| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Standard +0.3 This is a straightforward exponential model question requiring: (a) substitution of t=0, (b) differentiation and solving a simple exponential equation using logarithms, (c) identifying the horizontal asymptote, and (d) stating a model limitation. All techniques are standard AS-level procedures with clear scaffolding, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06b Gradient of e^(kx): derivative and exponential model1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| £18 000 | B1 | No requirement to have units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dV}{dt} = -3925e^{-0.25t}\) | M1 | Award for making link between gradient and rate of change; attempt to differentiate \(V\) to \(\frac{dV}{dt} = ke^{-0.25t}\) required on both sides |
| \(\frac{dV}{dt} = -3925e^{-0.25t}\) correct | A1 | Achieves \(\frac{dV}{dt} = -3925e^{-0.25t}\) or \(\frac{dV}{dt} = 15700 \times -0.25e^{-0.25t}\) |
| Sets \(-3925e^{-0.25T} = -500 \Rightarrow 3925e^{-0.25T} = 500\) | A1* | Given answer; all aspects must be correct; \(t\) must be changed to \(T\) at some point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-0.25T} = 0.127... \Rightarrow -0.25T = \ln 0.127...\) | M1 | Proceeds from \(e^{-0.25T} = A,\ A > 0\) using ln's to \(\pm 0.25T = \ldots\) |
| \(T = 8.24\) (awrt) | A1 | \(T = \text{awrt } 8.24\) or \(-\frac{1}{0.25}\ln\left(\frac{20}{157}\right)\); allow \(t = \text{awrt } 8.24\) |
| 8 years 3 months | A1 | Correct answer and solution only; answers obtained numerically score 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 2300 | B1 | Condone £2 300 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Any suitable reason e.g. other factors affect price (condition/mileage); accident damage; price may rise as car becomes rare; £2300 too large a scrap value | B1 | Accept "Scrappage schemes may pay more (or less) than £2 300"; do not accept "does not take into account inflation"; cannot score by suggesting alternative models |
## Question 14:
### Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| £18 000 | B1 | No requirement to have units |
### Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dV}{dt} = -3925e^{-0.25t}$ | M1 | Award for making link between gradient and rate of change; attempt to differentiate $V$ to $\frac{dV}{dt} = ke^{-0.25t}$ required on both sides |
| $\frac{dV}{dt} = -3925e^{-0.25t}$ correct | A1 | Achieves $\frac{dV}{dt} = -3925e^{-0.25t}$ or $\frac{dV}{dt} = 15700 \times -0.25e^{-0.25t}$ |
| Sets $-3925e^{-0.25T} = -500 \Rightarrow 3925e^{-0.25T} = 500$ | A1* | Given answer; all aspects must be correct; $t$ must be changed to $T$ at some point |
### Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-0.25T} = 0.127... \Rightarrow -0.25T = \ln 0.127...$ | M1 | Proceeds from $e^{-0.25T} = A,\ A > 0$ using ln's to $\pm 0.25T = \ldots$ |
| $T = 8.24$ (awrt) | A1 | $T = \text{awrt } 8.24$ or $-\frac{1}{0.25}\ln\left(\frac{20}{157}\right)$; allow $t = \text{awrt } 8.24$ |
| 8 years 3 months | A1 | Correct answer and solution only; answers obtained numerically score 0 |
### Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| 2300 | B1 | Condone £2 300 |
### Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| Any suitable reason e.g. other factors affect price (condition/mileage); accident damage; price may rise as car becomes rare; £2300 too large a scrap value | B1 | Accept "Scrappage schemes may pay more (or less) than £2 300"; do not accept "does not take into account inflation"; cannot score by suggesting alternative models |
---\begin{enumerate}
\item The value of a car, $\pounds V$, can be modelled by the equation
\end{enumerate}
$$V = 15700 \mathrm { e } ^ { - 0.25 t } + 2300 \quad t \in \mathbb { R } , t \geqslant 0$$
where the age of the car is $t$ years.\\
Using the model,\\
(a) find the initial value of the car.
Given the model predicts that the value of the car is decreasing at a rate of $\pounds 500$ per year at the instant when $t = T$,\\
(b) (i) show that
$$3925 \mathrm { e } ^ { - 0.25 T } = 500$$
(ii) Hence find the age of the car at this instant, giving your answer in years and months to the nearest month.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
The model predicts that the value of the car approaches, but does not fall below, $\pounds A$.\\
(c) State the value of $A$.\\
(d) State a limitation of this model.
\hfill \mbox{\textit{Edexcel AS Paper 1 2019 Q14 [9]}}