅ w + w ⋅ u = − 25, , A, , a, , = OA × BM, = ab sinθ =|a × b|, , 2, , (ix) Area of triangle, 1, (a) Area of ∆ ABC = | AB × AC|, 2, A, , Vector Product (or Cross Product), The vector product of two vectors a and b is a vector, and is given by, $, a × b = ab sin θ n, B, B, n, θ, O, , b, a, , A, , $ is a, where θ be the angle between a and b , and n, perpendicular unit vector to plane of a and b ., ⇒, |a × b|=|a||b|sin θ, , Properties of Vector Product, (i) Vector product is not commutative. i.e.,, a× b≠b×a, but, a× b= − b×a, (ii) If m and n be two scalars and a and b be two vector, quantities, then, ( m a) × ( n b ) = mn(a × b ) = ( n a) × ( m b ), (iii) a × ( b + c) = a × b + a × c (distributive law), , C, , (b) If a, b and c are position vectors of A, B and C, respectively, then, 1, area of ∆ ABC = |a × b + b × c + c × a|, 2, (x) Moment, (a) About a point Moment = r × F, Where r be the position vector of any point P, and F be the force about the point O., (b) About a line The moment of a force F acting, at a point P about a line L is a scalar given by, ( r × F) ⋅ a$ ., Where, a$ is a unit vector in the direction of the, line, and OR = R, where O is any point on the, line., , Example 12. If a × b = c × d and a × c = b × d, then, a − d is parallel to (where a ≠ d and b ≠ c.), , (a) ( b − c), (c) ( b ⋅ c), , (b) ( b + c), (d) None of these
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574, , NDA/NA Mathematics, , Solution (a) ( a − b) × ( b − c), = a ×b−a ×c−d×b+ d×c, = c×d−b×d+ b×d−c×d, =0, So, (a − d) is parallel to (b − c)., , (vi) [a b c + d ] = [a b c] + [a b d ], (vii) [a + b b + c c+ a] = 2 [a b c], (viii) [a − b b − c c − a] = 0, (ix) [a × b b × c c × a] = [a b c]2, (x) Volume of parallelopiped = [a b c] where a, b, and c are adjacent sides of parallelopiped., , Example 13. Find the moment about the point i + 2 j + 3k of, a force represented by i + j + k acting through the point, 2i + 3j + k., (a) 3i + 3 j, (b) 3i − 3j, (c) 2i + 2 j, (d) None of these, , Solution (b) Here, r = (2i + 3j + k) − ( i + 2j + 3k), r = i + j − 2k, and, F = i + j + k., Then, the required moment is given by, r × F = (i + j − 2 k ) × (i + j + k ), i j k, , Tetrahedron, A tetrahedron is a three-dimensional figure formed by, four triangle OABC is a tetrahedron with ∆ ABC as the, base. OA, OB, OC , AB, BC and CA are known as edges of the, tetrahedron. OA, BC ; OB, CA and OC , AB are known as the, pairs of opposite edges. A tetrahedron in which all edges, are equal is called a regular tetrahedron. Any two edge of, regular tetrahedron are perpendicular to each other., A(a), , = 1 1 −2 = 3 i − 3 j, 1 1 1, , a, D, , ∴ Moment about a point = 3i − 3j, b, , Scalar Triple Product, The scalar triple product of three vectors a, b and c is, [a b c ]., or, [a b c] = a ⋅ ( b × c), If, a = a1i + a2 j + a3 k, b = b1i + b2 j + b3 k, and, c = c1i + c2 j + c3 k, then, a1 a2 a3, [a b c] = b1 b2 b3, c1 c2 c3, , Properties of Scalar Triple Product, (i) The value of scalar triple product does not depend, upon the position of dot and cross., a ⋅ ( b × c) = (a × b ) ⋅ c, (ii) If a, b , c are cyclically permuted the value of scalar, product remains same., The change of cyclic order of vectors in scalar triple, product changes the sign of the scalar but not the, magnitude., i.e.,, [a b c] = [ b c a] = [c a b ], and, [a b c] = − [b a c] = − [c b a], = − [a c b ], (iii) The scalar triple product of three vectors is zero, if, any two of them are equal., (iv) The scalar triple product of three vectors is zero, if, two of them are parallel or collinear., (v) If three vectors a, b and c are collinear, then, [a b c] = 0., , c, , B(b), , C(c), , Volume of Tetrahedron, (i) The volume of a tetrahedron, 1, = (area of the base) (corresponding altitude), 3, 1, = [ AB BC AD ], 6, (ii) If a, b, c, are position vectors of vertices A, B and C, with respect to O, then volume of tetrahedron, 1, OABC = [a b c], 6, (iii) If a, b, c, d, are position vectors of vertices A, B, C , D, of a tetrahedron ABCD, then its volume, 1, = [b − a c − a d − a], 6, , Example 13. If the volume of the parallelopiped with a, b, and c as coterminus edges is 40 cm units, then the volume of, the parallelopiped having b + c, c + a and a + b as coterminus, edges in cubic units is, (a) 80, (b) 120, (c) 160, (d) 40, , Solution (a) Given, volume of parallelopiped, [ a b c] = 40, ∴ Volume of parallelopiped, = [ b + c c + a a + b], = 2 [ a b c], = 2 × 40 = 80 cu units
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575, , Vector Algebra, , Example 14. If the vectors 2i − 3j + 4k, i + 2 j − k and, mi − j + 2 k are coplanar, then the value of m is, 5, 8, 2, 7, (b), (c) −, (a), (d), 8, 5, 3, 4, , Solution (b) For coplanar, 2, , −3, , 4, , 1, , 2, , −1 = 0, , m, , −1, , 2, , ⇒, , 2 ( 4 − 1) + 3 (2 + m) + 4 ( −1 − 2m) = 0, , ⇒, , m=, , 8, 5, , Vector Triple Product, If a, b and c are three vector quantities, then, a × ( b × c)represents the vector triple product and is given, by, a × ( b × c) = (a ⋅ c)b − (a ⋅ b ) c, , (iii) The formula a × ( b × c) = (a ⋅ c)b − (a ⋅ b ) c is true, only, when the vector outside the bracket is on the, left most side. If it is not, we first shift on left by using, the properties of cross product and then apply the, same formula., Thus, ( b × c) × a = − { a × ( b × c)}, = − {(a ⋅ c)b − (a ⋅ b ) c}, = (a ⋅ b ) c − (a ⋅ c)b, (iv) Vector triple product is a vector quantity., (v) a × ( b × c) ≠ (a × b ) × c, , Example 16. If a and b are unit vectors, then the vector, ( a + b) × ( a × b) is parallel to the vector, (a) a − b, (b) a + b, (c) 2a − b, (d) 2a + b, Solution (b) Now, ( a + b) × ( a × b), = a × ( a × b) + b × ( a × b), = ( a ⋅ a) a − ( a ⋅ a) b + ( b ⋅ b) a − ( b ⋅ a) b, = ( a ⋅ b) a − b + a − ( b ⋅ a) b, = ( a − b)( a ⋅ b − 1), Hence, given vector is parallel to ( a − b)., , Properties of Vector Triple, Product, (i) The vector triple product a × ( b × c) is a linear, combination of those two vectors, which are within, brackets., (ii) The vector r = a × ( b × c) is perpendicular to a and, lies in the plane of b and c ., , Comprehensive Approach, n, , n, , The vector along the bisector of angle between vectors a and b is, λ(a + b), where λ is a parameter., The unit vector along the bisector of angle between a and b is, a+b, , where θ is angle between a and b., 2 cos θ / 2, , θ /2, θ /2, , n, , a⋅ a, , n, , n, n, , n, , n, , n, , n, , For any three vectors a, b and c,, [ a − b b − c c − a] = 0., If diagonals of a parallelogram is along the vectors a and b, then, 1, area of parallelogram = |a × b|., 2, If a, b and c are position vectors of A, B and C of ∆ ABC, then, a + b+ c, position vector of centroid G will be, and, 3, GA + GB + GC = 0., If D, E and F are mid-points of AB, BC and CA respectively of, ∆ ABC, then AD + BE + CF = 0., If circumcentre, centroid and orthocentre of triangle ABC are O, G, and H, then, , OA + OB + OC = 3OG = OH, and, HA + HB + HC = 3HG = 3HO, [a × b b × c c × a] = [a b c] 2, , n, , n, , n, , a⋅ b a ⋅ c, , = b⋅ a, , b⋅ b, , b⋅ c, , c⋅ a, , c⋅ b, , c⋅ c, , 1, Area of quadrilateral ABCD = |AC × BD|,, 2, where AC and BD are diagonals of quadrilateral., Volume of Tetrahedron If A, B, C, D are vertices of tetrahedron, 1, ABCD, then its volume = [ AB AC AD]., 6, If F be a force and d be a displacement, then, Work done = F ⋅ d, Incentre formula The position vector of the incentre of ∆ABC is, aa + bb + cc, a+ b+c, Orthocentre formula The position vector of the orthocentre of, a tan A + b tanB + c tanC, ∆ABC is, tan A + tanB + tanC
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Exercise, Level I, 1. Three non-zero, non-parallel coplanar vectors are, always, (a) linearly dependent, (b) linearly independent, (c) Either ‘a’ or ‘b’, (d) None of these, 2. Projection of 2i + j + k along i − j + k is, (a) 2 / 3, (b) 1 / 3, (c) 3, (d) 2 3, 3. Area of parallelogram ABCD, when AB = i + j + k, and BD = − i + j is, (a) 1 sq unit, (b) 6 sq units, (c) 3 sq units, (d) None of these, 4. If a and b are two vectors such that a ⋅ b = 0 and, a × b = 0,which one of the following is correct?, (NDA 2011 II), , (a), (b), (c), (d), , a is parallel to b, a is perpendicular to b, Either a or b is a null vector, None of the above, , 5. Angle between the vectors a = − i − 2 j + k and, b = xi + j + ( x + 1)k, (a) is obtuse angle, (b) is acute angle, (c) is right angle, (d) depends on x, 6. Which of the following expression is meaningless?, a, (b) (a × b ) ⋅ ( c × d ), (a), b, (c) a × ( b × c), (d) (a × b ) ⋅ c, 7. Which of the following is wrong?, (a)|a|2 = a 2, ‘a’ is a real number., (b)|a|2 = − a 2, (c)|a|2 = a 2, ‘a’ is a complex number., (d) None of the above, , 11. If the points A, B and C with position vectors 2i + 2 j,, λi + 8 j and 8i + 32 j are collinear, then λ is equal to, 8, 16, (a), (b), 5, 5, (c) 4, (d) None of these, 12. If a, b and c are three unit vectors, then, a ⋅ b + b ⋅ c + c ⋅ a is equal to (a + b + c = 0), (a) 0, (b) 1, 3, (c), (d) None of these, 2, 13. For what value of m are the points with position, vectors 10i + 3 j,12i − 5 j and mi + 11 j collinear?, (NDA 2011 II), , (a) – 8, (c) 8, , (b) 4, (d) 12, , 14. Angle between the vectors 3 (a × b ) and b − (a ⋅ b )a is, π, (a), (b) 0, 2, π, π, (c), (d), 4, 3, 15. If a, b are two unit vectors at an angle θ, then, magnitude of a + b is, (a) 2, (b) 2, θ, (d) None of these, (c) 2 cos, 2, 16. A unit vector in xy-plane makes an angle 45° with the, vector i + j is, i+ j, (a) i, (b), 2, i− j, (d) None of these, (c), 2, , 8. If a and b are position vectors of the points A and B, respectively, what is the position vector of a point C, on AB produced such that AC = 2 AB ?, (NDA 2007 I), (a) 2a − b (b) 2 b − a (c) a − 2 b (d) a − b, , 17. If G is the centroid of ∆ABC, then GA + GB + GC is, equal to, (a) 0, (b) 3 GA, (c) 3 GB, (d) 3 GC, , 9. If c is the unit vector perpendicular to both the vectors, a and b, what is another unit vector perpendicular to, both the vectors a and b?, (NDA 2011 II), (a) c × a, (b) c × b, (a × b ), (a × b ), (c) −, (d), |a × b|, |a × b|, , 18. If (a × b )2 + (a ⋅ b )2 = 144 and|a| = 4, then|b| is equal, to, (a) 16, (b) 8, (c) 3, (d) 12, , 10. If x, a, b and c are non-zero vectors such that, a ⋅ x = 0, b ⋅ x = 0 and c ⋅ x = 0, then [a b c] is, (a) 0, (b)|a| |b| |c|, 1, (d) None of these, (c) |a| |b| |c|, 3, , 19. Let p, q, r and s be respectively the magnitude of the, vectors 3i + 2 j, 2i + 2 j + k, 4i − j + k and 2i + 2 j + 3k., Which one of the following is correct?, (NDA 2011 I), (a) r > s > q > p, (b) s > r > p > q, (c) r > s > p > q, (d) s > r > q > p
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577, , Vector Algebra, 20. The position vector of three points A, B and C are, i + j, i − j and l i + m j + n k , respectively. The points, are collinear, if, (a) l = m = n = 1, (b) l = 1, m , n ∈ R, (c) l = 1, n = 0, m ∈ R, (d) m = 0, n = 1, l ∈ R, , 30. ABCD is a quadrilateral. Forces AB, CB, CD and DA, act along its sides. What is their resultant?, , 21. The magnitude, the displacement of vector from, position (2, 4, 2) to position (6, 1, 12), is, (b) 2 3, (a) 5 5, (c) 3 7, (d) 3 8, , 31. The magnitude of the projection of the vector, A = i − 2 j + k on the vector B = 4i − 4 j + 7k lies, between, (a) 1 and 2, (b) 2 and 3, (c) 3 and 4, (d) 4 and 5, , 22. if a, b are two unit vectors and θ is the angle between, θ, them, then the value of cos is equal to, 2, 1, 1, (a) |a − b|, (b) |a ⋅ b|, 2, 2, |a × b|, 1, (d) |a + b|, (c), 2|a||b|, 2, 23. If xi + yj + zk is a unit vector and x : y : z = 3 : 2 : 3,, what is the value of z?, (NDA 2011 I), 3, (b) 3, (a), 16, 3, (c), (d) 2, 4, 24. Position vector of a point P is r from origin of, coordinate axes. A force F passes through the point P., The moment of the force about the orign is, (a) r × F, (b) r ⋅ F, (c) F ⋅ r, (d) zero, 25. The projection of the vector a = i − 2 j + k on the vector, b = 4i − 4 j + 7k is equal to, 6, 19, (a), (b), 9, 9, 9, 6, (c), (d), 19, 19, 26. If the vectors i − xj − yk and i + xj + yk are, orthogonal to each other, then what is the locus of the, point ( x , y )?, (NDA 2012 I), (a) A parabola, (b) An ellipse, (c) A circle, (d) A straight line, 27. If a = i − k,, , (NDA 2010 II), , b = xi + j + (1 − x)k and, c = yi + xj + (1 + x − y)k,, then a ⋅ (b × c) depends upon, , (a) x only, (c) both x and y, , (b) y only, (d) Neither x nor y, , 28. The area of the parallelogram with a = 2i + 3 j − 5k, and b = i + j − k as consecutive sides is equal to, (b) 2 14 sq units, (a) 61 sq units, (c) 2 7 sq units, (d) 14 sq units, 29. If|a + b|=|a − b|, then, (a) a is parallel to b, (c)|a|=|b|, , (b) a is perpendicular to b, (d) None of these, , (NDA 2010 II), , (a) 2 CD, (c) 2 BC, , (b) 2 DA, (d) 2 CB, , 32. The moment about the point A ( 3, − 1, 3) of a force, F = 2i + j + 4k through the point B ( 5, 1, 4) is, (a) 3i + 2 j − k, (b) 7i − 6 j − 2k, (c) 5i + j + 3k, (d) i + 2 j − k, 33. If|a × b|2 +|a ⋅ b|2 = 144 and|a|= 4, then|b| is equal, to, (a) 3, (b) 8, (c) 12, (d) 16, 34. Let a and b be two unit vectors and α be the angle, between them. If (a + b ) is also the unit vectors what, is the value of α ?, (NDA 2010 I), π, π, (a), (b), 4, 3, 2π, π, (d), (c), 3, 2, 35. If C is the mid-point of AB and P is any point not lying, on AB, then, (a) PA + PB = 2 PC, (b) PA + PB = PC, (c) PA + PB = − PC, (d) PA + PB = − 2 PC, 36. The position vectors of points A, B, C , D are, a, b , 2 a + 3 b and a − 2 b, respectively. The vector AC, is, (a) a − 3 b, (b) 3b − a, (c) a + 3b, (d) − a − b, 37. What are the unit vectors parallel to xy-plane and, perpendicular to the vector 4i − 3 j + k ? (NDA 2009 II), (a) ± ( 3i + 4 j) / 5, (b) ± ( 4i + 3 j) / 5, (c) ± ( 3i − 4 j) / 5, (d) ± ( 4i − 3 j) / 5, 38. If P is a point on the circumference of a semi-circle of, radius a being bounded by the diameter BC, then, which one of the following is correct?, (a) BP ⋅ PC = 1, (b) BP ⋅ PC = a, (c) BP ⋅ PC = a 2, (d) BP ⋅ PC = 0, 39. What is the vector in the xy-plane through origin and, perpendicular to the vector r = ai + bj and of the, same length?, (NDA 2009 II), (a) − ai − bj, (b) ai − bj, (c) − ai + bj, (d) bi − aj, 40. What is the area of the parallelogram having, diagonal a = 3i + j − 2k and b = i − 3 j + 4k ?, (a) 5 2 sq units, (b) 4 3 sq units, (c) 5 3 sq units, (d) 10 3 sq units
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578, , NDA/NA Mathematics, , 41. The magnitude of the vectors a and b are equal and, the angle between them is 60°. If the vectors λa + b, and a − λb are perpendicular to each other them,, what is the value of λ?, (NDA 2009 II), (a) 1, (b) 2, (c) 3, (d) 4, 42. For any vector a, what is the value of, (a)|a|, (c) 3|a|2, , 2, , (b) 2|b|, (d) 4|a|2, , 43. Consider the diagonals of a quadrilateral formed by, the vectors 3i + 6 j − 2k and 4i − j + 3k. The, quadrilateral must be a, (NDA 2009 II), (a) square, (b) rhombus, (c) rectangle, (d) None of these, 44. What is the area of the rectangle having vertices A,B, C, 1, 1, and D with position vectors − i + j + 4k, i + j + 4k ,, 2, 2, 1, 1, i − j + 4k and i − j + 4k ?, (NDA 2012 I), 2, 2, (a) 1/2 sq unit, (b) 1 sq unit, (c) 2 sq units, (d) 4 sq units, 45. If, ABCDEF is a regular hexagon and, AB + AC + AD + AE + AF = λ AD, then λ is equal to, (a) 2, (b) 3, (c) 4, (d) 6, 46. Let a, b and c be the position vectors of points A, B, and C, respectively. Under which one of the following, conditions are the points A, B and C collinear?, (NDA 2009 I), , (a), (b), (c), (d), , 51. What is the value of m, if the vectors, 2 i − j + k, i + 2 j − 3k and 3 i + mj + 5k are coplanar?, (NDA 2012 I), , | a × i|2 + |a × j|2 + |a × k|2 ?, 2, , 50. If r1 = λi + 2 j + k and r2 = i + ( 2 − λ ) j + 2k are such, that | r1|>| r2|, then λ satisfies, which one of the, followng?, (NDA 2008 II), (a) λ = 0, (b) λ = 1, (c) λ < 1, (d) λ > 1, , a × b is equal to 0, b × c is parallel to a × b, a × b is perpendicular to b × c, (a × b ) + ( b × c) + ( c × a) is equal to 0, , 47. What is the angle between the yz-plane and the, vector i + j + k?, π, 1, 1, (a) − cos−1, (b) cos−1, 2, 3, 3, −1 1, (c) 90°, (d) π − cos, 3, 48. Let a = (1, − 2 , 3) and b = ( 3, 1, 2) be two vectors and, c be a vector of length l and parallel to (a + b ). What is, the value of c?, (NDA 2008 II), 1, 1, (b), (a), ( −2, − 3, 1), (1, 0, 1), 14, 2, 1, (d) None of these, (c), ( −5, − 4, − 1), 42, 49. If c is normal to a and b, then which one of the, following statement is not correct ?, (a) c is normal to a + b, (b) c is normal to a − b, (c) c is normal to a × b, (d) c is normal to a + 2 b, , (a) −2, (b) 2, (c) −4, (d) 4, 52. What is the value of λ, if the triangle whose vertices, are i, j and i + j + λk,will be right angled?, (NDA 2008 II), , (a) 2, , (b) 0, , (c) –1, , (d) 1, , 53. What is the volume of the rectangular parallelopiped, formed by the vectors i, 2 j, 3k?, (a) 14 cu units, (b) 6 cu units, (c) 1 cu unit, (d) 14 cu units, 54. If in a ∆OAC, B is the mid-point of AC and OA = a, and OB = b , then what is the value of OC ?, (a) (a + b ) / 2, (b) 2 b − 2 a, (c) 2 b − a, (d) 3 a − 2 b, 55. The scalar triple product ( A × B) ⋅ C of three vectors, (NDA 2008 II), A , B and C determines, (a) volume of a parallelopiped, (b) volume of a tetrahedron, (c) volume of an ellipsoid, (d) None of the above, 56. OAB is a given triangle such that OA = a, OB = b., Also, C is a point on AB such that AB = 2 BC. What is, the value of AC?, 1, 1, (a) ( b − a), (b) ( b + a), 2, 2, 3, 3, (c) (a − b ), (d) ( b − a), 2, 2, 57. If a and b are two unit vectors inclined at an angle, 60° to each other, which one of the following is, correct?, (NDA 2008 II), (a) |a + b|< 1, (b) |a + b|> 1, (c) |a − b|< 1, (d) |a − b|> 1, 58. If a, b , c are the position vectors of corners A, B, C of a, parallelogram ABCD, then what is the position, vector of the corner D?, (a) a + b + c, (b) a + b − c, (c) a − b + c, (d) − a + b + c, 59. If a is a position vector of a point (1, − 3) and A is, another point ( −1, 5), what are the coordinates of the, point B such that AB = a ?, (NDA 2008 I), (a) ( 2, 0), (b) ( 0, 2), (c) ( −2, 0), (d) ( 0, − 2)
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579, , Vector Algebra, 60. In a ∆ABC, angle B is obtuse and D , E , F are the, middle points of sides BC , CA , AB, respectively., Which one of the following vectors has the greatest, magnitude?, (a) BC, (b) CA, (c) AB, (d) AD, 61. Let a and b be the position vectors of A and B,, respectively. If C is the point 3a − 2b , which one of, the following is correct?, (NDA 2008 I), (a) C is in between A and B., (b) A is in between C and B., (c) B is in between A and C., (d) A, B and C are not collinear., π, 62. If two unit vectors pand q make an angle with each, 3, 1, other, what is the magnitude of p − q?, 2, 3, 1, (a) 0, (b), (c) 1, (d), 2, 2, 63. What is the locus of the point ( x , y ) for which the, vectors ( i − xj − 2k) and ( 2i + j + yk) are orthogonal?, (NDA 2008 I), , (a) A circle, (c) A parabola, , (b) An ellipse, (d) A straight line, , 64.What are the values of x for which the two vectors, ( x 2 − 1) i + ( x + 2) j + x 2 k and 2i − x j + 3 k are, orthogonal?, 1, (a) No real value of x, (b) x = and x = − 1, 2, 1, (d) x = − 1 and x = 2, (c) x = − and x = 1, 2, 65. What is the area of the rectangle of which r = ai + bj, is a semi-diagonal?, (NDA 2008 I), , (a) a 2 + b2, (c) 4( a + b2 ), , (b) 2 ( a 2 + b2 ), (d) 4 ab, , 66. Which one of the following is correct? If the vector c is, normal to the vectors a and b, then c is (NDA 2007 II), (a) parallel to both a + b and a − b., (b) normal to a − b and parallel to a + b., (c) normal to a + b and parallel to a − b., (d) normal of both a + b and a − b., 67. Let O be the origin and P , Q and R be the points such, that PO + OQ = QO + OR. Then, which one of the, following is correct?, (a) P , Q and R are the vertices of an equilateral, triangle, (b) P , Q and R are the vertices of an isosceles, triangle, (c) P , Q and R are collinear, (d) None of the above, 68. If a = i + 2 j − 3k and b = 3i − j + λk and (a + b ) is, perpendicular to a − b , what is the value of λ ?, (a) – 2, (b) ± 2, (NDA 2007 II), (c) 3, (d) ± 3, 69. Two vectors a and b are non-zero and non-collinear., What is the value of x for which the vectors, p = ( x − 2) a + b and q = ( x + 1) a − b are collinear?, (NDA 2007 I), , (a) 1, , 1, (b), 2, , 2, (c), 3, , (d) 2, , 70. What is the magnitude of the moment of the couple, consisting of the force F = 3i + 2 j − k acting through, the point i − j + k and −F acting through the point, (NDA 2007 I), 2i − 3 j − k ?, (b) 3 5, (c) 5 5, (d) 7 5, (a) 2 5, , Level II, 1. i × (a × i) + j × (a × j) + k × (a × k) is always equal to, (a) a, (b) − a, (c) 2 a, (d) −2 a, 2. If a, b and c are three non-coplanar vectors, then, [a × b b × c c × a] is equal to, (a) [a b c], (b) 0, (c) [a b c]2, (d) None of these, 3. If vectors a, b and c are non-coplanar, then the vector, a × b , b × c and c × a are, (a) linearly dependent, (b) linearly independent, (c) depends on vectors, (d) None of these, 4. If the vectors − i − 2xj − 3 yk and i − 3xj − 2 yk are, orthogonal to each other, what is the locus of the, point ( x , y )?, (NDA 2011 II), (a) A straight line, (b) An ellipse, (c) A parabola, (d) A circle, , 5. If vectors a, b and c are mutually perpendicular, vectors such that |a| =|b| = 10,|c| = 1, then the, length of vector 2 a + 2 b + 40 c is, (a) 20, (b) 20 6, (c) 40 6, (d) None of these, 6. If the points A, B, C and D with position vectors, i + j + k, 2i + 3 j + k, i + 2 j + 5k and λi + 3 j + 4k are, coplanar, then λ is equal to, (a) 5, (b) 7, (c) 2, (d) None of these, 7. Consider the following statements in respect of the, vectors u1 = (1, 2, 3), u2 = (2, 3, 1), u3 = (1, 3, 2) and, (NDA 2011 II), u4 = (4, 6, 2), I. u1 is parallel to u4., II. u2 is parallel to u4., III. u2 is parallel to u3 .
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580, , NDA/NA Mathematics, Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Only III, (d) Both I and III, , 8. If the magnitudes of two vectors a and b are equal,, then which one of the following is correct?, (NDA 2012 I), (a) (a + b ) is parallel to (a − b ), (b) (a + b ) ⋅ (a − b ) = 1, (c) (a + b ) is perpendicular to (a − b ), (d) None of the above, 9. If a, b and c are three non-coplanar vectors, [a + b b + c c + a] is equal to, (a) [a b c], (b) 2 [a b c], (d) None of these, (c) [a b c]2, 10. If i, j and k are unit orthonormal vectors and a is a, vector, if a × r = j, then a ⋅ r is, (a) 1, (b) 0, (c) 1/ 2, (d) None of these, 11. What is the sine of angle between the vectors, (NDA 2011 I), i + 2 j + 3k and − i + 2 j + 3k ?, (a), (c), , 13, 7, 13, 7, , (b), , 13, 7, , (d) None of these, , 12. Unit vectors perpendicular to 2i − j + 2k and, i + j + 4k are, 3i + 3 j − 6k, − 6i − 6 j + 3k, (b) ± , (a) ± , , , , , , , 9, 52, ( i + j + k), (c) ±, (d) None of these, 3, 13. A tetrahedron has vertices at O (0, 0, 0), B (2,1,3) and, C (−1, 1, 2). Then, the angle between the faces OAB, and ABC will be, 19 , 17 , (b) cos−1 , (a) cos−1 , 35, 31, (c) 30°, (d) 90°, 14. |a| = 10,|b| = 2 and a ⋅ b = 12, then what is the value, of|a × b|?, (NDA 2012 I), (a) 12, (b) 16, (c) 20, (d) 24, 15. If the position vector of a point P with respect to, origin O is i + 3 j − 2k and that of a point Q is, 3i + j − 2k, what is the position vector of the bisector, of the ∠ POQ?, (NDA 2010 II), (a) i − j − k, (b) i + j − k, (c) i + j + k, (d) None of these, 16. If|a| = 5,|b| = 12 and|a + b|= 13, then|a − b|is equal, to, (a) 13, (b) 5, (c) 12, (d) 15, 17. If a, b , c are unit vectors, such that a + b + c = 0 and, m = a ⋅ b + b ⋅ c + c ⋅ a, then, (a) m < 0, (b) m > 0, (c) m = 0, (d) None of these, , 18. Given, the points A( −2, 3,−3), B( 3, 2, 5), C(1, − 1, 2),, D( 3, 2, − 4). The projection of the vector AB on the, axis of CD is, 41, (a) −, 7, 41, (b), 7, 44, (c), 7, (d) None of the above, 19. If vectors ax i + 3 j − 5k and xi + 2 j + 2ax k make an, acute angle with each other, for all x ∈ R, then a, belongs to the interval, 1 , (b) ( 0, 1), (a) − , 0, 4 , 6, 23 , (d) −, (c) 0, , , 0, 25, 25 , vectors and |a|= 7. If, 1, a × ( b × c) + b × ( c × a) = a, then the angle between, 2, a and c is, (a) π / 3, (b) π / 6, (c) π / 2, (d) None of these, , 20. b, , and, , c, , are, , unit, , 21. Let α , β , γ be distinct real numbers. The points with, position vectors, α i + β j + γ k, β i + γ j + α k, γ i + α j + β k, (a) are collinear, (b) form an equilateral triangle, (c) form a scalene triangle, (d) form a right angled triangle, 22. Let p and q be the position vectors of P and Q,, respectively, with respect to O and | p|= p,|q|= q., The points R and S divide line segment PQ internally, and externally in the ratio 2 : 3 , respectively. If OR, and OS are perpendicular, then, (a) 9 p2 = 4q 2, (b) 4 p2 = 9q 2, (c) 9 p = 4q, (d) 4 p = 9q, 23. Let a, b and c be the distinct non-negative numbers., If the vectors ai + aj + ck,i + k and ci + cj + bk lie on a, plane, which one of the following is correct?, (NDA 2010 II), , (a), (b), (c), (d), , c is the arithmetic mean of a and b., c is the geometric mean of a and b., c is the harmonic mean of a and b., c is equal to zero., , 24. If a + b + c = αd , b + c + d = β a, then a + b + c + d is, equal to, (a) (β + 1) a, (b) α a, (c) (α + 1) b, (d) α a + β b, 25. If A = (1, 1, 1) and C = ( 0, 1, − 1) are given vectors, then, a vector B satisfying the equation A × B = C and, A ⋅ B = 3 is, 1 2 2, 2 2, (a) , , , (b) 1, , , 3 3 3, 3 3, 2 2 2, 5 2 2, (c) , , , (d) , , , 3 3 3, 3 3 3
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581, , Vector Algebra, 26. PQRS is a parallelogram, where PQ = 3i + 2 j − mk,, PS = i + 3 j + k and the area of the parallelogram is, 90.What is the value of m?, (NDA 2010 II), (a) 1, (c) 2, , (b) –1, (d) –2, , 27. If A(a), B( b ) and C( c) be the vertices of a triangle, whose circumcentre is the origin, then orthocentre is, given by, a+b+c, (a) a + b + c, (b), 3, a+ b+ c, (c), (d) None of these, 2, 28. Let a be a real number and, α = i + 2 j, β = 2 i + a j + 10 k, γ = 12 i + 2a j + a k be, three vectors, then α , β and γ are, , (a) linearly dependent for all a., (b) linearly independent for all a., (c) linearly dependent for a < 0., (d) linearly dependent for a > 0., 29. If a = ( 2, 1, − 1), b = (1, − 1, 0), c = ( 5, − 1, 1), then, what is the unit vector parallel to a + b − c in the, opposite direction?, (NDA 2012 I), i − j − 2k, i − 2 j + 2k, (a), (b), 3, 3, 2i − j + 2k, (c), (d) None of these, 3, 30. What is the value of b such that the scalar product of, the vector i + j + k with the unit vector parallel to the, sum of the vectors 2i + 4 j − 5k and bi + 2 j + 3k is, unity?, (NDA 2010 II), (a) –2, (b) –1, (c) 0, (d) 1, 31. If A, B, C and D are four points with vectors 3i, 3 j, 3k, and i + j + k, respectively, then D is the, (a) orthocentre of ∆ ABC, (b) lies on a circle passing through A, B and C, (c) is collinear with A, B, (d) None of the above, 32. The vector 2 i + 2 j + 3 k is rotated about origin, through an angle θ and becomes 2 i + 3 j + 2 k. Then,, angle θ is, 43, 16, (a) θ = cos−1 , (b) θ = sin−1, 17, 17, (c) θ = sin−1, , 33, 27, , 37 , (d) θ = tan−1 , , 16 , , 33. If a is non-zero vector of modulus a and m is a, non-zero scalar, then m a is a unit vector, if the value, of a is equal to, 1, 1, (d), (a) m, (b)|m|, (c), |m|, m, , 34. If the vector a = i + aj + a 2 k, b = i + bj + b2 k,, c = i + cj + c2 k are three non-coplanar vectors and, a a 2 1 + a3, b b2 1 + b3 = 0, then the value of abc is equal to, c, , c2, , 1 + c3, , (a) 2, (c) 0, , (b) 1, (d) – 1, , 35. Consider the following statements, I. For any three vectors a, b and c,, a ⋅ {( b + c) × (a + b + c)} = 0, II. For any three coplanar unit vectors d, e and f,, ( d × e) ⋅ f = 1, Which of the statements given above is/are correct?, (NDA 2010 I), , (a) Only I, (c) Both I and II, , (b) Only II, (d) Neither I nor II, , 36. If |a|= 4,|b|= 4,|c|= 5 such that a ⊥ ( b + c),, b ⊥ ( c + a) and c ⊥ (a + b ), then the value of, |a + b + c| is equal to, (a) 5, (b) 7, (d) 57, (c) 13, 37. A force is represented in magnitude and direction by, the line joining the point A(1, − 2, 4) to the point, B( 5, 2, 3). The moment about the point ( −2, 3, 5) is, equal to, (a) 9i − j + 32k, (b) 9i + j + 32k, (c) 9i − j − 32k, (d) 9i − j − 8k, 38. What is the geometric interpretation of the identity, (NDA 2010 I), (a − b ) × (a + b ) = 2(a × b )?, I. If the diagonals of a given parallelogram are used, as sides of a second parallelogram, then the area, of the second parallelogram is twice that of the, given parallelogram., II. If the semi-diagonals of a given parallelogram are, used as sides of a second parallelogram, then the, area of the second parallelogram is half that of, the given parallelogram., Select the correct answer using the codes given below, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 39. If r × b = c × b and r ⋅ a = 0, where a = 2i + 3 j − k,, b = 3i − j + k, c = i + j + 3k, then the value of r is, equal to, 1, (a) ( i + j + k), (b) 2 ( i + j + k), 2, 1, (c) 2 ( − i + j + k), (d) ( i − j + k), 2, 40. If in a right angled ∆ABC, the hypotenuse AB = p,, then the value of AB ⋅ AC + BC ⋅ BA + CB ⋅ CA is equal, to, (b) p2, (a) 2 p2, 1 2, (c) p, (d) p, 2
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582, , NDA/NA Mathematics, , 41. If p = a − b , q = a + b and|a|=|b|= 2, then the value, of| p × q| is equal to, (a) 2 16 − (a ⋅ b )2, , (b) 16 − (a ⋅ b )2, , (c) 2 4 − (a ⋅ b )2, , (d) 4 − (a ⋅ b )2, , π, 42. If a and b are two unit vectors inclined at an angle ,, 3, then the value of|a + b| is, (a) greater than 1, (b) less than 1, (c) equal to 1, (d) equal to 0, 43. Forces of magnitudes 3 and 2 units acting in the, directions 5i + 3 j + 4k and 3i + 4 j − 5k, respectively, act on a particle which is displaced from the point, (1, 1, − 1) to ( 3, 3, 1). The work done by the forces is, equal to, (a) 80 2 units, (b) 40 2 units, (d) 8 2 units, (c) 16 2 units, 44. A vector b is collinear with the vector a = (2, 1, −1), and satisfies the condition a ⋅ b = 3. What is b equal, to?, (NDA 2010 I), (a) (1, 1 / 2, − 1 / 2), (b) ( 2 / 3, 1 / 3, − 1 / 3), (c) (1 / 2, 1 / 4, − 1 / 4), (d) (1, 1, 0), 45. If a a vector of magnitude 50 is collinear with the, 15, vector b = 6i − 8 j −, k and makes an acute angle, 2, with the positive direction of z-axis, then the vector a, is equal to, (a) 24i − 32 j − 30k, (b) −24i + 32 j + 30k, (c) 16i − 16 j − 15k, (d) − 12i + 16 j − 30k, 46. A particle moves along a circular path of radius r in, xy-plane. The position vector R of this particle as a, function of its y-coordinate is, (a) x 2 − y 2 i + yj, (b) r 2 − y 2 i + yj, (c), , y 2 − r 2 i − yj, , (d) r 2 − y 2, , i − yj, , 47. If a + 2 b + 3 c = 0 and, (a × b) + (b × c) + (c × a ) = λ (b × c), then the value of λ, is equal to, , (a) 2, (c) 4, , (b) 3, (d) 6, , 48. Which one of the following vectors of magnitude 51, makes, equal, angles, with, three, vectors, i − 2j + 2k, −4i − 3k, and c = j ?, ,b =, a=, 3, 5, (NDA 2009 I), (a) 5i − j − 5k, (b) 5i + j + 5k, (c) −5i − j + 5k, (d) 5i + 5 j − k, 49. If a + b + c = 0 and|a|= 6,|b|= 8 and|c|= 10, then the, value of a ⋅ b + b ⋅ c + c ⋅ a is equal to, (a) 100, (b) – 100, (c) 200, (d) – 200, , 50. If the sides of a parallelogram are 2i + 4 j − 5k and, i + 2 j + 3k, then the unit vector parallel to one of the, diagonals is equal to, 1, (a) ( 3i + 6 j − 2k), 7, 1, (b) ( 3i − 6 j − 2k), 7, 1, (c) ( −3i + 6 j − 2k), 7, 1, (d) ( 3i + 6 j + 2k), 7, 51. EFGHIJ is a regular hexagon in which vector, EJ = i + j and IH = 2i + j + 3k, then the vector JI is, equal to, E, , i+j, , J, , I, , F, , G, , H, , (a) i + 3k, (b) j + 3k, (c) 3i + 2 j + 3k, (d) 2i + 3k, 52. The position vectors of P, Q, R and S are, and, i + j + k, 2i + 5 j, 3i + 2 j − 3k, i − 6j − k, respectively, then which one of the following forms a, parallel pair?, (a) PQ and RS, (b) PR and OS, (c) PR and OR, (d) None of these, 53. The projection of the vector 4i − 3 j + k on the line, passing through the points ( 2, 3, − 1) and (−2, − 4, 3) is, (a) 4, (b) 3, (c) 2, (d) 1, 54. If P, Q and R are the mid-points of the sides AB, BC, and CA , respectively of a ∆ ABC and if a, p and q are, the position vectors of A, P and Q respectively, what, is the position vector of R?, (NDA 2008 II), (a) 2a − ( p − q ), (b) ( p − q ) − 2a, a ( p − q), (c) a − ( p − q ), (d), −, 2, 2, 55. What is the work done by a force of 3 Newton whose, line of action has direction cosines 2/3, 1/3 and 2/3 in, a displacement from the point (1, 3, 5) to the point, (7, 9, 2), the unit of length being a metre?, (a) 6 Nm, (b) 9 Nm, (c) 12 Nm, (d) 18 Nm, 56. If A ≠ O and both the conditions, (i) A ⋅ B = A ⋅ C and, (ii) A × B = A × C hold simultaneously, then, (a) B = C = O, (b) B = C, (c) B ≠ C, (d) B ≠ O , C ≠ O
