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MATHEMATICS, , Daily Practice Problems, Class:XI, Q.1, , Time: 20 Min., , [SINGLE CORRECT CHOICE TYPE], [5 × 3 = 15], If the cubic equation x 3 + 2x 2 – 4x + 5 = 0 has roots ,  and  , then the value of, , , , 3, , , , , ,  5 3  5  3  5, 13 , , (A) 13, Q.2, , Discussion Date: 26-27/08/2015, , Target JEE, M.M.: 21, Dpp. No.-56, , , , is equal to, , (B) 9, , (C) 8, , The value of x satisfying the equation, is equal to, (B) , , , , (A), ,  , , log  x, , ·, , (D) 5, ,  , , log 2 x, , , ·, ,  , , (C) 3, , log 4 x, , , (D), , ·, ,  , , log 8 x, , , ……  = 3, , 1, 3, , Q.3, , If two prime numbers are the roots of the equation 5x2 – 25x + 2 –  = 0 then sum of all possible, values of , is, (A) 1, (B) –1, (C) 11, (D) –11, , Q.4, , Number of solution satisfying the equations tan 4 = cot 5 and sin 2 = cos  in[0, 2], is, (A) 2, (B) 3, (C) 4, (D) 1, , Q.5, , The arithmetic mean of  and  is x2 + 1 and geometric mean of  and  is 2x + 1., If harmonic mean of  and is 4, then x equals, (A), , 1, 4, , (B), , 1, 2, , 3, 4, [MATRIX TYPE], , (C), , (D) 1, [2+2+2=6], , n, ,  cos(4r  3), Q.6, , Let fn () =, , r 1, n, ,  sin(4r  3), , ., , r 1, , Column-I, , Column-II, , (A), ,   , f 3   is equal to,  40 , , (P), , 2 1, , (B), ,  5 , f5 ,  is equal to,  108 , , (Q), , 2 1, , (C), ,   , f7 ,  is equal to,  156 , , (R), , 2 3, , (S), , 2 3, , PAGE # 1

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MATHEMATICS, , Daily Practice Problems, Class:XI, , Time: 20 Min., , Discussion Date: 28-29/08/2015, , Target JEE, M.M.: 22, Dpp. No.-57, , [SINGLE CORRECT CHOICE TYPE], Q.1, , The maximum value of the function f (x) =, (A), , Q.2, , If, , 1, 4, , (B), , (B), , (C) 1, , (D) 2, , 1, 2, , (C), , 3, 4, , (D) 1, , 1, 1, , 1, a2, a3, 4 is equal to, a, a, (C) 4, (D) 5, , If a > 0, then the minimum value of sum of numbers, (A) 2, , Q.4, , 1, , is, x 2| x | 2, 2, , 1, 1, 1, , , , then sin2 7 is equal to, cos  cos 3 cos 5, , (A) 0, , Q.3, , 1, 2, , [3 × 3 = 9], , (B) 3, , [REASONING TYPE], [1 × 3 = 3], 2, Statement -1 : The number of common solutions of the trigonometric equations 2 sin  – cos 2 = 0, and 2 cos2 – 3 sin  = 0 in the interval [0, 2] is two., Statement-2 : The number of solutions of the equation, 2cos2 – 3 sin  = 0 in the interval [0, ], is two., (A) Statement-1 is false, statement-2 is true., (B) Statement-1 is true, statement-2 is false., (C) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1., (D) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1., [INTEGER TYPE / SUBJECTIVE], , , , then find the value of 4 cos 2 cos   2 cos 2 ., 7, , Q.5, , If  =, , Q.6, , Let d bet the minimum value of f (x) = 5x2 – 2x +, , , If, ,  1  (n  1)d  r n 1, n 1, , equals, , [2 × 5 = 10], , 26, and f (x) is symmetric about x = r.., 5, , p, , where p and q are relative prime, then find the value, q, , of (3q – p)., , PAGE # 2

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MATHEMATICS, , Daily Practice Problems, , Class:XI, , Target JEE, Discussion Date: 31-01/08-09/2015 M.M.: 30, Dpp. No.-58, , Time: 30 Min., , [SINGLE CORRECT CHOICE TYPE], Q.1, , 1º , 1º, , The value of  sin 7  cos 7  is equal to, 2 , 2, , , 2 2  3 1, 4 6  2, (B) –, 4, 2 2, , (A) –, Q.2, , 2 2  3 1, 2 2, , (C), , (D), , 4 6  2, 4, , Let 45°   < 90°. If tan  + cot  = tan2 + cot2 then the value of (sin  + cos ), is, (A), , Q.3, , [5 × 3 = 15], , 2, , (B), , 1, 2, , 3 1, 2, , (C), , The smallest positive solution of the equation 2sin  · 4sin, (A), , , 4, , (B), , , 6, , (C), , 2, , , , (D), , · 8sin, , 3, , , , ·16sin, , , 3, , 4, , , , 2, 3 1, , ……  = 4, is equal to, , (D), , 5, 12, , Q.4, , The number of integral values of k for which the equation 7cosx + 5sinx = 2k + 1 has a solution is, (A) 4, (B) 8, (C) 10, (D) 12, , Q.5, , Number of distinct real solutions of the equation x2 – 3 x 2 + 2 = 0 is equal to, (A) 1, (B) 2, (C) 3, (D) 4, , Q.6, , [INTEGER TYPE / SUBJECTIVE], [3 × 5 = 15], 2, A quadratic function y = px + qx + r (where p, q and r are distinct) satisfies 3 conditions., (i), The graph of y passes through the point (1, 6)., 1 1 1, , , form arithmetic progression., p q r, (iii), p, r, q form geometric progression., Find the value of (p2 + q2 + r2)., (ii), , Q.7, , The quadratic polynomial P(x) = ax 2 + bx + c has two different zeroes including –2., The quadratic polynomial Q(x) = ax 2 + cx + b has two different zeroes including 3., If  and  be the other zeroes of P(x) and Q(x) respectively, thenfind the value of, , Q.8, , , ., , , Find all values of  between 0° & 180° satisfying the equation;, cos 6  + cos 4 + cos 2 + 1 = 0., , PAGE # 3

