CAIE P2 2011 November — Question 1 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2011
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| > |linear|
DifficultyStandard +0.3 This requires squaring both sides to eliminate moduli, then solving a quadratic inequality—a standard technique for |expression| > |expression| problems. It's slightly above average difficulty due to the algebraic manipulation required, but follows a well-established method taught in P2 with no novel insight needed.
Spec1.02l Modulus function: notation, relations, equations and inequalities

1 Solve the inequality \(| x + 2 | > \left| \frac { 1 } { 2 } x - 2 \right|\).

AnswerMarks Guidance
State or imply non-modular inequality \((x + 2)^2 > \left(\frac{1}{2}x - 2\right)^2\), or corresponding equation or pair of linear equationsM1
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equationsM1
Obtain critical values \(-8\) and \(0\)A1
State correct answer \(x < -8\) or \(x > 0\)A1 [4]
OR
AnswerMarks Guidance
Obtain one critical value, e.g. \(x = -8\), by solving a linear equation (or inequality) or from a graphical method or by inspectionB1
Obtain the other critical value similarlyB2
State correct answer \(x < -8\) or \(x > 0\)B1 [4]
State or imply non-modular inequality $(x + 2)^2 > \left(\frac{1}{2}x - 2\right)^2$, or corresponding equation or pair of linear equations | M1 |
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations | M1 |
Obtain critical values $-8$ and $0$ | A1 |
State correct answer $x < -8$ or $x > 0$ | A1 | [4] |

**OR**

Obtain one critical value, e.g. $x = -8$, 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 $x < -8$ or $x > 0$ | B1 | [4] |
1 Solve the inequality $| x + 2 | > \left| \frac { 1 } { 2 } x - 2 \right|$.

\hfill \mbox{\textit{CAIE P2 2011 Q1 [4]}}