2- Question Report (2)

Target : JEE (Main + Advanced) 2017/21-03-2017 50. You have two C6H10O ketones, I and II. Both are 50. C6H10O III. optically active, but I is racemized by treatment I  with acid and II is not. Wolff kishner reduction of II both ketones gives the same achiral hydrocarbon,  C6H12  formula C H . What reasonable structures may III ? 6 12 be assigned to I and II respectively ? (1) I is 3-Methyl-4-Penten-2-one (1) I, 3--4--2- II is 4-Methyl-1-Penten-3-one II, 4--1--3- (2) I is 2-Methyl cyclopentanone (2) I , 2- II is 3-Methyl cyclopentanone II is 3- (3) I is 3-Methyl cyclopentanone (3) I is 3- II is 2-Methyl cyclopentanone II is 2- (4) I is 2-Ethyl cyclobutanone (4) I is 2- II is 3-Ethyl cyclobutanone II is 3- 51. Which of the following does not have the correct 51.  order of given property ? (1) Ga < Al < In < Tl ()  (1) Ga < Al < In < Tl (Atomic size) (2) I < F < Cl < Br ( ) (2) I2 < F2 < Cl2 < Br2 (Bond energy) 22 2 2 (3) PH3 < NH3 < HF < H2O (Boiling point) (3) PH < NH < HF < H O ()2 33 (4) BF3 < NF3 < NH3 (Dipole moment) (4) BF3 < NF3 < NH3 () 52. The distance betweentwo adjuscent carbon atoms 52.  is maximumin : (2) Graphite (1)  (2)  (1) Diamond (3) Benzene (4) Ethene (3)  (4)   H-28/40 1001CT102116064

Leader Course (Score-I) & Enthusiast Course (Score-II)/21-03-2017 53. Which of the following does not liberate a brown 53. ? gas ? (1) LiNO  3 (1) Action of heat on LiNO3 (2) Action of heat on KNO (2) KNO  3 3 (3) Zn HNO3 (3) Reaction of zinc with conc HNO (4) NaNO3 H2SO4  3 (4) Addition of conc H2SO4 on NaNO3 54. Self reduction process is used in the extraction 54.  of -  (1) Iron (2) Zinc (1) Iron (2) Zinc (3) Aluminium (4) Lead (3) Aluminium (4) Lead 55. The ammine complex of metal ions Cu2+, Ni2+ and 55.  Cu2+, Ni2+ Zn2+   Zn2+ have shapes respectively -  (1) Octahedral, Square planar, Tetrahedral (1)  (2) Square planar, Octahedral, Tetrahedral (2)  (3) Square planar, Tetrahedral, Octahedral (3)  (4) Tetrahedral, Square planar, Octahedral (4)  56. Geometrical as wellas opticalisomerismis shown 56.  by : (1) Cr C2O4 3 3 (2) Co  NH3 3 Cl3  (1) Cr  C2O4  3 3  (2) Co  NH3 3 Cl3  (3) Cr H2O2 C2O4 2  (3) Cr H2O2 C2O4 2  (4) CoenCl4  (4) CoenCl4   1001CT102116064 H-29/40

Target : JEE (Main + Advanced) 2017/21-03-2017 57. The reaction of white phosphorus with sodium 57.  hydroxide solutiongives :  (1) Phosphine and sodium salt of a dibasic acid (1) PH  3 (2) Phosphine and sodiumsalt ofamonobasic acid (2) PH  3 (3) Phosphine and sodium salt of a tribasic acid (3) PH3  (4) None of these (4)  58. The qualitative distinction of ZnSO4 and Al2(SO4)3 58. ZnSO Al (SO )      4 2 43 can be done by using the reagent :  (1) NH4OH (2) NaOH (1) NH OH (2) NaOH (3) Any of these (4) none of these 4 (3) (4) 59. KI is oxidised into I2 by using the reagent : 59. KI I (1) KMnO4 (neutral or slightly alkaline solution) 2 (2) Ozone (alkaline solution) (1) KMnO4 () (2) Ozone () (3) CuSO solution (3) CuSO  4 4 (4) All of these (4)  60. Ammonia is liberated in the reaction of : 60.  (1) Mg3N2 + H2O (1) Mg N + H O (2) NaNO + Zn + NaOH 32 2 3 (2) NaNO + Zn + NaOH 3 (3) CaNCN + H O 2 (3) CaNCN + H2O (4) All of these (4)   H-30/40 1001CT102116064

