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

Answer 1

Answer: (C)

$2f\left(x\right)$

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

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

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

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

$=2\left(f\left(x\right)\right)$