If `a, b in R` , and the equation `x^(2) + (a - b) x - a - b + 1 = 0 ` has real roots for all `b in R`, then a lies in the in terval

Options

A. `(1, infty)`
B. `(0, infty)`
C. `(- infty, 1)`
D. `(-1, 1)`

Answer

To determine the interval in which \( a \) lies such that the equation

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

has real roots for all \( b \in \mathbb{R} \), we need to analyze the discriminant of the quadratic equation. The general form of a quadratic equation is \( Ax^2 + Bx + C = 0 \), where the discriminant \( D \) is given by:

\[ D = B^2 - 4AC \]

In our case, we can identify:

- \( A = 1 \)
- \( B = a - b \)
- \( C = -(a + b) + 1 = -a - b + 1 \)

Thus, the discriminant \( D \) becomes:

\[ D = (a - b)^2 - 4 \cdot 1 \cdot (-a - b + 1) \]

Expanding this, we have:

\[ D = (a - b)^2 + 4(a + b - 1) \]

Now, we need to ensure that \( D \geq 0 \) for all \( b \in \mathbb{R} \).

### Step 1: Expand the Discriminant

First, we expand \( D \):

\[ D = (a^2 - 2ab + b^2) + 4a + 4b - 4 \]

Combining terms, we get:

\[ D = a^2 - 2ab + b^2 + 4a + 4b - 4 \]

### Step 2: Rearranging the Discriminant

We can rearrange this as:

\[ D = b^2 + (-2a + 4)b + (a^2 + 4a - 4) \]

### Step 3: Condition for Real Roots

For the quadratic in \( b \) to have real roots for all \( b \), the discriminant of this new quadratic must be non-positive:

\[ (-2a + 4)^2 - 4 \cdot 1 \cdot (a^2 + 4a - 4) \leq 0 \]

### Step 4: Calculate the Discriminant

Calculating the discriminant:

\[ (-2a + 4)^2 = 4a^2 - 16a + 16 \]

Now, we compute:

\[ 4(a^2 + 4a - 4) = 4a^2 + 16a - 16 \]

### Step 5: Set Up the Inequality

Setting up the inequality:

\[ 4a^2 - 16a + 16 - (4a^2 + 16a - 16) \leq 0 \]

This simplifies to:

\[ -32a + 32 \leq 0 \]

### Step 6: Solve the Inequality

Solving for \( a \):

\[ -32a \leq -32 \implies a \geq 1 \]

### Step 7: Determine the Interval

Since we want the discriminant to be non-positive, we also need to ensure that the leading coefficient of the quadratic in \( b \) is positive, which it is since \( A = 1 > 0 \).

Thus, the final condition for \( a \) is:

\[ a \leq 1 \]

### Conclusion

Therefore, the interval for \( a \) is:

\[ a \in [1, \infty) \]