| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | January |
| Marks | 12 |
| 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 with straightforward application to a real-world context. Part (a) is routine bookwork using R = √(a² + b²) and tan α = b/a. Parts (b) and (c) follow directly from part (a) using standard techniques. While multi-part with a modeling context, it requires only mechanical application of well-practiced methods with no novel insight or complex problem-solving. |
| Spec | 1.02z Models in context: use functions in modelling1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| VIIIV SIHI NI JIIHM 10 N OC | VIIV 5141 NI 3114 M I ON OC | VI4V SIHIL NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = \sqrt{4+16} = \sqrt{20}\) or \(2\sqrt{5}\) | B1 | No working needed; condone \(R = \pm\sqrt{20}\) |
| \(\tan\alpha = \frac{4}{2}\) | M1 | Or \(\tan\alpha = \pm\frac{4}{2}\) or \(\tan\alpha = \pm\frac{2}{4}\); if \(R\) used accept \(\sin\alpha = \pm\frac{4}{"R"}\) or \(\cos\alpha = \pm\frac{2}{"R"}\) |
| \(\alpha = 1.11\) (awrt) | A1 | Also accept \(\sqrt{20}\sin(x-1.11)\); answers in degrees are A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Maximum is \(12+2R\); minimum is \(12-2R\) | M1 | Uses maximum \(12+2R\) or minimum \(12-2R\) with their value of \(R\) |
| Maximum \(= 20.9\) hours (20h 57m); minimum \(= 3.06\) hours (3h 3m) | A1 A1 | Maximum and minimum value awrt 20.9 and 3.06; \(12 \pm 2\sqrt{20}\) or \(12 \pm 4\sqrt{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(17 = 12 + k"R"\sin\left(\frac{2\pi t}{365} \pm "\alpha"\right)\) | M1 | Attempt to interpret model in terms of (a); allow \(k=1\); condone slips |
| \(\sin\left(\frac{2\pi t}{365} \pm "\alpha"\right) = \ldots\) | dM1 | Attempting to make \(\sin\left(\frac{2\pi t}{365} \pm "\alpha"\right)\) the subject |
| For proceeding to one value of \(t\) from \(17 = 12 + 2"R"\sin\left(\frac{2\pi t}{365} \pm "\alpha"\right)\) | M1 | Method for finding at least one value of \(t\), \(0 < t < 365\), from correct starting point with \(2\times\) their \(R\) |
| \(t = 99\) (days) or 212 or 213 (days) | A1 | One correct value awrt 99 or awrt 212/213 |
| For finding two values of \(t\) | dM1 | Attempting second value; dependent on previous M; usually moving from \(\frac{2\pi t}{365} \pm "\alpha" = \pi - \beta\) |
| \(t = 99\) (days) and 212 or 213 (days) | A1 | awrt 99 and awrt 212 or 213 only, \(0 < t < 365\) |
# Question 12:
## Part (a):
| $R = \sqrt{4+16} = \sqrt{20}$ or $2\sqrt{5}$ | B1 | No working needed; condone $R = \pm\sqrt{20}$ |
| $\tan\alpha = \frac{4}{2}$ | M1 | Or $\tan\alpha = \pm\frac{4}{2}$ or $\tan\alpha = \pm\frac{2}{4}$; if $R$ used accept $\sin\alpha = \pm\frac{4}{"R"}$ or $\cos\alpha = \pm\frac{2}{"R"}$ |
| $\alpha = 1.11$ (awrt) | A1 | Also accept $\sqrt{20}\sin(x-1.11)$; answers in degrees are A0 | **(3)** |
## Part (b):
| Maximum is $12+2R$; minimum is $12-2R$ | M1 | Uses maximum $12+2R$ or minimum $12-2R$ with their value of $R$ |
| Maximum $= 20.9$ hours (20h 57m); minimum $= 3.06$ hours (3h 3m) | A1 A1 | Maximum and minimum value awrt 20.9 and 3.06; $12 \pm 2\sqrt{20}$ or $12 \pm 4\sqrt{5}$ | **(3)** |
## Part (c):
| $17 = 12 + k"R"\sin\left(\frac{2\pi t}{365} \pm "\alpha"\right)$ | M1 | Attempt to interpret model in terms of (a); allow $k=1$; condone slips |
| $\sin\left(\frac{2\pi t}{365} \pm "\alpha"\right) = \ldots$ | dM1 | Attempting to make $\sin\left(\frac{2\pi t}{365} \pm "\alpha"\right)$ the subject |
| For proceeding to one value of $t$ from $17 = 12 + 2"R"\sin\left(\frac{2\pi t}{365} \pm "\alpha"\right)$ | M1 | Method for finding at least one value of $t$, $0 < t < 365$, from correct starting point with $2\times$ their $R$ |
| $t = 99$ (days) or 212 or 213 (days) | A1 | One correct value awrt 99 or awrt 212/213 |
| For finding two values of $t$ | dM1 | Attempting second value; dependent on previous M; usually moving from $\frac{2\pi t}{365} \pm "\alpha" = \pi - \beta$ |
| $t = 99$ (days) **and** 212 or 213 (days) | A1 | awrt 99 **and** awrt 212 or 213 only, $0 < t < 365$ | **(6)** |
**(12 marks)**\begin{enumerate}
\item (a) Express $2 \sin x - 4 \cos x$ in the form $R \sin ( x - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$
\end{enumerate}
Give the exact value of $R$ and give the value of $\alpha$, in radians, to 3 significant figures.
In a town in Norway, a student records the number of hours of daylight every day for a year. He models the number of hours of daylight, $H$, by the continuous function given by the formula
$$H = 12 + 4 \sin \left( \frac { 2 \pi t } { 365 } \right) - 8 \cos \left( \frac { 2 \pi t } { 365 } \right) , \quad 0 \leqslant t \leqslant 365$$
where $t$ is the number of days since he began recording.\\
(b) Using your answer to part (a), or otherwise, find the maximum and minimum number of hours of daylight given by this formula. Give your answers to 3 significant figures.\\
(c) Use the formula to find the values of $t$ when $H = 17$, giving your answers to the nearest integer.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\
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VIIIV SIHI NI JIIHM 10 N OC & VIIV 5141 NI 3114 M I ON OC & VI4V SIHIL NI JIIYM ION OC \\
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\hfill \mbox{\textit{Edexcel C34 2018 Q12 [12]}}