Question

Solve  \[\frac{1}{\left| x \right| - 3} \leq \frac{1}{2}\]

Answer 1

\[\text{ As }, \frac{1}{\left| x \right| - 3} \leq \frac{1}{2}\]
\[ \Rightarrow \frac{1}{\left| x \right| - 3} - \frac{1}{2} \leq 0\]
\[ \Rightarrow \frac{2 - \left( \left| x \right| - 3 \right)}{2\left( \left| x \right| - 3 \right)} \leq 0\]
\[ \Rightarrow \frac{2 - \left| x \right| + 3}{2\left( \left| x \right| - 3 \right)} \leq 0\]
\[ \Rightarrow \frac{5 - \left| x \right|}{\left| x \right| - 3} \leq 0\]
\[\text{ Case I: When } x \geq 0, \left| x \right| = x, \]
\[\frac{5 - x}{x - 3} \leq 0\]
\[ \Rightarrow \left( 5 - x \leq 0 \text{ and } x - 3 > 0 \right) \text{ or } \left( 5 - x \geq 0 \text{ and } x - 3 < 0 \right)\]
\[ \Rightarrow \left( x \geq 5 \text{ and } x > 3 \right) \text{ or } \left( x \leq 5 \text{ and } x < 3 \right)\]
\[ \Rightarrow x \geq 5 or x < 3\]
\[ \Rightarrow x \in \left( 0, 3 \right) \cup [5, \infty )\]
\[\text{ Case II: When } x < 0, \left| x \right| = - x, \]
\[\frac{5 + x}{- x - 3} \leq 0\]
\[ \Rightarrow \frac{x + 5}{x + 3} \geq 0\]
\[ \Rightarrow \left( x + 5 > 0 \text{ and } x + 3 > 0 \right) or \left( x + 5 < 0 \text{ and } x + 3 < 0 \right)\]
\[ \Rightarrow \left( x > - 5 \text{ and } x > - 3 \right) \text{ or } \left( x < - 5 \text{ and } x < - 3 \right)\]
\[ \Rightarrow x > - 3 \text{ or } x < - 5\]
\[ \Rightarrow x \in \left( - \infty , - 5 \right) \cup \left( - 3, \infty \right)\]
\[\text{ So, from both the cases, we get }\]
\[x \in \left( - \infty , - 5 \right) \cup \left( - 3, \infty \right) \cup \left( 0, 3 \right) \cup [5, \infty )\]
\[ \therefore x \in ( - \infty , - 5] \cup \left( - 3, 3 \right) \cup [5, \infty )\]