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583, , Vector Algebra, 57. If n forces PA1 , PA 2 ,... , PA n diverge from point P, and n other forces A1Q , A 2Q,.....,A n Q converge to a, point Q, then the resultant of the 2n forces is, represented in magnitude and direction by, (a) n PQ, (b) n QP, (c) 2n PQ, (d) n 2PQ, 58. If A + t B is perpendicular to C, where, A = i + 2 j + 3k, B = − i + 2 j + k and C = 3i + j, then t, is equal to, (a) 4, (b) 5, (c) 6, (d) 7, 59. If a = a1i + a2 j + a3 k, where a1 , a2 , a3 are scalar, quantities and i, j, k are unit vectors in three, mutually, perpendicular, directions,, then, |a × i|2 +|a × k|2 is equal to, (b) a12 + a22 + a32, (a) a12 + a32, (c) 2a12 + a22 + a32, (d) a12 + 2a22 + a32, 60. Two forces are equal to 2 OA and 3 BO, their, resultant being λ OG, where G is the point on AB, 2, BG, such that, = − . What is the value of λ?, (NDA 2008 II), 3, AG, (a) 1, (b) –1, (c) 2, (d) None of these, 61. If the work done by a force i + j + 8 k along a given, vector in the xy-plane is 8 units and the magnitude of, the given vector is 4 3, then the given vector is, represented as, (a) ( 4 + 2 2) i + ( 4 − 2 2 ) j, (b) ( 4i + 4 2 j), (c) ( 4 2 i + 4 j), (d) ( 4 + 2 2 ) ( i + j), 62. The, coordinates, of, A,, B,, C,, D, are, ( 3, 5, − 3), ( 2, 3, − 1), (1, 2, 3) and ( 3, 5, 7), respectively., Then,, (a) AB is perpendicular to CD., (b) AB is parallel to CD., (c) the angle between AB and CD is π/ 3., (d) the angle between AB and CD is 2π/ 3., 63. OP and OQ are two lines in the xy-plane each of, length l and making angles of 30° and 60° with the, x-axis, mk is a vector. The cross product of vectors mk, and ( OP + OQ ) is equal to, 1, 1, (a) − ml ( 3 + 1) ( i − j) (b) ml ( 3 + 1) ( i − j), 2, 2, 1, 1, (d) − ml ( 3 − 1) ( i − j), (c) ml ( 3 − 1) ( i + j), 2, 2, 64. Consider the following, If a and b are the vectors forming consecutive sides of, a regular hexagon ABCDEF, then, (NDA 2008 I), I. CE = b − 2a, II. AE = 2b − a, III. FA = a − b, , Which of the above are correct?, (a) I and II, (b) II and III, (c) I and III, (d) I, II and III, 65. For what value of k, the points with position vector, 60 i + 3 j, 40 i − 8 j and k i − 52 j are collinear?, (a) k = 40, (b) k = − 40, (c) k = − 30, (d) k = 20, 66. If a, b and c are three vectors of which every pair is, non-collinear. If the vectors a + b and b + c are, collinear with the vectors c and a respectively, then, which one of the following is correct?, (a) a + b + c is a null vector., (b) a + b + c is a unit vector., (c) a + b + c is a vector of magnitude of 2 units., (d) a + b + c is a vector of magnitude of 3 units., 67. If a, b and c are unit vectors such that a is, perpendicular to the plane of b , c and the angle, π, between b and c is, . What is the value of, 3, (NDA 2008 I), |a + b + c|?, (a) 1, (b) 2, (c) 3, (d) 4, 68. If a = i + j, b = 2 j − k and r × a = b × a, r × b = a × b,, r, then what is the value of ?, | r|, ( i + 3 j − k), ( i − 3 j + k), (b), (a), 11, 11, (i + 3 j + k ), ( i − 3 j − k), (c), (d), 11, 11, 69. Which one of the following vectors represents the, unit vector parallel to the yz-plane and perpendicular, to the vector 3i + 4 j − 2k?, ( −2i + j − k), ( j + 2k), (a), (b), 6, 5, ( i + j), ( 2i + 3 j + 9k), (c), (d), 2, 94, 70. What is the number of, perpendicular to the, b = ( 0, 1, 1)?, (a) 1, (c) 3, , vectors of length 1 unit, vectors a = (1, 1, 0) and, (NDA 2008 I), , (b) 2, (d) 4, , 71. A vector v of magnitude 4 units is equally inclined to, the vectors i + j, k + j, k + i. Which one of the, following is correct ?, 4, 4, (a) v =, (b) v =, ( i + j − k), (i − j − k), 3, 3, 4, (d) v = 4( i + j + k), (c) v =, ( i + j + k), 3, 72. If a = (1, 2, − 3) and b = ( 3, − 1, 2), then which one of the, following vectors is perpendicular to a + b?, (a) 2a − b, (b) 2a + b (c) a + 2 b, (d) a − b, 73. How many unit vectors are there perpendicular to, both the vectors 2i + 3 j + 4k and i − 2 j + 3k?, (a) 1, (b) 2, (c) zero, (d) ∞
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584, , NDA/NA Mathematics, , 74. If a , b and c are non-zero vectors and, |(a × b ) ⋅ c| =|a || b||c|, which one of the following is, correct?, (NDA 2007 II), (a) a ⋅ b = b ⋅ c = c ⋅ a ≠ 0, (b) a ⋅ b = 0, (c) b ⋅ c = 0, (d) a ⋅ b = b ⋅ c = c ⋅ a = 0, 75. OA is a diagonal of a cube. The vector sum of the, three vectors determined by the diagonals of three, adjacent faces of the cube passing through the same, corner O has magnitude λ., Which one of the following is correct?, (a) λ = OA, (b) λ < OA, (c) λ = 3 ⋅ OA, (d) λ = 2 ⋅ OA, 76. A, B, C and D are four points, and E and F are the, middle points of AC and BD, respectively. What is, the value of the vector sum AB + CB + CD + AD ?, (a) 2 EF, (b) 3 EF, (c) 4 EF, (d) None of these, 77. The vectors AB = c, BC = a and CA = b are the sides, of a ∆ABC. Which of the following vectors, represent(s) the median AD?, 1, 1, 1, 1, III. a + b, II. − b + c, I. a + c, 2, 2, 2, 2, Select the correct answer using the codes given below, (a) I and II, (b) I and III, (c) I only, (d) II only, 78. A force 5i at the point O, P is a point such that OP is of, length l and is inclined to the direction of the force at, an angle of 30°. The moment of the force about the, point P is 10 units. What is the value of l ?, 5 3, units, (b) 4 units, (a), 4, 3, unit, (d) 3 units, (c), 4, 79. In a ∆PQR, the position vectors of the points Q and R, are 2 i + 3 j + 5k and i + j + k, respectively. S and T, are the middle points of PQ and PR, respectively., What is the length of ST ?, 51, (a) 51 units, (b), units, 2, 21, (c), units, (d) 21 units, 2, 80. If |a × b| 2 = k|a ⋅ b|2 +|a|2| b|2 , then what is the, value of k?, (a) 1, (b) 2, (c) 0, (d) –1, 81. Two coterminus vectors a and b include an angle of, 60°. If |a|=|b|= c, then what is the length of the, diagonal (of the parallelogram formed by a and b), that does not include the coterminus point?, (c) 3 c, (d) 2 c, (a) c, (b) 2 c, , 82. For any two vectors a and b, consider the following, statements, (NDA 2007 I), I. |a + b| =|a − b| ⇔ a and b are orthogonal., II. |a + b| =|a| +| b| ⇔ a and b are orthogonal., III.|a + b|2 =|a|2 +| b|2 ⇔ a and b are orthogonal., Which of the above statements are correct?, (a) I and II, (b) I and III, (c) II and III, (d) I, II and III, 83. The direction ratios of a vector PQ are 4, 1, x,, respectively and the magnitude of the vector PQ is, 42. What are the direction cosines of the vector PQ?, (a) 4, 1, 5, (b) 4/ 42 , 1/ 42, 5/ 42, (c) 2/21, 1/42, – 5/42, (d) 2/21, 1/42, 5/42, 84. The position vectors of the points A, B, C, D are, i + j + mk, 2i + 3 j, 3i + 5 j − 2k, − j + k, respectively, and AB and CD are parallel. What is the value of m?, (a) – 1, (b) 1, (c) – 3, (d) 3, 85. Consider the following statements, If u = a − b + c, v = 2 va − 3 b and w = a + 3 c, then, I. u, v, w are coplanar, if [a, b, c] ≠ 0, II. u, v, w are not coplanar, if [a, b, c] ≠ 0, III. u, v, w are coplanar, if there exist scalars,, α , β , γ not all zero such that α u, β v , γ w = 0., Which of the above is/are correct?, (a) I only, (b) II only, (c) III only, (d) I and III, 86. Two vectors 2i + mj − 3nk and 5i + 3mj + nk are such, that their magnitudes are, respectively 14 and 35,, where m and n are integers. Which one of the, following is correct?, (NDA 2007 I), (a) m takes 1 value and n takes 1 value, (b) m takes 1 value and n takes 2 values, (c) m takes 2 values and n takes 1 value, (d) m takes 2 values and n takes 2 values, 87. Which one of the following is correct?, If a and b are unit vectors, then, (a) it is not possible that both a + b and a − b are, unit vectors., (b) both a + b and a − b are unit vectors., (c) a + b is a unit vector but a − b is zero vector, if a, and b are parallel., (d) both a + b and a − b are unit vectors only when a, and b are perpendicular to each other., 88. Consider the following statements, Vectors which make an angle of π /4 with the vector, i + j + 2 k are given by, I. i + j, II. i + 2 k, III. 2 k
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585, , Vector Algebra, Which of the statements given above are correct?, (a) All I, II and III, (b) I and II, (c) II and III, (d) I and III, 89. If a and b are unit vectors inclined at an angle of 30°, to each other, which one of the following is correct?, (NDA 2007 I), , (a) |a + b|> 1, (c) |a + b| = 2, , (b) 1 <|a + b|< 2, (d) |a + b|> 2, , 90. Let a, b, c be non-coplanar vectors and, p=, , b×c, c× a, a ×b, ,q =, ,r =, ., [a b c], [a b c], [a b c], , (b) −3, , (c) 3, , (d) −9, , 91. If p ≠ 0 and the conditions p ⋅ q = p ⋅ r and p × q = p × r, hold simultaneously, then which one of the following, is correct?, (a) q ≠ r, (b) q = − r, (c) q ⋅ r = 0 (d) q = r, 92. Let ABCD be a parallelogram whose diagonals, intersect at P and let O be the origin, then what is, OA + OB + OC + OD equal to?, (a) OP, (b) 2OP, (c) 3OP, (d) 4OP, 93. If the vectors i − 2xj − 3 yk and i + 3xj + 2 yk are, orthogonal to each other, then what is the locus of the, point ( x , y )?, (a) A circle, (b) An ellipse, (c) A parabola, (d) A hyperbola, 94. Consider the following statements, I. If a × b = c × d and a × c = b × d, then a − d is, parallel to b − c., II. For, any, three, vectors, (a − b ) ⋅ ( b + c) × ( c − a) = 0., III. Let ABCD be a quadrilateral, then, AB + DC = DB + AC, Which of the statements given above are correct?, , (a) I and II, (c) III and I, , (b) II and III, (d) All I, II and III, , 95. Consider the following statements, I. If a and b are two unit vectors inclined at an, angle π / 3, then|a + b|> 1., II. The vector a × ( b × c) is coplanar with a and b., Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, , Directions (Q. Nos. 96-101), , 96. Assertion (A) If a, b, c are non-coplanar vectors, and p ⋅ a = p ⋅ b = p ⋅ c = 0, then p is a zero vector., Reason (R) Since, p is perpendicular to both a and, b it is normal to the plane of a and b. Then, since, p ⋅ c = 0, c must lie in the plane of a and b. But this, contrary to the data. Hence, p must be zero., 97. Assertion (A) The work done when the force and, displacement are perpendicular to each other is zero., , What is the value of, (a – b – c) ⋅ p + (b – c – a ) ⋅ q + (c – a – b) ⋅ r?, , (a) 0, , (b) Both A and R are individually true but R is, not the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., , Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is, the correct explanation of A., , Reason (R) The dot product A ⋅ B vanishes, is the, vectors A and B are perpendicular., (NDA 2009 I), 98. Assertion (A) If a = i + j + k, b = 4i + 3 j + 4 k and, c = i + α j + β k are linearly dependent vectors and, ( c) = 3, then β = 1., Reason (R) If a, b and c are linearly dependent, vectors, then [a b c] = 0., 99. Assertion (A) The volume of the parallelopiped, whose coterminus edges are represented by, a = 2i − 3 j + k b = i − j + 2k and c = 2i + j − k is, 14 cu units., Reason (R) The volume of the parallelopiped whose, coterminus edges are a, b , c is [a b c]., 100. Assertion (A) Three points with position vectors a,, b and c are collinear provided there exist three, scalars x , y , z not all zero, such that x a + y b + z c = 0,, where x + y + z = 0., Reason (R) This is a necessary condition for the, three points to be collinear., 101. Assertion (A) The work done when force and, displacement are perpendicular to each other, is zero, Reason (R) The dot product A ⋅ B vasnishes, if the, vector A and B are perpendicular., 102. If a = i + j + k , b = i − j + k, c = i + j − k and d = i − j − k, then match the List I with List II, List I, (i), (ii), (iii), (iv), , a⋅b, b⋅ c, [a b c], b×c, , Codes, (i) (ii) (iii) (iv), (a) A B C, D, (b) B A D, C, (c) A B D, C, (d) B A C, D, , List II, A. a ⋅ d, B. b ⋅ d, C. 2 j − 2 k, D. 4
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586, , NDA/NA Mathematics, , Directions (Q. Nos. 103-105) Let us defined two, , vectors a and b whose position vectors are 3i − j + 5k, and 2i + 3 j + k, 103. Find the angle between a and b ., 8, 8, (a), (b), 7, 10, 8, (c), (d) None of these, 7 10, 104. The value of a ⋅ b is, (a) 8, (b) 7, , three vectors., a = i + j + k, b = i − j + k and c = i + 2 j − k, On the basis of this solve the following question., 106. Find the value of [a b c]., (a) 4, (b) 3, (c) 2, (d) 1, , (d) − 8, , (c) 6, , Directions (Q. Nos. 106-108) Suppose there are, , 105. If a and b are adjacent sides of a parallelogram, then, area is equal to, (a) 436, (b) 426, (d) None of these, (c) 425, , 107. Find the value of a ×( b × c), (a) i + 4 j + 2k, (b) i − 4 j + 2k, (c) i + 2 j + 3k, (d) None of the above, 108. The value of [a + b b + c c + a] is, (a) 3, (b) 8, (c) 2, (d) 1, , Answers, Level I, 1., 11., 21., 31., 41., 51., 61., , (a), (b), (a), (b), (a), (c), (b), , 2., 12., 22., 32., 42., 52., 62., , (a), (d), (d), (b), (b), (b), (b), , 3., 13., 23., 33., 43., 53., 63., , (d), (c), (c), (a), (b), (b), (d), , 4., 14., 24., 34., 44., 54., 64., , (c), (a), (a), (c), (c), (c), (c), , 5., 15., 25., 35., 45., 55., 65., , (a), (c), (b), (a), (b), (a), (d), , 6., 16., 26., 36., 46., 56., 66., , (a), (a), (c), (c), (d), (a), (d), , 7., 17., 27., 37., 47., 57., 67., , (c), (a), (d), (a), (a), (b), (c), , 8., 18., 28., 38., 48., 58., 68., , (b), (c), (d), (d), (d), (c), (b), , 9., 19., 29., 39., 49., 59., 69., , (c), (c), (b), (d), (c), (b), (b), , 10., 20., 30., 40., 50., 60., 70., , (a), (c), (d), (c), (d), (b), (c), , 2., 12., 22., 32., 42., 52., 62., 72., 82., 92., 102., , (c), (a), (a), (a), (a), (a), (a), (d), (b), (d), (b), , 3., 13., 23., 33., 43., 53., 63., 73., 83., 93., 103., , (b), (a), (b), (c), (d), (d), (a), (b), (b), (a), (c), , 4., 14., 24., 34., 44., 54., 64., 74., 84., 94., 104., , (d), (b), (a), (d), (a), (c), (c), (d), (b), (d), (a), , 5., 15., 25., 35., 45., 55., 65., 75., 85., 95., 105., , (b), (b), (d), (a), (b), (c), (b), (d), (c), (c), (b), , 6., 16., 26., 36., 46., 56., 66., 76., 86., 96., 106., , (d), (a), (a), (d), (b), (b), (a), (c), (d), (a), (a), , 7., 17., 27., 37., 47., 57., 67., 77., 87., 97., 107., , (b), (a), (a), (a), (d), (a), (b), (c), (a), (a), (b), , 8., 18., 28., 38., 48., 58., 68., 78., 88., 98., 108., , (c), (a), (b), (c), (a), (b), (a), (b), (d), (a), (b), , 9., 19., 29., 39., 49., 59., 69., 79., 89., 99., , (b), (c), (c), (c), (b), (d), (b), (c), (b), (a), , 10., 20., 30., 40., 50., 60., 70., 80., 90., 100., , (d), (c), (d), (b), (a), (a), (b), (d), (c), (a), , Level II, 1., 11., 21., 31., 41., 51., 61., 71., 81., 91., 101., , (c), (b), (b), (a), (a), (c), (a), (c), (a), (d), (a)
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Hints & Solutions, Level I, 1. If three vectors a, b and c are coplanar, then, x a + y b + z c = 0 holds for non-zero value of scalars x, y, and z. So, they are linearly dependent., 2. Let a = 2i + j + k, b = i − j + k, a⋅b, Projection of a along b =, |b|, (2i + j + k) ⋅ (i − j + k), =, 3, 2 −1 + 1 2, =, =, 3, 3, , (a × b), |a × b|, (a × b), –, |a × b|, , C, , a, , 10. x is perpendicular to a, b and c it means a, b and c are, coplanar, so [a b c] = 0., , B, , i, , j k, , =|AB × AD|= 1, −1, , 1 1, 1 0, , 11. Points (2, 2), (λ, 8), 2 2 1, λ 8 1 =0, , =|i (−1) − j(1) + k(2)|, = 1 + 1 + 4 = 6 sq units, , 5. a ⋅ b = (− i − 2 j + k) ⋅ (x i + j + (x + 1) k) (obtuse angle), , = − x − 2 + x + 1 = − 1< 0, 6. Option (a) is meaningless., 7. |a|2 = a ⋅ a = a 2 but|z|2 ≠ z 2, (As one side is real but other side is imaginary)., 8. Let c be the position vector of C., AB = OB − OA = b − a, and, AC = 2 AB = 2 b − 2 a, , (given), , B, , C, , b, a, , c, , O, , Now, in ∆ AOC ,, ⇒, ⇒, , and, , (8, 32), , are, , collinear,, , so, , 8 32 1, , 4. If a ⋅ b = 0 ⇒ a ⊥ b, and, a × b = 0 ⇒ a || b, But both conditions can’t exist simultaneously. The, one possible way for both conditions to exist, simultaneously is that either of a and b is a null, vector., , A, , b, , a× b, in the vertical upper direction and, | a× b |, the other perpendicular unit vector c on both vectors a, and b is, (a × b), ,, c=−, | a× b |, which is vertically below direction., , b, , A, , a, , O, , Then, c =, , 3. Area of parallelogram, D, , 9. If c is perpendicular to both vectors a and b, then, c= a× b, But c is the unit vector., , AC = OC − OA, 2 b −2 a = c − a, c =2 b −2 a + a =2 b − a, , ⇒, ⇒, ⇒, , 2 (8 − 32) − 2 ( λ − 8) + 1(32λ − 64) = 0, −48 − 2λ + 16 + 32λ − 64 = 0, 16, 30λ = 96 ⇒ λ =, 5, , 12. ( a + b + c)2 = |a|2 + |b|2 + |c|2 + 2 (a ⋅ b + b ⋅ c + c ⋅ a ), ⇒ 0 = 1 + 1 + 1 + 2 (a ⋅ b + b ⋅ c + c ⋅ a ), 3, ⇒, a ⋅ b + b⋅ c + c⋅ a = −, 2, 13. Let AO = 10i + 3 j, OB = 12i − 5 j, and, OC = mi + 11 j, Since, A, B and C are collinear., We have,, AB = λBC, ⇒, (OB − OA ) = λ (OC − OB), ⇒, (2i − 8 j) = λ {(m − 12)i + 16 j}, On comparing the coefficients of i, j and k, we get, …(i), λ (m − 12) = 2, and, 16λ = − 8, 1, ⇒, λ=−, 2, From Eq. (i), we get, 1, − (m − 12) = 2, 2, ⇒, m − 12 = − 4, ⇒, m =8, Alternate Method If the given position vectors are, collinear, then the area of triangle should be zero.
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588, , NDA/NA Mathematics, 10, 12, m, 10, , 3, −5, 11, 3, , 1, 1 =0, 1, 1, , ⇒, , 2, m − 10, , −8, 8, , 0 =0, 0, , ⇒, ⇒, , 16 + 8(m − 10) = 0, 8m = 64 ⇒ m = 8, , R2 → R2 − R1 , , , R3 → R3 − R1, , 15. If OA = a , OB = b, , OM =, , and, , So, magnitude of a + b = 2|OM|= 2 cos, , (as|OB| = 1), θ, 2, , 16. Let r = xi + yj (vector in xy-plane), ⇒, x2 + y2 = 1, (xi + yj) ⋅ (i + j), ∴, cos 45° =, x2 + y2 12 + 12, ⇒, (x i + y j) ⋅ (i + j) = 2 cos 45°, ⇒, x + y =1, From Eqs. (i) and (ii), we get, (1 − y) 2+ y2 = 1 ⇒ 1 + y2 − 2 y + y2 = 1, ⇒, y = 0, 1, If y = 0, x = 1 and when y = 1, then x = 0., ∴, r = i or r = j, , …(i), , …(ii), , 17. We have, GB + GC = (1 + 1) GD = 2 GD, where D is the, mid-point of BC., ∴, GA + GB + GC = GA + 2 GD = GA − GA = 0, (G divides AD in the ratio 2 : 1, ∴ 2 GD = − GA), 18. We have, (a × b)2 + (a ⋅ b) 2 = 144, ⇒|a|2|b|2 sin 2 θ + |a|2|b|2 cos 2 θ = 144, |a|2|b|2 = 144, ⇒, |b| = 3, 19., , = 16 + 9 + 100, =5 5, 22. Given that a and b are two unit vectors and ‘ θ ’ is the, angle between them, Then,, |a + b|2 =| a |2 + | b|2 + 2| a || b|cos θ, [ Q|a|=|b|= 1], = (1 + 1 + 2 × 1 × cos θ ), = 2 (1 + cos θ ), 1, 2θ, = | a + b|2, cos, ⇒, 2 4, θ 1, ⇒, cos = | a + b|, 2 2, , 14. Let A = 3 (a × b), B = b − (a ⋅ b) a, ∴ A ⋅ B = 3 (a × b) ⋅ (b − ( a ⋅ b)a ), = 3 ([a b b ] − [a b a ] (a ⋅ b)) = 0, π, ∴ A ⊥ B i.e., θ =, 2, a+b, 2, θ, |OM| = cos, 2, , AB = OB − OA, = (6 − 2)i + (1 − 4) j + (12 − 2)k, = 4i − 3 j + 10k, And its magnitude = 42 + (−3)2 + 102, , 23., , Q xi + yj + zk is a unit vector., ∴, x2 + y2 + z 2 = 1, and, x: y: z = 3 :2 :3, ⇒, x = 3k, y = 2k and z = 3k, ∴ ( 3k)2 + (2k)2 + (3k)2 = 1, 1, 1, ⇒ k=, ⇒ 3k2 + 4k2 + 9k2 = 1 ⇒ k2 =, 16, 4, 1 3, Hence,, z = 3k = 3 × =, 4 4, , (given), , 24. We know that, if position vector of a point P is r from, origin of coordinate axes and a force F passes through, the point P, then moment of the force about the origin is, r× F, 25. We have, a = i − 2 j + k and b = 4i − 4 j + 7k, ∴ Projection of a on b is given by, a ⋅ b (i − 2 j + k) ⋅ (4i − 4 j + 7k), =, =, | b|, 16 + 16 + 49, 19 19, =, =, 81 9, 26. Since, the vectors i − xj − yk and (i + x j + y k) are, orthogonal, to, each, other,, then, (i − xj − yk) ⋅ (i + xj + yk) = 0, , [Q |a|= 4], , p = Magnitude of 3i + 2 j = 9 + 4 = 13, q = Magnitude of 2i + 2 j + k = 4 + 4 + 1 = 3, r = Magnitude of 4i − j + k = 16 + 1 + 1 = 18 = 3 2, and s = Magnitude of 2i + 2 j + 3k = 4 + 4 + 9 = 17, ∴, r >s> p>q, , 20. Since, points are collinear, Q, AB = λ AC, ⇒, −2 j = λ [(l − 1 ) i + (m − 1) j + nk], On comparing both sides, we get, λ (l − 1) = 0, λ (m − 1) = − 2, λn = 0, ⇒, l = 1, n = 0, m ∈ R, 21. Let the position vector be OA = (2, 4, 2) and, OB = (6, 1, 12)., ∴ The displacement vector is given by, , ⇒ 1 − x2 − y2 = 0 ⇔ x2 + y2 = 1, which represent a circle with centre at ogigin and, having radius is 1., 27. Q, a = i − k, b = xi + j + (1 − x) k, and, c = yi + xj + (1 + x − y) k, 1 0 −1, ∴, a ⋅ (b × c) = x 1 1 − x, y x 1+ x− y, Expanding along R1, = 1 (1 + x − y − x + x2) − 1 (x2 − y), = 1 − y + x2 − x 2 + y = 1, which shows that a ⋅ (b × c) does not depend on x and y., 28. We have, a = 2i + 3 j − 5k and b = i + j − k, ∴ The area of parallelogram is given by, A = | a × b|
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589, , Vector Algebra, i j k, a × b = 2 3 −5, , Now,, , 1 1 −1, ⇒, ⇒, ⇒, , a × b = 2i − 3 j − k, | a × b| = 4 + 9 + 1, A = 14 sq units, , 29. We have,| a + b| = | a − b|, ⇒, | a |2 + | b|2 + 2|a || b|cos θ, | a |2 + | b|2 − 2|a || b|cos θ, ⇒, 4| a || b|cos θ = 0, ⇒, cos θ = 0 ⇒ θ = 90°, 30. In ∆ ACD,, D, , C, , B, , A, , |a |= 4, 12, ∴, | b| =, =3, 4, 34. Q, a ⋅ b =| a || b|cos α, …(i), ⇒, cos α = a ⋅ b, (Q | a | = | b| = 1, a and b are unit vectors), [Q (a + b) is unit vector], Now,, | a + b| = 1, ⇒ | a |2 + | b|2 + 2a ⋅ b = 1, [from Eq. (i)], ⇒, 1 + 1 + 2 cos α = 1, ⇒, 2 cos α = −1, 1, 2π, cos α = − = cos, ⇒, 2, 3, 2π, α=, ⇒, 3, Since,, , …(i), CD + DA = CA, Now in ∆ ABC,, …(ii), CA + AB = CB, From Eqs. (i) and (ii), we get, (CD + DA ) + AB = CB, ⇒ CB + CD + DA + AB = 2 CB, A⋅B, 31. Projection of A on B is given by , ×B, 2, |B | , 1 × 4 + (−2) × (−4) + 1 × 7, × (4i − 4 j + 7k), =, 42 + (−4)2 + (7)2, 4+8+ 7, =, (4i − 4 j + 7k), (16 + 16 + 49), 19, =, (4i − 4 j + 7k), 81, , 35. Since, C is mid-point of AB, ∴, AC = CB, Now, in ∆ APC,, PA + AC = PC, and in ∆ BPC,, PC + CB = PB, ⇒, PC = PB − CB, On adding Eqs. (i) and (ii), we get, 2PC = PA + PB + AC − CB, ⇒, 2 PC = PA + PB, , …(i), , …(ii), , 36. We have, OA = a, OB = b, OC = 2a + 3b ,, OD = a − 2 b, ∴, AC = OC − OA, =2 a + 3 b − a, = a + 3b, , 19, ∴ Magnitude = (16 + 16 + 49), 81, , 37. By taking option (a), condition of perpendicularity a ⋅ b = 0, (3i + 4 j), 1, ±, ⋅ (4i − 3 j + k) = (12 − 12), 5, 5, =0, , = 2 .11, Magnitude of the projection of A and B lies between, 2 and 3., , 38. Since, P is a point on circumference of a semi-circle of, radius a, which is bounded by the diameter BC., In ∆PBC,, , 2, , P, , 32. Here, we have,, F = 2i + j + 4k, and the points are A (3, −1, 3) and B (5, 1, 4), ∴, AB = (5i + j + 4k) − (3i − j + 3k), = 2i + 2 j + k, Now, moment of F about A is given by AB × F, = (2i + 2 j + k) × (2i + j + 4k), i j k, = 2 2 1, 2 1 4, = i (2 × 4 − 1 × 1) − j(2 × 4 − 2 × 1) + k (2 × 1 − 2 × 2), = 7 i − 6 j − 2k, 33. We have,| a × b|2 + | a ⋅ b|2 = 144, ⇒, |a|2|b|2 sin 2 θ + |a|2|b|2 cos 2 θ = 144, ⇒, | a |2| b|2 = 144, ⇒, | a || b|= 12, , BP, a, B, , BC, , PC, C, , BP ⋅ PC = |BP| ⋅ |PC| cos 90°, BP ⋅ PC = 0, 39. Let r1 = bi − aj, Condition of perpendicularity a ⋅ b = 0, Now,, r1 ⋅ r = (bi − aj) ⋅ (ai + bj), = ab − ab = 0, 40. Given that,, The diagonals of a parallelogram are, a = 3 i + j − 2k and b = i − 3 j + 4 k, k, i j, 3, 1, 2, ∴ a ×b = , −, , , 1 −3 4 , , (by option ‘d’)
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591, , Vector Algebra, 51. We know that, if three vectors a, b, c are coplanar, then, [a b c] = 0, Let a = 2i − j + k, b = i + 2 j − 3k and c = 3i + m j + 5k, 2 −1 1, Then,, [a b c] = 1 2 −3 = 0, 3, , m, , 5, , ⇒, , 2 (10 + 3m) + 1 (5 + 9) + 1 (m − 6) = 0, , ⇒, , 20 + 6m + 14 + m − 6 = 0, , ⇒, , 7m + 28 = 0, , ⇒, , 57. | a + b| = | a + b|2, = | a |2 + | b|2 + 2| a || b|cos 60°, , m = −4, , 52. Given that, the vertices of ∆ABC are i, j and i + j + λk., Then,, AB = 2 , BC = 1 + λ2, and, , ⇒ C is mid-point of AB, 1, AC = AB, 2, 1, = (b − a ), 2, , ∴, , ∴, , (Q θ = 60° ), [Q a and b are unit vectors ⇒| a | = | b| = 1], 1, = (1)2 + (1)2 + 2 ⋅ 1 ⋅ 1 ⋅ = 3, 2, | a + b| > 1, , 58. In ∆ AOB,, , AB = OB − OA, =b− a, CD = − AB = a − b, , CA = 1 + λ2, ∴, In ∆ ODC,, , A (1, 0, 0), , B, (0, 1, 0), , OD = OC + CD, = c+ a− b, , 59. Given that, a = i − 3 j and OA = − i + 5 j, Let the coordinates of B are (x, y), then OB = xi + yj, ∴, AB = a = OB − OA, ⇒ (x + 1) i + ( y − 5) j = i − 3 j, (after comparing), ⇒ x + 1 = 1 and y − 5 = − 3, ⇒, x = 0 and y = 2, ∴ Coordinates of B are (0, 2)., , C, (1, 1, λ), , Q ∆ABC is right angled triangle., ∴, AB2 = BC 2 + CA 2, ⇒, 2 = 1 + λ2 + 1 + λ2, ⇒, λ =0, 53. Volume of rectangular parallelopiped having sides as, i , 2 j, 3k, 0, 0, 1, 0, 2, 0, =, , , 3, 0, 0, , 0, 2, = 6 cu units, =, 3, 0, , 60. It is clear that CA has greatest magnitude, since side, opposite to greatest angle is longest and ∠B is greatest, angle in ∆ ABC. Thus, CA has greatest magnitude., A, , E, , F, , 54. Since, B is the mid-point of AC, then, O, D, , B, , b, , a, , A, , ⇒, ⇒, , B, , 61. Q a and b are position vectors of A and B, respectively, and position vector of C is 3a − 2b., From above, it is clear that A is in between C and B., C, , OA + OC, OB =, 2, a + OC, b=, 2, OC = 2b − a, , 55. (A × B).C represents the volume of a parallelopiped., 56. In ∆ OAB,, OA + AB = OB, ⇒, AB = OB − OA, =b− a, Q, AB = 2BC, , C, , (0, 1), B, , (1, 0), λ, , A, , (3, –2), 1, , C, , By section formula,, 3λ + 0 −2λ + 1 , OA = , ,, = (1, 0), λ+1 , λ+1, On comparing,, 3λ + 0 = 1 + λ, 1, λ=, ⇒, 2, Q So, point A divides BC internally.
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592, , NDA/NA Mathematics, , 62. Q p and q are unit vectors, which makes an angle, , π, 3, , 67. Given condition is,, PO + OQ = QO + OR, , with each other., ∴, Now,, , ∴, 63., , π 1, =, 3 2, 2, 1, 1, 2, p − q = |p|2 + |q|2 − p ⋅ q, 2, 4, 2, 1 1, =1 + −, 4 2, 5 1 5 −2, = − =, 4 2, 4, 3, =, 4, 1, 3, p− q =, 2, 2, p ⋅ q = |p||q|cos, , 66., , Q c is normal to the vectors a and b, then, c ⋅ a = 0 and c ⋅ b = 0, Now,, c ⋅ (a + b) = c ⋅ a + c ⋅ b, =0 + 0 =0, and, c ⋅ (a − b) = c ⋅ a − c ⋅ b, =0 −0 =0, ∴ c is normal to (a + b) and (a − b), , OQ + OQ = OP + OR, , ⇒, , 2 OQ = OP + OR, OP + OR, OQ =, 2, , Which represent Q is the mid-point of P and Q i.e., P, Q, and R are collinear., , 64. Two vectors (x2 − 1) i + (x + 2) j + x2k and, 2i − xj + 3k are orthogonal, if, 2 (x2 − 1) − x (x + 2) + 3x2 = 0, ⇒, 2 x2 − 2 − x2 − 2 x + 3x2 = 0, ⇒, 4 x2 − 2 x − 2 = 0, ⇒, (2 x2 − x − 1) = 0, ⇒, (2 x + 1) (x − 1) = 0, 1, and, x=−, x=1, ⇒, 2, Q Semi-diagonal is r = ai + bj, ∴ Sides of rectangle are 2a and 2b., Hence, area of rectangle = 2a × 2b = 4ab, , − OP + OQ = − OQ + OR, , ⇒, ⇒, , Q (i − xj − 2k) and (2i + j + yk) are orthogonal., ∴ (i − xj − 2k) ⋅ (2i + j + yk) = 0, ⇒, 2 − x − 2y = 0, ⇒, x + 2y = 2, Which is an equation of straight line., Thus, the locus of the point (x, y) is a straight line., , 65., , ⇒, , 68. Q, a = i + 2 j − 3k and b = 3i − j + λk, ∴ a + b = i + 2 j − 3 k + 3 i − j + λk, = 4i + j + (λ − 3) k, and a − b = i + 2 j − 3k − 3i + j − λk, = − 2i + 3 j − (3 + λ ) k, Q (a + b) is perpendicular to (a − b)., ∴, (a + b) ⋅ (a − b) = 0, ⇒ {4i + j + (λ − 3) k} { −2i + 3 j − (3 + λ ) k} = 0, ⇒, −8 + 3 + (32 − λ2) = 0, ⇒, 4 − λ2 = 0, ⇒, λ=±2, 69., , (given), , Since, p and q are collinear, then p = λq, ⇒, (x − 2) a + b = λ (x + 1) a − λ b, On equating the coefficients,, x − 2 = λ (x + 1) and − λ = 1, ⇒, x − 2 = − (x + 1), 1, ⇒, 2x = 1 ⇒ x =, 2, , 70. Here, F = 3 i + 2 j − k, and, r1 − r2 = i − j + k − 2i + 3 j + k, = − i + 2 j+ 2 k, Moment of couple = (r1 − r2) × F, = (− i + 2 j + 2 k) × (3 i + 2 j − k), i, j, k, 2, = −1 2, 3, , 2, , −1, , = (−2 − 4)i − (1 − 6) j + k(−2 − 6), = − 6i + 5 j − 8k, Magnitude of the moment = | − 6i + 5 j − 8k|, = 36 + 25 + 64 = 5 5, , Level II, 1. i × (a × i ) + j × (a × j) + k × (a × k), = (i ⋅ i ) a − (i ⋅ a )i + ( j ⋅ j) a − ( j ⋅ a ) j + (k ⋅ k)a − (k ⋅ a ) k, = 3a − {(i ⋅ a ) i + ( j ⋅ a ) j + (k ⋅ a ) k}, = 3a − a = 2a, 2. [ a × b b × c c × a ], = ( a × b) ⋅ {(b × c) × (c × a )}, = (a × b) ⋅ {(b × c ⋅a ) c − (b × c ⋅ c) a }, , θ, , c, , = (a × b ⋅c) (b × c ⋅ a ), = [a b c]2, , b, , a
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593, , Vector Algebra, 3. [a × b b × c c × a ] = [a b c]2 ≠ 0, (As a, b and c are non-coplanar), So, a × b, b × c and c × a are non-coplanar vectors or, they are linearly independent., 4. Let, a = − i − 2xj − 3 yk, and, b = i − 3 xj − 2 yk, If both the vectors are orthogonal to each other, then, a⋅b = 0, ⇒ (− i − 2xj − 3 yk) ⋅ (i − 3xj − 2 yk) = 0, ⇒, − 1 + 6 x2 + 6 y2 = 0, ⇒, 6 x2 + 6 y2 = 1, 1, ⇒, x2 + y 2 =, 6, which is the equation of a circle., 5. Since, a , b and c are mutually perpendicular vectors., Let, a = 10i , b = 10 j, c = k, ∴, |2a + 2b + 40c| = |20i + 20 j + 40k|, =|20(i + j + 2k)| = 20 6, 6. If the vectors AB, AC and AD are coplanar, then the, points A , B, C and D are coplanar., Now, AB = i + 2 j + 0k, AC = 0i + 1 j + 4k, and, AD = (λ − 1)i + 2 j + 3k, 1, 2 0, ∴, 0 1 4 =0, λ −1 2, ⇒, ⇒, ⇒, ⇒, , 7., , Since, u 4 = 2u 2 ⇒ (4, 6, 2) = 2(2, 3, 1), ∴ u 2 is parallel to u 4., , 8. Given that,|a| = |b|, If (a + b) is parallel to (a − b)., Then, (a + b) × (a − b) should be equal to zero., ∴, , a12, , + b12 + c12 a 22 + b22 + c22, , ∴ Angle between the vectors i + 2 j + 3k and, − i + 2 j + 3k is given by, cos θ =, =, Now,, , 1 × (−1) + 2 × 2 + 3 × 3, 1+4+9 1+4+9, −1 + 4 + 9 12 6, =, =, 14, 14 7, , 36, 49, 49 − 36, 13, 13, =, =, =, 49, 49, 7, , sin θ = 1 − cos 2 θ = 1 −, , 12. a = 2i − j + 2k, b = i + j + 4k, i, , j, , ∴ a × b = 2 −1, 1 1, , k, 2, 4, , = i (−4 − 2) − j (8 − 2) + k (2 + 1), = − 6i − 6 j + 3k, −6 i − 6 j + 3 k, Unit vector in this direction = ±, 36 + 36 + 9, =±, , 3, , 1(3 − 8) − 2 { −4 (λ − 1)} = 0, −5 − 2 { − 4 λ + 4 } = 0, −5 + 8 λ − 8 = 0, 8λ = 13, 13, λ=, 8, , ⇒, , a1a 2 + b1b2 + c1c2, , cos θ =, , (a + b) × (a − b) = a × a + b × a − a × b − b × b, , =0 − a × b − a × b −0, = −2 a × b ≠0, (a + b) ⋅ (a − b) = a ⋅ a + b ⋅ a − a ⋅ b − b ⋅ b, =1 + a⋅b − a⋅b −1, =0 ≠1, i.e., (a + b) is perpendicular (a − b), 9. (a + b) ⋅ (b + c) × (c + a ), = (a + b) ⋅ [b × c + b × a + c × a], = a ⋅ b × c +0 + 0 + 0 + 0 + b ⋅ c × a ], = 2 [a b c], 10. Since, a × r = j ⇒ a and r are perpendicular to j i.e.,, parallel to xz-plane but they can be of any magnitude, and at any angle, so a ⋅ r cannot be determined., 11. We know that, the angle between the vectors, a1i + b1 j + c1k and a 2i + b2j + c2k is given by, , −6 i − 6 j + 3 k, 9, , 13. Angle between the faces OAB and ABC is same as angle, between normals of the faces OAB and ABC. Vector, along the normal of OAB, i, j k, (say), = 1 2, 1 = 5i − j − 3k = a, 2, , 1, , 3, , Vector along the normal of ABC., i, j k, = 1 −1 2 = i − 5 j − 3 k = b, −2 −1 1, a ⋅b, 5+5+9, =, ∴ cos θ =, |a||b| 35 35, 19, ⇒, θ = cos −1 , 35, , (say), , 14. Given that,|a| = 10,|b| = 2, and, , a ⋅ b = 12, , ⇒, , |a||b|cos θ = 12, , ⇒, , 10 ⋅ 2 ⋅ cos θ = 12, , ⇒, , cos θ = 3 / 5, , ... (i), , sin θ = 1 − cos θ, 2, , = 1 − 9 /5 = 4 /5, Now,, , $, |a × b| = |a||b||sin θn|, $|, =|a||b||sin θ|n, = 10 ⋅ 2 ⋅ 1 ⋅|sin θ|, = 10 ⋅ 2 ⋅ 1 ⋅|4 / 5|, 4, = 20 × = 4 × 4 = 16, 5
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594, 15., , NDA/NA Mathematics, (given), Q OP = i + 3 j − 2k and OQ = 3i + j − 2k, By option, let i + j − k be required position vector of the, bisector of the ∠ POQ. Then, it will make equal angles, with OP and OQ., ∴ Angle between i + 3 j − 2k and i + j − k,, , , 1+3+2, 6 , = cos −1 , θ = cos −1 , , 14 3 , 1 + 9 + 4 1 + 1 + 1, and angle between 3i + j − 2k and i + j − k, , , , 1+3+2, 6 , = cos −1 , φ = cos −1 , , 14 3 , 9 + 1 + 4 1 + 1 + 1, , Q(q), , S, , OS = 3p − 2q, OR ⋅ OS = 0, 9 p2 = 4 q 2, , 23. Since, vectors ai + aj + ck, i + k and ci + cj + bk lie on a, plane i.e., are coplanar., a a c, ∴, 1 0 1 =0, c, ., , 18. Since,, AB = 5i − j + 8k and CD = 2i + 3 j − 6k, Projection of vector AB along CD, AB ⋅ CD, =, |CD|, (5i − j + 8k) ⋅ (2i + 3 j − 6k), =, 4 + 9 + 36, 10 − 3 − 48, 41, =, =−, 7, 7, 19. Since, vectors make an acute angle with each other, so, their dot product must be positive., i.e.,, ax2 − 10ax + 6 > 0 ,, ∀ x ∈R, ⇒, − ax2 + 10ax − 6 < 0 ,, ∀ x ∈R, 6, 2, ⇒, − a < 0 and 100a < 24a ⇒ 0 < a <, 25, 1, 20. a × (b × c) + b × (c × a ) = a, 2, 1, ⇒ (a ⋅ c) b − (a ⋅ b) c + (b ⋅ a ) c − (b ⋅ c) a = a, 2, 1, , (a ⋅ b) b − b ⋅ c + a = 0, ⇒, , 2, 1, ⇒, a⋅ c = 0 , b⋅ c = −, 2, π, ∴ Angle between a and c is ., 2, , = QR = RP, ∴ P , Q and R are the vertices of an equilateral triangle., , R, , O, , 17. a ⋅ b + b ⋅ c + c ⋅ a, 1, = − (|a|2 + |b|2 + |c|2 ) < 0, 2, So,, m <0, , 21. Let the points are P (α i + β j + γ k), Q (βi + γ j + α k), and, R(γ i + α j + β k)., Here, PQ = (α − β )2 + (β − γ )2 + (γ − α )2, , 2q + 3p, 5, P(p), , and, Now,, ⇒, , Hence, θ = φ, 16. |a + b| = 13, ⇒, |a|2 + |b|2 + 2a ⋅b = 169, ⇒, 25 + 144 + 2a ⋅ b = 169, ⇒, a ⋅ b=0, Now,, |a − b|2 = |a|2 + |b|2 − 2a ⋅ b, = 25 + 144 = 169, ⇒, |a − b| = 13, a+ b+ c =0, ⇒, (a + b + c) ⋅(a + b + c) = 0, ⇒ |a|2 + |b|2 + |c|2+2 (a ⋅ b + b ⋅ c + a ⋅ c) = 0, , 22. OR =, , c, , b, , Expanding along R2, ⇒, −1 (ab − c2) − 1 (ac − ac) = 0 ⇒ ab − c2 = 0, ⇒ c is the geometric mean of a and b., 24. a + b + c = αd , b + c + d = βa, ⇒, (a + b + c + d ) = (α + 1)d, and, also, a + b + c + d = (β + 1) a, 25. Let B = x i + y j + z k, Now,, A× B= C, i j k, 1 1 1 =j−k, ⇒, x, , y, , …(i), , z, , and, x+ y+ z=3, (Q A ⋅ B = 3) …(ii), From, Eq. (i), we get, i (z − y) − j (z − x) + k ( y − x) = j − k, ⇒, y = z , x − z = 1, y − x = − 1, From Eq. (ii), we get, x + 2z = 3, ⇒, 3z = 2, 2, y=z=, ⇒, 3, 5, and, x=, 3, 26. Q PQ = 3i + 2 j − mk and PS = i + 3 j + k, i j k, ∴ Area of parallelogram = 3 2 − m = | PQ × PS|, 1 3 1, = | i (2 + 3m) − j (3 + m) + k (9 − 2)|, = (2 + 3m)2 + (3 + m)2 + 72, ⇒ 90 = 4 + 9m2 + 12m + 9 + m2 + 6m + 49, ⇒, 10 m2 + 18 m − 28 = 0, ⇒, 5m2 + 9m − 14 = 0, 2, ⇒, 5m + 14 m − 5m − 14 = 0, ⇒ m ( 5m + 14 ) − 1 ( 5m + 14 ) = 0, ⇒, ( 5m + 14 ) (m − 1 ) = 0, 14, ⇒, m = 1 or −, 5, , (given)
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595, , Vector Algebra, , 27. Centroid G is given by, , a + b +c, . The line joining the, 3, A (a), , H, , 2, , 1, O, G, , B (b), , 33. Given that, a is a non-zero vector of modulus a and m is, a non-zero scalar., (given), ∴, | a|= a, Let, a = ma, $ |= 1], ⇒, 1 =|m|| a |, [Q |b, 1, ∴, a=, |m|, a = i + a j + a2 k, b = i + b j + b2 k, and, c = i + c j + c2 k, are three non-coplanar vectors,, i.e.,, [a b c] ≠ 0 [by properties of vectors], 1 a a2, ∴, 1 b b2 ≠ 0, , 34. We have,, C(c), , orthocentre and the circumcentre is divided by centroid, in 2 : 1 ratio internally. Therefore, H will be given by, a + b + c., → →, →, 28. Determinant corresponding to α , β and γ is, 1 2 0, 2 a 10 = a 2 − 24a + 240, which is never zero as its, 12 2a a, discriminant < 0. Hence, vectors are always linearly, independent for all, values of a., , 1, Now,, , 29. Given, a = (2, 1, − 1), b = (1, − 1, 0) and c = (5 − 1, 1), Now, a + b − c = (2 + 1 − 5, 1 − 1 + 1, − 1 + 0 − 1), = (− 2, 1, − 2) = d, d, (− 2, 1, − 2), $, ∴ Unit vector (d ) =, =, |d|, (− 2)2 + (1)2 + (− 2)2, , (say), , (− 2, 1, − 2) 1, = (− 2, 1, − 2), =, 4+1+4 3, But in opposite direction d = − d$, =, , 1, 2i − j + 2 k, (2, − 1, 2) =, 3, 3, , 30. Let A = i + j + k, B = 2i + 4 j − 5k, and, C = bi + 2 j + 3 k, ∴, B + C = 2 i + 4 j − 5 k + bi + 2 j + 3 k, = (2 + b) i + 6 j − 2k, Unit vector parallel to B + C, (2 + b) i + 6 j − 2k, $ =, n, (2 + b)2 + 62 + (− 2)2, (2 + b) i + 6 j − 2k, $ =, n, b2 + 4 b + 44, $ =1, Now, (i + j + k) ⋅ n, (according to question), ⇒, ⇒, ⇒, ⇒, ⇒, , 2 + b + 6 − 2 = b2 + 4 b + 44, (b + 6)2 = b2 + 4b + 44, b + 36 + 12b = b2 + 4b + 44, 8b = 8, b =1, 2, , 31. Obviously, D is the centroid of ∆ ABC. But ABC is an, equilateral triangle, so D is orthocentre, circumcentre, and incentre also., 4 + 6 + 6 16, 32. cos θ =, =, 17 17 17, ∴, , 33, 33, sin θ =, , tan θ =, 17, 16, , ⇒, , c, , c2, , a, , a 2 1 + a3, , b, c, , b2 1 + b3 = 0, c2 1 + c3, , a a 2 a3, 1 a a2, 1 b b2 + b b2 b3 = 0, c c2 c3, 1 c c2, 1 a, , ⇒ 1 b, 1 c, , a2, , 1 a, , b2 + abc 1 b, c2, 1 c, , a2, b2 = 0, c2, , ⇒, , 1 a a2, 1 b b2 (1 + abc) = 0, 1 c c2, , Q, , 1 a a2, 1 b b2 ≠ 0, 1 c c2, , ∴, ⇒, , (1 + abc) = 0, abc = − 1., , 35. a {(b + c) × (a + b + c)} = 0, = a ⋅ { b × a + b × b + b × c + c × a + c × b + c × c}, = 0 + 0 + a ⋅ (b × c) + 0 + a ⋅ (c × b) + 0, , Q a ⋅ (b × a ) = 0, , , a ⋅ (b × b) = 0, , , a ⋅ (c × a ) = 0, = a ⋅ (b × c) − a ⋅ (b × c) = 0, , , , , a ⋅ (c × c) = 0, , , and c × b = − b × c, and for any three coplanar vectors d, e and f,, (d × e) ⋅ f = 0, 36. Given that,, | a | = 4,| b| = 4,| c| = 5, Then, a ⊥ (b + c), b ⊥ (c + a ) and c ⊥ (a + b), i.e.,, a ⋅ (b + c) = 0 ⇒ a ⋅ b + a ⋅ c = 0, and, b ⋅ (c + a ) = 0, ⇒, b⋅ c + b⋅ a = 0, , …(i), …(ii)
