WBJEE 2016 Maths Question Answerkey Solutions PDF

WB-JEE-2016 (Mathematics)
__________________________________________________________________________________________
1 31 7
1. Let A and B be two events such that P(A  B) = , P(A  B) = and P ( B) = then
6 45 10
a. A and B are independent
b. A and B are mutually exclusive
A 1
c. P  <
B 6
B 1
d. P  <
A 6
1° 1°
2. The value of cos 15° cos 7 sin7 is
2 2
1
a.
2
1
b.
8
1
c.
4
1
d.
16
3. The smallest positive root of the equation tan x – x = 0 lies in

a. (0,/2)
b. (/2,)
 3 
c.   , 
 2 
 3 
d.  ,2 
 2 
4. If in a triangle ABC, AD, BE and CF are the altitudes and R is the circumradius, then the
radius of the circumcircle of EF is
R
a.
2
2R
b.
3
1
c. R
3
d. None of these
WB-JEE-2016 (Mathematics) Page | 1

5. The points (–a, –b), (a, b), (0, 0) and (a2, ab), a  0, b  0 always lie on this line. Hence,
collinear
a. collinear
b. vertices of a parallelogram
c. vertices of a rectangle
d. lie on a circle
6. The line AB cuts off equal intercepts 2a from the axes. From any point P on the line AB
perpendiculars PR and PS are drawn on the axes. Locus of mid-point of RS is
a
a. x–y=
2
b. x+y=a
c. x2 + y2 = 4a2
d. x2 – y2 = 2a2
7. x + 8y – 22 = 0, 5x + 2y – 34 = 0, 2x – 3y + 13 = 0 are the three sides of a triangle. The

area of the triangle is
a. 36 square unit
b. 19 square unit
c. 42 square unit
d. 72 square unit
8. The line through the points (a, b) and (–a, –b) passes through the point
a. (1, 1)
b. (3a, –2b)
c. (a2, ab)
d. (a, b)
x y x y 1
9. The locus of the point of intersection of the straight lines + = K and – = , where
a b a b k
k is a non-zero real variable, is given by
a. a straight line
b. an ellipse
c. a parabola
d. a hyperbola

10. The equation of a line parallel to the line 3x + 4y = 0 and touching the circle x2 + y2 = 9 in
the first quadrant is
a. 3x + 4y = 15
b. 3x + 4y = 45
c. 3x + 4y = 9
d. 3x + 4y = 27
11. A line passing through the point of intersection of x+y=4 and x–y=2 makes an angle
3
tan–1   with the x-axis. It intersects the parabola y2=4(x–3) at points (x1, y1) and (x2,
4
y2) respectively. Then |x1–x2| is equal to
16
a.
9
32
b.
9
40
c.
9
80
d.
9
12. Then equation of auxiliary circle of the 16x2+25y2+32x–100y=284 is

a. x2 + y2 + 2x – 4y – 20 = 0
b. x2 + y2 + 2x – 4y = 0
c. (x+1)2+(y–2)2 = 400
d. (x+1)2+(y–2)2=225
x2 y 2
13. If PQ is a double ordinate of the hyperbola – = 1 such that OPQ is equilateral, O
a2 b2
being the centre. Then the eccentricity e satisfies
2
a. 1 < e <
3
2
b. e =
2
3
c. e =
2
2
d. e >
3

14. If the vertex of the conic y2 – 4y = 4x – 4a always lies between the straight lines; x+y=3
and 2x+2y–1=0 then
a. 2<a<4
1
b. – < a < 2
2
c. 0<a<2
1 3
d. – < a <
2 2
15. A straight line joining the points (1,1,1) and (0,0,0) intersects the plane 2x+2y+z=10 at
a. (1, 2, 5)
b. (2, 2, 2)
c. (2, 1, 5)
d. (1, 1, 6)
16. Angle between the planes x+y+2z=6 and 2x–y+z=9 is


a.
4

b.
6

c.
3

d.
2
 dy 
17. If y =(1+x)(1+x2)(1+x4)........(1+x2n) then the value of   at x = 0 is
 dx 
a. 0
b. –1
c. 1
d. 2
18. If f(x) is an odd differentiable function defined on (–,) such that f' (3) = 2, then f' (–3)
equal to
a. 0
b. 1
c. 2
d. 4

(1– x )
 1+ x  (1–x )
19. lim  
x →1 2+ x
 
a. is 1
b. does not exist
2
c. is
3
d. is/n 2
  e 
 log  x2  
20. If f ( x ) = tan–1     + tan–1  3+2log x  then the value of f'' (x) is
 1– 6log x 
 log ex2  ( )  
 
a. x2
b. x
c. 1
d. 0
log x
21.  3x
dx is equal to
1
( ) +c
2
a. log x
3
2
( ) +c
2
b. log x
3
2
( log x ) + c
2
c.
3
1
( log x ) + c
2
d.
3
22.  2x (f'(x)+ f(x)log2) is equal to

a. 2xf'(x)+c
b. 2xlog2+c
c. 2xf(x) +c
d. 2x+c

1
1 
23.  log  –1  dx =
0 x 
a. 1
b. 0
c. 2
d. None of these
 n + 1 + n + 2 + .... + 2n –1 
 
24. The value of lt  3  is
n →

 n 2 

a.
2
3
(
2 2 –1 )
b.
2
3
( 2 –1 )
c.
2
3
( 2 +1 )
d.
2
3
(
2 2 +1 )
dy
25. If the solution of the differential equation x + y = xex be, xy = ex (x) + c then (x) is
dx
equal to
a. x+1
b. x–1
c. 1–x
d. x
26. The order of the differential equation of all parabolas whose axis of symmetry along x-
axis is
a. 2
b. 3
c. 1
d. None of these

27. The line y = x +  is tangent to the ellipse 2x2 + 3y2 = 1. Then  is
a. –2
b. 1
5
c.
6
2
d.
3
28. The area enclosed by y = 5 – x 2 and y = |x–1| is

 5 
a.  4 – 2  sq.units
 
5 – 2
b. sq.units
2
 5 1 
c.  4 – 2  sq.units
 
 
d.  2 –5  sq.units
 
29. Let S be the set of points whose abscissas and ordinates are natural numbers. Let P  S
such that the sum of the distance of P from (8,0) and (0,12) is minimum among all
elements in S. Then the number of such points P in S is
a. 1
b. 3
c. 5
d. 11
l
30. Time period T of a simple pendulum of length l is given by T = 2 . If the length is
g
increased by 2%, then an approximate change in the time period is
a. 2%
b. 1%
1
c. %
2
d. None of these

31. The cosine of the angle between any two diagonals of a cube is
1
a.
3
1
b.
2
2
c.
3
1
d.
3
32. If x is a positive real number different from 1 such that logax, logbx, logcx are in A.P., then
a+c
a. b=
2
b. b = ac
c2 = ( ac )
log a b
c.
d. None of (A), (B), (C) are correct
33. If a, x are real numbers and |a| < 1, |x| < 1, then 1 + (1+a)x + (1+a+a2)x2 + .... is equal to
1
a. (1– a)(1– ax)
1
b. (1– a)(1– x)
1
c. (1– x)(1– ax)
1
d. (1+ ax)(1– a)
34. if log0.3 (x–1) < log0.09 (x–1), then x lies in the interval
a. (2, )
b. (1, 2)
c. (–2, –1)
d. None of these

