<p>The number of ordered pairs (<i>a</i>, <i>b</i>), where <i>a</i>, <i>b</i> are integers satisfying the inequality \(\min\left(x^2 + (a-b)x + (1-a-b)\right) > \max\left(-x^2 + (a+b)x - (1+a+b)\right)\) for all \(x \in \mathbb{R}\), is:</p>
Step-by-Step Solution
Key Concept: For the inequality to hold for all x ∈ ℝ, the minimum value of the first parabola must exceed the maximum value of the second parabola. Since the first parabola opens upward, its minimum exists; since the second opens downward, its maximum exists. We need: min(f) > max(g).
<p><strong>Step 1: Identify the parabolas</strong></p><p>Let f(x) = x² + (a-b)x + (1-a-b) and g(x) = -x² + (a+b)x - (1+a+b).</p><p>f(x) opens upward (coefficient = 1 > 0) and g(x) opens downward (coefficient = -1 < 0).</p><p><strong>Step 2: Find the minimum of f(x)</strong></p><p>The vertex occurs at x = -(a-b)/2. The minimum value is:</p><p>f_min = (1-a-b) - [(a-b)²/4] = (1-a-b) - (a-b)²/4</p><p><strong>Step 3: Find the maximum of g(x)</strong></p><p>The vertex occurs at x = (a+b)/2. The maximum value is:</p><p>g_max = -(1+a+b) + [(a+b)²/4] = (a+b)²/4 - (1+a+b)</p><p><strong>Step 4: Set up the inequality f_min > g_max</strong></p><p>(1-a-b) - (a-b)²/4 > (a+b)²/4 - (1+a+b)</p><p>(1-a-b) + (1+a+b) > (a+b)²/4 + (a-b)²/4</p><p>2 > [(a+b)² + (a-b)²]/4</p><p>8 > (a+b)² + (a-b)²</p><p><strong>Step 5: Expand and simplify</strong></p><p>(a+b)² + (a-b)² = a² + 2ab + b² + a² - 2ab + b² = 2a² + 2b²</p><p>So: 8 > 2a² + 2b² → 4 > a² + b²</p><p><strong>Step 6: Find integer solutions</strong></p><p>We need a² + b² < 4 where a, b are integers.</p><p>Possible values: a² + b² ∈ {0, 1, 2, 3}</p><p>• a² + b² = 0: (a,b) = (0,0) → 1 pair</p><p>• a² + b² = 1: (a,b) ∈ {(±1,0), (0,±1)} → 4 pairs</p><p>• a² + b² = 2: (a,b) ∈ {(±1,±1)} → 4 pairs</p><p>• a² + b² = 3: No integer solutions</p><p>Total: 1 + 4 + 4 = 9 pairs</p><p><strong>∴ Answer: 9</strong></p>
Correct Answer: 9