Applications of Derivatives
Minima of Functions
Grade 12
Question:
<p>The least value of \(\alpha \in R\) for which \(4\alpha x^2 + \dfrac{1}{x} \ge 1\), for all \(x > 0\), is</p>
<p>(1) \(\dfrac{1}{64}\)</p>
<p>(2) \(\dfrac{1}{32}\)</p>
<p>(3) \(\dfrac{1}{27}\)</p>
<p>(4) \(\dfrac{1}{25}\)</p>
Step-by-Step Solution
Key Concept: Find the minimum value of the function f(x) = 4αx² + 1/x for x > 0, then ensure this minimum is ≥ 1. The critical point occurs where f'(x) = 0, and the least α is found by analyzing the boundary condition where the minimum exactly equals 1.
<p><strong>Step 1:</strong> For the inequality 4αx² + 1/x ≥ 1 to hold for all x > 0, the minimum value of f(x) = 4αx² + 1/x must be ≥ 1.</p><p><strong>Step 2:</strong> Find the critical point by taking the derivative: f'(x) = 8αx - 1/x² = 0, which gives 8αx³ = 1, so x = (1/8α)^(1/3).</p><p><strong>Step 3:</strong> The least value of α occurs when the minimum value of f(x) equals exactly 1. At the critical point x = (1/8α)^(1/3):</p><p>f(x) = 4α · (1/8α)^(2/3) + (8α)^(1/3) = 4α · (1/8α)^(2/3) + (8α)^(1/3)</p><p><strong>Step 4:</strong> Simplifying: f(x) = 4α · 1/(2α)^(2/3) + (8α)^(1/3) = 4α^(1/3)/2^(2/3) + 2α^(1/3) = 2^(4/3)α^(1/3) + 2α^(1/3) = α^(1/3)(2^(4/3) + 2)</p><p><strong>Step 5:</strong> Let this equal 1: α^(1/3) · 3·2^(1/3) = 1, giving α^(1/3) = 1/(3·2^(1/3))</p><p><strong>Step 6:</strong> Therefore α = 1/(3³·2) = 1/54</p><p>∴ Answer: C</p>
Correct Answer: C