If $\frac{1}{|x|-3}≤\frac{1}{2}$, then value of $x$:

$x ∈(-∞,-5] ∪ (-3,3) ∪ [5,∞)$

The correct answer is Option (4) → $x ∈(-∞,-5] ∪ (-3,3) ∪ [5,∞)$ **

Solve the inequality:

$\displaystyle \frac{1}{|x|-3} \le \frac12$

1. Domain

$|x| - 3 \neq 0 \;\Rightarrow\; |x| \neq 3$ So $x \neq 3$ and $x \neq -3$.

2. Consider two cases based on the sign of the denominator

Case 1: $|x| - 3 < 0$ → $|x| < 3$

Then the denominator is negative.

$\displaystyle \frac{1}{\text{negative}}$ is always negative.

RHS is positive ($\frac12$).

A negative number is always $\le$ a positive number.

So all $x$ with $|x| < 3$ satisfy the inequality:

$(-3,\,3)$

Case 2: $|x| - 3 > 0$ → $|x| > 3$

Denominator positive → multiply without changing inequality:

$\displaystyle \frac{1}{|x|-3} \le \frac12$

Multiply both sides by $(|x|-3)$:

$1 \le \frac12(|x|-3)$

Multiply by 2:

$2 \le |x|-3$

Add 3:

$|x| \ge 5$

Thus:

$x \le -5$ or $x \ge 5$

Final Answer

${(-\infty,-5] \;\cup\; (-3,3) \;\cup\; [5,\infty)}$