Questions - A-Level Maths

OCR Further Mechanics 2021 June Q1

9 marks Standard +0.3

1 A car of mass 800 kg is driven with its engine generating a power of 15 kW .

The car is first driven along a straight horizontal road and accelerates from rest. Assuming that there is no resistance to motion, find the speed of the car after 6 seconds.
The car is next driven at constant speed up a straight road inclined at an angle $\theta$ to the horizontal. The resistance to motion is now modelled as being constant with magnitude 150 N . Given that $\sin \theta = \frac { 1 } { 20 }$, find the speed of the car.
The car is now driven at a constant speed of $30 \mathrm {~ms} ^ { - 1 }$ along the horizontal road pulling a trailer of mass 150 kg which is attached by means of a light rigid horizontal towbar. Assuming that the resistance to motion of the car is three times the resistance to motion of the trailer, find
- the resistance to motion of the car,
- the magnitude of the tension in the towbar.

OCR Further Mechanics 2021 June Q2

11 marks Challenging +1.2

2 One end of a light inextensible string of length 0.8 m is attached to a fixed point, $O$. The other end is attached to a particle $P$ of mass $1.2 \mathrm {~kg} . P$ hangs in equilibrium at a distance of 1.5 m above a horizontal plane. The point on the plane directly below $O$ is $F$.
$P$ is projected horizontally with speed $3.5 \mathrm {~ms} ^ { - 1 }$. The string breaks when $O P$ makes an angle of $\frac { 1 } { 3 } \pi$ radians with the downwards vertical through $O$ (see diagram).
\includegraphics[max width=\textwidth, alt={}, center]{0428f2f2-12c4-4e89-93ab-8cfe2c5aca4a-02_757_889_1482_251}

Find the magnitude of the tension in the string at the instant before the string breaks.
Find the distance between $F$ and the point where $P$ first hits the plane.

OCR Further Mechanics 2021 June Q3

17 marks Standard +0.3

3 This question is about modelling the relation between the pressure, $P$, volume, $V$, and temperature, $\theta$, of a fixed amount of gas in a container whose volume can be varied. The amount of gas is measured in moles; 1 mole is a dimensionless constant representing a fixed number of molecules of gas. Gas temperatures are measured on the Kelvin scale; the unit for temperature is denoted by K . You may assume that temperature is a dimensionless quantity. A gas in a container will always exert an outwards force on the walls of the container. The pressure of the gas is defined to be the magnitude of this force per unit area of the walls, with $P$ always positive. An initial model of the relation is given by $P ^ { \alpha } V ^ { \beta } = n R \theta$, where $n$ is the number of moles of gas present and $R$ is a quantity called the Universal Gas Constant. The value of $R$, correct to 3 significant figures, is $8.31 \mathrm { JK } ^ { - 1 }$.

Show that $[ P ] = \mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 2 }$ and $[ R ] = \mathrm { ML } ^ { 2 } \mathrm {~T} ^ { - 2 }$.
Hence show that $\alpha = 1$ and $\beta = 1$. 5 moles of gas are present in the container which initially has volume $0.03 \mathrm {~m} ^ { 3 }$ and which is maintained at a temperature of 300 K .
Find the pressure of the gas, as predicted by the model. An improved model of the relation is given by $\left( P + \frac { a n ^ { 2 } } { V ^ { 2 } } \right) ( V - n b ) = n R \theta$, where $a$ and $b$ are constants.
Determine the dimensions of $b$ and $a$. The values of $a$ and $b$ (in appropriate units) are measured as being 0.14 and $3.2 \times 10 ^ { - 5 }$ respectively.
Find the pressure of the gas as predicted by the improved model. Suppose that the volume of the container is now reduced to $1.5 \times 10 ^ { - 4 } \mathrm {~m} ^ { 3 }$ while keeping the temperature at 300 K .

By considering the value of the pressure of the gas as predicted by the improved model, comment on the validity of this model in this situation. \section*{Total Marks for Question Set 5: 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

AOs

Guidance

1

(a)

\(\begin{aligned}

15000 \times 6 = \frac { 1 } { 2 } \times 800 v ^ { 2 }

v = \sqrt { 225 } = 15 \mathrm {~m} \mathrm {~s} ^ { - 1 } \end{aligned}\)

M1

A1

[2]

1.1a

1.1

Finding energy input using $W = P t$ and equating to KE gained

\(\begin{aligned}

\text { Or } \frac { 15000 } { v } = 800 \frac { \mathrm {~d} v } { \mathrm {~d} t }

\Rightarrow \frac { 75 } { 2 } \int _ { 0 } ^ { 6 } \mathrm {~d} t = \int _ { 0 } ^ { v } 2 v \mathrm {~d} v \text { oe } \end{aligned}\)

1

(b)

Considering forces along the road: $800 g \sin \theta + 150 = D$ $15000 = D v$ $v = \frac { 15000 } { 800 g \times \frac { 1 } { 20 } + 150 } = \text { awrt } 27.7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$

M1

A1

[3]

1.1a

1.1

Where $D$ is the driving force

Use of $P = F v$

27.675276....

Or $W = 800 g d \sin \theta + 150 d$, where $d$ is distance travelled …

$\ldots$ in time $t$ and $W = 15000 t$ and $v = \frac { d } { t }$

1

(c)

\(\begin{aligned}	D = \frac { 15000 } { 30 } \quad ( = 500 )
	\frac { 15000 } { 30 } = R _ { \text {total } }
	R _ { \mathrm { C } } = \frac { 3 } { 4 } \times 500 = 375 \mathrm {~N} \end{aligned}\)
Car: $500 = T + 375 \Rightarrow T = 125 \mathrm {~N}$

M1

A1

[4]

1.1a

2.2a

1.1

Driving force ...

... equals total resistance

Or Trailer: $T = R _ { \mathrm { T } } = \frac { 1 } { 4 } \times 500 = 125 \mathrm {~N}$

Question		Answer	Marks	AOs	Guidance
\multirow[t]{5}{*}{2}	\multirow[t]{5}{*}{(a)}	$\frac { 1 } { 2 } m \times 3.5 ^ { 2 } = \frac { 1 } { 2 } m v ^ { 2 } + m g \times 0.8 \left( 1 - \cos \frac { 1 } { 3 } \pi \right)$	M1	3.1b	Conservation of energy (could be general angle for RHS)
		$v ^ { 2 } = 3.5 ^ { 2 } - 0.8 g = 4.41$	A1	1.1	Or $v = 2.1$
		$T - 1.2 g \cos \frac { 1 } { 3 } \pi = \frac { 1.2 v ^ { 2 } } { 0.8 }$	M1	3.1b	Resolving weight and use of NII with correct centripetal acceleration
		So tension when string breaks is awrt 12.5 N	A1	3.2a
			[4]

Question

Answer

Marks

AOs

Guidance

\multirow[t]{3}{*}{2}

\multirow[t]{3}{*}{(b)}

$v _ { y } = 2.1 \sin \frac { 1 } { 3 } \pi$ upwards

$P$ starts freefall $1.5 + 0.8 \left( 1 - \cos \frac { 1 } { 3 } \pi \right)$ above plane $- \left( 1.5 + 0.8 \left( 1 - \cos \frac { 1 } { 3 } \pi \right) \right) = \left( 2.1 \sin \frac { 1 } { 3 } \pi \right) t - \frac { 1 } { 2 } g t ^ { 2 }$

$- 1.9 = ( 1.05 \sqrt { 3 } ) t - 4.9 t ^ { 2 }$ or $4.9 t ^ { 2 } - 1.8186 \ldots t - 1.9 = 0$

$t$ = awrt 0.835

Horizontal distance in freefall is $2.1 \cos \frac { 1 } { 3 } \pi \times 0.835 \ldots$

$0.8 \sin \frac { 1 } { 3 } \pi + 2.1 \cos \frac { 1 } { 3 } \pi \times 0.835 \ldots =$ awrt 1.57 m

M1

B1

M1

A1

М1

A1

3.1b

1.1

2.2a

Direction can be implied by eg use of -9.8

(=1.9)

Setting up 3 term quadratic equation in $t$ with adjustment to $\pm 1.5$ as displacement

Correct equation soi

BC

0.877...

horizontal distance while attached plus horizontal distance in freefall.

