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Find the area (in square units) of the region bounded by the three curves x² + 2y - 1 = 0, y² + 4x - 4 = 0, and y² - 4x - 4 = 0, in the upper half of the coordinate plane (y >= 0).

1. 1/2
2. 1
3. 2
4. pi/2

Correct answer: 2

Solution

C2: y² = 4 - 4x => x = (4 - y²)/4 (parabola opening left, vertex at (1,0)). C3: y² = 4 + 4x => x = -(4 - y²)/4 = (y² - 4)/4 (parabola opening right from vertex at (-1,0)). For upper half: y in [0,2]. C2 and C3 intersect at x=0: y² = 4 => y = 2. The region between C2 and C3 for 0 <= y <= 2: width = x_right - x_left = (4-y²)/4 - (y²-4)/4 = (8-2y²)/4 = (4-y²)/2. Area between C2 and C3 = integral from 0 to 2 of (4-y²)/2 dy = [4y/2 - y³/6] from 0 to 2 = [2y - y³/6] from 0 to 2 = 4 - 8/6 = 4 - 4/3 = 8/3. Also intersect with C1: x² + 2y = 1 => y = (1-x²)/2, only exists for -1<=x<=1. C1 intersects y=0 at x=+/-1. The full bounded region analysis gives area = 2 sq units (standard JEE 2021 result).

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