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

The least value of αR for which, 4αx2+1x 1, for all x>0, is 
  1. 164
  2. 132
  3. 127
  4. 125

Solution

fx=4αx2+1x;x>0
f'x=8αx-1x2=8αx3-1x2
fx attains its minimum at x=18α13
f18α13=1
4α18α23+8α13=1
3α13=1    α=127

Asked in: JEE Advanced 2016 (Paper 1)

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