Question

If $f:R\to R,g:R\to R$ 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. ${\left(\begin{array}{c}\frac{x+7}{2}\end{array}\right)}^{\frac{1}{3}}$
• B. ${\left(\begin{array}{c}\frac{x-7}{2}\end{array}\right)}^{\frac{1}{3}}$
• C. ${\left(\begin{array}{c}x-\frac{7}{2}\end{array}\right)}^{\frac{1}{3}}$
• D. ${\left(\begin{array}{c}x+\frac{7}{2}\end{array}\right)}^{\frac{1}{3}}$

Answer 1

The correct option is B ${\left(\begin{array}{c}\frac{x-7}{2}\end{array}\right)}^{\frac{1}{3}}$

Given,

$f\left(x\right)=2x-3$ and $g\left(x\right)=x^3+5$

Now, $\left(fog\right)\left(x\right)=f\left(g\right(x\left)\right)=2(x^3+5)-3$

$=2x^3+10-3=2x^3+7$

Let, $\left(fog\right)\left(x\right)=y,$ then,

$y=2x^3+7$

$⟹x={\left(\frac{y-7}{2}\right)}^{\frac{1}{3}}$

$\therefore fo{g}^{-1}\left(x\right)={\left(\frac{x-7}{2}\right)}^{\frac{1}{3}}$

Hence, B is the correct option.