| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove hyperbolic identity from exponentials |
| Difficulty | Standard +0.8 Part (a) is a straightforward proof from exponential definitions (routine for Further Maths students). Part (b) requires careful parametric differentiation with hyperbolic functions. Part (c) involves second derivative calculation and solving a non-trivial equation involving sech²t, requiring multiple steps and algebraic manipulation. Overall, this is a standard Further Maths question with moderate computational demand but no exceptional insight required. |
| Spec | 1.07s Parametric and implicit differentiation4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{sech}\,t = \frac{2}{e^t+e^{-t}} \quad \tanh t = \frac{e^t - e^{-t}}{e^t+e^{-t}}\) | B1 | |
| \(1 - \left(\frac{2}{e^t+e^{-t}}\right)^2 = \frac{(e^t+e^{-t})^2 - 4}{(e^t+e^{-t})^2} = \frac{e^{2t}+2+e^{-2t}-4}{(e^t+e^{-t})^2} = \frac{(e^t-e^{-t})^2}{(e^t+e^{-t})^2}\) | M1 A1 | Expands, gets to \(\frac{e^{2t}+2+e^{-2t}-4}{(e^t+e^{-t})^2}\) for M1. AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dt} = 4\tanh^3 t\,\text{sech}^2 t\) | B1 | |
| \(\frac{dx}{dt} = \tanh t\,\text{sech}^2 t - \tanh t \;(= \tanh t(\text{sech}^2 t - 1) = -\tanh^3 t)\) | M1 A1 | M1 sensible attempt at derivative of \(x\) |
| \(\frac{dy}{dx} = \frac{dy}{dt}\times\frac{dt}{dx} = \frac{4\tanh^3 t\,\text{sech}^2 t}{-\tanh^3 t} = -4\,\text{sech}^2 t\) | M1 A1 | Applies chain rule, must substitute their \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) for M1. AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = \frac{d}{dt}\!\left(\frac{dy}{dx}\right)\times\frac{dt}{dx} = \frac{8\,\text{sech}^2 t\,\tanh t}{-\tanh^3 t} = -8\frac{\text{sech}^2 t}{\tanh^2 t}\) | M1 A1 | Finds \(\frac{d^2y}{dx^2}\). For M1: \(\frac{d}{dt}\!\left(\frac{dy}{dx}\right) = c\,\text{sech}^2 t\tanh t\). AEF. Accept \(-8\,\text{cosech}^2 t\) |
| \(-8\frac{\text{sech}^2 t}{\tanh^2 t} = -\frac{9}{2} \Rightarrow 8(1-\tanh^2 t) = \frac{9}{2}\tanh^2 t \Rightarrow \tanh^2 t = \frac{16}{25}\) | M1 A1 | Sets equal to \(-\frac{9}{2}\), uses identity from (a). Accept \(\sinh^2 t = \frac{16}{9}\) or \(\cosh^2 t = \frac{25}{9}\) |
| \((x,y) = \left(\frac{8}{25} + \ln\frac{3}{5},\, \frac{881}{625}\right)\) | A1 A1 | A1 for each correct coordinate. \(t = \ln 3\) |
# Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{sech}\,t = \frac{2}{e^t+e^{-t}} \quad \tanh t = \frac{e^t - e^{-t}}{e^t+e^{-t}}$ | B1 | |
| $1 - \left(\frac{2}{e^t+e^{-t}}\right)^2 = \frac{(e^t+e^{-t})^2 - 4}{(e^t+e^{-t})^2} = \frac{e^{2t}+2+e^{-2t}-4}{(e^t+e^{-t})^2} = \frac{(e^t-e^{-t})^2}{(e^t+e^{-t})^2}$ | M1 A1 | Expands, gets to $\frac{e^{2t}+2+e^{-2t}-4}{(e^t+e^{-t})^2}$ for M1. AG |
---
# Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dt} = 4\tanh^3 t\,\text{sech}^2 t$ | B1 | |
| $\frac{dx}{dt} = \tanh t\,\text{sech}^2 t - \tanh t \;(= \tanh t(\text{sech}^2 t - 1) = -\tanh^3 t)$ | M1 A1 | M1 sensible attempt at derivative of $x$ |
| $\frac{dy}{dx} = \frac{dy}{dt}\times\frac{dt}{dx} = \frac{4\tanh^3 t\,\text{sech}^2 t}{-\tanh^3 t} = -4\,\text{sech}^2 t$ | M1 A1 | Applies chain rule, must substitute their $\frac{dy}{dt}$ and $\frac{dx}{dt}$ for M1. AG |
---
# Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dx^2} = \frac{d}{dt}\!\left(\frac{dy}{dx}\right)\times\frac{dt}{dx} = \frac{8\,\text{sech}^2 t\,\tanh t}{-\tanh^3 t} = -8\frac{\text{sech}^2 t}{\tanh^2 t}$ | M1 A1 | Finds $\frac{d^2y}{dx^2}$. For M1: $\frac{d}{dt}\!\left(\frac{dy}{dx}\right) = c\,\text{sech}^2 t\tanh t$. AEF. Accept $-8\,\text{cosech}^2 t$ |
| $-8\frac{\text{sech}^2 t}{\tanh^2 t} = -\frac{9}{2} \Rightarrow 8(1-\tanh^2 t) = \frac{9}{2}\tanh^2 t \Rightarrow \tanh^2 t = \frac{16}{25}$ | M1 A1 | Sets equal to $-\frac{9}{2}$, uses identity from (a). Accept $\sinh^2 t = \frac{16}{9}$ or $\cosh^2 t = \frac{25}{9}$ |
| $(x,y) = \left(\frac{8}{25} + \ln\frac{3}{5},\, \frac{881}{625}\right)$ | A1 A1 | A1 for each correct coordinate. $t = \ln 3$ |8
\begin{enumerate}[label=(\alph*)]
\item Starting from the definitions of sech and tanh in terms of exponentials, prove that
$$1 - \operatorname { sech } ^ { 2 } t = \tanh ^ { 2 } t$$
\includegraphics[max width=\textwidth, alt={}]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-14_77_1547_360_347} ......................................................................................................................................... ......................................................................................................................................... . ........................................................................................................................................ ........................................................................................................................................ .......................................................................................................................................\\
\includegraphics[max width=\textwidth, alt={}, center]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-14_72_1573_911_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-14_67_1570_1005_324}
The curve $C$ has parametric equations
$$\mathrm { x } = \frac { 1 } { 2 } \tanh ^ { 2 } \mathrm { t } + \text { Insecht } , \quad \mathrm { y } = 1 + \tanh ^ { 4 } \mathrm { t } , \quad \text { for } t > 0$$
\item Show that $\frac { d y } { d x } = - 4 \operatorname { sech } ^ { 2 } t$.
\item Find the coordinates of the point on $C$ with $\frac { d ^ { 2 } y } { d x ^ { 2 } } = - \frac { 9 } { 2 }$, giving your answer in the form $( a + \ln b , c )$ where $a , b$ and $c$ are rational numbers.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q8 [14]}}