Question

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

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

Answer 1

Answer: (C)

odd and is strictly increasing in $g:(-\infty ,\infty )$

$g\left(u\right)+g(-u)$
$=2ta{n}^{-1}\left(e^u\right)-\frac{\pi }{2}+2ta{n}^{-1}\left(\frac{1}{e^u}\right)-\frac{\pi }{2}$.
$=2ta{n}^{-1}\left(e^u\right)+2co{t}^{-1}\left(e^u\right)-\pi$
$=2\left(\frac{\pi }{2}\right)-\pi$
$=0$.
Hence
$g\left(u\right)+g(-u)=0$
Hence $g\left(u\right)$ is an odd function.
$g^{\prime }\left(u\right)=\frac{e^u}{1+e^{2u}}$
Now
$f\left(x\right)=e^x$ is strictly increasing for all real x.
Hence
$g^{\prime }\left(u\right)>0$ for all real x.
Hence
$g\left(u\right)$ is increasing for all real x.