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)

$y = \frac{ t - \frac{1}{t}}{1 + \frac{1}{t}}$ 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} $
$\Rightarrow 4x\, log_e e = log_e (\frac{1 + y}{1 - y})$
$\Rightarrow x = \frac{1}{4} log_e(\frac{1 + y}{ 1 -y})$
$\therefore f^{-1}(y) = \frac{1}{4} log_e (\frac{1 + y}{1 - y})$
$\therefore f^{-1} (x) = \frac{1}{4} log_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.