CAIE P2 2018 June — Question 1 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2018
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| < |linear|
DifficultyStandard +0.3 This is a standard modulus inequality requiring consideration of critical points (x = 2/3 and x = -5) and testing intervals, but follows a routine procedure taught in P2. It's slightly above average difficulty due to requiring systematic case analysis and algebraic manipulation across multiple regions, but remains a textbook exercise with no novel insight required.
Spec1.02l Modulus function: notation, relations, equations and inequalities

1 Solve the inequality \(| 3 x - 2 | < | x + 5 |\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
Either
State or imply non-modular inequality \((3x-2)^2 < (x+5)^2\) or corresponding equation or pair of linear equationsB1
Attempt solution of 3-term quadratic equation or of 2 linear equationsM1
Obtain critical values \(-\frac{3}{4}\) and \(\frac{7}{2}\)A1
State answer \(-\frac{3}{4} < x < \frac{7}{2}\)A1
Or
Obtain critical value \(\frac{7}{2}\) from graph, inspection, equationB1
Obtain critical value \(-\frac{3}{4}\) similarlyB2
State answer \(-\frac{3}{4} < x < \frac{7}{2}\)B1
Total4
**Question 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| **Either** | | |
| State or imply non-modular inequality $(3x-2)^2 < (x+5)^2$ or corresponding equation or pair of linear equations | B1 | |
| Attempt solution of 3-term quadratic equation or of 2 linear equations | M1 | |
| Obtain critical values $-\frac{3}{4}$ and $\frac{7}{2}$ | A1 | |
| State answer $-\frac{3}{4} < x < \frac{7}{2}$ | A1 | |
| **Or** | | |
| Obtain critical value $\frac{7}{2}$ from graph, inspection, equation | B1 | |
| Obtain critical value $-\frac{3}{4}$ similarly | B2 | |
| State answer $-\frac{3}{4} < x < \frac{7}{2}$ | B1 | |
| **Total** | **4** | |
1 Solve the inequality $| 3 x - 2 | < | x + 5 |$.\\

\hfill \mbox{\textit{CAIE P2 2018 Q1 [4]}}