Applications of Derivatives
Rolle's Theorem / Mean Value Theorem
Grade 12
Question:
<p>If \(a_0, a_1, a_2, a_3\) are all positive, then \(4a_0x^3 + 3a_1x^2 + 2a_2x + a_3 = 0\) has atleast one root in \((-1, 0)\), then</p>
<p>(a) \(a_0 + a_2 = a_1 + a_3\)</p>
<p>(b) \(4a_0 + 2a_2 > 3a_1 + a_3\)</p>
<p>(c) \(4a_0 + 2a_2 < 3a_1 + a_3\)</p>
<p>(d) \(4a_0 + 2a_2 = 3a_1 + a_3\)</p>
Step-by-Step Solution
Key Concept: Recognize that 4a₀x³ + 3a₁x² + 2a₂x + a₃ is the derivative of a₀x⁴ + a₁x³ + a₂x² + a₃x, then apply Rolle's theorem to the parent function on the interval [-1, 0].
<p><strong>Step 1:</strong> Recognize the pattern. Let f(x) = a₀x⁴ + a₁x³ + a₂x² + a₃x</p><p><strong>Step 2:</strong> Then f'(x) = 4a₀x³ + 3a₁x² + 2a₂x + a₃ (the given equation)</p><p><strong>Step 3:</strong> Evaluate f at the boundary points:</p><p>• f(-1) = a₀(-1)⁴ + a₁(-1)³ + a₂(-1)² + a₃(-1) = a₀ - a₁ + a₂ - a₃</p><p>• f(0) = 0</p><p><strong>Step 4:</strong> Since all aᵢ > 0, we have f(-1) = a₀ - a₁ + a₂ - a₃, which is generally not zero. However, f is continuous and differentiable on [-1, 0].</p><p><strong>Step 5:</strong> The given equation 4a₀x³ + 3a₁x² + 2a₂x + a₃ = 0 is exactly f'(x) = 0. By Rolle's theorem, if f(-1) and f(0) have opposite signs (or one is zero), then ∃ at least one c ∈ (-1, 0) where f'(c) = 0.</p><p><strong>Step 6:</strong> Alternatively, observe: f(0) = 0 and f'(x) = 4a₀x³ + 3a₁x² + 2a₂x + a₃. Since all coefficients are positive, f'(x) > 0 for x > 0 and f is increasing on (0, ∞). For x ∈ (-1, 0), if f has a local extremum, then f'(x) = 0 at that point.</p><p>∴ The equation 4a₀x³ + 3a₁x² + 2a₂x + a₃ = 0 has at least one root in (-1, 0)</p>
Correct Answer: D