BITSAT 2011 Paper

f(x) = 2x3–3x2 – 12x + 4
⇒ f ' (x) = 6x2 – 6x – 12 = 6(x2 – x – 2)
= 6(x – 2) (x + 1)
For maxima and minima f ' (x) = 0
∴ 6(x – 2)(x + 1) = 0 ⇒ x = 2, – 1
Now, f '' (x) = 12x - 6
At x = 2; f '' (x) = 24 - 6 =18 > 0
∴ x = 2 , local min. point
At x = – 1; f '' (x) = 12(-1) - 6 = -18 < 0
∴ x = –1 local max. point