| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Surface area of revolution with hyperbolics |
| Difficulty | Challenging +1.8 This is a multi-step Further Maths question combining surface area of revolution formula, substitution, and integration involving hyperbolic functions. While it requires several techniques and careful algebraic manipulation, the question is highly scaffolded with clear guidance at each stage (showing specific forms, given substitution). The integration itself, once set up, uses standard hyperbolic results that FP2 students would recognize. More challenging than typical A-level due to the Further Maths content and length, but the structured parts make it accessible to well-prepared students. |
| Spec | 1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| S.A. \(= 2\pi\int_0^9 y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = 2\pi\int_0^9\left(\frac{9}{2}e^{\frac{1}{9}x}\right)\sqrt{1+\left(\frac{1}{2}e^{\frac{1}{9}x}\right)^2}\,dx\) | B1; M1 | 1.1a / 3.4 |
| \(= 9\pi\int_0^9 e^{\frac{1}{9}x}\sqrt{1+\frac{1}{4}e^{\frac{2}{9}x}}\,dx = \frac{9\pi}{2}\int_0^9 e^{\frac{1}{9}x}\sqrt{4+e^{\frac{2}{9}x}}\,dx\) | A1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = e^{\frac{1}{9}x} \Rightarrow du = \frac{1}{9}e^{\frac{1}{9}x}\,dx\) | B1 | 2.2a |
| \(\Rightarrow \int e^{\frac{1}{9}x}\sqrt{4+e^{\frac{2}{9}x}}\,dx = \int e^{\frac{1}{9}x}\sqrt{4+e^{\frac{2}{9}x}}\cdot 9e^{-\frac{1}{9}x}\,du = 9\int\sqrt{4+u^2}\,du\) | M1 | 1.1b |
| \(= 9\int\frac{4+u^2}{\sqrt{4+u^2}}\,du = 9\int\frac{2+u^2}{\sqrt{4+u^2}}\,du + 9\int\frac{2}{\sqrt{4+u^2}}\,du\) | M1 | 3.1a |
| \(= 9\int\frac{2u+u^3}{u\sqrt{4+u^2}}\,du + 18\int\frac{1}{\sqrt{4+u^2}}\,du\) | M1 | 2.1 |
| \(\{x=0\Rightarrow u=1\}\); \(\{x=9\Rightarrow u=e\}\) \(\Rightarrow \int_0^9 e^{\frac{1}{9}x}\sqrt{4+e^{\frac{2}{9}x}}\,dx = 9\int_1^e \frac{2u+u^3}{\sqrt{4u^2+u^4}}\,du + 18\int_1^e\frac{1}{\sqrt{4+u^2}}\,du\) | A1\* | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(9\int\frac{2u+u^3}{\sqrt{4u^2+u^4}}\,du = \frac{9}{4}\int\frac{8u+4u^3}{\sqrt{4u^2+u^4}}\,du = \alpha\sqrt{4u^2+u^4}\) | M1 | 1.1b |
| \(18\int\frac{1}{\sqrt{4+u^2}}\,du = \beta\,\text{arsinh}\!\left(\frac{u}{2}\right)\) or \(\beta\ln\!\left(u+\sqrt{u^2+a^2}\right)\) | M1 | 1.1b |
| \(9\int_1^e\frac{2u+u^3}{\sqrt{4u^2+u^4}}\,du + 18\int_1^e\frac{1}{\sqrt{4+u^2}}\,du = \left[\frac{9}{2}\sqrt{4u^2+u^4}+18\,\text{arsinh}\!\left(\frac{u}{2}\right)\right]_1^e\) | A1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses a correct formula for the surface area | B1 | Correct formula required |
| Applies surface area formula with \(\frac{dy}{dx} = Ke^{\frac{1}{9}x}\) | M1 | Must use correct form of derivative |
| Simplifies correctly with \(K = \frac{9\pi}{2}\) | A1 | As shown in working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct connecting equation: \(du = \frac{1}{9}e^{\frac{1}{9}x}\,dx\) or \(\frac{du}{dx} = \frac{1}{9}e^{\frac{1}{9}x}\) | B1 | Accept any equivalent form |
| Full substitution to obtain integral in terms of \(u\) only, with \(dx\) replaced appropriately | M1 | Limits not needed at this stage |
| Writes integral as \(9\int\frac{4+u^2}{\sqrt{4+u^2}}\,du\) | M1 | Then either splits to extract \(\frac{1}{\sqrt{4+u^2}}\) term, or multiplies numerator/denominator appropriately |