Leader Course (Score-I) & Enthusiast Course (Score-II)/21-03-2017 PART C - MATHEMATICS 61. The solution of D.E. 61.  (x cot y + ln cos x) dy+ (ln sinyy tan x)dx = 0 (x cot y+ ln cos x) dy+ (ln sinyy tan x)dx = 0 (1) (sin x)y (cos y)x = c  (2) (sin y)x (cos x)y = c (1) (sin x)y (cos y)x = c (3) (sin x)y (sin y)x = c (2) (sin y)x (cos x)y = c (4) (cot x)y (cot y)x = c (3) (sin x)y (sin y)x = c (4) (cot x)y (cot y)x = c 62. If f(x) be such that f(x) = max (|2x|, 2x3), x  R 62.  f(x)  (1) f(x) is discontinuous at one point f(x) = max (|2x|,2x3), x  R (2) f(x) is differentiable  x  R (1) f(x)  (2) f(x)  x  R (3) f(x) is non differentiable at one point only (3) f(x)  (4) f(x) is non differentiable at 4 points only (4) f(x) 4  63. 2x4 14x2  8x  49 63.  f(x)  2x4 14x2  8x  49   If the range of f(x)  x4  7x2  4x  23 is x4  7x2  4x  23 (a, b], then (a +b) is (a, b] (a +b) = (1) 3 (2) 4 (1) 3 (2) 4 (3) 5 (4) 6 (3) 5 (4) 6 64. Let f(x) = x3  3x2 + 3x + 1 and g be inverse of 64. f(x) = x3  3x2 + 3x + 1 g it, then area bounded by the curve y = g(x) with  y = g(x), x x= 1, x = 2  x axis between x = 1, x = 2 is (in square units) ()  1 1 1 1 (1) 2 (2) 4 (1) 2 (2) 4 (4) 1 (4) 1 3 3 (3) 4 (3) 4  1001CT102116064 H-31/40

Target : JEE (Main + Advanced) 2017/21-03-2017 65. If f(x) satisfies the relation 65.  f(x)  f  5x  3y   5 f  x 3 f  y  x, y R f  5x  3y   5 f  x 3 f  y  x, y R  2  2  2  2 f(0) = 1, f '(0) = 2 then period of sin (f(x)) is    f(0)=1, f '(0) = 2  (1) 2 (2)  sin (f(x))  (3) 3 (4) 4 (1) 2 (2)  (3) 3 (4) 4 12 12 66. If 12K. 12CK .11CK 1 is equal to 66. 12K. 12CK .11CK 1 = K 1 K 1 12  2119 17 ..........3  212  p then p is 12  2119 17 ..........3  212  p p  11! 11! (1) 2 (2) 4 (1) 2 (2) 4 (3) 8 (4) 6 (3) 8 (4) 6 67. The line x2  y 1  z 1 intersects the curve 67. x  2  y 1  z 1  xy = c2, z = 0  3 2 1 , 3 2 1 xy = c2, z = 0 if c is equal to c= (1) ± 1 1 (1) ± 1 1 (2) ± 3 (2) ± 3 (3) ± 5 (4) ± 2 (3) ± 5 (4) ± 2   f  x  1  dx    1   x    x  68. If  f  x dx  1  f  x dx  1  is equal to  then  68.  f x  dx   (1) 0 (2) 1 (1) 0 (2) 1 (3) 1 (4) 2 (3) 1 (4) 2  H-32/40 1001CT102116064

Leader Course (Score-I) & Enthusiast Course (Score-II)/21-03-2017 69. ABCD is a rhombus. The circumradii of ABD 69. ABCD ABD ACD  25 225 25       and ACD are 2 and 25. Then the area of  rhombus is (1) 400 sq.unit (1) 400 sq.unit (2) 600 sq. unit (2) 600 sq. unit (3) 200 sq. unit (3) 200 sq. unit (4) 800 sq. unit (4) 800 sq. unit 70. If z is a complex number satisfying 70.  z |z|2  |z|  2 <0  |z|2  |z|  2 <0, then the value of z2  z sin , z2zsin  for all values of  , is  (1) equal to 4 (2) equal to 6 (1) 4  (2) 6  (3) more than 6 (4) less than 6 (3) 6  (4) 6  71. If the graph of y = ax3 + bx2 + cx + d is 71.  y = ax3 + bx2 + cx + d x= k  symmetric about the line x = k then  (1) k = c (2) k   c (1) k = c (2) k   c b b (3) a  c k  0 (4) none of these (3) a c  k  0 (4)  2b 2b 72. The solution set of inequality 72.      tan1 x 1    2  1    sec1       tan1 x 1    cot 1 2  1    sec1   cot1 x  tan1 x 2   2 cot 1 x  2   lim x  2  cot1 x  tan1 x 2   2 x  2   lim x  2  x   x   is (where [.] denotes the greatest integer ([.]  ) function) (1) (tan 1, tan 2) (2) (cot 1, cot 2) (1) (tan 1, tan 2) (2) (cot 1, cot 2) (3) (tan 1, tan2) (4) (tan1,  ) (3) (tan 1, tan2) (4) (tan1,  )  1001CT102116064 H-33/40

