Question

If ${\sin }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\log a$, then $\frac{d^{2}y}{d{x}^2}$ equals

• A. $\frac{x}{y}$
• B. $\frac{y}{x^2}$
• C. $\frac{y}{x}$
• D. $0$

Answer 1

Answer: (D)

$0$

We have ${\sin }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\log a$
$\Rightarrow \frac{x^2-y^2}{x^2+y^2}=\sin \left(\log a\right)$
$\Rightarrow \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\sin \left(\log a\right)$
(on putting $y=x\tan \theta$)
$\Rightarrow \cos 2\theta =\sin \left(\log a\right)$
$\Rightarrow 2\theta ={\cos }^{-1}\left(\sin \log a\right)$
$\Rightarrow \theta =\frac{1}{2}{\cos }^{-1}\left\{\sin \left(\log a\right)\right\}$
${\tan }^{-1}\left(\frac{y}{x}\right)=\frac{1}{2}{\cos }^{-1}\left\{\sin \log a\right\}$
$\Rightarrow \frac{1}{1+\frac{y^2}{x^2}}\cdot \frac{x\frac{dy}{dx}-y}{x^2}=0$
$\Rightarrow x\frac{dy}{dx}-y=0$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$ ...(i)
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-y}{x^2}+\frac{1}{x}\cdot \frac{dy}{dx}=\frac{-y}{x^2}+\frac{1}{x}\left(\frac{y}{x}\right)$ ....{from (i)}
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-y}{x^2}+\frac{y}{x^2}=0$