Edexcel C3 2017 June — Question 8 9 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2017
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeApplied rate of change
DifficultyStandard +0.8 This question requires quotient rule differentiation of a complex exponential function, solving a transcendental equation (setting derivative to zero), and understanding asymptotic behavior. While the individual techniques are C3 standard, the algebraic complexity of the quotient rule application and solving for T when dP/dt = 0 elevates this above a routine question. Part (d) requires conceptual understanding of limits, making this a solid mid-to-upper difficulty C3 question.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives

8. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f0a633e3-5c63-4d21-8ffa-d4e7dc43a536-26_663_1454_210_242} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} The number of rabbits on an island is modelled by the equation $$P = \frac { 100 \mathrm { e } ^ { - 0.1 t } } { 1 + 3 \mathrm { e } ^ { - 0.9 t } } + 40 , \quad t \in \mathbb { R } , t \geqslant 0$$ where \(P\) is the number of rabbits, \(t\) years after they were introduced onto the island.
A sketch of the graph of \(P\) against \(t\) is shown in Figure 3.
  1. Calculate the number of rabbits that were introduced onto the island.
  2. Find \(\frac { \mathrm { d } P } { \mathrm {~d} t }\) The number of rabbits initially increases, reaching a maximum value \(P _ { T }\) when \(t = T\)
  3. Using your answer from part (b), calculate
    1. the value of \(T\) to 2 decimal places,
    2. the value of \(P _ { T }\) to the nearest integer.
      (Solutions based entirely on graphical or numerical methods are not acceptable.) For \(t > T\), the number of rabbits decreases, as shown in Figure 3, but never falls below \(k\), where \(k\) is a positive constant.
  4. Use the model to state the maximum value of \(k\).

Question 8:
Part (a):
AnswerMarks Guidance
AnswerMark Guidance
\(P_0 = \frac{100}{1+3} + 40 = 65\)B1
Part (b):
AnswerMarks Guidance
AnswerMark Guidance
\(\frac{d}{dt}e^{kt} = Ce^{kt}\)M1 Allow \(C=1\); may be within incorrect product/quotient rule
\(\frac{dP}{dt} = \frac{(1+3e^{-0.9t})\times -10e^{-0.1t} - 100e^{-0.1t}\times -2.7e^{-0.9t}}{(1+3e^{-0.9t})^2}\)M1 A1 Full application of quotient rule; denominator must be present; correct unsimplified answer
Part (c)(i):
AnswerMarks Guidance
AnswerMark Guidance
Sets \(\frac{dP}{dt} = 0\); factorises to reach \(Ae^{\pm 0.9t} = B\)M1 Condone double error on \(e^{-0.1t} \times e^{-0.9t} = e^{-0.1t-0.9t}\); look for correct indices
\(e^{-0.9t} = \frac{10}{240}\), i.e. \(e^{0.9t} = 24\)M1 Correct order of operations taking ln; cannot be awarded from impossible equations
\(t = \frac{10}{9}\ln(24) = 3.53\)M1 A1 Accept \(t = \frac{10}{9}\ln(24)\) or exact equivalent; awrt 3.53
Part (c)(ii):
AnswerMarks Guidance
AnswerMark Guidance
Sub \(t = 3.53 \Rightarrow P_t = 102\)A1 awrt 102 following 3.53; M marks must have been awarded
Part (d):
AnswerMarks Guidance
AnswerMark Guidance
\(40\)B1 Condone \(P \to 40\), \(k\ldots 40\) or likewise
# Question 8:

## Part (a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $P_0 = \frac{100}{1+3} + 40 = 65$ | B1 | |

## Part (b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d}{dt}e^{kt} = Ce^{kt}$ | M1 | Allow $C=1$; may be within incorrect product/quotient rule |
| $\frac{dP}{dt} = \frac{(1+3e^{-0.9t})\times -10e^{-0.1t} - 100e^{-0.1t}\times -2.7e^{-0.9t}}{(1+3e^{-0.9t})^2}$ | M1 A1 | Full application of quotient rule; denominator must be present; correct unsimplified answer |

## Part (c)(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| Sets $\frac{dP}{dt} = 0$; factorises to reach $Ae^{\pm 0.9t} = B$ | M1 | Condone double error on $e^{-0.1t} \times e^{-0.9t} = e^{-0.1t-0.9t}$; look for correct indices |
| $e^{-0.9t} = \frac{10}{240}$, i.e. $e^{0.9t} = 24$ | M1 | Correct order of operations taking ln; cannot be awarded from impossible equations |
| $t = \frac{10}{9}\ln(24) = 3.53$ | M1 A1 | Accept $t = \frac{10}{9}\ln(24)$ or exact equivalent; awrt 3.53 |

## Part (c)(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| Sub $t = 3.53 \Rightarrow P_t = 102$ | A1 | awrt 102 following 3.53; M marks must have been awarded |

## Part (d):

| Answer | Mark | Guidance |
|--------|------|----------|
| $40$ | B1 | Condone $P \to 40$, $k\ldots 40$ or likewise |

---
8.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f0a633e3-5c63-4d21-8ffa-d4e7dc43a536-26_663_1454_210_242}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

The number of rabbits on an island is modelled by the equation

$$P = \frac { 100 \mathrm { e } ^ { - 0.1 t } } { 1 + 3 \mathrm { e } ^ { - 0.9 t } } + 40 , \quad t \in \mathbb { R } , t \geqslant 0$$

where $P$ is the number of rabbits, $t$ years after they were introduced onto the island.\\
A sketch of the graph of $P$ against $t$ is shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Calculate the number of rabbits that were introduced onto the island.
\item Find $\frac { \mathrm { d } P } { \mathrm {~d} t }$

The number of rabbits initially increases, reaching a maximum value $P _ { T }$ when $t = T$
\item Using your answer from part (b), calculate
\begin{enumerate}[label=(\roman*)]
\item the value of $T$ to 2 decimal places,
\item the value of $P _ { T }$ to the nearest integer.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)

For $t > T$, the number of rabbits decreases, as shown in Figure 3, but never falls below $k$, where $k$ is a positive constant.
\end{enumerate}\item Use the model to state the maximum value of $k$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C3 2017 Q8 [9]}}