CAIE P2 2012 June — Question 1 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2012
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 = -3 and x = -1/2) and testing regions, but follows a routine algorithmic approach taught in P2. It's slightly above average difficulty due to the need to handle multiple cases systematically, but requires no novel insight beyond applying the standard technique.
Spec1.02l Modulus function: notation, relations, equations and inequalities
1 Solve the inequality \(| x + 3 | < | 2 x + 1 |\).
AnswerMarks Guidance
Either: State or imply non-modular inequality \((x + 3)^2 < (2x + 1)^2\) or corresponding equation or pair of linear equationsB1
Attempt solution of 3-term quadratic or of 2 linear equationsM1
Obtain critical values \(-\frac{4}{3}\) and \(2\)A1
State answer \(x < -\frac{4}{3}, x > 2\)A1 [4]
Or: Obtain critical value \(x = 2\) from graphical method, inspection, equationB1
Obtain critical value \(x = -\frac{4}{3}\) similarlyB2
State answer \(x < -\frac{4}{3}, x > 2\)B1 [4] 
Either: State or imply non-modular inequality $(x + 3)^2 < (2x + 1)^2$ or corresponding equation or pair of linear equations | B1 |
Attempt solution of 3-term quadratic or of 2 linear equations | M1 |
Obtain critical values $-\frac{4}{3}$ and $2$ | A1 |
State answer $x < -\frac{4}{3}, x > 2$ | A1 | [4]

Or: Obtain critical value $x = 2$ from graphical method, inspection, equation | B1 |
Obtain critical value $x = -\frac{4}{3}$ similarly | B2 |
State answer $x < -\frac{4}{3}, x > 2$ | B1 | [4]

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1 Solve the inequality $| x + 3 | < | 2 x + 1 |$.

\hfill \mbox{\textit{CAIE P2 2012 Q1 [4]}}
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Q1 4 Q2 5 Q3 5 Q4 8 Q5 8 Q6 9 Q7 11