A) \[2f({{x}^{2}})\]
B) \[f({{x}^{2}})\]
C) \[2f(2x)\]
D) \[2f(x)\]
Correct Answer: D
Solution :
Given that,\[f(x)=\log \left( \frac{1+x}{1-x} \right)\] \[\therefore \]\[f\left( \frac{2x}{1+{{x}^{2}}} \right)=\log \left( \frac{1+\frac{2x}{1+{{x}^{2}}}}{1-\frac{2x}{1+{{x}^{2}}}} \right)\] \[=\log {{\left( \frac{1+x}{1-x} \right)}^{2}}\] \[=2\log \left( \frac{1+x}{1-x} \right)=2f(x)\]You need to login to perform this action.
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