Question
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
Mathematics
Inverse Trigonometric Functions
2026
JEE Main 2026 (21 January Shift 2)
JEE Main Mathematics Question (2026) — Solution
Options
- A. A
- B. B
- C. C
- D. D
Answer
A. A
Step-by-step solution
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
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