| 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 requiring algebraic manipulation. Part (b) involves standard parametric differentiation with hyperbolic functions. Part (c) requires computing the second derivative using the chain rule, solving for t, then substituting back—this is multi-step but follows standard Further Maths techniques. Overall, this is a solid Further Maths question requiring competence across several techniques but no exceptional insight. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| 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) or equivalent. 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) or equivalent. 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={}]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-14_77_1547_360_347} ......................................................................................................................................... ......................................................................................................................................... . ........................................................................................................................................ ........................................................................................................................................ .......................................................................................................................................\\
\includegraphics[max width=\textwidth, alt={}, center]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-14_72_1573_911_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-14_67_1573_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]}}