Question

If $f\left(x\right)=\log \left(\frac{1+x}{1-x}\right)$, then the value of $f\left(\frac{2x}{1+x^2}\right)$ is

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

Answer 1

Answer: (A)

$2f\left(x\right)$

Given that,

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

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

$=\log \left\{\frac{1+x^2+2x}{1+x^2-2x}\right\}$

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

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

$=2\log \left\{\frac{1+x}{1-x}\right\}$

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