CAIE P2 2014 June — Question 1 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2014
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| > |linear|
DifficultyStandard +0.3 This is a standard modulus inequality requiring systematic case analysis (squaring both sides or considering critical points x = 2/3 and x = -4), but the algebra is straightforward and the technique is well-practiced in P2. Slightly above average difficulty due to the two-modulus comparison, but remains a routine textbook exercise.
Spec1.02l Modulus function: notation, relations, equations and inequalities
1 Solve the inequality \(| 3 x - 2 | \geqslant | x + 4 |\).
Either path:
AnswerMarks Guidance
State or imply non-modular inequality \((3x - 2)^2 > (x + 4)^2\) or corresponding equation or pair of linear equationsB1
Attempt solution of 3-term quadratic equation or of 2 linear equationsM1
Obtain critical values \(-\frac{1}{4}\) and \(3\)A1
State answer \(x < -\frac{1}{4}, x > 3\)A1 [4]
Or path:
AnswerMarks Guidance
Obtain critical value \(x = 3\) from graphical method, inspection, equationB1
Obtain critical value \(x = -\frac{1}{4}\) similarlyB2
State answer \(x < -\frac{1}{4}, x > 3\)B1 [4] 
**Either path:**

State or imply non-modular inequality $(3x - 2)^2 > (x + 4)^2$ or corresponding equation or pair of linear equations | B1 |
Attempt solution of 3-term quadratic equation or of 2 linear equations | M1 |
Obtain critical values $-\frac{1}{4}$ and $3$ | A1 |
State answer $x < -\frac{1}{4}, x > 3$ | A1 | [4]

**Or path:**

Obtain critical value $x = 3$ from graphical method, inspection, equation | B1 |
Obtain critical value $x = -\frac{1}{4}$ similarly | B2 |
State answer $x < -\frac{1}{4}, x > 3$ | B1 | [4]
1 Solve the inequality $| 3 x - 2 | \geqslant | x + 4 |$.

\hfill \mbox{\textit{CAIE P2 2014 Q1 [4]}}
This paper (7 questions)
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Q1 4 Q2 6 Q3 6 Q4 8 Q5 8 Q6 8 Q7 10