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 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 .

JEE Main 202623 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 23 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Differential Equations

Let f be a twice differentiable non-negative function such that...

Let

f

be a twice differentiable non-negative function such that

(f(x))^{2}=25+\int_{0}^{x}\left((f(\mathrm{t}))^{2}+\left(f^{\prime}(\mathrm{t})\right)^{2}\right) \mathrm{dt}

. Then the mean of

f\left(\log _{\mathrm{e}}(1)\right), f\left(\log _{\mathrm{e}}(2)\right), \ldots.., f\left(\log _{\mathrm{e}}(625)\right)

is equal to

\_\_\_\_

Correct answer

Step-by-step solution

Differentiating

(f(x))^2 = 25 + \int_0^x((f(t))^2 + (f'(t))^2)\,dt

2ff' = f^2 + (f')^2 \Rightarrow (f - f')^2 = 0 \Rightarrow f' = f

f(x) = Ce^x

. At

x = 0

f(0)^2 = 25 \Rightarrow f(0) = 5

(non-negative), so

f(x) = 5e^x

f(\log_e k) = 5k

for

k = 1, 2, \ldots, 625

.
Mean

= \frac{5(1+2+\cdots+625)}{625} = \frac{5 \times 625 \times 626}{2 \times 625} = \frac{5 \times 626}{2} = 1565

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