QJEEM24Ma5a54: JEE Main 2024 Mathematics - Class 12 JEEM24Ma5a54 Solution: If the solution curve of the differential equation passes…

$\begin{aligned}& \int\left(\frac{1}{x}+\frac{\ln x}{x}\right) d x+\int \frac{d y}{1+y^2}=0 \\& \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=C\end{aligned}$

Put $x=y=1$

$\begin{aligned}& \therefore C=\frac{\pi}{4} \\& \Rightarrow \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=\frac{\pi}{4}\end{aligned}$

Put $x=e$

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