Page 1 :
CHAPTER, , 1, , Function, 1.1, , Function, , 1.2, , Types of Function, , 1.3, , Inverse of a Function, , 1.4, , Composite Function, , 1.5, , Simple Algebraic Function, , 1.6, , Trigonometric Function, , Sir Isaac Newton was one of the greatest, mathematician and physicist, the world, has ever produced. His contributions in, mathematics are spread in almost every, field. However, he developed Calculus, both, differential and integral, that are, still the, most important aspects of higher mathematics., Using Calculus, he made it easier to find the, area bounded by closed curves. In physics, his, most famous and world changing contribution, was the discovery of existence of gravitation, force in our surrounding., , 1, , Speedy Optional Maths – Book 10

Page 2 :
Function, , 1.1 Function, Let A and B be two non empty sets. Then a relation from set A to Set B is said to be a, function if every element of the first set has a unique element in second set. , , 1.2 Type of Function, One to one function, , ¦, , A function from set A to set B is said to be one to one, function if every element of the domain (i.e. the first, set) has a unique/ different image in the co-domain, (second set where it is mapped). Here, the function, shown alongside is one to one function, , 1, 2, 3, , a, b, c, d, , A, , B, , Many to one function, , ¦, , A function f: A→B is said to be many to one function, if at least two elements of domain have a unique (same), image in co-domain. In diagram, elements 1, 2 and 3 of, domain A have a unique image 'a' in co-domain. So , it, is a many to one function., , a, b, , 1, 2, 3, A, , B, ¦, , Onto function, A function f: A→ B is said to be an onto function if the, range is equals to the co - domain i.e. no elements are, left over in the second set after mapping., , 1, 2, , a, b, , Here,, , A, , B, , f: A→B is an onto function., ¦, , Into function, , 1, 2, 3, , a, b, c, , A function f: A→B is said to into function if the range is, not equal to the co-domain. Instead the range is proper, subset of co-domain. In other words, if all the elements of, A, B, co-domain are not mapped with the elements of domain, then it is known as into function. In the diagram the element 'c' of co-domain is left to be mapped., So , it is an into function., , 1.3 Inverse of a function, Let f : A→B be one to onto function. Than the function defined form B to A on the same, time such that f-1 (y) =x is called inverse of a function where x ∈A and y ∈B. It is denoted by, f-1 and read as inverse., , Speedy Optional Maths – Book 10, , 2

Page 3 :
Function, In figure,, , ¦, , ¦, , 1, 2, 3, , 1, 2, 3, , a, b, , a, b, c, d, , A, A, B, , fig. i fig. ii, Here, in figure no. (i), inverse function does not exit as it, is many to one onto function and in figure no.(ii)inverse, function does not exist as it is one to one into and the, 1, inverse of f exists only in the case of one to one and onto, 2, function., 3, -1, In figure (iii) f(1)=a , f (a)=1, A, , f(2)=b f-1(b)=2, , f(3)=c f-1(c)=3, , B, ¦, , a, b, c, ¦ -1, , B, , fig. iii, , Procedural steps in finding inverse of algebraic function, , (i) Put y in place of f(x) as f(x)=y, (ii) Interchange the roles of x and y., (iii) Get new y which is f-1(x), , , , Worked out examples, , Example 1:, , If f= {(1,2), (2,3), (4,5) }find the inverse of f., Solution:, , , , , Here, , , f, , = {(1,2), (2,3), (4,5)}, , In arrow diagram,, \ , , f-1, , = {(2,1), (3,2), (5,4)}, , Example 2: a) If f(x) = 7x – 2, find f–1(x)., b) If g: x ® 5x–3 , find g–1(2)., 2x+1, a) Solution:, , , , , Here, f(x), , =7x–2, , y, , =7x–2, , Interchanging the roles of x and y,, , x, , , =7y–2, , , , or, x + 2 =7y, x+2, x+2, \ y, = 7 ,f–1(x) = 7, b) Solution:, 5x–3, Here, g: x → 2x+1, , , , or,, , , , , or,, , 5x–3, y = 2x+1, , 5x–3, g(x) = 2x+1, 3, , Speedy Optional Maths – Book 10

Page 4 :
Function, , , Interchanging the roles of x and y, , , , 5y–3, x = 2y+1, , or,, or,, , x+3, = 5–2x, 2+3, 5, 5, g -1(2) = 5-2(2) = 5-4 = 1 = 5, y, , , , or,, , x(2y + 1) = 5y – 3, , , , or,, , 2xy + x, , , , or,, , Example 3:, , If f = {(x:y) : y = 3x - k} and f -1(5)=4, find the value of k., , , , , , , , Solution:, Here, f , \ f(x) , Also, f -1(5), Now, for f(x), , = 5y – 3, , x + 3 = 5y – 2xy, , = {x:y= 3x – k}, = 3x – k, = 4 ......................................... (i), , f(x) , , = 3x – k, , , , = 3x – k, , or, y , , Interchanging the roles of x and y, or,, or,, \, \, , x = 3y – k, , From (i) and (ii), 5+k, =4, 3, or,, 5+k, = 12, , x+k, 3 = y, x+k, f -1(x) = 3, 5+k, f -1(5) = 3, , \, , ........... (ii), , k , , = 12 – 5 = 7, , Example 4: Given a function f(x + 3) = 3 x + 5, find f -1(x), = 3x + 5, = 3x + 9 – 4, = 3(x + 3) – 4, = 3a – 4, = 3x – 4, , 1.4 Composite function, Let f: A®B be a function and, g: B®C be another function., Then a function defined, from A to C is known as, composite function of f and, g. It is denoted by gof and, read as g knot f, g circle f or, f followed by g., , A, , y, , Interchanging the values of x and y,, , or,, , x, x+4, 3, , \, , f -1(x), , ¦, , x, , = 3x – 4, = 3y – 4, =y, x+4, = 3, , B, , g, , ¦{x}, , gof, , This also tells us that f is the first function operated and g is the second., Speedy Optional Maths – Book 10, , 4, , C, g ¦{(x)}, , Solution:, Here,, f(x + 3), , or, , f(x + 3), , or, , f(x + 3), , or, , f(a), Now,, , f(x), , for, f -1(x)

Page 5 :
Function, Example 5: If f = {(1,3), (0,0), (–1, –3)} and g= {(0,2), (–3,5)} then show the function, , , (gof) (x) in an arrow diagram and hence find it in ordered pair form and gof (1)., , , Solution:, , Here, f = {(1,3), (0,0), ( –1, –3)}, g = {(0,2), (–3, –1), (3,5)}, , , So, (gof) in arrow diagram,, , ¦, , A, , g, , B, , C, , 1, , 3, , 5, , 0, , 0, , 2, , -1, , -3, , -1, , go¦, , (gof) is ordered pair form (gof) = {(1,5), (0,2), ( –1, –1)}, (gof) (1) = 5, , Example 6: If f(x) = 2 x and g(x) = x + 1, Find gof(x), gof( –2), fog(x) and fog(2)., , , Solution:, , Here,, , f(x), , = 2x, , , , g(x), , =x+1, , , , (gof)(x),, , For, , , (gof)(x), , , , (gof)( -2), , , , Here,, , f(x), , g(x), , , For, , , y, , = 3x – 7, = x+2, 5, , Now,, By the question,, , x+2, = 5, x+2, = 5 [ g(x) = y], , , , Interchanging the roles of x and y,, , , , or, , , x, , = y+2, 5, , , , or, , , 5x, , =y+2, , y, , = g -1(x) = 5x – 2, , \ , , = f{g(x)}, = f(x + 1), = 2(x + 1), = 2x + 2, = 2.2 + 2, =4+2, =6, , x+2, and g -1 of(x) =f(x), find the value of x., 5, , g –1(x),, , g(x), or, , , fog(x), , , , and fog(2), , , , = g{f(x)}, = g{2x}, = 2x + 1, = 2( –2) + 1, = –4 + 1, = –3, , Example 7: If f(x) = 3x – 7, g(x) =, Solution:, , Again,, , 5, , g -10 f(x) , , = f(x), , or,, or,, or,, or,, or,, , g {f(x)}, g -1{3x – 7}, 5(3x – 7) – 2, 15x – 35 – 2, 12x , , or,, , x , , \, , x , , = f(x), = 3x – 7, = 3x – 7, =3x – 7, = 30, = 30, 12, 5, =, 2, , -1, , 5, 2, , Speedy Optional Maths – Book 10

Page 6 :
Function, Example 8: If f(x) = 2x + 3 and g(x) = x + 2. Then find (fog) -1(X)., Solution:, Here,, , f(x), , = 2x + 3, , , , g(x), , =x+2, , Now, For (fog) -1 (x), , Now,, , , For, , , or,, , fof (x), = f{g(x)}, , , , = f(x + 2), , , , = 2(x + 2) + 3, , , , = 2x + 4 + 3, , , , = 2x + 7, , Example 9: If f(x) = 4x – 17 and g(x) =, Solution:, Here,, , = 2x + 7, , f(x), , y, , = f{4x – 17}, , , , = 4(4x – 17) – 17, , , , = 16x – 68 – 17, , , , = 5x–8, 2, g-1(x), , = 5x–8, 2, According to the question,, , \, , = 16x – 85, , fog(x) , , for g (x), , or,, , 16x – 85 , , = g -1(x), 5x–8, = 2, , or,, , 32x – 170, , = 5x – 8, , or,, , 27x , , = 162, , x , , = 162, 27, =6, , -1, , 2x+8, 5, 2x+8, , or, , y, = 5, , Interchanging the roles of x and y,, x, = 2y+8, 5, , or, , 5x, = 2y + 8, , , , y , , 2x+8, and fof(x) = g-1 (x) then find the value of x., 5, , , , Also,, , = 2x + 7, , Interchanging the value of x and y,, x–7, or,, =y, 2 , \, (fog) -1 (x), = x–7, 2, , = 4x – 17, 2x+8, g(x), = 5, Now,, (fog)(x) = f{f(x)}, , , , fog(x) , , g(x), , =, , \, , x , , =6, , Example 10: If f(x) = 2x – 4 then prove that fof -1(x) is an identity function., Solution:, Here,, , f(x), , \ f -1(x) , , = 2x – 4, , Now, (fof -1) (x), , Now,, , , For , , f -1(x),, , , , f(x), , = 2x – 4, , , , y, , = 2x – 4, , or, , , = x+4, 2, , , , =f, , x+4, 2, , , , =f, , x+4, 2, , , , =x+4–4, , Interchanging the roles of x and y,, , , , =x, , , , x, , = 2y – 4, , \, , =x, , , , y, , = x+4, 2, , It is an identity function., , or, , , Speedy Optional Maths – Book 10, , 6, , fof (x) , -1, , –4

Page 7 :
Function, , EXERCISE 1.1, General Section, 1., , Define inverse of a function with an arrow diagram., , 2., , a), , Write the range and inverse of a, , function f =, b), , , {(, , 1, 2, 2, , )(, , 1, , 3, 3, , )(, , 1, , 4, 4, , ¦, , )}, , 1, 2, 3, , Find the inverse of a function f from the, arrow diagram given aside., , a, b, c, , 3., , Find the inverse of the following functions:, , 4., , f(x)=5x – 1 , b) g:x → 5x + 2, 3x+7, , d) g(x)= 2x+5 , x∈R, c) f(x)=, 2, 2, x–5, x, 5x-3 , x∈R, e) h(x)=, , f) f =, 2x+1, 2x, If f(x) = x2 – 5 whose range is 4. Find the show f(x) in an arrow diagram and find, a), , {(, , 5., , ), , }, , the inverse of f(x) if possible., a) If f(x)= 3x–4 , find f –1(x) and f –1(3), 5, 3x–2, b) If g(x)=, , find g –1(– 2), 4, c) If h(x)= 1 , find h –1 –2, 3, 5x+4, , ( ), , 6., , Define composite function with an example., , 7., , a), , , b), , c), , If f={(1,2), (3,4), (5,6)} and g= {(2,3), (4,6), (6,8)}. Find the composite of gof., Does fog exist?, If f= {(1,3), (0,0), (–1, –3)} and g= {(0,2), (–3, –1), (3,5)} then find gof in an, arrow diagram and also present it in an ordered pair., If g= {(1,2), (2,3), (3,4)} and h= {(2,3), (3,4), (4,5)} then calculate hog (1) and, , , , hog (2)., , 8., , If f(x)=x and g(x)=x + 1 then find, , a), , , , (i) (fog) (x) , , (ii) (gof) (x), , , , (iii) (fof) (x) , , (iv) (gog) (x), , , , (v) (fog) (1) , , (vi) (gof), , b), , ( –12 ), , If f(x)= x – 5 and g(x)= 5 – x, check if (fog) (x) = (gof) (x)., , 7, , Speedy Optional Maths – Book 10

Page 8 :
Function, 9., , a), b), , If g:x → 5x – 2 and h:x → 5+x , find (hog) -1(x)., 2, x+3, If f(x)=, and g(x)=3x + 4, find fof -1(x) and fog -1(x). Also find out, 2, , f -1 og -1(1)., , Creative Section, 10. a), b), c), 11. a), b), , If f(x)=2x + 3 and f -1(7)=k, then find out value of k., If f:x → 4x + 3 and if f(x)=f -1(x), find the value of x., If g= {(x, 5x – k)} and if g -1(2)= 4 , find the value of k., 5, If f(x)=4x – 7 and g(x)=3x – 5. If fg -1(x)=15, find the value of x., If f(x)= x–2 and g(x)= 1 and if f -1(x)=gof(x) find the value of x., x, 2x+1, , x, If the functions f(x)= x–2 and g(x)=bx – 2. Also gf(4)= – 8, find the values of, f -1( – 2) and b., , 12. a), , b), , If f(x)= 3x + a and if ff(6)= 10, find the value of a and f -1(4)., , c), , If f(x)=(2x + k) and ff(4)= 10, find the value of k., , 13. a), b), 14. a), , If f(x)= 2x+5 ; x ≠ – 2, find f -1(x) and show that ff -1(x) is an identity function., x+2, If the function f(x)= 2x+1 , prove that f -1 o f(x) is an identity function., 4, If f(x + 3)=x + 6, find f -1(x)., , b), , If f(2x + 5)=4x + 13, find f -1(x)., , c), , If f(3x + 4)=5x + 8, find f -1(x)., , 15. a), b), 16. a), b), , If f -1(x)= x–1 , find the function f(x)., 3, If f -1(x)= x – 2, find f(x) and f(3)., If (fog)(x)= 7x–1 and f(x)= 3x + 5, find g(x), where g(x) is linear., 3, If g(x)=2x and (fog)(x)= 6x – 2, find f(x), where f(x) is a linear function., , 1.5 Simple algebraic functions, The Identity Function, Let ƒ be a function from a set A to the set A itself defined by ƒ(x) = x. Then ƒ is, called an identity function., Speedy Optional Maths – Book 10, , 8

Page 9 :
Function, Y, , If ƒ:R → R is an identity function, its graph is a straight line, bisecting the angle between the axes of coordinates., , ¦ (x) =x, , X', , X, , Y', ¦, , The Constant Function, , 1, , Let ƒ be a function from A to B and B is a singleton, set. Then, ƒ is called a constant function., , 2, , a, , 3, , Y, , ¦ (x) =c, , X', , X, , If ƒ: R→ R be defined by ƒ(x) = c where c is, constant, then the function ƒ is a constant function, and its graph is a straight line parallel to the, x – axis., , Y', , Y, , The Linear Function, Let A and B be two sets and ƒ be a function from, A to B. Then ƒ is called a linear function if it is, defined as follows:, ƒ(x) = ax + b, where a and b are real constants., For example,, Let ƒ:R →R be defined by ƒ(x) = x + 1. There, order pairs of this function are given in the, following table:, , x, , 0, , 2, , –2, , y = ƒ(x), , 1, , 3, , –1, , ¦ (x)=x + 1, (2,3), (1,3), (0,1), , X', , (-1,0), , X, , (-2,-1), , Y', , The graph of ƒ (x) is a straight line as shown in the following figure:, 9, , Speedy Optional Maths – Book 10

Page 10 :
Function, , The Quadratic Function, , Y, , Let ƒ be a function from A to B. Then ƒ is, called a quadratic if ƒ is defined as follows:, ƒ (x) = ax2 + bx + c, where a, b, c are, real constants., For example,, Let ƒ: R→R be defined by, ƒ(x) = 4x2 + 8x + 5. Then ƒ is a quadratic, function. Some of its ordered pairs are:, , x, , –1, , –2, , 0, , y = ƒ(x), , 1, , 5, , 5, , 4x, , 2, , + 8x + 3, , (0,5), , (-2,3), , (-1,1), , X', , X, , The graph of ƒ(x) is as shown in the figure., Y', , The Cubic Function, , Let ƒ be a function from A to B. Then ƒ is called a cubic function if it is defined as, ƒ(x) = ax3 + bx2 + cx + d where, a, b, c, and d are real numbers or constants., For example,, Y, Let ƒ: R→R be defied by ƒ(x) = x3., Some ordered pairs of this function, ¦ (x)=x, are given in the following table., , 3, , (1,1), , x, , –2, , –1, , 0, , 1, , 2, , y = ƒ(x), , –8, , –1, , 0, , 1, , 8, , X', , X, , (0 , 0, (-1,-1), , The graph of ƒ(x) is as shown in the figure., , (-2,-8), , 1.6 Trigonometric functions, , Y', , Trigonometric function is the function which associates each angle with the definite real, number. So, the domain of trigonometric function is the set of angles and its co – domain, is the set of real numbers. Traditionally trigonometric functions are defined for angles of, a triangle. But these trigonometric function can be defined for angle of any magnitude., Y, , Let us consider an angle q placed in the standard, position. Draw circle with center at O and radius r., Let the circle intersect the terminal arm at some point, P(x, y) as shown in the figure. Draw PM ⊥ OX., For any angle q the six trigonometric ratios: the sine,, cosine, tangent, cosecant, secant and cotangent may be, defined in terms of x - coordinate 'x', y – coordinate, 'y' and the radius 'r' by the formula:, , P(x,y), r, X', , q, O, , Y', , Speedy Optional Maths – Book 10, , 10, , y, M, , X

Page 11 :
Function, y, x, = r ,, cos, = r ,, r, r, =y,, sec, =x, cosec, , The definition given readily gives, y y, =0, (ii), (i) sin0°, = r = r, r, x, =1, cos0°, = r = r, y 0, = 0 , tan0°, =x= r, y 0, =0, (iv), (iii) sin180°, = r = r, x -r, = –1 , cos180°, = r = r, y 0, = 0 etc. , tan180°, = x = -r, , y, =x,, x, = y, , tan, , sin, , And, , sin90°, cos90°, tan90°, sin270°, cos270°, tan270°, , cot, y, = r, x, = r, y, =x, y, = r, x, = r, y, = r, , y, = r =1, 0, = r =0, r, = 0 (Undefined), -r, = r = –1, 0, = r =0, -r, = 0 (Undefined) etc., , , , , The trigonometric ration for angles such as 30°, 45°, 60° etc. can be calculated, with the help of elementary plane geometry. Following table shows the values of, trigonometrically rations from 0° to 180°., 0°, , , , 30°, 1, 2, , 45°, 1, 2, , 60°, , 90°, , 120°, , 150°, 1, 2, , 180°, , 3, 2, , 135°, 1, 2, , 3, 2, , 1, , 3, –2, , –1, , sin, , 0, , cos, , 1, , 3, 2, , 1, 2, , 1, 2, , 0, , – 1, 2, , – 1, 2, , tan, , 0, , 1, 3, , 1, , 3, , ∞, , – 3, , –1, , –, , 1, 3, , 0, , 0, , The function defined as follows are trigonometric functions:, , ƒ( ) = sin , , , ƒ( ) = cos ,, , ƒ( ) = tan , , ƒ( ) = cosec ,, , ƒ( ) = sec ,, , ƒ( ) = cot, , 1.7 Graphs of Trigonometric Functions, To each angle x there corresponds a unique value for the trigonometric ratios and hence the, trigonometric ratios such as sinx, cosx, tanx including the reciprocal ratios are functions of, the angle x. A function can be represented in various ways. One of the important ways by, which a function can be represented is the graph. In a trigonometric function, the variable, is the angle. For different values of the angle, the values of the trigonometric ratios will be, different. To draw the graph of trigonometric function, we take the angle x along X-axis and, the value of trigonometric ratio sinx, cosx, tanx, etc. along Y-axis. Taking different values of, the angle as the x-coordinates, and the corresponding values of the trigonometric ratio, as y-coordinates, we plot the points (x, sinx), etc. on the plane with rectangular axes., Then we join these points freely to get required graph of the trigonometric function., Consider a circle with radius r placed in standard position. Let a revolving line OP start from, OX and trace out an angle XOP = x°. Draw perpendicular PM from P to the X-axis., 11, , Speedy Optional Maths – Book 10

