Question

If $(x)=\log \left( \frac{1+x}{1-x} \right),-1$.

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

Answer 1

Answer: (D)

$f(x)$

Answer 2

The correct option is(D): f(x).

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