Modeling context with interpretation

1. Two species of insects, \(X\) and \(Y\), co-exist on an island. The populations of the species at time \(t\) years are \(x\) and \(y\) respectively, where \(x\) and \(y\) are measured in millions. The situation can be modelled by the differential equations

$$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = 3 x + 10 y \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = x + 6 y \end{aligned}$$

1. 1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 9 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 8 x = 0\).
   2. Find the general solution for \(x\) in terms of \(t\).
2. Find the corresponding general solution for \(y\).
3. When \(t = 0 , \frac { \mathrm {~d} x } { \mathrm {~d} t } = 5\) and the population of species \(Y\) is 4 times the population of species \(X\). Find the particular solution for \(x\) in terms of \(t\).

10. The following simultaneous equations are to be solved. $$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = 4 x + 2 y + 6 \mathrm { e } ^ { 3 t } \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = 6 x + 8 y + 15 \mathrm { e } ^ { 3 t } \end{aligned}$$

1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 12 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 20 x = 0\).
2. Given that \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 9\) and \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 10\) when \(t = 0\), find the particular solution for \(x\) in terms of \(t\). Additional page, if required. Write the question number(s) in the left-hand margin. \section*{PLEASE DO NOT WRITE ON THIS PAGE} \section*{PLEASE DO NOT WRITE ON THIS PAGE} \section*{PLEASE DO NOT WRITE ON THIS PAGE}

1. A scientist is studying the effect of introducing a population of white-clawed crayfish into a population of signal crayfish.
   At time \(t\) years, the number of white-clawed crayfish, \(w\), and the number of signal crayfish, \(s\), are modelled by the differential equations

$$\begin{aligned} & \frac { \mathrm { d } w } { \mathrm {~d} t } = \frac { 5 } { 2 } ( w - s ) \\ & \frac { \mathrm { d } s } { \mathrm {~d} t } = \frac { 2 } { 5 } w - 90 \mathrm { e } ^ { - t } \end{aligned}$$

1. Show that $$2 \frac { \mathrm {~d} ^ { 2 } w } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} w } { \mathrm {~d} t } + 2 w = 450 \mathrm { e } ^ { - t }$$
2. Find a general solution for the number of white-clawed crayfish at time \(t\) years.
3. Find a general solution for the number of signal crayfish at time \(t\) years. The model predicts that, at time \(T\) years, the population of white-clawed crayfish will have died out. Given that \(w = 65\) and \(s = 85\) when \(t = 0\)
4. find the value of \(T\), giving your answer to 3 decimal places.
5. Suggest a limitation of the model.

1. Two compounds, \(X\) and \(Y\), are involved in a chemical reaction. The amounts in grams of these compounds, \(t\) minutes after the reaction starts, are \(x\) and \(y\) respectively and are modelled by the differential equations

$$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = - 5 x + 10 y - 30 \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = - 2 x + 3 y - 4 \end{aligned}$$

1. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 50$$
2. Find, according to the model, a general solution for the amount in grams of compound \(X\) present at time \(t\) minutes.
3. Find, according to the model, a general solution for the amount in grams of compound \(Y\) present at time \(t\) minutes. Given that \(x = 2\) and \(y = 5\) when \(t = 0\)
4. find
   1. the particular solution for \(x\),
   2. the particular solution for \(y\). A scientist thinks that the chemical reaction will have stopped after 8 minutes.
5. Explain whether this is supported by the model.

1. A scientist is studying the effect of introducing a population of type \(A\) bacteria into a population of type \(B\) bacteria.

At time \(t\) days, the number of type \(A\) bacteria, \(x\), and the number of type \(B\) bacteria, \(y\), are modelled by the differential equations $$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = x + y \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = 3 y - 2 x \end{aligned}$$

1. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 0$$
2. Determine a general solution for the number of type \(A\) bacteria at time \(t\) days.
3. Determine a general solution for the number of type \(B\) bacteria at time \(t\) days. The model predicts that, at time \(T\) hours, the number of bacteria in the two populations will be equal. Given that \(x = 100\) and \(y = 275\) when \(t = 0\)
4. determine the value of \(T\), giving your answer to 2 decimal places.
5. Suggest a limitation of the model.

1. A company plans to build a new fairground ride. The ride will consist of a capsule that will hold the passengers and the capsule will be attached to a tall tower. The capsule is to be released from rest from a point half way up the tower and then made to oscillate in a vertical line.

The vertical displacement, \(x\) metres, of the top of the capsule below its initial position at time \(t\) seconds is modelled by the differential equation, $$m \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = 200 \cos t , \quad t \geqslant 0$$ where \(m\) is the mass of the capsule including its passengers, in thousands of kilograms.
The maximum permissible weight for the capsule, including its passengers, is 30000 N .
Taking the value of \(g\) to be \(10 \mathrm {~ms} ^ { - 2 }\) and assuming the capsule is at its maximum permissible weight,

1. 1. explain why the value of \(m\) is 3
   2. show that a particular solution to the differential equation is $$x = 40 \sin t - 20 \cos t$$
   3. hence find the general solution of the differential equation.
2. Using the model, find, to the nearest metre, the vertical distance of the top of the capsule from its initial position, 9 seconds after it is released.

1. A patient is treated by administering an antibiotic intravenously at a constant rate for some time.

Initially there is none of the antibiotic in the patient.
At time \(t\) minutes after treatment began

• the concentration of the antibiotic in the blood of the patient is \(x \mathrm { mg } / \mathrm { ml }\)
• the concentration of the antibiotic in the tissue of the patient is \(y \mathrm { mg } / \mathrm { ml }\)

The concentration of antibiotic in the patient is modelled by the equations $$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.025 y - 0.045 x + 2 \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = 0.032 x - 0.025 y \end{aligned}$$

1. Show that $$40000 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2800 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 13 y = 2560$$
2. Determine, according to the model, a general solution for the concentration of the antibiotic in the patient's tissue at time \(t\) minutes after treatment began.
3. Hence determine a particular solution for the concentration of the antibiotic in the tissue at time \(t\) minutes after treatment began. To be effective for the patient the concentration of antibiotic in the tissue must eventually reach a level between \(185 \mathrm { mg } / \mathrm { ml }\) and \(200 \mathrm { mg } / \mathrm { ml }\).
4. Determine whether the rate of administration of the antibiotic is effective for the patient, giving a reason for your answer.

1. At the start of the year 2000, a survey began of the number of foxes and rabbits on an island.

At time \(t\) years after the survey began, the number of foxes, \(f\), and the number of rabbits, \(r\), on the island are modelled by the differential equations $$\begin{aligned} & \frac { \mathrm { d } f } { \mathrm {~d} t } = 0.2 f + 0.1 r \\ & \frac { \mathrm {~d} r } { \mathrm {~d} t } = - 0.2 f + 0.4 r \end{aligned}$$

1. Show that \(\frac { \mathrm { d } ^ { 2 } f } { \mathrm {~d} t ^ { 2 } } - 0.6 \frac { \mathrm {~d} f } { \mathrm {~d} t } + 0.1 f = 0\)
2. Find a general solution for the number of foxes on the island at time \(t\) years.
3. Hence find a general solution for the number of rabbits on the island at time \(t\) years. At the start of the year 2000 there were 6 foxes and 20 rabbits on the island.
4. 1. According to this model, in which year are the rabbits predicted to die out?
   2. According to this model, how many foxes will be on the island when the rabbits die out?
   3. Use your answers to parts (i) and (ii) to comment on the model.

9 The quantity of grass on an island at time \(t\) years is \(x\), in appropriate units. At time \(t = 0\) some rabbits are introduced to the island. The population of rabbits on the island at time \(t\) years is \(y\), in units of 100s of rabbits. An ecologist who is studying the island suggests that the following pair of simultaneous first order differential equations can be used to model the population of rabbits and quantity of grass for \(t \geqslant 0\). $$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = 3 x - 2 y , \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = y + 5 x \end{aligned}$$

1. (a) Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = a \frac { \mathrm {~d} x } { \mathrm {~d} t } + b x\) where \(a\) and \(b\) are constants which should be found.
   (b) Find the general solution for \(x\) in real form.
2. Find the corresponding general solution for \(y\). At time \(t = 0\) the quantity of grass on the island was 4 units. The number of rabbits introduced at this time was 500 .
3. Find the particular solutions for \(x\) and \(y\).
4. The ecologist finds that the model predicts that there will be no grass at time \(T\), when there are still rabbits on the island. Find the value of \(T\).
5. State one way in which the model is not appropriate for modelling the quantity of grass and the population of rabbits for \(0 \leqslant t \leqslant T\). \section*{OCR} \section*{Oxford Cambridge and RSA}

10 In a predator-prey environment the population, at time \(t\) years, of predators is \(x\) and prey is \(y\). The populations of predators and prey are measured in hundreds. The populations are modelled by the following simultaneous differential equations. $$\frac { \mathrm { d } x } { \mathrm {~d} t } = y \quad \frac { \mathrm {~d} y } { \mathrm {~d} t } = 2 y - 5 x$$

1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } - 5 x\).
2. 1. Find the general solution for \(x\).
   2. Find the equivalent general solution for \(y\). Initially there are 100 predators and 300 prey.
3. Find the particular solutions for \(x\) and \(y\).
4. Determine whether the model predicts that the predators will die out before the prey.

10 A swing door is a door to a room which is closed when in equilibrium but which can be pushed open from either side and which can swing both ways, into or out of the room, and through the equilibrium position. The door is sprung so that when displaced from the equilibrium position it will swing back towards it. The extent to which the door is open at any time, \(t\) seconds, is measured by the angle at the hinge, \(\theta\), which the plane of the door makes with the plane of the equilibrium position. See the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{e792797e-6d20-4fc3-9733-43db10f764d7-5_367_1116_625_246} In an initial model of the motion of a certain swing door it is suggested that \(\theta\) satisfies the following differential equation. \(4 \frac { \mathrm {~d} ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + 25 \theta = 0\)

1. 1. Write down the general solution to (\textit{).
   2. With reference to the behaviour of your solution in part (a)(i) explain briefly why the model using (}) is unlikely to be realistic. In an improved model of the motion of the door an extra term is introduced to the differential equation so that it becomes \(4 \frac { \mathrm {~d} ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + \lambda \frac { \mathrm { d } \theta } { \mathrm { d } t } + 25 \theta = 0 \quad ( \dagger )\) where \(\lambda\) is a positive constant.
2. In the case where \(\lambda = 16\) the door is held open at an angle of 0.9 radians and then released from rest at time \(t = 0\).
   1. Find, in a real form, the general solution of ( \(\dagger\) ).
   2. Find the particular solution of ( \(\dagger\) ).
   3. With reference to the behaviour of your solution found in part (b)(ii) explain briefly how the extra term in ( \(\dagger\) ) improves the model.
3. Find the value of \(\lambda\) for which the door is critically damped.

16

1. Show that $$\int _ { 0.5 } ^ { 4 } \frac { 1 } { t } \ln t \mathrm {~d} t = a ( \ln 2 ) ^ { 2 }$$ where \(a\) is a rational number to be found.
   16
2. A curve \(C\) is defined parametrically for \(t > 0\) by $$x = 2 t \quad y = \frac { 1 } { 2 } t ^ { 2 } - \ln t$$ The arc formed by the graph of \(C\) from \(t = 0.5\) to \(t = 4\) is rotated through \(2 \pi\) radians about the \(x\)-axis to generate a surface with area \(S\) Find the exact value of \(S\), giving your answer in the form $$S = \pi \left( b + c \ln 2 + d ( \ln 2 ) ^ { 2 } \right)$$ where \(b , c\) and \(d\) are rational numbers to be found.
   [0pt] [7 marks]
   
   \includegraphics[max width=\textwidth, alt={}]{a9f88195-e545-43f2-a13a-6459d14e1cda-25_2488_1719_219_150}
   Question number Additional page, if required.
   Write the question numbers in the left-hand margin.

