then what is \[f(x)\] equal to?
A) \[\ln \left( \frac{1-x}{1+x} \right)\]
B) \[\ln \left( \frac{2+x}{1-x} \right)\]
C) \[{{\tan }^{-1}}\left( \frac{1-x}{1+x} \right)\]
D) \[{{\tan }^{-1}}\left( \frac{1+x}{1-x} \right)\]
Correct Answer: A
Solution :
\[f({{x}_{1}})-f({{x}_{2}})=f\left( \frac{{{x}_{1}}-{{x}_{2}}}{1-{{x}_{1}}{{x}_{2}}} \right)\] \[{{x}_{1}},{{x}_{2}}\in (-1,1)\] then \[f(x)=\log \frac{(1-x)}{(1+x)}\] \[f({{x}_{1}})=\log \frac{1-{{x}_{1}}}{1+{{x}_{1}}}\] \[f({{x}_{2}})=\log \frac{1-{{x}_{2}}}{1+{{x}_{2}}}\] \[f({{x}_{1}})-f({{x}_{2}})=\log \frac{1-{{x}_{1}}}{1+{{x}_{1}}}-\log \frac{1-{{x}_{2}}}{1+{{x}_{2}}}\] \[=\log \,\frac{(1-{{x}_{1}})}{(1+{{x}_{1}})}\times \frac{(1+{{x}_{2}})}{(1-{{x}_{2}})}\] \[=\log \,\frac{(1-{{x}_{1}}+{{x}_{2}}-{{x}_{1}}{{x}_{2}})}{(1+{{x}_{1}}-{{x}_{2}}-{{x}_{1}}{{x}_{2}})}\] \[=\log \,\frac{(1-{{x}_{1}}{{x}_{2}})-({{x}_{1}}-{{x}_{2}})}{(1-{{x}_{1}}{{x}_{2}})+({{x}_{1}}-{{x}_{2}})}\] \[=\log \frac{1-\left( \frac{{{x}_{1}}-{{x}_{2}}}{1-{{x}_{1}}{{x}_{2}}} \right)}{1+\left( \frac{{{x}_{1}}-{{x}_{2}}}{1-{{x}_{1}}{{x}_{2}}} \right)}\] \[f({{x}_{1}})-f({{x}_{2}})=f\left( \frac{{{x}_{1}}-{{x}_{2}}}{1-{{x}_{1}}{{x}_{2}}} \right)\]
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