| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Surface area of revolution: Cartesian curve |
| Difficulty | Challenging +1.8 This is a structured two-part surface area of revolution question requiring two successive substitutions. Part (a) involves a straightforward substitution with exponentials, while part (b) requires the less common hyperbolic substitution u=sinh v and knowledge of the integral of cosh²v. The question is well-scaffolded with both substitutions given, making it a challenging but guided exercise rather than requiring novel insight. The algebraic manipulation and hyperbolic identities push this above average difficulty for Further Maths. |
| Spec | 4.07e Inverse hyperbolic: definitions, domains, ranges4.08e Mean value of function: using integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = 2\pi\int_{\ln\frac{4}{3}}^{\ln\frac{12}{5}} e^x\sqrt{1+e^{2x}}\,dx = 2\pi\int_{\frac{4}{3}}^{\frac{12}{5}} u\sqrt{1+u^2}\left(\frac{1}{u}\right)du\) | M1A1 | Correct formula and applies substitution, AG |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u = \sinh v \Rightarrow du = \cosh v\,dv\) | B1 | |
| \(A = 2\pi\int_{\ln 3}^{\ln 5}\sqrt{1+\sinh^2 v}\cosh v\,dv \quad A = 2\pi\int_{\ln 3}^{\ln 5}\cosh^2 v\,dv\) | M1A1 | Applies substitution and \(\cosh^2 x = 1+\sinh^2 x\). Need limits for A1 |
| \(\pi\int_{\ln 3}^{\ln 5}(\cosh 2v+1)\,dv\) | M1 | Applies \(2\cosh^2 x = \cosh 2x + 1\) |
| \(= \pi\left[\frac{1}{2}\sinh 2v + v\right]_{\ln 3}^{\ln 5}\) | A1 | Correct integration |
| \(\pi\left(\frac{312}{50}+\ln 5-\frac{40}{18}-\ln 3\right) = \pi\left(\frac{904}{225}+\ln\frac{5}{3}\right)\) | A1 | AG |
| 6 |
# Question 3:
## Part 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = 2\pi\int_{\ln\frac{4}{3}}^{\ln\frac{12}{5}} e^x\sqrt{1+e^{2x}}\,dx = 2\pi\int_{\frac{4}{3}}^{\frac{12}{5}} u\sqrt{1+u^2}\left(\frac{1}{u}\right)du$ | M1A1 | Correct formula and applies substitution, AG |
| | **2** | |
## Part 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u = \sinh v \Rightarrow du = \cosh v\,dv$ | B1 | |
| $A = 2\pi\int_{\ln 3}^{\ln 5}\sqrt{1+\sinh^2 v}\cosh v\,dv \quad A = 2\pi\int_{\ln 3}^{\ln 5}\cosh^2 v\,dv$ | M1A1 | Applies substitution and $\cosh^2 x = 1+\sinh^2 x$. Need limits for A1 |
| $\pi\int_{\ln 3}^{\ln 5}(\cosh 2v+1)\,dv$ | M1 | Applies $2\cosh^2 x = \cosh 2x + 1$ |
| $= \pi\left[\frac{1}{2}\sinh 2v + v\right]_{\ln 3}^{\ln 5}$ | A1 | Correct integration |
| $\pi\left(\frac{312}{50}+\ln 5-\frac{40}{18}-\ln 3\right) = \pi\left(\frac{904}{225}+\ln\frac{5}{3}\right)$ | A1 | AG |
| | **6** | |
---3 A curve has equation $y = \mathrm { e } ^ { x }$ for $\ln \frac { 4 } { 3 } \leqslant x \leqslant \ln \frac { 12 } { 5 }$. The area of the surface generated when the curve is rotated through $2 \pi$ radians about the $x$-axis is denoted by $A$.
\begin{enumerate}[label=(\alph*)]
\item Use the substitution $u = \mathrm { e } ^ { x }$ to show that
$$A = 2 \pi \int _ { \frac { 4 } { 3 } } ^ { \frac { 12 } { 5 } } \sqrt { 1 + u ^ { 2 } } \mathrm {~d} u$$
\item Use the substitution $u = \sinh v$ to show that
$$A = \pi \left( \frac { 904 } { 225 } + \ln \frac { 5 } { 3 } \right) .$$
\includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-06_2716_38_109_2012}\\
\includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-07_2726_35_97_20}
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q3 [12]}}