CAIE P2 2008 November — Question 1 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2008
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| > |linear|
DifficultyStandard +0.3 This is a straightforward modulus inequality requiring consideration of critical points (x=0 and x=3) and testing intervals, which is a standard technique taught in P2. While it requires systematic case analysis, it's a routine application of the method with no conceptual surprises, making it slightly easier than average.
Spec1.02l Modulus function: notation, relations, equations and inequalities

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

AnswerMarks Guidance
State or imply non-modular inequality \((x-3)^2 > (2x)^2\) or corresponding quadratic equation or pair of linear equations \((x-3) = \pm 2x\)M1
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equationsM1
Obtain critical values \(x = 1\) and \(x = -3\)A1
State answer \(-3 < x < 1\)A1 [4]
OR:
AnswerMarks Guidance
Obtain critical value \(x = -3\) from a graphical method, or by inspection, or by solving a linear inequality or linear equationB1
Obtain the critical value \(x = 1\) similarlyB2
State answer \(-3 < x < 1\)B1 [4]
State or imply non-modular inequality $(x-3)^2 > (2x)^2$ or corresponding quadratic equation or pair of linear equations $(x-3) = \pm 2x$ | M1 |
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations | M1 |
Obtain critical values $x = 1$ and $x = -3$ | A1 |
State answer $-3 < x < 1$ | A1 | [4]

**OR:**

Obtain critical value $x = -3$ from a graphical method, or by inspection, or by solving a linear inequality or linear equation | B1 |
Obtain the critical value $x = 1$ similarly | B2 |
State answer $-3 < x < 1$ | B1 | [4]
1 Solve the inequality $| x - 3 | > | 2 x |$.

\hfill \mbox{\textit{CAIE P2 2008 Q1 [4]}}