Let ƒ(x) = (1 + cos 8 x) (ax 2 + bx + c)

We are given dx = dx

⇒ dx = dx + dx

⇒ dx = 0

If ƒ(x) > 0 (ƒ(x) < 0) ∀ x ∈ [1, 2],

Then dx > 0

But

∴ ƒ(x) is partly positive and partly negative on [1, 2].

⇒ there exist α , β ∈ [1, 2] such that

ƒ( α ) > 0 and ƒ( β ) < 0.

As ƒ is continuous on [1, 2] there exists γ lying between α and β (and hence between 1 and 2) such that ƒ( γ ) = 0

⇒ (1 + cos 8 γ ) (a γ 2 + b γ + c) = 0

⇒ a γ 2 + b γ + c = 0 [  1 + cos 8 γ ≥ 1]

Thus, ax 2 + bx + c = 0 has at least one root in [1, 2].