| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.8 This is a straightforward AS-level exponential modelling question requiring only direct substitution (parts a,b), solving a simple exponential equation using logarithms (part b), understanding asymptotic behaviour (part c), and reading an asymptote from a graph (part d). All techniques are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure and need for logarithms. |
| Spec | 1.02z Models in context: use functions in modelling1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Temperature \(= 83°C\) | B1 | Must include units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(18+65e^{-\frac{t}{8}}=35 \Rightarrow 65e^{-\frac{t}{8}}=17\) | M1 | Uses information, proceeds to \(Pe^{\pm\frac{t}{8}}=Q\); condone slips |
| \(t = -8\ln\left(\frac{17}{65}\right)\) or \(\ln65 - \frac{t}{8}=\ln17 \Rightarrow t=\ldots\) | dM1 | Full method using correct log laws; \(e^x\) and \(\ln x\) are inverses; condone one error |
| \(t = 10.7\) | A1 | awrt \(10.7\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As \(t\to\infty\), \(\theta\to18\) from above; minimum temperature is \(18°C\) | B1 | States suitable reason with minimal conclusion; e.g. substituting \(\theta=15\) shows \(e^{-t/8}\) cannot be negative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A+B=94\) or \(A+Be^{-1}=50\) | M1 | Uses model to form at least one equation |
| \(A+B=94\) and \(A+Be^{-1}=50\) | A1 | Both equations correct |
| Full method to find at least one value for \(A\) | dM1 | Dependent on first M |
| Deduces \(\mu = \frac{50e-94}{e-1}\) or accept \(\mu=\) awrt \(24.4\) | A1 |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Temperature $= 83°C$ | B1 | Must include units |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $18+65e^{-\frac{t}{8}}=35 \Rightarrow 65e^{-\frac{t}{8}}=17$ | M1 | Uses information, proceeds to $Pe^{\pm\frac{t}{8}}=Q$; condone slips |
| $t = -8\ln\left(\frac{17}{65}\right)$ or $\ln65 - \frac{t}{8}=\ln17 \Rightarrow t=\ldots$ | dM1 | Full method using correct log laws; $e^x$ and $\ln x$ are inverses; condone one error |
| $t = 10.7$ | A1 | awrt $10.7$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $t\to\infty$, $\theta\to18$ from above; minimum temperature is $18°C$ | B1 | States suitable reason with minimal conclusion; e.g. substituting $\theta=15$ shows $e^{-t/8}$ cannot be negative |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A+B=94$ or $A+Be^{-1}=50$ | M1 | Uses model to form at least one equation |
| $A+B=94$ and $A+Be^{-1}=50$ | A1 | Both equations correct |
| Full method to find at least one value for $A$ | dM1 | Dependent on first M |
| Deduces $\mu = \frac{50e-94}{e-1}$ or accept $\mu=$ awrt $24.4$ | A1 | |\begin{enumerate}
\item The temperature, $\theta ^ { \circ } \mathrm { C }$, of a cup of tea $t$ minutes after it was placed on a table in a room, is modelled by the equation
\end{enumerate}
$$\theta = 18 + 65 \mathrm { e } ^ { - \frac { t } { 8 } } \quad t \geqslant 0$$
Find, according to the model,\\
(a) the temperature of the cup of tea when it was placed on the table,\\
(b) the value of $t$, to one decimal place, when the temperature of the cup of tea was $35 ^ { \circ } \mathrm { C }$.\\
(c) Explain why, according to this model, the temperature of the cup of tea could not fall to $15 ^ { \circ } \mathrm { C }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bcbd842f-b2e2-4587-ab4c-15a57a449e5d-16_675_951_973_573}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The temperature, $\mu ^ { \circ } \mathrm { C }$, of a second cup of tea $t$ minutes after it was placed on a table in a different room, is modelled by the equation
$$\mu = A + B \mathrm { e } ^ { - \frac { t } { 8 } } \quad t \geqslant 0$$
where $A$ and $B$ are constants.\\
Figure 2 shows a sketch of $\mu$ against $t$ with two data points that lie on the curve.\\
The line $l$, also shown on Figure 2, is the asymptote to the curve.\\
Using the equation of this model and the information given in Figure 2\\
(d) find an equation for the asymptote $l$.
\hfill \mbox{\textit{Edexcel AS Paper 1 2020 Q8 [9]}}