<p>Find the set of all possible real values of <em>a</em> such that the inequality \((x - (a-1))(x - (a^2 + 2)) < 0\) holds for all \(x \in (-1, 3)\).</p>
Step-by-Step Solution
Key Concept: For the product (x - (a-1))(x - (a²+2)) to be negative for all x in (-1,3), the roots must bracket this interval: one root must be ≤ -1 and the other must be ≥ 3. This requires a-1 ≤ -1 and a²+2 ≥ 3 (or vice versa, but checking which is the smaller root).
<p><strong>Step 1:</strong> Identify the roots. The expression equals zero when x = a-1 or x = a²+2.</p><p><strong>Step 2:</strong> Verify root ordering. Compare a-1 and a²+2: a²+2 - (a-1) = a² - a + 3 = (a - 1/2)² + 11/4 > 0 always. So a-1 < a²+2 for all real a.</p><p><strong>Step 3:</strong> Apply the interval condition. For (x - (a-1))(x - (a²+2)) < 0 to hold for ALL x ∈ (-1,3), we need the product to be negative exactly when a-1 < x < a²+2. This means the interval (-1,3) must be contained in (a-1, a²+2).</p><p><strong>Step 4:</strong> Set up inequalities. We require: a-1 ≤ -1 AND a²+2 ≥ 3.</p><p><strong>Step 5:</strong> Solve both conditions.</p><p>From a-1 ≤ -1: a ≤ 0</p><p>From a²+2 ≥ 3: a² ≥ 1, so a ≤ -1 or a ≥ 1</p><p><strong>Step 6:</strong> Find intersection. The conditions a ≤ 0 AND (a ≤ -1 or a ≥ 1) give: a ≤ -1.</p><p>∴ Answer: <strong>a ≤ -1</strong></p>
Correct Answer: a ≤ -1