Aspire Faculty ID #18056 · Topic: JEE Main 2026 (24 January Morning Shift) · Just now
JEE Main 2026 (24 January Morning Shift)

Let a differentiable function $f$ satisfy the equation $\int_0^{36} f\left(\frac{tx}{36}\right)dt=4\alpha f(x)$ If $y=f(x)$ is a standard parabola passing through the points $(2,1)$ and $(-4,\beta)$, then $\beta^\alpha$ is equal to ______.

Solution

Put $\frac{tx}{36}=y$

$\int_0^x f(y)\frac{36}{x}dy=4\alpha f(x)$

$\Rightarrow \int_0^x f(y)dy=\frac{\alpha f(x)x}{9}$

Differentiate

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

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

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

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

For standard parabola ⇒ power = 2

$\frac{9}{\alpha}-1=2 \Rightarrow \alpha=3$

$f(2)=1 \Rightarrow c=\frac{1}{4}$

$f(x)=\frac{x^2}{4}$

$f(-4)=\beta=4$

$\beta^\alpha=4^3=64$

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