| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Logistic/bounded growth |
| Difficulty | Standard +0.8 This is a multi-part logistic differential equation question requiring exponential modeling, partial fractions, separation of variables, and solving for a specific value. While the techniques are A-level standard (partial fractions, integration, logarithms), the extended multi-step nature, application context, and need to manipulate the logistic equation solution make it moderately challenging—above average but not requiring exceptional insight. |
| Spec | 1.02y Partial fractions: decompose rational functions1.06i Exponential growth/decay: in modelling context1.08k Separable differential equations: dy/dx = f(x)g(y) |
|
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 25 \times 1.32^t\) | B1 (AO 3.3) | Forms correct model, or applies repeated percentage increase 4 times; PI by AWRT 75.9 |
| Substitutes \(t = 5\) into their model: \(x = 25 \times 1.32^5 = 100.18\ldots\) | M1 (AO 3.4) | Or applies repeated percentage increase 5 times |
| \(x = 101\) (condone 100) | A1 (AO 3.2a) | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The value predicted by the exponential model will grow without limit | E1 (AO 3.5b) | Explains that the model grows exponentially; must refer to model and exponential |
| This can't be true as there are only 900 tomato plants in the greenhouse | E1 (AO 3.5a) | Refers to 900 plants; condone reference to "tomato(es)" or "plants" in place of "tomato plants" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = \frac{x(900-x)}{2700}\), rearranges to \(\frac{2700}{x(900-x)}\frac{dx}{dt} = 1\) | B1 (AO 3.1a) | Rearranges to obtain one of the required forms; where \(P \times Q = 2700\); may include integral signs |
| \(\frac{2700}{x(900-x)} \equiv \frac{A}{x} + \frac{B}{900-x}\); \(2700 \equiv A(900-x) + Bx\); \(x=0 \Rightarrow A = 3\); \(x=900 \Rightarrow B = 3\) | M1 (AO 3.1a) | Forms partial fraction equation with correct denominators and uses appropriate method to find numerators; PI by correct \(A\) and \(B\) without incorrect working |
| \(\int\!\left(\frac{3}{x} + \frac{3}{900-x}\right)dx = \int dt\) | R1 (AO 2.1) | Obtains correct \(A\) and \(B\) and concludes; accept \(\int 1\, dt\); condone missing brackets |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Integrates to obtain \(\ln x\) or \(\pm\ln(900-x)\) | M1 (3.1a) | Condone missing brackets |
| Integrates to obtain \(\ln x\) and \(\pm\ln(900-x)\) | M1 (1.1a) | Condone missing brackets |
| \(3(\ln x - \ln(900-x)) + c = t\) | A1 (1.1b) | OE; condone missing \(+c\) |
| Uses \(t=0\), \(x=25\) to obtain value for \(c\); \(3(\ln 25 - \ln(900-25)) + c = 0\), \(c = 10.67\) | M1 (3.4) | |
| \(t = 3(\ln x - \ln(900-x)) + 10.67\) e.g. \(t = 3\ln\left(\dfrac{35x}{900-x}\right)\) | A1 (1.1b) | ACF; if \(c\) given as decimal accept AWRT 11 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3(\ln 450 - \ln(450)) + 10.67 = 10.67\ldots\) | M1 (3.4) | Substitutes \(x=450\) into model from (b)(ii) |
| \(t = 11\) days | A1 (3.2a) | CAO; "It takes 11 days from when damage is first noticed until half the plants are damaged" |
## Question 10(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 25 \times 1.32^t$ | B1 (AO 3.3) | Forms correct model, or applies repeated percentage increase 4 times; PI by AWRT 75.9 |
| Substitutes $t = 5$ into their model: $x = 25 \times 1.32^5 = 100.18\ldots$ | M1 (AO 3.4) | Or applies repeated percentage increase 5 times |
| $x = 101$ (condone 100) | A1 (AO 3.2a) | CAO |
**Subtotal: 3 marks**
---
