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

Given ^ -1 (1- ) + ^ -1 (1- ) = 4 Applying the formula ^ -1 x + ^ -1 y = ^ -1 ( x+y 1-xy ), we get: (1- ) + (1- ) 1 - (1- )(1- ) = ( 4 ) = 1 2 - - = 1 - (1 - - + ) 2 - - = + - 2( + ) = 2 + Given = 1 3 , we have = 1 3 . Substituting this into the equation: 2( + ) = 2 + 1 3 2( + ) = 7 3 + = 7 6 Therefore, 6( + ) = 6 7 6 = 7 Answer: 7

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

The general term of the given series is T_p = ^ -1 ( 2^ p-1 1 + 2^ 2p-1 ). This can be rewritten by expressing the numerator and denominator in terms of 2^p and 2^ p-1 : T_p = ^ -1 ( 2^p - 2^ p-1 1 + 2^p 2^ p-1 ) Using the inverse trigonometric identity ^ -1 ( x - y 1 + xy ) = ^ -1 x - ^ -1 y, we get: T_p = ^ -1 (2^p) - ^ -1 (2^ p-1 ) Summing T_p from p = 1 to 11 gives a telescoping series: _ p=1 ^ 11 T_p = ( ^ -1 (2^1) - ^ -1 (2^0) ) + ( ^ -1 (2^2) - ^ -1 (2^1) ) + + ( ^ -1 (2^ 11 ) - ^ -1 (2^

Q: 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:

Let y = y_1 + y_2 y_1 = ^ -1 ( 3 x - 4 x 4 x + 3 x ) Dividing the numerator and denominator by 4 x: y_1 = ^ -1 ( 3 4 - x 1 + 3 4 x ) Using the formula ^ -1 ( A - B 1 + AB ) = ^ -1 A - ^ -1 B: y_1 = ^ -1 ( 3 4 ) - ^ -1 ( x) = ^ -1 ( 3 4 ) - x Differentiating with respect to x: dy_1 dx = -1 Now, y_2 = 2 ^ -1 ( x 1+ 1-x^2 ) Substituting x = : y_2 = 2 ^ -1 ( 1+ ) y_2 = 2 ^ -1 ( 2 ( /2) ( /2) 2 ^2( /2) ) y_2 = 2 ^ -1 ( 2 ) = 2 ( 2 ) = = ^ -1 x Differentiating with respect to x: dy_2 dx = 1 1-x^2 At x

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

Let = ^ -1 ( 2 13 ). Then = 2 13 , = 3 13 , = 2 3 . So (2 ) = 2 2 3 1 - 4 9 = 12 5 . Let = ^ -1 ( 3 10 ). Then = 3 10 , = 1 10 , = 1 3 . So (2 ) = 2 1 3 1 - 1 9 = 3 4 . Therefore (2 - 2 ) = 12 5 - 3 4 1 + 12 5 3 4 = 33 20 56 20 = 33 56

Q: 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 \_\_\_\_

Let = 1 2 ^ -1 2 3 , then 1 2 ^ -1 1 3 = ( 4 - ) k = + = 1 = 2 2 k = 2 2 3 = 3 ^ -1 (3x - 1) = ^ -1 x - ^ -1 x ^ -1 (3x - 1) = 2 - 2 ^ -1 x 3x - 1 = ( 2 - 2 ^ -1 x ) 3x - 1 = 2x^2 - 1 x = 0, 3 2 (rejected) No. of solution = 1

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

For ^ -1 ( 1 x^2-2x-2 ) to be defined, we need | 1 x^2-2x-2 | 1, i.e., |x^2-2x-2| 1. Case 1: x^2-2x-2 1 x^2-2x-3 0 (x-3)(x+1) 0 x -1 or x 3. Case 2: x^2-2x-2 -1 x^2-2x-1 0 (x-1)^2 2 1- 2 x 1+ 2 . Domain = (- , -1] [1- 2 , 1+ 2 ] [3, ). So = -1,\ = 1- 2 ,\ = 1+ 2 ,\ = 3. + + + = -1 + 1 - 2 + 1 + 2 + 3 = 4.

