If from the Mean Value Theorem,$f'({x_1}) = \frac{f(b) - f(a)}{b - a}$,then

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

If $f(x)=x^3+p x^2+q x$ is defined on $[0,2]$ such that $f(0)=f(2)$ and $f^{\prime}\left(1+\frac{1}{\sqrt{3}}\right)=0$,then $p^2+q^2=$

If the equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x = 0$,where $a_1 \neq 0$ and $n \geq 2$,has a positive root $x = \alpha$,then the equation $n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1 = 0$ has a positive root which is:

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 $x \in [-2, 2]$

Let $f(x)=x^3+2x^2-x$ be a real-valued function. Then,the value of Lagrange's constant $C$ in $(-1,2)$ is