For the function $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-6{\mathrm{x}}^2+\mathrm{ax}+\mathrm{b},$ if Rolle's theorem holds in [1, 3] with $\mathrm{c}=2+\frac{1}{\sqrt{3}}$ then (a, b) is equal to

$$\begin{gathered} \mathrm{f}\left(1\right)=\mathrm{f}\left(3\right) \Rightarrow 1-6+\mathrm{a}+\mathrm{b}=27-54+3\mathrm{a}+\mathrm{b} \\ \Rightarrow \mathrm{a}-5=-27+3\mathrm{a} \\ \Rightarrow 2\mathrm{a}=22 \Rightarrow \mathrm{a}=11 \\ {\mathrm{f}}^1\left(\mathrm{x}\right)=3{\mathrm{x}}^2-12\mathrm{x}+\mathrm{a} \\ {\mathrm{f}}^1\left(\mathrm{c}\right)=0 \Rightarrow 3{\mathrm{x}}^2-12\mathrm{x}+11=0 \\ \mathrm{x}=\frac{12\pm \sqrt{144-132}}{6} \\ =2\pm \frac{1}{\sqrt{3}} \end{gathered}$$