CAIE P3 2013 June — Question 1 3 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2013
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| = |linear| (both linear inside)
DifficultyModerate -0.8 This is a straightforward modulus equation requiring students to consider cases based on critical points (x=0 and x=2), then solve linear equations in each region. It's a standard textbook exercise testing basic understanding of modulus definitions with minimal algebraic complexity, making it easier than average but not trivial since it requires systematic case analysis.
Spec1.02l Modulus function: notation, relations, equations and inequalities1.02t Solve modulus equations: graphically with modulus function
Solve the equation \(|x - 2| = |\frac{1}{3}x|\). [3]
AnswerMarks
\((x-2)^2 = \left(\frac{1}{3}x\right)^2\) or pair of equations \(x - 2 = \pm\frac{1}{3}x\)M1
EITHER:
AnswerMarks
Obtain answer \(x = 3\)A1
Obtain answer \(x = \frac{3}{2}\), or equivalentA1
OR:
AnswerMarks Guidance
Obtain answer \(x = 3\) by solving an equation or by inspectionB1
State or imply the equation \(x - 2 = -\frac{1}{3}x\), or equivalentM1
Obtain answer \(x = \frac{3}{2}\), or equivalentA1 [3] 
$(x-2)^2 = \left(\frac{1}{3}x\right)^2$ or pair of equations $x - 2 = \pm\frac{1}{3}x$ | M1 |

**EITHER:**
Obtain answer $x = 3$ | A1 |
Obtain answer $x = \frac{3}{2}$, or equivalent | A1 |

**OR:**
Obtain answer $x = 3$ by solving an equation or by inspection | B1 |
State or imply the equation $x - 2 = -\frac{1}{3}x$, or equivalent | M1 |
Obtain answer $x = \frac{3}{2}$, or equivalent | A1 | [3]
Solve the equation $|x - 2| = |\frac{1}{3}x|$. [3]

\hfill \mbox{\textit{CAIE P3 2013 Q1 [3]}}
This paper (10 questions)
View full paper
Q1 3 Q2 5 Q3 5 Q4 6 Q5 6 Q6 8 Q7 9 Q8 11 Q9 11 Q10 11