Question

$\text{If}f\left(x\right)={\sin }^{-1}\left(\frac{4x}{4+x^2}\right),$ then

• A. $f(\sqrt{8}-2)=\pi -2{\tan }^{-1}(\sqrt{2}-1)$
• B. $f(\sqrt{12}-\sqrt{3})=\frac{2\pi }{3}$
• C. $f(-\left(\frac{3}{\sqrt{8}-2}\right))=-\pi +2{\tan }^{-1}\left(\frac{3}{\sqrt{8}-2}\right)$
• D. $f\left(\pi \right)=\pi -2{\tan }^{-1}\left(\frac{\pi }{2}\right)$

Answer 1

The correct option is D $f\left(\pi \right)=\pi -2{\tan }^{-1}\left(\frac{\pi }{2}\right)$

$f\left(x\right)={\sin }^{-1}\left(\frac{4x}{4+x^2}\right)$
$\text{Put}x=2\tan \theta$
$\begin{array}{cc}\text{Then}f\left(x\right)& ={\sin }^{-1}\left(\sin 2\theta \right)\\ & =\{\begin{array}{c}-\pi -2\theta ,\frac{-\pi }{2}<\theta <\frac{-\pi }{4}\\ 2\theta ,\frac{-\pi }{4}\le \theta \le \frac{\pi }{4}\\ \pi -2\theta ,\frac{\pi }{4}<\theta <\frac{\pi }{2}\end{array}\\ & \\ & =\{\begin{array}{c}-\pi -2{\tan }^{-1}\left(\frac{x}{2}\right),x<-2\\ 2{\tan }^{-1}\left(\frac{x}{2}\right),-2\le x\le 2\\ \pi -2{\tan }^{-1}\left(\frac{x}{2}\right),x>2\end{array}\end{array}$

$-2<\sqrt{8}-2<2$
$\therefore f(\sqrt{8}-2)=2{\tan }^{-1}\left(\frac{\sqrt{8}-2}{2}\right)=2{\tan }^{-1}(\sqrt{2}-1)$

$-2<\sqrt{12}-\sqrt{3}<2$
$\therefore f(\sqrt{12}-\sqrt{3})=2{\tan }^{-1}\left(\frac{\sqrt{12}-\sqrt{3}}{2}\right)$

$\frac{-3}{\sqrt{8}-2}=\frac{-3(1+\sqrt{2})}{2}<-2\therefore f(-\left(\frac{3}{\sqrt{8}-2}\right))=-\pi +2{\tan }^{-1}\left(\frac{3}{2(\sqrt{8}-2)}\right)$

$\pi >2$
$\therefore f\left(\pi \right)=\pi -2{\tan }^{-1}\left(\frac{\pi }{2}\right)$