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JEE Main202621 January 2026Evening ShiftMathematicsInverse Trigonometric FunctionsActual

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 _ _ _ _ .

Correct answer

0

Step-by-step solution

Let = ⁻¹x and = ⁻¹x . 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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