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596, , NDA/NA Mathematics, Similarly, c ⋅ (a + b) = 0, …(iii), ⇒, c⋅ a + c⋅ b = 0, ∴ On adding Eqs. (i), (ii) and (iii), we get, 2 (a ⋅ b) + 2 (b ⋅ c) + 2 (c ⋅ a ) = 0, ∴| a + b + c|2 = |a|2 + |b|2 + | c|2 + 2 (a ⋅ b + b ⋅ c + c ⋅ a ), = 16 + 16 + 25 + 0 = 57, ⇒| a + b + c| = 57, , 37. Given that, the position vectors of A and B are, OA = i − 2 j + 4k ,, OB = 5i + 2 j + 3k, and, OP = − 2i + 3 j + 5k, ∴ Force = OB − OA, ⇒, F = 4i + 4 j − k = AB, Now, moment about any point ‘P’ is given by, PA = OA − OP = 3i − 5 j − k, Moment of the force = r × F, = PA × AB, [Q torque = r × F], = (3i − 5 j − k) × (4i + 4 j + k), i j, k, = 3 −5 −1, 4 4 −1, = 9i − j + 32k, 38. If a and b are the adjacent sides of any parallelogram,, then its area = | a × b|, ∴ Area of ABCE = | a × b|, Now,, AC = a + b, Then,, CS = a − b, So, area of ACSR = |(a + b) × (a − b)|, = 2| a × b|, (given), Q (a + b) × (a − b) = 2(a × b), So, statement I is true., a + b, Now, side, AD = , , 2 , a − b, Then,, AQ = , , 2 , R, , Q, (a – b), 2, A, , P, (a + b), 2, , a, (a, , S, (a – b), , B, , b, , C, , a + b a − b, area of AQPD = , , ×, 2 2 , 1, = | (a + b) × (a − b)|, 4, 1, = × 2| a × b|, 4, 1, = | a × b|, 2, So, statement II is also true., So,, , ∴, , b = c × b and r ⋅ a = 0, a = 2i + 3 j − k, b = 3i − j + k, c = i + j+ 3k, r = xi + y j + zk, i j k, r× b= x y z, , …(i), , 3 −1 1, = ( y + z )i − (x − 3z ) j + (− x − 3 y)k, i j k, and, , c× b = 1, , 1, , 3 = 4i + 8 j − 4k, , 3 −1 1, Now,, r× b= c× b, ⇒ ( y + z ) i − (x − 3z ) j + (− x − 3 y)k = 4i + 8 j − 4k, Equating the coefficient of i, j and k, we get, …(i), y+ z =4, …(ii), −x + 3z = 8, and, …(iii), −x − 3 y = − 4, Since,, r⋅a = 0, …(iv), ⇒, 2x + 3 y − z = 0, On adding Eqs. (i), (ii) and (iii), we get, …(v), −2 x − 2 y + 4 z = 8, Again, adding Eqs. (iv) and (v), we get, …(vi), y + 3z = 8, Now, from Eq (vi), put 3z = 8 − y in Eq (ii), we get, −x + 8 − y = 8, ⇒, x=− y, Put x = − y in Eq. (iii), we get y = 2 and similarly from, (vi), z = 2, ∴, r = xi + y j + zk, ⇒, r = 2 (− i + j + k), 40. Let the position vectors of A and B be a and b with, respect to C as origin., B( b ), ∴, AB = b − a, and, a⋅b = 0, Similarly,, AC = − a, b, BC = b, BA = a − b, CB = b, and, ∴, , E, , ), +b, , 39. We have, r ×, Where,, and, Let, , C (c), A(a), a, CA = a, AB ⋅ AC + BC ⋅ BA + CB ⋅ CA, = (b − a ) ⋅ (− a ) + (− b) ⋅ (a − b) + b ⋅ a, = − a ⋅ b + a 2 − a ⋅ b + b2 + b ⋅ a, (because a ⋅ b = 0), = a 2 + b2, = CA 2 + CB2, [Q AB = p (given)], = AB2 = p2, , 41. We have, p = a − b, q = a + b, and, | a | = |b| = 2, ∴, p ⋅ q = (a − b) ⋅ (a + b), = a − b ⋅ a + a ⋅ b − b2, = a 2 − b2, =0, [Q |a|=|b|= 2], Angle between p and q is 90°., ∴, | p × q|=| p|| q|sin θ, =| p|q|sin 90° =|p|| q|, ⇒, | p × q|=| a − b|| a + b|
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597, , Vector Algebra, ⇒, ⇒, , | p × q | = (a 2 + b2 )2 − (2a ⋅ b)2, |p × q| = (4 + 4) − 4 (a ⋅ b), 2, , ⇒, , | p × q | = 64 − 4 (a ⋅ b)2, , ⇒, , | p × q | = 2 16 − (a ⋅ b)2, , 2, , 42. Given that, a and b are two unit vectors inclined at an, π, angle ., 3, π, i.e.,, | a | = | b| = 1 and θ =, 3, π, 2, 2, ∴ | a + b| = | a | + |b| + 2| a || b|cos, 3, 1, = 1 + 1 + 2 ×1 ×1 ×, 2, = 3, ⇒ | a + b| = 3 > 1, ∴, | a + b|>1, 43. Unit vector in the direction of 5i + 3 j + 4k, 5i + 3 j + k, a× b , , $, =, Q n =| a × b |, 25 + 9 + 16, Which has forces of magnitude 3, 3 × (5i + 3 j + 4k), i.e.,, F1 =, 5 2, 2 × (3i + 4 j − 5k), Similarly,, F2 =, 5 2, Total force, F = F1 + F2, 1, =, (21i + 17 j + 2k), 5 2, and displacement is given by, d = r2 − r1 = (3i + 3 j + k) − (i + j − k), = 2 (i + j + k), Total work done = force × displacement, = F ⋅d, 1, (21i + 17 j + 2k) ⋅ 2 (i + j + k), =, 5 2, 1 ×2, =, × 40 = 8 2 units, 5 2, 44. Let vector b = xi + yj + zk, and, a =2i + j − k, Given that,, …(i), a⋅b = 3, ⇒, (xi + yj + zk) ⋅ (2i + j − k) = 3, ⇒, 2x + y − z = 3, Q Vectors a and b are collinear, i.e., Angle between, both the vectors should be 0°., Then, a ⋅ b =| a || b|cos 0, ⇒, a ⋅ b = 4 + 1 + 1 x2 + y 2 + z 2 × 1, ⇒, , a⋅b = 6, , x2 + y2 + z 2, , …(ii), , From Eqs. (i) and (ii), we get, ⇒, , 3 = 6 x2 + y 2 + z 2, 3, x2 + y2 + z 2 =, ⇒, 2, 1, 1, Here, b = 1, , − will satisfy Eq. (iii)., 2, 2, , …(iii), , 45. We have, magnitude of vector a = 50., Which is collinear with vector, 15, b = 6i − 8 j −, k, 2, Let, a =xi + y j + zk, ⇒, | a | = | x i + y j + z k|, ⇒, 50 = x2 + y2 + z 2, a and b are collinear, a = λb, 15 , , ⇒ xi + yj + zk = λ 6i − 8 j −, k, , 2 , 625, 2500 = λ2 , ⇒, , 4 , , Q, ∴, , ⇒, λ2 = 16, ⇒, λ=±4, Since, a makes an acute angle with the positive, direction of z-axis, this is possible, when λ = − 4, 15 , , (Q z-axis +ve), a = − 4 6i − 8 j −, k, ∴, , 2 , = − 24i + 32 j + 30k, 46. When a particle moves along a circular path of radius r, in xy-plane, the position vector R of this particle is, (xi + yj) and particle moves in circular path, so equation, of circle is, x2 + y2 = r 2, ⇒, x = r 2 − y2, ∴ Coordinate is r 2 − y2i + yj., 47. We have, a + 2b + 3c = 0, and (a × b) + (b × c) + (c × a ) = λ (b × c), From Eq. (i),, a × b = − 3 (c × b), ⇒, a × b = 3 (b × c), and, a × c = − 2 (b × c), ⇒, c × a = 2 (b × c), C, , …(i), …(ii), , D, , ( i + 2 j + 3 k), , A, , B, (2i+ 4j + 5k ), , ∴ a× b+ b× c+ c× a, = 3 (b × c) + (b × c) + 2 (b × c), From Eq. (ii), = 6 (b × c), 6 (b × c) = λ (b × c), ⇒, λ =6, 48. From option (a),, Let d = 5i − j − 5k ⇒ |d | = 51, (i − 2 j + 2k), ⋅ (5i − j − 5k), a ⋅d, 3, Then, cos θ1 =, =, 3, | a| | d|, 1 ⋅ 51, |5 /3 + 2 /3 − 10 /3|, 1, =, =, 51, 51
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598, , NDA/NA Mathematics, , b⋅d, Similarly, cos θ 2 =, =, | b| | d|, , (−4i − 3k), ⋅ (5i − j − 5k), 5, 1 ⋅ 51, , −4 + 3, =, 51, c ⋅d, cos θ3 =, =, | c| | d|, 1, = −, =, 51, , 1, 51, j ⋅ (5i − j − 5k), And, 1 ⋅ 51, 1, 51, 1 , Here,, θ1 = θ 2 = θ3 = cos −1 , , 51 , So, the vector 5i − j − 5k makes and equal angles with, three vectors a, b and c., =, , 49. Given that,|a|= 6,|b|= 8 and|c|= 10, Also,, a + b + c =0, ⇒, (a + b + c)2 = 0, ⇒, | a |2 + | b|2 + | c|2 + 2 (a ⋅ b + b ⋅ c + c ⋅ a) = 0, ⇒ 36 + 64 + 100 + 2 (a ⋅ b + b ⋅ c + c ⋅ a ) = 0, 200, ⇒, a ⋅ b + b⋅ c + c⋅ a = − , = − 100, 2 , 50. Here, given that sides of a parallelogram are, 2i + 4 j − 5k, and, i + 2 j + 3k, Let AB and AD be the two adjacent sides of the, parallelogram ABCD., [Q BC = AD], ∴, AC = AB + BC = AB + AD, ⇒, AC = 2i + 4 j − 5k + i + 2 j + 3k, ⇒, AC = 3i + 6 j − 2k, and unit vector parallel to AC, 3i + 6 j − 2k 3i + 6 j − 2k, =, =, 9 + 36 + 4, 49, 1, = (3i + 6 j − 2k), 7, 51. From the adjoining figure, A, , B, , C, , AB represents,, EJ = i + j, BC represents,, IH = 2i + j + 3k, and, AC represents JI., By triangle law of vector addition, AC = AB + BC, = (i + j) + (2i + j + 3k), Thus,, JI = 3i + 2 j + 3k, 52. Position vectors of a points are, OP = i + j + k, OQ = 2i + 5 j, OR = 3i + 2 j − 3k and OS = i − 6 j − k, ∴, PQ = OQ − OP = i + 4 j − k, and, RS = OS − OR = − 2i − 8 j + 2k, , PQ ⋅ RS, |PQ||RS|, −2 − 32 − 2 −36, =, =, = −1, 36, 18 72, ⇒, cos θ = cos π, ⇒, θ=π, Hence, PQ and RS form a parallel pair., , ∴, , cos θ =, , 53. Let a = 4i − 3 j + k, P = 2i + 3 j − k, and Q = − 2i − 4 j + 3k, PQ = − 4i − 7 j + 4k, ∴ Projection of the vector a on PQ is given by, a ⋅ PQ, =, |PQ|, (4i − 3 j + k) ⋅ (−4i − 7 j + 4k), =, (−4)2 + (−7)2 + 42, −16 + 21 + 4, 16 + 49 + 16, , =, =, , 9, =1, 9, , 54. Let the position vectors of B, C and R are b, c and r,, respectively., A, a, O, p, q, , (b) B, , ∴ In ∆AOB,, ⇒, , a+ b, 2, b = 2p − a, , p=, , r, P, , R, , Q, , C (c), , (by section formula), , …(i), b+ c, Now, in ∆BOC ,, (by section formula), q=, 2, ⇒, c = 2q − b, [from Eq. (i)], ⇒, c = 2q − (2p − a ), …(ii), c = 2q − 2p + a, a+ c, and in ∆AOC , r =, (by section formula), 2, (2q − 2p + a ) + a, [from Eq. (ii)], =, 2, = q − p + a = a − (p − q ), 2 1 2, 55. Given DC’s of F are , , ., 3 3 3, Then, the direction ratios of F = 2i + j + 2k, 3 (2i + j + 2k), If the forces is of 3N, then F =, 4+1+4, = 2i + j + 2k, Let points P = (1, 3, 5) and Q = (7, 9, 2), Position vector of, P, OP = i + 3 j + 5k, Position vector of, Q, OQ = 7i + 9 j + 2k
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599, , Vector Algebra, Displacement from P to Q, PQ = (7i + 9 j + 2k) − (i + 3 j + 5k), = 6i + 6 j − 3k, Work done,, W = F ⋅ PQ, = (2i + j + 2k) ⋅ (6i + 6 j − 3k), = 12 + 6 − 6, ⇒, W = 12 Nm, 56. Given, conditions are, …(i), A ⋅ B = A ⋅ C but A ≠ 0, …(ii), A × B = A × C but A ≠ 0, From Eq. (i), we get, A ⋅ B = A ⋅ C ⇒ A ⋅ (B − C) = 0, Either A = 0 or B = C or A ⊥ (B − C), ⇒, From Eq. (ii), we get, A× B= A× C, ⇒, A × (B − C) = 0, ⇒ Either A = 0 or B = C or A||B − C, But it is given for both Eq. (i) and (ii) that A ≠ 0, then, clearly, B=C, , ∴ | a × i |2 + | a × k|2 = a 22 + a32 + a12 + a 22, = a12 + 2a 22 + a32., BG, 2, 60. Given,, =−, AG, 3, A, , –3, λ, , O, , G, , In ∆AOG,, , 2 × BG − (−3) × AG, (by section formula), AG − BG, BG, +3, 2×, 2 (−2 /3) + 3, AG, [from Eq. (i)], =, =, BG, 1 − (− 2 /3), 1−, AG, 4, 5, − +3, 3, =, = 3 =1, 2, 5, 1+, 3, 3, 1 ⋅ OG = λOG ⇒ λ = 1, OG =, , Q, , 61. Let r be the vector in xy-plane, then, r = xi + y j, , A2, A3, Q, , P, , B, , 2, , 57. According to question, we see that, Resultant of PA1 + A1Q = PQ, A1, , …(i), , ⇒, , x2 + y 2 = 4 3, , (given), , x + y = 48, Work done = F ⋅ r = 8, 2, , 2, , (given), P, , An, , Resultant of PA 2 + A 2Q = PQ, Resultant of PA3 + A3 Q = PQ, Similarly,, Resultant of PA n + A nQ = PQ, ∴, PA1 + A1Q + K+ PA n + A nQ, = PQ + PQ + . . . + PQ = n PQ, ∴ The resultant of 2n forces = nPQ., 58. We have, A = i + 2 j + 3k, B = − i + 2 j + k, and, C = 3i + j, Now, A + t B = i + 2 j + 3k + t (− i + 2 j + k), = (1 − t )i + (2 + 2 t ) j + (3 + t )k, Since, A + t B is perpendicular to C, then, 3 (1 − t ) + 1 (2 + 2 t ) + (3 + t )0 = 0, ⇒, 3 −3t+2 + 2t =0, ⇒, − t = −5, ⇒, t =5, 59. Here, we have,, a = a1i + a 2j + a3 k, Now,, a × i = (a1 i + a 2j + a3 k) × i, = − a 2k + a3 j, and, a × k = (a1i + a 2j + a3 k) × k, = − a1 j + a 2i, , A, , C, , B, , ⇒, (i + j + 8 k) ⋅ (xi + yj) = 8, ⇒, x+ y=8, ⇒, (x + y)2 = 64, ⇒, x2 + y2 + 2xy = 64, ⇒, 48 + 2xy = 64, ⇒, xy = 8, Now,, (x − y)2 = x2 + y2 − 2xy, ⇒, (x − y)2 = 48 − 16, x− y=± 4 2, ⇒, On solving Eqs. (i) and (ii), we get, x = 4 + 2 2 and y = 4 − 2 2, ∴, ∴ Required vector,, r = (4 + 2 2 )i + (4 − 2 2) j, , …(i), , …(ii), , 62. Given, points are A (3, 5, − 3), B (2, 3, − 1), C (1, 2, 3) and, D (3, 5, 7)., ∴, AB = − i − 2 j + 2k, CD = 2i + 3 j + 4k, Now, AB ⋅ CD = (−1) × 2 + (−2) × 3 + 2 × 4, = −2 −6 + 8= −8 + 8 =0, ∴, AB ⊥ CD
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600, , NDA/NA Mathematics, Since, AB and BC are collinear., k − 40, = − 20, 4, ⇒, k = − 80 + 40, k = − 40, , 63. According to question, we have, y, , ⇒, Q, l, , l, , P, , °, 60, 30°, , x, , O, , 3, 1 , OP = l , i + j, 2 , 2, 1, 3 , OQ = l i +, j, 2 , 2, l, Now,, OP + OQ = [( 3 + 1) i + ( 3 + 1) j], 2, , i, j, k, , , 0, 0, m, ∴ mk × (OP + OQ) = , l, l, ( 3 + 1) ( 3 + 1) 0, , 2, 2, ml, , ml, , =i −, ( 3 + 1) + j, ( 3 + 1), 2, , 2, , 1, = − ml ( 3 + 1) (i − j), 2, 64., E, , D, , F, , C, b, A, , In ∆ABC ,, ⇒, and, ∴, In ∆ACD,, ⇒, In ∆CDE ,, ⇒, In ∆AEF,, ⇒, , a, , B, , AB + BC = AC, …(i), AC = a + b, (by property), AD = 2BC, AD = 2b, AC + CD = AD, CD = 2b − (a + b) = b − a, CE = CD + DE = b − a − a, = b − 2a, FA − CD = − (b − a ) = a − b, AE = EF + FA = − BC + FA, AE = − b − (b − a ), = a − 2b, , 65. Let the position vectors be OA = 60i + 3 j, OB = 40i − 8 j and OC = ki − 52j, Now,, AB = (40 − 60) i + (− 8 − 3) j = − 20i − 11 j, and, BC = (k − 40) i + (− 52 + 8) j, = (k − 40) i − 44 j, k − 40, , = 4 , i − 11 j, , , 4, , , , 66. Since, a + b is collinear with c, ⇒, a + b = xc, x ∈ R, ⇒, a + b + c = (x + 1) c, and b + c is collinear with a, ⇒, b + c = ya , y ∈ R, ⇒, , a + b + c = ( y + 1) a, , ⇒, , (x + 1) c = ( y + 1) a, , ⇒, , x + 1 = 0 and y + 1 = 0, , ⇒, , x = − 1 and y = − 1, , ∴, , ...(i), , [from Eq. (i)], , a+ b+ c= 0, , Hence, a + b + c is a null vector., 67. Since, a is perpendicular to b and c, ∴, a ⋅ b = | a || b|cos 90°, =0, and, a ⋅ c = | a || c|cos 90° = 0, π, and angle between b and c =, 3, π, ∴, b ⋅ c =| b|| c|cos, 3, 1, (Q b and c are unit vectors), =, 2, (| b| = | c| = 1), Now, | a + b + c|2 = | a |2 + | b|2 + | c|2, + 2 (a ⋅ b + b ⋅ c + c ⋅ a ), 1, , , = 1 + 1 + 1 + 2 ⋅ 0 + + 0, , , 2, [Q a, b and c are unit vectors ⇒| a | = | b| = | c| = 1], =1 + 1 + 1 + 1 =4, ⇒, | a + b + c| = 2, 68. We have,, r × a = b × a ⇒ (r − b) × a = 0, ⇒, (r − b )|| a ⇒ r − b = λa, ...(i), ⇒, r = b + λa, Similarly, the equation of the line r × b = a × b can be, rewritten as, ...(ii), r= a + µ b, From Eqs. (i) and (ii), we get, r = i + 3j – k, |r| = 1 + 9 + 1 = 11, ⇒, ∴, , r, i + 3j – k, =, |r|, 11, , 69. Let a vector parallel to yz-plane, ⇒, a = yj + z k, and, b = 3i + 4 j − 2k, Since, a is perpendicular to b., ∴, a ⋅ b = ( yj + zk) ⋅ (3i + 4 j − 2k) = 0, , ...(i)
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601, , Vector Algebra, ⇒, 4 y − 2z = 0, ⇒, z = 2y, On Putting the value of z in Eq. (i), we get, a = yj + 2 y k = y ( j + 2k), a, y ( j + 2k) j + 2k, =, =, ∴ Unit vector =, |a|, 5, y2 + 4 y2, 70., , Q, , 74. Since,|(a × b) ⋅ c |is the volume (V ) of the parallelopiped, whose adjacent edges are a, b and c., i.e.,, V = | a × b|OL, ⇒, V =| a || b|(sin θ ) (cos φ )| c|, ^, n, , 1, 1, , E, , b, a, , = i− j+ k, , 72. We have, a = (1, 2, − 3) and b = (3, − 1, 2), ∴, (a + b) ⋅ (a + λb) = 0, ⇒, (4i + j – k)⋅[ i + 2j – 3k + λ (3i – j + 2k)] = 0, ⇒, 4 (1 + 3λ ) + 1 (2 − λ ) − 1 (− 3 + 2λ ) = 0, ⇒, 4 + 12λ + 2 − λ + 3 − 2λ = 0, ⇒, 9λ = − 9 ⇒ λ = − 1, ∴ Required vector is (a + (− 1) b) i. e. , (a − b)., a = 2i + 3 j + 4 k, b = i – 2j + 3k, j, k, i, 2, 3, 4, ∴, a × b =, , , 1, 2, 3, −, , = i [9 + 8] − j [6 − 4 ] + k [ − 4 − 3 ], = 17i – 2 j – 7k, This is perpendicular to both a and b., 17i – 2 j – 7k, ∴ Unit vector = ±, 172 + 22 + 72, 17i – 2 j – k, =±, 342, ∴ Number of unit vectors are 2., , 73. Let, and, , G, , φ, , 0, 1, , and, | a × b| = 1 + 1 + 1 = 3, Unit vector perpendicular to a and b, (a × b), =±, | a × b|, i− j+ k, =±, 3, Thus, the number of vectors of length 1 unit, perpendicular to the vectors a and b is 2., i + j j + k k + i, 71. Let v = λ , +, +, , 2, 2, 2 , λ, ...(i), v=, [2i + 2 j + 2k], ⇒, 2, λ2, ⇒, |v|2 =, (4 + 4 + 4), 2, 16 × 2 16 8, 2 2, ⇒, =, = ⇒, = λ [Q |v|= 4], λ2 =, 12, 6 3, 3, 2 2, in Eq. (i), we get, Put, λ=, 3, 2 2, (2i + 2 j + 2k), v=, 3 2, 4, (i + j + k), v=, ⇒, 3, , F, , c, , a = i + j and b = j + k, i j k, a×b= 1, 0, , C, , B, , O, , A, , D, , Where, 0 ≤ θ ≤ π is the angle between a and b., Now,, | (a × b) ⋅ c|=| a| | b| | c|, ⇒, |sin θ cos φ | = 1 ⇒ sin θ = 1, cos φ = 1, π, θ = , φ = 0 ⇒ a ⋅ b = b⋅ c = c⋅ a = 0, ⇒, 2, ∴There are two unit vectors perpendicular to both, vectors., 75. Let a be the sides of the cube, then the coordinates of, A , D , B and C are, z, k, B, A, , C, , i, O, , y, , x, , D, , j, , (a , a , a ), (a , a , 0), (a , 0, a ) and (0, a , a ), respectively., Now, OD = ai + aj,OB = ai + ak, OC = aj + ak and OA = ai + aj + ak, Given that,, OD + OB + OC = λ, ∴ ai+ a j+ ai+ ak+ a j+ ak=λ, ⇒, ⇒, , 2ai + 2aj + 2 ak = λ, 2 (ai + aj + ak) = λ, , ⇒, , 2 OA = λ, , 76. Given that, F is mid-point of BD., ∴ In ∆ ABD,, C, , D, , E, , F, A, , B, , 1 ⋅ AB + 1 ⋅ AD, AF =, 1+1, ⇒, , AB + AD = 2AF, , ...(i)
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602, , NDA/NA Mathematics, , CF =, , In ∆BCD,, , 1 ⋅ CB + 1 ⋅ CD, 1+1, , P, , ...(ii), ⇒, CB + CD = 2 CF, On adding Eqs. (i) and (ii), we get, AB + AD + CB + CD = 2AF + 2 CF, = − 2 (FA + FC), = − 2 (2 FE), (Q E is the mid-point of the AC), = − 4 FE, = 5 EF, 77. In ∆ABD,, A, , b, , c, , B, , D, , C, , a, , In ∆ABD,, , AB = AD + DB, 1, c = AD − a, 2, (Q D is the mid-point of BC), 1, AD = a + c, 2, Also, in ∆ACD,, AD + DC = CA, 1, AD + a = b, ⇒, 2, 1, ⇒, AD = b − a, 2, Hence, only statement I represents median AD., , 78. Moment = r × F, ⇒, 10 =|r||F|sin θ, ⇒, 10 = 5 l sin 30°, 10, 10, ⇒, l=, =, =4, 5 sin 30° 5 × 1, 2, P, , ∴, , 30°, , x, F, , l = 4 units, , 79. Position vector of Q, OQ = 2i + 3 j + 5k., and position vector of R, OR = i + j + k, QS and T are the mid-points of PQ and PR,, respectively., 1, ST = QR, ∴, 2, Now,, QR = OR − OQ, = (i + j + k) − (2i + 3 j + 5k), = − i − 2 j − 4k, , Q, , ⇒, , ...(i), , T, , R, , |QR| = (−1)2 + (−2)2 + (−4)2, , = 1 + 4 + 16 = 21, 1, [from Eq. (i)], ∴, ST = QR, 2, 1, 21, ST =, 21 =, units, 2, 2, $ sin θ, 80. Q a × b = | a | | b | n, 2, 2, ...(i), ⇒, |a × b| = |a| |b|2 sin 2 θ, Now,, a ⋅ b =|a||b|cos θ, ...(ii), ⇒, |a ⋅ b|2 = |a|2|b|2 cos 2 θ, On adding Eqs. (i) and (ii), we get, |a × b|2 + |a ⋅ b|2 = |a|2|b|2 (sin 2 θ + cos 2 θ ), ⇒, |a × b|2 = − | a ⋅ b|2 + | a |2|b|2, But, (given), |a × b|2 = k|a ⋅ b|2 + |a|2|b|2, ⇒, k = −1, 81. Let AC = k, and, ∴, ⇒, ⇒, , |AC| = k, AB = AC + CB, AC = AB − CB ⇒ k = a − b, |k|2 = |a|2 + |b|2 − 2|a||b|cos 60°, (Q angle between a and b is 60°), 1, = c2 + c2 − 2c2 ⋅ = c2, 2, [Q|a| = |b| = c given], ⇒, k=c, ∴ Length of required diagonal is c., , 82., , l, , O, , S, , I. | a + b| = | a − b|, On squaring both sides, we get, | a + b|2 = | a − b|2, 2, ⇒, | a | + | b|2 + 2 a ⋅ b = | a |2 + | b|2 − 2 a ⋅ b, ⇒, 4a ⋅ b = 0 ⇒ a ⋅ b = 0, ⇒ a and b are orthogonal to each other., II. | a + b| = | a | + | b|, On squaring both sides, we get, | a + b|2 = (| a | + | b|)2, 2, 2, ⇒, | a | + | b| + 2a ⋅ b = | a |2 + | b|2 + 2| a || b|, ⇒, 2|a || b|cos θ = 2| a || b|, ⇒, cos θ = 1 = cos 0 ⇒ θ = 0, ⇒ a and b are parallel to each other., III. | a + b|2 = | a |2 + | b|2, ⇒ | a | 2 + | b|2 + 2 a ⋅ b = | a |2 + | b|2, ⇒, a⋅b = 0, ⇒ a and b are orthogonal.
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603, , Vector Algebra, , ⇒, 8n 2 = 8m2, …(v), ⇒, n=±m, ∴ m takes 2 values and n takes 2 values from Eqs. (iii), and (iv)., , 83. Given that, direction ratios of a vector PQ are 4, 1, x and, magnitude is 42., ∴, 16 + 1 + x2 = 42, ⇒, x2 = 25 ⇒ x = ± 5, ∴Direction cosines of vectors PQ are, 4, 1, 5, ,, ,, 42, 42, 42, 84. The position vector of points A , B, C and D are, i + j + mk, 2i + 3 j, 3i + 5 j − 2k and − j + k, respectively., Now,, AB = OB − OA, = 2i + 3 j – i – j − mk = i + 2 j − mk, and, CD = OD − OC, = j − k − 3i − 5j + 2k, = − 3i − 6 j + 3k, Since, AB and CD are parallel, j, k , i, m, ∴, AB × (1), 1, 2, −, = 0, , −, 3, −, 6, 3, , , ⇒, ⇒, ⇒, , i (6 − 6m) − j (3 − 3m) + k (− 6 + 6) = 0, 6 − 6m = 0, m =1, , 85. We have,, u = a − b + c, v = 2a − 3b and, w = a + 3c, I. Now, u ⋅ (v × w), = (a − b + c) ⋅ ((2a − 3b) × (a + 3b)), = (a − b + c) ⋅ [2(0) + 6(a × c), − 3(b × a ) − 9 (b × c)], = 6 [a a c] − 3 [a ba ] − 9 [ a b c], − 6 [b a c] + 3 [b b a ] + 9 [b b c], + 6 [c a c] − 3 [c b a ] − 9 [c b c], = − 9 [a b c] − 6 [b a c] − 3 [c b a ], = − 9 [a b c] + 6 [a b c] + 3 [a b c], =0, II. If u, v, w are coplanar, then it is not necessary that, [a b c] ≠ 0., III. It is true that, if u, v, w are coplanar, then there, exist three scalars α , β , γ not all zero such that, αu + βv + γw = 0., ∴ Only statement III is correct., 86. QGiven, |2 i + m j − 3 nk| + 14, and, |5i + 3mj + nk| = 35, From Eq. (i), we get, |2i + mj − 3nk|2 = 14, ⇒, 4 + m2 + 9n 2 = 14, ⇒, m2 + 9n 2 = 10, and from Eq. (ii), we get, 25 + 9m2 + n 2 = 35, ⇒, 9m2 + n 2 = 10, From Eqs. (iii) and (iv), we get, m2 + 9n 2 = 9m2 + n 2, , …(i), …(ii), , 87. If a and b are unit vectors, then it is not possible that, both a + b and a − b are unit vectors., π, 88. Let a = i + j + 2k and θ = ; |a| = 1 + 1 + 2 = 2, 4, I. Let b = i + j ⇒ |b| = 2, a⋅b, cos θ =, |a| |b|, (i + j + 2k) ⋅ (i + j) 1 + 1, ⇒, cos θ =, =, 2 2, 2 2, π, ⇒, θ=, 4, ∴ It is correct statement., II. Let b = i + 2k ⇒ |b| = 1 + 2 = 3, a⋅b, ∴, cos θ =, |a||b|, (i + j + 2k)⋅ (i + 2k) 1 + 2, =, =, 2 3, 2 3, 3, 3, =, =, 2, 2 3, π, θ=, ⇒, 6, ∴ It is not correct statement., III. Let, b = 2k, |b| = 2, ⇒, a⋅ b, (i + j+ 2 k) ⋅ 2k, ∴, cos θ =, =, |a| |b|, 2 2, 2, 1, =, =, 2 2, 2, π, θ=, ⇒, 4, ∴ It is correct statement., ∴ Statements I and III are correct., 89., , We know that,, | a + b|2 = | a |2 + | b|2 + 2 a ⋅ b, (Q a ⋅ b = | a|| b|cos θ ), 2, ⇒, | a + b| = 1 + 1 + 2| a || b|cos 30°, (Q a and b are unit vectors ⇒| a | = | b| = 1), 3, =1 + 1 + 2 ×, 2, ⇒, | a + b|2 = 2 + 3, ⇒, ⇒, , …(iii), , …(iv), , 90. Q, , | a + b| = 2 + 3 = 1.93, 1 < | a + b| < 2, b× c, c× a, p=, , q=, [abc], [abc], , and, , r=, , a× b, [abc], , ∴ (a − b − c) ⋅ p + (b − c − a ) ⋅ q + (c − a − b) ⋅ r, a ⋅ (b × c) b ⋅ (c × a ) c ⋅ (a × b), =, +, +, [a b c], [a b c], [a b c], [a b c] [a b c] [a b c], =, +, +, =3, [a b c] [a b c] [a b c]
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604, , NDA/NA Mathematics, , 91. Q, p⋅ q = p⋅ r, ⇒, p ⋅ (q − r ) = 0, ⇒ p is perpendicular to q − r, and, p× q= p× r, ⇒, p × (q − r ) = 0, ⇒, p is parallel to q − r., Which is not possible simultaneously. Thus, the given, conditions hold simultaneously, if q = r., 92. Since, the diagonals of a parallelogram bisect each, other, p is the mid-point of AC and BD both,, ∴, OA + OC = 2OP and OB + OD = 2 OP, ⇒ OA + OB + OC + OD = 4 OP, 93. Vectors, are i − 2xj − 3 yk and i + 3xj + 2 yk, Since, they are orthogonal., Dot product of vector = 0, ⇒, ⇒ (i − 2xj − 3 yk) ⋅ (i + 3xj + 2 yk) = 0, ⇒, 1 − 6 x 2 − 6 y2 = 0, 1, x2 + y2 =, ⇒, 6, Which is equation of circle, Hence, locus of point (x, y) is a circle., 94. I. Given, a × b = c × d and a × c = b × d, Now,, (a − d ) × (b − c) = a × b − a × c − d × b + d × c, = c × d −b × d + b × d − c × d, =0, ∴ a − d is parallel to b − c., Hence, statement I is correct., II. (a − b) ⋅ (b + c) × (c + a ) = (a − b) ., (b × c + b × a + c × c + c × a ), = (a − b) ⋅(b × c + b × a + c × a ) [Q c × c = 0 ], = a . ( b × c) + a ⋅ ( b × a ) + a ⋅ (c × a ) − b ⋅ ( b × a ), − b ⋅ (b × a ) − b ⋅ (c × a ), = [a b c] − [a b c], [∴ [a a b] = 0], =0, D, III., C, A, B, , We know, AB + BC = AC, and, DC + CB = DB, On adding Eqs. (i) and (ii), we get, AB + BC + DC + CB = AC + DB, or, AB + BC + DC − BC = AC + DB, ∴, AB + DC = AC + DB, Hence, all statements are correct., 95. I. Now,| a + b|2 = | a |2 + | b|2 + 2| a || b|cos, , …(i), ...(ii), , π, 3, , 1, = 3 ⇒ | a + b|= 3 > 1, 2, II. We have, a × (b × c) = (a ⋅ c) b − (a ⋅ b)c, = xb + y c,, where, x = a ⋅ c, y = − a ⋅ b, Therefore, a × (b × c) is coplanar with b and c., = 1 + 1 + 2 ⋅1 ⋅1 ⋅, , 96. p ⋅ a = 0 and p ⋅ i = 0, ⇒, p ⊥ a and p ⊥ b, ⇒ p is perpendicular to the plane of a and b, Also,, p⋅ c = 0 ⇒ p ⊥ c, c lies in the plane of a and b., But, a, b and c are non-coplanar., ⇒ c must not lie in the plane of a and b, ⇒, p =0, Hince, A and R are true and R is the correct explantion, of A., 97. (A) We know that,, Work done = F ⋅ d =| F| ⋅ | d|cos θ, Since,, θ = 90°, = F ⋅d =| F| ⋅ | d|cos 90° = 0, (R) A ⋅ B = 0, ⇒ A and B are perpendicular., Both A and R are true and R is the correct, explanation of A., 98. Since, a, b and c are linearly dependent vectors, then, [a b c] = 0, 1 1 1, ⇒, 4 3 4 =0, 1, ⇒, , 1 α β, 0, 0, , 0 =0, −1, 1 α −1 β −1, 4, , ⇒, ⇒, , −β + 1 =0, β =1, , 99. Volume of parallelopiped = [a b c], 2 −3 1, ⇒, , 1 −1, 2, , 1, , 2 = 2 (1 − 2) + 3 (−1 − 4) + 1 (1 + 2), −1, = − 14, = 14 cu units, , 100. (A) It is true that, if a , b, c are collinear and three, scalars x, y, z not all zero, then x a + y b + z c = 0, where x + y + z = 0, (R) It is the necessary condition., ∴Both A and R are individually true and R is the, correct explanation of A., 101. (A) Force and displacement both are vector quantities., The work done = (Force) ⋅ (Displacement), = F ⋅d, Now, F ⋅ d = | F | | d | cos θ, where θ is the angle between F and d when F and d, are perpendicular to each other, ⇒, θ = 90°, ⇒, cos θ = cos 90° = 0, ⇒, F ⋅d = 0, ⇒ The work done is zero., (R) It is true that, if A and B are perpendicular then, A ⋅ B = 0., ∴ Both A and R are true and R is the correct, explanation of A.
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605, , Vector Algebra, a ⋅ b = (i + j + k) ⋅ (i − j + k) = 1 − 1 + 1 = 1, b ⋅ d = (i − j + k) ⋅ (i − j − k) = 1 + 1 − 1 = 1, ∴ a⋅b= b ⋅d, (ii), b ⋅ c = (i − j + k) ⋅ (i + j − k), =1 −1 −1 = −1, a ⋅ d = (i + j + k) ⋅ (i − j − k) = 1 − 1 − 1 = − 1, ∴ b⋅ c = a⋅d, 1 1, 1, (iii) [a b c ]= 1 −1 1, , 102. (i), , 1, , −1, , 1, , = 1(1 − 1) − 1 (−1 − 1) + 1 (1 + 1), =0 + 2 + 2 =4, i, , j, , k, , (iv) b × c = 1 −1 1 = 2 j + 2 k, 1 1 −1, , Solutions (Q. Nos. 103-105), 103. Given, a = 3i − j + 5k and b = 2i + 3 j + k, a1a 2 + b1b2 + c1c2, ∴ cos θ =, a12 + b12 + c12 a 22 + b22 + c22, =, =, , 3 ×2 −1 ×3 + 5 ×1, 6 −3 + 5, =, 9 + 1 + 25 4 + 9 + 1, 35 14, 8, 7 10, , 104. Q a ⋅ b = (3i − j + 5k) ⋅ (2i + 3 j + k), =6 −3 + 5 =8, i j k, 105. Now, a × b = 3 −1 5, 2, , 3, , 1, , = i (− 1 − 15) − j(3 − 10) + k(9 + 2), = − 16i + 7 j + 11k, ∴ Area of parallelogram = |a × b|, =|− 16 i + 7 j + 11k|, = 256 + 49 + 121 = 426, , Solutions (Q. Nos. 106-108), 1 1, 1, 106. ∴ [a b c] = 1 −1 1 = 1(1 − 2) − 1(−1 − 1) + 1(2 + 1), 1 2 −1, = −1 + 2 + 3 =4, 107. a × [b × c] = (a ⋅ c)b − (a ⋅ b)c, = [(i + j + k) ⋅ (i + 2j − k)] (i − j + k), − [(i + j + k) ⋅ (i − j + k)](i +2j − k), = (1 + 2 − 1)(i − j + k) − (1 − 1 + 1)(i + 2 j − k), = (2i − 2 j + k) − (i + 2 j − k), = i − 4 j + 2k, 108. [a + b b + c c + a ], = 2 [a b c], = 2(4) = 8
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29, Three Dimensional, Geometry, PQ = ( x2 − x1 )2 + ( y2 − y1 )2 + ( z 2 − z1 )2, , Coordinates of a Point in a, Space, , y, P(x1, y1, z1), , Three mutually perpendicular lines in space divide the, space in 8 octants. Let the lines be XOX ′ , YOY ′ and ZOZ ′, intersecting at O. If P ( x , y , z ) is a point in space., Y, , O, Z, , z, , P (x, y, z), X´, , %, , X, , O, , Y´, , | x|= distance from Y Z–plane,, | y| = distance from Z X–plane,, and, | z| = distance from X Y–plane., The following table show the signs of coordinates of, points in various octants, Here,, , OXYZ, OX ′ YZ, OXY ′ Z, OXYZ ′, OX ′ Y ′ Z, OX ′ YZ ′, OXY ′ Z ′, OX ′ Y ′ Z ′, , y, , x, , +, −, +, +, −, −, +, −, , +, +, −, +, −, +, −, −, , two, , y2 + z 2, , points, , Solution (a) Since, coordinates of P(1, 2, 3) and Q(7, 5, 8), then, PQ = (7 − 1) 2 + (5 − 2) 2 + (8 − 3) 2, = (6 2 + 3 2 + 5 2), = (36 + 9 + 25) = 70 units, , z, , Section Formula, , +, +, +, –, +, −, −, −, , Let P ( x1 , y1 , z1 ) and Q( x2 , y2 , z 2 ) be two points. Let R be, a point on the line segment joining P and Q internally in, the ratio m : n. Then, the coordinates of R are, mx2 + nx1 my2 + ny1 mz 2 + nz1 , ,, ,, , , m+n, m+n, m+n , , Distance Formula, Distance between, Q ( x2 , y2 , z 2 ) is, , Distance of a point from x-axis =, , Example 1. Find the distance between P(1, 2, 3) and, Q( 7, 5, 8)., (b) 55 units, (a) 70 units, (c) 60 units, (d) 72 units, , Z´, , Octant, Coordinate, , Q(x2, y2, z2), x, , P ( x1 , y1 , z1 ), , and, , And if P and Q are such that R divides the join of P and, Q externally in the ratio m : n. Then, the coordinates of R, are, mx2 − nx1 my2 − ny1 mz 2 − nz1 , ,, ,, , , m−n, m−n, m−n
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607, , Three Dimensional Geometry, , Example 2. Find the coordinates of a point, which divides, , the joining of P(1, 2, 3) and Q(5, − 2, 3) internally in the ratio, 1 : 2., 1 2 1, 7 2 , (a) , , , (b) , , 3, 3 3 3, 3 3 , 7 −2 , (c) − ,, (d) None of these, , 3, 3 3 , , Solution (b) Let R( x, y , z) be the point, which divides the, joining of P(1, 2, 3) and Q(5, − 2, 3) internally in the ratio 1 : 2., 1× 5 + 2 ×1 7, ∴, x=, =, 1+ 2, 3, , and, , y=, , 1 × ( −2) + 2 × 2 2, =, 1+ 2, 3, , z=, , 1× 3 + 2 × 3 9, = =3, 1+ 2, 3, , Let G( x, y , z) be the centroid of a triangle, then, 1+ 2 + 3, x=, =2, 3, 1+ 2 + 3, 1+ 2 + 3, y=, = 2 and z =, =2, 3, 3, ∴ Coordinates of the centroid of a triangle are (2, 2, 2)., , Example 4. The area of ∆ABC, if the coordinates of vertices, A, B and C are (1, 1, 1,, ) (1, − 1, 2) and (1, 3, 4) respectively is, (a) 5 sq units, (b) 3 sq units, (c) 4 sq units, (d) None of these, , Solution (c) Since, the coordinates of vertices of a ∆ ABC are, A(1, 1, 1), B(1, − 1, 2) and C(1, 3, 4), respectively., i j k, 1, Area of ∆ABC =, 0 −2 1, 2, 0 2 3, 1, = | i ( −6 − 2) − j(0) + k(0)|, 2, 1 2 1, =, 8 = × 8 = 4 sq units, 2, 2, , 7 2 , ∴ Coordinates of R are , , 3 ., 3 3 , , Triangle, , Tetrahedron, , Let A( x1 , y1 , z1 ), B ( x2 , y2 , z 2 ) and C( x3 , y3 , z3 ) be the, vertices of a ∆ABC., , and, Let, C( x3 , y3 , z3 ), A( x1 , y1 , z1 ), B( x2 , y2 , z 2 ),, D( x4 , y4 , z 4 ) be the vertices of tetrahedron., (i) Centroid of tetrahedron is given by, x1 + x2 + x3 + x4 y1 + y2 + y3 + y4 z1 + z 2 + z3 + z 4 , ,, ,, G=, , , , 4, 4, 4, (ii) Volume of tetrahedron is given by, x1 y1 z1 1, 1 x 2 y2 z 2 1, V =, 6 x3 y3 z3 1, x 4 y4 z 4 1, , Centroid of Triangle, The intersection point of all three medians of a triangle, is called centroid of triangle, whose coordinate is, x + x2 + x3 y1 + y2 + y3 z1 + z 2 + z3 , G= 1, ,, ,, ., , , 3, 3, 3, y, A(x1, y1, z1), C(x3, y3, z3), B(x2, y2, z2), x, , Example 5. If A(1, − 2, 3), B(3, 1, 2), C(1, 5, 7) and D(1, − 1, 1) are, the vertices of a tetrahedron, then find the centroid and, volume of tetrahedron., 3 3 13, 3 −3 13, (a) , , , 6, (b) ,, , , 6, 2 4 4 , 2 4 4 , 1 2 13, (d) None of these, (c) , , ,6, 2 3 4 , , z, , Area of Triangle, 1, Area of triangle = ||AB × AC||, 2, 1, =, 2, , Solution (a) Centroid of tetrahedron, , i, , j, , k, , x2 − x1, x3 − x1, , y2 − y1, y3 − y1, , z 2 − z1, z3 − z1, , Example 3. The coordinates of centroid of a ∆ABC, whose, vertices are A(1, 1, 1), B(2, 2, 2) and C(3, 3, 3) is, (a) (2, 2, 2), (b) (1, 2, 2), (c) (0, 2, 1), (d) None of these, , Solution (a) Since, the vertices of a triangle are A(1, 1, 1,) B(2, 2, 2), and C(3, 3, 3)., , 1 + 3 + 1 + 1 −2 + 1 + 5 − 1 3 + 2 + 7 + 1 3 3 13, G=, ,, ,, = , , , , 2 4 4 , 4, 4, 4, , 1 −2, 1 3 1, Volume of tetrahedron =, 6 1 5, 1 −1, =, , 3 1, 2 1, 7 1, 1 1, , 3 2 1, 3 1 1, 3 1 2, 1 2 1, 1, 5 7 1 + 2 1 7 1 + 3 1 5 1 −1 1 5 7 , , 6, 1 1 1, 1 −1 1, 1 −1 1 , −1 1 1
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608, , NDA/NA Mathematics, P ≡ ( x1 , y1, z1 ) andQ ≡ ( x2 , y2, z 2 ) and ABbe a given line with, DC’s as l, m, n. If the line segment PQ makes angle θ with, the line AB, then, , 1, [6 + 2(12) + 3(12) − 30 ], 6, 1, = [6 + 24 + 36 − 30 ] = 6 cu units, 6, =, , Q, Z, , Direction Ratios and Direction, Cosines, Direction cosines of AB are cos α , cos β and cos γ, where, α , β and γ angles made by X-axis, Y-axis and Z-axis,, respectively, where the point A is ( a1 , a2 , a3 ) and the point, B is ( b1 , b2 , b3 )., b − a1, b − a2, b − a3, cos α = 1, , cos β = 2, , cos γ = 3, r, r, r, where, r =|AB|., Normally, direction cosines are denoted by l, m and n, and always l 2 + m 2 + n 2 = 1., If direction cosines of AB are l , m and n,then direction, cosines of line AB are l , m , n and also − l , − m and – n., Direction cosines of X-axis, Y-axis and Z-axis are,, respectively (1, 0, 0), (0, 1, 0) and (0, 0, 1)., If a , b, c are numbers proportional to l , m , n (direction, cosines of line), then a , b, c are called direction ratios or, direction numbers., a, b, where,, l=, ,m =, 2, 2, 2, 2, a +b +c, a + b2 + c2, and, , %, %, , n=, , c, a + b2 + c2, 2, , 1 1 1, Example 6. If the direction cosines of a line are , , ,, c c c , , (a) 0 < c<1 (b) c > 2, , (c) c = ± 2, , (d) c = ± 3, , Solution (d) Since, DC’s of a line are , , ., , , 1 1 1, c c c, , Q, , l 2 + m2 + n 2 = 1, 2, 2, 1, 1, 1, + + =1, c, c, c, 3, =1, c2, c2 = 3, 2, , ∴, ⇒, ⇒, ⇒, , c=± 3, , Projection, Projection of a line joining the points P (x1, y1, z1 ), and Q (x2 , y2 , z2 ) on another line, whose direction, cosines are l, m and n Let PQ be a line segment, where, , P, , M, , N', , θ, , Q, , P, , N, K', , O M, , X, , A, , P', , Q', , B, , Y, , Projection of PQ is P ′ Q ′ = PQ cosθ, = ( x2 − x1 ) cos α + ( y2 − y1 ) cos β + ( z 2 − z1 ) cos γ, = ( x2 − x1 )l + ( y2 − y1 )m + ( z 2 − z1 )n, For X-axis, l = 1, m = 0, n = 0., Hence, projection of PQ on X-axis = x2 − x1, Similarly, projection of PQ on Y-axis = y2 − y1, and projection of PQ on Z-axis = z 2 − z1, , Angle between Two Lines, If direction cosines of two lines are ( l1 , m1 , n1 ) and, ( l2 , m2 , n 2 ), then angle θ between them is given by, cos θ =| l1l2 + m1m2 + n1n 2|., If direction ratios are given ( a1 , b1 , c1 ) and ( a2 , b2 , c2 ), respectively, then, cosθ =, , The direction ratios of a line joining the points A(x1 , y1 , z1 ) and, B(x 2 , y2 , z 2 ) be x 2 − x1 , y2 − y1 and z 2 − z1 ., Projection of a line obtained by joining two given points to, another line whose D C’s are given, is given by, (x 2 − x1 )l + ( y2 − y1 )m + (z 2 − z1 ) n., , then, , K, , a1a2 + b1b2 + c1c2, a12, , + b12 + c12, , a22 + b22 + c22, , ., , Now, if l1l2 + m1m2 + n1n 2 = 0, lines are perpendicular, and if l1 = l2 , m1 = m2 and n1 = n 2 lines are parallel., Similarly, if a1a2 + b1b2 + c1c2 = 0, then lines are, perpendicular and if, a1 b1 c1, , then lines are parallel., =, =, a2 b2 c2, , Example 7. If direction cosines of two lines are (1, 0, 1) and, ( −1, 0, 1), then the angle between these two lines is, (a) 45°, (b) 90°, (c) 60°, (d) 30°, Solution (b) Since, direction cosines of two lines are (1, 0 , 1) and, ( −1, 0 , 1)., Let θ be the angle between them, ∴, cos θ = |1 × ( −1) + 0 + (1 × 1)|, = | − 1 + 1| = 0, ⇒, θ = 90 °, , Straight Line, Every equation of the first degree represents a, plane.Two equations of the first degree are satisfied by the, coordinates of every point on the line of intersection of the, planes represented by them.
