JEE MAIN - Mathematics (2024 - 9th April Evening Shift - No. 8)
If $$\log _e y=3 \sin ^{-1} x$$, then $$(1-x^2) y^{\prime \prime}-x y^{\prime}$$ at $$x=\frac{1}{2}$$ is equal to
$$9 e^{\pi / 2}$$
$$9 e^{\pi / 6}$$
$$3 e^{\pi / 2}$$
$$3 e^{\pi / 6}$$
Explanation
$$\begin{aligned} &\log _e y=3 \sin ^{-1} x\\ &\begin{aligned} & y=e^{3 \sin ^{-1} x} \\ & \frac{d y}{d x}=e^{3 \sin ^{-1} x} \cdot \frac{3}{\sqrt{1-x^2}} \end{aligned} \end{aligned}$$
$$\sqrt{1-x^2} \frac{d y}{d x}=3 y$$
Again differentiate
$$\begin{aligned} & \sqrt{1-x^2} \cdot y^{\prime \prime}-\frac{2 x}{2 \sqrt{1-x^2}} y^{\prime}=3 y^{\prime} \\ & (1-x)^2 y^{\prime \prime}-x y^{\prime}=3 y^{\prime}\left(\sqrt{1-x^2}\right) \end{aligned}$$
So value of $$3 y^{\prime}\left(\sqrt{1-x^2}\right)$$ at $$x=\frac{1}{2}$$
$$\begin{aligned} & 3 \cdot \frac{3}{\sqrt{1-x^2}} e^{\sin ^{-1} x}\left(\sqrt{1-x^2}\right) \\ & =9 e^{3 \frac{\pi}{6}}=9 e^{\frac{\pi}{2}} \end{aligned}$$
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