Question

The inverse of the function $y=\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}$ is

• A. ${\log }_e\left(\frac{1+2x}{1-2x}\right)$
• B. $\frac{1}{4}{\log }_e\left(\frac{1-x}{1+x}\right)$
• C. $\frac{1}{4}{\log }_e\left(\frac{1+x}{1-x}\right)$
• D. an odd function

Answer 1

Answer: (C), (D)

$\frac{1}{4}{\log }_e\left(\frac{1+x}{1-x}\right)$
an odd function

$y=\frac{t-\frac{1}{t}}{t+\frac{1}{t}}$ where $t=e^{2x}$
$\therefore y(t^2+1)=t^2-1$
$\Rightarrow t^2=\frac{1+y}{1-y}$
$\Rightarrow {\left(e^{2x}\right)}^2=\frac{1+y}{1-y}(\therefore t=e^{2x})$
$\Rightarrow e^{4x}=\frac{1+y}{1-y}$
$\Rightarrow 4x={\log }_e\left(\frac{1+y}{1-y}\right)$
$x=\frac{1}{4}{\log }_e\left(\frac{1+y}{1-y}\right)$
$\therefore f^{-1}\left(x\right)=\frac{1}{4}{\log }_e\left(\frac{1+x}{1-x}\right)=p\left(x\right)$
$\therefore p(-x)=-p\left(x\right)$
$\therefore f^{-1}(-x)=-f^{-1}\left(x\right)$
$\Rightarrow$ Thus $f^{-1}\left(x\right)$ is an odd function.