Question

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

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

Answer 1

Answer: (B)

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