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If the maximum area bounded by y 2 = 4 x and the line y = m x ∀ m ∈ 1 , 3 is k square units, then the smallest prime number greater than 3 k is

Options

  1. A3
  2. B5
  3. C7
  4. D11

Correct answer

D. 11

Step-by-step solution

The area covered is maximum if ‘ m ’ is minimum, i.e. m = 1 ∴ Required area ∫ 0 4 4 x - x d x = 2 ⋅ x 3 2 3 2 − x 2 2 0 4 = 4 3 x 3 2 - x 2 2 0 4 = 32 3 - 8 = 8 3 sq. units ∴ k = 8 3 Thus required prime number = 11

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