Suppose `a in R, "If" 3x^(2) + 2(a^(2) +1) x + (a^(2) - 3a +2) = 0` possesses roots of opposite signs, then a lies in t he interval :

Options

A. `(- infty, -1)`
B. `(-1,1)`
C. (1,2)
D. (2,3)

Answer

To solve the problem, we need to determine the interval of \( a \) such that the quadratic equation

\[ 3x^2 + 2(a^2 + 1)x + (a^2 - 3a + 2) = 0 \]

has roots of opposite signs.

### Step 1: Identify the coefficients
The quadratic equation can be represented in the standard form \( Ax^2 + Bx + C = 0 \), where:
- \( A = 3 \)
- \( B = 2(a^2 + 1) \)
- \( C = a^2 - 3a + 2 \)

### Step 2: Condition for opposite signs
For the roots of the quadratic equation to be of opposite signs, the product of the roots must be negative. The product of the roots of a quadratic equation is given by:

\[ \text{Product of roots} = \frac{C}{A} \]

Thus, we have:

\[ \frac{C}{A} = \frac{a^2 - 3a + 2}{3} < 0 \]

### Step 3: Set up the inequality
We need to solve the inequality:

\[ a^2 - 3a + 2 < 0 \]

### Step 4: Factor the quadratic expression
To factor \( a^2 - 3a + 2 \), we look for two numbers that multiply to \( 2 \) and add to \( -3 \). The factors are \( -1 \) and \( -2 \):

\[ a^2 - 3a + 2 = (a - 1)(a - 2) \]

### Step 5: Solve the inequality
Now we solve the inequality:

\[ (a - 1)(a - 2) < 0 \]

### Step 6: Determine the critical points
The critical points are \( a = 1 \) and \( a = 2 \). We will test the intervals defined by these points: \( (-\infty, 1) \), \( (1, 2) \), and \( (2, \infty) \).

1. **For \( a < 1 \)** (e.g., \( a = 0 \)):
\[ (0 - 1)(0 - 2) = 1 \cdot 2 = 2 > 0 \]
2. **For \( 1 < a < 2 \)** (e.g., \( a = 1.5 \)):
\[ (1.5 - 1)(1.5 - 2) = 0.5 \cdot (-0.5) = -0.25 < 0 \]
3. **For \( a > 2 \)** (e.g., \( a = 3 \)):
\[ (3 - 1)(3 - 2) = 2 \cdot 1 = 2 > 0 \]

### Step 7: Conclusion
The inequality \( (a - 1)(a - 2) < 0 \) holds true in the interval \( (1, 2) \).

Thus, the interval of \( a \) for which the quadratic equation has roots of opposite signs is:

\[ \boxed{(1, 2)} \]