| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Applied rate of change |
| Difficulty | Standard +0.8 This question requires quotient rule differentiation of an exponential function with different powers (0.6t and 0.8t), then solving dP/dt = 0 by manipulating exponential equations. While the individual techniques are standard C3/C4 content, the algebraic complexity of simplifying the derivative and solving the resulting exponential equation elevates this above a routine exercise. The multi-step nature and need for careful algebraic manipulation make it moderately challenging but still within expected A-level scope. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=0 \Rightarrow (P=)200 - \frac{160}{15+1} = 190 \Rightarrow 190\,000\) | M1A1 | Sets \(t=0\); award if candidate attempts \(200 - \frac{160}{15+1}\); correct answer only, accept 190 000 or \((P=)\) 190; no decimals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^{kt} \rightarrow ae^{kt}\) | M1 | For showing \(e^{kt} \rightarrow ae^{kt}\) where \(a\) is constant; may be embedded in product/quotient rule |
| \(\frac{dP}{dt} = -\frac{(15+e^{0.8t})\times 96e^{0.6t} - 160e^{0.6t} \times 0.8e^{0.8t}}{(15+e^{0.8t})^2}\) | M1A1 | Applying quotient rule correctly; correct un-simplified or simplified \(\frac{dP}{dt}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(\pm\frac{(15+e^{0.8t})\times 96e^{0.6t} - 160e^{0.6t}\times 0.8e^{0.8t}}{(15+e^{0.8t})^2} = 0 \Rightarrow e^{0.8t} = 45\) | M1A1 | Sets \(\frac{dP}{dt}=0\) to obtain \(pe^{0.8t}=q\) or equivalent; \(e^{0.8t}=45\) or equivalent correct equation; allow recovery if signs reversed |
| \(T = \frac{\ln 45}{0.8} = 4.76\) | M1A1 | Having set \(\frac{dP}{dt}=0\), correct order of operations taking ln; cannot be awarded from impossible equations e.g. \(e^{0.8t}=-45\); accept \(\frac{\ln 45}{0.8}\) or awrt \(= 4.76\) |
# Question 13:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=0 \Rightarrow (P=)200 - \frac{160}{15+1} = 190 \Rightarrow 190\,000$ | M1A1 | Sets $t=0$; award if candidate attempts $200 - \frac{160}{15+1}$; correct answer only, accept 190 000 or $(P=)$ 190; no decimals |
**(2 marks)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{kt} \rightarrow ae^{kt}$ | M1 | For showing $e^{kt} \rightarrow ae^{kt}$ where $a$ is constant; may be embedded in product/quotient rule |
| $\frac{dP}{dt} = -\frac{(15+e^{0.8t})\times 96e^{0.6t} - 160e^{0.6t} \times 0.8e^{0.8t}}{(15+e^{0.8t})^2}$ | M1A1 | Applying quotient rule correctly; correct un-simplified or simplified $\frac{dP}{dt}$ |
**(3 marks)**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $\pm\frac{(15+e^{0.8t})\times 96e^{0.6t} - 160e^{0.6t}\times 0.8e^{0.8t}}{(15+e^{0.8t})^2} = 0 \Rightarrow e^{0.8t} = 45$ | M1A1 | Sets $\frac{dP}{dt}=0$ to obtain $pe^{0.8t}=q$ or equivalent; $e^{0.8t}=45$ or equivalent correct equation; allow recovery if signs reversed |
| $T = \frac{\ln 45}{0.8} = 4.76$ | M1A1 | Having set $\frac{dP}{dt}=0$, correct order of operations taking ln; cannot be awarded from impossible equations e.g. $e^{0.8t}=-45$; accept $\frac{\ln 45}{0.8}$ or awrt $= 4.76$ |
**(4 marks)**
---13.
Figure 5
A colony of ants is being studied. The number of ants in the colony is modelled by the equation
$$P = 200 - \frac { 160 \mathrm { e } ^ { 0.6 t } } { 15 + \mathrm { e } ^ { 0.8 t } } \quad t \in \mathbb { R } , t \geqslant 0$$
where $P$ is the number of ants, measured in thousands, $t$ years after the study started. A sketch of the graph of $P$ against $t$ is shown in Figure 5
\begin{enumerate}[label=(\alph*)]
\item Calculate the number of ants in the colony at the start of the study.
\item Find $\frac { \mathrm { d } P } { \mathrm {~d} t }$
The population of ants initially decreases, reaching a minimum value after $T$ years, as shown in Figure 5
\item Using your answer to part (b), calculate the value of $T$ to 2 decimal places.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q13 [9]}}