The number of integral values of a for which `x^(2) - (a-1) x+3 = 0` has both roots positive and `x^(2) + 3x + 6 - a = 0` has both roots negative is

Options

A. 0
B. 1
C. 2
D. infinite

Explanation

To solve the problem, we need to analyze the two quadratic equations given:

1. \( f(x) = x^2 - (a-1)x + 3 = 0 \)
2. \( g(x) = x^2 + 3x + 6 - a = 0 \)

We need to find the integral values of \( a \) such that the first equation has both roots positive and the second equation has both roots negative.

### Step 1: Analyze the first equation \( f(x) \)

For the quadratic equation \( f(x) \) to have both roots positive, we need to satisfy the following conditions:

1. **Condition 1**: \( f(0) > 0 \)
\[ f(0) = 3 > 0 \quad \text{(This condition is always satisfied)} \]

2. **Condition 2**: The discriminant \( D \) must be positive:
\[ D = (a-1)^2 - 4 \cdot 1 \cdot 3 > 0 \]
\[ (a-1)^2 - 12 > 0 \]
\[ (a-1)^2 > 12 \]
Taking square roots:
\[ a - 1 > 2\sqrt{3} \quad \text{or} \quad a - 1 < -2\sqrt{3} \]
Thus:
\[ a > 1 + 2\sqrt{3} \quad \text{or} \quad a < 1 - 2\sqrt{3} \]

3. **Condition 3**: The vertex of the parabola must be positive:
\[ x = \frac{-(a-1)}{2 \cdot 1} = \frac{1-a}{2} > 0 \]
\[ 1 - a > 0 \implies a < 1 \]

### Step 2: Analyze the second equation \( g(x) \)

For the quadratic equation \( g(x) \) to have both roots negative, we need to satisfy the following conditions:

1. **Condition 1**: \( g(0) > 0 \)
\[ g(0) = 6 - a > 0 \implies a < 6 \]

2. **Condition 2**: The discriminant \( D \) must be positive:
\[ D = 3^2 - 4 \cdot 1 \cdot (6 - a) > 0 \]
\[ 9 - 24 + 4a > 0 \]
\[ 4a > 15 \implies a > \frac{15}{4} \]

3. **Condition 3**: The vertex must be negative:
\[ x = \frac{-3}{2} < 0 \quad \text{(This condition is always satisfied)} \]

### Step 3: Combine the conditions

From the analysis of \( f(x) \):
- \( a > 1 + 2\sqrt{3} \) or \( a < 1 - 2\sqrt{3} \)
- \( a < 1 \)

From the analysis of \( g(x) \):
- \( a < 6 \)
- \( a > \frac{15}{4} \)

### Step 4: Calculate \( 1 + 2\sqrt{3} \)

Calculating \( 2\sqrt{3} \):
\[ 2\sqrt{3} \approx 2 \times 1.732 \approx 3.464 \]
Thus:
\[ 1 + 2\sqrt{3} \approx 4.464 \]

### Step 5: Determine the range for \( a \)

Combining the conditions:
- From \( f(x) \): \( a > 4.464 \)
- From \( g(x) \): \( \frac{15}{4} = 3.75 < a < 6 \)

Thus, the combined range for \( a \) is:
\[ 4.464 < a < 6 \]

### Step 6: Find integral values of \( a \)

The only integral value of \( a \) in the range \( (4.464, 6) \) is:
\[ a = 5 \]

### Final Answer

The number of integral values of \( a \) is:
\[ \boxed{1} \]