, and 
, then (fog)(x) is equal to
A) 3f(x)
B)
C) 2f(x)
D)
Correct Answer: A
Solution :
\[f(x)=lo{{g}_{e}}\left( \frac{1+x}{1-x} \right)\] \[\therefore \,\,\,(fog)\,(x)=lo{{g}_{e}}\left[ \frac{1+\frac{3x+{{x}^{3}}}{1+3{{x}^{2}}}}{1-\frac{3x+{{x}^{3}}}{1+3{{x}^{2}}}} \right]\] \[=\,\,{{\log }_{e}}=\left[ \frac{1+3{{x}^{2}}+3x+{{x}^{3}}}{1+3{{x}^{2}}-3x-{{x}^{3}}} \right]={{\log }_{e}}\,{{\left[ \frac{1+x}{1-x} \right]}^{3}}\], \[=\,\,3\,{{\log }_{e}}\left[ \frac{1+x}{1-x} \right]=3\,f\,(x)\]
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