JEE Main 2024 — Mathematics Calculus

$\begin{aligned} & \ln (y)=3 \sin ^{-1} x \\ & \frac{1}{y} \cdot y^{\prime}=3\left(\frac{1}{\sqrt{1-x^2}}\right) \\ & \Rightarrow y^{\prime}=\frac{3 y}{\sqrt{1-x^2}} \text { at } x=\frac{1}{2} \\ & \Rightarrow y^{\prime}=\frac{3 e^{3\left(\frac{\pi}{6}\right)}}{\frac{\sqrt{3}}{2}}=2 \sqrt{3} e^{\frac{\pi}{2}} \\ & \Rightarrow y^{\prime \prime}=3\left(\frac{\sqrt{1-x^2} y^{\prime}-y \frac{1}{2 \sqrt{1-x^2}}(-2 x)}{\left(1-x^2\right)}\right)\end{aligned}$​ln(y)=3sin−1xy1​⋅y′=3(1−x2​1​)⇒y′=1−x2​3y​ at x=21​⇒y′=23​​3e3(6π​)​=23​e2π​⇒y′′=3((1−x2)1−x2​y′−y21−x2​1​(−2x)​)​ $\begin{aligned} & \Rightarrow\left(1-x^2\right) y^{\prime \prime}=3\left(3 y+\frac{x y}{\sqrt{1-x^2}}\right) \\ & \downarrow \text { at } x=\frac{1}{2}, y=e^{3 \sin ^{-1}\left(\frac{1}{2}\right)}=\mathrm{e}^{3\left(\frac{\pi}{6}\right)}=\mathrm{e}^{\frac{\pi}{2}}\end{aligned}$​⇒(1−x2)y′′=3(3y+1−x2​xy​)↓ at x=21​,y=e3sin−1(21​)=e3(6π​)=e2π​​ $\begin{aligned} & \left.\left(1-x^2\right) y^{\prime \prime}\right|_{\text {at } x=\frac{1}{2}}=3\left(3 \mathrm{e}^{\frac{\pi}{2}}+\frac{\frac{1}{2}\left(e^{\frac{\pi}{2}}\right)}{\frac{\sqrt{3}}{2}}\right) \\ & =3 \mathrm{e}^{\frac{\pi}{2}}\left(3+\frac{1}{\sqrt{3}}\right) \\ & \left(1-x^2\right) y^{\prime \prime}-\left.x y^{\prime}\right|_{\text {atx } x=\frac{1}{2}} \\ & =3 \mathrm{e}^{\frac{\pi}{2}}\left(3+\frac{1}{\sqrt{3}}\right)-\frac{1}{2}\left(2 \sqrt{3} \mathrm{e}^{\frac{\pi}{2}}\right)=9 \mathrm{e}^{\frac{\pi}{2}}\end{aligned}$​(1−x2)y′′​at x=21​​=3(3e2π​+23​​21​(e2π​)​)=3e2π​(3+3​1​)(1−x2)y′′−xy′∣atx x=21​​=3e2π​(3+3​1​)−21​(23​e2π​)=9e2π​​