$f:[1,3] \rightarrow R$ is a function defined as $f(x)=x^3+a x^2+b x$. If $f(1)-f(3)=0$ and $f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$, then $a-b$ is equal to

Consider a quadratic equation $ax^2 + bx + c = 0,$ where $2a + 3b + 6c = 0$ and let $g(x) = a\frac{x^3}{3} + b\frac{x^2}{2} + cx.$
Statement $1:$ The quadratic equation has at least one root in the interval $(0, 1).$
Statement $2:$ The Rolle's theorem is applicable to function $g(x)$ on the interval $[0, 1].$

If a function $f$ is differentiable on $R$ such that $f^{\prime}(x) \leq 4$ for all $x \in R$; and if $f(2)=-6$ and $f(6)=8$,then the value of $f(4)$ belongs to the interval

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 $x \in [-2, 2]$

$A$ value of $c$ according to the Lagrange's mean value theorem for $f(x)=(x-1)(x-2)(x-3)$ in $[0,4]$ is

For all $x \in [0, 2024]$,assume that $f(x)$ is differentiable,$f(0) = -2$,and $f^{\prime}(x) \geq 5$. Then the least possible value of $f(2024)$ is: