Question

Solve  \[\left| x - 1 \right| + \left| x - 2 \right| + \left| x - 3 \right| \geq 6\]

Answer 1

\[\text{ We have }, \left| x - 1 \right| + \left| x - 2 \right| + \left| x - 3 \right| \geq 6 . . . . . \left( i \right)\]

\[\text{ As }, \left| x - 1 \right| = \binom{x - 1, x \geq 1}{1 - x, x < 1}; \]

\[\left| x - 2 \right| = \binom{x - 2, x \geq 2}{2 - x, x < 2}\text{ and }\]

\[\left| x - 3 \right| = \binom{x - 3, x \geq 3}{3 - x, x < 3}\]

\[\text{ Now }, \]

\[\text{ Case I: When } x < 1, \]

\[1 - x + 2 - x + 3 - x \geq 6\]

\[ \Rightarrow 6 - 3x \geq 6\]

\[ \Rightarrow 3x \leq 0\]

\[ \Rightarrow x \leq 0\]

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

\[\text{ Case II: When } 1 \leq x < 2, \]

\[x - 1 + 2 - x + 3 - x \geq 6\]

\[ \Rightarrow 4 - x \geq 6\]

\[ \Rightarrow x \leq 4 - 6\]

\[ \Rightarrow x \leq - 2\]

\[\text{ So }, x \in \phi\]

\[\text{ Case III : When } 2 \leq x < 3, \]

\[x - 1 + x - 2 + 3 - x \geq 6\]

\[ \Rightarrow x \geq 6\]

\[\text{ So }, x \in \phi\]

\[\text{ Case IV : When } x \geq 3, \]

\[x - 1 + x - 2 + x - 3 \geq 6\]

\[ \Rightarrow 3x - 6 \geq 6\]

\[ \Rightarrow 3x \geq 12\]

\[ \Rightarrow x \geq \frac{12}{3}\]

\[ \Rightarrow x \geq 4\]

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

\[\text{ So, from all the four cases, we get }\]

\[x \in ( - \infty , 0] \cup [4, \infty )\]