\(\begin{aligned}

\frac { \sqrt { 331 } + 3 \sqrt { 3 } } { 28 }

\frac { 3 \sqrt { 331 } + 9 \sqrt { 3 } } { 80 } \end{aligned}\)

Alternative solution

Projection velocity is $2.1 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $\frac { 1 } { 3 } \pi$ above horizontal

$P$ starts freefall $1.5 + 0.8 \left( 1 - \cos \frac { 1 } { 3 } \pi \right)$ above plane

Equation of trajectory: $y = x \tan \frac { 1 } { 3 } \pi - \frac { g x ^ { 2 } } { 2 \times 2.1 ^ { 2 } \cos ^ { 2 } \left( \frac { 1 } { 3 } \pi \right) }$ $- 1.9 = x \sqrt { 3 } - \frac { 40 } { 9 } x ^ { 2 } \text { or } 4.444 \ldots x ^ { 2 } - 1.732 \ldots x - 1.9 = 0$

$x =$ awrt 0.877

$0.8 \sin \frac { 1 } { 3 } \pi + 0.877 \ldots =$ awrt 1.57 m

B1ft

B1

М1

A1ft М1

A1

soi; ft their 2.1

(=1.9)

Or suitably adjusted if not using $P$ 's position when string breaks as the origin

Correct equation (with their 2.1) soi oe: formulation of explicit quadratic equation in $x$

BC

$\frac { 3 \sqrt { 331 } + 9 \sqrt { 3 } } { 80 }$

[7]

Question

Answer

Marks

AOs

Guidance

3

(a)

\(\begin{aligned}

{ [ F ] = [ m a ] = \mathrm { MLT } ^ { - 2 } }

{ [ P ] = \frac { [ F ] } { [ A ] } = \mathrm { MLT } ^ { - 2 } \mathrm {~L} ^ { - 2 } = \mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 2 } }

{ [ R ] = [ F ] [ d ] }

{ [ R ] = \mathrm { MLT } ^ { - 2 } \mathrm {~L} = \mathrm { ML } ^ { 2 } \mathrm {~T} ^ { - 2 } } \end{aligned}\)

M1

A1

M1

A1

[4]

1.2

2.2a

3.1b

3.2a

Determining dimensions of energy

3

(b)

\(\begin{aligned}

\left( \mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 2 } \right) ^ { \alpha } \mathrm { L } ^ { 3 \beta } = \mathrm { ML } ^ { 2 } \mathrm {~T} ^ { - 2 }

\mathrm { M } ^ { \alpha } = \mathrm { M } \Rightarrow \alpha = 1

\mathrm {~L} : - \alpha + 3 \beta = 2 \Rightarrow \beta = 1 \text { (AGG) } \end{aligned}\)

M1

A1

[3]

3.3

3.4

1.1

Using their dimensions or similarly with T

$n$ and $\theta$ dimensionless

3

(c)

\(\begin{aligned}

P = \frac { 5 \times 8.31 \times 300 } { 0.03 }

415500 \mathrm { Nm } ^ { - 2 } \end{aligned}\)

M1

A1

[2]

3.4

1.1

Correctly substituting values

With their $\alpha$ and $\beta$

3

(d)

\(\begin{aligned}

{ [ b ] = \mathrm { L } ^ { 3 } }

{ [ a ] = \mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 2 } \left( \mathrm {~L} ^ { 3 } \right) ^ { 2 } }

{ [ a ] = \mathrm { ML } ^ { 5 } \mathrm {~T} ^ { - 2 } } \end{aligned}\)

B1

M1

A1

[3]

2.2a

1.1

Using equality of $\left[ a / V ^ { 2 } \right]$ and $[ P ]$

With $n$ dimensionless

3

(e)

$P + \frac { 0.14 \times 5 ^ { 2 } } { 0.03 ^ { 2 } } = \frac { 5 \times 8.31 \times 300 } { 0.03 - 5 \times 3.2 \times 10 ^ { - 5 } }$

awrt $414000 \mathrm { Nm } ^ { - 2 }$

M1

A1

[2]

3.4

1.1

Correctly substituting values

With their $\alpha$ and $\beta$

3

(f)

$P = \frac { 5 \times 8.31 \times 300 } { 1.5 \times 10 ^ { - 4 } - 5 \times 3.2 \times 10 ^ { - 5 } } - \frac { 0.14 \times 5 ^ { 2 } } { \left( 1.5 \times 10 ^ { - 4 } \right) ^ { 2 } } \approx - 1.4 \times 10 ^ { 9 }$

A negative value is not a valid value for pressure...

...and so the model is not valid for these values

B1

E1

[3]

3.4

3.5b

2.2a

Ignore units

OR from $V - n b < 0$, so since $R , \theta , n$ and $V$ are all positive $P < 0$

OCR Further Mechanics 2021 June Q2

8 marks Challenging +1.2

2 Three particles, $A , B$ and $C$, of masses $2 \mathrm {~kg} , 3 \mathrm {~kg}$ and 5 kg respectively, are at rest in a straight line on a smooth horizontal plane with $B$ between $A$ and $C$. Collisions between $A$ and $B$ are perfectly elastic. The coefficient of restitution for collisions between $B$ and $C$ is $e$.
$A$ is projected towards $B$ with a speed of $5 u \mathrm {~ms} ^ { - 1 }$ (see diagram).
\includegraphics[max width=\textwidth, alt={}, center]{709f3a7a-d857-4813-98ab-de6b41a3a8dc-02_190_885_1151_260} Show that only two collisions occur.

OCR Further Mechanics 2021 June Q3

9 marks Challenging +1.2

3 A particle $P$ of mass 8 kg moves in a straight line on a smooth horizontal plane. At time $t \mathrm {~s}$ the displacement of $P$ from a fixed point $O$ on the line is $x \mathrm {~m}$ and the velocity of $P$ is $v \mathrm {~ms} ^ { - 1 }$. Initially, $P$ is at rest at $O$.
$P$ is acted on by a horizontal force, directed along the line away from $O$, with magnitude proportional to $\sqrt { 9 + v ^ { 2 } }$. When $v = 1.25$, the magnitude of this force is 13 N .

Show that $\frac { 1 } { \sqrt { 9 + v ^ { 2 } } } \frac { \mathrm {~d} v } { \mathrm {~d} t } = \frac { 1 } { 2 }$.
Find an expression for $v$ in terms of $t$ for $t \geqslant 0$.
Find an expression for $x$ in terms of $t$ for $t \geqslant 0$.

OCR Further Mechanics 2021 June Q4

13 marks Standard +0.8

4 Particles $A , B$ and $C$ of masses $2 \mathrm {~kg} , 3 \mathrm {~kg}$ and 5 kg respectively are joined by light rigid rods to form a triangular frame. The frame is placed at rest on a horizontal plane with $A$ at the point $( 0,0 )$, $B$ at the point ( $0.6,0$ ) and $C$ at the point ( $0.4,0.2$ ), where distances in the coordinate system are measured in metres (see Fig. 1). \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{709f3a7a-d857-4813-98ab-de6b41a3a8dc-03_311_661_338_258} \captionsetup{labelformat=empty} \caption{Fig. 1}

\end{figure} $G$, which is the centre of mass of the frame, is at the point $( \bar { x } , \bar { y } )$.

- Show that $\bar { x } = 0.38$.
- Find $\bar { y }$.
- Explain why it would be impossible for the frame to be in equilibrium in a horizontal plane supported at only one point.
A rough plane, $\Pi$, is inclined at an angle $\theta$ to the horizontal where $\sin \theta = \frac { 3 } { 5 }$. The frame is placed on $\Pi$ with $A B$ vertical and $B$ in contact with $\Pi . C$ is in the same vertical plane as $A B$ and a line of greatest slope of $\Pi . C$ is on the down-slope side of $A B$. The frame is kept in equilibrium by a horizontal light elastic string whose natural length is $l \mathrm {~m}$ and whose modulus of elasticity is $g \mathrm {~N}$. The string is attached to $A$ at one end and to a fixed point on $\Pi$ at the other end (see Fig. 2). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{709f3a7a-d857-4813-98ab-de6b41a3a8dc-03_605_828_1525_248} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} The coefficient of friction between $B$ and $\Pi$ is $\mu$.
Show that $l = 0.3$.

Show that $\mu \geqslant \frac { 14 } { 27 }$. \section*{Total Marks for Question Set 6: 38} \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

AOs

Guidance

1

(a)

\(\begin{aligned}

\mathrm { KE } = \frac { 1 } { 2 } \times 2 \times \binom { - 19.5 } { - 60 } \cdot \binom { - 19.5 } { - 60 } = 19.5 ^ { 2 } + 60 ^ { 2 }

\text { awrt } 3980 \mathrm {~J} \end{aligned}\)

M1 \(\begin{aligned}

\text { A1 }

{ [ 2 ] }

\end{aligned}\)

1.1a

1.1

Using KE $= \frac { 1 } { 2 } m \mathbf { v } . \mathbf { v }$ or $\frac { 1 } { 2 } m v ^ { 2 }$ and attempting to find dot product or magnitude

1

(b)

\(\begin{aligned}	\binom { 4 t } { - 2 } = 2 \mathbf { a } = 2 \frac { \mathrm {~d} \mathbf { v } } { \mathrm {~d} t }
	\Rightarrow \mathbf { v } = \binom { t ^ { 2 } } { - t } + \mathbf { u } \end{aligned}\)
$t = 0 \Rightarrow \mathbf { u } = \binom { - 19.5 } { - 60 }$ so $\mathbf { v } = \binom { t ^ { 2 } - 19.5 } { - t - 60 }$ oe $P = \mathbf { F } \cdot \mathbf { v } = \binom { 4 t } { - 2 } \cdot \binom { t ^ { 2 } - 19.5 } { - t - 60 } = 4 t ^ { 3 } - 76 t + 120$
$4 t ^ { 3 } - 76 t + 120 = 0$
$t = 2,3$

M1

A1

М1

A1

[6]

1.1

2.2a

Using $\mathbf { F } = m \mathbf { a }$ with $\mathbf { a } = \frac { \mathrm { d } \mathbf { v } } { \mathrm { d } t }$ soi

Integrating to find $\mathbf { v }$ (u may be missing or may look like a scalar at this stage)

Using $P =$ F.v and attempt at dot product

$\mathbf { B C }$ If $t = - 5$ included then A0

Must be reasonable attempt at integration

Question

Answer

Marks

AOs

Guidance

\multirow[t]{7}{*}{2}

M1

1.1

Conservation of momentum

\multirow{3}{*}{}

A1

1.1

Both

$A$ colliding with $B$ : $2 \times 5 u = 2 v _ { A } + 3 v _ { B }$ $\frac { v _ { B } - v _ { A } } { 5 u } = 1$

$v _ { A } = - u$ and $v _ { B } = 4 u$

M1

1.1

Conservation of momentum

\multirow{4}{*}{}

$V _ { B } = \frac { u ( 3 - 5 e ) } { 2 }$

A1

1.1

Correct $V _ { B }$

$V _ { C } = \frac { 3 u ( 1 + e ) } { 2 }$, but this need not be stated; ignore any wrong value seen, provided not used further

$e \leq 1 \Rightarrow V _ { B } \geq \frac { u ( 3 - 5 ) } { 2 } = - u$

M1

2.1

oe; eg stating that the greatest leftwards speed of $B$ occurs when $e = 1$ and evaluating this

Or could see eg assumption $V _ { B } < v _ { A }$ leading to $e > 1$ stated as a contradiction

So $A$ and $B$ are both travelling in the negative direction, but $\left| v _ { A } \right| \geq \left| v _ { B } \right|$ hence $B$ and $A$ do not collide again, ie there are only two collisions (AG)

E1

2.2a

Conclusion correctly drawn

Allow statement such as ' $B$ is not catching up with $A$ ' in place of a formal inequality, provided correct working is seen

[8]

Question

Answer

Marks

AOs

Guidance

3

(a)

\(\begin{aligned}

F = k \sqrt { 9 + 1.25 ^ { 2 } } = 13

k = 4

4 \sqrt { 9 + v ^ { 2 } } = 8 \frac { \mathrm {~d} v } { \mathrm {~d} t } \Rightarrow \frac { 1 } { \sqrt { 9 + v ^ { 2 } } } \frac { \mathrm {~d} v } { \mathrm {~d} t } = \frac { 1 } { 2 } \quad ( \mathbf { A G } ) \end{aligned}\)

M1

A1

[3]

3.3

1.1

3.3

Substituting $v = 1.25$ and $F = 13$

Clear use of $F = m a$ leading to AG

3

(b)

\(\begin{aligned}

\int \frac { 1 } { \sqrt { 9 + v ^ { 2 } } } \mathrm {~d} v = \frac { 1 } { 2 } t + c

\sinh ^ { - 1 } \left( \frac { v } { 3 } \right) = \frac { 1 } { 2 } t + c

c = 0

v = 3 \sinh \left( \frac { 1 } { 2 } t \right) \end{aligned}\)

M1

B1

A1

[4]

3.4

1.1

3.4

Separating the variables and correctly integrating to obtain bt

Integrating other side to obtain $p \sinh ^ { - 1 } q v$

Substituting $t = 0 , v = 0$ into a solution of the DE to obtain $c$

Condone omission of $c$ for these two M marks

Or using 0 as limits on integrals

3

(c)

\(\begin{aligned}

\frac { \mathrm { d } x } { \mathrm {~d} t } = 3 \sinh \left( \frac { 1 } { 2 } t \right) \Rightarrow x = C + 6 \cosh \left( \frac { 1 } { 2 } t \right)

t = 0 , x = 0 \Rightarrow C = - 6 \Rightarrow x = 6 \left( \cosh \left( \frac { 1 } { 2 } t \right) - 1 \right) \text { oe } \end{aligned}\)

