Question

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

Answer 1

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