<p>Let \(f(x) = ax^2 + 2bx - 3c\) has no real root and \(\dfrac{3c}{4} < a + b\), then:</p>
Step-by-Step Solution
Key Concept: Since f(x) has no real roots, its discriminant must be negative: 4b² + 12ac < 0, which gives b² < -3ac. Combined with the inequality 3c/4 < a, we can establish bounds on the relationship between coefficients and determine which statements must be true.
<p><strong>Step 1:</strong> Since f(x) = ax² + 2bx - 3c has no real roots, the discriminant must be negative:</p><p>Δ = (2b)² - 4(a)(-3c) < 0</p><p>4b² + 12ac < 0</p><p>∴ <strong>b² < -3ac</strong> ... (i)</p><p><strong>Step 2:</strong> From inequality (i), since b² ≥ 0, we need -3ac > 0, which means <strong>ac < 0</strong>.</p><p>This tells us a and c have opposite signs.</p><p><strong>Step 3:</strong> Given: 3c/4 < a</p><p>If c > 0, then 3c/4 > 0, so a > 3c/4 > 0, contradicting ac < 0.</p><p>Therefore <strong>c < 0</strong> (and consequently <strong>a > 0</strong>).</p><p><strong>Step 4:</strong> From b² < -3ac and a > 0, c < 0: we get b² < 3a|c|</p><p>This bounds |b| and shows <strong>b can be positive or negative</strong>, but |b| is restricted.</p><p><strong>Step 5:</strong> From 3c/4 < a with c < 0: we have a > 3c/4. Since c < 0, this is automatically satisfied for a > 0, giving us information about how negative c can be relative to a.</p><p>∴ Statements typically true: <strong>a > 0, c < 0</strong> (and depending on options: ac < 0, b² < -3ac)</p>
Correct Answer: A,B