The inverse of the function f:R x :x<1 given by f(x)= e^ x -e^ -x e^x+e^ -x is:

The inverse of the function f:R x :x<1 given by f(x)= e^ x -e^ -x…

MCQ

The inverse of the function $\text{f}:\text{R}\rightarrow\{\text{x}\in\text{R}:\text{x}<1\}$ given by $\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$ is:

✓
$\frac{1}{2}\log\frac{1+\text{x}}{1-\text{x}}$
B
$\frac{1}{2}\log\frac{2+\text{x}}{2-\text{x}}$
C
$\frac{1}{2}\log\frac{1-\text{x}}{1+\text{x}}$
D
None of these

✓

Answer

Correct option: A.

$\frac{1}{2}\log\frac{1+\text{x}}{1-\text{x}}$

Let $f^{-1}(x) = y .....(1)$
$\Rightarrow \text{f(y)}=\text{x}$
$\Rightarrow \frac{\text{e}^{\text{y}}-\text{e}^{-\text{y}}}{\text{e}^{\text{y}}+\text{e}^{-\text{y}}}=\text{x}$
$\Rightarrow \frac{\text{e}^{-\text{y}}(\text{e}^{2\text{y}}-1)}{\text{e}^{-\text{y}}(\text{e}^{2\text{y}}+1)}=\text{x}$
$\Rightarrow (\text{e}^{2\text{y}}-1)=\text{x}(\text{e}^{2\text{y}}+1)$
$\Rightarrow \text{e}^{2\text{y}}-1=\text{xe}^{2\text{y}}+\text{x}$
$\Rightarrow \text{e}^{2\text{y}}=\frac{1+\text{x}}{1-\text{x}}$
$\Rightarrow 2\text{y}=\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
$\Rightarrow \text{y}=\frac{1}{2}\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
$\Rightarrow \text{f}^{-1}(\text{x})=\frac{1}{2}\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big) [$From $(1)]$