| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Rate of change in exponential model |
| Difficulty | Moderate -0.3 This is a straightforward exponential growth question requiring basic substitution (parts a,b) and standard differentiation of an exponential function (part c). While part (c) involves finding dN/dt at a specific point, it's a direct application of the chain rule with no conceptual challenges—slightly easier than the average A-level question due to its routine nature. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(N=5000(1.04)^t, t \in \mathbb{R}, t \geq 0\) | ||
| \(\{t=0\} \Rightarrow N=5000\) (bacteria) | B1 cao | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\pm\left(\frac{5000(1.04)^2-5000}{5000}\right)100\) or \(\pm\left(\frac{5408-5000}{5000}\right)100\) | M1 | A full method for finding a percentage increase. |
| \(=8.16(\%)\) | A1 | \(8\) or \(8.1\) or \(8.16\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dN}{dt}=5000(1.04)^t\ln(1.04)\) or \(\frac{dN}{dt}=5000(e^{\ln(1.04)})\ln(1.04)\) | M1 A1 | |
| or \(\frac{dN}{dt}=N\ln(1.04)\) or \(\frac{1}{N}\frac{dN}{dt}=\ln(1.04)\) | ||
| At \(t=T, 15000=5000(1.04)^T \Rightarrow 3=(1.04)^T \Rightarrow T=\frac{\ln 3}{\ln 1.04}=28.01...\) | ||
| \(\{At t=T,\}\frac{dN}{dt}=5000(3)\ln(1.04)\) | dM1 | Substitutes their found \((1.04)^T\) or their found \(T\) (or \(t\)) into \(\frac{dN}{dt}\) or their found \(N\) into \(\frac{dN}{dt}=N\ln(1.04)\) |
| or \(\frac{dN}{dt}=5000(1.04)^{28.01}\ln(1.04)\) | or \(N\) into \(\frac{dN}{dt}=N\ln(1.04)\) | |
| \(=588.3106973...\left(\frac{\text{bacteria}}{\text{hour}}\right)\) | A1 | \(590\) or awrt \(588\) |
| [4] |
**(a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $N=5000(1.04)^t, t \in \mathbb{R}, t \geq 0$ | | |
| $\{t=0\} \Rightarrow N=5000$ (bacteria) | B1 cao | [1] |
**(b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\pm\left(\frac{5000(1.04)^2-5000}{5000}\right)100$ or $\pm\left(\frac{5408-5000}{5000}\right)100$ | M1 | A full method for finding a percentage increase. |
| $=8.16(\%)$ | A1 | $8$ or $8.1$ or $8.16$ |
| | | [2] |
**Notes for (b)**
B1: 5000 cao.
M1: A full method for finding a percentage increase.
A1: $8$ or $8.1$ or $8.16$
Note: $(1.04)^2$ or $1.0816$ or $0.0816$ by itself is M0; but followed by either $8$ or $8.2$ or $8.16$ is M1A1.
Note: Applying $\left(\frac{5000(1.04)^2-5000}{5408}\right)100$ or equivalent (answer of $7.54(\%)$) is M0A0.
**(c)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dN}{dt}=5000(1.04)^t\ln(1.04)$ or $\frac{dN}{dt}=5000(e^{\ln(1.04)})\ln(1.04)$ | M1 A1 | |
| or $\frac{dN}{dt}=N\ln(1.04)$ or $\frac{1}{N}\frac{dN}{dt}=\ln(1.04)$ | | |
| At $t=T, 15000=5000(1.04)^T \Rightarrow 3=(1.04)^T \Rightarrow T=\frac{\ln 3}{\ln 1.04}=28.01...$ | | |
| $\{At t=T,\}\frac{dN}{dt}=5000(3)\ln(1.04)$ | dM1 | Substitutes their found $(1.04)^T$ or their found $T$ (or $t$) into $\frac{dN}{dt}$ or their found $N$ into $\frac{dN}{dt}=N\ln(1.04)$ |
| or $\frac{dN}{dt}=5000(1.04)^{28.01}\ln(1.04)$ | | or $N$ into $\frac{dN}{dt}=N\ln(1.04)$ |
| $=588.3106973...\left(\frac{\text{bacteria}}{\text{hour}}\right)$ | A1 | $590$ or awrt $588$ |
| | | [4] |
**Notes for (c)**
M1: Award M1 for $\frac{dN}{dt}=\pm\lambda(1.04)^t$ or $\frac{dN}{dt}=\pm\lambda N$ or $\frac{dN}{dt}=\pm\lambda e^{\ln(1.04)t}$ or $\frac{1}{N}\frac{dN}{dt}=\pm\lambda$ where $\lambda \neq 0$ is a constant.
**EXCEPTION:** Award M0, however, for $\frac{dN}{dt}=...(1.04)^{t-1}$ or $\frac{dN}{dt}=...(1.04)^{t-1}$ or equivalent.
Note: Award M0 for expressions such as $\frac{dN}{dt}=5000(1.04)^{t-1}$ or $\frac{dN}{dt}=5000t(1.04)^{t-1}$
Note: You can award M1 for $\frac{dN}{dt}=5000(1.04)^t$
A1: $\frac{dN}{dt}=5000(1.04)^t\ln(1.04)$ or $\frac{dN}{dt}=5000(e^{\ln(1.04)t})\ln(1.04)$ or $\frac{dN}{dt}=N\ln(1.04)$
or $\frac{1}{N}\frac{dN}{dt}=\ln(1.04)$ or equivalent.
dM1: (dependent on the first M mark) For substituting their found $(1.04)^T$ (or $(1.04)^t$) or their found $T$ (or $t$) into their $\frac{dN}{dt}=f(t)$; or their found $N$ or $N=15000$ into their $\frac{dN}{dt}=f(N)$.
A1: $590$ or anything that rounds to 588
---\begin{enumerate}
\item The number of bacteria, $N$, present in a liquid culture at time $t$ hours after the start of a scientific study is modelled by the equation
\end{enumerate}
$$N = 5000 ( 1.04 ) ^ { t } , \quad t \geqslant 0$$
where $N$ is a continuous function of $t$.\\
(a) Find the number of bacteria present at the start of the scientific study.\\
(b) Find the percentage increase in the number of bacteria present from $t = 0$ to $t = 2$
Given that $N = 15000$ when $t = T$,\\
(c) find the value of $\frac { \mathrm { d } N } { \mathrm {~d} t }$ when $t = T$, giving your answer to 3 significant figures.\\
\hfill \mbox{\textit{Edexcel C4 2014 Q3 [7]}}