Inverse Trigonometric Functions

1. If x, y, z are in A.P. and tan–1x, tan–1y and tan–1z

are also in A.P., then [JEE (Main)-2013] 146 145
(1) (2)
12 12
(1) x = y = z (2) 2x = 3y = 6z
(3) 6x = 3y = 2z (4) 6x = 4y = 3z 145 145
(3) (4)
10 11
dy
2. If y  sec(tan1 x ), then at x = 1 is equal to 6. If x = sin –1 (sin10) and y = cos –1 (cos10), then
dx y – x is equal to [JEE (Main)-2019]
[JEE (Main)-2013]
(1) 7 (2) 10
1 1 (3) 0 (4) 
(1) (2)
2 2
 19  n 
(3) 1 (4) 2 7. The value of cot   cot –1  1   2 p   is
 n 1  p 1  
  
 2x 
3. Let tan1 y  tan1 x  tan1  2 [JEE (Main)-2019]
 1 x 
1 19 23
where | x |  . Then a value of y is (1) (2)
3 21 22
[JEE (Main)-2015]
22 21
(3) (4)
3x  x3 3x  x3 23 19
(1) (2)
1  3x 2 1  3x 2 8. All x satisfying the inequality (cot–1x)2 – 7 (cot–1x)
+ 10 > 0, lie in the interval [JEE (Main)-2019]
3x  x3 3x  x3
(3) (4) (1) (cot 2, )
1  3x 2 1 3x2
(2) (cot 5, cot 4)
  (3) (–, cot 5)  (cot 4, cot 2)
4. Consider f  x   tan1  1  sin x  , x  0,   .
 1  sin x   2 (4) (–, cot 5)  (cot 2, )

A normal to y = f(x) at x  also passes 3  1
6 9. If  = cos –1   ,  = tan–1   , where 0 < ,
5 3
through the point [JEE (Main)-2016]

< , then  –  is equal to
 2     2
(1)  0,  (2)  ,0 
 3   6 
[JEE (Main)-2019]
    9 
(3)  , 0  (4) (0, 0) 1  9  cos 1 
 4  (1) tan   (2) 
 14   5 10 
–1  2  –1  3   3
5. If cos    cos     x   , then x is 1  9   9 
 3x   4x  2  4 (3) sin   (4) tan1  
equal to [JEE (Main)-2019]  5 10   5 10 
MATHEMATICS ARCHIVE - JEE (Main)

y 15. If S is the sum of the first 10 terms of the series

10. If cos–1x – cos–1 = , where –1  x  1, –2  y
2
 1  1  1  1
y tan1   + tan–1   + tan –1   + tan –1   +...,
 2, x  , then for all x, y, 4x2 – 4xy cos + y2 3
  7
  13
   21
2 then tan(S) is equal to [JEE (Main)-2020]
is equal to : [JEE (Main)-2019]
(1) 2 sin2 (2) 4 sin2 – 2x2y2 6 5
(1) – (2)
5 11
(3) 4 cos2 + 2x2y2 (4) 4 sin2
10 5
–1  12  –1  3  (3) (4)
11. The value of sin   – sin   is equal to 11 6
 13  5
[JEE (Main)-2019]  1  x 2  1
16. The derivative of tan1   with respect to
  56   63   x 
– sin–1    – sin–1    
(1) (2)
2  65   65 
 2x 1  x 2  1
tan1   at x  is
–1  33    9   1  2x 2  2
(3)  – cos   (4) – cos –1    
 65  2  65 
[JEE (Main)-2020]
 
12. If f(x) = (secx + tanx),   x  , and
tan –1
2 2 2 3 2 3
f(0) = 0, then f(1) is equal to [JEE (Main)-2020] (1) (2)
3 5

 1 1
(1) (2) 3 3
4 4 (3) (4)
10 12
2  1
(3) (4) 6
3 4 
4 4 17. If y   k cos –1  5 cos kx – 5 sin kx  , then
k =1
13. The domain of the function
dy
at x = 0 is ________. [JEE (Main)-2020]
1  |
x | 5  dx
f x  sin  2  is
 x 1 
 n  1  
, a ]  [a,  . Then a is equal to 18. lim tan   tan –1   is equal to
n
r 1  1 r  r2  
[JEE (Main)-2020] ______. [JEE (Main)-2021]

1  17 17 1 1 63 
(1) (2) 1 19. A possible value of tan  sin  is :
2 2 4 8 
[JEE (Main)-2021]
17  1 17
(3) (4) (1) 2 2  1 (2)
2 2 7 1

