The number of ordered pairs (x, y) of real number satisfying `4x^2 - 4x + 2 = sin^2 y and x^2 + y^2 <= 3` equal to

Options

A. 0
B. 2
C. 4
D. 8

Answer

To find the number of ordered pairs \((x, y)\) of real numbers satisfying the equations \(4x^2 - 4x + 2 = \sin^2 y\) and \(x^2 + y^2 \leq 3\), we can follow these steps:

### Step 1: Analyze the first equation
We start with the equation:
\[ 4x^2 - 4x + 2 = \sin^2 y \]
We can rewrite the left-hand side:
\[ 4x^2 - 4x + 2 = 4\left(x^2 - x + \frac{1}{2}\right) \]
To simplify further, we complete the square:
\[ x^2 - x + \frac{1}{2} = \left(x - \frac{1}{2}\right)^2 + \frac{1}{4} \]
Thus, we have:
\[ 4\left(\left(x - \frac{1}{2}\right)^2 + \frac{1}{4}\right) = 4\left(x - \frac{1}{2}\right)^2 + 1 \]
So, we can rewrite the equation as:
\[ 4\left(x - \frac{1}{2}\right)^2 + 1 = \sin^2 y \]

### Step 2: Determine the range of \(\sin^2 y\)
The function \(\sin^2 y\) has a range of \([0, 1]\). Therefore, we set up the inequality:
\[ 4\left(x - \frac{1}{2}\right)^2 + 1 \leq 1 \]
This simplifies to:
\[ 4\left(x - \frac{1}{2}\right)^2 \leq 0 \]
Since \(4\left(x - \frac{1}{2}\right)^2\) is always non-negative, the only solution occurs when:
\[ 4\left(x - \frac{1}{2}\right)^2 = 0 \implies x - \frac{1}{2} = 0 \implies x = \frac{1}{2} \]

### Step 3: Substitute \(x\) into the second equation
Now that we have \(x = \frac{1}{2}\), we substitute this into the second equation:
\[ x^2 + y^2 \leq 3 \]
This becomes:
\[ \left(\frac{1}{2}\right)^2 + y^2 \leq 3 \]
\[ \frac{1}{4} + y^2 \leq 3 \implies y^2 \leq 3 - \frac{1}{4} = \frac{12}{4} - \frac{1}{4} = \frac{11}{4} \]
Thus, we have:
\[ y^2 \leq \frac{11}{4} \implies -\sqrt{\frac{11}{4}} \leq y \leq \sqrt{\frac{11}{4}} \implies -\frac{\sqrt{11}}{2} \leq y \leq \frac{\sqrt{11}}{2} \]

### Step 4: Determine the possible values of \(y\)
The values of \(y\) can take any value in the interval \([- \frac{\sqrt{11}}{2}, \frac{\sqrt{11}}{2}]\). Since \(\sin^2 y\) can take any value from \(0\) to \(1\), we can find specific values of \(y\) that correspond to \(\sin^2 y = 0\) and \(\sin^2 y = 1\).

### Step 5: Find the ordered pairs
The specific values of \(y\) that satisfy \(\sin^2 y = 0\) are:
\[ y = n\pi \quad (n \in \mathbb{Z}) \]
The specific values of \(y\) that satisfy \(\sin^2 y = 1\) are:
\[ y = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
However, we need to ensure these values of \(y\) lie within the interval \([- \frac{\sqrt{11}}{2}, \frac{\sqrt{11}}{2}]\).

From the range of \(y\), we find two specific values:
1. \(y = -\frac{\pi}{2}\)
2. \(y = \frac{\pi}{2}\)

Thus, the ordered pairs are:
1. \(\left(\frac{1}{2}, -\frac{\pi}{2}\right)\)
2. \(\left(\frac{1}{2}, \frac{\pi}{2}\right)\)

### Final Answer
The number of ordered pairs \((x, y)\) satisfying the given conditions is:
\[ \boxed{2} \]