If \(\displaystyle\frac{1}{|x|-3} \le \frac{1}{2}\), then value of \(x\):
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(x \in (-\infty,-5] \cup (-3,3) \cup [5,\infty)\)
The expression is not defined when \(|x| = 3\), so \(x \ne \pm 3\). The sign of the denominator decides the two cases.
Case 1: \(|x| \lt 3\), that is \(-3 \lt x \lt 3\).
\(|x| - 3 \lt 0\)
The left side is negative, so it is always less than \(\frac{1}{2}\). Every x in \((-3, 3)\) satisfies the inequality.
Case 2: \(|x| \gt 3\). The denominator is positive, so multiplying both sides by \(2(|x| - 3)\) keeps the sign:
\(2 \le |x| - 3\)
\(|x| \ge 5\)
\(x \le -5 \text{ or } x \ge 5\)
Combining the two cases:
\(x \in (-\infty,-5] \cup (-3,3) \cup [5,\infty)\)
Why the other options are incorrect
Option (1) \(x \in (-\infty,-6) \cup (-3,3] \cup (5,\infty)\) — Incorrect. It includes \(x = 3\), where the expression is not defined, and leaves out \(x = 5\), where \(\frac{1}{5 - 3} = \frac{1}{2}\) satisfies the inequality.
Option (2) \(x \in (-\infty,5) \cup (5,\infty)\) — Incorrect. It includes \(x = 4\), where \(\frac{1}{4 - 3} = 1 \gt \frac{1}{2}\).
Option (3) \(x \in (-\infty,-3) \cup [3,\infty)\) — Incorrect. It includes \(x = 4\), where the left side is \(1 \gt \frac{1}{2}\), and leaves out \(x = 0\), where the left side is \(-\frac{1}{3} \lt \frac{1}{2}\).