CAIE P2 2018 June — Question 4 8 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2018
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeDerive stationary point equation
DifficultyStandard +0.3 This is a straightforward calculus question requiring quotient rule differentiation, setting dy/dx=0, and applying a given iterative formula. Part (ii) involves routine algebraic manipulation to reach the required form, and part (iii) is mechanical iteration. While multi-step, it requires only standard A-level techniques with no novel insight or challenging problem-solving.
Spec1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

4 \includegraphics[max width=\textwidth, alt={}, center]{873a104f-e2e2-49bb-b943-583769728fbb-06_355_839_260_653} The diagram shows the curve with equation \(y = \frac { 5 \ln x } { 2 x + 1 }\). The curve crosses the \(x\)-axis at the point \(P\) and has a maximum point \(M\).
  1. Find the gradient of the curve at the point \(P\).
  2. Show that the \(x\)-coordinate of the point \(M\) satisfies the equation \(x = \frac { x + 0.5 } { \ln x }\).
  3. Use an iterative formula based on the equation in part (ii) to find the \(x\)-coordinate of \(M\) correct to 4 significant figures. Show the result of each iteration to 6 significant figures.

Question 4(i):
AnswerMarks Guidance
AnswerMark Guidance
Use quotient rule or equivalentM1 Obtaining two terms in numerator and \((2x+1)^2\) in denominator for a quotient
Obtain correct \(\dfrac{\frac{5}{x}(2x+1) - 10\ln x}{(2x+1)^2}\) or equivalent, or \(\frac{5}{x}(2x+1)^{-1} - 10\ln x(2x+1)^{-2}\) or equivalentA1 Obtaining one term with \((2x+1)^{-1}\) and a second term with \((2x+1)^{-2}\); condone poor use of brackets if recovered later
Substitute \(x=1\) to obtain \(\frac{15}{9}\) or \(\frac{5}{3}\) or equivalent, wwwA1
Total3
Question 4(ii):
AnswerMarks Guidance
AnswerMark Guidance
Equate numerator to zero and attempt relevant arrangementM1 Need to see at least one line of working after either \(10 + \frac{5}{x} - 10\ln x = 0\) or their numerator (which must have at least 2 terms, one involving \(\ln x\)) \(= 0\)
Confirm \(x = \dfrac{x + 0.5}{\ln x}\)A1 AG; necessary detail needed
Total2
Question 4(iii):
AnswerMarks Guidance
AnswerMark Guidance
Use iteration process correctly at least onceM1
Obtain final answer 3.181A1
Show sufficient iterations to 6 sf to justify answer or show sign change in interval \((3.1805, 3.1815)\)A1
Total3
## Question 4(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| Use quotient rule or equivalent | M1 | Obtaining two terms in numerator and $(2x+1)^2$ in denominator for a quotient |
| Obtain correct $\dfrac{\frac{5}{x}(2x+1) - 10\ln x}{(2x+1)^2}$ or equivalent, or $\frac{5}{x}(2x+1)^{-1} - 10\ln x(2x+1)^{-2}$ or equivalent | A1 | Obtaining one term with $(2x+1)^{-1}$ and a second term with $(2x+1)^{-2}$; condone poor use of brackets if recovered later |
| Substitute $x=1$ to obtain $\frac{15}{9}$ or $\frac{5}{3}$ or equivalent, www | A1 | |
| **Total** | **3** | |

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## Question 4(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| Equate numerator to zero and attempt relevant arrangement | M1 | Need to see at least one line of working after either $10 + \frac{5}{x} - 10\ln x = 0$ or their numerator (which must have at least 2 terms, one involving $\ln x$) $= 0$ |
| Confirm $x = \dfrac{x + 0.5}{\ln x}$ | A1 | AG; necessary detail needed |
| **Total** | **2** | |

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## Question 4(iii):

| Answer | Mark | Guidance |
|--------|------|----------|
| Use iteration process correctly at least once | M1 | |
| Obtain final answer 3.181 | A1 | |
| Show sufficient iterations to 6 sf to justify answer or show sign change in interval $(3.1805, 3.1815)$ | A1 | |
| **Total** | **3** | |

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\includegraphics[max width=\textwidth, alt={}, center]{873a104f-e2e2-49bb-b943-583769728fbb-06_355_839_260_653}

The diagram shows the curve with equation $y = \frac { 5 \ln x } { 2 x + 1 }$. The curve crosses the $x$-axis at the point $P$ and has a maximum point $M$.\\
(i) Find the gradient of the curve at the point $P$.\\

(ii) Show that the $x$-coordinate of the point $M$ satisfies the equation $x = \frac { x + 0.5 } { \ln x }$.\\

(iii) Use an iterative formula based on the equation in part (ii) to find the $x$-coordinate of $M$ correct to 4 significant figures. Show the result of each iteration to 6 significant figures.\\

\hfill \mbox{\textit{CAIE P2 2018 Q4 [8]}}