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

For x (- 1 2 6 , 1 2 6 ), 24x^2 So ^ -1 (4x) + ^ -1 (6x) = ^ -1 ( 10x 1-24x^2 ) = 6 . 10x 1-24x^2 = 1 3 24x^2 + 10 3 x - 1 = 0. x = -10 3 300+96 48 = -5 3 3 11 24 . x_1 = -5 3 +3 11 24 0.054 (valid, lies in range). x_2 = -5 3 -3 11 24 -0.776 (rejected, outside range). Number of solutions = 1.

Question 1

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: ( ) In the light

Accepted Answer

We are given = 3 ^ -1 ( 6 11 ) and = 3 ^ -1 ( 4 9 ). First, we find the range of . Since 1 2 6 Multiplying by 3, we get 2 Since lies in the second quadrant, ( ) Next, we find the range of . Since 0 3 Multiplying by 3, we get Thus, lies in the third quadrant. Now, we determine the sign of ( + ). Adding the inequalities for and : 2 + 3 2 The angle ( + ) lies in the interval ( 3 2 , 9 4 ), which covers the fourth quadrant and a part of the first quadrant. In both of these regions, the cosine functi

Question 2

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

Accepted Answer

Let = ^ -1 (x 2 ) = x 2 From the right-angled triangle, = x 2 1+2x^2 Let = ^ -1 1-x^2 = 1-x^2 From the right-angled triangle, = 1 - (1-x^2) = x = = x 1-x^2 Given = , substituting the values: x 2 1+2x^2 = x 1-x^2 Since x (0,1), x 0. Dividing by x and squaring both sides: 2 1+2x^2 = 1 1-x^2 2(1-x^2) = 1+2x^2 2 - 2x^2 = 1 + 2x^2 4x^2 = 1 x^2 = 1 4 Since x (0,1), x = 1 2 Answer: 1 2

Question 3

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:

Accepted Answer

For the function f(x) = ^ -1 ( x+[x] 3 ) to be defined, the argument of the inverse sine function must lie in the interval [-1, 1]. -1 x+[x] 3 1 -3 x + [x] 3 Using the fractional part function \ x\ , we can write x = [x] + \ x\ , where 0 \ x\ -3 2[x] + \ x\ 3 We analyze the possible integer values for [x]: If [x] -2, then 2[x] + \ x\ -4 + \ x\ If [x] = -1, then 2[x] + \ x\ = -2 + \ x\ . Since 0 \ x\ If [x] = 0, then 2[x] + \ x\ = \ x\ . Since 0 \ x\ If [x] = 1, then 2[x] + \ x\ = 2 + \ x\ . Sinc

Question 4

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

Accepted Answer

Given ^ -1 (1- ) + ^ -1 (1- ) = 4 Applying the formula ^ -1 x + ^ -1 y = ^ -1 ( x+y 1-xy ), we get: (1- ) + (1- ) 1 - (1- )(1- ) = ( 4 ) = 1 2 - - = 1 - (1 - - + ) 2 - - = + - 2( + ) = 2 + Given = 1 3 , we have = 1 3 . Substituting this into the equation: 2( + ) = 2 + 1 3 2( + ) = 7 3 + = 7 6 Therefore, 6( + ) = 6 7 6 = 7 Answer: 7

Question 5

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

Accepted Answer

The general term of the given series is T_p = ^ -1 ( 2^ p-1 1 + 2^ 2p-1 ). This can be rewritten by expressing the numerator and denominator in terms of 2^p and 2^ p-1 : T_p = ^ -1 ( 2^p - 2^ p-1 1 + 2^p 2^ p-1 ) Using the inverse trigonometric identity ^ -1 ( x - y 1 + xy ) = ^ -1 x - ^ -1 y, we get: T_p = ^ -1 (2^p) - ^ -1 (2^ p-1 ) Summing T_p from p = 1 to 11 gives a telescoping series: _ p=1 ^ 11 T_p = ( ^ -1 (2^1) - ^ -1 (2^0) ) + ( ^ -1 (2^2) - ^ -1 (2^1) ) + + ( ^ -1 (2^ 11 ) - ^ -1 (2^

Question 6

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:

