Answer :

Subtracting 6 from both the sides, we get–


|x–1|+|x–2|+|x–3|–6≥0


Here, we have 4 cases:


Case 1: –∞ <x<1


For this case, |x–1|=–(x–1), |x–2|=–(x–2) and |x–3|=–(x–3)


–(x–1+x–2+x–3+6)≥0


x–1+x–2+x–3+6<0


3x<0


x<0


xϵ (–∞ , 0) …(1)


Case 2:1<x<2


For this case, |x–1|=x–1, |x–2|=–(x–2) and |x–3|=–(x–3)


x–1–x+2–x+3–6≥0


–x–2≥0


x+2<0


x<–2


Which doesn’t signify the interval


Case 3:2<x<3


For this case, |x–1|=x–1, |x–2|=x–2 and x–3=–(x–3)


x–1+x–2–x+3–6≥0


x–6≥0


x≥ 6


Which doesn’t signify the interval


Case 4:3<x<∞


For this case, |x–1|=x–1, |x–2|=x–2 and |x–3|=x–3


x–1+x–2+x–3–6≥0


x–1+x–2+x–3–6≥0


3x–12>0


x>4


xϵ (4 , ∞ ) …(2)


x ϵ (–∞ , 0) (4 , ∞ ) (from 1 and 2)


We can verify the answers using graph as well.



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