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Inverse Trigonometric Functions — JEE Main & Advanced Mathematics PYQs

266 previous year questions from Inverse Trigonometric Functions with answers and solutions. Numbered list, year tags, and one-tap solutions — built for serious JEE / NEET practice.

266 questionsMathematicsSolutions on every page
1

Considering only the principal values of the inverse trigonometric functions, the value of ⁻¹( (-11)) + 10 (2 ⁻¹ ( 1 2 ) ) + 10 (2 ⁻¹(2)) is

JEE Advanced 2026 Solution View → 2

Let = 3 ⁻¹ ( 6 11 ) and = 3 ⁻¹ ( 4 9 ) , where inverse trigonometric functions take only the principal values. Given below are two statements: Statement I: ( + ) > 0 . Statement II

JEE Main 2026 Solution View → 3

If ( ⁻¹(x 2 )) = ( ⁻¹ 1-x^2 ) , x (0,1) , then the value of x is :

JEE Main 2026 Solution View → 4

Let [ ] denote the greatest integer function. If the domain of the function f(x) = ⁻¹ ( x+[x] 3 ) is [ , ) , then ^2 + ^2 is equal to:

JEE Main 2026 Solution View → 5

Let 0 < < 1 , = 1 3 and ⁻¹(1- ) + ⁻¹(1- ) = 4 . Then 6( + ) is equal to:

JEE Main 2026 Solution View → 6

If 4 + _ p=1 ¹¹ ⁻¹ ( 2^ p-1 1 + 2^ 2p-1 ) = , then is equal to __________.

JEE Main 2026 Solution View → 7

If y = ⁻¹ ( 3 x - 4 x 4 x + 3 x ) + 2 ⁻¹ ( x 1+ 1-x^2 ) , then dy dx at x = 3 2 is equal to:

JEE Main 2026 Solution View → 8

Considering the principal values of inverse trigonometric functions, the value of the expression (2 ⁻¹ ( 2 13 )-2 ⁻¹ ( 3 10 ) ) is equal to :

JEE Main 2026 Solution View → 9

If k= ( 4 + 1 2 ⁻¹ ( 2 3 ) )+ ( 1 2 ⁻¹ ( 2 3 ) ) , then the number of solutions of the equation ⁻¹(k x-1)= ⁻¹ x- ⁻¹ x is _ _ _ _

JEE Main 2026 Solution View → 10

If the domain of the function f(x)= ⁻¹ ( 1 x²-2 x-2 ) , is (- , ] [ , ] [ , ) , then + + + is equal to

JEE Main 2026 Solution View → 11

The number of solutions of ⁻¹ 4x + ⁻¹ 6x = 6 , where - 1 2 6 < x < 1 2 6 , is equal to

JEE Main 2026 Solution View → 12

Let the maximum value of ( ⁻¹ x )²+ ( ⁻¹ x )² for x [- 3 2 , 1 2 ] be m n ² , where gcd ( m , n )=1 . Then m + n is equal to _ _ _ _ .

JEE Main 2026 Solution View → 13

If the domain of the function f(x)= ⁻¹ ( 2 x-5 11-3 x )+ ⁻¹ (2 x²-3 x+1 ) is the interval [ , ] , then +2 is equal to :

JEE Main 2026 Solution View → 14

The total number of real solutions of the equation = ⁻¹(2 )- 1 2 ⁻¹ ( 6 9+ ^2 ) is (Here, the inverse trigonometric functions ⁻¹ x and ⁻¹ x assume values in [- 2 , 2 ] and (- 2 , 2

JEE Advanced 2025 Solution View → 15

The value of ⁻¹ ( 1+ ^2(2) -1 (2) )- ⁻¹ ( 1+ ^2 ( 1 2 ) +1 ( 1 2 ) ) is equal to

JEE Main 2025 Solution View → 16

The sum of the infinite series ⁻¹ ( 7 4 )+ ⁻¹ ( 19 4 )+ ⁻¹ ( 39 4 )+ ⁻¹ ( 67 4 )+ . is :-

JEE Main 2025 Solution View → 17

Considering the principal values of the inverse trigonometric functions, ⁻¹ ( 3 2 x+ 1 2 1-x^2 ),- 1 2 x 1 2 , is equal to

JEE Main 2025 Solution View → 18

If y= ( 3 + ⁻¹ x 2 ) , then (x-y)^2+3 y^2 is equal to _____.

JEE Main 2025 Solution View → 19

Let ( S = x: ⁻¹ x= + ⁻¹ x+ ⁻¹(2 x+1) ). Then ( _ x ~S (2 x-1)^2 ) is equal to ______.

JEE Main 2025 Solution View → 20

( ⁻¹ 3 5 + ⁻¹ 5 13 + ⁻¹ 33 65 ) is equal to:

JEE Main 2025 Solution View → 21

If 0 , then the expression ⁻¹ + (1+ ^2 ) ( - ) + ⁻¹ + (1+ ^2 ) ( - ) + ⁻¹ + (1+ ^2 ) ( - ) is equal to :

JEE Main 2025 Solution View → 22

If for some , ; , + -8 and ^2 ( ⁻¹ )+ cosec ^2 ( ⁻¹ )-36 , then ^2+ is_______.

JEE Main 2025 Solution View → 23

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of 16 ( ( ⁻¹ x )^2+ ( cosec ⁻¹ x )^2 ) is :

JEE Main 2025 Solution View → 24

Considering only the principal values of the inverse trigonometric functions, the value of ( ⁻¹ ( 3 5 )-2 ⁻¹ ( 2 5 ) ) is

JEE Advanced 2024 Solution View → 25

The integral _ 1 / 4 ^ 3 / 4 (2 ⁻¹ 1-x 1+x ) d x is equal to

JEE Main 2024 Solution View → 26

Let the inverse trigonometric functions take principal values. The number of real solutions of the equation 2 ⁻¹ x+3 ⁻¹ x= 2 5 , is _______

JEE Main 2024 Solution View → 27

For n N , if ⁻¹ 3+ ⁻¹ 4+ ⁻¹ 5+ ⁻¹ n= 4 , then n is equal to_____

JEE Main 2024 Solution View → 28

Given that the inverse trigonometric function assumes principal values only. Let x , y be any two real numbers in [-1,1] such that ⁻¹ x- ⁻¹ y= , - 2 . Then, the minimum value of x^

JEE Main 2024 Solution View → 29

If a = sin − 1 sin 5 and b = cos − 1 cos 5 , then a 2 + b 2 is equal to

JEE Main 2024 Solution View → 30

For α , β , γ ≠ 0 . If sin − 1 α + sin − 1 β + sin − 1 γ = π and α + β + γ α − γ + β = 3 α β , then γ equal to

JEE Main 2024 Solution View → 31

Let x = m n ( m , n are co-prime natural numbers) be a solution of the equation cos 2 sin - 1 x = 1 9 and let α , β ( α > β ) be the roots of the equation m x 2 - n x - m + n = 0 .

JEE Main 2024 Solution View → 32

Considering only the principal values of inverse trigonometric functions, the number of positive real values of x satisfying tan - 1 ( x ) + tan - 1 ( 2 x ) = π 4 is :

JEE Main 2024 Solution View → 33

For x &#8712; ( - 1 , 1 ] , the number of solutions of the equation sin - 1 x = 2 tan - 1 x is equal to

JEE Main 2023 Solution View → 34

If S = x &#8712; &#8477; : sin - 1 x + 1 x 2 + 2 x + 2 - sin - 1 x x 2 + 1 = &#960; 4 then &#8721; x &#8712; S sin x 2 + x + 5 &#960; 2 - cos x 2 + x + 5 &#960; is equal to _ _ _ _

JEE Main 2023 Solution View → 35

Let S = x &#8712; R : 0 &#60; x &#60; 1 and 2 tan - 1 1 - x 1 + x = cos - 1 1 - x 2 1 + x 2 . If n ( S ) denotes the number of elements in S then :

JEE Main 2023 Solution View → 36

Let S be the set of all solutions of the equation cos - 1 2 x - 2 cos - 1 1 - x 2 = &#960; , x &#8712; - 1 2 , 1 2 . Then &#8721; x &#8712; S 2 sin - 1 x 2 - 1 is equal to

JEE Main 2023 Solution View → 37

Let a , b &#8834; 0 , 2 &#960; be the largest interval for which sin - 1 sin &#952; - cos - 1 sin &#952; &#62; 0 , &#952; &#8712; 0 , 2 &#960; , holds . If &#945; x 2 + &#946; x +

