JEE Main 2026 — 23 January, Morning Shift — Question 7

$\left(x^{4} d y+4 x^{3} y d x\right)=-2 \sin x d x$

$\Rightarrow \int \mathrm{d}\left(\mathrm{x}^{4} \mathrm{y}\right)=\int-2 \sin \mathrm{xdx}$

$\Rightarrow \mathrm{x}^{4} \mathrm{y}=2 \cos \mathrm{x}+\mathrm{c}$

$\Rightarrow \mathrm{x}^{4} \mathrm{f}(\mathrm{x})=2 \cos \mathrm{x}+\mathrm{c}$

As $\mathrm{f}\left(\frac{\pi}{2}\right)=0$

So, $\mathrm{c}=0$

$$\begin{aligned} & \left(\frac{\pi}{3}\right)^{4} \mathrm{f}\left(\frac{\pi}{3}\right)=2 \cos \frac{\pi}{3} & \end{aligned}$$

$\pi^{4} \mathrm{f}\left(\frac{\pi}{3}\right)=81$