The abscissa of the points of the curve $y = x^3$ in the interval $[-2, 2]$,where the slope of the tangents is equal to the slope of the secant line as per the Mean Value Theorem for the interval $[-2, 2]$,are:

The value of $C$ in the Mean Value Theorem for the function $f(x) = x^{2}$ in the interval $[2, 4]$ is:

Let $f$ be a polynomial function defined on $[2,7]$. If $f(2)=3$ and $f^{\prime}(x) \leq 5$ for all $x$ in $(2,7)$,then the maximum possible value attained by $f$ at $x=7$ is

Consider the following functions:
$I) f(x) = \begin{cases} \frac{1}{2}-x, & x < \frac{1}{2} \\ (\frac{1}{2}-x)^2, & x \geq \frac{1}{2} \end{cases}$
$II) f(x) = |3x-1|$
$III) f(x) = x|x|$
$IV) f(x) = |x|$
Then on $[0, 1]$,Lagrange's Mean Value Theorem $(LMVT)$ is applicable to which of the functions?

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

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?