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JEE Advanced2016MathematicsApplication of DerivativesActual

The least value of α ∈ R for which, 4 α x 2 + 1 x ≥ 1 , for all x > 0 , is

Options

  1. A1 64
  2. B1 32
  3. C1 27
  4. D1 25

Correct answer

C. 1 27

Step-by-step solution

f x = 4 α x 2 + 1 x ; x > 0 ⇒ f ' x = 8 α x - 1 x 2 = 8 α x 3 - 1 x 2 ⇒ f x attains its minimum at x = 1 8 α 1 3 f 1 8 α 1 3 = 1 ⇒ 4 α 1 8 α 2 3 + 8 α 1 3 = 1 ⇒ 3 α 1 3 = 1     ⇒     α = 1 27

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