13
35. The value of  (in + in+1 ),i = –1, is
n=1
a. i
b. i–1
c. 1
d. 0
1 1 1
36. If z1, z2, z3 are imaginary numbers such that |z1| = |z2| = |z3| = + + = 1 then
z1 z2 z3
|z1 +z2+z3| is
a. Equal to 1
b. Less than 1
c. Greater than 1
d. Equal to 3
37. If p, q are the roots of the equation x2 + px + q = 0, then

a. p = 1, q = –2
b. p = 0, q = 1
c. p = –2, q = 0
d. p = –2, q = 1
38. The number of values of k for which the equation x2 – 3x + k = 0 has two distinct roots
lying in the interval (0, 1) are
a. Three
b. Two
c. Infinitely many
d. No values of k satisfies the requirement
39. The number of ways in which the letters of the word ARRANGE can be permuted such
that the R’s occur together is
7
a.
22
7
b.
2
6
c.
2
d. 5 2

1 1 1
40. If 5
+ 6 = 4 , then the value of r equals to
Cr Cr Cr
a. 4
b. 2
c. 5
d. 3
41. For +ve integer n, n3 + 2n is always divisible

a. 3
b. 7
c. 5
d. 6
42. In the expansion of (x – 1) (x – 2) .... (x – 18), the coefficient of x17 is
a. 684
b. –171
c. 171
d. –342
43. 1+ nC1 cos + nC2 cos 2+.......+ nCn cos n equals

n
 θ nθ
a.  2cos 2  cos 2
 
nθ
b. 2cos2
2
θ
c. 2cos2n
2
n
d.
 2θ
 2cos 2 
 
1 log x y log x z
44. If x, y and z be greater than 1, then the value of log y x 1 log y z is
log z x log z y 1
a. log x. logy. log z
b. log x + logy + log z
c. 0
d. 1 – {(log x).(logy).(logz)}

45. Let A is a 3 × 3 matrix and B is its adjoint matrix. If |B| = 64, then |A| =
a. ±2
b. ±4
c. ±8
d. ±12
    1 
 cos 4 – sin 4   
2
46. Let Q =   and  then Q3x is equal to
 sin  cos    1 
   
 4 4  2
0
a.  
1
 1 
 
b.  2
 1 
 
 2
 –1 
c.  
0
 1 
 
 2
d.
 1 
 
 2
47. Let R be a relation defined on the set Z of all integers and xRy when x + 2y is divisible by
3. Then
a. R is not transitive
b. R is symmetric only
c. R is an equivalence relation
d. R is not an equivalence relation
48. If A = 5n – 4n –1 : n  N and B = 16 ( n –1 ) : n  N , then

a. A=B
b. AB=
c. AB
d. BA

49. If the function f : R→R is defined by f(x) = (x2+1)35  xR, then f is
a. one-one but not onto
b. onto but not one-one
c. neither one-one nor onto
d. both one-one and onto
50. Standard Deviation of n observations a1, a2, a3 .....an is . Then the standard deviation of
the observations a1 ,a2 …… an is
a. 
b. –
c.  
d. n
51. The locus of the midpoints of chords of the circle x2+y2 =1 which subtends a right angle
at the origin is
(0,0)
/4
M (h,k)
/4
1
a. x2 + y 2 =
4
1
b. x2 + y 2 =
2
c. xy = 0
d. x2 – y2 = 0
52. The locus of the midpoints of all chords of the parabola y2 = 4ax through its vertex is
another parabola with directrix
P(at2,2at)
M (h,k)
(0,0)
a. x=–a
b. x=a
c. x=0
a
d. x=–
2

53. If [x] denotes the greatest integer less than or equal to x, then the value of the integral
2
 x  x dx equals
2
5
a.
3
7
b.
3
8
c.
3
4
d.
3
54. The number of points at which the function f(x) = max {a – x, a + x, b}, –  x  , 0 < a <
b cannot be differentiable
y
b
x
–a a
a. 0
b. 1
c. 2
d. 3
55. For non-zero vectors aandb if a + b < a – b , then aandb are
a. Collinear
b. Perpendicular to each other
c. Inclined at an acute angle
d. Inclined at an obtuse angle
dy
56. General solution of y + by 2 = acosx, 0 < x < 1 is Here c is an arbitrary constant
dx
a. y2 = 2a(2b sinx + cosx) + ce–2bx
b. (4b2 + 1)y2 = 2a(sinx + 2bcosx) + ce–2bx
c. (4b2 + 1)y2 = 2a(sinx + 2bcosx) + ce2bx
d. y2 = 2a(2bsinx + cosx) + ce–2bx

57. The points of the ellipse 16x2 + 9y2 = 400 at which the ordinate decreases at the same
rate at which the abscissa increases is/are given by
 16   –16 
a.  3, 3  &  –3, 3 
   
 –16   16 
b.  3, 3  &  –3, 3 
   
 1 1  1 1
c.  16 , 9  &  – 16 , 9 
   
 1 1  1 1
d.  16 , 9  &  – 16 , 9 
   
58. The letters of the word COCHIN are permuted and all permutation are arranged in an
alphabetical order as in an English dictionary. The number of words that appear before
the word COCHIN is
a. 96
b. 48
c. 183
d. 267
2 0 0  a 0 0
   
59. If the matrix A=  0 2 0 , then A =  0 a 0  , n  N where
n
2 0 2 b 0 a
   
a. a = 2n, b = 2n
b. a = 2n, b = 2n
c. a = 2n, b = n2n–1
d. a = 2n, b = n2n
60. The sum of n terms of the following series; 13 +33 + 53 +73 +…. is
a. n2(2n2 – 1)
b. n3(n – 1)
c. n3 + 8n + 4
d. 2n4 + 3n2
61. If  and  are roots of ax2 + bx + c = 0 then the equation whose roots are 2 and 2 is
a. a2x2 – (b2 – 2ac)x + c2 = 0
b. a2x2 + (b2 – ac)x + c2 = 0
c. a2x2 + (b2 + ac)x + c2 = 0
d. a2x2 + (b2 + 2ac)x + c2 = 0

62. If  is an imaginary cube root of unity, then the value of
(2 – )(2 – 2) + 2(3 – )(3 – 2) + ..... + (n – 1)(n – )(n – 2) is
n2
( n +1) – n
2
a.
4
n2
( n +1) + n
2
b.
4
n2
( n +1)
2
c.
4
n2
d. (n + 1)– n
4
63. If nCr–1 = 36, nCr = 84 and nCr+1 = 126 then the value of nC8 is
a. 10
b. 7
c. 9
d. 8
64. In a group 14 males and 6 females, 8 and 3 of the males and females respectively are
aged above 40 years. The probability that a person selected at random from the group is
aged above 40 years, given that the selected person is female, is
2
a.
7
1
b.
2
1
c.
4
5
d.
6
65. The equation x3 –yx2 + x – y = 0 represents
a. a hyperbola and two straight lines
b. a straight line
c. a parabola and two straight lines
d. a straight line and a circle

