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

$f(x) \leq 1$

$f(x)$ satisfies Longrages theorem in $\{0,2\}$

$f(0)=0$      $f'(x) \leq 1 / 2$

for atieast some $x_0$

$f'\left(x_0\right) =\frac{f(2)-f(0)}{2-0}$ (as per LMVT theorem)

$f'\left(x_0\right) =\frac{f(2)-0}{2}$

$f'\left(x_0\right) =\frac{f(2)}{2}$  as  $f'\left(x_0\right) \leq \frac{1}{2}$

So LMVT holds for [0, x] as well where 0 < x < 2

we get $f'(x) = \frac{f(x) - f(0)}{x - 0} = \frac{f(x)}{x}$

So $x f'(x_0) = f(x)$

$f(x) = x f'(x_0) ≤ 2 × \frac{1}{2} = 1$

f(x) ≤ 1