6- Question Report (6)

Leader Course/Phase-III to VII/Score-I/11-03-2017/Paper-2 58. Select the correct match among following : 58.  (1) OHC NO2 < CHO (1) OHC NO2 < CHO NO2 NO2 (Order of dipole moment)  (2) H3C Cl < CH3 (2) H3C Cl < CH3 Cl Cl (Order of boiling point)  (3) < (3) < (Order of heat of combustion)   (4) F F (4) F F F> F> F F (Order of melting point)   1001CT103516018 E-29/39

59. Select the incorrect order among following Target : JEE (Main + Advanced) 2017/11-03-2017/Paper-2 Cl 59.  Cl (1) > Cl > (1) > Cl > O Cl O O O Cl O O (Order of reactivity towards S 1-reaction)  S1- N N Cl Cl Cl Cl (2) > Cl > (2) > Cl > (Order of reactivity towards E –reaction)  E – 1 1 Cl Cl Cl Cl (3) > Cl > (3) > Cl > (Order of reactivity towards E –reaction)  E – 2 2 Cl Cl Cl Cl (4) > Cl > (4) > Cl > (Order of reactivity towards S 1-reaction)  S1- N N 60. (1) NaSH (l eq) 60. (1) NaSH (l eq) Br (2) NaOH Br (2) NaOH Cl Cl The major product obtained in above  reaction is : (1) (2) (1) (2) SH OH OH SH SH OH OH SH (3) (4) Me Et (3) (4) Me Et S SH SH OH S OH  E-30/39 1001CT103516018

Leader Course/Phase-III to VII/Score-I/11-03-2017/Paper-2 PART C - MATHEMATICS 61. If cot 2x  tan x  cosec kx , then the value 61.  cot 2x  tan x  cosec kx  33 3 33 3 of tan–1(tank) equals - tan–1(tank)  (1) 2 (2) 2 –  (3)  – 2 (4) 2 – 2 (1) 2 (2) 2 –  (3)  – 2 (4) 2 – 2 62. Let   are the roots of the quadratic 62.  ax2 + bx + c = 0  equation ax2 + bx + c = 0. If a, b, c are in A.P. a, b, c =15 and  = 15, then  equals   (1) –21 (2) –29 (3) –31 (4) –39 (1) –21 (2) –29 (3) –31 (4) –39 63. ƒ(x)  1  x; 0  x 1 2 63.  ƒ(x)  1  x; 0  x 1 2 1x2 1  x 2 If , ƒ(x)dx ƒ(x )dx   6)1 / 3 then   6)1 / 3 0 (7x 0 (7x  ;  ; is equal to  55 31 1 31 55 31 1 31 (1) (2) (3) (4) (1) (2) (3) (4) 42 12 42 21 42 12 42 21 64. Let ƒ(x) = x x x x....... (x > 0), then ƒ'(3) 64. ƒ(x) = x x x x....... (x > 0) ƒ'(3) is equal to  (1) 4 (2) 6 (3) 8 (4) 10 (1) 4 (2) 6 (3) 8 (4) 10 65. Suppose ƒ '(x) ƒ(x)  0 , where ƒ(x) is 65. ƒ '(x) ƒ(x)  0 ,ƒ(x)  ƒ ''(x) ƒ '(x) ƒ ''(x) ƒ '(x) continuously differentiable function with  ƒ '(x)  0  ƒ(0 ) = 1  ƒ'(x)  0 and satisfy ƒ(0) = 1 and ƒ'(0) = 2, then the number of solution(s) of equation ƒ'(0) = 2 ƒ(x)=x2 ƒ(x) = x2 is equal to-  (1) 0 (2) 1 (3) 2 (4) 3 (1) 0 (2) 1 (3) 2 (4) 3  1001CT103516018 H-31/39

Target : JEE (Main + Advanced) 2017/11-03-2017/Paper-2 66. If n((e – 1)exy + x2) = x2 + y2, then dy 66. n((e – 1)exy + x2) = x2 + y2 ddyx(1,0) dx (1,0) is  (1) 0 (2) 1 (3) 2 (4) 4 (1) 0 (2) 1 (3) 2 (4) 4 1 1 67. The value of eex (1  x.ex )dx is 67.  eex (1  x.ex )dx  0 0 (1) e (2) ee (3) ee – e (4) ee – 1 (1) e (2) ee (3) ee – e (4) ee – 1 68. If  is the interior angle of a regular octagon, 68.          then lim tan   1 is equal to lim tan  1   [sin   cos ]  [sin   cos ] (Note : [k] denotes greatest integer less than (:[.] )  or equal to k) (1) 0 (2) –1 (1) 0 (2) –1 (3) 1 (4) 2 (3) 1 (4) 2 69. Let ƒ(x) and g(x) be differentiable functions 69. ƒ(x) g(x)    R on R. If h(x) = ƒ(g(ƒ(x))), where ƒ(2) = 1, h(x)= ƒ(g(ƒ(x))), ƒ(2) = 1, g(1) = 2 and ƒ'(2) = g'(1) = 4, then h'(2) is g(1) = 2 ƒ'(2) = g'(1) = 4 h'(2)  equal to -  (1) 8 (2) 16 (3) 64 (4) 36 (1) 8 (2) 16 (3) 64 (4) 36 70. The value of  3k2  3k  1 is equal to 70.  3k2  3k  1  k1 (k2  k)3 k1 (k2  k)3 1 1 1 (4) 1 1 1 1 (4) 1 (1) (2) (3) (1) (2) (3) 8 4 2 8 4 2  H-32/39 1001CT103516018

