CAIE P2 2024 June — Question 6 9 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2024
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeRearrange to iterative form
DifficultyStandard +0.3 This is a structured multi-part question with clear scaffolding. Part (a) requires quotient rule differentiation (routine). Parts (b)-(c) involve algebraic manipulation and sign-checking (standard). Part (d) applies a given iterative formula repeatedly—mechanical calculation rather than problem-solving. All techniques are standard P2 content with no novel insight required, making it slightly easier than average.
Spec1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams
\includegraphics{figure_6} The diagram shows the curve with equation \(y = \frac{\ln(2x + 1)}{x + 3}\). The curve has a maximum point M.
  1. Find an expression for \(\frac{dy}{dx}\). [2]
  2. Show that the x-coordinate of M satisfies the equation \(x = \frac{x + 3}{\ln(2x + 1)} - 0.5\). [2]
  3. Show by calculation that the x-coordinate of M lies between 2.5 and 3.0. [2]
  4. Use an iterative formula based on the equation in part (b) to find the x-coordinate of M correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]
Question 6:
AnswerMarks Guidance
6(a)Attempt use of quotient rule M1
(x3) 2  ln(2x1)
Obtain 2x1
AnswerMarks Guidance
(x3)2A1 OE
2
AnswerMarks Guidance
6(b)Equate first derivative to zero and arrange as far as 2x1... M1
x3
Confirm x 0.5
AnswerMarks Guidance
ln(2x1)A1 Answer given – necessary detail needed.
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(c)x3
Consider sign of x 0.5 or equivalent for 2.5 and 3.0
AnswerMarks
ln(2x1)M1
Obtain 0.07 0.0696... and 0.4 0.4166... and justify
AnswerMarks Guidance
conclusionA1 Answer given – necessary detail needed.
2
Alternative Method for Question 6(c)
x3
Consider the values of fx 0.5and obtain
ln2x1
AnswerMarks Guidance
f2.52.57 (2.5696…) and f32.58 (2.58339…)(M1)
Conclude f2.53 and f32.5 so root lies in given interval(A1) Answer given – necessary detail needed.
2
AnswerMarks Guidance
6(d)Use iterative process correctly at least once M1
Obtain final answer 2.569A1 Answer required to exactly 4sf.
Show sufficient iterations to 6 sf to justify answer or show sign
AnswerMarks
change in interval [2.5685, 2.5695]A1
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(a) ---
6(a) | Attempt use of quotient rule | M1 | Or equivalent method.
(x3) 2  ln(2x1)
Obtain 2x1
(x3)2 | A1 | OE
2
--- 6(b) ---
6(b) | Equate first derivative to zero and arrange as far as 2x1... | M1
x3
Confirm x 0.5
ln(2x1) | A1 | Answer given – necessary detail needed.
2
Question | Answer | Marks | Guidance
--- 6(c) ---
6(c) | x3
Consider sign of x 0.5 or equivalent for 2.5 and 3.0
ln(2x1) | M1
Obtain 0.07 0.0696... and 0.4 0.4166... and justify
conclusion | A1 | Answer given – necessary detail needed.
2
Alternative Method for Question 6(c)
x3
Consider the values of fx 0.5and obtain
ln2x1
f2.52.57 (2.5696…) and f32.58 (2.58339…) | (M1)
Conclude f2.53 and f32.5 so root lies in given interval | (A1) | Answer given – necessary detail needed.
2
--- 6(d) ---
6(d) | Use iterative process correctly at least once | M1
Obtain final answer 2.569 | A1 | Answer required to exactly 4sf.
Show sufficient iterations to 6 sf to justify answer or show sign
change in interval [2.5685, 2.5695] | A1
3
Question | Answer | Marks | Guidance
\includegraphics{figure_6}

The diagram shows the curve with equation $y = \frac{\ln(2x + 1)}{x + 3}$. The curve has a maximum point M.

\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\frac{dy}{dx}$. [2]
\item Show that the x-coordinate of M satisfies the equation $x = \frac{x + 3}{\ln(2x + 1)} - 0.5$. [2]
\item Show by calculation that the x-coordinate of M lies between 2.5 and 3.0. [2]
\item Use an iterative formula based on the equation in part (b) to find the x-coordinate of M correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2024 Q6 [9]}}
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