If \[f({{x}_{1}})-f({{x}_{2}})=f\left( \frac{{{x}_{1}}-{{x}_{2}}}{1-{{x}_{1}}{{x}_{2}}} \right)\] for \[{{x}_{1}},{{x}_{2}}\in [-1,\,1]\], then \[f(x)\] is [Roorkee 1998]

\[{{\tan }^{-1}}\frac{(1-x)}{(1+x)}\]

\[{{\tan }^{-1}}\frac{(1+x)}{(1-x)}\]

If \[f({{x}_{1}})-f({{x}_{2}})=f\left( \frac{{{x}_{1}}-{{x}_{2}}}{1-{{

1.	If \[f({{x}_{1}})-f({{x}_{2}})=f\left( \frac{{{x}_{1}}-{{x}_{2}}}{1-{{x}_{1}}{{x}_{2}}} \right)\] for \[{{x}_{1}},{{x}_{2}}\in [-1,\,1]\], then \[f(x)\] is [Roorkee 1998]
A.	\[\log \frac{(1-x)}{(1+x)}\]
B.	\[{{\tan }^{-1}}\frac{(1-x)}{(1+x)}\]
C.	\[\log \frac{(1+x)}{(1-x)}\]
D.	\[{{\tan }^{-1}}\frac{(1+x)}{(1-x)}\]
Answer» C. \[\log \frac{(1+x)}{(1-x)}\]

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If \[f({{x}_{1}})-f({{x}_{2}})=f\left( \frac{{{x}_{1}}-{{x}_{2}}}{1-{{x}_{1}}{{x}_{2}}} \right)\] for \[{{x}_{1}},{{x}_{2}}\in [-1,\,1]\], then \[f(x)\] is [Roorkee 1998]

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