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JEE Main202623 January 2026Morning ShiftMathematicsDifferential EquationsActual

Let f be a twice differentiable non-negative function such that (f(x))²=25+ ₀^ x ((f( t ))²+ (f^ ( t ) )² ) dt . Then the mean of f ( _ e (1) ), f ( _ e (2) ), .., f ( _ e (625) ) is equal to _ _ _ _ .

Correct answer

0

Step-by-step solution

Differentiating (f(x))^2 = 25 + ₀^x((f(t))^2 + (f'(t))^2) ,dt : 2ff' = f^2 + (f')^2 (f - f')^2 = 0 f' = f . f(x) = Ce^x . At x = 0 : f(0)^2 = 25 f(0) = 5 (non-negative), so f(x) = 5e^x . f( _e k) = 5k for k = 1, 2, , 625 . Mean = 5(1+2+ +625) 625 = 5 625 626 2 625 = 5 626 2 = 1565 .

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