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JEE MainMathematicsArea Under Curves

Let a region in the first quadrant be bounded by the parabola y^2 = 6x and the line y = x . The area of the portion of this region that lies strictly inside the circle x^2 + y^2 = 16 is equal to :

Options

  1. A2 3 + 4 3 3
  2. B2 3 + 2 3 3
  3. C2
  4. D2 3 + 2 3 3 + 2

Correct answer

B. 2 3 + 2 3 3

Step-by-step solution

The region is bounded by the parabola y^2 = 6x (upper boundary) and the line y = x (lower boundary), and is restricted inside the circle x^2 + y^2 = 16 . Find the intersection of the parabola y^2 = 6x and the circle x^2 + y^2 = 16 in the first quadrant: x^2 + 6x - 16 = 0 (x+8)(x-2) = 0 x = 2 Find the intersection of the line y = x and the circle x^2 + y^2 = 16 in the first quadrant: x^2 + x^2 = 16 2x^2 = 16 x = 2 2 The upper boundary of the required region changes at x = 2 . For x [0, 2] , the upper boundary is the

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