Applications of Derivatives
Rolle's Theorem
Grade 12
Question:
<p>If the equation \(A_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x = 0\), \(a_1 \neq 0\), \(n \geq 2\), has a positive root \(x = \alpha\), then the equation \(n a_n x^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1 = 0\) has a positive root, which is</p>
<p>greater than \(\alpha\)</p>
<p>smaller than \(\alpha\)</p>
<p>greater than or equal to \(\alpha\)</p>
<p>equal to \(\alpha\)</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 on [0, α], f'(x) must have a root in (0, α).
<p><strong>Step 1:</strong> Let f(x) = A_n x^n + a_{n-1} x^{n-1} + ... + a_1 x. Note that f(0) = 0 (no constant term) and f(α) = 0 (given).</p><p><strong>Step 2:</strong> Since f is continuous on [0, α] and differentiable on (0, α), by Rolle's Theorem, ∃ c ∈ (0, α) such that f'(c) = 0.</p><p><strong>Step 3:</strong> f'(x) = n·A_n x^{n-1} + (n-1)a_{n-1} x^{n-2} + ... + a_1</p><p><strong>Step 4:</strong> The equation n·a_n x^{n-1} + (n-1)a_{n-1} x^{n-2} + ... + a_1 = 0 has a positive root c where 0 < c < α.</p><p><strong>Step 5:</strong> Since a_1 ≠ 0, f'(0) = a_1 ≠ 0, so the root c is strictly between 0 and α.</p><p>∴ <strong>Answer: B</strong> (The positive root lies between 0 and α)</p>
Correct Answer: B