 4 5 16  1 1
14. 2   sin1  sin1  sin1  is equal to (3) (4)
 5 13 65  7 2 2
 4 
[JEE (Main)-2020] 20. cosec  2 cot 1 5  cos 1    is equal to :
  5 
 7 [JEE (Main)-2021]
(1) (2)
2 4
65 65
(1) (2)
3 5 56 33
(3) (4)
2 4 75 56
(3) (4)
56 33
ARCHIVE - JEE (Main) MATHEMATICS

26. The number of solutions of the equation

 1  2
sin 1 x cos 1 x tan 1 y sin1  x2    cos1  x2    x2 , for x  [  1,1]
21. If   ; 0 < x < 1, then the  3  3
a b c and [x] denotes the greatest integer less than or
 c  equal to x, is : [JEE (Main)-2021]
value of cos   is : [JEE (Main)-2021]
ab (1) Infinite (2) 2
(3) 4 (4) 0
1 y2
(1) 1 – y2 (2)
y y cos ec –1x
27. The real valued function f  x   , where [x]
x  x
1  y2 1  y2 denotes the greatest integer less than or equal to x,
(3) 2 (4) is defined for all x belonging to :
1 y 2y
[JEE (Main)-2021]
 (1) all non-integers except the interval [–1, 1]
22. If 0 < a, b < 1, and tan–1a + tan–1 b = , then the
4 (2) all integers except 0, –1, 1
value of (3) all reals except integers

 a2  b2   a3  b3   a 4  b 4  (4) all reals except the interval [–1, 1]

(a + b) –   + 
 
 – 
 
 +
 28. The number of real roots of the equation
 2   3   4  
... is : [JEE (Main)-2021] tan1 x(x  1)  sin1 x2  x  1  is :
4
e [JEE (Main)-2021]
(1) e2 – 1 (2) loge   (1) 2 (2) 1
2
(3) 4 (4) 0
(3) e (4) loge2
23. Given that the inverse trigonometric function take  3  5 
29. The value of tan  2 tan1    sin1    is equal to
principal values only. Then, the number of real  5
   13  
values of x which satisfy [JEE (Main)-2021]
 3x   4x 
sin1    sin 1  1
  sin x is equal to : (1)
220
(2)
151
 5   5  21 63
[JEE (Main)-2021] 181 291
(3) (4)
(1) 3 (2) 1 69 76
(3) 0 (4) 2 1  1  x 
30. The domain of the function cosec   is :
24. If cot–1() = cot–12 + cot–18 + cot–118 + cot–132  x 
+... upto 100 terms, then  is : [JEE (Main)-2021]
[JEE (Main)-2021]  1   1 
(1)   ,0   1,  (2)  ,    0
(1) 1.01 (2) 1.02  2   2 
(3) 1.03 (4) 1.00  1  1 
(3)  1,    0,  (4)   ,    0
25. The sum of possible values of x for  2  2 
 1  1  8 
tan–1(x  1)  cot 1    tan  31  is
2 2
 x  1    31. If sin1 x  cos1 x  a; 0 < x < 1, a  0,
[JEE (Main)-2021] then the value of 2x2 – 1 is [JEE (Main)-2021]

30 31  2a   4a 
(1)  (2)  (1) cos   (2) sin  
4 4      

32 33  4a   2a 
(3)  (4)  (3) cos   (4) sin  
4 4      
MATHEMATICS ARCHIVE - JEE (Main)

38. Considering only the principal values of inverse

1  1  sin x  1  sin x   
32. If y ( x )  cot   , x   ,   , functions, the set [JEE (Main)-2021]
 1  sin x  1  sin x  2 
dy 5 
then at x  is [JEE (Main)-2021] A  x  0: tan–1(2 x )  tan–1(3 x ) 
dx 6 4
1 (1) Is a singleton
(1)  (2) 0
2
(2) Contains two elements
1
(3) –1 (4) (3) Contains more than two elements
2
33. Let M and m respectively be the maximum and (4) Is an empty set
minimum values of the function f(x) = tan–1(sinx +
39. T he set of all values of k for which
 
cosx) in 0,  . Then the value of tan(M – m) is 3 3
 2 tan1 x  cot 1 x  k 3 , x  R , is the interval
equal to [JEE (Main)-2021]

(1) 2  3 (2) 2  3 [JEE (Main)-2022]

(3) 3  2 2 (4) 3  2 2
34. The domain of the function  1 7  1 13 
(1)  ,  (2)  24 , 16 
 32 8   
 3 x 2  x – 1  x  1
f ( x )  sin–1    cos –1   is
 ( x – 1)2   x  1
   1 13   1 9
(3)  ,  (4)  32 , 8 
[JEE (Main)-2021]  48 16   

