Q46 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
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: 1565
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 $= \frac{5(1+2+\cdots+625)}{625} = \frac{5 \times 625 \times 626}{2 \times 625} = \frac{5 \times 626}{2} = 1565$.
$2ff' = f^2 + (f')^2 \Rightarrow (f - f')^2 = 0 \Rightarrow f' = f$.
$f(x) = Ce^x$. 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 $= \frac{5(1+2+\cdots+625)}{625} = \frac{5 \times 625 \times 626}{2 \times 625} = \frac{5 \times 626}{2} = 1565$.
Hence, the correct answer is 1565.