CAIE P2 2014 November — Question 1 3 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2014
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| = |linear| (both linear inside)
DifficultyModerate -0.8 This is a straightforward modulus equation requiring consideration of cases where expressions inside the modulus are positive or negative. It's a standard textbook exercise with clear methodology (squaring both sides or case analysis) and involves only basic algebraic manipulation, making it easier than average but not trivial since students must handle the modulus concept correctly.
Spec1.02l Modulus function: notation, relations, equations and inequalities

1 Solve the equation \(| 3 x - 1 | = | 2 x + 5 |\).

AnswerMarks Guidance
Square both sides obtaining 3 terms on each sideM1
Solve 3-term quadratic equationM1
Obtain \(-\frac{4}{3}\) and 6A1 [3]
Or
AnswerMarks Guidance
Obtain value 6 from graphical method, inspection, linear equation, ...B1
Obtain value \(-\frac{4}{3}\) similarlyB2 [3]
Square both sides obtaining 3 terms on each side | M1 |
Solve 3-term quadratic equation | M1 |
Obtain $-\frac{4}{3}$ and 6 | A1 | [3]

**Or**

Obtain value 6 from graphical method, inspection, linear equation, ... | B1 |
Obtain value $-\frac{4}{3}$ similarly | B2 | [3]
1 Solve the equation $| 3 x - 1 | = | 2 x + 5 |$.

\hfill \mbox{\textit{CAIE P2 2014 Q1 [3]}}