Question

If $f\left(x\right)={\log }_e\left(\frac{1-x}{1+x}\right),\left|x\right|<1,$ then $f\left(\frac{2x}{1+x^2}\right)$ is equal to :

• A. $2f(x)$
• B. $2f(x^2)$
• C. $(f(x){)}^2$
• D. $-2f(x)$

Answer 1

Answer: (A)

$f\left(x\right)={\log }_e\left(\frac{1-x}{1+x}\right),\left|x\right|<1$
$f\left(\frac{2x}{1+x^2}\right)=\ell n\left(\frac{1-\frac{2x}{1+2x^2}}{1+\frac{2x}{1+x^2}}\right)$
$=\ell n\left(\frac{{\left(x-1\right)}^2}{{\left(x+1\right)}^2}\right)=2\ell n\left|\frac{1-x}{1+x}\right|=2f\left(x\right)$