JEE MAIN - Mathematics (2024 - 29th January Morning Shift - No. 30)
If the solution curve $$y=y(x)$$ of the differential equation $$\left(1+y^2\right)\left(1+\log _{\mathrm{e}} x\right) d x+x d y=0, x > 0$$ passes through the point $$(1,1)$$ and $$y(e)=\frac{\alpha-\tan \left(\frac{3}{2}\right)}{\beta+\tan \left(\frac{3}{2}\right)}$$, then $$\alpha+2 \beta$$ is _________.
Answer
3
Explanation
$$\begin{aligned} & \int\left(\frac{1}{x}+\frac{\ln x}{x}\right) d x+\int \frac{d y}{1+y^2}=0 \\ & \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=C \end{aligned}$$
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 $$x=e$$
$$\begin{aligned} & \Rightarrow \mathrm{y}=\tan \left(\frac{\pi}{4}-\frac{3}{2}\right)=\frac{1-\tan \frac{3}{2}}{1+\tan \frac{3}{2}} \\ & \therefore \alpha=1, \beta=1 \\ & \Rightarrow \alpha+2 \beta=3 \end{aligned}$$
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