66. If the first and the (2n+1)th terms of an AP, GP and HP are equal and their nth terms are
respectively a, b, c then always
a. a=b=c
b. abc
c. a+c=b
d. ac – b2 = 0
1
67. The coordinates of a point on the line x + y + 1 = 0 which is at a distance unit from the
5
line 3x + 4y + 2 = 0 are
a. (2, – 3)
b. (–3, 2)
c. (0, –1)
d. (–1, 0)
68. If the parabola x2 = ay makes an intercept of length 40 unit on the line y – 2x = 1 then
a is equal to
a. 1
b. – 2
c. –1
d. 2
69. if f(x) is a function such that f(x) = (x –1)2(4 – x), then

a. f(0) = 0
b. f(x) is increasing in (0, 3)
c. x = 4 is a critical point of f(x)
d. f(x) is decreasing in (3, 5)
70. On the ellipse 4x2 + 9y2 = 1, the points at which the tangents are parallel to the line 8x =
9y are
2 1
a.  , 
5 5
 2 1
b.  – , 
 5 5
 2 1
c.  – ,– 
 5 5
2 1
d.  ,– 
5 5

3000 1,for0  t < 1 3000  2016 
71. If  φ( t ) =  then    ( t – r')( t – 2016)  dt =
–3000  0otherwise –3000  –3000 
a. a real number
b. 1
c. 0
d. does not exist
72. If the equation x2 + y2 –10x + 21 = 0 has real roots x = a and y =  then

a. 3  x  7
b. 3  y  7
c. – 2  y  2
d. – 2  x  2
73. If z = sin  – i cos  then for any integer n,

1  n 
a. zn + n = 2cos  – nθ 
z  2 
1  n 
b. zn + n = 2sin  – nθ 
z  2 
1  n 
c. zn – n = 2i cos  – nθ 
z  2 
1  n 
d. zn – n = 2i sin  nθ– 
z  2 
74. Let f : X → X be such that f(f(x)) = x for all x  X and X  R, then

a. f is one-to-one
b. f is onto
c. f is one-to-one but not onto
d. f is onto but not one-to-one
3 3 3
75. If A, B are two events such that P ( A  B ) ³ and  P ( A  B )  then
4 8 8
11
a. P ( A ) + P ( B ) 
8
3
b. P ( A ) .P ( B ) 
8
7
c. P ( A ) + P ( B) 
8
d. None of these

ANSWER KEYS
1. (a) 2. (b) 3. (c) 4. (a) 5. (a) 6. (b) 7. (b) 8. (c,d) 9. (d) 10. (a)
11. (b) 12. (a) 13. (d) 14. (b) 15. (b) 16. (c) 17. (c) 18. (c) 19. (c) 20. (d)
21. (a) 22. (c) 23. (b) 24. (a) 25. (b) 26. (a) 27. (c) 28. (c) 29. (b) 30. (b)
31. (a) 32. (c) 33. (c) 34. (a) 35. (b) 36. (a) 37. (a) 38. (c) 39. (c) 40. (b)
41. (a) 42. (b) 43. (a) 44. (c) 45. (c) 46. (c) 47. (c) 48. (c) 49. (c) 50. (c)
51. (b) 52. (d) 53. (b) 54. (c) 55. (d) 56. (b) 57. (a) 58. (a) 59. (d) 60. (a)
61. (a) 62. (a) 63. (c) 64. (b) 65. (b) 66. (b,d) 67. (b,d) 68. (a,b) 69. (b,c) 70. (b,d)
71. (a,b) 72. (a,c) 73. (a,c) 74. (a,b) 75. (a,c)

Solution
1. (a)
1 31 7
P(A B) = , P(A B) = P(B) = (given)
6 45 10
7 3
P(B) = 1 – =
10 10
 P(A B) = P(A) + P(B) – P(A B)
31 3 1 5
 = P(A) + –  P(A) =
45 10 6 9
5 3 1
 P(A) × P(B) =  = = P(A B)
9 10 6
 A, B are independent events
2. (b)
1 1
Cos 15° cos 7 sin 7
2 2
1º 1
2cos 7 sin 7 cos15º
 2 2
2
sin15º cos15º 2
 
2 2
sin 30º 1
 =
4 8
3. (c)
 tan x = x
y = tan x ……….(1)
y = x ………….. (2)
y = tanx
y=x
  3
2 2
 3 
It is clearly visible that solution lies in  , 
 2 

4. (a)
Here, DEF is a pedal triangle of ABC
We know
A
F E
B C
D
EF = a cosA = R sinA (side of pedal )
FDE = 180º –2A
Let, circumradius of DEF be R.
Now by sine rule in DEF
EF R sin 2A R
2R = =  R'=
sin  FDE sin(180 – 2 A) 2
5. (a)
–a –b
a b
1
= 0 0
2 2
a ab
–a –b
 = ( –ab + ab + 0 – 0 + 0 – 0 – a 2 b + a 2 b )
1
2
=0
 Hence, points are collinear
6. (b)
Equation of line which cuts equal intercepts from axes is x + y = 2a
B
P
S
h,k m
R A
x+y=2a
Let, co-ordinates of the midpoint be m (h, k).

So, R and S are (2h, 0) and (0, 2k)
Therefore, p must be (2h, 2k)
 P lies on AB
 2h + 2k = 2a
x+y=a
7. (b)

L1: x + 8y = 22
L2: 5x + 2y = 34
L3: 2x – 3y = –13
On solving L1, L2 & L3 we get
A = (–2, 3), B = (6, 2) and C = (4, 7)
–2 3
1 6 2
Area =
2 4 7
–2 3
1
= (–4 –18 + 42 – 8 + 12 + 14) = 19 square units
2
8. (c,d)
Equation of line passes through points (a, b) & (–a, –b) is
–2b
 (y – b) = (x– a)
–2a
 ay – ab = bx – ab
 ay = bx
Now check options  (C) & (D) are correct
9. (d)
x y
L1 = + =k
a b
x y 1
L2 = – =
a b k
Let point of intersection be ( )
    1
So, + = k and – =
a b a b k
      
  +  –  = 1
 a b  a b 
 2 2
 – =1
a 2 b2
x 2 y2
Locus: – = 1 which is hyperbola
a 2 b2
10. (a)

Let, equation of line which is parallel to given line is
3x + 4y = k
This line is tangent to circle
d=r
0+0–k 0
 =3 (0, 0)
5 3x + 4y = k
 k =  15
So, equation of tangent in first quadrant is 3x + 4y = 15
11. (b)
L1: x + y = 4
L2: x – y = 2
 line passing through the point of intersection is
 L = L1 + L2 = 0
 L = (x + y – 4) +  (x – y – 2) =0
 L = x(1 + ) + y (1 – ) – 4 – 2 = 0
3
 ML = (given)
4
1+  3
 =
–1 +  4
  = –7
Equation of line is L = –6x + 8y + 10 = 0
3x – 5
y= (put in equation of parabola)
4
2
 3x – 5 
  = 4(x– 3)  9x – 94x + 217 = 0
2
 4 
94 217
 x1 + x2 = , x1 x 2 =
9 9
32
 |x1 – x2| = (x1 + x 2 ) 2 – 4x1x 2 =
9
12. (a)

