Let the function $\mathrm{g}:(-\infty, \infty) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ be…
Let the function $\mathrm{g}:(-\infty, \infty) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ be given by $\mathrm{g}(\mathrm{u})=2 \tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)-\frac{\pi}{2}$. Then g is
even and is strictly increasing in $(0, \infty)$.
odd and is strictly decreasing in $(-\infty, \infty)$.
odd and is strictly increasing in $(-\infty, \infty)$.
neither even nor odd, but is strictly increasing in $(-\infty, \infty)$.
Solution
$\begin{aligned}
& \mathrm{g}(\mathrm{u})=2 \tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)-\frac{\pi}{2} \\
& \Rightarrow \mathrm{~g}(\mathrm{u})=\tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)-\left(\frac{\pi}{2}-\tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)\right) \\
& \Rightarrow \mathrm{g}(\mathrm{u})=\tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)-\cot ^{-1}\left(e^{\mathrm{u}}\right) \\
& \Rightarrow \mathrm{g}(-\mathrm{u})=\tan ^{-1}\left(\mathrm{e}^{-\mathrm{u}}\right)-\cot ^{-1}\left(\mathrm{e}^{-\mathrm{u}}\right) \\
& \Rightarrow \mathrm{g}(-\mathrm{u})=\tan ^{-1}\left(\frac{1}{\mathrm{e}^{\mathrm{u}}}\right)-\cot ^{-1}\left(\frac{1}{\mathrm{e}^u}\right) \\
& \Rightarrow \mathrm{g}(-\mathrm{u})=\cot ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)-\tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)=-\mathrm{g}(\mathrm{u})
\end{aligned}$ Now, $\dot{g(\mathrm{u}})=2 \tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)-\frac{\pi}{2}$
$\Rightarrow \mathrm{g}^{\prime}(\mathrm{u})=\frac{2 \mathrm{e}^{\mathrm{u}}}{1+\mathrm{e}^{2 \mathrm{u}}}\gt0 \text { for all } \mathrm{u} \in(-\infty, \infty)$
$\Rightarrow \mathrm{g}$ is strictly increasing function in $(-\infty, \infty)$. Hence, $\mathrm{g}(\mathrm{u})$ is odd and is strictly increasing in $(-\infty, \infty)$.