A) \[{{\log }_{10}}\left[ \frac{x}{x-2} \right]\]
B) \[{{\log }_{e}}\left[ \frac{x}{x-2} \right]\]
C) \[\frac{1}{2}{{\log }_{e}}\left[ \frac{y}{2-y} \right]\]
D) None of these
Correct Answer: C
Solution :
| Given function, \[f(x)=\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}+1\] |
| Let \[f(x)=y\] |
| \[\therefore \,\,\,y=\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}+1\Rightarrow y-1=\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}\] |
| \[\Rightarrow \,\,y-1=\frac{{{e}^{x}}-\frac{1}{{{e}^{x}}}}{{{e}^{x}}+\frac{1}{{{e}^{x}}}}=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}\]\[\Rightarrow \,\,\,\frac{y-1}{1}=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}\]\[\Rightarrow \,\,\,\,\frac{y-1+1}{y-1-1}=\frac{{{e}^{2x}}-1+{{e}^{2x}}+1}{{{e}^{2x}}-1-{{e}^{2x}}-1}\] |
| \[\Rightarrow \,\,\,\frac{y}{y-2}=\frac{2{{e}^{2x}}}{-2}\]\[\Rightarrow \,\,\,\frac{y}{y-2}=-{{e}^{2x}}\]\[\Rightarrow \,\,\,\,\frac{y}{2-y}={{e}^{2x}}\Rightarrow \,\,2x={{\log }_{e}}\left( \frac{y}{2-y} \right)\] |
| \[\therefore \,\,\,\,x=\frac{1}{2}{{\log }_{e}}\left( \frac{y}{2-y} \right)\] |
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