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

In the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(c)$. If $a = 4$,$b = 9$,and $f(x) = \sqrt{x}$,then the value of $c$ is:

$f:[1,3] \rightarrow R$ 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

If $a, b, c$ are real numbers such that $a + b + c = 0$,then the quadratic equation $3ax^2 + 2bx + c = 0$ has

The number of admissible values of $C$ obtained when the Lagrange's mean value theorem is applied for the function $f(x)=x$ on $[2,5]$ is

Consider the 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-$2$: Rolle's theorem can be applied to the function $g(x)$ in the interval $[0, 1]$.