| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Linear transformation to find constants |
| Difficulty | Moderate -0.3 This is a structured multi-part question on exponential modeling with logarithmic linearization. Parts (a)-(d) are routine applications of standard techniques (interpreting constants, taking logs, solving simultaneous equations, substitution), while part (e) requires basic contextual reasoning. Slightly easier than average due to heavy scaffolding and straightforward calculations. |
| Spec | 1.02z Models in context: use functions in modelling1.06g Equations with exponentials: solve a^x = b1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Time (minutes) | 10 | 20 |
| Temperature (degrees Celsius) | 30 | 12 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = 0\) gives \(\theta = A\); \(A\) is the starting temperature of the water | R1 | Substitutes \(t=0\) to obtain \(\theta = A\), states \(10^{-kt}=1\) at \(t=0\), and infers \(A\) is initial temperature |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\log_{10}\theta = \log_{10}A + \log_{10}10^{-kt} = \log_{10}A - kt\) | B1 | Uses logarithms correctly. Must see clear evidence of use. \(\log_{10}A \times \log_{10}10^{-kt}\) scores B0. AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=10, \theta=30\); \(t=20, \theta=12\) substituted to form at least one correct equation | M1 | |
| \(\log_{10}30 = \log_{10}A - 10k\) and \(\log_{10}12 = \log_{10}A - 20k\) | A1 | Two correct equations |
| \(k = \frac{1}{10}\log_{10}2.5 = 0.0398\) | M1 | Solves equations for exact \(k\). ACF or AWFW 0.039 to 0.04 |
| \(A = 75\) | A1 | AWRT 75 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(75 \times 10^{-0.039 \times 45}\) | M1 | Substitutes their \(k\), \(A\) and \(t=45\) into equation |
| \(= 1.2\); \(1.2 > 1\) | A1 | \(\theta\) AWFW 1.18 to 1.32, or \(t = 47.1\) AWFW 46.8 to 48.1 |
| Model does not support Zena's statement | R1 | Compares AWFW 1.18 to 1.32 with 1, or AWFW 46.8 to 48.1 with 45 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. After 45 minutes the outside temperature may have changed; model implies water never cools to \(0°C\); other factors (e.g. air currents); not enough measurements taken; water behaves differently near \(0°C\); we don't know what happens after \(t=20\) | E1 | States a valid problem with the model. Do not accept reference to rates of change unless fully qualified |
## Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = 0$ gives $\theta = A$; $A$ is the starting temperature of the water | R1 | Substitutes $t=0$ to obtain $\theta = A$, states $10^{-kt}=1$ at $t=0$, and infers $A$ is initial temperature |
## Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_{10}\theta = \log_{10}A + \log_{10}10^{-kt} = \log_{10}A - kt$ | B1 | Uses logarithms correctly. Must see clear evidence of use. $\log_{10}A \times \log_{10}10^{-kt}$ scores B0. AG |
## Question 10(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=10, \theta=30$; $t=20, \theta=12$ substituted to form at least one correct equation | M1 | |
| $\log_{10}30 = \log_{10}A - 10k$ and $\log_{10}12 = \log_{10}A - 20k$ | A1 | Two correct equations |
| $k = \frac{1}{10}\log_{10}2.5 = 0.0398$ | M1 | Solves equations for exact $k$. ACF or AWFW 0.039 to 0.04 |
| $A = 75$ | A1 | AWRT 75 |
## Question 10(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $75 \times 10^{-0.039 \times 45}$ | M1 | Substitutes their $k$, $A$ and $t=45$ into equation |
| $= 1.2$; $1.2 > 1$ | A1 | $\theta$ AWFW 1.18 to 1.32, or $t = 47.1$ AWFW 46.8 to 48.1 |
| Model does not support Zena's statement | R1 | Compares AWFW 1.18 to 1.32 with 1, or AWFW 46.8 to 48.1 with 45 |
## Question 10(e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. After 45 minutes the outside temperature may have changed; model implies water never cools to $0°C$; other factors (e.g. air currents); not enough measurements taken; water behaves differently near $0°C$; we don't know what happens after $t=20$ | E1 | States a valid problem with the model. Do **not** accept reference to rates of change unless fully qualified |
---10 As part of an experiment, Zena puts a bucket of hot water outside on a day when the outside temperature is $0 ^ { \circ } \mathrm { C }$.
She measures the temperature of the water after 10 minutes and after 20 minutes. Her results are shown below.
\begin{center}
\begin{tabular}{ | l | l | l | }
\hline
Time (minutes) & 10 & 20 \\
\hline
Temperature (degrees Celsius) & 30 & 12 \\
\hline
\end{tabular}
\end{center}
Zena models the relationship between $\theta$, the temperature of the water in ${ } ^ { \circ } \mathrm { C }$, and $t$, the time in minutes, by
$$\theta = A \times 10 ^ { - k t }$$
where $A$ and $k$ are constants.
10
\begin{enumerate}[label=(\alph*)]
\item Using $t = 0$, explain how the value of $A$ relates to the experiment.
10
\item Show that
$$\log _ { 10 } \theta = \log _ { 10 } A - k t$$
10
\item Using Zena's results, calculate the values of $A$ and $k$.\\
10
\item Zena states that the temperature of the water will be less than $1 ^ { \circ } \mathrm { C }$ after 45 minutes. Determine whether the model supports this statement.\\
10
\item Explain why Zena's model is unlikely to accurately give the value of $\theta$ after 45 minutes.
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2019 Q10 [10]}}