The number of points in the interval [2,4], at which the function f(x)=[x^2-x-(1)/(2)], wh…

Question

The number of points in the interval $[2,4]$, at which the function $f(x)=[{x}^{2}-x-\frac{1}{2}]$, where $[\cdot ]$ denotes the greatest integer function, is discontinuous, is \_\_\_\_\_\_\_\_.

Answer: 10

Accepted Answer

For the greatest integer function

$f(x)=[{x}^{2}-x-\frac{1}{2}],$

the function is discontinuous at those points where

${x}^{2}-x-\frac{1}{2}$

takes an integer value.

So, let

${x}^{2}-x-\frac{1}{2}=n,$

where $n$ is an integer.

We need to find how many such points lie in the interval $[2,4]$.

First, consider

$g(x)={x}^{2}-x-\frac{1}{2}.$

Now check its values at the endpoints:

At $x=2$,

$g(2)=4-2-\frac{1}{2}=\frac{3}{2}.$

At $x=4$,

$g(4)=16-4-\frac{1}{2}=\frac{23}{2}=11.5.$

Also,

${g}^{'}(x)=2x-1.$

For $x\in [2,4]$, we have $2x-1>0$, so $g(x)$ is strictly increasing on $[2,4]$.

Hence, as $x$ goes from $2$ to $4$, $g(x)$ takes every value from $\frac{3}{2}$ to $\frac{23}{2}$ exactly once.

Now count the integers in the interval

$[\frac{3}{2},\frac{23}{2}]=[1.5,11.5].$

These integers are

$2,3,4,5,6,7,8,9,10,11.$

There are $10$ integers.

Since each integer value corresponds to exactly one $x\in [2,4]$, the function $f(x)$ is discontinuous at exactly $10$ points.

Therefore, the required number of points is

$10.$

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