A) \[\frac{1}{2}{{\log }_{10}}\left( \frac{1+x}{1-x} \right)\]
B) \[\frac{1}{2}{{\log }_{10}}\left( \frac{1-x}{1+x} \right)\]
C) \[\frac{1}{4}{{\log }_{10}}\left( \frac{2x}{2-x} \right)\]
D) None of these
Correct Answer: A
Solution :
\[y=\frac{{{10}^{x}}-{{10}^{-x}}}{{{10}^{x}}+{{10}^{-x}}}\Rightarrow x=\frac{1}{2}{{\log }_{10}}\left( \frac{1+y}{1-y} \right)\] Let \[y=f(x)\] Þ \[x=\pi ,\,\,f(\pi )=-\tan \frac{\pi }{4}=-1\] Þ \[{{f}^{-1}}(y)=\frac{1}{2}{{\log }_{10}}\left( \frac{1+y}{1-y} \right)\] Þ \[g({{\sin }^{2}}x)=\,|\sin x|\].
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