CAIE P2 2013 June — Question 2 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2013
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=4 and x=8) and testing regions, which is a routine technique taught in P2. While it requires systematic case analysis across three intervals, the algebraic manipulation within each case is straightforward, making it slightly easier than average but not trivial.
Spec1.02l Modulus function: notation, relations, equations and inequalities

2 Solve the inequality \(| x - 8 | > | 2 x - 4 |\).

Either
AnswerMarks Guidance
State or imply non-modular inequality \((x - 8)^2 > (2x - 4)^2\), or corresponding equation or pair of linear equationsM1
Make reasonable solution attempt at a quadratic, or solve two linear equationsM1
Obtain critical values 4 and -4A1
State correct answer \(-4 < x < 4\)A1 [4]
Or
AnswerMarks Guidance
Obtain one critical value, e.g. \(x = 4\), by solving a linear equation (or inequality) or from a graphical method or by inspectionB1
Obtain the other critical value similarlyB2
State correct answer \(-4 < x < 4\)B1 [4]
**Either**

State or imply non-modular inequality $(x - 8)^2 > (2x - 4)^2$, or corresponding equation or pair of linear equations | M1 |
Make reasonable solution attempt at a quadratic, or solve two linear equations | M1 |
Obtain critical values 4 and -4 | A1 |
State correct answer $-4 < x < 4$ | A1 | [4]

**Or**

Obtain one critical value, e.g. $x = 4$, by solving a linear equation (or inequality) or from a graphical method or by inspection | B1 |
Obtain the other critical value similarly | B2 |
State correct answer $-4 < x < 4$ | B1 | [4]
2 Solve the inequality $| x - 8 | > | 2 x - 4 |$.

\hfill \mbox{\textit{CAIE P2 2013 Q2 [4]}}