Consider the following inequality:
(×): a|x| + 1 ≤ |x − 2|
where a is a real constant.
Which one of the following describes the complete set of values of a such that (×) is true for all real x?
- Aa ≤ 32
- Ba ≤ 1
- Ca ≤ 12
- Da ≤ 0
- Ea ≤ −12
- Fa ≤ −1
- Ga ≤ −32
- HThere are no such values of a.
Show the answer and worked solution
answer · E
- Aa ≤ 32
- Ba ≤ 1
- Ca ≤ 12
- Da ≤ 0
- Ea ≤ −12
- Fa ≤ −1
- Ga ≤ −32
- HThere are no such values of a.
Test the value that makes the right-hand side smallest before doing the general work: at x = 2 the inequality reads 2a + 1 ≤ 0, so a ≤ −12 is forced. Now check that a ≤ −12 is enough, splitting at x = 0 and x = 2. For x < 0 it becomes x(1 − a) ≤ 1, and the left side is negative, so it holds. For 0 ≤ x ≤ 2 it becomes (a+1)x ≤ 1, whose worst case is x = 2, giving a ≤ −12 again. For x ≥ 2 it becomes (a − 1)x ≤ −3, and since a − 1 < 0 the worst case is the smallest x, namely x = 2, giving 2 − 2a ≥ 3, once more a ≤ −12.