JEE Main 2025 — 23 January, Evening Shift — Question 11

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

$\frac{d y}{y}=\sin \left(\frac{x}{y}\right) d\left(-\frac{x}{y}\right)$

$\Rightarrow \quad \ell \mathrm{ny}=\cos \frac{\mathrm{x}}{\mathrm{y}}+\mathrm{C}$

$$\begin{aligned} & x(1)=\frac{\pi}{2}\\ \Rightarrow 0=\cos \frac{\pi}{2}+C \Rightarrow C=0 \\& \ell n y=\cos \frac{x}{y} \end{aligned}$$

but $y=2 \Rightarrow \cos \frac{x}{2}=\ln 2$

$$\begin{aligned} \cos x & =2 \cos ^{2} \frac{x}{2}-1 \\& =2(\ln 2)^{2}-1 \end{aligned}$$