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JEE Main201910 Jan 2019Evening ShiftMathematicsDeterminantsActual

Let A = 2 b 1 b b 2 + 1 b 1 b 2 , where b > 0 . Then the minimum value of det A b is:

Options

  1. A2 3
  2. B- 2 3
  3. C3
  4. D- 3

Correct answer

A. 2 3

Step-by-step solution

A = 2 b 1 b b 2 + 1 b 1 b 2 = 2 2 b 2 + 2 - b 2 - b 2 b - b + 1 b 2 - b 2 - 1 = 2 b 2 + 2 - b 2 - 1 = 2 b 2 + 4 - b 2 - 1 = b 2 + 3 A b = b + 3 b = b - 3 b 2 + 2 3 Hence, the minimum value of A b is 2 3 . ∵ b - 3 b 2 ≥ 0

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