| Multiplies through to get numerator of first integral correct | M1 | Or splits to extract \(\frac{u}{\sqrt{4u^2+u^4}}\) term |
| Deduces correct limits and achieves given answer — must have split the integral correctly | A1 | Fully correct, limits correctly deduced from substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(9\int\frac{2u+u^3}{\sqrt{4u^2+u^4}}\,du = \alpha\sqrt{4u^2+u^4}\) | M1 | Integrates to correct form |
| \(18\int\frac{1}{\sqrt{4+u^2}}\,du = \beta\,\text{arsinh}\!\left(\frac{u}{2}\right)\) or \(\beta\ln\!\left(u+\sqrt{u^2+a^2}\right)\) | M1 | Integrates to correct form |
| Fully correct integration (need not be simplified; may be in terms of a different variable if substitution used; no constant of integration required; \(\frac{9\pi}{2}\) not required for this mark) | A1 | |
| Applies limits from (b) to integrated expression in \(u\) and subtracts correct way round. If returned to \(x\): limits 9 and 0 must be used. Needs \(\frac{9\pi}{2}\) for this mark. | ddM1 | If limits not shown, follow through on \(\frac{9\pi}{2} \times 42.6089\ldots\) or correct answer |
| \(= \frac{9\pi}{2}(42.6089\ldots) \approx\) 602 cm² | A1 | awrt 602, units required. Answer only with no integration shown scores no marks. |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| S.A. $= 2\pi\int_0^9 y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = 2\pi\int_0^9\left(\frac{9}{2}e^{\frac{1}{9}x}\right)\sqrt{1+\left(\frac{1}{2}e^{\frac{1}{9}x}\right)^2}\,dx$ | **B1; M1** | 1.1a / 3.4 |
| $= 9\pi\int_0^9 e^{\frac{1}{9}x}\sqrt{1+\frac{1}{4}e^{\frac{2}{9}x}}\,dx = \frac{9\pi}{2}\int_0^9 e^{\frac{1}{9}x}\sqrt{4+e^{\frac{2}{9}x}}\,dx$ | **A1** | 2.1 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = e^{\frac{1}{9}x} \Rightarrow du = \frac{1}{9}e^{\frac{1}{9}x}\,dx$ | **B1** | 2.2a |
| $\Rightarrow \int e^{\frac{1}{9}x}\sqrt{4+e^{\frac{2}{9}x}}\,dx = \int e^{\frac{1}{9}x}\sqrt{4+e^{\frac{2}{9}x}}\cdot 9e^{-\frac{1}{9}x}\,du = 9\int\sqrt{4+u^2}\,du$ | **M1** | 1.1b |
| $= 9\int\frac{4+u^2}{\sqrt{4+u^2}}\,du = 9\int\frac{2+u^2}{\sqrt{4+u^2}}\,du + 9\int\frac{2}{\sqrt{4+u^2}}\,du$ | **M1** | 3.1a |
| $= 9\int\frac{2u+u^3}{u\sqrt{4+u^2}}\,du + 18\int\frac{1}{\sqrt{4+u^2}}\,du$ | **M1** | 2.1 |
| $\{x=0\Rightarrow u=1\}$; $\{x=9\Rightarrow u=e\}$ $\Rightarrow \int_0^9 e^{\frac{1}{9}x}\sqrt{4+e^{\frac{2}{9}x}}\,dx = 9\int_1^e \frac{2u+u^3}{\sqrt{4u^2+u^4}}\,du + 18\int_1^e\frac{1}{\sqrt{4+u^2}}\,du$ | **A1\*** | 3.4 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $9\int\frac{2u+u^3}{\sqrt{4u^2+u^4}}\,du = \frac{9}{4}\int\frac{8u+4u^3}{\sqrt{4u^2+u^4}}\,du = \alpha\sqrt{4u^2+u^4}$ | **M1** | 1.1b |
| $18\int\frac{1}{\sqrt{4+u^2}}\,du = \beta\,\text{arsinh}\!\left(\frac{u}{2}\right)$ or $\beta\ln\!\left(u+\sqrt{u^2+a^2}\right)$ | **M1** | 1.1b |
| $9\int_1^e\frac{2u+u^3}{\sqrt{4u^2+u^4}}\,du + 18\int_1^e\frac{1}{\sqrt{4+u^2}}\,du = \left[\frac{9}{2}\sqrt{4u^2+u^4}+18\,\text{arsinh}\!\left(\frac{u}{2}\right)\right]_1^e$ | **A1** | 2.1 |
# Surface Area Question (Final Part):
---
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses a correct formula for the surface area | B1 | Correct formula required |
| Applies surface area formula with $\frac{dy}{dx} = Ke^{\frac{1}{9}x}$ | M1 | Must use correct form of derivative |
| Simplifies correctly with $K = \frac{9\pi}{2}$ | A1 | As shown in working |
---
