Let f be a differentiable function satisfying f(x) = (2/√3) ∫_{0}^{√3} f(λ²x/3) dλ, x 0 and f(1) =

Q: Wait, why do we need to differentiate? Can't we just solve the integral directly?

That is a brilliant question! The problem is that $f(x)$ is inside the integral, and we do not know its form yet. We cannot integrate a function we have not defined. By differentiating, we transform the 'integral equation' into a 'differential equation,' which is much easier to solve.

Q: How did you know to substitute $t = \frac{\lambda^2 x}{3}$?

Look at the argument of $f$ inside the integral: $\frac{\lambda^2 x}{3}$. When we want to use the Leibniz rule, we need the upper limit of the integral to match the variable $x$ in the function. By setting $t$ equal to the entire argument, we force the upper limit to become $x$, making the differentiation step straightforward.

Q: Why does the constant $C$ become a multiplier $k$?

When we integrate both sides, we get $\ln f(x) = \frac{1}{2} \ln x + C$. Exponentiating both sides gives $f(x) = e^{\frac{1}{2} \ln x + C} = e^C \cdot e^{\ln \sqrt{x}}$. Since $e^C$ is just a constant, we rename it $k$ to keep our algebra clean. It is a standard practice to simplify constants early!

Q: Is it always valid to use the Leibniz Rule here?

Yes, because the problem states that $f$ is a differentiable function. The Leibniz Rule requires the integrand to be continuous and the limits to be differentiable, which are satisfied here. Always check the differentiability condition before applying the rule!

Q: What if I forgot the $d\lambda$ term during substitution?

That is a common trap! When you change variables, you must also change the differential $d\lambda$ to $dt$. If you miss the Jacobian factor (the derivative of the substitution), your equation will not balance, and you will end up with an incorrect differential equation. Always be meticulous with your differentials.

Question

Let $f$ be a differentiable function satisfying $f(x) = \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} f \left( \frac{\lambda^2 x}{3} \right) d\lambda, x > 0$ and $f(1) = \sqrt{3}$. If $y = f(x)$ passes through the point $(\alpha, 6)$, then $\alpha$ is equal to \_\_\_\_.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing before a complex integral equation. It looks like a labyrinth, with the function $f(x)$ trapped both outside and inside the integral sign:

$$f(x) = \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} f \left( \frac{\lambda^2 x}{3} \right) d\lambda$$

At first glance, it feels impossible. How can we find a function that is defined by its own average behavior? This is the beauty of JEE Advanced mathematics—it is not about brute force; it is about finding the hidden symmetry that allows us to simplify the complex into the elegant.

### The Art of Substitution

To solve this, we must first liberate the variable $x$ from the argument of $f$. We perform a change of variables. Let $t = \frac{\lambda^2 x}{3}$.

This is our key. By rearranging for $\lambda$, we find $\lambda = \sqrt{\frac{3t}{x}}$, and consequently, the differential $d\lambda = \frac{\sqrt{3}}{2\sqrt{tx}} dt$.

As we transform the limits, we see the magic happen: when $\lambda = 0$, $t = 0$, and when $\lambda = \sqrt{3}$, $t = x$. Our integral now looks like this:

$$f(x) = \frac{2}{\sqrt{3}} \int_{0}^{x} f(t) \left( \frac{\sqrt{3}}{2\sqrt{tx}} \right) dt$$

Notice how the constants cancel out beautifully, leaving us with the simplified form:

$$\sqrt{x} f(x) = \int_{0}^{x} \frac{f(t)}{\sqrt{t}} dt$$

### The Power of Leibniz

Now, we stand at the threshold of the most powerful tool in our arsenal: the Leibniz Integral Rule. We have an equation where the right side is an integral with a variable limit $x$.

If we differentiate both sides with respect to $x$, the integral sign will vanish. Applying the product rule to the left side, we get:

$$\frac{d}{dx}(\sqrt{x} f(x)) = \frac{1}{2\sqrt{x}} f(x) + \sqrt{x} f'(x)$$

On the right side, the fundamental theorem of calculus tells us that the derivative of the integral is simply the integrand evaluated at the upper limit:

$$\frac{d}{dx} \int_{0}^{x} \frac{f(t)}{\sqrt{t}} dt = \frac{f(x)}{\sqrt{x}}$$

Equating these two, we arrive at a simple, clean differential equation:

$$\sqrt{x} f'(x) = \frac{f(x)}{2\sqrt{x}}$$

### The Final Descent

We are almost there. By separating the variables, we get $\frac{f'(x)}{f(x)} = \frac{1}{2x}$. Integrating both sides yields $\ln f(x) = \frac{1}{2} \ln x + C$.

Using the properties of logarithms, this simplifies to $f(x) = k\sqrt{x}$. We are given the boundary condition $f(1) = \sqrt{3}$, which immediately reveals that $k = \sqrt{3}$.

Thus, our function is $f(x) = \sqrt{3x}$.

To find $\alpha$ such that $f(\alpha) = 6$, we simply solve $\sqrt{3\alpha} = 6$. Squaring both sides gives $3\alpha = 36$, leading us to the final, satisfying result:

**$\alpha = 12$**

You have navigated the labyrinth, transformed the integral, and emerged with the answer. This is the essence of mathematics—taking a daunting problem and, through logical steps, revealing the simple truth hidden beneath.

Animated Solution for Mathematics - Differential Equations: Let $f$ be a differentiable function satisfying $f (x) = \frac{2}{3} \int_{0}^{3} f (\frac{λ ^{2} x}{3}) d λ, x > 0$ and $f (1) = 3$ . If $y = f (x)$ passes through the point $(α, 6)$ , then $α$ is equal to \_\_\_\_.

Enter Numerical Value:

Visualized Solution

Understanding the Integral Equation

Change of Variables: Substitution

Transforming the Limits

Simplifying the Equation

Applying Newton-Leibniz Rule

Forming the Differential Equation

Solving by Variable Separation

Finding the General Solution

Applying the Boundary Condition

Solving for $α$

Conclusion and Key Takeaways

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Substitution

The Power of Leibniz

The Final Descent

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