Indian Institute of Foreign Trade 2011 Solved Paper

Given that $\left|\mathit{x}-1\right|+|\mathit{x}-2|+|\mathit{x}-3|\ge 6$
Case 1: When$\mathit{x}>3$
$\left(\mathit{x}-1\right)+\left(\mathit{x}-2\right)+\left(\mathit{x}-3\right)\ge 6$
$\mathit{x}\ge 4$
Therefore, the value of $\mathit{x}∊\left[4,\mathit{\infty }\right)$
Case 2: When
$\left(\mathit{x}-1\right)+\left(\mathit{x}-2\right)-\left(\mathit{x}-3\right)\ge 6$
$\mathit{x}\ge 6$
Therefore, no possible value of x in this domain
Case 3: When
$\left(\mathit{x}-1\right)-\left(\mathit{x}-2\right)-\left(\mathit{x}-3\right)\ge 6$
$\mathit{x}\le -2$
Therefore, no possible value of x in this domain.
Case 3: When
$-\left(\mathit{x}-1\right)-\left(\mathit{x}-2\right)-\left(\mathit{x}-3\right)\ge 6$
$\mathit{x}\le 0$
Therefore, the value of $\mathit{x}∊\left(-\mathit{\infty },0\right]$
Therefore, the value of x that will satisfy this inequality: $\mathit{x}∊\left(-\mathit{\infty },0\right]\cup \left[4,\mathit{\infty }\right)$
Hence, option B is the correct answer.