 1  1 40. T he domain of the function

(1) 0,  (2) 0, 
 4  2
 x2 – 5x  6 
 1 1  1 1 cos–1  
(3)  ,   0 (4) [–2, 0]   ,   x2 – 9 
4 2 4 2 f (x)    is [JEE (Main)-2022]
loge ( x 2 – 3 x  2)
35. cos –1(cos(–5))  sin–1(sin(6)) – tan –1(tan(12))
is equal to (The inverse trigonometric functions take
the principal values) [JEE (Main)-2021] (1) (– , 1)  (2,  )
(1) 3 + 1 (2) 3 – 11
(2) (2, )
(3) 4 – 11 (4) 4 – 9
 1 
k  6r  (3)  – , 1  (2, )
36. Let Sk   tan1  . Then lim Sk is  2 
 22r 1  3 2r 1  k 
r 1  
equal to : [JEE (Main)-2021]
3  1   3  5 3 – 5 
(1) tan–1(3) (2) tan1   (4)  – , 1  (2,  ) –  , 
2  2   2 2 

3 
(3) cot 1   (4) 41. Let x * y = x2 + y3 and (x * 1) * 1 = x * (1 * 1).
2 2
 x 4  x2  2 
50
–1 1
Then a value of 2 sin1   is
 x 4  x2  2 
37. If  tan  p, then the value of tan p is  
r 1 2r 2
[JEE (Main)-2021] [JEE (Main)-2022]

50  
(1) 100 (2) (1) (2)
51 4 3

101 51  
(3) (4) (3) (4)
102 50 2 6
ARCHIVE - JEE (Main) MATHEMATICS

  15    n
 cos   –1   1  
 4   is equal to 47. The value of lim 6 tan   tan–1  
42. The value of tan –1  n  r  1  r 2  3r  3  
   
 sin  4  
    is equal to [JEE (Main)-2022]
[JEE (Main)-2022] (1) 1 (2) 2

  (3) 3 (4) 6
(1) – (2) –
4 8 48. The domain of the function

5 4  1 
(3) – (4) – 1 
12 9  2 sin  2 
cos1   4x  1 
is [JEE (Main)-2022]
  
43. Let f ( x )  2 cos –1 x  4 cot –1 x – 3 x 2 – 2 x  10,  –1, 1 ,  
If [a, b] is the range of the function f, then  
4a – b is equal to [JEE (Main)-2022]  1 1
(1) R    , 
(1) 11 (2) 11 –   2 2
(3) 11 +  (4) 15 –  (2) ,  1  1,   0
44. If the inverse trigonometric functions take principal
 1   1 
values, then (3)  ,  ,  0
 2   2 
 3   4  2   4   1   1 
cos –1  cos  tan –1     sin  tan–1     (4)  ,  ,   0
 10   3  5   3   2  2 
is equal to [JEE (Main)-2022]   1  1 
49. 50 tan  3 tan1    2 cos 1  
  2  5 
(1) 0 (2)
4 1 
 4 2 tan  tan 1 2 2  is equal to ______.
  2 
(3) (4)
3 6 [JEE (Main)-2022]

 2   7   3   1 5 1
45. sin1  sin   cos1  cos   tan1  tan  is 50. tan  2 tan1  sec 1  2 tan1  is equal to
 3   6   4   5 2 8 

equal to [JEE (Main)-2022]
[JEE (Main)-2022]
11 17 (1) 1 (2) 2
(1) (2)
12 12
1 5
(3) (4)
31 3 4 4
(3) (4) 
12 4 51. For k  R, let the solution of the equation

 50 

46. The value of cot   tan–1 
1 cos  sin1 x cot tan1 cos sin1 

  is  
n 1  1  n  n2  
   1
 k,0  x 
[JEE (Main)-2022] 2
Inverse trigonometric functions take only principal
26 25
(1) (2) values. If the solutions of the equation x2 – bx – 5 =
25 26
1 1  b
0 are  and , then 2 is equal to ______.
50 52  2
 2  k
(3) (4)
51 51 [JEE (Main)-2022]
MATHEMATICS ARCHIVE - JEE (Main)

52. Considering only the principal values of the inverse

trigonometric functions, the domain of the function 1 sin–1 x cos –1 x
56. If 0  x  and  , then a value
2  
1
 x2  4x  2 
f ( x )  cos   is [JEE (Main)-2022]
 x2  3   2 
 
of sin   is
 
 1  1 
(1)   ,  (2)   ,  
 4  4  2 2 2 2
(1) 4 1  x 1  2 x (2) 4x 1  x 1  2 x
 1   1
(3)  ,   (4)   , 
 3   3 
(3) 2 x 1  x 2 1  4x 2 2
(4) 4 1  x 1  4 x
2
53. Considering the principal values of the inverse
trigonometric functions, the sum of all the solutions
[JEE (Main)-2022]
of the equation cos–1(x) – 2sin–1(x) = cos–1(2x) is equal
57. The domain of the function
to [JEE (Main)-2022]

