Question

If $f\left(x\right)=[x-2]$, then (where $[.]$ denotes the greatest integer function)

• A. $f^{\prime }\left(2.5\right)=\frac{1}{2}$ and $f^{\prime }\left(5\right)=3$
• B. $f^{\prime }\left(2.5\right)=0$ and $f^{\prime }\left(5\right)=3$
• C. $f^{\prime }\left(2.5\right)=0$ and $f^{\prime }\left(5\right)$ does not exist
• D. Both $f^{\prime }\left(2.5\right)$ and $f^{\prime }\left(5\right)$ do not exist.

Answer 1

The correct option is C $f^{\prime }\left(2.5\right)=0$ and $f^{\prime }\left(5\right)$ does not exist

Given $f\left(x\right)=[x-2]$

$L{f}^{\prime }\left(2.5\right)={lim}_{h\to 0}\frac{f(2.5-h)-f\left(2.5\right)}{-h}$
$={lim}_{h\to 0}\frac{[2.5-h-2]-[2.5-2]}{-h}$
$={lim}_{h\to 0}\frac{0}{-h}=0$
Now, $R{f}^{\prime }\left(2.5\right)={lim}_{h\to 0}\frac{(2.5+h)-f\left(2.5\right)}{h}$
$={lim}_{h\to 0}\frac{[2.5+h-2]-[2.5-2]}{h}={lim}_{h\to 0}\frac{0}{h}=0$
$\therefore f^{\prime }\left(2.5\right)=0$
$f\left(x\right)=[x-2]$ is not continuous at integer $(x=5)$
$\Rightarrow$ It is not differentiable at $x=5$
Therefore, $f^{\prime }\left(5\right)$ does not exist.