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(B) By the Mean Value Theorem,there exists some $c \in (0, 2)$ such that $\frac{f(2) - f(0)}{2 - 0} = f'(c)$.Given $f(0) = 0$,we have $\frac{f(2)}{2}

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$|f(x)| \le 1$

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(B) By the Mean Value Theorem,there exists some $c \in (0, 2)$ such that $\frac{f(2) - f(0)}{2 - 0} = f'(c)$.Given $f(0) = 0$,we have $\frac{f(2)}{2} = f'(c)$.For any $x \in [0, 2]$,applying the Mean Value Theorem on $[0, x]$,there exists $c_x \in (0, x)$ such that $\frac{f(x) - f(0)}{x - 0} = f'(c_x)$.Since $f(0) = 0$,this simplifies to $\frac{f(x)}{x} = f'(c_x)$,which implies $f(x) = x \cdot f'(c_x)$.Taking the absolute value,$|f(x)| = |x| \cdot |f'(c_x)|$.Given $|f'(x)| \le \frac{1}{2}$ for all $x \in [0, 2]$,we have $|f(x)| \le |x| \cdot \frac{1}{2}$.Since the maximum value of $x$ in $[0, 2]$ is $2$,we get $|f(x)| \le 2 \cdot \frac{1}{2} = 1$.Thus,$|f(x)| \le 1$.

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:

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