(1) 0 (2) 1  
f x  sin1 2 x 2  3   log2  log 1 x 2  5 x  5 ,
   
1 1  2 
(3) (4) 
2 2
where [t] is the greatest integer function, is
54. The sum of the absolute maximum and absolute
minimum values of the function [JEE (Main)-2022]
f x  tan1 sin x  cos x in the interval [0, ] is
 5 5 5   
(1)   ,  (2)  5  5 , 5  5 
 1    2 2   2 2 
 
(1) 0 (2) tan 1  
 2 4

 1     5 5  5 5
(3) cos 1   (4) (3)  1,  (4) 1, 
 3 4 12  2   2 
 
[JEE (Main)-2022]

55. The domain of the function 1 1 4 

58. Let x = sin(2tan –1 ) and y  sin  tan . If
2 3 
 x2  3x  2 
f ( x )  sin–1   is
 x 2  2x  7 
  S     : y 2  1  x , then  163
S
is equal to
(1) [1, ) (2) [–1, 2] _______.
(3) [–1, ) (4) (–, 2] [JEE (Main)-2022]
[JEE (Main)-2022]

  
Inverse Trigonometric Functions

1. Answer (1)
1 x
= tan tan
–1 x z 4 2
2y = x + z and 2tan–1y = tan
1 xz
x
=
4 2
2y x z
1 y2 1 xy 1
f (x) and at x , f ( x)
2 6 3
y2 xz So, equation of normal is
x, y, z are in GP 2
y 2 x y 2x
x=y=z 3 6 3
2. Answer (1) 5. Answer (2)

y sec sec –1 1 x2 1 x2 1 2 1 3 3
cos cos x
3x 4x 2 4
dy x
dx 1 2 1 3
1 x2 cos cos
3x 2 4x
dy 1
dx x 1 2 sin 1 x cos 1 x
2
3. Answer (1)
1 2 1 3
2x cos sin
tan y 1
tan x 1
tan 1 3x 4x
1 x2
2 16 x 2 9
1 3x x 3 cos 1
cos 1
3tan–1 x = tan 3x 4x
1 3x 2

3x x3 3 16 x 2 9
y sin 1
cos 1
1 3x2 4x 4x
4. Answer (1)
2 16 x 2 9
1 1 sin x
f (x) tan 3x 4x
1 sin x
64 81
2 x2
x x 9 16
cos sin
1 2 2
= tan
x x
2 145 3
cos sin x x
2 2 12 4
ARCHIVE - JEE (Main) MATHEMATICS

6. Answer (4) 9. Answer (3)

x= sin–1(sin10) 3
cos
5
3 10 3 4
x = 3 – 10 2 2 tan
3 x 3
1
3 10 4 and tan
and y = cos–1(cos10) 3
4 x
tan tan
y = 4 – 10 tan
1 tan · tan
y – x = (4 – 10) – (3 – 10) =
7. Answer (4) 4 1
3 3 1
19 n 4 13
cot cot –1 1 2p 1
9 9
n 1 p 1
9
19
cot cot –1
1 n n 1 13
n 1
1 9 1 9 1 13
tan sin cos
19 13 5 10 5 10
n 1 n
cot tan–1
n 1 1 n 1n 10. Answer (4)

19 y
cos–1x – cos–1
cot tan–1 n 1 – tan–1 n 2
n 1
xy y2
cot tan–1 20 – tan–1 1 cos–1 1 x2 . 1
2 4
20 – 1
cot tan –1
1 20 1 xy 1 x2 4 y 2
cos
2 2
19 21 21
cot tan –1 cot cot –1
21 19 19 xy + 1 x2 4 y 2 2cos
8. Answer (1)
(xy – 2 cos )2 = (1 – x2) (4 – y2)
x2y2 + 4cos2 – 4xycos = 4 – y2 – 4x2 + x2y2
2 4x2 – 4xycos + y2 = 4sin2
2 11. Answer (1)
cot 2 3 12
1 1
sin sin
5 13

1 3 5 12 4
sin
(cot–1x – 5) (cot–1 – 2) > 0 5 13 13 5
cot–1x (– , 2) (5, ) ...(i) ( xy 0 and x2 + y2 1)
But cot–1x lies in (0, )
1 33 1 33
From equation (i) sin sin
65 65
So, cot–1x (0, 2)
By graph, 1 56 1 56
cos sin
x (cot 2, ) 65 2 65
MATHEMATICS ARCHIVE - JEE (Main)

