Question

If \( \log_e y = 3 \sin^{-1}x \), then \( (1 - x)^2 y'' - xy' \) at \( x = \frac{1}{2} \) is equal to:

• A. \( 9e^{\pi/6} \)
• B. \( 3e^{\pi/6} \)
• C. \( 3e^{\pi/2} \)
• D. \( 9e^{\pi/2} \)

Answer 1

The Correct Option is D

Provided:

\[\ln(y) = 3 \sin^{-1}(x)\]

Differentiating both sides with respect to \(x\):

\[\frac{1}{y} \cdot y' = 3 \left( \frac{1}{\sqrt{1 - x^2}} \right)\]

\[\Rightarrow y' = \frac{3y}{\sqrt{1 - x^2}}\]}

At \(x = \frac{1}{2}\):

\[y' = \frac{3e^{3\sin^{-1}\left(\frac{1}{2}\right)}}{\sqrt{1 - \left(\frac{1}{2}\right)^2}}\]}

\[= \frac{3e^{\frac{\pi}{2}}}{\frac{\sqrt{3}}{2}}\]}

\[= 2\sqrt{3} e^{\frac{\pi}{2}}\]}

Differentiating again to find \(y''\):

\[y'' = 3 \left( \frac{\sqrt{1 - x^2} \, y' - y \cdot \frac{1}{\sqrt{1 - x^2}}(-2x)}{1 - x^2} \right)\]

Therefore:

\[(1 - x^2)y'' = 3 \left( 3y + \frac{xy}{\sqrt{1 - x^2}} \right)\]

At \(x = \frac{1}{2}\):

\[y = e^{3\sin^{-1}\left(\frac{1}{2}\right)} = e^{\frac{\pi}{2}}\]}

Substituting into the equation:

\[(1 - x^2)y''\big|_{x=\frac{1}{2}} = 3e^{\frac{\pi}{2}} \left( 3 + \frac{1}{\sqrt{3}} \right)\]

Computing \((1 - x^2)y'' - xy'\) at \(x = \frac{1}{2}\):

\[(1 - x^2)y'' - xy' = 3e^{\frac{\pi}{2}}\left( 3 + \frac{1}{\sqrt{3}} \right) - \frac{1}{2}\left( 2\sqrt{3} e^{\frac{\pi}{2}} \right)\]

\[= e^{\frac{\pi}{2}} \left( 9 + \sqrt{3} - \sqrt{3} \right)\]}

\[= 9e^{\frac{\pi}{2}}\]}

Final Answer:

\[\boxed{9e^{\frac{\pi}{2}}}\]}