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'\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) $