Applications of Derivatives
Rolle's Theorem
Grade 12
Question:
<p>Given that \(a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x = 0\), \(a_1 \neq 0\), \(x \geq 2\) has a positive root \(x = \alpha\), and \(na_n x^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1 = 0\) has a positive root, say \(\beta\). Then which of the following is true?</p>
<p>\(\beta > \alpha\)</p>
<p>\(\beta < \alpha\)</p>
<p>\(\beta = \alpha\)</p>
<p>\(\beta \geq \alpha\)</p>
Step-by-Step Solution
Key Concept: The second equation is the derivative of the first equation divided by x. By Rolle's theorem applied to f(x) = a_n x^n + a_{n-1}x^{n-1} + ⋯ + a_1 x, if f(α) = 0 and f(0) = 0, then f'(ξ) = 0 for some ξ ∈ (0,α), which means β < α.
<p><strong>Step 1:</strong> Define f(x) = a_n x^n + a_{n-1}x^{n-1} + ⋯ + a_1 x. Note that f(0) = 0 (constant term is 0) and f(α) = 0 (given, α is a positive root with x ≥ 2).</p><p><strong>Step 2:</strong> Compute f'(x) = na_n x^{n-1} + (n-1)a_{n-1}x^{n-2} + ⋯ + a_1. The second equation states f'(β) = 0.</p><p><strong>Step 3:</strong> Apply Rolle's Theorem: Since f is differentiable on [0,α], f(0) = 0, and f(α) = 0, there exists at least one point β ∈ (0,α) where f'(β) = 0.</p><p><strong>Step 4:</strong> Since β is a root of f'(x) = 0 with β ∈ (0,α) and α ≥ 2, we have β < α.</p><p>∴ Answer: B (The relationship is β < α)</p>
Correct Answer: B