| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Critique single model appropriateness |
| Difficulty | Moderate -0.3 This is a straightforward application of exponential modelling with standard depreciation context. Part (a) requires simple substitution to find the decay factor (0.8), part (b) asks for basic model evaluation by comparing predicted vs actual values, and part (c) tests conceptual understanding of the decay parameter. While it involves multiple parts and some interpretation, it requires no novel insight—just routine application of V = V₀r^t and comparison of numerical outputs. |
| Spec | 1.02z Models in context: use functions in modelling1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses model \(V = Ae^{\pm kt}\) | M1 | Do not allow if \(k\) is fixed; condone different variable names |
| Substitutes \(t=0, V=20000 \Rightarrow A=20000\) | M1 | Candidates writing \(V=20000e^{kt}\) directly score M1 M1 |
| Substitutes \(t=1, V=16000 \Rightarrow 16000 = 20000e^{-k} \Rightarrow k = \ldots\) | dM1 | Dependent on both previous M's; must use logs correctly |
| \(V = 20000e^{-0.223t}\) | A1 | Accuracy to at least 3sf; or \(V=20000e^{t\ln 0.8}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(t=10\) into \(V = 20000e^{-0.223t} \Rightarrow V = (\pounds 2150)\) | M1 | Or substitutes \(V=2000\) and finds \(t\) |
| Model is reliable as \(\pounds 2150 \approx \pounds 2000\) | A1 | Must compare to \(\pounds 2000\) with reason; do not allow "not reliable because it is not the same" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Make the \(-0.223\) less negative; or adapt model e.g. \(V = 18000e^{-0.223t}+2000\) | B1ft | Follow through on their constant; condone not stating \(-0.223\) lies outside \((-0.223, 0)\) |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses model $V = Ae^{\pm kt}$ | M1 | Do not allow if $k$ is fixed; condone different variable names |
| Substitutes $t=0, V=20000 \Rightarrow A=20000$ | M1 | Candidates writing $V=20000e^{kt}$ directly score M1 M1 |
| Substitutes $t=1, V=16000 \Rightarrow 16000 = 20000e^{-k} \Rightarrow k = \ldots$ | dM1 | Dependent on both previous M's; must use logs correctly |
| $V = 20000e^{-0.223t}$ | A1 | Accuracy to at least 3sf; or $V=20000e^{t\ln 0.8}$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $t=10$ into $V = 20000e^{-0.223t} \Rightarrow V = (\pounds 2150)$ | M1 | Or substitutes $V=2000$ and finds $t$ |
| Model is reliable as $\pounds 2150 \approx \pounds 2000$ | A1 | Must compare to $\pounds 2000$ with reason; do not allow "not reliable because it is not the same" |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Make the $-0.223$ less negative; or adapt model e.g. $V = 18000e^{-0.223t}+2000$ | B1ft | Follow through on their constant; condone not stating $-0.223$ lies outside $(-0.223, 0)$ |\begin{enumerate}
\item In a simple model, the value, $\pounds V$, of a car depends on its age, $t$, in years.
\end{enumerate}
The following information is available for $\operatorname { car } A$
\begin{itemize}
\item its value when new is $\pounds 20000$
\item its value after one year is $\pounds 16000$\\
(a) Use an exponential model to form, for car $A$, a possible equation linking $V$ with $t$.
\end{itemize}
The value of car $A$ is monitored over a 10-year period.\\
Its value after 10 years is $\pounds 2000$\\
(b) Evaluate the reliability of your model in light of this information.
The following information is available for car $B$
\begin{itemize}
\item it has the same value, when new, as car $A$
\item its value depreciates more slowly than that of $\operatorname { car } A$\\
(c) Explain how you would adapt the equation found in (a) so that it could be used to model the value of car $B$.
\end{itemize}
\hfill \mbox{\textit{Edexcel Paper 1 2019 Q7 [7]}}