| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Linear transformation to find constants |
| Difficulty | Moderate -0.3 This is a standard C3 exponential modelling question requiring linearisation via logarithms, plotting points, reading gradient/intercept from a graph, and basic substitution. The techniques are routine (taking ln of both sides, using y=mx+c form) with no novel problem-solving required, though it involves multiple steps across several parts making it slightly more substantial than the most basic exercises. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form2.02c Scatter diagrams and regression lines |
| Time \(( t\) years \()\) | 0 | 1 | 2 | 3 | 4 |
| Number of bats, \(P\) | 100 | 170 | 300 | 340 | 360 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P = P_0 e^{kt} \Rightarrow \ln P = \ln P_0 + kt\) | M1A1 | |
| So if \(y\) is identified with \(\ln P\), \(m\) with \(k\) and \(x\) with \(t\), we have \(y = mx + c\); gradient \(k\), intercept \(\ln P_0\) | E1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Table of values: \(t=0, P=100, \ln P = 4.61\); \(t=1, P=170, \ln P=5.14\); \(t=2, P=300, \ln P=5.70\) | B1 | For table of values |
| Points plotted | B1 | Points plotted |
| Straight line drawn | B1 | Straight line |
| Gradient estimate \(k \approx 0.55\) | M1, A1 | Gradient |
| Intercept about \(4.61\); so \(P_0 = \exp(4.61) = 100\), to 2sf | B1 | \(P_0\) |
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Table and curve for 1st model | M1, A1 | For 1st curve |
| Table and curve for 2nd model | M1, A1 | For 2nd curve |
| Values: \(t=0: 52.2\); \(t=1: 170\); \(t=2: 287.8\); \(t=3: 336.1\); \(t=4: 357.4\) | ||
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(375 = 150\arctan(t-1) + 170\) | B1 | |
| \(\Rightarrow t = 1 + \tan\frac{205}{150}\) | M1, A1 | |
| \(= 5.83\ldots\) So the population will exceed 375 in 6 years | B1 | |
| Total: 4 |
## Question 9:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P = P_0 e^{kt} \Rightarrow \ln P = \ln P_0 + kt$ | M1A1 | |
| So if $y$ is identified with $\ln P$, $m$ with $k$ and $x$ with $t$, we have $y = mx + c$; gradient $k$, intercept $\ln P_0$ | E1 | |
| | **Total: 3** | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Table of values: $t=0, P=100, \ln P = 4.61$; $t=1, P=170, \ln P=5.14$; $t=2, P=300, \ln P=5.70$ | B1 | For table of values |
| Points plotted | B1 | Points plotted |
| Straight line drawn | B1 | Straight line |
| Gradient estimate $k \approx 0.55$ | M1, A1 | Gradient |
| Intercept about $4.61$; so $P_0 = \exp(4.61) = 100$, to 2sf | B1 | $P_0$ |
| | **Total: 6** | |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Table and curve for 1st model | M1, A1 | For 1st curve |
| Table and curve for 2nd model | M1, A1 | For 2nd curve |
| Values: $t=0: 52.2$; $t=1: 170$; $t=2: 287.8$; $t=3: 336.1$; $t=4: 357.4$ | | |
| | **Total: 5** | |
### Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $375 = 150\arctan(t-1) + 170$ | B1 | |
| $\Rightarrow t = 1 + \tan\frac{205}{150}$ | M1, A1 | |
| $= 5.83\ldots$ So the population will exceed 375 in 6 years | B1 | |
| | **Total: 4** | |9 Answer parts (ii) and (iii) of this question on the Insert provided.
The bat population of a colony is being investigated and data are collected of the estimated number of bats in the colony at the beginning of each year.
It is thought that the population may be modelled by the formula
$$P = P _ { 0 } \mathrm { e } ^ { k t }$$
where $P _ { 0 }$ and $k$ are constants, $P$ is the number of bats and $t$ is the number of years after the start of the collection of data.\\
(i) Explain why a graph of $\ln P$ against $t$ should give a straight line. State the gradient and intercept of this line.\\
(ii) The data collected are as follows.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | }
\hline
Time $( t$ years $)$ & 0 & 1 & 2 & 3 & 4 \\
\hline
Number of bats, $P$ & 100 & 170 & 300 & 340 & 360 \\
\hline
\end{tabular}
\end{center}
Using the first three pairs of data in the table, plot $\ln P$ against $t$ on the axes given on the Insert, and hence estimate values for $P _ { 0 }$ and $k$.\\
(Work to three significant figures.)
This model assumes exponential growth, and assumes that once born a bat does not die, continuing to reproduce. This is unrealistic and so a second model is proposed with formula
$$P = 150 \arctan ( t - 1 ) + 170$$
(You are reminded that arctan values should be given in radians.)\\
(iii) Plot on a single graph on the Insert the curves $P = P _ { 0 } \mathrm { e } ^ { k t }$ for your values of $P _ { 0 }$ and $k$ and $P = 150 \arctan ( t - 1 ) + 170$. The data pairs in the table above have been plotted for you.\\
(iv) Using the second model calculate an estimate of the number of years it is before the bat population exceeds 375.
\section*{Insert for question 3.}
(i) Sketch the graph of $y = 2 \mathrm { f } ( x )$\\
\includegraphics[max width=\textwidth, alt={}, center]{3853d1e7-ae1f-4eca-93c7-96f03b6d31c3-6_641_1431_541_354}\\
(ii) Sketch the graph of $y = \mathrm { f } ( 2 x )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{3853d1e7-ae1f-4eca-93c7-96f03b6d31c3-6_691_1539_1468_374}
\section*{Insert for question 9.}
(ii) Plot $\ln P$ against $t$.\\
\includegraphics[max width=\textwidth, alt={}, center]{3853d1e7-ae1f-4eca-93c7-96f03b6d31c3-7_704_1442_443_338}\\
(iii) Plot the curves $P = P _ { 0 } \mathrm { e } ^ { k t }$ and $P = 150 \arctan ( t - 1 ) + 170$ for your values of $P _ { 0 }$ and $k$. The data pairs are plotted on the graph.\\
\includegraphics[max width=\textwidth, alt={}, center]{3853d1e7-ae1f-4eca-93c7-96f03b6d31c3-7_780_1399_1546_333}
\hfill \mbox{\textit{OCR MEI C3 Q9 [18]}}