M1

A1

[2]

3.4

2.2a

Replacing $v$ with $\frac { \mathrm { d } x } { \mathrm {~d} t }$ and integrating to obtain $r \cosh s t$

Condone omission of $C$ (or use of ' $c$ ' again)

Question

Answer

Marks

AOs

Guidance

4

(a)

\(\begin{aligned}

\text { Use of } \bar { x } = \frac { \Sigma m x } { \Sigma m } \text { or } \bar { y } = \frac { \Sigma m y } { \Sigma m }

\bar { x } = \frac { 2 \times 0 + 3 \times 0.6 + 5 \times 0.4 } { 2 + 3 + 5 } = \frac { 3.8 } { 10 } = 0.38

\bar { y } = \frac { 2 \times 0 + 3 \times 0 + 5 \times 0.2 } { 2 + 3 + 5 } = \frac { 1 } { 10 } = 0.1 \end{aligned}\)

M1

A1

[3]

1.2

3.3

Values must be substituted

Intermediate step must be seen

4

(b)

$G$ is not on the frame so the weight would have an unbalanced moment about the support point

E1

[1]

2.4

4

(c)

Moments about $B : 0.6 T = 10 g \times 0.1$ $T = \frac { g ( 0.8 - l ) } { l }$

Eliminate $T : \frac { 5 g } { 3 } = \frac { g ( 0.8 - l ) } { l } \Rightarrow 8 l = 2.4 \Rightarrow l = 0.3$ (AG)

M1

A1

[3]

3.4

1.1

Each term must be of the form $F \times d$

Correct form for Hooke's law

Intermediate working must be seen

4

(d)

Any two resolving equations oe, eg: (vertically): $R \cos \theta + F \sin \theta = 10 g$

(horizontally): $R \sin \theta = F \cos \theta + T$ (parallel to plane): $F + T \cos \theta = 10 g \sin \theta$ (perpendicular to plane): $R = T \sin \theta + 10 g \cos \theta$

Use $T = \frac { 5 } { 3 } g , \sin \theta = \frac { 3 } { 5 } , \cos \theta = \frac { 4 } { 5 }$ to derive equations $4 R + 3 F = 50 g$ and $9 R = 12 F + 25 g$ oe $R = 9 g$ and $F = \frac { 14 } { 3 } g$

$F \leq \mu R \Rightarrow \frac { 14 } { 3 } g \leq \mu \times 9 g$

$\therefore \mu \geq \frac { 14 g } { 3 \times 9 g } = \frac { 14 } { 27 } \quad$ (AG)

M1

A1

M1

A1

[6]

3.4

1.1a

1.1

3.4

1.1

Or taking moments eg about the point where the string is attached to the plane $( 1 \times N = ( 0.8 + 0.1 ) \times 10 g )$ or the point where the lines of action of $R$ and the weight intersect: $\frac { 0.1 } { \sin \theta } \times F = \left( 0.6 - \frac { 0.1 } { \tan \theta } \right) \times T$

Forming equations that lead to numerical values for $F$ and $R$

Both

$F \leq \mu R$ with values substituted in

Intermediate working must be seen

$R$ is the normal contact force, $F$ is the frictional force (up the slope)

OCR Further Pure Core 1 2021 June Q2

4 marks Standard +0.3

2 In this question you must show detailed reasoning.
You are given that $z = \sqrt { 3 } + \mathrm { i }$.
$n$ is the smallest positive whole number such that $z ^ { n }$ is a positive whole number.

Determine the value of $n$.
Find the value of $z ^ { n }$.

OCR Further Pure Core 1 2021 June Q3

5 marks Moderate -0.3

3 Prove by induction that, for all positive integers $n , 7 ^ { n } + 3 ^ { n - 1 }$ is a multiple of 4.

OCR Further Pure Core 1 2021 June Q4

7 marks Challenging +1.2

4

Determine an expression for $\sum _ { r = 1 } ^ { n } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }$ giving your answer in the form $\frac { 1 } { 4 } - \frac { 1 } { 2 } \mathrm { f } ( n )$.
Find the value of $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }$.

OCR Further Pure Core 1 2021 June Q5

16 marks Challenging +1.2

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.
$\frac { \mathrm { d } x } { \mathrm {~d} t } = y \quad \frac { \mathrm {~d} y } { \mathrm {~d} t } = 2 y - 5 x$

Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } - 5 x$.
1. Find the general solution for $x$.
2. Find the equivalent general solution for $y$. Initially there are 100 predators and 300 prey.
Find the particular solutions for $x$ and $y$.

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

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

[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

OCR Further Pure Core 1 2021 June Q1

9 marks Standard +0.8

1 The equation of the curve shown on the graph is, in polar coordinates, $r = 3 \sin 2 \theta$ for $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.
\includegraphics[max width=\textwidth, alt={}, center]{4282a136-2ad0-43ce-929a-3d154a4c4af1-02_481_675_440_264}

The greatest value of $r$ on the curve occurs at the point $P$.
1. Show that $\theta = \frac { 1 } { 4 } \pi$ at the point $P$.
2. Find the value of $r$ at the point $P$.
3. Mark the point $P$ on a copy of the graph.
In this question you must show detailed reasoning. Find the exact area of the region enclosed by the curve.

OCR Further Pure Core 1 2021 June Q2

7 marks Standard +0.3

2 You are given that $\mathrm { f } ( x ) = \ln ( 2 + x )$.

Determine the exact value of $\mathrm { f } ^ { \prime } ( 0 )$.
Show that $\mathrm { f } ^ { \prime \prime } ( 0 ) = - \frac { 1 } { 4 }$.
Hence write down the first three terms of the Maclaurin series for $\mathrm { f } ( x )$. You are given that $\mathbf { A } = \left( \begin{array} { c c c } 1 & 2 & 1 \\ 2 & 5 & 2 \\ 3 & - 2 & - 1 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { c c c } 1 & 0 & 1 \\ - 8 & 4 & 0 \\ 19 & - 8 & - 1 \end{array} \right)$.
Find $\mathbf { A B }$.
Hence write down $\mathbf { A } ^ { - 1 }$.
You are given three simultaneous equations $$\begin{array} { r } x + 2 y + z = 0 \\ 2 x + 5 y + 2 z = 1 \\ 3 x - 2 y - z = 4 \end{array}$$
1. Explain how you can tell, without solving them, that there is a unique solution to these equations.
2. Find this unique solution.

OCR Further Pure Core 1 2021 June Q4

9 marks Standard +0.8

4

Given that $u = \tanh x$, use the definition of $\tanh x$ in terms of exponentials to show that $$x = \frac { 1 } { 2 } \ln \left( \frac { 1 + u } { 1 - u } \right)$$
Solve the equation $4 \tanh ^ { 2 } x + \tanh x - 3 = 0$, giving the solution in the form $a \ln b$ where $a$ and $b$ are rational numbers to be determined.
Explain why the equation in part (b) has only one root.

