| Exam Board | WJEC |
|---|---|
| Module | Unit 3 (Unit 3) |
| Year | 2018 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve k|linear| compared to |linear| |
| Difficulty | Standard +0.3 This is a modulus equation requiring case analysis (typically 2-3 cases based on critical points x = -1/2 and x = 2), but the algebraic manipulation is straightforward once cases are identified. It's slightly above average difficulty as it requires systematic case work and checking solutions, but remains a standard A-level technique without requiring deep insight. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities |
| Answer | Marks |
|---|---|
| 1 | 0 |
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 1 | 4 |
| 1 | 5 |
| 1 | 6 |
| 1 | 7 |
Question 1:
1 | 0
1 | 1
1 | 2
1 | 3
1 | 4
1 | 5
1 | 6
1 | 7Solve the equation
$$|2x + 1| = 3|x - 2|.$$ [4]
\hfill \mbox{\textit{WJEC Unit 3 2018 Q1 [4]}}