<p>Let \(z\) be the set of integers. If \(A = \{x \in z : 2^{(x+2)(x^2 - 5x + 6)} = 1\}\) and \(B = \{x \in z : -3 < 2x - 1 < 9\}\), then the number of subsets of the set \(A \times B\) is:</p>
Step-by-Step Solution
Key Concept: For 2^k = 1, we need k = 0. Factor the exponent (x+2)(x²-5x+6) = 0 and find integer solutions, then intersect with the constraint on B.
<p><strong>Step 1:</strong> Find set A. We need 2^((x+2)(x²-5x+6)) = 1, which means (x+2)(x²-5x+6) = 0.</p><p><strong>Step 2:</strong> Factor: (x+2)(x-2)(x-3) = 0, giving x ∈ {-2, 2, 3}.</p><p><strong>Step 3:</strong> Since x ∈ ℤ, we have A = {-2, 2, 3}.</p><p><strong>Step 4:</strong> Set B is defined as B = {x ∈ ℤ : -3 < x < 3}, so B = {-2, -1, 0, 1, 2}.</p><p><strong>Step 5:</strong> Find A ∩ B by taking common elements: A ∩ B = {-2, 2}.</p><p>∴ Answer: B</p>
Correct Answer: B