Question

If a₀, a₁, a₂, a₃ are all positive, then

4a₀x³ + 3a₁x² + 2a₂x + a₃ = 0 has at least one root in (–1, 0) if

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

Answer 1

Answer: (A)

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

Answer 2

P(x) ≡ 4a₀x³ + 3a₁x² + 2a₂ x + a₃ is a polynomial and hence is continuous for all x. P(x) = 0 has a root in (–1, 0) if it takes both positive and negative values in (–1, 0) as continuity implies that P(x) = 0 at least one point. This will happen if either P(–1).P(0) < 0 or the area enclosed by the graph of P(x), the x-axis and the ordinates at x = – 1 and x = 0 is zero.

As 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.