Question

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

• A. $\sqrt{\frac{x^{1/3}+b}{a}}$
• B. $\sqrt{\frac{x^3-b}{a}}$
• C. $\sqrt{\frac{x^{1/3}-b}{a}}$
• D. $\sqrt{\frac{x^3+b}{a}}$

Answer 1

The correct option is C $\sqrt{\frac{x^{1/3}-b}{a}}$

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

$f\left(g\right(x\left)\right)=g\left(f\right(x\left)\right)=x$ is possible only when $g$ is the inverse of $f$ and vice-versa.
$f\left(x\right)=(a{x}^2+b{)}^3$
If $g\left(x\right)=f^{-1}\left(x\right)$
Let $y=f\left(x\right)=(a{x}^2+b{)}^3$
$\Rightarrow \pm \sqrt{\frac{y^{1/3}-b}{a}}=x$
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 ${}^{\prime }{+}^{\prime }$sign
$\Rightarrow g\left(x\right)=\sqrt{\frac{x^{1/3}-b}{a}}$