<p>70. The set of values of \(a\) for which \((a-1)x^2 - (a+1)x + a - 1 \geq 0\) is true for all \(x \geq 2\) is</p>
<p>(1) \((-\infty, 1)\)</p>
<p>(2) \(\left[1, \dfrac{7}{3}\right)\)</p>
<p>(3) \(\left(\dfrac{7}{3}, \infty\right)\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: For the inequality to hold for all x ≥ 2, analyze the quadratic's behavior on [2, ∞) by checking the vertex location, discriminant sign, and boundary value f(2). The critical insight is that if a = 1, the expression becomes linear; otherwise, we need the parabola to be non-negative throughout [2, ∞).
<p><strong>Step 1:</strong> Consider the expression (a-1)x² - (a+1)x + a - 1. If a = 1, it becomes -2x ≥ 0, which is false for x ≥ 2. So a ≠ 1.</p><p><strong>Step 2:</strong> For a ≠ 1, rewrite as (a-1)[x² - x] - (1-a) - x ≥ 0, or factor: (a-1)(x² - 1) - (a+1)x + (a-1) ≥ 0. At x = 2: (a-1)(4) - (a+1)(2) + (a-1) = 4a - 4 - 2a - 2 + a - 1 = 3a - 7 ≥ 0, giving a ≥ 7/3.</p><p><strong>Step 3:</strong> Check that for a > 1, the parabola opens upward. The vertex is at x = (a+1)/(2(a-1)). For a ≥ 7/3 > 1, we have vertex x = (a+1)/(2(a-1)) ≤ 2 when (a+1) ≤ 4(a-1), i.e., a ≥ 5/3. Since 7/3 > 5/3, the vertex lies left of x = 2, so f is increasing on [2, ∞). The minimum on [2, ∞) is at x = 2.</p><p><strong>Step 4:</strong> Therefore f(2) ≥ 0 is sufficient: 3a - 7 ≥ 0 ⟹ a ≥ 7/3.</p><p>∴ Answer: a ∈ [7/3, ∞) or a ≥ 7/3</p>
Correct Answer: 3