CAIE P2 2004 November — Question 1 3 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2004
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| > |linear|
DifficultyModerate -0.5 This is a straightforward modulus inequality requiring students to consider cases based on critical values (x = 0 and x = -1), then solve linear inequalities in each region. While it requires systematic case analysis, the algebraic manipulation is simple and the technique is standard for P2, making it slightly easier than average.
Spec1.02l Modulus function: notation, relations, equations and inequalities
1 Solve the inequality \(| x + 1 | > | x |\).
Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
State or imply non-modular inequality \((x+1)^2 > x^2\) or corresponding quadratic or linear equation \(x+1=-x\)B1 EITHER method
Obtain critical value \(-\frac{1}{2}\)B1
State answer \(x > -\frac{1}{2}\)B1
Obtain critical value \(-\frac{1}{2}\) by linear inequality or graphical method or inspectionB2 OR method
State answer \(x > -\frac{1}{2}\)B1 Total: 3
[For \(2x+1>0\), \(x > +\frac{1}{2}\), or similar reasonable method]M1 
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply non-modular inequality $(x+1)^2 > x^2$ or corresponding quadratic or linear equation $x+1=-x$ | B1 | EITHER method |
| Obtain critical value $-\frac{1}{2}$ | B1 | |
| State answer $x > -\frac{1}{2}$ | B1 | |
| Obtain critical value $-\frac{1}{2}$ by linear inequality or graphical method or inspection | B2 | OR method |
| State answer $x > -\frac{1}{2}$ | B1 | **Total: 3** |
| [For $2x+1>0$, $x > +\frac{1}{2}$, or similar reasonable method] | M1 | |

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1 Solve the inequality $| x + 1 | > | x |$.

\hfill \mbox{\textit{CAIE P2 2004 Q1 [3]}}
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Q1 3 Q2 3 Q3 4 Q4 5 Q5 6 Q6 8 Q7 11 Q8 10