Question

If $f\left(x\right)=\log \left(\frac{1+x}{1-x}\right)\mathrm{and}g\left(x\right)=\frac{3x+x^3}{1+3x^2}$, then f(g(x)) is equal to
(a) f(3x)
(b) {f(x)}³
(c) 3f(x)
(d) −f(x)

Answer 1

(c) 3f(x)

$f\left(x\right)=\log \left(\frac{1+x}{1-x}\right)\mathrm{and}g\left(x\right)=\frac{3x+x^3}{1+3x^2}$
$$\begin{gathered} \text{Now}, \\ \frac{1+g\left(x\right)}{1-g\left(x\right)}=\frac{1+{\displaystyle \frac{3x+x^3}{1+3x^2}}}{1-\frac{3x+x^3}{1+3x^2}} \\ =\frac{1+3x^2+3x+x^3}{1+3x^2-3x-x^3} \\ =\frac{(1+x{)}^3}{(1-x{)}^3} \\ \mathrm{Then},f\left(g\right(x\left)\right)=\log \left(\frac{1+g\left(x\right)}{1-g\left(x\right)}\right) \\ =\log {\left(\frac{1+x}{1-x}\right)}^3 \\ =3\log \left(\frac{1+x}{1-x}\right) \\ =3f\left(x\right)) \end{gathered}$$