JEE-Mains 2014 (13/30)

Calculus Level 3

If x = − 1 x=-1 and x = 2 x=2 are extreme points of f ( x ) = α log ⁡ ∣ x ∣ + β x 2 + x f(x)=\alpha \log|x|+\beta x^2+x , then :

α = 2 , β = 1 2 \alpha=2,\beta=\frac{1}{2} α = 2 , β = − 1 2 \alpha=2,\beta=-\frac{1}{2} α = − 6 , β = 1 2 \alpha=-6,\beta=\frac{1}{2} α = − 6 , β = − 1 2 \alpha=-6,\beta=-\frac{1}{2}

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Since x = − 1 & x = 2 x=-1 \text{ \& } x=2 are extreme points, it is implied that

f ′ ( − 1 ) = f ′ ( 2 ) = 0 f'(-1) = f'(2) = 0

f ′ ( − 1 ) = − α − 2 β + 1 = 0 → α + 2 β = 1 f'(-1) = -\alpha - 2\beta + 1 = 0 \rightarrow \alpha + 2\beta = 1

f ′ ( 2 ) = α 2 + 4 β + 1 = 0 f'(2) = \frac{\alpha}{2} + 4\beta + 1 = 0

Solving these two equations we get,

α = 2 & β = − 1 2 \alpha = 2 \text{ \& } \beta = \frac{-1}{2}

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