Solve: |x-1|+|x-2|+|x-3| > (or) = 6
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Here the change point are x = 1 , 2 , 3
Hence we consider the following case
(I) x < 1
(II) 1 < x < 2
(III) 2 < x < 3
(IV) x > 3
case (I) x < 1
-(x - 1) - ( x - 2) - (x - 3) ≥ 3
-3x + 6 ≥6 or -3x ≥ 0 ∴ x ≥\, 0
Which is < 1 and hence the solution
case (II) 2 ≥ x <3
(x - 1) - ( x - 2) - (x - 3) ≥ 3
-3x + 6 ≥6 or -x ≥ 2 x ≥\, -2
This does not satisfy given condition of case (II) Hence no solution
case (III) 2 ≥ x <3
(x - 1) - ( x - 2) - (x - 3) ≥ 3
x ≥\, 6
This does not satisfy given condition of case (III) Hence no solution
case (IV) x ≥3
(x - 1) - ( x - 2) - (x - 3) ≥ 3
x ≥\, 14 or x ≥\, 4
This does not satisfy given condition of case (III) Hence no solution
Thus the required solution by case I are IV are x ≥\, 0 or x ≥\, 4
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