Let f be a twice differentiable non-negative function such that (f(x))² = 25 + ∫₀ˣ ((f(t))² + (f'(t)

Q: Wait, why can we just differentiate the integral? Is it always that simple?

Great question! The Leibniz Rule for differentiation under the integral sign is a powerful tool in your JEE arsenal. Whenever you see an integral with a variable limit, differentiating is the standard 'key' to unlock the functional form hidden inside. It transforms a global property (the integral) into a local property (the derivative).

Q: Why did we reject the negative root when solving $(f(0))^2 = 25$?

The problem explicitly states that $f$ is a non-negative function. In mathematics, we must always respect the constraints given in the problem statement. Since $f(x) \ge 0$, we must choose $f(0) = 5$ instead of $-5$.

Q: How did you know $f(\ln k) = 5k$ so quickly?

Once we found $f(x) = 5e^x$, we simply substituted $x = \ln k$. Using the fundamental property of logarithms, $e^{\ln k} = k$. Thus, $f(\ln k) = 5e^{\ln k} = 5k$. It is just a beautiful simplification that makes the summation much easier!

Q: Is there a shortcut for the sum of the sequence?

Yes! The sequence is $5, 10, 15, \dots, 5(625)$. This is an Arithmetic Progression. The mean of an AP is simply the average of the first and last terms: $\frac{5(1) + 5(625)}{2} = \frac{5(626)}{2} = 5(313) = 1565$. Both methods lead to the same elegant result.

Q: Why did the integral term $(f'(t))^2$ appear in the derivative?

When we differentiate $\int_0^x ((f(t))^2 + (f'(t))^2)dt$ with respect to $x$, the Fundamental Theorem of Calculus tells us that the derivative is simply the integrand evaluated at the upper limit $x$. That is why $(f(x))^2 + (f'(x))^2$ pops out!

Question

Let $f$ be a twice differentiable non-negative function such that $(f(x))^2 = 25 + \int_0^x ((f(t))^2 + (f'(t))^2)dt$. Then the mean of $f(\log_e(1)), f(\log_e(2)), \dots, f(\log_e(625))$ is equal to ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex, intimidating integral equation: $(f(x))^2 = 25 + \int_0^x ((f(t))^2 + (f'(t))^2)dt$. It looks like a fortress, but in the world of JEE Advanced, we dismantle these structures by peeling back the layers to reveal the elegant simplicity hidden underneath.

### The Power of Differentiation

Our first goal is to liberate $f(x)$ from the integral. Whenever you see an integral with a variable limit $x$, your intuition should immediately suggest differentiation. By applying the Leibniz Rule, we determine how the total area under the curve changes as we push the boundary $x$ further.

When we differentiate both sides with respect to $x$, the constant $25$ vanishes. The integral, via the Fundamental Theorem of Calculus, yields the integrand evaluated at $x$:

$$\frac{d}{dx}(f(x))^2 = (f(x))^2 + (f'(x))^2$$

Applying the chain rule to the left side, we obtain $2f(x)f'(x) = (f(x))^2 + (f'(x))^2$. This is not just a random collection of terms; it is a perfect square waiting to be recognized.

### The Perfect Square Revelation

If we rearrange the terms, we get $(f(x))^2 - 2f(x)f'(x) + (f'(x))^2 = 0$. This is the expansion of the following identity:

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

For the square of a real expression to be zero, the expression itself must be zero. Therefore, we have discovered the core identity: $f'(x) = f(x)$.

This is a fundamental differential equation indicating that the function is its own rate of change. Solving $\frac{df}{dx} = f$ leads to $\int \frac{df}{f} = \int dx$, which results in $\ln|f(x)| = x + C$, or simply $f(x) = Ae^x$.

### Finding the Identity

We return to our original equation to find the constant $A$. By setting $x = 0$, the integral term becomes zero because the limits of integration are identical. We are left with $(f(0))^2 = 25$.

Since the problem defines $f$ as non-negative, we conclude $f(0) = 5$. Substituting this into $f(x) = Ae^x$, we find $A = 5$. Our function is revealed as:

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

### The Final Calculation

The problem asks for the mean of $f(\ln 1), f(\ln 2), \dots, f(\ln 625)$. Using our function, $f(\ln k) = 5e^{\ln k} = 5k$. The sequence is simply $5(1), 5(2), 5(3), \dots, 5(625)$.

We are calculating the mean of an arithmetic progression. The sum is $\sum_{k=1}^{625} 5k = 5 \cdot \frac{625(626)}{2}$. Dividing by the number of terms, $625$, we get:

$$	ext{Mean} = \frac{5 \cdot 625 \cdot 626}{2 \cdot 625} = \frac{5 \cdot 626}{2} = 5 \cdot 313 = 1565$$

Through the power of calculus and logical deduction, we have navigated the complexity and arrived at the elegant result of **1565**. Remember, every complex problem is just a series of simple, logical steps waiting for you to connect them.

Enter Numerical Value:

Visualized Solution

Analyze the Integral Equation

Differentiate using Leibniz Rule

Simplify to a Perfect Square

Solve the Differential Equation

Find the Initial Condition $f (0)$

Evaluate the Sequence Terms

Set Up the Arithmetic Mean

Calculate the Sum of Natural Numbers

Final Computation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Power of Differentiation

The Perfect Square Revelation

Finding the Identity

The Final Calculation

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