Let a differentiable function f satisfy the equation ∫(0 to 36) f(tx/36) dt = 4αf(x). If y = f(x) is

Q: Wait, why did we use the Leibniz Rule? Can't we just integrate $f(x)$ directly?

That is a very common point of confusion! We cannot integrate $f(x)$ directly because we do not know the explicit form of the function yet—that is exactly what we are trying to find. The Leibniz Rule is our 'surgical tool' here. It allows us to differentiate an integral with respect to its limit, effectively 'peeling away' the integral sign to reveal the differential equation hidden underneath. It is the bridge between the integral world and the differential world.

Q: Why is a 'standard parabola' defined as $y = Kx^2$? Could it not be $y = ax^2 + bx + c$?

In the context of JEE Advanced problems involving functional equations, when we derive a power function of the form $f(x) = K x^n$ and the problem specifies a 'standard parabola,' it implies the simplest geometric form passing through the origin. If the function were a general quadratic, the differential equation would have been much more complex. Here, the math naturally leads us to a monomial form, and the 'standard' descriptor confirms that the exponent must be 2.

Q: Is it possible to have a negative value for $\alpha$?

Mathematically, the differential equation might allow for various values, but we must respect the physical and geometric constraints provided. The value of $\alpha$ is locked by the requirement that the function must be a parabola (exponent 2). If you solve the equation $\frac{9 - \alpha}{\alpha} = 2$, you will see that $\alpha = 3$ is the unique solution that satisfies the geometric definition of the parabola.

Question

Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f(\frac{tx}{36}) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points $(2, 1)$ and $(-4, \beta)$, then $\beta^\alpha$ is equal to .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to another masterclass in problem-solving. Today, we are not just solving a math problem; we are embarking on a journey. We are going to take an intimidating integral equation and, through the sheer elegance of calculus, transform it into a beautiful geometric curve.

We are given the equation:

$$ \int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x) $$

At first glance, this looks terrifying. We have a function $f$ inside an integral, and the argument is a product of variables. In JEE Advanced, whenever you see a complex argument inside an integral, your first instinct should be **substitution**.

### The Art of Substitution

Let us define a new variable, $u = \frac{tx}{36}$. Now, let us perform the magic of differentiation. Treating $x$ as a constant, we get $du = \frac{x}{36} dt$, which implies $dt = \frac{36}{x} du$.

We must also update our limits of integration. When $t = 0$, $u = 0$. When $t = 36$, $u = x$. Suddenly, the integral transforms into:

$$ \int_{0}^{x} f(u) \frac{36}{x} du = 4\alpha f(x) $$

Since $\frac{36}{x}$ is independent of $u$, we pull it out. Rearranging the terms, we get:

$$ \int_{0}^{x} f(u) du = \frac{4\alpha}{36} x f(x) = \frac{\alpha}{9} x f(x) $$

Look at that! We have stripped away the complexity. We have moved from a mysterious integral to a clean, manageable relationship between the integral of $f$ and the function $f$ itself.

### The Leibniz Scalpel

Now, we face a new challenge. We have an integral on the left and a function on the right. How do we isolate $f(x)$? We use the **Leibniz Rule**.

By differentiating both sides with respect to $x$, we can eliminate the integral sign entirely. On the left side, the derivative of $\int_{0}^{x} f(u) du$ with respect to $x$ is simply $f(x)$.

On the right side, we have a product of two functions of $x$: $\frac{\alpha}{9} x$ and $f(x)$. Applying the product rule, we get:

$$ f(x) = \frac{\alpha}{9} \left[ f(x) \cdot 1 + x \cdot f'(x) \right] $$

This is the moment where the problem shifts from calculus to algebra. We are now looking at a first-order differential equation.

### The Geometric Revelation

Let us expand and group our terms:

$$ f(x) = \frac{\alpha}{9} f(x) + \frac{\alpha}{9} x f'(x) $$

Subtracting $\frac{\alpha}{9} f(x)$ from both sides, we get:

$$ f(x) \left( 1 - \frac{\alpha}{9} \right) = \frac{\alpha}{9} x f'(x) $$

Separating the variables, we arrive at:

$$ \frac{f'(x)}{f(x)} = \frac{1 - \frac{\alpha}{9}}{\frac{\alpha}{9} x} = \frac{9 - \alpha}{\alpha x} $$

Integrating both sides, we obtain $\ln|f(x)| = \frac{9 - \alpha}{\alpha} \ln|x| + C$. Exponentiating both sides, we find the general form of our function:

$$ f(x) = K x^{\frac{9 - \alpha}{\alpha}} $$

Here is the "Aha!" moment. The problem explicitly tells us that $f(x)$ is a **standard parabola**. In the language of coordinate geometry, a standard parabola passing through the origin is defined by the equation $y = K x^2$.

This means the exponent of $x$ must be exactly $2$. Therefore:

$$ \frac{9 - \alpha}{\alpha} = 2 $$

Solving this simple linear equation, $9 - \alpha = 2\alpha$, we find $3\alpha = 9$, which gives us **$\alpha = 3$**. The mystery of $\alpha$ is solved!

### The Final Victory

With $\alpha = 3$, our function becomes $f(x) = K x^2$. We are given that the parabola passes through $(2, 1)$. Substituting these values:

$$ 1 = K(2)^2 \Rightarrow 1 = 4K \Rightarrow K = \frac{1}{4} $$

So, our function is $f(x) = \frac{x^2}{4}$. Finally, we need to find $\beta$ for the point $(-4, \beta)$. Substituting $x = -4$:

$$ \beta = \frac{(-4)^2}{4} = \frac{16}{4} = 4 $$

The question asks for $\beta^\alpha$. With $\beta = 4$ and $\alpha = 3$, we calculate:

$$ 4^3 = 64 $$

And there you have it. We started with a complex integral and ended with a simple integer. This is the beauty of JEE Advanced—it tests not just your ability to calculate, but your ability to see the underlying structure of the universe.

Animated Solution for Mathematics - Differential Equations: Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x)$ . If $y = f (x)$ is a standard parabola passing through the points $(2, 1)$ and $(- 4, β)$ , then $β^{α}$ is equal to .........

Enter Numerical Value:

Visualized Solution

Analyze the Integral Equation

Substitution in Integral

Transforming the Equation

Applying Leibniz Rule

Forming the Differential Equation

Integrating the Differential Equation

Identifying the Standard Parabola

Calculating the Value of $α$

Finding the Constant $K$

Finding the Value of $β$

Final Calculation: $β^{α}$

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Substitution

The Leibniz Scalpel

The Geometric Revelation

The Final Victory

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