A) \[{{\tan }^{-1}}\frac{(1+x)}{(1-x)}\]
B) \[{{\tan }^{-1}}\frac{(1-x)}{(1+x)}\]
C) \[\log \frac{(1+x)}{(1-x)}\]
D) none of these
Correct Answer: B
Solution :
We have.\[f({{x}_{1}})-f({{x}_{2}})=f\left( \frac{{{x}_{1}}-{{x}_{2}}}{1-{{x}_{1}}{{x}_{2}}} \right)\] Since, \[{{x}_{1}},{{x}_{2}}\in [-1,1].\] Let \[{{x}_{1}}=-1\]and \[{{x}_{2}}=1,\]then \[f(-1)-f(1)=f\left[ \frac{-1-1}{1+1} \right]\] \[=f(-1)\] \[\Rightarrow \] \[f(1)=0\] Now take, \[f(x)={{\tan }^{-1}}\left( \frac{1-x}{1+x} \right)\] \[f(1)={{\tan }^{-1}}\left( \frac{1-1}{1+1} \right)={{\tan }^{-1}}0\] = 0 Which is satisfied.
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