The value of $c$ for the Lagrange's mean value theorem for $f(x)=\sqrt{x^2-x}, x \in[1,4]$ is

If $2a + 3b + 6c = 0$,then at least one root of the equation $ax^2 + bx + c = 0$ lies in the interval:

Let $f(x) = \begin{cases} \frac{x^p}{(\sin x)^q} & \text{if } 0 < x \leq \frac{\pi}{2} \\ 0 & \text{if } x = 0 \end{cases}$ where $p, q \in \mathbb{R}$. Then, Lagrange's Mean Value Theorem is applicable to $f(x)$ in the closed interval $[0, \frac{\pi}{2}]$ if:

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 $b c=$

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 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