15

1. Find the value of \(r\). 15
2. Show that \(\mu = 3\) 15
3. Solve the coupled differential equations to find both \(x\) and \(y\) in terms of \(t\).
   [0pt] [9 marks]
   
   \includegraphics[max width=\textwidth, alt={}]{f1ec515d-184a-4462-a6d2-5876d3e19117-27_2493_1721_214_150}
   Additional page, if required.
   Write the question numbers in the left-hand margin.

14 On an isolated island some rabbits have been accidently introduced. In order to eliminate them, conservationists have introduced some birds of prey.
At time \(t\) years \(( t \geq 0 )\) there are \(x\) rabbits and \(y\) birds of prey.
At time \(t = 0\) there are 1755 rabbits and 30 birds of prey.
When \(t > 0\) it is assumed that:

• the rabbits will reproduce at a rate of \(a \%\) per year
• each bird of prey will kill, on average, \(b\) rabbits per year
• the death rate of the birds of prey is \(c\) birds per year
• the number of birds of prey will increase at a rate of \(d \%\) of the rabbit population per year.

This system is represented by the coupled differential equations: $$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.4 x - 13 y \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = 0.01 x - 1.95 \end{aligned}$$ 14

1. State the value of \(a\), the value of \(b\), the value of \(c\) and the value of \(d\) [0pt] [2 marks]
   14
2. Solve the coupled differential equations to find both \(x\) and \(y\) in terms of \(t\)

9 A curve passes through the point (-2, 4.73) and satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 2 } - x ^ { 2 } } { 2 x + 3 y }$$ Use Euler's step by step method once, and then the midpoint formula $$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right) , \quad x _ { r + 1 } = x _ { r } + h$$ once, each with a step length of 0.02 , to estimate the value of \(y\) when \(x = - 1.96\) Give your answer to five significant figures.
[0pt] [4 marks]

5 In a predator-prey environment the population, at time \(t\) years, of predators is \(x\) and prey is \(y\). The populations of predators and prey are measured in hundreds. The populations are modelled by the following simultaneous differential equations. \(\frac { \mathrm { d } x } { \mathrm {~d} t } = y \quad \frac { \mathrm {~d} y } { \mathrm {~d} t } = 2 y - 5 x\)

1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } - 5 x\).
2. 1. Find the general solution for \(x\).
   2. Find the equivalent general solution for \(y\). Initially there are 100 predators and 300 prey.
3. Find the particular solutions for \(x\) and \(y\).
4. Determine whether the model predicts that the predators will die out before the prey. \section*{Total Marks for Question Set 1: 37} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
   b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
   A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
   c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
   d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
   e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.
   • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
   • When a value is not given in the paper accept any answer that agrees with the correct value to \(\mathbf { 3 ~ s } . \mathbf { f }\). unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.
   Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
   Candidates using a value of \(9.80,9.81\) or 10 for \(g\) should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
   f Rules for replaced work and multiple attempts:
   • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
   • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
   • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.
   For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
   If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
   h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Abbreviations}

   Abbreviations used in the mark scheme	Meaning
   dep*	Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark
   cao	Correct answer only
   ое	Or equivalent
   rot	Rounded or truncated
   soi	Seen or implied
   www	Without wrong working
   AG	Answer given
   awrt	Anything which rounds to
   BC	By Calculator
   DR	This question included the instruction: In this question you must show detailed reasoning.

   \end{table}

   Question			Answer	Marks	AO	Guidance
   1	(a)		\(\begin{aligned}	\operatorname { un } _ { A B } = \left( \begin{array} { l } 1
   0
   2 \end{array} \right) , \operatorname { un } _ { A C } = \left( \begin{array} { l } 3
   1
   9 \end{array} \right)
   	\Rightarrow A B \times A C = \left( \begin{array} { c } - 2
   - 3
   1 \end{array} \right) \end{aligned}\)	M1 A1	1.1a 1.1	Either correct Cross product BC
   				[3]
   	(b)		\(\begin{aligned}	- 2 x - 3 y + z = d
   	\Rightarrow \text { e.g. } - 2 \times 0 - 3 \times 1 - 1 \times 4 = - 7
   	\Rightarrow 2 x + 3 y - z = 7 \text { oe } \end{aligned}\)	M1 A1 [2]	3.1a 1.1	Use of their vector product and substitution of one point
   2	(a)		\(\begin{aligned}	\operatorname { DR }
   	\arg ( z ) = \frac { \pi } { 6 }
   	\Rightarrow \arg \left( z ^ { n } \right) = \frac { n \pi } { 6 } = 2 \pi
   	\Rightarrow n = 12 \end{aligned}\)	B1 M1 A1 [3]	1.1 2.1 1.1	Equating arg to \(2 \pi\)
   	(b)		4096 or \(2 ^ { 12 }\)	B1 [1]	1.1

