Question

Given $f\left(x\right)=\log \left(\frac{1+x}{1-x}\right)$ and $g\left(x\right)=\frac{3x+x^3}{1+3x^2},fog\left(x\right)$ equals

• A. $-f\left(x\right)$
• B. $3f\left(x\right)$
• C. ${\left[f\left(x\right)\right]}^3$
• D. none of these

Answer 1

Answer: (B)

$3f\left(x\right)$

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

$\therefore fog\left(x\right)=\log \left(\frac{1+\frac{3x+x^3}{1+3x^2}}{1-\frac{3x+x^3}{1+3x^2}}\right)$

$=\log \left(\frac{(1+x{)}^3}{(1-x{)}^3}\right)=3\log \left(\frac{1+x}{1-x}\right)=3f\left(x\right)$