If from the Mean Value Theorem,$f'({x_1}) = \frac{f(b) - f(a)}{b - a}$,then

If the Mean Value Theorem holds for the function $f(x)=(x-1)(x-2)(x-3)$ on the interval $x \in [0, 4]$,then the values of $c$ as per the theorem are:

Let $y = f(x)$ and $y = g(x)$ be two differentiable functions in $[0, 2]$ such that $f(0) = 3$,$f(2) = 5$,$g(0) = 1$,and $g(2) = 2$. If there exists at least one $c \in (0, 2)$ such that $f'(c) = k g'(c)$,then $k$ must be:

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

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

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