CAIE P2 2013 November — Question 1 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2013
SessionNovember
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 = -1 and x = -5/3) and testing regions, but follows a routine method taught in P2. It's slightly above average difficulty due to the algebraic manipulation needed across multiple cases, but remains a textbook-style question with no novel insight required.
Spec1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities

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

AnswerMarks
State or imply non-modular inequality \((x+1)^2 < (3x+5)^2\), 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 \(-2\) and \(-\frac{3}{2}\)A1
State correct answer \(x < -2\) or \(x > -\frac{3}{2}\)A1
Or
AnswerMarks Guidance
Obtain one critical value, e.g. \(x = -2\), by solving a linear equation (or inequality) or from a graphical method or by inspectionB1
Obtain the other critical value similarlyB2
State correct answer \(x < -2\) or \(x > -\frac{3}{2}\)B1 [4]
State or imply non-modular inequality $(x+1)^2 < (3x+5)^2$, 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 $-2$ and $-\frac{3}{2}$ | A1 |
State correct answer $x < -2$ or $x > -\frac{3}{2}$ | A1 |

**Or**

Obtain one critical value, e.g. $x = -2$, 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 $x < -2$ or $x > -\frac{3}{2}$ | B1 | [4]

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1 Solve the inequality $| x + 1 | < | 3 x + 5 |$.

\hfill \mbox{\textit{CAIE P2 2013 Q1 [4]}}