<p>Two sets <em>A</em> and <em>B</em> are as under:<br>\(A = \{(a,b) \in \mathbb{R} \times \mathbb{R} : |a-5| < 1 \text{ and } |b-5| < 1\}\);<br>\(B = \{(a,b) \in \mathbb{R} \times \mathbb{R} : 4(a-6)^2 + 9(b-5)^2 \leq 36\}\)<br>Then</p>
Step-by-Step Solution
Key Concept: Set A represents the interior and boundary of a square centered at (5,5) with side length 2√2, while Set B represents the interior and boundary of an ellipse centered at (6,5). The key is finding the intersection by comparing these two regions geometrically.
<p><strong>Step 1: Identify Set A</strong></p><p>From |a-5| < 1 and |b-5| < 1, we get:<br>4 < a < 6 and 4 < b < 6<br>This is the interior of a square with vertices at (4,4), (6,4), (6,6), (4,6) centered at (5,5).</p><p><strong>Step 2: Identify Set B</strong></p><p>From 4(a-6)² + 9(b-5)² ≤ 36, divide by 36:<br>(a-6)²/9 + (b-5)²/4 ≤ 1<br>This is an ellipse centered at (6,5) with semi-major axis a=3 (along a-axis) and semi-minor axis b=2 (along b-axis).</p><p><strong>Step 3: Find A ∩ B</strong></p><p>The square A has vertices (4,4), (6,4), (6,6), (4,6). The ellipse B is centered at (6,5).<br>• Right boundary of square: a = 6 (where ellipse is centered)<br>• At a = 6: 9(b-5)² ≤ 36 → (b-5)² ≤ 4 → 3 ≤ b ≤ 7<br>• Intersection at a = 6: 4 ≤ b ≤ 6 (within square's b-range)<br><br>Check left edge a = 4: 4(4-6)² + 9(b-5)² ≤ 36 → 16 + 9(b-5)² ≤ 36 → (b-5)² ≤ 20/9<br>This gives b ≈ 3.06 to 6.94, overlapping with [4,6].<br><br>The intersection A ∩ B is non-empty and forms a lens-shaped region where the ellipse cuts through the right portion of the square.</p><p>∴ Answer: A</p>
Correct Answer: A