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MATHEMATICS, , Daily Practice Problems, Class:XI, Q.1, , Time: 30 Min., , [SINGLE CORRECT CHOICE TYPE], Maximum value of f() = sin14 + cos20 for all  R, is, (A) 1, , Q.2, , (B), , (C), , 2, , The smallest integral value c of x such that, (A) c2 + 3c – 4 = 0, , Q.3, , Discussion Date: 02-03/09/2015, , (B) c2 – 7c + 6 = 0, , (A) 9, , (B) 12, , If sin  =, , 3, , , 0<<, then the value of, 7, 2, , (A) –1, , (B) 0, , 3, , Q.5, , 3, 2, , [5 × 3 = 15], , (D) cannot be determined, , x 5, > 0, satisfies, x  5x  14, (C) c2 – 5c + 4 = 0, (D) c2 + 5c – 6 = 0, 2, , If sum of all possible values of x  (0, 2) satisfying the equation, 2 cos x · cosec x – 4 cos x – cosec x = – 2, is, , Q.4, , Target JEE, M.M.: 30, Dpp. No.-59, , 2r, and P =, Let S =  cos, 7, r 1, , (A) S + P < 0, , k, , then k is equal to, 4, (C) 16, (D) 32, , 1, cos 2 , , is equal to, tan 2   1, cosec2   1, (C) 1, , 3, ,  cos, r 1, , (B) S + P > 0, , (D), , 3, 4, , 2r, , then, 7, (C), , S, =4, P, , (D) S – P > 0, , Q.6, , [INTEGER TYPE], Find the sum of the values of m for which the quadratic equations, x2 – 11x + m = 0 and x2 – 14x + 2m = 0 may have common root., , Q.7, , Consider an equation with x as a variable 7 sin 3x – 2 sin 9x = sec2  + 4 cosec2 ,, then find the value of, , Q.8, , , , [3 × 5 = 15], , , , 15, (minimum positive root)  (maximum negative root) ., , , There are two sets M1 and M2 each of which consists of three numbers in arithmetic sequence, whose sum is 15. Let d1 and d2 be the common differences such that d1 = 1 + d2 and 8p1 = 7p2, where p1 and p2 are the product of the numbers respectively in M1 and M2. If d2 > 0 then find, p p , the value of  2 1  .,  d1  d 2 , , PAGE # 4

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MATHEMATICS, , Daily Practice Problems, Class:XI, , Time: 20 Min., , Discussion Date: 04-05/09/2015, , Target JEE, M.M.: 24, Dpp. No.-60, , [SINGLE CORRECT CHOICE TYPE], Q.1, , In a triangle XYZ, let a, b and c be the lengths of the sides opposite to the angles X,Yand Z respectively., If 1 + cos 2X – 2cos 2Y = 2sin X sin Y, then possible value(s) of, (A) 1, , (B) 2, , (C) 3, , a, is (are), b, (D) 5, , [MULTIPLE CORRECT CHOICE TYPE], Q.2, , [1 × 2 = 2], , [1 × 2 = 2], , In a triangle XYZ, let a, b and c be the lengths of the sides opposite to the angles X,Yand Z respectively., sin (X  Y ), , then possible values of n for which cos (n) = 0 is (are), sin Z, (B) 2, (C) 3, (D) 5, , If 2 (a2 – b2) = c2 and  =, (A) 1, , [INTEGER TYPE / SUBJECTIVE], , [4 × 5 = 20], , ab, bc, , b,, are in A.P. If ,  are the roots of the quadratic equation, 1  ab, 1  bc, 2ac x2 + 2abc x + (a + c) = 0, then find the value of (1 + )(1 + )., , Q.3, , For a, b, c  R – {0}, let, , Q.4, , Let the quadratic equation x2 + 3x – k = 0 has roots a, b and x2 + 3x – 10 = 0 has roots c, d such that, modulus of difference of the roots of the first equation is equal to twice the modulus of the difference of, the roots of the second equation. If the value of 'k' can be expressed as rational number in the lowest, form as m n then find the value of (m + n)., , Q.5, , If 24a – b = log7(2401) and log4(98a + 138b) = 5, then find the value of the product (ab)., , Q.6, , Find the general solution of the equation , 2 + tan x · cot, , x, x, + cot x · tan = 0, 2, 2, , PAGE # 5