| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Applied context modeling |
| Difficulty | Standard +0.3 This is a standard C3 harmonic form question with routine application to a real-world context. Part (a) requires the standard R cos(x-α) conversion using R²=a²+b² and tan α=b/a. Parts (b-d) apply this result straightforwardly: finding intercepts, identifying max/min from amplitude, and solving a simple trigonometric equation. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.05f Trigonometric function graphs: symmetries and periodicities1.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = \sqrt{6^2 + 2.5^2} = 6.5\) | B1 | Accept \(R = \text{awrt } 6.50\), \(\frac{13}{2}\). Do not accept \(R = \pm 6.50\) |
| \(\tan\alpha = \frac{2.5}{6} \Rightarrow \alpha = \text{awrt } 0.395\) | M1A1 | M1 for \(\tan\alpha = \pm\frac{2.5}{6}\) or \(\pm\frac{6}{2.5}\). Answer in degrees \(22.6°\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((0, 6)\) | B1 | Accept \(y=6\), \((0,6)\), awrt \(y=6.00\), \(f(0)=6\) or marked on curve. Do not accept \((6,0)\) |
| awrt \((1.97, 0)\) and \((5.11, 0)\) | M1A1 | M1 for attempt at either x-intercept from \(\frac{\pi}{2} + \text{their } 0.395\) or \(\frac{3\pi}{2} + \text{their } 0.395\). Both answers correct for A1. Accept awrt \((1.97,0)\) and \((5.11,0)\). In degrees: \((112.6, 0)\) and \((292.6, 0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempts \(12 + {'}R{'}\) or \(12 - {'}R{'}\) | M1 | |
| Either \(H_{\max} = 18.5\) or \(H_{\min} = 5.5\) | A1 | |
| Both \(H_{\max} = 18.5\) and \(H_{\min} = 5.5\) | A1 | Accept answers just written down with limited working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Sub \(H=16\), proceed to \(6.5\cos\!\left(\frac{2\pi t}{52} \pm {'}0.395{'}\right) = 4\) | M1 | |
| \(\left(\frac{2\pi t}{52} - {'}0.395{'}\right) = \text{awrt } 0.91\) | A1 | Accept in terms of \(x\): \((x - {'}0.395{'}) = \text{awrt } 0.91\) or \(= \text{invcos}\frac{4}{6.5}\) |
| \(t = \left(\text{awrt } 0.908 \pm {'}0.395{'}\right) \times \frac{52}{2\pi} = 11\ (10.78)\) | dM1A1 | Dependent on previous M. Accept awrt \(10.7/10.8\) or \(11\). The \(0.395\), \(2\pi\) and \(52\) must be used |
| \(\left(\frac{2\pi t}{52} \pm {'}0.395{'}\right) = \text{awrt } 2\pi - 0.908 \Rightarrow t = 48\ (47.75)\) | ddM1A1 | Dependent on both previous M's. Accept \(47.7/47.8\) or \(48\). Both values must be correct and rounded from values correct to 1 d.p. |
# Question 7:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = \sqrt{6^2 + 2.5^2} = 6.5$ | B1 | Accept $R = \text{awrt } 6.50$, $\frac{13}{2}$. Do not accept $R = \pm 6.50$ |
| $\tan\alpha = \frac{2.5}{6} \Rightarrow \alpha = \text{awrt } 0.395$ | M1A1 | M1 for $\tan\alpha = \pm\frac{2.5}{6}$ or $\pm\frac{6}{2.5}$. Answer in degrees $22.6°$ is A0 |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(0, 6)$ | B1 | Accept $y=6$, $(0,6)$, awrt $y=6.00$, $f(0)=6$ or marked on curve. Do not accept $(6,0)$ |
| awrt $(1.97, 0)$ and $(5.11, 0)$ | M1A1 | M1 for attempt at either x-intercept from $\frac{\pi}{2} + \text{their } 0.395$ or $\frac{3\pi}{2} + \text{their } 0.395$. Both answers correct for A1. Accept awrt $(1.97,0)$ and $(5.11,0)$. In degrees: $(112.6, 0)$ and $(292.6, 0)$ |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempts $12 + {'}R{'}$ or $12 - {'}R{'}$ | M1 | |
| Either $H_{\max} = 18.5$ or $H_{\min} = 5.5$ | A1 | |
| Both $H_{\max} = 18.5$ and $H_{\min} = 5.5$ | A1 | Accept answers just written down with limited working |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Sub $H=16$, proceed to $6.5\cos\!\left(\frac{2\pi t}{52} \pm {'}0.395{'}\right) = 4$ | M1 | |
| $\left(\frac{2\pi t}{52} - {'}0.395{'}\right) = \text{awrt } 0.91$ | A1 | Accept in terms of $x$: $(x - {'}0.395{'}) = \text{awrt } 0.91$ or $= \text{invcos}\frac{4}{6.5}$ |
| $t = \left(\text{awrt } 0.908 \pm {'}0.395{'}\right) \times \frac{52}{2\pi} = 11\ (10.78)$ | dM1A1 | Dependent on previous M. Accept awrt $10.7/10.8$ or $11$. The $0.395$, $2\pi$ and $52$ must be used |
| $\left(\frac{2\pi t}{52} \pm {'}0.395{'}\right) = \text{awrt } 2\pi - 0.908 \Rightarrow t = 48\ (47.75)$ | ddM1A1 | Dependent on both previous M's. Accept $47.7/47.8$ or $48$. Both values must be correct and rounded from values correct to 1 d.p. |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{be00fdaa-2fe3-4f06-a710-08ec67fb911e-13_456_881_214_534}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the curve $C$, with equation $y = 6 \cos x + 2.5 \sin x$ for $0 \leqslant x \leqslant 2 \pi$
\begin{enumerate}[label=(\alph*)]
\item Express $6 \cos x + 2.5 \sin x$ in the form $R \cos ( x - \alpha )$, where $R$ and $\alpha$ are constants with $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$. Give your value of $\alpha$ to 3 decimal places.
\item Find the coordinates of the points on the graph where the curve $C$ crosses the coordinate axes.
A student records the number of hours of daylight each Sunday throughout the year. She starts on the last Sunday in May with a recording of 18 hours, and continues until her final recording 52 weeks later.
She models her results with the continuous function given by
$$H = 12 + 6 \cos \left( \frac { 2 \pi t } { 52 } \right) + 2.5 \sin \left( \frac { 2 \pi t } { 52 } \right) , \quad 0 \leqslant t \leqslant 52$$
where $H$ is the number of hours of daylight and $t$ is the number of weeks since her first recording.
Use this function to find
\item the maximum and minimum values of $H$ predicted by the model,
\item the values for $t$ when $H = 16$, giving your answers to the nearest whole number.\\[0pt]
[You must show your working. Answers based entirely on graphical or numerical methods are not acceptable.]\\
\includegraphics[max width=\textwidth, alt={}, center]{be00fdaa-2fe3-4f06-a710-08ec67fb911e-14_40_58_2460_1893}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2014 Q7 [15]}}