Q47 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{t x}{36}\right) d t=4 \alpha f(x)$. If $y=f(x)$ is a standard parabola passing through the points $(2,1)$ and $(-4, \beta)$, then $\beta^{\alpha}$ is equal to $\_\_\_\_$.
Correct answer: 64
Solution
From $\int_0^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$, substitute $u = \frac{tx}{36}$ to get $\frac{36}{x} \int_0^x f(u) du = 4\alpha f(x)$.
Differentiating: $f(x) = \frac{\alpha}{9}[f(x) + xf'(x)]$, which simplifies to $(9-\alpha)f(x) = \alpha x f'(x)$.
This gives $\frac{f'(x)}{f(x)} = \frac{9-\alpha}{\alpha x}$, integrating to $f(x) = Kx^{\frac{9-\alpha}{\alpha}}$.
For f to be a standard parabola: $\frac{9-\alpha}{\alpha} = 2 \Rightarrow \alpha = 3$.
Thus $f(x) = Kx^2$. Using point $(2,1)$: $K = \frac{1}{4}$, so $f(x) = \frac{x^2}{4}$.
At $(-4, \beta)$: $\beta = \frac{16}{4} = 4$.
Therefore $\beta^\alpha = 4^3 = 64$
Differentiating: $f(x) = \frac{\alpha}{9}[f(x) + xf'(x)]$, which simplifies to $(9-\alpha)f(x) = \alpha x f'(x)$.
This gives $\frac{f'(x)}{f(x)} = \frac{9-\alpha}{\alpha x}$, integrating to $f(x) = Kx^{\frac{9-\alpha}{\alpha}}$.
For f to be a standard parabola: $\frac{9-\alpha}{\alpha} = 2 \Rightarrow \alpha = 3$.
Thus $f(x) = Kx^2$. Using point $(2,1)$: $K = \frac{1}{4}$, so $f(x) = \frac{x^2}{4}$.
At $(-4, \beta)$: $\beta = \frac{16}{4} = 4$.
Therefore $\beta^\alpha = 4^3 = 64$
Hence, the correct answer is 64.