| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.3 This is a straightforward exponential model question requiring standard techniques: substitution to find A, solving a logarithmic equation, differentiation for rate of change, and interpretation of the model's limitations. All parts are routine P3 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.07j Differentiate exponentials: e^(kx) and a^(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = 93\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(100 = 125 - \text{"93"}e^{-0.109T} \Rightarrow Ae^{-0.109T} = ...\) | M1 | Sets \(100 = 125 - \text{"93"}e^{-0.109T}\) and proceeds. Condone slips. Candidates using 100 000 score no marks |
| \(Ae^{kT} = B \Rightarrow T = \dfrac{\ln\!\left(\dfrac{B}{A}\right)}{k}\) | dM1 | Uses correct order of operations and correct log work from \(Ae^{kT}=B\), \(AB>0\) |
| \(T = 12.05\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dN}{dt} = 0.109 \times \text{"93"}e^{-0.109\times7}\) | M1 | Differentiates to form \(\lambda e^{-0.109t}\), \(\lambda\) constant, substitutes \(t=7\) |
| \(4730\) (total sales per month) | A1 | awrt 4730. Accept 4.73 thousand. "Sales per month" may be omitted but A0 if incorrect unit given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The limit is 125 000 / the model has a limit below 150 000 | B1 | A correct reason. Accept: limit is 125 000; \(N\) cannot reach 150; asymptote at 125. Score B0 if "negative time" reached and used as reason, or if 150 000 used in equation |
# Question 4:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = 93$ | B1 | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $100 = 125 - \text{"93"}e^{-0.109T} \Rightarrow Ae^{-0.109T} = ...$ | M1 | Sets $100 = 125 - \text{"93"}e^{-0.109T}$ and proceeds. Condone slips. Candidates using 100 000 score no marks |
| $Ae^{kT} = B \Rightarrow T = \dfrac{\ln\!\left(\dfrac{B}{A}\right)}{k}$ | dM1 | Uses correct order of operations and correct log work from $Ae^{kT}=B$, $AB>0$ |
| $T = 12.05$ | A1 | cao |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dN}{dt} = 0.109 \times \text{"93"}e^{-0.109\times7}$ | M1 | Differentiates to form $\lambda e^{-0.109t}$, $\lambda$ constant, substitutes $t=7$ |
| $4730$ (total sales per month) | A1 | awrt 4730. Accept 4.73 thousand. "Sales per month" may be omitted but A0 if incorrect unit given |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| The limit is 125 000 / the model has a limit below 150 000 | B1 | A correct reason. Accept: limit is 125 000; $N$ cannot reach 150; asymptote at 125. Score B0 if "negative time" reached and used as reason, or if 150 000 used in equation |
---\begin{enumerate}
\item A new mobile phone is released for sale.
\end{enumerate}
The total sales $N$ of this phone, in thousands, is modelled by the equation
$$N = 125 - A \mathrm { e } ^ { - 0.109 t } \quad t \geqslant 0$$
where $A$ is a constant and $t$ is the time in months after the phone was released for sale.\\
Given that when $t = 0 , N = 32$\\
(a) state the value of $A$.
Given that when $t = T$ the total sales of the phone was 100000\\
(b) find, according to the model, the value of $T$. Give your answer to 2 decimal places.\\
(c) Find, according to the model, the rate of increase in total sales when $t = 7$, giving your answer to 3 significant figures.\\
(Solutions relying entirely on calculator technology are not acceptable.)
The total sales of the mobile phone is expected to reach 150000\\
Using this information,\\
(d) give a reason why the given equation is not suitable for modelling the total sales of the phone.
\hfill \mbox{\textit{Edexcel P3 2023 Q4 [7]}}