Question

Solve for x, |x + 1| + |x| > 3.

Answer 1

On L.H.S. of the given inequality, We have two terms both containing modulus. By equating the expression within the modulus to zero, we get x = –1, 0 as critical points. These critical points divide the real line in three parts as (– `oo`, – 1), [–1, 0), [0, `oo`).

Case I: When `–oo < x < –1`

|x + 1| + |x| > 3

⇒ –x – 1 – x > 3

⇒ x < –2

Case II: When –1 ≤ x < 0,

|x + 1| + |x| > 3

⇒ x + 1 – x > 3

⇒ 1 > 3    ....(Not possible)

Case III: When 0 ≤ x < `oo`,

|x + 1| + |x| > 3

⇒ x + 1 + x > 3

⇒ x > 1

Combining the results of cases (I), (II) and (III), We get x ∈ `(–oo , –2) ∪ (1, oo)`.