JEE Main 2026 — 24 January, Morning Shift — Question 20

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

Put $\frac{\mathrm{tx}}{36}=\mathrm{y}$ $\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{x}}{36}$

$\int_{0}^{\mathrm{x}} \frac{f(\mathrm{y}) 36 \mathrm{dy}}{\mathrm{x}}=4 \alpha f(\mathrm{x})$

$\int_{0}^{\mathrm{x}} f(\mathrm{y}) \mathrm{dy}=\frac{\alpha f(\mathrm{x}) \mathrm{x}}{9}$

$f(\mathrm{x})=\frac{\alpha}{9}\left(f(\mathrm{x})+\mathrm{x} f^{\prime}(\mathrm{x})\right)$

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

$\Rightarrow(9-\alpha) f(\mathrm{x})=\alpha \mathrm{x} f^{\prime}(\mathrm{x})$

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

$\log _{\mathrm{e}} f(\mathrm{x})=\left(\frac{9}{\alpha}-1\right) \log _{\mathrm{e}} \mathrm{x}+\log _{\mathrm{e}} \mathrm{c}$

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

for standard parabola $\frac{9}{\alpha}-1=2$

$\alpha=3$

$f(\mathrm{x})=\mathrm{cx}^{2}$

passing through $(2,1)$

$1=4 c \Rightarrow c=1 / 4$

$y=\frac{x^{2}}{4}$ passing through $(-4, \beta)$

$\beta=4$

$\beta^{x}=4^{3}=64$