Question

If $2{\tan }^{-1}x+{\sin }^{-1}\frac{2x}{1+x^2}$ is independent of $x$ then

• A. $xϵ(1,+\infty )$
• B. $xϵ(-1,1)$
• C. $xϵ(-\infty ,-1)$
• D. none of these

Answer 1

The correct option is A $xϵ(1,+\infty )$

$2{\tan }^{-1}x+{\sin }^{-1}\frac{2x}{1+x^2}$
We know that for |x|>1 , $2{\tan }^{-1}x=\pi -2{\sin }^{-1}\frac{2x}{1+x^2}$
$=\pi -{\sin }^{-1}\frac{2x}{1+x^2}+{\sin }^{-1}\frac{2x}{1+x^2}$
$=\pi$ (independent of x)