Question
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 \_\_\_\_.
JEE Main
Mathematics
Differential Equations
2026
JEE Main 2026 (23 January Shift 1)
JEE Main Mathematics Question (2026) — Solution
Options
- A. A
- B. B
- C. C
- D. D
Answer
A. A
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.
Related: Mathematics — Differential Equations · All PYQ Banks