| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | October |
| Marks | 7 |
| 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. Part (a) uses the routine R sin(x + α) formula with R = √(1² + 2²) = √5 and tan α = 2. Parts (b) and (c) directly apply this result to find maximum temperature (5 + √5) and time by solving a simple equation. All steps are algorithmic with no novel insight required, making it slightly easier than average. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = \sqrt{5}\) | B1 | \(R=\sqrt{5}\) only |
| \(\tan\alpha = 2 \Rightarrow \alpha = \ldots\) | M1 | Proceeds to value of \(\alpha\) from \(\tan\alpha=\pm 2\), \(\tan\alpha=\pm\frac{1}{2}\), \(\sin\alpha=\pm\frac{2}{"R"}\) or \(\cos\alpha=\pm\frac{1}{"R"}\). Implied by awrt 1.11 (radians) or 63.4 (degrees). |
| \(\alpha = 1.107\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\theta = 5+\sqrt{5}\sin\!\left(\frac{\pi t}{12}+1.107-3\right)\), maximum is \((5+\sqrt{5})\ ^\circ\text{C}\) or awrt \(7.24\ ^\circ\text{C}\) | B1ft | Deduces maximum temperature is \((5+\sqrt{5})\ ^\circ\text{C}\). Follow through on their \(R\), so allow \((5+\text{"R"})\ ^\circ\text{C}\). Condone lack of units. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{\pi t}{12}+1.107-3=\dfrac{\pi}{2} \Rightarrow t=\ldots\) | M1 | Complete strategy to find \(t\). Follow through on their 1.107 but angle must be in radians. |
| \(t = \text{awrt } 13.2\) | A1 | |
| Either 13:14 or 1:14 pm or 13 hours 14 minutes after midnight | A1 | Accept 13:14, 1:14 pm, 13h 14, or 1 hour 14 minutes after midday. |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = \sqrt{5}$ | B1 | $R=\sqrt{5}$ only |
| $\tan\alpha = 2 \Rightarrow \alpha = \ldots$ | M1 | Proceeds to value of $\alpha$ from $\tan\alpha=\pm 2$, $\tan\alpha=\pm\frac{1}{2}$, $\sin\alpha=\pm\frac{2}{"R"}$ or $\cos\alpha=\pm\frac{1}{"R"}$. Implied by awrt 1.11 (radians) or 63.4 (degrees). |
| $\alpha = 1.107$ | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\theta = 5+\sqrt{5}\sin\!\left(\frac{\pi t}{12}+1.107-3\right)$, maximum is $(5+\sqrt{5})\ ^\circ\text{C}$ or awrt $7.24\ ^\circ\text{C}$ | B1ft | Deduces maximum temperature is $(5+\sqrt{5})\ ^\circ\text{C}$. Follow through on their $R$, so allow $(5+\text{"R"})\ ^\circ\text{C}$. Condone lack of units. |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{\pi t}{12}+1.107-3=\dfrac{\pi}{2} \Rightarrow t=\ldots$ | M1 | Complete strategy to find $t$. Follow through on their 1.107 but angle must be in radians. |
| $t = \text{awrt } 13.2$ | A1 | |
| Either 13:14 or 1:14 pm or 13 hours 14 minutes after midnight | A1 | Accept 13:14, 1:14 pm, 13h 14, or 1 hour 14 minutes after midday. |\begin{enumerate}
\item (a) Express $\sin x + 2 \cos x$ in the form $R \sin ( x + \alpha )$ where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$\\
Give the exact value of $R$ and give the value of $\alpha$ in radians to 3 decimal places.
\end{enumerate}
The temperature, $\theta ^ { \circ } \mathrm { C }$, inside a room on a given day is modelled by the equation
$$\theta = 5 + \sin \left( \frac { \pi t } { 12 } - 3 \right) + 2 \cos \left( \frac { \pi t } { 12 } - 3 \right) \quad 0 \leqslant t < 24$$
where $t$ is the number of hours after midnight.\\
Using the equation of the model and your answer to part (a),\\
(b) deduce the maximum temperature of the room during this day,\\
(c) find the time of day when the maximum temperature occurs, giving your answer to the nearest minute.
\hfill \mbox{\textit{Edexcel Paper 1 2020 Q6 [7]}}