JEE Main 2026 (Online) 24th January Morning ShiftQuestion 1
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 $\_\_\_\_\_\_$ .
Definite Integration
Answer$64$
Solution
Concept. Substituting $u=\frac{tx}{36}$ turns the condition into $\int_0^xf=\frac{\alpha x}{9}f(x)$; differentiating gives $f=c\,x^k$. Formulas used. (1) $\dfrac{d}{dx}\int_0^xf(u)\,du=f(x)$(2) $\dfrac{f'}{f}=\dfrac{k}{x}\Rightarrow f=c\,x^k$Step 1 - Change variable. $u=\frac{tx}{36}$, $dt=\frac{36}{x}du$, $t:0\to36$ gives $u:0\to x$: $$\frac{36}{x}\int_0^xf(u)\,du=4\alpha f(x)\ \Longrightarrow\ \int_0^xf(u)\,du=\frac{\alpha x}{9}f(x).$$ Step 2 - Differentiate. By (1), $$f=\frac\alpha9\big(f+xf'\big)\ \Longrightarrow\ \frac{f'}{f}=\frac{9-\alpha}{\alpha}\cdot\frac1x.$$ By (2), $f=c\,x^{(9-\alpha)/\alpha}$. Step 3 - Standard parabola. $y=cx^2$ needs $\frac{9-\alpha}{\alpha}=2$, so $\alpha=3$. Through $(2,1)$: $c=\frac14$, so $f=\frac{x^2}{4}$ and $\beta=f(-4)=4$. Step 4 - Value. $\beta^\alpha=4^3=64$. Answer: 64