Question

Solve  \[\frac{\left| x - 2 \right| - 1}{\left| x - 2 \right| - 2} \leq 0\] 

Answer 1

\[\text{ As }, \frac{\left| x - 2 \right| - 1}{\left| x - 2 \right| - 2} \leq 0\]

\[\text{ Case I: When } x \geq 2, \left| x - 2 \right| = x - 2, \]

\[\frac{x - 2 - 1}{x - 2 - 2} \leq 0\]

\[ \Rightarrow \frac{x - 3}{x - 4} \leq 0\]

\[ \Rightarrow \left( x - 3 \leq 0 \text{ and } x - 4 > 0 \right) \text{ or } \left( x - 3 \geq 0 \text{ and } x - 4 < 0 \right)\]

\[ \Rightarrow \left( x \leq 3 \text{ and } x > 4 \right) \text{ or } \left( x \geq 3 \text{ and } x < 4 \right)\]

\[ \Rightarrow \phi \text{ or } \left( 3 \leq x < 4 \right)\]

\[ \Rightarrow 3 \leq x < 4\]

\[\text{ So }, x \in [3, 4)\]

\[\text{ Case II: When } x \leq 2, \left| x - 2 \right| = 2 - x, \]

\[\frac{2 - x - 1}{2 - x - 2} \leq 0\]

\[ \Rightarrow \frac{1 - x}{- x} \leq 0\]

\[ \Rightarrow \frac{x - 1}{x} \leq 0\]

\[ \Rightarrow \left( x - 1 \leq 0 \text{ and } x > 0 \right) or \left( x - 1 \geq 0 \text{ and } x < 0 \right)\]

\[ \Rightarrow \left( x \leq 1 \text{ and }x > 0 \right) \text{ or } \left( x \geq 1 \text{ and } x < 0 \right)\]

\[ \Rightarrow \left( 0 < x \leq 1 \right) \text{ or } \phi\]

\[ \Rightarrow 0 < x \leq 1\]

\[\text{ So }, x \in (0, 1]\]

\[ \therefore \text{ From both the cases, we get }\]

\[x \in (0, 1] \cup [3, 4)\]