Let the function $g:(-\infty, \infty) \rightarrow(-\pi / 2, \pi / 2)$ be given by $g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}$. Then, g is
Answer & explanation
Correct answer: option 3
We have,
$g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}$
$\Rightarrow g'(u)=\frac{2 e^u}{1+e^{2 u}}>0$ for all $u \rightarrow(-\infty, \infty)$
⇒ g is strictly increasing function in $(-\infty, \infty)$.
Now,
$g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}$
$\Rightarrow g(u)=\tan ^{-1}\left(e^u\right)-\left(\frac{\pi}{2}-\tan ^{-1}\left(e^u\right)\right)$
$\Rightarrow g(u)=\tan ^{-1}\left(e^u\right)-\cot ^{-1}\left(e^u\right)$
$\Rightarrow g(-u)=\tan ^{-1}\left(e^{-u}\right)-\cot ^{-1}\left(e^{-u}\right)$
$\Rightarrow g(-u)=\tan ^{-1}\left(e^{1 / u}\right)-\cot ^{-1}\left(e^{1 / u}\right)$
$\Rightarrow g(-u)=\cot ^{-1}\left(e^u\right)-\tan ^{-1}\left(e^u\right)=-g(u)$
Hence, g(u) is odd and is strictly increasing in $(-\infty, \infty)$