If x = - 1 and x = 2 are extreme points of f x = α log x + β x 2 + x , then

If x=-1 and x=2 are extreme points of fx=αlogx+βx2+x, then 
  1. α = 2 ,  β = - 1 2
  2. α = 2 ,  β = 1 2
  3. α = - 6 ,  β = 1 2
  4. α = - 6 ,  β =- 1 2

Solution

fx=αlogx+βx2+x

If x < 0

f⁡x=αlog-x+βx2+x

∴   f⁡'x=-α-x+2βx+1

If x > 0

f⁡x=αlogx+βx2+x

∴   f⁡'x=αx+2βx+1

f⁡ ′ - 1 = - α - 2 β + 1 = 0

f⁡ ′ 2 = α 2 + 4 β + 1 = 0

2 f⁡ ′ - 1 = - 2 α - 4 β + 2 = 0

and f'2=α2+4β+1=0

Adding,

- 3 α 2 + 3 = 0

∴    α = 2

∴    β = - 1 2

Asked in: JEE Main 2014 (06 Apr)

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