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609, , Three Dimensional Geometry, and will be perpendicular, if, a1a2 + b1b2 + c1c2 = 0., , Therefore, the two equations of that line, ax + by + cz + d = 0, and a ′ x + b′ y + c′ z + d ′ = 0 together represent a, straight line., , (i) Equation of a Line Passing Through a Given Point, Cartesian equation of a straight line passing through, a fixed point ( x1 , y1 , z1 ) and having direction ratios a,, b, c is, x − x1 y − y1 z − z1, =, =, a, b, c, (symmetrical form of a straight line), , (ii) Equation of a Line Passing Through Two Given, points, If A( x1 , y1 , z1 ), B( x2 , y2 , z 2 ) be two given points, then, equations to the line AB are, x − x1, y − y1, z − z1, =, =, x2 − x1 y2 − y1 z 2 − z1, %, , %, , %, , Equation of x-axis, x −0, y−0 z −0, or y = 0 and z = 0, =, =, 1, 0, 0, Equation of y-axis, x −0, y−0 z −0, or x = 0 and z = 0, =, =, 0, 1, 0, Equation of z-axis, x −0, y−0 z −0, or x = 0 and y = 0, =, =, 0, 0, 1, , Example 8. The equation of a straight line passing through, A(1, 2, 3) and B(3, 5, − 1) is, x −1 y − 2 z − 3, (a), =, =, 2, 3, −4, x −1 y − 2 z − 3, (c), =, =, 4, −3, −2, , (b), , x +1 y − 2 z − 3, =, =, 2, 4, −3, , (d) None of these, , Solution (a) The equation of a straight line passing through, A(1, 2, 3) and B(3, 5, − 1) is given by, x −1 y − 2 z − 3, =, =, 3 − 1 5 − 2 −1 − 3, x −1 y − 2 z − 3, =, =, ⇒, 2, 3, −4, , Angle between Two Lines, , %, , (x 2 , y2 , z 2 ) is, [(x 2 − x1 ) 2 + ( y2 − y1 ) 2 + (z 2 − z1 ) 2, − { l (x 2 − x1 ) + m ( y2 − y1 ) + n (z 2 − z1 )}]1 / 2, or, , Then,, , y2 − y1, , l, , m, , 2, , +, , y2 − y1, , z 2 − z1, , m, , n, , 2, , z 2 − z1, , x 2 − x1, , n, , l, , 2, , Example 9. The lines x = ay + b, z = cy + d and, x = a ′ y + b ′, z = c ′ y + d ′ are perpendicular, if aa ′ + cc ′ + 1is, equal to, (a) 0, (b) 3, (c) 2, (d) 5, Solution (a) We can write the equations of straight line as, x−b, z −d, = y, y =, a, c, x− b y −0 z −d, =, =, ⇒, a, 1, c, x − b′, z − d′, and, = y, y =, a′, c′, x − b′ y − 0 z − d ′, =, =, ⇒, a′, c′, 1, x − x1 y − y1 z − z1, We know,, =, =, a1, b1, c1, x − x2 y − y 2 z − z 2, and, =, =, a2, b2, c2, , are perpendicular, if, a1a2 + b1b2 + c1c2 = 0, ⇒, aa′ + 1 + cc′ = 0, , Skew-Lines, Two non-parallel, non-intersecting straight lines are, Skew-lines., Let, the, straight, lines, are, x − x1 y − y1 z − z1, x − x2 y − y2 z − z 2, and, and d, =, =, =, =, a1, b1, c1, a2, b2, c2, is the shortest distance between them., L, , x − x1 y − y1 z − z1, =, =, a1, b1, c1, x − x2 y − y2 z − z 2, ., =, =, a2, b2, c2, a1a2 + b1b2 + c1c2, cos θ =, 2, a1 + b12 + c12 a22 + b22 + c22, , Two lines will be parallel, if, a1 b1 c1, =, =, a2 b2 c2, , x 2 − x1, , +, , Let θ be the angle between, and, , x − x1 y − y1, z − z1, and point, =, =, l, m, n, , Distance between a line, , B, , A, , C, , ∴ d=, , M, , D, , x2 − x1, , y2 − y1, , z 2 − z1, , l1, , m1, , n1, , l2, , m2, , n2, , ( l1m2 − l2m1 ) + ( m1n 2 − m2n1 )2 + ( l1n 2 − l2n1 )2, 2
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610, , NDA/NA Mathematics, , Example 10. Find the shortest distance between the lines, , x−1 y− 2 z− 3, x− 2 y− 4 z −5, and, ., =, =, =, =, 2, 3, 4, 3, 4, 5, (b) 3 / 6, (a) 1 / 6, (d) None of these, (c) 2 / 6, , Solution (a) Given, lines are, , x −1 y − 2 z − 3, …(i), =, =, 2, 3, 4, x−2 y − 4 z −5, …(ii), and, =, =, 3, 4, 5, Here, x1 = 1, y1 = 2, z1 = 3, x2 = 2, y 2 = 4, z 2 = 5,, l1 = 2, m1 = 3, n1 = 4, l2 = 3, m2 = 4, n2 = 5, Shortest distance between the lines (i) and (ii) are modulus of, x2 − x1 y 2 − y1 z 2 − z1, l1, m1, n1, l2, , m2, , n2, , (l1m2 − l2m1) 2 + (m1n2 − m2n1) 2 + (l1n2 − l2n1) 2, x2 − x1 y 2 − y1 z 2 − z1, Now,, , l1, l2, , m1, m2, , n1, n2, , …(iii), , −1 −2 −2, = 2 3, 4, 3 4 5, , Equation of the planes bisecting the angles between, two, planes, and, a1x + b1 y + c1z + d1 = 0, a2x + b2 y + c2 z + d2 = 0 is, a1x + b1 y + c1z + d1, a x + b2 y + c2z + d2, =± 2, 2, 2, 2, a1 + b1 + c1, a22 + b22 + c22, , Condition, (i) If a1a2 + b1b2 + c1c2 is negative, then origin lies in, the acute angle between., (ii) If a1a2 + b1b2 + c1c2 is positive, then origin lies in, the obtuse angle between the given planes., , Example 11. Find the equation of the bisector planes of the, , Solution (a) The two given planes are, 2x − y + 2z + 3 = 0 and 3x − 2y + 6z + 8 = 0, , Plane, A plane is a surface such that, if any two points are, taken on it, the line segment joining them lies completely, on the surface., , Normal to Plane, A line perpendicular to a plane is called normal to the, plane. It is clear that every line lie in a plane perpendicular, to the normal to the plane., , Equation of a plane through a point ( x1 , y1 , z1 ) is, A ( x − x1 ) + B ( y − y1 ) + C ( z − z1 ) = 0., Here, A, B, C are direction ratios of a line normal to, plane., , Equation of the Planes Bisecting, the Angles between Two Planes, , (a) (14x − 7y + 14z + 21) = ± (9x − 6y + 18z + 24), (b) (14x + 7y + 14z + 21) = ± (9x − 6y + 18z + 24), (c) (14x − 7y + 14z + 21) = ± (9x + 6y + 18z + 24), (d) None of the above, , ⇒, , Plane Through a Point ( x 1, y 1, z 1 ), , From Eq. (iii), shortest distance between lines (i) and (ii), 1 1, = −, =, 6, 6, , 2x − y + 2z + 3 = 0, , ⇒, , 2x − y + 2z + 3, 3x − 2y + 6z + 8, =±, 4 + 1+ 4, 9 + 4 + 36, 2x − y + 2z + 3, 3x − 2y + 6z + 8, =±, 3, 7, (14x − 7y + 14z + 21) = ± (9x − 6y + 18z + 24), , General equation of a plane is, Ax + By + Cz + D = 0 (Linear equation in x , y and z ), , = (8 − 9) 2 + (15 − 16) 2 + (10 − 12) 2 = 6, , planes, , i.e.,, , General Equation of a Plane, , = − 1 (15 − 16) + 2 (10 − 12) − 2 (8 − 9) = − 1, Also, (l1m2 − l2m1) 2 + (m1n2 − m2n1) 2 + (n1l2 − n21, l )2, , angles between the, 3x − 2y + 6 z + 8 = 0., , where, d1, d 2 > 0, a1x + b1y + c1z + d1, a x + b2y + c2z + d 2, =± 2, ∴, 2, 2, 2, a1 + b1 + c1, a22 + b22 + c22, , and, , Example 12. Let P( −7, 1, − 5) be a point on a plane and let O, be the origin. If OP is normal to the plane, then the equation, of the plane is, (a) 7x − y + 5 z + 75 = 0, (b) 7x + y − 5 z + 73 = 0, (c) 7x + y + 5 z + 73 = 0, (d) None of these, Solution (a) Equation of any plane passing through ( −7,1, − 5) is, a( x + 7) + b(y − 1) + c( z + 5) = 0, The DR’s of normal to the above plane are, a = −7, b = 1, c = −5, ∴From Eq.(i), we get, −7( x + 7) + 1(y − 1) − 5( z + 5) = 0, ⇒, 7x − y + 5z + 75 = 0, , ...(i), , Example 13. A plane meets the coordinate axes in A, B, C, such that the centroid of the ∆ ABC is the point ( p, q, r), then, the equation of the plane is, x y z, x y z, (b) + + = 3, (a) + + = 2, p q r, p q r, x y z, (c) + + = 4, (d) None of these, p q r, , Solution (b) Let the required equation of plane be, x y z, + + =1, a b c, , …(i)
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612, , NDA/NA Mathematics, ax1 + by1 + cz1 + d, , Example 15. The normals to the planes x − y + z = 1,, 3x + 2y − z + 2 = 0 are inclined at an angle of, (a) 0°, (b) 30°, (c) 45°, , a 2 + b2 + c2, (d) 90°, , Solution (d)The direction cosines of the normal to the plane, ⇒, , x − y + z = 1are proportional to (1, − 1, 1), 1, 1, 1, DC’s of normal to the plane are , ,−, ,, ., 3, 3, 3, , Similarly, DC’s of normal to 3x + 2y − z + 2 = 0 are, 2, 1 , 3, ,, ,−, ., , 14 14, 14 , ∴ If θ be an angle between the normals of the plane, 1 , 1 3 1 2 1 , ⇒ cos θ = , + −, , + −, , 3 14 , 3 14 3 14 , 3 − 2 −1, =, = 0 ⇒ θ = 90 °, 42, Thus, the normals are inclined to each other at 90°., , Equation of Plane Parallel to Coordinate, Planes or Perpendicular to Coordinate Axes, (i) Equation of plane parallel to YOZ-plane, (or perpendicular to x-axis) and at a distance ‘a’ from, it is x = a., (ii) Equation of plane parallel to ZOX-plane, (or perpendicular to y-axis) and at a distance ‘b’ from, it is y = b., (iii) Equation of plane parallel to XOY-plane, (or perpendicular to z-axis) and at a distance ‘c’ from, it is z = c., , Equation of Plane Perpendicular to Coordinate, Planes or Parallel to Coordinate Axes, (i) Equation of plane perpendicular to YOZ-plane or, parallel to x-axis is By + Cz + D = 0., (ii) Equation of plane perpendicular to ZOX-plane or, parallel to y-axis is Ax + Cz + D = 0., (iii) Equation of plane perpendicular to XOY-plane or, parallel to z- axis is Ax + By + D = 0., (iv) Equation of plane parallel to a given plane, ax + by + cz + d = 0 is ax + by + cz + d ′ = 0 i.e., only, constant term is changed., , Foot of Perpendicular from a Point A (α , β, γ ) to, a Given Plane ax + by + cz + d = 0, If AP be the perpendicular from A to the given plane,, then it is parallel to the normal, so that its equation is, x−α y−β z − γ, =, =, = r (say), a, b, c, Any point P on it is ( ar + α , br + β , cr + γ ) , It lies on the, given plane and we find the value ofr and hence, the point P., , Perpendicular Distance, The length of the perpendicular from the point, P ( x1 , y1 , z1 ) to the plane ax + by + cz + d = 0 is, , ., , Distance, between, two, parallel, planes, Ax + By + Cz + D1 = 0 and Ax + By + Cz + D2 = 0 is, D2 ~ D1, A2 + B2 + C 2, , Position of Two Points wrt a Plane, Two points P ( x1 , y1 , z1 ) and Q( x2 , y2 , z 2 ) lie on the same, or opposite sides of a plane ax + by + cz + d = 0according to, ax1 + by1 + cz1 + d and ax2 + by2 + cz 2 + d are of same or, opposite signs. The plane divides the line joining the, points P and Q externally or internally according to P and, Q are lying on same or opposite sides of the plane., , Example 16. Find the distance of the point (2, 1, 0) from, the plane 2 x + y + 2 z + 5 = 0., (a) 10/3, (b) 5/3, (c) 7/3, (d) None of these, Solution (a) Required distance, =, , 2 × 2 + 1× 1+ 2 × 0 + 5, 2 +1 + 2, 2, , 2, , 2, , =, , 10, 3, , Example 17. Find the distance between the parallel planes, 2 x − y + 2 z + 3 = 0 and 4x − 2y + 4z + 5 = 0., (a) 1/6, (b) 5/6, (c) 4/6, (d) None of these, Solution (a) Let P( x1, y1, z1) be any point on 2x − y + 2z + 3 = 0., Then,, 2x1 − y1 + 2z1 + 3 = 0, The length of perpendicular from P( x1, y1, z1) to, 4x − 2y + 4z + 5 = 0, 4x1 − 2y1 + 4z1 + 5, =, 4 2 + ( −2) 2 + 4 2, 2(2x1 − y1 + 2z1) + 5, 2( −3) + 5, 1, =, =, =, 6, 6, 36, , …(i), , Family of Planes, 1. Let P1 ≡ a1x + b1 y + c1z + d1 = 0 and P2 ≡ a2x + b2 y, + c2z + d2 = 0 be two planes, then P1 + λP2 = 0 (where,, λ is a parameter) represents family of planes passing, through line of intersection of the planes P1 = 0 and, P2 = 0., 2. ax + by + cz = k represents family of planes parallel, to the plane ax + by + cz + d = 0. (where, k is a, parameter)., , Example 18. Find the equation of plane containing the line, , of intersection of the planes x + y + z − 6 = 0, 2 x + 3y + 4z + 5 = 0 and passing through (1, 1, 1., ), (a) 20 x + 23y + 26 z + 69 = 0, (b) 20 x + 23y + 26 z − 69 = 0, (c) 20 x + 23y − 26 z − 69 = 0, (d) None of the above, , and
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613, , Three Dimensional Geometry, , Solution (b) The Equation of the plane through the line of, intersection of the given planes is,, ( x + y + z − 6) + λ (2x + 3y + 4z + 5) = 0, If it passes through (1, 1, 1), ⇒, (1 + 1 + 1 − 6) + λ (2 + 3 + 4 + 5) = 0, 3, λ=, ⇒, 14, 3, Putting λ =, in Eq. (i), we get, 14, 3, ( x + y + z − 6) +, (2x + 3y + 4z + 5) = 0, 14, ⇒, 20 x + 23y + 26z − 69 = 0, , The fixed point is called the centre and the constant, distance is called the radius of the sphere., P (r), , …(i), , Angle between Line and Plane, x−α y−β z − γ, and, =, =, l, m, n, the plane ax + by + cz + d = 0 is given by, al + bm + cn, sin θ =, 2, 2, ( a + b + c2 ) ( l 2 + m 2 + n 2 ), The angle θ between the line, , (i) The line is perpendicular to the plane if and only if, a, b, c, =, = ., l m n, (ii) The line is parallel to the plane if and only if, al + bm + cn = 0., (iii) The line lies in the plane if and only if, al + bm + cn = 0 and aα + bβ + cγ + d = 0., , Example 19. The angle between the line, 3x − 1 y + 3 5 − 2 z, and the plane 3x − 3y − 6 z = 0, =, =, 3, −1, 4, is equal to, π, π, π, π, (a), (b), (c), (d), 6, 4, 3, 2, , Solution (d) Given, line and plane can be written as, 1, 5, z−, 3 = y+3 =, 2, 1, −1, −2, and, x − y − 2z = 0, Here, a1 = 1, b1 = −1, c1 = −2 and a2 = 1, b2 = − 1, c2 = −2, a1a2 + b1b2 + c1c2, Sin θ =, ∴, a12 + b12 + c12 a22 + b22 + c22, 1 × 1 + ( −1) × ( −1) + ( −2)( −2), =, 1+ 1+ 4 1+ 1+ 4, 6, π, =, = 1⇒ θ =, 2, 6 6, x−, , Sphere, A sphere is the locus of a point, which moves in space in, such a way that its distance from a fixed point always, remains constant., , C (a ), , General Equation of Sphere, The, general, equation, of, a, sphere, is, with, centre, x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0, ( − u , − v , − w) i.e., ( −1 / 2) coefficient of x, − (1 / 2) coefficient, 1, of y, − coefficient of z) and radius = u 2 + v 2 + w2 − d ., 2, , Equation of Sphere in Various Forms, (i) Equation of sphere with given centre and, radius The equation of a sphere with centre ( a , b, c) and, radius R is, ( x − a )2 + ( y − b)2 + ( z − c)2 = R 2, , …(i), , If the centre is at the origin, then Eq.(i) takes the form, x 2 + y 2 + z 2 = R 2, which is known as the standard form of, the equation of the sphere., (ii) Diameter form of the equation of a, sphere If ( x1 , y1 , z1 ) and ( x2 , y2 , z 2 ) are the coordinates of, the extremities of a diameter of a sphere, then its equation, is ( x − x1 )( x − x2 ) + ( y − y1 )( y − y2 ) + ( z − z1 )( z − z 2 ) = 0., , Example 20. Find the equation of sphere, if (1, 2, 3) and, (1, − 1, 0) are the vertices of diameter., (a) x 2 + y 2 + z 2 − 2 x − y − 3z − 1 = 0, (b) x 2 + y 2 + z 2 + 2 x + y − 3z − 1 = 0, (c) x 2 + y 2 + z 2 − 2 x + y − 3z − 1 = 0, (d) None of the above, Solution (a) If (1, 2, 3) and (1, −1, 0) are the vertices of diameter,, then equation of sphere is, ( x − 1) ( x − 1) + (y − 2) (y + 1) + ( z − 3) ( z − 0) = 0, ⇒, x2 − 2x + 1 + y 2 − y − 2 + z 2 − 3z = 0, ⇒, , x2 + y 2 + z 2 − 2x − y − 3z − 1 = 0, , Example 21. Find the equation of sphere, whose centre is, C (5, −2, 3) and which passes through the point P (8, − 6, 3)., (a) x 2 + y 2 + z 2 − 10 x + 4y + 6 z + 13 = 0, (b) x 2 + y 2 + z 2 − 10 x + 4y − 6 z + 13 = 0, (c) x 2 + y 2 + z 2 + 10 x + 4y − 6 z + 13 = 0, (d) None of the above, Solution (b) We have, radius, = (8 − 5) 2 + ( −6 + 2) 2 + (3 − 3) 2 = 9 + 16 + 0 = 5, Equation of required sphere is, ( x − 5) 2 + (y + 2) 2 + ( z − 3) 2 = 5 2, ⇒, , x2 + y 2 + z 2 − 10 x + 4y − 6z + 13 = 0
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614, , NDA/NA Mathematics, , Section of a Sphere by a Plane, Consider a sphere intersected by a plane. The set of, points common to both sphere and plane is called a plane, section of a sphere. The plane section of a sphere is always, a circle. The equations of the sphere and the plane taken, together represent the plane section., , Example 22. The radius of the circle in which the sphere, , x 2 + y 2 + z 2 + 2 x − 2y − 4z − 19 = 0 is cut by the plane, x + 2y + 2 z + 7 = 0 is, (a) 1, (b) 2, (c) 3, (d) 4, , Solution (c) The centre and radius of given sphere are C( −1,1, 2), and R = ( −1) 2 + (1) 2 + (2) 2 + 19 = 5, Length of perpendicular from centre C on the plane,, −1 × 1 + 1 × 2 + 2 × 2 + 7, d=, =4, 12 + 2 2 + 2 2, , C, , P, , M, , ∴ Radius of circle = R 2 − d 2 = 25 − 16 = 3, , Q, , Let C be the centre of the sphere and M be the foot of, the perpendicular from C on the plane. Then, M is the, centre of the circle and radius of the circle is given by, PM = CP 2 − CM 2 ., The centre M of the circle is the point of intersection of, the plane and line CM which passes through C and is, perpendicular to the given plane., Centre The foot of the perpendicular from the centre, of the sphere to the plane is the centre of the circle., (Radius of circle)2 = (Radius of sphere)2− (Perpendicular, from centre of spheres on the plane)2, , Condition of Tangency of a Plane to a Sphere, Plane ax + by + cz + d = 0 touches a sphere, x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + k = 0,, If distance of plane from centre of sphere is equal to, radius of sphere., − au − bv − cw + d, i.e.,, = u 2 + v 2 + w2 − k., 2, 2, 2, a +b +c, %, , The conditions of orthogonal intersection of two sphere is, 2 u1 u 2 + 2v1v2 + 2w1 w2 = d1 + d 2, , Comprehensive Approach, n, n, n, , n, , n, , n, , n, , Any plane parallel to x-axis is of the form by + cz = d ., Intersection point of planes x = a, y = b and z = c is ( a, b , c)., Image ( x, y , z) (or reflection) of a point ( x1 , y1 , z1) in a plane, ax + by + cz + d = 0 is given by, x − x1 y − y1 z − z1, =, =, a, b, c, −2 ( ax1 + by1 + cz1 + d ), =, a2 + b 2 + c 2, Foot ( x, y , z) of a point ( x1 , y1 , z1) in a plane ax + by + cz + d = 0, is given by, x − x1 y − y1 z − z1, =, =, a, b, c, ( ax1 + by1 + cz1 + d ), =−, a2 + b 2 + c 2, Any plane parallel to X-Y plane is z = constant, similarly plane, parallel to Y-Z plane is x = constant and plane parallel to Z - X, plane is y = constant. x = 0, y = 0 and z = 0 are, respectively Y-Z,, Z-X and X-Y planes., Points P ( x1 , y1 , z1) and Q ( x2 , y 2 , z 2) are on same side of plane, if, and, ax + by + cz + d = 0,, ax1 + by1 + cz1 + d, ax2 + by 2 + cz 2 + d are of same sign. If they are of opposite sign,, then the points are on the opposite sides., Equation of plane parallel to planes ax + by + cz + d1 = 0 and, ax + by + cz + d 2 = 0 and equidistance from them is, d + d2, ax + by + cz + 1, =0, 2 , , n, , The, equation, of, x − x1 y − y1 z − z1, =, =, a1, b1, c1, x − x2 y − y 2 z − z 2, is, =, =, a2, b2, c2, , n, , n, , n, , n, , n, , n, , plane, and, , containing, parallel, to, , the, the, , line, line, , x − x1 y − y1 z − z1, , a1, b1, c1 = 0, a2, b2, c2, x − x1 y − y1 z − z1, lies in the plane, The line, =, =, a1, b1, c1, a2x + b2y + c2z + d 2 = 0, then a2x1 + b2y1 + c2z1 + d 2 = 0, and, a1a2 + b1b2 + c1c2 = 0, There are four lines, which are equally inclined to the coordinate, axes., 1, The angle between any two diagonals of a cube is cos−1 ., 3, The angle between a diagonal of a cube and the diagonal of a, 2, faces of the cube is cos−1 , ., 3, When two spheres touch each other the common tangent plane is, S1 − S2 = 0 and when they cut in a circle, the plane of the circle is, S1 − S2 = 0 ; coefficient of x2 ,y 2 , z 2 being unity in both the cases., (i) The plane cuts the sphere in a circle, iff p < r and in this case,, the radius of circle is r 2 − p 2 ., (ii) The plane touches the sphere; iff p = r., (iii) The plane does not meet the sphere, iff p > r., Equation of concentric sphere Any sphere concentric with, the, sphere, is, x2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0, x2 + y 2 + z 2 + 2ux + 2vy + 2wz + λ = 0 where λ is some real,, which makes it a sphere.
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Exercise, Level I, 1. The coordinates of the point in which the line joining, the points ( 3, 5, − 7) and ( −2, 1, 8) is intersected by the, plane yz are given by, 13, 2, , − 13, , (b) 0, −, (a) 0,, , − 2, ,− , , , , 5, 5, 5, 13 2, 13 , (c) 0,, (d) 0,, , , , 2, 5 5, 5 , 2. The projection of any line on coordinate to be,, respectively 3, 4, 5, then its length is, (a) 50, (b) 12, (d) None of these, (c) 5 2, 3. The direction cosines of the normal to the plane, x + 2 y − 3z + 4 = 0 are, 1, 2, 3, 1, 2, 3, ,, ,, (a) −, ,−, (b), ,−, 14 14 14, 14, 14, 14, 2, 3, 1, 2, 1, 3, ,, (d), ,−, (c) −, ,, ,, 14 14, 14 14, 14, 14, 4. The angle between the straight lines, x+1 y− 2 z + 3, x−1 y+ 2 z − 3, and, is, =, =, =, =, 2, 5, 4, 1, 2, −3, (a) 45°, (b) 30°, (c) 60°, (d) 90°, 5. The length of the perpendicular from the origin to the, plane 3x + 4 y + 12z = 52 is, (a) 3, (b) – 4, (c) 5, (d) None of these, 6. What is the locus of points of intersection of a sphere, and a plane?, (NDA 2011 II), (a) Circle, (b) Ellipse, (c) Parabola, (d) Hyperbola, 7. The intercepts of the plane 5x − 3 y + 6z = 60 on the, coordinate axes are, (a) (10, 20, − 10), (b) (10, − 20, 12), (c) (12, − 20, 10), (d) (12, 20, − 10), 8. The direction cosines of the line joining the points, ( 4, 3, − 5) and ( −2, 1, − 8) are, 6 2 3, 2 3 6, (b) , , , (a) , , , 7 7 7, 7 7 7, 6 3 2, (c) , , , (d) None of these, 7 7 7, 9. If a line lies in the octant OXYZ and it makes equal, angles with the axes, then, 1, 1, (a) l = m = n =, (b) l = m = n = ±, 3, 3, 1, 1, (d) l = m = n = ±, (c) l = m = n = −, 3, 2, , 10. What is the acute angle, x + y + 2z = 3 and −2x + y − z, π, π, (a), (b), (c), 5, 4, , between the planes, (NDA 2011 I), = 11 ?, π, π, (d), 6, 3, , 11. If α , β , γ be the angles, which a line makes with the, coordinate axes, then, (a) sin2 α + cos2 β + sin2 γ = 1, (b) cos2 α + cos2 β + cos2 γ = 1, (c) sin2 α + sin2 β + sin2 γ = 1, (d) cos2 α + cos2 β + sin2 γ = 1, 12. If a line makes the angles α , β and γ with the axes,, then what is the value of 1 + cos 2α + cos 2β + cos 2γ ?, (NDA 2012 I), , (a) –1, , (b) 0, , (c) 1, , (d) 2, , 13. Equation of x-axis is, x y z, x y z, x y z, x y z, (a) = = (b) = = (c) = = (d) = =, 1 1 1, 0 1 1, 1 0 0, 0 0 1, 14. The, two, planes, and, ax + by + cz + d = 0, ax + by + cz + d1 = 0, where d ≠ d1 , have (NDA 2010 II), (a) one point only in common, (b) three points in common, (c) infinite points in common, (d) no points in common, 15. If direction cosines of two lines are proportional to, ( 2, 3, − 6) and ( 3, − 4, 5), then the acute angle between, them is, 18 2 , 49, (b) cos−1 , (a) cos−1 , , 36, 35 , (c) 96°, , 18 , (d) cos−1 , 35, , 16. The value of aa ′ + bb ′ + cc ′ being negative. The, origin will lie in the acute angle between the planes, ax + by + cz + d = 0 and a ′ x + b ′ y + c ′ z + d ′ = 0, if, (a) a = a ′ = 0, (b) d and d′ are of same sign, (c) d and d′ are of opposite sign, (d) None of the above, 17. The equation to the straight line passing through the, points ( 4, − 5, − 2) and ( −1, 5, 3) is, x−4 y+5 z+2, x+1 y− 5 z − 3, (b), (a), =, =, =, =, 1, −1, −2, 1, 2, −1, x, y z, x, y, z, (c), (d) =, = =, =, −1 5 3, 4 −5 −2, 18. Let O ( 0, 0, 0), P ( 3, 4, 5), Q ( m , n , r ) and R (1, 1, 1) be, the vertices of a parallelogram taken in order. What, is the value of m + n + r ?, (NDA 2010 I), (a) 6, (b) 12, (c) 15, (d) More than 15
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616, , NDA/NA Mathematics, , 19. If a line makes an angles α , β , γ , δ with four diagonals, of, a, cube,, then, the, value, of, sin2 α + sin2 β + sin2 γ + sin2 δ is, (a) 4/3, (b) 8/3, (c) 7/3, (d) 1, 20. The xy-plane divides the line joining the points, ( −1, 3, 4) and ( 2, − 5, 6), (a) internally in the ratio 2 : 3, (b) internally in the ratio 3 : 2, (c) externally in the ratio 2 : 3, (d) externally in the ratio 3 : 2, 21. The projection of the line joining the points (3, 4, 5), and (4, 6, 3) on the line joining the points ( −1, 2, 4), and (1, 0, 5) is, (a) 4/3, (b) 2/3, (c) – 4/3, (d) 1/2, 22. What is the equation of the plane through z-axis and, x−1 y+ 2 z − 3, parallel to the line, ?, =, =, cos θ sin θ, 0, (NDA 2010 I), (a) x cot θ + y = 0, (c) x + y cot θ = 0, , (b) x tan θ − y = 0, (d) x − y tan θ = 0, , 23. In the space the equation by + cz + d = 0 represents a, plane perpendicular to the plane, (a) YOZ, (b) Z = k, (c) ZOX, (d) XOY, 24. The points A( 4, 5, 1), B( 0, − 1, − 1), C( 3, 9, 4), D( −4, 4, 4) are, (a) collinear, (b) coplanar, (c) non-coplanar, (d) non-collinear and non-coplanar, , and, , 25. The direction cosines of a line are proportional to, ( 2, 1, 2) and the line intersects a plane, perpendicularly at the point (1, − 2, 4). What is the, distance of the plane from the point (3, 2, 3)?, (NDA 2009 II), , (a), , 3, , (b) 2, , (c) 2 2, , (d) 4, , 26. What are the direction ratios of normal to the plane, (NDA 2012 I), 2 − y + 2z + l = 0 ?, 1, (a) < 2, 1, 2 >, (b) < 1 , − , 1 >, 2, (c) < 1, − 2, 1 >, (d) None of these, 27. The point of intersection of the line joining the points, ( −3, 4 , − 8) and ( 5, − 6, 4) with the xy-plane is, (a) (7/3, – 8/3, 0), (b) (– 7/3, – 8/3, 0), (c) (– 7/3, 8/3, 0), (d) (7/3, 8/3, 0), 28. If the angle between the lines, whose direction ratios, π, are ( 2, − 1, 2) and ( x , 3, 5) is , then the smallest value, 4, of x is, (a) 52, (b) 4, (c) 2, (d) 1, 29. The foot of the perpendicular drawn from the origin, to a plane is the point (1, − 3, 1). What is the intercept, cut on the x-axis by the plane?, (NDA 2009 II), (d) 11, (a) 1, (b) 3, (c) 11, 30. The equation of the sphere, whose centre is (1, 1, 1), and which passes through (3, 3, 2), is, , (a) x 2 + y 2 + z 2 + 2x + 2 y + 2z = 6, (b) x 2 + y 2 + z 2 − 2x − 2 y − 2z = 0, (c) x 2 + y 2 + z 2 − 2x − 2 y − 2z = 6, (d) x 2 + y 2 + z 2 + 2x + 2 y + 2z = 38, 31. A straight line with direction cosines (0, 1, 0) is, (a) parallel to the x-axis., (b) parallel to the y-axis., (c) parallel to the z-axis., (d) equally inclined to all the axes., 32. What is the diameter of the sphere, x 2 + y 2 + z 2 − 4x + 6 y − 8z − 7 = 0 ? (NDA 2012 I), (a) 4 units (b) 5 units (c) 6 units (d) 12 units, 33. Under what condition does the equation, x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0 represent a, real sphere?, (NDA 2009 I), 2, 2, 2, 2, 2, 2, (b) u + v + w2 > d, (a) u + v + w = d, 2, 2, 2, (d) u 2 + v 2 + w2 < d 2, (c) u + v + w < d, 34. If a line makes angles of 60° and 45° with the positive, direction of the axes of x and y respectively, then the, angle made by the line with positive direction of the, z-axis is equal to, (a) 60°, (b) 120°, (c) Either 60° or 120°, (d) Neither 60° nor 120°, 35. ABC is a triangle and AD is the median. If the, coordinates of A are (4, 7, –8) and the coordinates of, centroid of the ∆ABC are (1, 1, 1). What are the, coordinates of D?, 11, , 1, 1, (b) − , − 2 , , (a) − , 2 , 11, , 2, 2, 2, (c) ( −1, 2 , 11), (d) ( −5 , − 11, 19), 36. If the points (5, –1, 1), (–1, –3, 4) and (1, – 6, 10) are, the three vertices of a rhombus taken in order, then, which one of the following is the fourth vertex?, 7 11, , (a) ( 7, − 4, 11), (b) 3, − , , , 2 2, (c) ( 7, − 4, 7), (d) ( 7, 4, 11), 37. What is the equation of a plane through the x-axis, and passing through the point (1, 2, 3)? (NDA 2009 I), (a) x + y + z = 6, (b) x = 1, (c) y + z = 5, (d) z + y = 1, 38. The planes bx − ay = n , cy − bz = l and az − cx = m, intersect in a line. Which one of the following is, correct?, (a) a + b + c = 0, (b) al + bm + cn = 0, (c) a/ l + b/ m + c/ n = 0, (d) l + m + n = 0, 39. What, is, the, condition, for, the, plane, ax + by + cz + d = 0 to be perpendicular to xy-plane?, (a) a = 0, (b) b = 0, (c) c = 0, (d) a + b + c = 0, 40. The equation by + cz + d = 0 represents a plane, parallel to which one of the following? (NDA 2008 II), (a) x-axis, (b) y-axis, (c) z-axis, (d) None of these
