Question

If $f(x)=(ax^2+b)^3 ,b\in \mathbb R, a\in \mathbb R -\{0\}$ and $g(x)$ is a function such that $f(g(x))=g(f(x))=x,$ then $g(x)=$
(Given that $f$ and $g$ are bijective functions)

Answer 1

\(\because\) It is given \(f(x)\) is bijective function, which is possible iff domain and range are restricted.

$f(g(x))=g(f(x))=x$ is possible only when $g $ is the inverse of $f$ and vice-versa.
$f(x)=(ax^2+b)^3$
If $g(x)=f^{-1}(x)$
Let $y=f(x)=(ax^2+b)^3$
$ \Rightarrow \pm\sqrt{\dfrac{y^{1/3}-b}{a}}=x$
Clearly taking positive or negative sign will give us the inverse function. But we have to select exactly one of them, because we know that inverse of a function is unique. Based on options we have to take $'+'$sign
$ \Rightarrow g(x)=\sqrt{\dfrac{x^{1/3}-b}{a}}$