JEE MAIN - Mathematics (2024 - 29th January Morning Shift - No. 19)
A function $$y=f(x)$$ satisfies $$f(x) \sin 2 x+\sin x-\left(1+\cos ^2 x\right) f^{\prime}(x)=0$$ with condition $$f(0)=0$$. Then, $$f\left(\frac{\pi}{2}\right)$$ is equal to
2
1
$$-$$1
0
Explanation
$$\begin{aligned}
& \frac{d y}{d x}-\left(\frac{\sin 2 x}{1+\cos ^2 x}\right) y=\sin x \\
& \text { I.F. }=1+\cos ^2 x \\
& y \cdot\left(1+\cos ^2 x\right)=\int(\sin x) d x \\
& =-\cos x+C \\
& x=0, C=1 \\
& y\left(\frac{\pi}{2}\right)=1
\end{aligned}$$
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