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ExamsJEE AdvancedMaths

The area of the region {(x, y): y² <= 4x, x + y <= 1, x >= 0, y >= 0} equals a*sqrt(2) + b. Find a - b.

  1. 8/3
  2. 10/3
  3. 6
  4. -2/3

Correct answer: 6

Solution

The intersection gives y = -2 + 2*sqrt(2); integrating (1 - y) - y²/4 in y yields (8/3)*sqrt(2) - 10/3, so a = 8/3, b = -10/3.

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