Q. Let the function $g:( - \infty,\ \infty) \rightarrow \left( - \frac{\pi}{2},\frac{\pi}{2} \right)$ be given by$g(u) = 2\tan^{- 1}{\left( e^{u} \right) - \frac{\pi}{2}}$. 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 $( - \infty,\ \infty)$
Explanation
(c)
Given, $g(u) = 2\tan^{- 1}\left( e^{u} \right) - \frac{\pi}{2},\ for\ u \in ( - \infty,\ \infty)$
Now, $g( - u) = 2\tan^{- 1}{\left( e^{- u} \right) - \frac{\pi}{2}}$
$= 2\left( \cot^{- 1}\left( e^{u} \right) \right) - \frac{\pi}{2}$
$= 2\left( \frac{\pi}{2} - \tan^{- 1}{(e^{u})} \right) - \frac{\pi}{2}$
$= - g(u)$
$\therefore\ \ \ g(u)$ Is an odd function.
Also, $g^{'}(u) = 2\frac{1}{1 + \left( e^{u} \right)^{2}}.e^{u} - 0 > 0$
Which is strictly increasing in $( - \infty,\ \infty)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.