| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Interpret or complete given sketch of two |linear| functions |
| Difficulty | Moderate -0.3 This is a straightforward modulus function question requiring sketching two linear modulus functions and finding their intersection. Part (a) is simple substitution, (b) requires reading the graph's turning point, (c) involves solving |t-3|+4 > 8-|2t-6| by considering cases (standard technique), and (d) asks for model critique. While multi-part, each component uses routine A-level techniques without requiring novel insight, making it slightly easier than average. |
| Spec | 1.02h Express solutions: using 'and', 'or', set and interval notation1.02l Modulus function: notation, relations, equations and inequalities1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(N_A - N_B = (3+4)-(8-6) = \ldots\) | M1 | Uses models at \(t=0\); allow slips evaluating \(N_A\) and \(N_B\) |
| 5000 (subscribers) | A1 | 5000 or 5 thousand (subscribers); 5 is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((T=)3\) | B1 | Just look for number 3; e.g. \(t>3\) or "just after 3" acceptable |
| This was the point when company A had the lowest number of subscribers | B1 | Any acceptable reason e.g. minimum, vertex, gradient becomes positive, graph changes direction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-t+7=2t+2\) o.e. or \(t+1=14-2t\) o.e. | B1 | Forms one valid equation |
| \(-t+7=2t+2 \Rightarrow t=\ldots\) or \(t+1=14-2t \Rightarrow t=\ldots\) | M1 | Attempts to solve one valid equation |
| One of the two critical values \(t=\frac{5}{3}\) or \(t=\frac{13}{3}\) | A1 | |
| Both of \(t < ``\frac{5}{3}"\), \(t > ``\frac{13}{3}"\) | A1ft | Chooses outside region for their two values of \(t\) |
| \(\left\{t \in \mathbb{R} : t < \frac{5}{3}\right\} \cup \left\{t \in \mathbb{R} : t > \frac{13}{3}\right\}\) | A1 | Fully correct solution in set notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((-t+7)^2=(2t+2)^2\) o.e. or \((t+1)^2=(14-2t)^2\) o.e. | B1 | |
| Solves to give \(t=\ldots\) (gives \(-9\) and \(\frac{5}{3}\), or \(15\) and \(\frac{13}{3}\)) | M1 | |
| One of \(t=\frac{5}{3}\) or \(t=\frac{13}{3}\) | A1 | |
| Both \(t<``\frac{5}{3}"\), \(t>``\frac{13}{3}"\) | A1ft | |
| \(\left\{t \in \mathbb{R} : t < \frac{5}{3}\right\} \cup \left\{t \in \mathbb{R} : t > \frac{13}{3}\right\}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ | t-3\ | +4=8-\ |
| \((3t-9)^2=4^2 \Rightarrow 9t^2-54t+65=0 \Rightarrow t=\ldots\) (gives \(\frac{5}{3}\) and \(\frac{13}{3}\)) | M1 | |
| One of \(t=\frac{5}{3}\) or \(t=\frac{13}{3}\) | A1 | |
| Both \(t<``\frac{5}{3}"\), \(t>``\frac{13}{3}"\) | A1ft | |
| \(\left\{t \in \mathbb{R} : t < \frac{5}{3}\right\} \cup \left\{t \in \mathbb{R} : t > \frac{13}{3}\right\}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The number of subscribers will become negative (when \(t>7\)) | B1 | Any indication subscribers become negative e.g. "allows negative subscribers (which isn't possible)"; not "subscribers will become zero" |
# Question 12:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $N_A - N_B = (3+4)-(8-6) = \ldots$ | **M1** | Uses models at $t=0$; allow slips evaluating $N_A$ and $N_B$ |
| 5000 (subscribers) | **A1** | 5000 or 5 thousand (subscribers); 5 is A0 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(T=)3$ | **B1** | Just look for number 3; e.g. $t>3$ or "just after 3" acceptable |
