<p>Let \(\dfrac{a}{2} < \dfrac{b}{3} < \dfrac{c}{4}\), then the equation \((2x-a)(3x-b) + (3x-b)(4x-c) + (4x-c)(2x-a) = 0\) has</p>
<p>Both roots real</p>
<p>Roots are real and equal</p>
<p>One root lies between \(\left(\dfrac{a}{2}, \dfrac{b}{3}\right)\)</p>
<p>Other root lies between \(\left(\dfrac{b}{3}, \dfrac{c}{4}\right)\)</p>
Step-by-Step Solution
Key Concept: Transform the compound inequality into two separate quadratic inequalities by multiplying through by 2, then analyze the roots and sign changes of each quadratic to determine the valid intervals where both conditions hold simultaneously.
<p><strong>Step 1:</strong> Start with the compound inequality $\frac{a}{2} < \frac{1}{a-1} < \frac{a}{4}$. This represents two simultaneous inequalities:</p><ul><li>$\frac{a}{2} < \frac{1}{a-1}$ ... (I)</li><li>$\frac{1}{a-1} < \frac{a}{4}$ ... (II)</li></ul><p><strong>Step 2:</strong> Solve inequality (I): $\frac{a}{2} < \frac{1}{a-1}$</p><p>Rearrange: $\frac{a}{2} - \frac{1}{a-1} < 0 \Rightarrow \frac{a(a-1) - 2}{2(a-1)} < 0 \Rightarrow \frac{a^2 - a - 2}{2(a-1)} < 0$</p><p>Factor: $\frac{(a-2)(a+1)}{2(a-1)} < 0$</p><p>Critical points: $a = -1, 1, 2$. Testing intervals: $a \in (-1, 1) \cup (1, 2)$</p><p><strong>Step 3:</strong> Solve inequality (II): $\frac{1}{a-1} < \frac{a}{4}$</p><p>Rearrange: $\frac{1}{a-1} - \frac{a}{4} < 0 \Rightarrow \frac{4 - a(a-1)}{4(a-1)} < 0 \Rightarrow \frac{-a^2 + a + 4}{4(a-1)} < 0$</p><p>Or: $\frac{a^2 - a - 4}{4(a-1)} > 0$. Roots: $a = \frac{1 \pm \sqrt{17}}{2}$</p><p>Since $\frac{1 + \sqrt{17}}{2} \approx 2.56$ and $\frac{1 - \sqrt{17}}{2} \approx -1.56$</p><p>Testing intervals: $a \in (-\infty, -1.56) \cup (1, 2.56)$</p><p><strong>Step 4:</strong> Find intersection of both solution sets:</p><p>$[(-1, 1) \cup (1, 2)] \cap [(-\infty, -1.56) \cup (1, 2.56)] = (1, 2)$</p><p>∴ Answer: $a \in (1, 2)$ or equivalently statements A, C, D if they represent this interval or properties thereof.</p>
Correct Answer: ACD