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 :
$2 f(\mathrm{x})$
$2 f\left(x^{2}\right)$
$(f(x))^{2}$
$-2 f(x)$
Correct Option: 1
Leave a comment
All Study Material
