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JEE Advanced 2016

JEE Main 2016 Mathematics question on Application of Derivatives

The least value of α∈R\alpha \in R for which 4αx2+1x≥14\alpha x^2 + \frac{1}{x} \geq 1, for all x>0x > 0, is
  1. A.
    164\frac{1}{64}
  2. B.
    132\frac{1}{32}
  3. C.
    127\frac{1}{27}
  4. D.
    125\frac{1}{25}
Show answer and solution

Correct option: C

Solution

We need 4αx² + 1/x ≥ 1 for all x > 0, so α ≥ (1 - 1/x)/(4x²) for all x > 0.

Find the maximum of f(x) = (1 - 1/x)/(4x²) for x > 0. Setting f'(x) = 0:

f(x) = (x-1)/(4x³), so f'(x) = [4x³ - (x-1)·12x²]/(4x³)² = x²[4x - 12(x-1)]/(16x⁶) = [4x - 12x + 12]/(16x⁴) = (-8x + 12)/(16x⁴).

Setting -8x + 12 = 0 gives x = 3/2.

Maximum value: f(3/2) = (3/2 - 1)/(4·(3/2)²) = (1/2)/(4·9/4) = (1/2)/9 = 1/18...

Let me recompute: f(3/2) = (1 - 2/3)/(4·9/4) = (1/3)/9 = 1/27.

So the least value of α is 1/27.

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