If x = – 1 and x = 2 are extreme points of f(x) = α log | x

Engineering

Mathematics

Maxima and Minima

Question

If x = – 1 and x = 2 are extreme points of f(x) = α log | x | + βx² + x, then

$α = - 6, β = \frac{- 1}{2}$

$α = 2, β = \frac{- 1}{2}$

$α = 2, β = \frac{1}{2}$

$α = - 6, β = \frac{1}{2}$

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$f' (x) = \frac{α}{x} + 2 β x + 1$

Now, f '(– 1) = 0

⇒ – α – 2β + 1 = 0 …… (1)

and f '(2) = 0

$\Rightarrow \frac{α}{2} + 4 β + 1 = 0$ …… (2)

$∴$ On solving, we get

α + 2β = 1

α + 8β = – 2

—————

6β = – 3 ⇒ β = $\frac{- 1}{2}$ ,

Now, α = 1 – 2β = 2

Hence, α = 2, $β = \frac{- 1}{2}$