12. Answer (1) 15. Answer (4)

1 sin x x 1 1 1 1 1 1
f '( x ) tan 1
tan 1
tan S tan tan tan .........
cos x 4 2 3 7 13
1 1 1 1 1
x tan tan tan
, 1 1 2 1 2 3
in the neighbourhood of x = 1
4 2 2 2 1
.........
x 1 3 4
f '( x )
4 2 1 2 1 1 3 2
tan tan
1 2 1 1 3 2
x2
f (x) x c 4 3
1 11 10
4 4 tan .......... tan 1
1 3 4 1 11 10
f(0) = 0 c=0 = (tan–12 – tan–11) + (tan–1 3 – tan–12) +
1 1 (tan–14 – tan–13) + ..... + (tan–111 – tan–110)
So f (1) = tan–111 – tan–11
4 4 4
1 11 1
13. Answer (1) tan
1 11 1
x 5
f (x) sin–1 1 5
2 tan
x 1 6

x 5 5
–1 1 tan(S) =
2 6
x 1
16. Answer (3)
x2 – x – 4 0
1 x2 – 1
2 d tan–1
1 x –1 x
1– 17 1 17 d tan–1
x – x – 0 x
2 2 dx
2
2 x 1– x 2 x 1– x 2
1 17 d tan–1 d tan–1
x 1– 2x 2 1– 2 x 2
2
dx
1 17 1 17 2
x – ,– , 1 x –1
2 2 Simplifying tan–1 Put x = tan
x
14. Answer (3)
sec – 1 1– 1– 2sin2 / 2
–1 4 5 16 tan–1 tan–1
2 – sin + sin–1 + sin–1 tan 2sin / 2 cos / 2 2
5 13 65

4 5 16 1 x2 – 1 tan–1 x
=2 – tan–1 + tan–1 + tan–1 tan–1
3 12 63 x 2
45
2 x 1– x 2
–1 3
12 16 & similarly tan–1 Put x = sin
=2 – tan + tan–1 1– 2 x 2
4 5 63
1–
3 12 sin2
tan–1 2 = 2sin–1x
63 16 cos 2
=2 – tan –1 + tan–1
16 63 Hence required derivative
1
63 63
=2 – tan –1 + cot –1 2(1 x 2 ) 1– x 2 3
16 16
2 4(1 x )2 10
3 1– x 2 1
=2 – = x
2 2 2
ARCHIVE - JEE (Main) MATHEMATICS

17. Answer (91)

6
3 4
y k cos –1 cos kx – sin kx
k =1 5 5

3 4
Let cos and sin
5 5
6 3 7
2cos2 1 2cos2
y k cos –1 {cos coskx – sin sinkx} 4 4 4 4
k =1
7 7
cos2 cos
6 4 8 4 2 2
k cos –1 cos kx
k =1 1
tan
4 7
6 6
k kx k x 2
k 20. Answer (1)
k 1 k 1
1 1 4
cos ec 2 cot 5 cos
dy 6
6 7 13 5
k2 91
dx k 1 6 1 5 1 3
cos ec tan tan
12 4
18. Answer (1)
1 56
cos ec tan
1 r 1 r 33
tan –1 tan 1
1 r r2 1 r 1r 1 65 65
cos ec cos ec
56 56
tan –1 r 1 – tan 1 r 21. Answer (3)
sin –1x cos –1 x tan –1 y
n
1 k (say)
tan 1 a b c
So (tan–12 – tan–11)
r 1 1 r r2 sin–1 x = ak, cos–1x = bk and tan–1y = ck
+ (tan–13 – tan–12) + .......... + (tan–1(n + 1) – Now,
tan–1n)
= tan–1(n + 1) – tan–11 sin –1 x cos –1 x
2
n
1
lim tan tan –1
n r 1 1 r r2 (a b)x
2
–1 –1
lim tan tan n 1 – tan 1
n k
2(a b)
= tan 1
2 4 c
Now tan –1 y
19. Answer (3) 2(a b)

1 63
tan sin 1 tan ? c
4 8 4 cos cos(2 tan –1 y)
ab
Let

sin 1
63 1 63 1 – y2
8
and cos
8
sin
8
tan 63 and 6
cos cos –1 [if y > 0]
1 y2
1 9 9
2cos2 1 2cos2 cos2
2 8 2 8 2 16
1– y 2
3
cos
2 4
1 y2
MATHEMATICS ARCHIVE - JEE (Main)

