$\mathrm{If} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3+{\mathrm{px}}^2+\mathrm{qx}$ is defined on [0, 2] such that $\mathrm{f}\left(0\right)=\mathrm{f}\left(2\right) \mathrm{and} {\mathrm{f}}^1\left[1+\frac{1}{\sqrt{3}}\right]=0$ then  ${\mathrm{p}}^2+{\mathrm{q}}^2=$

$$\begin{gathered} \mathrm{f}\left(0\right)=\mathrm{f}\left(2\right)\Rightarrow 0+0+0=8+4\mathrm{p}+2\mathrm{q} \\ \Rightarrow 2\mathrm{p}+\mathrm{q}=-4 \\ {\mathrm{f}}^1\left(\mathrm{x}\right)=3{\mathrm{x}}^2+2\mathrm{px}+\mathrm{q} \\ {\mathrm{f}}^1\left(1+\frac{1}{\sqrt{3}}\right)=0\Rightarrow 3{\left(1+\frac{1}{\sqrt{3}}\right)}^2+2\mathrm{p}\left(1+\frac{1}{\sqrt{3}}\right)-4-2\mathrm{p}=0 \\ 3+1+\frac{6}{\sqrt{3}}+\frac{2\mathrm{p}}{\sqrt{3}}-4=0 \\ \Rightarrow 2\mathrm{p}+6=0 \Rightarrow \mathrm{p}=-3 \\ q=2 \\ {p}^2+q^2=13 \end{gathered}$$