 16x2 + 25y2 + 32x – 100y = 284
 16(x2 + 2x) + 25(y2 + 4y) = 284
 16 (x + 1)2 + 25(y – 2)2 = 284 + 16 +100
 16 (x + 1)2 + 25(y – 2)2 = 400
(x + 1) 2 (y– 2) 2
 + =1
25 16
So, the auxiliary circle is (x + 1)2 + (y – 2)2 = 25
 x2 + y2 + 2x – 4y – 20 = 0
13. (d)
 OPQ is equilateral
(OP)2 = (PQ)2
P (asec, btan)
Q(asec, – btan)
 a2sec2 + b2 tan2 = (2b tan)2
 a2sec2 = 3b2 tan2
a2
 sin2 = 2
3b
 Now, sin2 < 1
a2
 2 1
3b
b2 1
 2 
a 3
On adding 1 both sides
b2 1
l+ 2 >1+
a 3
4
 e2 >
3
On taking root both sides
2
e>
3
14. (b)

 y2 – 4y = 4x – 4a
 y2 – 4y + 4 = 4x – 4a + 4
 (y – 2)2 = 4(x–(a–1))
 Vertex: (a–1, 2)
Vertex lies in between lines
 L1 × L2 < 0
 (a – 1 + 2 – 3) × (2(a–1) + 4 – 1) < 0
 (a – 2) (2a + 1) < 0
 1 
 a  – , 2 
 2 
15. (b)
Equation of line joining points (1, 1, 1) & (0, 0, 0) is
x–0 y–0 z–0
 L: = =
1– 0 1– 0 1– 0
i.e. L : x = y = z = k (let) ……..(1)
Any point on L is p(k, k, k)
Since line & plane intersect at a point so, P lies on 2x + 2y + z = 10
 2k + 2k + k = 10
k=5
So, point is (2, 2, 2)
16. (c)
a1b1 + a 2 b 2 + a 3b3
Angle between planes = cos =
a + a 22 + a 32 · b12 + b 22 + b32
2
1
1 2 + 1 (–1) + 2 1
 cos =
12 + 12 + 22 · 22 + (–1) 2 + 12
3
 cos =
6
1
 cos =
2

=
3
17. (c)

y = (1 + x) (1 + x2) (1+ x4) ………..(1 + x2n)
Taking log both sides
logy = log(1 + x) + log (1 + x)2 + ……… + log(1 + x2n)
On differentiating both sides
1 dy 1 2x 2n x 2n –1
= + + .......... +
y dx 1 + x 1 + x 2 1 + x 2n
dy  1 2x 2n x 2n –1 
= y + + .......... +   at x = 0 y = 1
 1+ x 1+ x 1 + x 2n 
2
dx
dy
= 1·1 =1
dx x =0
18. (c)
 f(x) is odd differentiable
 f(–x) = –f(x)
On differentiating both sides
 –f (–x) = – f (x)
 f (x) = f (–x) ……..(1)
 put x = 3 in eq. (1)
 f (3) = f(–3)
 f (–3) = 2  f (3) = 2 (given)
19. (c)
 1– x 
 
 1 + x  1–x 
lim  
x →1 2 + x
 
 1– x 
 
 1 + x  (1– x )(1– x ) 
 lim  
x →1 2 + x
 
 1 
 
 1 + x  1+ x
 lim  
x →1 2 + x
 
1
 1 + 1 1+1
 
 2 +1 
1
 2 2
  .
3
20. (d)

  e 
 log  x 2  
f(x) = tan–1     + tan –1  3 + 2 log x 
 
 log(ex )   1 – 6 log x 
2
 
 
 1 – 2 log x  –1  3 + 2 log x 
f(x) = tan–1   + tan  
 1 + 2 log x   1 – 3.2 log x 
f(x) = tan–1 1 – tan–1 ( 2 log x ) + tan –1 3 + tan –1 ( 2 log x )
f(x) = tan–11 + tan–13 = constant
On differentiating with respect to x
f (x) = 0
Again differentiating with respect to x
f (x) = 0
21. (a)
log x
I=  3x
dx
Let log x =t
1 1 dx
 . dx = dt  = 2dt
x 2 x x
t
I=  3 2dt
2  t2  1
( )
2
 I =   + c  I = log x +c
3 2  3
22. (c)
I =  2x (f '(x) + f(x) log 2) dx
Let g(x) = 2xf(x)  g(x) = 2x(f(x) + f(x)n2)

I =  g '(x) dx
I = g (x) + C
I = 2x f(x) + C
23. (b)

1 
1
I =  log  –1 dx
0 x 
1– x 
1
 I =  log   dx
0  x 
On applying king property
 x 
1
 I =  log   dx
0  1 – x 
1– x 
1
= –  log   dx = –I
0  x 
I=–I
 2I = 0
I=0
24. (a)
 n + 1 + n + 2 + ....... + 2n –1 
lim  
n →
 n 3/2 
1  1 2 n –1 
 lim  1 + + 1 + + ..... + 1 + 
n → n
 n n n 
n –1
1 r
 lim  1+
n → n n
r =1
1
n –1
 
0
1 + x dx  upper limit = lim
n → n
=1
1
2 1
 (1 + x)3/2 lower limit = lim =0
3 n → n
0

2
3
2 2 –1( )
25. (b)

dy
x + y = xe x
dx
dy y
+ = e x …… linear differential equation
dx x
1
I.F = e  x = enx = x
dx
Solution of L.D.E
 y. x =  e x .x dx
 y . x = xex –  e x dx
 y.x = xex – ex + c
 xy = ex(x – 1) + c
On comparing with given relation xy = ex  (x) + c
(x) = (x–1)
26. (a)
Let, equation of parabola y2 = 4a (x – b) ……..(1)
On differentiating with respect to x
 2yy = 4a
Again differentiating with respect to x
 2yy + 2(y)2 = 0
Order of differential equation = 2
27. (c)
x 2 y2 1
E: + = ……....(1)
3 2 6
L: y = x +  …………(2)
Line is tangent to ellipse
 2x2 + 3 (x + )2 = 1
 2x2 + 3 (x2 + 2x + ) =1
 5x2 + 6x + 32 – 1 = 0
D=0
 362 – 20 (32–1) = 0
5
 – 242 + 20 = 0  62 – 5 = 0   = 
6
28. (c)

y = 5 – x2 …… (1)
y = |x – 1| …… (2)
From equation (1) & (2)
 5 – x2 = (x– 1)2
 5 – x2 = x2 –2x + 1
 2x2 – 2x – 4 = 0
 x2 – x – 2 = 0  x = –1, 2
y = |x – 1|
y= 5 – x 2
–1 2
( )
2
Required area = A = 5 – x 2 – | x –1| dx
–1
( ) dx −  (1 – x) dx − (x–1) dx
2 1 2
= 5–x 2
–1 –1 1
5 5
=2+ –
4 2
5 1
= –
4 2
29. (b)
Equation of line AB =
x y
 + =1 B (0, 12)
8 12
 3x + 2y = 2y
P (x, y)
 Sum of distance of P is minimum,
so P will be on line AB
O (0,0) A (8, 0)
AB: 3x + 2y = 24
x y
2 9
Possible points 4
 6
6
 3
 (x, y)  (2, 9), (4, 6), (6, 3)
30. (b)

