9709 P33 - Jun 2022 - Q1
1489
Find, in terms of a, the set of values of x satisfying the inequality:
2|3x + a| < |2x + 3a|, where a is a positive constant.
Solution
First, remove the modulus by considering the cases for the inequality:
1. Solve the inequality without modulus: \(2(3x + a)^2 < (2x + 3a)^2\).
2. Expand both sides: \(2(9x^2 + 6ax + a^2) < 4x^2 + 12ax + 9a^2\).
3. Simplify: \(18x^2 + 12ax + 2a^2 < 4x^2 + 12ax + 9a^2\).
4. Rearrange: \(14x^2 - 7a^2 < 0\).
5. Factorize: \(14(x^2 - \frac{a^2}{2}) < 0\).
6. Solve for critical points: \(x = \pm \frac{a}{\sqrt{2}}\).
7. Simplify critical points: \(x = \frac{a}{2}\) and \(x = -\frac{5}{8}a\).
8. Determine the interval: \(-\frac{5}{8}a < x < \frac{1}{4}a\).
