CAIE P2 2024 November — Question 2 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2024
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInequalities
TypeSolve absolute value inequality
DifficultyStandard +0.3 This is a straightforward absolute value inequality requiring case analysis (x ≥ 7 and x < 7), solving two linear inequalities, and checking validity of solutions. It's slightly above average difficulty due to the need for systematic case work and careful attention to which solutions are valid in each case, but remains a standard textbook exercise with no novel insight required.
Spec1.02l Modulus function: notation, relations, equations and inequalities

Solve the inequality \(|x - 7| > 4x + 3\). [4]

Question 2:
AnswerMarks Guidance
2Attempt solution of equation or inequality, where signs of x and 4x are different M1
Obtain 4 …
AnswerMarks Guidance
5A1 OE
… and finally no other valueA1
Conclude x 4
AnswerMarks Guidance
5A1  4
Allow −, .
 5
Alternative Method for Question 2
AnswerMarks
State or imply non-modulus equation (x−7)2 =(4x+3)2 or inequalityB1
Attempt solution of three-term quadratic equation or inequalityM1
Obtain finally 4 only
AnswerMarks
5A1
Conclude x 4
AnswerMarks Guidance
5A1  4
Allow −, 
 5
4
AnswerMarks Guidance
QuestionAnswer Marks
Question 2:
2 | Attempt solution of equation or inequality, where signs of x and 4x are different | M1
Obtain 4 …
5 | A1 | OE
… and finally no other value | A1
Conclude x 4
5 | A1 |  4
Allow −, .
 5
Alternative Method for Question 2
State or imply non-modulus equation (x−7)2 =(4x+3)2 or inequality | B1
Attempt solution of three-term quadratic equation or inequality | M1
Obtain finally 4 only
5 | A1
Conclude x 4
5 | A1 |  4
Allow −, 
 5
4
Question | Answer | Marks | Guidance
Solve the inequality $|x - 7| > 4x + 3$. [4]

\hfill \mbox{\textit{CAIE P2 2024 Q2 [4]}}