OCR Further Pure Core 1 2021 June Q5

7 marks Standard +0.8

5 In this question you must show detailed reasoning. Find $\int _ { - 1 } ^ { 11 } \frac { 1 } { \sqrt { x ^ { 2 } + 6 x + 13 } } \mathrm {~d} x$ giving your answer in the form $\ln ( p + q \sqrt { 2 } )$ where $p$ and $q$ are integers to be determined. \section*{Total Marks for Question Set 2: 38} \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}

\includegraphics[max width=\textwidth, alt={}]{4282a136-2ad0-43ce-929a-3d154a4c4af1-07_1717_2605_141_242}

Question

Answer

Marks

AO

Guidance

2

(a)

$\mathrm { f } ( x ) = \ln ( 2 + x ) \Rightarrow \mathrm { f } ^ { \prime } ( x ) = \frac { 1 } { 2 + x } \Rightarrow \mathrm { f } ^ { \prime } ( 0 ) = \frac { 1 } { 2 }$

M1

A1

[2]

1.1

Differentiation

(b)

\(\begin{aligned}

\mathrm { f } ^ { \prime } ( x ) = \frac { 1 } { 2 + x } \Rightarrow \mathrm { f } ^ { \prime \prime } ( x ) = - \frac { 1 } { ( 2 + x ) ^ { 2 } }

\Rightarrow \mathrm { f } ^ { \prime \prime } ( 0 ) = \frac { - 1 } { ( 2 + 0 ) ^ { 2 } } = - \frac { 1 } { 4 } \end{aligned}\)

M1

A1

[2]

1.1

2.1

Differentiation

AG

(c)

\(\begin{aligned} \mathrm { f } ( 0 )

= \ln 2

\mathrm { f } ( x )

= \mathrm { f } ( 0 ) + x \mathrm { f } ^ { \prime } ( 0 ) + \frac { x ^ { 2 } } { 2 } \mathrm { f } ^ { \prime \prime } ( 0 )

= \ln 2 + \frac { 1 } { 2 } x - \frac { 1 } { 8 } x ^ { 2 } \end{aligned}\)

B1

B1FT

[3]

1.1

soi

2 two terms correct $3 ^ { \text {rd } }$ term also correct.

FT their values

3

(a)

\(\left( \begin{array} { l l l } 4

0

4

0

4 \end{array} \right)\)

B1

[1]

1.1

(b)

\(\frac { 1 } { 4 } \left( \begin{array} { c c c } 1

0

1

- 8

4

0

19

- 8

- 1 \end{array} \right)\) oe

B1

[1]

1.1

(c)

(i)

E.g. Because expressed in matrix form the system of equations is $\mathbf { A X } = \mathbf { C }$ where $\mathbf { A }$ is the matrix is $\mathbf { A }$ from above

E.g. And it has been established that an inverse exists.

B1

[2]

2.4

Question

Answer

Marks

AO

Guidance

(c)

(ii)

\(\begin{aligned}

\left( \begin{array} { c c c } 1

2

1

2

5

2

3

- 2

- 1 \end{array} \right) \left( \begin{array} { l } x

y

z \end{array} \right) = \left( \begin{array} { l } 0

1

4 \end{array} \right)

\Rightarrow \left( \begin{array} { l } x

y

z \end{array} \right) = \frac { 1 } { 4 } \left( \begin{array} { c c c } 1

0

1

- 8

4

0

19

- 8

- 1 \end{array} \right) \left( \begin{array} { l } 0

1

4 \end{array} \right) = \left( \begin{array} { c } 1

1

- 3 \end{array} \right)

\Rightarrow x = 1 , y = 1 , z = - 3 \end{aligned}\)

M1

A1

[2]

3.1a

1.1

Use B and multiply

May be seen in part (i)

BC

4

(a)

\(\begin{aligned}

\tanh x = \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } = u

\Rightarrow \mathrm { e } ^ { x } ( 1 - u ) = \mathrm { e } ^ { - x } ( 1 + u )

\Rightarrow \mathrm { e } ^ { 2 x } = \left( \frac { 1 + u } { 1 - u } \right) \Rightarrow x = \frac { 1 } { 2 } \ln \left( \frac { 1 + u } { 1 - u } \right) \end{aligned}\)

M1

A1

M1 A1

[4]

3.1a

2.1

3.1a

1.1

Use of exponentials

Attempt to find $\mathrm { e } ^ { 2 x }$

(b)

\(\begin{aligned}

4 \tanh ^ { 2 } x + \tanh x - 3 = 0

\Rightarrow ( 4 u - 3 ) ( u + 1 ) = 0 \Rightarrow u = \frac { 3 } { 4 } , ( - 1 )

\Rightarrow x = \frac { 1 } { 2 } \ln \left( \frac { 1 + \frac { 3 } { 4 } } { 1 - \frac { 3 } { 4 } } \right) = \frac { 1 } { 2 } \ln 7 \quad \text { oe }

\text { i.e. } a = \frac { 1 } { 2 } , b = 7 \quad \text { oe } \end{aligned}\)

M1

A1

M1

A1

[4]

1.1a

1.1

Solve quadratic

Substitute into the standard form

(c)

E.g. Because - $1 < \tanh x < 1$

B1

[1]

2.4

e.g. One root has been rejected as outside range.

Question

Answer

Marks

AO

Guidance

5

DR \(\begin{aligned}	\int _ { - 1 } ^ { 11 } \frac { 1 } { \sqrt { x ^ { 2 } + 6 x + 13 } } \mathrm {~d} x
	x ^ { 2 } + 6 x + 13 \equiv ( x + 3 ) ^ { 2 } + 4
	\Rightarrow \int _ { - 1 } ^ { 11 } \frac { 1 } { \sqrt { x ^ { 2 } + 6 x + 13 } } \mathrm {~d} x = \int _ { - 1 } ^ { 11 } \frac { 1 } { \sqrt { ( x + 3 ) ^ { 2 } + 4 } } \mathrm {~d} x
	= \left[ \ln \left( ( x + 3 ) + \sqrt { ( x + 3 ) ^ { 2 } + 4 } \right) \right] _ { - 1 } ^ { 11 } \end{aligned}\)
i.e. $p = 3 , q = 2$

M1

3.1a

Attempt to complete the square

OCR Further Pure Core 1 2021 June Q1

4 marks Standard +0.3

1 In this question you must show detailed reasoning.
The quadratic equation $x ^ { 2 } - 2 x + 5 = 0$ has roots $\alpha$ and $\beta$.

Write down the values of $\alpha + \beta$ and $\alpha \beta$.
Hence find a quadratic equation with roots $\alpha + \frac { 1 } { \beta }$ and $\beta + \frac { 1 } { \alpha }$. Using the formulae for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$, show that $\sum _ { r = 1 } ^ { 10 } r ( 3 r - 2 ) = 1045$.

OCR Further Pure Core 1 2021 June Q3

6 marks Standard +0.8

3 The equation of a plane is $4 x + 2 y + z = 7$.
The point $A$ has coordinates $( 9,6,1 )$ and the point $B$ is the reflection of $A$ in the plane.
Find the coordinates of the point $B$. You are given the matrix $\mathbf { A }$ where $\mathbf { A } = \left( \begin{array} { l l l } a & 2 & 0 \\ 0 & a & 2 \\ 4 & 5 & 1 \end{array} \right)$.

