Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$, defined by $f(x) = (x - \sqrt{3})sgn(x^2-3)$. Then the number of points of discontinuity of $f$ is: (where "sgn" denotes signum function)

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

1

Answer 2

To determine the number of points of discontinuity of the function $f(x) = (x - \sqrt{3})sgn(x^2-3)$, we need to analyze the behavior of the function at points where the signum function's argument changes sign or becomes zero.

The signum function $sgn(u)$ is defined as: $sgn(u) = 1$ if $u > 0$ $sgn(u) = 0$ if $u = 0$ $sgn(u) = -1$ if $u < 0$

The signum function is discontinuous at $u=0$. In our case, $u = x^2 - 3$. So, potential points of discontinuity for $f(x)$ are where $x^2 - 3 = 0$, which means $x^2 = 3$, so $x = \sqrt{3}$ or $x = -\sqrt{3}$.

Let's analyze the continuity of $f(x)$ at these two points.

Case 1: At $x = \sqrt{3}$

1. Value of the function at $x = \sqrt{3}$: $f(\sqrt{3}) = (\sqrt{3} - \sqrt{3})sgn((\sqrt{3})^2 - 3) = (0)sgn(3 - 3) = 0 \cdot sgn(0) = 0 \cdot 0 = 0$.

2. Left-hand limit at $x = \sqrt{3}$: $\lim_{x \to \sqrt{3}^-} f(x) = \lim_{x \to \sqrt{3}^-} (x - \sqrt{3})sgn(x^2 - 3)$ As $x \to \sqrt{3}^-$, $x$ is slightly less than $\sqrt{3}$. This means $x^2$ is slightly less than $3$, so $x^2 - 3 < 0$. Therefore, $sgn(x^2 - 3) = -1$. $\lim_{x \to \sqrt{3}^-} (x - \sqrt{3})(-1) = (\sqrt{3} - \sqrt{3})(-1) = 0 \cdot (-1) = 0$.

3. Right-hand limit at $x = \sqrt{3}$: $\lim_{x \to \sqrt{3}^+} f(x) = \lim_{x \to \sqrt{3}^+} (x - \sqrt{3})sgn(x^2 - 3)$ As $x \to \sqrt{3}^+$, $x$ is slightly greater than $\sqrt{3}$. This means $x^2$ is slightly greater than $3$, so $x^2 - 3 > 0$. Therefore, $sgn(x^2 - 3) = 1$. $\lim_{x \to \sqrt{3}^+} (x - \sqrt{3})(1) = (\sqrt{3} - \sqrt{3})(1) = 0 \cdot (1) = 0$.

Since $f(\sqrt{3}) = \lim_{x \to \sqrt{3}^-} f(x) = \lim_{x \to \sqrt{3}^+} f(x) = 0$, the function $f(x)$ is continuous at $x = \sqrt{3}$.

Case 2: At $x = -\sqrt{3}$

1. Value of the function at $x = -\sqrt{3}$: $f(-\sqrt{3}) = (-\sqrt{3} - \sqrt{3})sgn((-\sqrt{3})^2 - 3) = (-2\sqrt{3})sgn(3 - 3) = (-2\sqrt{3}) \cdot sgn(0) = (-2\sqrt{3}) \cdot 0 = 0$.

2. Left-hand limit at $x = -\sqrt{3}$: $\lim_{x \to -\sqrt{3}^-} f(x) = \lim_{x \to -\sqrt{3}^-} (x - \sqrt{3})sgn(x^2 - 3)$ As $x \to -\sqrt{3}^-$, $x$ is slightly less than $-\sqrt{3}$. For example, $x = -1.8$. Then $x^2 = (-1.8)^2 = 3.24$, which is greater than $3$. So $x^2 - 3 > 0$. Therefore, $sgn(x^2 - 3) = 1$. $\lim_{x \to -\sqrt{3}^-} (x - \sqrt{3})(1) = (-\sqrt{3} - \sqrt{3})(1) = -2\sqrt{3}$.

3. Right-hand limit at $x = -\sqrt{3}$: $\lim_{x \to -\sqrt{3}^+} f(x) = \lim_{x \to -\sqrt{3}^+} (x - \sqrt{3})sgn(x^2 - 3)$ As $x \to -\sqrt{3}^+$, $x$ is slightly greater than $-\sqrt{3}$. For example, $x = -1.7$. Then $x^2 = (-1.7)^2 = 2.89$, which is less than $3$. So $x^2 - 3 < 0$. Therefore, $sgn(x^2 - 3) = -1$. $\lim_{x \to -\sqrt{3}^+} (x - \sqrt{3})(-1) = (-\sqrt{3} - \sqrt{3})(-1) = (-2\sqrt{3})(-1) = 2\sqrt{3}$.

Since $\lim_{x \to -\sqrt{3}^-} f(x) = -2\sqrt{3}$ and $\lim_{x \to -\sqrt{3}^+} f(x) = 2\sqrt{3}$, the left-hand limit and the right-hand limit are not equal. Therefore, the limit of $f(x)$ as $x \to -\sqrt{3}$ does not exist, and $f(x)$ is discontinuous at $x = -\sqrt{3}$.

Combining both cases, the function $f(x)$ is discontinuous only at $x = -\sqrt{3}$. Thus, there is only 1 point of discontinuity.

The final answer is $\boxed{1}$.

Explanation: The function $f(x) = (x - \sqrt{3})sgn(x^2-3)$ is a product of a continuous function $g(x) = x - \sqrt{3}$ and a function $h(x) = sgn(x^2-3)$ which is discontinuous at $x^2-3=0$, i.e., $x=\sqrt{3}$ and $x=-\sqrt{3}$.

1. At $x = -\sqrt{3}$: $g(-\sqrt{3}) = -2\sqrt{3} \neq 0$. When a continuous function (non-zero at the point) is multiplied by a discontinuous function, the product is discontinuous. Thus, $f(x)$ is discontinuous at $x = -\sqrt{3}$.
2. At $x = \sqrt{3}$: $g(\sqrt{3}) = 0$. When a continuous function that is zero at a point is multiplied by a function with a finite jump discontinuity, the product often becomes continuous. In this case, $\lim_{x \to \sqrt{3}^-} f(x) = \lim_{x \to \sqrt{3}^-} (x-\sqrt{3})(-1) = 0$, $\lim_{x \to \sqrt{3}^+} f(x) = \lim_{x \to \sqrt{3}^+} (x-\sqrt{3})(1) = 0$, and $f(\sqrt{3})=0$. Since all three values are equal, $f(x)$ is continuous at $x = \sqrt{3}$. Therefore, there is only one point of discontinuity.