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Let $f (x)$ satisfies the conditions 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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$| f (x) | \leq 2$

$f (x) \leq 1$

$f (x) = 2 x$

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

answer is B.

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Detailed Solution

$\frac{f (x) - f (0)}{x - 0} = f^{'} (c) for some c \in (0, 2)$

$\Rightarrow \frac{f (x)}{x} = f^{'} (c) \Rightarrow f^{'} (c) \leq \frac{1}{2} \Rightarrow f (x) \leq \frac{1}{2} x \leq \frac{1}{2} \times 2 \Rightarrow f (x) \leq 1$

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