Question

Let $f:R\to R$ and $g:R\to R$ be two functions given by $f\left(x\right)=2x-3$ and $g\left(x\right)=x^3+5$. Then, $\left(fog{)}^{-1}\right(x)$ is equal to

• A. ${\left(\frac{x+7}{2}\right)}^{1/3}$
• B. ${\left[\frac{x-7}{2}\right]}^{1/3}$
• C. ${\left(\frac{x-7}{7}\right)}^{1/3}$
• D. None of these

Answer 1

The correct option is B ${\left[\frac{x-7}{2}\right]}^{1/3}$

Given, $f:R\to R,g:R\to R$ such that
$f\left(x\right)=2x-3,g\left(x\right)=x^3+5$
It is a bijective function.
$\therefore f^{-1}\left(x\right)$ and $g^{-1}\left(x\right)$ exist.
Let $y=f\left(x\right)=2x-3\Rightarrow 2x=y+3$
$\Rightarrow x=\frac{y+3}{2}\Rightarrow f^{-1}\left(x\right)=\frac{x+3}{2}$
Again, let $y=g\left(x\right)=x^3+5$
$\Rightarrow x^3=y-5\Rightarrow x=(y-5{)}^{1/3}$
$\therefore g^{-1}\left(x\right)=(x-5{)}^{1/3}$
Now, $\left(fog{)}^{-1}\right(x)=(g^{-1}o{f}^{-1}\left)\right(x)=g^{-1}(f^{-1}\left(x\right))$
$=g^{-1}\left(\frac{x+3}{2}\right)=[\frac{x+3}{2}-5]={\left[\frac{x-7}{2}\right]}^{1/3}$.