CAIE P2 2010 June — Question 1 3 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2010
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| > constant (greater than)
DifficultyEasy -1.2 This is a straightforward application of the basic modulus inequality rule, requiring only splitting into two cases (2x-3 > 5 or 2x-3 < -5) and solving two simple linear inequalities. It's a standard textbook exercise testing recall of a fundamental technique with minimal steps, making it easier than average.
Spec1.02l Modulus function: notation, relations, equations and inequalities

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

AnswerMarks Guidance
EITHER: State or imply non-modular inequality \((2x - 3)^2 > 5^2\), or corresponding equation or pair of linear equationsM1
Obtain critical values \(-1\) and \(4\)A1
State correct answer \(x < -1, x > 4\)A1 [3]
OR: State one critical value, e.g. \(x = 4\), having solved a linear equation (or inequality) or from a graphical method or by inspectionB1
State the other critical value correctlyB1
State correct answer \(x < -1, x > 4\)B1 [3]
EITHER: State or imply non-modular inequality $(2x - 3)^2 > 5^2$, or corresponding equation or pair of linear equations | M1 |
Obtain critical values $-1$ and $4$ | A1 |
State correct answer $x < -1, x > 4$ | A1 | [3]

OR: State one critical value, e.g. $x = 4$, having solved a linear equation (or inequality) or from a graphical method or by inspection | B1 |
State the other critical value correctly | B1 |
State correct answer $x < -1, x > 4$ | B1 | [3]
1 Solve the inequality $| 2 x - 3 | > 5$.

\hfill \mbox{\textit{CAIE P2 2010 Q1 [3]}}