Wwe have
T = 2
g
Taking log both sides

1
log T = log 2 + (log  – log g)
2
Differentiating both sides we get
1 11
 dT = · d
T 2
dT 1d  1
 100 =  100  =  2 = 1%
T 2  2
31. (a)
 1 1 1 
Direction cosine of diagonal OD =  , 
 3 3 3
 1 1 –1 
Direction cosine of diagonal FB =  , 
 3 3 3
Z
F(0,0,1) E(1,0,1)
G(0,1,1) D(1,1,1)
O A(1,0,0)
x
(0,0,0)
C
(0,1,0) B(1,1,0)
1 1 1 1 1  –1  1
cos = . + . +  =
3 3 3 3 3  3 3
32. (c)

loga x , log b x , log c x → A.P
Then,
 2 log b x = log a x + log c x
2 1 1
 = +
log x b log x a log x c
 2 log x a log x c = log x ac·log x b
 log x c 2 = log x ac·log x b
 c2 = (ac)loga b
33. (c)
1 + (1 + a) x + (1 + a + a2) x2 + ………
Multiply & divide by (1 – a) we get
1
= [(1– a) + (1 – a2) x + (1 – a3)x2 + ……..  ]
1– a
1
= [(1 + x + x2 + ……….  ) – (a + a2 x + a3 x2 + ………  )
1– a
1  1 a 
= –

1 – a 1 – x 1 – ax 
1 1– ax – a + ax 
=
(1– a)  (1– x)(1– ax) 
1
=
(1 – x)(1 – a x)
34. (a)
log.3 (x–1)  log.09 (x–1)  x –1 > 0  x > 1 ………(1)
 log.3 (x–1)  log (.3)2 (x–1)
 2 log.3 (x–1)  log.3 (x–1)
 log.3 (x–1) 2  log.3 (x–1)
 (x – 1)2 > (x – 1)
 x2 – 3x + 2 > 0
x (–,1)  (2, ) ……..(2)
From (1) & (2)
x (2, )
35. (b)

13
 (i
n =1
n
+ i n +1 )
 we know,
i + i2 + i3 + i4 = 0
13 13
  in +  in +1
n =1 n =1
 i13 + i14
 i + i2
i–l
36. (a)
1 1 1 1
+ + =1 zz =| z |2 = 1  z =
z z 2 z3 z
 z1 + z 2 + z 3 = 1
 | z1 + z 2 + z 3 | = 1 | z |=| z |
 | z1 + z2 + z3 | = 1
37. (a)

x2 + px + q=0

p+q=–p  2p + q = 0 …..(1)
 pq = q  q(p–1) = 0 … (2)
From equation (1) & (2)
 (–2p)(p–1) = 0
 p=0 or p = 1
q=0 or q = –2

38. (c)
(1) D > 0
9–4k>0
9
k< …..(1)
4 /2
–/2
1
0 1 y = cosx
(2) f(0) > 0

k>0 …… (2)
(3) f(1) > 0
1–3+k>0
k>2 …..(3)
From (1), (2) & (3)
 9
k   2, 
 4
39. (c)
A RR ANGE
6
Number of ways when R’s occurs together =
2
120.6
=
2
= 360
40. (b)
1 1 1
 5
+ 6 = 4
Cr Cr Cr
r!(5 – r)! r!(6 – r)! r!(4 – r)!
 + =
5! 6! 4!
 6(5 – r) + (6 – r) (5 – r) = 6 × 5
 30 – 6r + r2 – 11 r + 30 = 30
 r2 – 17r + 30 = 0
 r = 2, 15 (not possible)
r=2

41. (a)
n3 + 2n, where n is positive integer i.e. n = 1, 2, 3
When
n = 1, 13 + 2.1 = 3 = 3 × 1
n = 2, 23 + 2.2 = 12 = 3 × 4
n = 3, 33 + 2×3 = 33 = 3 × 11
n = 4, 43 + 2×4 = 72 = 3 × 24
n = 5, 53 + 2×5 = 135 = 3 × 45
Hence, n3 + 3n is divisible by 3
Option (a) is correct
42. (b)
(x –1) (x–2) (x–3) ….. (x–18)
(x –1) (x–2) = x2–3x + 2, coefficient of x = –3 = –(1+2)
(x –1) (x–2) (x–3) = (x2–3x + 2) (x – 3)
= x3 – 6x2 + 11x –6, coefficient of x2 = – 6 = –(1 + 2 + 3)
As, similarly,
Coefficient of x17 = –(1+2+3+……+18)
 18(18 + 1) 
=–  
 2 
= – 9 × 19 = –171
Hence option (b) is correct

43. (a)
= 1 + nC1 cos + nC2 cos2 + ……+ nCn cosn
 (1+x)n = nC0 + nC1x + nC2x2 + ….. + nCnxn
Put x = ei
(1+ ei) = 1+ nC1 ei + nC2 (ei)2 + ….. nCn (ei)n
= 1+ nC1 ei + nC2 e2i + ….. nCn (eni)
(ei = cos + isin)
 (1+cos + isin)n = 1 + nC1 ei + nC2 e2i + nC3 e3i + ….. + nCn eni
n
   
  1 + 2cos2 –1 + 2isin cos  = 1 + nC1 (cos + isin) + nC2 (cos2 + isin2)+….+nCn
 2 2 2
(cosn + isinn)
n
   
  2cos2 + 2isin cos  = (1 + nC1 cos + nC2cos2 +….. + nC2 cosn) + i(nC1sin + nC2
 2 2 2
sin2 +….+ nCn sinn)
  
 2ncosn (cos + i sin )n = (1 + nC1 cos + nC2cos2 +….. + nCn cosn) + i(nC1sin + nC2
2 2 2
sin2 +….+ nCn sinn)
n
  i 2 
 2 cos
n n
 e  = (1 + nC1 cos + nC2cos2 +….. + nCn cosn) + i(nC1sin + nC2 sin2
2  
+….+ nCn sinn)
  n n 
 2ncosn  cos 2 + isin 2  = (1 + C1 cos + C2cos2 +….. + Cn cosn) + i( C1sin +
n n n n
2  
nC2 sin2 +….+ nCn sinn)
  n n 
 2ncosn  cos + isin  = (1 + nC1 cos + nC2cos2 +….. + nCn cosn) + i(nC1sin +
2  2 2 
nC2 sin2 +….+ nCn sinn)
n n
  n   n
  2cos  .cos + i  2cos 2  .sin 2 = (1 + C1 cos + C2cos2 +….. + Cn cosn) +
n n n
 2 2  
i(nC1sin + nC2 sin2 + nC3 sin3 + ….. + nCn sinn)
Comparing both side with real part we get
n
  n
 2cos 2  cos 2 = 1 + C1cos + C2 cos2 + ….. + Cn cosn)
n n n
 
Hence option (a) is correct.

