Question

If $f(x)=\log \left(\frac{1+x}{1-x}\right),-1<x<1,$ then $f\left(\frac{3x+x^3}{1+3x^2}\right)-f\left(\frac{2x}{1+x^2}\right)$ is

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

Answer 1

Answer: (D)

Given, $f(x)=\log \left(\frac{1+x}{1-x}\right)$
$\therefore$ $f\left(\frac{3x+x^3}{1+3x^2}\right)-f\left(\frac{2x}{1+x^2}\right)$
$=\log \left(\frac{1+\left(\frac{3x+x^3}{1+3x2}\right)}{1-\left(\frac{3x+x^3}{1+3x^2}\right)}\right)-\log \left(\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1+x^2}}\right)$
$=\log {\left(\frac{1+x}{1-x}\right)}^3-\log {\left(\frac{1+x}{1-x}\right)}^2$
$=\log \left(\frac{1+x}{1-x}\right)$
$=f(x)$