Q: 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 ), respect

Let = 1 2 ^ -1 ( 6 9+ ^2 ) aligned & = ^ -1 (2 )- \\ & + = ^ -1 (2 ) \\ & ( + )=2 \\ & + 1- =2 ....(1)\\ & 2 = 6 9+ ^2 = 2 1+ ^2 \\ & 3 +3 ^2 =9 + ^2 \\ & 3( -3 )= ( -3 ) \\ & = 3 or =3 aligned aligned & Case-I : =3 \\ & + 3 1- ^2 3 =2 aligned =0, 2 3 =1- ^2 3 =1,-1 =0,-1,1 aligned & Case-II : = 3 \\ & + 3 -2 =2 aligned

Q: Considering only the principal values of the inverse trigonometric functions, the value of 3 2 cos - 1 2 2 + π 2 + 1 4 sin - 1 2 2 π 2 + π 2 + tan - 1 2 π is _______. (

3 2 cos - 1 2 2 + π 2 + 1 4 sin - 1 2 2 π 2 + π 2 + tan - 1 2 π = 3 2 tan - 1 π 2 + 1 4 tan - 1 2 2 π π 2 - 2 + tan - 1 2 π = π 2 + 1 2 tan - 1 π 2 - 1 4 tan - 1 2 2 π 2 - π 2 = π 2 + 1 2 tan - 1 π 2 - 1 4 tan - 1 2 · π 2 1 - π 2 2 = π 2 + 1 2 tan - 1 π 2 - 1 4 - π + 2 tan - 1 π 2 = π 2 + π 4 = 3 π 4 = 2 . 36

Q: For any positive integer n , let S n : ( 0 , ∞ ) → R be defined by S n x = ∑ k = 1 n cot - 1 1 + k ( k + 1 ) x 2 x where for any x ∈ R , cot - 1 ( x ) ∈ ( 0 , π ) an

S n x = ∑ k = 1 n cot - 1 1 + k ( k + 1 ) x 2 x = ∑ k = 1 n cot - 1 1 + k x ( k + 1 ) x ( k + 1 ) x - k x = ∑ k = 1 n tan - 1 ( k + 1 ) x - k x 1 + k x ( k + 1 ) x = ∑ k = 1 n tan - 1 ( k + 1 ) x - tan - 1 k x = tan - 1 ( n + 1 ) x - tan - 1 n x + … + tan - 1 3 x - tan - 1 2 x + tan - 1 2 x - tan - 1 x = tan - 1 ( n + 1 ) x - tan - 1 x = tan - 1 ( n + 1 ) x - x 1 + ( n + 1 ) x 2 = tan - 1 n x 1 + ( n + 1 ) x 2 1. S 10 x = tan - 1 10 x 1 + 11 x 2 = π 2 - c

Question 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

Accepted Answer

For the first term, ^ -1 ( (-11)): The principal value branch of ^ -1 (x) is (0, ). Since the cotangent function is periodic with period , (-11) = (-11 + 4 ). Using 3.14, we have 4 12.56, so -11 + 4 1.56, which lies in the interval (0, ). Thus, ^ -1 ( (-11)) = 4 - 11. For the second term, 10 (2 ^ -1 ( 1 2 ) ): Since ^ -1 ( 1 2 ) = 4 , we get: 10 (2 4 ) = 10 ( 2 ) = 10(1) = 10. For the third term, 10 (2 ^ -1 (2)): Let = ^ -1 (2) ( ) = 2. Using the multiple angle formula (2 ) = 2 ( ) 1 + ^2( ) , w

Question 2

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 ), respect

Accepted Answer

Let = 1 2 ^ -1 ( 6 9+ ^2 ) aligned & = ^ -1 (2 )- \ & + = ^ -1 (2 ) \ & ( + )=2 \ & + 1- =2 ....(1)\ & 2 = 6 9+ ^2 = 2 1+ ^2 \ & 3 +3 ^2 =9 + ^2 \ & 3( -3 )= ( -3 ) \ & = 3 or =3 aligned aligned & Case-I : =3 \ & + 3 1- ^2 3 =2 aligned =0, 2 3 =1- ^2 3 =1,-1 =0,-1,1 aligned & Case-II : = 3 \ & + 3 -2 =2 aligned

Question 3

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

Accepted Answer

Let E = ( ^ -1 ( 3 5 ) - 2 ^ -1 ( 2 5 ) ). Evaluate the inverse trigonometric expressions: Let = ^ -1 ( 3 5 ), so = 3 5 and = 3 4 . Let = ^ -1 ( 2 5 ), so = 2 5 and = 1 2 . Compute the double angle: Using (2 ) = 2 1 - ^2 , we obtain (2 ) = 2 1 2 1 - ( 1 2 )^2 = 4 3 . Apply the tangent subtraction formula: E = ( - 2 ) = - (2 ) 1 + (2 ) = 3 4 - 4 3 1 + 3 4 4 3 = - 7 12 2 = - 7 24 . Final result: - 7 24

Question 4

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 _______. (

Accepted Answer

3 2 cos - 1 2 2 + &#960; 2 + 1 4 sin - 1 2 2 &#960; 2 + &#960; 2 + tan - 1 2 &#960; = 3 2 tan - 1 &#960; 2 + 1 4 tan - 1 2 2 &#960; &#960; 2 - 2 + tan - 1 2 &#960; = &#960; 2 + 1 2 tan - 1 &#960; 2 - 1 4 tan - 1 2 2 &#960; 2 - &#960; 2 = &#960; 2 + 1 2 tan - 1 &#960; 2 - 1 4 tan - 1 2 &#183; &#960; 2 1 - &#960; 2 2 = &#960; 2 + 1 2 tan - 1 &#960; 2 - 1 4 - &#960; + 2 tan - 1 &#960; 2 = &#960; 2 + &#960; 4 = 3 &#960; 4 = 2 . 36

Question 5

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 ) &#8712; ( 0 , &#960; ) an

Accepted Answer

S n x = &#8721; k = 1 n cot - 1 1 + k ( k + 1 ) x 2 x = &#8721; k = 1 n cot - 1 1 + k x ( k + 1 ) x ( k + 1 ) x - k x = &#8721; k = 1 n tan - 1 ( k + 1 ) x - k x 1 + k x ( k + 1 ) x = &#8721; k = 1 n tan - 1 ( k + 1 ) x - tan - 1 k x = tan - 1 ( n + 1 ) x - tan - 1 n x + &#8230; + tan - 1 3 x - tan - 1 2 x + tan - 1 2 x - tan - 1 x = tan - 1 ( n + 1 ) x - tan - 1 x = tan - 1 ( n + 1 ) x - x 1 + ( n + 1 ) x 2 = tan - 1 n x 1 + ( n + 1 ) x 2 1.  S 10 x = tan - 1 10 x 1 + 11 x 2 = &#960; 2 - c

Inverse Trigonometric Functions — Previous Year Questions & Solutions