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617, , Three Dimensional Geometry, , 41. For what value of k are the lines, , z −1, x, and, = 2− y =, 2, k, , x+1 y−1, =, = z + 2 perpendicular to each other?, 3, k, (a) –2, (b) –1, (c) 1, (d) 2, , 42. If, the, radius, of, the, sphere, x 2 + y 2 + z 2 − 6x − 8 y + 10z + λ = 0 is unity, what is, the value of λ ?, (NDA 2008 II), (a) 49, (b) 7, (c) –49, (d) –7, 43. Under which one of the following conditions will the, two planes x + y + z = 7 and αx + βy + γz = 3, be, parallel (but not coincident)?, (NDA 2008 I), 3, (a) α = β = γ = 1 only, (b) α = β = γ = only, 7, (c) α = β = γ, (d) None of these, 44. How many arbitrary constants are there in the, equation of a plane?, (NDA 2008 I), (a) 2, (b) 3, (c) 4, (d) Any finite number, x−3 y−4 z−5, is parallel to, 45. The straight line, =, =, 2, 3, 4, which one of the following?, (NDA 2008 I), (a) 4x + 3 y − 5z = 0, (b) 4x + 5 y − 4z = 0, (c) 4x + 4 y − 5z = 0, (d) 5x + 4 y − 5z = 0, 46. If θ is the acute angle between the diagonals of a cube,, then which one of the following is correct?, (a) θ < 30°, (b) θ = 60°, (NDA 2008 I), (c) 30° < θ < 60°, (d) θ > 60°, 47. What is the equation of the plane passing through, ( x1 , y1 , z1 ) and normal to the line with < a , b, c > as, direction ratios?, (NDA 2007 II), , (a), (b), (c), (d), , ax + by + cz = ax1 + by1 + cz1, a( x + x1 ) + b( y + y1 ) + c( z + z1 ) = 0, ax + by + cz = 0, ax + by + cz = x1 + y1 + z1 = 0, , 48. If the sum of the squares of the distances of the point, ( x , y , z ) from the points ( a , 0, 0) and ( −a , 0, 0) is 2c2 ,, then which one of the following is correct? (NDA 2007 I), (a) x 2 + a 2 = 2c2 − y 2 − z 2 (b) x 2 + a 2 = c2 − y 2 − z 2, (c) x 2 − a 2 = c2 − y 2 − z 2 (d) x 2 + a 2 = c2 + y 2 + z 2, 49. What is the angle between the two lines, whose, direction ratios are, ( 3 − 1, − 3 − 1, 4) and (− 3 − 1, 3 − 1, 4)?, π, π, π, π, (a), (b), (c), (d), 6, 4, 3, 2, 50. If α , β , γ be angles, which the vector r = λi + 2 j − k, makes with the coordinate axes, then what is the, value of sin2 α + sin2 β + sin2 γ ?, (a) 2, (b) 1, (c) λ 2 + 1, (d) 1 − λ 2, 51. Which one of the following is correct?, The, three, planes, 2x + 3 y − z − 2 = 0,, 3x + 3 y + z − 4 = 0, x − y + 2z − 5 = 0 intersect, (NDA 2007 I), , (a) at a point, (c) at three points, , (b) at two points, (d) in a line, , 52. What, is, the, centre, of, the, sphere, ax 2 + by 2 + cz 2 − 6x = 0, if the radius is 1 unit?, (a), (b), (c), (d), , (NDA 2007 I), ( 0, 0, 0), (1, 0, 0), ( 3, 0, 0), Cannot be determined as values of a, b, c are, unknown, , Level II, 1. The equation of the plane in which the lines, x−5 y−7 z+3, x−8 y−4 z−5, and, lie, is, =, =, =, =, 4, 4, −5, 7, 1, 3, (a) 17x − 47 y − 24z + 172 = 0, (b) 17x + 47 y − 24z + 172 = 0, (c) 17x + 47 y + 24z + 172 = 0, (d) 17x − 47 y + 24z + 172 = 0, , 4. What is the equation of the plane passing through, the point (1, − 1, − 1) and perpendicular to each of the, planes x − 2 y − 8z = 0 and 2x + 5 y − z = 0 ?, , 2. What is the sum of the squares of direction cosines of, the line joining the points (1, 2, –3) and (–2, 3, 1) ?, , 5. The equation of the plane passing through the line, x−1 y+ 2 z − 3, and the point (4, 3, 7) is, =, =, 5, 6, 4, (a) 4x + 8 y + 7z = 41, (b) 4x − 8 y + 7z = 41, (c) 4x − 8 y − 7z = 41, (d) 4x − 8 y + 7z = 39, , (NDA 2012 I), , (a) 0, , (b) 1, , (c) 3, , (d), , 2, 26, , x y z, + + = 1 cut the axes in A, B, C, then, 2 3 4, the area of the ∆ ABC is, (b) 41 sq units, (a) 29 sq units, (c) 61 sq units, (d) None of these, , 3. The plane, , (NDA 2011 II), , (a) 7x − 3 y + 2z = 14, (c) x − 7 y + 3z = 4, , (b) 2x + 5 y − 3z = 12, (d) 14x − 5 y + 3z = 16, , 6. The centre of sphere passes through four points, (0, 0, 0), (0, 2, 0), (1, 0, 0) and (0, 0, 4) is, , 1, , 1, (b) − , 1, 2, (a) , 1, 2, , 2, , 2, , 1, 1 , (c) , 1, − 2, (d) 1, , 2, 2 , , 2
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618, , NDA/NA Mathematics, , 7. The shortest distance between the lines, x−3 y−8 z−3, x+3 y+7 z−6, and, is, =, =, =, =, 3, 1, 2, 4, −1, −3, (d) 3 30, (b) 2 30, (c) 5 30, (a) 30, 8. The equation to sphere passing through origin and, the points ( −1, 0, 0), ( 0, − 2, 0) and ( 0, 0, − 3) is, x 2 + y 2 + z 2 + f ( x , y , z ) = 0.What is f ( x , y , z )equal to?, (NDA 2011 II), , (a) − x − 2 y − 3z, (c) x + 2 y + 3z − 1, , (b) x + 2 y + 3z, (d) x + 2 y + 3z + 1, x y z, 9. The angle between the line = = and the plane, 2 3 4, 3x + 2 y − 3z = 4 is, (a) 45°, (b) 0°, 24 , −1 , (d) 90°, (c) cos , , 29 22 , 10. The equation of a line of intersection of planes, 4x + 4 y − 5z = 12 and 8x + 12 y − 13z = 32 can be, written as, x−1 y+ 2 z, x−1 y− 2 z, (b), (a), =, =, =, =, 2, 4, −3, 2, 3, 4, x y+1 z − 2, x y z−2, (c) =, (d) = =, =, 2, 3, 4, 2 3, 4, 11. Consider the following relations among the angles, α , β and γ made by a vector with the coordinate axes, (NDA 2011 I), , I. cos 2α + cos 2β + cos 2γ = − 1, II. sin2 α + sin2 β + sin2 γ = 1, Which of the above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 12. The equation of the plane, which makes with, coordinate axes a triangle with its centroid (α , β , γ ), is, (a) αx + βy + γz = 3, (b) αx + βy + γz = 1, x y z, x y z, (d) + + = 1, (c) + + = 3, α β γ, α β γ, 13. A variable plane moves, so that the sum of the, reciprocals of its intercepts on the coordinates axes is, 1/2 . Then, the plane passes through, 1 1 1, (a) , ,− , (b) ( −1, 1, 1), 2 2 2, (c) ( 2, 2, 2), (d) ( 0, 0, 0), 14. The direction cosines l , m , n of two lines are, connected by the relations l + m + n = 0, lm = 0, then, the angle between them is, (a) π/ 3, (b) π/ 4, (c) π / 2, (d) 0, 15. Equation of the plane passing through the line of, intersection of the planes P ≡ ax + by + cz + d = 0,, P ′ ≡ a ′ x + b ′ y + c ′ z + d ′ = 0 and parallel to x-axis is, (a) Pa − P ′ a ′ = 0, (b) P / a + P ′/ a ′ = 0, (c) Pa + P ′ a ′ = 0, (d) P / a = P ′/ a ′, , 16. Under what condition do the planes bx − ay = n ,, cy − bz = l , az − cx = m intersect in a line?, (NDA 2010 I), , (a) a + b + c = 0, (c) al + bm + cn = 0, , (b) a = b = c, (d) l + m + n = 0, , 17. What is the cosine of angle between the planes, x + y + z + 1 = 0 and 2x − 2 y + 2z + 1 = 0? (NDA 2012 I), 1, 1, (a), (b), 2, 3, 2, (d) None of these, (c), 3, 18. The equation of the plane passing through (2, 3, 4), and parallel to the plane 5x − 6 y + 7z = 3 is, (a) 5x − 6 y + 7z + 20 = 0 (b) 5x − 6 y + 7z − 20 = 0, (c) −5x + 6 y − 7z + 3 = 0 (d) 5x + 6 y + 7z + 3 = 0, 19. If the plane x + 2 y + 2z − 15 = 0 cuts the circle, x 2 + y 2 + z 2 − 2 y − 4z − 11 = 0, then radius of circle is, (b) 5, (c) 7, (d) 3, (a) 3, 20. What is the area of the triangle with vertices ( 0, 2, 2),, (NDA 2010 I), ( 2, 0, − 1) and ( 3, 4, 0)?, 15, (a), sq units, (b) 15 sq units, 2, 7, (c) sq units, (d) 7 sq units, 2, 21. The points A( 5, − 1, 1), B( 7, − 4, 7), C(1, − 6, 10) and, D( −1, − 3, 4) are vertices of a, (a) square, (b) rhombus, (c) rectangle, (d) None of these, 22. If P = ( 0, 1, 0), Q = ( 0, 0, 1), then projection of PQ on, the plane x + y + z = 3 is, (a) 3, (b) 3, (d) 2, (c) 2, 23. Coordinate of a point equidistant from the points, (0, 0, 0), ( a , 0, 0), ( 0, b, 0), ( 0, 0, c) is, a b c, a b c, (a) , , , (b) , , , 4 4 4, 2 4 4, a, b, c, , , (c) , , , (d) ( a , b, c), 2 2 2, 24. What is the equation of the sphere, which has its, centre at ( 6, − 1, 2) and touches the plane, (NDA 2009 II), 2x − y + 2z − 2 = 0 ?, 2, 2, 2, (a) x + y + z + 12x − 2 y + 4z + 16 = 0, (b) x 2 + y 2 + z 2 + 12x − 2 y + 4z − 16 = 0, (c) x 2 + y 2 + z 2 − 12x + 2 y − 4z + 16 = 0, (d) x 2 + y 2 + z 2 − 12x + 2 y − 4z + 25 = 0, 25. The equation of the sphere touching the three, coordinates planes is, (a) x 2 + y 2 + z 2 + 2a( x + y + z ) + 2a 2 = 0, (b) x 2 + y 2 + z 2 − 2a( x + y + z ) + 2a 2 = 0, (c) x 2 + y 2 + z 2 ± 2a( x + y + z ) + 2a 2 = 0, (d) x 2 + y 2 + z 2 ± 2ax ± 2ay ± 2az + 2a 2 = 0
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619, , Three Dimensional Geometry, 26. The plane 2x − (1 + λ ) y + 3λz = 0 passes through the, intersection of the planes as 2x − y = 0 andy + zy = 0 is, (a) 2x − y = 0 and y + 3z = 0, (b) 2x − y = 0 and y − 3z = 0, (c) 2x + 3z = 0 and y = 0, (d) None of the above, , 33. The points A( 5, 9, 9), B ( 0, − 1, − 6) and C( 5, − 11, − 1), are the vertices of, (a) an acute angled triangle, (b) a right angled triangle with unequal sides, (c) an isosceles right angled triangle, (d) an obtuse angled triangle, , 27. The radius of the circle in which the sphere, x 2 + y 2 + z 2 + 2x − 2 y − 4z − 19 = 0 is cut by the plane, x + 2 y + 2z + 7 = 0, is, (a) 1, (b) 2, (c) 3, (d) 4, x−2 y−3 z−4, and, 28. The, line, =, =, 1, 1, −k, x−1 y− 4 z − 5, are coplanar, if, =, =, k, 2, 1, (a) k = 0 or −1, (b) k = 0 or 1, (c) k = 0 or −3, (d) k = 3 or −3, , 34. The line passing through the points (1, 2, − 1) and, ( 3, − 1, 2) meets the yz-plane at the points is, 7 5, , 7 5, (b) 0, , , (a) 0, − , , , 2 2, , 2 2, 7, 5, 5, , , 7, (d) 0, , − , (c) 0, − , − , , 2, 2, 2, 2, , 29. What are the direction ratios of the line determined, by the planes x − y + 2z = 1 and x + y − z = 3?, (a) ( −1, 3, 2), (b) ( −1, − 3 2) (NDA 2009 II), (c) ( 2, 1, 3), (d) ( 2, 3, 2), 30. Consider the following statements, I. If two spheres touch each other externally, then, the distance between their centres is r1 + r2., II. Any plane parallel to x-axis is of the form, bx + cz = d., x − x1 y − y1 z − z1, lies in the plane, III. The line, =, =, a1, b1, c1, a2x + b2 y + c2z + d2 = 0, if a1a2 + b1b2 + c1c2 = d2., Which of the statements given above are correct?, (a) I and II, (b) II and III, (c) I and III, (d) All I, II and III, 31. Consider the following statements, I. Foot ( x , y , z ) of a point ( x1 , y1 , z1 ) in a plane, ax + by + cz + d = 0 is given by, x − x1 y − y1 z − z1, =, =, a, b, c, ( ax1 + by1 + cz1 + d ), =−, a 2 + b2 + c2, II. Image ( x , y , z ) of a point ( x1 , y1 , z1 ) in a plane, ax + by + cz + d = 0 is given by, x − x1 y − y1 z − z1 −( ax1 + by1 + cz1 + d ), =, =, =, a, b, c, a 2 + b2 + c2, Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 32. The point of intersection of the lines, x+1, y+3 z+5, x−2 y−4 z−6, and, is, =, =, =, =, 3, 5, 7, 1, 3, 5, (a) (1/2, 1/2, –3/2), (b) (– 1/2, – 1/2, 3/2), (c) (1/2, – 1/2, – 3/2), (d) (– 1/2, 1/2, 3/2), , 35. Find the coordinate of the point, which is equidistant, from the points ( 0, 0, 0), ( 2, 0, 0), ( 0, 4, 0), ( 0, 0, 6)., respectively., (NDA 2008 II), (a) (1, 2, 3), (b) ( 2, 3, 1), (c) ( 3, 1, 2), (d) (1, 3, 2), 36. In three dimensional space, the path of a point,, whose distance from the x-axis is 3 times its distance, from the yz-plane is, (a) y 2 + z 2 = 9x 2, (b) x 2 + y 2 = 3z 2, 2, 2, 2, (c) x + z = 3 y, (d) y 2 − z 2 = 9x 2, 37. The equation of the sphere through x 2 + y 2 + z 2 = 4,, 2x + 3 y + 4z = 7 and (1, 2, 0) is given by, (a) x 2 + y 2 + z 2 − 2x − 3 y + 4 = 0, (b) x 2 + y 2 + z 2 − 2x − 3 y − 4z + 3 = 0, (c) x 2 + y 2 + z 2 − 2x − 3 y − 6z + 3 = 0, (d) x 2 + y 2 + z 2 − 2x − 3 y − 8z + 5 = 0, 38. The equation of sphere, which passes through the, origin and makes intercepts 3, 4 and 5 on the, coordinate axes, is given by, (a) x 2 + y 2 + z 2 − 3x − 4 y + 5z = 0, (b) x 2 + y 2 + z 2 − 3x − 4 y − 5z = 0, (c) x 2 + y 2 + z 2 − 4x − 5 y − 3z = 0, (d) x 2 + y 2 + z 2 − 3x + 4 y − 5z = 0, 39. The line passing through (1, 2, 3) and having, direction ratios given by < 1, 2, 3 > cuts the x-axis at a, distance k from origin. What is the value of k?, (NDA 2008 II), , (a) 0, , (b) 1, , (c) 2, , (d) 3, , 40. The equation of the plane containing the points, (1, 0, 0), (0, 2, 0) and (0, 0, 3) is given by, (a) x + 2 y + 3z = 1, (b) 3x + 2 y + z = 2, (c) 6x + 3 y + 2z = 6, (d) 6x + 3 y + 2z = 8, 41. The direction cosines of the line perpendicular to the, line with direction ratios (1, –2, –2) and (0, 2, 1) are, 1 2, 2 1 2, 2, (b) − , , , (a) , − , , 3 3 3, 3, 3 3, 1, 2, 2, 2 1, 2, (c) , , − , (d) − , − , − , 3 3, 3, 3, 3, 3
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620, , NDA/NA Mathematics, , 42. Match List I (Equations of spheres) with List II, (Their centres) and select the correct answer using, the codes given below the lists., List I, (Equations of spheres), A., B., C., D., , List II, (Their centres), , x2 + y 2 + z 2 + 3 x, −3 y + 3z − 49 = 0, x2 + y 2 + z 2 − 4 x, −6 y − 2 z + 9 = 0, , 1., , (2, 3, 1), , 2., , 2 x2 + 2 y 2 + 2 z 2, −x − y − z = 0, x2 + y 2 + z 2 − 2 x, −2 y − 2 z = 0, , 3., , 3, 3 3, − , , − , 2 2, 2, (1, 1, 1), , Codes, A B, (a) 2 4, (c) 2 1, , C, 1, 4, , D, 3, 3, , 1 1 1, , , , 4 4 4, , 4., , A, (b) 3, (d) 3, , B, 1, 4, , C, 4, 1, , D, 2, 2, m, 43. Under what condition are the two lines y =, x + α,, l, n, m′, n′, and, x + α′ ,, x + β′, z = x+β, y=, z=, l, l′, l′, orthogonal?, (NDA 2008 I), (a) αα ′ + ββ ′ + 1 = 0, (b) (α + α ′ )(β + β ′ ) = 0, (c) ll ′ + mm ′ + nn ′ = 1, (d) ll ′ + mm ′ + nn ′ = 0, 44. The condition for the lines x = az + b, y = cz + d and, x = a1z + b1 , y = c1z + d1 to be perpendicular is, (a) ac1 + a1c + 1 = 0, (b) aa1 + cc1 + 1 = 0, (c) ac1 + a1c − 1 = 0, (d) aa1 + cc1 − 1 = 0, 45. The lines 2x = 3 y = − z and 6x = − y = − 4z, (a) are parallel, (b) are perpendicular, (c) intersect at an angle of 45°, (d) intersect at an angle of 60°, 46. The distance between the parallel planes, 4x − 2 y + 4z + 9 = 0 and 8x − 4 y + 8z + 21 = 0 is, 1, 1, (b), (a), 4, 2, 3, 7, (c), (d), 2, 4, 47. The locus of a point such that the sum of the squares, of its distances from the planes x + y + z = 0, x − z = 0, and x − 2 y + z = 0 is 9, is, (b) x 2 + y 2 + z 2 = 6, (a) x 2 + y 2 + z 2 = 3, 2, 2, 2, (d) x 2 + y 2 + z 2 = 12, (c) x + y + z = 9, 48. Which of the following conditions such that the line, x− p y− q z −r, lies, on, the, plane, =, =, l, m, n, Ax + By + Cz + D = 0 is/are correct?, I. lp + mq + nr + D = 0, , II. Ap + Bq + Cr + D = 0, III. Al + Bm + Cn = 0, Select the correct answer using the codes given below, (a) Only I, (b) I and II, (c) I and III, (d) II and III, 49. The equation of a sphere is x 2 + y 2 + z 2 − 10z = 0. If, one end point of a diameter of the sphere is, ( −3, − 4, 5), what is the other end point? (NDA 2007 II), (a) ( −3, − 4, − 5), (b) ( 3, 4, 5), (c) ( 3, 4, − 5), (d) ( −3, 4, − 5), 50. The plane x − 1 = 0 intersects the sphere, x 2 + y 2 + z 2 = 4 in a circle, whose centre and radius r, are given by, (a) ( 0, 0, 1), r = 3, (b) ( 0, 1, 0), r = 1, (d) (1, 0, 0), r = 1, (c) (1, 0, 0), r = 3, 51. The ratio in which the joining of ( 2, 1, 5) and ( 3, 4, 3), 1, divided by the plane x + y − z = , is, 2, (a) 3 : 5, (b) 5 : 7, (c) 1 : 3, (d) 4 : 5, 52. The direction ratios of the line OP are equal and the, length OP = 3. Then, the coordinates of the point P, are, (a) ( −1, − 1, − 1), (b) ( 3 , 3 , 3 ), (d) ( 2, 2, 2), (c) ( 2 , 2 , 2 ), 53. What is the area of the triangle, whose vertices are, (0, 0, 0), (3, 4, 0) and (3, 4, 6)?, (a) 12 sq units, (b) 15 sq units, (c) 30 sq units, (d) 36 sq units, 54. The planes x = 2 y − 3 and z = 2 y − 1 intersect along a, line. What are the direction ratios of the line?, (a) ( 3, 2, 0) (b) ( 2, 3, 0) (c) ( 2, 1, 2) (d) (1, 2, 2), 55. Given, the sphere x 2 + y 2 + z 2 − 4x + 6 y − 8z − 71 = 0, and the two points A (1, − 1, 2) and B ( 2, − 3, 4). Which, one of the following is correct?, (a) A, B are inside the sphere., (b) A, B are outside the sphere., (c) A is inside the sphere and B is outside the sphere., (d) A is outside the sphere and B is inside the sphere., 56. What is the angle between the line 6x = 4 y = 3z and, the plane 3x + 2 y − 3z = 4?, (a) 0°, (b) π/ 6, (c) π/ 3, (d) π / 2, 57. What is the locus of a point, which is equidistant, from the points (1, 2, 3) and ( 3, 2, − 1)?, (a) x + z = 0, (b) x − 3z = 0, (c) x − z = 0, (d) x − 2z = 0, 58. What is the locus of a point, which moves such that, the difference of the squares of its distances from two, given points in space is constant?, (a) A straight line, (b) A plane, (c) A sphere, (d) Any surface other than a plane or a sphere
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621, , Three Dimensional Geometry, 59. P ( a , b, c), Q ( a + 2, b + 2, c − 2), and R ( a + 6, b + 6, c − 6) are collinear, Consider the following statements, I. R divides PQ internally in the ratio 3 : 2., II. R divides PQ externally in the ratio 3 : 2., III. Q divides PR internally in the ratio 1 : 2., Which of the statements given above is/are correct?, (a) I only, (b) II only, (c) I and III, (d) II and III, 60. Consider the points ( a − 1, a , a + 1), ( a , a + 1, a − 1), and ( a + 1, a − 1, a )., I. These points always form the vertices of an, equilateral triangle for any real value of a., II. The area of the triangle formed by these points is, independent of a., Which of the statement(s) given above is/are correct ?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 61. Consider the following statements, I. Equations ax + by + cz + d = 0,, a ′ x + b′ y + c′ z + d ′ = 0, represent a straight line., II. Equation of the form, x−α y−β z − γ, =, =, l, m, n, represent a straight line passing through the point, (α , β , γ ) and having direction ratios proportional to, l , m , n., Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 62. If the direction ratios of the normal to a plane are, < l , m , n > and the length of the normal is p, then, what is the sum of intercepts cut-off by the plane, from the coordinate axes?, 1, 1 1, (a) p +, + , l m n, (b) p ( l 2 + m 2 + n 2 ), 1, 1 1, (c) p ( l 2 + m 2 + n 2 ) +, + , l m n, p, 1, 1 1, (d), + , +, 2, 2, 2 l, m n, (l + m + n ), 63. If O , P are the points (0, 0, 0,), (2, 3, –1) respectively,, then what is the equation to the plane through P at, right angles to OP ?, (a) 2x + 3 y + z = 16, (b) 2x + 3 y − z = 14, (c) 2x + 3 y + z = 14, (d) 2x + 3 y − z = 0, , Directions (Q. Nos. 64-66) Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 64. Assertion (A) The image of the point P(1, 3, 4) in, the plane 2x − y + z + 3 = 0 is ( −3, 5, 2)., Reason (R) The image ( x , y , z ) of a point ( x1 , y1 , z1 ), in a plane ax + by + cz + d = 0 is given by, x − x1 y − y1 z − z1, 2 ( ax1 + by1 + cz1 + d ), ., =, =, =−, a, b, c, a 2 + b2 + c2, 65. Assertion (A) Two spheres x 2 + y 2 + z 2 + 2x + 2 y, + 2z + 5 = 0 and x 2 + y 2 + z 2 + 3x + 3 y + 7z + 6 = 0, touch each other, then the equation of common, tangent is x + y + 5z + 1 = 0., Reason (R) If two spheres S1 and S 2 touch each, other, then S1 − S 2 = 0 is common tangent plane., 66. Assertion (A) If < l , m , n > are direction cosines of, a line, there can be a line, whose direction cosines are, l2 + m2 m2 + n 2 n 2 + l2, ,, ,, ., 2, 2, 2, , (NDA 2007 I), , Reason (R) The sum of direction cosines of a line is, unity., , Directions (Q. Nos. 67-69), , Let us consider two, straight lines, x+4 y−3 z+2, x y −1 z, and L2 = =, L1 =, =, =, =, 3 −2 1, 1, 2, 3, , 67. The DC’s of L1 is, 1, 2, 3, (a) <, ,, ,, >, 14 14 14, 1, 2, 1, (c) <, ,, ,, >, 14 14 14, , (b) <, , −1, 14, , ,, , −2, 14, , ,, , 3, 14, , >, , (d) None of these, , 68. The angle between L1 and L2 is, 1, (a) cos− 1 , (b) cos− 1 ( 7), 7, 2, (d) None of these, (c) cos− 1 , 7, 69. Find the equation of plane, which line L1 lies in the, end whose normal to the plane is (1, 3, 2)., (a) x + 3 y + 2z + 1 = 0, (b) x + 3 y + 2z − 1 = 0, (c) x − 3 y + 2z − 1 = 0, (d) None of these
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Answers, Level I, 1., 11., 21., 31., 41., 51., , (d), (b), (c), (b), (b), (d), , 2., 12., 22., 32., 42., 52., , (c), (b), (b), (d), (a), (b), , 3., 13., 23., 33., 43., , (d), (c), (a), (b), (c), , 4., 14., 24., 34., 44., , (d), (d), (b), (a), (c), , 5., 15., 25., 35., 45., , (d), (b), (b), (b), (c), , 6., 16., 26., 36., 46., , (a), (b), (b), (c), (d), , 7., 17., 27., 37., 47., , (c), (a), (a), (b), (a), , 8., 18., 28., 38., 48., , (a), (c), (b), (b), (b), , 9., 19., 29., 39., 49., , (a), (b), (d), (c), (c), , 10., 20., 30., 40., 50., , (d), (c), (d), (a), (a), , 2., 12., 22., 32., 42., 52., 62., , (b), (c), (c), (c), (c), (a), (c), , 3., 13., 23., 33., 43., 53., 63., , (c), (c), (c), (b), (d), (b), (b), , 4., 14., 24., 34., 44., 54., 64., , (d), (a), (c), (d), (b), (c), (a), , 5., 15., 25., 35., 45., 55., 65., , (b), (d), (b), (a), (b), (a), (a), , 6., 16., 26., 36., 46., 56., 66., , (a), (c), (b), (a), (a), (a), (c), , 7., 17., 27., 37., 47., 57., 67., , (d), (b), (c), (b), (c), (d), (a), , 8., 18., 28., 38., 48., 58., 68., , (b), (b), (c), (b), (d), (b), (a), , 9., 19., 29., 39., 49., 59., 69., , (b), (c), (a), (a), (b), (d), (b), , 10., 20., 30., 40., 50., 60., , (b), (a), (a), (c), (c), (c), , Level II, 1., 11., 21., 31., 41., 51., 61., , (a), (a), (b), (a), (a), (b), (c), , Hints & Solutions, Level I, 1. Let the joining the points (3, 5, − 7) and (−2, 1, 8), which, divide the plane in k :1., Coordinate of that point is, −2 k + 3 k + 5 8 k − 7 , ,, ,, ., , k+1 k+1 k+1 , Since, these are the coordinates of yz-plane, −2 k + 3, ∴, =0, k+1, 3, k=, ⇒, 2, 3, , 3, + 5 8 × − 7, , 2, 2, ,, ∴ Point is 0,, , 3 +1 3 +1 , , 2, 2, 13 10, 13 , or, , or 0,, , 2, 0,, 5 5, 5 , 2. It is clear from the figure, that the coordinate of P is, (3, 4, 5)., y, P, , R, 4, O, 5, S, z, , 3, , Q, , x, , OP = (3 − 0)2 + (4 − 0)2 + (5 + 0)2, = 9 + 16 + 25 = 5 2, 3. Equation of plane is x + 2 y − 3z + 4 = 0. Its direction, ratios are (1, 2, −3)., ∴Direction cosines are, , , 1, 2, 3, , , −, ,, ,, 2, , 1 + 22 + (−3)2 12 + 22 + (−3)2, 12 + 22 + (−3)2 , ∴, , 2, 3 , 1, or , ,, ,−, , 14 14, 14 , 4. The direction ratio of the lines are, a1 = 2, b1 = 5, c1 = 4, and, a 2 = 1, b2 = 2, c2 = − 3, a1a 2 + b1b2 + c1c2, ∴, cos θ =, a12 + b12 + c12 a 22 + b22 + c22, 2 ⋅ 1 + 5 ⋅ 2 + 4(−3), =, 22 + 52 + 42 12 + 22 + (−3)2, (2 + 10 − 12), ⇒, θ = cos −1, 4 + 25 + 16 1 + 4 + 9, = cos −1 (0), ⇒, θ = 90°, 5. The length of the perpendicular from origin to the plane, is, 0 + 0 + 0 − 52 , , p =, 9 + 16 + 144 , −52 , =4, =, 13
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623, , Three Dimensional Geometry, 6. If a sphere is intersected by a plane, then the locus of set, of points common to both sphere and plane is always a, circle., Sphere, , C, Circle, A, , B, Plane, , 7. The given equation can be rewritten as, 5x 3 y 6z, x, y, z, −, +, =1⇒, −, +, =1, 60 60 60, 12 20 10, Hence, the intercepts are (12, − 20, 10)., 8. Let the points be P = (4, 3, − 5) and Q = (−2, 1, − 8), Now, | PQ| = (−2 − 4)2 + (1 − 3)2 + (−8 + 5)2, , ⇒, 2 cos 2 α + 2 cos 2 β + 2 cos 2 γ = 2, ⇒ (1 + cos 2α ) + (1 + cos 2β ) + (1 + cos 2γ ) = 2, 1 + cos 2α + cos 2β + cos 2γ = 2 − 2 = 0, 13. The DC’s of x-axis is (1, 0, 0)., The equation of x-axis is, x−0 y−0 z−0, =, =, 1, 0, 0, x y z, ⇒, = =, 1 0 0, 14. Two planes ax + by + cz + d = 0 and ax + by + cz + d1 = 0, are parallel to each other., (Q d ≠ d1 ), ∴ There have no common point., |a1a 2 + b1b2 + c1c2|, 15. We know, cos θ =, 2, a1 + b12 + c12 a 22 + b22 + c22, cos θ =, , = 36 + 4 + 9 = 7, DC’s of line PQ is, x − x1, y − y1, z − z1, ,m = 2, ,n= 2, l= 2, | PQ |, | PQ |, | PQ |, 6, 2, 3, ∴, l= ,m = , n =, 7, 7, 7, 9. Since, it is given that line makes equal angle with the, coordinate axes, ∴, l=m = n, We know,, l2 + m 2 + n 2 = 1, 1, ⇒, 3 l2 = 1 ⇒ l2 =, 3, 1, (neglect negative sign), ⇒, l=, 3, 1, l=m = n =, ∴, 3, 10. The equation of planes are x + y + 2z = 0 and, −2x + y − z = 11., We know that, the angle between the planes, a1x + b1 y + c1z + d1 = 0 and a 2x + b2y + c2z + d2 = 0 is, given by, cos θ =, , a1a 2 + b1b2 + c1c2, a12, , + b12 + c12 a 22 + b22 + c22, , Here, a1 = 1, b1 = 1, c1 = 2, a 2 = − 2, b2 = 1, c2 = − 1, 1 × (−2) + 1 × 1 + 2 × (−1), cos θ =, ∴, 1+1+4 4+1+1, −2 + 1 − 2, 3 1, π, =, = = = cos, 6 2, 3, 6 6, π, θ=, ⇒, 3, 11. We know, if l, m, n are DC’s of a line, then, l2 + m 2 + n 2 = 1, ⇒, cos 2 α + cos 2 β + cos 2 γ = 1, where, l = cos α, m = cos β , n = cos γ, 12. Given that α, β and γ are angles, which makes a line, with the axes, then, cos 2 α + cos 2 β + cos 2 γ = 1, , =, cos θ =, , |(2)(3) + (3)(−4) + (−6)(5)|, 2 + 32 + (−6)2 32 + (−4)2 + (5)2, 2, , |6 − 12 − 30|, 4 + 9 + 36 9 + 16 + 25, 36, 18 2, =, 35, 7⋅ 5 2, , 18 2 , θ = cos −1 , , 35 , , ⇒, , 16. The d and d′ are of same sign., 17. We know, if the line passing through (x1 , y1 , z1 ) and, (x2, y2, z2), then equation of line is, x − x1, y − y1, z − z1, =, =, x2 − x1 y2 − y1 z2 − z1, Since, the line passing through (4, − 5, − 2) and (−1, 5, 3)., , ∴ The equation of straight line is, , x−4 y+ 5 z + 2, =, =, −2, 1, −1, Which is the required equation of straight line., 18. We know that, in a parallelogram, diagonals bisect each, other. Mid-point of OQ = Mid-point of PR, 0 + m 0 + n 0 + r 1 + 3 1 + 4 1 + 5, ,, ,, ,, ,, ∴ , , ≡, 2, 2, 2 2, 2, 2 , ⇒, m = 4, n = 5, r = 6, Hence, m + n + r = 4 + 5 + 6 = 15, 19. We know,, cos 2 α + cos 2 β + cos 2 γ + cos 2 δ =, , 4, 3, , … (i), , where, α , β , γ and δ are the angles with diagonals of, cube, then from Eq. (i), we get, 4, 3, 8, ⇒, sin 2 α + sin 2 β + sin 2 γ + sin 2 δ =, 3, 20. Let xy-plane divides the line joining the points (−1, 3, 4), and (2, − 5, 6) in λ : 1., 6λ + 4, The z-coordinate is, λ+1, 1 − sin 2 α + 1 − sin 2 β + 1 − sin 2 γ + 1 − sin 2 δ =
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624, , NDA/NA Mathematics, , For xy-plane, z-coordinate = 0, 6λ + 4, 2, ⇒, 0=, ⇒λ = −, 3, λ+1, Required ratio = − 2 : 3, Its externally in the ratio 2 : 3., (Q ‘–’ sign show that the point divide the line.), 21. Let A ≡ (3, 4, 5), B ≡ (4, 6, 3), C ≡ (−1, 2, 4), D ≡ (1, 0, 5), For AB, x2 − x1 = 4 − 3 = 1, y2 − y1 = 6 − 4 = 2 ,, z2 − z1 = 3 − 5 = − 2, 2 −2 1, Let DC’s l, m, n for CD are ,, ,, 3 3 3, ∴ Projection of AB on CD = Σl (x2 − x1 ), 4, 1, 2, 2, = 1 + − 2 + (−2) = −, 3, 3, 3, 3, 22. Equation of plane through z-axis is, …(i), ax + by = 0, This plane is parallel to the line, x−1 y+ 2 z −3, =, =, cos θ sin θ, 0, ⇒, a cos θ + b sin θ = 0, ⇒, a = − b tan θ, ∴, − b tan θx + by = 0, ⇒, x tan θ − y = 0, Which is the required equation of plane., 23. Given equation is by + cz + d = 0,, which is in yz-plane, i. e. , x = 0, or, x + 0⋅ y + 0⋅ z = 0, Clearly, given plane is perpendicular to yz-plane., 24. Direction ratios of AB = (−4, − 6, − 2), AC = (−1, 4, 3),, AD = (−8, − 1, 3), Points A, B, C, D are coplanar, if, −4 −6 −2, [ AB BC CD ] = −1, , 4, , −8 −1, , 3, 3, , = − 4 (12 + 3) + 6 (−3 + 24) − 2 (1 + 32), = − 4 (15) + 6 (21) − 2 (33), = − 60 + 126 − 66 = 0, 25. Equation of plane passing through (1, − 2, 4) and the, direction cosines of whose normal (2, 1, 2), is, 2(x − 1) + 1( y + 2) + 2(z − 4) = 0, ⇒, 2x + y + 2z − 8 = 0, ∴ Required distance, 2(3) + 1(2) + 2(3) − 8 , ax + by + c , =, Q Distance =, , 4+1+4, a 2 + b2 , , 6, = =2, 3, 26. Equation of plane,, 2x − y + 2z + 1 = 0, 1, 1, ⇒, x− y+ z + =0, 2, 2, Here, direction ratios of normal to the plane, 1, = < 1, − ,1 >, 2, 27. Given, points are (−3, 4, − 8) and (5, − 6, 4)., ∴ The equation of the line joining giving points is, , x+3, y −4, z+8, =, =, = λ (say), −6 − 4 4 + 8, 5+3, x+ 3 y−4 z + 8, =, =, =λ, ⇒, −10, 8, 12, Since, it cuts xy-plane, i.e., z = 0, z+8, Now,, =λ, 12, 0+8, 2, ⇒, =λ ⇒ λ=, 12, 3, ∴, x + 3 = 8λ and y − 4 = − 10λ, 7, 8, and y = −, x=, ⇒, 3, 3, 7 8 , ∴ Point of intersection = , − , 0, 3 3 , 28. Given that, the direction ratio of the lines are (2, –1, 2), π, and (x, 3, 5), and angle between them is ., 4, We know that,, a1a 2 + b1b2 + c1c2, cos θ =, 2, a1 + b12 + c12 a 22 + b22 + c22, π, 2 ⋅ x − 1 ⋅3 + 2 ⋅5, cos =, ⇒, 4, 4 + 1 + 4 x2 + 9 + 25, ⇒, , 1, 2x + 7, =, 2 3 x2 + 34, , On squaring both sides, we get, 1 4x2 + 49 + 28x, =, 2, 9 (x2 + 34), 1 4x2 + 49 + 28x, =, ⇒, 2, 9 (x2 + 34), 2, ⇒, 9x + 306 = 8x2 + 98 + 56x, ⇒, x2 − 56x + 208 = 0, ⇒, (x − 52) (x − 4) = 0, ⇒, x = 4, 52, ∴ The smallest value of x = 4., 29. Equation of plane passing through (1, − 3, 1) and the, direction cosines of whose normal (1, − 3, 1) is, 1(x − 1) − 3( y + 3) + 1(z − 1) = 0, ⇒, x − 3 y + z − 11 = 0, x, y, z, −, +, = 0 (intercept form), ⇒, 11 11 /3 11, The above plane intercept the x-axis at a distance of 11., 30. Let the equation of the required sphere be, …(i), x2 + y2 + z 2 + 2 ux + 2vy + 2wz + d = 0, The centre is given to be (1, 1, 1), ∴ From Eq. (i), we get, …(ii), x2 + y2 + z 2 + 2x + 2 y + 2z + d = 0, Since, Eq. (ii) passes through (3, 3, 2), ∴, 9 + 9 + 4 + 6 + 6 + 4 + d =0, ⇒, d = − 38, ∴ From Eq. (i), the equation of sphere is, x2 + y2 + z 2 + 2x + 2 y + 2z − 38 = 0, ⇒, x2 + y2 + z 2 + 2x + 2 y + 2z = 38, 31. As we know, direction cosines of y-axis are (0, 1, 0). If the, line is parallel to y-axis, then the direction cosines of the, line are (0, 1, 0).