Page 12 :
Function, Y, , Y, , 2nd, , 1st, , quadrant, , quadrant, , P, , P, x°, X', , M, , X', , X, , O, , x°, O, , 3rd, , X', , 4th, , Y, , Y, , M, , X, , Y', , Y', quadrant, , M, , x°, , x°, X', , X, , O, , quadrant, , M, O, , X, , P, , P, Y', , First quadrant, , : OP revolves from OX to OY, x varies from 0° to 90°, MP varies from 0 to, r, MP is positive, OM varies from r to 0 and OM is positive., , Second quadrant : OP revolves from OY to OX’, x varies from 90° to 180°, MP varies from, r to 0, MP is positive, OM varies from 0 to – r and OM is negative., Third quadrant, , : OP revolves from OX’ to OY’, x varies from 90° to 270°, MP varies from 0, to – r, MP is negative, OM varies from – r to 0 and OM is negative., , Fourth quadrant : OP revolves from OY’ to OX, x varies from 270° to 360°, MP varies from – r, to 0, MP is negative, OM varies from 0 to r and OM is positive., , Graph of sine function, The sine function is defined by y = sinx =, , MP MP, =, ., OP, r, , When x varies from 0° to 90°, y = sin x varies from 0 to 1 because MP varies from 0 to r., When x varies from 90° to 180°, y = sinx varies from 1 to 0 because MP varies from r to 0., When x varies from 180° to 270°, y= sin x varies from 0 to –1 because MP varies from 0 to –r., When x varies from 270° to 360°, y = sin x varies from – 1 to 0., Some standard values of x and the corresponding values of sin x are given below:, x, sinx, , 0°, 0, , 30°, 0.5, , 60°, 0.56, , 90°, 1.0, , 120°, 0.86, , Speedy Optional Maths – Book 10, , 150° 180° 210°, 0.5, 0, –0.5, 12, , 240°, –0.86, , 270°, –1.0, , 300°, –0.86, , 330°, –0.5, , 360°, 0

Page 13 :
Function, Plotting these values in the graph paper, we get the following graph., , If we study this sine curve, we will get the following facts:, y = sin x attains the maximum value of 90° and minimum value at 270°. The maximum, and the minimum value of sin x are 1 and –1 respectively. So, y = sin x oscillates in the, limit – 1 to 1., , (i), , (ii) y = sin x has positive values in the first and the second quadrants where the graph, is above X-axis and negative values in the third and the fourth quadrants where the, graph is below X-axis., , Graph of Cosine function, The cosine function is defined by, OM OM, =, ., OP, r, When x increases from 0° to 90°, cos x decreases from 1 to 0 as OM decreases from r to 0. When, x increased from 90° to 180°, cos x decreases from 0 to –1 because OM changes from 0 to – r., When x, changes from 180° to 270°, cos x changes from – 1 to 0 similarly, when x increases from 270° to, 360°, cos x increases from 0 to 1., Some standard values of x and the corresponding values of cos x are given below:, y= cos x =, , x, , 0°, , 30°, , 60°, , 90°, , 120°, , 150°, , 180°, , 210°, , 240°, , 270°, , 300°, , 330°, , 360°, , cosx, , 1.0, , 0.86, , 0.5, , 0, , – 0.5, , –0.86, , – 1. 0, , –0.86, , –0.5, , 0, , 0.5, , –0.86, , 1.0, , 13, , Speedy Optional Maths – Book 10

Page 14 :
Function, Plotting these values in the graph paper, we get the following graph., , (i), , Y = cosx attains the maximum value at x = 0° and 360° and minimum value at 180°., The maximum and minimum value of cosx are 1 and –1 respectively. So, y =, cosx oscillates in the limit – 1 to 1., , (ii), , y = cosx has positive values in the first and fourth quadrants where the graph is, above X-axis and negative values in the second and third quadrants where the, graph is below X-axis., , Graphs of Tangent Function, The tangent function is defined by y = tanx = MP ., OM, , When x increases from 0° to 90°, tanx increases from 0 to ∞ (undefined) because MP changes, from 0 to r and OM changes from r to 0. When x increases from 90° to 180°, tanx changes, from – ∞ to 0 because MP changes from r to 0 and OM changes from 0 to –r. When x, increases from 180° to 270°, tanx changes from 0 to ∞ because MP changes from 0 to –r and, OM changes from – r to 0. When x increases from 270° to 360°, tanx changes from – ∞ to 0, because MP changes from – r to 0 and OM changes from 0 to r., , Speedy Optional Maths – Book 10, , 14

Page 15 :
Function, Some standard values of x and the corresponding values of tanx are given below:, x, , 0°, , 30°, , 60°, , 90°, , 120°, , 150°, , 180°, , 210°, , 240°, , 270°, , 300°, , 330°, , 360°, , Tanx, , 0, , 0.58, , 1.73, , ∞, , –1.73, , –0.58, , 0, , 0.58, , 1.73, , ∞, , –1.73, , –0.58, , 0, , Plotting these values in the graph paper, we get the following graph., Y, 4, 3, , y=3, , (1.249, 3), , 2, 1, X', , - 3p, 2, , -p, , -p, , 2, , p, , p, , 2, , -1, , 3p, 2, , 2p, , X, , -2, -3, Y', Y', , y = tan x has positive values in the first and the third quadrants where the graph is above, X-axis and negative values in the second and the fourth quadrants where the graph is, below X-axis. The maximum and minimum value of tan x cannot be defined., , EXERCISE 1.2, 1., , Draw the graphs of the following trigonometric functions:, (a), f(x) = sinx, (–p ≤ x ≤ p), (b), f(x) = cosx, (–p ≤ x ≤ p), (c), f(x) = tanx, (–p ≤ x ≤ p), , 2., , Draw the graphs of the following trigonometric functions:, (a), f(x) = sinx, (–2p ≤ x ≤ 2p), (b), f(x) = cosx, (–2p ≤ x ≤ 2p), (c), f(x) = tanx, (–2p ≤ x ≤ 2p), , 15, , Speedy Optional Maths – Book 10

Page 16 :
CHAPTER, , 2, , Polynomial, , Shrinivasa Ramanujan Iyengar, the greatest, Indian mathematician of 20th century,, contributed immensely in fields like number, theory, mathematical analysis, string theory, and crystallography. His genius has been, admired by some greatest contemporary, mathematicians of his time. He is hailed to be, one of the most famous mathematicians in the, field of number theory. Although he lived for, a short span of 32 years, he compiled nearly, 3900 phenomenal results that leave even the, best mathematical brains of today in sheer awe, and wonder!, Speedy Optional Maths – Book 10, , 16, , 2.1, , Polynomial, , 2.2, , Multiplication of Polynomials, , 2.3, , Division of Polynomials, , 2.4, , Synthetic Division, , 2.5, , Remainder Theorem, , 2.6, , Factor Theorem, , 2.7, , Factorization of Polynomials, , 2.8, , Polynomial Equation

Page 18 :
Polynomial, , Properties of Multiplication of polynomials, Multiplication of the polynomials follow the following properties:, a), , b), , c), , Closure property, If ƒ(x) and g(x) be two polynomials then their product ƒ(x). g(x) is also a , polynomials., Commutative property, If ƒ(x) and g(x) be two polynomials, then ƒ(x).g(x) = g(x).ƒ(x)., Associative property, , , , If ƒ(x). {g(x) and h(x)} be three polynomials. then, f(x) .{g(x).h(x)} = {f(x).g(x)}.h(x), d), , e), , , 2.3, , Multiplicative identity, For any polynomial ƒ(x) there is the unit polynomial I(x) = 1xº = 1 such that, ƒ(x).1 = ƒ(x). Here, 1 is said to be the multiplicative identity., Distribute property, If f(x), g(x) and h(x) be three polynomials then, f(x) be three polynomials then,, f(x) [g(x) + h(x)] = f(x).g(x) + f(x).h(x), , Division of polynomials, Let us consider the following examples, Divide x2 + 5x + 6 by x + 3, Solution:, , ), , (, , x + 3 x2 + 5x + 6, x+2, x2 + 3x, – –, 2x + 6, 2x + 6, –, –, , 0, Hence, Remainder (R) = 0 and Quotient Q(x)=x + 2, , Theoretical Generalization, If a polynomial ƒ(x) is divided by a non – zero polynomial g(x) then there exist, unique polynomials Q(x) and R(x) such that ƒ(x) = g(x).Q(x) + R(x), Where, ƒ(x) is dividend, g(x) is divisor, Q(x) is quotient and R(x) is remainder., If R(x) = 0, then the divisor g(x) is a factor of the dividend ƒ(x). The other factor of, ƒ(x) is the quotient Q(x)., The relation ƒ(x) = g(x).Q(x) + R(x), or, Dividend = Divisor × Quotient + Remainder. It is known as division algorithm., , The following steps are to be used to divide a polynomial by the other:, a), , Arrange the divided ƒ(x) and divisor g(x) in standard form i.e. generally descending, powers of variable x., , Speedy Optional Maths – Book 10, , 18

Page 19 :
Polynomial, b), , Divide the first term of divided ƒ(x) by the first term of divisor g(x) to get the first term, of quotient Q(x)., , c), , Multiply each term of divisor g(x) by the first term of quotient Q(x) obtained in step(ii), and subtract the product so obtained from the dividend ƒ(x)., , d), , Take the remainder obtained in step (iii) as new dividend and continue the above, process until the degree of the remainder is less then of the divisor., Example:, , Divide ƒ(x) = 4x3 – 3x2 + x + 7 by g(x) = x – 2., , Solution:, , ), , (, , 4x2 + 5x + 11, x - 2 4x3 – 3x2 + x + 7, 3, 4x, – 8x2, –, +, 5x2 + x + 7, 2, 5x, – 10x, –, +, 11x + 7, 11x – 22, , –, +, 29, Hence, Dividend ƒ(x) = 4x3 – 3x2 + x + 7 Divisor g(x) = x – 2, Quotient Q(x) = 4x2 + 5x + 11 Remainder R = 29, We can write in the form of ƒ(x) = g(x) Q(x) + R, 4x3 + 3x2 + x + 7 = (x – 2) (4x2 + 5x + 11) + 29., , EXERCISE 2.1, 1., , 2., , 3., , Find the product of:, a), , 3x3 + 5x2 + 9x + 8, , and, , 3x + 4., , b), , 4x2 + 5x + 7, , and, , x + 1., , c), , 3x3 – 2x2 + 5x – 1, , and, , 2x – 3., , Divide the polynomial ƒ(x) by g(x) if, a), , ƒ(x) = 8x4 + 2x3 – 3x – 8, , g(x) = 2x2 – 3, , b), , ƒ(x) = 10x2 – 3 – 6x + 3x3, , g(x) = 4 – x, , c), , ƒ(x) = 4x2 + 32x + 15 , , g(x) = 4x + 3., , a), , If 2x3 – 9x2 + 5x – 5 = (2x – 3) Q(x) + R, find the quotient and remainder., , b), , Find Q(x) and R if 3x4 – 8x2 – 20 = (x2 – 2) Q(x) + R., , 2.4 Synthetic division, Synthetic division is the process which helps us to find the quotient and remainder, when a polynomial ƒ(x) is divided by x – a., 19, , Speedy Optional Maths – Book 10

Page 20 :
Polynomial, Let us divide 4x3 – 3x2 + x + 7 by x – 2., x – 2) 4x3 – 3x2 + x + 7 (4x2 + 5x + 11, 3, –4x, – 8x2, – +, 5x2 + x + 7, 5x2 – 10x, 11x + 7, – 11x +, – 22, 29, Here, quotient is 4x2 + 5x + 11 and remainder is 29. Then we can write, 4x3 – 3x2 + x + 7 = (x – 2) (4x2 + 5x + 11) + 29., Same quotient and remainder can be obtained by applying the following process., Here, ƒ(x) = 4x3 – 3x2 + x + 7, and, x – a = x – 2, \, Now,, , a = 2., , Here, quotient Q(x) = 4x2 + 5x + 11 and remainder R = 29., This process for division is called Synthetic division., , Following steps are performed for synthetic division:, •, , Change the sign of the constant term in the divisor. – 2 is replaced by 2 in the above, example., , •, , Write down the coefficients of divided ƒ(x) with their signs., coefficients of x3, x2, x, x° respectively in the above example., , •, , Write '0' as the coefficient for any missing term., , •, , Bring down the first coefficient 4., , •, , Multiply 4 by the constant term of the divisor; with sign changed, here it is 2 after, change of sign., , •, , Write the result under the next coefficient – 3 and add., , •, , Multiply the result obtained (i.e., coefficient 1, and add., , •, , Repeat the process., , 5) by, , 2, , 4,, , – 3,, , 1,, , 7 are, , and write the result under the next, , The last number so obtained is the remainder. The number just before the last is, constant term; the third number from the end is the coefficient of x and so on of the, quotient., Speedy Optional Maths – Book 10, , 20

Page 21 :
Polynomial, Example:, Use synthetic division to divide x3 – 6x2 + 11x – 6 by x – 2., , Solution:, , Let ƒ(x) = x3 – 6x2 + 11x – 6, , And x – a = x – 2, , \ a = 2, , The constant term in the divisor with sign changed =, coefficients in order, we have, , 2. Writing the, , , , Hence, Quotient Q(x) = x2 – 4x + 3, Remainder R = 0., , Example:, , , , , , , Find the quotient and remainder when 2x3 + 7x2 – 5 is divided by x + 3., Solution:, Let ƒ(x) = 2x3 + 7x2 – 5, And x – a = x + 3 = x – ( – 3), The constant term in the divisor with sign changed = – 3., Writing the coefficients in order, we have, , , , \ Quotient Q(x) = 2x2 + x – 3 Remainder R = 4., , Application of synthetic Division, Let Q(x) and R be the quotient and remainder when a polynomial ƒ(x) is divided, by ax - b., Then, ƒ(x) = (ax – b). Q(x) + R = a x – b . Q(x) + R = x – b . g(x) + R, a, a, Where a. Q(x) = g(x), Or, = 1 g(x)., a, Here g(x) and R are the quotient and remainder when ƒ(x) is divided x – b ., a, This result leads us to conclude that the process of synthetic division discussed, earlier is also useful to find out the quotient and remainder when ƒ(x) is divided by, ax – b., , (, , ), , ), , (, , (, , 21, , ), , Speedy Optional Maths – Book 10

Page 22 :
Polynomial, , (, , ), , (, , ), , Since ax – b = a x – b , we first divide ƒ(x) by x – b, to get the quotient g(x) and, a, a, remainder R. The quotient g(x) is again divided by a to get the required quotient Q(x)., We proceed similarly to get the quotient and remainder when ƒ(x) is divided by, (ax + b)., Example:, , , , , , , , Find the quotient and remainder when 2x2 – 9x2 + 5x – 5 is divided by, 2x – 3., Solution:, Let ƒ(x) = 2x3 – 9x2 + 5x – 5, Here, 2x – 3 = 2 x – 3, 2, 3, The constant term in x – 3, with sign changed = 2 ., 2, Writing the coefficients in order, we have, , (, , (, , ), , ), , Now, the required quotient = 3 (2x2 – 6x – 4) = x2 – 3x – 2 and remainder = – 11., 2, Example:, , , , , , , , Find the quotient and remainder when 4x3 + 2x2 – 4x + 3 is divided by 2x + 3., Solution:, Let ƒ(x) = 4x3 + 2x2 – 4x + 3, Here, 2x + 3 = 2 x + 3, =2 x– x– 3, 2, 2, 3, The constant term in the divider x + 3, with sign changed = – 2 ., 2, Writing the coefficients in order, we have, , Speedy Optional Maths – Book 10, , (, , ) { (, ( ), , 22, , )}

Page 23 :
Polynomial, ,  Worked out examples, Example 1: Divide x3 – 6x2 + 11x – 6 by x – 3., , Solution:, , Here; Let, f(x) = x3 – 6x2 + 11x – 6 and, , g(x) = x – 3, , Comparing g(x) with x – a, we get a=3, , So, writing the coefficients of f(x) in descending order of x., , , 3, , , , , , Hence,, Quotient,, Remainder,, , 3, 3, , 9, 2, , Q(x) = x2 – 3x + 2, R, =0, , Example 2: Divide f(y) = y3 – 19y – 30 by g(y) = y + 1 and hence find the quotient and the, remainder., , Solution:, Here,, f(y), = y3 – 19y – 30, , g(y), =y+1, , Comparing, g(y) with y – a, a = – 1, , Writing the coefficients in descending order of y,, , , , , Example 3:, , , , , Hence,, Quotient,, Q(y), = y2 – y – 18, Remainder,, R, = – 12, Find the quotient and remainder when 2x6 – 9x4 + 5x2 – 5 is divided by 2x2 – 3., Solution:, Let, f(x), = 2x6 – 9x4 + 5x2 – 5 and g(x) = 2x2 – 3, 2, Suppose, x = y then our problem changes to f(y) = 2y3 – 9y2 + 5y – 5 and, g(y) = 2y – 3, , , , So, 2y – 3, , , , , (y – 32 ), The constant term in (y – 3 ) with sign changed =, 2, =2, , 3 ., 2, , Writing the coefficients in order, we have, , 23, , Speedy Optional Maths – Book 10

Page 24 :
Polynomial, , Now, the required quotient = 1 (2x2 – 6x – 4) = x2 – 3x – 2 and, 2, remainder = – 11, Lastly, putting the value of y, we get, Quotient = x4 – 3x2 – 2, Example 4:, , , , , , , , Find quotient and remainder when y5 – 1 is divided by y + 1., Solution:, Let, ƒ(y) = y5 – 1, and, g(y) = y + 1, Comparing, g(y) with y – a, a = – 1, writing the coefficients in descending order of y,, We get,, 4, , R, , , , , 2.5, , Hence,, Quotient,, Remainder,, , Q(y) = y4 – y3 + y2 – y + 1, (R) = 0, , Remainder theorem, , Statement: If a polynomial ƒ(x) is divided by x – a, then the remainder is ƒ(a)., Proof: If, we divided ƒ(x) by x - a, then we get Q(x) as quotient and R as remainder., Then, ƒ(x) = (x - a). Q(x) + R, put x, = a., Then. ƒ(a) = (a - a) Q(a) + R, or,, R, = ƒ(a), Hence, remainder = ƒ(a) = the value of polynomial ƒ(x) at x = a., Corollary 1: If a polynomial ƒ(x) is divided by (x + a), then remainder R = ƒ( –a)., Proof: If we divide ƒ(x) by (x + a), then we get Q(x) as quotient and R as remainder., Then, ƒ(x), = (x + a). Q(x) = R, Put x, = – a. Then,, , ƒ(– a), = ( – a + a). Q( –a) + R, or,, R, = ƒ( –a)., Hence, remainder = ƒ( – a) = the value of the polynomial at x = – a., Speedy Optional Maths – Book 10, , 24

