A) even
B) \[f({{x}_{1}})f({{x}_{2}})=f({{x}_{1}}+{{x}_{2}})\]
C) \[\frac{f({{x}_{1}})}{f({{x}_{2}})}=f({{x}_{1}}-{{x}_{2}})\]
D) odd
E) neither even nor odd
Correct Answer: D
Solution :
\[\because \]\[f(-x)=\log \left( \frac{1-x}{1+x} \right)\] \[=-\log \left( \frac{1+x}{1-x} \right)\] \[\Rightarrow \]\[f(-x)=-f(x)\] \[\therefore \]Given function is an odd function.You need to login to perform this action.
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