CAIE P3 2014 November — Question 1 4 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2014
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| < |linear|
DifficultyStandard +0.3 This is a standard modulus inequality requiring consideration of critical points (x = 1/3 and x = -5/2) and testing intervals, but follows a routine method taught explicitly in P3. It's slightly above average difficulty due to the algebraic manipulation across multiple cases, but remains a textbook exercise with no novel insight required.
Spec1.02l Modulus function: notation, relations, equations and inequalities

1 Solve the inequality \(| 3 x - 1 | < | 2 x + 5 |\).

Either route:
AnswerMarks Guidance
State or imply non-modular inequality \((3x-1)^2 < (2x+5)^2\) or corresponding quadratic equation or pair of linear equations \(3x-1 = \pm(2x+5)\)B1
Solve a three-term quadratic or two linear equations \(5x^2 - 26x - 24 < 0\)M1
Obtain \(-\frac{4}{5}\) and \(6\)A1
State \(-\frac{4}{5} < x < 6\)A1 [4]
Or route:
AnswerMarks Guidance
Obtain value \(6\) from graph, inspection or solving linear equationB1
Obtain value \(-\frac{4}{5}\) similarlyB2
State \(-\frac{4}{5} < x < 6\)B1 [4]
**Either route:**
State or imply non-modular inequality $(3x-1)^2 < (2x+5)^2$ or corresponding quadratic equation or pair of linear equations $3x-1 = \pm(2x+5)$ | B1 |
Solve a three-term quadratic or two linear equations $5x^2 - 26x - 24 < 0$ | M1 |
Obtain $-\frac{4}{5}$ and $6$ | A1 |
State $-\frac{4}{5} < x < 6$ | A1 | [4]

**Or route:**
Obtain value $6$ from graph, inspection or solving linear equation | B1 |
Obtain value $-\frac{4}{5}$ similarly | B2 |
State $-\frac{4}{5} < x < 6$ | B1 | [4]
1 Solve the inequality $| 3 x - 1 | < | 2 x + 5 |$.

\hfill \mbox{\textit{CAIE P3 2014 Q1 [4]}}