Question

Let $f:[-1,2] \to \mathbb{R}$ be defined by $f(x) = [x^2-3]$ where $[.]$ denotes greatest integer function, then the number of points of discontinuity for the function $f$ in $(-1,2)$ is

• A. 5
• B. 4
• C. 3
• D. 2

Hint:

The greatest integer function $[f(x)]$ is discontinuous whenever $f(x)$ is an integer, with a key exception: if $f(x)$ merely touches an integer value at a local minimum or maximum without crossing it, the function may be continuous at that point.

Answer 1

The Correct Option is C

Step 1: Identify the condition for discontinuity.
The greatest integer function $[u]$ is discontinuous whenever its argument $u$ takes on an integer value. So, we need to find the values of $x$ in the interval $(-1,2)$ for which the inner function, $h(x) = x^2-3$, is an integer.

Step 2: Find the range of the inner function over the given interval.
The interval for $x$ is $(-1,2)$. For $x \in (-1,2)$, the values of $x^2$ are in the interval $[0, 4)$. Therefore, the range of $h(x) = x^2-3$ is $[0-3, 4-3)$, which is $[-3, 1)$.

Step 3: Find the integer values that the inner function takes.
The integer values in the range $[-3, 1)$ are $-3, -2, -1, 0$.

Step 4: Find the corresponding x-values and check for continuity.
We solve $x^2-3 = n$ for each integer $n$, which is $x^2 = n+3$. For $n=-3$: $x^2 = 0 \implies x=0$. At $x=0$, the function $h(x)=x^2-3$ has a local minimum. For $x$ in a small neighborhood around 0, $h(x) \ge -3$. For example, if $x \in (-0.1, 0.1)$, then $x^2-3 \in [-3, -2.99)$. In this neighborhood, $[x^2-3]$ is always $-3$. Thus, the function is continuous at $x=0$. For $n=-2$: $x^2 = 1 \implies x=\pm 1$. Only $x=1$ is in the interval $(-1,2)$. At $x=1$, the function $h(x)$ is strictly increasing, so it crosses the integer value, causing a discontinuity. For $n=-1$: $x^2 = 2 \implies x=\pm \sqrt{2}$. Only $x=\sqrt{2}$ is in the interval $(-1,2)$. This is a point of discontinuity. For $n=0$: $x^2 = 3 \implies x=\pm \sqrt{3}$. Only $x=\sqrt{3}$ is in the interval $(-1,2)$ (since $\sqrt{3} \approx 1.732$). This is a point of discontinuity.

Step 5: Count the points of discontinuity.
The points of discontinuity in the open interval $(-1,2)$ are $1, \sqrt{2}, \sqrt{3}$. There are 3 such points.