22. Answer (4) 100

1 1
tan 2k 1 tan 2k 1
1 1 1 a b
tan a tan b tan k 1
a ab 4
= tan–1201 – tan–1 1
a + b + ab = 1
(1 + a)(1 + b) = 2 1 200
tan
Given 202
= cot–1(1.01)
a2 a3 b2 b3
a ... b ... Hence = 1.01
2 3 2 3
25. Answer (3)
ln(1 + a) + ln(1 + b)
1 8
ln(1 + a)(1 + b) = ln 2 tan 1 (x 1) cot 1
tan 1

x 1 31
23. Answer (1)
8
3x 4x tan 1 (x 1) tan 1(x 1) tan 1
1 1 1 31
sin sin sin x
5 5
1 (x 1) (x 1) 1 8
tan tan
1 (x 1)(x 1) 31
1 3x 16x 2 4x 9x 2 1
sin 1 1 sin x x 4
5 25 5 25
2 x2 31
1
4x2 + 31x – 8 = 0 x or x 8
1 3x 25 16x2 4x 25 9x2 1 4
sin sin x
25 1
x does not satisfy
4
3x 25 16x 2 4x 25 9x 2 25x 32
Hence, sum of possible values of x 8
4
x = 0 or 3 25 16x 2 4 25 9x 2 25 26. Answer (4)
Squaring both sides
1 1 1
sin x2 cos 1
x2 1 x2
2 1 3 3
x
2
1 1 4 1
x2 , ; so x 2 0 or 1
1 3 3 3 3
x
2 Hence L.H.S. is always equal to .
1 and x2 = has no solution in [–1, 1].
x 0,
2 27. Answer (1)
24. Answer (1) cosec–1x defined for x (– , –1] [1, )

also {x} 0 x Z
1 1 1
cot cot 2 cot 8 cot 1 18 cot 1
32 ...100 terms f(x) is defined for all non-integer except interval
[–1, 1]
28. Answer (4)
11 11 1 1 1 1
tan tan tan tan ...100 term x(x + 1) 0 ...(i)
2 8 18 32
x2 + x + 1 1
100
1 1 x2 + x 0 ...(ii)
tan
2
k 1 2k x2 +x=0 x = 0, –1
When x = 0 or –1
100 n 2k 1 2k 1
1 2 1
tan tan
2 1 2k 1 2k 1 LHS
k 1 4k k 1 No solution
2
ARCHIVE - JEE (Main) MATHEMATICS

29. Answer (1) 32. Answer (1)

x x
6 6
1 sin x sin cos and
1 3 1 1 15 2 2
2tan tan 5
tan 5
tan 1
5 1 9 16 8
5
2 25 x x
1 sin x sin cos
2 2
x
3 5 15 5 2 sin
2tan 1 sin 1 tan 1 tan 1 y(x) cot 1 2 cot 1 tan
x x
5 13 8 12 x 2 2 2
2cos
2
15 5
dy 1
tan 1 8 12
15 5 dx 2
1 , 33. Answer (4)
8 12
Range of sinx + cosx for x 0, is [1, 2]
180 40 220 2
tan 1 tan 1
21 21 So, M tan 1 2 and m = tan–11
30. Answer (2)
For domain 2 1
M m tan 1
2 1
1 x 1 x
1 or 1 2 1
x x tan(M m) 3 2 2
2 1
1 2x 1 34. Answer (3)
0 or 0
x x 3x 2 x –1 x –1
[–1, 1] and [–1, 1]
2 x 1
1 ( x – 1)
x ,0 or x (0, )
2 1 1
x –2, and x – ,0 , – 1 and
2 4
1 x 0,
domian x ,0 0,
2
1 1
finally x 0 ,
1 4 2
i.e x , – 0 35. Answer (3)
2
cos –1(cos(–5)) –5 2 a (say)
31. Answer (4)
(sin–1 x)2 – (cos–1 x)2 = (sin–1 x + cos–1 x) (sin–1 x sin –1(sin 6) 6–2 b (say)
–cos–1 x) = a –1
tan (tan12) 12 – 4 c (say)
a + b – c = 2 – 5 + 6 – 2 – 12 + 4
2cos 1 x a = 4 – 11
2 2
36. Answer (3)
2r
1 2a r
2cos x 6 3r 1
2 Let Tk tan 1 tan 1
2r 1 2r 1 2r 1
2 3 2
take sine both sides 1
3
r r 1
2a 2 2
sin 2cos 1 x sin 3 3 2
r
2
r 1
2 tan 1
tan 1
tan 1
2r 1 3 3
2
2a 1
cos 2cos 1 x sin 3

k k 1
2a 2 2
2cos2 cos 1 x 1 sin then Sk Tk tan 1 tan 1
r 1 3 3

2a 2
2 x 2 1 sin lim Sk tan 1 .
k 3
MATHEMATICS ARCHIVE - JEE (Main)

