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 :
Let\[y=f(x)\] \[\Rightarrow \] \[y=\frac{{{10}^{x}}-{{10}^{x}}}{{{10}^{x}}+{{10}^{-x}}}\] \[\Rightarrow \] \[y({{10}^{x}}+{{10}^{-x}})={{10}^{x}}-{{10}^{-x}}\] \[\Rightarrow \] \[y{{.10}^{2x}}-{{10}^{+2x}}=-1-y\] \[\Rightarrow \] \[{{10}^{2x}}(1-y)=1+y\] \[\Rightarrow \] \[{{10}^{2x}}=\frac{1+y}{1-y}\] \[\Rightarrow \] \[2x={{\log }_{10}}\frac{1+y}{1-y}\] \[\Rightarrow \] \[x=\frac{1}{2}{{\log }_{10}}\left( \frac{1+y}{1-y} \right)\] \[\Rightarrow \] \[{{f}^{-1}}(y)=\frac{1}{2}{{\log }_{10}}\left( \frac{1+y}{1-y} \right)\] \[[\because y=f(x)\Rightarrow {{f}^{-1}}(y)=x]\] \[\Rightarrow \] \[{{f}^{-1}}(x)=\frac{1}{2}{{\log }_{10}}\left( \frac{1+x}{1-x} \right)\]
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