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प्रश्न
Solve \[\left| x - 1 \right| + \left| x - 2 \right| + \left| x - 3 \right| \geq 6\]
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उत्तर
\[\text{ We have }, \left| x - 1 \right| + \left| x - 2 \right| + \left| x - 3 \right| \geq 6 . . . . . \left( i \right)\]
\[\text{ As }, \left| x - 1 \right| = \binom{x - 1, x \geq 1}{1 - x, x < 1}; \]
\[\left| x - 2 \right| = \binom{x - 2, x \geq 2}{2 - x, x < 2}\text{ and }\]
\[\left| x - 3 \right| = \binom{x - 3, x \geq 3}{3 - x, x < 3}\]
\[\text{ Now }, \]
\[\text{ Case I: When } x < 1, \]
\[1 - x + 2 - x + 3 - x \geq 6\]
\[ \Rightarrow 6 - 3x \geq 6\]
\[ \Rightarrow 3x \leq 0\]
\[ \Rightarrow x \leq 0\]
\[\text{ So }, x \in ( - \infty , 0]\]
\[\text{ Case II: When } 1 \leq x < 2, \]
\[x - 1 + 2 - x + 3 - x \geq 6\]
\[ \Rightarrow 4 - x \geq 6\]
\[ \Rightarrow x \leq 4 - 6\]
\[ \Rightarrow x \leq - 2\]
\[\text{ So }, x \in \phi\]
\[\text{ Case III : When } 2 \leq x < 3, \]
\[x - 1 + x - 2 + 3 - x \geq 6\]
\[ \Rightarrow x \geq 6\]
\[\text{ So }, x \in \phi\]
\[\text{ Case IV : When } x \geq 3, \]
\[x - 1 + x - 2 + x - 3 \geq 6\]
\[ \Rightarrow 3x - 6 \geq 6\]
\[ \Rightarrow 3x \geq 12\]
\[ \Rightarrow x \geq \frac{12}{3}\]
\[ \Rightarrow x \geq 4\]
\[\text{ So }, x \in [4, \infty )\]
\[\text{ So, from all the four cases, we get }\]
\[x \in ( - \infty , 0] \cup [4, \infty )\]
