| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Half-life and doubling time |
| Difficulty | Moderate -0.5 This is a straightforward application of exponential models with logarithmic transformation. Part (a) requires recognizing that ln P = ln a + kt gives gradient k and intercept ln a, both directly readable from the graph. Part (b) is a standard doubling time calculation using logarithms. The question involves routine algebraic manipulation with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply \(\ln P = \ln a + kt\) or \(\ln P = \ln a + k(\ln e)t\) | B1 | Can be implied by both \(a\) and \(k\) correct; \(P = e^3 e^{\frac{1}{20}t}\) gets B1B1 |
| \(\ln P = \dfrac{1}{20}t + 3\), B0 until associated with \(a\) and/or \(k\) | ||
| State \(k = \dfrac{1}{20}\), not from \(\dfrac{dP}{dt} = k\) | B1 | OE. Can be embedded in \(P = ae^{kt}\) |
| \(\ln a = 3 \Rightarrow a = 20\) to 2 sf | B1 | Must be 2 sf; can be embedded in \(P = ae^{kt}\) |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Form a correct equation in \(t\) using \(a\) and \(k\), or their \(a\) and \(k\) where \(a\) will cancel (or are both numerical) | M1 | E.g. \(2a = ae^{kt}\), \(2 = e^{kt}\), \(kt = \ln 2\) |
| Obtain \(t = 14\) [hours] | A1 | Allow 13.75 [hrs] (13 hrs 45 min) to 14 [hrs]. ISW |
| Total | 2 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $\ln P = \ln a + kt$ or $\ln P = \ln a + k(\ln e)t$ | B1 | Can be implied by both $a$ and $k$ correct; $P = e^3 e^{\frac{1}{20}t}$ gets B1B1 |
| $\ln P = \dfrac{1}{20}t + 3$, B0 until associated with $a$ and/or $k$ | | |
| State $k = \dfrac{1}{20}$, not from $\dfrac{dP}{dt} = k$ | B1 | OE. Can be embedded in $P = ae^{kt}$ |
| $\ln a = 3 \Rightarrow a = 20$ to 2 sf | B1 | Must be 2 sf; can be embedded in $P = ae^{kt}$ |
| **Total** | **3** | |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Form a correct equation in $t$ using $a$ and $k$, or their $a$ and $k$ where $a$ will cancel (or are both numerical) | M1 | E.g. $2a = ae^{kt}$, $2 = e^{kt}$, $kt = \ln 2$ |
| Obtain $t = 14$ [hours] | A1 | Allow 13.75 [hrs] (13 hrs 45 min) to 14 [hrs]. ISW |
| **Total** | **2** | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-04_527_634_255_717}
The number of bacteria in a population, $P$, at time $t$ hours is modelled by the equation $P = a \mathrm { e } ^ { k t }$, where $a$ and $k$ are constants. The graph of $\ln P$ against $t$, shown in the diagram, has gradient $\frac { 1 } { 20 }$ and intersects the vertical axis at $( 0,3 )$.
\begin{enumerate}[label=(\alph*)]
\item State the value of $k$ and find the value of $a$ correct to 2 significant figures.
\item Find the time taken for $P$ to double. Give your answer correct to the nearest hour.\\
\includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-05_2723_33_99_22}
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q3 [5]}}