44. (c)
1 log x y log x z
= log y x 1 log y z
log z x log z y 1
log x x log x y log x z

= log y x log y y log y z ( log aa = 1)
log z x log z y log z z
log x log y log z

 R1
log x log x log x
log x log y log z
=  R2
log y log y log y
log x log y log z
 R3
log z log z log z
1 1 1
Taking , , Common R1, R2, R3 we get
log x log y log z
log x log y log z
= log x log y log z
log x log y log z
=0
Since all rows are identical.
45. (c)
  B is adjoint matrix of A
 B = adj(A)
 |B| = |adj(A)|
 64 = |adj(A)| ( |B| = 64)
 |A|(n–1) = 64 (n = 3)
 |A|(3–1) = 64
 |A|2 = 82
 |A| = ±8
Hence option (c) is correct.
46. (c)

  
cos 4 –sin 4 
Given Q =  
 sin  cos  
 4 4 
cos  –sin 
Let Q() =  
 sin  cos  
cos  –sin  cos  –sin 
Q(). Q() =  .  
 sin  cos    sin  cos  
 cos2  – sin2  –cos  sin  – sin  cos 
= 
sin  cos  + cos  sin  –sin2  + cos2  
cos2 –sin2
Q2() =  
 sin2 cos2 
cos2 –sin2 cos  –sin 
Now Q2(). Q() =  .  
 sin2 cos2   sin  cos  
 cos2.cos  – sin2.sin  –cos2 sin  – sin2.cos  
= 
cos .sin2 + cos2.sin  –sin .sin2 + cos2.cos  
cos3 –sin3
 Q3() =  
 sin3 cos3 
       1 1 
cos  3 4  –sin  3 4   – –
    2 2
 Q3() =  = 
       1 1 
 sin  3  cos  3    –
  4  4   2 2 
 1 1   1   1 1 1 1 
– 2 –
2  2 – 2 . 2 – 2 . 2 
Now Q3(). x     =  
 1 –
1   1   1 1 1 1 
 2 2   2   2. 2– 2. 2 
 
 1 1
– 2 – 2  –1
= =  
– 1 – 1   0 
 2 2 
Hence option (c) is correct
47. (c)

R = {x, y : x, y  z, x + 2y is divisible by 3}
Reflexive: Let x, y  z
x=y
x + 2x = 3x
it is divisible by 3
(x, x)  R
So it is reflexive
Symmetric: If x R y  x + 2y is divisible by 3.
Now, y + 2x = 3x + 3y – (x + 2y) is divisible by 3.
yRx
i.e. it is symmetric.
Transitive: x R y  x + 2y is divisible by 3.
y R z  y + 2z is divisible by 3.
 x + 2y + y + 2z is divisible by 3.
 x + 3y + 2z is divisible by 3.
 x + 2z is divisible by 3.
xRz
Hence transitive
Therefore, R is equivalence relation.
48. (c)
A = {5n – 4n–1 : n  N}
When n = 1, 51 – 4 × 1–1 = 0
n = 2, 52 – 4 × 2–1 = 16
n = 3, 53 – 4 × 3–1 = 112
n = 4, 54 – 4 × 4–1 = 608
...................
 A = {0, 16, 112, 608}
While, B = {16(n–1), n  N}
B = {0, 16, 32, 48……}
Hence it is clear that A  B
49. (c)

f(x) = (x2 + 1)35.  x  R
f(–x) = ((–x)2 +1)35
= (x2 +1)35 = f(x)
Hence f(x) is an even function
So its range  R (f(x) > 0  x  R)
And also it is not one-one and not onto.
50. (c)
Observations are a1, a2, a3 ……. an
a1 + a2 + .... + an
Mean ( x )=
n
x i2
 = –(x)2
n
2
a12 + a22 + a32 + ....a2n  a + a + ... + a n 
= – 1 2 
n  n 
a1 , a2, a3, ……., an
2
( a1 )2 + ( a2 )2 + ..... + ( an )2   a1 +  a2 + ..... +  an 
1 = – 
n  n 
2
 a2 + a2 + ..... + a2n  2  a1 + a2 + ..... + an 
=   1 2
2
–  
 n   n 
2
a12 + a22 + ..... + a2n  a + a + ..... + an 
= || – 1 2 
n  n 
1 = || 
51. (b)

OM = (0– h)2 + (0– k)2
OM = h2 + k 2
 In OPM ()
   Perpendicular h2 + k 2 
sin   = =
4 hypotenuse 1 /4
P
1 h +k
2 2 M(h,k)
 =
2 1
Squaring both side, we get
1 h2 + k 2
 =
2 1
 2(h + k2) = 1
2
(h, k) → (x, y)
2(x2+y2) = 1
1
 x2+y2 =
2
Hence option (b) is correct.
52. (d)
Let midpoint of chord = (h, k)
at 2 + 0
h=  at2 = 2h …..(1)
2
2at – 0
K=  at = k
2 P(at2, 2at)
k
t=
a
Putting value of t in (1) we get M(h, k)
2
k
 a   = 2h (0,0)
a
 k2 = 2ah  y2 = 2ax
Directrix of parabola y2 = 2ax is
–a
x=
2
Hence option (d) is correct
53. (b)

2
Let I =  0
x 2[x]dx
1 2
I= 
0
x 2[x]dx + 1
x 2[x]dx
  0  x < 1  [x] = 0
1  x < 2  [x] = 1
1 2
I=  0
x 2 .0dx +  1
x 2 .1dx
2
=0+ 1
x 2 dx
2
 x3 
=0+  
3
23 13
= –
3 3
8 1 7
= – =
3 3 3
54. (c)
f(x) = a – x,a + x.b

y=b
–a < x < , a < a < b
y=a–x y=a–x
y=a+x
y=a+x
(–a,0) (a,0)
y=b
Possible graph of f(x) is shown.
There are two sharp turn. Hence f(x) is not differentiable at two points.
55. (d)
| a + b|| a – b|

Squaring both sides we get
| a + b|2  | a – b|2
 | a |2 + | b|2 +2 | a |. | b| cos  < | a |2 + | b|2 – 2|a |.| b| cos 
( is the angle between a & b )
 2 |a | | b| cos  < – 2 | a | | b| cos 
 4 |a | | b| cos  < 0
 |a | | b| cos  < 0
 cos  < 0
  is an obtuse angle.
Hence (d) option is correct.
56. (b)
dy
y + by2 = acosx, 0 < x < 1
dx
dy dt
Put y2 = t  2y =
dx dx
dy 1 dt
y =
dx 2 dx
1 dt
+ bt = a cosx
2 dx
dt
 + 2bt = 2a cosx
dx
It is linear in ‘t’ we get
 IF = e 
2bdx
= e2bx
Solution is
t. IF =  2acosx . IF dx
 t. e2bx =  2acosx . e2bx dx

