Basic Mathematics & Logarithm
Inequalities involving absolute values
Grade 11
Question:
<p>If \(\dfrac{1}{|x|-3}\leq\dfrac{1}{2}\). Then,</p>
<p>\(x\in(-10,-5)\cup(-3,3)\cup[5,10)\)</p>
<p>\(x\in(-\infty,-5)\cup(-3,3)\cup[5,\infty)\)</p>
<p>\(x\in(-\infty,-10]\cup(-3,3)\cup[10,\infty)\)</p>
<p>\(x\in(-15,-10]\cup(-3,3)\cup[10,15)\)</p>
Step-by-Step Solution
Key Concept: When taking reciprocals of positive numbers in an inequality, the inequality sign flips. You must also ensure the denominator is positive and handle the absolute value by considering cases based on its definition.
<p><strong>Step 1:</strong> For the inequality to be defined, we need |x| - 3 ≠ 0, so |x| ≠ 3.</p><p><strong>Step 2:</strong> We have two cases based on the sign of (|x| - 3).</p><p><strong>Case 1:</strong> If |x| - 3 > 0 (i.e., |x| > 3), then multiplying both sides by (|x| - 3) preserves the inequality:<br>1 ≤ ½(|x| - 3)<br>2 ≤ |x| - 3<br>|x| ≥ 5</p><p><strong>Case 2:</strong> If |x| - 3 < 0 (i.e., |x| < 3), then multiplying both sides by (|x| - 3) reverses the inequality:<br>1 ≥ ½(|x| - 3)<br>2 ≥ |x| - 3<br>|x| ≤ 5</p><p>Combined with |x| < 3, this gives |x| < 3.</p><p><strong>Step 3:</strong> Combining both cases: |x| < 3 or |x| ≥ 5<br>Which gives: -3 < x < 3 or x ≤ -5 or x ≥ 5</p><p><strong>Step 4:</strong> In interval notation: x ∈ (-∞, -5] ∪ (-3, 3) ∪ [5, ∞)</p><p>∴ Answer: B</p>
Correct Answer: B