CAIE P3 2018 November — Question 1 4 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2018
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve k|linear| compared to |linear|
DifficultyModerate -0.3 This is a straightforward modulus inequality requiring systematic case analysis (typically 3 cases based on critical points x = 0.5 and x = -4), then solving linear inequalities in each region. While it requires careful organization and checking multiple cases, the algebraic manipulation is routine and the technique is standard for A-level, making it slightly easier than average.
Spec1.02l Modulus function: notation, relations, equations and inequalities

Solve the inequality \(3|2x - 1| > |x + 4|\). [4]

Question 1:
AnswerMarks
1State or imply non-modular inequality
32 ( 2x−1 )2 > ( x+4 )2 , or corresponding quadratic equation, or pair
AnswerMarks Guidance
of linear equations/inequalities 3 ( 2x−1 )= ±( x+4 )B1 35x2 −44x−7=0
Make reasonable attempt at solving a 3-term
AnswerMarks Guidance
quadratic, or solve two linear equations for xM1 Allow for reasonable attempt at factorising
e.g. (5x – 7)(7x + 1)
7 1
Obtain critical values x = and x = −
AnswerMarks Guidance
5 7A1 Accept 1.4 and -0.143 or better for penultimate A mark
7 1
State final answer x> , x<−
AnswerMarks Guidance
5 7A1 ‘and’ is A0, 7 < x<−1 is A0. Must be exact values.
5 7
Must be strict inequalities in final answer
Alternative
7
Obtain critical value x = from a graphical method
AnswerMarks Guidance
5B1 or by inspection, or by solving a linear equation or an inequality
1
Obtain critical value x=− similarly
AnswerMarks
7B2
7 1
State final answer x> or x<− or equivalent
AnswerMarks Guidance
5 7B1 [Do not condone ⩾ for >, or ⩽ for <.]
4
AnswerMarks Guidance
QuestionAnswer Marks
Question 1:
1 | State or imply non-modular inequality
32 ( 2x−1 )2 > ( x+4 )2 , or corresponding quadratic equation, or pair
of linear equations/inequalities 3 ( 2x−1 )= ±( x+4 ) | B1 | 35x2 −44x−7=0
Make reasonable attempt at solving a 3-term
quadratic, or solve two linear equations for x | M1 | Allow for reasonable attempt at factorising
e.g. (5x – 7)(7x + 1)
7 1
Obtain critical values x = and x = −
5 7 | A1 | Accept 1.4 and -0.143 or better for penultimate A mark
7 1
State final answer x> , x<−
5 7 | A1 | ‘and’ is A0, 7 < x<−1 is A0. Must be exact values.
5 7
Must be strict inequalities in final answer
Alternative
7
Obtain critical value x = from a graphical method
5 | B1 | or by inspection, or by solving a linear equation or an inequality
1
Obtain critical value x=− similarly
7 | B2
7 1
State final answer x> or x<− or equivalent
5 7 | B1 | [Do not condone ⩾ for >, or ⩽ for <.]
4
Question | Answer | Marks | Guidance
Solve the inequality $3|2x - 1| > |x + 4|$. [4]

\hfill \mbox{\textit{CAIE P3 2018 Q1 [4]}}