If the roots of the quadratic equation `ax^(2)+bx-b=0`, where `a`, `b in R` such that `a*b gt 0`, are `alpha` and `beta`, then the value of `log_(|(beta-1)|)|(alpha-1)|` is

Options

A. (a) `1`
B. (b) `-1`
C. (c) `0`
D. (d) none of these

Explanation

To solve the problem, we need to analyze the given quadratic equation and its roots. Let's break down the solution step by step.

### Step 1: Identify the roots of the quadratic equation
The given quadratic equation is:
\[ ax^2 + bx - b = 0 \]

Let the roots of this equation be \( \alpha \) and \( \beta \).

### Step 2: Use Vieta's formulas
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{-b}{a} \)

### Step 3: Set up the equation involving the roots
We can express the relationship between the roots as follows:
\[ \alpha \beta = \alpha + \beta \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \alpha \beta - \alpha - \beta = 0 \]
Adding 1 to both sides, we have:
\[ \alpha \beta - \alpha - \beta + 1 = 1 \]

### Step 5: Factor the left-hand side
Factoring the left-hand side gives:
\[ (\alpha - 1)(\beta - 1) = 1 \]

### Step 6: Analyze the conditions on \( a \) and \( b \)
Given that \( ab > 0 \), we have two cases:
1. Both \( a > 0 \) and \( b > 0 \)
2. Both \( a < 0 \) and \( b < 0 \)

In both cases, we can conclude that:
- The sum \( \alpha + \beta < 0 \)
- The product \( \alpha \beta < 0 \)

### Step 7: Determine the signs of \( \alpha \) and \( \beta \)
From the conditions, we can deduce:
- One root is negative and the other is positive.

### Step 8: Evaluate \( | \alpha - 1 | \) and \( | \beta - 1 | \)
Since one root is negative and the other is positive, we analyze:
- If \( \alpha < 0 \), then \( \alpha - 1 < -1 \) and \( | \alpha - 1 | = -(\alpha - 1) = 1 - \alpha \)
- If \( \beta > 0 \) and \( \beta < 1 \), then \( \beta - 1 < 0 \) and \( | \beta - 1 | = 1 - \beta \)

### Step 9: Substitute into the logarithm
Using the relationship \( (\alpha - 1)(\beta - 1) = 1 \), we have:
\[ | \alpha - 1 | \cdot | \beta - 1 | = 1 \]

### Step 10: Take logarithm
Taking the logarithm gives:
\[ \log_{| \beta - 1 |} | \alpha - 1 | = \log_{| \beta - 1 |} \left( \frac{1}{| \beta - 1 |} \right) = -1 \]

### Final Answer
Thus, the value of \( \log_{|( \beta - 1 )|} |( \alpha - 1 )| \) is:
\[ \boxed{-1} \]