Page 25 :
Polynomial, , ( ), , Corollary 2: If a polynomial ƒ(x) is divided by (ax + b), then remainder R = ƒ –b, a, Proof: When we divide ƒ(x) by ax + b, then we get Q(x) as quotient and R as remainder., Then, ƒ(x) , , = (ax + b). Q(x) + R, , (, , ), , = a x + b . Q(x) + R, a, b, =–a ., b, =a –b – a, .Q – b +R, a, a, = ƒ – b = the value of ƒ(x) at x = – b ., a, a, , , Put x , , ( ), , { ( )} ( ), ( ), , Then ƒ – b, a, \ R , , ( ba ) ., , Corollary 3: If a polynomial ƒ(x) is divided by ax – b, then remainder R = ƒ, , Proof: When we divide ƒ(x) by ax – b, then we get Q(x) as quotient and R as remainder., Then, ƒ(x) , , = (ax – b). Q(x) + R, , (, , ), , = a x – b . Q(x) + R, a, b, Put x = a ., b, Then ƒ b, =a b – a .Q b +R, a, a, a, b, or, ƒ, =0+R, a, \ R , = ƒ b = the value of ƒ(x) at x = b ., a, a, , , , ( ), ( ), , (, , ) ( ), , ( ), , Example 5:, , , , , , , , If ƒ(x) = x3 + 6x2 – x – 30 is divided by x – 2, find the remainder., Solution:, Here, ƒ(x) = x3 + 6x2 – x – 30, And g(x) = x – 2, Comparing g(x) with x – a, we get, a = 2., By remainder theorem,, R = ƒ(a) = ƒ(2) = 23 + 6.22 – 2 – 30 = 8 + 24 – 32 = 0, , Example 6: Using remainder theorem, find the remainder on dividing 3x2 + 5x – 11 by, 2x + 5., , Solution:, , Let, ƒ(x) = 3x2 + 5x – 11., , , , (, , Here g(x) = 2x + 5 = 2 x + 5, 2, Comparing g(x) with x – a,, , ) = 2{x –( 52 )}, , 25, , Speedy Optional Maths – Book 10

Page 26 :
Polynomial, we get,, , a, , -5, = 2, , ( ), ( ), , =ƒ – 5, 2, 2, = 3 – 5 + 5 – 5 – 11, 2, 2, 25 25, = 3. 4 – 2 – 11, 75 25, 19, = 4 – 2 – 11 = 75 – 50 – 44 = – 4, 4, So, Remainder R, , ( ), , Example 7: Find the remainder on dividing 2x2 – 7x – 1 by x + 3., , Solution:, , Let, ƒ(x) = 2x2 – 7x – 1, , g(x) = x + 3, , Comparing g(x) with x – a, we get, a= – 3, , Now, Remainder R= ƒ( – 3) = 2( – 3)2 – 7 ( – 3) – 1 = 18 + 21 – 1 = 38., Example 8: Find the remainder when 2x3 + 13x2 – 36 is divided by (x – 3)., , Solution:, , Here, ƒ(x) = 2x3 + 13x2 – 36, , g(x) = x – 2, , Comparing g(x) with x – a, a=3, , Using remainder theorem,, , \, Remainder, R, = f(a), , = 2(3)3 + 13(3)2 – 36, , = 2.27 + 13.9 – 36 = 54 + 117 – 36 = 135, Example 9: When x3 + 3x2 – kx + 4 is divided by x – 2, the remainder is k. Find the value, of k., , Solution:, Let, ƒ(x) = x3 + 3x2 – kx + 4, By question,, , g(x) = x – 2, Remainder, R , =k, Comparing g(x) with x – a, we get, a = 2., i.e. ƒ(2) , =k, or, 8 + 3.4 – 2k + 4, =k, or, 24 –2k , =k, or, 3k , = 24, \, k , = 8., Example 10:Find the value of k when kx3 + 3x2 + 4x – 10 leaves a remainder of – 5 when, divided by x + 3., Solution:, , Let, f(x), = kx3 + 3x2 + 4x – 10 and g(x) = x + 3, , Comparing g(x) with (x – a), we get, a = – 3, Speedy Optional Maths – Book 10, , 26

Page 27 :
Polynomial, , Using remainder theorem,, , Remainder, R , = f(x), , = k( – 3)3 + 9( – 3)2 + 4( – 3) – 10, , = – 27k + 9 × 9 – 12 – 10, , = – 27k – 59, , But, by the question,, , Remainder , =–5, , or,, – 27k , = – 5 + 59, , or,, – 27k , = 54, , \, k , =–2, , 2.6 Factor theorem, Statement: If ƒ(x) is a polynomial and a is a real number, then (x – a) is a factor of, ƒ(x) if ƒ(a) = 0., Proof: If we divide ƒ(x) by x – a, then we get Q(x) as quotient and R as remainder., Then, , ƒ(x) = (x – a). Q(x) + R, ......................................... (i), Put , x, = a. Then, , ƒ(a) = (a – a). Q(a) + R, or, , ƒ(a) = R, When , ƒ(a) = 0, then R = 0., Putting the value of R in (i), we get, , ƒ(x) = (x – a). Q(x), so, (x – a) is a factor of ƒ(x)., Hence, (x – a) is a factor of ƒ(x) if ƒ(a) = 0., Corollary 1: If ƒ(x) is a polynomial and a is a real number, then (x + a) is a factor of ƒ(x) if, ƒ(–a) = 0., Corollary 2: If ƒ(x) is a polynomial and a, b are real number, then (ax + b) is a factor of ƒ(x), , ( ), , if ƒ -b = 0. Converse of the factor theorem: If (x - a) is a factor of ƒ(x), ƒ(a) = 0, a, Example 11:Show that (x – 5) is a factor of 2x2 – 11x + 5., Solution:, , Let, ƒ(x) = 2x2 – 11x + 5, , And x – a = x – 5, , \, a, = 5., , Now, remainder R = ƒ(a) = 2.52 – 11.5 + 5 = 50 – 55 + 5 = 0, , \, By factor theorem, x – 5 is a factor of 2x2 – 11x + 5., , 27, , Speedy Optional Maths – Book 10

Page 28 :
Polynomial, Example 12:Show that (x + 3) is a factor of 2x2 – x – 21., Solution:, , Let, ƒ(x) = 2x2 – x – 21, , Here, x + 3 = x – (–3)., , Now, remainder R = ƒ(–3) = 2(–3)2 – (–3) – 21 = 18 + 3 – 21 = 0, , \, By factor theorem, (x + 3) is a factor of 2x2 – x – 21., Example 13:Show that (2x – 1) is a factor of 2x2 – 11x + 5., Solution:, , Let, ƒ(x), = 2x2 – 11x + 5., , (, ), ( ), ( ) ( ), , =2 x – 1, 2, , Now, remainder R = ƒ 1, 2, 2, =2 1, – 11 1 + 5, 2, 2, 11, , = 2. 1 – 2 + 5 = 1 – 11 + 5 = 1 – 11 + 10 = 0, 2, 4, 2, 2, 2, , \, By factor theorem, (2x – 1) is a factor of 2x – 11x + 5., , , , Here, 2x – 1, , , , Example 14: Show that (2x + 7) is a factor of 2x3 + 7x2 – 4x – 14., , Solution:, , Let ƒ(x) = 2x3 + 7x2 – 4x – 14, , { ( – 72 )} = 2{x – (– 72 )}, = ƒ(– 7 ) = 2(– 7 ) + 7(– 7 ) – 4(– 7 ) – 14, 2, 2, 2, 2, = 2.(- 343 ) +7.(- 49 ) + 14 – 14 = – 343 + 343 + 14 – 14 = 0, 8, 4, 4, 4, , , , Here, 2x + 7 = 2 x +, , , , Now, R, , , , , \, , 3, , 2, , By factor theorem, (2x + 7) is a factor of 2x3 + 7x2 – 4x – 14., , Example 15: If (x – 2) is a factor of the polynomial x2 – 3x + 5k, find the value of k., , Solution:, , Let, ƒ(x), = x2 – 3x + 5k, , And x – a, = x2 – 2, , \, a, =2, , Here, (x – 2) is a factor ƒ(x)., , Then, by the converse of factor theorem,, , ƒ(2), =0, , or,, (2)2 – 3.(2) + 5k, =0, , or,, 4 – 6 + 5k, =0, , or,, 5k, =2, , or,, k, = 2, 5, Speedy Optional Maths – Book 10, , 28

Page 29 :
Polynomial, Example 16: If 2x + 7 is a factor of 2x3 + 7x2 – kx – 14, find the value of k., Solution:, , Here,, , Let, f(x), = 2x3 + 7x2 – kx – 14, , , , , , (, , = 2x + 7 = 2. x + 7, 2, –7, Comparing g(x) with x – a, a = 2, So, Remainder is given by, Remainder, R = f(a), , , , g(x), , ), , ( ) ( ) ( ), ( ) ( ), 3, , 2, , = 2 –7 + 7 –7 – k –7 – 14, 2, 2, 2, –343, = 2, + 7 49 + 7k – 14 = –343 + –343 + 7k – 14, 8, 4, 4, 2, 2, 4, , = 7k – 14, 2, , , \, , (2x + 7) is a factor of f(x), R = 0, , , , or,, , 7k – 14 = 0, 2, , , , or,, , k = 14×2 = 4, 7, , Example 17: If (x + 1) and (x – 2) are factor of x3 + ax2 – bx – 6, find the values of, a and b., Solution:, , Let, ƒ(x) = x3 + ax2 – bx – 6., , (x + 1) is a factor of x3 + ax2 – bx – 6., , So,, ƒ(–1) = 0, , or,, (–1)3 + a(–1)2 – b(–1) – 6, =0, , or,, a + b – 7 , = 0 .............................. (i), 3, 2, , Again, (x – 2) is a factor of x + ax – bx – 6., , So,, ƒ(2) , =0, 3, 2, , or,, (2) + a.(2) – b(2) – 6, =0, , or,, 4a – 2b + 2 , =0, , or,, 2a – b + 1 , = 0 ............................. (ii), , Adding (i) and (ii), we get, , 3a – 6, =0, , or,, a, =2, , Putting a = 2 in (i), we get, , 2+b–7, =0, , or,, b, =5, , Hence, a = 2 and b = 5., , 29, , Speedy Optional Maths – Book 10

Page 30 :
Polynomial, Example 18: Prove that the polynomial f(x) = x(x – 1) (x – 2) (x – 3) – 120 has a factor, (x – 5)., Solution:, , Let, f(x) = x(x – 1) (x – 2) (x – 3) – 120, , g(x) = x – 5, , Comparing g(x) with x – a, a = 5, , Now, Remainder, R, = f(a), , = 5(5 – 1) (5 – 2) (5 – 3) – 120, , = 5. 4. 3. 2 – 120, , = 120 – 120 = 0, , \ Remainder, R, = 0, x – 5 is a factor of x (x – 1) (x – 2) (x – 3) – 120., , 2.7 Factorization of a polynomial, The splitting of a polynomial in term of product of its factor is known as, factorization. Factor theorem and synthetic division can be very good tools for, factorization of a polynomial., Example 19: Factorize: x3 – 4x2 + x + 6., Solution:, , Constant term of this polynomial is 6 and the possible factor of 6 are:, , ± 1, ± 2, ± 3, ± 6., , Let ƒ(x) = x3 – 4x2 + x + 6, , Since the degree of ƒ(x) is 3, so there will be at most three factor., , When x = 1, ƒ(1) = 1 – 4 + 1 + 6 = 4, , \, (x – 1) is not a factor., , When x = 1, ƒ(–1) = 1 – 4 – 1 + 6 = 0, , \, (x + 1) is a factor., , When x = 2, ƒ(2) = 8 – 4.4 + 2 + 6 = 0, , \, (x – 2) is a factor., , When x = –2, ƒ(–2) = –8 – 4.4 – 2 + 6 = 20, , \, (x + 2) is not factor., , When x = 3, ƒ(3) = 27 – 4.9 + 3 + 6 = 0, , \, (x – 3) is a factor., , \, (x + 1), (x – 2) and (x – 3) are three factor., , \, x3 – 4x2 + x + 6 = (x + 1) (x – 2) (x – 3)., But instead of finding all the factors by using factor theorem, the synthetic, division can be used after getting one factor with the help of factor, theorem., Example 20: Factorize: x3 – 6x2 + 11x – 6, , Solution:, , Let ƒ(x) = x3 – 6x2 + 11x – 6, Speedy Optional Maths – Book 10, , 30

Page 31 :
Polynomial, , , , , Here constant term of the polynomial ƒ(x) is 6 and the possible factors of 6 are:, ± 1, ± 2, ± 3, ± 6., When x = 1, ƒ(1) = (1)3 – 6(1)2 + 11 (1) – 6 1 – 6 + 11 – 6 = 0. So, (x – 1) is a, factor ƒ(x)., Now, we use synthetic division to find other factor., , , , 1, , 1, 1, , , , , , , -6, 1, -5, , 11, -5, 6, , -6, 6, 0, , \, Quotient is x2 – 5x + 6., So, the remaining factors of x3 – 6x2 + 11x – 6 are the factors of x2 – 5x + 6., Now, x2 – 5x + 6 = x2 – 3x – 2x + 6 = x(x – 3) – 2(x – 3) = (x – 3) (x – 2), Hence, x3 – 6x2 + 11x – 6 = (x – 1) (x – 2) (x – 3)., , Alternative Method, Example 21: Factorize: x3 – 4x2 + x + 6., , Solution:, , Here, constant term is 6. Its factors are: ± 1, ± 2, ± 3, ± 6., , Let, ƒ(x), = x3 – 4x2 + x + 6, , When x, = 1, ƒ(1) = 1 – 4 + 1 + 6 = 4, , When x, = –1, ƒ(–1) = –1 – 4 – 1 + 6 = 0, , \ (x – 1) is not a factor. , \ (x + 1) is a factor of f(x)., , , Now splitting the second and third terms of the expression in such a way that, each of the pairs of terms so formed has a factor (x + 1), we have, = x3 + x2 – 5x2 – 5x + 6x + 6, , x3 – 4x2 + x + 6, , = x2(x + 1) – 5x(x + 1) + 6(x + 1), , = (x + 1) (x2 – 5x + 6), , = (x + 1) (x2 – 3x – 2x + 6), , = (x + 1) {x (x – 3) – 2(x – 3)}, , = (x + 1) (x – 2) (x – 3)., The general rule of factor theorem leads to the following cases which are helpful to, the factorization:, a), If there is any polynomial containing integral powers of x the sum of whose, coefficients is zero, then (x – 1) is a factor of that polynomial., b), If there is any polynomial containing integral power of x, the sum of the, coefficients of odd power of x is equal to the sum of the remaining coefficients,, then (x + 1) is a factor of that polynomial., 31, , Speedy Optional Maths – Book 10

Page 32 :
Polynomial, Example 22: Factorize: x3 – 11x2 + 31x – 21., Solution:, , Here, sum of the coefficients = 1 – 11 + 31 – 21 = 0, , So, (x – 1) is a factor of the polynomial x3 – 11x2 + 31x – 21. Now splitting the, second and the third terms so that the terms can be grouped to have a factor, (x – 1), we get, , x3 – 11x2 + 31x – 21, = x3 – x2 – 10x2 + 10x + 21x – 21, , = x2(x – 1) – 10x (x – 1) + 21 (x – 1), , = (x – 1) (x2 – 10x + 21), , = (x – 1) (x2 – 7x – 3x + 21), , = (x – 1) {x(x – 7) – 3 (x – 7)} = (x – 1) (x – 7) (x – 3)., Example 23: Factorize: x3 – 4x2 + x + 6, , Solution:, , Sum of the coefficients of odd powers of x = 1 + 1 = 2, , Sum of the coefficients of even powers of x = –4 + 6 = 2, , So, (x + 1) is a factor of x3 – 4x2 + x + 6., , , , , , , Now, x3 – 4x2 + x + 6 = x3 + x2 – 5x2 – 5x + 6x + 6, = x2(x + 1) – 5x (x + 1) + 6 (x + 1), = (x + 1) (x2 – 5x + 6), = (x + 1) {x (x – 3) – 2 (x – 3)}, = (x + 1) (x – 3) (x – 2)., , 2.8 Polynomial Equation, Let ƒ(x) = anxn + an–1xn–1 + ... + a0 be a polynomial in x. Then ƒ(x) = 0 is called a, polynomial equation in x., ax + b = 0 is a linear equation, ax2 bx + c = 0 is a quadratic equation, ax3 + bx2 + cx + d = 0 is a cubic equation, ax4 + bx3 + cx2 + dx + e = 0 is a big bi-quadratic equation, (Here, a, b, c, d, e are real numbres.), If a is a number such that ƒ(a) = 0, then a is called a root of the polynomial, equation, ƒ(x) = 0., For example:, Consider a quadratic equation 3x2 + 5x – 2 = 0, When, x = – 2, then, , 3x2 + 5x – 2 = 3(–2)2 + 5(–2) – 2 = 12 – 10 – 2 = 0, So,, x = – 2 is a root of polynomial equation 3x2 + 5x – 2 = 0., Speedy Optional Maths – Book 10, , 32

Page 33 :
Polynomial, Example 24: Solve: 2x3 + 13x2 – 36 = 0, Solution:, 2x3 + 13x2 – 36 = 0, , or,, 2x3 + 4x2 + 9x2 + 18x – 18x – 36, , or,, 2x2 (x + 2) + 9x (x + 2) – 18 (x + 2), , or,, (x + 2) (2x2 + 9x – 18) , , or,, (x + 2) {2x (x + 6) – 3 (x + 6)}, , or,, (x + 2) (x + 6) (2x – 3) , , , , , , , Either x + 2 , or,, x + 6 , or,, 2x – 3 , From (i), x, From (ii), x, , , , From (iii), , x, , , , Hence,, , x, , =0, =0, =0, =0, =0, , = 0 ................................... (i), = 0 ................................... (ii), = 0 ................................... (iii), = –2, = –6, 3, = 2, = –2, –6, 3, 2, , Example 25: Solve: x3 – 4x2 + 5x – 2 = 0, , Solution:, , Let, ƒ(x) = x3 – 4x2 + 5x – 2, , When x = 1, ƒ(1) = 1 – 4 + 5 – 2 = 0, , \, x – 1 is a factor of ƒ(x)., , Let us use Synthetic division to get other factor., , 1, , 1, 1, , \, , -4, 1, -3, , 5, -3, 2, , -2, 2, 0, , Quotient is x2 – 3x + 2., , , Now, x2 – 3x + 2 = x2 – x + 2 = x(x – 2) – 1(x – 1) = (x – 2) (x – 1), , So, ƒ(x), = x3 – 4x2 + 5x – 2, , = (x – 1) (x – 1) (x – 2), , Now, (x – 1) (x – 1) (x – 2) = 0, From (i) x = 1, , Either, x – 1 = 0 ................................... (i), From (ii) x = 1, , or,, x – 1 = 0 ................................... (ii), From (iii) x = 2, , or,, x – 2 = 0 ................................... (iii), Hence, x = 1, 1, 2., 33, , Speedy Optional Maths – Book 10

Page 34 :
Polynomial, Example 26: If a polynomial kx3 – 8x2 + px + 6 is exactly divisible by (x2 – 2x – 3) then, find the values of k and p., Solution:, , Here, Let the given polynomial be, , ƒ(x) = kx3 – 8x2 + px + 6 is divided by g(x) = x2 – 2x – 3, , So,, x2 – 2x – 3, , = x2 – (3 – 1) x – 3, , = x2 – 3x + x – 3, , = (x – 3) (x + 1), , Hence, (x – 3) and (x + 1) are factor of f(x), , So, comparing x – 3 with x – a, a = 3, , Using remainder theorem,, , Remainder, R , = f(a), , = k(3)3 – 8(3)2 + p(3) + 6, , = 27k – 72 + 3p + 6, , = 27k + 3p – 66, , \, (x + 3) is a factor of f(x), , R=0, , \, 27k + 3p – 66 = 0 ................................................. (i), , Again,, , (x + 1) is also a factor of f(x), , So, comparing x + 1 with x – a, a = –1, , \, R, = f(a), , or,, 0, = k( –1)3 – 8( –1)2 + p( –1) + 6, , or,, 0, =–k–8–p+6, , or,, k + p = – 2 ................................................. (ii), , Put the value of p in eqn (i), , 27 k + 3(–2 – k) – 66 = 0, , or,, 27 k – 6 – 3k – 66 = 0, , or,, 24 k = 72, , \, k=3, , Put the value of k in eqn (ii), we get, , p=2–3=–5, , \, p = –5, , k=3, , , Speedy Optional Maths – Book 10, , 34

