A) \[{{\log }_{10}}(2-x)\]
B) \[\frac{1}{2}{{\log }_{10}}\left( \frac{1+x}{1-x} \right)\]
C) \[\frac{1}{2}{{\log }_{10}}(2x-1)\]
D) \[\frac{1}{4}{{\log }_{10}}\left( \frac{2x}{2-x} \right)\]
Correct Answer: B
Solution :
\[f(x)=\frac{{{10}^{x}}-{{10}^{-x}}}{{{10}^{x}}+{{10}^{-x}}}=y\](suppose) \[\therefore \] \[x={{f}^{-1}}(y)\] \[\Rightarrow \] \[\frac{{{10}^{x}}-\frac{1}{{{10}^{x}}}}{{{10}^{x}}+\frac{1}{{{10}^{x}}}}=y\] \[\Rightarrow \] \[{{10}^{2x}}-1=y({{10}^{2x}}+1)\] \[\Rightarrow \] \[{{10}^{2x}}-y{{10}^{2x}}=y+1\] \[\Rightarrow \] \[{{10}^{2x}}=\frac{y+1}{1-y}\] \[\Rightarrow \] \[{{10}^{x}}=\sqrt{\frac{y+1}{1-y}}\] \[\Rightarrow \] \[{{\log }_{10}}{{10}^{x}}={{\log }_{10}}{{\left( \frac{y+1}{1-y} \right)}^{1/2}}\] \[\Rightarrow \] \[x=\frac{1}{2}{{\log }_{10}}\left( \frac{1+y}{1-y} \right)\] \[\Rightarrow \] \[{{f}^{-1}}(y)=\frac{1}{2}{{\log }_{10}}\left( \frac{1+y}{1-y} \right)\] \[\Rightarrow \] \[{{f}^{-1}}(x)=\frac{1}{2}{{\log }_{10}}\left( \frac{1+y}{1-y} \right)\]
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