If `alpha,beta` be the roots of `x^(2)-a(x-1)+b=0`, then the value of `(1)/(alpha^(2)-a alpha)+(1)/(beta^(2)-a beta)+(2)/(a+b)` is

Options

A. `(4)/(a+b)`
B. `(1)/(a+b)`
C. 0
D. `-1`

Answer

To solve the problem, we need to evaluate the expression

\[ \frac{1}{\alpha^2 - a\alpha} + \frac{1}{\beta^2 - a\beta} + \frac{2}{a+b} \]

given that \(\alpha\) and \(\beta\) are the roots of the quadratic equation

\[ x^2 - a(x - 1) + b = 0. \]

### Step 1: Rewrite the quadratic equation
The given quadratic equation can be rewritten as:

\[ x^2 - ax + a + b = 0. \]

### Step 2: Use the property of roots
Since \(\alpha\) and \(\beta\) are the roots of the equation, they satisfy:

\[ \alpha^2 - a\alpha + (a + b) = 0 \quad \text{and} \quad \beta^2 - a\beta + (a + b) = 0. \]

From these equations, we can express \(\alpha^2 - a\alpha\) and \(\beta^2 - a\beta\):

\[ \alpha^2 - a\alpha = - (a + b) \quad \text{and} \quad \beta^2 - a\beta = - (a + b). \]

### Step 3: Substitute into the expression
Now substituting these results into the expression we want to evaluate:

\[ \frac{1}{\alpha^2 - a\alpha} = \frac{1}{-(a + b)} \quad \text{and} \quad \frac{1}{\beta^2 - a\beta} = \frac{1}{-(a + b)}. \]

Thus, we have:

\[ \frac{1}{\alpha^2 - a\alpha} + \frac{1}{\beta^2 - a\beta} = \frac{1}{-(a + b)} + \frac{1}{-(a + b)} = \frac{-2}{a + b}. \]

### Step 4: Combine with the third term
Now we can combine this with the third term in the expression:

\[ \frac{-2}{a + b} + \frac{2}{a + b} = 0. \]

### Final Answer
Thus, the value of the expression

\[ \frac{1}{\alpha^2 - a\alpha} + \frac{1}{\beta^2 - a\beta} + \frac{2}{a+b} \]

is

\[ \boxed{0}. \]