Page 35 :
Polynomial, , EXERCISE 2.2, General Section, 1., , Divide the following and hence find Q(x) & R, a) x2 + 5x + 6 by x + 2, b) 3x3 – 2x2 – 13x + 10 by x – 2, c) 8x4 + 2x3 – 3x – 8 by 2x2 – 3, , 2., , Using synthetic division, a) divide f(x) = 3x2 – 4x + 6 by g(x) = x – 3 and hence calculate the quotient and the, remainder., b) Divide f(y) = 2y4 + 5y3 + 6 by g(y) = y + 2 and hence obtain the quotinte and the, remainder., c) Divide f(z) = z3 – 19z – 30 by g(z) = z – 5 and hence calculate the quotient and the, remainder., , 3., , Using synthetic division, find the quotient and the remainder when, a) 4x3 + 2x2 – 4x + 3 is divided by 2x + 3, b) 2x3 – 9x2 + 4x – 6 is divided by 2x – 3, c) 12x6 – 9x2 – 5 is divided by 2x2 + 1, , 4., , a) Define remainder theorem and find the remainder when 3x3 – 5x2 + 2x – 3 is, divided by x – 2 with the help of remainder theorem., b) Find the remainder when 5x3 + 7x2 – 3x + 2 is divided by x + 3 with the help of, remainder theorem., , 5., , a) Define factor theorem. Use factor theorem to show (x – 3) is a factor of, x3 – 4x2 + x + 6., b) Prove that (2x + 1) is a factor of 6x3 + 13x2 + 17x + 6., c) Use factor theorem to determine whether g(x) = x + 3 is a factor of the polynomial, f(x) = x3 – 4x2 – 18x + 9., d) Prove that f(x) = x(x + 1) (x + 3) (x + 4) – 4 has a factor (x + 2)., , 6., , a) If x3 + ax2 – x + 7 leaves a remainder 4 when divided by x – 3 find the value of a., b) The polynomial f(x) = kx3 + 9x2 + 4x – 8 when divided by (x + 3) leaves a, remainder – 20. Find the value of k., c) If 2x3 – 4x2 + 6x – k leaves a remainder 18 when divided by (x – 2) find the value, of k., , 7., , a), b), c), d), , Use factor theorem to find the factor when f(x) = x3 – 2x2 + 1., If (x – 2) is a factor of the polynomial x2 – 3x + 5a, find the value of a., If (x + 1) is a factor of 2x3 – kx2 – 8x + 5, find the value of k., If 2x + 1 is a factor of 2x3 + ax2 + x + 2, find a., 35, , Speedy Optional Maths – Book 10

Page 37 :
CHAPTER, , 3, , Sequence, and Series, , Leonardo of Pisa, the greatest European Italian, mathematician, popularly known as Fibonacci,, was the first mathematician to introduce, Hindu-Arabic system in Europe, that is the, positional system of using ten digits with a, decimal point and zero. He is popular for using, Fibonacci number sequence, that is 1, 1, 2, 3,, 5, 11... in the book authored by him, known as, Book of Calculation, the Liber Abaci., , 37, , 3.1, , Sequence, , 3.2, , Series, , 3.3, , Arithmetic Progression, , 3.4, , Sum of Arithmetic Series, , 3.5, , Arithmetic Mean, , 3.6, , Geometric Progression, , 3.7, , Sum of Geometric Series, , 3.8, , Geometric Mean, , 3.9, , Relation between A.M and G.M, , 3.10. Natural Numbers, , Speedy Optional Maths – Book 10

Page 38 :
Sequence and Series, , 3.1 , , Sequence, , , A sequence is a set of numbers written in a particular order., , E.g., 1, 3, 5, 7, 9, ........ is a set of odd numbers., , 1, 4, 9, 16, ......... is a set of square numbers., , 3.2 , , Series, , , , A series is a sum of the terms of a sequence. If there are n terms in the sequence, and we evaluate the sum then we often write, , Sn = u1 + u2 + u3 + ....................... + un, , 3.3 , , Arithmetic progression, , , , , , Let's consider the two common sequences:, 1, 3, 5, 7, ........ and 0, 10, 20, 30, .......... ., It is easy to see how these sequences are formed. They start with a particular first, term and then to get successive terms we just add a fixed value to the previous, term. In the first sequence, we add 2 to get next term where as in the second, sequence we add 10. So, the difference between consecutive term in each, sequence is constant. we could also subtract a constant instead because that is, just the same as adding a negative constant. e.g. in a sequence 8, 5, 2, – 1, – 4,, ......... the difference between the consecutive terms is – 3. Any sequence wih this, property is called an arithmetic progression or A.P. for short., , , , If 'a' stands for the first term of the sequence and 'd' stands for the common, difference between the successive terms then,, , , First term (t1) , , Second term (t2) , , Third term (t3) , , Fourth term (t4) , , . , , . , , . , nth term (tn) , , , , , =a, =a+d, = a + d + d = a + 2d, = a + 3d, ., ., ., = a + (n – 1) d, , Hence, tn = a + (n – 1) d, Sometimes, we also write 'l' for the last term of a finite sequence. So, we also write, l = a + (n – 1) d., , Speedy Optional Maths – Book 10, , 38

Page 39 :
Sequence and Series, , , , Worked out examples, , Example 1: Write down the first five terms of an A.P with first term 8 and common, difference 7., Solution:, , Here, rirst term (t1) = 8 = a, , Common difference (cd) = 7, , So, second term (t2), =a+d, , = 8 + 7 = 15, , Third term (t3), = a + 2d, , =8+2×7, , = 8 + 14 = 22, , Fourth term (t4), = a + 3d = 8 + 3 × 7, , = 8 + 21 = 29, , Fifth term (t5) , = a + 4d = 8 + 4 × 7, , = 8 + 28 = 36, , Hence, the first five terms of an A.P. with first term 8 and common difference, 7 are 8, 15, 22, 29 and 36., Example 2: Find the eighth term of the sequence 7, 11, 15, 19 ........., , Solution:, , Here, first term (a) = 7 common difference (cd) = 11 – 7 = 4, , Now, eighth term (t8), = a + 7d, , = 7 + 7 × 4 = 35, Example 3: If the 4th term and the common difference of an A.P. are 27 and 5, respectively, find the 1st term., , Solution:, = 27, , Here, 4th term, , i.e., a + 3d , = 27, [ tn = a + (n – 1)d], , or,, a + 3 × 5 = 27, , or,, a = 27 – 15, , ∴, a = 12, Example 4: If there be 60 terms in A.P. of whose the first term is 8 and last term 185,, find the 21st term., , Solution:, , Here last term (tn) = 185, , i.e., a + (n – 1) d = 185, , or,, 8 + (60 – 1) d = 185, , or,, d = 185–8 = 177 = 3, 59, 59, , Now, 21st term (t21) = a + 20d = 8 + 20 × 3 = 8 + 60 = 68, 39, , Speedy Optional Maths – Book 10

Page 40 :
Sequence and Series, Example 5: How many terms are there in the series 1 + 4 + 7 + ..... + 64?, Solution:, , First term (a) = 1 common difference (cd) = 4 – 1 = 3, , Here, last term (tn), = 64, , i.e., a + (n – 1) d, = 64, , or,, 1 + (n – 1) × 3, = 64, , or,, 3n – 2 , = 64, , or,, 3n , = 66, , or,, n , = 66 = 22, 3, , Hence, 64 is a 22th term., Example 6: An A.P is given by k, 2k , k , 0, ........, 3, 3, , a), Find the sixth term., , b), Find the nth term, , c), If the 20th term is 15, find k., , Solution:, Here, the given sequence is k, 2k , k , 0, ........, 3, 3, , So, common difference (d) = t2 – t1, 2k–3k, –k, , = 2k –k =, = 3, 3, 3, , a), Now, sixth term (t6) = a + 5d, , ( –3k ), , , , =k+5, , , , c), , 20th term (t20), , = k – 5k = -2k, 3, 3, = a + (n – 1) d, –k, = k + (n – 1) 3, = k – kn + k = 4k–kn = k(4–n), 3, 3, 3, 3, = 15, , , , or, a + 19d , , = 15, , , , or, k + 19 – k, 3, or, – 16k , 3, or, k , , b), , nth term (tn) , , , , , , , , ( ), , = 15, = 15, = -15×3 = –45, 16, 16, , Example 7: If the 5th term of an A.P. is 19 and the 8th term is 31, which term is 67?, Solutions : Since, tn = a + (n – 1)d, 5th term, = 19, , i.e., a + 4d , = 19 ....................................... (i), Speedy Optional Maths – Book 10, , 40

Page 41 :
Sequence and Series, 8th term, , i.e., a + 7d , , = 31, = 31 ....................................... (ii), , , Subtracting (i) from (ii), we get, , 3d , = 12, , or,, d , = 12 = 4, 3, , Putting the value of d in equation (i), we get, , a + 4d , = 19, , or,, a+4×4, = 19, , or,, a , = 19 – 16 = 3, , , , , , , , Let,, Then, or,, or,, , nth term, a + (n – 1) d, 3 + (n – 1) × 4, 3 + 4n – 4, , = 67, = 67, = 67, = 67, or,, n , = 68 = 17, 3, Hence, 67 is the 17th term., , Example 8: The 5th term of an A.P. is 28 and the 9th term is 48. Are 63 and 95 terms of, this A.P. ?, , Solution:, , Here, 5th term = 28, , i.e., a + 4d = 28, ....................................... (i), th, , Again, 9 term = 48, , i.e., a + 8d = 48, ....................................... (ii), , , Subtracting (i) from (ii), we get , , , , = 20 , , or,, d, = 20 = 5 , 4, , Putting the value of d in equation, , (i) we get , , a + 4d , = 28 , , or,, a+4×5, = 28 , , or,, a = 28 – 20 = 8 , , , , , , , , 4d, , or,, or,, \, , If possible, let 95 is the pth term of A.P., Then,, a + (p – 1) d, = 95, or,, 8 + (p – 1) × 5, = 95, or,, 5p + 3 , = 95, or,, 5p , = 92, , If possible, let 63 is the nth term of or,, A.P., Then, a + (n – 1) d, = 63 \, or,, 8 + (n – 1) × 5, = 63 , or,, 5n + 3 , = 63 , 41, , 5n , = 60, 12, == 12, n=, 3, 63 is the 12th term of A.P., , p , , = 92, 5, , 95 is not any term of the A.P., because p is fraction, which is, not possible., Speedy Optional Maths – Book 10

Page 42 :
Sequence and Series, Example 9: The nth term of series 9 + 7 + 5 + ... is same as the nth term of the series, 15 + 12 + 9 + ... find n., , , , Solution:, , , , For the series 9 + 7 + 5 + ...,, , , , First term (a) = 9, , , , Common difference (d) = 7 – 9 = – 2, , , , \ nth term, , , , For the series 15 + 12 + 9 + ...,, , , , First term (a) , , = 15, , , , Common difference (d), , = 12 – 15 = – 3, , , , \ nth term , , = a + (n – 1) d = 15 + (n – 1) (– 3) = 18 – 3n, , , , By question,, , = a + (n – 1) d = 9 + (n – 1) (– 2) = 11 – 2n, , , , 11 – 2n, , = 18 – 3n, , , , or,, , 3n – 2n, , = 18 – 11, , , , or,, , n , , =7, , EXERCISE 3.1, General Section, 1., , Write down the first five terms of an A.P. whose first term and common difference are, as below:, a) First term (a) = 1, Common difference = 3, b) First term (a) = – 2, Common difference = – 2, c) First term (a) = 3 , Common difference = 1, 2, 5, , 2., , Find the 7th term and 9th term of an A.P. where, a) First term (a) = 9, Common difference = 5, b) First term (a) = 1 , Common difference = –1, 3, 3, , 3., , Find the first term for each of the following A.P.'s, a) d = – 3, 8th term = 30, b) 4th term = 27, common difference = 5, , 4., , Find the common difference of the following Arithmetic sequences., a) First term = 3, 7th term = 27 b) a = 13 and t31 = – 77, , Speedy Optional Maths – Book 10, , 42

Page 43 :
Sequence and Series, 5., , Determine the number of terms for each of the following A.P.'s., a) First term = 1, Common difference = 6, Last term = 49, b) tn = 1, a = 15, d = – 2, c) 44, 36, 28, ....., – 380, , 6., , a) Which term of the sequence 2, 4, 6, ...., 512 is 256?, b) Which term of the series 3 + 6 + 9 + 12 + ..... is 96?, c) How many terms are there in the series 5 + 9 + 13 + ..... + 77?, d) Is 65 a term of the series 9 + 11 + 13 + 15 + ....?, e) Is – 42 a term of the series 7 + 2 – 3 + ....? Give reason., , Creative Section, 7., , a) If fourth and the 10th term of an A.P are 75 and 117 respectively. Find the first, term and the common difference., b) The 3rd term and the 13th term of an arithmetic sequence are –40 and 0, respectively. What is the 28th term of the sequence?, c) If the 5th and 10th term of an Arithmetic sequence are 17 and 42 respectively., Find its 20th term., , 8., , a) If 6th term and 9th term of an A.P. are 23 and 35 which term is 67?, b) The 5th term of an A.P. is 28 and the 9th term is 48. Is 93 a term of the sequence?, Give reason., c) Is 40 a term of an arithmetic sequence whose 11th and 15th terms are 23 and 35, respectively?, , 9., , a) If the nth term of the A.P. 7, 12, 17, 22, ... is equal to the nth term of another A.P., 27, 30, 33, 36, .... Find the value of n., b) If the twelveth, 85th and the last term of an arithmetic sequence are 38, 257 and, 395 respectively. Calculate how many terms are there in the sequence., , 10. a) Ram earns a monthly salary of Rs. 20, 000 per month. This salary increases by, Rs. 5000 per year. How much will his salary be in the 10th year?, b) A man repays a loan of Rs. 3250 by paying Rs. 20 in the first day then increases, the payment by Rs. 15 every day. How long will it take him to clear his debt?, , 43, , Speedy Optional Maths – Book 10

Page 44 :
Sequence and Series, , 3.4 Sum of an Arithmetic Series, Let us consider an arithmetic series, a + (a + d) + (a + 2d) + (a + 3d) + .... + (l – 2d) + (l – d) + l., Here, first term = a, common difference = d, number of terms = n,, last term (tn) = l and the term before last term = l – d, If the sum of n terms is denoted by Sn, then, Sn = a + (a + d) + (a + 2d) + ... + (l – 2d) + (l – d) + l, ..................................... (i), Writting the terms in reverse order,, Sn = l + (l – d) + (l – 2d) + ... + (a + 2d) + (a + d) + a, ..................................... (ii), Adding the corresponding terms of (i) and (ii), Sn, = a + (a + d) + (a + 2d) + ... +(l – 2d) + (l – d) + l, Sn, = l + (l – d) + (l – 2d) + ... + (a + 2d) + (a + d) + a, 2Sn, , , , = (a + l) + (a + l) + (a + l) + ... + (a + l) + (a + l) + (a + l), = n times (a + l), = n (a + l), \, Sn = n [a + l], 2, , But, the last term l = a + (n – 1) d, So, Sn = n (a + l) = n [a + a + (n – 1) d] = n [2a + (n – 1) d], 2, 2, 2, \, Sn = n [2a + (n – 1) d], 2, , Condition, , Sum formula, , If l, a, n are known, , Sn = n (a + l), 2, , If d, a, n of an A.P. are known, , Sn = n [2a + (n – 1) d], 2, , , , , Sum of first n natural numbers, The numbers 1, 2, 3, 4, ..., n are called the first n natural numbers. They form an, arithmetic sequence., , First term (a) = 1, Common difference (d) = 2 – 1 = 1, number of term (n) = n, , If Sn denoted the sum of these first n natural numbers, then, Sn = n [2a + (n – 1) d] = n [2.1 + (n – 1) .1] = n (2 + n – 1) = n (n + 1), 2, 2, 2, 2, Sum of first n odd natural numbers, 1, 3, 5, 7, ..., (2n – 1) are the first n odd numbers. They also form an arithmetic, sequence., , Here, first term (a) = 1, Common difference (d) = 3–1 = 2, number of terms (n) = n, , If Sn denotes the sum of these first n odd numbers, then, Sn = n [2a + (n – 1) d] = n [2.1 + (n – 1) d] = n (2 + 2n – 2) = n × 2n = n2, 2, 2, 2, 2, Sum of first n even natural numbers, , , 2, 4, 6, ..., 2n are the first n even numbers. This is also an arithmetic sequence., , , , Here, first term (a) = 2, common difference (d) = 4 – 2 = 2, number of term (n) = n, Then, Sn = n [2a + (n–1)d] = n [2.2 + (n–1).2] = n [4 + 2n–2] = n (2n + 2) = n(n + 1), 2, 2, 2, 2, Speedy Optional Maths – Book 10, , 44

Page 45 :
Sequence and Series, , , Note: Some problems in A.P. can be solved easily if we denote the terms as follows:, , No. of terms in A.P., Three, , , , , , Terms, (a – d), a, (a + d), , Five, , (a – 2d), (a – d), a, (a + d), (a + 2d), , Four, , (a – 3d), (a – d), (a + d), (a + 3d), , Six, (a – 5d), (a – 3d), (a – d), (a + d), (a + 3d), (a + 5d), Note that if the number of terms is odd, we take the middle term as a and common, difference as d. If the number of terms is even, we take the two middle terms as, (a – d) and (a + d) and common difference as 2d., , Worked out examples, , Example 1: Find the sum of first 50 terms of the sequence 1, 3, 5, 7, 9, ...., Solution:, , This is an arithmetic progression as the common difference is 2 through, over. So, a = 1, d = 2 and n= 50., , Now, we know,, , , Sn, , , , S50, , , , ∴ S50, Example 2:, , , , , , , , , , = n {2a + (n – 1) d}, 2, 50, =, {2.1 + (50 – 1) 2}, 2, = 25 {2 + 49.2}, = 25.100 = 2500, , From the series 5 + 10 + 15 + .... + 45, find n then find Sn., Solution:, Here, a = 5, d = 10 - 5 = 5, tn = 45, We have,, tn, = a + (n – 1) d, or,, 45, = 5 + (n – 1) 5, or,, 40, = (n – 1) 5, or,, (n – 1) = 8, \, n, =8+1=9, Now, Sn, = n (a + 1) = 9 {5 + 45} = 9 × 25 = 225, 2, 2, , Example 3: Find the first term of an A.P. whose common difference is – 7 and the sum, of 8 terms is – 76., , Solution:, , Sum of 8 terms (S8) = – 76, , Common difference (d) = – 7, , First term (a) = ?, , 45, , Speedy Optional Maths – Book 10

Page 46 :
Sequence and Series, , , We know,, , , , , , , , or, , or, , or, , \ , , = n {2a + (n – 1) d}, 2, S8, = 8 {2a + (8 – 1) 7}, 2, – 76 = 4 {2a + (– 49)}, – 19 = 2a – 49, 30, = 2a, a, = 15, , Sn, , Example 4: If the 1st term of an A.P. is 100 and the sum of its 10 term is 550. Find the, common difference., , Solution:, , Here, 1st term (a) = 100, , Sum of its first 10 term (S10) = 550, , We know that,, , , , or,, , S10, , = n {2a + (n – 1) d}, 2, 10, = 2 {2.100 + (10 – 1) d}, , , , , , , or,, or,, or,, \, , 550, 110, 9d, d, , = 5{200 + 9d}, = 200 + 9d, = – 90, = – 10, , Sn, , Example 5: Find the number of terms in an arithmetic series if its first terms is 3,, common difference is 4 and the sum of terms is 210., , Solution:, , Here, first term (a) = 3,, , Common difference (d) = 4, , Sum of the terms (Sn) = 210, , , i.e., , n [2a + (n – 1) d] , 2, , , , , , , , , , , or,, or,, or,, or,, or,, or,, or,, or,, , n [2.3 + (n – 1).4] , n (6 + 4n – 4) , n (4n + 2) , n (2n + 1) , 2n2 + n – 210 , 2n2 + 21n – 20n – 210, n (2n + 21) – 10 (2n + 21), (2n + 21) (n – 10) , , , , , , = 210, , = 420, = 420, = 420, = 210, =0, =0, =0, =0, \, n , = – 21 or 10, 2, But, n cannot be negative and fraction as it denotes number of terms., So, n = 10, , Speedy Optional Maths – Book 10, , 46

