<p>One of the roots of \(ax^2 + bx + c = 0\) is greater than 2 and the other is less than −1. If the roots of \(cx^2 + bx + a = 0\) are <em>α</em> and <em>β</em>, then</p>
<p>\(0 < \alpha < \dfrac{1}{2}\) and \(-1 < \beta < 0\)</p>
<p>\(\alpha < \dfrac{1}{2}\) and \(\beta < -1\)</p>
<p>\(\alpha > \dfrac{1}{2}\) and \(\beta > -1\)</p>
<p>\(\alpha < 2\) and \(\beta > -1\)</p>
Step-by-Step Solution
Key Concept: If roots of ax² + bx + c = 0 lie on opposite sides of 2 and -1, then evaluating the quadratic at these points reveals sign conditions. These conditions on a, b, c transfer to constraints on the roots α, β of the reciprocal coefficient equation cx² + bx + a = 0.
<p><strong>Step 1:</strong> Let roots of ax² + bx + c = 0 be r₁ > 2 and r₂ < -1.</p><p><strong>Step 2:</strong> Since r₁ > 2 and r₂ < -1, we have f(2)·a and f(-1)·a both have specific signs. Calculate: f(2) = 4a + 2b + c and f(-1) = a - b + c.</p><p><strong>Step 3:</strong> One root > 2 and other < -1 means the parabola crosses x-axis on opposite sides of [-1, 2]. Thus: a·f(2) < 0 and a·f(-1) < 0 (both points inside the root interval have opposite sign to 'a').</p><p><strong>Step 4:</strong> For cx² + bx + a = 0, note that if r is a root of ax² + bx + c = 0, then 1/r satisfies c(1/r)² + b(1/r) + a = 0, giving cr² + br + a = 0 after multiplying by r². The roots α, β of cx² + bx + a = 0 are reciprocals: α = 1/r₁ and β = 1/r₂.</p><p><strong>Step 5:</strong> Since r₁ > 2: then 0 < α = 1/r₁ < 1/2. Since r₂ < -1: then -1 < β = 1/r₂ < 0.</p><p><strong>Step 6:</strong> Therefore both roots α and β lie in the interval (-1, 1/2), or more precisely: -1 < β < 0 < α < 1/2.</p><p>∴ Answer: A</p>
Correct Answer: A