Question

Let a, b, c be non-zero real number such that

$\int_{0}^{1}{(1 + \cos^{8}x)(ax^{2} + bx + c)dx}$= $\int_{0}^{2}{(1 + \cos^{8}x)(ax^{2} + bx + c)dx}$

Then the quadratic equation ax² + bx + c = 0 has –

• A. No root in (0, 2)
• B. At least one root in (1, 2)
• C. A double root (0, 2)
• D. None of these

Answer 1

Answer: (B)

At least one root in (1, 2)

Answer 2

Let ƒ(x) = (1 + cos⁸ x) (ax² + bx + c)

We are given $\int_{0}^{1}{ƒ(x)}$dx = $\int_{0}^{2}{ƒ(x)}$dx

⇒ $\int_{0}^{1}{ƒ(x)}$dx = $\int_{0}^{1}{ƒ(x)}$dx + $\int_{1}^{2}{ƒ(x)}$dx

⇒$\int_{1}^{2}{ƒ(x)}$dx = 0

If ƒ(x) > 0 (ƒ(x) < 0) ∀ x ∈ [1, 2],

Then $\int_{1}^{2}{ƒ(x)}$dx > 0 $\left( \int_{1}^{2}{ƒ(x)dx < 0} \right)$

But $\int_{1}^{2}{ƒ(x)dx = 0}$

∴ 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 + cos⁸ γ) (aγ² + bγ + c) = 0

⇒ aγ² + bγ + c = 0 [Q 1 + cos⁸ γ ≥ 1]

Thus, ax² + bx + c = 0 has at least one root in [1, 2].