| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Linear transformation to find constants |
| Difficulty | Moderate -0.3 This is a standard C2 logarithmic transformation question with routine steps: verifying a percentage decrease, converting to log form (direct application of log laws), plotting points, reading intercept and gradient from a graph, and making a prediction. All techniques are textbook exercises requiring no problem-solving insight, though the multi-part structure and context make it slightly more substantial than the most basic questions. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context |
| Year | 1980 | 1985 | 1990 | 1995 | 2000 | 2005 |
| Population | 25000 | 22000 | 18750 | 16250 | 13500 | 12000 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 50% of 25 000 is 12 500 and the population [in 2005] is 12 000 [so consistent] | B1 | or 12 000 is 48% of 25 000 so less than 50% [so consistent] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\log_{10} P = \log_{10} a - kt\) or \(\log_{10}\frac{P}{a} = -kt\) o.e. www | B2 | condone omission of base; M1 for \(\log_{10} P = \log_{10} a + \log_{10} 10^{-kt}\) or better www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Values: \(4.27, 4.21, 4.13, 4.08\) | B1 | Accept \(4.273\ldots, 4.2108\ldots, 4.130\ldots, 4.079\ldots\) rounded to 2 or more dp |
| Plots | B1 | 1 mm tolerance; f.t. if at least two calculated values correct |
| Ruled line of best fit drawn | B1 | f.t. their values if at least 4 correct values are correctly plotted; must have at least one point on or above and at least one point on or below the line and must cover \(0 \leq t \leq 25\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(a = 25000\) to \(25400\) | B1 | Allow \(10^{4.4\ldots}\) |
| \(0.01 \leq k \leq 0.014\) | B2 | M1 for \(-k = \frac{\Delta y}{\Delta x}\) using values from table or graph; condone \(+k\); A1 for \(k\); A1 for \(a\) |
| \(P = a \times 10^{-kt}\) or \(P = 10^{\log a - kt}\) with values in acceptable ranges | B1 | B0 if left in logarithmic form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(P = a \times 10^{-0.35k}\) | M1 | Their \(a\) and \(k\) |
| \(8600\) to \(9000\) | A1 | |
| Comparing their value with \(9375\) o.e. and reaching the correct conclusion for their value | A1 | f.t. |
**Question 12(i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| 50% of 25 000 is 12 500 and the population [in 2005] is 12 000 [so consistent] | **B1** | or 12 000 is 48% of 25 000 so less than 50% [so consistent] | |
---
**Question 12(ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\log_{10} P = \log_{10} a - kt$ or $\log_{10}\frac{P}{a} = -kt$ o.e. www | **B2** | condone omission of base; **M1** for $\log_{10} P = \log_{10} a + \log_{10} 10^{-kt}$ or better www | |
## Question 12 (iii):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Values: $4.27, 4.21, 4.13, 4.08$ | **B1** | Accept $4.273\ldots, 4.2108\ldots, 4.130\ldots, 4.079\ldots$ rounded to 2 or more dp |
| Plots | **B1** | 1 mm tolerance; f.t. if at least two calculated values correct |
| Ruled line of best fit drawn | **B1** | f.t. their values if at least 4 correct values are correctly plotted; must have at least one point on or above and at least one point on or below the line and must cover $0 \leq t \leq 25$ |
## Question 12 (iv):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $a = 25000$ to $25400$ | **B1** | Allow $10^{4.4\ldots}$ | **M1** for a correct first step in solving a pair of valid equations in either form |
| $0.01 \leq k \leq 0.014$ | **B2** | **M1** for $-k = \frac{\Delta y}{\Delta x}$ using values from table or graph; condone $+k$; **A1** for $k$; **A1** for $a$ |
| $P = a \times 10^{-kt}$ or $P = 10^{\log a - kt}$ with values in acceptable ranges | **B1** | **B0** if left in logarithmic form | **A1** for $P = a \times 10^{-kt}$ |
## Question 12 (v):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $P = a \times 10^{-0.35k}$ | **M1** | Their $a$ and $k$ | Allow $\log P = \log a - 35k$ |
| $8600$ to $9000$ | **A1** | |
| Comparing their value with $9375$ o.e. and reaching the correct conclusion for their value | **A1** | f.t. |12 The table shows the size of a population of house sparrows from 1980 to 2005.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Year & 1980 & 1985 & 1990 & 1995 & 2000 & 2005 \\
\hline
Population & 25000 & 22000 & 18750 & 16250 & 13500 & 12000 \\
\hline
\end{tabular}
\end{center}
The 'red alert' category for birds is used when a population has decreased by at least $50 \%$ in the previous 25 years.\\
(i) Show that the information for this population is consistent with the house sparrow being on red alert in 2005.
The size of the population may be modelled by a function of the form $P = a \times 10 ^ { - k t }$, where $P$ is the population, $t$ is the number of years after 1980, and $a$ and $k$ are constants.\\
(ii) Write the equation $P = a \times 10 ^ { - k t }$ in logarithmic form using base 10, giving your answer as simply as possible.\\
(iii) Complete the table and draw the graph of $\log _ { 10 } P$ against $t$, drawing a line of best fit by eye.\\
(iv) Use your graph to find the values of $a$ and $k$ and hence the equation for $P$ in terms of $t$.\\
(v) Find the size of the population in 2015 as predicted by this model.
Would the house sparrow still be on red alert? Give a reason for your answer.
\hfill \mbox{\textit{OCR MEI C2 2011 Q12 [13]}}