The inverse of the function f : R→{x ∈ R : x < 1} given by

Question:

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

Solution:

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).

 

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