<p>The area enclosed by the curves \(|y + x| \leq 1, |y - x| \leq 1\) and \(2x^2 + 2y^2 = 1\) is</p>
<p>(a) \(\left(2 + \frac{\pi}{2}\right)\) sq units</p>
<p>(b) \(\left(2 - \frac{\pi}{2}\right)\) sq units</p>
<p>(c) \(\left(3 + \frac{\pi}{4}\right)\) sq units</p>
<p>(d) \(\left(3 - \frac{\pi}{4}\right)\) sq units</p>
Step-by-Step Solution
Key Concept: The area enclosed is the area of the square defined by the absolute value inequalities minus the area of the circle that lies within the square. The square |y+x| ≤ 1, |y-x| ≤ 1 is a diamond shape, and we subtract the circular region 2x² + 2y² = 1 that overlaps with it.
<p><strong>Step 1: Identify the square region.</strong> Let u = y+x and v = y-x. Then |u| ≤ 1 and |v| ≤ 1 defines a square in the uv-plane with area 4. Converting back to xy-coordinates: this is a square with vertices at (1,0), (0,1), (-1,0), (0,-1), which has area 2 sq units (diagonal = 2, area = ½ × 2 × 2 = 2).</p><p><strong>Step 2: Identify the circle.</strong> The curve 2x² + 2y² = 1 can be rewritten as x² + y² = ½, which is a circle with radius r = 1/√2 centered at origin. Its area is πr² = π(½) = π/2 sq units.</p><p><strong>Step 3: Verify the circle lies within the square.</strong> The circle has radius 1/√2 ≈ 0.707. Check if circle points satisfy the square constraints: At x = 1/√2, y = 0: |y+x| = 1/√2 < 1 ✓ and |y-x| = 1/√2 < 1 ✓. The circle is entirely contained within the square.</p><p><strong>Step 4: Calculate enclosed area.</strong> The area enclosed by both curves is the square area minus the circular area: Area = 2 - π/2 sq units.</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B