Question

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

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

Answer 1

The correct option is C $2f\left(x\right)$

As, $f\left(x\right)={\log }_e\left(\frac{1-x}{1+x}\right)$
$\therefore f\left(\frac{2x}{1+x^2}\right)={\log }_e\left(\frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}\right)={\log }_e\left(\frac{1+x^2-2x}{1+x^2+2x}\right)=\ln {\left(\frac{1-x}{1+x}\right)}^2=2f\left(x\right)$