CAIE P2 2012 November — Question 1 3 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2012
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| < |linear|
DifficultyStandard +0.3 This is a straightforward modulus inequality requiring students to square both sides (since both are positive) to eliminate the modulus signs, then solve the resulting linear inequality. It's slightly above average difficulty as it requires knowing the squaring technique and careful algebraic manipulation, but it's a standard textbook exercise with no novel insight required.
Spec1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities

1 Solve the inequality \(| 2 x + 1 | < | 2 x - 5 |\).

AnswerMarks
State or imply non-modular inequality \(2x + 1)^2 < (2x - 5)^2\), or corresponding equation or pair of linear equationsM1
Obtain critical value 1A1
State correct answer \(x < 1\)A1
[3]
OR
AnswerMarks
State the critical value \(x = 1\), by solving a linear equation (or inequality) or from a graphical method or by inspectionB2
State correct answer \(x < 1\)B1
[3]
State or imply non-modular inequality $2x + 1)^2 < (2x - 5)^2$, or corresponding equation or pair of linear equations | M1 |
Obtain critical value 1 | A1 |
State correct answer $x < 1$ | A1 |
| | [3] |

**OR**

State the critical value $x = 1$, by solving a linear equation (or inequality) or from a graphical method or by inspection | B2 |
State correct answer $x < 1$ | B1 |
| | [3] |
1 Solve the inequality $| 2 x + 1 | < | 2 x - 5 |$.

\hfill \mbox{\textit{CAIE P2 2012 Q1 [3]}}