NTA Abhyas JEE Main2020MathematicsApplication of DerivativesPractice

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

Options

Aα = 2 , β = - 1 2
Bα = 2 , β = 1 2
Cα = - 6 , β = 1 2
Dα = - 6 , β = - 1 2

Correct answer

A. α = 2 , β = - 1 2

Step-by-step solution

f x = α log x + βx 2 + x If x < 0 f ⁡ x = α log - x + β x 2 + x ∴ f ⁡ ′ x = - α - x + 2 β x + 1 If x > 0 f ⁡ x = α log x + β x 2 + x ∴ f ⁡ ′ x = α x + 2 β x + 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 = α 2 + 4 β + 1 = 0 Adding, - 3 α 2 + 3 = 0 ∴ α = 2 ∴ β = - 1 2

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