Question:

<p><strong>Step 1:</strong> Rearrange the inequality to one side:</p><p>$$\frac{x+3}{x^2-x-2} - \frac{1}{x-4} \geq 0$$</p><p><strong>Step 2:</strong> Factor the denominator on the left:</p><p>$$x^2 - x - 2 = (x-2)(x+1)$$</p><p>So the inequality becomes:</p><p>$$\frac{x+3}{(x-2)(x+1)} - \frac{1}{x-4} \geq 0$$</p><p><strong>Step 3:</strong> Find a common denominator $(x-2)(x+1)(x-4)$:</p><p>$$\frac{(x+3)(x-4) - (x-2)(x+1)}{(x-2)(x+1)(x-4)} \geq 0$$</p><p><strong>Step 4:</strong> Expand the numerator:</p><p>$(x+3)(x-4) = x^2 - x - 12$</p><p>$(x-2)(x+1) = x^2 - x - 2$</p><p>Numerator: $x^2 - x - 12 - (x^2 - x - 2) = -10$</p><p><strong>Step 5:</strong> The inequality becomes:</p><p>$$\frac{-10}{(x-2)(x+1)(x-4)} \geq 0$$</p><p><strong>Step 6:</strong> For this fraction to be non-negative, the denominator must be negative (since numerator is -10):</p><p>$$(x-2)(x+1)(x-4) < 0$$</p><p><strong>Step 7:</strong> Analyze sign using critical points: $x = -1, 2, 4$</p><p>Test intervals:</p><p>• $x < -1$: $(-)(-)(-) = -$ ✓ (denominator negative)</p><p>• $-1 < x < 2$: $(-)(-)(+) = +$ ✗</p><p>• $2 < x < 4$: $(+)(+)(-) = -$ ✓ (denominator negative)</p><p>• $x > 4$: $(+)(+)(+) = +$ ✗</p><p><strong>Step 8:</strong> Exclude points where denominator = 0: $x ≠ -1, 2, 4$</p><p>Solution: $x < -1$ or $2 < x < 4$</p><p>∴ Answer: C</p>