<p>Let \(g(x_2) \leq f(x_1)\) for all \(x \in [0,1]\), where \(f(x)\) is an increasing function on \([0,1]\) with minimum value \(f(0) = -1\). Given \(h(x) = x^2 - 2ax + 5\), find the minimum value of \(a\) such that \(h(x) \leq 0\) for at least one \(x \in [1, 2]\).</p>
Step-by-Step Solution
Key Concept: For h(x) = x² - 2ax + 5 ≤ 0 to have at least one solution in [1,2], we need the minimum value of h(x) on this interval to be non-positive. Since h is a parabola opening upward with vertex at x = a, we must find when min h(x) on [1,2] equals zero, then solve for the minimum a.
<p><strong>Step 1:</strong> Analyze h(x) = x² - 2ax + 5. This is a parabola with vertex at x = a and opens upward.</p><p><strong>Step 2:</strong> For h(x) ≤ 0 to have at least one solution in [1,2], we need min{h(x) : x ∈ [1,2]} ≤ 0.</p><p><strong>Step 3:</strong> Consider cases based on vertex location:</p><ul><li><strong>Case 1:</strong> If a < 1, minimum is at x = 1: h(1) = 1 - 2a + 5 = 6 - 2a ≤ 0 ⟹ a ≥ 3 (contradicts a < 1)</li><li><strong>Case 2:</strong> If a ∈ [1,2], minimum is at x = a: h(a) = a² - 2a² + 5 = 5 - a² ≤ 0 ⟹ a² ≥ 5 ⟹ a ≥ √5 ≈ 2.236 (contradicts a ≤ 2)</li><li><strong>Case 3:</strong> If a > 2, minimum is at x = 2: h(2) = 4 - 4a + 5 = 9 - 4a ≤ 0 ⟹ a ≥ 9/4 = 2.25</li></ul><p><strong>Step 4:</strong> The minimum value of a for which h(x) ≤ 0 has at least one solution in [1,2] is a = <strong>9/4 or 2.25</strong>.</p><p>∴ Answer: A</p>
Correct Answer: A