<p>Let \(f(x) = x - \dfrac{1}{x+1}\) and \(g(x) = x^2 - 2ax + 4\), where \(a\) is a parameter. If \(\forall\, x_1 \in [0, 1]\) there exists some \(x_2 \in [1, 2]\), such that \(f(x_1) \geq g(x_2)\). Then the minimum value of \(a\) is:</p>
Step-by-Step Solution
Key Concept: The condition ∀x₁∈[0,1] ∃x₂∈[1,2] such that f(x₁)≥g(x₂) is equivalent to min f(x) on [0,1] ≥ min g(x) on [1,2]. Find both minimum values and set up an inequality in terms of parameter a.
<p><strong>Step 1: Find minimum of f(x) on [0,1]</strong></p><p>f(x) = x - 1/(x+1), so f'(x) = 1 + 1/(x+1)² > 0 for all x.</p><p>Thus f is strictly increasing on [0,1], so min f = f(0) = 0 - 1/1 = <strong>-1</strong></p><p><strong>Step 2: Find minimum of g(x) on [1,2]</strong></p><p>g(x) = x² - 2ax + 4, so g'(x) = 2x - 2a = 0 gives x = a.</p><p>The vertex is at x = a with g(a) = a² - 2a² + 4 = 4 - a²</p><p><strong>Case 1:</strong> If a < 1, then g is increasing on [1,2], so min g = g(1) = 1 - 2a + 4 = 5 - 2a</p><p><strong>Case 2:</strong> If 1 ≤ a ≤ 2, then min g = g(a) = 4 - a²</p><p><strong>Case 3:</strong> If a > 2, then g is decreasing on [1,2], so min g = g(2) = 4 - 4a + 4 = 8 - 4a</p><p><strong>Step 3: Apply the condition min f ≥ min g</strong></p><p>We need -1 ≥ min g(x) on [1,2].</p><p><strong>Case 1:</strong> -1 ≥ 5 - 2a ⟹ 2a ≥ 6 ⟹ a ≥ 3 (contradicts a < 1)</p><p><strong>Case 2:</strong> -1 ≥ 4 - a² ⟹ a² ≥ 5 ⟹ a ≥ √5 (contradicts 1 ≤ a ≤ 2)</p><p><strong>Case 3:</strong> -1 ≥ 8 - 4a ⟹ 4a ≥ 9 ⟹ a ≥ 9/4</p><p>Since we need a > 2 for Case 3 and 9/4 = 2.25 > 2, this is valid.</p><p>∴ The minimum value of a is <strong>9/4</strong> (Answer: B)
Correct Answer: B