JEE Mathematics 2026 Question with Solution

Given:

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

with $f$ twice differentiable and non-negative.

Find: The mean of

$$f(\log_2 1), f(\log_2 2), \ldots, f(\log_2 625)$$

Differentiate both sides with respect to $x$ using the Leibniz rule:

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

Rearranging,

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

So,

$$f'(x) = f(x)$$

This differential equation gives

$$f(x) = Ce^x$$

Now put $x = 0$ in the original equation:

$$f(0)^2 = 25$$

Since $f$ is non-negative,

$$f(0) = 5$$

Hence,

$$C = 5$$

and therefore

$$f(x) = 5e^x$$

the solution then evaluates the required mean by taking the logarithm in the arguments as natural logarithm, so that

$$f(\ln k) = 5k$$

Thus the mean is

$$\frac{1}{625}\sum_{k=1}^{625} 5k = \frac{5}{625} \cdot \frac{625 \cdot 626}{2} = 1565$$

Therefore, the mean is $1565$.

Given: The integral equation and the non-negativity of $f$.

Find: The required mean.

The extracted working states:

1. Differentiate the given integral equation.
2. Obtain

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

3. Recognize this as a perfect square:

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

4. Hence,

$$f'(x) = f(x)$$

5. Solve the differential equation:

$$f(x) = Ce^x$$

6. Use $x=0$ in the original equation to get $f(0)^2 = 25$, hence $f(0)=5$ because $f$ is non-negative.
7. Therefore,

$$f(x) = 5e^x$$

8. The solution explicitly notes a base issue: it first writes $f(\log_2 k)=5e^{\log_2 k}$, then corrects itself and proceeds with the assumption that the logarithm is natural logarithm for the final evaluation.
9. Using that evaluation,

$$\text{Mean} = \frac{1}{625}\sum_{k=1}^{625} 5k = 1565$$

So the final answer concluded on the solution is $1565$.