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625, , Three Dimensional Geometry, 32. The equation of sphere is, x2 + y2 + z 2 − 4x + 6 y − 8z − 7 = 0, Compare with,, ax2 + by2 + cz 2 + 2ux + 2vy + 2wz + d = 0, Here, u = − 2, v = 3, w = − 4 and d = − 7, ∴Radius of sphere = u 2 + v2 + w2 − d, = (− 2)2 + (3)2 + (−4)2 + 7, = 4 + 9 + 16 + 7 = 36 = 6, ∴Diameter = 2 × Radius, = 2 × 6 = 12 units, 33. The given equation represents a real sphere, if, (by definition), u 2 + v2 + w2 > d, 34. Let α = 60° , β = 45°, We know that,, cos 2 α + cos 2 β + cos 2 γ = 1, ∴, cos 2 60° + cos 2 45° + cos 2 γ = 1, 2, 2, 1, 1, 2, ⇒, + + cos γ = 1, 2, 2, 1 1, ⇒, + + cos 2 γ = 1, 4 2, 1, cos 2 γ =, ⇒, 4, 1, ⇒, cos γ = ±, 2, ⇒, γ = 60° , 120°, ∴, γ = 60°, (Q positive direction only), 35. Let the coordinates of D are (x, y, z)., A(4, 7, –8), , 2, G (1,1,1), 1, B, , D ( x , y, z), , For x-coordinate,, , C, , 2 × x+ 1 ×4, 1+2, 1, x=−, 2, , 1=, , ⇒, ⇒, For y-coordinate,, , 2× y+1×7, ⇒ y= −2, 1+2, and for z-coordinate,, 2 × z + 1 × (−8), 11, 1=, ⇒ z=, 1+2, 2, 11, 1, ∴ Coordinates of D are − , − 2, ., 2, 2, 1=, , 36. Given that, the points of rhombus are A (5, − 1, 1),, B (−1, − 3, 4), and C (1, − 6, 10)., Let coordinates of fourth vertex D be (x, y, z )., Coordinates of the mid-point of AC = Coordinates of the, mid-point of BD, 5 + 1 −1 − 6 1 + 10 −1 + x −3 + y 4 + z , ⇒, ,, ,, ,, ,, , , =, 2, 2, 2 2, 2, 2 , , x − 1 y − 3 z + 4, −7 11, , = , ,, ,, , 3,, 2, 2 2, 2, 2 , x−1, y−3, 7, ⇒, = 3,, =− ,, 2, 2, 2, z + 4 11, =, 2, 2, ⇒, x = 7, y = − 4, z = 7, Let coordinates of fourth vertex are, (x, y, z ) = (7, − 4, 7)., 37. The equation of the plane passing through x-axis is, x = a., This also passes through (1, 2, 3), ∴, x=1, (Q a = 1), Which is the required equation of plane., 38. The planes bx − ay = n , cy − bz = l and az − cx = m, intersect in a line. This planes satisfies the option (b), ∴, al + bm + cn, = a (cy − bz ) + b (az − cx) + c (bx − ay), = acy − abz + abz − bcx + bcx − acy = 0, ⇒, , ⇒, , al + bm + cn = 0, , 39. The condition for the plane ax + by + cz + d = 0 to be, perpendicular to xy-plane, if the coefficient of z is equal, to zero., i.e.,, c=0, 40. The equation of plane, which is parallel to x-axis is, by + cz = d , where x = 0., x 2 − y z −1, 41. Given, lines are =, =, k, 1, 2, x+ 1 y−1 z + 2, and, =, =, 3, k, 1, The direction ratios of these lines are (k, 1, 2) and, (3, k, 1). These lines will be perpendicular, if, l1l2 + m1m2 + n1n2 = 0, ⇒, 3k − k + 2 = 0, ⇒, 2k + 2 = 0, ⇒, k = −1, 42. Radius of sphere, 2, , 2, , 2, , 10, 8, 6, = − + − + − λ, 2, 2, 2, (given), 1 = 9 + 16 + 25 − λ, ⇒, λ = 49, 43. The given equation of planes are, …(i), x+ y+ z=7, and, …(ii), αx + βy + γz = 3, These planes will be parallel, if the coefficients of x, y, and z are the same, i.e., α = β = γ, 44. General equation of a plane is ax + by + cz + d = 0, This equation consists 4 arbitrary constants., x−3 y−4 z −5, 45. Straight line, is parallel to the plane, =, =, 2, 3, 4, but perpendicular to its normal., ∴ From option (c),, Plane, 4x + 4 y − 5z = 0, Now, by condition of perpendicularity,, a1a 2 + b1b2 + c1c2 = 0, So,, (2)(4) + (3)(4) + (4)(−5) = 0, 0 =0
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626, , NDA/NA Mathematics, , 46. We know that, the angle between the diagonals of cube, is, 1, θ = cos −1 ., 3, If θ is an acute angle, then θ > 60° ., 47. The equation of the plane passing through (x1 , y1 , z1 ) and, normal to the line with < a , b, c > as direction ratios, is, a (x − x1 ) + b ( y − y1 ) + c (z − z1 ) = 0, ⇒, ax − ax1 + by − by1 + cz − cz1 = 0, ⇒, ax + by + cz = ax1 + by1 + cz1, Which is required equation of plane., 48. Since, the sum of squares of the distances of the point, P (x, y, z ) from the points A (a , 0, 0) and B(− a , 0, 0) is 2c2., ∴, PA 2 + PB2 = 2c2, ⇒, (x − a )2 + y2 + z 2 + (x + a )2 + y2 + z 2 = 2c2, ⇒ x2 + a 2 − 2xa + 2 y2 + 2z 2 + x2 + a 2 + 2xa = 2c2, ⇒, ⇒, , x2 + a 2 + y2 + z 2 = c2, x + a = c − y2 − z 2, 2, , 2, , 2, , 49. Direction ratios of two lines are, ( 3 − 1, − 3 − 1, 4) and (− 3 − 1, 3 − 1, 4), Now, angle between two lines is, ( 3 − 1) (− 3 − 1) + (− 3 − 1) ( 3 − 1) + 4 × 4, cos θ =, ( 3 − 1)2 + (− 3 − 1)2 + 42 (− 3 − 1)2 + ( 3 − 1)2 + 42, =, , − (3 − 1) − (3 − 1) + 16, 4 − 2 3 + 4 + 2 3 + 16 4 + 2 3 + 4 − 2 3 + 16, , 12, 24 24, 12 1, ⇒, cos θ =, =, 24 2, π, ∴, θ=, 3, 50. We know,, cos 2 α + cos 2 β + cos 2 γ = 1, ⇒, 1 − sin 2 α + 1 − sin 2 β + 1 − sin 2 γ = 1, =, , ⇒, , sin 2 α + sin 2 β + sin 2 γ = 2, , 51. We know that, planes never intersect at points, they, intersect in a line., 52. For the sphere,, Coefficient of x = Coefficient of y = Coefficient of z, ⇒, a=b=c, So,, ax2 + by2 + cz 2 − 6x = 0, 6x, ⇒, x2 + y 2 + z 2 −, =0, a, , 3, Centre = , 0, 0, ∴, , a, radius = 1, , Given that,, 2, , 3, + 0 + 0 =1, a, 3, =1, a, a =3, ∴ Centre = (1, 0, 0), , Level II, 1. The equation of plane, in which the line, x−5 y− 7 z + 3, =, =, 4, 4, −5, lies, is A (x − 5) + B( y − 7) + C (z + 3) = 0, , …(i), , where, A, B and C are the direction ratios of the, plane. Since, the first line lies on the plane., ∴ Direction ratios of normal to the plane is, perpendicular to the direction ratios of line, i. e. ,, 4 A + 4B − 5 C = 0, , …(ii), , x−8 y−4 z−5, Also, since line, lies in this, =, =, 7, 1, 3, plane. The direction ratios is also perpendicular to, this line, ∴, , 7 A + B + 3C = 0, , …(iii), , From Eqs. (ii) and (iii), we get, A, B, C, =, =, 17 −47 −24, , ∴ The required equation of plane is, , 17 (x − 5) − 47( y − 7) + (−24)(z + 3) = 0, ⇒, 17x − 47 y − 24z + 172 = 0, 2. The equation of line passing through (x1 , y1 , z1 ) and, (x2, y2, z2) is, x − x1, y − y1, z − z1, …(i), =, =, x2 − x1 y2 − y1 z2 − z1, , Here, (x2 − x1 ), ( y2 − y1 ) and (z2 − z1 ) are direction ratios, of that line., Then, its direction cosines are, (x2 − x1 ), ( y2 − y1 ), (z2 − z1 ), and n =, l=, ,m =, 2, 2, Σ (x2 − x1 ), Σ (x2 − x1 ), Σ (x2 − x1 )2, Given, (x1 , y1 , z1 ) = (1, 2, 3), and, (x2, y2, z2) = (− 2, 3, 1), (− 2 − 1), l=, ∴, (− 3)2 + (1)2 + (4)2, (3 − 2), m=, 2, (− 3) + (1)2 + (4)2, (1 + 3), and, n=, (− 3)2 + (1)2 + (4)2, −3, 1, 4, ⇒, l=, ,m =, ,n =, 26, 26, 26, 2, 2, 2, −3 , 1 , 4 , 2, 2, 2, l +m +n =, ∴, +, +, , 26 , 26 , 26 , 9, 1, 16 26, =, +, +, =, =1, 26 26 26 26, x y z, 3. The given equation of plane is + + = 1., 2 3 4, x y z, On comparing with + + = 1, a b c
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627, , Three Dimensional Geometry, α − α ′ β − β′, l, m, , ⇒, , a = 2, b = 3, c = 4, 1, Area of ∆ ABC =, a 2b2 + b2c2 + c2a 2, 2, 1, ∆=, 4 × 9 + 9 × 16 + 16 × 4, 2, 1, =, 36 + 144 + 64, 2, 1, =, 244 = 61 sq units, 2, 4. Let the equation of the plane which passes through the, point (1, − 1, − 1) be, a (x − 1) + b ( y + 1) + c (z + 1) = 0, … (i), , where, a, b, c are the direction ratios of the normal of, the plane (i)., Now, plane (i) is perpendicular to the planes, x − 2 y − 8z = 0 and 2x + 5 y − z = 0, then, , ...(ii), a − 2b − 8c = 0, and, ...(iii), 2a + 5b − c = 0, Now, solving Eqs. (ii) and (iiii) by cross multiplication, method., a, b, c, =, =, (2 + 40) (− 16 + 1) (5 + 4), a, b, c, ⇒, =, =, 42 − 15 9, Now, from Eq. (i), we get, 42 (x − 1) − 15 ( y + 1) + 9 (z + 1) = 0, ⇒, 42x − 15 y + 9z = 42 + 15 − 9, ⇒, 42x − 15 y + 9z = 48, ⇒, 14x − 5 y + 3z = 16, 5. Any plane through given line is, A (x − 1) + B( y + 2) + C (z − 3) = 0, … (i), , Where, A, B and C are the DR’s of the normal to the, plane. Since, the straight line lies on the plane., ∴ DR’s of plane is perpendicular to the line, i.e.,, , 5 A + 6B + 4C = 0, … (ii), Since, it passes through (4, 3, 7), we get, 3 A + 5B + 4C = 0, …(iii), On solving Eqs. (ii) and (iii), we get, A B C, = =, −4 8 −7, ∴ Equation of required plane is, 4x − 8 y + 7z = 41., 6. Let the equation of sphere passing through origin is, x2 + y2 + z 2 + 2 ux + 2vy + 2wz = 0, It passes through (0, 2, 0), ∴, 4 + 4v = 0 ⇒ v = − 1, Also, it passes through (1, 0, 0), 1, ∴, 1 + 2u =0 ⇒ u = −, 2, and it passes through (0, 0, 4), ∴, 16 + 8w ⇒, w = −2, 1, , ∴ Centre of sphere is (− u , − v, − w) = , 1, 2 ., 2, , 7. The shortest distance between the lines, x−3 y−8 z −3, x+ 3 y+ 7 z −6, and, is, =, =, =, =, −1, −3, 3, 1, 2, 4, , SD =, , =, , l′, , γ − γ′, n, , m′, , n′, , (Σ mn′ + nm′ )2, 6 15 − 3, 3 −1 1, −3, , 2, , 4, , (− 4 − 2) + (12 + 3)2 + (6 − 3)2, 270, =, = 270 = 3 30, 270, 2, , 8. The general equation of sphere is, x2 + y2 + z 2 + 2ux + 2vy + 2wz + d = 0, , …(i), , Which passes through the origin and the points, ( −1, 0, 0), ( 0, − 2, 0) and ( 0, 0, − 3), then, , 0 + 0 + 0 + 0 + 0 + 0 + d =0, d =0, 1 + 0 + 0 − 2u + 0 + 0 + d = 0, 2u = d + 1, 1, ⇒, 2u = 0 + 1 ⇒ u =, 2, and, 0 + 4 + 0 + 0 − 4v + 0 + d = 0, ⇒, 4v = 4 + d, ⇒, 4v = 4 + 0 ⇒ v = 1, and, 0 + 0 + 9 + 0 + 0 − 6w + d = 0, ⇒, 6w = 9 + d, 3, ⇒, 6w = 9 + 0 ⇒ w =, 2, On putting these values in Eq. (i), we get, 1, 3, x2 + y2 + z 2 + 2 x + 2 (1) y + 2 z + 0 = 0, 2, 2, 2, 2, 2, ⇒, x + y + z + x + 2 y + 3z = 0, Comparing it with,, x2 + y2 + z 2 + f (x, y, z ) = 0, ⇒, f (x, y, z ) = x + 2 y + 3z, 9. Angle between the plane and line is, aa ′ + bb ′ + cc ′, sin θ =, a 2 + b2 + c2 a ′ 2 + b ′ 2 + c ′ 2, 3, 2, 3, 2 × + 3 × −4 ×, 4, 4, 4, ∴, sin θ =, ⇒, and, ⇒, , 2, , 2, , 2, , 3, 2, 3, + + − , 4, 4, 4, 6 6 12, + −, 4 4 4, =, =0, 9, 4, 9, 4 + 9 + 16, +, +, 16 16 16, ⇒, sin θ = sin 0, ⇒, θ = 0°, 10. Given that, equation of planes are, 4x + 4 y − 5z = 12, … (i), and, 8x + 12 y − 13z = 32, … (ii), Let (l, m, n ) be the direction ratios of the line., ∴ Eqs. (i) and (ii) becomes, 4l + 4m − 5n = 0, … (iii), and, 8l + 12m − 13n = 0, …(iv), 22 + 32 + 42
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628, , NDA/NA Mathematics, , l m, n, l m n, =, =, ⇒ =, =, 8 12 16 2 3 4, Now, we take intersection point with z = 0 is given by, 4x + 4 y = 12, …(v), and, 8x + 12 y = 32, …(vi), On solving Eqs. (v) and (vi), we get (1, 2, 0), x−1 y−2 z, =, = ., ∴ Required line is, 2, 3, 4, 11. Q, cos 2 α + cos 2 β + cos 2 γ = 1, ⇒, 2 cos 2 α + 2 cos 2 β + 2 cos 2 γ = 2, 2, ⇒ 2 cos α − 1 + 2 cos 2 β − 1 + 2 cos 2 γ − 1 = 2 − 3, ⇒, cos 2α + cos 2β + cos 2γ = − 1, and now,, 1 − sin 2 α + 1 − sin 2 β + 1 − sin 2 γ = 1, ⇒, sin 2 α + sin 2 β + sin 2 γ = 2, Hence, only statement I is correct., 12. Let the vertices of a triangle are, A (a , 0, 0), B(0, b, 0) and C (0, 0, c), and the equation of plane is, x y z, + + =1, …(A), a b c, Q Centroid of ∆ ABC is (α , β , γ ), a+0+0, ∴, = α ⇒ a = 3α, …(i), 3, 0+ b+0, Similarly,, = β ⇒ b = 3β, …(ii), 3, 0+0+ c, and, =γ ⇒, c = 3γ, …(iii), 3, ⇒, , Putting the values of α , β , γ from Eqs. (i), (ii) and (iii), x, y, z, in Eq. (A), we get, +, +, =1, 3 α 3β 3γ, ⇒, , x y z, + + =3, α β γ, , Which is the equation of required plane., 1 1 1 1, 13. Given that + + =, a b c 2, and equation of plane is, x y z, + + =1, a b c, , 15. Given that, equation of planes are, P ≡ ax + by + cz + d = 0, … (i), and, P ′ ≡ a ′ x + b ′ y + c′ z + d ′ = 0, … (ii), Condition for any plane parallel to x-axis is, P + λP ′ = 0, … (iii), ⇒, ax + by + cz + d + λ (a ′ x + b ′ y + c ′ z + d ) = 0, For parallel to x-axis, coefficient of x = 0, a, ⇒, a + λa ′ = 0 ⇒ λ = −, a′, ∴ From Eq. (iii), we get, a, P P′, P−, P′ =0 ⇒ =, a′, a a′, 16. The given equation of planes are, bx − ay = n, cy − bz = l, az − cx = m, Let z = 0 and solving Eqs. (i) and (ii), we get, , x=, , n al, and, +, b cb, , y=, , …(i), …(ii), …(iii), , l, c, , If all three planes intersect in a line, then the point, n al l , , , 0 satisfying the Eq. (iii),, +, b cb c , n al , a ×0 − c +, ⇒, =m, b cb , ⇒, − nc − al = bm, ⇒, al + bm + cn = 0, , The planes bx − ay = n , cy − bz = l and az − cx = m, intersect in a line, if al + bm + cn = 0., …(i), …(ii), , From Eqs. (i) and (ii), we get, x = 2, y = 2, z = 2, ∴, (x, y, z ) ≡ (2, 2, 2), 14. Given that, l + m + n = 0, and, lm =0, From Eq. (i),, l = − (m + n ), and from Eq. (ii), − (m + n ) ⋅ m = 0, ⇒, − (m2 + mn ) = 0, ⇒, m2 + mn = 0, ⇒, m + n = 0, m = 0, If m = 0,, l+ m + n =0, l1 m1 n1, Then,, =, =, −1, 0, 1, and if m + n = 0, l + m + n = 0, l2 m2 n2, Then,, =, =, 0 −1, 1, ∴, (l1 , m1, n1 ) = (−1, 0, 1), , (l2, m2, n2) = (0, − 1, 1), , and, , ∴ Angle between them, 0+0+1, 1 π, = =, cos θ =, 1+0+1 0+1+1 2 3, , …(i), …(ii), , [from Eq. (i)], , 17. Given equation of two planes,, x+ y+ z + 1 =0, and, 2x − 2 y + 2z + 1 = 0, Here,, a1 = 1, b1 = 1 and c1 = 1, a 2 = 2, b2 = − 2 and ⇒ c2 = 2, Let θ be the angle between them., |a1a 2 + b1b2 + c1c2|, Then, cos θ =, a12 + b12 + c12 a 22 + b22 + c22, ⇒, , cos θ =, , ⇒, , cos θ =, , |(1) (2) + (1) (−2) + (1)(2)|, (1) + (1)2 + (1)2 + (2)2 + (–2)2 + (2)2, 2, , |2 − 2 + 2|, ||, 1, 2, =, ⇒ cos θ =, 3, 3 12, 3 (2 3 ), , 18. Equation of any plane passing through (2, 3, 4) is, A (x − 2) + B( y − 3) + C (z − 4) = 0, … (i), Plane (i) is parallel to 5x − 6 y + 7z = 3, ∴ DR's of this plane is same as the Eq. (i), i. e. ,, A = 5, B = − 6, C = 7, ∴, 5(x − 2) − 6( y − 3) + 7(z − 4) = 0, ⇒, 5x − 6 y + 7z − 20 = 0 is the required plane., 19. Since, the centre of sphere x2 + y2 + z 2 − 2 y − 4z − 11 = 0, is (0, 1, 2) and radius is 4., , Distance of a plane x + 2 y + 2z − 15 = 0 from point, (0,1,2)
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629, , Three Dimensional Geometry, , =, , |0 + 2 + 4 − 15| 9, = =3, 3, 1+4+4, , | NQ | =, , |0 + 0 + 1 − 3|, 1 +1 +1, 2, , 2, , 2, , =, , 2, 3, , | PQ | = (0 − 0)2 + (0 − 1)2 + (1 − 0)2 = 2, | RP | = | MP | − | MR| = | MP | − | NQ | = 0, , O, , ∴, N, , P, , Now, NP = OP 2 − ON 2 = 42 − 32 = 16 − 9 = 7, , ∴, , Radius of circle = 7, , 20. Let A = (0, 2, 2), B = (2, 0, − 1) and C = (3, 4, 0), Now, AB = (2, − 2, − 3) and AC = (3, 2, − 2), 1, ∴ Area of triangle = ||AB × AC||, 2, i, j, k, 1, =, 2 −2 −3, 2, 3, 2, −2, , 1, = | [i ( 4 + 6) + j ( −4 + 9) + k ( 4 + 6)]|, 2, 1, = |10i + 5 j + 10k|, 2, 1, 1, =, (10)2 + (5)2 + (10)2 =, 225, 2, 2, 15, sq units, =, 2, 21. Given, points are A (5, − 1, 1), B(7, − 4, 7) , C (1, − 6, 10), and D(−1, − 3, 4) ., Now,, AB = (7 − 5)2 + (−4 + 1)2 + (7 − 1)2, = 4 + 9 + 36 = 7, BC = (1 − 7)2 + (−6 + 4)2 + (10 − 7)2, = 36 + 4 + 9 = 7, CD = (−1 − 1)2 + (−3 + 6)2 + (4 − 10)2, = 4 + 9 + 36 = 7, DA = (−1 − 5)2 + (−3 + 1)2 + (4 − 1)2, = 36 + 4 + 9 = 7, ∴, AB = BC = CD = DA = 7, Also, ( it is not perpendicular), AB ⋅ AC ≠ 0, ⇒ ABCD is not square., ∴ It is a rhombus., 22. Given, plane is x + y + z − 3 = 0. From point P and Q, draw PM and QN perpendicular on the given plane and, QR ⊥ MP., P, , Q, , | MP | =, , 1 +1, 2, , 2, , +1, , 2, , Q Radius = Perpendicular distance to the plane from, the centre, ∴, , Radius =, , 2(6) − 1(−1) + 2(2) − 2, 15, =, =5, 3, 4+1+4, , ∴ Equation of sphere is, (x − 6)2 + ( y + 1)2 + (z − 2)2 = 52, 2, ⇒, x + y2 + z 2 − 12x + 2 y − 4z + 16 = 0, 25. Since, the given sphere touching the three coordinates, planes. So, it is clear that, if radius is a, then centre is, (a , a , a )., , ∴ The equation of sphere at the centre ( a , a , a ) and, radius a is, (x − a )2 + ( y − a )2 + (z − a )2 = a 2, x2 + y2 + z 2 − 2ax − 2ay − 2az + 3a 2 = a 2, ∴, x2 + y2 + z 2 − 2a (x + y + z ) + 2a 2 = 0, is the required equation of sphere., 26. We know that, the equation of a plane passing through, the intersection of the planes, a1x + b1 y + c1z + d1 = 0 and a 2x + b2y + c2z + d2 = 0 is, (a1x + b1 y + c1z + d1 ) + λ (a 2x + b2y + c2z + d2) = 0, where, λ is constant., Thus, the equation of plane, 2x − (1 + λ ) y + 3λz = 0 can be rewritten as, (2x − y) + λ (− y + 3z ) = 0, , So, it is clear that the equation of plane passes, through the intersection of planes, 2x − y = 0, , x 2 + y 2 + z 2 + 2x − 2 y − 4z − 19 = 0, and equation of plane is, , M, , |0 + 1 + 0 − 3|, , 23. The required point is the centre of the sphere through, the given points., Let the equation of sphere passing through (0,0, 0) is, x2 + y2 + 2 ux + 2vy + 2wz = 0, …(i), Sphere (i) is passing through (a , 0, 0), (0, b, 0), and (0, 0, c), a 2 + 2 ua = 0 ⇒ u = − a /2, b2 + 2 v b = 0 ⇒ v = − b /2, c2 + 2 w c = 0 ⇒ w = − c/2, a b c, Therefore, centre of sphere is , , ., 2 2 2, Which is the required point., 24. Given, the centre of sphere to be (6, − 1, 2)., , and, y − 3z = 0, 27. Given, equation of sphere is, , R, , N, , | NM | = |QR| = PQ 2 − RP 2 = ( 2 )2 − 0 = 2, , =, , 2, 3, , x + 2 y + 2z + 7 = 0, , The centre of sphere is C (−1, 1, 2), and radius = 12 + 12 + 22 + 19 = 25 = 5
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630, , NDA/NA Mathematics, The perpendicular distance from centre of sphere to, the plane x + 2 y + 2z + 7 = 0, BC =, , −1 + 2 + 4 + 7 12, =, =4, 3, 1+4+4, , Let r be the radius of the circle., (–1, 1, 2), Now, in ∆ABC,, C, r 2 = BA 2 = AC 2 − BC 2, r 2 = 52 − 42 = 25 − 16, ⇒, r2 = 9, A, B, ⇒, r =3, 28. We know, if two lines are coplanar, then, x2 − x1 y2 − y1 z2 − z1, l1, l2, ⇒, , m1, n1, =0, m2, n2, 1 − 2 4 −3 5 −4, 1, k, , 3λ − 3 5λ − 7, =, 1, 3, , ⇒, , 9λ − 9 = 5λ − 7 ⇒ λ =, , 7, 5, , 3, ∴ The point of intersection is − 1, − 3, − 5 or, , 2, 2, 2, 1, 1, 3, , , ,− ,− ., 2 2, 2, 33. Given that, the coordinates of points A, B, and C are, (5, 9, 9), (0, − 1, − 6) and (5, − 11, − 1), respectively., Now,, AB = (0 − 5)2 + (−1 − 9)2 + (−6 − 9)2, = 350, , −1 1, , 1, , BC = (5 − 0)2 + (−11 + 1)2 + (−1 + 6)2, , 1 −k = 0, 2 1, , = 25 + 100 + 25 = 150, and, , 0, ⇒, , 0, , 1, , Since, the line formed intersected by planes and the, normal of the plane are perpendicular, then, by taking option (a), −1 (1 + λ ) + 3(λ − 1) + 2(2 − λ ) = 0, ⇒, − 1 − λ + 3λ − 3 + 4 − 2λ = 0, 0 =0, 30. I. When two spheres touch externally, the distance, between their centres is equal to the sum of their, radii. i.e., r1 + r2., , II. Any plane parallel to x is of the form by + cz = d., III. Line lies in a plane, if dot product of DR’s of a line, and DR’s of normal to the plane is zero., , i.e.,, a1a 2 + b1b2 + c1c2 = 0, Hence, statements I and II are correct., 31. Only statement I is correct., 32. Given, lines are, x+1 y+3 z+5, =, =, = λ (say ), 3, 5, 7, x−2 y−4 z −6, and, =, =, 1, 3, 5, , …(i), …(ii), , Any point on line (i) is ( 3λ − 1, 5λ − 3, 7λ − 5). Let, this point be the point of intersection of Eqs. (i) and, (ii), then it will also satisfy Eq. (ii), 3λ − 1 − 2 5λ − 3 − 4 7λ − 5 − 6, =, =, 1, 3, 5, , AC = (5 − 5)2 + (−11 − 9)2 + (−1 − 9)2, = 400 + 100 = 500, , 2, 1 + k −k = 0, k+2, 1, 1, , c1 → c1 + c2, c2 → c2 − c3, ⇒, k2 + 3k = 0, ⇒, k(k + 3) = 0, ⇒, k = 0 or −3, 29. The intersection of the given plane is, x − y + 2z − 1 + λ (x + y − z − 3) = 0, ⇒ x(1 + λ ) + y(λ − 1) + z (2 − λ ) − 3λ − 1 = 0, Direction ratios of normal to the above plane is, (1 + λ , λ − 1, 2 − λ ), , ∴, , 1, 2, , = 25 + 100 + 225 = 350, , −k = 0 ⇒ 1, 1, k, , 1, 2, , ⇒, , ∴ From above, we see that, AB2 + BC 2 = AC 2, and, AB ≠ BC ≠ AC, ∴ ∆ ABC is right angled triangle with unequal sides., 34. The equation of line passing through the points (1, 2, –1), and (3, –1, 2) is, x−1, y−2, z+1, =, =, 3 − 1 −1 − 2 2 + 1, x−1 y−2 z + 1, or, …(i), =, =, −3, 2, 3, Since, the line meets the yz-plane, therefore x = 0, 0 −1 y−2 z + 1, =, =, ⇒, 2, −3, 3, y−2, z+1, 1, 1, and, =−, =−, ⇒, −3, 3, 2, 2, 7, 5, and z = −, ⇒, y=, 2, 2, 7 5, ∴ Required point is 0, , − ., 2 2, 35. Let the equation of sphere, which passes through the, given four points be, x2 + y2 + z 2 + 2ux + 2vy + 2wz + d = 0, ∴, 0 + 0 + 0 + 0 + 0 + 0 + d =0, …(i), ⇒, d =0, So, (2)2 + 0 + 0 + 4u + 0 + 0 + d = 0, 4u + d = − 4, …(ii), ⇒, 4u + 0 = − 4 ⇒ u = − 1, Therefore,, 0 + (4)2 + 0 + 0 + 8v + 0 + d = 0, ⇒, 16 + 8v + d = 0, …(iii), ⇒, 8v + 0 = − 16 ⇒ v = − 2, Now,, 0 + 0 + (6)2 + 0 + 0 + 12w + d = 0, ⇒, 36 + 12w + d = 0, …(iv), ⇒, 12w + 0 = − 36 ⇒ w = − 3, , ∴ Centre of sphere = (1, 2, 3), , So, the required point is the centre of sphere, which is, equidistant from given four points.
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631, , Three Dimensional Geometry, 36. Let P (x1 , y1 , z1 ) be the point., Then, distance of P from x-axis =, , y12, , +, , 1 0 1, 1 0 1, 0 0 1, 1 0 0, ⇒ x 2 0 1 − y 0 0 1 + z 0 2 1 −1 0 2 0 = 0, 0 0 1, 0 3 1, 0 3 1, 0 0 3, , z12, , In yz-plane, x = 0, , Given that distance of P ( x1 , y1 , z1 ) from x = 0 is, , x1, , ., 1, Distance of P from x-axis = 3 × distance of P from, yz-plane., y12 + z12 = 3x1, On squaring both sides, we get, y12 + z12 = 9x12, Thus, path of P (x1 , y1 , z1 ) is, y2 + z 2 = 9x2., 37. The equation of required sphere is given by, (x2 + y2 + z 2 − 4) + k (2x + 3 y + 4z − 7) = 0, Since, it passes through (1, 2, 0), ∴, (1 + 4 − 4) + k (2 + 6 − 7) = 0, ⇒, 1 + k =0, ⇒, k = −1, ∴ Equation of sphere from Eq. (i) is, x2 + y2 + z 2 − 4 − 2x − 3 y − 4z + 7 = 0, ⇒, x2 + y2 + z 2 − 2x − 3 y − 4z + 3 = 0, , …(i), , 38. Let the equation of sphere be, …(i), x2 + y2 + z 2 + 2ux + 2vy + 2wz + d = 0, Since, it passes through (0, 0, 0), ∴, d =0, It meets x-axis at (3, 0, 0), 3, ∴, 9 + 6u =0, ⇒ u=−, 2, It meets y-axis at (0, 4, 0), ∴, 16 + 8v = 0, ⇒ v= −2, It meets z-axis at (0, 0, 5), ∴, 25 + 10w = 0, 5, w=−, ⇒, 2, ∴ Equation of sphere from Eq. (i) is, x2 + y2 + z 2 − 3x − 4 y − 5z = 0, 39. The equation of line, which passes through (1, 2, 3) and, is having direction ratios (1, 2, 3), is, x−1 y−2 z −3, (say), =, =, =a, 1, 2, 3, ∴, x−1 = a, y − 2 = 2a, and, z − 3 = 3a, ⇒, x = a + 1,, y = 2a + 2 and z = 3a + 3, At, x-axis, y = 0 and z = 0, ⇒, 2a + 2 = 0 and 3a + 3 = 0, ⇒, a = − 1 and a = − 1, ∴, x = (−1) + 1 = 0, 40. We know that, the equation of plane containing three, points, is given by, x y z 1, x y z 1, x1 y1 z1 1, 1 0 0 1, =0 ⇒, =0, x2 y2 z2 1, 0 2 0 1, 0 0 3 1, x3 y3 z3 1, , ⇒, , x [1 (6 − 0)] − y [1 (0 − 3)] + z [1 (2 − 0)], − 1 [1(6 − 0)] = 0, ⇒, 6x + 3 y + 2z − 6 = 0, ⇒, 6x + 3 y + 2z = 6, 41. Let a , b, c are direction ratios of line. If this line is, perpendicular to the line, whose direction ratios are, (1, − 2, − 2) and (0, 2, 1)., Then,, 1⋅ a − 2⋅ b − 2⋅ c = 0, …(i), ⇒, a − 2b − 2c = 0, and, 0⋅ a + 2⋅ b + 1⋅ c = 0, …(ii), ⇒, 0 + 2b + c = 0, On solving Eqs. (i) and (ii), we get, a, b, c, =, =, −2 + 4 0 − 1 2 + 0, a, b, c, =, =, ⇒, 2 (−1) 2, ∴ Direction cosines of the required line are, 2, 1, 2, ,−, ,, 2, 2, 2, 2, 2, 2, 2, 2 + (−1) + 2, 2 + (−1) + 2, 2 + (−1)2 + 22, 1 2, 2, or, , − , ., 3, 3 3, 42. A. Centre of sphere, x2 + y2 + z 2 + 3x − 3 y + 3z − 49 = 0 is, 3 3 3, − , ,− ., 2 2 2, B. Centre of sphere, x2 + y2 + z 2 − 4x − 6 y − 2z + 9 = 0 is (2, 3, 1)., C. Centre of sphere, 2 x2 + 2 y2 + 2 z 2 − x − y − z = 0, x y z, or, x2 + y2 + z 2 − − − = 0 is, 2 2 2, 1 1 1, , , ., 4 4 4, D. Centre of sphere, x2 + y2 + z 2 − 2x − 2 y − 2z = 0 is, (1, 1, 1)., ∴ The correct option is (c)., m, n, m′, 43. The two lines y = x + α , z = x + β and y =, x + α′ ,, l, l, l′, n′, z=, x + β′ can be written as,, l′, α, β, z−, y−, x−0, l, l, =, =, n, m, 1, l, l, α′, β′, y−, z−, x−0, l′, l′, and, =, =, m, n, ′, ′, 1, l′, l′, Both lines are perpendicular, if, m m′ n n′ , (1)(1) + + = 0, l l′ l l′ , ⇒, , ll′ + mm′ + nn′ = 0
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632, , NDA/NA Mathematics, , 44. Given, lines are x = az + b, y = cz + d, and, x = a1z + b1 , y = c1z + d1, or it can be rewritten as, x − b y−d, =, =z, a, c, x − b1 y − d1, and, =, =z, a1, c1, , ⇒, ⇒, …(i), … (ii), , Let us consider l1 , m1 , n1 and l2 , m2 , n 2 are DC ′ s of, Eqs. (i) and (ii), then, From Eq. (i), we get, l1 m1 n1, =, =, 1, a, c, Similarly,, l2 m2 n2, =, =, a1, c1, 1, If two lines are perpendicular, then, l1l2 + m1m2 + n1n2 = 0, From Eqs. (iii) and (iv), we get, ⇒, aa1 + cc1 + 1 = 0, 45. Given, two lines are, 2x = 3 y = − z, x, y, z, =, =, ⇒, 1 / 2 1 / 3 1 / (−1), and, ⇒, , Since,, , …(iii), , …(iv), , 6x = − y = − 4z, x, y, z, =, =, 1, 1, 1, 6 (−1), (−4), , …(ii), , x2 + y2 + z 2 = 9, , 48. We have, equation of line is, x− p y−q z−r, =, =, l, m, n, Ax + By + Cz + D = 0, , …(ii), , Since, the line lies in the plane, therefore its point, ( p, q , r) lies in the plane., ∴, , Ap + Bq + Cr + D = 0, , And the normal to the plane is perpendicular to the, line (i), ∴, Al + Bm + Cn = 0, ∴ Statements II and III are correct., 49. The equation of sphere is, x2 + y2 + z 2 − 10z = 0, ∴ The centre of sphere is C (0, 0, 5)., , C, , A, , B, , (0,0,5), , Coordinates of one end point of a diameter of the, sphere is A( −3, − 4, 5)., Let coordinates of another end point of this diameter, is B ( x1 , y1 , z1 )., , 1 1, 1 1, 1, 1, 1 1 1, ⋅ +, ⋅ +, ⋅, =, − + =0, 6 2 (−1) 3 (−4) (−1) 12 3 4, , 46. Here, two parallel planes are, 4x − 2 y + 4z + 9 = 0, , …(i), , 8x − 4 y + 8z + 21 = 0, , …(ii), , Distance of plane (i) from origin, 4 ×0 −2 ×0 + 4 ×0 +9 9 3, = =, =, 6 2, 16 + 4 + 16, , −3 + x1, = 0 ⇒ x1 = 3, 2, −4 + y1, = 0 ⇒ y1 = 4, 2, 5 + z1, and, = 5 ⇒ z1 = 5, 2, ∴ Required coordinates are (3, 4, 5)., 50. Let C be the centre of this sphere and let M be the foot of, perpendicular from C on plane., Let P be any point on the circle., ∴, , Distance of plane (ii) from origin, 8 × 0 − 4 × 0 + 8 × 0 + 21 21 7, =, =, =, 12 4, 64 + 16 + 64, ∴ Distance between the given two planes, 7 3 1, = − =, 4 2 4, 47. Let the locus of a point be (x, y, z ). From the given, condition,, 2, 2, 2, x − 2 y + z, x − z, x + y + z, =9, +, +, , , , 2 , , 3 , 6, ⇒, , …(i), , and equation of plane is, , …(i), , ∴ Angle between both lines is 90°. Thus, both lines, are perpendicular to each other., , and, , 6x2 + 6 y2 + 6z 2 = 54, , 2 (x + y + z )2 + 3 (x − z )2 + (x − 2 y + z )2 = 54, , ⇒ 2 (x2 + y2 + z 2 + 2xy + 2 yz + 2zx) + 3 (x2 + z 2 − 2xz ), + (x2 + 4 y2 + z 2 − 4xy − 4 yz + 2xz ) = 54, , C, , P, , M, , Q, , Then, CP = Radius of sphere., The coordinates of centre of sphere x2 + y2 + z 2 = 4 are, (0, 0, 0) and radius = 4 = 2 ., Now, equation of line CM normal to given plane, x − 1 = 0 through centre C (0, 0, 0) of sphere is, x−0 y−0 z −0, =, =, = λ (say), 1, 0, 0, , …(i)
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633, , Three Dimensional Geometry, ⇒, , x=λ, , Any point on Eq. (i) is ( λ , 0, 0). If this point lies on, given plane, then we have,, λ −1 =0, λ =1, ∴ Centre of circle = (1, 0, 0), , Now, CM = length of perpendicular from C ( 0, 0, 0) on, plane x − 1 = 0, | x ⋅ 0 − 1| 1, =, = =1, 1, 12, , Radius of circle, PM = PC 2 − MC 2, , ⇒, x2 = 1, ⇒, x=±1, ⇒, x = − 1, y = − 1, z = − 1, ∴ Coordinates of P = (−1, − 1, − 1)., 53. Let A (0, 0, 0), B (3, 4, 0) and C (3, 4, 6) be the vertices of a, ∆ABC., ∴, ∆ = (∆2x + ∆2y + ∆2z ), Now,, , 0 0 1, y1 z1 1, 1, 1, 1, y2 z2 1 = 4 0 1 = [1 (24)] = 12, ∆x =, 2, 2, 2, 4 6 1, y3 z3 1, , Similarly,, 0, 1, 0, 2, 6, 0, 1, ∆z = 3, 2, 3, , = 4 −1 = 3, , ∆y =, , ∴ Radius of circle is 3 and centre is (1, 0, 0)., 1, 51. Let plane x + y − z = divides the line joining the points, 2, A (2, 1, 5) and B (3, 4, 3) at a point C in the ratio λ : 1 ., Then, coordinates of C are, 3λ + 2 4λ + 1 3λ + 5, ,, ,, , ., λ + 1 λ + 1 λ +1 , Since, point C lies on the plane., , Therefore, coordinates of C must satisfy the equation, of plane., 3λ + 2 4λ + 1 3λ + 5 1, , +, −, =, λ +1 λ +1 λ +1 2, 1, ⇒, 3λ + 2 + 4λ + 1 − 3λ − 5 = (λ + 1), 2, 1, 4λ − 2 = (λ + 1), ⇒, 2, ⇒, 8λ − 4 = λ + 1, ⇒, 7λ = 5, 5, ⇒, λ=, 7, ∴ Required ratio is λ : 1 = 5 : 7., 52. If O be origin, then let coordinates of P are (x, y, z )., ∴, , and, , 4 1, , 1, [12 − 12] = 0, 2, , = 10 units, CA = (1 − 2)2 + (−1 + 3)2 + (2 − 4)2, = (−1)2 + (−2)2 + (−2)2, = 1 + 4 + 4 = 9 =3, CB = (2 − 2)2 + (−3 + 3)2 + (4 − 4)2, = 0 + 0 + 0 =0, , x, , This shows that points A and B are inside the sphere., , Let a , b, c are direction ratios of OP., Then, a = 0 − x, b = 0 − y, c = 0 − z, ⇒, x = − a , y = − b, z = − c, But a, b, c are equal ., ∴, − x, − y, − z are also equal., We have, distance of OP = 3, ⇒, (0 − x)2 + (0 − y)2 + (0 − z )2 = 3, 3 x2 = 3, 3 x2 = 3, , 4 1 =, , x2 + y2 + z 2 − 4x + 6 y − 8z − 71 = 0, whose centre is, C (2, − 3, 4) and, radius = 22 + (−3)2 + 42 + 71 = 100, , and, , ⇒, ⇒, , 3 1, 0 1, , 1, [1 (−18) ] = − 9, 2, , 55. Equation of given sphere is, , P (x, y, z), , z, O (0, 0, 0), , 3 1 =, , Area of ∆ = 122 + (−9)2 + (0)2 = 144 + 81, ∴, = 225, = 15 sq units, 54. The planes x = 2 y − 3 and z = 2 y − 1 intersect along a, line. Equation of planes can be rewritten as, x+ 3 y−0, …(i), =, 2, 1, z + 1 y −0, and, …(ii), =, 2, 1, From Eqs. (i) and (ii), we get, x+ 3 y−0 z + 1, =, =, 2, 1, 2, ∴ Direction ratios of the line are (2, 1, 2)., , Now,, , y, , 0 1, , (as x, y, z are equal), , 56. The equation of the given lines is, 6x = 4 y = 3z, x−0 y−0 z −0, =, =, ⇒, 1 /6, 1 /4, 1 /3, and equation of the plane is, 3x + 2 y − 3z − 4 = 0, Let θ be the angle between line and plane, then, sin θ =, , a1a 2 + b1b2 + c1c2, a12, , + b12 + c12 a 22 + b22 + c22
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634, , NDA/NA Mathematics, , 1, 1, 1, × 3 + × 2 + (−3) ×, 6, 4, 3, =, 1, 1, 1, 9+4+9, +, +, 36 16 9, , , , , 1 −1, =, = 0°, 1, 1, 1, , +, +, 22, 36 16 9, , ∴, θ = 0°, 57. Let (h , k, l) be the point, which is equidistant from the, points (1, 2, 3) and (3, 2, –1), ⇒ (h − 1)2 + (k − 2)2 + (l − 3)2, = (h − 3)2 + (k − 2)2 + (l + 1)2, ⇒, (h − 1) + (l − 3)2 = (h − 3)2 + (l + 1)2, 2, ⇒ h + 1 − 2 h + l2 − 6 l + 9 = h 2 − 6 h + 9 + l2 + 2 l + 1, ⇒, − 2h − 6l = − 6h + 2l, ⇒, 6h − 2h − 6l − 2l = 0, ⇒, 4h − 8l = 0, ⇒, h − 2l = 0, ∴ Locus of point (h , k, l) is, x−2 z =0, 58. Let the two points A (a , b, c) and B(a1 , b1 , c1 )., Let the locus of a point be P (x, y, z )., According to the given condition, (where, d is constant), PA 2 − PB2 = d, ⇒, (x − a )2 + ( y − b)2 + (z − c)2 − [(x − a1 )2, + ( y − b1 )2 + (z − c1 )2] = d, 2, 2, ⇒, x + a − 2ax + y2 + b2 − 2by + z 2 + c2 − 2cz, 2, − (x + a12 − 2a1x + y2 + b12 − 2b1 y + z 2 + c12 − 2c1z ) = d, ⇒, a 2 − a12 + b2 − b12 + c2 − c12 + 2x (a1 − a ), + 2 y (b1 − b) + 2z (c1 − c) = d, This represents an equation of plane., 59. We, have,, and, P (a , b, c); Q (a + 2, b + 2, c − 2), R(a + 6, b + 6, c − 6) are collinear., I. If R divides PQ internally in the ratio 3 : 2, then, coordinates of, 3a + 6 + 2a 3b + 6 + 2b 3c − 6 + 2c, R=, ,, ,, , , , 5, 5, 5, 5a + 6 5b + 6 5c − 6, ,, ,, =, , 5, 5, 5 , 2, , ∴ It is not true., II. If R divides PQ externally in the ratio 3 : 2, then, coordinates of, 3a + 6 − 2a 3b + 6 − 2b 3c − 6 − 2c, R=, ,, ,, , , , 1, 1, 1, = (a + 6, b + 6, c − 6), ∴ It is true statement., III. If Q divides PR internally in the ratio 1 : 2, then, coordinates of, a + 6 + 2a b + 6 + 2b c − 6 + 2c, ,, ,, Q=, , , , 3, 3, 3, = (a + 2, b + 2, c − 2), ∴ It is true., ∴ Statements II and III are true., , 60. Let A (a − 1, a , a + 1), B (a , a + 1, a − 1) and, C (a + 1, a − 1, a ) are the vertices of a ∆ABC., ∴, , AB = (a − a + 1)2 + (a + 1 − a )2 + (a − 1 − a − 1)2, = 1+1+4= 6, BC = (a + 1 + a )2 + (a − 1 − a − 1)2 + (a − a + 1)2, = 1+4+1= 6, , and CA = (a − 1 − a − 1)2 + (a − a + 1)2 + (a + 1 − a )2, = 4+1+1= 6, ∴, , AB = BC = CA, , ∴ ∆ ABC is an equilateral triangle and these given, points are vertices of an equitaeral triangle for any, real value of a., Now, area of a ∆ABC, i j k, 1, 1, = |AB × AC| = 1 1 −2, 2, 2, 2 −1 −1, 1, = |−3i − 3 j − 3 k|, 2, 1, =, 9+ 9 + 9, 2, 27, sq units, =, 2, Thus, the area of triangle formed by these points is, independent of a., 61. Clearly, both the statements are correct., 62. DR’s of normal to the plane is l, m, n ,i.e., normal vector, = li + mj + nk, li + mj + nk, $ =, ⇒ Normal unit vector n, l2 + m 2 + n 2, Distance from origin (d ) = p, $ =d, ⇒ Equation of a plane is r ⋅ n, (li + mj + nk ), ⇒, (x i + yj + 2k ), =p, l2 + m 2 + n 2, ⇒, , lx + my + nz = p l2 + m2 + n 2, , ⇒ Intercepts on x, y and z-axes are, p 2, x=, l + m2 + n 2, l, p 2, y=, l + m2 + n 2, m, p 2, z=, l + m2 + n 2, n, , 1 1 1, ∴Sum of intercepts = p l2 + m2 + n 2 + + , l m n, 63. DR’s of OP = (2 − 0, 3 − 0, − 1 − 0), , = ( 2, 3, − 1), , Equation of a plane passing through a point, ( 2, 3, − 1) and normal to a vector OP is, 2 ( x − 2) + 3 ( y − 3) − 1 ( z + 1) = 0, ⇒, , 2x + 3 y − z − 4 − 9 − 1 = 0, 2x + 3 y − z = 14
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635, , Three Dimensional Geometry, 64. The image of the point P(1, 3, 4) in the plane, 2x − y + z + 3 = 0 is given by, x−1 y−3 z −4, 2 (2 − 3 + 4 + 3), =, =, =−, −1, 2, 1, 4+1+1, x−1 y−3 z −4, =, =, = −2, ⇒, −1, 2, 1, x−1 y−3 z −4, ⇒, =, =, = −2, −1, 2, 1, ⇒, x = − 3, y = 5, z = 2, , ∴ Coordinates of image of the point P (1, 3, 4) is, ( −3, 5, 2)., 65. Equation of the spheres are, …(i), x2 + y2 + z 2 + 2x + 2 y + 2z + 5 = 0, and, …(ii), x2 + y2 + z 2 + 3x + 3 y + 7z + 6 = 0, Thus, equation of common tangent is, (x2 + y2 + z 2 + 2x + 2 y + 2z + 5), − (x2 + y2 + z 2 + 3x + 3 y + 7z + 6) = 0, ⇒, −x − y − 5z − 1 = 0, ⇒, x + y + 5z + 1 = 0, 66. If the direction cosines of any line is < l, m, n > , then, l2 + m 2 + n 2 = 1, So, for the direction cosines,, , 2, , 2, , =, , l +m, m +n, n +l, +, +, 2, 2, 2, 2, , 2, , 2, , 2, , 2, , 2, , 68. Angle between two lines is, a1a 2 + b1b2 + c1c2, cos θ =, 2, a1 + b12 + c12 a 22 + b22 + c22, Here, DR’s of L1 (a1 , b1 , c1 ) = (1, 2, 3), DR’s of L 2, (a 2, b2, c2) = (3, − 2, 1), ∴, , ⇒, , l2 + m 2 m 2 + n 2 n 2 + l2, ,, ,, 2, 2, 2, 2, 2, 2, 2, l2 + m 2 , , , + m +n + n +l , =, , , , , , 2 , 2, 2 , , , , , , 1, (2l2 + 2m2 + 2n 2), 2, = l2 + m2 + n 2 = 1,, while l + m + n ≠ 1, 67. DR’s of line L1 is (1, 2, 3)., ∴DR’s of, 1, 2, 3, ,, ,, >, L1 = <, 2, 2, 2, 2, 2, 2, 2, 1 +2 +3, 1 +2 +3, 1 + 22 + 32, 1, 2, 3, ,, ,, =<, >, 14 14 14, =, , 2, , cos θ =, , 1 × 3 + 2 × (−2) + 3 × 1, , 12 + 22 + 32 32 + (−2)2 + 12, 2, 2, 1, =, =, =, 14 14 14 7, 1, θ = cos −1 , 7, , 69. Equation of any plane through L1 is, a (x + 4) + b( y − 3) + c(z + 2) = 0, , Also, normal of a plane is (a , b, c) = (1, 3, 2), ∴, ⇒, , 1(x + 4) + 3( y − 3) + 2(z + 2) = 0, x + 3 y + 2z − 1 = 0
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30, , Statistics, , (ii) Continuous Frequency Distribution, , Statistics, Statistics is the branch of science, which deals with the, collection, analysis and interpretation of numerical data., , Some Important Terms, , Classification of Data, Classification of data is the first step in statistics, towards achieving the goal on conclusion. There are, different sources of collection of data., Sources of data, , Internal, , External, , Primary, , A frequency distribution in which data are arranged in, classes or groups which are not exactly measurable., , Secondary, , The two main type of data on the basis of collection are, (i) Primary data It is the data collected actually in, the process of investigation by the investigator. It is, original and is first hand information., (ii) Secondary data Data which is already collected, by other persons is called as secondary data., , Presentation of data, Raw ou Ungrouped data When the data are, presented in random and is not prepared according to some, order. It does not give us a clear picture of the class., Grouped data When the data is arranged in any, manner like ascending or descending etc. It can also be, presented in the form of a table called frequency, distribution table., , Types of Frequency, Distribution, Frequency distribution are of two types, , (i) Discrete Frequency Distribution, A frequency distribution is called discrete frequency, distribution if data are presented in a way that exact, measurements of the units are clearly shown., , Frequency Number of observations falling in a, particular class is called frequency of that class., Class marks It is the mid-point of the class interval., Lower limit of class + Upper limit of class, Class mark =, 2, Cumulative, frequency The, cumulative, frequency of a class interval is the sum of frequencies, of all classes upto that class. (including the frequency, of that particular class)., Example 1 Consider the following statements, The appropriate number of classes while constructing a, frequency distribution should be chosen such that, I. The class frequency first increases to peak and then, declines., II. The class frequency should cluster around the class, mid-point., Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, , Solution (b) The appropriate number of classes while, constructing a frequency distribution should be chosen such that, the class frequency should cluster around the class mid-point., , Graphical Representation of, Frequency Distribution, Bar Diagrams, In bar diagrams, only the length of the bars are taken, into consideration. The width of each bar can be any, but, widths of all the bars is same and space between these bars, should be same. The width of the bar has no special, meaning., e.g., The bar diagram of the following data is
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637, , Statistics, , Registration of, vehicles in 2011, No. of vehicles, , Car, , Bus, , Scooters, , Bikes, , 40, , 20, , 25, , 35, , 40, , Frequency Polygon, , 30, , To draw the frequency polygon of an ungrouped, frequency distribution, we plot the points with abscissa as, the variate values and the ordinate as the corresponding, frequencies. These plotted points are joined by straight, lines to obtain the frequency polygon., , Bikes, , Cars, , 10, , Scooters, , 20, Bus, , Frequency, , y, , x-axis on a suitable scale. On each class interval erect, rectangles with heights proportional to the frequency of, the corresponding class interval, so that the area of the, rectangle is proportional to the frequency of the class., , x, , 0, , Cumulative Frequency Curve, (Ogive), , Registered Vehicles, , Pie Diagrams, Pie diagram is used to represent a relative frequency, distribution. A pie diagram consists of a circle divided into, as many sectors or there are classes in a frequency, distribution. Sum of all the angles of sectors is 360°, , When we plot the upper class limit along x-axis and, cumulative frequencies along y-axis. And on joining, then, we get a curve called an ogive., y, , Temperature, , 85, Cars, 120º, 60º, , 70, 65, , Scooters, , frequency × 360° , Central angle = , , total frequency , e.g., The pie diagram of the above data is, 40, Central angle for cars =, × 360° = 120°, 120, 20 × 360°, Central angle for bus =, = 60°, 120, 25 × 360°, Central angle for scooters =, = 75°, 120, 35 × 360°, Central angle for bikes =, = 105°, 120, , 15:00, , 14:00, , 13:00, , 12:00, , x, , 11:00, , 0, , 10:00, , Bikes, , 75º, , 75, , 9:00, , 105º, , Bus, , 80, , Time, , Measures of Central Tendency, , Histogram, To draw a histogram of a given continuous frequency, distribution, we first mark off all the class intervals along, , Generally average value of a distribution in the middle, part of the distribution such type of values are known as, measures of central tendency., An average of a distribution is the value of the variable, which is representative of the entire distribution., The following are the five measures of central, tendency., 1. Arithmetic Mean, 2. Geometric Mean, 3. Harmonic Mean, 4. Median, 5. Mode, , y, , 1. Arithmetic Mean, , Frequency, , 50, 40, 30, 20, 10, 0, , 10 20, , 30 40 50 60, Class interval, , x, , Let x1 , x2 , x3 , K , xn are n individual data, then, x + x 2 + K + xn, 1 n, x= 1, =, Σ xi, n, n i =1, 1 n, or, x = A+, Σ di, n i =1, where, A = assumed mean and di = xi − A
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638, , NDA/NA Mathematics, , For Frequency Distribution, , Example 3. Arithmetic mean of 9 observations is 100 and, , Let x1 , x2 , K , xn are n observations,, corresponding frequencies are f1 , f2 , K , fn , then, , where, , n, , x=, , x1 f1 + x2 f2 + K + xn fn, =, f1 + f2 + K + fn, , Σ fi xi, , i =1, n, , Σ fi, , i =1, n, , x = A+, , or, , arithmetic mean of 6 observations is 80, then the arithmetic, mean of 5 observations is, (a) 90, (b) 91, (c) 92, (d) 93, , ∴, , Σ fi di, , i =1, n, , where, A = assumed mean and di = xi − A, (c) 24.56, , Class Interval, , Frequency, , 0–10, 10–20, 20–30, 30–40, 40–50, , 22, 38, 46, 35, 20, , 9 × 100 + 6 × 80, 9+6, , 900 + 480, 15, 1380, =, = 92, 15, , (d) 25.56, , 2. Geometric Mean (GM), (i) If x1 , x2 , K , xn are n non-zero observations, then, GM = ( x1 , x2 , K , xn )1/ n, 1 n, , Σ log xi , = antilog , n i =1, , , Solution (c), Class interval, , xi, , fi, , f i xi, , 0–10, 10–20, 20–30, 30–40, 40–50, , 5, 15, 25, 35, 45, , 22, 38, 46, 35, 20, Σfi = 161, , 110, 570, 1150, 1225, 900, Σfi xi = 3955, , Mean =, , n1x1 + n2x2, n1 + n2, , =, , Example 2. The mean for following distribution is, (b) 23.24, , x12 =, =, , Σ fi, , i =1, , (a) 22.33, , n1 = 9, x1 = 100 , n2 = 6 and x2 = 80, , Solution (c) Given,, , Σfi xi 3955, =, Σfi, 161, , = 24.56 (approx), , (ii) For frequency distribution, 1 n, , GM = antilog , Σ ( fi log xi ), i =1, Σ, f, i, , , Example 4. The geometric mean of 1, 2, 2 2, …, 2 n is, (a) 2 n/ 2, (c) 2 n, , (b) 2( n+1)/ 2, (d) 2 n+1, 1, , Solution (a) Geometric mean = (1⋅ 2 ⋅ 22 K 2n) n + 1, 1, , Weighted Arithmetic Mean, , = (2(1 + 2 + 3 + K + n)) n + 1, , If w1 , w2 , w3 , K , wn are corresponding weight to the, observations x1 , x2 , x3 , K , xn , then, w x + w2x2 + K + wn xn, Weighted arithmetic mean = 1 1, w1 + w2 + K + wn, , = (2, , Combined Mean, If two sets of observations are given, then combined, mean for the two sets can be calculated with the help of, following formula, n x + n 2x2, x12 = 1 1, n1 + n 2, where, x12 = combined mean of two sets of observations, x1 = mean of first set of observations, n1 = number of observations in first set, x2 = mean of second set of observations, n 2 = number of observations in second set, , n ( n + 1) 1, 2 )n+1, , n, , = 22, , 3. Harmonic Mean (HM), (i) If x1 , x2 , K , xn are n observations, then, 1, HM =, n, 1, 1, Σ , n i = 1 xi , (ii) For frequency distribution, 1, HM =, n, f , 1, Σ i, Σfi i = 1 xi , Relation Among AM, GM and HM, AM ≥ GM ≥ HM
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639, , Statistics, , 1 1 1, 2 3 4, 2, (b), 19, 19, (d), 16, , Example 5. The harmonic mean of , , , K,, 19, 2, 2, (c), 16, , (a), , 1, is, 17, , 1, 1 n 1, Σ , n i = 1 xi , 2, 1, =, =, 1, (2 + 3 + K + 17) 19, 16, , Solution (b) Harmonic mean =, , Example 6. The median for the following distribution is, , (a) 20, , (a) If n is an odd number, then Median = Value of, n + 1, , th term, 2 , (b) If n is an even number, then, Median, n, n, , Value of th term + Value of + 1 th term, 2, , 2, =, 2, (ii) Median of a discrete series First, arrange the, value of given observations (or variables) in ascending, order, then find the cumulative frequency., (a) If n is an odd number, then Median = value of, n + 1, , th term, 2 , (b) If n is an even number, then, Median, n, n, , value of th term + value of + 1 th term, 2, 2, , =, 2, (iii) Median of a continuous series First find the, cumulative frequency table of given observations, then, n, find the group (median group) of th observation. Then,, 2, n, , − c, 2, , Median = l +, ∴, ×h, f, , Frequency, , 0–10, 10–20, 20–30, 30–40, 40–50, , 22, 38, 46, 35, 20, , (b) 22.46, , (c) 24.46, , (d) 25, , Solution (c) Cumulative frequency table, Class Interval, , f, , cf, , 0–10, 10–20, 20–30, 30–40, 40–50, , 22, 38, 46, 35, 20, , 22, 60, 106, 141, 161, , 4. Median, (i) Median of an individual series Let the number, of observations is n. First arrange the value of the, observation in ascending or descending order., , Class Interval, , n 161, =, = 80.5, 2, 2, Hence, median group is 20 − 30, , n, − c, 80.5 − 60 , 2, Median, =l + , ∴, × 10, × h = 20 + , , , f, 46, , , , , Here,, , = 20 + 4.46 = 24.46, , 5. Mode, (i) Mode of an individual series The mode of an, individual series is that value of variable, which is, repeated more than the other variables of the series., (ii) Mode of a discrete series The mode of a discrete, series is that value of variable for which the frequency is, maximum., (iii) Mode of a continuous series First find the, modal group, which has maximum frequency, then, f1 − f0, Mode = l +, ×h, 2 f1 − f0 − f2, where, l = lower limit of modal group, h = size of modal group, f1 = frequency of modal group, f0 = frequency of a group before to modal group, f2 = frequency of a group next to modal group, Relation among Mean, Median and Mode, Mode = 3 (Median) − 2 (Mean), , Example 7. The mode of the following distribution is, , where, l = lower limit of median group, f = frequency of median group, h = size of median group, c = cumulative frequency of a group before to, median group, (a) 40, , Class Interval, , Frequency, , 0–20, 20–40, 40–60, 60–80, 80–100, , 17, 28, 32, 24, 19, , (b) 42.67, , (c) 46.67, , (d) 7
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640, , NDA/NA Mathematics, , Solution (c) From the given table, it is clear that modal group is, 40–60., Here,, , l = 40 , f0 = 28, f1 = 32, f2 = 24, f1 − f0, Mode = l +, ×h, 2f1 − f0 − f2, 32 − 28, = 40 +, × 20, 64 − 28 − 24, 4 × 20, = 40 +, 12, = 40 + 6.67, = 46. 67, , ∴, , Solution (a), x, , f, , cf, , 2, 3, 4, 5, 6, , 3, 4, 8, 4, 1, , 3, 7, 15, 19, 20, , N = ∑ fi = 20, , Here,, , N 20, 3N, =, =5 ,, = 15, 4 4, 4, , Measures of Dispersion, , Q cf just greater than 5 is 7., , The measure of the scatteredness of the mass of figures, in a series about an average is called the measure of, variation or dispersion., The measures of dispersion commonly used are, 1. Range, 2. Quartile deviation, 3. Mean deviation, 4. Standard deviation, , and cf just greater than 15 is 19., Q3 = 5, Q − Q1 5 − 3, ∴Quartile deviation = 3, =, =1, 2, 2, Hence,, , 3. Mean Deviation, For frequency distribution mean deviation from the, average A (usually mean, median or modes) is given by, n, , 1. Range, , MD =, , The range is the difference of maximum and minimum, observation of observations of a distribution. If L and S are, maximum and minimum observation of distribution then,, Coefficient of range =, , L−S, L+S, , 2. Quartile Deviation, Quartile deviation or semi-interquartile range Q is, given by, 1, Q = (Q3 − Q1 ), ∴, 2, Q − Q1, and coefficient of, Q= 3, Q3 + Q1, where, Q1 and Q3 are the first and third quartiles of the, distribution, respectively., , Example 8. The quartile deviation for the following data is, , (a) 1, , x, , f, , 2, 3, 4, 5, 6, , 3, 4, 8, 4, 1, , (b) 2, , (c) 3, , Σ fi|xi − A|, , i =1, , n, , Σ fi, , i =1, , where, A = mean or median or mode, Coefficient of mean deviation, Mean deviation, =, Average from which it is calculated, , Range = L − S, and, , Q1 = 3, , Hence,, , 4. Standard Deviation, Usually denoted by Greek letter small sigma (σ)., , (i) Standard Deviation for Ungrouped Data, n, , i =1, , SD (σ ) =, , n, , 1 n 2 1 n, Σ xi − Σ xi , i, =, 1, =, 1, i, n, n, , , σ=, , or, , 2, , (ii) Standard Deviation for Grouped Data, n, , σ=, or, (d) 4, , Σ ( xi − x )2, , where,, , σ=, , Σ fi ( xi − x )2, , i =1, , N, 1 n, , 1, Σ fi xi2 − , Σ fi xi , i, =, 1, =, 1, i, N, N, , n, , N = Σfi, , 2
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641, , Statistics, Variance = σ 2, , Greater is the value of coefficient of variation of a, distribution, there is more variability in that distribution., , σ, and Coefficient of dispersion = ., x, , Example 11. The following is the record of goals scored by, team A in a football session., , Example 9. The mean deviation of 3, 4, 5, 6, 7 is, (a) 1, , (b) 1.2, , Solution (b) Mean, x =, Mean deviation =, , (c) 0, , (d) 2, , 3+ 4+5+6+7, =5, 5, , 1 n, Σ | xi − x|, n i=1, , 1, (| 3 − 5 | + | 4 − 5 | + |5 − 5 | + | 6 − 5 | + | 7 − 5 |), 5, 1, 6, = (2 + 1 + 0 + 1 + 2) = = 1.2, 5, 5, =, , Example 10. The variance of the following distribution is, 2, 1/3, , xi, fi, , (a) 10, , 3, 1/2, , (b) 16, , 11, 1/6, , (c) 22, , (d) 32, , fi, , fi x i, , fi x i2, , 2, 3, 11, , 1/3, 1/2, 1/6, Σfi = 1, , 2/3, 3/2, 11/6, Σfi xi = 4, , 4/3, 9/2, 121/6, , 2, , Variance ( σ 2) =, , 1, 9, , 2, 7, , 3, 5, , 4, 3, , For the team B, mean number of goals scored per match was, 2 with standard deviation 1.25 goals. Which team may be, considered more consistent., (a) Team A, (b) Team B, (c) both teams have same coefficient of variation, (d) None of the above, , Solution (a) For team A,, Number of goals x i, Number of matches, fi, , 0, , 1, , 2, , 3, , 4, , Total, , 1, , 9, , 7, , 5, , 3, , 25, , x 2i, , 0, , 1, , 4, , 9, , 16, , fi x i, , 0, , 9, , 14, , 15, , 12, , 50, , fi x 2 i, , 0, , 9, , 28, , 45, , 48, , 130, , Solution (a), xi, , 0, 1, , Number of goals scored, Number of matches, , ∑ f x 50, x= i i =, =2, ∑ fi, 25, Standard deviation,, 1, 5, σ=, 25 × 130 − (50) 2 =, 130 − 100 = 1.095, 25, 25, 1095, σ, ., = 100 = 54.75, ∴Coefficient of variation = × 100 =, x, 2, , Σfi xi2 = 26, 2, , Σfi xi2 Σfi xi , 26 4, −, − = 26 − 16 = 10, =, 1 1, Σfi, Σfi , , For Team B,, Mean,, , Comparison of Variability of, Distribution, , x=2, , σ =1. 25, σ, Coefficient of variation = × 100, x, 1. 25, =, × 100 = 62 . 5, 2, Coefficient of variation of goals of team A is less than that of B., Therefore, team A is more consistent than B, SD,, , To compare the variability of two or more, distributions, we calculate the coefficient of variation of, each distribution i.e.,, Deviation, σ, Coefficient of variation =, × 100 = × 100, x, Mean, , Comprehensive Approach, n, , σ1 and σ 2 are standard deviation respectively, then, Combined standard deviation, n σ 2 + n2σ 22 + n1d12 + n2d 22, σ12 = 1 1, n1 + n2, , In statistical observations, the sum of deviations of individual, values from arithmetic mean is zero, i.e.,, n, , Σ fi ( xi − x) = 0, , i =1, n, , n, , n, , n, n, , In statistical observations, the sum of squares of deviations of, individual values from arithmetic mean is least., If each of n given observations is doubled, then the mean is also, become twice., 1/ 2, 1, Standard deviation of n natural numbers = (n2 − 1) , 12, , Combined Standard Deviation, If A1 and A2 are two series in which number of observations are n1, and n2, respectively. Let x1 and x2 are their arithmetic mean and σ1, , d1 = ( x1 − x12), d 2 = ( x2 − x12), n1x1 + n2x2, n1 + n2, σ, Coefficient of variation = × 100, x, where,, , and, n, , n, , x12 =, , If each item of a data is increased or decreased by the same, constant. The standard devition of the data remains unchanged.
