| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch rational with linear numerator |
| Difficulty | Challenging +1.2 Part (a) is straightforward identification of asymptotes from rational function form. Parts (b)(i) and (b)(ii) require standard volumes of revolution techniques, but the rational function makes integration moderately challenging—students must handle partial fractions or substitution carefully. The two-axis rotation adds some complexity, but these are well-practiced A-level Further Maths techniques without requiring novel insight. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| 16(a) | Deduces that a = 3 | 2.2a |
| D educes that b = 2 | 2.2a | B1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 16(b)(i) | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| PI by correct answer. | 2.2a | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI by correct answer. | 1.1a | M1 |
| Obtains AWRT 21.2 | 1.1b | A1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 16(b)(ii) | Deduces that y-intercept = 2.5 | |
| PI correct answer. | 2.2a | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI correct answer. | 2.2a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI by correct answer. | 1.1a | M1 |
| Obtains AWRT 14.1 | 1.1b | A1 |
| Subtotal | 4 | |
| Question total | 9 | |
| Q | Marking Instructions | AO |
Question 16:
--- 16(a) ---
16(a) | Deduces that a = 3 | 2.2a | B1 | a = 3, b = 2
D educes that b = 2 | 2.2a | B1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 16(b)(i) ---
16(b)(i) | 5
Deduces that x-intercept is – .
3
PI by correct answer. | 2.2a | B1 | 3x+5
f(x) =
x+2
5
x-intercept = –
3
0 3x+5 2
V = π∫ dx
x+2
5
–
3
= 21.2040…
= 21.2
Uses π∫y2dx
Condone missing dx and
missing/incorrect limits.
PI by correct answer. | 1.1a | M1
Obtains AWRT 21.2 | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 16(b)(ii) ---
16(b)(ii) | Deduces that y-intercept = 2.5
PI correct answer. | 2.2a | B1 | y-intercept = 2.5
3x+5
y =
x+2
yx + 2y = 3x + 5
x(y – 3) = 5 – 2y
5–2y
x =
y–3
2.55–2y 2
V = π∫ dy
y–3
0
= 14.1360…
= 14.1
Deduces an expression for x in
terms of y
PI correct answer. | 2.2a | M1
Uses π∫x2dy
Condone missing dy, use of dx
and missing/incorrect limits.
PI by correct answer. | 1.1a | M1
Obtains AWRT 14.1 | 1.1b | A1
Subtotal | 4
Question total | 9
Q | Marking Instructions | AO | Marks | Typical SolutionThe function f is defined by
$$f(x) = \frac{ax + 5}{x + b}$$
where $a$ and $b$ are constants.
The graph of $y = f(x)$ has asymptotes $x = -2$ and $y = 3$
\begin{enumerate}[label=(\alph*)]
\item Write down the value of $a$ and the value of $b$
[2 marks]
\item The diagram shows the graph of $y = f(x)$ and its asymptotes.
The shaded region $R$ is enclosed by the graph of $y = f(x)$, the $x$-axis and the $y$-axis.
\includegraphics{figure_16}
\begin{enumerate}[label=(\roman*)]
\item The shaded region $R$ is rotated through $360°$ about the $x$-axis to form a solid.
Find the volume of this solid.
Give your answer to three significant figures.
[3 marks]
\item The shaded region $R$ is rotated through $360°$ about the $y$-axis to form a solid.
Find the volume of this solid.
Give your answer to three significant figures.
[4 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2024 Q16 [9]}}