| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve mixed sinh/cosh linear combinations |
| Difficulty | Challenging +1.2 This is a multi-part applied hyperbolic functions question requiring substitution of coordinates into cosh equations, solving simultaneous equations with hyperbolic identities, and comparison of models. While it involves several steps and hyperbolic functions (a Further Maths topic), the solution path is relatively straightforward: substitute known points, use cosh²-sinh²=1, and evaluate. The algebraic manipulation is standard for FM students, making this moderately above average difficulty but not requiring deep insight. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07b Hyperbolic graphs: sketch and properties |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (a) | Min value of cosh is 1 (and point on ground |
| Answer | Marks |
|---|---|
| (so 0 = k×1 – 1 =>) k = 1 | M1 |
| Answer | Marks |
|---|---|
| [2] | 2.2a |
| 2.2a | Using minimum point of curve |
| and knowledge of cosh graph | Could be derived by |
| Answer | Marks |
|---|---|
| (b) | Passes through (0, 3) => 3 = cosh(–b) – 1 |
| Answer | Marks |
|---|---|
| => a = ½ln(4 + √15) | *M1 |
| Answer | Marks |
|---|---|
| [5] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Use of (0, 3) to derive an |
| Answer | Marks |
|---|---|
| Use of (2, 0) to derive b = 2a | accept |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | (By symmetry of both;) (4, 3) | B1 |
| [1] | 2.2a |
Question 9:
9 | (a) | Min value of cosh is 1 (and point on ground
is at the minimum)
(so 0 = k×1 – 1 =>) k = 1 | M1
A1
[2] | 2.2a
2.2a | Using minimum point of curve
and knowledge of cosh graph | Could be derived by
differentiation
If zero scored then sc1 for k=1
www
(b) | Passes through (0, 3) => 3 = cosh(–b) – 1
=> b = –cosh–1(3 + 1)
b = (±)ln(4 + √(42 – 1))
=> b = ln(4 + √15)
Passes through (2, 0) => 0 = cosh(2a – b) –
1
=> b = 2a
=> a = ½ln(4 + √15) | *M1
dep*M
1
A1
M1
A1
[5] | 3.3
3.1a
1.1
3.3
1.1 | Use of (0, 3) to derive an
expression for b
Correct numerical use of formula
Use of (2, 0) to derive b = 2a | accept
4
Or rearcraonshge(s−. 𝑏𝑏)=𝑘𝑘
Could be from (a). Allow ft
(c) | (By symmetry of both;) (4, 3) | B1
[1] | 2.2a
PMT
Y541/01 Mark Scheme November 2020
Jofra’s model: d = cosh(5a – b) – 1 M1 3.4 Use of x = 5 with their values of a = 1.0317..., b = 2.0634..., d =
J J
a and b to predict d. Must have – 10.067... Condone 10.07 only if
1 clear evidence of production
AG A1 1.1 From correct values
d – d = 10.067... – 6.75 = 3.32 (3 sf)
J H
[3]
159 Two thin poles, $O A$ and $B C$, are fixed vertically on horizontal ground. A chain is fixed at $A$ and $C$ such that it touches the ground at point $D$ as shown in the diagram.
On a coordinate system the coordinates of $A$, $B$ and $D$ are $( 0,3 ) , ( 5,0 )$ and $( 2,0 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{c07ba83a-75fa-42dc-9bfd-6fc2f9226a23-5_805_1554_452_258}
It is required to find the height of pole $B C$ by modelling the shape of the curve that the chain forms.\\
Jofra models the curve using the equation $\mathrm { y } = \mathrm { k } \cosh ( \mathrm { ax } - \mathrm { b } ) - 1$ where $k , a$ and $b$ are positive constants.
\begin{enumerate}[label=(\alph*)]
\item Determine the value of $k$.
\item Find the exact value of $a$ and the exact value of $b$, giving your answers in logarithmic form.
Holly models the curve using the equation $y = \frac { 3 } { 4 } x ^ { 2 } - 3 x + 3$.
\item Write down the coordinates of the point, $( u , v )$ where $u$ and $v$ are both non-zero, at which the two models will agree.
\item Show that Jofra's model and Holly's model disagree in their predictions of the height of pole $B C$ by 3.32 m to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2020 Q9 [11]}}