Basic Mathematics & Logarithm
Inequalities
Grade 11
Question:
<p>The solution of the inequality \(\dfrac{x+7}{x-5} + \dfrac{3x+1}{2} \geq 0\) is</p>
<p>\([1, 3] \cup (5, \infty)\)</p>
<p>\((1, 3) \cup (5, \infty)\)</p>
<p>\((-\infty, 1) \cup (5, \infty)\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Convert the inequality to a single rational expression by finding a common denominator, then determine the sign of the expression using a sign chart with critical points (zeros and undefined points).
<p><strong>Step 1:</strong> Find common denominator (2(x-5)) and combine:</p><p>$$\frac{2(x+7) + (3x+1)(x-5)}{2(x-5)} \geq 0$$</p><p><strong>Step 2:</strong> Expand the numerator:</p><p>$$2(x+7) + (3x+1)(x-5) = 2x + 14 + 3x^2 - 15x + x - 5 = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$$</p><p><strong>Step 3:</strong> The inequality becomes:</p><p>$$\frac{3(x-1)(x-3)}{2(x-5)} \geq 0$$</p><p><strong>Step 4:</strong> Critical points are x = 1, 3 (zeros of numerator) and x = 5 (undefined). Create sign chart:</p><p>• For x < 1: (−)(−)/(−) = negative ✗</p><p>• For 1 ≤ x < 3: (+)(−)/(−) = positive ✓</p><p>• For 3 ≤ x < 5: (+)(+)/(−) = negative ✗</p><p>• For x > 5: (+)(+)/(+) = positive ✓</p><p><strong>Step 5:</strong> Include boundary points where numerator = 0 (x = 1, 3) but exclude x = 5 (undefined):</p><p>∴ Answer: x ∈ [1, 3] ∪ (5, ∞)</p>
Correct Answer: A