| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Surface area of revolution with hyperbolics |
| Difficulty | Challenging +1.8 This is a Further Maths question combining hyperbolic function manipulation, parametric curves, and surface area of revolution. Part (a) requires proving an identity from exponential definitions (standard FM technique). Part (b) involves applying the surface area formula to parametric equations with hyperbolics, then integrating using the double angle identity from part (a). While technically demanding with multiple steps, the techniques are all standard Further Maths material with clear signposting—harder than typical A-level but not requiring exceptional insight. |
| Spec | 4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cosh A = \frac{1}{2}(e^A + e^{-A})\) | B1 | Writes in exponential form |
| \(2\cosh^2 A = \frac{1}{2}(e^{2A} + 2 + e^{-2A}) = \frac{1}{2}(e^{2A} + e^{-2A}) + 1 = \cosh 2A + 1\) | M1 A1 | Expands, AG |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dx}{dt} = 4\sinh 2t + 3\), \(\quad \frac{dy}{dt} = 3\sinh 2t - 4\) | B1 | |
| \((4\sinh 2t + 3)^2 + (3\sinh 2t - 4)^2 = 25(\sinh^2 2t + 1) = 25\cosh^2 2t\) | M1 A1 | Expands and applies \(\cosh^2 A = \sinh^2 A + 1\) |
| \(2\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}(2\cosh 2t + 3t)(5\cosh 2t)\,dt \left[= \pi\int_{-\frac{1}{2}}^{\frac{1}{2}} 20\cosh^2 2t + 30t\cosh 2t\,dt\right]\) | A1 | Correct formula for surface area, AG |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(20\int_{-\frac{1}{2}}^{\frac{1}{2}}\cosh^2 2t\,dt = 10\int_{-\frac{1}{2}}^{\frac{1}{2}}\cosh 4t + 1\,dt\) | M1 | Applies \(\cosh^2 A = \frac{1}{2}(\cosh 2A + 1)\) |
| \(10\left[\frac{1}{4}\sinh 4t + t\right]_{-\frac{1}{2}}^{\frac{1}{2}} = 10\left(\frac{1}{2}\sinh 2 + 1\right)\) | M1 A1 | Integrates |
| \(30\int_{-\frac{1}{2}}^{\frac{1}{2}} t\cosh 2t\,dt = 30\left(\frac{1}{2}\left[t\sinh 2t\right]_{-\frac{1}{2}}^{\frac{1}{2}} - \frac{1}{2}\int_{-\frac{1}{2}}^{\frac{1}{2}}\sinh 2t\,dt\right)\) | M1 A1 | Integrates by parts |
| \(15\left[t\sinh 2t - \frac{1}{2}\cosh 2t\right]_{-\frac{1}{2}}^{\frac{1}{2}} = 0\) | A1 | Accept \(\int_{-\frac{1}{2}}^{\frac{1}{2}} t\cosh 2t\,dt = 0\) since \(t\cosh 2t\) is odd |
| \(10\pi\left(\frac{1}{4}(e^2 - e^{-2}) + 1\right)\) | A1 | 64.82457... |
| Total: 7 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(10\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}(2\cosh 2t + 3t)\cosh 2t\,dt = 10\pi\left[\cosh 2t \sinh 2t + \frac{3}{2}t\sinh 2t\right]_{-\frac{1}{2}}^{\frac{1}{2}} - 5\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\sinh 2t(4\sinh 2t + 3)\,dt\) | M1 A1 | Integrates by parts: \(u = 2\cosh 2t + 3t\), \(u' = 4\sinh 2t + 3\), \(v' = \cosh 2t\), \(v = \frac{1}{2}\sinh 2t\) |
| \(10\pi\left[\frac{1}{2}\sinh 4t\right]_{-\frac{1}{2}}^{\frac{1}{2}} - 20\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\sinh^2 2t\,dt - \left(15\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\sinh 2t\,dt\right) = 10\pi\left[\frac{1}{2}\sinh 4t\right]_{-\frac{1}{2}}^{\frac{1}{2}} - 10\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\cosh 4t - 1\,dt - \left(15\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\sinh 2t\,dt\right)\) | M1 A1 | Applies \(2\sinh^2 2t = \cosh 4t - 1\) |
| \(\pi\left[\frac{5}{2}\sinh 4t - \frac{10}{4}\sinh 4t + 10t - \left(\frac{15}{2}\cosh 2t\right)\right]_{-\frac{1}{2}}^{\frac{1}{2}}\) | M1 A1 | Integrates |
| \(\pi\left[\frac{5}{2}\sinh 4t + 10t - \left(\frac{15}{2}\cosh 2t\right)\right]_{-\frac{1}{2}}^{\frac{1}{2}} = \pi(5\sinh 2 + 10) = 10\pi\left(\frac{1}{4}(e^2 - e^{-2}) + 1\right)\) | A1 | 64.82457... |
| Total: 7 marks |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cosh A = \frac{1}{2}(e^A + e^{-A})$ | B1 | Writes in exponential form |
