<p>The number of integral values of \(m\) for which the quadratic expression, \((1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)\), \(x \in R\), is always positive, is __________.</p>
Step-by-Step Solution
Key Concept: For a quadratic expression to be always positive, the coefficient of x² must be positive AND the discriminant must be negative (ensuring no real roots). This creates a system of inequalities in m that must all be satisfied simultaneously.
<p><strong>Step 1:</strong> For the quadratic (1+2m)x² - 2(1+3m)x + 4(1+m) to always be positive, we need:</p><ul><li>Condition A: 1 + 2m > 0 ⟹ m > -1/2</li><li>Condition B: Discriminant < 0</li></ul><p><strong>Step 2:</strong> Calculate the discriminant:</p><p>D = [-2(1+3m)]² - 4(1+2m)·4(1+m)</p><p>D = 4(1+3m)² - 16(1+2m)(1+m)</p><p>D = 4[(1+3m)² - 4(1+2m)(1+m)]</p><p>D = 4[1 + 6m + 9m² - 4(1 + m + 2m + 2m²)]</p><p>D = 4[1 + 6m + 9m² - 4 - 12m - 8m²]</p><p>D = 4[m² - 6m - 3]</p><p><strong>Step 3:</strong> For D < 0:</p><p>4(m² - 6m - 3) < 0</p><p>m² - 6m - 3 < 0</p><p><strong>Step 4:</strong> Solve m² - 6m - 3 = 0:</p><p>m = (6 ± √(36+12))/2 = (6 ± √48)/2 = (6 ± 4√3)/2 = 3 ± 2√3</p><p>Since √3 ≈ 1.732, we get: 3 - 2√3 ≈ -0.464 and 3 + 2√3 ≈ 6.464</p><p>So m² - 6m - 3 < 0 when: -0.464 < m < 6.464</p><p><strong>Step 5:</strong> Combine both conditions:</p><p>m > -1/2 AND -0.464 < m < 6.464</p><p>This gives: -0.464 < m < 6.464 (since -1/2 = -0.5 is already in this range)</p><p><strong>Step 6:</strong> Integral values of m in this range: m ∈ {0, 1, 2, 3, 4, 5, 6}</p><p>∴ Answer: <strong>7</strong></p>
Correct Answer: 7