<p>If the function \(f(x) = (x^2 + ax + 2a)e^x\) is a strictly increasing function in \((-\infty, \infty)\). Then the number of integral values of \(a\) is:</p>
Step-by-Step Solution
Key Concept: For f(x) to be strictly increasing on ℝ, we need f'(x) ≥ 0 for all x ∈ ℝ, with equality only at isolated points. Since f'(x) involves a product with e^x (always positive), the quadratic factor must be non-negative everywhere.
<p><strong>Step 1:</strong> Find f'(x) using the product rule.</p><p>f(x) = (x² + ax + 2a)e^x</p><p>f'(x) = (2x + a)e^x + (x² + ax + 2a)e^x</p><p>f'(x) = e^x[(2x + a) + (x² + ax + 2a)]</p><p>f'(x) = e^x[x² + (a+2)x + (a + 2a)]</p><p>f'(x) = e^x[x² + (a+2)x + 3a]</p><p><strong>Step 2:</strong> Since e^x > 0 for all x, f(x) is strictly increasing ⟺ x² + (a+2)x + 3a ≥ 0 for all x ∈ ℝ.</p><p><strong>Step 3:</strong> For a quadratic ax² + bx + c to be non-negative for all x, the discriminant must be ≤ 0.</p><p>Discriminant: Δ = (a+2)² - 4(1)(3a) ≤ 0</p><p>a² + 4a + 4 - 12a ≤ 0</p><p>a² - 8a + 4 ≤ 0</p><p><strong>Step 4:</strong> Solve a² - 8a + 4 = 0 using the quadratic formula:</p><p>a = (8 ± √(64-16))/2 = (8 ± √48)/2 = (8 ± 4√3)/2 = 4 ± 2√3</p><p><strong>Step 5:</strong> Since the coefficient of a² is positive, a² - 8a + 4 ≤ 0 when:</p><p>4 - 2√3 ≤ a ≤ 4 + 2√3</p><p>With √3 ≈ 1.732: 4 - 3.464 ≤ a ≤ 4 + 3.464</p><p>0.536 ≤ a ≤ 7.464</p><p><strong>Step 6:</strong> The integral values of a in this range are: {1, 2, 3, 4, 5, 6, 7}</p><p>∴ Answer: D (7 integral values)</p>
Correct Answer: D