Page 47 :
Sequence and Series, Example 6:, , , , , , , , If the sum of first 12 terms of an A.P. whose 5th term is 17 and 10th term is 42., Solution:, Here, 5th term (t5), = 17, i.e., a + 4d , = 17 ......................................... (i), Again, 10th term (t10), i.e., a + 9d , , = 42, = 42 ......................................... (ii), , , Subtracting (i) from (ii), we get, , 5d, = 25, = 25 = 5, 5, , Putting the value of d in (i), we get, , a + 4.5, = 17, , or,, a , = 17 – 20 = – 3, , , or,, , d, , , , Now, sum of first 12 terms = n [2a + (n – 1) d] = 12 [2.(– 3) + (12 – 1).5], 2, 2, , = 6 [– 6 + 55] = 6 × 49 = 294, Example 7: Find the sum of the numbers from 1 to 100 which are exactly divisible by 3., , Solution:, , The sequence of the number from 1 to 100 which are divisible by 3 is 3, 6, 9,, 12, ..., 99., , Here,, , , , i.e., , or,, , or,, , first term (a) = 3,, common difference (d) = 6 – 3 = 3, Last term (l) = 99, a + (n – 1) d = 99, 3 + (n – 1).3 = 99, 3n , = 99, , , , \, , , , Now, required sum, , n , , , , = 99 = 33, 3, n, = (a + l), 2, = 33 (3 + 99) = 33 × 102 = 33 × 102, 2, 2, 2, , (, , ) = 1683, , Example 8: The sum of the three numbers in A.P. is 15 and the sum of their squares is, 83. Find the numbers., , Solution:, , Let the three numbers in A.P. be a – d, a and (a + d), , Now, sum of the numbers = 15, , i.e., (a – d) + a + (a + d) = 15, 47, , Speedy Optional Maths – Book 10

Page 48 :
Sequence and Series, , , or,, , 3a, , = 15, , , , , , , = 15 = 5, 3, Again, (a – d)2 + a2 + (a + d)2 , or,, (5 – d)2 + 52 + (5 + d)2 , or,, 25 – 10d + d2 + 25 + 25 + 10d + d2, or,, 2d2 + 75 , , , , or,, , , , , , , , , , , , , \, d , When a = 5 and d = 2,, First number , =a–d=5–2=3, Second number, =a=5, Third number, =a+d=5+2=7, When a = 5 and d = – 2,, First number , =a–d=5+2=7, Second number, =a=5, Third number, =a+d=5–2=3, Hence, three numbers in A.P. are 3, 5, 7 or 7, 5, 3., , a, , d2 , , = 83, = 83, = 83, = 83, =8 = 4, 2, =±2, , EXERCISE 3.2, General Section, 1., , Find the sum of the following series., a) 4 + 7 + 10 + ...... up to 12 terms., c) 8 + 13 + 18 + ...... + 63 , , b), d), , e) 1 + 1 + 0 + ...... – 17 f), 2, 2, , 44 + 39 + 34 + ...... up to 50 terms., 2 – 9 – 20 – ...... – 130, 13 + 21 + 4 + ...... + up to 8 terms., 2, 4, , 2., , a) The first term of an arithmetic series is 32 with common difference 8. Find the, sum of first 8 terms., b) If the fourth term of an A.P. is 4 and the common difference is 6. Find the sum, of first 16 terms., , 3., , a) The sum of 15 terms of an A.P. is 435 and the common difference is 4. Find, the first term., b) Find the first term of an arithmetic series whose sum of its first seven terms is, 0 and the common difference is – 3., c) An arithmetic series has 10 terms. If its last term is 50 and the sum of its terms, 275, find the first term., d) An arithmetic series with 9 terms has the sum 135. If the first term is 3, find, the last term., , Speedy Optional Maths – Book 10, , 48

Page 49 :
Sequence and Series, 4., , a) Find the number of term in an arithmetic progression which has its first term, 16, common difference 4 and the sum 120., b) Find the number of term in an arithmetic series which has its first term 7,, common difference 4 and the sum 900., c) Find the number of term in an arithmetic series which has its first term 49,, last term -31 and the sum 153., d) The first and the last terms of an A.P. are 7 and 23 respectively. Find the, number of terms in the series if its sum is 90., , 5., , a) The first term of an A.P. is 5, number of terms is 30 and the sum of terms is, 1455. Find the common difference., b) The sum of an A.P. of 10 terms is 210. Find the common difference if the first, term is 3., c) The first term of an A.P. is 9, the number of terms 7 and their sum 0. Find the, common difference., d) What is the common difference of an A.P. when the first term is 1, the last term 50 and, the sum 204., , Creative Section, 6. a) If the 9th and 29th terms of A.P. are 40 and 60 respectively, find the sum of first, 30 terms., b) If the fourth term of an A.P. is 1 and the sum of its first eight terms is 18, find, the tenth term of the series., c) If the third term of an arithmetic sequence is 1 and the fifth term is 7, find the, sum of first ten terms of a sequence., d) The sum of the ten terms of an A.P. is 50 and its fifth term is treble the, second term. Calculate the first term and the sum of first twenty terms., 7., , a) If the sum of the first three terms of an arithmetic series is 42 and that of the, first five terms is 80, find the 20th term of the series., b) If the sum of first seven terms of an A.P. is 14 and the sum of the first ten terms, is 125 then find the fourth term of the series., c) The sum of first 4 terms of an A.P. is 26 and the sum of first 8 terms is 100., Find the sum of first 12 terms., , 8., , a) Find three numbers in A.P. such that their sum is 36 and their product is 1140., b) Find three numbers in A.P. such that their sum is 21 and their product is 280., c) Find three numbers in A.P. such that their sum is 12 and the sum of their, squares is 50., , 9., , a) If the 6th term of an A.P. is 46, find the sum of first 11 terms., , 49, , Speedy Optional Maths – Book 10

Page 50 :
Sequence and Series, b), c), 10., , a), , b), , In an arithmetic sequence, the sixth term is equal to 3 times the fourth term, and the sum of the first three terms is – 12. Find A.P., If the 8th term of an A.P is 45, find the sum of first 15 terms., The first and last term of an A.S are – 24 and 72 respectively. If the sum of all, terms of the series is 600. Find the number of terms and the common, difference of the series., In an A.S, the sum of first ten terms is 520. If the seventh term is double of its, third term, calculate the first term and the common difference of the series., , 11., , a), b), c), , Find the sum of the first 50 natural numbers., Find the sum of the first 100 odd numbers., Find the sum of first 50 even numbers., , 12., , a), b), , Find the sum of all numbers between 200 and 400 divisible by 7., Find the sum of all numbers between 1 and 100 divisible by 3., , 3.5 Arithmetic Mean, If three numbers are in arithmetic progression, then the middle term is called the, arithmetic mean of first and third term. Since the three numbers 2, 4, 6 are in, arithmetic progression (with common difference d = 2), 4 is the arithmetic mean, between 2 and 6., If the three numbers a, b, c are in arithmetic progression, then by definition,, , b–a=c–b, or, 2b = a + c, , ( ), ( ), , a+c, 2, a+c, Hence the arithmetic mean between a and c is, 2, Again, when any numbers of quantities are in A.P., all the terms in between the, first and the last terms are called the arithmetic means between these two terms., For example, since, 5, 8, 11, 14, 17, 20, 23 are in A.P., so, 8, 11, 14, 17, 20 are the, arithmetic means between 5 and 23., or,, , b+b=a+c, , or, b, , =, , Arithmetic Means between two numbers a and b, Let a and b be two given numbers. Again, Let m1, m2, m3, ..., mn be n arithmetic, means between a and b. Then, a, m1, m2, m3, ..., mn, be are in A.P., Here, number of arithmetic means = n, So,, number of terms of A.P. = n + 2, It means,, , b, = (n + 2)th term of A.P., or,, b, = a + (n + 2 – 1) d, where d is common difference, or,, b, = a + (n + 1) d, Speedy Optional Maths – Book 10, , 50

Page 51 :
Sequence and Series, or, (n + 1) d, , =b–a, , , \ d, , b–a, = n+1, , Now,, , arithmetic means are given by, , , , m1 =, , a+d, , , , m2 =, , a + 2d, , , , m3 =, , a + 3d, , b–a, = a + n+1, = a + 2(b – a), n+1, 3(b, – a), =a+, n+1, , , , ., , ., , ., , , , ., , ., , ., , , , ., , ., , ., , , , mn =, , a + nd, , = a + n(b – a), n+1, , , , Worked out examples, , Example 1: Find the A.M. between 2 and 4., Solution:, , Let m be the A.M between 2 and 4, , So, a = 2, b = 4, m = ?, , , We have,, , A.M. = a+b, 2, , or,, m, = 2+4 = 3, 2, Example 2: If 3, x, y, –9, are in A.P., find the values of x and y., , Solution:, , Here, first term (a) = 3, , Number of terms (n) = 4, , Last term (t4) = – 9, , i.e., a + (n – 1) d = – 9, , or,, 3 + (4 – 1) d = – 9, , or,, 3d = – 9 – 3, , , or,, , d = –12 = – 4, 3, , , Now, x = a + d = 3 – 4 = – 1 and, , y = a + 2d = 3 + 2 (– 4) = 3 – 8 = – 5, , 51, , Speedy Optional Maths – Book 10

Page 52 :
Sequence and Series, Example 3: Insert 5 arithmetic means between 11 and 29., , Solution:, , Here, first term (a) = 11, last term (b) = 29, number of means (n) = 5, \ Common difference (d) = b–a = 29–11 = 18 = 3, 6, 5+1, n+1, , Now, m1, =a+d, = 11 + 3, = 14, m2, = a + 2d, = 11 + 2 × 3, = 17, m3, = a + 3d, = 11 + 3 × 3, = 20, m4, = a + 4d, = 11 + 4 × 3, = 23, m5, = a + 5d, = 11 + 5 × 3, = 26, , , , , Here,, , , i.e., , or,, , Second Method:, First term (a), Last term (t7), a + 6d , 11 + 6d, , = 11 number of terms (n) = 5 + 2 = 7, = 29, = 29, = 29, , = 29–11 = 18 = 3, 6, 6, , Now, m1, =a+d, = 11 + 3 = 14, m2, = a + 2d, = 11 + 2 × 3 = 17, m3, = a + 3d, = 11 + 3 × 3 = 20, m4, = a + 4d, = 11 + 4 × 3 = 23, m5, = a + 5d, = 11 + 5 × 3 = 26, , Thus 5 arithmetic means are: 14, 17, 20, 23 and 26., , , \, , d , , Example 4: The product of two numbers is 45 and their arithmetic means is 7. Find the, numbers., , Solution:, , Let a and b be the two required numbers. Then., , a . b , = 45, ............................................... (i), , or,, , , , Again, a + b , =7, 2, or,, a + b , = 17, ............................................... (ii), Putting the value of b in (ii), we get, , , , , b , , = 45, a, , , , , , a + 45, a, , = 14, , , , , , , a2 + 45, a2 – 9a – 5a + 45, (a – 9) (a – 5), a , , = 14a, =0, =0, = 5 or 9, , or,, or,, or,, \, , Speedy Optional Maths – Book 10, , 52

Page 53 :
Sequence and Series, If a = 5, then b = 45 = 45 = 9, 5, a, 45, 45, If a = 9, then b =, =, = 5., a, 9, Hence, required two numbers are 5 and 9 or 9 and 5., , , , , , Example 5: There are n arithmetic means between 20 and 80 such that the ratio of the, first mean to the last mean is 1:3. Find n., , Solution:, , Here, the sequence is 20, ......., 80., , Since there are n A.M's, the total number of terms = n + 2 (including the first, and last term), , Since they are in A.P. we can write 20, 20 + d, ........, 80 – d, 80, , where, 20 + d is the second term or first mean and 80 – d is the second last, term or last mean., , By the question,, Now, we know, tn, = a + (n – 1)d, First mean = 1, 3, Last mean, or, 80, = 20 + (n – 1)5, 1, 20, +, d, or,, 60, = 5n – 5, , or,, =, 3, 80 – d, or, 65, = 5n, , or,, 60 + 3d, = 80 – d, \ n, = 65 = 13, , or,, 4d , = 20, 5, , , \, , d , , \ n, = 13, There are n – 2 = 13 – 2 = 11, A.M.'s between 20 and 80 is 11, , =5, , EXERCISE 3.3, General Section, 1., , Find the arithmetic mean between, a) a and b, b) 5 and 15, , c), , – 9 and 15, , f), , (a – b)2 and (a + b)2, , d), , 1 and 1, 2, 4, , 2., , a), b), c), d), e), f), , Find the value of x if 7, x and 11 are in A.P., Find the value of x if 13, x, 27 are in A.P., If 6, p, q and 18 are in A.P. find the value of p and q., 5, x, y, 11 are in an arithmetic series, find the value of x and y., If 96, p, q, r, 68 are in A.P. find the values of p, q and r., If k, k2 + 1 and k + 6 are in A.P., find k., , 3., , a) One number is reciprocal to another and arithmetic mean between them is 5 ., 4, , e), , 5 and 5, 3, 2, , find them., b) One number is 3 times of the other and their arithmetic mean is 10. Find them., 53, , Speedy Optional Maths – Book 10

Page 54 :
Sequence and Series, 4., , , , a), b), c), , Insert 4 arithmetic means between 2 and – 18, Insert 5 arithmetic means between 2 and 20, Insert 6 arithmetic means between – 3 and 32., , , , d) Insert 3 arithmetic means between 3 and 13, 2, 2, , 5., , If the third mean between 6 and 16 is 12, find the number of arithmetic means., , 6., , a), , , , b), , , , c), , 7., , , 8., , , a), b), , a), b), , There are n arithmetic means between 1 and 31 such that the 7th mean: (n – 1)th, mean = 5:9. Find n., There are n arithmetic means between 5 and 35. If the second mean: the last, mean = 1:4, fins n., There are n arithmetic means between 1 and 70. If the first mean: the last, mean = 4:67, find n., Insert 6 arithmetic means between 4 and b so that the 5th mean may be 14. Also, find b., There are 3 arithmetic means between a and b. If 1st and 3rd means are 10 and, 20 respectively find a and b., The product of two numbers is 45 and their arithmetic mean is 9. Find the, numbers., The product of two number is 48 and their arithmetic mean is 7. Find the, numbers., , 3.6 Geometric Progression, , , , Let us consider the sequence, 2, 6, 18, 54, .........., Here, each term in the sequence is 3 times the previous term. Similarly, in the, sequence, 1, – 2, 4, – 8, ... each term is 2 times the previous term. Sequences, such that where the ratio between to consecutive terms is constant is called a, geometric progression or G.P. for short., , , , If a stands for the first term of the sequence and r stands for the common ratio of, the sequence, then, , , , , , , , , , , First term (t1), Second term (t2), Third term (t3), Fourth term (t4), . , . , . , nth term (tn) , , Speedy Optional Maths – Book 10, , =, =, =, =, , a, ar, ar2, ar3, ., ., ., = arn – 1, , 54

Page 55 :
Sequence and Series, , , , Worked out examples, , Example 1: Write down the first five terms of geometric progression which has first, term 1 and common ratio 1 ., 2, Solution:, , Here, first term (a) = 1, , , , Common ratio (r) = 1, 2, Second term (t2) = a.r = 1. 1 = 1, 2, 2, , ( 12 ) = 14, = ar = 1. ( 1 ) = 1, 2, 8, = ar = 1. ( 1 ) = 1, 2, 16, 2, , , , Third term (t3), , = ar2 = 1., , , , Fourth term (t4), , , , Fifth term (t5) , , , , Hence, the terms in order are 1, 1 , 1 , 1 , 1 ., 2 4 8 16, , 3, , 3, , 4, , 4, , Example 2: Find the 10th term of G.P. with first term 3 and common ratio 2., , Solution:, , Here, first term (a) = 3, , Common ratio (r) = 2, , 10th term = ?, , we know,, tn, = arn – 1, , \, t10, = ar9, , So,, t10, = 3.29, , = 3. 512, , = 1536, Example 3: Find the common ratio and 7th term of the G.P. 2, – 6, 18, ........., , Solution:, , Here, the sequence is 2, – 6, 18, ........, , , So, common ratio (r) =, , , \ 7th term, , , , , t2, = -6 = – 3, t1, 2, , = ar6, = 2 (– 3)6, = 2 × 729, = 1458, , ( tn = arn – 1), , 55, , Speedy Optional Maths – Book 10

Page 56 :
Sequence and Series, Example 4: Find the number of terms in the series 6 + 12 + 24 + ... + 768, Solution:, , Here, first term (a) = 6, common ratio (r) = 12 = 2, 6, , Let the no. of terms be n., , Then, nth term, = 768, n–1, , or,, ar , = 768, , or,, 6(2)n – 1 , = 768, , , or,, , 2n – 1 , , 768, = 6, , , , , , , or,, or,, or,, \, , 2n – 1 , 2n – 1 , n – 1 , n , , = 128, = 27, =7, =7+1=8, , Example 5: find the common ratio of a G.P. whose 4th term is 54 and the first, term 2., , Solution:, , Here, first term (a) = 2 , or, r3, = 54 = 27, 2, 4th term (t4) = 54 , or, r3, = 33, , i.e., ar3 , = 54 , \, r, =3, , or,, 2. r3 , = 54, Example 6: Find the 1st term of a G.P. whose 3rd term is 18 and the common, ratio is 1 ., 2, , Solution:, 2, , Here, third term (t3) = 18, or, , a 3, = 18, 2, 3, , Common ratio (r) =, = 18, or, a. 9, 2, 4, , So,, or, a, = 18×4 = 8, t3, = 18, 9, \ a, =8, , or,, ar2, = 18, , ( ), , Example 7: Which term of the series 16 + 24 + 36 + ... is 81?, , Solution:, , , , , Here, first term (a) = 16, common ratio (r) = 24 = 3, 16, 2, Let 81 be the nth term of the series., Then, arn – 1 = 81, , , , or,, , ( ), , 16. 3, 2, , n–1, , = 81, , Speedy Optional Maths – Book 10, , 56

Page 57 :
Sequence and Series, , , or,, , , , or,, , , , or,, \, So,, , , , , ( 32 ), ( 32 ), , n–1, , n–1, , = 81, 16, =, , ( 32 ), , 4, , n – 1 = 4, n = 5, 81 is the 5th term of the series., , Example 8: Is 1 a term of the series 2 + 1 + 1 + 1 + .... ?, 128, 2, 4, , Solution:, , , , Here, first term (a) = 2, Common ratio (r) = 1, 2, 1, If possible, let 128 be the nth term of the series., 1, Then, arn – 1 = 128, , , , or,, , , , or,, , , , or,, , n–1, , , , \, , n = 9 which is natural number, , , , So,, , 1, th, 128 is a term of the series and it is the 9 term., , , , , ( 12 ), ( 12 ), 2., , n–1, , n–1, , 1, = 256, =, , ( 12 ), , 8, , =8, , Example 9: If the 4th and 7th terms of a G.P. are 24 and 192 respectively, find the, 12th term., , Solution:, , Here, fourth term = 24, , i.e., ar3 = 24, .................................... (i), , Seventh term = 192, , i.e., ar6 = 192, .................................... (ii), , , Dividing (ii) by (i), we get, , 6, 192, = ar3 = 24, ar, , , , , , or,, or,, \, , r3, r3, r, , =8, = 23, =2, , 57, , Speedy Optional Maths – Book 10

