JEE Main | 2014Math miscellaneousHard
Question
If x = -1 and x = 2 are extreme points of f(x) = α log |x| + βx2 + x then
Options
A.α = 2, β = - 
B.α = 2, β =
C.α = - 6, β =
D.α = - 6, β = -
Solution
f(x) = α log |x| + βx2 + x
f′(x) =
+ 2 βx + 1 = 0 at x = - 1, 2
- α -2β + 1 = 0 ⇒ α + 2β = 1 .....(i)
+ 4β + 1 = 0 ⇒ α + 8β = - 2 .....(ii)
6β = - 3 ⇒ β = -
∴ α = 2
f′(x) =
+ 2 βx + 1 = 0 at x = - 1, 2- α -2β + 1 = 0 ⇒ α + 2β = 1 .....(i)
+ 4β + 1 = 0 ⇒ α + 8β = - 2 .....(ii) 6β = - 3 ⇒ β = -
∴ α = 2
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