The function $f(x) = x(x + 3)e^{-\frac{1}{2}x}$ satisfies all the conditions of Rolle's theorem in $[-3, 0]$. Find the value of $c$ such that $f'(c) = 0$.

Let $f:[1,3] \rightarrow R$ be a continuous function that is differentiable in $(1,3)$ and $f^{\prime}(x)=|f(x)|^{2}+4$ for all $x \in(1,3).$ Then,

Let $f:(a, b) \rightarrow R$ be a twice differentiable function such that $f(x) = \int_{a}^{x} g(t) \, dt$ for a differentiable function $g(x)$. If $f(x) = 0$ has exactly five distinct roots in $(a, b)$,then $g(x) g'(x) = 0$ has at least:

If the function $f(x) = ax^3 + bx^2 + 11x - 6$, defined on $[1, 3]$, satisfies all the conditions of Rolle's theorem for $c = 2 + \frac{1}{\sqrt{3}}$, then

Let $f(x)$ be a differentiable function in $[0, 2]$,$f(0) = 0$ and $f'(x) \le \frac{1}{2}$ for all $x \in [0, 2]$. Then:

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