| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Time to reach target in exponential model |
| Difficulty | Moderate -0.8 This is a straightforward exponential model question requiring only basic substitution (part a) and solving a simple exponential equation using logarithms (part b). Both parts are routine C3 techniques with no problem-solving insight needed, making it easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(20 \text{ (mm}^2)\) | B1 | Sight of 20 relating to value of \(A\) at \(t=0\); accept its sight in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(40 = 20\,e^{1.5t} \rightarrow e^{1.5t} = \frac{40}{20} = 2\) | A1 | Substitutes \(A=40\) or twice answer to (a) and proceeds to \(e^{1.5t} = \text{constant}\); accept non-numerical answers |
| \(1.5t = \ln 2 \rightarrow t = \frac{\ln c}{1.5}\) | M1 | Correct ln work; order must be correct; accept non-numerical answers |
| \(t = \frac{\ln 2}{1.5}\) (awrt \(0.46\)) | A1 | Achieves \(\frac{\ln(2)}{1.5}\) or awrt 0.46 (2sf) |
| \(12.28\) or \(28\) (minutes) | A1 | Either 12:28 or 28 minutes; cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln(40) = \ln 20 + 1.5t\) | M1, A1 | Substitutes \(A=40\) and takes ln of both sides; proceeds to \(\ln(40) = \ln 20 + \ln e^{1.5t}\) |
| \(\ln(40) - \ln 20 = 1.5t\) or \(\ln\!\left(\frac{40}{20}\right) = 1.5t \rightarrow t =\) | M1 | Make \(t\) the subject with correct \(\ln\) work |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $20 \text{ (mm}^2)$ | B1 | Sight of 20 relating to value of $A$ at $t=0$; accept its sight in (b) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $40 = 20\,e^{1.5t} \rightarrow e^{1.5t} = \frac{40}{20} = 2$ | A1 | Substitutes $A=40$ or twice answer to (a) and proceeds to $e^{1.5t} = \text{constant}$; accept non-numerical answers |
| $1.5t = \ln 2 \rightarrow t = \frac{\ln c}{1.5}$ | M1 | Correct ln work; order must be correct; accept non-numerical answers |
| $t = \frac{\ln 2}{1.5}$ (awrt $0.46$) | A1 | Achieves $\frac{\ln(2)}{1.5}$ or awrt 0.46 (2sf) |
| $12.28$ or $28$ (minutes) | A1 | Either 12:28 or 28 minutes; cao |
**Alt 1 – taking ln of both sides on line 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln(40) = \ln 20 + 1.5t$ | M1, A1 | Substitutes $A=40$ and takes ln of both sides; proceeds to $\ln(40) = \ln 20 + \ln e^{1.5t}$ |
| $\ln(40) - \ln 20 = 1.5t$ or $\ln\!\left(\frac{40}{20}\right) = 1.5t \rightarrow t =$ | M1 | Make $t$ the subject with correct $\ln$ work |
---3. The area, $A \mathrm {~mm} ^ { 2 }$, of a bacterial culture growing in milk, $t$ hours after midday, is given by
$$A = 20 \mathrm { e } ^ { 1.5 t } , \quad t \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Write down the area of the culture at midday.
\item Find the time at which the area of the culture is twice its area at midday. Give your answer to the nearest minute.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2012 Q3 [6]}}