Question

Let the function $g:\left(-\infty ,\infty \right)\to \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ be given by $g(u)=2{\tan }^{-1}(e^u)-\frac{\pi }{2}$. then , g is

• A. even and is strictly increasing in $(0,\infty )$
• B. odd and is strictly decreasing in $(-\infty ,\infty )$
• C. odd and is strictly increasing in $(-\infty ,\infty )$
• D. neither even nor odd, but is strictly increasing in $(-\infty ,\infty )$

Answer 1

Answer: (C)

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