JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let y = x 2 . Since 0 x , we have 0 y 2 . The given expression becomes f(y) = 16 y ^3 y. Differentiating with respect to y: f'(y) = 16( y ^3 y + y 3 ^2 y(- y)) f'(y) = 16 ^2 y( ^2 y - 3 ^2 y) For maximum value, f'(y) = 0: ^2 y - 3 ^2 y = 0 ^2 y = 1 3 Since y [0, 2 ], y = 1 3 y = 6 . Substituting y = 6 in f(y): f ( 6 ) = 16 ( 6 ) ^3 ( 6 ) f ( 6 ) = 16 1 2 ( 3 2 )^3 f ( 6 ) = 8 3 3 8 = 3 3 The maximum value is 3 3 . Answer: 3 3