Leader Course/Phase-III to VII/Score-I/11-03-2017/Paper-2 71. If1  ƒ(x))ƒ(x)dx  4 , then the area of 71.  1 (4x3  ƒ( x ))ƒ( x )dx  4   y= ƒ(x), 7 (4x3  07 0 region bounded by y = ƒ(x), x-axis and x-x= 1x = 2  ordinate x = 1 and x = 2, is -  17 15 17 15 (1) (2) (1) (2) 2 2 2 2 13 11 13 11 (3) 2 (4) 2 (3) 2 (4) 2 72. The general solution y(x) of the differential 72.  y  y  d  dt d  dt     equation x  x , is x x y(x)  dy dy (1) y = n|1 – x| + C (1) y = n|1 – x| + C (2) y = –n|1 – x| + C (2) y = –n|1 – x| + C (3) y = –n|1 + x| + C (3) y = –n|1 + x| + C (4) y = n|1 + x| + C (4) y = n|1 + x| + C (Note : C is constant of integration) (:C ) 73. If the equation tan4x – 2sec2x + [a]2 = 0 has 73. tan4x – 2sec2x + [a]2 = 0  atleast one solution, then the complete range     'a'     of 'a' (where a  R) is - (a  R) (Note : [k] denotes greatest integer less than (:[.] )  or equal to k) (2) [–2, 1] (1) [–1, 1] (2) [–2, 1] (1) [–1, 1] (4) [–2, 2) (3) [–1, 2) (4) [–2, 2) (3) [–1, 2)  1001CT103516018 H-33/39

Target : JEE (Main + Advanced) 2017/11-03-2017/Paper-2 74. Let ƒ(x) = Ax3 – Bx – tanx.sgn(x) be an – (2n , n  I    2 (2n  n  I 74.  x  R  1)   2  even function  x  R –  1) , , ƒ(x) = Ax3 – Bx – tanx.sgn(x)  1 1  where A = sin2 – sin + 4 A = sin2 – sin + 4 and B  tan2   2 tan   1 , then the B  tan2   2 tan   1    33 33 number of value(s) of  in  3 ,2 is -  3 ,2 - 2 2 (where sgnx denotes signum function of x) (sgnx, x ) (1) 0 (2) 1 (3) 2 (4) 4 (1) 0 (2) 1 (3) 2 (4) 4 75. Let sec1[ sin2 x]dx  ƒ(x)  C , (valid x  0) 75. sec1[sin2 x]dx  ƒ(x)  C , (x  0 where [k] denotes greatest integer less than )[.]  or equal to k and ƒ(0) = 0, then the value of  ƒ  8   at x = 2 is (where () dash denotes ƒ(0) = 0 x=2   ƒ  8      x     x       the derivative) (() ) (1) 2 (2) 4 (3) 8 (4) 16 (1) 2 (2) 4 (3) 8 (4) 16  n    n    76.  2 k  1    2 k  1       n1     The sum n1 cot1   k 1   is equal to 76. cot1   k 1        3 3  (1) 3  cot1 2 (2)   cot1 3 (1) 3  cot1 2 (2)   cot1 3 4 2 4 2 (3)  (4)   tan1 2 (3)  (4)   tan1 2 2 2  H-34/39 1001CT103516018

Leader Course/Phase-III to VII/Score-I/11-03-2017/Paper-2 77. Let ƒ(x) = x3 + bx2 + cx + d, 0 < b2 < c, 77. ƒ(x) = x3 + bx2 + cx + d, 0 < b2 < c then ƒ ƒ (1) is bounded (1)  (2) has a local maxima (2)  (3) has a local minima (3)  (4) is strictly increasing (4)  ee 78. For n  N, let Pn  (nx)n dx , then 78. n  N   Pn  (nx)n dx    11 (P10 – 90P8) is equal to (P10 – 90P8)  (1) 10e (2) –9e (1) 10e (2) –9e (3) –9 (4) 10 (3) –9 (4) 10 79. The variance of 20 observation is 5. If each 79. 20 52 observation is multiplied by 2, then the new         variance of the resulting observations, is  (1) 5 (2) 10 (1) 5 (2) 10 (3) 20 (4) 40 (3) 20 (4) 40 80. A flag-staff 5m high stands on a building of 80.  5m  25m   height 25m . At an observer who is at height 30m  of 30m, the flag-staff and the building  subtend equal angles. The distance of the  observer from the top of the flag staff is - (1) 5 3 3 53 (2) 5 3 2 2 (1) 2 (2) 5 2 (3) 5 2 (4) none of these (3) 5 2 (4)  3 3  1001CT103516018 H-35/39

