| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Integrate using hyperbolic substitution |
| Difficulty | Challenging +1.8 Part (a) is a routine proof from exponential definitions (standard Further Maths exercise). Part (b) requires surface of revolution formula, executing a non-trivial hyperbolic substitution, applying the identity from (a), and integrating hyperbolic functions—multiple sophisticated steps but follows a guided path with the substitution provided. Challenging for Further Maths but not requiring exceptional insight. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.08e Mean value of function: using integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cosh 2A = \frac{1}{2}(e^{2A} + e^{-2A})\), \(\sinh A = \frac{1}{2}(e^A - e^{-A})\) | B1 | |
| \(2\sinh^2 A = \frac{1}{2}(e^A - e^{-A})^2 = \frac{1}{2}(e^{2A} - 2 + e^{-2A}) = \cosh 2A - 1\) | M1 A1 | Expands, AG. A0 for mixing variables e.g. \(\sinh A = \frac{1}{2}(e^x - e^{-x})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S = \int_0^3 2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = 2\pi\int_0^3 x^2\sqrt{1+4x^2}\,dx\) | M1 A1 | Correct formula with correct limits. Correct limits for M1. (Limits may be recovered.) |
| \(S = \pi\int_0^{\sinh^{-1}\frac{4}{3}} \frac{1}{4}\sinh^2 u\sqrt{1+\sinh^2 u}\cosh u\,du\) | M1 | Applies given substitution to their expression with correct limits. (Limits may be recovered.) |
| \(= \frac{1}{4}\pi\int_0^{\sinh^{-1}\frac{4}{3}} \sinh^2 u\cosh^2 u\,du\) | A1 | Must be simplified. |
| \(= \frac{1}{16}\pi\int_0^{\sinh^{-1}\frac{4}{3}} \sinh^2 2u\,du\) | M1 | Applies \(\sinh 2u = 2\sinh u\cosh u\). May use double angle formulae for cosh and sinh instead. |
| \(= \frac{1}{32}\pi\int_0^{\sinh^{-1}\frac{4}{3}} \cosh 4u - 1\,du\) | M1 | Applies \(\sinh^2 A = \frac{1}{2}(\cosh 2A - 1)\). \(S\) must have the form \(a\int\sinh^2 u\cosh^2 u\,du\) |
| \(= \frac{1}{32}\pi\left[\frac{1}{4}\sinh 4u - u\right]_0^{\sinh^{-1}\frac{4}{3}}\) | A1 | |
| \(\sinh^{-1}\frac{4}{3} = \ln 3\) | B1 | |
| \(S = \frac{1}{32}\pi\left(\frac{1}{8}(e^{\ln 81} - e^{-\ln 81}) - \ln 3\right) = \frac{1}{32}\pi\left(\frac{1}{8}\left(81 - \frac{1}{81}\right) - \ln 3\right) = \frac{1}{32}\pi\left(\frac{820}{81} - \ln 3\right)\) | A1 | AG |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cosh 2A = \frac{1}{2}(e^{2A} + e^{-2A})$, $\sinh A = \frac{1}{2}(e^A - e^{-A})$ | B1 | |
| $2\sinh^2 A = \frac{1}{2}(e^A - e^{-A})^2 = \frac{1}{2}(e^{2A} - 2 + e^{-2A}) = \cosh 2A - 1$ | M1 A1 | Expands, AG. A0 for mixing variables e.g. $\sinh A = \frac{1}{2}(e^x - e^{-x})$ |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S = \int_0^3 2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = 2\pi\int_0^3 x^2\sqrt{1+4x^2}\,dx$ | M1 A1 | Correct formula with correct limits. Correct limits for M1. (Limits may be recovered.) |
| $S = \pi\int_0^{\sinh^{-1}\frac{4}{3}} \frac{1}{4}\sinh^2 u\sqrt{1+\sinh^2 u}\cosh u\,du$ | M1 | Applies given substitution to their expression with correct limits. (Limits may be recovered.) |
| $= \frac{1}{4}\pi\int_0^{\sinh^{-1}\frac{4}{3}} \sinh^2 u\cosh^2 u\,du$ | A1 | Must be simplified. |
| $= \frac{1}{16}\pi\int_0^{\sinh^{-1}\frac{4}{3}} \sinh^2 2u\,du$ | M1 | Applies $\sinh 2u = 2\sinh u\cosh u$. May use double angle formulae for cosh and sinh instead. |
| $= \frac{1}{32}\pi\int_0^{\sinh^{-1}\frac{4}{3}} \cosh 4u - 1\,du$ | M1 | Applies $\sinh^2 A = \frac{1}{2}(\cosh 2A - 1)$. $S$ must have the form $a\int\sinh^2 u\cosh^2 u\,du$ |
| $= \frac{1}{32}\pi\left[\frac{1}{4}\sinh 4u - u\right]_0^{\sinh^{-1}\frac{4}{3}}$ | A1 | |
| $\sinh^{-1}\frac{4}{3} = \ln 3$ | B1 | |
| $S = \frac{1}{32}\pi\left(\frac{1}{8}(e^{\ln 81} - e^{-\ln 81}) - \ln 3\right) = \frac{1}{32}\pi\left(\frac{1}{8}\left(81 - \frac{1}{81}\right) - \ln 3\right) = \frac{1}{32}\pi\left(\frac{820}{81} - \ln 3\right)$ | A1 | AG |
---7
\begin{enumerate}[label=(\alph*)]
\item Starting from the definitions of cosh and sinh in terms of exponentials, prove that
$$2 \sinh ^ { 2 } A = \cosh 2 A - 1$$
\includegraphics[max width=\textwidth, alt={}, center]{1fa404d4-5e14-4356-9b6d-f176d5a9f6db-12_79_1556_358_347}\\
\includegraphics[max width=\textwidth, alt={}]{1fa404d4-5e14-4356-9b6d-f176d5a9f6db-12_69_1575_466_328} ....................................................................................................................................... ........................................................................................................................................
\item A curve has equation $\mathrm { y } = \mathrm { x } ^ { 2 }$, for $0 \leqslant x \leqslant \frac { 2 } { 3 }$. The area of the surface generated when the curve is rotated through $2 \pi$ radians about the $x$-axis is denoted by $S$.\\
Use the substitution $\mathrm { X } = \frac { 1 } { 2 } \operatorname { sinhu }$ to show that $S = \frac { 1 } { 32 } \pi \left( \frac { 820 } { 81 } - \ln 3 \right)$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q7 [12]}}