Question

Let $f:R\to R,g:R\to R$, be two functions, such that f(x) =2x – 3, g (x) = $x^3$ + 5.
The function $(fog{)}^{-1}$ (x) is equal to

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

Answer 1

The correct option is D ${\left(\frac{x-7}{2}\right)}^{\frac{1}{3}}$

We have, f : R $\to$ R, g: R $\to$ R defined by f(x) = 2x - 3 and g (x) = $x^3$ + 5
It can be checked that f(x) and g(x) are bijective functions
$\therefore$ fo g is also bijective and (fog) = f(g(x)) = f $(x^3+5)=2(x^3+5)-3=2x^3+7$
$\left(fog\right)\left(x\right)=y\Rightarrow 2x^3+7=y\Rightarrow x={\left(\frac{y-7}{2}\right)}^{\frac{1}{3}}$
$\therefore \left(fog{)}^{-1}\right(x)={\left(\frac{x-7}{2}\right)}^{\frac{1}{3}},xϵR$
$\therefore$ The correct answer is (d).