The inverse of the function $f: R \rightarrow\{x \in R: x<1\}$ given by $f(x)=\frac{e^{z}-e^{-x}}{e^{x}+e^{-x}}$ 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
Let $f^{-1}(x)=y \quad \ldots(1)$
$\Rightarrow f(y)=x$
$\Rightarrow \frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}=x$
$\Rightarrow \frac{e^{-y}\left(e^{2 y}-1\right)}{e^{-y}\left(e^{2 y}+1\right)}=x$
$\Rightarrow\left(e^{2 y}-1\right)=x\left(e^{2 y}+1\right)$
$\Rightarrow e^{2 y}-1=x e^{2 y}+x$
$\Rightarrow e^{2 y}(1-x)=x+1$
$\Rightarrow e^{2 y}=\frac{1+x}{1-x}$
$\Rightarrow 2 y=\log _{e}\left(\frac{1+x}{1-x}\right)$
$\Rightarrow y=\frac{1}{2} l \operatorname{og}_{e}\left(\frac{1+x}{1-x}\right)$
$\Rightarrow f^{-1}(x)=\frac{1}{2} \log _{e}\left(\frac{1+x}{1-x}\right)$ [from (1)]
So, the answer is (a).
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.