| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve differential equations with hyperbolics |
| Difficulty | Challenging +1.2 Part (a) is a routine proof from definitions requiring straightforward algebraic manipulation of exponentials. Part (b) is a standard first-order linear ODE requiring integrating factor method with hyperbolic functions—more involved than typical A-level but follows a well-established procedure. The hyperbolic context and Further Maths setting elevate it slightly above average difficulty. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cosh x = \frac{1}{2}(e^x + e^{-x})\) | B1 | |
| \(\frac{2}{4}(e^x + e^{-x})^2 = \frac{1}{2}(e^{2x} + e^{-2x} + 2) = \cosh 2x + 1\) | M1 A1 | Expands, AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{2\int \tanh x\, dx} = e^{2\ln\cosh x} = \cosh^2 x\) | M1 A1 | Finds integrating factor |
| \(\frac{d}{dx}(y\cosh^2 x) = \cosh^2 x\) | M1 | Correct form on LHS, \(\frac{d}{dx}(yI)\) for their integrating factor \(I\), and attempt to integrate their RHS |
| \(y\cosh^2 x = \int \cosh^2 x\, dx = \frac{1}{2}\int \cosh 2x + 1\, dx = \frac{1}{2}\left(\frac{1}{2}\sinh 2x + x\right) + C\) | M1 A1 | Uses \(\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)\). M1 A1 is for RHS |
| \(1 = C\) | M1 | Finds \(C\) |
| \(y = \text{sech}^2 x\left(\frac{1}{4}\sinh 2x + \frac{1}{2}x + 1\right)\) | M1 A1 | Divides through by their coefficient of \(y\) |
## Question 5(a):
| $\cosh x = \frac{1}{2}(e^x + e^{-x})$ | B1 | |
|---|---|---|
| $\frac{2}{4}(e^x + e^{-x})^2 = \frac{1}{2}(e^{2x} + e^{-2x} + 2) = \cosh 2x + 1$ | M1 A1 | Expands, AG |
## Question 5(b):
| $e^{2\int \tanh x\, dx} = e^{2\ln\cosh x} = \cosh^2 x$ | M1 A1 | Finds integrating factor |
|---|---|---|
| $\frac{d}{dx}(y\cosh^2 x) = \cosh^2 x$ | M1 | Correct form on LHS, $\frac{d}{dx}(yI)$ for their integrating factor $I$, and attempt to integrate their RHS |
| $y\cosh^2 x = \int \cosh^2 x\, dx = \frac{1}{2}\int \cosh 2x + 1\, dx = \frac{1}{2}\left(\frac{1}{2}\sinh 2x + x\right) + C$ | M1 A1 | Uses $\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)$. M1 A1 is for RHS |
| $1 = C$ | M1 | Finds $C$ |
| $y = \text{sech}^2 x\left(\frac{1}{4}\sinh 2x + \frac{1}{2}x + 1\right)$ | M1 A1 | Divides through by their coefficient of $y$ |
---5
\begin{enumerate}[label=(\alph*)]
\item Starting from the definitions of cosh and sinh in terms of exponentials, prove that
$$2 \cosh ^ { 2 } x = \cosh 2 x + 1$$
\includegraphics[max width=\textwidth, alt={}, center]{d421652f-576d-4843-abbf-54404e225fec-08_67_1550_374_347}
\item Find the solution of the differential equation
$$\frac { d y } { d x } + 2 y \tanh x = 1$$
for which $y = 1$ when $x = 0$. Give your answer in the form $y = f ( x )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q5 [11]}}