CAIE P2 2010 June — Question 3 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2010
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 = 0.5 and x = -4) and testing regions, but the algebraic manipulation is straightforward. Slightly above average difficulty as it requires systematic case analysis rather than just routine manipulation, but this is a well-practiced technique at A-level.
Spec1.02l Modulus function: notation, relations, equations and inequalities

3 Solve the inequality \(| 2 x - 1 | < | x + 4 |\).

AnswerMarks Guidance
EITHER State or imply non-modular inequality \((2x-1)^2 < (x+4)^2\), 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 \(-1\) and \(5\)A1
State correct answer \(-1 < x < 5\)A1 [4]
OR Obtain one critical value, e.g. \(x = 5\), by solving a linear equation (or inequality) or from a graphical method or by inspectionB1
Obtain the other critical value similarlyB2
State correct answer \(-1 < x < 5\)B1
**EITHER** State or imply non-modular inequality $(2x-1)^2 < (x+4)^2$, 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 $-1$ and $5$ | A1 |

State correct answer $-1 < x < 5$ | A1 | [4]

**OR** Obtain one critical value, e.g. $x = 5$, 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 $-1 < x < 5$ | B1 |
3 Solve the inequality $| 2 x - 1 | < | x + 4 |$.

\hfill \mbox{\textit{CAIE P2 2010 Q3 [4]}}