37. Answer (2)

Max will occur around t
2
1 2
tan 1 tan–1
2r 2 1 (4r 2 – 1) 3
7 3
Range of ƒ t ,
32 8
1 (2r 1) – (2r 1)
= tan 1 (2r 1)(2r 1)
1 7
k ,
32 8
= tan 1 2r 1 – tan–1(2r 1)
40. Answer (4)
50 50
1 1
tan tan –1(2r 1) – tan–1(2r – 1)
r 1 2r 2 r 1 x2 – 5x 6
–1 1 and x2 – 3x + 2 > 0, 1
2
x 9
p = tan–1(101) – tan–1 1
( x – 3)(2 x 1) 5( x – 3)
0 0
2
–1 101– 1 x –9 x2 – 9
= tan
1 101 1
Solution to this inequality is
50 –1
tan p = x , – 3
51 2
38. Answer (1)
For x2 – 3x + 2 > 0 and 1
tan–1(2x) + tan–1(3x) =
4 3– 5 3 5
x (– ,1) (2, ) – ,
5x 5x 2 2
tan–1 1
1 6x 2 4 1 6x 2
Com bining the two s olution sets (taking
i.e. 6x2 + 5x – 1 = 0
intersection)
(6x–1)(x + 1) = 0
1 1 3– 5 3 5
x= (as x 0) x – ,1 (2, )– ,
6 2 2 2
Hence A is a singleton set
41. Answer (2)
39. Answer (1)
Given x * y = x2 + y3 and (x * 1) * 1 = x * (1 * 1)
So, (x2 + 1) * 1 = x * 2
Let tan–1x = t ,
2 2 (x2 + 1)2 + 1 = x2 + 8
x4 + 2x2 + 2 = x2 + 8
1
cot x t
2 (x2)2 + x2 – 6 = 0
(x2 + 3)(x2 – 2) = 0
3 2
ƒ(t ) t3 t ƒ (t ) 3t 2 3 t
2 2 x2 2

x4 x2 2 4
ƒ (t) = 0 at t = Now, 2sin 1 2sin 1
4 x4 x 2
2 8

3 3 3
ƒ t min 2·
64 64 32 6 3
ARCHIVE - JEE (Main) MATHEMATICS

42. Answer (2) 46. Answer (1)

15 50
1
cos 1 cot tan–1
–1 4 1 n n2
tan n 1
sin
4
50
(n 1) n
= cot tan–1
n 1 1 (n 1) n
1
1
tan–1 2
1 50
2 = cot tan–1(n 1) tan–1 n
n 1

tan–1 1 2 tan–1 2 1
= cot tan –1 51 tan–1 1

8
51 1
= cot tan –1
43. Answer (2) 1 51

f (x ) 2cos–1 x 4cot –1 x – 3x2 – 2x 10 x [–1, 1]

52
= cot cot –1
2 4 50
f (x ) – – – 6x – 2 0 x [–1, 1]
1– x 2 1 x2
26
So, f(x) is decreasing function and range of f(x) is =
25
f (1), f (–1) , which is 5, 5 9
47. Answer (3)
Now 4a – b = 4( + 5) – (5 + 9)
n
= 11 – 1
lim 6 tan tan–1
2
44. Answer (3)
n r 1 r 3r 3

3 4 2 4
cos –1 cos tan–1 sin tan –1 n
(r 2) (r 1)
10 3 5 3 = lim 6 tan tan –1
n r 1 1 (r 2)(r 1)

3 3 2 4
cos–1 · ·
10 5 5 5 n
= lim 6 tan (tan–1(r 2) tan–1(r 1))
n r 1
1
cos –1
2 3
–1 –1
= lim 6 tan tan ( n 2) tan 2
45. Answer (1) n

3 – 3
sin–1 cos –1 tan–1(–1) 1
2 2 = 6 tan cot –1
2 2

5
– 1
3 6 4 = 6 tan tan –1
2
4 10 – 3 11
12 12 =3
MATHEMATICS ARCHIVE - JEE (Main)

48. Answer (4) 1 1

1
50 tan tan 4 2 tan tan 1 2 2
2 2
2 1 1
1 sin 1
2
4x 1 1
50 4 2 tan , where 2 tan 1 2 2
2
1
sin 1 2 tan
2 2 2
4x 1 2 2 … (i)
1 tan2
1 2 2 tan2 2 tan 2 2 0
1 1
2
4x 1
2 2 tan2 4 tan 2 tan 2 2 0

