Solve for x, the inequalities in

Let y = |x – 2|, then

Now, if y < 1, then

y – 1 < 0 and y – 2 < 0

and

This is not required

if y > 2, then

y – 1 > 0 and y – 2 > 0

and

This is not required

if 1 ≤ y < 2, then

y – 1 ≥ 0 and y – 2 < 0

and

This is required answer,

Hence

1 ≤ y < 2

⇒ 1 ≤ |x – 2| < 2

Therefore, there are two cases

⇒ 1 ≤ x – 2 < 2

⇒ 3 ≤ x < 4

and

⇒ 1 ≤ -(x – 2) < 2

⇒ 1 ≤ - x + 2 < 2

Multiplying with -1 all sides

⇒ -2 ≤ x – 2 < -1

Adding 2 both side

⇒ 0 ≤ x < 1

∴ Hence, solution is [0, 1) ∪ [3, 4)