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct connecting equation: $du = \frac{1}{9}e^{\frac{1}{9}x}\,dx$ or $\frac{du}{dx} = \frac{1}{9}e^{\frac{1}{9}x}$ | B1 | Accept any equivalent form |
| Full substitution to obtain integral in terms of $u$ only, with $dx$ replaced appropriately | M1 | Limits not needed at this stage |
| Writes integral as $9\int\frac{4+u^2}{\sqrt{4+u^2}}\,du$ | M1 | Then either splits to extract $\frac{1}{\sqrt{4+u^2}}$ term, or multiplies numerator/denominator appropriately |
| Multiplies through to get numerator of first integral correct | M1 | Or splits to extract $\frac{u}{\sqrt{4u^2+u^4}}$ term |
| Deduces correct limits and achieves given answer — must have split the integral correctly | A1 | Fully correct, limits correctly deduced from substitution |
---
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $9\int\frac{2u+u^3}{\sqrt{4u^2+u^4}}\,du = \alpha\sqrt{4u^2+u^4}$ | M1 | Integrates to correct form |
| $18\int\frac{1}{\sqrt{4+u^2}}\,du = \beta\,\text{arsinh}\!\left(\frac{u}{2}\right)$ or $\beta\ln\!\left(u+\sqrt{u^2+a^2}\right)$ | M1 | Integrates to correct form |
| Fully correct integration (need not be simplified; may be in terms of a different variable if substitution used; no constant of integration required; $\frac{9\pi}{2}$ not required for this mark) | A1 | |
| Applies limits from (b) to integrated expression in $u$ and subtracts correct way round. If returned to $x$: limits 9 and 0 must be used. Needs $\frac{9\pi}{2}$ for this mark. | ddM1 | If limits not shown, follow through on $\frac{9\pi}{2} \times 42.6089\ldots$ or correct answer |
| $= \frac{9\pi}{2}(42.6089\ldots) \approx$ **602 cm²** | A1 | awrt 602, units required. Answer only with no integration shown scores no marks. |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c20a4592-74c6-4f58-b63b-984b171b1bfd-28_552_380_264_468}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c20a4592-74c6-4f58-b63b-984b171b1bfd-28_524_446_274_1151}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 1 shows a French horn with a detachable bell section.\\
The shape of the bell section can be modelled by rotating an exponential curve through $360 ^ { \circ }$ about the $x$-axis, where units are centimetres.
The model uses the curve shown in Figure 2, with equation
$$y = \frac { 9 } { 2 } e ^ { \frac { 1 } { 9 } x } \quad 0 \leqslant x \leqslant 9$$
\begin{enumerate}[label=(\alph*)]
\item Show that, according to this model, the external surface area of the bell section is given by
$$K \int _ { 0 } ^ { 9 } \mathrm { e } ^ { \frac { 1 } { 9 } x } \sqrt { 4 + \mathrm { e } ^ { \frac { 2 } { 9 } x } } \mathrm {~d} x$$
where $K$ is a real constant to be determined.
\item Use the substitution $u = e ^ { \frac { 1 } { 9 } x }$ to show that
$$\int _ { 0 } ^ { 9 } \mathrm { e } ^ { \frac { 1 } { 9 } x } \sqrt { 4 + \mathrm { e } ^ { \frac { 2 } { 9 } x } } \mathrm {~d} x = 9 \int _ { a } ^ { b } \frac { 2 u + u ^ { 3 } } { \sqrt { 4 u ^ { 2 } + u ^ { 4 } } } \mathrm {~d} u + 18 \int _ { a } ^ { b } \frac { 1 } { \sqrt { 4 + u ^ { 2 } } } \mathrm {~d} u$$
where $a$ and $b$ are constants to be determined.
Hence, using algebraic integration,
\item determine, according to the model, the external surface area of the bell section of the horn, giving your answer to 3 significant figures.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2024 Q8 [13]}}