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 $b c=$

The value of $c$ for which the Mean Value Theorem holds for the function $f(x) = \log_{e} x$ on the interval $[1, 3]$ is:

If for $f(x) = 2x - x^2$,Lagrange's Mean Value Theorem satisfies in $[0, 1]$,then the value of $c \in [0, 1]$ is

Let $S$ be the set of all functions $f:[0,1] \rightarrow \mathbb{R}$ which are continuous on $[0,1]$ and differentiable on $(0,1)$. Then for every $f \in S$,there exists a $c \in (0,1)$,depending on $f$,such that:

If $f(x)$ is a differentiable function,$f^{\prime}(x) \geq 5$ for all $x \in [2, 6]$,$f(2) = 4$ and $f(3) = 15$,then a possible value of $f(6)$ is:

If $f(x) = x^{3}$ and $g(x) = x^{3} - 4x$ in the interval $[-2, 2]$,consider the following statements:
$(a)$ $f(x)$ and $g(x)$ satisfy the Mean Value Theorem.
$(b)$ $f(x)$ and $g(x)$ both satisfy Rolle's Theorem.
$(c)$ Only $g(x)$ satisfies Rolle's Theorem.
Which of these statements is correct?