Question

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

• A. $f\left(x\right)$
• B. $2f\left(x\right)$
• C. $3f\left(x\right)$
• D. $4f\left(x\right)$

Answer 1

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

Given:

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

Substitute $x$ as $\frac{2x}{1+x^2}$, in the above function, we get

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

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

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

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

Using $\log x^m=m\cdot \log x$, we get

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

Hence, option B.