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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Differentiation

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Differentiation

If _e y=3 ^ -1 x , then (1-x^2) y^ -x y^ at x=(1)/(2) is equal to

If log⁡ey=3sin⁡−1x\log _e y=3 \sin ^{-1} x, then \left(1-x^2\right) y^{\prime \prime}-x y^{\prime}atatx=\frac{1}{2} is equal to

Options

  1. A3eπ/63 e^{\pi / 6}
  2. B9eπ/29 e^{\pi / 2}
  3. C3eπ/23 e^{\pi / 2}
  4. D9eπ/69 e^{\pi / 6}

Correct answer

B. 9eπ/29 e^{\pi / 2}

Step-by-step solution

ln⁡(y)=3sin⁡−1x1y⋅y′=3(11−x2)⇒y′=3y1−x2 at x=12⇒y′=3e3(π6)32=23eπ2⇒y′′=3(1−x2y′−y121−x2(−2x)(1−x2))\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}⇒(1−x2)y′′=3(3y+xy1−x2)↓ at x=12,y=e3sin⁡−1(12)=e3(π6)=eπ2\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′′∣at x=12=3(3eπ2+12(eπ2)32)=3eπ2(3+13)(1−x2)y′′−xy′∣atx x=12=3eπ2(3+13)−12(23eπ2)=9eπ2\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}$
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