| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential growth/decay model setup |
| Difficulty | Moderate -0.8 This is a straightforward exponential model question requiring only direct substitution (part a), routine logarithm manipulation (part b), and standard differentiation of an exponential function (part c). All three parts are textbook exercises with no problem-solving or novel insight required, making it easier than average for A-level. |
| Spec | 1.06b Gradient of e^(kx): derivative and exponential model1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3500\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3500(1.035)^t > 10000 \Rightarrow (1.035)^t > \frac{20}{7}\) (awrt 2.86) | M1 A1 | M1: Substitutes \(N=10000\), proceeds to \((1.035)^t \ldots A\) |
| \(\Rightarrow t > \frac{\log\frac{20}{7}}{\log 1.035} = 30.516 =\) 30 hrs 31 mins or 30 hrs 32 mins | M1 A1 | M1: Proceeds correctly to find value of \(t\); A1: 30 hrs 31 mins or 30 hrs 32 mins (not 1831 minutes) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dN}{dt} = 3500(1.035)^t \ln 1.035 \Rightarrow \frac{dN}{dt}\bigg | _{t=8} = 3500(1.035)^8 \ln 1.035 =\) awrt 159 | B1 M1 A1 |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3500$ | B1 | |
**(1 mark)**
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3500(1.035)^t > 10000 \Rightarrow (1.035)^t > \frac{20}{7}$ (awrt 2.86) | M1 A1 | M1: Substitutes $N=10000$, proceeds to $(1.035)^t \ldots A$ |
| $\Rightarrow t > \frac{\log\frac{20}{7}}{\log 1.035} = 30.516 =$ 30 hrs 31 mins or 30 hrs 32 mins | M1 A1 | M1: Proceeds correctly to find value of $t$; A1: 30 hrs 31 mins or 30 hrs 32 mins (not 1831 minutes) |
**(4 marks)**
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dN}{dt} = 3500(1.035)^t \ln 1.035 \Rightarrow \frac{dN}{dt}\bigg|_{t=8} = 3500(1.035)^8 \ln 1.035 =$ awrt 159 | B1 M1 A1 | B1: Correct derivative; M1: Substitutes $t=8$ into their $\frac{dN}{dt}$; A1: awrt 159 |
**(3 marks, 8 marks total)**
---3. The number of bacteria in a liquid culture is modelled by the formula
$$N = 3500 ( 1.035 ) ^ { t } , \quad t \geqslant 0$$
where $N$ is the number of bacteria $t$ hours after the start of a scientific study.
\begin{enumerate}[label=(\alph*)]
\item State the number of bacteria at the start of the scientific study.\\
(1)
\item Find the time taken from the start of the study for the number of bacteria to reach 10000\\
Give your answer in hours and minutes, to the nearest minute.
\item Use calculus to find the rate of increase in the number of bacteria when $t = 8$ Give your answer, in bacteria per hour, to the nearest whole number.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q3 [8]}}