If the function $f(x)=\sqrt{x^2-4}$ satisfies the Lagrange's mean value theorem on $[2, 4]$,then the value of $C$ is

Consider the following statements:
Statement $I$: If $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\ldots+\frac{a_n}{n+1}=0$,where $a_0, a_1, \ldots, a_n$ are real numbers,then the polynomial $P(x) = a_0+a_1 x+a_2 x^2+\ldots+a_n x^n$ has a zero in the interval $(0,1)$.
Statement $II$: If $f:[a, b] \rightarrow R$ is continuous on $[a, b]$ and $f$ is differentiable in $(a, b)$,where $a>0$ and if $\frac{f(a)}{a}=\frac{f(b)}{b}$,then there exists $c \in(a, b)$ such that $c f^{\prime}(c)=f(c)$.
Which one of the following options is true?

If the function $f(x)=x^3+b x^2+c x-6$ satisfies all the conditions of Rolle's theorem in $[1,3]$ and $f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$,then $b c=$

If Rolle's theorem holds for the function $f(x) = 2x^3 + ax^2 + bx$ in the interval $[-1, 1]$ for the point $c = \frac{1}{2}$,then the value of $2a + b$ is

Let $f^{\prime}(0)=-3$ and $f^{\prime}(x) \leq 5$ for all real values of $x$. The $f(2)$ can have possible maximum value as

The value $C$ of the Lagrange's mean value theorem for the function $f(x)=x(x-1)(x-2)$ in the interval $[0, 1/2]$ is