<p>Given inequality is \(-3 \leq \dfrac{x^2 - \lambda x - 2}{x^2 + x + 1} \leq 2\). Find the number of integral values of \(\lambda\).</p>
Step-by-Step Solution
Key Concept: Split the compound inequality into two separate inequalities and recognize that the denominator x² + x + 1 is always positive (Δ < 0), so inequality direction is preserved. The solution requires finding the range of λ where both inequalities hold for all real x.
<p><strong>Step 1:</strong> Since x² + x + 1 = (x + ½)² + ¾ > 0 for all real x, we can multiply through without flipping inequalities.</p><p><strong>Step 2:</strong> <strong>Left inequality:</strong> -3 ≤ (x² - λx - 2)/(x² + x + 1)</p><p>-3(x² + x + 1) ≤ x² - λx - 2</p><p>-3x² - 3x - 3 ≤ x² - λx - 2</p><p>4x² + (λ - 3)x + 1 ≥ 0</p><p>For all real x: Δ ≤ 0 → (λ - 3)² - 16 ≤ 0 → (λ - 3)² ≤ 16</p><p>-4 ≤ λ - 3 ≤ 4 → <strong>-1 ≤ λ ≤ 7</strong></p><p><strong>Step 3:</strong> <strong>Right inequality:</strong> (x² - λx - 2)/(x² + x + 1) ≤ 2</p><p>x² - λx - 2 ≤ 2(x² + x + 1)</p><p>x² - λx - 2 ≤ 2x² + 2x + 2</p><p>-x² - (λ + 2)x - 4 ≤ 0</p><p>x² + (λ + 2)x + 4 ≥ 0</p><p>For all real x: Δ ≤ 0 → (λ + 2)² - 16 ≤ 0 → (λ + 2)² ≤ 16</p><p>-4 ≤ λ + 2 ≤ 4 → <strong>-6 ≤ λ ≤ 2</strong></p><p><strong>Step 4:</strong> Both conditions must hold: [-1, 7] ∩ [-6, 2] = <strong>[-1, 2]</strong></p><p>Integral values: -1, 0, 1, 2</p><p>∴ Answer: <strong>4 integral values</strong></p>
Correct Answer: C