A) \[f(x)\]
B) \[2f(x)\]
C) \[-2f(x)\]
D) \[{{(f(x))}^{2}}\]
Correct Answer: B
Solution :
\[f(x)=\ell n\left( \frac{1-x}{1+x} \right)\] |
\[f\left( \frac{2x}{1+{{x}^{2}}} \right)=\ell n\left( \frac{1-\frac{2x}{1+{{x}^{2}}}}{1+\frac{2x}{1+{{x}^{2}}}} \right)=\ell n\left( \frac{1+{{x}^{2}}-2x}{1+{{x}^{2}}+2x} \right)\] |
\[=\ell n\left( {{\left( \frac{1-x}{1+x} \right)}^{2}} \right)=2\ell n\left( \frac{1-x}{1+x} \right)=2f(x).\] |
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