BITSAT 2013 Paper

Dividing R at

2

3

,

4

3

and 2, analyse 4 cases
When x ≤

2

3

, the inequality becomes
2 – 3x + 4 – 3x + 6 – 3x ≥ 12.
implying – 9x ≥ 0 ⇒ x ≤ 0.
when x ≥ 2 the ineqality becomes
3x – 2 + 3x – 4 + 3x – 6 ≥ 12,
Implying 9x ≥ 24 ⇒ x ≥ 8/3
The inequality in invalid in the other two sections.
∴ either x ≤ 0 or x ≥ 8/3