Question

f a₀, a₁, a₂, a₃ are all positive, then 4a₀ x³ + 3a₁ x² + 2 a₂ x + a₃ = 0 has at least one root in (–1, 0) if –

• A. a₀ + a₂ = a₁ + a₃ and 4a₀ + 2a₂> 3a₁ + a₃
• B. 4a₀ + 2a₂< 3a₁ + a₃
• C. 4a₀ + 2a₂ = 3a₁ + a₃ and a₀ + a₂< a₁ + a₃
• D. None of these

Answer 1

Answer: (A)

a₀ + a₂ = a₁ + a₃ and 4a₀ + 2a₂> 3a₁ + a₃

Answer 2

P(x) = 4a₀ x³ + 3a₁ x² + 2a₂ x + a₃ is a polynomial, so it is continuous for all x.

P(x) = 0 has a root in (–1, 0).

P(–1) P(0) < 0 or the area enclosed by the graph of P(x), x-axis and the ordinates at x = – 1, and

x = 0 is zero.

P(0) = a₃ > 0, P(– 1) = – 4a₀ + 3a₁ – 2a₂ + a₃ < 0 or

4a₀ + 2a₂ > 3a₁ + a₃

$\int_{- 1}^{0}{}$P(x) dx = 0 gives a₀ + a₂ = a₁ + a₃