The number of ordered pairs `(a,b)` where a,b are integers satisfying the inequality min `(x ^(2) +(a-b) x + (1-a-b)) gtmax (-x ^(2) +(a+b)x-(1+a+b)AA x in R,` is :

Answer

To solve the problem of finding the number of ordered pairs \((a, b)\) where \(a\) and \(b\) are integers satisfying the inequality

\[ \min(x^2 + (a-b)x + (1-a-b)) > \max(-x^2 + (a+b)x - (1+a+b)) \]

for all \(x \in \mathbb{R}\), we can follow these steps:

### Step 1: Identify the Quadratic Functions
We have two quadratic functions:
1. \(f(x) = x^2 + (a-b)x + (1-a-b)\)
2. \(g(x) = -x^2 + (a+b)x - (1+a+b)\)

### Step 2: Determine the Nature of the Quadratics
- The function \(f(x)\) is a **concave up** parabola (since the coefficient of \(x^2\) is positive).
- The function \(g(x)\) is a **concave down** parabola (since the coefficient of \(x^2\) is negative).

### Step 3: Find the Minimum of \(f(x)\) and Maximum of \(g(x)\)
- The minimum value of \(f(x)\) occurs at the vertex, given by:

\[ x = -\frac{b}{2a} = -\frac{(a-b)}{2} \quad \text{(where \(a = 1\))} \]

The minimum value is:

\[ f\left(-\frac{(a-b)}{2}\right) = \left(-\frac{(a-b)}{2}\right)^2 + (a-b)\left(-\frac{(a-b)}{2}\right) + (1-a-b) \]

Calculating this gives:

\[ = \frac{(a-b)^2}{4} - \frac{(a-b)^2}{2} + (1-a-b) = \frac{(a-b)^2}{4} - \frac{2(a-b)^2}{4} + (1-a-b) = -\frac{(a-b)^2}{4} + (1-a-b) \]

- The maximum value of \(g(x)\) also occurs at the vertex, given by:

\[ x = -\frac{b}{2a} = \frac{(a+b)}{2} \quad \text{(where \(a = -1\))} \]

The maximum value is:

\[ g\left(\frac{(a+b)}{2}\right) = -\left(\frac{(a+b)}{2}\right)^2 + (a+b)\left(\frac{(a+b)}{2}\right) - (1+a+b) \]

Calculating this gives:

\[ = -\frac{(a+b)^2}{4} + \frac{(a+b)^2}{2} - (1+a+b) = \frac{(a+b)^2}{4} - (1+a+b) \]

### Step 4: Set Up the Inequality
We need to satisfy the inequality:

\[ -\frac{(a-b)^2}{4} + (1-a-b) > \frac{(a+b)^2}{4} - (1+a+b) \]

### Step 5: Simplify the Inequality
Rearranging gives:

\[ -\frac{(a-b)^2}{4} + 1 - a - b > \frac{(a+b)^2}{4} - 1 - a - b \]

This simplifies to:

\[ -\frac{(a-b)^2}{4} + 2 > \frac{(a+b)^2}{4} \]

Multiplying through by 4 (to eliminate the fractions):

\[ - (a-b)^2 + 8 > (a+b)^2 \]

Rearranging gives:

\[ 8 > (a+b)^2 + (a-b)^2 \]

### Step 6: Expand and Simplify
Expanding the squares:

\[ 8 > (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2) = 2a^2 + 2b^2 \]

This simplifies to:

\[ 4 > a^2 + b^2 \]

### Step 7: Find Integer Solutions
We need to find integer pairs \((a, b)\) such that:

\[ a^2 + b^2 < 4 \]

The possible integer pairs \((a, b)\) are:
- \( (0, 0) \)
- \( (0, 1) \)
- \( (0, -1) \)
- \( (1, 0) \)
- \( (1, 1) \)
- \( (1, -1) \)
- \( (-1, 0) \)
- \( (-1, 1) \)
- \( (-1, -1) \)
- \( (2, 0) \)
- \( (0, 2) \)
- \( (-2, 0) \)
- \( (0, -2) \)

However, since \(a^2 + b^2\) must be less than 4, the only valid pairs are:
- \((0, 0)\)
- \((0, 1)\)
- \((0, -1)\)
- \((1, 0)\)
- \((1, 1)\)
- \((1, -1)\)
- \((-1, 0)\)
- \((-1, 1)\)
- \((-1, -1)\)

### Final Count
Counting these valid pairs gives us a total of **9 ordered pairs**.