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

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