Question

Let $f:(-\infty ,\infty )\to \left(\frac{-\pi }{2},\frac{\pi }{2}\right)$ 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, $f^{\prime }(x)=\frac{2}{1+e^{2x}}{e}^x>0\Rightarrow$ strictly increasing