If the function

The given function is $f(x)=x^{3}-6 x^{2}+a x+b$.

It is given that $f(x)$ defined on $[1,3]$ satisfies Rolle's theorem for $c=2+\frac{1}{\sqrt{3}}$.

$\therefore f(1)=f(3)$ and $f^{\prime}(c)=0$

Now,

$f(1)=f(3)$

$\Rightarrow 1-6+a+b=27-54+3 a+b$

$\Rightarrow-5+a=-27+3 a$

$\Rightarrow 2 a=22$

$\Rightarrow a=11$

Also,

$f(x)=x^{3}-6 x^{2}+a x+b$

$\Rightarrow f^{\prime}(x)=3 x^{2}-12 x+a$

$\therefore f^{\prime}(c)=0$

$\Rightarrow 3 c^{2}-12 c+a=0$

$\Rightarrow 3 c^{2}-12 c+11=0 \quad(a=11)$

Since both equations $f(1)=f(3)$ and $f^{\prime}(c)=3 c^{2}-12 c+11=0$ are independent of $b$, so $b$ can taken any real value.

$\therefore a=11$ and $b \in \mathrm{R}$

Thus, if the function $f(x)=x^{3}-6 x^{2}+a x+b$ defined on [1, 3] satisfies Roll's theorem for $c=2+\frac{1}{\sqrt{3}}$, then $a=11$ and $b \in \mathrm{R}$.

If the function $f(x)=x^{3}-6 x^{2}+a x+b$ defined on $[1,3]$ satisfies Roll's theorem for $c=2+\frac{1}{\sqrt{3}}$, then a = ___11___, b = ___R___.