| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.3 This is a standard exponential decay modeling question with shifted asymptote. Part (a) requires simple substitution at t=0, part (b) involves logarithms to solve for B, part (c) requires differentiation and evaluation (routine calculus), and part (d) tests understanding of horizontal asymptotes. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.06i Exponential growth/decay: in modelling context1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((A=)8\) | B1 | Award for \((T=)10+8e^{-Bt}\); "A=" not required, just look for 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(16 = 10 + \text{"8"}e^{-Bt} \Rightarrow e^{-Bt} = \ldots\) | M1 | Sets up equation and rearranges to form \(e^{-Bt}=\ldots\) with or without \(t=45\) |
| \(e^{-45B} = \frac{3}{4} \Rightarrow -45B = \ln\frac{3}{4} \Rightarrow B = \ldots\) | dM1 | Uses \(t=45\), takes ln of both sides; depends on previous M |
| \(B = \text{awrt } 0.00639\) | A1 | e.g. \(\frac{1}{45}\ln\frac{4}{3}\) or \(-\frac{1}{45}\ln\frac{3}{4}\); correct answer only scores no marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dT}{dt} = -\text{"8"}\times\frac{1}{45}\ln\!\left(\frac{4}{3}\right)\times e^{-\frac{1}{45}\ln\left(\frac{4}{3}\right)\times 2} = \ldots\) | M1 | Differentiates using chain rule to obtain \(\ldots e^{-Bt}\) (not \(\ldots e^{Bt}\)) and substitutes \(t=2\) |
| \(= -0.0505\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The temperature has a (lower) limit of \(10\,°C\), or: \(5 = 10 + \text{"8"}e^{-Bt} \Rightarrow e^{-Bt} = -\frac{5}{\text{"8"}}\) which is not possible / cannot find log of negative number | B1 | Either interpretation accepted |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(A=)8$ | B1 | Award for $(T=)10+8e^{-Bt}$; "A=" not required, just look for 8 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $16 = 10 + \text{"8"}e^{-Bt} \Rightarrow e^{-Bt} = \ldots$ | M1 | Sets up equation and rearranges to form $e^{-Bt}=\ldots$ with or without $t=45$ |
| $e^{-45B} = \frac{3}{4} \Rightarrow -45B = \ln\frac{3}{4} \Rightarrow B = \ldots$ | dM1 | Uses $t=45$, takes ln of both sides; depends on previous M |
| $B = \text{awrt } 0.00639$ | A1 | e.g. $\frac{1}{45}\ln\frac{4}{3}$ or $-\frac{1}{45}\ln\frac{3}{4}$; correct answer only scores no marks |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dT}{dt} = -\text{"8"}\times\frac{1}{45}\ln\!\left(\frac{4}{3}\right)\times e^{-\frac{1}{45}\ln\left(\frac{4}{3}\right)\times 2} = \ldots$ | M1 | Differentiates using chain rule to obtain $\ldots e^{-Bt}$ (not $\ldots e^{Bt}$) and substitutes $t=2$ |
| $= -0.0505$ | A1 | |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The temperature has a (lower) limit of $10\,°C$, or: $5 = 10 + \text{"8"}e^{-Bt} \Rightarrow e^{-Bt} = -\frac{5}{\text{"8"}}$ which is not possible / cannot find log of negative number | B1 | Either interpretation accepted |\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying entirely on calculator technology are not acceptable.}
The temperature, $T ^ { \circ } \mathrm { C }$, of the air in a room $t$ minutes after a heat source is switched off, is modelled by the equation
$$T = 10 + A \mathrm { e } ^ { - B t }$$
where $A$ and $B$ are constants.\\
Given that the temperature of the air in the room at the instant the heat source was switched off was $18 ^ { \circ } \mathrm { C }$,\\
(a) find the value of $A$
Given also that, exactly 45 minutes after the heat source was switched off, the temperature of the air in the room was $16 ^ { \circ } \mathrm { C }$,\\
(b) find the value of $B$ to 3 significant figures.
Using the values for $A$ and $B$,\\
(c) find, according to the model, the rate of change of the temperature of the air in the room exactly two minutes after the heat source was switched off.\\
Give your answer in ${ } ^ { \circ } \mathrm { C } \min ^ { - 1 }$ to 3 significant figures.\\
(d) Explain why, according to the model, the temperature of the air in the room cannot fall to $5 ^ { \circ } \mathrm { C }$
\hfill \mbox{\textit{Edexcel P3 2024 Q5 [7]}}