Quadratic Equations
Quadratic Inequalities
Grade 11
Question:
<p><strong>Ex. 78:</strong> Given \(f(x) = (x + 2a)(x + a - 4)\) where \(a \in \mathbb{R}\). If \(f(x) < 0\) for \(-1 < x < 1\), then 'a' satisfies:</p>
<p>(a) \(-1 < a < 3\)</p>
<p>(b) \(-\frac{1}{2} < a < \frac{3}{2}\)</p>
<p>(c) \(-3 < a < -\frac{1}{2}\)</p>
<p>(d) \(-3 < a < -\frac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: For a quadratic to be negative on an interval, evaluate it at the endpoints and set both values negative.
<p>Given \(f(x) = (x + 2a)(x + a - 4) = x^2 + (3a - 4)x + 2a(a - 4)\).</p><p>For \(f(x) < 0\) when \(-1 < x < 1\), we need:</p><p>\(f(-1) < 0\) and \(f(1) < 0\)</p><p>\(f(-1) = (-1 + 2a)(-1 + a - 4) = (2a - 1)(a - 5) < 0\)</p><p>\(f(1) = (1 + 2a)(1 + a - 4) = (1 + 2a)(a - 3) < 0\)</p><p>From these conditions: \(-1 < a < 3\)</p>
Correct Answer: a