Find, in terms of $a$, the determinant of $\mathbf { A }$, simplifying your answer.
Hence find the values of $a$ for which $\mathbf { A }$ is singular. You are given the following equations which are to be solved simultaneously. $$\begin{aligned} a x + 2 y & = 6 \\ a y + 2 z & = 8 \\ 4 x + 5 y + z & = 16 \end{aligned}$$
For each of the values of $a$ found in part (b) determine whether the equations have
- a unique solution, which should be found, or
- an infinite set of solutions or
- no solution.
A particle is suspended in a resistive medium from one end of a light spring. The other end of the spring is attached to a point which is made to oscillate in a vertical line. The displacement of the particle may be modelled by the differential equation
$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 10 \sin t$
where $x$ is the displacement of the particle below the equilibrium position at time $t$.
When $t = 0$ the particle is stationary and its displacement is 2 .
Find the particular solution of the differential equation.

Write down an approximate equation for the displacement when $t$ is large. \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}

\includegraphics[max width=\textwidth, alt={}]{5392d682-c940-41bb-bcaf-10e7a7440c80-06_1740_2523_141_242}

Question

Answer

Marks

AO

Guidance

4

(a)

$\operatorname { det } \mathbf { A } = a ^ { 2 } - 10 a + 16$

M1

A1

[2]

1.1a

1.1

Attempt to work out the determinant

(b)

$a ^ { 2 } - 10 a + 16 = 0 \Rightarrow ( a - 2 ) ( a - 8 ) = 0 \Rightarrow a = 2,8$

M1

A1

[2]

1.1a

1.1

Solving their quadratic soi

(c)

For both values there is no unique solution as $\operatorname { det } \mathbf { A } = 0$
For $a = 2$, equations are: \(\begin{aligned}	p _ { 1 } : 2 x + 2 y = 6
	p _ { 2 } : 2 y + 2 z = 8
	p _ { 3 } : 4 x + 5 y + z = 16
	2 p _ { 1 } + \frac { 1 } { 2 } p _ { 2 } = p _ { 3 } \end{aligned}\)
So there is an infinite set of solutions.
For $a = 8$, equations are: \(\begin{aligned}	p _ { 1 } : 8 x + 2 y = 6
	p _ { 2 } : 8 y + 2 z = 8
	p _ { 3 } : 4 x + 5 y + z = 16
	\frac { 1 } { 2 } p _ { 1 } + \frac { 1 } { 2 } p _ { 2 } \neq p _ { 3 } \text { as it gives } 4 x + 5 y + z = 7 \end{aligned}\)
and $16 \neq 7$ so no solution

B1

М1

A1

М1

A1

[7]

2.4

2.1

1.1

2.2a

Soi by correct answers

Substitute one of their values and solve

"correct answers" means solns are either infinite or nonexistent.

\includegraphics[max width=\textwidth, alt={}]{5392d682-c940-41bb-bcaf-10e7a7440c80-09_1653_2523_141_242}

(b)

\(\begin{aligned}

x = \mathrm { e } ^ { - t } ( A \cos 2 t + B \sin 2 t ) + 2 \sin t - \cos t

\text { As } t \rightarrow \infty , \quad \mathrm { e } ^ { - t } \rightarrow 0

\Rightarrow x \approx 2 \sin t - \cos t \end{aligned}\)

М1

A1

[2]

3.4

Consider behaviour of $\mathrm { e } ^ { - \mathrm { t } }$

Accept = ft their equation from (a)

Accept this final line for 2 marks

OCR Further Pure Core 1 2021 June Q1

3 marks Standard +0.3

1 Indicate by shading on an Argand diagram the region $$\{ z : | z | \leqslant | z - 4 | \} \cap \{ z : | z - 3 - 2 i | \leqslant 2 \} .$$

OCR Further Pure Core 1 2021 June Q2

4 marks Standard +0.3

2 In this question you must show detailed reasoning. You are given that $x = 2 + 5 \mathrm { i }$ is a root of the equation $x ^ { 3 } - 2 x ^ { 2 } + 21 x + 58 = 0$.
Solve the equation.

OCR Further Pure Core 1 2021 June Q3

7 marks Standard +0.3

3 The diagram shows part of the curve $y = 5 \cosh x + 3 \sinh x$.
\includegraphics[max width=\textwidth, alt={}, center]{ef967953-70b5-4dd1-a342-ad488b5fa79f-02_426_661_906_260}

Solve the equation $5 \cosh x + 3 \sinh x = 4$ giving your solution in exact form.
In this question you must show detailed reasoning. Find $\int _ { - 1 } ^ { 1 } ( 5 \cosh x + 3 \sinh x ) \mathrm { d } x$ giving your answer in the form $a \mathrm { e } + \frac { b } { \mathrm { e } }$ where $a$ and $b$ are integers to be determined.

OCR Further Pure Core 1 2021 June Q4

6 marks Challenging +1.2

4 You are given that $y = \tan ^ { - 1 } \sqrt { 2 x }$.

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
Show that $\int _ { \frac { 1 } { 6 } } ^ { \frac { 1 } { 2 } } \frac { \sqrt { x } } { \left( x + 2 x ^ { 2 } \right) } \mathrm { d } x = k \pi$ where $k$ is a number to be determined in exact form.

OCR Further Pure Core 1 2021 June Q5

6 marks Standard +0.8

5 The function $\operatorname { sech } x$ is defined by $\operatorname { sech } x = \frac { 1 } { \cosh x }$.

Show that $\operatorname { sech } x = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 }$.
Using a suitable substitution, find $\int \operatorname { sech } x \mathrm {~d} x$.

OCR Further Pure Core 1 2021 June Q6

12 marks Challenging +1.2

6 In this question you must show detailed reasoning.
You are given the complex number $\omega = \cos \frac { 2 } { 5 } \pi + i \sin \frac { 2 } { 5 } \pi$ and the equation $z ^ { 5 } = 1$.

Show that $\omega$ is a root of the equation.
Write down the other four roots of the equation.
Show that $\omega + \omega ^ { 2 } + \omega ^ { 3 } + \omega ^ { 4 } = - 1$.
Hence show that $\left( \omega + \frac { 1 } { \omega } \right) ^ { 2 } + \left( \omega + \frac { 1 } { \omega } \right) - 1 = 0$.

Hence determine the value of $\cos \frac { 2 } { 5 } \pi$ in the form $a + b \sqrt { c }$ where $a , b$ and $c$ are rational numbers to be found. Total Marks for Question Set 4: 38 \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.
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. 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.
g 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
oe	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

\multirow[t]{2}{*}{3

}

\(\begin{aligned}	5 \cosh x + 3 \sinh x = 4
	\Rightarrow 5 \left( \frac { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } { 2 } \right) + 3 \left( \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { 2 } \right) = 4
	\Rightarrow 4 \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } = 4 \end{aligned}\)
Alternatively: \(\begin{aligned}	5 \cosh x + 3 \sinh x \equiv R \cosh ( x + \alpha )
	\text { where } R = \sqrt { 25 - 9 } = 4
	\tanh \alpha = \frac { 3 } { 5 } \Rightarrow \alpha = \tanh ^ { - 1 } \frac { 3 } { 5 } = \frac { 1 } { 2 } \ln \left( \frac { 1 + \frac { 3 } { 5 } } { 1 - \frac { 3 } { 5 } } \right) = \frac { 1 } { 2 } \ln 4 = \ln 2 \text { M1 A1 }
	\Rightarrow 4 \cosh ( x + \alpha ) = 4 \Rightarrow \cosh ( x + \alpha ) = 0
	\Rightarrow x = - \alpha = - \ln 2 \end{aligned}\)