## Question 10(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The value predicted by the exponential model will grow without limit | E1 (AO 3.5b) | Explains that the model grows exponentially; must refer to model and exponential |
| This can't be true as there are only 900 tomato plants in the greenhouse | E1 (AO 3.5a) | Refers to 900 plants; condone reference to "tomato(es)" or "plants" in place of "tomato plants" |
**Subtotal: 2 marks**
---
## Question 10(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = \frac{x(900-x)}{2700}$, rearranges to $\frac{2700}{x(900-x)}\frac{dx}{dt} = 1$ | B1 (AO 3.1a) | Rearranges to obtain one of the required forms; where $P \times Q = 2700$; may include integral signs |
| $\frac{2700}{x(900-x)} \equiv \frac{A}{x} + \frac{B}{900-x}$; $2700 \equiv A(900-x) + Bx$; $x=0 \Rightarrow A = 3$; $x=900 \Rightarrow B = 3$ | M1 (AO 3.1a) | Forms partial fraction equation with correct denominators and uses appropriate method to find numerators; PI by correct $A$ and $B$ without incorrect working |
| $\int\!\left(\frac{3}{x} + \frac{3}{900-x}\right)dx = \int dt$ | R1 (AO 2.1) | Obtains correct $A$ and $B$ and concludes; accept $\int 1\, dt$; condone missing brackets |
**Subtotal: 3 marks**
## Question 10(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Integrates to obtain $\ln x$ or $\pm\ln(900-x)$ | M1 (3.1a) | Condone missing brackets |
| Integrates to obtain $\ln x$ and $\pm\ln(900-x)$ | M1 (1.1a) | Condone missing brackets |
| $3(\ln x - \ln(900-x)) + c = t$ | A1 (1.1b) | OE; condone missing $+c$ |
| Uses $t=0$, $x=25$ to obtain value for $c$; $3(\ln 25 - \ln(900-25)) + c = 0$, $c = 10.67$ | M1 (3.4) | |
| $t = 3(\ln x - \ln(900-x)) + 10.67$ e.g. $t = 3\ln\left(\dfrac{35x}{900-x}\right)$ | A1 (1.1b) | ACF; if $c$ given as decimal accept AWRT 11 |
## Question 10(b)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3(\ln 450 - \ln(450)) + 10.67 = 10.67\ldots$ | M1 (3.4) | Substitutes $x=450$ into model from (b)(ii) |
| $t = 11$ days | A1 (3.2a) | CAO; "It takes 11 days from when damage is first noticed until half the plants are damaged" |10 A gardener has a greenhouse containing 900 tomato plants.
The gardener notices that some of the tomato plants are damaged by insects.\\
Initially there are 25 damaged tomato plants.\\
The number of tomato plants damaged by insects is increasing by $32 \%$ each day.\\
10
\begin{enumerate}[label=(\alph*)]
\item The total number of plants damaged by insects, $x$, is modelled by
$$x = A \times B ^ { t }$$
where $A$ and $B$ are constants and $t$ is the number of days after the gardener first noticed the damaged plants.
10 (a) (i) Use this model to find the total number of plants damaged by insects 5 days after the gardener noticed the damaged plants.\\
10 (a) (ii) Explain why this model is not realistic in the long term.\\
10
\item A refined model assumes the rate of increase of the number of plants damaged by insects is given by
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { x ( 900 - x ) } { 2700 }$$
10 (b) (i) Show that
$$\int \left( \frac { A } { x } + \frac { B } { 900 - x } \right) \mathrm { d } x = \int \mathrm { d } t$$
where $A$ and $B$ are positive integers to be found.\\
\begin{center}
\begin{tabular}{|l|}
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\begin{tabular}{l}
10 (b) (iii) Hence, find the number of days it takes from when the damage is first noticed until half of the plants are damaged by the insects. \\[0pt]
[2 marks] $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ \\
\end{tabular} \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2022 Q10 [15]}}