Question

Let $f(x)=[x]$, $x\in(0,6)$, where $[x]$ is the greatest integer function. Then the number of discontinuities of $f(x)$ is:

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5

Hint:

For an open interval $(a, b)$, the number of discontinuities of $[x]$ is $\lfloor b - \epsilon \rfloor - \lfloor a \rfloor$, effectively counting integers between $a$ and $b$.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The greatest integer function, \( f(x) = [x] \), also known as the floor function, gives the greatest integer less than or equal to x. This function is known to be discontinuous at every integer value. We need to find how many integers are within the given interval.
Step 2: Key Formula or Approach:
The function \( f(x) = [x] \) is discontinuous at all points \( x \in \mathbb{Z} \) (the set of integers). We need to count the number of integers in the open interval \( (0, 6) \).
Step 3: Detailed Explanation:
The function is defined on the interval \( (0, 6) \). This means \( 0<x<6 \).
The greatest integer function \( [x] \) has a "jump" discontinuity whenever x is an integer. At these points, the left-hand limit is different from the right-hand limit and the function value.
For example, at x=2:
- \( \lim_{x \to 2^-} [x] = 1 \)
- \( \lim_{x \to 2^+} [x] = 2 \)
- \( f(2) = [2] = 2 \)
Since the left and right limits are not equal, the function is discontinuous at x=2.
We need to identify all the integer values within the interval \( (0, 6) \). The integers in this interval are:
1, 2, 3, 4, 5
The endpoints 0 and 6 are not included in the interval.
The function \( f(x) = [x] \) will be discontinuous at each of these integer points.
Counting these points, we find there are 5 points of discontinuity.
Step 4: Final Answer:
The number of discontinuities of f(x) in the interval (0,6) is 5.