In the function$f\left(x\right)=a{x}^3+b{x}^2+11x-6$  satisfies conditions of Rolle’s Theorem in $[1,3]$  and$f\text{'}\left(2+\frac{1}{\sqrt{3}}\right)=0,$  then value of a and b are respectively

$\begin{array}{c}\because f\left(1\right)=f\left(3\right)\\ \Rightarrow a+b+11-6=27a+9b+33-6\end{array}$

$\Rightarrow 13a+4b=-11$

And       $f\text{'}\left(x\right)=3a{x}^2+2bx+11\mathrm{.....}\left(i\right)$

$\begin{array}{c}\Rightarrow f\text{'}\left(2+\frac{1}{\sqrt{3}}\right)=3a{\left(2+\frac{1}{\sqrt{3}}\right)}^2+2b\left(2+\frac{1}{\sqrt{3}}\right)+11=0\\ \Rightarrow 3a\left(4+\frac{1}{3}+\frac{4}{\sqrt{3}}\right)+2b\left(2+\frac{1}{\sqrt{3}}\right)+11=0\mathrm{......}\left(ii\right)\end{array}$

From Eqs, (i) and (ii), we get $a=1,b=-6.$