Question

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

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

Answer 1

The correct option is D $\pi$

$f\left(x\right)=2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\text{Put}x=\tan \theta ,\theta \in (\frac{\pi }{4},\frac{\pi }{2})\text{as}x\in (1,\infty )f\left(x\right)=2{\tan }^{-1}\tan \theta +{\sin }^{-1}(\frac{2\tan \theta }{1+{\tan }^2\theta })=2\theta +{\sin }^{-1}\left(\sin 2\theta \right)....\left(1\right)\text{Since,}{\sin }^{-1}\sin x=x,x\in [-\frac{\pi }{2},\frac{\pi }{2}]\text{Now,}\frac{\pi }{4}<\theta <\frac{\pi }{2}\Rightarrow \frac{\pi }{2}<2\theta <\pi \Rightarrow -\pi <-2\theta <-\frac{\pi }{2}\Rightarrow 0<\pi -2\theta <\frac{\pi }{2}\therefore f\left(x\right)=2\theta +\pi -2\theta =\pi \therefore f\left(5\right)=\pi$