<p>If \(-3 \leq \dfrac{x^2 - \lambda x - 2}{x^2 + x + 1} \leq 2\) for all \(x \in R\), then how many integral values of \(\lambda\) exist?</p>
Step-by-Step Solution
Key Concept: Convert the compound inequality into two separate conditions, then use the fact that the denominator x² + x + 1 is always positive (discriminant < 0) to clear denominators and rearrange into quadratic forms that must be non-negative for all real x.
<p><strong>Step 1:</strong> Note that x² + x + 1 = (x + 1/2)² + 3/4 > 0 for all x ∈ ℝ.</p><p><strong>Step 2:</strong> From -3 ≤ (x² - λx - 2)/(x² + x + 1), multiply by denominator (positive, so inequality preserves):<br/>-3(x² + x + 1) ≤ x² - λx - 2<br/>-3x² - 3x - 3 ≤ x² - λx - 2<br/>4x² + (λ - 3)x + 1 ≥ 0 for all x</p><p><strong>Step 3:</strong> For this to hold for all x: Δ₁ = (λ - 3)² - 16 ≤ 0<br/>(λ - 3)² ≤ 16<br/>|λ - 3| ≤ 4<br/>-1 ≤ λ ≤ 7</p><p><strong>Step 4:</strong> From (x² - λx - 2)/(x² + x + 1) ≤ 2, multiply by denominator:<br/>x² - λx - 2 ≤ 2(x² + x + 1)<br/>x² - λx - 2 ≤ 2x² + 2x + 2<br/>-x² - (λ + 2)x - 4 ≤ 0<br/>x² + (λ + 2)x + 4 ≥ 0 for all x</p><p><strong>Step 5:</strong> For this to hold for all x: Δ₂ = (λ + 2)² - 16 ≤ 0<br/>(λ + 2)² ≤ 16<br/>|λ + 2| ≤ 4<br/>-6 ≤ λ ≤ 2</p><p><strong>Step 6:</strong> Intersection of both conditions: -1 ≤ λ ≤ 2<br/>Integral values: λ ∈ {-1, 0, 1, 2}</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: 4