Question

The interval(s) in which the value of $2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ is independent of $x$ is

• A. $(-1,1)$
• B. $(-\infty ,1]$
• C. $[1,\infty )$
• D. $\varphi$

Answer 1

The correct option is C $[1,\infty )$

${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
$\text{Let}x=\tan \theta ,\theta ϵ(-\frac{\pi }{2},\frac{\pi }{2}),2\theta ϵ(-\pi ,\pi )$
$\therefore {\sin }^{-1}\left(\sin 2\theta \right)$
$=\left\{\begin{array}{cc}2\theta ;& -\frac{\pi }{2}<2\theta <\frac{\pi }{2}\\ \pi -2\theta ;& \frac{\pi }{2}\le 2\theta <\pi \\ -\pi -2\theta ;& -\pi <2\theta \le -\frac{\pi }{2}\end{array}\right\}$
$=\left\{\begin{array}{cc}2{\tan }^{-1}x;& -1<x<1\\ \pi -2{\tan }^{-1}x;& x\ge 1\\ -\pi -2{\tan }^{-1}x;& x\le -1\end{array}\right\}$
So $2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ is independent of $x$ in $(-\infty ,-1]\cup [1,\infty )$.