JEE Main 2023 Solution View → 38

If sin - 1 &#945; 17 + cos - 1 4 5 - tan - 1 77 36 = 0 , 0 &#60; &#945; &#60; 13 , then sin - 1 sin &#945; + cos - 1 cos &#945; is equal to

JEE Main 2023 Solution View → 39

Let y = f x represent a parabola with focus - 1 2 , 0 and directrix y = - 1 2 . Then S = x &#8712; &#8477; : tan - 1 f x + sin - 1 f x + 1 = &#960; 2 :

JEE Main 2023 Solution View → 40

Let a 1 = 1 , a 2 , a 3 , a 4 , &#8230; . . be consecutive natural numbers. Then tan - 1 1 1 + a 1 a 2 + tan - 1 1 1 + a 2 a 3 + &#8230; . . + tan - 1 1 1 + a 2021 a 2022 is equal

JEE Main 2023 Solution View → 41

If the sum of all the solutions of tan - 1 2 x 1 - x 2 + cot - 1 1 - x 2 2 x = &#960; 3 , - 1 &#60; x &#60; 1 , x &#8800; 0 , is &#945; - 4 3 , then &#945; is equal to _____ .

JEE Main 2023 Solution View → 42

tan - 1 1 + 3 3 + 3 + sec - 1 8 + 4 3 6 + 3 3 =

JEE Main 2023 Solution View → 43

Considering only the principal values of the inverse trigonometric functions, the value of 3 2 cos - 1 2 2 + &#960; 2 + 1 4 sin - 1 2 2 &#960; 2 + &#960; 2 + tan - 1 2 &#960; is __

JEE Advanced 2022 Solution View → 44

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos - 1 x - 2 sin - 1 x = cos - 1 2 x is equal to

JEE Main 2022 Solution View → 45

For k &#8712; &#8477; , let the solutions of the equation cos sin - 1 x cot tan - 1 cos sin - 1 x = k , 0 &#60; x &#60; 1 2 be &#945; and &#946; , where the inverse trigonometric f

JEE Main 2022 Solution View → 46

If 0 &#60; x &#60; 1 2 and sin - 1 x &#945; = cos - 1 x &#946; , then a value of sin 2 &#960; &#945; &#945; + &#946; is

JEE Main 2022 Solution View → 47

tan 2 tan - 1 1 5 + sec - 1 5 2 + 2 tan - 1 1 8 is equal to:

JEE Main 2022 Solution View → 48

Let x = sin 2 tan - 1 &#945; and y = sin 1 2 tan - 1 4 3 . If S = &#945; &#8712; &#8477; : y 2 = 1 - x , then &#8721; &#945; &#8712; S 16 &#945; 3 is equal to _______.

JEE Main 2022 Solution View → 49

50 tan 3 tan - 1 1 2 + 2 cos - 1 1 5 + 4 2 tan 1 2 tan - 1 2 2 is equal to ______.

JEE Main 2022 Solution View → 50

The value of lim n &#8594; &#8734; 6 tan &#8721; r = 1 n tan - 1 1 r 2 + 3 r + 3 is equal to

JEE Main 2022 Solution View → 51

The value of cot &#8721; n = 1 50 tan - 1 1 1 + n + n 2 is

JEE Main 2022 Solution View → 52

sin - 1 sin 2 &#960; 3 + cos - 1 cos 7 &#960; 6 + tan - 1 tan 3 &#960; 4 is equal to

JEE Main 2022 Solution View → 53

If the inverse trigonometric functions take principal values, then cos - 1 3 10 cos tan - 1 4 3 + 2 5 sin tan - 1 4 3 is equal to

JEE Main 2022 Solution View → 54

The value of tan - 1 cos 15 &#960; 4 - 1 sin &#960; 4 is equal to

JEE Main 2022 Solution View → 55

Let x &#215; y = x 2 + y 3 and x &#215; 1 &#215; 1 = x &#215; 1 &#215; 1 . Then a value of 2 sin - 1 x 4 + x 2 - 2 x 4 + x 2 + 2 is

JEE Main 2022 Solution View → 56

The set of all values of k for which tan - 1 x 3 + cot - 1 x 3 = k &#960; 3 , x &#8712; R , is the interval

