| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.3 This is a standard Newton's Law of Cooling application requiring substitution to find A, solving a logarithmic equation, and model evaluation. All techniques are routine for A-level: finding constants from initial conditions, rearranging exponentials, and basic critical thinking about model validity. Slightly easier than average due to straightforward structure and clear signposting. |
| Spec | 1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{t=0,\ \theta=75 \Rightarrow 75=25+A \Rightarrow A=50\} \Rightarrow \theta = 25 + 50e^{-0.03t}\) | B1 | AO3.3; Applies \(t=0\), \(\theta=75\) to give complete model \(\theta = 25 + 50e^{-0.03t}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{\theta=60\} \Rightarrow 60 = 25 + \text{"50"}e^{-0.03t} \Rightarrow e^{-0.03t} = \frac{60-25}{\text{"50"}}\) | M1 | AO3.4; Applies \(\theta=60\) and their value of \(A\), rearranges to make \(e^{-0.03t}\) the subject. Note: later working can imply this mark |
| \(t = \frac{\ln(0.7)}{-0.03} = 11.8891... = 11.9\) minutes (1 dp) | A1 | AO1.1b; Obtains 11.9 (minutes) with no errors in manipulation seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| A valid evaluation of the model relating to large values of \(t\). E.g. As \(20.3 < 25\) the model is not true for large values of \(t\); or \(e^{-0.03t} = \frac{20.3-25}{\text{"50"}} = -0.094\) does not have any solutions; or \(t=120 \Rightarrow \theta = 25 + 50e^{-0.03(120)} = 26.36...\) which is not approximately equal to 20.3 | B1 | AO3.5a |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{t=0,\ \theta=75 \Rightarrow 75=25+A \Rightarrow A=50\} \Rightarrow \theta = 25 + 50e^{-0.03t}$ | B1 | AO3.3; Applies $t=0$, $\theta=75$ to give complete model $\theta = 25 + 50e^{-0.03t}$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{\theta=60\} \Rightarrow 60 = 25 + \text{"50"}e^{-0.03t} \Rightarrow e^{-0.03t} = \frac{60-25}{\text{"50"}}$ | M1 | AO3.4; Applies $\theta=60$ and their value of $A$, rearranges to make $e^{-0.03t}$ the subject. Note: later working can imply this mark |
| $t = \frac{\ln(0.7)}{-0.03} = 11.8891... = 11.9$ minutes (1 dp) | A1 | AO1.1b; Obtains 11.9 (minutes) with no errors in manipulation seen |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| A valid evaluation of the model relating to large values of $t$. E.g. As $20.3 < 25$ the model is not true for large values of $t$; or $e^{-0.03t} = \frac{20.3-25}{\text{"50"}} = -0.094$ does not have any solutions; or $t=120 \Rightarrow \theta = 25 + 50e^{-0.03(120)} = 26.36...$ which is not approximately equal to 20.3 | B1 | AO3.5a |
---\begin{enumerate}
\item A cup of hot tea was placed on a table. At time $t$ minutes after the cup was placed on the table, the temperature of the tea in the cup, $\theta ^ { \circ } \mathrm { C }$, is modelled by the equation
\end{enumerate}
$$\theta = 25 + A \mathrm { e } ^ { - 0.03 t }$$
where $A$ is a constant.
The temperature of the tea was $75 ^ { \circ } \mathrm { C }$ when the cup was placed on the table.\\
(a) Find a complete equation for the model.\\
(b) Use the model to find the time taken for the tea to cool from $75 ^ { \circ } \mathrm { C }$ to $60 ^ { \circ } \mathrm { C }$, giving your answer in minutes to one decimal place.
Two hours after the cup was placed on the table, the temperature of the tea was measured as $20.3 ^ { \circ } \mathrm { C }$.
Using this information,\\
(c) evaluate the model, explaining your reasoning.
\hfill \mbox{\textit{Edexcel Paper 2 Q3 [4]}}