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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
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Verified Answer
The correct answer is:
odd and is strictly increasing in $(-\infty, \infty)$
odd and is strictly increasing in $(-\infty, \infty)$
$\therefore \quad g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}$
for $\quad u \in(-\infty, \infty)$
and
$$
\begin{aligned}
g(-u) & =2 \tan ^{-1}\left(e^{-u}\right)-\frac{\pi}{2} \\
& =2\left(\cot ^{-1}\left(e^u\right)\right)-\frac{\pi}{2} \\
& =2\left(\frac{\pi}{2}-\tan ^{-1}\left(e^u\right)\right)-\frac{\pi}{2} \\
& =\frac{\pi}{2}-2 \tan ^{-1}\left(e^u\right)=-g(u)
\end{aligned}
$$
$\therefore \quad g(-u)=-g(u)$
$\Rightarrow g(u)$ is an odd function.
for $\quad u \in(-\infty, \infty)$
and
$$
\begin{aligned}
g(-u) & =2 \tan ^{-1}\left(e^{-u}\right)-\frac{\pi}{2} \\
& =2\left(\cot ^{-1}\left(e^u\right)\right)-\frac{\pi}{2} \\
& =2\left(\frac{\pi}{2}-\tan ^{-1}\left(e^u\right)\right)-\frac{\pi}{2} \\
& =\frac{\pi}{2}-2 \tan ^{-1}\left(e^u\right)=-g(u)
\end{aligned}
$$
$\therefore \quad g(-u)=-g(u)$
$\Rightarrow g(u)$ is an odd function.
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