Mathematics - Functions Question with Solution | TestHub

Given that $g(u)=2{\tan }^{-1}\left(e^u\right)-\frac{\pi }{2}$ $\therefore g(-u)=2{\tan }^{-1}\left(e^{-u}\right)-\frac{\pi }{2}=$ $2{\tan }^{-1}\left(\frac{1}{e^u}\right)-\frac{\pi }{2}$ $=2{\cot }^{-1}\left(e^u\right)-\frac{\pi }{2}=2\left[\frac{\pi }{2}-{\tan }^{-1}\left(e^u\right)\right]-\frac{\pi }{2}$ $=\pi -2{\tan }^{-1}\left(e^u\right)-\frac{\pi }{2}$ $=\frac{\pi }{2}-2{\tan }^{-1}\left(e^u\right)=-g(u)$ $\therefore g$ is an odd function. Also $g^{\prime }(u)=\frac{2e^u}{1+e^{2u}}>0,\forall u\in (-\infty ,\infty )$ $\therefore g$ is strictly increasing on $(-\infty ,\infty )$.