CAIE P2 2006 June — Question 1 3 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2006
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: |expression| > k splits into expression > k or expression < -k. Requires only algebraic manipulation of two simple linear inequalities with no conceptual difficulty beyond recalling the standard technique.
Spec1.02l Modulus function: notation, relations, equations and inequalities

1 Solve the inequality \(| 2 x - 7 | > 3\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
EITHER: State or imply non-modular inequality \((2x-7)^2 > 3^2\), or corresponding equationM1
Obtain critical values 2 and 5A1
State correct answer \(x < 2,\ x > 5\)A1
OR: State one critical value, e.g. \(x = 5\), by solving a linear equation/inequality or graphical method or inspectionB1
State the other critical value correctlyB1
State correct answer \(x < 2,\ x > 5\)B1 Total: 3
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| **EITHER:** State or imply non-modular inequality $(2x-7)^2 > 3^2$, or corresponding equation | M1 | |
| Obtain critical values 2 and 5 | A1 | |
| State correct answer $x < 2,\ x > 5$ | A1 | |
| **OR:** State one critical value, e.g. $x = 5$, by solving a linear equation/inequality or graphical method or inspection | B1 | |
| State the other critical value correctly | B1 | |
| State correct answer $x < 2,\ x > 5$ | B1 | Total: 3 |

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1 Solve the inequality $| 2 x - 7 | > 3$.

\hfill \mbox{\textit{CAIE P2 2006 Q1 [3]}}