Accepted Answer

Let y = y_1 + y_2 y_1 = ^ -1 ( 3 x - 4 x 4 x + 3 x ) Dividing the numerator and denominator by 4 x: y_1 = ^ -1 ( 3 4 - x 1 + 3 4 x ) Using the formula ^ -1 ( A - B 1 + AB ) = ^ -1 A - ^ -1 B: y_1 = ^ -1 ( 3 4 ) - ^ -1 ( x) = ^ -1 ( 3 4 ) - x Differentiating with respect to x: dy_1 dx = -1 Now, y_2 = 2 ^ -1 ( x 1+ 1-x^2 ) Substituting x = : y_2 = 2 ^ -1 ( 1+ ) y_2 = 2 ^ -1 ( 2 ( /2) ( /2) 2 ^2( /2) ) y_2 = 2 ^ -1 ( 2 ) = 2 ( 2 ) = = ^ -1 x Differentiating with respect to x: dy_2 dx = 1 1-x^2 At x

Question 7

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

Accepted Answer

Let = ^ -1 ( 2 13 ). Then = 2 13 , = 3 13 , = 2 3 . So (2 ) = 2 2 3 1 - 4 9 = 12 5 . Let = ^ -1 ( 3 10 ). Then = 3 10 , = 1 10 , = 1 3 . So (2 ) = 2 1 3 1 - 1 9 = 3 4 . Therefore (2 - 2 ) = 12 5 - 3 4 1 + 12 5 3 4 = 33 20 56 20 = 33 56

Question 8

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 \_\_\_\_

Accepted Answer

Let = 1 2 ^ -1 2 3 , then 1 2 ^ -1 1 3 = ( 4 - ) k = + = 1 = 2 2 k = 2 2 3 = 3 ^ -1 (3x - 1) = ^ -1 x - ^ -1 x ^ -1 (3x - 1) = 2 - 2 ^ -1 x 3x - 1 = ( 2 - 2 ^ -1 x ) 3x - 1 = 2x^2 - 1 x = 0, 3 2 (rejected) No. of solution = 1

Question 9

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

Accepted Answer

For ^ -1 ( 1 x^2-2x-2 ) to be defined, we need | 1 x^2-2x-2 | 1, i.e., |x^2-2x-2| 1. Case 1: x^2-2x-2 1 x^2-2x-3 0 (x-3)(x+1) 0 x -1 or x 3. Case 2: x^2-2x-2 -1 x^2-2x-1 0 (x-1)^2 2 1- 2 x 1+ 2 . Domain = (- , -1] [1- 2 , 1+ 2 ] [3, ). So = -1,\ = 1- 2 ,\ = 1+ 2 ,\ = 3. + + + = -1 + 1 - 2 + 1 + 2 + 3 = 4.

Question 10

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

Accepted Answer

For x (- 1 2 6 , 1 2 6 ), 24x^2 So ^ -1 (4x) + ^ -1 (6x) = ^ -1 ( 10x 1-24x^2 ) = 6 . 10x 1-24x^2 = 1 3 24x^2 + 10 3 x - 1 = 0. x = -10 3 300+96 48 = -5 3 3 11 24 . x_1 = -5 3 +3 11 24 0.054 (valid, lies in range). x_2 = -5 3 -3 11 24 -0.776 (rejected, outside range). Number of solutions = 1.

Question 11

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 \_\_\_\_.

Accepted Answer

Let = ^ -1 x and = ^ -1 x. Using + = 2 : f(x) = ^2 + ^2 = ^2 + ( 2 - )^2 = 2 ^2 - + ^2 4 This equals 2( - 4 )^2 + ^2 8 , which is minimized at = 4 . For x [- 3 2 , 1 2 ], we have [- 3 , 4 ]. Distance from - 3 to vertex: 4 + 3 = 7 12 . Distance from 4 to vertex: 0. Maximum occurs at x = - 3 2 where = - 3 : f(- 3 2 ) = 2( ^2 9 ) + ^2 3 + ^2 4 = ^2( 8 + 12 + 9 36 ) = 29 36 ^2 Since (29, 36) = 1, we have m = 29, n = 36, so m + n = 65

Question 12

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

Accepted Answer

For ^ -1 ( 2x-5 11-3x ): -1 2x-5 11-3x 1 gives x 16 5 or x > 6. For ^ -1 (2x^2 - 3x + 1): -1 2x^2 - 3x + 1 1. Lower bound always satisfied (discriminant Upper bound gives x(2x-3) 0, so 0 x 3 2 . Intersection of both conditions: [0, 3 2 ]. Thus = 0, = 3 2 . + 2 = 0 + 3 = 3.

Inverse Trigonometric Functions — Previous Year Questions & Solutions