М1

3.1a

Use of exponentials

Multiply by $\mathrm { e } ^ { x }$

Alternatively make cosh the subject, square and use Pythagoras to give quadratic in cosh

Alternatively use compound angle formula

[4]

(b)

DR \(\begin{aligned}

\int _ { - 1 } ^ { 1 } ( 5 \cosh x + 3 \sinh x ) \mathrm { d } x = [ 5 \sinh x + 3 \cosh x ] _ { - 1 } ^ { 1 }

= \left( 5 \frac { \mathrm { e } ^ { 1 } - \mathrm { e } ^ { - 1 } } { 2 } + 3 \frac { \mathrm { e } ^ { 1 } + \mathrm { e } ^ { - 1 } } { 2 } \right) - \left( 5 \frac { \mathrm { e } ^ { - 1 } - \mathrm { e } ^ { 1 } } { 2 } + 3 \frac { \mathrm { e } ^ { - 1 } + \mathrm { e } ^ { 1 } } { 2 } \right)

= \left( 4 \mathrm { e } ^ { 1 } - \mathrm { e } ^ { - 1 } \right) - \left( 4 \mathrm { e } ^ { - 1 } - \mathrm { e } ^ { 1 } \right)

= 5 \mathrm { e } - \frac { 5 } { \mathrm { e } } \end{aligned}\)

М1

1.1

Attempt at integral (i.e. one function changed)

Alternatively:

M1 convert (including possibly using result from (a))

M1 integrate and use limits correctly

\includegraphics[max width=\textwidth, alt={}]{ef967953-70b5-4dd1-a342-ad488b5fa79f-09_1333_2525_159_242}

Question

Answer

Marks

AO

Guidance

4

(b)

\(\begin{aligned}

\int _ { 1 / 6 } ^ { 1 / 2 } \frac { 1 } { ( 1 + 2 x ) \sqrt { x } } \mathrm {~d} x = \sqrt { 2 } \int _ { 1 / 6 } ^ { 1 / 2 } \frac { 1 } { ( 1 + 2 x ) \sqrt { 2 x } } \mathrm {~d} x

= \sqrt { 2 } \left[ \tan ^ { - 1 } \sqrt { 2 x } \right] _ { 1 / 6 } ^ { 1 / 2 } = \sqrt { 2 } \left( \tan ^ { - 1 } 1 - \tan ^ { - 1 } \frac { 1 } { \sqrt { 3 } } \right)

= \sqrt { 2 } \left( \frac { \pi } { 4 } - \frac { \pi } { 6 } \right) = \frac { \sqrt { 2 } } { 12 } \pi

\text { So } k = \frac { \sqrt { 2 } } { 12 } \end{aligned}\)

M1

A1

М1

A1

[4]

3.1a

1.1

Get into form of (a). Ignore limits

Correct form

Use (a) and correct limits in correct order.

ое

Alternatively: \(\begin{aligned}

\text { Let } u = \sqrt { x }

\mathrm {~d} u = \frac { 1 } { 2 \sqrt { x } } \mathrm {~d} x \Rightarrow \mathrm {~d} x = 2 \sqrt { x } \mathrm {~d} u = 2 u \mathrm {~d} u

\int _ { \frac { 1 } { 6 } } ^ { \frac { 1 } { 2 } } \frac { \sqrt { x } } { \left( x + 2 x ^ { 2 } \right) } \mathrm { d } x = \int _ { x = \frac { 1 } { 6 } } ^ { x = \frac { 1 } { 2 } } \frac { u } { \left( u ^ { 2 } + 2 u ^ { 4 } \right) } 2 u \mathrm {~d} u = 2 \int _ { x = \frac { 1 } { 6 } } ^ { x = \frac { 1 } { 2 } } \frac { 1 } { \left( 1 + 2 u ^ { 2 } \right) } \mathrm { d } u

= \sqrt { 2 } \left[ \tan ^ { - 1 } u \sqrt { 2 } \right] _ { x = \frac { 1 } { 6 } } ^ { x = \frac { 1 } { 2 } }

= \sqrt { 2 } \left[ \tan ^ { - 1 } \sqrt { 2 x } \right] _ { x = \frac { 1 } { 6 } } ^ { x = \frac { 1 } { 2 } } = \sqrt { 2 } \left( \tan ^ { - 1 } 1 - \tan ^ { - 1 } \frac { 1 } { \sqrt { 3 } } \right) = \sqrt { 2 } \left( \frac { \pi } { 4 } - \frac { \pi } { 6 } \right)

= \frac { \pi \sqrt { 2 } } { 12 } \end{aligned}\)

Make a substitution

Get into correct form

Use standard result with correct limits in correct order

Question

Answer

Marks

AO

Guidance

5

(a)

\(\begin{aligned}

\cosh x = \frac { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } { 2 } = \frac { \mathrm { e } ^ { 2 x } + 1 } { 2 \mathrm { e } ^ { x } }

\Rightarrow \operatorname { sech } x = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 } \quad \text { AG } \end{aligned}\)

М1

A1

[2]

1.1

2.1

Use of $\cosh x$ in exponentials

(b)

\(\begin{aligned}

u = \mathrm { e } ^ { x } \Rightarrow \mathrm {~d} u = \mathrm { e } ^ { x } \mathrm {~d} x

\Rightarrow \mathrm {~d} x = \frac { \mathrm { d } u } { u }

\Rightarrow \int \operatorname { sech } x \mathrm {~d} x = \int \left( \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 } \right) \mathrm { d } x

= \int \frac { 2 u } { u ^ { 2 } + 1 } \cdot \frac { \mathrm {~d} u } { u }

= 2 \tan ^ { - 1 } ( u ) + c = 2 \tan ^ { - 1 } \left( \mathrm { e } ^ { x } \right) + c

\text { Alternatively: }

u = \sinh x \Rightarrow \mathrm {~d} u = \cosh x \mathrm {~d} x

\Rightarrow \int \operatorname { sech } x \mathrm {~d} x = \int \frac { 1 } { \cosh x } \cdot \frac { \mathrm {~d} u } { \cosh x } = \int \frac { \mathrm { d } u } { \cosh ^ { 2 } x }

= \int \frac { \mathrm { d } u } { 1 + \sinh ^ { 2 } x } = = \int \frac { \mathrm { d } u } { 1 + u ^ { 2 } }

= \tan ^ { - 1 } u + c

= \tan ^ { - 1 } ( \sinh x ) + c

\text { Alternatively: }

\int \operatorname { sech } x \mathrm {~d} x = \int \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 } \mathrm {~d} x

\text { Let } \mathrm { e } ^ { x } = \tan u \Rightarrow \mathrm { e } ^ { x } \mathrm {~d} x = \sec ^ { 2 } u \mathrm {~d} u \Rightarrow \mathrm {~d} x = \frac { \sec ^ { 2 } u } { \tan u } \mathrm {~d} u

\Rightarrow \int \operatorname { sech } x \mathrm {~d} x = \int \frac { 2 \tan u } { \tan ^ { 2 } u + 1 } \cdot \frac { \sec ^ { 2 } u } { \tan u } \mathrm {~d} u = 2 \int \mathrm {~d} u

= 2 u + c

= 2 \tan ^ { - 1 } \left( \mathrm { e } ^ { x } \right) + c \end{aligned}\)

