If $f(x)=2\left(\tan ^{-1}\left(e^x\right)-\frac{\pi}{4}\right)$, then $f(x)$ is:

odd and is strictly increasing in $(-\infty, \infty)$

The correct answer is Option (3) → odd and is strictly increasing in $(-\infty, \infty)$

$f(x)=2\left(\tan ^{-1}\left(e^x\right)-\frac{\pi}{4}\right)$

$f(-x)=2\left(\tan^{-1}\left(\frac{1}{e^x}\right)-\frac{\pi}{4}\right)$

$f(-x)=2\left(\cot^{-1}e^x-\frac{\pi}{4}\right)$

$=2\left(\frac{\pi}{2}-\tan^{-1}e^x-\frac{\pi}{4}\right)$

$=-2\left(\tan^{-1}e^x-\frac{\pi}{4}\right)$

$f(-x)=-f(x)$ odd function

$f(-x)=\frac{2e^x}{1+e^{2x}}>0$ always

$f(x)$ → strictly increasing