JEE Main 2022 Solution View → 57

For any positive integer n , let S n : ( 0 , &#8734; ) &#8594; R be defined by S n x = &#8721; k = 1 n cot - 1 1 + k ( k + 1 ) x 2 x where for any x &#8712; R , cot - 1 ( x ) &#871

JEE Advanced 2021 Solution View → 58

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 Solution View → 59

If sin - 1 x 2 - cos - 1 x 2 = a ; 0 &#60; x &#60; 1 , a &#8800; 0 , then the value of 2 x 2 - 1 is

JEE Main 2021 Solution View → 60

If &#8721; r = 1 50 tan - 1 1 2 r 2 = p , then the value of tan p is :

JEE Main 2021 Solution View → 61

The value of tan 2 tan - 1 3 5 + sin - 1 5 13 is equal to:

JEE Main 2021 Solution View → 62

The number of real roots of the equation tan - 1 x x + 1 + sin - 1 x 2 + x + 1 = &#960; 4 is:

JEE Main 2021 Solution View → 63

The number of solutions of the equation sin - 1 x 2 + 1 3 + cos - 1 x 2 - 2 3 = x 2 for x &#8712; [ - 1 , 1 ] , and [ x ] denotes the greatest integer less than or equal to x , is

JEE Main 2021 Solution View → 64

If cot - 1 ( &#945; ) = cot - 1 2 + cot - 1 8 + cot - 1 18 + cot - 1 32 + &#8230; . upto 100 terms, then &#945; is:

JEE Main 2021 Solution View → 65

The sum of possible values of x for tan - 1 x + 1 + cot - 1 1 x - 1 = tan - 1 8 31 is:

JEE Main 2021 Solution View → 66

Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin - 1 3 x 5 + sin - 1 4 x 5 = sin - 1 x is equal to:

JEE Main 2021 Solution View → 67

Let S k = &#8721; r = 1 k tan - 1 6 r 2 2 r + 1 + 3 2 r + 1 , then lim k &#8594; &#8734; S k is equal to :

JEE Main 2021 Solution View → 68

If sin - 1 x a = cos - 1 x b = tan - 1 y c ; 0 &#60; x &#60; 1 , then the value of cos &#960; c a + b is:

JEE Main 2021 Solution View → 69

cosec 2 cot - 1 5 + cos - 1 4 5 is equal to:

JEE Main 2021 Solution View → 70

A possible value of tan 1 4 sin - 1 63 8 is:

JEE Main 2021 Solution View → 71

lim n &#8594; &#8734; tan &#8721; r = 1 n tan - 1 1 1 + r + r 2 is equal to_______.

JEE Main 2021 Solution View → 72

If S is the sum of the first 10 terms of the series, tan - 1 1 3 + tan - 1 1 7 + tan - 1 1 13 + tan - 1 1 21 + &#8230; &#8230; then tan ( S ) is equal to :

JEE Main 2020 Solution View → 73

2 &#960; - sin - 1 4 5 + sin - 1 5 13 + sin - 1 16 65 is equal to :

JEE Main 2020 Solution View → 74

If y = &#8721; k = 1 6 k cos - 1 3 5 cos k x - 4 5 sin k x then d y d x at x = 0 is

JEE Main 2020 Solution View → 75

If sin - 1 ⁡ 1 3 + sin - 1 ⁡ 2 3 = sin - 1 ⁡ x , then the value of x is

NTA Abhyas JEE Main 2020 Solution View → 76

If x takes all permissible negative values, then sin - 1 ⁡ x is equal to

NTA Abhyas JEE Main 2020 Solution View → 77

In a Δ A B C , if ∠ A = ∠ B = 1 2 sin - 1 6 + 1 2 3 + sin - 1 1 3 and length of the side opposite to ∠ C is c = 6 ⋅ 3 1 4 , then the area of Δ A B C is

NTA Abhyas JEE Main 2020 Solution View → 78

If α = sin - 1 ⁡ 3 2 + sin - 1 ⁡ 1 3 and β = cos - 1 ⁡ 3 2 + cos - 1 ⁡ 1 3 , then

NTA Abhyas JEE Main 2020 Solution View → 79

The complete solution set of the inequality cos - 1 ⁡ ( cos 4 ) > 3 x 2 - 4 x is

NTA Abhyas JEE Main 2020 Solution View → 80

The value of sin ⁡ cot - 1 ⁡ x is

NTA Abhyas JEE Main 2020 Solution View → 81

Solution set of [ sin - 1 ⁡ x ] > [ cos - 1 ⁡ x ] , where [.] denotes the greatest integer function, is

