CAIE P3 2019 November — Question 1 4 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2019
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve k|linear| compared to |linear|
DifficultyStandard +0.3 This is a standard modulus inequality requiring case-by-case analysis based on critical points x = -2 and x = 1/3. Students must consider three intervals, square both sides or use sign analysis, then combine solution sets. Slightly above average difficulty due to the algebraic manipulation required, but follows a well-practiced technique with no novel insight needed.
Spec1.02l Modulus function: notation, relations, equations and inequalities

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

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
State or imply non-modular inequality \((x+2)^2 > (3x-1)^2\), or corresponding quadratic equation, or pair of linear equations \(2(x+2) = \pm(3x-1)\)B1
Make reasonable attempt at solving a 3-term quadratic, or solve two linear equations for \(x\)M1
Obtain critical values \(x = -\frac{3}{5}\) and \(x = 5\)A1
State final answer \(-\frac{3}{5} < x < 5\)A1
Alternative method:
Obtain critical value \(x = 5\) from graphical method, or by inspection, or by solving a linear equation or inequalityB1
Obtain critical value \(x = -\frac{3}{5}\) similarlyB2
State final answer \(-\frac{3}{5} < x < 5\)B1
Total4
**Question 1:**

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

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

\hfill \mbox{\textit{CAIE P3 2019 Q1 [4]}}