If Rolle's theorem holds for the function \(f(x)=x^3+b x^2+a x+5\) on \([1,3]\) with \(c=2+\frac{1}{\sqrt{3}}\), then the values of \(a\) and \(b\) respectively are

Since \(\mathrm{f}(x)\) satisfies the Rolle's theorem,
\(f(1)=f(3)\)
\(\therefore \quad 1+b+a+5=27+9 b+3 a+5\)
\(\Rightarrow 2 a+8 b=-26\)
\(\Rightarrow a+4 b=-13\) ...(i)
\(\vec{f}(x)=x^3+b x^2+a x+5\)
\(\therefore \quad \mathrm{f}^{\prime}(x)=3 x^2+2 \mathrm{~b} x+\mathrm{a}\)
Now, \(\mathrm{f}^{\prime}(\mathrm{c})=0\)
\(
\begin{aligned}
& \Rightarrow \mathrm{f}^{\prime}\left(2+\frac{1}{\sqrt{3}}\right)=0 \\
& \Rightarrow 3\left(2+\frac{1}{\sqrt{3}}\right)^2+2 \mathrm{~b}\left(2+\frac{1}{\sqrt{3}}\right)+\mathrm{a}=0 \\
& \Rightarrow 3\left(4+\frac{4}{\sqrt{3}}+\frac{1}{3}\right)+4 \mathrm{~b}+\frac{2 \mathrm{~b}}{\sqrt{3}}+\mathrm{a}=0 \\
& \Rightarrow \mathrm{a}+4 \mathrm{~b}+\frac{2 \mathrm{~b}+12}{\sqrt{3}}+13=0 \\
& \Rightarrow-13+\frac{2 \mathrm{~b}+12}{\sqrt{3}}+13=0 \\
& \Rightarrow \frac{2 \mathrm{~b}+12}{\sqrt{3}}=0 \\
& \Rightarrow \mathrm{b}=-6
\end{aligned}
\)
Substituting \(b=-6\) in (i), we get \(a=11\)