A) \[{{\log }_{e}}{{\left( \frac{x-2}{x-1} \right)}^{1/2}}\]
B) \[{{\log }_{e}}{{\left( \frac{x-1}{3-x} \right)}^{1/2}}\]
C) \[{{\log }_{e}}{{\left( \frac{x}{2-x} \right)}^{1/2}}\]
D) \[{{\log }_{e}}{{\left( \frac{x-1}{x+1} \right)}^{-2}}\]
Correct Answer: B
Solution :
\[y=\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}+2\,\,\Rightarrow \,\,y=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}+2\,\,\] \[\Rightarrow \,\,{{e}^{2x}}=\frac{1-y}{y-3}=\frac{y-1}{3-y}\,\,\Rightarrow \,\,x=\frac{1}{2}{{\log }_{e}}\,\left( \frac{y-1}{3-y} \right)\] \[\Rightarrow {{f}^{-1}}(y)={{\log }_{e}}\,{{\left( \frac{y-1}{3-y} \right)}^{1/2}}\Rightarrow {{f}^{-1}}(x)={{\log }_{e}}{{\left( \frac{x-1}{3-x} \right)}^{1/2}}\].
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