Question

If $f(x)=(a{x}^2+b{)}^3$, then the function $g$ such that $f(g(x))=g(f(x))$ is given by

• A. $g\left(x\right)=\left(\frac{b-x^{1/3}}{a}\right)$
• B. $g\left(x\right)=\frac{1}{{\left(a{x}^2+b\right)}^3}$
• C. $g(x)=(a{x}^2+b{)}^{1/3}$
• D. $g\left(x\right)={\left(\frac{x^{1/3}-b}{a}\right)}^{1/2}$

Answer 1

Answer: (D)

$y=f\left(x\right)={\left(a{x}^2+b\right)}^3$
$\Rightarrow a{x}^2+b=y^{1/3}$
$\Rightarrow x^2=\frac{y^{1/3}-b}{a}$
$\Rightarrow x={\left(\frac{y^{1/3}-b}{a}\right)}^{1/2}$
$=f^{-1}\left(y\right)$
Since, $f(g(x))=g(f(x))$
$\therefore$ We have, $g\left(x\right)={\left(\frac{x^{1/3}-b}{a}\right)}^{1/2}$.