Question

The interval for which $2{\tan }^{-1}x+{\sin }^{-1}\frac{2}{1+x}$ is independent of $x$ is

• A. $\left|x\right|<1$
• B. $\left|x\right|>1$
• C. $\left|x\right|=1$
• D. $\varphi$

Answer 1

Answer: (B)

$\left|x\right|>1$

We know that,
if $\left|x\right|\le 1$, then $2{\tan }^{-1}x={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
and if $\left|x\right|>1$, then $\pi -2{\tan }^{-1}x={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
Thus, if $\left|x\right|>1$,
$2{\tan }^{-1}x+{\sin }^{-1}\frac{2x}{1+x^2}=2{\tan }^{-1}x+\pi -2{\tan }^{-1}x=\pi$
which is independent of $x$.