Question

Let $f :(-\infty, \infty) \rightarrow\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ be given by $f ( x )=\frac{\pi}{2}-2 \cot ^{-1}\left( e ^{ x }\right)$, then $f$ is

• A. odd and is strictly decreasing in $(0,-\infty)$.
• B. even and is strictly increasing in $(-\infty, \infty)$.
• C. odd and is strictly increasing in $(-\infty, \infty)$.
• D. neither odd nor even but is strictly increasing in $(-\infty, \infty)$.

Answer 1

Answer: (C)

$f (- x )=\frac{\pi}{2}-2 \cot ^{-1}\left( e ^{- x }\right)=\frac{\pi}{2}-2 \cot ^{-1}\left(\frac{1}{ e ^{ x }}\right)=\frac{\pi}{2}-2 \tan ^{-1} e ^{ x } $
$=\frac{\pi}{2}-2\left(\frac{\pi}{2}-\cot ^{-1} e ^{ x }\right)=2 \cot ^{-1} e ^{ x }-\frac{\pi}{2}=- f ( x )$
$\Rightarrow$ odd function, $f ^{\prime}( x )=\frac{2}{1+ e ^{2 x }} e ^{ x }>0 \Rightarrow$ strictly increasing