| $2\cosh^2 A = \frac{1}{2}(e^{2A} + 2 + e^{-2A}) = \frac{1}{2}(e^{2A} + e^{-2A}) + 1 = \cosh 2A + 1$ | M1 A1 | Expands, AG |
| **Total: 3 marks** | | |
## Question 8(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dx}{dt} = 4\sinh 2t + 3$, $\quad \frac{dy}{dt} = 3\sinh 2t - 4$ | B1 | |
| $(4\sinh 2t + 3)^2 + (3\sinh 2t - 4)^2 = 25(\sinh^2 2t + 1) = 25\cosh^2 2t$ | M1 A1 | Expands and applies $\cosh^2 A = \sinh^2 A + 1$ |
| $2\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}(2\cosh 2t + 3t)(5\cosh 2t)\,dt \left[= \pi\int_{-\frac{1}{2}}^{\frac{1}{2}} 20\cosh^2 2t + 30t\cosh 2t\,dt\right]$ | A1 | Correct formula for surface area, AG |
| **Total: 4 marks** | | |
## Question 8(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $20\int_{-\frac{1}{2}}^{\frac{1}{2}}\cosh^2 2t\,dt = 10\int_{-\frac{1}{2}}^{\frac{1}{2}}\cosh 4t + 1\,dt$ | M1 | Applies $\cosh^2 A = \frac{1}{2}(\cosh 2A + 1)$ |
| $10\left[\frac{1}{4}\sinh 4t + t\right]_{-\frac{1}{2}}^{\frac{1}{2}} = 10\left(\frac{1}{2}\sinh 2 + 1\right)$ | M1 A1 | Integrates |
| $30\int_{-\frac{1}{2}}^{\frac{1}{2}} t\cosh 2t\,dt = 30\left(\frac{1}{2}\left[t\sinh 2t\right]_{-\frac{1}{2}}^{\frac{1}{2}} - \frac{1}{2}\int_{-\frac{1}{2}}^{\frac{1}{2}}\sinh 2t\,dt\right)$ | M1 A1 | Integrates by parts |
| $15\left[t\sinh 2t - \frac{1}{2}\cosh 2t\right]_{-\frac{1}{2}}^{\frac{1}{2}} = 0$ | A1 | Accept $\int_{-\frac{1}{2}}^{\frac{1}{2}} t\cosh 2t\,dt = 0$ since $t\cosh 2t$ is odd |
| $10\pi\left(\frac{1}{4}(e^2 - e^{-2}) + 1\right)$ | A1 | 64.82457... |
| **Total: 7 marks** | | |
### Alternative Method for 8(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $10\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}(2\cosh 2t + 3t)\cosh 2t\,dt = 10\pi\left[\cosh 2t \sinh 2t + \frac{3}{2}t\sinh 2t\right]_{-\frac{1}{2}}^{\frac{1}{2}} - 5\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\sinh 2t(4\sinh 2t + 3)\,dt$ | M1 A1 | Integrates by parts: $u = 2\cosh 2t + 3t$, $u' = 4\sinh 2t + 3$, $v' = \cosh 2t$, $v = \frac{1}{2}\sinh 2t$ |
| $10\pi\left[\frac{1}{2}\sinh 4t\right]_{-\frac{1}{2}}^{\frac{1}{2}} - 20\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\sinh^2 2t\,dt - \left(15\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\sinh 2t\,dt\right) = 10\pi\left[\frac{1}{2}\sinh 4t\right]_{-\frac{1}{2}}^{\frac{1}{2}} - 10\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\cosh 4t - 1\,dt - \left(15\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\sinh 2t\,dt\right)$ | M1 A1 | Applies $2\sinh^2 2t = \cosh 4t - 1$ |
| $\pi\left[\frac{5}{2}\sinh 4t - \frac{10}{4}\sinh 4t + 10t - \left(\frac{15}{2}\cosh 2t\right)\right]_{-\frac{1}{2}}^{\frac{1}{2}}$ | M1 A1 | Integrates |
| $\pi\left[\frac{5}{2}\sinh 4t + 10t - \left(\frac{15}{2}\cosh 2t\right)\right]_{-\frac{1}{2}}^{\frac{1}{2}} = \pi(5\sinh 2 + 10) = 10\pi\left(\frac{1}{4}(e^2 - e^{-2}) + 1\right)$ | A1 | 64.82457... |
| **Total: 7 marks** | | |8
\begin{enumerate}[label=(\alph*)]
\item Starting from the definition of cosh in terms of exponentials, prove that
$$2 \cosh ^ { 2 } A = \cosh 2 A + 1$$
The curve $C$ has parametric equations
$$\mathrm { x } = 2 \cosh 2 \mathrm { t } + 3 \mathrm { t } , \quad \mathrm { y } = \frac { 3 } { 2 } \cosh 2 \mathrm { t } - 4 \mathrm { t } , \quad \text { for } - \frac { 1 } { 2 } \leqslant t \leqslant \frac { 1 } { 2 }$$
The area of the surface generated when $C$ is rotated through $2 \pi$ radians about the $y$-axis is denoted by $A$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that $A = 10 \pi \int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } ( 2 \cosh 2 t + 3 t ) \cosh 2 t d t$.
\item Hence find $A$ in terms of $\pi$ and e.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q8 [14]}}