JEE Main 2026 — 28 January, Evening Shift — Question 14

$x d y-y d x=x^{2} \cot x d x$

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

$\mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\cot \mathrm{x} \mathrm{dx}$

$\int \mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\int \cot \mathrm{x} \mathrm{dx}$

$\frac{\mathrm{y}}{\mathrm{x}}=\log _{\mathrm{e}} \sin \mathrm{x}+\mathrm{C}$

given $\mathrm{y}\left(\frac{\pi}{2}\right)=\frac{\pi}{2}$

$\Rightarrow \mathrm{c}=1$

$\mathrm{y}=\mathrm{x}\left(\log _{\mathrm{e}} \sin \mathrm{x}+1\right)$

$y\left(\frac{\pi}{6}\right)=\frac{\pi}{6}\left[-\log _{e} 2+1\right]$

$y\left(\frac{\pi}{4}\right)=\frac{\pi}{4}\left[-\frac{1}{2} \log _{e} 2+1\right]$

$6 y\left(\frac{\pi}{6}\right)-8 y\left(\frac{\pi}{4}\right)$

$=\pi\left[\left(-\log _{\mathrm{e}} 2+1\right)+2\left(\frac{1}{2} \log _{\mathrm{e}} 2-1\right)\right]$ $=\pi[1-2]=-\pi$