Differential Equations – Integral Equations – JEE Main 23 Jan 2026 Shift 1

Solution:

Differentiate the given equation with respect to $x$ using Leibniz rule:
$$ \frac{d}{dx} (f(x))^2 = \frac{d}{dx} \left( 25 + \int_{0}^{x}((f(t))^{2}+(f^{\prime}(t))^{2})dt \right) $$

$$ 2f(x)f'(x) = 0 + (f(x))^2 + (f'(x))^2 $$

Rearrange the terms:
$$ (f(x))^2 + (f'(x))^2 – 2f(x)f'(x) = 0 $$
$$ (f(x) – f'(x))^2 = 0 $$
$$ f(x) = f'(x) $$

This is a standard differential equation:
$$ \frac{f'(x)}{f(x)} = 1 \implies \ln f(x) = x + C $$
$$ f(x) = A e^x $$

To find $A$, put $x=0$ in the original integral equation:
$$ (f(0))^2 = 25 + \int_{0}^{0} (…) dt $$
$$ (f(0))^2 = 25 \implies f(0) = 5 $$
(Since $f$ is non-negative).

Substitute $f(0)=5$ into $f(x) = A e^x$:
$$ 5 = A e^0 \implies A = 5 $$
So, $f(x) = 5e^x$.

We need the mean of the sequence:
$S = \{ f(\ln 1), f(\ln 2), \dots, f(\ln 625) \}$
Number of terms $n = 625$.

The $k$-th term is:
$$ f(\ln k) = 5e^{\ln k} = 5k $$

The sum of the sequence is:
$$ \Sigma = \sum_{k=1}^{625} 5k = 5 \times \frac{625 \times (625+1)}{2} $$
$$ \Sigma = 5 \times \frac{625 \times 626}{2} = 5 \times 625 \times 313 $$

The mean is $\frac{\Sigma}{n}$:
$$ \text{Mean} = \frac{5 \times 625 \times 313}{625} $$
$$ \text{Mean} = 5 \times 313 = 1565 $$

Ans. 1565