Question

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

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

Answer 1

The correct option is C $|f\left(x\right)|\le 1$

By Lagrange's mean value theorem in $[0,2]$, there exists $c\in (0,2)$ such that
$f^{\prime }\left(c\right)=\frac{f\left(2\right)-f\left(0\right)}{2}$
$\Rightarrow 2f^{\prime }\left(c\right)=f\left(2\right)-f\left(0\right)$
$\Rightarrow 2f^{\prime }\left(c\right)=f\left(2\right)$ , since $f\left(0\right)=0$
Now, since $|f^{\prime }\left(x\right)|\le \frac{1}{2}$ for all $x\in [0,2]$,

$\Rightarrow |f^{\prime }\left(c\right)|\le \frac{1}{2}$
$\Rightarrow 2|f^{\prime }\left(c\right)|\le 1$
$\Rightarrow |f\left(2\right)|\le 1$

Also,

$|f^{\prime }\left(k\right)|=\left|\frac{f\left(k\right)-f\left(0\right)}{k}\right|\le \frac{1}{2}$

$\therefore |f\left(k\right)|\le \frac{k}{2}$ for $k\in [0,2]$
$\therefore |f\left(x\right)|\le 1$