Question

If $f\left(x\right)=\ln \frac{1+x}{1-x}$ and $g\left(x\right)=\frac{3x+x^3}{1+3x^2}$, then $f\left[g\right(x\left)\right]$ equals.

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

Answer 1

Answer: (C)

$3f\left(x\right)$

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

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

$=\ln \frac{1+3x^2+3x+x^3}{1+3x^2-3x-x^3}$

$=\ln {\left[\frac{1+x}{1-x}\right]}^3$

$=3\ln \frac{1+x}{1-x}$

$\Rightarrow f\left[g\right(x\left)\right]=3f\left(x\right)$

Hence, option C.