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

For which interval does the function $f(x) = \frac{x^2 - 3x}{x - 1}$ satisfy all the conditions of Rolle's theorem?

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 $[0, 1].$

The value of $c$ in the Lagrange's mean value theorem for $f(x)=\sqrt{x-2}$ in the interval $[2,6]$ is

Let $f(x)=x^3+2x^2-x$ be a real-valued function. Then, the value of Lagrange's constant $C$ in $(-1,2)$ is

If the function $f(x)=a x^3+b x^2+11 x-6$ satisfies the conditions of Rolle's theorem in $[1,3]$ and $f^{\prime}\left(2+\frac{1}{\sqrt{3}}\right)=0$,then $a+b=$