A) \[0\le 0\le 4\]
B) \[x\le 0\,\,or\,\,x\ge 4\]
C) \[x\le -2\,\,or\,\,x\ge 4\]
D) \[x\ge -2\,\,or\,\,x\le 4\]
Correct Answer: B
Solution :
\[|x-\alpha |=x-\alpha \]if\[x\ge \alpha \] \[=\alpha -x\]if\[x\le \alpha \] Now mark \[1,\,\,2\] and \[3\] as shown and the regions as I. \[x\le 1\] II. \[1\le x\le 2\] III. \[2\le x\le 3\] IV. \[x\ge 3\] and discuss the inequation in all these regions. Region I. Since\[x<1\], therefore given inequation is \[(1-x)+(2-x)+(3-x)\le 6\] or \[-3x\ge 0\] or \[x\le 0\] \[\therefore \] \[x\le 1\] and \[x\le 0\] or \[x\le 0\] Region II. Since \[1\le x\le 2\] therefore given inequation is \[(x-1)+(2-x)+(3-x)\ge 6\] or \[4-x\ge 6\] or \[x\le -2\] It is impossible to find\[x\]such that \[1\le x\le 2\] and \[x\le -2\]. Region III. Since\[2\le x\le 3\], therefore given inequation is \[(x-1)+(x-2)+(3-x)\ge 6\] or \[x\ge 6\] Here also it is impossible to find \[x\] such that \[2\le x\le 3\] and\[x\ge 6\] Region IV. Since\[x\ge 3\], therefore given inequation is \[(x-1)+(x-2)+(x-3)\ge 6\] or \[x\ge 4\] \[\therefore \] \[x\ge 3\] and \[x\ge 4\] or \[x\ge 4\] \[\therefore \] \[x\]will satisfy\[x<0\] or \[x\ge 4\]
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