2a
 te2bx = (sinx + 2b cos x). e2bx + c
4b2 + 1
2a(sinx + 2bcosx) 2bx
 y2. e2bx = e +c
4b2 + 1
 (4b2 + 1)y2 = 2a(sinx + 2bcosx) + ce–2bx
57. (a)
Given, 16x2 + 9y2 = 400

16x 2 9y 2
 + =1
400 400
x2 y2
 + =1
25 400
9
Here a2 = 25  a = 5
400 20
b2 = b=
9 3
Any point on ellipse is (a cos, b sin )
 20 
  5cos , sin  
 3 
20
x = 5cos, y = sin 
3
dx dy 20
 = –5sin  , = cos 
d d 3
dx dy
 =– (Since ordinate is decreasing)
d d
20
 –5sin = – cos
3
4
 tan  =
3
3 3 5
cos  = or – 4
5 5
4 4
sin  = or – 
5 5
3
 16   –16 
 Points are  3,  and  –3,
 3   3 
58. (a)
Arranging in alphabetical order → C, C, H, I, N, O
Number of words that appear before the word COCHIN is
CC …….. → 4!
CH …….. → 4!
CI …….. → 4!
CN …….. → 4!
COCHIN …….. → 1
 No. of words before …….. COCHIN = 4! + 4! + 4! + 4! = 96
59. (d)

2 0 0
A = 0 2 0
2 0 2
2 0 0 2 0 0 4 0 0   22 0 0
0 4 0  =  
A2 = A. A = 0 2 0 0 2 0 =  
2
 0 2 0
   
2 0 2 2 0 2 8 0 4  2.22 0 0
 
 4 0 0  2 0 0
A3 = A2. A =  0 4 0  0 2 0
   
 8 0 4  2 0 2
 8 0 0  23 0 0
 
=  0 8 0 = 3
 0 2 0
 
24 0 8 3.23 0 23 
 
............
 2n 0 0   a 0 0
 
An =  0 2n 0  = 0 a 0
 
n.2n 0 2n   b 0 a 

 On comparing both matrix we get
a = 2n, b = n.2n
Hence option (d) is correct.
60. (a)
Given series is
Sn = 13 + 33 + 53 + 73 + . . . . . . . .
tr = (2r–1)3 = r3 – 1 – 3.2r.1(2r–1)
= r3–1–12r2 + 6r
n n
 Sn =  tr = (8r –1–12r
r =1 r =1
3 2
+ 6r)
n n n n
= 8 r3 –  1–12 r2 + 6 r
r=1 r =1 r =1 r=1
2
 n(n+ 1)  n(n+ 1)(2n+ 1) n(n + 1)
= 8  – n – 12 +6
 2  6 2
= 2n (n+1) – n –2n(n+1) (2n+1) + 3n(n+1)
2 2
= 2n2(n2+1+2n) – n –(2n2+2n) (2n+1) + 3n2 + 3n

= 2n4 + 2n2 + 4n3 – n –4n3 – 4n2 –2n2–2n + 3n2 + 3n
= 2n4 –n2
Sn = n2(2n2–1)
Hence option (a) is correct
61. (a)

   &  are roots of ax2 + bx + c = 0
Let y = x2  x = y
Putting y in the given equation, we get
 a( y )2 + b( y ) + c = 0
 ay + b y + c = 0
 b( y ) = –ay – c
 b2y = a2y2 + c2 +2acy
 a2y2 – (b2–2ac) y + c2 = 0
So the required equation is
a2x2–(b2–2ac) x + c2 = 0
Hence option (a) is correct
62. (a)
  (2–) (2–) + 2(3–) (3–) + ….+ (n–1) (n–)(n–2)
n
 Tr = (r–1)(r– )(r–  )
r =2
2
n
= (r – r  – r + )(r–  )
r =2
2 2
n
(
=  r3 – r2 – r2 + r – 2r2 + r3 + r2 – 3 )
r=2
n
=  ((r – r ( + 1 +  ) + r( + 
r=2
3 2 2 3
+ 2 )– 3 )
( 1 +  +  =   = )
( ) )
n
=  r3 – r2  0 + r  + 1 + 2 –1 (
r=2
n
= (r –1)
r =2
3
n n
=  r3 –1
r =2 r =2
 n(n+ 1)2  
=   –1 – (n–1)
  2  
n2(n+ 1)2
= –n
4
63. (c)

Given nCr–1 = 36 nCr = 84, nCr+1 = 126
n n n
……(i), = 84…..(ii), = 126 ….(iii)
r –1 n – r + 1 r n–r r + 1 n – r –1
Dividing (i) (ii) we get
n n 36
 =
r –1 n – r + 1 r n – r 84
n r n–r 3
× =
r –1 n – r + 1 n 7
r r –1 n – r 3
=
r –1(n– r + 1) n – r 7
r 3
 =
n– r +1 7
7r = 3n – 3r + 3
10r – 3n = 3 …..(iv)
Now dividing eq. (ii)  (iii) we get
n n 84
 =
r n – r r + 1 n – r –1 126
n r + 1 n – r –1 84
 =
r n–r n 126
r +1 2
=  5r–2n = – 3 ….(v)
n–r 3
Solving eq. (iv) & (v) we get r = 3 & n = 9
 nC8 = 9C8 = 9
64. (b)
Since there are total 14 males and 6 females in a group. In which 8 males and 3 females
are aged above 40 years.
Here out of 6 females 3 are above 40 and 3 are aged below 40. So probability of person
1
aged above 40 given female person =
2
65. (b)

Given equation, x3 – yx2 + x – y = 0
 x2(x – y) + (x – y) = 0
 (x – y) (x2 + 1) = 0
So only possibility is x – y = 0 or x2 + 1 = 0
 x = y or x2 + 1 = 0 (not possible)
Hence, given equation represents straight line. Hence option (B) is correct.
66. (b, d)
There is mistake in question.
If there are (2n – 1)th terms instead of (2n + 1) terms then nth terms of the AP, GP and
HP are the AM, GM & HM of the 1st and the last terms.
So, a  b  c&ac – b2 (B,D)
Otherwise if there are (2n + 1) terms then the nth terms should be in decreasing order of
AP. GP & HP.
i.e. a  b  c (B)
67. (b,d)
Let any parametric point on the line x + y + 1 = 0 is (t, –1 –t ).
Distance of (t, –1 – t) from 3x + 4y + 2 = 0 is
3  t + 4(–1 – t) + 2 1
 =
32 + 42 5
3t – 4 – 4 t + 2 1
 =
5 5
–t – 2 1
 =
5 5
 | t + 2 |= 1
 t + 2 = 1
 t + 2 = 1
 t + 2 = 1 or t + 2 = – 1
t = – 1 or t = – 3
 Possible points on the line x + y + 1 = 0
(t, –1–t)  (–1, –1 + 1) = (–1, 0)
(t, –1–t)  (–3, –1 + 3) = (–3, 2)
Hence option (B, D) are correct
68. (a, b)