Target : JEE (Main + Advanced) 2017/21-03-2017 xx 73. Let f  x   t2  2t  2 dt, where x is set of 73. f  x   t2  2t  2 dt, x  0 0 real numbers satisfying the inequation        log 1  log 1 2 2 6x  x2 8  0 . If range of f(x) is 6x  x2 8  0   [a, b] then (a + b) is f(x)  [a, b] (a + b)  (1) 50 (2) 56 (1) 50 (2) 56 (3) 72 (4) 32 (3) 72 (4) 32 74. The equation of the plane through the 74. x+ 2y + z  1 = 0 intersection of the planes x + 2y + z  1 = 0 2x + y + 3z  2 = 0     and 2x + y + 3z  2 = 0 and perpendicular to     x+y+z 1=0  the plane x+y+z1=0 and x + ky + 3z  1 = 0. x + ky + 3z  1 = 0   k  Then the value of k is  (1) 5 (2)  3 (1) 5 (2)  3 2 2 2 2 5 3 5 3 (3) 2 (4) 2 (3) 2 (4) 2   1   x p sin  1   x x3 , x0   x    x  x p sin  x x3 , x0 75. f x   75. Let f x    0, x0  0, x0 then complete set of values of p for which f \"(x) p f\"(x), x = 0  is continuous at x = 0 is  (1) [2, ) (2) [3,  ) (1) [2, ) (2) [3,  ) (3) (4,) (4) [2,) (3) (4,) (4) [2,)  H-34/40 1001CT102116064

Leader Course (Score-I) & Enthusiast Course (Score-II)/21-03-2017 76. If lim ax2  bx  c  1 76.  lim ax2  bx  c  1 2 2 x 1 2x 12 x 1 2x 12 2 2 then lim x ax bx c is  x ax bx c  lim x2 x  2 x2 x  2 1 1 (1) 0 (2) 2 (1) 0 (2) 2 (3) 2 (4) 6 (3) 2 (4) 6 d 2x3  3x2  x  3 BC 77.  d  2x3  3x2  x  3  A  B  C dx  x2  x  2  2 77. If dx  x2  x  2   A   12  x 22    x 12  x 2   x  then (A  B + C) is  (A  B + C)  (1) 4 (2) 7 (1) 4 (2) 7 (3) 2 (4) 0 (3) 2 (4) 0 78. Lines are drawn from a point P (1, 3) to a 78. P(1, 3) x2+y22x + 4y  8 = 0.  circle x2 + y2  2x + 4y  8 = 0. Which meets 2AB  the circle at 2 points A & B, then the minimum , PA + PB  value of PA + PB is (1) 6 (2) 8 (1) 6 (2) 8 (3) 10 (4) 12 (3) 10 (4) 12    79. If Tn  n2 1 n! & Sn= T1 + T2 + T3 +......Tn n2 79.  Tn  1 n! S =T +T + T +......T 3 n n12 Let T10 a where a & b are realtively prime TS1100  a a b  S10 b b natural numbers, then the value of (b  a) is (b  a)  (1) 8 (2) 9 (1) 8 (2) 9 (3) 10 (4) 11 (3) 10 (4) 11  1001CT102116064 H-35/40

Target : JEE (Main + Advanced) 2017/21-03-2017 80. If g(x) = 2f (2x3  3x2) + f(6x2  4x3  3), 80.  g(x) = 2f (2x3  3x2) + f(6x2  4x3  3),  x R and f\"(x) > 0,  x R , then  x  R f\"(x) > 0,  x R ,  g'(x) > 0 for x belonging to g'(x) > 0,  x  (1)  ,  1   0,1 (1)  ,  1   0,1  2   2  (2)   1 , 0   1,   (2)   1 , 0   1,    2   2  (3) 0,  (3) 0,  (4) ,1 (4) ,1 81. Let I   /6 cos x dx, J   /2 cos x dx . Which of 81. I   /6 cos x dx, J   /2 cos x dx       0 x  /3 x 0x  /3 x the following is CORRECT ? ? (1) I   ,J   (2) I   ,J   (1) I   ,J   (2) I   ,J   6 6 6 6 6 6 6 6 (3) I   ,J   (4) I   ,J   (3) I   ,J   (4) I   ,J   6 6 6 6 6 6 6 6 82. A variable line ax + by + c = 0, where a, b, c 82. ax + by + c = 0, a, b, c  are in A.P., is normal to a circle x 2   y   2     x  2   y   2   , which is orthogonal to x2 + y2 4x 4y1 = 0        = circle x2 + y2 4x 4y1 = 0. The value of      is equal to (1) 3 (2) 5 (1) 3 (2) 5 (3) 10 (4) 7 (3) 10 (4) 7  H-36/40 1001CT102116064

