CAIE P2 2011 November — Question 1 3 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2011
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| < constant (pure inequality)
DifficultyEasy -1.2 This is a straightforward application of the basic modulus inequality rule: |expression| < k means -k < expression < k. Students simply need to recall this definition, set up -3 < 4-5x < 3, and solve two linear inequalities. It requires minimal problem-solving and is a standard textbook exercise testing routine technique.
Spec1.02l Modulus function: notation, relations, equations and inequalities

1 Solve the inequality \(| 4 - 5 x | < 3\).

AnswerMarks
State or imply non-modular inequality \((4-5x)^2 < 3^2\), or corresponding equation or pair of linear equationsM1
Obtain critical values \(\frac{1}{5}\) and \(\frac{7}{5}\)A1
State correct answer \(\frac{1}{5} < x < \frac{7}{5}\)A1
OR
AnswerMarks Guidance
State one critical value, e.g. \(x = \frac{1}{5}\), by solving a linear equation (or inequality) or from a graphical method or by inspectionB1
State the other critical value correctlyB1
State correct answer \(\frac{1}{5} < x < \frac{7}{5}\)B1 [3]
State or imply non-modular inequality $(4-5x)^2 < 3^2$, or corresponding equation or pair of linear equations | M1 |

Obtain critical values $\frac{1}{5}$ and $\frac{7}{5}$ | A1 |

State correct answer $\frac{1}{5} < x < \frac{7}{5}$ | A1 |

**OR**

State one critical value, e.g. $x = \frac{1}{5}$, by solving a linear equation (or inequality) or from a graphical method or by inspection | B1 |

State the other critical value correctly | B1 |

State correct answer $\frac{1}{5} < x < \frac{7}{5}$ | B1 | [3]
1 Solve the inequality $| 4 - 5 x | < 3$.

\hfill \mbox{\textit{CAIE P2 2011 Q1 [3]}}