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MATHEMATICS, , Daily Practice Problems, Class:XI (Phase-1), , Time: 24 Min. Discussion Date: 20-21/07/2015, , Target JEE, M.M.: 24, Dpp. No.-45, , Q.1, , [SINGLE CORRECT CHOICE TYPE], [8 × 3 = 24], 6, 6, Difference between the maximum and minimum values of sin x + cos x for all x  R, is, (A) 0.80, (B) 0.75, (C) 0.60, (D) 0.50, , Q.2, , If p(q – r)x2 + q(r – p)x + r(p – q) = 0 has equal roots, then, (A), , Q.3, , 1 1, , p r, , (B) p + r, , If sec x + tan x = 3 then value of tan, (A) 2, , Q.4, , (B), , 1, 2, , 2, is equal to, q, , (C), , p r, , r p, , (D) p2 + r2, , (C), , 1, 3, , (D), , x, is, 2, 2, 3, , The best graph of the quadratic expression y = ax2 + bx + c, where a > 0, b < 0 and c < 0, is, y, , y, , (A), , x, , O, , (B), x, , O, , y, , y, , (C), , (D), , O, , x, , x, , O, , Q.5, , If tan 51º – 2 tan 12º = tan xº and tan  · tan(60º – ) · tan(60º + ) = tan xº, then the minimum positive, value of  is equal to, (A) 9º, (B) 13º, (C) 17º, (D) 21º, , Q.6, , The difference between the corresponding real roots of the equations, x2 + ax + b = 0 and x2 + bx + a = 0 (a b) is same, then, (A) a + b – 4 = 0, (B) a + b + 4 = 0, (C) a – b = 4, , Q.7, , Q.8, , (D) none of these, , If a, b and c (where c is the largest of the three numbers) are the sides of a right triangle, then, (A) log c2 (a 2  b 2 )  0, , (B) log a 2 (c 2  b 2 )  0, , (C) logc (a 2  b 2 )  2, , (D) loga (b 2  c 2 )  2, , If 4 sin  cos  – 8 sin3 cos  =, Then the value of log, (A) 0, , , cos, 2, , 1, 3, and 8 cos4 – 8 cos2 + 1 =, , where  , 2, 2, ,  ,  0,  .,  2, ,  sin   cos  , ,  is equal to, 2, , , , (B) 1, , (C) 2, , (D) 3, , PAGE # 1

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MATHEMATICS, , Daily Practice Problems, Class:XI (Phase-1), , Time: 20 Min. Discussion Date: 22-23/07/2015, , Target JEE, M.M.: 21, Dpp. No.-46, , [SINGLE CORRECT CHOICE TYPE], Q.1, , Product of all the solution of equation x log10  100  2, , x, , (A), Q.2, , (B) 1, , 3, , (C) 10, , log3 2, ,  x is, , , (D) 100, , If S = (sin2  + cos4 ) for all  R, then, (A), , Q.3, , 1, 10, , log 2 3, , [3 × 3 = 9], , 13, S1, 16, , (B) 1  S  2, , (C), , 3, 13, S, 4, 16, , (D), , 3, S1, 4, , If , ,  are the roots of 2x3 – x + 1 = 0 then the value of 3 + 3 + 3 is, (A) –, , 1, 2, , (B), , 1, 2, , (C) –, , 3, 2, , (D) –, , 5, 4, , [MULTIPLE CORRECT CHOICE TYPE], 17, , Q.4, , Let C =  cos (10r )º and S =, r 1, , (A) C – S = 0, Q.5, , Let, , [3 × 4 = 12], , 17, ,  cot (10r )º , then, r 1, , (B) C + S = 0, , (C) C S, , (D) C S, , a = log 3 log 2 3log  2  ,, log 576, b = 3 log 2  log 3 the base of the logarithm being 10,, , c = 2  sum of the solution of the equation (3) 4 x  (3) 2 x  log3 (12)   27 = 0, and, , , , d = 7 (log7 2  log7 3), , then a  b  c  d  simplifies to, (A) rational which is not natural., (B) natural but not prime., (C) irrational, (D) even but not composite, , Q.6, , ,  2 , 2, The value of x satisfying the equation log 2 2 x   log 2 x  log 2    = 2, is,  x , , (A) a prime number, (B) a composite number, (C) an even number, (D) an odd number, , PAGE # 2

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MATHEMATICS, , Daily Practice Problems, Class:XI (Phase-1), , Time: 24 Min. Discussion Date: 24-25/07/2015, , Target JEE, M.M.: 24, Dpp. No.-47, , [SINGLE CORRECT CHOICE TYPE], Q.1, , 1, 1,  1, ,   is equal to, If 2a = 4b = 8c and abc = 288 then ,  2a 4b 8c , , (A), Q.2, , [2 × 3 = 6], , 11, 48, , (B), , 11, 24, , (C), , 11, 96, , (D), , 13, 96, , If ,  ( < ) are the real roots of equation x2 – (k + 4)x + k2 – 12 = 0 such that 4 (), then, the number of integral values of k is equal to, (A) 4, (B) 5, (C) 6, (D) 7, [MULTIPLE CORRECT CHOICE TYPE], , Q.3, , Let x1 and x2 satisfies the equation x2 + 6 = x1logx 5 (x1 > x2), then, (A) x12 + x 22 = 13, (C) x1 and x2 are relatively prime, , Q.4, , [2 × 4 = 8], , (B) 2x1 + 3x2 = 12, (D) x1 and x2 are both rational number, , If a sec  – c tan  = d and b sec  + d tan  = c, where a, b, c, d are distinct, non-zero real numbers, then, (A) a2 + b2 = c2 + d2, (C) tan  =, , ac  bd, bc  ad, , (B) sec  =, , c2  d 2, ad  bc, , (D) a2 + d2 = b2 + c2, , Q.5, , [INTEGER TYPE], [2 × 5 = 10], 2, Find the greatest integral value of c so that both the roots of the equation (c – 5)x – 2cx + (c – 4) = 0, are positive, one root is less than 2 and other root is lying between 2 and 3., , Q.6, , Let f (n) = log1155(n3) for all n  N, then find the value of f (7) + f (11) + f (15)., , PAGE # 3

