Question

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

Answer 1

\[\text{ As }, \frac{\left| x + 2 \right| - x}{x} < 2\]
\[ \Rightarrow \frac{\left| x + 2 \right| - x}{x} - 2 < 0\]
\[ \Rightarrow \frac{\left| x + 2 \right| - x - 2x}{x} < 0\]
\[ \Rightarrow \frac{\left| x + 2 \right| - 3x}{x} < 0\]
\[\text{ Case I: When } x \geq - 2, \left| x + 2 \right| = \left( x + 2 \right), \]
\[\frac{\left( x + 2 \right) - 3x}{x} < 0\]
\[ \Rightarrow \frac{2 - 2x}{x} < 0\]
\[ \Rightarrow \frac{- 2\left( x - 1 \right)}{x} < 0\]
\[ \Rightarrow \frac{x - 1}{x} > 0\]
\[ \Rightarrow \left( x - 1 > 0 \text{ and } x > 0 \right) or \left( x - 1 < 0 \text{ and } x < 0 \right)\]
\[ \Rightarrow \left( x > 1 \text{ and } x > 0 \right) \text{ or } \left( x < 1 \text{ and} x < 0 \right)\]
\[ \Rightarrow x > 1 \text{ or } x < 0\]
\[ \Rightarrow x \in [ - 2, 0) \cup \left( 1, \infty \right)\]
\[\text{ Case II }: \text{ When }x \leq - 2, \left| x + 2 \right| = - \left( x + 2 \right), \]
\[\frac{- \left( x + 2 \right) - 3x}{x} < 0\]
\[ \Rightarrow \frac{- x - 2 - 3x}{x} < 0\]
\[ \Rightarrow \frac{- 4x - 2}{x} < 0\]
\[ \Rightarrow \frac{- 2\left( 2x + 1 \right)}{x} < 0\]
\[ \Rightarrow \frac{2x + 1}{x} > 0\]
\[ \Rightarrow \left( 2x + 1 > 0 \text{ and } x > 0 \right) \text{ or } \left( 2x + 1 < 0 \text{ and } x < 0 \right)\]
\[ \Rightarrow \left( x > \frac{- 1}{2} and x > 0 \right) \text{ or } \left( x < \frac{- 1}{2} \text{ and } x < 0 \right)\]
\[ \Rightarrow x > 0 or x < \frac{- 1}{2}\]
\[ \Rightarrow x \in ( - \infty , - 2] \cup \left( 0, \infty \right)\]
\[\text{ So, from both the cases, we get }\]
\[x \in [ - 2, 0) \cup \left( 1, \infty \right) \cup ( - \infty , - 2] \cup \left( 0, \infty \right)\]
\[ \therefore x \in \left( - \infty , 0 \right) \cup \left( 1, \infty \right)\]