A) \[\log \frac{(1-x)}{(1+x)}\]
B) \[{{\tan }^{-1}}\frac{(1-x)}{(1+x)}\]
C) \[\log \frac{(1+x)}{(1-x)}\]
D) \[{{\tan }^{-1}}\frac{(1+x)}{(1-x)}\]
Correct Answer: B
Solution :
When \[{{x}_{1}}=-1\] and \[{{x}_{2}}=1\] , then \[f(-1)-f(1)=f\left[ \frac{-1-1}{1+1(1)} \right]=f(-1)\Rightarrow f(1)=0\] Which is satisfied when \[f(x)={{\tan }^{-1}}\left( \frac{1-x}{1+x} \right)\] When \[{{x}_{1}}={{x}_{2}}=0\], then \[f(0)-f(0)=f\left[ \frac{0-0}{1-0} \right]=f(0)\Rightarrow f(0)=0\] When \[{{x}_{1}}=-1\] and \[{{x}_{2}}=0\] then \[f(-1)-f(0)=f\left( \frac{-1-0}{1-0} \right)=f(-1)\Rightarrow f(0)=0\] Which is satisfied when \[f(x)=\log \left( \frac{1-x}{1+x} \right)\]and \[f(x)=\log \left( \frac{1+x}{1-x} \right)\].
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