The range of the function `y=sqrt(2{x}-{x}^2-3/4)` (where, denotes the fractional part) is

Options

A. `[-1/4,1/4]`
B. `[0,1/2)`
C. `[0,1/4]`
D. `[1/4,1/2]`

Answer

To find the range of the function \( y = \sqrt{2 \{x\} - \{x^2\} - \frac{3}{4}} \), where \( \{x\} \) denotes the fractional part of \( x \), we can follow these steps:

### Step 1: Understand the fractional part
The fractional part \( \{x\} \) of a number \( x \) is defined as \( x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \). Thus, \( 0 \leq \{x\} < 1 \).

### Step 2: Substitute \( t \) for the fractional part
Let \( t = \{x\} \). Therefore, the function can be rewritten as:
\[ y = \sqrt{2t - t^2 - \frac{3}{4}} \]

### Step 3: Find the expression under the square root
We need to ensure that the expression under the square root is non-negative:
\[ 2t - t^2 - \frac{3}{4} \geq 0 \]

### Step 4: Rearranging the inequality
Rearranging gives us:
\[ -t^2 + 2t - \frac{3}{4} \geq 0 \]
Multiplying through by -1 (and flipping the inequality) gives:
\[ t^2 - 2t + \frac{3}{4} \leq 0 \]

### Step 5: Factor the quadratic
To factor \( t^2 - 2t + \frac{3}{4} \), we can find the roots using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot \frac{3}{4}}}{2 \cdot 1} \]
Calculating the discriminant:
\[ = \frac{2 \pm \sqrt{4 - 3}}{2} = \frac{2 \pm 1}{2} \]
This gives us:
\[ t = \frac{3}{2} \quad \text{and} \quad t = \frac{1}{2} \]

### Step 6: Analyze the intervals
The roots \( t = \frac{1}{2} \) and \( t = \frac{3}{2} \) divide the number line into intervals. We need to test the sign of the quadratic in these intervals:
- For \( t < \frac{1}{2} \): positive
- For \( \frac{1}{2} < t < \frac{3}{2} \): negative
- For \( t > \frac{3}{2} \): positive

Thus, the quadratic is non-positive in the interval:
\[ \frac{1}{2} \leq t \leq \frac{3}{2} \]

### Step 7: Restrict \( t \) to the fractional part
Since \( t = \{x\} \) must satisfy \( 0 \leq t < 1 \), we restrict our interval:
\[ \frac{1}{2} \leq t < 1 \]

### Step 8: Find the range of \( y \)
Now substituting the endpoints back into the function:
1. For \( t = \frac{1}{2} \):
\[ y = \sqrt{2 \cdot \frac{1}{2} - \left(\frac{1}{2}\right)^2 - \frac{3}{4}} = \sqrt{1 - \frac{1}{4} - \frac{3}{4}} = \sqrt{0} = 0 \]
2. For \( t \) approaching 1:
\[ y = \sqrt{2 \cdot 1 - 1^2 - \frac{3}{4}} = \sqrt{2 - 1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Conclusion
Thus, the range of the function \( y \) is:
\[ [0, \frac{1}{2}) \]