If the function $f(x) = ax^3 + bx^2 + 26x - 24$ satisfies the conditions of Rolle's theorem in $[2, 4]$ and $f^{\prime}\left(3 + \frac{1}{\sqrt{3}}\right) = 0$,then the value of $ab$ is equal to

Examine if Rolle's Theorem is applicable to the following function. Can you say something about the converse of Rolle's Theorem from this example?
$f(x) = [x]$ for $x \in [-2, 2]$

For the Mean Value Theorem $f(b) - f(a) = (b - a) f'(x_1)$ where $a < x_1 < b$,if $f(x) = 1/x$,then $x_1 = ?$

For all real values of $a_{0}, a_{1}, a_{2}, a_{3}$ satisfying $a_{0}+\frac{a_{1}}{2}+\frac{a_{2}}{3}+\frac{a_{3}}{4}=0$, the equation $a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}=0$ has a real root in the interval

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?

Let $f$ be a twice differentiable function on $(1,6)$. If $f(2)=8$,$f'(2)=5$,$f'(x) \geq 1$ and $f''(x) \geq 4$ for all $x \in (1,6)$,then: