JEE Main 2024 — 29 January, Shift 1 — Question 25

Sol. $\int\left(\frac{1}{x}+\frac{\ln x}{x}\right) d x+\int \frac{d y}{1+y^{2}}=0$

$\ln x+\frac{(\ln x)^{2}}{2}+\tan ^{-1} y=C$ Put $x=y=1$

$\therefore \mathrm{C}=\frac{\pi}{4}$

$\Rightarrow \ln x+\frac{(\ln x)^{2}}{2}+\tan ^{-1} y=\frac{\pi}{4}$

Put $\mathrm{x}=\mathrm{e}$

$\Rightarrow \mathrm{y}=\tan \left(\frac{\pi}{4}-\frac{3}{2}\right)=\frac{1-\tan \frac{3}{2}}{1+\tan \frac{3}{2}}$ $\therefore \alpha=1, \beta=1$

$\Rightarrow \alpha+2 \beta=3$