Question

Let $f\left(x\right)$ satisfy all the conditions of mean value theorem in $\left[0,2\right]$ if $f\left(0\right)=0$ and $\left|f\text{'}\left(x\right)\right|\le \frac{1}{2},\forall xin\left[0,2\right]$ then:

• A. $f\left(x\right)\le 2$
• B. $f\left(x\right)\le 1$
• C. $f\left(x\right)=2x$
• D. $f\left(x\right)=3$ for at least on $xin\left[0,2\right]$

Answer 1

The correct option is B

$f\left(x\right)\le 1$

Explanation for the correct answer:

Given that: $f\left(0\right)=0$

$\left|f\text{'}\left(x\right)\right|\le \frac{1}{2},\forall xin\left[0,2\right]$

It is given that $f\left(x\right)$ satisfy the mean value theorem in $\left[0,2\right]$ which means that there exists a point $c$ such that:

$$\begin{gathered} f\text{'}\left(c\right)=\frac{f\left(x\right)-f\left(0\right)}{x-0},0<c<x<2 \\ \Rightarrow f\text{'}\left(c\right)=\frac{f\left(x\right)-0}{x} \\ \Rightarrow x.f\text{'}\left(c\right)=f\left(x\right) \\ \Rightarrow f\left(x\right)=x.f\text{'}\left(c\right) \\ \left|f\left(x\right)\right|=\left|x.f\text{'}\left(c\right)\right| \\ \le \left|x\right|.\left|f\text{'}\left(c\right)\right| \\ \le 2\times \frac{1}{2} \\ \le 1 \end{gathered}$$

Therefore, $f\left(x\right)\le 1$.

Hence, the correct answer is Option (B).