The value $C$ of the Lagrange's mean value theorem for the function $f(x)=x(x-1)(x-2)$ in the interval $[0, 1/2]$ is

If $a, b, c \in \mathbb{R}$ and satisfy $3a + 5b + 15c = 0$,then the equation $ax^4 + bx^2 + c = 0$ has:

If $f:[a, b] \rightarrow [c, d]$ is a continuous and strictly increasing function, then $\frac{d-c}{b-a}$ is

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

$A$ value of $c$ according to the Lagrange's mean value theorem for $f(x)=(x-1)(x-2)(x-3)$ in $[0,4]$ is

Let $f(x)$ satisfy the requirements of Lagrange's Mean Value Theorem in $[0, 2]$. If $f(0) = 0$ and $|f'(x)| \leqslant \frac{1}{2}$ for all $x \in [0, 2]$,then-