CAIE P2 2023 March — Question 4 7 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2023
SessionMarch
Marks7
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Mark schemeDownload PDF ↗
TopicInequalities
TypeSolve absolute value inequality
DifficultyModerate -0.3 This is a straightforward modulus inequality question with standard techniques. Part (a) requires basic graph sketching of linear and modulus functions. Part (b) involves solving |2x-1| < 3x-3 by considering cases or using the graph, which is routine. Part (c) applies the same inequality with a substitution (x = ln N), requiring one additional step to find N = e^2 ≈ 7.39, so N = 8. The question is slightly easier than average as it follows a predictable structure with no conceptual surprises, though it does require careful execution across multiple parts.
Spec1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities1.02m Graphs of functions: difference between plotting and sketching1.02s Modulus graphs: sketch graph of |ax+b|
  1. Sketch, on the same diagram, the graphs of \(y = |2x - 1|\) and \(y = 3x - 3\). [2]
  2. Solve the inequality \(|2x - 1| < 3x - 3\). [3]
  3. Find the smallest integer \(N\) satisfying the inequality \(|2 \ln N - 1| < 3 \ln N - 3\). [2]
Question 4:
AnswerMarks
4Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
AnswerMarks Guidance
4(a)Draw V-shaped graph with vertex on positive x-axis B1
Draw approximately correct graph of y=3x−3 with greater
AnswerMarks Guidance
gradientB1 Crossing x-axis between origin and vertex of first graph.
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
4(b)Attempt solution of linear equation where signs of 2x and 3x are
differentM1
Solve −2x+11=3x−3 to obtain x=14
AnswerMarks Guidance
5A1 OE
Conclude x14
AnswerMarks Guidance
5A1 OE
Alternative method for Question 4(b)
Attempt solution of 3-term equation (2x−11)2 =(3x−3)2 to
AnswerMarks Guidance
obtain at least one value of xM1 Or equivalent inequality.
Obtain at least x=14
AnswerMarks Guidance
5A1 OE
Conclude x14
AnswerMarks Guidance
5A1 OE
3
AnswerMarks
4(c)Attempt value of N (maybe non-integer at this stage) using
logarithms and their answer to part (b).M1
Conclude with single integer 17A1
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | Draw V-shaped graph with vertex on positive x-axis | B1
Draw approximately correct graph of y=3x−3 with greater
gradient | B1 | Crossing x-axis between origin and vertex of first graph.
2
Question | Answer | Marks | Guidance
--- 4(b) ---
4(b) | Attempt solution of linear equation where signs of 2x and 3x are
different | M1
Solve −2x+11=3x−3 to obtain x=14
5 | A1 | OE
Conclude x14
5 | A1 | OE
Alternative method for Question 4(b)
Attempt solution of 3-term equation (2x−11)2 =(3x−3)2 to
obtain at least one value of x | M1 | Or equivalent inequality.
Obtain at least x=14
5 | A1 | OE
Conclude x14
5 | A1 | OE
3
--- 4(c) ---
4(c) | Attempt value of N (maybe non-integer at this stage) using
logarithms and their answer to part (b). | M1
Conclude with single integer 17 | A1
2
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\alph*)]
\item Sketch, on the same diagram, the graphs of $y = |2x - 1|$ and $y = 3x - 3$. [2]
\item Solve the inequality $|2x - 1| < 3x - 3$. [3]
\item Find the smallest integer $N$ satisfying the inequality $|2 \ln N - 1| < 3 \ln N - 3$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2023 Q4 [7]}}
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