Target : JEE (Main + Advanced) 2017/11-03-2017/Paper-2 81. Which of the following is injective but not 81.         surjective ? ? (1) ƒ : N  N, ƒ(x) = 2x + 3 (1) ƒ : N  N, ƒ(x) = 2x + 3 4x  3 4x  3 (2) ƒ : R  R, ƒ(x) = 5 (2) ƒ : R  R, ƒ(x) = 5 (3) ƒ : R  R, ƒ(x) = x3 – x (3) ƒ : R  R, ƒ(x) = x3 – x (4) ƒ : R  R, ƒ(x) = n(|x| + 1) (4) ƒ : R  R, ƒ(x) = n(|x| + 1) 82. A function ƒ satisfies the relation 82.  ƒ  ƒ(x) = ƒ(x) + ƒ(x) + ....... where ƒ(x) is ƒ(x) = ƒ(x) + ƒ(x) + .......  a differentiable function indefinitely. If ƒ(x)    ƒ(1) = 5, then the value of ƒ(1) + ƒ(1) is ƒ(1) = 5 ƒ(1) + ƒ(1)  equal to (1) 0 (2) –5 (1) 0 (3) 5 (2) –5 (3) 5 (4)  (4) cannot be determined 83. Let g(x) = ||x + 2| – 3|. If 'a' denotes the 83. g(x) = ||x + 2| – 3| 'a'  number of relative minima, 'b' denotes the 'b'  number of relative maxima and 'c' denotes  'c',g(x)      the product of the zeroes of g(x), then the (a+2b– c)  value of (a + 2b – c) is - (1) –1 (2) –2 (1) –1 (2) –2 (3) 8 (4) 9 (3) 8 (4) 9 84. Let b1, b2,......, bn be a geometric sequence 84. b1, b2,......, bn    such that b1 + b2 = 1 and bk  2 . Given b1 + b2 = 1  bk  2 b2< 0  k 1 k 1 that b2 < 0, then the value of b1 is (1) 2  2 (2) 1  2 b1  (1) 2  2 (2) 1  2 (3) 2  2 (4) 4  2 (3) 2  2 (4) 4  2  H-36/39 1001CT103516018

Leader Course/Phase-III to VII/Score-I/11-03-2017/Paper-2 85. For the primitive integral equation 85. y dx + y2dy = xdy; ydx + y2dy = xdy; x  R, y > 0, y = y(x), y(1) = 1, x  R, y > 0, y = y(x), y(1) = 1 y(–3)  then y(–3) is - (1) 3 (2) 2 (1) 3 (2) 2 (3) 1 (4) 5 (3) 1 (4) 5 86. The number of solutions that the equation 86.       0,          si n 5  c o s 3  4  sin5cos3 = sin9cos7 has in 0,  is = sin9cos7 has  4  (1) 4 (2) 5 (1) 4 (2) 5 (3) 6 (4) 7 (3) 6 (4) 7 87. If n arithmetic means a1,a2,......an are 87. n a1,a2,......an 50 100  inserted between 50 and 100 and n harmonic nh1, h2, ...... hn  means h1, h2, ...... hn are inserted between          the same two numbers, then a2hn–1  a2hn–1 is equal to 10000 (1) 5000 10000 (1) 5000 (2) n (2) n (3) 10000 250 (3) 10000 250 (4) n (4) n 88. If the difference between the roots of the 88. x2 + ax + b = 0  equation x2 + ax + b = 0 is equal to the x2 + bx + a = 0 (a  b)  difference between the roots of the equation - x2 + bx + a = 0 (a  b), then - (1) a + b = 4 (2) a + b = –4 (1) a + b = 4 (2) a + b = –4 (3) a – b = 4 (4) a – b = –4 (3) a – b = 4 (4) a – b = –4  1001CT103516018 H-37/39

Target : JEE (Main + Advanced) 2017/11-03-2017/Paper-2 89. The set of values of 'a' such that 89. 'a'      [1, 2] x2 – 2ax + a2 – 6a < 0 in [1, 2] is - x2 – 2ax + a2 – 6a < 0 - (1) [4  15, 4  15] (1) [4  15, 4  15] (2) [5  21, 4  15] (2) [5  21, 4  15] (3) [5  21, 5  21] (3) [5  21, 5  21] (4) [4  15, 5  21] (4) [4  15, 5  21] 90. Statement-1 : ~(p  ~q) is equivalent 90. -1 : ~(p  ~q), p  q  to p  q. -2 : ~(p  ~q)  Statement-2 : ~(p  ~q) is a tautology. (1) -I   -II  -II -I  (1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) -I   -II  -II (2) Statement-1 is True, Statement-2 is True; -I    Statement-2 is NOT a correct explanation (3) -I  -II for Statement-1. (3) Statement-1 is True, Statement-2 is False. (4) -I -II  (4) Statement-1 is False, Statement-2 is True.  H-38/39 1001CT103516018

Leader Course/Phase-III to VII/Score-I/11-03-2017/Paper-2   1001CT103516018 H-39/39