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MATHEMATICS, , Daily Practice Problems, Class:XI (Phase-1), , Time: 20 Min. Discussion Date: 27-28/07/2015, , Target JEE, M.M.: 22, Dpp. No.-48, , [SINGLE CORRECT CHOICE TYPE], Q.1, , Range of the function f (x) =, (A), , 1, 2, , (B), , [4 × 3 = 12], , x 2  9x  20, , x  R does not contain, x 2  6x  8, , 1, 4, , (C) 3, , (D) 4, , Q.2, , Let f () = sin2 + cos2 + tan2 + sec2 + cosec2 + cot2, then the minimum value of f () is, (A) 3, (B) 6, (C) 7, (D) not possible to determine, , Q.3, , If ,  be the roots of 4x2 – 17x +  = 0,  R such that 1 <  < 2 and 2 <  < 3, then the number, of integral values of  is, (A) 1, (B) 2, (C) 3, (D) 4, , Q.4, , Number of real values of x satisfying the equation log 2, (A) 0, (B) 1, (C) 2, , , , , , , , x  1 = log3 1  x, (D) 3, , [INTEGER TYPE], , Q.5, ,  is equal to, [2 × 5 = 10], ,  b3 , log 2  ,  8   1 and log  9  = log  b  . If the largest single digit number which can divide, Given, 3, 2,  27 , 4, a, log3  2 , a , a, the value of   is m, then find the value of m., b, , Q.6, , The value of (tan218°) (tan254°) can be expressed as a rational number, , p, in lowest terms, q, , where p, q  N. Find the value of (p + q)., , PAGE # 4

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MATHEMATICS, , Daily Practice Problems, Class:XI (Phase-1), Q.1, Q.2, , Q.3, , Q.4, , Time: 20 Min. Discussion Date: 29-30/07/2015, , Target JEE, M.M.: 24, Dpp. No.-49, , [SINGLE CORRECT CHOICE TYPE], If + 2ax + 10 – 3a > 0, x R, then range of a is equal to, (A) – 5 < a < 2, (B) a < – 5, (C) a > 5, (D) 2 < a < 5, , [3 × 3 = 9], , x2, , The values of p for which one root of the equation x2 – (p + 1)x + p2 + p = 8 exceeds 2 and the other, is lesser than 2 are given by, (A) 3 < p < 10, (B) p  –2, (C) p  10, (D) –2 < p < 3, 2,  1  cot x  cot 2 x ,  1  cos 2 x  sin 2 x , , , , Let P (x) = , ,  1  tan x  tan 2 x  , then the minimum value of P(x) equals,  1  cos 2 x  sin 2 x , , , (A) 1, (B) 2, (C) 4, (D) 16, , [INTEGER TYPE], [3 × 5 = 15], Given that log 2 = 0.301, find the number of digits before decimal in the solution to the equation, log 5 log 4 log 3 (log 2 x   0 ., , Q.5, , Q.6, , , 3, 5 , Find the absolute value of  sec  sec,  sec  ., 7, 7, 7 , , , Suppose one of the roots of the equation ax2 + bx + c = 0 is 3 + 5 , where a, b, c  I, then find the, least value of |a + b + c|., , PAGE # 5

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MATHEMATICS, , Daily Practice Problems, , Class:XI (Phase-1), , Target JEE, Time: 20 Min. Discussion Date: 31-01/07-08/2015 M.M.: 20, Dpp. No.-50, [SINGLE CORRECT CHOICE TYPE], , Q.1, , Number of integral values of x satisfying the inequality, (A) 1, , (B) 2, , (C) 3, , [4 × 3 = 12], , x 2 ( x  1)(x 2  1)(x  2), 0 , is, x 2  2x  3, (D) 4, , Q.2, , If log2 N = 4 + b1 and log3N = 2 + b2 where b1, b2 [0, 1) , then largest integral value of N is, (A) 9, (B) 11, (C) 16, (D) 26, , Q.3, , The expression x2 + 2xy + ky2 + 2x + k = 0 can be resolved into two linear factors, then k , (A) {0, 1}, (B) {1, 2}, (C) {0, 2}, (D) {1, 4}, , Q.4, ,  | x 1| , Given that x is a real number, then the domain of f (x) = log  2,  is,  x  3x  2 , (A) x  {– 2, – 1}, (B) x  (– , – 2)  (– 1, ), (C) x  (– 2, – 1), (D) x  (– , ), , [MULTIPLE CORRECT CHOICE TYPE], Q.5, , Q.6, , [2 × 4 = 8], , , , , , If the equation 4sin  x  · cos x   = a2 + 3 sin 2x – cos 2x has a solution, then the value of a, 3, 6, , , can be, (A) –3, (B) 0, (C) 2, (D) 3, , Let ,  and  satisfies 0 <  <  <  < 2. If cos (x + ) + cos (x + ) + cos (x + ) = 0, for allx  R. Then ( – ) can be equal to, (A), , , 3, , (B), , 2, 3, , (C), , 4, 3, , (D), , 5, 3, , PAGE # 6