JEE Main 2026 — Jan 23, Shift 1 — Mathematics solution

JEE Main2026Jan 23, Shift 1Mathematics

Let f be a twice differentiable non-negative function such that $(f (x))^{2} = 25 + \int_{0}^{x} ((f (t))^{2} + (f^{'} (t))^{2}) d t$ . Then the mean of $f (lo g_{e} 1), f (lo g_{e} 2), \dots, f (lo g_{e} 625)$ is equal to ______.

Your answer

Solution

Differentiate both sides:

$2 f (x) f^{'} (x) = f (x)^{2} + f^{'} (x)^{2} \Rightarrow (f (x) - f^{'} (x))^{2} = 0 \Rightarrow f^{'} (x) = f (x)$

So $f (x) = c e^{x}$ . From the equation at $x = 0$ : $f (0)^{2} = 25$ , and since $f \geq 0$ , $f (0) = 5$ , giving $c = 5$ and $f (x) = 5 e^{x}$ .

Compute the mean. $f (ln k) = 5 k$ , so:

Mean $= \frac{5 ( 1 + 2 + \dots + 625 )}{625} = \frac{5 \cdot \frac{625 \cdot 626}{2}}{625} = \frac{5 \cdot 626}{2} = 1565$ .

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