If e y = 3 ^-1 x, then (1 - x^2) y'' - x y' at x = 12 is equal to… — Solution | JEE Mathematics

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If $\log_e y = 3 \sin^{-1} x$ , then $\left(1 - x^2\right) y'' - x y'$ 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

Correct answer:D

Step-by-step solution

Standard Method

Given: $\ln(y) = 3\sin^{-1}(x)$

Find: $\left(1-x^2\right)y''-xy'$ at $x=\frac{1}{2}$ .

Differentiate both sides with respect to $x$ :

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

So,

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

At $x=\frac{1}{2}$ ,

y = e^{3\sin^{-1}(1/2)} = e^{\pi/2}

Therefore,

y' = \frac{3e^{\pi/2}}{\sqrt{1-\left(\frac{1}{2}\right)^2}} = \frac{3e^{\pi/2}}{\frac{\sqrt{3}}{2}} = 2\sqrt{3}e^{\pi/2}

Differentiate $y' = \frac{3y}{\sqrt{1-x^2}}$ again:

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

Hence,

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

Now put $x=\frac{1}{2}$ and $y=e^{\pi/2}$ :

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

Therefore,

\begin{aligned} (1-x^2)y'' - xy' &= 3e^{\pi/2}\left(3 + \frac{1}{\sqrt{3}}\right) - \frac{1}{2}\left(2\sqrt{3}e^{\pi/2}\right) \\ &= e^{\pi/2}\left(9 + \sqrt{3} - \sqrt{3}\right) \\ &= 9e^{\pi/2} \end{aligned}

Therefore, the correct option is D, and the value is $9e^{\pi/2}$ .

Using explicit form of $$y$$

From $\log_e y = 3\sin^{-1}x$ , we get

y = e^{3\sin^{-1}x}

Then,

y' = e^{3\sin^{-1}x}\cdot 3\cdot \frac{1}{\sqrt{1-x^2}} = \frac{3y}{\sqrt{1-x^2}}

At $x=\frac{1}{2}$ ,

\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}

y = e^{3\pi/6} = e^{\pi/2}

This confirms the exponent used in the final evaluation.

The second provided approach contains an intermediate appearance of $e^{\pi/6}$ in places where $e^{\pi/2}$ should be used, but its final conclusion still matches the first approach and the correct option. Using the consistent substitution gives $9e^{\pi/2}$ .

Common mistakes

Using $\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}$ is wrong. The correct value is $\frac{\pi}{6}$ , so $3\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}$ . Evaluate the inverse sine first, then multiply by $3$ .
Differentiating $\ln y$ as if it were just $y'$ is incorrect. Since $y$ is a function of $x$ , use chain rule: $\frac{d}{dx}(\ln y)=\frac{1}{y}y'$ .
While differentiating $\frac{3y}{\sqrt{1-x^2}}$ , treating $y$ as a constant is incorrect. Both $y$ and $\sqrt{1-x^2}$ depend on $x$ , so product or quotient rule must be applied carefully.

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