<p>If <span>\(x^2 + 3x + 1 + \lambda(x+1) > -10\)</span> for all <span>\(x \in \mathbb{R}\)</span>, and <span>\(\lambda\)</span> is an integer, find the sum of all integer values of <span>\(\lambda\)</span>.</p>
Step-by-Step Solution
Key Concept: Rearrange as a quadratic in x: x² + (3+λ)x + (1+λ+10) > 0 for all x∈ℝ. For a quadratic to be always positive, its discriminant must be negative: (3+λ)² - 4(11+λ) < 0.
<p><strong>Step 1:</strong> Rearrange the inequality:</p><p>x² + 3x + 1 + λ(x+1) > -10</p><p>x² + 3x + 1 + λx + λ + 10 > 0</p><p>x² + (3+λ)x + (11+λ) > 0</p><p><strong>Step 2:</strong> For this quadratic to be positive for all x∈ℝ, the discriminant must be strictly negative (since the coefficient of x² is positive):</p><p>Δ = (3+λ)² - 4(1)(11+λ) < 0</p><p>9 + 6λ + λ² - 44 - 4λ < 0</p><p>λ² + 2λ - 35 < 0</p><p><strong>Step 3:</strong> Factor: (λ+7)(λ-5) < 0</p><p>This gives: -7 < λ < 5</p><p><strong>Step 4:</strong> Find all integers in this range:</p><p>λ ∈ {-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4}</p><p><strong>Step 5:</strong> Sum = -6-5-4-3-2-1+0+1+2+3+4 = -15 + 25 = 10</p><p>∴ Answer: 10</p>
Correct Answer: 10