A) \[\frac{1}{2}{{\log }_{a}}\left( \frac{1+x}{1-x} \right)\]
B) \[\frac{1}{2}{{\log }_{a}}\left( \frac{1-x}{1+x} \right)\]
C) \[\frac{1}{2}{{\log }_{e}}\left( \frac{1+x}{1-x} \right)\]
D) \[\frac{1}{2}{{\log }_{e}}\left( \frac{1-x}{1+x} \right)\]
Correct Answer: A
Solution :
[a] \[\because y=f(x)=\frac{{{a}^{x}}-{{a}^{-x}}}{{{a}^{x}}+{{a}^{-x}}}=\frac{{{a}^{x}}-\frac{1}{{{a}^{x}}}}{{{a}^{x}}+\frac{1}{{{a}^{x}}}}\,\Rightarrow \frac{y}{1}=\frac{{{a}^{2x}}-1}{{{a}^{2x}}+1}\] \[\Rightarrow \frac{y+1}{y-1}=\frac{{{a}^{2x}}-1+{{a}^{2x}}+1}{{{a}^{2x}}-1-{{a}^{2x}}-1}\] [by componendo & dividendo] \[\Rightarrow \frac{y+1}{y-1}=\frac{2{{a}^{2x}}}{-2}=-{{a}^{2x}}\Rightarrow {{a}^{2X}}=\frac{1+y}{1-y}\] Taking log on base a on both sides, we have \[\Rightarrow 2x.lo{{g}_{a}}a=lo{{g}_{a}}\left( \frac{1+y}{1-y} \right)\,\Rightarrow {{a}^{2x}}=\log a\left( \frac{1+y}{1-y} \right)\]\[\Rightarrow x=\frac{1}{2}.\log a\left( \frac{1+y}{1-y} \right)\] \[\therefore {{f}^{-1}}(x)=\frac{1}{2}.\log a\left( \frac{1+x}{1-x} \right)\] Hence, option [a] is correct.
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