NTA Abhyas JEE Main 2020 Solution View → 82

The relation tan - 1 ⁡ 1 + x 1 - x = π 4 + tan - 1 ⁡ x holds true for all

NTA Abhyas JEE Main 2020 Solution View → 83

The domain set of the function f ( x ) = tan − 1 x − cot − 1 x + cos − 1 ( 2 − x ) is

NTA Abhyas JEE Main 2020 Solution View → 84

Let x + 1 x = 2, y + 1 y = - 2 and sin &#8722; 1 x + cos &#8722; 1 y = m &#960;, &#160; then the value of m is

NTA Abhyas JEE Main 2020 Solution View → 85

The value of the expression cot -1 1 2 + cot -1 9 2 + cot -1 2 5 2 + cot -1 4 9 2 + ....... upto n terms is

NTA Abhyas JEE Main 2020 Solution View → 86

The value of a for which a x 2 + sin - 1 &#8289; x 2 - 2 x + 2 + cos - 1 &#8289; x 2 - 2 x + 2 = 0 has a real solution, is

NTA Abhyas JEE Main 2020 Solution View → 87

The real solutions of the equation tan -1 x x + 1 + sin -1 x 2 + x + 1 = π 2 are

NTA Abhyas JEE Main 2020 Solution View → 88

If x , y , z are in arithmetic progression and tan - 1 ⁡ x , tan - 1 ⁡ y and tan - 1 ⁡ z are also in arithmetic progression, then

NTA Abhyas JEE Main 2020 Solution View → 89

Let S 1 is the complete solution set of the inequality c o s - 1 x &#62; c o s - 1 x 2 and S 2 is the complete solution set of the inequality c o t - 1 x 2 - 5 c o t - 1 x + 6 &#62

NTA Abhyas JEE Main 2020 Solution View → 90

Consider the function f x = c o s - 1 2 x + s i n - 1 2 x - 1 , then (where [.] represents the greatest integer part function)

NTA Abhyas JEE Main 2020 Solution View → 91

If cot - 1 &#8289; x 2 - 7 &#160; cot - 1 &#8289; x + 10 &#62; 0 , then the range of x will be

NTA Abhyas JEE Main 2020 Solution View → 92

The range of the function f x = s i n - 1 x 2 1 + x 2 , x ∈ R is

NTA Abhyas JEE Main 2020 Solution View → 93

The value(s) of x satisfying the equation sin - 1 ⁡ 1 - x - 2 sin - 1 ⁡ x = π 2 is/are

NTA Abhyas JEE Main 2020 Solution View → 94

If x satisfies the inequality t a n - 1 x 2 + 3 t a n - 1 x - 4 > 0 , then the complete set of values of x is

NTA Abhyas JEE Main 2020 Solution View → 95

The value of x for which sin ⁡ cot - 1 ⁡ 1 + x = cos ⁡ ( tan - 1 ⁡ x ) is

NTA Abhyas JEE Main 2020 Solution View → 96

The area bounded by the curve y = c o s - 1 sin &#8289; x + &#960; 2 - c o s - 1 cos &#8289; x and the x -axis, where &#960; 2 &#8804; x &#8804; &#960; , is equal to

NTA Abhyas JEE Main 2020 Solution View → 97

If a 1 , a 2 , a 3 are in arithmetic progression and d is the common difference, then tan - 1 ⁡ d 1 + a 1 a 2 + tan - 1 ⁡ d 1 + a 2 a 3 =

NTA Abhyas JEE Main 2020 Solution View → 98

If f x = t a n - 1 x 2 + 4 x + s i n - 1 x 2 + 4 x + 1 , then

NTA Abhyas JEE Main 2020 Solution View → 99

If the value of the expression tan ⁡ 1 2 cos - 1 ⁡ 2 5 is in the form of a + b where a , b ∈ Z , then the value of a + b b is

NTA Abhyas JEE Main 2020 Solution View → 100

Let f x = s i n - 1 x 1 - x - x 1 - x 2 , &#160; &#8704; 0 &#8804; x &#8804; 1 , then f x is

