Question 6 - A-Level Maths

CAIE P2 2024 June — Question 6 9 marks

Exam Board	CAIE
Module	P2 (Pure Mathematics 2)
Year	2024
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Product & Quotient Rules
Type	Show derivative equals given algebraic form
Difficulty	Standard +0.3 This is a standard quotient rule differentiation followed by routine algebraic manipulation and iterative numerical methods. Part (a) is straightforward application of quotient rule with ln; part (b) requires setting derivative to zero and rearranging (shown result guides the algebra); parts (c) and (d) are mechanical substitution and iteration. Slightly above average due to the multi-part nature and algebraic manipulation required, but all techniques are standard P2 content with no novel insight needed.
Spec	1.07l Derivative of ln(x): and related functions 1.07n Stationary points: find maxima, minima using derivatives 1.07q Product and quotient rules: differentiation 1.09a Sign change methods: locate roots 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

6 \includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-10_417_700_310_685} The diagram shows the curve with equation \(y = \frac { \ln ( 2 x + 1 ) } { x + 3 }\). The curve has a maximum point \(M\).

Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
Show that the \(x\)-coordinate of \(M\) satisfies the equation \(x = \frac { x + 3 } { \ln ( 2 x + 1 ) } - 0.5\).
Show by calculation that the \(x\)-coordinate of \(M\) lies between 2.5 and 3.0 .
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.

Show mark scheme Show mark scheme source

Question 6(a):

Answer	Marks	Guidance
Answer	Marks	Guidance
Attempt use of quotient rule	M1	Or equivalent method
Obtain \(\frac{(x+3)\frac{2}{2x+1} - \ln(2x+1)}{(x+3)^2}\)	A1	OE

Question 6(b):

Answer	Marks	Guidance
Answer	Marks	Guidance
Equate first derivative to zero and arrange as far as \(2x + 1 = \ldots\)	M1
Confirm \(x = \frac{x+3}{\ln(2x+1)} - 0.5\)	A1	Answer given – necessary detail needed

Question 6(c):

Answer	Marks	Guidance
Answer	Marks	Guidance
Consider sign of \(x - \frac{x+3}{\ln(2x+1)} + 0.5\) or equivalent for \(2.5\) and \(3.0\)	M1
Obtain \(-0.07\) \((0.0696\ldots)\) and \(0.4\) \((0.4166\ldots)\) and justify conclusion	A1	Answer given – necessary detail needed

Alternative Method 6(c):

Answer	Marks	Guidance
Answer	Marks	Guidance
Consider values of \(f(x) = \frac{x+3}{\ln(2x+1)} - 0.5\) and obtain \(f(2.5) = 2.57\ (2.5696\ldots)\) and \(f(3) = 2.58\ (2.58339\ldots)\)	(M1)
Conclude \(f(2.5) < 3\) and \(f(3) > 2.5\) so root lies in given interval	(A1)	Answer given – necessary detail needed

Question 6(d):

Answer	Marks	Guidance
Answer	Marks	Guidance
Use iterative process correctly at least once	M1
Obtain final answer \(2.569\)	A1	Answer required to exactly 4 sf
Show sufficient iterations to 6 sf to justify answer or show sign change in interval \([2.5685,\ 2.5695]\)	A1

## Question 6(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt use of quotient rule | M1 | Or equivalent method |
| Obtain $\frac{(x+3)\frac{2}{2x+1} - \ln(2x+1)}{(x+3)^2}$ | A1 | OE |

## Question 6(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Equate first derivative to zero and arrange as far as $2x + 1 = \ldots$ | M1 | |
| Confirm $x = \frac{x+3}{\ln(2x+1)} - 0.5$ | A1 | Answer given – necessary detail needed |

## Question 6(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Consider sign of $x - \frac{x+3}{\ln(2x+1)} + 0.5$ or equivalent for $2.5$ and $3.0$ | M1 | |
| Obtain $-0.07$ $(0.0696\ldots)$ and $0.4$ $(0.4166\ldots)$ and justify conclusion | A1 | Answer given – necessary detail needed |

**Alternative Method 6(c):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Consider values of $f(x) = \frac{x+3}{\ln(2x+1)} - 0.5$ and obtain $f(2.5) = 2.57\ (2.5696\ldots)$ and $f(3) = 2.58\ (2.58339\ldots)$ | (M1) | |
| Conclude $f(2.5) < 3$ and $f(3) > 2.5$ so root lies in given interval | (A1) | Answer given – necessary detail needed |

## Question 6(d):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use iterative process correctly at least once | M1 | |
| Obtain final answer $2.569$ | A1 | Answer required to exactly 4 sf |
| Show sufficient iterations to 6 sf to justify answer or show sign change in interval $[2.5685,\ 2.5695]$ | A1 | |

Show LaTeX source

6\\
\includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-10_417_700_310_685}

The diagram shows the curve with equation $y = \frac { \ln ( 2 x + 1 ) } { x + 3 }$. The curve has a maximum point $M$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Show that the $x$-coordinate of $M$ satisfies the equation $x = \frac { x + 3 } { \ln ( 2 x + 1 ) } - 0.5$.
\item Show by calculation that the $x$-coordinate of $M$ lies between 2.5 and 3.0 .
\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.
\end{enumerate}

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

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