Question

If x satisfies |x – 1| + |x – 2| + |x – 3| ≥ 6, then (1) 0 ≤ x ≤ 4 (2) x ≤ 2 or x ≥ 4 (3) x ≤ 0 or x ≥ 4 (4) None of these

Answer 1

Answer:

verified✔

*⃣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

Step-by-step explanation:

HOPE IT'S HELPFUL......

Answer 2

Step-by-step explanation: