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 

 

पर्याय

• \[\frac{1}{2} \log \frac{1 + x}{1 - x}\]

•  \[\frac{1}{2} \log \frac{2 + x}{2 - x}\]

• \[\frac{1}{2} \log \frac{1 - x}{1 + x}\]

• none of these

Answer 1

\[\text{Let} f^{- 1} \left( x \right) = y . . . \left( 1 \right)\] 
\[ \Rightarrow f\left( y \right) = x\] 
\[ \Rightarrow \frac{e^y - e^{- y}}{e^y + e^{- y}} = x\] 
\[ \Rightarrow \frac{e^{- y} \left( e^{2y} - 1 \right)}{e^{- y} \left( e^{2y} + 1 \right)} = x\] 
\[ \Rightarrow \left( e^{2y} - 1 \right) = x\left( e^{2y} + 1 \right)\] 
\[ \Rightarrow e^{2y} - 1 = x e^{2y} + x\] 
\[ \Rightarrow e^{2y} \left( 1 - x \right) = x + 1\] 
\[ \Rightarrow e^{2y} = \frac{1 + x}{1 - x}\] 
\[ \Rightarrow 2y = \log_e \left( \frac{1 + x}{1 - x} \right)\] 
\[ \Rightarrow y = \frac{1}{2}l {og}_e \left( \frac{1 + x}{1 - x} \right)\] 
\[ \Rightarrow f^{- 1} \left( x \right) = \frac{1}{2}l {og}_e \left( \frac{1 + x}{1 - x} \right) [\text{from}\left( 1 \right)]\] 

So, the answer is (a).