| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Applied context modeling |
| Difficulty | Standard +0.3 This is a standard harmonic form question from Further Maths. Part (a) is routine application of R cos(θ+α) formula using standard identities. Parts (b) and (c) follow directly from the harmonic form with straightforward max/min and inequality solving. Part (d) is a simple modeling comment. While it's Further Maths content, it's a textbook application with no novel insight required, making it slightly easier than average overall. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(R = \sqrt{5}\) | B1 | \(R = \sqrt{5}\) only |
| \(\tan\alpha = \frac{1}{2}\) or \(\sin\alpha = \frac{1}{\sqrt{5}}\) or \(\cos\alpha = \frac{2}{\sqrt{5}} \Rightarrow \alpha = ...\) | M1 | Proceeds to a value for \(\alpha\) from \(\tan\alpha = \pm\frac{1}{2}\) or \(\sin\alpha = \pm\frac{1}{``R"}\) or \(\cos\alpha = \pm\frac{2}{``R"}\) |
| \(\alpha = 0.464\) | A1 | \(\alpha =\) awrt \(0.464\) (radians) or awrt \(26.6\) (degrees) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(3 + 2\sqrt{5}\) | B1ft | For \(\left(3+2\sqrt{5}\right)\) m or awrt \(7.47\) m. Follow through on their \(R\): allow \(3 + 2\times\) Their \(R\). Allow in decimals with at least 3sf accuracy. Condone lack of units. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\cos(0.5t + 0.464) = 1 \Rightarrow 0.5t + 0.464 = 2\pi \Rightarrow t = ...\) | M1 | Uses \(0.5t \pm\text{"0.464"} = 2\pi\) to obtain a value for \(t\). Follow through on their \(0.464\) but angle must be in radians. Possible in degrees using \(0.5t \pm\text{"26.6"} = 360\) |
| \(t = 11.6\) | A1 | Awrt \(11.6\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(3 + 2\sqrt{5}\cos(0.5t + 0.464) = 0\) leading to \(\cos(0.5t + 0.464) = -\dfrac{3}{2\sqrt{5}}\) | M1 | Uses the model, sets \(3 + 2\text{"}\sqrt{5}\text{"}\cos(...) = 0\) and proceeds to \(\cos(...) = k\) where \(\ |
| \(0.5t + 0.464 = \cos^{-1}\!\left(-\dfrac{3}{2\sqrt{5}}\right) \Rightarrow t = 2\!\left(\cos^{-1}\!\left(-\dfrac{3}{2\sqrt{5}}\right) - 0.464\right)\) | dM1 | Solves \(\cos(0.5t \pm \text{"0.464"}) = k\) where \(\ |
| \(2(3.977... - 0.464) - 2(2.306... - 0.464)\) | dM1 | Fully correct strategy to find required duration (e.g. finds 2 consecutive values of \(t\) when \(H=0\) and subtracts). Depends on previous M1. |
| \(= 3.34\) | A1 | Correct value. Must be \(3.34\) (not awrt) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| e.g. the "\(3\)" would need to vary | B1 | Correct refinement e.g. as in scheme. If a specific function is suggested to replace the "\(3\)" it must be sensible e.g. a trigonometric function rather than a quadratic/linear one. |
# Question 15:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $R = \sqrt{5}$ | B1 | $R = \sqrt{5}$ only |
| $\tan\alpha = \frac{1}{2}$ or $\sin\alpha = \frac{1}{\sqrt{5}}$ or $\cos\alpha = \frac{2}{\sqrt{5}} \Rightarrow \alpha = ...$ | M1 | Proceeds to a value for $\alpha$ from $\tan\alpha = \pm\frac{1}{2}$ or $\sin\alpha = \pm\frac{1}{``R"}$ or $\cos\alpha = \pm\frac{2}{``R"}$ |
| $\alpha = 0.464$ | A1 | $\alpha =$ awrt $0.464$ (radians) or awrt $26.6$ (degrees) |
**(3 marks)**
---
