| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Show constant equals specific form |
| Difficulty | Moderate -0.3 This is a standard exponential modelling question requiring substitution of initial conditions to find A, algebraic manipulation with logarithms to show k equals a specific form, and interpretation of the model's asymptotic behaviour. All techniques are routine for C3/C4 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=0, \theta=38 \Rightarrow 38 = 20 + Ae^{-k \times 0}\) | M1 | For substituting \(t=0\) and \(\theta=38\) into \(\theta = 20 + Ae^{-kt}\) |
| \(\Rightarrow A = 18\) | A1 | Correct value for \(A\). \(A=18\) with no working scores both marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=16, \theta=24.5 \Rightarrow 24.5 = 20 + \text{"18"}e^{-k\times16}\) | M1 | For substituting \(t=16\) and \(\theta=24.5\) into \(\theta=20+\text{their "A"}e^{-kt}\) |
| \(\Rightarrow 18e^{-k\times16} = 4.5\) or \(e^{-k\times16} = \frac{1}{4}\) | A1 | This mark is for a correct equation with the constants combined. Allow equivalent correct equations e.g. \(e^{16k}=4\) |
| \(\Rightarrow e^{16k}=4 \Rightarrow 16k = \ln 4\) or \(\Rightarrow \ln18e^{-k\times16} = \ln4.5 \Rightarrow \ln18 + \ln e^{-k\times16} = \ln4.5 \Rightarrow \ln e^{-k\times16} = \ln\frac{1}{4} \Rightarrow -16k = \ln\frac{1}{4}\) | M1 | Uses correct log or exponential work to move from: \(e^{\pm nk}=C\) to \(\pm nk = \alpha \ln C\) or \(pe^{\pm nk}=q\) to \(\pm nk = \alpha \ln \beta\) |
| \(-16k = \ln\frac{1}{4} \Rightarrow k = -\frac{1}{16}\ln\frac{1}{4} = \frac{1}{8}\ln 2\)* | A1* | Shows that \(k=\frac{1}{8}\ln 2\). There must be at least one intermediate line between their \(\pm nk = \alpha \ln C\) or their \(\pm nk = \alpha \ln \beta\) and the printed answer. So for example \(-16k = \ln\frac{1}{4} \Rightarrow k=\frac{1}{8}\ln 2\)* scores A0 as there is no intermediate line. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=40 \Rightarrow \theta = 20 + \text{"18"}e^{\frac{1}{8}\ln2\times40}\) | M1 | Substitutes \(t=40\) into the given equation with their \(A\) and the given value of \(k\) to obtain a value for \(\theta\) |
| \(\Rightarrow \theta = \text{awrt } 20.6\,(°C)\) | A1 | Awrt 20.6. Correct answer only scores both marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. The lower limit is 20; \(\theta > 20\); As \(t\) tends to infinity temperature tends to 20; The temperature cannot go below 20; \(e^{-kt}\) tends towards zero so the temperature tends to 20; \(e^{-kt}\) is always positive so the temperature is always bigger than 20; Substitutes \(\theta=19\) in \(\theta=20+\text{"18"}e^{-kt}\) (may be implied by e.g. \(e^{-kt}=-\frac{1}{18}\)) and states e.g. that you cannot find the log of a negative number or "which is not possible". Do not accept \(e^{-kt}\) cannot be negative without reference to the "20" | B1 |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=0, \theta=38 \Rightarrow 38 = 20 + Ae^{-k \times 0}$ | M1 | For substituting $t=0$ and $\theta=38$ into $\theta = 20 + Ae^{-kt}$ |
| $\Rightarrow A = 18$ | A1 | Correct value for $A$. **$A=18$ with no working scores both marks** |
---
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=16, \theta=24.5 \Rightarrow 24.5 = 20 + \text{"18"}e^{-k\times16}$ | M1 | For substituting $t=16$ and $\theta=24.5$ into $\theta=20+\text{their "A"}e^{-kt}$ |
| $\Rightarrow 18e^{-k\times16} = 4.5$ or $e^{-k\times16} = \frac{1}{4}$ | A1 | This mark is for a correct equation with the **constants combined.** Allow equivalent correct equations e.g. $e^{16k}=4$ |
| $\Rightarrow e^{16k}=4 \Rightarrow 16k = \ln 4$ or $\Rightarrow \ln18e^{-k\times16} = \ln4.5 \Rightarrow \ln18 + \ln e^{-k\times16} = \ln4.5 \Rightarrow \ln e^{-k\times16} = \ln\frac{1}{4} \Rightarrow -16k = \ln\frac{1}{4}$ | M1 | Uses correct log or exponential work to move from: $e^{\pm nk}=C$ to $\pm nk = \alpha \ln C$ **or** $pe^{\pm nk}=q$ to $\pm nk = \alpha \ln \beta$ |
| $-16k = \ln\frac{1}{4} \Rightarrow k = -\frac{1}{16}\ln\frac{1}{4} = \frac{1}{8}\ln 2$* | A1* | Shows that $k=\frac{1}{8}\ln 2$. There must be **at least one intermediate line** between their $\pm nk = \alpha \ln C$ or their $\pm nk = \alpha \ln \beta$ and the printed answer. So for example $-16k = \ln\frac{1}{4} \Rightarrow k=\frac{1}{8}\ln 2$* scores A0 as there is no intermediate line. |
---
## Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=40 \Rightarrow \theta = 20 + \text{"18"}e^{\frac{1}{8}\ln2\times40}$ | M1 | Substitutes $t=40$ into the given equation with their $A$ and the given value of $k$ to obtain a value for $\theta$ |
| $\Rightarrow \theta = \text{awrt } 20.6\,(°C)$ | A1 | Awrt 20.6. **Correct answer only scores both marks** |
---
## Question 5(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. The lower limit is 20; $\theta > 20$; As $t$ tends to infinity temperature tends to 20; The temperature cannot go below 20; $e^{-kt}$ tends towards zero so the temperature tends to 20; $e^{-kt}$ is always positive so the temperature is always bigger than 20; Substitutes $\theta=19$ in $\theta=20+\text{"18"}e^{-kt}$ (may be implied by e.g. $e^{-kt}=-\frac{1}{18}$) and states e.g. that you cannot find the log of a negative number or "which is not possible". **Do not accept** $e^{-kt}$ cannot be negative without reference to the "20" | B1 | |
---5. A bath is filled with hot water. The temperature, $\theta ^ { \circ } \mathrm { C }$, of the water in the bath, $t$ minutes after the bath has been filled, is given by
$$\theta = 20 + A \mathrm { e } ^ { - k t }$$
where $A$ and $k$ are positive constants.
Given that the temperature of the water in the bath is initially $38 ^ { \circ } \mathrm { C }$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $A$.
The temperature of the water in the bath 16 minutes after the bath has been filled is $24.5 ^ { \circ } \mathrm { C }$.
\item Show that $k = \frac { 1 } { 8 } \ln 2$
Using the values for $k$ and $A$,
\item find the temperature of the water 40 minutes after the bath has been filled, giving your answer to 3 significant figures.
\item Explain why the temperature of the water in the bath cannot fall to $19 ^ { \circ } \mathrm { C }$.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2019 Q5 [9]}}