   Question		Answer	Marks	AO	Guidance
   \multirow[t]{5}{*}{3}				2.1	\multirow[b]{2}{*}{ Alternatively: \(\begin{aligned} \mathrm { f } ( k + 1 ) = 7 ^ { k + 1 } + 3 ^ { k } = 7.7 ^ { k } + 3.3 ^ { k - 1 } = 7 \left( 4 k - 3 ^ { k - 1 } \right) + 3.3 ^ { k - 1 } = \ldots \end{aligned}\) Alternatively: \(\begin{aligned} \mathrm { f } ( k + 1 ) = 7 ^ { k + 1 } + 3 ^ { k } = 7.7 ^ { k } + 3.3 ^ { k - 1 } = 7 \left( 4 k - 3 ^ { k - 1 } \right) + 3.3 ^ { k - 1 } = \ldots \end{aligned}\) }
   		Base case: For \(n = 1,7 ^ { n } + 3 ^ { n - 1 } = 7 + 1 = 8\) which is a multiple of 4. Assume that \(\mathrm { f } ( k ) = 7 ^ { k } + 3 ^ { k - 1 } = 4 \lambda\) for some integer, \(\lambda\) \(\begin{aligned} \mathrm { f } ( k + 1 ) = 7 ^ { k + 1 } + 3 ^ { k } = 7.7 ^ { k } + ( 7 - 4 ) 3 ^ { k - 1 } = 7 \mathrm { f } ( k ) - 4.3 ^ { k - 1 } = 7.4 \lambda - 4.3 ^ { k - 1 } = 4 \left( 7 \lambda - 3 ^ { k - 1 } \right) = 4 \lambda ^ { \prime } \end{aligned}\)	M1	2.1
   		Where \(\lambda ^ { \prime }\) is an integer because \(\lambda\) is and so rhs is a multiple of 4 .	A1	2.5
   		So if true for \(n = k\), true also for \(n = k + 1\). But it is true for \(n = 1\)	A1	2.2a
   			[5]

   Question			Answer	Marks	AO	Guidance
   \multirow[t]{7}{*}{4}	\multirow[t]{7}{*}{(a)}		\(\begin{aligned}	\frac { 1 } { r ( r + 1 ) ( r + 2 ) } = \frac { A } { r } + \frac { B } { r + 1 } + \frac { C } { r + 2 }
   	\Rightarrow A ( r + 1 ) ( r + 2 ) + B r ( r + 2 ) + C r ( r + 1 ) = 1
   	\Rightarrow A = \frac { 1 } { 2 } \cdot B = - 1 , C = \frac { 1 } { 2 }
   	\Rightarrow \frac { 1 } { r ( r + 1 ) ( r + 2 ) } = \frac { 1 } { 2 } \left( \frac { 1 } { r } - \frac { 2 } { r + 1 } + \frac { 1 } { r + 2 } \right) \end{aligned}\)	M1	3.1a	Attempt to find constants by comparing 3 coefficients or substituting 3 values for \(r\)
   			\(\sum _ { r = 1 } ^ { n } \frac { 1 } { r ( r + 1 ) ( r + 2 ) } =\)
   				M1	1.1	Expand sum using their partial fractions
   				M1	2.1	Cancel terms
   			\(\begin{aligned}	= \frac { 1 } { 2 } \left[ \left( \frac { 1 } { 1 } - \frac { 2 } { 2 } + \frac { 1 } { 2 } \right) + \left( - \frac { 1 } { n + 1 } + \frac { 1 } { n + 2 } \right) \right]
   	= \frac { 1 } { 4 } - \frac { 1 } { 2 ( n + 1 ) } + \frac { 1 } { 2 ( n + 2 ) } \end{aligned}\)
   				A1	1.1	Answer in the form \(\frac { 1 } { 4 } - \frac { 1 } { 2 } \mathrm { f } ( n )\)
   				[6]
   	(b)		\(\frac { 1 } { 4 }\)	B1	2.2a

