Question

The inverse of the function $f:R\to \left\{x\in R:x<1\right\}$ 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) none of these

Answer 1

$$\begin{gathered} \text{Let}{f}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow f\left(y\right)=x \\ \Rightarrow \frac{e^y-e^{-y}}{e^y+e^{-y}}=x \\ \Rightarrow \frac{e^{-y}\left(e^{2y}-1\right)}{e^{-y}\left(e^{2y}+1\right)}=x \\ \Rightarrow \left(e^{2y}-1\right)=x\left(e^{2y}+1\right) \\ \Rightarrow e^{2y}-1=x{e}^{2y}+x \\ \Rightarrow e^{2y}\left(1-x\right)=x+1 \\ \Rightarrow e^{2y}=\frac{1+x}{1-x} \\ \Rightarrow 2y={\log }_e\left(\frac{1+x}{1-x}\right) \\ \Rightarrow y=\frac{1}{2}l{\mathrm{og}}_e\left(\frac{1+x}{1-x}\right) \\ \Rightarrow f^{-1}\left(x\right)=\frac{1}{2}l{\mathrm{og}}_e\left(\frac{1+x}{1-x}\right)\text{[from}\left(1\right)\text{]} \end{gathered}$$
So, the answer is (a).