Mathematics - Functions Question with Solution | TestHub

Given:

$f\left(x\right)=ax-3$ 

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

Now, let

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

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

Let

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

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

$\Rightarrow x=2y^3+7$

So, inverse of $h^{-1}\left(x\right)$ is $2x^3+7$ i.e., 

$h\left(x\right)=fog\left(x\right)=2x^3+7$

Also,

$fog\left(x\right)=a\left(x^b+c\right)-3$

$\Rightarrow a{x}^b+ac-3=2x^3+7$

On comparing, we get

$a=2, b=3,c=5$

So,

$$\begin{gathered} f\left(x\right)=2x-3 \\ g\left(x\right)=x^3+5 \end{gathered}$$

Now,

$fog \left(ac\right)=fog\left(10\right)=f\left(1005\right)=2007$

$\left(gof\right)\left(b\right)=gof\left(3\right)=g\left(3\right)=32$

Therefore,

$fog \left(ac\right)+\left(gof\right)\left(b\right)=2007+32=2039$