Question

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

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

Answer 1

Answer: (C)

We have, $f(x)={\log }_e\left(\frac{1+x}{1-x}\right)$
$f\left(\frac{2x}{1+x^2}\right)={\log }_e\left(\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1+x^2}}\right)$
$={\log }_e\left(\frac{1+x^2+2x}{1+x^2-2x}\right)$
$={\log }_e{\left(\frac{1+x}{1-x}\right)}^2$
$=2{\log }_e\left(\frac{1+x}{1-x}\right)$
$\left(\because \log a^b=b\log a\right)$
$=2f(x)$