Page 58 :
Sequence and Series, , Putting the value of r in (i), ar3, = 24, , or,, a.23, = 24, , or,, a.8, = 24, , or,, a, = 24 = 3, 8, , Now, the 12th term = ar11 = 3.(2)11 = 3 × 2048 = 6144., Example 10: In a geometric sequence, if the seventh term is 27 times the fourth term, and the 2nd term is 6, find the sequence., , Solution:, , Here, seventh term = 27 × fourth term, , i.e., ar6, = 27 × ar3, 6, ar, , or,, = 27, ar3, , or,, r3, = 27, , or,, r3, = 33, , \, r, =3, , , , , , , Again, second term = 6, i.e., ar, =6, or,, a.3, =6, \, a, = 6 =2, 3, Now, the required G.S. is a, ar, ar2, ar3, ... i.e. 2, 6, 18, 54, ..., , EXERCISE 3.4, General Section, 1., , Find the first five terms of a G.P. whose first term and common ratio are as below:, a), First term (a) = 1, Common ratio (r) = 2, b), First term (a) = 3, Common ratio (r) = 1, 3, c), First term (a) = 2, Common ratio (r) = 2, d), First term (a) = 2, Common ratio (r) = – 2, , 2., , Find the tenth and 13th term of the sequence where, , c), , t1 = 3, r = 1 , 5, First term = 25, Second term = 125, , a), , In the sequence 1, – 3, 9, ......, find the common ratio and the 8th term., , b), , In the series 4 – 1 + 1 – 1 + ..., find the common ratio and the 9th term., 4 16, , a), , 3., , Speedy Optional Maths – Book 10, , 58, , b), , First term = – 1, Common ratio = 3, , d), , First term = 7, t2 = 14

Page 59 :
Sequence and Series, c), 4., , 5., , Find the common ratio of a G.P. whose first term 1 and 9th term is 2., 8, , Find the number of terms in each of the following series., 8 + 12 + 18 + ....... + 91 1, 8, 3 + 6 + 12 + ...... + 768, , a), , 2 + 4 + 8 + 16 + ....... + 512, , b), , c), , 27 + 9 + 3 + ....... + 1 , 81, , e), , a), , What will be the first term of a geometric sequence with common ratio 3 and, the third term 36 ?, The fourth term of the series in G.P. is 81 and the common ratio is 3. Find the, first term., The third of the G.P. is 18 and its common ratio is 3. Find the fifth term., Find the 6th term of a G.P. whose first term is 2 and the 4th term is 54., , b), c), d), , Creative Section, 6., a), Which term of the sequence 7, 14, 28, 56, ... is 1792 ?, b), , Which term of the series 1 + 1 + 1 + 2 + 4 + ... is 128 ?, 4, 2, , c), , Which term of the G.P. is 64 whose first term is 27 and the second term, 27, is 18 ?, Are 70 and 80 terms of the sequence, 5, 10, 20, ... ?, , d), 7., , a), b), , 8., , a), b), , 9., , a), b), , 10., , a), , First and second terms of a geometrical sequence are 32 and 8 respectively., What will be the sixth term?, If the 6th and 10th terms of a geometrical sequence rae 64 and 1024 respectively., Find its 8th term., If 6th and 9th term of a geometrical sequence are 1 and 8 respectively which, term of the sequence is 64 ?, Which term of the sequence 1 if its 3rd and 5th terms are 27 and 3, 9, respectively?, The 4th term of a G.P. is 54 and the 6th term is 24. Is 64 a term of the G.P. ?, 9, 1, nd, th, The 2 term and 6 term of a G.P. are, and 4 respectively. Is 32 a term of, 9, the G.P. ?, The third term of a geometric series is 27 times the 6th term and the 4th term, is 9. Find the series., , 59, , Speedy Optional Maths – Book 10

Page 60 :
Sequence and Series, b), , In a geometric series, 7th term is 16 times the third term and the fifth term is, 1 . Find the 2nd term., 16, , 3.7 Sum of a Geometric Series, Let us consider a geometric series, a + ar + ar2 + ar3 + ... + arn – 3 + arn – 2 + arn – 1, Here, first term = a, common ratio = r, number of terms = n, last term (l) = arn – 1, If the sum of n terms is denoted by Sn, then, Sn = a + ar + ar2 + ar3 + ... + arn – 3 + arn – 2 + arn – 1, , ................................ (i), , Multiplying both sides of (i) by r, we get, rSn = ar + ar2 + ar3 + ar4 + ... + arn – 2 + arn – 1 + arn, , ................................ (ii), , Subtracting (i) from (ii), we get, rSn = ar + ar2 + ar3 + ar4 + ... + arn – 2 + arn – 1 + arn, Sn = a + ar + ar2 + ar3 + ... + arn – 3 + arn – 2 + arn – 1, –, –, –, –, –, –, –, –, , rSn – Sn = arn – a, or, (r – 1) Sn = a (rn – 1), n, Sn = a(r – 1), r–1, n, Again, Sn = ar – a, r–1, , \, , n–1, or, Sn = ar .r – 1, r–1, , \ Sn = lr – a, r–1, , Condition, , Sum formula, , If r, a, n are known, , Sn = a(r – 1), r–1, , If r, a, l are known, , Sn = lr – a, r–1, , n, , Note: Some problems in G.P. can be solved easily if we denote the terms as follows:, , Speedy Optional Maths – Book 10, , 60

Page 61 :
Sequence and Series, , No. of terms in G.P., , Terms, , Three, , a, r, a, r2, a, r3, a, r5, , Five, , , , , , , a, ar, , , a , a, ar, ar2, r, a, Four, , , ar, ar3, r, a, Six, , 3 , a , ar, ar3, ar5, r, r, If the number of terms is odd, we take the middle term as a and the common ratio, as r. If the number of terms is even, we take ' a ' and 'ar' as the middle terms and r2, r, as the common ratio., , Worked out examples, , Example 1: Find the sum of the geometric series 2 + 6 + 18 + 54 + ... where there are 6, terms in the series., Solution:, , Here, We have,, , 2 + 6 + 18 + 54 + ......., , So, we have,, First term (a) = 2, t2, = 6 =3, t1, 2, , , , Common ratio (r) =, , , , Number of terms (n) = 6, , , , So,, , Sn, , , , , n, = a(r – 1), r–1, 6, = 2(3 – 1), 3–1, 2(729 – 1), =, = 728, 2, , Example 2: Find the sum of the series 256 + 128 + ... + 2 + 1., , Solution:, , Here, first term (a) = 256, , Last term (l) = 1, , , , , \, , Common ratio (l) = 128 = 1, 256, 2, Sn, , –a, = lr, r–1 =, , – 511, 1. 1 – 256, 2, 2, =, = 511, 1, 1 –1, –, 2, 2, , 61, , Speedy Optional Maths – Book 10

Page 62 :
Sequence and Series, Example 3: The sum of first four terms of, G.P. with common ratio 3 is 80., Find the first term., Solution:, , Here,, , Sum of first four terms (S4) = 80, , Common ratio (r) = 3, , We know that,, n, = a(r – 1), r–1, 4, S4, = a(3 – 1), 3–1, a(80), , or,, 80, = 2, , \, a, =2, , \, The first term is 2., , , , Sn, , Example 4: Find the common ratio of a, geometric sequence whose, first term is 2 and 4th term is, 54?, Solution:, , Here, first term (a) = 2, 4th term (t4) = 54, , Common ratio (r) = ?, , We know that,, tn, = arn – 1, , \, t4, = 2(r)4 – 1, , or,, 54, = 2r3, , or,, 27, = r3, 3, , or,, (3), = (r)3, , \, r, =3, , Example 4: Find the number of terms in the G.P. 96, – 48, 24, ...., 3 ., 8, Solution:, , Here, first term (a) = 96, , , Last term (l) = 3, 8, , , , Common ratio (r) = –48 = -1, 96, 2, , , , No. of term (n) = ?, , , We know,, l, , , or, 3, 8, , , , or,, , 3, 8×96, , , , or,, , 1, 8×32, , = arn – 1, , ( ), = (– 1 ), 2, = (– 1 ), 2, = 96 1, 2, , n–1, , n–1, , or,, , (– 12 ) , , =, , \, \, , n – 1 , n , , =8, =9, , 8, , (– 12 ), , n–1, , n–1, , Example 5: Find the sum of first 6 terms of G.P. whose second term is 48 and fifth term is 6., , Solution:, , Second term , = 48, , or,, a r , = 48, ............................................. (i), , fifth term = 6, , or,, ar4 , =6, ............................................. (ii), Speedy Optional Maths – Book 10, , 62

Page 63 :
Sequence and Series, , , Dividing (i) by (ii), we get , , , , , or,, , , , \, , Putting the value of r in (i), we get, , ar, ar4, 1, r3, , = 48, 6, = 8 , , a, , = 1 = 48, 2, , r, , = 1 , 2, , a, , = 96, , or,, , {( 12 ) –1} = (96 641 –1), 6, , , , Now, Sum of first 6 terms =, , a(r – 1), r–1 =, 6, , 96, , 1 –1, 2, , – 1, 2, , ( ), , 96 – 63, 64, =, – 1, 2, = 96 × 63 × 2 = 189, 1, 64, , , , , Example 6: Three numbers are in G.P. If their sum is 21 and their product is 64, find them., Solution:, , Let three numbers in G.P. be a , a and ar., r, , Here, product of the numbers = 64, a × a × ar = 64, , or,, r, , or,, a3 = 64, , or,, a3 = 43, , or,, a=4, When r = 4 and a = 4, , Again, sum of the numbers = 21, a = 4 =1, a, , or,, + a + ar = 21, r, 4, r, a, = 4 and, , or, a a + 1 + r, = 21, ar = 4 × 4 = 16, r, , , , (, , , , , , , , , or,, or,, or,, or,, or,, or,, , , , \, , ), , 4 (1 + r + r2), 4r2 + 4r + 4, 4r2 – 17r + 4, 4r2 – 16r – r + 4, 4r (r – 4) –1(r – 4), (r – 4) (4r – 1), r, , = 21r, = 21r, =0, =0, =0, =0, , Hence,, , = 4 or 1, 4, , a, , r, , required three numbers are 1, 4,, 16 or 16, 4, 1, When r, , a, ar, , 63, , = 1 and a = 4, 4, 4, =, = 16, 1, 4, =4, =4× 1 =1, 4, , Speedy Optional Maths – Book 10

Page 64 :
Sequence and Series, , EXERCISE 3.5, General Section, 1., , Find the sum of the following series, a), 2 + 4 + 8 + ... till 8 terms , c), e), , 2., , 3., , b), , 3 + 6 + 12 + ... till 20 terms, 81 – 27 + 9 – ... till 8 terms, , d), , 81 + 27 + 9 + ... till 5 terms, 1 + 1 + 1 + ... till 7 terms, 2, 4, , Find the sum of the series., 1 + 1 + 1 + ... + 1, 128, 2, 4, 81 + 27 + 9 + ... + 1, 81, , a), , 3 + 12 + 48 + ... + 768 , , b), , c), , 1 + 2 + 4 + ... + 144 , 9, 3, , d), , a), , What will be the first term of a geometric series which has its sum 280,, common ratio 3 and the last term 189 ?, Find the first term of a GP whose sum is 728, common ratio is 3 and the last, term is 486., The first term, common ratio and sum of its terms of a geometric series are 7, 3, and 280 respectively. Find the last term., The first term, common ratio and sum of the term of a geometric series are 27,, , b), c), d), , 1 and 40 40 respectively. Find the last term., 81, 2, 4., , a), , b), c), d), , 5., , a), b), c), d), , How many terms of the series 64 + 32 + 16 + ... must be taken so that the, sum may be 127 1 ?, 2, How many terms of the series 3 – 6 + 12 ... must be taken so that the sum may, be – 63 ?, Find the numbers of terms of the geometric series whose first term is 3, common, ratio 4 and the sum 255., How many terms are there in the G.P whose first terms is 2, common ratio 3, and the sum 728 ?, The first term of a G.P. is 3 and sum of the first three terms is 39. Find, common ratio., In a geometric series first term is 1 and sum of the first three term is 7., Find r., Sum of the first 6 terms of a GP is 28 and the common ratio is 3. Find the first, term., Sum of the first 8 terms of a GP is 1530 and the common ratio is 2. Find the, first term., , Speedy Optional Maths – Book 10, , 64

Page 65 :
Sequence and Series, Creative Section, 6., a), The third and 6th terms of a GP are 8 and 64 respectively. Find the sum of, the first 4 terms., b), Find the sum of the first 4 terms of a geometrical series whose second and fifth, terms are respectively 6 and 162., 7., , a), , b), c), , 8., , a), b), , 9., , a), b), c), , 10., , a), b), c), d), , The sum of the two terms of a GP is 6 and that of the first four terms is, 15 . Find the sum of the first six terms., 2, The sum of the first 2 terms of a GP is 8 and that of first 4 terms is 80. Find the, sum of the first 6 terms., In a geometric series, the sum of the first three terms is 1 and the sum of its first, 6 terms is 28. Find the common ratio and the first term., The sum of the first 8 terms of a GP is 5 times the sum of first 4 terms. Find the, common ratio., The sum of the first 6 terms of a GP is 9 times the sum of the first 3 terms. Find r., In a GP, the sum of the three numbers is 14 and their product is 64. Find them., The product of three terms of a GP is 216 and their sum is 19. Find the, numbers., The sum of three consecutive terms in G.P is 62 and their product is 1000, find, the terms., The sum of the first 2 terms of a GP is 9 and the sum of the last 2 terms is, 8, 1152. There are 12 terms altogether. Find the common ratio., In a geometric series, if the seventh term is 27 times the fourth term and the, sum of the first two terms is 8, find the series., The sum of first four terms is a G.P. is 30 and that of the last four terms is 960., if its first and last terms are 2 and 512 respectively, find the common ratio., The sum of first four terms is a G.P. is 30 and that of the last four terms is 960., if its first term is 2, find the common ratio., , 3.8 Geometric Mean, If the three numbers (or quantities) are in G.P., then the middle term is called the, geometric mean of the other two terms. In other words, the geometric mean of two, non zero numbers is defined as the square root of their product., Let a, G, b be three numbers in G.P., then the common ratio is the same i.e., G = b, a, G, or, G2 = a b, or,, , G = ab, 65, , Speedy Optional Maths – Book 10

Page 66 :
Sequence and Series, Hence the geometric mean of two numbers a and b is the square root of their, product i.e. ab, , , So, the geometric mean between 2 and 8 is G = ab = 2×8 = 16 = 4, , When any numbers of quantities are in G.P., all the terms in between the first and, last terms are called the geometric means between these two quantities., , , Let a, G1, G2, G3, ..., Gn, b be (n + 2) numbers (i.e. n number of means and two, extreme values) in G.P., then G1, G2, G3, ..., Gn are n geometric means between a, and b. If r is the common ratio of this G.P., Then, b = ar(n + 2) – 1, , or,, , b, , = a rn + 1, , , , rn + 1, , = b, a, , r, , b, = a, , or,, or,, , , , ( ), , 1, n+1, , ( ), , a, Now, 1 mean G1 = a r = a b, st, , ( ), b, mean G = ar = a ( a ), b, mean G = ar = a ( a ), , 2, , a, 2 mean G2 = ar = a b, nd, , 3, , rd, , n, , th, , , , n+1, , 2, , 3, , n, , 3, , n, , Let a, ar, ar2, ar3, .......,, , 1, n+1, , 3, n+1, , n, n+1, , l l l, ,, ,, , l be a sequence in G.P. then,, r 3 r2 r, , a, , first term, , ar, , second term, , – 1st mean, , ar2, , third term, , – 2nd mean, , ar3, , fourth term, , – 3rd mean, , fourth last term, , – third last mean, , third last term, , – second last mean, , second last term, , – last mean, , l, , r3, l, , r2, l, , r, , l, , last term, , Speedy Optional Maths – Book 10, , 66

Page 67 :
Sequence and Series, , Worked out examples, , , , Example 1: Find a G.M. between 25 and 625., Solution:, , Here, first term (a) = 25, last term (b) = 625, , Now, GM = ab = 25×625 = 125, 32, Example 2: Insert 4 G.M.'s between 81 and 3 ., , Solution:, , , Here, first term (a) = 81, last term (b) = 32 , no. of means (n) = 4, 3, , ( ), , b, Now, common ratio (r) = a, , , , , , 1, n+1, , =, , 1, 5, , ( ) ( ), 32, 3, , 81, , = 32, 243, , ( ), , 2, = 3, , 5×, , 1, 5, , = 2, 3, , First mean G1 = ar = 81 × 2 = 54, 3, , ( 23 ) = 36, Third mean G = ar = 81 × ( 2 ) = 24, 3, Fourth mean G = ar = 81 × ( 2 ) = 16, 3, 2, , Second mean G2 = ar2 = 81 ×, , 3, , 3, , 3, , 4, , 4, , 4, , , , Alternative Method:, , , , , First term (a) = 81, last term (tn) = 32 ,, 3, No. of terms (n) = 4 + 2 = 6, , , , Now, by the formula, tn, , = arn – 1, , 32, 3, , = 81.r6 – 1, , or,, , 32, 3×81, , = r5, , or,, , r5, , =, , , or,, , G.M. are, ar, ar2, , ( 23 ), , 5, , ar3, , = 81 × 2, 3, = 81 × 4, 9, = 81 × 8, 27, = 81 × 16, 81, , ar4, = 2, 3, Then the required geometric means are 54, 36, 24 and 16., , \, , , 1, 5, , r, , = 54, = 36, = 24, = 16, , Example 3: If 5, x, y, 40 are in G.P. find x and y., 67, , Speedy Optional Maths – Book 10

Page 68 :
Sequence and Series, , , , , , Solution:, Here,, The sequence is 5, x, y, 40, a = 5 and tn = 40, , , We have,, tn, , or,, 40, , or,, 8, , or,, (r)3, , \, r, , x = r, 5, or, x = 2 × 5 = 10, , Now,, = arn – 1, = 5r4 – 1, = r3, = (2)3, =2, , 40 = r, y, or, y = 40 = 20, 2, \ x = 10 and y = 20, Also,, , Example 4: Some geometric means are inserted between 5 and 80. Find the number, of means between the two numbers if the second mean is 20. Also find the, values of remaining means., , Solution:, , Here, the second mean = 20, , We also know, second mean = 3rd term, , So,, ar2, = 20, , or,, 5.r2 = 20, , or,, r2, =4, , \, r, =2, , Now,, tn, = arn – 1, So, the number of mean = n – 2 = 3, n–1, , or,, 80, = 5.2, Now, first mean (m1) = a.r, , or,, 16, = 2n – 1 , = 5 . 2 = 10, 4, n–1, or,, 2, = 2 , third mean (m2) = ar3 = 5 . 23, \, n, = 5 , = 5 . 8 = 40, , 3.9 Relation between A.M. and G.M., "Arithmetic mean (A.M.) is a always greater then Geometric mean (G.M.) between, two positive real unequal numbers"., Let us consider two numbers 2 and 8, 2+8, Here,, A.M. between 2 and 8 = 2 = 5, , G.M. between 2 and 8 = 2×8 = 4, \, A.M. > G.M., Now, to prove the above statement mathematically, let us consider two real, positive unequal numbers a and b., Speedy Optional Maths – Book 10, , 68

Page 69 :
Sequence and Series, Then,, , , A.M. between a and b =, , a+b, 2, , G.M. between a and b = ab, , Now, , , a+b–2 ab, = a + b – ab =, 2, 2, , A.M. – G.M., 2, , , , =, , ( a ) –2 a b +( b), 2, , 2, , (, =, , a – b, , ), , 2, , 2, , Square of any real number though it is negative is always positive., So,, A.M. – G.M. > 0, or,, A.M. – G.M. + G.M. > 0 + G.M., \, A.M. > G.M., Example 5: the A.M. of two number is 15 and their G.M. is 9. Find the numbers., , Solution:, , Let a and b be rhe two numbers whose A.M. is 15 and the G.M. is 9., , , , Then, a + b = 15, 2, or,, a + b = 30, , , , , And ab, or,, ab, , , , or,, , , , b, , =9, = 81, 81, = a, , ............................................... (i), , ............................................... (ii), , Putting the value of b in (i), we get, 81, a + a , = 30, or, a2 + 81 , = 30a, or, a2 – 30a + 81 , =0, 2, or, a – 27a – 3a + 81, =0, or, a(a – 27) –3 (a – 27) = 0, or, (a – 27) (a – 3) , =0, a , = 27 or 3, , When a = 27, then b = 81 = 81 = 3, a, 27, When a = 3, then b = 81 = 81 = 27, a, 3, Hence, required two numbers are, 3 and 27 or 27 and 3., , 3.10 Relation between A.M and G.M., , Telescopic Sum Method, a), Sum of the first n natural numbers, , Let, Sn = 1 + 2 + 3 + ... + n, , We know that, r2 – (r – 1)2 = r2 – r2 + 2r – 1, , or,, r2 – (r – 1)2 = 2r – 1, 69, , Speedy Optional Maths – Book 10

