Question:
If $f(x)=\log _{e}\left(\frac{1-x}{1+x}\right),|x|<1$, then $f\left(\frac{2 x}{1+x^{2}}\right)$ is equal to:
Correct Option: 1
Solution:
$f(x)=\log \left(\frac{1-x}{1+x}\right),|x|<1$
$f\left(\frac{2 x}{1+x^{2}}\right)=\log \left(\frac{1-\frac{2 x}{1+x^{2}}}{1+\frac{2 x}{1+x^{2}}}\right)$
$=\log \left(\frac{1+x^{2}-2 x}{1+x^{2}+2 x}\right)$
$=\log \left(\frac{1-x}{1+x}\right)^{2}$
$=2 \log \left(\frac{1-x}{1+x}\right)$
$=2 f(x)$
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