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Consider A = ∫ 0 1 d x 1 + x 3 , then A satisfies

Options

  1. AA > π 4
  2. BA < π 4
  3. CA = π 4
  4. DA = π 6

Correct answer

A. A > π 4

Step-by-step solution

As, x ∈ 0,1 ⇒ x 2 > x 3 ⇒ 1 + x 2 > 1 + x 3 or 1 1 + x 2 < 1 1 + x 3 Thus, ∫ 0 1 d x 1 + x 2 < ∫ 0 1 d x 1 + x 3 i.e., t a n - 1 x 0 1 < A ⇒ A > t a n - 1 1 - t a n 1 0 ⇒ A > π 4

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