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
Answer & explanation
Correct answer: option 2
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 $x \in\{0,2\}$