Question

Which of the following represents an odd function?

• A. $f\left(x\right)=\frac{{(1+e^x)}^2}{e^x}$
• B. $g\left(x\right)={\sec }^{-1}\left(\sec x\right)$
• C. $h\left(x\right)=\cos \left({\cos }^{-1}x\right)$
• D. $k\left(x\right)={\cot }^{-1}\left(\cot x\right)$

Answer 1

The correct option is C $h\left(x\right)=\cos \left({\cos }^{-1}x\right)$

$f\left(x\right)=\frac{(1+e^x{)}^2}{e^x}$
$f(-x)=\frac{(1+e^{-x}{)}^2}{e^{-x}}$
$=\frac{(1+e^x{)}^2}{e^{-x}{e}^{2x}}$
$=\frac{(1+e^x{)}^2}{e^x}$
$=f\left(x\right)$
Hence $f(-x)=f\left(x\right)$. Therefore, $f\left(x\right)$ is an even function.
$g\left(x\right)=se{c}^{-1}\left(sec\right(x\left)\right)$
$g(-x)=se{c}^{-1}\left(sec\right(-x\left)\right)$
$=se{c}^{-1}\left(sec\right(x\left)\right)$
$=g\left(x\right)$
Hence $g(-x)=g\left(x\right)$. Therefore, $g\left(x\right)$ is an even function.
$h\left(x\right)=cos\left(co{s}^{-1}\right(x\left)\right)$
$h(-x)=cos\left(co{s}^{-1}\right(-x\left)\right)$
$=cos(\pi -co{s}^{-1}(x\left)\right)$
$=-cos\left(co{s}^{-}\right(x\left)\right)$
$=-h\left(x\right)$
Hence $h(-x)=-h\left(x\right)$. Therefore, $h\left(x\right)$ is an odd function.
$k\left(x\right)=co{t}^{-1}\left(cot\right(x\left)\right)$
$k(-x)=co{t}^{-1}\left(cot\right(-x\left)\right)$
$=co{t}^{-1}(-cot(x\left)\right)$
$=\pi -co{t}^{-1}\left(cot\right(x\left)\right)$
Hence $k\left(x\right)$ is neither odd nor even.