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MATHEMATICS, , Daily Practice Problems, Class:XI, , Time: 24 Min., , Discussion Date: 07-08/09/2015, , Target JEE, M.M.: 24, Dpp. No.-61, , [SINGLE CORRECT CHOICE TYPE], , [8 × 3 = 24], , Q.1, , The value of expression 2(sin6  + cos6 ) – 3 (sin4  + cos4 ) + 1 is equal to, (A) 2, (B) 0, (C) 4, (D) 6, , Q.2, , The value of x satisfying the equation 2(eln (x +3)) (ln e(2x – 3)) =, (A) 4, , (B), , 7, 4, , (C), , 9, 4, , 1 ln (– 6x + 20), e, , is, 2, , (D), , 98 7, 8, , Q.3, , Let S be the sum of the first n terms of the arithmetic sequence 8, 12, 16, .....……., and, T be the sum of first n terms arithmetic sequence 17, 19, 21, ….......….. ., If S – T = 0, then the value of n is equal to, (A) 8, (B) 10, (C) 18, (D) 22, , Q.4, , A cubic polynomial P(x) is such that P(1) = 1, P(2) = 2, P(3) = 3 and P(4) = 5, then P(6), is equal to, (A) 7, (B) 10, (C) 13, (D) 16, , Q.5, , The smallest integral value of p for which the polynomial P(x) = (p – 3)x2 – 2px + 5p is positive for, atleast one real x, is, (A) 0, (B) 1, (C) 2, (D) 3, , Q.6, , The number of solutions of the equation sin 2 – 2cos  + 4 sin  = 4 in [0, 5] is equal to, (A) 3, (B) 4, (C) 5, (D) 6, , Q.7, , If harmonic mean of numbers, (A) 5 (210), , Q.8, , , 1 1 1, 1, , 2 , 3 , , 10 is 10, then  is equal to, 2 2 2, 2 1, 2, (B) 10 (210), (C) 5, (D) 10, , If the quadratic equations in x, (2 sin2) x2 – 3x + 2 cos2  = 0 and (cos 2)x2 – 3x + 3 sin 2 = 0,    5 , ,, have both roots common, then number of possible values of  in , is equal to,  2 2 , , (A) 3, , (B) 5, , (C) 6, , (D) 7, , PAGE # 1

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MATHEMATICS, , Daily Practice Problems, Class:XI, , Time: 24 Min., , Discussion Date: 09-10/09/2015, , Target JEE, M.M.: 24, Dpp. No.-62, , [SINGLE CORRECT CHOICE TYPE], Q.1, , Q.2, , [8 × 3 = 24], ,  , In the interval 0,  , the equation cos2 x – cos x – x = 0 has,  2, (A) no solution, (B) exactlyone solution, (C) exactlytwo solutions, (D) more than two solutions, , Let a, b, c are distinct positive real numbers such that a > b > c. If 2 log(a – c), log (a2 – c2),, log (a2 + 2b2 + c2 ) taken in that order are in A.P., then, (A), , a,, , b,, , c are in A.P.., , (C) a, b, c are in G.P., , (B) a, b, c are in A.P., a,, , (D), , b,, , .P., c are in G.P., , Q.3, , Let   are roots of x2 – bx = b (b > 0) and ,  are roots of, The minimum value of (p2 – 8q) is equal to, (A) – 6, (B) – 4, (C) 0, (D) 4, , Q.4, , For real values of x and y, the minimum value of f (x, y) = x2 + 2xy + 2y2 + 6y + 10, is, (A) 0, (B) 1, (C) 3, (D) 10, , Q.5, , If the first term of geometric progression g1, g2, g3, g4, .......... is unity, then the minimum value of, 4g2 + 5g3 is equal to, (A), , Q.6, , 8, 5, , (B), , 2, 5, , (C), , 4, 5, , Let a1, a2, a3, ……, an are in A.P. such that an = 100, a40 – a39 =, , (D), , x2 + px + q = 0., , 1, 5, , 3, then 15th term of A.P. from end, 5, , is, (A), , Q.7, , Q.8, , 448, 5, , (B), , 452, 5, , (C), , 454, 5, , (D), , 458, 5, , 2, cos 2 x,  3  2cos x   4 which are lying in the interval, If the sum of all the solutions of the equation 2, , , [0, 315] is k, then the value of k, is, (A) 4950, (B) 5050, (C) 5151, (D) 5055, , If a, b, c are in A.P. then the numerical value of tan, (A) 1/4, , (B) 1/3, , C, A, tan is, 2, 2, , (C) 1/2, , (D) 1, , PAGE # 2

