CAIE P2 2003 November — Question 1 3 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2003
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInequalities
TypeSolve absolute value inequality
DifficultyModerate -0.8 This is a straightforward absolute value inequality requiring students to split into two cases (8-3x < 2 and 8-3x > -2) and solve two linear inequalities. It's a standard textbook exercise testing basic understanding of absolute values with minimal steps, making it easier than average but not trivial since it requires correct handling of inequality direction when dividing by negative coefficients.
Spec1.02l Modulus function: notation, relations, equations and inequalities

Find the set of values of \(x\) satisfying the inequality \(|8 - 3x| < 2\). [3]

AnswerMarks
EITHER: State or imply non-modular inequality e.g. \(-2 < 8-3x < 2\), or \((8-3x)^2 < 2^2\), or corresponding equation or pair of equationsM1
Obtain critical values \(2\) and \(3\frac{1}{3}\)A1
State correct answer \(2 < x < 3\frac{1}{3}\)A1
OR: State one critical value (probably \(x = 2\)), from a graphical method or by inspection or by solving a linear equality or equationB1
State the other critical value correctlyB1
State correct answer \(2 < x < 3\frac{1}{3}\)B1
Total: [3]
**EITHER:** State or imply non-modular inequality e.g. $-2 < 8-3x < 2$, or $(8-3x)^2 < 2^2$, or corresponding equation or pair of equations | M1 |

Obtain critical values $2$ and $3\frac{1}{3}$ | A1 |

State correct answer $2 < x < 3\frac{1}{3}$ | A1 |

**OR:** State one critical value (probably $x = 2$), from a graphical method or by inspection or by solving a linear equality or equation | B1 |

State the other critical value correctly | B1 |

State correct answer $2 < x < 3\frac{1}{3}$ | B1 |

**Total: [3]**
Find the set of values of $x$ satisfying the inequality $|8 - 3x| < 2$. [3]

\hfill \mbox{\textit{CAIE P2 2003 Q1 [3]}}