| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Range from trigonometric functions |
| Difficulty | Moderate -0.5 This question involves straightforward application of transformations to find constants in given models. Part (a) requires solving two simultaneous equations from given conditions (max height 60m, starting height 2m). Part (b) is direct substitution. Part (c) uses amplitude and vertical shift properties of cosine (max/min values give amplitude 29 and shift β=31, starting condition gives α). Part (d) requires recognizing that cosine is periodic while quadratic is not. All steps are standard techniques with no novel problem-solving required, though the multi-part structure and context make it slightly more substantial than pure recall. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 60\) | B1 | May be implied by 60 substituted for \(a\) in final model |
| \(2 = \text{"60"} - b(-20)^2 \Rightarrow b = ...\) | M1 | Substitute \(t=0\), \(H=2\) and their \(a\), proceed to value for \(b\) |
| \(H = 60 - 0.145(t-20)^2\) | A1 | Must be in terms of \(H\) and \(t\); equivalent forms accepted e.g. \(H = -\frac{29}{200}t^2 + 5.8t + 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Height \(= 2\) m | B1 | Condone lack of units; can score even if model in (a) is incorrect |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\alpha = 180\) or \(\beta = 31\) | M1 | Condone \((\alpha =) \pi\) |
| \(H = 29\cos(9t + 180)^\circ + 31\) | A1 | Equivalent e.g. \(H = -29\cos(9t) + 31\); must be in terms of \(H\) and \(t\); does not require degree symbol |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. "The model allows for more than one circuit" | B1 | Any valid reason referencing: repetition/multiple cycles; at 2 mins carriage back at start; original model gives negative heights after 40s; original model would not return to start after 2 mins |
## Question 13:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 60$ | B1 | May be implied by 60 substituted for $a$ in final model |
| $2 = \text{"60"} - b(-20)^2 \Rightarrow b = ...$ | M1 | Substitute $t=0$, $H=2$ and their $a$, proceed to value for $b$ |
| $H = 60 - 0.145(t-20)^2$ | A1 | Must be in terms of $H$ and $t$; equivalent forms accepted e.g. $H = -\frac{29}{200}t^2 + 5.8t + 2$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Height $= 2$ m | B1 | Condone lack of units; can score even if model in (a) is incorrect |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha = 180$ or $\beta = 31$ | M1 | Condone $(\alpha =) \pi$ |
| $H = 29\cos(9t + 180)^\circ + 31$ | A1 | Equivalent e.g. $H = -29\cos(9t) + 31$; must be in terms of $H$ and $t$; does not require degree symbol |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. "The model allows for more than one circuit" | B1 | Any valid reason referencing: repetition/multiple cycles; at 2 mins carriage back at start; original model gives negative heights after 40s; original model would not return to start after 2 mins |
---\begin{enumerate}
\item On a roller coaster ride, passengers travel in carriages around a track.
\end{enumerate}
On the ride, carriages complete multiple circuits of the track such that
\begin{itemize}
\item the maximum vertical height of a carriage above the ground is 60 m
\item a carriage starts a circuit at a vertical height of 2 m above the ground
\item the ground is horizontal
\end{itemize}
The vertical height, $H \mathrm {~m}$, of a carriage above the ground, $t$ seconds after the carriage starts the first circuit, is modelled by the equation
$$H = a - b ( t - 20 ) ^ { 2 }$$
where $a$ and $b$ are positive constants.\\
(a) Find a complete equation for the model.\\
(b) Use the model to determine the height of the carriage above the ground when $t = 40$
In an alternative model, the vertical height, $H \mathrm {~m}$, of a carriage above the ground, $t$ seconds after the carriage starts the first circuit, is given by
$$H = 29 \cos ( 9 t + \alpha ) ^ { \circ } + \beta \quad 0 \leqslant \alpha < 360 ^ { \circ }$$
where $\alpha$ and $\beta$ are constants.\\
(c) Find a complete equation for the alternative model.
Given that the carriage moves continuously for 2 minutes,\\
(d) give a reason why the alternative model would be more appropriate.
\hfill \mbox{\textit{Edexcel Paper 1 2023 Q13 [7]}}