CAIE P2 2009 November — Question 1 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2009
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| < |linear|
DifficultyStandard +0.3 This requires squaring both sides to eliminate moduli, expanding and simplifying a quadratic inequality, then factorizing to find the solution set. It's slightly above average difficulty as it involves multiple algebraic steps and understanding that squaring preserves the inequality direction for positive expressions, but it's a standard technique taught in P2 with no novel insight required.
Spec1.02l Modulus function: notation, relations, equations and inequalities

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

AnswerMarks Guidance
EITHER: Obtain a non-modular inequality from \(2x + 3)^2 < (x - 3)^2\), or corresponding quadratic equation, or pair of linear equations \(2x + 3 = \pm(x - 3)\)M1
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equationsM1
Obtain critical values \(x = -6\) and \(x = 0\)A1
State answer \(-6 < x < 0\)A1
OR: Obtain the critical value \(x = -6\) from a graphical method or by inspection, or by solving a linear equation or inequalityB1
Obtain the critical value \(x = 0\) similarlyB2
State answer \(-6 < x < 0\)B1 [4 marks]
**EITHER:** Obtain a non-modular inequality from $2x + 3)^2 < (x - 3)^2$, or corresponding quadratic equation, or pair of linear equations $2x + 3 = \pm(x - 3)$ | M1 |
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations | M1 |
Obtain critical values $x = -6$ and $x = 0$ | A1 |
State answer $-6 < x < 0$ | A1 |

**OR:** Obtain the critical value $x = -6$ from a graphical method or by inspection, or by solving a linear equation or inequality | B1 |
Obtain the critical value $x = 0$ similarly | B2 |
State answer $-6 < x < 0$ | B1 | [4 marks]
1 Solve the inequality $| 2 x + 3 | < | x - 3 |$.

\hfill \mbox{\textit{CAIE P2 2009 Q1 [4]}}