Page 70 :
Sequence and Series, , , , , Put r = 1 then,, Put r = 2 then,, Put r = 3 then,, , 12 – 02 = 2.1 – 1, 22 – 12 = 2.2 – 1, 32 – 22 = 2.3 – 1, , , , Put r = n then,, , n2 – (n – 1)2 = 2.n – 1, , , , or, , , or, , , , , , , n2 – 02 = 2 (1 + 2 + 3 + ... + n) – n.1, n2 = 2Sn – n, 2Sn = n2 + n, , Sn = n (n + 1), 2, Thus, the sum of the first n natural number is Sn = n (n + 1), 2, The general term of the series 1 + 2 + 3 + ... + n is tn = n, So, the sum of the first n natural number can be written in sigma notation as, follows:, \ , , n, , , , 1 + 2 + 3 + ... + n = ∑n = n (n+1), 2, n=1, , b), , , , Sum of the first n odd numbers, Let, Sn = 2 + 4 + 6 + ... + 2n, The general term of this series is tn = 2n, , , , n, n, 2 + 4 + 6 + ... + 2n = ∑2n = 2∑n = 2 × n (n+1) = n(n + 1), 2, n=1, n=1, Thus, the sum of the first n natural even number is Sn = n (n + 1), , c), , , , Sum of the first n odd numbers, Let, Sn = 1 + 3 + 5 + ... + 2n – 1, The general term of this series is tn = 2n – 1, , , , So,, , , , So,, , n, , 1 + 3 + 5 + ... + 2n – 1 = ∑(2n – 1), n=1, , , , n, , n, , n=1, , n=1, , = ∑2n – ∑1, , n, n, = 2∑n – ∑1 = 2 × n (n + 1) – n, 2, n=1, n=1, , = n2 + n – n = n2, , \, Sum of the first n odd natural numbers is Sn = n2, , , , d), Sum of the squares of the first n natural numbers, , Let, Sn = 12 + 22 + 32 + ... + n2, , We know that:, r3 – (r – 1)3 = r3 – (r3 – 3r2 + 3r – 1), , or,, r3 – (r – 1)3 = 3r2 – 3r + 1, , Putting r = 1, 2, 3, ..., n successively and adding them we get,, Speedy Optional Maths – Book 10, , 70

Page 71 :
Sequence and Series, 13 – 03 = 3.12 – 3.1 + 1, 23 – 13 = 3.22 – 3.2 + 1, 33 – 23 = 3.32 – 3.3 + 1, n3 – (n–1)3 = 3.n2 – 3.n + 1, , , n3 – 03 = 3 (12 + 22 + 32 + ... + n2) – 3 (1 + 2 + 3 + ... + n) + n, , , , or,, , , , or,, , , , or,, , , , or,, , , , or,, , , , or,, , , , So,, , n3 = 3 Sn – 3 × n (n + 1) + n, 2, 3, 3, 3Sn = n + n(n + 1) – n, 2, 3Sn = n ( n + 1) (n – 1) + 3 n(n + 1), 2, 3, 3Sn = n(n + 1) (n – 1) + n(n + 1), 2, 3, 3Sn = n(n + 1) (n – 1 + ), 2, n(n+1) (2n+1), Sn =, 6, n, n(n+1) (2n+1), 12 + 22 + 32 + ... + n2 = ∑n2 =, 6, n=1, , e), Sum of the cubes of the first n natural numbers, , Let, Sn = 13 + 23 + 33 + ... + n3, , We know that,, r4 – (r–1)4 = 4r3 – 6r2 + 4r – 1, , Putting r = 1, 2, 3, ..., n successively and adding them we get,, , 14 – 04 = 4.13 – 6.12 + 4.1 – 1, , 24 – 14 = 4.23 – 6.22 + 4.2 – 1, , 34 – 24 = 4.33 – 6.32 + 4.3 – 1, , , n4 – (n–1)4 = 4.n3 – 6.n2 + 4.n – 1, , , , n4 – 04 = 4(13 + 23 + 33 + .. + n3) –6(12 + 22 + 32 + .. + n2) +4(1 + 2 + .. + n) –n, , , , or,, , n4, , = 4Sn – 6 ×, , , , , , , , or,, or,, or,, or,, or,, , 4Sn, 4Sn, 4Sn, 4Sn, 4Sn, , = n4 + n + n(n + 1) (2n + 1) – 2n (n + 1), = n (n + 1) (n2 – n + 1) + n (n + 1) (2n + 1) –2n (n + 1), = n (n + 1) (n2 – n + 1 + 2n + 1 – 2), = n(n + 1) (n2 + n), = n2 (n + 1)2, , , , or,, , Sn, , =, , [, , n(n+1), 2, , n(n+1) (2n+1), n(n+1), +4×, –n, 6, 2, , ], , 2, , 71, , Speedy Optional Maths – Book 10

Page 72 :
Sequence and Series, , , [, , n(n+1), 2, The formulae derived above are summarized below: , Thus, the sum of the cubes of the n natural numbers is Sn =, , Sn = n [2a + (n – 1)d], 2, n, , ∑n2 =, , n=1, , n, , ∑n = n (n+1), 2, n=1, , n(n+1) (2n+1), 6, , n, , ∑n3 =, , n=1, , [ n(n+1), ], 2, , tn = a + (n–1)d, , Sn = n (n + 1), 2, , Example 6: Find the sum of, , a), 1 + 2 + 3 + 4 + ....... 100 terms, , b), 12 + 22 + 32 + 42 + ....... 92, , c), 13 + 23 + 33 + 43 + ....... (20)3, , Solution:, , a), Here,, , This is the sum of squares of first 100 natural numbers, , \, n , = 100, n(n+1), , \, Sn, =, 2, 100(100+1), =, 6, = 5050, , b), Here,, , , This is the sum of squares of first 9 natural numbers, , , , \, , n , , =9, , \, Sn, = n (n + 1) (2n + 1), 6, = 9 (9 + 1) (2 × 9 + 1), 6, = 9 × 10 × 19 = 285, 6, , c), Here,, , This is the sum of cubes of first 20 natural numbers., , \, n , = 20, n2(n+1)2, 4, 2, (20) (21)2, =, 4, = 44100, , , \, , Sn, , Speedy Optional Maths – Book 10, , =, , 72, , 2, , ], , 2

Page 73 :
Sequence and Series, , EXERCISE 3.6, General Section, 1., , 2., , 3., , 4., , Find the geometric mean between the two numbers., a), , 9 and 16, , b), , 8 and 4, , c), , d), , a and b, , e), , – 9 and – 81, , f), , 8 and 1, 2, – 4 and – 25, , Find the values of x and y if, a), , 64, x, y, 8 are in G.P. , , c), , 5, x, y, 40 are in G.P., , 2 , x, 3 are in G.P., 3, 2, , b), , Find the values of the variables if they are in G.P., a), k – 5, k + 1, k + 5, b) 3k, k + 6, 3k + 8, d), , 1 , x, y, z, 2, 8, , a), , Insert 3 geometric means between 1 and 9., 9, Insert 4 G.M.'s between 2 and 162., 3, Insert 5 G.M.'s between 3 and 2187., , b), c), , c) 3, p, q, r, 243, , e) x + 4, x – 2 and x – 6, , Creative Section, 5., , a), b), , Third mean between two numbers 27 and 1 is 1. Find the number of means., 27, There are three geometrical means between a and b. If the first and the third, means are 25 and 125 respectively, Find a and b., , 6., , a), , There are 4 geometrical means between a and b. If the 2nd and 4th means are, , b), , 1 and 1 respectively, find a and b., 2, 8, Some geometrical means are inserted between 5 and 160. If the third mean, between them is 40, find the number of means., , 7., , a), , There are 'n' GM's between 1 and 64. If 1st mean: last mean = 1:16, find n., , b), , There are 'n' GM's between 1 and 25. If the ratio of the first mean to third, 25, mean is 1:25. Find n., 73, , Speedy Optional Maths – Book 10

Page 74 :
Sequence and Series, 8., , a), , If A.M. and G.M. between two numbers are 10 and 8 respectively, find the two, numbers., , b), , Find two numbers whose arithmetic mean is 34 and geometric mean is 16., , c), , If the arithmetic mean of two numbers is 45 and their geometric mean is 27,, find the numbers., , d), , The arithmetic mean of two numbers is 25 and their geometric mean is 15. Find, the numbers., , 9., , e), , Find two numbers whose arithmetic mean is 5 and geometric mean 4., , f), , Find two numbers whose A.M. is 50 and G.M. is 40., , Find the sum of the following., a), , 1 + 2 + 3 + 4 + .... up to 50 terms., , b), , 12 + 22 + 32 + 42 + 52 + .... up to 100 terms., , c), , 13 + 23 + 33 + 43 + .... up to 100 terms., , , , Speedy Optional Maths – Book 10, , 74

Page 75 :
CHAPTER, , 4, , Linear, Programming, 4.1, , Linear Inequality in two variables, , 4.2, , System of Linear Inequalities, , 4.3, , Linear programming with convex, polygonal region, , 4.4, , Graph of Linear Function, , 4.5, , Graph of Quadratic Equation, , 4.6, , Graph of Cubic Equation, , 4.7, , Solution of quadratic and linear, equation, , Euler, a Swiss mathematician and physicist is, widely acclaimed for his contributions in the, field of function notation. His mathematical, genius has been proved by his contribution in, the fields of infinitesimal calculus and graph, theory. His mathematical works and research, in optics, mechanics, fluid dynamics have, been hailed to be very influential., , 75, , 4.8, , Solving the quadratic equation, by graphical method, , 4.9, , Finding the Equation of Parabola, , Speedy Optional Maths – Book 10

Page 76 :
Linear Programming, Two numbers a and b are to be equal if their difference a -b is zero, otherwise they are said to, be unequal. If is equal to b, we write a = b and we call it an equality. If a and b are unequal,, then there are two possibilities:, (i) the difference a-b is positive or , (ii) the difference a-b is negative., If the difference a -b is positive a is said to be greater then b and we a > b(a is greater then, b),If the difference a -b is negative, a is said to be less then b and we write a < b (a is less, than b), In either case, we call it an inequality., Graphically , such inequalities may be represented by two points on a number line. In, such a representation, the greater number is represented by a point to the right of the point, representing , the smaller number. In the same way. The smaller number is represented by a, point on the left side of the point representing the greater number. Given a number 3, there, are infinite number of numbers greater....or less then 3. The numbers 4567,....etc. are all, greater than 3, while the numbers......-4, -3,..-@, -1. 0, 1,2, are all less then 3. Given the set, of real numbers, a statement such as' x denotes any one of the real numbers greater than 3", is written symbolically as x > 3. Similarly, the symbol x <3 represents the statement " x is, a real number less than 3"., The graph (number line) of the inequalities x > 3 and x< 3 are as follows:, , The graph (number line ) of the first inequality x > 3 is the half - line or ray to the right of, the point x = 3: where as that of the second inequality x , 3 is the half line or ray to the left of, the point x = 3. In each case, the point x = 3 is excluded. If the point x = 3 is also included,, we write it as x . 3 and x , 3 respectively., The symbol x ≤3 is real as's is greater than or equal to 3' and the symbol x ≤ 3 is read as ', is less or equal to 3'. The graph (number line) of x 3 and x ≤ 3 are shown in the following, figures:, , The graph (number line) of x, , 3 and x ≤ 3 are called closed half lines or closed- rays., , 4.1 Linear Inequality in two variables, The general form of linear equation in x and y is ax + by + c =0. A relation represented by, ax +by + c < 0 or ax + by +c< 0 or ax + by + c > 0 or ax + by + c> 0 is known as the, liner inequality in variables x and y., , Speedy Optional Maths – Book 10, , 76

Page 77 :
Linear Programming, Graph of liner Inequalities:, (a) Graph of x ≤ 0, First of all we draw the line x=0. It is Y- axis., It is called boundary line. Since x ≤ 0,, the half plane representing x ≤ 0 lies to the, left of Y-axis including Y-axis., , (b) Graph of x ≥ 0, Equation of boundary line is x = 0, which is Y -axis. Since x ≥ 0, the half, plane representing x ≥ 0 line to the, right of Y-axis including Y-axis., , O, , O, , (c) The graphs of y ≤ 0 and y ≥ 0 are as follows:, , O, , O, , (d) Graph of x ≥ 1 and x ≤ 1, Equation of boundary line in each case is x = 1 which is parallel to Y - axis. Since, x ≥1, the half plane representing x ≥1 lies right of the line x = 1 and since x ≤1 the, half plane representing x ≤1 lies left of the line x =1., , O, O, , 77, , Speedy Optional Maths – Book 10

Page 78 :
Linear Programming, (e) Graph of y ≥ 1 and y ≤ 1, Equation of boundary line in each case is y = 1 which is parallel to X - axis. Since y, ≥1, the half plane representing y ≥1 lies above the line y = 1 and since y ≤1 the half, plane representing y ≥1 lies below the line y =1., , O, , y ≥1, , y ≤1, , O, , (f) Graph of 2x + 3y ≤ 6., Equation of boundary line is 2x + 3y =6. when x, = 0, then y = 2 and when y =, then x = 3 So, the, boundary line passes through the points (0,2) and, (3,0) . putting x = 0 and y = 0 as the testing point, in 2x + 3y <6 we get 0<6 (which is true). So the, half plane representing 2x + 3y 6 contains origin., , (g) Graph of 3x + 4y ≥ 12., Equation of boundary line is 3x + 4y = 12., when x = 0, then y = 3 and when y = 0, then, x = 4 So, the boundary line passes through the, points (0, 3) and (4, 0). putting x = 0 and y = 0 as, the testing point in 3x + 4y ≥ 12, we get 0 ≥12, (which is false). so, the half plane representing., 3x + 4y ≥ 12 does not contain origin., , O, , O, , 4.2 System of Linear Inequalities, A set of two or more linear inequalities having a common solution set is said to be the, system of linear inequalities. In this system, the plane regions determined by the set of, inequalities are shown on the same graph paper., , Speedy Optional Maths – Book 10, , 78

Page 79 :
Linear Programming, (a) Solution set of 2x + 5y ≤ 10 and x ≥ 0., Equation of boundary line of inequality 2x =, 5y =10 when x =, then y = 2 when y = 0,then, x =5 So the boundary line passes through the, points(0,2) and (5,0) putting x = 0 and y = 0 as, the testing point in 2x + 5y < 10 we get 0 < 10, O, (which is true). So, the half plane representing, 2x + 5y ≤ 10 contains the origin. Again, equation of boundary line of inequality x ≥ 0, is x = 0 which is y-axis. Since x ≥ 0, the half, plane representing x ≥ 0 lies to the right of y -axis. The shaded region represents, the solution set., , Solution set of y ≤ 2 x + 4, y≥- x-2 and y< 4- 4 x., (i) First inequality y ≤ 2x + 4, , Equation of boundary line is y = 2x + 4, , when x = 0, then y = 4, when y = 0, then x = -2, , So, the boundary line passes through the points (0,4) and (-2,0), Putting x = 0 andy = 0 as the testing point in y <2 x + 4 , we get 0 < 4, (which is true), , So, the half plane representing y ≤ 2 x + 4 contains origin., (b), , (ii), , , , , , , Second inequality y ≥ -x-2., Equation of boundary line is y = -x-2, when x = 0, then y = -2, when y = 0, then x = -2, So, the boundary line passes through the points (0,-2) and (-2,0), Putting x = 0 and y = 0 as the testing point in y > -x-2 we gent 0> -2, (which is true)., So, the half plane representing y ≥ -x- 2 contains origin., , (iii) Third inequality y ≤ 4 - 4 x, , Equation of boundary line is y = 4-4 x., , when x = 0, then y = 4, , when y = 0, then x = 1, So, the boundary line passes through the, points (0,4) and (1,0)., , Putting x = 0 and y = 0 as the testing point in, y < 4-4x we get 0 < 4 (which is true). So, the half, plane representing y ≤ 4 - 4x contains the origin., The shaded region represents the solution set of given system of linear, inequalities., 79, , Speedy Optional Maths – Book 10

Page 80 :
Linear Programming, Note: The shaded region in the above example is a triangular region which is the common, region of three closed half - planes determined by the inequalities y ≤ 2x + 4,, y ≥ -x-2 and y ≤ 4 -4x., Here, triangular region ABC is called a convex polygonal region. Any plane region, which is the interaction of a finite number of closed half- planes is called a convex, polygonal region. A plane region is convex when it contains every line segment joining, any two points in the region., , 4.3 Liner Programming with convex polygonal region, Convex polygonal regions in the plane are important in studying the values of linear, expression in the from F = ax + by + c where a, b and c are real numbers and x and y, are variable. Any liner function F = ax + by + c define on system of linear attains its, maximum and minimum values at the vertices of that system of linear qualities. Such, linear function F = ax + by + c whose value is to be maximized or minimized is sad, to be an objective function. The conditions (liner inequalities) to be satisfied by the, variables x and y of the objective function are said to be the constraints. The convex, polygonal region determined by the constraints is also called as feasible region., Liner Programming, Liner programming is a mathematical technique for solving problems by maximization, or minimization (optimization)to certain constrains. If we think of a business, organization then the next thing comes to our mind is profit. So having maximum, profit from the constraints like number of workers, wages paid to the workers, type, of goods used to produce any etc is an example of Linear Programming. This linear, programming is also called constrained optimizations problem., The main elements of any constraints optimization problem are:, a) Variable, The variable usually represent things that we can adjust and control., E.g the rate at which to manufacture., b) Objective function, This is a mathematical expression that combines the variable to express our goal., It may represent profit, for example: We will either be required to either minimize, or maximize the objective function., Speedy Optional Maths – Book 10, , 80

Page 81 :
Linear Programming, c) Constraints:, , These are mathematical expressions that combine the variable express limits, on the possible solutions . E.g, they may express the idea that the numbers of, workers available to operate a particular machine is limited., d) Variable bounds, , The variable are usually bounds and all the variables are not permitted to, take, my, value from minus infinity. E.g 0 and 100 might found the, production rate of a particular machine., , , , Worked out examples, , Examples 1 : Maximize: P= 5x+9y subject to constraints x + y ≤ 10, 2x + 3y ≤ 24, x>0, y > 0., Solution:, The constraints given are x + y ≤ 10, 2x + 3y ≤ 24, x>0, y>0, Considering the inequations as equations, x + y = 10 , ..................................................(i), 2x + 3y= 24 , ..................................................(ii), x = 0 , ...................................................(iii), y = 0 , ....................................................(iv), Taking equation (i), x + y =10, y= 10 - x, , x, , 10, , 0, , y, , 0, , 10, , Taking origin (0, 0) as the testing point, in equation (i)., we get, ) + 0≤10 (True), So the region including the orign represents the feasible region., Taking equation (ii), y= 24-2x, 3, , x, , 0, , 6, , y, , 8, , 4, , Taking origin (0, 0)as the testing point, o≥24 (True ). So the region including, the represents the feasible region., Taking equation (iii), x= 0, x = 0 is the line of y axis, so, x ≥ 0 represents the region on the right side of, y axis., Taking equation (iv), y = 0, y = 0 is the line of x axis so, y ≥ 0 represents the region above x axis., , 81, , Speedy Optional Maths – Book 10

