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Let f (x) satisfy the requirements of Lagrange’s mean value theorem in [0,2]. If $f (0) = 0 and |f^{1} (x)| \leq \frac{1}{2} for all x in [0, 2] then$

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$| f (x) | \leq 1$

$f (x) = 2 x$

$f (x) \leq 2$

$f (x) = 3 for at least one x in [0, 2]$

answer is B.

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Detailed Solution

$\begin{array}{l} |\frac{f (2) - f (0)}{2}| \leq \frac{1}{2} \\ | f (2) | \leq 1 \\ |\frac{f (x) - f (0)}{x}| \leq \frac{1}{2} \\ | f (x) | \leq \frac{x}{2} \leq 1 . \end{array}$

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