| This was the point when company A had the lowest number of subscribers | **B1** | Any acceptable reason e.g. minimum, vertex, gradient becomes positive, graph changes direction |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-t+7=2t+2$ o.e. or $t+1=14-2t$ o.e. | **B1** | Forms one valid equation |
| $-t+7=2t+2 \Rightarrow t=\ldots$ or $t+1=14-2t \Rightarrow t=\ldots$ | **M1** | Attempts to solve one valid equation |
| One of the two critical values $t=\frac{5}{3}$ or $t=\frac{13}{3}$ | **A1** | |
| Both of $t < ``\frac{5}{3}"$, $t > ``\frac{13}{3}"$ | **A1ft** | Chooses outside region for their two values of $t$ |
| $\left\{t \in \mathbb{R} : t < \frac{5}{3}\right\} \cup \left\{t \in \mathbb{R} : t > \frac{13}{3}\right\}$ | **A1** | Fully correct solution in set notation |
### Squaring Method - Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(-t+7)^2=(2t+2)^2$ o.e. or $(t+1)^2=(14-2t)^2$ o.e. | **B1** | |
| Solves to give $t=\ldots$ (gives $-9$ and $\frac{5}{3}$, or $15$ and $\frac{13}{3}$) | **M1** | |
| One of $t=\frac{5}{3}$ or $t=\frac{13}{3}$ | **A1** | |
| Both $t<``\frac{5}{3}"$, $t>``\frac{13}{3}"$ | **A1ft** | |
| $\left\{t \in \mathbb{R} : t < \frac{5}{3}\right\} \cup \left\{t \in \mathbb{R} : t > \frac{13}{3}\right\}$ | **A1** | |
### Squaring Method - Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\|t-3\|+4=8-\|2t-6\| \Rightarrow \|t-3\|+\|2t-6\|=4 \Rightarrow 3t-9=4$ o.e. | **B1** | |
| $(3t-9)^2=4^2 \Rightarrow 9t^2-54t+65=0 \Rightarrow t=\ldots$ (gives $\frac{5}{3}$ and $\frac{13}{3}$) | **M1** | |
| One of $t=\frac{5}{3}$ or $t=\frac{13}{3}$ | **A1** | |
| Both $t<``\frac{5}{3}"$, $t>``\frac{13}{3}"$ | **A1ft** | |
| $\left\{t \in \mathbb{R} : t < \frac{5}{3}\right\} \cup \left\{t \in \mathbb{R} : t > \frac{13}{3}\right\}$ | **A1** | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The number of subscribers will become negative (when $t>7$) | **B1** | Any indication subscribers become negative e.g. "allows negative subscribers (which isn't possible)"; not "subscribers will become zero" |12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3f6f3f19-a1d0-488b-a1a4-302cc4cf5a1e-34_643_652_210_708}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The number of subscribers to two different music streaming companies is being monitored.
The number of subscribers, $N _ { \mathrm { A } }$, in thousands, to company $\mathbf { A }$ is modelled by the equation
$$N _ { \mathrm { A } } = | t - 3 | + 4 \quad t \geqslant 0$$
where $t$ is the time in years since monitoring began.\\
The number of subscribers, $N _ { \mathrm { B } }$, in thousands, to company B is modelled by the equation
$$N _ { \mathrm { B } } = 8 - | 2 t - 6 | \quad t \geqslant 0$$
where $t$ is the time in years since monitoring began.\\
Figure 2 shows a sketch of the graph of $N _ { \mathrm { A } }$ and the graph of $N _ { \mathrm { B } }$ over a 5-year period.\\
Use the equations of the models to answer parts (a), (b), (c) and (d).
\begin{enumerate}[label=(\alph*)]
\item Find the initial difference between the number of subscribers to company $\mathbf { A }$ and the number of subscribers to company B.
When $t = T$ company A reduced its subscription prices and the number of subscribers increased.
\item Suggest a value for $T$, giving a reason for your answer.
\item Find the range of values of $t$ for which $N _ { \mathrm { A } } > N _ { \mathrm { B } }$ giving your answer in set notation.
\item State a limitation of the model used for company B.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2023 Q12 [10]}}