| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Moderate -0.3 This is a standard log-linear modelling question requiring routine manipulation of logarithms and exponentials. Part (a) involves taking log₁₀ of both sides (straightforward application of log laws), parts (b-c) are direct substitution/calculation, and part (d) requires basic contextual reasoning about exponential growth limitations. Slightly easier than average due to being highly procedural with clear scaffolding. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\log_{10} N = \log_{10} A + kt\log_{10} 2\) | M1 | |
| Equation above is of the form \(y = mx + c\) [with \(\log_{10} N\) as \(y\) and \(t\) as \(x\)] | E1 | |
| Gradient \(= k\log_{10} 2\) | A1 | |
| Intercept \(= \log_{10} A\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k\log_{10} 2 = 0.2 \Rightarrow k = 0.66[438...]\) | B1 | |
| \(\log_{10} A = 2 \Rightarrow A = 100\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(N = 100 \times 2^{0.66...\times 24} = 6\,300\,000\) FT their \(A\), \(k\) | B1 | Answer in range \(5\,860\,000\) to \(6\,400\,000\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| E.g. the piece of bread may not be sufficient to support the number of bacteria | E1 | OR bacterial growth may obey different rules for large values of \(t\) |
# Question 9:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\log_{10} N = \log_{10} A + kt\log_{10} 2$ | M1 | |
| Equation above is of the form $y = mx + c$ [with $\log_{10} N$ as $y$ and $t$ as $x$] | E1 | |
| Gradient $= k\log_{10} 2$ | A1 | |
| Intercept $= \log_{10} A$ | A1 | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k\log_{10} 2 = 0.2 \Rightarrow k = 0.66[438...]$ | B1 | |
| $\log_{10} A = 2 \Rightarrow A = 100$ | B1 | |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $N = 100 \times 2^{0.66...\times 24} = 6\,300\,000$ FT their $A$, $k$ | B1 | Answer in range $5\,860\,000$ to $6\,400\,000$ |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| E.g. the piece of bread may not be sufficient to support the number of bacteria | E1 | OR bacterial growth may obey different rules for large values of $t$ |
---9 A biologist is investigating the growth of bacteria in a piece of bread.\\
He believes that the number, $N$, of bacteria after $t$ hours may be modelled by the relationship $N = A \times 2 ^ { k t }$, where $A$ and $k$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Show that, according to the model, the graph of $\log _ { 10 } N$ against $t$ is a straight line.
Give, in terms of $A$ and $k$,
\begin{itemize}
\item the gradient of the line
\item the intercept on the vertical axis.
\end{itemize}
The biologist measures the number of bacteria at regular intervals over 22 hours and plots a graph of $\log _ { 10 } N$ against $t$. He finds that the graph is approximately a straight line with gradient 0.20 . The line crosses the vertical axis at 2.0 .
\item Find the values of $A$ and $k$.
\item Use the model to predict the number of bacteria after 24 hours.
\item Give a reason why the model may not be appropriate for large values of $t$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 Q9 [8]}}