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Exercise, Level I, (a) 17, , (b) 18, , (c) 19, , (d) 20, , 1. Which one of the following is a source of data for, primary investigations?, (a) News papers, (b) Magazines, (c) Questionnaires, (d) All of these, , 10. What is the geometric mean of 10, 40 and 60?, , 2. What is the cumulative frequency curve of statistical, data commonly called?, (NDA 2011 I), (a) Cartogram, (b) Histogram, (c) Pictogram, (d) Ogive, , 11. What is the median of the distribution 3, 7, 6, 9, 5, 4, and 2?, (NDA 2011 II), (a) 5, (b) 6, (c) 7, (d) 8, , 3. What is the mode for the data 20, 20, 20, 21, 21, 21,, 21, 21, 22, 22, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 24,, 24, 25?, (NDA 2012 I), (a) 7, (b) 21, (c) 22, (d) 25, 4. In statistics, a suitable graph for representing the, partitioning of total into subparts is, (a) an ogive, (b) a pictograph, (c) a histogram, (d) a pie chart, 5. Which one of the following represents statistical data?, (a) The name of all owners of shops located in a, shopping complex., (b) A list giving the names of all states of India., (c) A list of all European countries and their, respective capital cities., (d) The volume of a rainfall in certain geographical, area, recorded every month for 24 consecutive, months., 6. The frequency distribution of some given numbers is, Value, , Frequency, , 1, 2, 3, 4, , (NDA 2011 II), , (a) 10 3 3, , (b) 20 3 3, , (c) 40 3 3, , (d) 70 3 3, , 12. The mean of 30 given numbers, when it is given that, the mean of 10 of them is 12 and the mean of the, remaining 20 is 9, is equal to, (a) 11, (b) 10, (c) 9, (d) 5, 13. Mean of 100 observations is 45. It was later found, that two observations 19 and 31 were incorrectly, recorded as 91 and 13. The correct mean is, (a) 44, (b) 45, (c) 44.46, (d) 45.54, 14. If n = 20, x = 50 and Σx 2 = 84000, then the variance is, equal to, (a) 1500, (b) 1700, (c) 1750, (d) 1800, 15. In a class of 100 students, the average amount of, pocket money is ` 35 per student. If the average is, ` 25 for girls and ` 50 for boys, then the number of, girls in the class is, (a) 20, (b) 40, (c) 60, (d) 80, 16. The standard deviation of the observations 22, 26,, 28, 20, 24, 30 is, (a) 2, (b) 2.4, (c) 3, (d) 3.42, 17. The SD of 15 items is 6 and if each item is decreased, by 1, then standard deviation will be, 91, (a) 5, (b) 7, (c), (d) 6, 15, , 5, 4, 6, f, , If the mean is known to be 3, then the value of f is, (a) 3, (b) 7, (c) 10, (d) 14, 7. The average of the squares of the numbers 0, 1, 2, 3,, 4,..., n is, 1, 1, (a) n ( n + 1), (b) n ( 2n + 1), 2, 6, 1, 1, (d) n ( n + 1), (c) ( n + 1)( 2n + 1), 6, 6, 8. A variate X takes values 2, 3, 4, 2, 5, 4, 3, 2 and 1., What is the mode?, (NDA 2011 II), (a) 2, (b) 3, (c) 4, (d) 5, 9. Which one of the following is the mean of the data, given below?, (NDA 2011 II), xi, , 6, , 10, , 14, , 18, , 24, , 28, , 30, , fi, , 2, , 4, , 7, , 12, , 8, , 4, , 3, , 18. A variate X takes values 2, 9, 3, 7, 5, 4, 3, 2, 10. What, is the median?, (NDA 2012 II), (a) 2, (b) 4, (c) 7, (d) 9, 19. What is the geometric mean of the data 2, 4, 8, 16 and, 32?, (NDA 2011 II), (a) 2, (b) 4, (c) 8, (d) 16, 20. What is the mean deviation of the data 2, 9, 9, 3, 6, 9, and 4?, (NDA 2011 II), (a) 2.23, (b) 2.57, (c) 3.23, (d) 3.57, 21. A set of n values x1 , x2 ,..... , xn has standard, deviation σ. What is the standard deviation of n, (NDA 2011 I), values x1 + k, x2 + k,........ , xn + k?, (a) σ, (b) σ + k, (c) σ − k, (d) kσ, 22. The observations 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95, are arranged in ascending order. What is the value of, x, if the median of the data is 63?, (a) 61, (b) 62, (c) 62.5, (d) 63
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643, , Statistics, 23. The median of 19 observations is 30. Two more, observations are made and the values of these are 8, and 32. The median of the 21 observations taken, together is equal to, (a) 28, (b) 30, (c) 32, (d) 34, , respectively. The arithmetic mean of the marks, secured by students of section C, who appeared for a, test in the same subject, which carried 75 marks is, 51. What is the average percentage of marks secured, by all the 100 students of the three sections?, , 24. The adjoining diagram gives a pie chart representing, the units of electricity sold to different classes in a, month. The angle of the sector corresponding to, supply for industries is, , (NDA 2009 II), , 13%, Street, light, , 16%, Others, 25%, For, Industries, , (b) 46°, , (c) 60°, , (d) 90°, , 26. What is the arithmetic mean of first 16 natural, numbers with weights begin the number itself?, (NDA 2012 I), , (b) 33/2, , (c) 11, , (c) 65, , (d) 67.5, , 31. What is the least value of the standard deviation of, 5 integers, no two of which are equal? (NDA 2009 II), (a) 5, (b) 2, (c) 2, (d) No such least value can be computed., , Month 1 Month 2 Month 3 Month 4, , 25. Which one of the following is not correct in respect of, mean?, (a) The sum of the deviations of individual, observations from the mean is zero., (b) If each observation in the data is replaced by, mean x, then the sum remains unaltered., (c) If x is the mean of x1 ,... , xn , then the mean of, x1 ± a , x2 ± a ,... , xn ± a is equal to x ± a., (d) If x is the mean of x1 ,.... , xn , then the mean of, log x1, log x2 ,.... , log xn is equal to log x., , (a) 17/2, , (b) 70.8, , 32. The average sales and standard deviation of sales for, four months for a company are as follows., , 46%, Domestic, supply, , (a) 25°, , (a) 70, , (d) 187/2, , 27. In a batch of 15 students, IF the marks of 10, students, who passed are 70, 50, 95, 40, 60, 70, 80, 90,, 75, 80, then the nedian marks of the all 15 students is, (a) 40, (b) 50, (c) 60, (d) 70, 28. The distributions X and Y with total number of, observations 36, 64 and mean 4, 3, respectively are, combined. What is the mean of the resulting, distribution X + Y ?, (NDA 2010 II), (a) 3.26, (b) 3.32, (c) 3.36, (d) 3.42, 29. The geometric mean of three numbers was computed, as 6. It was subsequently found that in this, computation, a number 8 was wrongly read as 12., What is the correct geometric mean?, (NDA 2010 I), (a) 4, (b) 3 5, (c) 2 3 18, (d) None of these, 30. A class consists of 3 sections A, B and C with 35, 35, and 30 students, respectively. The arithmetic means, of the marks secured by students of sections A and B,, who appeared for a test of 100 marks are 74 and 70,, , Average, Sales, , 30, , 57, , 82, , 28, , Standard, Deviation of, Sales, , 2, , 3, , 4, , 2, , During which month are the sales most consistent?, (NDA 2009 I), , (a) Month 1, (c) Month 3, , (b) Month 2, (d) Month 4, , 33. The harmonic mean of two numbers is 21.6. If one of, the numbers is 27, what is the other number?, (NDA 2009 I), , (a) 16.2, , (b) 17.3, , (c) 18, , (d) 20, , 34. What is the standard deviation of numbers 7, 9, 11,, 13 and 15?, (NDA 2009 I), (a) 2.2, (b) 2.4, (c) 2.6, (d) 2.8, 35. For a set of discrete numbers, three measures of, central tendency are given below., I. Arithmetic mean II. Median III. Geometric mean, Which of the above measures may not have a, meaningful definition?, (NDA 2008 II), (a) Only I, (b) Only II, (c) Only III, (d) All of them are meaningfully defined., 36. If the three observations are 3, − 6 and − 6, then what, is their harmonic mean?, (NDA 2011 II), (a) 0, (b) ∞, (c) − 1 / 2, (d) −3, 37. What is the arithmetic mean of the series, n, C1 , n C2 , n C3 ,... ,n Cn ?, (NDA 2008 II), n, n, ( 2 − 1), 2, (a), (b), n, ( n + 1), ( 2n ), 2( n + 1), (c), (d), n, ( n + 1)
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644, , NDA/NA Mathematics, , 38. The average age of 20 students in a class is 15 yr. If, the teacher’s age is included, the average increases, by one. What is the teacher’s age?, (NDA 2008 II), (a) 30 yr, (b) 21 yr, (c) 42 yr, (d) 36 yr, , 40. The data below records the itemwise quarterly, expenditure of a private organization., Items of expenditure, , 39. If n1 and n 2 are the sizes, G1 and G 2 are the geometric, means of two series respectively, which one of the, following expresses the geometric mean (G) of the, combined series?, (NDA 2008 I), n1G1 + n 2 G2, (a) logG =, n1 + n 2, n 2 log G1 + n1 log G2, (b) log G =, n1 + n 2, n1 log G1 + n 2 log G2, (c) G =, n1 + n 2, (d) None of the above, , Amount (in ` lakh), , 1., , Salaries, , 6, , 2., , TA & DA, , 4.9, , 3., , House rent and postage, , 3.6, , 4., , All other expenses, , 5.5, , Total, , 20, , The data is represented by a pie diagram. What is the, sectorial angle of the sector with largest area?, (NDA 2007 II), , (a) 120°, (c) 100°, , (b) 108°, (d) 90°, , Level II, 1. The weighted AM of first n natural numbers whose, weights are equal to the corresponding numbers is, equal to, 1, (a) 2n +1, (b) ( 2n + 1), 2, 1, 2n + 1, (c) ( 2n + 1), (d), 3, 6, 2. The mean of a set of abservations is x. If each, observation is divided by α ≠ 0 and then is increased, by 10, then the mean of the new set is, x, x + 10, (b), (a), α, α, x + 10α, (c), (d) α x + 10, α, 3. Some measures of central tendency for n discrete, observations are given below., I. Arithmetic mean, II. Geometric mean, III. Harmonic mean, IV. Median, A desirable property of a measure of central tendency, is, if every observation is multiplied by c, then the, measure of central tendency is also multiplied by c,, where c > 0. Which of the above measures satisfy the, property?, (NDA 2011 II), (a) I, II and III, (b) I, II and IV, (c) III and IV, (d) I, II, III and IV, n, , n, , i =1, , i =1, , 4. If ∑ ( xi − 2) = 110, ∑ ( xi − 5) = 20,what is the mean?, (a) 11/2, (c) 17/3, , (NDA 2009 I), , (b) 2/11, (d) 17/9, , 5. The arithmetic mean of data with observations, a, a + d, a + 2d,.....,a + 2md is, (a) a + md, (b) a + (m − 1)d, 1, 1, (d) a + ( m + 1) d, (c) a + md, 2, 2, , 6. What is the value of n for which the numbers 1, 2,, 3,....,n have variance 2?, (NDA 2008 II), (a) 4, (b) 5, (c) 6, (d) 8, 7., , X, , 1, , 2, , 3, , 4, , Frequency, , 2, , 3, , f, , 5, , The frequency distribution of a discrete variable X, with one missing frequency f is given above. If the, 23, arithmetic mean of X is, , what is the value of the, 8, missing frequency?, (NDA 2008 II), (a) 5, (b) 6, (c) 8, (d) 10, 8. Consider the two series of observations A and B as, follows., Series A, , 1019, , 1008, , 1015, , 1006, , 1002, , Series B, , 1.9, , 0.8, , 1.5, , 0.6, , 0.2, , If the standard deviation of the series A is 38, then, what is the standard deviation of the Series B?, (NDA 2008 I), , (a) 3.8, , (b) 0.38, , (c) 0.38, , (d) 38, , 9. Consider the following three methods of collecting, data, I. Collecting data from government offices., II. Collecting data from public libraries., III. Collecting data by telephonic interview., (NDA 2008 II), , Select the correct answer using the codes given, below., (a) All the three methods gives secondary data., (b) I and II give secondary and III gives primary data., (c) I and III give secondary and II gives primary data., (d) II and III give secondary and I gives primary, data.
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645, , Statistics, 10. Consider the following statements, I. A continuous random variable can take all values, in an interval., II. A random variable which takes a finite number of, values is necessarily discrete., III. Construction of a frequency distribution is based, on data which are discrete., (NDA 2008 II), Which of the above statements are correct?, (a) Both I and II, (b) Both II and III, (c) Both I and III, (d) All of these, 11. The populations of four towns A, B, C and D as on, 2001 are as follows., (NDA 2008 II), Town, , Population, , A, B, C, D, , 6863, 519, 12185, 1755, , What is the most appropriate diagram to present the, above data?, (a) Pie diagram, (b) Bar chart, (c) Cubic chart, (d) Histogram, 12. Let x be the mean of n observations x1 , x2 ,.... , xn . . If, ( a − b) is added to each observation, what is the mean, of new set of observations?, (NDA 2008 I), (a) 0, (b) x, (d) x + ( a − b), (c) x − ( a − b), 13. The mean weight of all students in a class is 60 kg., The mean weight of boys in a class is 70 kg while, mean weight of girls is 55 kg, what is the ratio of the, number of girls?, (NDA 2007 I), (a) 2:1, (b) 1:2, (c) 1:4, (d) 4:1, 14. Students of two schools appeared for a common test, carrying 100 marks. The arithmetic means of their, marks for schools I and II are 82 and 86, respectively., If the number of students of school II is 1.5 times the, number of students of school I, what is the arithmetic, mean of the marks of all the students of both the, schools?, (NDA 2007 II), (a) 84, (b) 84.2, (c) 84.4, (d) Cannot be determined, 15. If the median of x/ 5, x, x/4, x/2 and x/3, where, x > 0, is, 8, then the value of x is equal to, (a) 26, (b) 24, (c) 20, (d) 32, 16. Each observation of a raw data whose variance is σ 2,, is multiplied by n. What is the variance of the new, set?, (a) σ 2, (b) n 2 σ 2, σ2, (c) nσ 2, (d), n, , 17. In the following frequency distribution, class limits, of some of the class intervals and mid-value of a class, are missing. However, the mean of the distribution is, known to be 46.5., Class, Intervals, , Mid-values, , Frequency, , x1 − x2, , 15, , 10, , x2 − x3, , 30, , 40, , x3 − x4, , M, , 30, , x4 − x5, , 75, , 10, , x5 − 100, , 90, , 10, , The values of x1 , x2 , x3 , x4 , x5 , respectively will be, (a) (0, 20, 40, 60, 80), (b) (40, 50, 60, 70, 80), (c) (10, 20, 40, 70, 80), (d) (0, 19.5, 39.5, 69.5, 80), 18. Two variables X and U are related by the, relationship X = 5 + 2U. The mean and coefficient of, variation of X are 10 and 2.6, respectively. The, coefficient of variation of the variable U is, (a) 5.2, (b) 2.6, (c) 1.3, (d) 52, 19. If in a frequency distribution table with 12 classes,, the width of each class is 2.5 and the lowest class, boundary is 6.1, what is the upper class boundary of, the highest class?, (NDA 2007 II), (a) 30.1, (b) 27.6, (c) 30.6, (d) 36.1, 20. Students of three sections of a class, having 30, 30, and 40 students appeared for a test of 100 marks., The arithmetic means of the marks of the three, sections are 72.2, 69 and 64.1 in that order. What is, the arithmetic mean of the marks of all the students, of the three sections?, (NDA 2011 II), (a) 66.6, (b) 67.3, (c) 68, (d) 70.6, 21. If the variance of the data 2, 4, 5, 6 and 17 is v, then, what is the variance of the data 4, 8, 10, 12 and 34?, (a) v, (b) 4v, (c) v 2, (d) 2v, 22. Consider the following data, , Mean Wages of workers, Standard Deviation of Wages, , Factory, A, , Factory, B, , ` 540, , ` 620, , ` 40.50, , ` 31, , What is the variability in the wages of the workers in, Factory A?, (NDA 2010 II), (a) 100% more than the variability in the wages of, the workers in Factory B., (b) 50% more than the variability in the wages of the, workers in Factory B., (c) 50% less than the variability in the wages of the, workers in Factory B., (d) 150% more than the variability in the wages of, the workers in Factory B.
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646, , NDA/NA Mathematics, , 23. The standard deviation of some consecutive integers, is found to be 2. Which of the following statements, best describes the nature of the consecutive integers?, (NDA 2010 II), , (a) The integers are any set of eight consecutive, integers., (b) The integers are any set of eight consecutive, positive integers., (c) The integers are any set of seven consecutive, integers., (d) None of the above, 24. In a factory, there are 30 men and 20 women, employees. If the average salary of men is ` 4050 and, the average salary of all the employees is ` 3550,, what is the average salary of women? (NDA 2009 I), (a) ` 3800, (b) ` 3300, (c) ` 3000, (d) ` 2800, 25. The marks scored by two students A and B in six, subjects are given below ., (NDA 2009 I), A, , 71, , 56, , 55, , 75, , 54, , 49, , B, , 55, , 74, , 83, , 54, , 38, , 52, , Which one of the following statements is most, appropriate?, (a) The average scores of A and B are same but A is, consistent., (b) The average scores of A and B are not same but A, is consistent., (c) The average scores of A and B are same but B is, consistent., (d) The average scores of A and B are not same but B, is consistent., 26. Following table gives the mean and variance of, monthly demand for four products A, B, C and D in a, supermarket, (NDA 2009 II), Product, , A, , B, , C, , D, , Mean demand, , 60, , 90, , 80, , 120, , Variance, , 12, , 25, , 36, , 16, , For which product the demand is consistent?, (a) Product A, (b) Product B, (c) Product C, (d) Product D, 27., , Class, Interval, , 1-5, , 6-10, , 11-15, , 16-20, , Frequency, , 3, , 7, , 6, , 5, , Consider the following statements in respect of the, above frequency distribution., (NDA 2011 I), I. The median is contained in the modal class., II. The distribution is bell-shaped., Which of the above statements is/are correct?, , (a) Only I, (c) Both I and II, , (b) Only II, (d) Neither I nor II, , 28. The mean of 7 observations is 10 and that of, 3 observations is 5. What is the mean of all the 10, observations?, (NDA 2011 II), (a) 15, (b) 10, (c) 8.5, (d) 7.5, 29. Following is the frequency distribution of life lenght, in hours of 100 electric bulbs., Life length of bulbs, (in hours), , 8.5–3.5, 13.5–18.5, , 18.5–23.5, 23.5–28.5, 28.5–33.5, 33.5–38.5, , No. of bulbs, , 7, x, 40, y, 10, 2, , If the median of life length is 20 h, then what are the, missing frequencies (xy)?, (a) (28, 13), (b) (23, 18), (c) (31, 10), (d) (25, 16), 30. Consider the following statements, I. The data, which are collected from the unit or, individual respondents directly for the purpose of, certain study or information are known as, primary data., II. The data obtained in a census study are primary, data., Which of the above statements is/are correct?, (NDA 2009 II), , (a) Only I, (c) Both I and II, , (b) only II, (d) Neither I nor II, , Directions (Q. Nos. 31-32) Each of these questions, contain two statements, one is Assertion (A) and other is, Reason (R). Each of these questions also has four, alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 31. Assertion (A) Data collected in decennial censuses, are not statistical data., Reason (R) Since, no probability is involved in this, data collection, it amounts of 100% collection of, existing data., (NDA 2008 II)
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647, , Statistics, 32. Assertion (A) For the sets of numbers 12, 6, 7, 3, 15,, 20, 18, 5 and 9, 3, 8, 8, 9, 8, 9, 18 the range is equal to, 15 for both. Hence, dispersion is also same for the two, sets of numbers., Reason (R) In general, when extreme values are, present, the range is poor measure of dispersion., , Directions (Q. Nos. 33-34), , The following table, gives the continuous frequency distribution of a, continuous variable X., (NDA 2011 I), Class, Interval, , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , Frequency, , 5, , 10, , 20, , 5, , 10, , 33. What is the median of the above frequency, distribution?, (a) 23, (b) 24, (c) 25, (d) 26, 34. What is the, distribution?, (a) 25, (c) 27, , mean, , of, , the, , above, , frequency, , (b) 26, (d) 28, , Directions (Q. Nos. 35-39) Some data is kept on a, computer disk but unfortunately some of it is lost, because of a virus. Only the following could be recovered., Performance, Average, , Male, Female, Total, , Good, , Excellent Total, , 10, 32, 30, , Directions (Q. Nos. 40-42), , The frequency, distribution of life of 90 TV tubes whose median life is, 17 months is as follow., (NDA 2010 II), Life of TV tubes, , 0–5, 5–10, 10–15, 15–20, 20–25, 25-30, , 3, 12, x, 35, y, 4, , 40. What is the lower limit of the median class?, (a) 10, (b) 15, (c) 20, (d) 25, 41. What is the missing frequency Y?, (a) 20, (b) 16, (c) 15, , 35. How many male students are average?, (a) 10, (b) 16, (c) 30, (d) 32, 36. How many students is are both female and excellent?, (a) 0, (b) 8, (c) 16, (d) 32, 37. What proportion of good students are male?, (a) 0, (b) 0.73, (c) 0.4, (d) 1.0, 38. What proportion of female students are good?, (a) 0, (b) 0.25, (c) 0.5, (d) 1.0, 39. How many students are both male and good?, (a) 10, (b) 16, (c) 22, (d) 48, , (d) 12, , 42. What is the cumulative frequency of the modal class?, (a) 31, (b) 35, (c) 66, (d) Cannot be determined, , Directions (Q. Nos. 43-46) Study the pie chart, given below and answer the questions that follow., The following pie chart gives the distribution of funds in, a five year plan under the major heads of development, expenditures., Agriculture (A), Industry (I), Education (E),, Employment (Em) and Miscellaneous (M)., The total allocation is 72000 (in crore of rupees)., M, , An expert committee was formed, which decided that, the following facts were self evident., Half the students were either excellent or good., 40% of the students were females., One-third of the male students were average., , Number of TV, Tubes, , (In Months), , 70º, A 90º, , 50º I, 110º 40º, E, Em, , 43. Which head is allocated the maximum funds?, (a) Agriculture, (b) Employment, (c) Industry, (d) Miscellaneous, 44. How much money (in crore) is allocated to, Education?, (a) 6000, (b) 8000, (c) 10000, (d) 10800, 45. How much money (in crore) is allocated to both, Agriculture and Employment?, (a) 40000, (b) 36000, (c) 42000, (d) 44000, 46. How much excess money (in crore) is allocated to, Miscellaneous over Education?, (a) 8000, (b) 7500, (c) 6000, (d) 6500
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Answers, Level I, 1., 11., 21., 31., , (c), (a), (a), (c), , 2., 12., 22., 32., , (d), (b), (b), (c), , 3., 13., 23., 33., , (c), (c), (b), (c), , 4., 14., 24., 34., , (d), (b), (d), (d), , 5., 15., 25., 35., , (d), (c), (d), (d), , 6., 16., 26., 36., , (d), (d), (c), (b), , 7., 17., 27., 37., , (b), (d), (c), (a), , 8., 18., 28., 38., , (a), (b), (c), (d), , 9., 19., 29., 39., , (c), (c), (c), (b), , 10., 20., 30., 40., , (b), (b), (b), (b), , 2., 12., 22., 32., 42., , (c), (d), ( b), (b), (c), , 3., 13., 23., 33., 43., , (d), (b), (c), (c), (b), , 4., 14., 24., 34., 44., , (c), (c), (d), (b), (b), , 5., 15., 25., 35., 45., , (a), (b), (b), (b), (a), , 6., 16., 26., 36., 46., , (b), (b), (d), (a), (c), , 7., 17., 27., 37., , (b), (c), (d), (b), , 8., 18., 28., 38., , (b), (c), (c), (b), , 9., 19., 29., 39., , (b), (d), (c), (c), , 10., 20., 30., 40., , (a), (c), (c), (b), , Level II, 1., 11., 21., 31., 41., , (c), (b), (b), (a), (a), , Hints & Solutions, Level I, 1. Primary investigation is done by questionnaires., 2. The cumulative frequency curve of statistical data is, called ogive., , 9., , 3. The given observation, 3 times, , 5 times, , 20, 20, 20, , 21, 21, 21, 21, 21,, 5 times, , 7 times, 22, 22, 22, 22, 22, 22, 22,, 2 times, , 1 time, , 24, 24,, , 25, , 23, 23, 23, 23, 23, , Mode = Higher frequency of observation = 22 (7 times), 4. In statistics, a suitable graph for representing the, partitioning of total into subparts is a pie chart., 5. The volume of rainfall in certain geographical area, recorded every month for 24 consecutives months., 6., f x, f, x, 1, 2, 3, 4, , ∴, , 5, 8, 18, 4f, , Σf = 15 + f, , Σfx = 31 + 4 f, , Σfx 31 + 4 f, =, Σf, 15 + f, 31 + 4 f, 3=, 15 + f, , Mean =, , ⇒, ⇒, ⇒, , 5, 4, 6, f, , 45 − 31 = f, f = 14, 12 + 22 + 32 + K + n 2, (n + 1), n (n + 1)(2n + 1) 1, =, = n (2n + 1), 6(n + 1), 6, , 7. Mean =, , 8. The value 2 is repeated more than the other values of, variate X . Hence, mode = 2, , ∴, , xi, , 6, , 10, , 14, , 18, , 24, , 28, , 30, , fi, , 2, , 4, , 7, , 12, , 8, , 4, , 3, , Σ fi xi, Σ fi, (12 + 40 + 98 + 216 + 192 + 112 + 90), =, ( 2 + 4 + 7 + 12 + 8 + 4 + 3), 760, =, = 19, 40, , Mean =, , 10. The geometric mean of the given observation, , = (10 ⋅ 40 ⋅ 60)1/ 3, = ( 24 × 1000)1/ 3, = ( 3)1/ 3 ( 8 × 1000)1/ 3, = ( 3)1/ 3 ( 2 × 10), = 20, , 3, , 3, , 11. Firstly, arrange the given observation in ascending, order 2, 3, 4, 5, 6, 7 and 9., , Total terms, n = 7 (odd), ( 7 + 1), n + 1, th term, So, Median = , th term =, 2 , 2, = 4 th term = 5, 12. Given, n1 = 10, x1 = 12, n2 = 20, x2 = 9, n x + n2x2 10 × 12 + 20 × 9, =, x= 1 1, ∴, n1 + n2, 10 + 20, 120 + 180 300, =, =, = 10, 30, 30, 13. Total of corrected observations, = 4500 − (91 + 13) + (19 + 31), = 4446, 4446, Mean =, ∴, = 44.46, 100
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649, , Statistics, 14. Given, n = 20, x = 50, Σ x2 = 84000, 1 2, 84000, ∴, Σ x − ( x )2 =, σ=, − (50)2 = 1700, n, 20, , ⇒, , 22. The observations are 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95., Number of observations = 10, ∴ Median, 10, , 10, Value of, th term + Value of , + 1 th term, , 2, 2, =, 2, Value of 5th term + Value of 6th term, =, 2, x + x + 2 2(x + 1), =, =, =x+1, 2, 2, But, Median = 63, (given), ∴, x = 62, , Variance = σ 2 = 1700, , 15. Let the number of girls in the class = y, ∴ Number of boys in the class = 100 − y, Now,, x = 25, n1 = y, x2 = 50, n2 = 100 − y, and, x = 35, n1 + n2 = 100, 25 × y + 50 × (100 − y), 35 =, ∴, 100, ⇒, 3500 = 25 y + 5000 − 50 y, ⇒, 25 y = 1500, ⇒, y = 60, , 23. The two more observations added are 8 and 32., ∴ One observation, i.e., 8 is less than given median 30, and second observation, i.e., 32 is more than given, median 30., So, the median remain same, i.e., 30., , ∴ Number of girls in the class = 60, , 16. x =, , 22 + 26 + 28 + 20 + 24 + 30 150, =, = 25, 6, 6, xi, , 22, , 26, , 28, , 20, , 24, , 30, , xi − x, , –3, , 1, , 3, , –5, , –1, , 5, , ( xi − x)2, , 9, , 1, , 9, , 25, , 1, , 25, , n, , SD =, , ∑, , i =1, , ( xi − x )2, , n, = 11 .67, , =, , 70, 6, , = 3.42 (approx), 17. If each item of a data is increased or decreased by the, same constant, the standard deviation of the data, remains unchanged. i.e., SD is 6., 18. The given observations are arranged in ascending order, 2, 2, 3, 3, 4, 5, 7, 9, 10., Here, total term = 9 (ood), 9 + 1, 10, ∴ Median = , th term = th term, 2 , 2, = 5th term = 4, 19. Required geometric mean = (2 ⋅ 4 ⋅ 8 ⋅ 16 ⋅ 32)1/5, = (21 + 2+ 3 + 4 + 5 )1/5, = (215 )1/5, = 23 = 8, 2+9+9+3+6+9+4, 20. Mean =, 7, 42, =, =6 = X, 7, Σ [ X − X], ∴ Mean deviation =, n, |2 − 6|+ 3|9 − 6|+ |3 − 6|+ |6 − 6|+ |4 − 6|, =, 7, 4 + 9 + 3 + 0 + 2 18, =, =, = 2 . 57, 7, 7, 21. We know that, if a number is added in values, then the, standard deviation remains unaltered., ∴ SD of new values = σ, , 24. From the pie chart,, 25 1, =, 100 4, Let θ be the angle corresponding to industries., 1, Then, θ = × 360° = 90°, 4, , The share of industries = 25% =, , 25. If x is the mean of x1 , x2, K , xn , then the mean of, log x1 , log x2, K , log xn is not equal to log x , therefore,, option (d) is not correct., 26. We know that, the arithmetic mean of n natural, number with weights being the number itself, n (n + 1) (2n + 1), Σn 2, 6, =, n (n + 1), Σn, 2, n (n + 1) (2n + 1), 2, (2n + 1), =, ×, =, 6, n (n + 1), 3, For 16 natural numbers, put n = 16, 2 × 16 + 1, 33, =, =, = 11, 3, 3, 27. Given, the marks of 10 students out of 15 in the, ascending order are 40, 50, 60, 70, 70, 75, 80, 80, 90, 95, n + 1, Median = value of , ∴, th term, 2 , 15 + 1, = value of , th term, 2 , = value of 8 term = 60, , (Qfirst five students are failed ), 36 × 4 + 64 × 3, 28. Required mean =, 36 + 64, 144 + 192 336, =, =, = 3.36, 100, 100, 29. When, we take 12 wrongly in place of 8, then geometric, mean = 6, ⇒, (x1 ⋅ x2 ⋅ 12)1/3 = 6, ...(i), ⇒, x1 ⋅ x2 ⋅ 12 = 216 ⇒ x1 ⋅ x2 = 18, Now, we take the right observation 8 in place of 12, then, the geometric mean = (x1 ⋅ x2 ⋅ 8)1/3 = (18 ⋅ 8)1/3 = 2 3 18
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650, , NDA/NA Mathematics, , 30. Since, section C carried 51 average marks of 75., 51, ∴ Section C carried =, × 100 = 68 average marks out, 75, of 100., Average percentage marks, 35 × 74 + 35 × 70 + 30 × 68, =, 100, 2590 + 2450 + 2040, =, = 70.80, 100, 31. Let us consider any five integers be 3, 4, 5, 6 and 7., 25, Its mean =, =5, 5, (5 − 3)2 + (5 − 4)2 + (5 − 5)2 + (5 − 6)2 + (5 − 7)2, SD =, 5, 4+1+0+1+4, =, = 2, 5, σ, 2, 32. Month 1,, CV = × 100 =, × 100 = 6.67, X, 30, 3, Month 2,, CV =, × 100 = 5.26, 57, 4, Month 3,, CV =, × 100 = 4.88, 82, 2, Month 4,, CV =, × 100 = 7.14, 28, Hence, in month 3 the sales are most consistent., 33. Given, HM = 21.6 and a = 27, We know that,, 2ab, 2 × 27 × b, HM =, ⇒ 21 .6 =, a+b, 27 + b, ⇒, ⇒, , 583.2 = 54b − 21.6b, 583.2, b=, = 18, 32 .4, , 7 + 9 + 11 + 13 + 15 55, =, = 11, 5, 5, Now, SD, , 34. X =, , ( 7 − 11)2 + ( 9 − 11)2 + (11 − 11)2 + (13 − 11)2, + (15 − 11)2, , =, =, , 5, 16 + 4 + 0 + 4 + 16, = 8 = 2 .8, 5, , (approx), , 35. For the set of discrete numbers, the arithmetic mean,, median, geometric means all of them are meaningfully, defined., 1, 1, 1, 36. Harmonic mean =, = =∞, =, 1 1, 1, 1 1 1 1 0, +, − , , +, 3 3 (−6) (−6) 3 3 3, 37. Arithmetic mean of the series, n, C + nC 2 + nC3 + ...+ nC n 2n − 1, = 1, =, n, n, (Q2n = nC 0 + nC1 + ...+ nC n ⇒ 2n − 1 = nC1 + nC 2 + ...+ nC n ), 38. Let the teacher’s age is x yr., 20 × 15 + x, 15 + 1 =, 21, ⇒, , 16 × 21 = 300 + x, , ⇒, , x = 336 − 300 = 36 yr, , 39. Required expression is, , log G =, , n 2 log G1 + n1 log G2, n1 + n 2, , 40. The largest amount occupies the largest area. Thus, the, salaries occupied the largest area, 6, × 360° = 108°, ∴ Sectorial angle =, 20, , Level II, 1. The required weighted mean is, , n (n + 1)(2n + 1), 1 ⋅1 + 2 ⋅2 + 3 ⋅3 + K + n ⋅ n, 6, x=, =, n (n + 1), 1+2+3+K+ n, 2, , =, , 2n + 1, 3, , 2. Let x1 , x2, x3 , …, xn be n observations. Then,, 1 n, Σ xi, x=, n i =1, 1 n xi, 1 1 n 1, ∴New mean, x =, Σ + 10 = Σ xi + ⋅ (10n ), α n i =1 n, n i =1 α, , =, , x + 10 α, 1, x + 10 =, α, α, , 3. Let a and b be two observations, then, a + b, I. Arithmetic mean, AM = , , 2 , Now, we multiply by c in every observation, then, , ac + bc, a + b, = c⋅ , , 2 , 2, II. Geometric mean, GM = ab, Now, we taken ac and bc, then, Then,, GM = ac. bc = c . ab, 2ab, III. Harmonic mean, HM =, a+b, AM =, , Now, we take, ac and bc, then, 2(ac) . (bc), 2c2 ab, HM =, =, ac + bc, c (b + a ), 2ab , HM = c . , , a + b, IV. Median, (i) Let a , b, d, f and h be five observations, then, total terms = 4 + 1 = 5 = n, 5 + 1, n + 1, Median = , th term = d, term = , 2 , 2
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651, , Statistics, 1, n (n + 1)2, ⇒ 2n = n (n + 1) (2n + 1) +, 6, 4, , Now, we take, ac, bc, dc, fc and hc, then, 5 + 1, Median = , th term = 3rd term, 2 , = c. d, (ii) Let a, b, d and f be four observations, then total, number of terms = 4 = n, , n, n, Median = th term + + 1 th term, , 2, 2, = 2nd term + 3 rd term = (b + d), Now, we take, ac, bc, dc, and fc, then, , n, n, Median = th term + + 1 th term, , 2, 2, = 2nd term + 3rd term, = bc + dc = c . (b + d ), Hence, all four satisfy the property., , ⇒, ⇒, , n, , 4. Q Given, ∑ (xi − 2) = 110, i =1, , ∴, ⇒, , x1 + x2 + ... + xn − 2n = 110, x1 + x2 + ... + xn = 2n + 110, , ...(i), , n, , and, , ∑, , (xi − 5) = 20, , i =1, , ⇒, x1 + x2 + ... + xn − 5n = 20, ⇒, x1 + x2 + ... + xn = 5n + 20, From Eqs. (i) and (ii), we get, 5n + 20 = 2n + 110, 3n = 90, ⇒, n = 30, ⇒, x1 + x2 + ... + xn, Now, Mean =, n, 5 × 30 + 20 170 17, =, =, =, 30, 30, 3, , (given), , 30 + 23f = 224 + 24f, f=6, , 8. Standard deviation of the series B, 1, =, (1 .92 + 0 .82 + 1 .52 + 0 .62 + 0 .22), 5, 2, 1 .9 + 0 .8 + 1 .5 + 0 .6 + 0.2, −, , , , 5, 6.9, − 1 = 1 .38 − 1, 5, = 0 .38, , ...(ii), , 9. Collection data from government offices and public, libraries is secondary and from telephonic interview is, primary., 10. Here, statement III is wrong because construction of a, frequency distribution is based on data which are both, discrete as well as continuous., [from Eq. (ii)], , = ( a + md ), n (n + 1), 2, 6. Mean of the numbers =, n, n+1, x =, ⇒, 2, 2, , n (n + 1) , , , 2, , , , =, , 5. S = a + (a + d ) + (a + 2d ) + K + (a + 2md ), 2m + 1, =, [2a + 2md ], 2, = (2m + 1)(a + md ), (2m + 1)(a + md ), ∴ Required arithmetic mean =, (2m + 1), , 2, , n + 1, −2 , , 2 , 2n + 1 n + 1 n + 1 , ⇒ 2n = n (n + 1), +, −, 4, 2 , 6, 4n + 2 − 3n − 3 , 2 = (n + 1) , ⇒, , 12, , ⇒ 24 = (n + 1) (n – 1), ⇒ n 2 − 1 = 24 ⇒, n 2 = 25, n = ±5, ⇒, 2 ×1 + 3 ×2 + 3f + 4 ×5, 7. Arithmetic mean =, 2+3+ f +5, 23 28 + 3 f, =, ⇒, 8, 10 + f, , 2, , n + 1, n + 1, n + 1, , , , 1 −, + 2 −, + 3 −, + ..., , , , 2 , 2 , 2 , Variance =, n, 2=, 2, n + 1, n + 1, (12 + 22 + 32 + .. ) + n , [1 + 2 + 3 ... ], −2 , 2 , 2 , n, , 11. Required diagram is a bar chart., 12. Since, x is the mean of n observations x1 , x2,... , xn, x = x1 + x2 + x3 + ... + xn, ∴, n, Now, (a − b) is added to each term., x + (a − b) + x2 + (a − b) + ... + xn + (a − b), ∴New mean = 1, n, x1 + x2 + ...+ xn n (a − b), =, +, = x + (a − b), n, n, 13. Let the number of boys in class (n1) = x, and let the number of girls in class (n2) = y, The mean weight of all students (w12) = 60 kg, Mean weight of boys (w1 ) = 70 kg, Mean weight of girls (w2) = 55 kg, w n1 + w2 n2, w12 = 1, n1 + n2, 70x + 55 y, 60 =, ⇒, x+ y, ⇒, ⇒, ⇒, , 60 x + 60 y = 70 x + 55y, 10 x = 5 y, x 1, = ⇒x : y = 1 : 2, y 2, , (given), (given), (given)
