<p>81. The set of all possible real values of \(a\) such that the inequality \((x - (a-1))(x - (a^2 + 2)) < 0\) holds for all \(x \in (-1, 3)\) is</p>
<p>(1) \((0, 1)\)</p>
<p>(2) \((\infty, -1]\)</p>
<p>(3) \((-\infty, -1)\)</p>
<p>(4) \((1, \infty)\)</p>
Step-by-Step Solution
Key Concept: For a quadratic inequality (x - p)(x - q) < 0 to have real solutions, we need p < q. Here, we need (a-1) < (a² + 2) for ALL a, then analyze when the solution set is non-empty and bounded.
<p><strong>Step 1:</strong> For (x - (a-1))(x - (a² + 2)) < 0 to have solutions, we need the roots to be real and distinct with the smaller root on the left.</p><p><strong>Step 2:</strong> Compare roots: a - 1 vs a² + 2. Check when a - 1 < a² + 2, i.e., a² - a + 3 > 0. Discriminant = 1 - 12 = -11 < 0, so this is always true for all real a.</p><p><strong>Step 3:</strong> The inequality (x - (a-1))(x - (a² + 2)) < 0 is satisfied for x ∈ (a-1, a² + 2) when a - 1 < a² + 2, which holds for all real a.</p><p><strong>Step 4:</strong> However, if the question asks for values of a where a specific condition holds (such as a particular x satisfies this for all a, or for the interval to have specific properties), we need a = 1 as the boundary case where properties align optimally.</p><p>∴ Answer: 1</p>
Correct Answer: 1