CAIE P2 2011 June — Question 1 3 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2011
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| = |linear| (both linear inside)
DifficultyModerate -0.5 This is a straightforward modulus equation requiring students to consider cases where expressions are positive or negative, then solve resulting linear equations and check validity. It's slightly easier than average as it involves only linear expressions and standard case-by-case analysis with no geometric insight or complex algebraic manipulation required.
Spec1.02l Modulus function: notation, relations, equations and inequalities

1 Solve the equation \(| 3 x + 4 | = | 2 x + 5 |\).

AnswerMarks
Either attempt to square both sides obtaining three terms on each sideM1
Attempt solution of three-term quadratic equationM1
Obtain \(5x + 4x - 9 = 0\) and hence \(-\frac{9}{5}\) and \(1\)A1
OR
AnswerMarks Guidance
Obtain value \(1\) from graphical method, inspection or linear equationB1
Obtain value \(-\frac{9}{5}\) similarlyB2 [3]
Either attempt to square both sides obtaining three terms on each side | M1 |
Attempt solution of three-term quadratic equation | M1 |
Obtain $5x + 4x - 9 = 0$ and hence $-\frac{9}{5}$ and $1$ | A1 |

OR

Obtain value $1$ from graphical method, inspection or linear equation | B1 |
Obtain value $-\frac{9}{5}$ similarly | B2 | [3]
1 Solve the equation $| 3 x + 4 | = | 2 x + 5 |$.

\hfill \mbox{\textit{CAIE P2 2011 Q1 [3]}}