A) \[2f(x)\]
B) \[2f({{x}^{2}})\]
C) \[{{(f(x))}^{2}}\]
D) \[-2f(x)\]
Correct Answer: A
Solution :
\[f(x)=lo{{g}_{e}}\left( \frac{1-x}{1+x} \right),|x|<1\] \[f\left( \frac{2x}{1+{{x}^{2}}} \right)=\ell n\left( \frac{1-\frac{2x}{1+2{{x}^{2}}}}{1+\frac{2x}{1+{{x}^{2}}}} \right)\] \[=\ell n\left( \frac{{{(x-1)}^{2}}}{{{(x+1)}^{2}}} \right)=2\ell n\left| \frac{1-x}{1+x} \right|=2f(x)\]You need to login to perform this action.
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