1 4x 2 2 2 tan 2 tan 2 0
1 0
2 (2n 1)(2x 1)
4x 1
1
tan 2 or
2

1
tan
2
1 1 tan 2 doesn't satisfy i
x , (0) , …(1)
2 2
1
25 4 2 29
2 2
1 1 4x 2
1 1 0 0
4x2 1 4 x2 1 4x2 1 50. Answer (2)

1 5 1
1 1 tan 2 tan 1 sec 1 2 tan 1
x x 5 2 8
2 2
0
1 1
x x 1 1
2 2
1 5 8 1 5
tan 2 tan sec
1 1 2
1
5 8

1 1
tan 2 tan 1 tan 1
3 2

1 1 1 1 2
x , , , …(2)
2 2 2 2 1 3 1
tan tan tan 1
1 2
From (1) & (2), 1
9

1 1 3 1
x , , {0} tan tan 1 tan 1
2 2 4 2
49. Answer (29)
3 1 5
tan tan 1 4 2 tan tan 1 4
1 1 1 1 1 3 5
50 tan tan 2 tan 2 tan 2 1
2 2 8 8
1
tan
4 2 tan 2 2 tan tan 1 2 2
2
ARCHIVE - JEE (Main) MATHEMATICS

51. Answer (12) 52. Answer (2)

x2 4x 2
cos sin 1
x cot tan 1
cos sin 1
k 1 1
2
x 3

x2 3 x2 4x 2 x2 3
1 1 2
cos sin x cot tan 1 x k
2x 2 4x 5 0 & –4x 1

1
x R & x
1 x 4
cos sin k
1 x2
1
So domain is , .
4

1 2x 2 53. Answer (1)

k
1 x2 cos–1x – 2sin–1x = cos–12x

1 1
For Domain : x ,
1 2x 2 2 2
k2
1 x2
1 1 1
cos x 2 cos x cos 2x
1 – 2x2 = k2 – k2x2 2
cos–1x + 2cos–1x = + cos–12x
2
k 1 cos(3cos–1x) = – cos(cos–12x)
x2 0
k2 2 4x3 = x

1
x = 0, ±
1 1 k2 2 2
2 2
2 …(1)
k2 1 54. Answer (3)
1
f x tan sin x cos x , 0,

and 1 …(2)
Let g x sin x cos x

3
2 2 sin x and x ,
k 2 4 4 4 4
2 1 5
k2 1
g x 1, 2

1 and tan–1x is an increasing function

k2
3 1 1
f x tan 1 ,tan 2

k2 2 1
and b = S.R = 2 1 4 ,tan 2
k2 1 4

1
Sum of fmax and fmin tan 2
b 4 4
12
k 2 1
3 1 1
cos
3 4
MATHEMATICS ARCHIVE - JEE (Main)

55. Answer (3)

cos 2 0 as 2 0,
–1 x2 3x 2 2
f (x) sin
2
x 2x 7
sin4 = 2 · 2 x 1 x 2 (1 2 x 2 )
x2 3x 2
–1 1 4 x 1 x 2 1 2x 2
x2 2x 7
57. Answer (3)
x2 3x 2
1
x2 2x 7 –1 2 x 2 – 3 2 log 1 ( x 2 – 5 x 5) 0
2
x2 3x 2 x2 2x 7
or 2 2x 2 5 0 x2 – 5x 5 1
5x –5
2 5
x –1 …(i) or 1 x x 2 – 5x 5 0 & x 2 – 5x 4 0
2
x 2 – 3x 2
–1 5 5– 5 5 5
2 x – , –1
x 2x 7 x – , ,
2 2 2
x2 3x 2 – x2 2x 7
5
2x 2 x 9 0 1, & x (– , 1) (4, )
2
x R …(ii)
Taking intersection
(i) (ii),
Domain [–1, ) 5 5
x 1,
2
56. Answer (2)
58. Answer (130)
sin–1 x cos–1 x –1 –1
Let k sin x cos x k( )
1 2
x sin 2 tan 2 …(i)
1
.
2k 1 4 1 1
and y sin tan 1 sin sin 1

2 3 5 5
2 2 –1
Now, 4k 4 sin x
Now, y 2 1 x
2k
1 2
2 1
–1 5 1 2
Here sin sin (4 sin x)
2 2
1 5 5 10
1
Let sin–1 x = x 0, 0, 2 2
5 2 0
2 4
x = sin 1
2,
2
cos 1 x2
3 1
16 16 23 16
sin2 2x · 1 x 2 S 23
= 130
cos 2 1 4 x 2 (1 x 2 ) (2x 2 1)2 1 2x 2