The value of $c$ for which the Lagrange's Mean Value Theorem $(LMVT)$ is applicable for the function $f(x) = x(x+3)(x-2)$ in the interval $[-1, 4]$ is:

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:

If $f(x)$ is differentiable on the interval $[2, 5]$ such that $f(2) = 1/5$ and $f(5) = 1/2$,then there exists a number $c$ such that $2 < c < 5$ and $f'(c) = \dots$

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 $f:[a, b] \rightarrow R$ be continuous in $[a, b]$, differentiable in $(a, b)$ and $f(a)=0=f(b)$. Then

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