Let $f(x)$ be a differentiable function in $[0, 2]$,$f(0) = 0$ and $f'(x) \le \frac{1}{2}$ for all $x \in [0, 2]$. Then:

According to the Mean Value Theorem,which of the following functions does not satisfy the conditions on the interval $[0, 1]$?

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:

Let $f(x)$ be a non-constant twice differentiable function defined on $(-\infty, \infty)$ such that $f(x)=f(1-x)$ and $f^{\prime}\left(\frac{1}{4}\right)=0$. Then
$(A)$ $f^{\prime \prime}(x)$ vanishes at least twice on $[0,1]$
$(B)$ $f^{\prime}\left(\frac{1}{2}\right)=0$
$(C)$ $\int_{-1 / 2}^{1 / 2} f\left(x+\frac{1}{2}\right) \sin x d x=0$
$(D)$ $\int_0^{1 / 2} f(t) e^{\sin \pi t} d t=\int_{1 / 2}^1 f(1-t) e^{\sin \pi t} d t$

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

Rolle's theorem is not applicable to the function $f(x) = |x|$ defined on $[-1, 1]$ because