if `alpha+(1)/(alpha)` and `2-beta-(1)/(beta)(alpha,betage0)` are the roots of the quadratic eqaution `x^(2)-2(a+1)x+a-3=0` then the integral values of a may be

Options

A. 3
B. 0
C. 5
D. `-1`

Explanation

To solve the problem, we need to analyze the roots of the quadratic equation given and find the integral values of \( a \) based on the conditions provided.

### Step-by-Step Solution:

1. **Identify the Roots**:
The roots of the quadratic equation \( x^2 - 2(a + 1)x + (a - 3) = 0 \) are given as:
- \( r_1 = \alpha + \frac{1}{\alpha} \)
- \( r_2 = 2 - \beta - \frac{1}{\beta} \)

2. **Use Vieta's Formulas**:
According to Vieta's formulas, for a quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots \( r_1 + r_2 = 2(a + 1) \)
- The product of the roots \( r_1 \cdot r_2 = a - 3 \)

3. **Calculate the Sum of Roots**:
\[ r_1 + r_2 = \left(\alpha + \frac{1}{\alpha}\right) + \left(2 - \beta - \frac{1}{\beta}\right) = 2(a + 1) \]

4. **Calculate the Product of Roots**:
\[ r_1 \cdot r_2 = \left(\alpha + \frac{1}{\alpha}\right)\left(2 - \beta - \frac{1}{\beta}\right) = a - 3 \]

5. **Analyze the Conditions**:
- Since \( \alpha \geq 0 \) and \( \beta \geq 0 \), we know that \( \alpha + \frac{1}{\alpha} \geq 2 \) (by AM-GM inequality).
- For \( r_2 \), we need to analyze \( 2 - \beta - \frac{1}{\beta} \). Since \( \beta \geq 0 \), it can be shown that \( 2 - \beta - \frac{1}{\beta} \) can take values less than or equal to 0.

6. **Set Up Inequalities**:
- From \( r_1 \geq 2 \):
\[ \alpha + \frac{1}{\alpha} \geq 2 \implies r_1 \geq 2 \]
- From \( r_2 \leq 0 \):
\[ 2 - \beta - \frac{1}{\beta} \leq 0 \implies \beta + \frac{1}{\beta} \geq 2 \implies \beta \geq 1 \]

7. **Substituting Values**:
- Substitute \( \beta = 1 \) into \( r_2 \):
\[ r_2 = 2 - 1 - 1 = 0 \]
- Substitute \( \beta > 1 \) into \( r_2 \) to find upper limits.

8. **Finding Integral Values of \( a \)**:
- From the product of the roots:
\[ r_1 \cdot r_2 = a - 3 \]
- Since \( r_1 \geq 2 \) and \( r_2 \leq 0 \), we can conclude:
- If \( r_1 \) is at least 2 and \( r_2 \) is less than or equal to 0, we can find values of \( a \) that satisfy the conditions.

9. **Possible Integral Values**:
- Check values for \( a \):
- If \( a = 3 \): \( r_1 \cdot r_2 = 0 \) (valid)
- If \( a = 5 \): \( r_1 \cdot r_2 = 2 \) (valid)
- If \( a = -1 \): \( r_1 \cdot r_2 = -4 \) (valid)

Thus, the integral values of \( a \) that satisfy the conditions are \( a = 3, 5, -1 \).