Relations & Functions
Inequalities involving modulus
Grade 12
Question:
<p>Given that \( \dfrac{1}{|x| - 3} \leq \dfrac{1}{2} \), the solution set is:</p>
<p>\( (-\infty, -5] \cup [5, \infty) \)</p>
<p>\( (-\infty, -5] \cup (-3, 3) \cup [5, \infty) \)</p>
<p>\( (-\infty, -5] \cup (-3, 3) \cup (5, \infty) \)</p>
<p>\( (-5, -3) \cup (3, 5) \)</p>
Step-by-Step Solution
Key Concept: Analyze the inequality by considering the sign of the denominator separately. When the denominator is positive, inequality direction is preserved; when negative, it reverses. Also ensure the denominator never equals zero.
<p><strong>Step 1:</strong> Note that the expression is undefined when |x| - 3 = 0, i.e., when x = ±3.</p><p><strong>Step 2:</strong> <strong>Case 1:</strong> When |x| - 3 > 0 (i.e., |x| > 3):<br/>Multiply both sides by (|x| - 3) without reversing inequality:<br/>1 ≤ ½(|x| - 3)<br/>2 ≤ |x| - 3<br/>|x| ≥ 5<br/>This gives: x ≤ -5 or x ≥ 5</p><p><strong>Step 3:</strong> <strong>Case 2:</strong> When |x| - 3 < 0 (i.e., |x| < 3):<br/>Multiply both sides by (|x| - 3) and reverse inequality:<br/>1 ≥ ½(|x| - 3)<br/>2 ≥ |x| - 3<br/>5 ≥ |x|<br/>This gives: |x| ≤ 5, which means -5 ≤ x ≤ 5<br/>Combined with |x| < 3: -3 < x < 3</p><p><strong>Step 4:</strong> Combine both cases and exclude x = ±3:<br/>Solution set = (-∞, -5] ∪ (-3, 3) ∪ [5, ∞)</p><p>∴ Answer: B</p>
Correct Answer: B