Solution:

$ \begin{aligned} \\ &\text { Here, } F \circ F=I \quad \text { (Given) } \\\\ &\quad F[F(x)]=I(x)=x \\\\ &\text { ie. } \quad F(p x+q)=x \\\\ &\Rightarrow \quad p(p x+q)+q=x \\\\ &\Rightarrow \quad p^2 x+p q+q=x \\\\ &\Rightarrow \quad p^2 x+p q+q-x=0 \\\\ &\Rightarrow \quad\left(p^2-1\right) x+q(p+1)=0\\ \end{aligned}\\ $

Therefore, $\quad p^2 - 1=0$

$ \begin{aligned}\\ &\Rightarrow \quad p^2=1 \\\\ &\Rightarrow \quad p=1\\ \end{aligned}\\ $

And, $\quad q(p+1)=0$

$ \begin{aligned}\\ &\Rightarrow \quad q(1+1)=0 \quad \text { [Putting value of } p] \\\\ &\Rightarrow \quad 2 q=0 \\\\ &\Rightarrow \quad q=0\\ \end{aligned}\\ $

So, $p=1$ and $q=0$ Ans