One of the roots of `ax^(2) + bx + c = 0` is greater than 2 and the other is less than -1. If the roots of `cx^(2) + bx + a = 0` are `alpha and beta`, then

Options

A. `0 lt alpha lt 1/2 and -1 lt beta lt 0`
B. `alpha lt 1/2 and beta lt -1`
C. `alpha gt 1/2 and beta gt -1`
D. `alpha lt 2 and beta gt -1`

Explanation

To solve the problem step by step, we will analyze the given quadratic equations and the relationships between their roots.

### Step 1: Understand the given equations
We have two quadratic equations:
1. \( ax^2 + bx + c = 0 \) with roots \( \gamma \) and \( \delta \).
2. \( cx^2 + bx + a = 0 \) with roots \( \alpha \) and \( \beta \).

We know from the problem that one root of the first equation is greater than 2 and the other is less than -1.

### Step 2: Set up the inequalities for the roots
Let’s denote the roots of the first equation as:
- \( \gamma > 2 \)
- \( \delta < -1 \)

### Step 3: Use the relationship between the roots
The roots of the second equation \( cx^2 + bx + a = 0 \) are the reciprocals of the roots of the first equation. Thus:
- \( \alpha = \frac{1}{\gamma} \)
- \( \beta = \frac{1}{\delta} \)

### Step 4: Analyze the implications of the inequalities
Since \( \gamma > 2 \):
- \( \alpha = \frac{1}{\gamma} < \frac{1}{2} \)

Since \( \delta < -1 \):
- \( \beta = \frac{1}{\delta} > -1 \)

### Step 5: Combine the results
From the analysis, we have:
- \( \alpha < \frac{1}{2} \)
- \( \beta > -1 \)

### Step 6: Determine the ranges for \( \alpha \) and \( \beta \)
Since \( \alpha \) must be positive (as \( \gamma > 2 \) implies \( \alpha < \frac{1}{2} \) and cannot be negative), we conclude:
- \( 0 < \alpha < \frac{1}{2} \)
- \( \beta > -1 \)

### Final Result
Thus, the ranges for the roots \( \alpha \) and \( \beta \) are:
- \( 0 < \alpha < \frac{1}{2} \)
- \( \beta > -1 \)