Question

For what values of a do the points of extremum of the function y = x³ – 3ax² + 3(a² – 1) x + 1 lie in the interval (– 2, 4)?

Answer 1

We have y = x³ – 3ax² + 3(a² – 1) x + 1

⇒ dy/dx = 3x² – 6ax + 3(a² – 1)

For max or min, dy/dx = 0 

⇒ 3x² – 6ax + 3(a² – 1) = 0

⇒  x² – 2ax + (a² – 1) = 0

Let g(x) = x² – 2ax + (a² – 1)

Clearly, g(– 2) > 0 and g(4) > 0

Now, g(– 2) > 0 gives

⇒  4 + 4a + a² – 1 > 0

⇒  a² + 4a + 3 > 0

⇒  (a + 1)(a + 3) > 0

⇒ a < –3, a > –1 ...(i)

Also, g(4) > 0 gives 16 – 8a + a² – 1 > 0

⇒  a² – 8a + 15 > 0

⇒ (a – 3)(a – 5) > 0

⇒  a < 3, a > 5 ...(ii)

Again, –2 < a < 4 ...(iii)

From (i), (ii) and (iii), we get,

–1 < a < 3

⇒ a ∈ (–1, 3)