WB-JEE-2016
t , 2t
(Mathematics) 1
2
1
t22, 2t2 r
Parabola: x2 = ay
Line: y – 2x = 1
(x2, y2)
y = 2x+1
Q
(x1, y1)
p
y = 2x+1
Solving parabola and line

x2 = a(1 + 2x)
 x2 = a + 2ax
 x2 – 2ax – a = 0
Let x1 and x2 are roots.
 x1 + x2 = 2a
x1 x2 = –a
 (x1 – x2)2 = (x1 + x2)2 – 4x1x2
= (2a)2 – 4(–a)
= 4a2 + 4a
(x1 – x2)2 = 4a(a + 1)
Point (x1, y1) lie on line y = 2 x + 1
 y1 = 2x1 + 1
Also point (x2, y2) line on line y = 2x + 1
 y2 = 2x2 + 1
 y1 – y2 = 2(x1 – x2)
(y1 – y2)2 = 4 (x1 – x2)2 = 4. 4a(a + 1)
(y1 – y2)2 = 16 a(a + 1)
Length PQ = (x1 – x 2 ) 2 + (y1 – y 2 ) 2
= 4a(a + 1) + 16a(a + 1)
PQ = 20a(a + 1)
 40 = 20 a(a + 1)
 40 = 20 a(a + 1)
 2 = a(a + 1)
 a2 + a – 2 =0
 (a + 2) (a – 1) =0
 a = –2, 1
Hence option (A, B) are correct.
69. (b, c)

 f (x) = (x – 1)2 (4 – x)
For critical point put f (x) = 0
 (x – 1)2 (4 – x) = 0 
 x = 1, 4
Therefore x = 1 & 4 are critical point of f(x)
Now sign scheme for f (x)
+ + – 3cos + 3cos
1 4
 f(x) is increasing in the interval (–, 4)
Hence also increasing in the interval (0, 3)
And f(x) is decreasing in the interval (4, )
Hence (B, C) option is correct.
From f (x), we can’t determine f(x) uniquely so f(o) can’t be predicted.
70. (b,d)
Given ellipse is 4x2 + 9y2 = 1
x2 y2
+ =1
1 1/ 9
1/ 4
1 1
Here, a2 = a=
4 2
1 1
b2 = b=
9 3
 Any point on ellipse is (acos, bsin)
 cos  sin  
 Point on ellipse is  , 
 2 3 
 cos  sin  
Equation of tangent at point  ,  is
 2 3 
xx1 yy1
+ 2 =1
a2 b
cos  sin 
x y
 2 + 3 =1
1 1
4 9

2x cos  3y sin 
 + =1
1 1
 2x cos + 3ysin = 1 …… (1)
8
Slope of line 9y = 8x  y = x ……. (2)
9
Equation slope of (1) and (2) we get
–2 cos  8 –3
 =  tan  =
3sin  9 4
C
5 3

A B
4
–4 3
Either, cos = ,sin  =
5 5
Or
4 3
Cos = , sin  = –
5 5
1 1 
So, a point on ellipse is (a cos, b sin) =  cos , sin  
2 3 
 1  –4  1  3    1  4  1  –3  
    ,    or    ,   
 2  5  3  5   2  5  3  5 
 –4 1   4 –1 
  ,  or  , 
 10 5   10 5 
 2 1 2 1
=  – ,  or  ,– 
 5 5 5 5
Hence option (B, D) are correct
71. (a,b)
1, 0  t 1
Given (t) = 
0, otherwise

 2016 
3000
So    (t– r') + (t– 2016)  dt
–3000  r ' = 2014 
3000
= –3000
[(t– 2014) + (t– 2015) + (t– 2016)](t– 2016)dt
2016
= [(t– 2014) + (t– 2015) + (t– 2016)](t– 2016)dt
–3000
2017
+ [(t– 2014) + (t– 2015) + (t– 2016)](t– 2016)dt
2016
3000
+ [(t– 2014) + (t– 2015) + (t– 2016)](t– 2016)](t– 2016)dt
2017
2016 2017 3000
= odt +  (0 + 0 + 1).1dt +  odt
3000 2016 2017
2017
=0+ 
2016
dt + 0
= [t] 2017
2016
= 2017 – 296
= 1.
Hence option (A, B) is correct
72. (a,c)
Given equation is x2 + y2 – 10x + 21 = 0
 x2 – 10x + (y2 + 21) = 0 have roots
x = a and y = b
 for real roots D  0
 (–10)2 – 4·1· (y2 + 21)  0
 100 – 4y2 – 84  0
 –4y2 + 16  0
 y2  4
 –2  y  2
Hence option (c) is correct
Now, y2 = –x2 + 10x – 2
For real roots of y
 –x2 + 10x – 21  0
 x2 + –10x + 21  0
 x2 –7x–3x + 21  0
 x(x–7)–3(x–7)  0
 (x – 7) (x – 3)  0
3 x7
Option (A) is correct
Hence option (A, C) are correct
73. (a,c)
 z = sin – icos

   
= cos   –  + i sin   – 
 2  2

i(  – )
=z e 2
n    
 i(  –  )   in   –  
So, zn =  e 2  = e  2  
 
     
zn = cos  n   –   + isin n   –  .....(1)
  2   2
1 1 1 1
Now = n= =  
( z )  i – 2   eni – 2 
n n
z 
e 
 
 

1 – ni(  – )  
n
= e 2
= cos(– n( – )) + isin(– n( – ))
z 2 2
       
= cos  n   –   – isin  n   –   .....(2)
  2    2 
Subtracting
1    
Zn – n
= cos n( – ) + i sin (n( – )) – cos (n(– ))+ i sin (n(– )
z 2 2 2 2
 
= 2 i sin n   – 
 2
1  n 
zn – = 2 i sin  n –  (option (c))
zn  2 
Adding (1) & (2) we get
1                
Zn + = cos  n   –   + i sin  n   –   + cos  n   –   – isin  n   –  
  2    2    2    2 
n
z
   
= 2 cos  n   –  
  2 
 n 
= 2 cos  n –  option (A)
 2 
74. (a,b)
f(f(x)) = x x X & x  R

So, f(x) = f–1(x)
 f(x) is self inverse
Hence f(x) is one-one and onto
Therefore, option (A, B) is correct.
75. (a,c)
3 1 3
P ( A  B)  and  P(A B) 
4 8 8
We know that
P(A B) = P(A) + P(B) – P(A B)
 P(A) + P(B) = P(A B) + P(A B) ……..(1)
3
  P(A B)  1
4
1 3
 P(A B) 
8 8
3 1 3
 +  P(A B) + P(A B)  1 +
4 8 8
7 6 +1 11
 =  P ( A  B ) + P(A B) 
8 8 8
From (1)
7 11
 P ( A) + P(B ) 
8 8
7
Hence P(A) + P(B)  option (c)
8
11
P(A) +P(B)  option (A)
8

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