| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Show constant equals specific form |
| Difficulty | Standard +0.3 This is a standard exponential modelling question requiring substitution of initial conditions, algebraic manipulation with logarithms, and differentiation. Part (a) is straightforward substitution, part (b) requires basic log manipulation to reach the required form, and parts (c)-(d) apply the found constants. All techniques are routine for P3 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(18 = A - 180 \times 1 \Rightarrow A = \ldots\) | M1 | Substitutes \(\theta=18\), \(t=0\), uses \(e^0=1\) |
| \(A = 198\) | A1 | Condone \(A=198°C\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(90 = 198 - 180e^{-5k} \Rightarrow 180e^{-5k} = 108\) | M1, A1 | Substitutes \(\theta=90\), \(t=5\) with their \(A\); correct equation |
| \(e^{-5k} = \frac{108}{180} \Rightarrow -5k = \ln 0.6 \Rightarrow k = \ldots\) | dM1 | Correct order of operations using ln; do not award if using \(\log_{10}\) |
| \(k = -\frac{1}{5}\ln\frac{3}{5}\) or \(k = \frac{1}{5}\ln\frac{5}{3}\) | A1 | Accept awrt \(0.102\); allow equivalent exact forms e.g. \(k = -0.2\ln 0.6\), \(k=\frac{1}{5}\ln\frac{180}{108}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\theta = 198 - 180e^{-9\times\frac{1}{5}\ln\frac{5}{3}} \Rightarrow \theta = \ldots\) | M1 | Substitutes their \(A\) and \(k\) with \(t=9\) |
| \(\theta = 126°C\) (awrt) | A1 | Condone lack of units; awrt 126 following \(k=\) awrt \(0.102\) sufficient |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{d\theta}{dt} = -180\left(-\frac{1}{5}\ln\frac{5}{3}\right)e^{-\frac{t}{5}\ln\frac{5}{3}} = \ldots\) | B1ft | Correct differentiation; follow through their \(k\); allow \(180ke^{-kt}\) |
| \(\frac{d\theta}{dt} = -180\left(-\frac{1}{5}\ln\frac{5}{3}\right)e^{-\frac{9}{5}\ln\frac{5}{3}} = \ldots\) | M1 | Substitutes \(t=9\) into \(\beta e^{-kt}\) form, following through their \(k\) |
| \(= 7.33\ °C\ \text{min}^{-1}\) (awrt) | A1 | Ignore units; award B0 M1 A1 for \(\pm\) awrt 7.33 following sign error in \(\frac{d\theta}{dt}\) |
# Question 5:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $18 = A - 180 \times 1 \Rightarrow A = \ldots$ | M1 | Substitutes $\theta=18$, $t=0$, uses $e^0=1$ |
| $A = 198$ | A1 | Condone $A=198°C$ |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $90 = 198 - 180e^{-5k} \Rightarrow 180e^{-5k} = 108$ | M1, A1 | Substitutes $\theta=90$, $t=5$ with their $A$; correct equation |
| $e^{-5k} = \frac{108}{180} \Rightarrow -5k = \ln 0.6 \Rightarrow k = \ldots$ | dM1 | Correct order of operations using ln; do not award if using $\log_{10}$ |
| $k = -\frac{1}{5}\ln\frac{3}{5}$ or $k = \frac{1}{5}\ln\frac{5}{3}$ | A1 | Accept awrt $0.102$; allow equivalent exact forms e.g. $k = -0.2\ln 0.6$, $k=\frac{1}{5}\ln\frac{180}{108}$ |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\theta = 198 - 180e^{-9\times\frac{1}{5}\ln\frac{5}{3}} \Rightarrow \theta = \ldots$ | M1 | Substitutes their $A$ and $k$ with $t=9$ |
| $\theta = 126°C$ (awrt) | A1 | Condone lack of units; awrt 126 following $k=$ awrt $0.102$ sufficient |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d\theta}{dt} = -180\left(-\frac{1}{5}\ln\frac{5}{3}\right)e^{-\frac{t}{5}\ln\frac{5}{3}} = \ldots$ | B1ft | Correct differentiation; follow through their $k$; allow $180ke^{-kt}$ |
| $\frac{d\theta}{dt} = -180\left(-\frac{1}{5}\ln\frac{5}{3}\right)e^{-\frac{9}{5}\ln\frac{5}{3}} = \ldots$ | M1 | Substitutes $t=9$ into $\beta e^{-kt}$ form, following through their $k$ |
| $= 7.33\ °C\ \text{min}^{-1}$ (awrt) | A1 | Ignore units; award B0 M1 A1 for $\pm$ awrt 7.33 following sign error in $\frac{d\theta}{dt}$ |
---5. The temperature, $\theta ^ { \circ } \mathrm { C }$, inside an oven, $t$ minutes after the oven is switched on, is given by
$$\theta = A - 180 \mathrm { e } ^ { - k t }$$
where $A$ and $k$ are positive constants.
Given that the temperature inside the oven is initially $18 ^ { \circ } \mathrm { C }$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $A$.
The temperature inside the oven, 5 minutes after the oven is switched on, is $90 ^ { \circ } \mathrm { C }$.
\item Show that $k = p \ln q$ where $p$ and $q$ are rational numbers to be found.
Hence find
\item the temperature inside the oven 9 minutes after the oven is switched on, giving your answer to 3 significant figures,
\item the rate of increase of the temperature inside the oven 9 minutes after the oven is switched on. Give your answer in ${ } ^ { \circ } \mathrm { C } \min ^ { - 1 }$ to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2021 Q5 [11]}}