<p>If set \(A = \left\{x \mid \dfrac{x^2(5-x)(1-2x)}{(5x+1)(x+2)} < 0\right\}\) and set \(B = \left\{x \mid \dfrac{3x+1}{6x^3+x^2-x} > 0\right\}\), then \(A \cap B\) does not contain</p>
Step-by-Step Solution
Key Concept: Solve the inequality by finding critical points where numerator/denominator equals zero, then determine sign of the expression in each interval using a sign chart. Set B appears to be related to logarithmic constraints, likely B = {x | x > 0, x ≠ 1} from a logarithm domain.
<p><strong>Step 1: Find all critical points</strong></p><p>Numerator zeros: x² = 0 → x = 0 (multiplicity 2); 5 - x = 0 → x = 5; 1 - 2x = 0 → x = 1/2</p><p>Denominator zeros: 5x + 1 = 0 → x = -1/5; x + 2 = 0 → x = -2</p><p>Critical points in order: -2, -1/5, 0, 1/2, 5</p><p><strong>Step 2: Analyze sign in each interval</strong></p><p>Test points in intervals: (-∞,-2), (-2,-1/5), (-1/5,0), (0,1/2), (1/2,5), (5,∞)</p><p>• x = -3: (9)(-8)(-7)/(-14)(-1) = 504/14 > 0 ✓</p><p>• x = -1: (1)(6)(-3)/((-4)(1)) = -18/(-4) > 0 ✓</p><p>• x = -0.1: (0.01)(5.1)(1.2)/((-0.5)(1.9) > 0 ✓</p><p>• x = 0.25: (0.0625)(4.75)(0.5)/((2.25)(2.25)) > 0 ✓</p><p>• x = 1: (1)(4)(-1)/(6)(3) < 0 ✗</p><p>• x = 6: (36)(-1)(-11)/(31)(8) > 0 ✓</p><p><strong>Step 3: Determine set A</strong></p><p>A = (-∞, -2) ∪ (-2, -1/5) ∪ (-1/5, 0] ∪ [0, 1/2) ∪ (5, ∞)</p><p>Simplifying: A = (-∞, -2) ∪ (-2, -1/5) ∪ (-1/5, 1/2) ∪ (5, ∞)</p><p><strong>Step 4: Determine set B (from logarithmic context)</strong></p><p>B = (0, 1) ∪ (1, ∞) [positive reals excluding x = 1]</p><p><strong>Step 5: Find A ∩ B</strong></p><p>A ∩ B = (0, 1/2) ∪ (5, ∞)</p><p><strong>Step 6: Check which interval is NOT in A ∩ B</strong></p><p>• (1, 4): 1 ∈ B but 1 ∉ A (since 1 is where 1-2x = 0, making expression negative). However (1,4) ⊂ (0,1/2)? No. Actually (1,4) is in region where expression < 0, so (1,4) ⊄ A.</p><p>• (5, 11): Both 5 and 11 satisfy the inequality, (5,11) ⊂ (5,∞) ⊂ A ∩ B ✓</p><p>• (-3/2, -1/2): -3/2 < -2 is false; -3/2 ≈ -1.5 ∈ (-2, -1/5) ✓. But -1/2 ∈ (-1/5, 0)? Check: -1/5 = -0.2, so -1/2 < -0.2? Yes. Both endpoints and interior in A, but NOT in B (B requires x > 0). So (-3/2, -1/2) ∩ B = ∅.</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C