| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Graphs & Exact Values |
| Type | Real-world modelling (tides, daylight, etc.) |
| Difficulty | Standard +0.3 This is a straightforward application of sine function properties to a real-world context. Part (a) requires finding minimum value (amplitude manipulation), part (b) involves solving a basic trigonometric inequality, and part (c) is qualitative interpretation. All parts use standard A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin\!\left(\dfrac{2\pi(t+101.75)}{365}\right) = -1\), giving \(-3.87 + 11.7 = 7.83\), i.e. 7 hours 50 mins | M1 (AO 3.4) | Uses \(\sin = -1\) in model to obtain \(-3.87 + 11.7\); if \(t\) value used then sine must evaluate to \(-1\); or differentiates, sets derivative to 0 and substitutes back |
| 470 minutes or \(\frac{47}{6}\) or \(7\frac{5}{6}\) hours | A1 (AO 3.2a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3.87\sin\!\left(\dfrac{2\pi(t+101.75)}{365}\right) + 11.7 = 14\) | M1 (AO 3.4) | Uses model to form equation/inequality with \(H=14\); condone incorrect inequality |
| \(t = 300.22\) or \(t = 408.77\) | A1 (AO 1.1b) | At least two correct values of \(t\); can be rounded or truncated e.g. \(-64.77\), \(43.779\), \(300.22\), \(408.77\) |
| \(408 - 300 = 108\) | A1 (AO 3.2a) | Subtracts appropriate pair of \(t\) values; accept 109 or 107; alternative: \(43 + (365-300) = 108\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sofia's refinement would increase the range of the graph | M1 (AO 3.3) | Accept: "range of the graph would increase"; "it would increase the fluctuation of the graph" |
| Sofia's graph suggests this is not the case, so the refinement is not appropriate | A1 (AO 3.5c) | Must explain refinement is not appropriate as her data/graph suggests a lower amplitude |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin\!\left(\dfrac{2\pi(t+101.75)}{365}\right) = -1$, giving $-3.87 + 11.7 = 7.83$, i.e. 7 hours 50 mins | M1 (AO 3.4) | Uses $\sin = -1$ in model to obtain $-3.87 + 11.7$; if $t$ value used then sine must evaluate to $-1$; or differentiates, sets derivative to 0 and substitutes back |
| 470 minutes or $\frac{47}{6}$ or $7\frac{5}{6}$ hours | A1 (AO 3.2a) | |
---
## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3.87\sin\!\left(\dfrac{2\pi(t+101.75)}{365}\right) + 11.7 = 14$ | M1 (AO 3.4) | Uses model to form equation/inequality with $H=14$; condone incorrect inequality |
| $t = 300.22$ or $t = 408.77$ | A1 (AO 1.1b) | At least two correct values of $t$; can be rounded or truncated e.g. $-64.77$, $43.779$, $300.22$, $408.77$ |
| $408 - 300 = 108$ | A1 (AO 3.2a) | Subtracts appropriate pair of $t$ values; accept 109 or 107; alternative: $43 + (365-300) = 108$ |
---
## Question 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sofia's refinement would increase the range of the graph | M1 (AO 3.3) | Accept: "range of the graph would increase"; "it would increase the fluctuation of the graph" |
| Sofia's graph suggests this is not the case, so the refinement is not appropriate | A1 (AO 3.5c) | Must explain refinement is not appropriate as her data/graph suggests a lower amplitude |
---8 Mike, an amateur astronomer who lives in the South of England, wants to know how the number of hours of darkness changes through the year.
On various days between February and September he records the length of time, $H$ hours, of darkness along with $t$, the number of days after 1 January.
His results are shown in Figure 1 below.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-08_940_1541_696_246}
\end{center}
\end{figure}
Mike models this data using the equation
$$H = 3.87 \sin \left( \frac { 2 \pi ( t + 101.75 ) } { 365 } \right) + 11.7$$
8
\begin{enumerate}[label=(\alph*)]
\item Find the minimum number of hours of darkness predicted by Mike's model. Give your answer to the nearest minute.\\[0pt]
[2 marks]
8
\item Find the maximum number of consecutive days where the number of hours of darkness predicted by Mike's model exceeds 14\\
8
\item Mike's friend Sofia, who lives in Spain, also records the number of hours of darkness on various days throughout the year.
Her results are shown in Figure 2 below.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-10_933_1537_561_248}
\end{center}
\end{figure}
Sofia attempts to model her data by refining Mike's model.\\
She decides to increase the 3.87 value, leaving everything else unchanged.\\
Explain whether Sofia's refinement is appropriate.\\
\includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-11_2488_1730_219_141}\\
$9 \quad$ Chloe is attempting to write $\frac { 2 x ^ { 2 } + x } { ( x + 1 ) ( x + 2 ) ^ { 2 } }$ as partial fractions, with constant numerators.
Her incorrect attempt is shown below.
Step 1
$$\frac { 2 x ^ { 2 } + x } { ( x + 1 ) ( x + 2 ) ^ { 2 } } \equiv \frac { A } { x + 1 } + \frac { B } { ( x + 2 ) ^ { 2 } }$$
Step 2
$$2 x ^ { 2 } + x \equiv A ( x + 2 ) ^ { 2 } + B ( x + 1 )$$
Step 3
$$\begin{aligned}
& \text { Let } x = - 1 \Rightarrow A = 1 \\
& \text { Let } x = - 2 \Rightarrow B = - 6
\end{aligned}$$
Answer
$$\frac { 2 x ^ { 2 } + x } { ( x + 1 ) ( x + 2 ) ^ { 2 } } \equiv \frac { 1 } { x + 1 } - \frac { 6 } { ( x + 2 ) ^ { 2 } }$$
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2020 Q8 [7]}}