Consider the function $f(x)=2x^3-3x^2-x+1$ and the intervals $I_1=[-1,0]$, $I_2=[0,1]$, $I_3=[1,2]$, $I_4=[-2,-1]$. Then,

Let $f: R \rightarrow R$ be a differentiable function such that $f(a)=0=f(b)$ and $f^{\prime}(a) f^{\prime}(b) > 0$ for some $a < b$. Then,the minimum number of roots of $f^{\prime}(x)=0$ in the interval $(a, b)$ is

For the curve $y = x^3$ in the interval $[-2, 2]$,find the abscissae of the points where the slope of the tangent is equal to the slope of the secant line passing through the endpoints of the interval,as per the Mean Value Theorem.

If $f:[a, b] \rightarrow [c, d]$ is a continuous and strictly increasing function, then $\frac{d-c}{b-a}$ is

If the function $f(x) = x(x+3) e^{-\frac{x}{2}}$ satisfies all the conditions of Rolle's theorem in $[-3, 0]$,then $c$ is

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?