Question

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

• A. ${\left(\frac{x+7}{2}\right)}^{1/3}$
• B. ${(x-\frac{7}{2})}^{1/3}$
• C. ${\left(\frac{x-2}{7}\right)}^{1/3}$
• D. ${\left(\frac{x-2}{2}\right)}^{1/3}$

Answer 1

The correct option is D ${\left(\frac{x-2}{2}\right)}^{1/3}$

$f\left(x\right)=2x-3,g\left(x\right)=x^3+5$ are bijections and hence $f^{-1}$ and $g^{-1}$ both exist.
$y=2x-3$
$\Rightarrow x=\frac{y+3}{2}$
$\therefore f^{-1}\left(y\right)=\frac{y+3}{2}\cdots \left(1\right)$
$y=x^3+5$
$\therefore x={(y-5)}^{1/3}=g^{-1}\left(y\right)$
$\therefore g^{-1}\left(x\right)={(x-5)}^{1/3}...\left(2\right)$
$\therefore {\left(fog\right)}^{-1}x=\left(g^{-1}o{f}^{-1}\right)x$
$=g^{-1}\left[f^{-1}\left(x\right)\right]=g^{-1}\left[\frac{x+3}{2}\right]$ by (1)
$={(\frac{x+3}{2}-5)}^{1/3}={\left(\frac{x-7}{2}\right)}^{1/3}$
Ans: D