The set of values of p for which the roots of the equation `3x^(2)+2x+p(p-1)=0` are of opposite signs is :

Options

A. `(-oo,0)`
B. `(0,1)`
C. `(1,oo)`
D. `(0,oo)`

Answer

To solve the problem, we need to find the set of values of \( p \) for which the roots of the quadratic equation

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

are of opposite signs.

### Step 1: Understand the condition for opposite signs
For the roots of a quadratic equation \( ax^2 + bx + c = 0 \) to be of opposite signs, the product of the roots must be negative. If we denote the roots by \( \alpha \) and \( \beta \), then we have:

\[ \alpha \cdot \beta < 0 \]

### Step 2: Find the product of the roots
Using Vieta's formulas, the product of the roots \( \alpha \cdot \beta \) is given by:

\[ \alpha \cdot \beta = \frac{c}{a} \]

In our equation, \( a = 3 \) and \( c = p(p-1) \). Therefore, we have:

\[ \alpha \cdot \beta = \frac{p(p-1)}{3} \]

### Step 3: Set up the inequality
For the roots to be of opposite signs, we need:

\[ \frac{p(p-1)}{3} < 0 \]

This simplifies to:

\[ p(p-1) < 0 \]

### Step 4: Analyze the inequality
The expression \( p(p-1) < 0 \) means that the product of \( p \) and \( (p-1) \) is negative. This occurs in the following scenarios:

1. \( p < 0 \) and \( p - 1 > 0 \) (which gives \( p > 1 \)) - This scenario is not possible.
2. \( p > 0 \) and \( p - 1 < 0 \) (which gives \( p < 1 \)) - This scenario is valid.

### Step 5: Combine the conditions
From the valid scenario, we have:

\[ 0 < p < 1 \]

### Conclusion
The set of values of \( p \) for which the roots of the equation are of opposite signs is:

\[ p \in (0, 1) \]

### Final Answer
Thus, the answer is:

\[ (0, 1) \]