Question

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

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

Answer 1

Answer: (C), (D)

$y=\frac{t-\frac{1}{t}}{1+\frac{1}{t}}$ where $t=e^{2x}$
$\therefore y(t^2+1)=t^2-1$
$\Rightarrow t^2=\frac{1+y}{1-y}$
$\Rightarrow (e^{2x}{)}^2=\frac{1+y}{1-y}(\because t=e^{2x})$
$\Rightarrow e^{4x}=\frac{1+y}{1-y}$
$\Rightarrow 4xlo{g}_{e}e=lo{g}_e(\frac{1+y}{1-y})$
$\Rightarrow x=\frac{1}{4}lo{g}_e(\frac{1+y}{1-y})$
$\therefore f^{-1}(y)=\frac{1}{4}lo{g}_e(\frac{1+y}{1-y})$
$\therefore f^{-1}(x)=\frac{1}{4}lo{g}_e(\frac{1+x}{1-x})=p(x)$(say)
$\therefore p(-x)=-p(x)$
$\therefore f^{-1}(-x)=-f^{-1}(x)$
which is an odd function.