Page 82 :
Hence from the graph,, O(0, 0). A (10, 0), B (6,4) and C (0, 8), represents the common feasible region., So under the objective function., , C, B, , p=5x + 9y,, O, , A, , Point, , x, , y, , z = 5x + 9y, , Volume, , Remark, , O(0, 0), , 0, , 0, , 5.0 + 9.0, , 0, , Minimum, , A (10, 0), , 10, , 0, , 5.10 + 9.0, , 50, , B (6, 4), , 6, , 4, , 5.6 + 9.4, , 66, , C (0, 8), , 0, , 6, , 5.0 + 9.8, , 72, , Maximum, , Hence the objective function P= 5x + 9y has maximum value 72 at C (0,8)., Example 2 : Find the maximum value of z = 3x + 4y under the constraints: x + y d 6,, x – y d 4, x t 0, y t 0, Solution:, The given inequalities are x + y d 6, x – y d 4, x t 0, y t 0., Considering these inequalities as equations, we get,, x+ y= 6, ............................................ (i), x- y=4, ............................................ (ii), x= 0, ............................................ (iii), y= 0, ............................................ (iv), Taking equation (i), we get,, x+y =6, or,, , y, , = 6 -x, , Table of values,, , x, , 0, , 6, , y, , 6, , 0, , Taking origin (0, 0) as the testing point, 0 + 0 d 6, or,, 0 d 6 [True], So, the region including the origin gives the feasible region., Taking equation (ii), we get,, x-y =4, or, y = x – 4, Speedy Optional Maths – Book 10, , 82

Page 83 :
Taking of values,, x, , 4, , 0, , y, , 0, , -4, , Taking origin (0,0) as the testing point,, , , 0+0 ≤4, , or 0 ≤ 4 (True), So, the region including the origin gives the feasible region., Taking equation (ii), we get ,, x= 0 is the line of y axis. So x ≥0 represents the line on the right side of y axis., Taking equation (iv) ie. y = 0, y = 0 is the line of x axis. So y ≥ 0 represents the line above the x axis., Hence from the graph,, O(0, 0), A (4, 0), B (5, 1), and C (0, 4) represents the common feasible region, So under the objective function, z = 3x + 4y,, , Point, , x, , y, , z = 3x + 4y, , Volume, , Remark, , O(0, 0), , 0, , 0, , 3.1 + 4.0, , 0, , Minimum, , A (4, 0), , 4, , 0, , 3.4 + 4.0, , 12, , B (5, 1), , 5, , 1, , 3.5 + 4.1, , 19, , C (0, 6), , 0, , 6, , 3,0 + 4.6, , 24, , Maximum, , Hence the objective function z= 3x + 4y has maximum value 24 at C(0. 6)., Example 3: R is the feasible region. Write down the inequalities that represent R and in, R maximize P = 4x -y, Solution:, , Here, x - intercept and y- intercept, of the boundary line AB are a = 3, and b = 2 respectively., So, equation of AB is, , or,, or,, , 2x + 3y, , =6, , The half plane with the boundary line AB does not contain origin. So the, inequality is 2x + 3y ≤ 6., 83, , Speedy Optional Maths – Book 10

Page 84 :
Again, the line BC passes through the points B(3,0) and C(6,2), So the, equation of BC is, , , y–0, , =, , 2 – 0 (x – 3), 6–3, , or,, , 2x – 6 = 6y, , or ,, , 2x – 3y = 6, , The half plane with the boundary line BC contains origin. So the inequality, is 2x - 3y ≤ 6, The boundary line AC is parallel to the x -axis and it passes through the point A (0,2)., ∴ Equation of AC is y = 2, The half plane with the boundary line AC contains origin. So, the inequality is y ≤ 2, Hence , the inequalities that represent R are 2x + 3y ≥6, 2x - 3y ≤ 6, y ≤ 2., Here, vertices of ∆ ABC are A (0, 2 ), B (3,0) and C (6, 2)., , Vertices, A (0, 2 ), , x, , y, , p = 4x-y, , Remarks, , 0, , 2, , -2, , Minimum, , B (3,0), , 3, , 0, , 12, , C (6, 2), , 6, , 2, , 22, , Maximum, , ∴ Maximum value of Pis 22 at the point C (6, 2), , EXERCISE 4.1, General Section, 1., Draw the graph of the following equations:, a) x = 0 , b) y= 0 , c) y= 2x + 1, 2., , 3., , Draw the graphs of the following inequalities., a) x ≥ 0 , b) y ≥ 0, c) x ≤ 0 , e) x ≥ 1 , f) x ≤ -2, g) y ≥ - 3, i) y ≥ 2x + 1, j) x + y ≥ 3, , d) x - y = 5, d) y ≤ 0, h) y ≤ 4, , Graph the following systems of inequalities and find vertices if they exist:, a) x + y ≥ 3 and 2x – y ≥ 2., b) x + 2y ≤ 8, 2y – x ≤ 6 and y ≥ 0., c) y + x + 2 ≥ 0, y ≤ 2x + 4 and y ≤ 4 – 4x., d) 2y – x ≤ 8, x – y ≤ 4, x ≥ 0 and y ≤ o., , Creative Section, 4. Maximize the objective function p under the following constraints., a) x + y ≤6, x – y ≥ – 2, x ≥ 0, y ≥ 0, P= 6x + 5y

Page 85 :
Linear Programming, b) 2y≥x – 1, x + y≤ 4, x≤0, y ≥0, P=3x + y, c) x + 2y ≤6, 3x + 2y≤12, x≥0, y ≥0, P= 5x + 6y, d) 2x + y ≥5, 3x + y ≤7, x≥0, P = 5x + 4y, 5., , a) Maximize the objective function under the following constraints x + y≥0,, x – y ≤0, y ≤2, x ≥ – 1, P=3x + 3y, b) Minimize Z = 2x – 3y, under the following constraints x – y ≤o,x ≥1, y ≤2., c) Find the maximum and minimum values of the objective function, , , , F= 16x – 2y + 40 subject to 3x + 5y ≤24, 0 ≤y ≤4, 0 ≤x ≤7., , 6. In the given diagram, the coordinates of B and C are (–2, –1), and (–2,8) respectively. The shaded region inside the triangle, ABC is represented by three inequalities. One of these is, x + Y≤6. Write down the coordinates of A and other two, inequalities. Also calculate the maximum value of x+ 2y, from the values which satisfy all three inequalities., , O, , O, , In the given figure, the co— ordinates of A, B and C are (3,, 2), (–3, 6) and (–3, –4) respectively. The shaded region inside, ∆ ABC is represented by three inequalities. Write down the, equation of these three inequalities and also calculate the, minimum value of 4x – 5y from the values which satisfy all, the three inequalities., , 8. In the given diagram, coordinates of the points A, E, c and (2,, 0), (5, 0), (0,5) respectively. OABC is the feasible region. Find, all the inequalities and hence minimize F = 6x + 10y + 20., , In the given diagram, ABC is the feasible region, AB is parallel, to the Y-axis and the coordinates of B are (5, 10). 2y + x≥5,, find other constraints and hence minimize C = 3x + 5y., O, 85, , Speedy Optional Maths – Book 10

Page 86 :
Linear Programming, , 10., , In the given diagram, OABC is the feasible region., Coordinates of B and C are (2,4) and (0,2) respectively,, If one of the constraints is x + y ≤ 6, find all the other, constraints and find the maximum value of P= 6x + 5y., , Graphs and Solutions of Linear and Quadratic Equations, 4.4 Graph of a Linear Function, An equation of first degree in the from of y = ax + b is known as a general equation, of first degree or linear function where a and b are constants., Draw the graph of y = 3x + 5, Solution:, Here, the given equation is y= 3x + 5, So, table of values,, , x, , 0, , 1, , -1, , y, , 5, , 8, , 2, , 4.5 Quadratic Equation, An equation of degree 2 is known as a quadratic equation which is the from of, y= ax2 + bx + c=0 where a#0. We get two values of x when the quadratic equations, are solved. They are also known as roots or solutions of the quadratic equation., Graph of quadratic equation, + Bx + c where a,b and c constants in the general form of a quadratic function., The nature of the graph is given by a parabola. The quadratic functions can classified, into the following categories., y= ax2, , a) y= ax2 , , , , b) y= ax2 +c , , c) y= ax2 + bx + c, , Worked out examples, Graph of y = ax2, Since y = ax2 or any other quadratic function gives a parabola (curve).We need to, have many points than a liner function for having a proper curve., , Speedy Optional Maths – Book 10, , 86

Page 87 :
Linear Programming, , Example 1: Draw the graph of y = x2, Solution:, x, , 0, , ±1, , ±2, , ±3, , y, , 0, , 1, , 4, , 9, , O, , Example 2: Draw the graph of y = –3x2, Solution:, y= – 3x2, x, , 0, , ±1, , ±2, , y, , 0, , –3, , – 12, , O, , Example 3: Draw the graph of y = 2x2 + 1, Solution:, Given, y = 2x2 + 1, , O, , x, , 0, , 1, , –1, , 2, , –2, , y, , 1, , 3, , 3, , 9, , 9, , Example 4: Draw the graph of y = 2x2 – 2, Solution:, Given, y = 2x2 – 2, x, , 0, , ±1, , ±2, , ±3, , y, , –2, , –1, , 2, , 7, , O, , Graph of y = ax2 + bx + c, While drawing the graph of y = ax2 + bx + c it doesn't become as easy as the graphs done, before as it will be hard to determine from where the curve turns. The turning point is, known, as the vertex of the parabola. So , we need to have the idea of finding the turning, point of a parabola. ., , 87, , Speedy Optional Maths – Book 10

Page 88 :
Linear Programming, , Finding the Vertex of Parabola, Given,, , ax2 + bx + c, , b, c, = a x2 + a x + a, b, b 2, b, or, y = a (x)2 +2.x. 2a + 2a – 2a, b 2 b2, c, or, y = a x + 2a – 4a2 + a, b 2 4ac – b2, or, y = a x + 2a + 4a2, b 2 4ac – b2, or, y = a x + 2a + 4a, –b, 4ac – b2, So,when, x= 2a , then y= 4a, , or,, , y, , 2, , c, +a, , –b 4ac – b2,  The vertex of the parabola y=ax2 + bx + c is ± 2a , 4a, , In a parabola, the vertical line which divides the parabola into two equal halves is, –b, known as line of symmetry. The equation of line is symmetry is given by x= 2a ., Example 5: Draw the graph of y= x2 – 2x – 3, Solution:, , , Here, the given equation is y= x2 – 2x – 3, which is in the form y= ax2 + bx + c. So,, , a=1, b= – 2, c= – 3, –b 4ac – b2, ∴Vertex of parabola, = 2a ,, 4a, (–2) 4.1 (–3) – (–2)2, , = 2.1 ,, 4.1, , = (1, – 4), So, taking 3 points on either side of x= 1, we get, , Example 6:, , O, , x, , 1, , 0, , –1, , –2, , 2, , 3, , 4, , y, , –4, , –3, , 0, , 5, , –3, , 0, , 5, , Draw the graph of y= x2 + 5x + 6., Solution:, , Here, the given equation is y= x2 + 5x + 6, which is in the form of y= ax2 + bx + c., So, a = 1, b= 5 and c = 6, –b 4ac – b2, , ∴ Vertex of parabola = 2a ,, 4a, , Speedy Optional Maths – Book 10, , 88, , O

Page 89 :
Linear Programming, (–5) 4.1 63) – (5)2, = 2.1 ,, 4.1, , , = ( – 2.5, – 0.25), , x, , – 2.5, , –1, , 0, , 1, , –3, , –4, , –5, , y, , – 0.25, , 2, , 6, , 12, , 2, , 6, , 12, , 4.6 Graph of Cubic Equations, Any equation of the form y= ax3 + bx2 + cx + d is a cubic function where a, b, c, and d are constants and a ≠ 0. We here shall not deal all but just deal about the graph, in the form y= ± ax3., Example 7:, , Draw the graph of y= x3., Solution:, Here, the equation is y= x3, , Example 8:, , O, , x, , 0, , 1, , –1, , 2, , –2, , y, , 0, , 1, , –1, , 8, , –8, , –x3, Draw the graph of y = 2 ., Solution:, –1, Here, the equation is y= 2 x3, x, , 0, , y, , 0, , 1, –1, 2, , –1, 1, 2, , O, 2, –4, , –2, 4, , 4.7 Solution of Quadratic and Linear equation, The point where the quadratic and the linear equation meets in known as its solution., The graphs of quadratic equation and linear equation meets/cuts at two points. The, co – ordinates of the point of intersection gives the solution. The solution is also called, the roots., Example 9: Solve graphically., y=x2 and y= 3 – 2x, Solution:, Let,, , , y= x2 , y=3 – 2x, , .................................... (i), .................................... (ii), O, 89, , Speedy Optional Maths – Book 10

Page 90 :
Linear Programming, , Taking equation (i), y= x2., x, , 0, , ±1, , ±2, , ±3, , , , y, 0, 1, 4, 9, This gives the parabola with its vertex at origin., , , , , Taking equation (ii), y= 3 – 2x, , , , , x, , 0, , 1, , 2, , 3, , –1, , –2, , –3, , y, , 3, , 1, , –1, , –3, , 5, , 1, , 9, , This gives a straight line., From graph, the parabola y=x2 and the straight line y=3 – 2x meets each other, at (1, 1) and ( – 3, 9). The solutions are (1, 1) and ( – 3, 9), , 4.8 Solving the Quadratic equation by Graphical method, As we have seen above the intersection of the straight line and the parabola gave its, solution. Similarly, the parabola/ quadratic equation can be solved and its roots can, be determined by the intersection of X axis and the parabola. Hence the intersection, of the parabola and X axis gives its solution., , Note: Sometimes the parabola may not be intersected by X-axis. In such, case the roots are expected to be imaginary., O, , Example 10: Solve graphically, x2 + 2x – 3= 0, Solution:, Let, y=x2 + 2x – 3= 0 .................................... (i), Comparing equation (i), with y=ax2 + bx + c, a=1, b=2 and c= – 3., –b 4ac – b2, ∴ Vertex of Parabola, = 2a ,, 4a, 2 4.1 (–3) – (–2)2, = 2.1,, 4.1, , = ( – 1, – 4), , Now, taking points around the vertex, we get, , , , x, , –1, , 0, , 1, , 2, , –2, , –3, , –4, , y, , –4, , –3, , 0, , 5, , –3, , 0, , 5, , As the parabola x2 + 2x – 3=0 cuts the X – axis at (1, 0) and ( –3, 0), x=1 and 3., , Example 11: Solve graphically: x2 – x – 2= 0 by using parabola and straight line., Speedy Optional Maths – Book 10, , 90

Page 91 :
O, , Linear Programming, The same problem of parabola can also be solved by splitting the given, parabola into and straight line., Solution:, Let,, y= x2 – x – 2= 0, or,, y= x2= x + 2, So, y= x2 , ............................................ (i), , y= x + 2 ............................................ (ii), , O, , , , , Equation (i) gives a parabola and equation (ii), gives a straight line. So the points of intersection gives the required solution., , , , Taking equation (i), y=x2, , , , , , x, , 0, , ±1, , ±2, , ±3, , y, , 0, , 1, , 4, , 9, , Taking equation (ii), y= x + 2, x, , 0, , 1, , –1, , 2, , –2, , y, , 2, , 3, , 1, , 4, , 0, , From graph as the parabola y=x2 and the straight line y= x + 2 meets at, (2, 4) and ( – 1, 1). x=2 and – 1 are the solutions., , 4.9 Finding the Equations of Parabola, The equation of parabola can also be calculated as its graph can be drawn through its, equation., Case 1: When the vertex is at origin and any other two points on parabola, are given., Since the vertex of this parabola is at origin its equation, will be y= ax2. Through the given points the value of ‘a’, will be calculated. Hence, the equation can be obtained., , O, 91, , Speedy Optional Maths – Book 10

Page 92 :
Linear Programming, Example 12: Find the equation of parabola through the points., Solution:, Here,, The equation of the parabola, whose vertex is at origin is given by, y= ax2. Since the parabola passes, through (1, 1)., , 1, = a (1)2, ∴, a, =1, ∴ The required equation is y =x2., , O, , Case 2: If the vertex is on Y axis at any other two points are given., , Since the vertex is on Y-axis the equation of parabola is, given by y= ax2 + c. Since c is the y–intercept it can, be simply read out from the graph or calculated using, simultaneous equations., , O, Example 13: Find the equation of parabola given below., , Solution:, Since the vertex of parabola lies on, the Y-axis its equation is given by, y=ax2 + c., As the y intercept (c)= 1 or the, vertex of parabola is at (1, 0),, c= 1., , O, , So, the equation reduces to y= ax2 + 1. Since the parabola passes through, (1, 3)., 3, =a 12 + 1, or,, a, =2, ∴ The required equation is y= 2x2 + 1, , Speedy Optional Maths – Book 10, , 92

Page 93 :
Linear Programming, , Case 3: If the vertex is anywhere and other two points along with the, vertex are given., Since the vertex is also not fixed the parabolas equation is formed by simply solving, the equations using simultaneous equations in the equation y=ax2 + bx + c., Example 14: Find the equation of parabola passing through the points (1, 0), (2, 0) and, (4, 6)., , Solution:, We know,, The equation of parabola is y=ax2 + bx + c. Since it passes through (1, 0), 0 = a (1)2 + b (1) + c, or,, a + b + c , = 0 ............................................ (i), Since it also passes through (2, 0), 0 = a (2)2 + b (2) + c, or,, 4a + 2b + c, = 0 ............................................ (ii), Subtracting (i) from (ii), we get, , 3a + b , = 0 ............................................ (iii), Again,, Since the parabola passes through (4, 6), 6 = a (4)2 + b (4) + c, or,, 16a + 4b + c, = 6 ............................................ (iv), Subtracting (ii) from (iv), , 12a + 2b , =6, or,, 6a + b , =3, , ............................................ (v), , Solving equation (iii) and equation (v), we get, 3a, =3, ∴, a, =1, Putting the value of a in equation (iii), 3.1 +b =0, ∴, b = –3, Again putting the value of a and b in equation (i), 1 + ( – 3) + c = 0, c = 2, ∴, The required equation is y= x2 – 3x + 2, 93, , Speedy Optional Maths – Book 10

Page 94 :
Linear Programming, , EXERCISE 4.2, General Section, 1., Draw the graphs of the following functions., a) y= x2 b) y= 2x2 c) y= 3x2, d) y= – x2 , , 1, e) y= – 2 x2, , 2., , Draw the graphs of following quadratic functions., a) y= 2x2 – 1 b) y= 4x2 + 5, c) y= x2 – 4 d) y= 4x2 – 3, , 3., , Draw the graphs of following quadratic functions., a) y= x2 + 2x – 8 , b) y= – x2 – 2x + 8, c) y= 4x2 – 8x + 3 , d) y= x2 – 2x + 3, , Creative Section, 4., Solve graphically and also indicate the point of intersection., a) y= x2 and y= 3 – 2x , b) y= x2 and y= 2x – 1, c) y= x2 – 2x and y= x , d) y= x2 – 5 and y= 2x + 3, e) y= x2 + 2x – 8 and y= – 5 , (f) y= x2 + 7x + 12 and x – y + 7= 0, , 5., Solve graphically., a) x2 + 2x – 3= 0 , b) x2 – 3x + 2= 0, c) x2 – x – 2= 0 , d) x2 – 5x + 6= 0 , e) x2 – x – 12= 0 , f) x2 + 7x + 12= 0, 6., , Solve the following equations with the help of a parabola and a straight line., a) x2 – 4x + 4= 0 , b) x2 – 2x – 8= 0, c) x2 + 7x + 12= 0 , d) – x2 – 2x + 3= 0, , 7., , a), , Find the equation of parabola passing through the point (0, 1), (1, 0) and, (2, 1)., b), Find the equation of parabola from the adjoining figures., , i. ii., , O, O, 8., , Draw the graphs of the following cubic functions., a), y= x3 b), , y= – x3, , c), , 1, y= – 3 x3, , y= 2x3 d), , , Speedy Optional Maths – Book 10, , 94