Leader Course (Score-I) & Enthusiast Course (Score-II)/21-03-2017 1 1 2 1 1 2 83. If A  0 2 1 and A3 = (aAI) (bAI), 83.  A  0 2 1 A 3 = (aAI) (bAI),  1 0 2 1 0 2 where a, b are integers and I is a 3 × 3 unit a, b  I, 3×3     matrix then value of (a + b) is equal to  (a + b)  (1) 4 (2) 5 (1) 4 (2) 5 (3) 6 (4) 7 (3) 6 (4) 7 84. The statement  p  q  p  q ~ q is 84.  p  q  p  q ~ q  (1) tautology (1)  (2) contradiction (2)  (3) open statement (3)  (4) neither tautology nor contradiction (4)  85. The average marks of 10 students in a class 85.  was 60 with a standard deviation 4, while the  average marks of other ten students was 40 with  a standard deviation 6. If all the 20 students  are taken together, their standard deviation  will be (1) 5 (2) 7.5 (1) 5 (2) 7.5 (3) 9.8 (4) 11.2 (3) 9.8 (4) 11.2 86. The number of ways in which 3 children can 86.          distribute 10 tickets out of 15 consecutively  numbered tickets themselves such that they get 5,32  consecutive blocks of 5, 3 and 2 tickets is  (1) 8C (2) 8C 3! (1) 8C5 (2) 8C5 3! 5 5 (3) 8C (3!)2 (4)  (3) 8C (3!)2 (4) none of these 5 5  1001CT102116064 H-37/40

Target : JEE (Main + Advanced) 2017/21-03-2017   87. Let b  iˆ  4 ˆj  6kˆ and   2iˆ 7 ˆj 10kˆ . If 87. iˆ  4 ˆj  6kˆ c  2iˆ  7 ˆj  10kˆ .   c b  a be a unit vector and the scalar triple product       a b c  a a b c  has the greatest value, then a is a equal to  (1)1  (1)1 3 iˆ  ˆj  kˆ 3 iˆ  ˆj  kˆ  1 (2) 5  1 2iˆ  ˆj  2kˆ (2) 5 2iˆ  ˆj  2kˆ  (3)1  (3)1 3 2iˆ  2 ˆj  kˆ 3 2iˆ  2 ˆj  kˆ 1 3iˆ  7 ˆj  kˆ 59 1 3iˆ  7 ˆj  kˆ  (4) 59  (4) 88. The locus of the orthocentre of the triangle 88.  y2 = 4ax  formed by the focal chord of the parabola  y2 = 4ax and the normals drawn at its extremeties is (1) y2 = a (x  3a) (1) y2 = a (x  3a) (2) y2 = a (x + 3a) (2) y2 = a (x + 3a) (3) y2 = a (x  4a) (3) y2 = a (x  4a) (4) y2 = a(x + 4a) (4) y2 = a(x + 4a)  H-38/40 1001CT102116064

Leader Course (Score-I) & Enthusiast Course (Score-II)/21-03-2017 89. In a tournament there are twelve players 89. P1,P2,P3,........ P12  P1, P2, P3,........ P12 and divided into six pairs at  random. From each game a winner is decided  on the basis of game played between the two  players of the pair. Assuming each player is of PP equal strength, then the probability that exactly 12 one out of P1 and P2 is among the losers is .  5 6 5 6 (1) (2) (1) 11 (2) 11 11 11 1 5 1 5 (3) 2 (4) 22 (3) 2 (4) 22 90. Point from which two distinct tangents can be 90.        drawn on two different branches of the x2  y2  1  hyperbola x2  y2  1 but no two different 25 16 x2 +y2=36 25 16 tangent can be drawn to the circle x2 + y2 = 36  is (1) (1, 6) (2) (1, 3) (1) (1, 6) (2) (1, 3) (3) (7, 1) 1 (3) (7, 1) 1 (4) (1, 2 ) (4) (1, 2 )  1001CT102116064 H-39/40

Target : JEE (Main + Advanced) 2017/21-03-2017   H-40/40 1001CT102116064