## Part (b)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $3 + 2\sqrt{5}$ | B1ft | For $\left(3+2\sqrt{5}\right)$ m or awrt $7.47$ m. Follow through on their $R$: allow $3 + 2\times$ Their $R$. Allow in decimals with at least 3sf accuracy. Condone lack of units. |
---
## Part (b)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\cos(0.5t + 0.464) = 1 \Rightarrow 0.5t + 0.464 = 2\pi \Rightarrow t = ...$ | M1 | Uses $0.5t \pm\text{"0.464"} = 2\pi$ to obtain a value for $t$. Follow through on their $0.464$ but angle must be in radians. Possible in degrees using $0.5t \pm\text{"26.6"} = 360$ |
| $t = 11.6$ | A1 | Awrt $11.6$ |
**(3 marks)**
---
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $3 + 2\sqrt{5}\cos(0.5t + 0.464) = 0$ leading to $\cos(0.5t + 0.464) = -\dfrac{3}{2\sqrt{5}}$ | M1 | Uses the model, sets $3 + 2\text{"}\sqrt{5}\text{"}\cos(...) = 0$ and proceeds to $\cos(...) = k$ where $\|k\| < 1$. Allow $3 + 2\text{"}\sqrt{5}\text{"}\cos(...) < 0$ |
| $0.5t + 0.464 = \cos^{-1}\!\left(-\dfrac{3}{2\sqrt{5}}\right) \Rightarrow t = 2\!\left(\cos^{-1}\!\left(-\dfrac{3}{2\sqrt{5}}\right) - 0.464\right)$ | dM1 | Solves $\cos(0.5t \pm \text{"0.464"}) = k$ where $\|k\|<1$ to obtain at least one value for $t$. Depends on previous M1. |
| $2(3.977... - 0.464) - 2(2.306... - 0.464)$ | dM1 | Fully correct strategy to find required duration (e.g. finds 2 consecutive values of $t$ when $H=0$ and subtracts). Depends on previous M1. |
| $= 3.34$ | A1 | Correct value. Must be $3.34$ (not awrt) |
**(4 marks)**
---
## Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| e.g. the "$3$" would need to vary | B1 | Correct refinement e.g. as in scheme. If a specific function is suggested to replace the "$3$" it must be sensible e.g. a trigonometric function rather than a quadratic/linear one. |
**(1 mark)**
**Total: 11 marks**\begin{enumerate}
\item (a) Express $2 \cos \theta - \sin \theta$ in the form $R \cos ( \theta + \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$
\end{enumerate}
Give the exact value of $R$ and the value of $\alpha$ in radians to 3 decimal places.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6c32000f-574f-473c-bd04-9cfe2c1bd715-44_440_1118_463_575}
\captionsetup{labelformat=empty}
\caption{Figure 6}
\end{center}
\end{figure}
Figure 6 shows the cross-section of a water wheel.\\
The wheel is free to rotate about a fixed axis through the point $C$.\\
The point $P$ is at the end of one of the paddles of the wheel, as shown in Figure 6.\\
The water level is assumed to be horizontal and of constant height.\\
The vertical height, $H$ metres, of $P$ above the water level is modelled by the equation
$$H = 3 + 4 \cos ( 0.5 t ) - 2 \sin ( 0.5 t )$$
where $t$ is the time in seconds after the wheel starts rotating.\\
Using the model, find\\
(b) (i) the maximum height of $P$ above the water level,\\
(ii) the value of $t$ when this maximum height first occurs, giving your answer to one decimal place.
In a single revolution of the wheel, $P$ is below the water level for a total of $T$ seconds. According to the model,\\
(c) find the value of $T$ giving your answer to 3 significant figures.\\
(Solutions based entirely on calculator technology are not acceptable.)
In reality, the water level may not be of constant height.\\
(d) Explain how the equation of the model should be refined to take this into account.
\hfill \mbox{\textit{Edexcel Paper 2 2021 Q15 [11]}}