Question

If $f\left(x\right)=\ln \left(\frac{1+x}{1-x}\right)$, then $f\left(\frac{2x}{1+x^2}\right)$ equals.

• A. $\left[f\right(x){]}^2$
• B. ${f\left(x\right)}^3$
• C. $2f\left(x\right)$
• D. $3f\left(x\right)$

Answer 1

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

$\left(x\right)=\log \left(\frac{1+x}{1-x}\right)$

$\therefore f\left(\frac{1+x}{1-x}\right)=\log \left[\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1-x^2}}\right]=\log \left[\frac{x^2+1+2x}{x^2+1-2x}\right]$

$=\log {\left[\frac{1+x}{1-x}\right]}^2=2\log \left[\frac{1+x}{1-x}\right]=2f\left(x\right)$