Sol. Since f(x) is positive and increasing function in I.

Therefore,

f(x) > 0 and f ′ (x) > 0 for all x ∈ I

Also, φ (x) is decreasing in I. Therefore,

φ′ (x) < 0 for all x ∈ I

Now,

φ (x) = 3f 4 –4f 3 –36f 2 –8

⇒ φ′ (x) = (12f 3 –12f 2 –72f) f ′

But, φ′ (x) < 0 for all x ∈ I. Therefore,

(12f 3 –12f 2 –72f) f ′ < 0 for all x ∈ I.

⇒ 12f (f 2 –f – 6) < 0 [  f ′ > 0]

⇒ f 2 – f – 6 < 0 [  f > 0]

⇒ (f –3) (f + 2) < 0

⇒ – 2 < f < 3

⇒ 0 ≤ f 2 < 9

Now,

G(x) = f 4 + f 2 –20

= (f 2 + 5) (f 2 –4)

Since 0 ≤ f 2 < 9. Therefore, f 2 – 4 can take both positive and negative values.

Hence, G(x) may not maintain the same sign in I