If the function $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^3+{\mathrm{bx}}^2+26\mathrm{x}-24$ satisfies conditions  of Rolle’s theorem in [2, 4] and ${\mathrm{f}}^{ \mathrm{l}} (3+\frac{1}{\sqrt{3}})=0$, then values of a and b are respectively

$\because \mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^3+{\mathrm{bx}}^2+26\mathrm{x}-24$ satisfies  Rolle’s theorem

$\begin{array}{l}\therefore \mathrm{f}\left(2\right)=\mathrm{f}\left(4\right)\Rightarrow 14\mathrm{a}+3\mathrm{b}+13=0\dots \dots .\left(\text{ i) and }{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=3{\mathrm{ax}}^2+2\mathrm{bx}+26\right.\\ \text{ Now, }{\mathrm{f}}^{\mathrm{\prime }}\left(3+\frac{1}{\sqrt{3}}\right)=0\Rightarrow 3\mathrm{a}{\left(3+\frac{1}{\sqrt{3}}\right)}^2+2\mathrm{b}\left(3+\frac{1}{\sqrt{3}}\right)+26=0\\ \Rightarrow 3\mathrm{a}\left(9+\frac{1}{3}+\frac{6}{\sqrt{3}}\right)+2\mathrm{b}\left(3+\frac{1}{\sqrt{3}}\right)+26=0\\ \Rightarrow \mathrm{a}(28+6\sqrt{3})+2\mathrm{b}\left(3+\frac{1}{\sqrt{3}}\right)+26=0\dots \dots .\left(\text{ ii }\right)\end{array}$

Solving (i) and (ii) we get $\mathrm{a}\left(\frac{26}{\sqrt{3}}\right)+26(-\frac{1}{\sqrt{3}})=0\Rightarrow \mathrm{a}=1,\mathrm{b}=-9$