NTA Abhyas JEE Main 2020 Solution View → 101

If f x = c o s - 1 x 3 2 - 1 - x - x 2 + x 3 , ∀ 0 ≤ x ≤ 1 , then the minimum value of f x is

NTA Abhyas JEE Main 2020 Solution View → 102

If f x = t a n - 1 2 x 1 + 2 2 x + 1 , then ∑ r = 0 9 f r is

NTA Abhyas JEE Main 2020 Solution View → 103

If c o t - 1 n 2 π > 2 π 3 , then the maximum value of the integer n is

NTA Abhyas JEE Main 2020 Solution View → 104

The maximum value of x that satisfies the equation s i n - 1 2 15 x = c o s - 1 14 x is

NTA Abhyas JEE Main 2020 Solution View → 105

If x = s i n 2 t a n - 1 3 and y = sin 1 2 tan - 1 4 3 , then

NTA Abhyas JEE Main 2020 Solution View → 106

For x ∈ 0 , π 2 , if c o s - 1 7 2 1 + cos ⁡ 2 x + s i n 2 x - 48 c o s 2 x sin ⁡ x = x - cos - 1 k cos ⁡ x , then the value of k is equal to

NTA Abhyas JEE Main 2020 Solution View → 107

The value of t a n - 1 1 - sin &#8289; x + 1 + sin &#8289; x 1 - sin &#8289; x - 1 + sin &#8289; x &#160; &#8704; x &#8712; 0 , &#960; 2 is equal to

NTA Abhyas JEE Main 2020 Solution View → 108

The number of roots of the equation s i n - 1 x - c o s - 1 x = s i n - 1 5 x - 3 is/are

NTA Abhyas JEE Main 2020 Solution View → 109

If s i n - 1 5 x + s i n - 1 12 x = s i n - 1 2 x + c o s - 1 2 x , then the value of x is equal to

NTA Abhyas JEE Main 2020 Solution View → 110

The range of the function f x = s i n - 1 x 2 - 1 3 - c o s - 1 x 2 + 2 3 is (where, x represents the greatest integer value of x )

NTA Abhyas JEE Main 2020 Solution View → 111

The minimum value of x which satisfies the inequality s i n - 1 x ≥ c o s - 1 x is λ , then the value of 2 λ is (use 2 = 1.41 )

NTA Abhyas JEE Main 2020 Solution View → 112

If t a n - 1 1 2 x + 1 + t a n - 1 1 4 x + 1 = c o t - 1 x 2 2 , then the number of all possible values of x is/are

NTA Abhyas JEE Main 2020 Solution View → 113

The number of solutions of the equation s i n - 1 x = sin ⁡ x - 1 is/are

NTA Abhyas JEE Main 2020 Solution View → 114

If c o s - 1 sin &#8289; x &#8805; s i n - 1 sin &#8289; x , then the number of integral values of x in the interval x &#8712; 0,3 &#960; are

NTA Abhyas JEE Main 2020 Solution View → 115

The minimum value of x which satisfies the inequality s i n - 1 x 2 ≥ c o s - 1 x 2 is

NTA Abhyas JEE Main 2020 Solution View → 116

If 3 t a n - 1 1 2 + 3 - t a n - 1 1 3 = t a n - 1 1 x , then the value of x is equal to

NTA Abhyas JEE Main 2020 Solution View → 117

If y = t a n - 1 1 1 + x + x 2 + t a n - 1 1 x 2 + 3 x + 3 + t a n - 1 1 x 2 + 5 x + 7 + . . . . + upto 2 n terms &#8704; x &#8805; 0 , then y 0 is

NTA Abhyas JEE Main 2020 Solution View → 118

If s i n - 1 x 2 - 2 s i n - 1 x + 1 ≤ 0 (where, . represents the greatest integral part of x ), then

NTA Abhyas JEE Main 2020 Solution View → 119

If f x = t a n - 1 ln e 2 / x 3 ln e 2 x 3 + t a n - 1 ln e 2 x 3 ln e / x 6 &#160; &#8704; x &#8805; e , then the incorrect statement is

NTA Abhyas JEE Main 2020 Solution View → 120

If the range of f x = t a n - 1 x + 2 s i n - 1 x + c o s - 1 x is a , b , then

NTA Abhyas JEE Main 2020 Solution View →
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