Question

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

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

Answer 1

The correct option is C

$\left\{2f\right(x\left)\right\}$

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