For some a, b, c ∈ ℕ , let f(x) = ax - 3 and g(x) = x ^ b + c, x ∈...

Given:

$f\left(x\right) = ax - 3$

$g\left(x\right) = {x}^{b} + c, x \in \mathbb{R} {\left(fog\right)}^{- 1}\left(x\right) = \left(\dfrac{x - 7}{2}\right)\dfrac{1}{3}$

Now, let

$h\left(x\right) = \left(f ∘ g\right)\left(x\right)$

$⇒ {h}^{- 1}\left(x\right) = \left(\dfrac{x - 7}{2}\right)\dfrac{1}{3}$

Let

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

$⇒ {y}^{3} = \left(\dfrac{x - 7}{2}\right)$

$⇒ x = 2{y}^{3} + 7$

So, inverse of ${h}^{- 1}\left(x\right) is$2{x}^{3} + 7$i.e., </p><p>$h\left(x\right)$= fog \left(x\right)$ = 2{x}^{3} + 7$</p><p>Also,</p><p>$fog \left(x\right)$= a \left({x}^{b} + c\right)$ - 3$</p><p>$⇒ a{x}^{b} + ac - 3 = 2{x}^{3} + 7$</p><p>On comparing, we get</p><p>$a = 2, b = 3, c = 5$</p><p>So,</p><p>$f\left(x\right)$= 2x - 3g \left(x\right)$ = {x}^{3} + 5$</p><p>Now,</p><p>$fog \left(ac\right)$= fog \left(10\right)$ = f \left(1005\right)$= 2007$

\left(gof\right)\left(b\right) = gof\left(3\right) = g\left(3\right) = 32$</p><p>Therefore,</p><p>$fog\left(ac\right) + \left(gof\right)\left(b\right) = 2007 + 32 = 2039$