<p>Let \(f(x) = (x^2 + ax + 2a)e^x\). If \(f(x)\) is an increasing function for all \(x \in \mathbb{R}\), find the number of integral values of \(a\).</p>
Step-by-Step Solution
Key Concept: For f(x) to be increasing on ℝ, we need f'(x) ≥ 0 for all x ∈ ℝ. Using the product rule on f(x) = (x² + ax + 2a)e^x, we get f'(x) = e^x[(x² + ax + 2a) + (2x + a)], which simplifies to e^x[x² + (a+2)x + (a+2a)]. Since e^x > 0, we need the quadratic x² + (a+2)x + 3a ≥ 0 for all x ∈ ℝ.
<p><strong>Step 1: Find f'(x) using the product rule.</strong></p><p>Given: 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 + (3a)]</p><p></p><p><strong>Step 2: Condition for f(x) to be increasing.</strong></p><p>For f(x) to be increasing on ℝ, we need f'(x) ≥ 0 for all x ∈ ℝ.</p><p>Since e^x > 0 for all x, we require:</p><p>x² + (a+2)x + 3a ≥ 0 for all x ∈ ℝ</p><p></p><p><strong>Step 3: Apply the discriminant condition.</strong></p><p>For a quadratic Ax² + Bx + C ≥ 0 for all x ∈ ℝ (with A > 0), the discriminant must satisfy Δ ≤ 0.</p><p>Here: A = 1 > 0, B = (a+2), C = 3a</p><p>Δ = (a+2)² - 4(1)(3a) ≤ 0</p><p>Δ = a² + 4a + 4 - 12a ≤ 0</p><p>Δ = a² - 8a + 4 ≤ 0</p><p></p><p><strong>Step 4: Solve the inequality a² - 8a + 4 ≤ 0.</strong></p><p>Using the quadratic formula for a² - 8a + 4 = 0:</p><p>a = (8 ± √(64 - 16))/2 = (8 ± √48)/2 = (8 ± 4√3)/2 = 4 ± 2√3</p><p></p><p><strong>Step 5: Find the range of a.</strong></p><p>We need: 4 - 2√3 ≤ a ≤ 4 + 2√3</p><p>Since √3 ≈ 1.732, we have 2√3 ≈ 3.464</p><p>Lower bound: 4 - 3.464 ≈ 0.536</p><p>Upper bound: 4 + 3.464 ≈ 7.464</p><p></p><p><strong>Step 6: Count integral values.</strong></p><p>The integral values of a in the range [0.536, 7.464] are:</p><p>a ∈ {1, 2, 3, 4, 5, 6, 7}</p><p>Number of integral values = 7</p><p></p><p>∴ Answer: C</p>
Correct Answer: C