Question

If f(x) = $2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right),x>1,$ then $f\left(5\right)$ is equal to:

• A. $\pi$
• B. $\frac{\pi }{3}$
• C. $4{\tan }^1\left(5\right)$
• D. ${\tan }^{-1}\left(\frac{64}{155}\right)$

Answer 1

Answer: (A)

$\pi$

$f\left(x\right)=2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
For $x>1$
${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=\pi -2{\tan }^{-1}x$
$\Rightarrow f\left(x\right)=2{\tan }^{-1}x+\pi -2{\tan }^{-1}x$
$\Rightarrow f\left(x\right)=\pi$
$\Rightarrow f\left(5\right)=\pi$