InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Solve for x, |x + 1| + |x| >3 . |
|
Answer» On LHS 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 (−∞, −1),[−1,0),[0. ∞). Case - I : When −∞ < 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 Case - III When 0 ≤ < ∞, x + 1| + |x| > 3 ⇒ x + 1 + x > 3 ⇒ x > 1. Combining the results of cases (I) (II) and (III), we get, x ∈ (−∞, −2),∪ [1, ∞) |
|