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 consideration of critical points at x=2 and x=-5, then testing regions or squaring both sides. While it requires systematic case analysis, it's a standard textbook exercise with no conceptual surprises, making it slightly easier than average.
Spec1.02l Modulus function: notation, relations, equations and inequalities

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

AnswerMarks Guidance
Either state or imply non-modular inequality \((x-2)^2 \geq (x+5)^2\), or corresponding equation or pair of linear equationsM1
Obtain critical value \(-\frac{3}{2}\)A1
State correct answer \(x \leq -\frac{3}{2}\)A1 [3]
Or state a correct linear equation for the critical value, e.g. \(x - 2 = -x - 5\), or corresponding correct linear inequality, e.g. \(x - 2 \geq -x - 5\)M1
Obtain critical value \(-\frac{3}{2}\)A1
State correct answer \(x \leq -\frac{3}{2}\)A1 [3]
Either state or imply non-modular inequality $(x-2)^2 \geq (x+5)^2$, or corresponding equation or pair of linear equations | M1 |
Obtain critical value $-\frac{3}{2}$ | A1 |
State correct answer $x \leq -\frac{3}{2}$ | A1 | [3]

Or state a correct linear equation for the critical value, e.g. $x - 2 = -x - 5$, or corresponding correct linear inequality, e.g. $x - 2 \geq -x - 5$ | M1 |
Obtain critical value $-\frac{3}{2}$ | A1 |
State correct answer $x \leq -\frac{3}{2}$ | A1 | [3]
1 Solve the inequality $| x - 2 | \geqslant | x + 5 |$.

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