Quadratic Equations
Sign of quadratic expression
Grade 11
Question:
<p>Set of all real values of \(a\) such that \(f(x) = \dfrac{(2a-1)x^2 + 2(a+1)x + (2a-1)}{x^2 - 2x + 40}\) is always negative is</p>
<p>(1) \((-\infty, 0)\)</p>
<p>(2) \((0, \infty)\)</p>
<p>(3) \((-\infty, 1/2)\)</p>
<p>(4) None</p>
Step-by-Step Solution
Key Concept: For f(x) to be always negative, the numerator must always be negative (since denominator x² - 2x + 40 is always positive with discriminant 4 - 160 < 0). This requires both the quadratic (2a-1)x² + 2(a+1)x + (2a-1) to have negative leading coefficient and no real roots.
<p><strong>Step 1:</strong> Verify denominator is always positive.</p><p>For x² - 2x + 40: Δ = 4 - 160 = -156 < 0 and leading coefficient = 1 > 0, so x² - 2x + 40 > 0 for all x ∈ ℝ. ✓</p><p><strong>Step 2:</strong> For f(x) < 0 for all x, numerator must be negative for all x.</p><p>Let N(x) = (2a-1)x² + 2(a+1)x + (2a-1)</p><p><strong>Step 3:</strong> Condition 1: Leading coefficient must be negative.</p><p>2a - 1 < 0 ⟹ a < 1/2</p><p><strong>Step 4:</strong> Condition 2: Numerator must have no real roots (parabola stays below x-axis).</p><p>Δ_N = [2(a+1)]² - 4(2a-1)(2a-1) < 0</p><p>= 4(a+1)² - 4(2a-1)² < 0</p><p>= (a+1)² - (2a-1)² < 0</p><p>= (a² + 2a + 1) - (4a² - 4a + 1) < 0</p><p>= -3a² + 6a < 0</p><p>= -3a(a - 2) < 0</p><p>= a(a - 2) > 0</p><p>⟹ a < 0 or a > 2</p><p><strong>Step 5:</strong> Find intersection of conditions: a < 1/2 AND (a < 0 or a > 2)</p><p>Since a > 2 contradicts a < 1/2, we need a < 0.</p><p>∴ Answer: a ∈ (-∞, 0)</p>
Correct Answer: C