Let f(x) satisfy the requirements of Lagrange | vedclass

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(B) According to Lagrange's Mean Value Theorem,for any $x \in (0, 2]$,there exists a $c \in (0, x)$ such that $f'(c) = \frac{f(x) - f(0)}{x - 0}$.Sinc

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$|f(x)| \leqslant 1$

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(B) According to Lagrange's Mean Value Theorem,for any $x \in (0, 2]$,there exists a $c \in (0, x)$ such that $f'(c) = \frac{f(x) - f(0)}{x - 0}$.Since $f(0) = 0$,we have $f'(c) = \frac{f(x)}{x}$.Given that $|f'(x)| \leqslant \frac{1}{2}$ for all $x \in [0, 2]$,it follows that $|f'(c)| \leqslant \frac{1}{2}$.Substituting this into our equation,we get $\left| \frac{f(x)}{x} \right| = |f'(c)| \leqslant \frac{1}{2}$.This implies $|f(x)| \leqslant \frac{|x|}{2}$.Since $x \in [0, 2]$,the maximum value of $|x|$ is $2$,so $|f(x)| \leqslant \frac{2}{2} = 1$.Thus,$|f(x)| \leqslant 1$.

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-

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