Sol. We have,

f(x) = x 3 + 3 (a –7)x 2 + 3(a 2 –9) x –1

∴ f ′ (x) = 3x 2 + 6(a –7) x + 3(a 2 –9)

= 3{x 2 + 2(a –7) x + (a 2 –9)}

For local maximum or minimum, we must have f ′ (x) = 0

⇒ x 2 + 2(a –7) x + a 2 –9 = 0

⇒ x = – (a –7) ±

Let x 1 = –(a –7) + and x 2 = – (a – 7) –

For x 1 , x 2 to be real, we must have

58 –14a > 0 ⇒ a < =

Now,

f ′′ (x) = 3{2x + 2(a –7)}

= 6 (x + a –7)

Clearly, f ′′ (x 2 ) = 6(x 2 + a –7) = –6 < 0

So, x 2 = – (a –7) – is a point of local maximum.

Now,

x 2 > 0

⇒ – (a –7) – > 0

⇒ 7 – a >

⇒ (7 – a) 2 > 58 – 14a

⇒ a 2 – 9 > 0

⇒ a < –3 or a > 3

Hence, a ∈ (– ∞ , – 3) ∪ (3, 29/7)