Correct Answer: 64

Explanation

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

Put $u=\dfrac{tx}{36}\Rightarrow t=\dfrac{36u}{x},\ dt=\dfrac{36}{x}du$.

Limits: $t=0\Rightarrow u=0$, $t=36\Rightarrow u=x$.

Hence, $$ \int_{0}^{36} f\!\left(\frac{tx}{36}\right)\,dt = \frac{36}{x}\int_{0}^{x} f(u)\,du. $$

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

Differentiate both sides with respect to $x$:

$$ \frac{36}{x}f(x)-\frac{36}{x^{2}}\int_{0}^{x}f(u)\,du = 4\alpha f'(x). $$

Using the original relation to eliminate the integral term, $$ \frac{36}{x}f(x)-\frac{4\alpha}{x}f(x)=4\alpha f'(x). $$

$$ (36-4\alpha)f(x)=4\alpha x f'(x). $$

Since $f(x)$ is a standard parabola, let $f(x)=ax^{2}$.

Then $f'(x)=2ax$.

Substitute: $$ (36-4\alpha)ax^{2}=4\alpha x(2ax). $$

$$ 36-4\alpha=8\alpha \Rightarrow \alpha=3. $$

Now $f(x)=ax^{2}$ and $f(2)=1$ gives $$ 4a=1 \Rightarrow a=\frac14. $$

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

Then $$ \beta=f(-4)=\frac{16}{4}=4. $$

Hence, $$ \beta^{\alpha}=4^{3}=\boxed{64}. $$

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