Let $f(x)$ satisfy the requirement of Lagrange's Mean value theorem in $[0,2]$. If $f(0)=0$ and $\left|f'(x)\right| \leq \frac{1}{2}$ for all $x \in[0,2]$, then
Answer & explanation
Correct answer: option 2
Let $x \in(0,2)$. Since $f(x)$ satisfies the requirements of Lagrange's mean value theorem in [0,2]. So, it also satisfies in $[0, x]$. Consequently, there exist $c \in(0, x)$ such that
$f'(c) =\frac{f(x)-f(0)}{x-0}$
$\Rightarrow f'(c)=\frac{f(x)}{x}$
$\Rightarrow \left|\frac{f(x)}{x}\right| =\left|f'(c)\right| \leq \frac{1}{2}$ $\left[∵\left|f^{\prime}(x)\right| \leq \frac{1}{2}\right]$
$\Rightarrow |f(x)| \leq \frac{|x|}{2}$
$\Rightarrow |f(x)| \leq \frac{x}{2}$ $[∵ x>0]$
$\Rightarrow |f(x)| \leq 1$ $[∵ x \in(0,2) \quad ∴|x|<2]$