| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Graphs & Exact Values |
| Type | Real-world modelling (tides, daylight, etc.) |
| Difficulty | Moderate -0.3 This is a standard A-level modelling question involving transformations of trigonometric graphs. Part (a) requires identifying amplitude, period, and phase shift from given information—routine application of formulas. Part (b) asks for a conceptual explanation about model appropriateness, which is straightforward given the context. The absolute value adds minor complexity but the overall problem follows a familiar template for Ferris wheel/circular motion questions, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.05f Trigonometric function graphs: symmetries and periodicities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces \(A = \pm 50\) or \(b = \frac{1}{4}\) | B1 | May be seen embedded within their equation |
| Deduces \(A = \pm 50\) and \(b = \frac{1}{4}\) | B1 | May be seen embedded within their equation |
| Uses \(t=0, H=1 \Rightarrow \alpha = \ldots\) e.g. \(1 = \text{"50"}\sin(\alpha)^\circ \Rightarrow \alpha = \ldots\) | M1 | Follow through on their value for \(A\). Allow for \(\pm 1 = \text{"50"}\sin(\alpha)^\circ \Rightarrow \alpha = \ldots\) where \(\alpha\) is in degrees or radians. Note in radians \(\sin^{-1}\!\left(\frac{1}{50}\right) \approx \frac{1}{50}\) (0.0200…) which may appear incorrect but is in fact ok. Also in degrees a value of e.g. 1.14 (truncated) would indicate the method. |
| \(H = \left\lvert \pm 50\sin\!\left(\frac{1}{4}t + 1.15\right)^\circ \right\rvert\) | A1 | Condone omission of degrees symbol and allow awrt 1.15 for \(\alpha\). Allow if correct equation seen anywhere in solution. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. the minimum height above the ground of the passenger on the original model was 0 m, or adding "\(d\)" means the passenger does not touch the ground | B1 | Gives a suitable explanation with no contradictory statements. Condone "so that pod/capsule/seat/passenger/ferris wheel/it etc. will not hit/touch the ground". Responses that focus on the starting point of the model are likely to score B0. |
## Question 9(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $A = \pm 50$ **or** $b = \frac{1}{4}$ | B1 | May be seen embedded within their equation |
| Deduces $A = \pm 50$ **and** $b = \frac{1}{4}$ | B1 | May be seen embedded within their equation |
| Uses $t=0, H=1 \Rightarrow \alpha = \ldots$ e.g. $1 = \text{"50"}\sin(\alpha)^\circ \Rightarrow \alpha = \ldots$ | M1 | Follow through on their value for $A$. Allow for $\pm 1 = \text{"50"}\sin(\alpha)^\circ \Rightarrow \alpha = \ldots$ where $\alpha$ is in degrees or radians. Note in radians $\sin^{-1}\!\left(\frac{1}{50}\right) \approx \frac{1}{50}$ (0.0200…) which may appear incorrect but is in fact ok. Also in degrees a value of e.g. 1.14 (truncated) would indicate the method. |
| $H = \left\lvert \pm 50\sin\!\left(\frac{1}{4}t + 1.15\right)^\circ \right\rvert$ | A1 | Condone omission of degrees symbol and allow awrt 1.15 for $\alpha$. Allow if correct equation seen anywhere in solution. |
**(4 marks)**
---
## Question 9(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. the minimum height above the ground of the passenger on the original model was 0 m, **or** adding "$d$" means the passenger does not touch the ground | B1 | Gives a suitable explanation with no contradictory statements. Condone "so that pod/capsule/seat/passenger/ferris wheel/it etc. will not hit/touch the ground". Responses that focus on the starting point of the model are likely to score B0. |
**(1 mark) — Total: 5 marks**
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{824d73c5-525c-4876-ad66-33c8f1664277-22_419_569_301_226}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{824d73c5-525c-4876-ad66-33c8f1664277-22_522_927_239_917}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 4 shows a sketch of a Ferris wheel.\\
The height above the ground, $H \mathrm {~m}$, of a passenger on the Ferris wheel, $t$ seconds after the wheel starts turning, is modelled by the equation
$$H = \left| A \sin ( b t + \alpha ) ^ { \circ } \right|$$
where $A$, $b$ and $\alpha$ are constants.\\
Figure 5 shows a sketch of the graph of $H$ against $t$, for one revolution of the wheel.\\
Given that
\begin{itemize}
\item the maximum height of the passenger above the ground is 50 m
\item the passenger is 1 m above the ground when the wheel starts turning
\item the wheel takes 720 seconds to complete one revolution
\begin{enumerate}[label=(\alph*)]
\item find a complete equation for the model, giving the exact value of $A$, the exact value of $b$ and the value of $\alpha$ to 3 significant figures.
\item Explain why an equation of the form
\end{itemize}
$$H = \left| A \sin ( b t + \alpha ) ^ { \circ } \right| + d$$
where $d$ is a positive constant, would be a more appropriate model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2022 Q9 [5]}}