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 $(0,\infty )$
• 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

The correct option is C odd and is strictly increasing in $(-\infty ,\infty )$

$g\left(u\right)=2{\tan }^{-1}\left(e^u\right)-\frac{\pi }{2}$

$g(-u)=2{\tan }^{-1}\left(e^{-u}\right)-\frac{\pi }{2}$

$=2{\tan }^{-1}\left(\frac{1}{e^u}\right)-\frac{\pi }{2}$

$=2{\cot }^{-1}\left(e^u\right)-\frac{\pi }{2}$

Now

$g\left(u\right)+g(-u)=2\left[{\tan }^{-1}\right(e^u)+{\cot }^{-1}(e^u\left)\right]-\pi$

$=2\left[\frac{\pi }{2}\right]-\pi$

$=0$

Hence, $g\left(u\right)$ is an odd function.

Now

$g^{\prime }\left(u\right)=\frac{2}{1+e^{2u}}.e^u$

$=\frac{2e^u}{1+e^{2u}}$

Therefore, $g^{\prime }\left(u\right)>0$ for all $uϵR$

Thus, $g\left(u\right)$ is an increasing function for all $uϵR$.

Hence, $g\left(u\right)$ is an odd and an increasing function.