   Question			Answer	Marks	AO	Guidance
   5	(a)		\(\begin{aligned}	\frac { \mathrm { d } x } { \mathrm {~d} t } = y \Rightarrow \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = \frac { \mathrm { d } y } { \mathrm {~d} t }
   	= 2 y - 5 x
   	\Rightarrow \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } - 5 x \end{aligned}\)	M1 \(\begin{aligned}	\text { A1 }
   	{ [ 2 ] }
   	\end{aligned}\)	3.4 2.1	Differentiate 1st DE and substitute back in AG
   	(b)	(i)	\(\begin{aligned}	\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 0
   	\text { A.E. is } n ^ { 2 } - 2 n + 5 = 0
   	\Rightarrow n = 1 \pm 2 \mathrm { i }
   	\Rightarrow x = \mathrm { e } ^ { t } ( A \sin 2 t + B \cos 2 t ) \end{aligned}\)	M1 \(\begin{aligned} \text { A1 } \text { A1 } \end{aligned}\) [3]	1.1 1.1 2.2a	For auxiliary equation BC
   	(b)	(ii)	\(\begin{aligned}	x = \mathrm { e } ^ { t } ( A \sin 2 t + B \cos 2 t ) \text { and } \frac { \mathrm { d } x } { \mathrm {~d} t } = y
   	\Rightarrow \frac { \mathrm {~d} x } { \mathrm {~d} t } = \mathrm { e } ^ { t } ( A \sin 2 t + B \cos 2 t ) + \mathrm { e } ^ { t } ( 2 A \cos 2 t - 2 B \sin 2 t )
   	\Rightarrow y = \mathrm { e } ^ { t } ( ( A - 2 B ) \sin 2 t + ( 2 A + B ) \cos 2 t ) \end{aligned}\)	M1 A1 [2]	1.1 2.2a
   	(c)		\(\begin{aligned}	x = \mathrm { e } ^ { t } ( A \sin 2 t + B \cos 2 t )
   	x = 1 , t = 0
   	\Rightarrow 1 = B
   	\Rightarrow y = \mathrm { e } ^ { t } ( ( A - 2 ) \sin 2 t + ( 2 A + 1 ) \cos 2 t )
   	y = 3 , t = 0
   	\Rightarrow 3 = 2 A + 1 \Rightarrow A = 1
   	\Rightarrow x = \mathrm { e } ^ { t } ( \sin 2 t + \cos 2 t ) , y = \mathrm { e } ^ { t } ( 3 \cos 2 t - \sin 2 t ) \end{aligned}\)	M1 A1 M1 A1 A1 [5]	3.3 1.1 3.3 1.1 3.3	Translate context into condition to find \(A\) Both stated with \(A\) and \(B\) substituted for

   Question		Answer	Marks	AO	Guidance
   \multirow[t]{4}{*}{(d)}		\multirow[t]{4}{*}{ \(x = 0 \Rightarrow x = \mathrm { e } ^ { t } ( \sin 2 t + \cos 2 t ) = 0\) So 1.12 years \(\begin{aligned} y = 0 \Rightarrow y = \mathrm { e } ^ { t } ( 3 \cos 2 t - \sin 2 t ) = 0 t = 0.624 \ldots \end{aligned}\) So 0.625 years \(0.625 < 1.12\) so the prey die out first. }	M1	3.1b	\multirow[t]{4}{*}{ Set either their \(x = 0\) or their \(y = 0\) and solve BC BC }
   			A1	3.4
   			A1	3.4
   			E1	3.4