М1

A1

М1

A1

3.1a

Must include $c$

Allow absence of $\mathrm { d } u$

Question

Answer

Marks

AO

Guidance

6

(a)

DR \(\begin{aligned}

\omega = \cos \frac { 2 \pi } { 5 } + \mathrm { i } \sin \frac { 2 \pi } { 5 }

\Rightarrow \omega ^ { 5 } = \left( \cos \frac { 2 \pi } { 5 } + \mathrm { i } \sin \frac { 2 \pi } { 5 } \right) ^ { 5 } = \cos 2 \pi + \mathrm { i } \sin 2 \pi = 1 + 0 \mathrm { i } = 1 \end{aligned}\)

M1 A1

[2]

2.1 1.1

Finding $\omega ^ { 5 }$ AG

Use of exponentials is satisfactory Could be argued backwards

(b)

$\omega ^ { 2 } , \omega ^ { 3 } , \omega ^ { 4 } , 1$

B1

[1]

1.1

Alternative:

Roots are $\cos \frac { 2 k \pi } { 5 } + \mathrm { i } \sin \frac { 2 k \pi } { 5 }$ for $k = 2,3,4$

and 1 ( or $k = 5$ )

Exponentials satisfactory

(c)

DR \(\begin{aligned}	\omega ^ { 5 } - 1 = 0
	\Rightarrow ( \omega - 1 ) \left( \omega ^ { 4 } + \omega ^ { 3 } + \omega ^ { 2 } + \omega + 1 \right) = 0
	\Rightarrow \omega ^ { 4 } + \omega ^ { 3 } + \omega ^ { 2 } + \omega = - 1 \end{aligned}\)
Alternatively: $1 + \omega + \omega ^ { 2 } + \omega ^ { 3 } + \omega ^ { 4 } = \frac { 1 - \omega ^ { 5 } } { 1 - \omega } = \frac { 0 } { 1 - \omega }$
M1
or $\omega + \omega ^ { 2 } + \omega ^ { 3 } + \omega ^ { 4 } = \omega \left( \frac { 1 - \omega ^ { 4 } } { 1 - \omega } \right) = \left( \frac { \omega - \omega ^ { 5 } } { 1 - \omega } \right) = \left( \frac { \omega - 1 } { 1 - \omega } \right) = - 1$ A1
Alternatively:
sum of roots $= = - \frac { b } { a }$ where $b = 0 \mathrm { M } 1 -$ needs
explanation - i.e. coefficient of $z ^ { 4 }$ term $= 0$

Use equation and $\omega$

AG

Question

Answer

Marks

AO

Guidance

6

(d)

\(\begin{aligned}

\mathbf { A G }

\left( \omega + \frac { 1 } { \omega } \right) ^ { 2 } + \left( \omega + \frac { 1 } { \omega } \right) - 1 = \omega ^ { 2 } + 2 + \frac { 1 } { \omega ^ { 2 } } + \omega + \frac { 1 } { \omega } - 1

= \frac { 1 } { \omega ^ { 2 } } \left( \omega ^ { 4 } + \omega ^ { 2 } + 1 + \omega ^ { 3 } + \omega \right) = 0

\text { Since } \frac { 1 } { \omega ^ { 2 } } \neq 0 , \omega ^ { 4 } + \omega ^ { 2 } + 1 + \omega ^ { 3 } + \omega = 0

\text { or from part (c) } \end{aligned}\)

M1

A1

2.1

1.1

2.2a

Multiply out

Alternatively: \(\begin{aligned}

\omega ^ { 4 } + \omega ^ { 3 } + \omega ^ { 2 } + \omega + 1 = 0

\Rightarrow \omega ^ { 2 } \left( \omega ^ { 2 } + \omega + 1 + \frac { 1 } { \omega } + \frac { 1 } { \omega ^ { 2 } } \right) = 0

\Rightarrow \omega ^ { 2 } \left( \left( \omega ^ { 2 } + 2 + \frac { 1 } { \omega ^ { 2 } } \right) + \left( \omega + \frac { 1 } { \omega } \right) + 1 - 2 \right) = 0

\Rightarrow \omega ^ { 2 } \left( \left( \omega + \frac { 1 } { \omega } \right) ^ { 2 } + \left( \omega + \frac { 1 } { \omega } \right) - 1 \right) = 0

\text { Since } \omega ^ { 2 } \neq 0 , \left( \omega + \frac { 1 } { \omega } \right) ^ { 2 } + \left( \omega + \frac { 1 } { \omega } \right) - 1 = 0 \quad \mathbf { A } \mathbf { 1 } \end{aligned}\)

[3]

For extraction of $\omega ^ { 2 }$

For dealing with the 2

Question		Answer	Marks	AO	Guidance
\multirow[t]{4}{*}{6}	(e)	\(\begin{aligned}	\frac { 1 } { \omega } = \cos \frac { 2 \pi } { 5 } - i \sin \frac { 2 \pi } { 5 }
	\Rightarrow \left( \omega + \frac { 1 } { \omega } \right) = 2 \cos \frac { 2 \pi } { 5 } \end{aligned}\)	B1	3.1a	$\omega + \frac { 1 } { \omega }$ may be seen in (d)
		\(\begin{aligned}	\text { From (iii) solving quadratic: } \left( \omega + \frac { 1 } { \omega } \right) = \frac { - 1 \pm \sqrt { 5 } } { 2 }
	\Rightarrow 2 \cos \frac { 2 \pi } { 5 } = \frac { \sqrt { 5 } - 1 } { 2 } \Rightarrow \cos \frac { 2 \pi } { 5 } = \frac { \sqrt { 5 } - 1 } { 4 }
	= - \frac { 1 } { 4 } + \frac { \sqrt { 5 } } { 4 } \text { or } - \frac { 1 } { 4 } + \frac { 1 } { 4 } \sqrt { 5 } \text { or } - 0.25 + 0.25 \sqrt { 5 } \end{aligned}\)	М1	3.1a	BC
			A1	2.3	For taking the valid value and presenting in correct form ое	No other forms acceptable
			[4]

OCR Further Pure Core 1 2021 June Q2

4 marks Moderate -0.8

2 In this question you must show detailed reasoning.

Determine the square roots of 25 i in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $0 \leqslant \theta < 2 \pi$.
Illustrate the number 25 i and its square roots on an Argand diagram.

OCR Further Pure Core 1 2021 June Q3

5 marks Challenging +1.2

3 By expanding $\left( z ^ { 2 } + \frac { 1 } { z ^ { 2 } } \right) ^ { 3 }$, where $z = \mathrm { e } ^ { \mathrm { i } \theta }$, show that $4 \cos ^ { 3 } 2 \theta = \cos 6 \theta + 3 \cos 2 \theta$.

OCR Further Pure Core 1 2021 June Q4

5 marks Standard +0.8

4 The equations of two non-intersecting lines, $l _ { 1 }$ and $l _ { 2 }$, are
$l _ { 1 } : \mathbf { r } = \left( \begin{array} { c } 1 \\ 2 \\ - 1 \end{array} \right) + \lambda \left( \begin{array} { c } 2 \\ 1 \\ - 2 \end{array} \right) \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { c } 2 \\ 2 \\ - 3 \end{array} \right) + \mu \left( \begin{array} { c } 1 \\ - 1 \\ 4 \end{array} \right)$.
Find the shortest distance between lines $l _ { 1 }$ and $l _ { 2 }$.

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