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