Question

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

• A. $\frac{1}{2}\log \frac{1+x}{1-x}$
• B. $\frac{1}{2}\log \frac{2+x}{2-x}$
• C. $\frac{1}{2}\log \frac{1-x}{1+x}$
• D. $noneofthese$

Answer 1

Answer: (A)

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

Given $f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$

So $y=\frac{e^x-e^{-x}}{e^x+e^{-x}}$

$⟹y=\frac{e^{2x}-1}{e^{2x}+1}$

$⟹y(e^{2x}+1)=e^{2x}-1$

$⟹e^{2x}(y-1)=-y-1$

$⟹e^{2x}=\frac{1+y}{1-y}$

$⟹2x=\log \frac{1+y}{1-y}$

$⟹x=\frac{1}{2}\log \frac{1+y}{1-y}$

$f^{-1}\left(x\right)=\frac{1}{2}\log \frac{1+x}{1-x}$