Basic Mathematics & Logarithm
Inequalities
Grade 11
Question:
<p>Complete solution set of inequality \(\dfrac{(x+2)(x+3)}{(x-2)(x-3)} \leq 1\) is</p>
<p>\((-\infty, 0]\)</p>
<p>\((-\infty, 0] \cup (2, 3)\)</p>
<p>\([2, 3]\)</p>
<p>\((-\infty, 2) \cup (3, \infty)\)</p>
Step-by-Step Solution
Key Concept: Rearrange the inequality to standard form by moving 1 to the left side, then find a common denominator to obtain a single rational expression. The sign analysis of this expression determines where the inequality holds.
<p><strong>Step 1:</strong> Rearrange to standard form:</p><p>$$\frac{(x+2)(x+3)}{(x-2)(x-3)} - 1 \leq 0$$</p><p><strong>Step 2:</strong> Find common denominator:</p><p>$$\frac{(x+2)(x+3) - (x-2)(x-3)}{(x-2)(x-3)} \leq 0$$</p><p><strong>Step 3:</strong> Expand numerator:</p><p>$(x+2)(x+3) = x^2 + 5x + 6$</p><p>$(x-2)(x-3) = x^2 - 5x + 6$</p><p>$(x+2)(x+3) - (x-2)(x-3) = (x^2 + 5x + 6) - (x^2 - 5x + 6) = 10x$</p><p><strong>Step 4:</strong> The inequality becomes:</p><p>$$\frac{10x}{(x-2)(x-3)} \leq 0$$</p><p><strong>Step 5:</strong> Sign analysis with critical points: $x = 0, 2, 3$</p><p>• $x < 0$: $\frac{(-)}{(+)} < 0$ ✓</p><p>• $0 < x < 2$: $\frac{(+)}{(-)} < 0$ ✓</p><p>• $2 < x < 3$: $\frac{(+)}{(-)} < 0$ ✓</p><p>• $x > 3$: $\frac{(+)}{(+)} > 0$ ✗</p><p>• At $x = 0$: expression equals 0 ✓</p><p><strong>Step 6:</strong> Exclude $x = 2$ and $x = 3$ (denominator = 0)</p><p>∴ Answer: $(-\infty, 0] \cup (2, 3)$</p>
Correct Answer: B