Question

If $f(x) = [x]$, for $x \in (-1, 2)$, then $f$ is discontinuous at (where $[x]$ represents floor function)

• A. $x = -1, 0, 1, 2$
• B. $x = -1, 0, 1$
• C. $x = 0, 1$
• D. $x = 2$

Hint:

The greatest integer floor function graph looks like a staircase. It snaps and breaks at every solid integer step! Just count the pure integers tucked inside the given interval limits. Since the interval is open at $-1$ and $2$, only $0$ and $1$ matter!

Answer 1

The Correct Option is C

Step 1: Recall how the floor function behaves.
The function $f(x)=[x]$ gives the greatest integer not bigger than $x$. It is smooth between integers but jumps at every integer.

Step 2: List the integers inside the interval.
The domain is the open interval $(-1,2)$. The end points $-1$ and $2$ are not included. The only integers sitting strictly inside are $0$ and $1$.

Step 3: Check the jump.
Near $x=1$, the value just to the left is $0$ and just to the right is $1$. Left and right limits differ, so there is a jump. The same thing happens at $x=0$.

Step 4: Conclusion.
The breaks occur only at $x=0$ and $x=1$. \[ \boxed{x=0,\ 1} \]