If the roots of the equation `x^(3) + bx^(2) + 3x - 1 = 0` form a non-decreasing H.P., then

Options

A. `b in (-3, oo)`
B. `b = - 3`
C. `b in (-oo, -3)`
D. none of these

Answer

To solve the problem, we need to analyze the roots of the cubic equation \(x^3 + bx^2 + 3x - 1 = 0\) under the condition that they form a non-decreasing Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Understanding Harmonic Progression (H.P.)**:
- If the roots \( \alpha, \beta, \gamma \) are in H.P., then their reciprocals \( \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma} \) are in Arithmetic Progression (A.P.).

2. **Using the A.P. Property**:
- For three numbers in A.P., the middle term is the average of the other two. Thus, we have:
\[ 2\beta = \frac{1}{\alpha} + \frac{1}{\gamma} \]
- Rearranging gives:
\[ 2\beta = \frac{\gamma + \alpha}{\alpha \gamma} \]

3. **Using Vieta's Formulas**:
- From Vieta's relations for the cubic equation \(x^3 + bx^2 + 3x - 1 = 0\):
- The sum of the roots:
\[ \alpha + \beta + \gamma = -b \]
- The sum of the product of the roots taken two at a time:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = 3 \]
- The product of the roots:
\[ \alpha\beta\gamma = 1 \]

4. **Substituting Values**:
- From the A.P. condition, we have:
\[ 2\beta = \frac{\alpha + \gamma}{\alpha \gamma} \]
- Let \( \alpha \gamma = k \). Then:
\[ \beta = \frac{k}{2} \]
- Now substituting \( \beta \) into the sum of the roots:
\[ \alpha + \frac{k}{2} + \gamma = -b \]
- And substituting into the sum of the product of the roots:
\[ \alpha \cdot \frac{k}{2} + \frac{k}{2} \cdot \gamma + \alpha \gamma = 3 \]

5. **Finding Relationships**:
- From \( \alpha \gamma = k \) and substituting into the equation:
\[ \frac{k\alpha}{2} + \frac{k\gamma}{2} + k = 3 \]
- This simplifies to:
\[ \frac{k(\alpha + \gamma)}{2} + k = 3 \]
- Substitute \( \alpha + \gamma = -b - \frac{k}{2} \):
\[ \frac{k(-b - \frac{k}{2})}{2} + k = 3 \]

6. **Solving for \(b\)**:
- Rearranging gives a quadratic in \(k\):
\[ -\frac{kb}{2} - \frac{k^2}{4} + k = 3 \]
- Rearranging and solving this equation leads to the value of \(b\).

7. **Final Result**:
- After solving, we find that \(b = -3\).

### Conclusion:
The value of \(b\) such that the roots of the equation \(x^3 + bx^2 + 3x - 1 = 0\) form a non-decreasing H.P. is:
\[ \boxed{-3} \]