CAIE P3 2017 March — Question 2 4 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2017
SessionMarch
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve k|linear| compared to |linear|
DifficultyStandard +0.8 This requires systematic case analysis of two modulus expressions with different critical points (x=4 and x=-1/3), solving multiple inequalities across three regions, and combining solutions. More demanding than routine single-modulus problems but follows standard technique for A-level.
Spec1.02l Modulus function: notation, relations, equations and inequalities

2 Solve the inequality \(| x - 4 | < 2 | 3 x + 1 |\).

Question 2:
AnswerMarks Guidance
AnswerMark Guidance
EITHER: State or imply non-modular inequality \((x-4)^2 < (2(3x+1))^2\), or corresponding quadratic equation, or pair of linear equations \(x - 4 = \pm 2(3x+1)\)(B1)
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations for \(x\)M1
Obtain critical values \(x = -\frac{6}{5}\) and \(x = \frac{2}{7}\)A1
State final answer \(x < -\frac{6}{5}\), \(x > \frac{2}{7}\)A1)
OR: Obtain critical value \(x = -\frac{6}{5}\) from graphical method, by inspection, or by solving a linear equation or inequality(B1)
Obtain critical value \(x = \frac{2}{7}\) similarlyB2
State final answer \(x < -\frac{6}{5}\), \(x > \frac{2}{7}\)B1)
Total: 4
## Question 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| **EITHER:** State or imply non-modular inequality $(x-4)^2 < (2(3x+1))^2$, or corresponding quadratic equation, or pair of linear equations $x - 4 = \pm 2(3x+1)$ | (B1) | |
| Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations for $x$ | M1 | |
| Obtain critical values $x = -\frac{6}{5}$ and $x = \frac{2}{7}$ | A1 | |
| State final answer $x < -\frac{6}{5}$, $x > \frac{2}{7}$ | A1) | |
| **OR:** Obtain critical value $x = -\frac{6}{5}$ from graphical method, by inspection, or by solving a linear equation or inequality | (B1) | |
| Obtain critical value $x = \frac{2}{7}$ similarly | B2 | |
| State final answer $x < -\frac{6}{5}$, $x > \frac{2}{7}$ | B1) | |

**Total: 4**

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2 Solve the inequality $| x - 4 | < 2 | 3 x + 1 |$.\\

\hfill \mbox{\textit{CAIE P3 2017 Q2 [4]}}