| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.8 This is a straightforward exponential model question requiring basic substitution (finding P when t=0), identifying the horizontal asymptote (long-term behavior as tââ), and solving a simple exponential equation for k. All steps are routine C3 techniques with no problem-solving insight needed, making it easier than average but not trivial since it requires understanding asymptotic behavior. |
| Spec | 1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(t=0\), \(P=7-2=5\), so 5 (million) | B1 | |
| In the long term \(e^{-kt}\to 0\) | M1 | allow substituting a large number for \(t\) (for both marks) |
| So long-term population is 7 (million) | A1 [3] | allow 7 unsupported |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P=7-2e^{-kt}\); when \(t=1\), \(P=5.5\) | ||
| \(\Rightarrow 5.5=7-2e^{-k}\) | M1 | |
| \(\Rightarrow e^{-k}=(7-5.5)/2=0.75\) | ||
| \(\Rightarrow -k=\ln((7-5.5)/2)\) | M1 | re-arranging and anti-logging â allow 1 slip (e.g. arith of \(7-5.5\), or \(k\) for \(-k\)) or \(\ln 2 - k = \ln 1.5\) o.e.; but penalise negative lns, e.g. \(\ln(-1.5)=\ln(-2)-k\) |
| \(\Rightarrow k=0.288\) (3 s.f.) | A1 [3] | 0.3 or better; allow \(\ln(4/3)\) or \(-\ln(3/4)\) if final ans; rounding from correct value \(k=0.2876820725\ldots\), penalise truncation and incorrect work with negatives |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t=0$, $P=7-2=5$, so 5 (million) | B1 | |
| In the long term $e^{-kt}\to 0$ | M1 | allow substituting a large number for $t$ (for both marks) |
| So long-term population is 7 (million) | A1 **[3]** | allow 7 unsupported |
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## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P=7-2e^{-kt}$; when $t=1$, $P=5.5$ | | |
| $\Rightarrow 5.5=7-2e^{-k}$ | M1 | |
| $\Rightarrow e^{-k}=(7-5.5)/2=0.75$ | | |
| $\Rightarrow -k=\ln((7-5.5)/2)$ | M1 | re-arranging and anti-logging â allow 1 slip (e.g. arith of $7-5.5$, or $k$ for $-k$) or $\ln 2 - k = \ln 1.5$ o.e.; but penalise negative lns, e.g. $\ln(-1.5)=\ln(-2)-k$ |
| $\Rightarrow k=0.288$ (3 s.f.) | A1 **[3]** | 0.3 or better; allow $\ln(4/3)$ or $-\ln(3/4)$ if final ans; rounding from correct value $k=0.2876820725\ldots$, penalise truncation and incorrect work with negatives |
---5 A termites' nest has a population of $P$ million. $P$ is modelled by the equation $P = 7 - 2 \mathrm { e } ^ { - k t }$, where $t$ is in years, and $k$ is a positive constant.\\
(i) Calculate the population when $t = 0$, and the long-term population, given by this model.\\
(ii) Given that the population when $t = 1$ is estimated to be 5.5 million, calculate the value of $k$.
\hfill \mbox{\textit{OCR MEI C3 2012 Q5 [6]}}