The range of the function `y=[x^2]-[x]^2` `x in [0,2]` (where [] denotes the greatest integer function), is

Options

A. [0]
B. [0,1]
C. [1,2]
D. [0,1,2]

Explanation

To find the range of the function \( y = [x^2] - [x]^2 \) for \( x \) in the interval \([0, 2]\), we will analyze the function step by step.

### Step 1: Understand the greatest integer function
The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\). For example:
- \([0] = 0\)
- \([1] = 1\)
- \([1.5] = 1\)
- \([2] = 2\)

### Step 2: Determine the intervals for \(x\)
We will consider the intervals of \(x\) in \([0, 2]\):
1. \(x \in [0, 1)\)
2. \(x \in [1, \sqrt{2})\)
3. \(x \in [\sqrt{2}, \sqrt{3})\)
4. \(x \in [\sqrt{3}, 2]\)

### Step 3: Calculate \(y\) for each interval

#### Interval 1: \(x \in [0, 1)\)
- Here, \(x^2 \in [0, 1)\) so \([x^2] = 0\)
- \([x] = 0\) since \(x < 1\)
- Therefore, \(y = [x^2] - [x]^2 = 0 - 0 = 0\)

#### Interval 2: \(x \in [1, \sqrt{2})\)
- Here, \(x^2 \in [1, 2)\) so \([x^2] = 1\)
- \([x] = 1\) since \(1 \leq x < \sqrt{2}\)
- Therefore, \(y = [x^2] - [x]^2 = 1 - 1 = 0\)

#### Interval 3: \(x \in [\sqrt{2}, \sqrt{3})\)
- Here, \(x^2 \in [2, 3)\) so \([x^2] = 2\)
- \([x] = 1\) since \(\sqrt{2} \approx 1.414\) and \(\sqrt{3} \approx 1.732\)
- Therefore, \(y = [x^2] - [x]^2 = 2 - 1 = 1\)

#### Interval 4: \(x \in [\sqrt{3}, 2]\)
- Here, \(x^2 \in [3, 4]\) so \([x^2] = 3\)
- \([x] = 1\) since \(\sqrt{3} \approx 1.732\) and \(x < 2\)
- Therefore, \(y = [x^2] - [x]^2 = 3 - 1 = 2\)

### Step 4: Collect the results
From the calculations above, we have:
- For \(x \in [0, 1)\): \(y = 0\)
- For \(x \in [1, \sqrt{2})\): \(y = 0\)
- For \(x \in [\sqrt{2}, \sqrt{3})\): \(y = 1\)
- For \(x \in [\sqrt{3}, 2]\): \(y = 2\)

### Conclusion
The possible values of \(y\) are \(0\), \(1\), and \(2\). Therefore, the range of the function \(y = [x^2] - [x]^2\) for \(x \in [0, 2]\) is \(\{0, 1, 2\}\).

### Final Answer
The range of the function is \( \{0, 1, 2\} \).

---