If 3x`-x^2ge2` and `y^2+yle2`, then

Options

A. `-1lexyle2`
B. `-2lexyle2`
C. `-4lexyle4`
D. `-4lexyle2`

Explanation

To solve the inequalities given in the question step by step, we will analyze each inequality separately and then find the relation between \(xy\).

### Step 1: Solve the first inequality \(3x - x^2 \geq 2\)

1. Rearrange the inequality:
\[ 3x - x^2 - 2 \geq 0 \]
This can be rewritten as:
\[ -x^2 + 3x - 2 \geq 0 \]

2. Multiply through by -1 (remember to reverse the inequality):
\[ x^2 - 3x + 2 \leq 0 \]

3. Factor the quadratic:
\[ (x - 1)(x - 2) \leq 0 \]

4. Determine the intervals where this inequality holds:
- The critical points are \(x = 1\) and \(x = 2\).
- Test intervals:
- For \(x < 1\), choose \(x = 0\): \((0 - 1)(0 - 2) = 2 > 0\) (not valid).
- For \(1 < x < 2\), choose \(x = 1.5\): \((1.5 - 1)(1.5 - 2) = 0.5 \cdot -0.5 < 0\) (valid).
- For \(x > 2\), choose \(x = 3\): \((3 - 1)(3 - 2) = 2 > 0\) (not valid).

5. Therefore, the solution for \(x\) is:
\[ 1 \leq x \leq 2 \]

### Step 2: Solve the second inequality \(y^2 + y \leq 2\)

1. Rearrange the inequality:
\[ y^2 + y - 2 \leq 0 \]

2. Factor the quadratic:
\[ (y - 1)(y + 2) \leq 0 \]

3. Determine the intervals where this inequality holds:
- The critical points are \(y = -2\) and \(y = 1\).
- Test intervals:
- For \(y < -2\), choose \(y = -3\): \((-3 - 1)(-3 + 2) = -4 < 0\) (valid).
- For \(-2 < y < 1\), choose \(y = 0\): \((0 - 1)(0 + 2) = -2 < 0\) (valid).
- For \(y > 1\), choose \(y = 2\): \((2 - 1)(2 + 2) = 4 > 0\) (not valid).

4. Therefore, the solution for \(y\) is:
\[ -2 \leq y \leq 1 \]

### Step 3: Find the range of \(xy\)

1. The maximum value of \(x\) is \(2\) and the maximum value of \(y\) is \(1\). Thus:
\[ \text{Maximum } xy = 2 \cdot 1 = 2 \]

2. The minimum value of \(y\) is \(-2\) and the minimum value of \(x\) is \(1\). Thus:
\[ \text{Minimum } xy = 1 \cdot (-2) = -2 \]

3. The minimum value of \(xy\) when \(x = 2\) and \(y = -2\) is:
\[ 2 \cdot (-2) = -4 \]

### Conclusion

Thus, the range of \(xy\) is:
\[ -4 \leq xy \leq 2 \]