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MATHEMATICS, , Daily Practice Problems, Class:XI, Q.1, , Q.2, , Time: 24 Min., , Target JEE, M.M.: 26, Dpp. No.-63, , Discussion Date: 11-12/09/2015, , [SINGLE CORRECT CHOICE TYPE], [6 × 3 = 18], Let f (x) = ax2 + bx + c, where a, b, c  R and a  0. If a2 + c2 – b2 + 2ac < 0 then the equation, f (x) = 0 has, (A) roots real and equal, (B) non-real roots, (C) both roots greater than unity, (D) roots real and unequal, 5, Let log b, =, where , 4, then the value of q is equal to, , (2sin 2 2), , (A), , 4, 5, , 1, sin 2 ,  , 0, ,, = . If b = 2q where q is rational,, ,  and, 2, 5, 2  cos ,  2, , 3, 5, , (B), , (C), , 3, 4, , (D), , 2, 3, , Q.3, , If (a3 – 3a – 2)x2 + (a2 – a – 2)x = (2a4 – 10a + b) is true for all real values of x where a, b  R, then b is equal to, (A) 6, (B) –12, (C) –6, (D) 12, , Q.4, , In a  ABC, the value of, (A), , Q.5, , (B), , R, 2r, , (C), , R, r, , In triangle ABC, BC = 3, CA = 5 and AB = 6, then the value of, , (A), Q.6, , r, R, , a cos A  b cos B  c cos C, is equal to, abc, , 5  2 14, 5, , (B), , 15  2 14, 15, , (C), , (D), , 2r, R, , 1  sin A, is equal to, A, A, , sin, cos, , , , 2, 2, , , 5  2 14, 5, , (D), , 15  2 14, 15, , Suppose that 'a' and 'b' are non zero real numbers and the equation x2 + ax + b = 0 has, solutions a and b. Then the pair (a, b), is, (A) (– 2, 1), (B) (– 1, 2), (C) (1, – 2), (D) (2, – 1), [MULTIPLE CORRECT CHOICE TYPE], , Q.7, , Q.8, , 3,     , then incorrect value of, 4, (A) 1 + cot , (B) 1 – cot , , If, , The value of the expression, (A) odd, (C) even composite, , [2 × 4 = 8], , cos ec 2   2 cot  is equal to, (C) – 1 + cot , , (D) –1 – cot , , 1, 1, 5, , , , is, log 4 (18) 2 log 6 (3)  log 6 (2) log 3 (18), , (B) an irrational, (D) twin prime with 5, PAGE # 3

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MATHEMATICS, , Daily Practice Problems, Class:XI, , Time: 24 Min., , Discussion Date: 14-15/09/2015, , Target JEE, M.M.: 24, Dpp. No.-64, , [SINGLE CORRECT CHOICE TYPE], Q.1, , In a triangle ABC , 2 a c sin, (A) a2 + b2  c2, , Q.2, , 1, (A  B + C) is equal to, 2, , (B) c2 + a2  b2, , (C) b2  c2  a2, , [6 × 3 = 18], , (D) c2  a2  b2, , 2n  1 , 1 3 5, The value of Lim     .......  n  is equal to, n   2, 4 8, 2 , , (A), , 5, 2, , (B) 3, , (C), , 7, 2, , (D) 4, , Q.3, , Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with, common difference 6. If first and last terms of this sequence are equal, then the last term is, (A) 2, (B) 4, (C) 8, (D) 16, , Q.4, , Angles A, B and C of a triangle ABC are in AP. If, (A) /6, , Q.5, , In triangle ABC if AB = 2, BC = 4 and AC = 5, then the value of, (A), , Q.6, , (B) /4, , b, 3, then  A is equal to, =, c, 2, (C) 5/12, (D) /2, , 1, 2, , (B), , 1, 2, , (C), , 2, 5, , sin A  sin B, is equal to, sin C, , (D), , 2, 5, , If the equation 22x + a · 2x+1 + a + 1 = 0 has roots of opposite sign then the exhaustive set of real values, of a, is, (A) (–, 0), , 2, , , (B)   1,, 3 , , , 2, , , (C)   ,, 3 , , , (D) (–1, ), , [COMPREHENSION TYPE], Paragraph for question nos. 7 to 8, , [2 × 3 = 6], , Consider the equation log2 (x2 – 5x + 5) · log5 (log2 ((x + 3) x + 4 ) ) = 0 whose roots, are , ,  and , where  <  <  < ., Q.7, Q.8, , The value of || + || + || + || is equal to, (A) 2, (B) 6, (C) 8, , (D) 10, , The quadratic equation whose roots are + and  +  is, (A) x2 – 2x – 15 = 0 (B) x2 – 3x + 10 = 0 (C) x2 – 5x + 10 = 0, , (D) x2 + 2x – 15 = 0, , PAGE # 4