Question

The number of discontinuities of the greatest integer function $f(x)=[x]$, $x\in(-\frac{7}{2},100)$

• A. 104
• B. 100
• C. 102
• D. 103

Hint:

Logic Tip: The number of integers in an inclusive range $[m, n]$ is calculated as $(n - m) + 1$. Here, the range of valid integers is $[-3, 99]$. Applying the formula yields $(99 - (-3)) + 1 = 102 + 1 = 103$.

Answer 1

The Correct Option is D

Concept:
The greatest integer function, denoted as $f(x) = [x]$ or $\lfloor x \rfloor$, returns the greatest integer less than or equal to $x$. This function has jump discontinuities at exactly every integer value of $x$. Thus, to find the number of discontinuities in a given open interval $(a, b)$, we just need to count the number of integers that lie strictly between $a$ and $b$.
Step 1: Define the interval boundaries clearly.
The given interval is $x \in \left(-\frac{7}{2}, 100\right)$. Converting the fraction to a decimal makes it easier to evaluate: $$-\frac{7}{2} = -3.5$$ So, the interval is $(-3.5, 100)$. Because it uses open parentheses, $-3.5$ and $100$ are not included in the interval.
Step 2: Identify the integers within this interval.
We need to find all integer values $k$ such that $-3.5<k<100$. The smallest integer greater than $-3.5$ is $-3$. The largest integer strictly less than $100$ is $99$. Therefore, the function is discontinuous at the points: $$x \in \{-3, -2, -1, 0, 1, 2, ......., 99\}$$
Step 3: Count the total number of integer points.
To count the total number of integers in this set, we can split them into negatives, zero, and positives:

• Positive integers: $1, 2, 3, ......., 99$ (Total = 99)
• Zero: $0$ (Total = 1)
• Negative integers: $-1, -2, -3$ (Total = 3)

Summing these up: $$\text{Total number of discontinuities} = 99 + 1 + 3 = 103$$