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652, , NDA/NA Mathematics, , 14. Let the number of students of school I = x, ∴ Number of students of school II = 1.5x, Mean of marks of school I = 82, and Mean of marks of school II = 86, x × 82 + 1.5x × 86, Combined mean =, ∴, x + 1.5x, x(82 + 129) 211, =, =, = 84.4, 2 .5x, 2 .5, , 17. Here, given x = 46.5, , Total, , 15, 30, M, 75, 90, , fx, 150, 1200, 30 M, 750, 900, , 100 3000 + 30 M, , Σfx, Mean, x =, Σf, 3000 + 30M, ⇒, 46.5 =, 100, ⇒, 4650 = 3000 + 30M, 4650 − 3000, M=, ⇒, 30, 1650, ⇒, M=, 30, ⇒, M = 55, Then, the value of x5 + 100 = 90 × 2, ⇒, x5 = 180 − 100 = 80, Similarly, x1 , x2, x3 , x4 are 10, 20, 40, 70, 80., , X = 5 + 2U, X −5, 2, X = 10, σ X2 = 2 .6, , U =, , Now, coefficient of variation of X − 5 = Coefficient of, variation of X., , ⇒, , 16. Suppose, we have a raw data, i.e.,, x1 , x2, K , xk, 1 k, Then,, Σ (xi − x )2, σ2 =, k i =1, If each value is multiplied by n, then the values are, nx1 , nx2, …, nxk, The AM of the new values is, nx1 + nx2 + K + nxk, = nx, k, Therefore, the variance of the new set of values is, 1 k, , 1 k, Σ (nxi − nx )2 = n 2 Σ (xi − x )2, 1, i, =, k i =1, k, , , = n 2σ 2, , x1 − x2, x2 − x3, x3 − x4, x4 − x5, x5 − 100, , Mean, , ⇒, , Now, arranging the terms in the increasing order, i.e.,, x, x x x x, , , , , x, the third term =, 3, 5 4 3 2, Given,, median = 8, x, =8, 3, x = 24, , f, 10, 40, 30, 10, 10, , ⇒, , ∴, , 15. Here, number of terms = 5 (which is odd), 5 + 1, ∴ Median = value of , th term, 2 , = value of 3rd term, , Class Interval Mid-values ( x), , 18. Given,, , [Q variance does not depend on change of origin], σ 2X − 5 = σ 2X = 2 .6, σ 2X − 5 1 2 1, = σ X = × 2 .6, 2, 2, 2, 2 .6, 2, σU =, = 1.3, 2, , 19. Upper class boundary of the highest class, = 6.1 + (2.5 × 12), = 6.1 + 30, = 36.1, 20. Let A, B and C be the sections of a class having 30, 30, and 40 students, respectively., Now, the total marks secured by the students of, section A = 30 × 72 . 2 = 2166, The total marks secured by the students of section, B = 30 × 69 = 2070, and the total marks secured by the students of section, C = 40 × 64.1 = 2564, So, the arithmetic mean of marks of all the students of, three sections, (2166 + 2070 + 2564), =, 100, 6800, =, = 68, 100, 21. We know that, var (λx) = λ2 var (x), Given data is, x = 2, 4, 5, 6, 17 and its variance,., var (x) = v, Now, multiply by 2 in above data numbers,, x = 4, 8, 10, 12, 34, Its variance, var (2x) = (2)2 var (x) = 4 v, , ...(i), , 22. The variability in the wages of the workers in factory A, is 50% more than the variability in the wages of the, workers in Factory B., SD, × 100, ∴ Coefficient of variation =, Mean, 40.50, For, factory A,, × 100 = 7.5, 540, 31, and for factory B,, × 100 = 5, 620, 23. Since, the standard deviation of some consecutive, integers is 2, then these integers are any set of seven, consecutive integers., 24. n = 50, x = 3550, n1 = 30, x1 = 4050 and n2 = 20, We know that, nx = n1x1 + n2x2, ⇒, 50 × 3550 = 30 × 4050 + 20x2, ⇒, 177500 − 121500 = 20x2, ⇒, x2 = 2800, Hence, average salary of women = ` 2800
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653, , Statistics, 25. Average of marks of A, 71 + 56 + 55 + 75 + 54 + 49 360, =, =, = 60, 6, 6, 121 + 16 + 25 + 225 + 36 + 121, 544, and SD =, =, = 9.52, 6, 6, Also, average of marks of B, 55 + 74 + 83 + 54 + 38 + 52 356, =, =, = 59.33 ≅ 59, 6, 6, 16 + 225 + 576 + 25 + 441 + 49, and, SD =, 6, 1332, =, = 222 = 15, 6, 9.52, Now, CV A =, × 100 = 15.87, 60, 15, and,, CVB =, × 100 = 25.42, 59, , and the mean of 3 observations = 5, 3, , 25 5, =, = 0.055, 90 90, , 7, , ∑, , i =1, , 16, 4, =, = 0.033, 120 120, , Class Interval, , f, , cf, , 0.5–5.5, 5.5–10.5, 10.5–15.5, 15.5–20.5, , 3, 7, 6, 5, , 3, 10, 16, 21, , 21, , 50, , 3, , 10, , xi + ∑ xi = 70 + 15 ⇒ ∑ xi = 85, i =1, , i =1, , ∴Mean of 10 observations =, , ∑, , xi, , i =1, , =, , 10, , 85, = 8 .5, 10, , 29. Let x = 31 and y = 10, then, CI, , x, , f, , cf, , 8.5–13.5, , 11, , 7, , 7, , 13.5–18.5, , 16, , 31, , 38, , 18.5–23.5, , 21, , 40, , 78, , 23.5–28.5, , 26, , 10, , 88, , 28.5–33.5, , 31, , 10, , 98, , 33.5–38.5, , 36, , 2, , 100, , N, = 50, 2, , N = 100, ∴, , Median group is 18.5–23.5, ∴, , Hence, we see that minimum coefficient of variance is, D, hence product is consistent., 27., , ....(ii), , i =1, , 10, , Q, , 36 6, Coefficient of variance C =, =, = 0.075, 80 80, Coefficient of variance D =, , 3, , 3, , = 5 ⇒ ∑ xi = 15, , On adding Eqs. (i) and (ii), we get, , 12 3.46, =, = 0.057, 60, 60, , Coefficient of variance B =, , xi, , i =1, , ⇒, , Thus, the average scores of A and B are not same but A, is consistent., SD, 26. Since, coefficient of variance =, Mean, Coefficient of variance A =, , ∑, , L1 = 18.5 , L 2 = 23.5 , C = 38, h = 5 , f = 40, 50 − 38, Median = 18.5 +, ×5, 40, 12 × 5, = 18.5 +, = 18.5 + 1 .5 = 20, 40, , Thus, our assumption is correct. Therefore, missing, numbers are 31 and 10, respectively., 30. Both statements are true., 31. A and R are correct, R is the correct explanation of A., 32. Dispersion is the limit of the deviation of the terms, about their mean., Dispersions which are mostly used are range, quartile, deviation, mean deviation, standard deviation., , N = 21, N 21, =, = 10.5, 2, 2, , Solutions (Q. Nos. 33-34), , Q Median class is 10.5 – 15.5, 10.5 − 10, Median = 10.5 +, ×5, 6, = 10.5 + 0.417 = 10.917, Thus, median is not contained in the modal class and, the distribution is not bell-shaped., Mean ≠ Median ≠ Mode, ∴, 28. Given, mean of 7 observations = 10, , Class, Interval, , f, , cf, , x, , fx, , 0-10, , 5, , 5, , 5, , 25, , 10-20, , 10, , 15, , 15, , 150, , 20-30, , 20, , 35, , 25, , 500, , 30-40, , 5, , 40, , 35, , 175, , 40-50, , 10, , 50, , 45, , 450, , 50, , 145, , 125, , 1300, , 7, , ⇒, , ∑ xi, , i =1, , 7, , = 10 ⇒, , 7, , ∑, , i =1, , xi = 70, , ...(i), , N 50, =, = 25, 2, 2
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654, , NDA/NA Mathematics, N = 90, N, = 45, 2, , 33. Median group is 20-30., 25 − 15, Median = 20 +, ⇒, × 10 = 20 + 5 = 25, 20, Σfx 1300, 34. Mean =, =, = 26, Σf, 50, , 40. Lower limit of median class is 15., 41. Given that, median = 17, , Solutions (Q. Nos. 35-39), Average, , Good, , Excellent, , Total, , Male, Female, , 16, 24, , 22, 8, , 10, —, , 48, 32, , Total, , 40, , 30, , 10, , 80, , 35. ∴, , 40% = 32, 32 × 100, Q 100% =, = 80, 40, ∴Number of boys = 80 − 32 = 48, 1, ∴Average male students = × 48 = 16, 3, , ⇒, ⇒, ⇒, ∴, ⇒, , , N, −C, 2, ×h, , M = l+, f, {45 − (15 + x)}, 17 = 15 +, ×5, 35, (30 − x), 17 = 15 +, 7, 119 = 105 + 30 − x, x = 16, x + y = 36, y = 36 − 16 = 20, , Unknown frequency = 20, , 36. 50% of the total student, i.e., 40 student were either, excellent or girl., ∴ No girl was excellent., 37. Number of males who are good = 48 − 16 − 10 = 22, 22, = 0.73, ∴ Required proportion =, 30, 38. Proportion of female students who are good, 8, =, = 0.25, 32, 39. Number of students who are both male and good, = 48 − 16 − 10 = 22, , Solutions (Q. Nos. 40-42), Class Interval, , Frequency, , cf, , 0–5, 5–10, 10–15, 15–20, 20–25, 25–30, , 3, 12, x, 35, y, 4, , 3, 15, 15 + x, 50 + x, 50 + x + y, 54 + x + y, , 42. Cumulative frequency of modal class, = 50 + x = 50 + 16 = 66, , Solutions (Q. Nos. 43-46), 43. Employment is allocated maximum funds., 44. Money allocated to education, 40°, =, × 72000, 360°, = 8000, 45. Money allocated to both Agriculture and Employment., 90° + 110°, =, × 72000, 360°, 200°, =, × 72000 = 40000, 360°, 70° − 40°, 46. Required value of money =, × 72000, 360°, 30°, =, × 72000, 360°, = 6000
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31, , Correlation, and Regression, Bivariate Frequency, Distribution, In a bivariate frequency distribution, we have two, variables of observation on which values are recorded for, each unit of observation. The distributions having more, than two variables of observation are called multivariate, distributions., Here, we confine ourselves to the study of bivariate, distribution in which two variables may be inter, dependent., , Correlation, The tendency of simultaneous variation between two, variables is called correlation. It denotes the degree of inter, dependence between variables., , Covariance, , Σ ( xi − x ) ( yi − y ), Σ ( xi − x )2 ⋅ Σ ( yi − y )2, , Modified Formula,, r=, where,, , n Σdxdy − Σdx Σdy, nΣdx − ( Σdx )2, 2, , nΣdy 2 − ( Σdy )2, , dx = x − x ,, dy = y − y, , Characteristics of Correlation, Coefficient, 1. − 1 ≤ r ≤ 1., 2. If r = 1, the correlation is perfect and positive., 3. If r = − 1, the correlation is perfect and negative., 4. The coefficient of correlation is a pure number, it is, independent of the unit of measurement., , The covariance between two variables x and y with n, pairs of observations ( x1 , y1 ), ( x2 , y2 ), .... , ( xn , yn ) is defined, as, Σ ( xi − x ) ( yi − y ), cov( x , y ) = σ xy =, n, , Σx y, = i i − xy, , n, where,, , =, , x=, , Σx i, Σy, and y = i, n, n, , Karl Pearson’s Correlation, Coefficient, The correlation coefficient r( x , y ), between two, variables x and y is given by, cov ( x , y ), r=, var ( x ) ⋅ var ( y ), , 5. The coefficient of correlation is independent of the, change of origin and scale., 6. If 0 < r < 1,there is a positive correlation between x, and y., 7. If − 1 < r < 0, there is a negative correlation, between x and y., 8. If r = 0, then x and y are said to be uncorrelated., , Example 1. Karl Pearson’s coefficient of correlation, between x and y for the data, x, , 65, , 66, , 67, , 67, , 68, , 72, , y, , 67, , 68, , 65, , 68, , 72, , 71, , is, , (a) 0.62, (c) 1, , (b) 0.82, (d) None of these
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656, , NDA/NA Mathematics, , Solution (a) Change the variables as ui = xi − 67 and vi = yi − 68,, we get, ui, , vi, , u i2, , v i2, , u iv i, , −2, , −1, , 4, , 1, , 2, , −1, , 0, , 1, , 0, , 0, , −3, , 0, , 9, , 0, , 0, , 0, , 0, , 0, , 0, , 1, , 4, , 1, , 16, , 4, , 5, , 3, , 25, , 9, , 15, , Σui2 = 31, , Σvi2 = 35, , Σuv, i i = 21, , Σvi = 3, , The term regression means “stepping back towards, the average”., , Lines of Regression, , n Σ uv, i i − ( Σui ) ( Σvi ), , r=, , [n Σ ui2 − ( Σui ) 2] [n Σvi2 − ( Σvi ) 2], 6 (21) − (3) (3), 117, =, =, = 0.62, 2, 2, × 201, 177, [6 (31) − (3) ] [6 (35) − (3) ], , The line of regression is the line which gives the best, estimate to the value of one variable for any specific value, of the other variable., , Equations of Regression Lines, The line of regression of y on x is, σy, (x − x ), y− y =r, σx, , Coefficient of Rank, Correlation, This formula is applied to the problems in which data, can not be measured quantitatively but qualitative, assessment is possible. In this case, the best individual is, given rank number 1, next rank 2, and so on. The, coefficient of rank correlation is given by the formula, 6 Σdi2, R=1−, n ( n 2 − 1), where, di is the difference of corresponding rank, and n is the number of pairs of observations., , Example 2. Eight students were ranked according to their, grades in both the practicals and theory examination of a, Physics course as given below, Practicals, , 4, , 6, , 3, , 5, , 7, , 2, , 8, , 1, , Theory, , 5, , 4, , 6, , 2, , 8, , 1, , 7, , 3, , Then, coefficient of rank correlation is, (a) 5 / 14, (b) 7 / 14, (c) 9 / 14, , 6 × 30, 6 Σdi2, =1−, 8 × 63, n (n 2 − 1), 5, 9, =1−, =, 14 14, , R =1−, , Regression, , 0, , Σui = 3, , ∴, , The line of regression of x on y is, σ, x − x = r x ( y − y), σy, , Properties of Regression Lines, 1. The two lines of regression pass through the point, ( x , y )., 2. Slope of the line of regression of y on x = byx ., 1, ., 3. Slope of the line of regression of x on y =, bxy, , Regression Coefficient, ‘b’ the slope of the line of regression of y on x is also, called the coefficient of regression of y on x., σy, Regression coefficient of y on x = byx = r, ., σx, Similarly,, , (d) 11/ 14, , Regression coefficient of x on y = bxy = r, , Solution (c), Practicals, , Theory, , di, , d i2, , 4, , 5, , 1, , 6, , 4, , −1, 2, , 3, , 6, , 5, , 2, , −3, 3, , 4, 9, 9, , 7, , 8, , −1, , 1, , 2, , 1, , 1, , 8, , 7, , 1, 1, , 1, , 3, , −2, , 1, 4, Σdi2 = 30, , σx, ., σy, , Properties of Regression, Coefficient, 1. Both regression coefficients and r have the same, sign., 2. Coefficient of correlation is the geometric mean, between the regression coefficients, r=±, , b yx × bxy, , 3. If one of the regression coefficients is greater than, unity, the other must be less than unity.
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657, , Correlation and Regression, 4. The modulus value of the arithmetic mean of the, regression coefficients is not less than the, modulus value of the correlation coefficient r. i.e.,, byx + bxy, 2, , Solution (b), , >|r|, , 5. Regression coefficients are independent of the, change of origin but not of scale., , Angle between Two Lines of, Regression, Ifθ be the angle between the two regression lines, then, (1 − r 2 ) σ x σ y , , ⋅ 2, tan θ = ±, σ + σ2 , |r|, x, y, π, , i. e. , if two variables are, 2, uncorrelated, then the lines of regression are, perpendicular to each other., If r = ± 1, then θ = 0 or π, i. e. , in the case of perfect, correlation, positive or negative the two lines of regression, coincide., If r = 0, then θ =, , Example 3. The equations of the lines of regression on the, basis of the following data, x, , 4, , 2, , 3, , 4, , 2, , y, , 2, , 3, , 2, , 4, , 4, , (a) x + y = 15, y + 4x = 15, (b) x + 4y = 15, 4x + y = 15, (c) x + 4y = 15, x + y = 15, (d) None of the above, , y2, , xy, , 16, , 4, , 8, , 4, , 9, , 6, , x, , y, , x2, , 4, , 2, , 2, , 3, , 3, , 2, , 9, , 4, , 6, , 4, , 4, , 16, , 16, , 16, , 2, , 4, , 4, , 16, , 8, , Σx = 15, , Σy = 15, , Σx = 49, , Σy = 49, , Σxy = 44, , 2, , n Σxy − ( Σx) ( Σy), n Σy 2 − ( Σy) 2, (5 × 44) − (15 × 15), =, 5 ( 49) − (15) 2, 220 − 225, 5, 1, =, =−, =−, 245 − 225, 20, 4, n Σxy − ( Σx) ( Σy), and, byx =, n Σx2 − ( Σx) 2, (5 × 44) − (15 × 15), =, 5 × 49 − (15) 2, 1, =−, 4, 15, Also,, x=, =3, 5, 15, and, y =, =3, 5, Lines of regression are, 1, y − 3 = − ( x − 3), 4, ⇒, x + 4y = 15, 1, and, x − 3 = − (y − 3), 4, ⇒, 4x + y = 15, , ∴, , bxy =, , 2
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Exercise, Level I, 1. When the correlation between two variables is, perfect, then the value of coefficient of correlation r is, (a) − 1, (b) + 1, (c) 0, (d) ± 1, , (c) Correlation coefficient is the harmonic mean of, the regression coefficient., (d) None of the above, , 2. If for the variables x and y, the two regression lines, are 3x + 2 y − 25 = 0 and 6x + y − 30 = 0, then the, coefficient of correlation r is equal to, (a) 0.5, (b) − 0.5, (c) 0.6, (d) − 0.6, , 11. If the correlation coefficient between X and Y is 0.7,, what is the correlation coefficient between, 3Y − 4, ?, U = 4X + 3 and V =, 2, (a) 0.6, (b) 0.7, (c) 0.8, (d) 1, , 3. If Σxi = 15, Σyi = 36, Σxi yi = 110 and n = 5, then cov, ( x , y ) is equal to, (a) 0.8, (b) 0.6, (c) 0.4, (d) 0.2, 4. The point of intersection of the regression lines, x + 2 y − 5 = 0 and 2x + 3 y − 8 = 0 will always be at, (a) mean, (b) median, (c) mode, (d) standard deviation, 5. If the regression equations of the variables x and y be, then, the, x = 1913, . − 0.83 y and y = 11.64 − 0.50x,, correlation coefficient between x and y is, (a) 0.66, (b) − 0.64, (c) 0.001, (d) − 0.001, 6. If var ( x ) = 8.25, var ( y ) = 33.96 and cov ( x , y ) = 10.2,, then the correlation coefficient is, (a) 0.89, (b) − 0.98, (c) 0.61, (d) − 016, ., 7. Consider the following data, x, , 5, , 7, , 8, , 4, , 6, , y, , 2, , 4, , 3, , 2, , 4, , What is the regression equation of y on x?, (NDA 2010 II), , (a) y = 0.6 + 0.4x, (c) y = 6 + 5x, , (b) y = 0.7 + 0.3x, (d) y = 4 + 9x, , 8. If the slopes of the line of regression of Y on X and of, X on Y are 30 and 60 respectively, then r( x , y ) is, (NDA 2007 I), , (a) − 1, , (b) 1, , (c), , 1, 3, , (d) −, , 1, 3, , 9. If the equation of the lines of regression of y on x and, that of x on y be y = ax + b and x = cy + d respectively,, then x and y are equal to, respectively, ab + c cd + a, bc + d ad + b, (b), (a), ,, ,, 1 − ab 1 − ad, 1 − ac 1 − ac, ad + c cd + d, (c), (d) None of these, ,, 1 − bc 1 − bc, 10. Which of the following statements is correct?, (a) Correlation coefficient is the arithmetic mean of, the regression coefficient., (b) Correlation coefficient is the geometric mean of, the regression coefficient., , 12. If x and y are independent variables, what is the, value of regression coefficient of y on x ?, (a) 0, (b) 1, (c) − 1, (d) ∞, 13. Two variables x and y are related by the linear, equation ax + by + c = 0. The coefficient of correlation, between the two is + 1, if, (a) a is positive, (b) b is positive, (c) a and b both are positive, (d) a and b are of opposite sign, 14. For the given data, the calculation corresponding to, all, values, of, pairs, is, following, (x, y), Σ ( x − x )2 = 36, Σ ( y − y )2 = 25, Σ ( x − x ) ( y − y ) = 20,, then the Karl Pearson's correlation coefficient is, (a) 0.2, (b) 0.5, (c) 0.66, (d) 0.33, 15. Two lines of regression are 3x + 4 y − 7 = 0 and, 4x + y − 5 = 0. Then, correlation coefficient between x, and y is, 3, 3, 3, 3, (c), (d) −, (b) −, (a), 4, 4, 16, 16, 16. Let X , Y be two variables with correlation coefficient, ρ( X , Y ) and variables U , V be related to X , Y by the, relation U = 2X , V = 3Y , then ρ(U , V ) is equal to, (a) ρ( X , Y ), (b) 6ρ( X , Y ), 3, (c) 6 ρ( X , Y ), (d) ρ( X , Y ), 2, 17. If the two lines of regression are 4x + 3 y + 7 = 0 and, 3x + 4 y + 8 = 0, then the means of x and y are, 4, 11, 4 11, (b) − ,, (a) − , −, 7, 7, 7 7, 4, 11, (c) , −, (d) 4, 7, 7, 7, 18. The ranks of some 10 students in Mathematics and, Physics are as follows. Two numbers within brackets, denote the ranks of the students in Mathematics and, Physics. (1, 1) (2, 10) (3, 3) (4, 4) (5, 5) (6, 7) (7, 2) (8, 6), (9, 8) (10,11). The rank correlation coefficient is, (a) 0.40, (b) 0.42, (c) 0.38, (d) 0.48.
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659, , Correlation and Regression, , Level II, 1. If byx and bxy are both positive ( where byx and bxy are, regression coefficients), then, 1, 1, 2, 1, 1, 2, (a), (b), +, <, +, >, byx bxy r, byx bxy r, 1, 1, r, (d) None of these, (c), +, <, byx bxy 2, 2. For, the, following, pairs, of, observations, (1, 10), ( 2, 9), ( 3, 9), ( 4, 8), ( 5, 6), ( 6, 12), ( 7, 4), ( 8, 3),, ( 9, 13), (10, 1), cov ( x , y ) is equal to, (a) 1.05, (b) 2.45, (c) − 4.45, (d) − 4.9, 3. In a partially destroyed laboratory record of an, analysis of correlation data, the following results are, legible, Variance of x = 9 and Regression equations are, 8x − 10 y + 66 = 0 and 40x − 18 y = 214, On the basis of above information, the coefficient of, correlation is, (a) 2/5, (b) 1/2, (c) 3/5, (d) 2/3, 4. In an experiment two variables X and Y are observed, on same units. It was recorded that σ X = 2σY . Which, one of the following statements is correct?, (a) The regression coefficient of Y on X is four times, the regression coefficient of X on Y, (b) The regression coefficient of Y on X is equal to the, regression coefficient of X on Y, (c) The regression coefficient of X on Y is four times, the regression coefficient of Y on X, (d) The value of correlation coefficient exceeds unity, 5. If X and Y are change into a + hU , b + kV, respectively, which of the correct relation between, the correlation coefficient bxy and bUV ? (NDA 2007 1), (b) kbXY = hbUV, (a) hbXY = kbUV, (d) k2bXY = h 2bUV, (c) bXY = bUV, 6. Consider the following statements, (NDA 2012 I), I. Two independent variables are always, uncorrelated., II. The coefficient of correlation between two, varibles X and Y is positive when X decrease,, then Y decreases., Which of the above statements is/are correct?, (a) Only I, (b) Only II, (c) Both I and II, (d) Neither I nor II, 7. In a study on the relationship between investment, ( X ) and profit (Y ), the following two regression, equations were obtained based on the data on X and, Y., 3X + Y − 12 = 0, X + 2Y − 14 = 0, What is the mean X ?, (a) 6, (b) 5, , (NDA 2009 II), , (c) 4, , (d) 2, , 8. a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0 are the lines of, regression of Y on X and X on Y, respectively. Which, one of the following is correct?, (a) a1a2 ≤ b1b2, (b) a1b2 ≤ a2b1, (d) b1b2 ≤ a1a2, (c) a2b1 ≤ a1b2, 9. If x1 and x2 are regression coefficients and r is the, coefficient of correlation, then, (b) x1 + x2 < r, (a) x1 − x2 > r, (d) None of these, (c) x1 + x2 ≥ 2r, 10. The two lines of regression are 8x − 10 y = 66 and, 40x − 18 y = 214 and variance of x series is 9. What is, the standard deviation of y series?, (NDA 2010 II), (a) 3, (b) 4, (c) 6, (d) 9, 11. Consider the following statements with regard to, correlation coefficient r between random variables x, and y., I. r = + 1 or − 1 means there is a linear relation, between x and y., II. − 1 ≤ r ≤ 1 and r 2 is a measure of the linear, relationship between the variables., Which of the statements given above is/are correct?, (a) Only I, (b) Only II, (NDA 2011 I), (c) Both I and II, (d) Neither I nor II, 12. Which of the following two sets of regression lines are, the true representative of the information from the, bivariate population?, I. x + 4 y = 15 and y + 3x = 12, x = 3, y = 3 ., 5, 30, ., II. 3x + 4 y = 9 and 4x + y = 1, x = −, ,y=, 10, 13, (a) Only I, (b) Only II, (c) Both I and II, (d) None of these, 13. If the two lines of regression are x + 4 y = 3 and, 3x + y = 15, then the value of x for y = 3 is, (a) − 4, (b) − 9, (c) 4, (d) 5, , Directions (Q. Nos. 14-16) Each of these, questions contain two statements, one is Assertion (A), and other is Reason (R). Each of these questions also has, four alternative choices, only one of which is the correct, answer. You have to select one of the codes (a), (b), (c) and, (d) given below., Codes, (a) Both A and R are individually true and R is the, correct explanation of A., (b) Both A and R are individually true but R is not, the correct explanation of A., (c) A is true but R is false., (d) A is false but R is true., 14. Assertion (A) We cannot find out the regression, equation of x on y from that of y on x ., Reason (R) In one equation x is dependent variable, and y is independent whereas in other equation y is, dependent variable and x is independent.
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660, , NDA/NA Mathematics, , 15. Assertion (A) If bxy is negative, then byx and r are, positive., Reason (R) Both regression coefficients and r have, the same sign., 16. Assertion (A) If one regression coefficient is greater, than 1, then the other will be less than one., Reason (R) The square root of the product of two, regression coefficients is the correlation coefficient, between the variables and the correlation coefficient, lies between − 1 and+ 1., , Directions (Q. Nos. 17-19), , For the following, , data, Mean, Standard deviation, Correlation coefficient, , x, 65, 5.0, 0.8, , y, 67, 2.5, , 17. The equation of line of regression of y on x is, 2, (a) y − 67 = ( x − 65), 5, 1, (b) y − 67 = ( x − 65), 5, 2, (c) x − 65 = ( y − 67), 5, 1, (d) x − 65 = ( y − 67), 5, 18. The equation of line of regression of x on y is, 4, 4, (b) x − 67 = ( y − 65), (a) x − 65 = ( y − 67), 5, 5, 8, 8, (c) x − 65 = ( y − 67), (d) x − 67 = ( y − 65), 5, 5, 19. Angle between the both lines of regression is, (b) tan− 1( 0.9), (a) tan− 1(1.8), −1, (d) tan− 1(1 .6), (c) tan ( 0.8), , Answers, Level I, 1. (d), 11. (b), , 2. (b), 12. (a), , 3. (c), 13. (d), , 4. (a), 14. (c), , 5. (b), 15. (b), , 6. (c), 16. (a), , 7. (a), 17. (a), , 8. (c), 18. (b), , 9. (b), , 10. (b), , 2. (c), 12. (c), , 3. (c), 13. (c), , 4. (c), 14. (a), , 5. (b), 15. (d), , 6. (a), 16. (a), , 7. (d), 17. (a), , 8. (b), 18. (c), , 9. (c), 19. (a), , 10. (b), , Level II, 1. (b), 11. (c)
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Hints & Solutions, Level I, 1. When correlation is perfect, then coefficient of, correlation is ± 1., 2. Given, regression lines are, 3x + 2 y − 25 = 0, and, 6x + y − 30 = 0, From Eq. (i), we have, 2 y = − 3x + 25, −3x 25, y=, +, 2, 2, −3, ∴, byx =, 2, From Eq. (ii), we have, 6x = − y + 30, −1, ⇒, x=, y+5, 6, 1, bxy = −, ∴, 6, r = − byx ⋅ bxy, ∴, , 7., , y2, , xy, , 25, , 4, , 10, , 49, , 16, , 28, , 3, , 64, , 9, , 24, , 2, , 16, , 4, , 8, , 6, , 4, , 36, , 16, , 24, , Σx = 30, , Σy = 15, , Σx2 = 190, , Σy2 = 49, , Σxy = 94, , ...(i), ...(ii), , 3. We have,, Σxi = 15, Σyi = 36, Σ xi yi = 110 and n = 5, We know that, (Σxi yi ), cov(x, y) =, −xy, n, x, 1, 15 36, , ⇒, cov(x, y) = (110) − , Q x = n , 5 5, 5, ⇒, cov(x, y) = 22 − 21.6 = 0.4, 4. The point of intersection of the regression lines passes, through mean., 5. Let us assume that line of regression of x on y is, x = − 0.83 y + 19.13 and that of y on x is, y = − 0.50x + 11.64, ⇒, bxy = − 0.83 and byx = − 0.50, Since, both regression coefficients are negative, ∴, , r = − bxy × byx, , ⇒, , r = − 0.83 × 0.50 = − 0.64, , 6. Given, var (x) = 8 .25, var ( y) = 33.96 and cov(x, y) = 10 .2, cov(x, y), 10.2, =, r=, ∴, var(x) ⋅ var( y), (8.25) ⋅ (33.96), =, , 10.2, 10.2, =, 280.17 16.74, , = 0.61, , y, , x2, , 5, , 2, , 7, , 4, , 8, 4, , 30, =6, 5, 15, y=, =3, 5, nΣxy − (Σx) (Σy), byx =, nΣx2 − (Σx)2, 5 × 94 − 30 × 15, =, 5 × 190 − (30)2, 470 − 450 20 2, =, =, = = 0.4, 950 − 900 50 5, x=, , and, ∴, , [Q bxy , byx < 0, so we take r negative], 3 1, =−, ×, 2 6, 1, =−, = − 0.5, 4, , x, , Hence, line of regression is, y − 3 = 0.4 (x − 6), ⇒, y = 0.4x + 3 − 2 .4, ⇒, y = 0.4x + 0.6, 1, 8. byx = tan 30° =, 3, 1, and, = tan 60° = 3, bxy, ∴, ⇒, , 1, 3, 1, 2, r =, 3, , bxy ⋅ byx =, , ⇒, , r=+, , 1, 3, , As, byx and bxy are positive., 9. The lines of regression are y = ax + b and x = cy + d., Since, the lines of regression passes through (x , y), …(i), y = ax + b, ⇒, …(ii), and, x = cy + d, On solving Eqs. (i) and (ii), we get, bc + d, ad + b, x=, ,y=, 1 − ac, 1 − ac, 10. Correlation coefficient is the geometric mean of the, regression coefficient., r = bxy ⋅ byx, 11. Since, correlation coefficient is independent of change of, origin and scale., ∴ It is remain same i.e., 0.7., 12. If x and y are independent variables, then value of, regression coefficient of y on x is 0.
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662, , NDA/NA Mathematics, , 13. The given relation is ax + by + c = 0, a, c, ⇒, by = − ax − c ⇒ y = − x −, b, b, Since, r = 1, if the slope of relation is positive., ∴ a and b are of opposite sign., , 16. Since, correlation coefficient is independent of change of, origin., ρ (U , V ) = ρ (X ,Y ), , 14. Given that,, Σ (x − x )2 = 36, Σ ( y − y)2 = 25, Σ (x − x ) ( y − y) = 20, Σ (x − x ) ( y − y), 20, 2, ∴, r=, =, =, 36 ⋅ 25 3, Σ (x − x )2 Σ ( y − y)2, ⇒, , r = 0.66, , 15. Let us assume that lines of regression of y on x is, 3x + 4 y − 7 = 0 and x on y is 4x + y − 5 = 0., or it can be rewritten as, 3x 7, y 5, and x = − +, y=−, +, 4 4, 4, 4, 3, 1, and bxy = −, ⇒, byx = −, 4, 4, r = − byx . bxy, ∴, 3 1, = − − − , 4 4, ⇒, , r=−, , 3, 4, , 17. Since, lines of regression pass through the means (x , y)., Hence, the equation will be 4 x + 3 y = − 7, and, 3x + 4 y = − 8, On solving both equations, we get, 4, 11, x=− ,y=−, 7, 7, 18., Rank in, Maths (X), , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , Rank in, Physics (Y), , 1, , 10, , 3, , 4, , 5, , 7, , 2, , 6, , 8, , 11, , d = X −Y, , 0, , –8, , 0, , 0, , 0, , –1, , 5, , 2, , 1, , –1, , 0, , 64, , 0, , 0, , 0, , 1, , 25, , 4, , 1, , 1, , 2, , d, , Total, , 96, , 6 Σd, n (n 2 − 1), 6 × 96, 6 × 96, =1 −, =1 −, 10 (100 − 1), 10 × 99, 32 23, =1 −, =, = 0.42, 55 55, , Rank correlation coefficient, R = 1 −, , 2, , Level II, Σ (x − x ) ( y − y), 44.5, =−, n, 10, = − 4.45, , 1. We know that,, AM ≥ GM, ⇒, , byx + bxy > 2 byx ⋅ bxy, , ⇒, , byx + bxy > 2r, byx + bxy 2, >, r, r2, byx + bxy 2, >, byx ⋅ bxy, r, , ⇒, ⇒, , 2., , cov(x, y) =, , ⇒, , 1, 1, 2, +, >, bxy byx r, , x, , y, , x−x, , y− y, , ( x − x) ( y − y), , 1, , 10, , –4.5, , 2.5, , –11.25, , 2, , 9, , –3.5, , 1.5, , –5.25, , 3, , 9, , –2.5, , 1.5, , –3.75, , 4, , 8, , –1.5, , 0.5, , –0.75, , 5, , 6, , – 0.5, , –1.5, , 0.75, , 6, , 12, , 0.5, , 4.5, , 2.25, , 7, , 4, , 1.5, , –3.5, , –5.25, , 8, , 3, , 2.5, , –4.5, , –11.25, , 9, , 13, , 3.5, , 5.5, , 19.25, , 10, , 1, , 4.5, , –6.5, , –29.25, , 55, , 75, , Here,, and, , –44.5, Σx 55, =, = 5 .5, n 10, Σy 75, y=, =, = 7.5, n 10, , x=, , 3. We have, variance of x = 9, and regression equations are, 8x − 10 y + 66 = 0, and, 40x − 18 y = 214, 8, 66, 18, 214, and x =, y=, x+, y+, ⇒, 10, 10, 40, 40, ∴ Regression coefficient of y on x is, 8 4, byx =, =, 10 5, 9, and x on y is bxy =, 20, 4 9, 9, 3, =, ⇒r =, ∴, r 2 = byx ⋅ bxy = ⋅, 5 20 25, 5, , ...(i), ...(ii), , 4. Here, two variables X andY are observed on same units, and, …(i), σ X = 2σY, σ, [from Eq. (i)], ∴, bXY = r X = 2r, σY, σY 1, and, bYX = r, = r, σX 2, ∴, 4bYX = bXY, ⇒ The regression coefficient of X on Y is four times the, regression coefficient of Y on X., 5. If X is changed to a + hU and Y to b + kV , then, h, bXY =, bUV, k, ⇒, kbXY = hbUV
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663, , Correlation and Regression, 6. I. If two variables are independent, then there is, uncorrelation i.e., r = 0, II. The coefficient of correlation between two variables, X and Y is positive. When X positive, then Y positive., 7. Since, lines of regression passes through (X , Y )., …(i), ∴, 3X + Y − 12 = 0, …(ii), and, X + 2Y − 14 = 0, On solving Eqs. (i) and (ii), we get, X = 2 and Y = 6, 8. Given, a1x + b1 y + c1 = 0 is the line of regression of Y on, X, so the slope, a, coefficient of x, m1 = −, =− 1, b1, coefficient of y, Similarly, a 2x + b2y + c2 = 0 is the line of regression of X, on Y, so the slope, b, coefficient of Y, m2 = −, =− 2, a2, coefficient of X, We know that, m1m2 ≤ 1, a1 b2 , So,, − × − ≤ 1 ⇒ a1b2 ≤ b1a 2, b1 a 2, 9. Given x1 = r, ∴, As,, ⇒, ∴, ⇒, , σy, σx, , , x2 = r, , σx, σy, , σy σx , +, x1 + x2 = r , , σ x σ y , AM ≥ GM, σy σx, σy σx, σy σx, +, ≥2, ⋅, ⇒, +, ≥2, σx σy, σx σy, σx σy, σy σx , r, +, ≥ 2r, σ x σ y , x1 + x2 ≥ 2r, , 10. For the line 8x − 10 y = 66, byx =, , 4, 5,, , and for the line 40x − 18 y = 214, bxy =, and, ⇒, ⇒, ∴, , 9, ,, 20, , σ 2x = 9 ⇒ σ x = 3, 36, r2 =, (Q r = byx ⋅ bxy ), 100, r = 0.6, 4, byx × σ x 5 × 3 4, 10, σy =, =, = ×3 ×, =4, r, 0.6, 5, 6, , 11. Both the statements I and II are correct, by property of, correlation coefficient., 12. Statement I Since, mean is satisfied the line of, regression, Here,, x = 3, y = 3, ∴, 3 + 4 (3) = 15, ⇒, 15 = 15, ∴ Statement I is true., Statement II, 5, 30, x=−, ,y=, ∴, 10, 13, , 5, 30, (3) +, (4) = 9, 10, 13, 15 120, −, +, =9, ⇒, 10, 13, ⇒, 1005 ≠ 1170, ∴ Statement II is false., In the second set of regression lines, the values of x, and y doesn’t satisfy the regression lines., ⇒, , −, , 13. The given equation of lines of regression are, …(i), x + 4y = 3, and, …(ii), 3x + y = 15, Now, we assume that Eq. (i) is the line of regression of y, on x and Eq. (ii) is the line of regression of x on y., 1, 1, ∴, byx = − , bxy = −, 4, 3, 1 1, 1, Now,, byx ⋅ bxy = − − =, 4 3 12, ∴ Our assumption is true., On putting y = 3 in Eq. (ii), we get, 3x + 3 = 15, ⇒, x=4, 14. Both are correct statements and R is the correct, explanation of A., 15. If bxy is negative, then byx and r are also negative, since, regression coefficients and r have the same sign., 16. A and R are individually true and R is the correct, explanation of A., x = 65, y = 67, σ x = 5.0, σ y = 2 .5, r = 0.8, The line of regression of y on x is, σy, (x − x ), y− y=r⋅, σx, 0.8 × 2 .5, y − 67 =, (x − 65), ⇒, 5, 2, ⇒, y − 67 = (x − 65), 5, , 17. Given,, , 18. The line of regression of x on y is, σ, x − x = r ⋅ x ( y − y), σy, 0.8 × 5, ⇒, x − 65 =, ( y − 67), 2 .5, 8, x − 65 = ( y − 67), ⇒, 5, 19. Angle between two regression lines is given by, 1 − r2 σx σy, ⋅ 2, tan θ =, r, σ x + σ y2, 1 − (0.8)2, 5 × 2 .5, ⋅ 2, 0.8, (5) + (2 .5)2, 0.36, 12 .5, =, ⋅, ⇒ tan θ = 1.8, 0.8 (25 + 6 .25), =, , ⇒, , θ = tan − 1 (1.8)
