Application of Derivatives 2 Question 7
####7. If the function $g:(-\infty, \infty) \rightarrow-\frac{\pi}{2}, \frac{\pi}{2}$ is given by $g(u)=2 \tan ^{-1}\left(e^{u}\right)-\frac{\pi}{2}$. Then, $g$ is
(2008, 3M)
(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 $(-\infty, \infty)$
Show Answer
Answer:
(a)
Solution:
- Given, $g(u)=2 \tan ^{-1}\left(e^{u}\right)-\frac{\pi}{2}$ for $u \in(-\infty, \infty)$
$ \begin{aligned} & 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 \frac{\pi}{2}-\tan ^{-1}\left(e^{u}\right)-\frac{\pi}{2} \\ &=\pi / 2-2 \tan ^{-1}\left(e^{u}\right)=-g(u) \\ & \therefore \quad g(-u)=-g(u) \\ & \Rightarrow \quad g(u) \text { is an odd function. } \end{aligned} $
We have, $g(u)=2 \tan ^{-1}\left(e^{u}\right)-\pi / 2$
$ \begin{aligned} & g^{\prime}(u)=\frac{2 e^{u}}{1+e^{2 u}} \\ & g^{\prime}(u)>0, \forall x \in R \end{aligned} $
$\left[\because e^{u}>0\right]$
So, $g^{\prime}(u)$ is increasing for all $x \in R$.
