Question

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

• A. $\frac{\pi}{2}$
• B. $\pi$
• C. $4\, \tan^{-1} \left(5\right)$
• D. $ \tan^{-1} \left(\frac{65}{156}\right)$

Answer 1

Answer: (B)

$f(x)=2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) x>1, f(5)=?$
We know that
$2 \tan ^{-1} x= \begin{cases} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \quad 1 \le x \le 1 \\ -\pi-\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \quad x < -1 \\ \pi-\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \quad x> 1 \end{cases} $
$\Rightarrow f(x)=2 \tan ^{-1} x+\left(\pi-2 \tan ^{-1} x\right)$
$\Rightarrow f(x)=\pi \forall x>1 \Rightarrow f(5)=\pi$