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

Answer 1

The correct option is B

$2f\left(x\right)$

Explanation for correct option:

Find the value of $f\left(\frac{2x}{1+x^2}\right)$:

Given,

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

Now,

$\begin{array}{rcl}f\left(\frac{2x}{1+x²}\right)& =& \log \left[\frac{1-\left({\displaystyle \frac{2x}{1+x^2}}\right)}{1+\left({\displaystyle \frac{2x}{1+x^2}}\right)}\right]\\ & =& \log \left(\frac{1+x^2-2x}{1+x^2+2x}\right)\\ & =& \log {\left(\frac{1-x}{1+x}\right)}^2\\ & =& 2\log \left(\frac{1-x}{1+x}\right)\left[log{a}^m=mloga\right]\\ & =& 2f\left(x\right)\end{array}$

Hence, the correct option is B.