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

f(x) ≤ 1

Given $f'(x) \leq \frac{1}{2}$

$\Rightarrow \frac{d}{d x}\left(f(x)-\frac{x}{2}\right) \leq 0$

$\Rightarrow h(x)=f(x)-\frac{x}{2}$ is a deceasing that.

Now h(0) = 0 and h(x) is decreasing

$\Rightarrow f(x) \frac{x}{2} \leq 0$

$\Rightarrow f(x) \leq \frac{x}{2}$

$\Rightarrow f(x) \leq 1$  since  $x \in\{0,2\}$