If the function $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^3+{\mathrm{bx}}^2+11\mathrm{x}-6$ satisfies the conditions of Rolle's theorem in [1, 3] and 
${\mathrm{f}}^{\text{'}}\left[2+\frac{1}{\sqrt{3}}\right]=0, \mathrm{then} \mathrm{a}+\mathrm{b}=$

$$\begin{gathered} \mathrm{f}\left(1\right)=\mathrm{f}\left(3\right)\Rightarrow \mathrm{a}+\mathrm{b}+11-6=27\mathrm{a}+9\mathrm{b}+33-6 \\ 26\mathrm{a}+8\mathrm{b}+22=0 \Rightarrow 13\mathrm{a}+4\mathrm{b}+11=0 \\ {\mathrm{f}}^1\left(\mathrm{x}\right)=3{\mathrm{ax}}^2+2\mathrm{bx}+11 \\ {\mathrm{f}}^1(2+\frac{1}{\sqrt{3}})=0 \\ \Rightarrow 3\mathrm{a}{\left(2+\frac{1}{\sqrt{3}}\right)}^2+2\mathrm{b}\left(2+\frac{1}{\sqrt{3}}\right)+11=0 \\ 3\mathrm{a}\left(4+\frac{1}{3}+\frac{4}{\sqrt{3}}\right)+2\mathrm{b}\left(2+\frac{1}{\sqrt{3}}\right)+11=0 \\ 12\mathrm{a}+\mathrm{a}+\frac{12\mathrm{a}}{\sqrt{3}}+\frac{(-11-13\mathrm{a})}{2}\left(2+\frac{1}{\sqrt{3}}\right)+11=0 \\ 26\mathrm{a}+\frac{24\mathrm{a}}{\sqrt{3}}-(11+13\mathrm{a})\left(2+\frac{1}{\sqrt{3}}\right)+22=0 \\ 26\mathrm{a}+\frac{24\mathrm{a}}{\sqrt{3}}-22-\frac{11}{\sqrt{3}}-26\mathrm{a}-\frac{13\mathrm{a}}{\sqrt{3}}+22=0 \\ \frac{11\mathrm{a}}{\sqrt{3}}-\frac{11}{\sqrt{3}}=0 \Rightarrow \mathrm{a}=1 \\ 4\mathrm{b}=-13\mathrm{a}-11=-24 \Rightarrow \mathrm{b}=-6 \\ \mathrm{a}+\mathrm{b}=-5 \end{gathered}$$