The number of points of discontinuity of the function f(x) = [x²/2] - [√x], x ∈ [0, 4], where [·] de

Q: Why do we only check where the internal functions are integers? Can't it be discontinuous elsewhere?

Great question! The Greatest Integer Function $[u]$ is continuous everywhere except at integers. Since our function is a combination of two such functions, the only places where the 'jumps' can originate are the points where the internal arguments $\frac{x^2}{2}$ or $\sqrt{x}$ hit an integer. Everywhere else, the function is locally constant or continuous, so no jumps occur.

Q: Wait, why did $x=4$ turn out to be continuous? I thought it was a suspect point!

That is the beauty of this problem! At $x=4$, both $\frac{x^2}{2}$ and $\sqrt{x}$ hit integers (8 and 2, respectively). This causes both terms to jump. However, the jump in the first term was exactly compensated by the jump in the second term, leading to a smooth transition at that specific point. Always verify the limits!

Q: Do I need to check the right-hand limit as well?

For a function to be continuous at $x=a$, we need $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$. In this problem, checking the left-hand limit and the function value is usually sufficient to detect the jump, but checking the right-hand limit is a great habit to ensure you haven't missed any subtle behavior.

Question

The number of points of discontinuity of the function $f(x) = [\frac{x^2}{2}] - [\sqrt{x}], x \in [0, 4]$, where $[\cdot]$ denotes the greatest integer function is ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The function is defined as $f(x) = [\frac{x^2}{2}] - [\sqrt{x}]$ within the interval $[0, 4]$. The Greatest Integer Function, denoted by $[\cdot]$, creates a "staircase" of discontinuities whenever its internal argument reaches an integer value.

To identify the points of discontinuity, we must find all $x \in [0, 4]$ such that either $\frac{x^2}{2} \in \mathbb{Z}$ or $\sqrt{x} \in \mathbb{Z}$.

### The Hunt for Suspects

First, consider the term $\frac{x^2}{2}$. As $x$ ranges from $0$ to $4$, the expression $\frac{x^2}{2}$ ranges from $0$ to $8$. The integers it hits are $\{1, 2, 3, 4, 5, 6, 7, 8\}$.

Solving $\frac{x^2}{2} = k$ for these integers yields the following candidate set:
$$x \in \{\sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4\}$$

Next, consider the term $\sqrt{x}$. In the interval $[0, 4]$, $\sqrt{x}$ ranges from $0$ to $2$. The integers it hits are $1$ and $2$.

Solving $\sqrt{x} = m$ gives us $x=1$ and $x=4$. Combining these, our master list of suspects is:
$$\{1, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4\}$$

### The Investigation: To Jump or Not to Jump?

We must verify if these points are actual points of discontinuity by checking the limits. Let us test $x=1$:
$$f(1) = [\frac{1}{2}] - [1] = 0 - 1 = -1$$

The left-hand limit is:
$$\lim_{x 	o 1^-} f(x) = [0.5^-] - [1^-] = 0 - 0 = 0$$
Since $-1 
eq 0$, $x=1$ is a point of discontinuity.

Moving to $x=\sqrt{2}$, we find:
$$f(\sqrt{2}) = [1] - [\sqrt{\sqrt{2}}] = 1 - 1 = 0$$
The left-hand limit is:
$$\lim_{x 	o \sqrt{2}^-} f(x) = [1^-] - [\sqrt{\sqrt{2}}^-] = 0 - 1 = -1$$
Because the values do not match, $x=\sqrt{2}$ is a point of discontinuity. This logic holds for $x=2$ and the cluster $\{\sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}\}$, where the first term jumps while the second remains constant.

### The Final Twist: The Case of $x=4$

Finally, we examine the endpoint $x=4$. The function value is:
$$f(4) = [\frac{16}{2}] - [\sqrt{4}] = 8 - 2 = 6$$

Now, we check the left-hand limit:
$$\lim_{x 	o 4^-} f(x) = [\frac{4^2}{2}^-] - [\sqrt{4}^-] = [8^-] - [2^-] = 7 - 1 = 6$$
Since the limit matches the function value, the jumps in the two terms cancel each other out. Therefore, the function is **continuous** at $x=4$.

### Conclusion

After our rigorous investigation, the points of discontinuity are $\{1, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}\}$.

There are a total of **8 points** of discontinuity. In the world of JEE, never assume; always verify.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The number of points of discontinuity of the function $f (x) = [\frac{x ^{2}}{2}] - [x], x \in [0, 4]$ , where $[\cdot]$ denotes the greatest integer function is ________.

Enter Numerical Value:

Visualized Solution

Defining the Function $f (x)$

Discontinuity of Greatest Integer Function

Candidate Points for $[\frac{x ^{2}}{2}]$

Candidate Points for $[x]$

The Master List of Candidates

Checking $x = 1$

Checking $x = 2$

Checking $x = 2$

Generalizing for $6$ to $14$

Checking the Endpoint $x = 4$

Final Conclusion

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Hunt for Suspects

The Investigation

The Final Twist

Conclusion

Similar Questions

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Enter Numerical Value:

Visualized Solution

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Discontinuity of Greatest Integer Function

Candidate Points for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [2x2​]

Candidate Points for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x​]

The Master List of Candidates

Checking .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Checking .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2​

Checking .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

Generalizing for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6​ to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 14​

Checking the Endpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4

Final Conclusion

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Hunt for Suspects

The Investigation

The Final Twist

Conclusion

Similar Questions

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Defining the Function $f (x)$

Candidate Points for $[\frac{x ^{2}}{2}]$

Candidate Points for $[x]$

Checking $x = 1$

Checking $x = 2$

Checking $x = 2$

Generalizing for $6$ to $14$

Checking the Endpoint $x = 4$

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