a level further mathematics - the maths orchard · the pearson edexcel level 3 advanced gce in...
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A Level Further Mathematics
Sample Assessment MaterialsPearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0)First teaching from September 2017First certifi cation from 2019 Issue 1
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All information in this document is correct at time of publication.
Original origami artwork: Mark Bolitho
Origami photography: Pearson Education Ltd/Naki Kouyioumtzis
ISBN 978 1 4469 3352 7
All the material in this publication is copyright
© Pearson Education Limited 2017
Contents
Introduction 1
General marking guidance 3
Paper 1 – sample question paper and mark scheme 5
Paper 2 – sample question paper and mark scheme 42
Paper 3A – sample question paper and mark scheme 71
Paper 4A– sample question paper and mark scheme 101
Paper 3B/4B – sample question paper and mark scheme 143
Paper 4E– sample question paper and mark scheme 177
Paper 3C/4C – sample question paper and mark scheme 213
Paper 4F – sample question paper and mark scheme 249
Paper 3D/4D – sample question paper and mark scheme 285
Paper 2G – sample question paper and mark scheme 325Mathematical formulae and statistical tables
Introduction
The Pearson Edexcel Level 3 Advanced GCE in Further Mathematics is designed for use in schools and colleges. It is part of a suite of AS/A Level qualifications offered by Pearson.
These sample assessment materials have been developed to support this qualification and will be used as the benchmark to develop the assessment students will take.
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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General marking guidance
All candidates must receive the same treatment. Examiners must mark the last
candidate in exactly the same way as they mark the first.
Mark schemes should be applied positively. Candidates must be rewarded for what
they have shown they can do rather than be penalised for omissions.
Examiners should mark according to the mark scheme – not according to their
perception of where the grade boundaries may lie.
All the marks on the mark scheme are designed to be awarded. Examiners should
always award full marks if deserved, i.e. if the answer matches the mark scheme.
Examiners should also be prepared to award zero marks if the candidate’s response is
not worthy of credit according to the mark scheme.
Where some judgement is required, mark schemes will provide the principles by
which marks will be awarded and exemplification/indicative content will not be
exhaustive. However different examples of responses will be provided at
standardisation.
When examiners are in doubt regarding the application of the mark scheme to a
candidate’s response, a senior examiner must be consulted before a mark is given.
Crossed-out work should be marked unless the candidate has replaced it with an
alternative response.
Specific guidance for mathematics
1. These mark schemes use the following types of marks:
M marks: Method marks are awarded for ‘knowing a method and attempting to
apply it’, unless otherwise indicated.
A marks: Accuracy marks can only be awarded if the relevant method (M) marks
have been earned.
B marks are unconditional accuracy marks (independent of M marks)
Marks should not be subdivided.
2. Abbreviations
These are some of the traditional marking abbreviations that may appear in the mark
schemes.
bod benefit of doubt
ft follow through
this symbol is used for
correct ft
cao correct answer only
cso correct solution only.
There must be no errors in
this part of the question to
obtain this mark
isw ignore subsequent working
awrt answers which round to
SC: special case
o.e. or equivalent (and
appropriate)
d… dependent
or dep
indep independent
dp decimal places
sf significant figures
The answer is printed on
the paper or ag- answer
given
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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or d… The second mark is
dependent on gaining the
first mark
3. All M marks are follow through.
All A marks are ‘correct answer only’ (cao.), unless shown, for example, as A1 ft to
indicate that previous wrong working is to be followed through. After a misread
however, the subsequent A marks affected are treated as A ft, but answers that don’t
logically make sense e.g. if an answer given for a probability is >1 or <0, should
never be awarded A marks.
4. For misreading which does not alter the character of a question or materially simplify
it, deduct two from any A or B marks gained, in that part of the question affected.
5. Where a candidate has made multiple responses and indicates which response they
wish to submit, examiners should mark this response. If there are several attempts at
a question which have not been crossed out, examiners should mark the final answer
which is the answer that is the most complete.
6. Ignore wrong working or incorrect statements following a correct answer.
7. Mark schemes will firstly show the solution judged to be the most common response
expected from candidates. Where appropriate, alternative answers are provided in the
notes. If examiners are not sure if an answer is acceptable, they will check the mark
scheme to see if an alternative answer is given for the method used. If no such
alternative answer is provided but deemed to be valid, examiners must escalate the
response to a senior examiner to review.
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Pearson Edexcel Level 3 GCE
Further Mathematics Advanced Paper 1: Core Pure Mathematics 1
Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes
Paper Reference(s)
9FM0/01
You must have: Mathematical Formulae and Statistical Tables, calculator
Candidates may use any calculator permitted by Pearson regulations. Calculators must not have the facility for algebraic manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them.
Instructions
Use black ink or ball-point pen.
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
Fill in the boxes at the top of this page with your name, centre number and candidate number.
Answer all questions and ensure that your answers to parts of questions are clearly labelled.
Answer the questions in the spaces provided – there may be more space than you need.
You should show sufficient working to make your methods clear. Answers without working may not gain full credit.
Answers should be given to three significant figures unless otherwise stated. Information
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
There are 9 questions in this question paper. The total mark for this paper is 75.
The marks for each question are shown in brackets – use this as a guide as to how much time to spend on each question.
Advice
Read each question carefully before you start to answer it.
Try to answer every question.
Check your answers if you have time at the end.
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Answer ALL questions. Write your answers in the spaces provided.
1. Prove that
1
1
1 3 12 2 3
n
r
n an b
r r n n
where a and b are constants to be found.
(5)
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Question 1 continued
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(Total for Question 1 is 5 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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2. Prove by induction that for all positive integers n,
3 1 2 1f 2 3 5n nn
is divisible by 17
(6)
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Question 2 continued
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(Total for Question 2 is 6 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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3.
4 3 2f ( ) 6 65z z az z bz
where a and b are real constants.
Given that 3 2iz is a root of the equation f(z) = 0, show the roots of f(z) = 0 on a
single Argand diagram.
(9)
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Question 3 continued
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(Total for Question 3 is 9 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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4.
Figure 1
The curve C shown in Figure 1 has polar equation
r = 4 + cos 2 0 2
At the point A on C, the value of r is 9
2
The point N lies on the initial line and AN is perpendicular to the initial line.
The finite region R, shown shaded in Figure 1, is bounded by the curve C, the initial line
and the line AN.
Find the exact area of the shaded region R, giving your answer in the form 3p q
where p and q are rational numbers to be found.
(9)
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Question 4 continued
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(Total for Question 4 is 9 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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5. A pond initially contains 1000 litres of unpolluted water.
The pond is leaking at a constant rate of 20 litres per day.
It is suspected that contaminated water flows into the pond at a constant rate of 25 litres
per day and that the contaminated water contains 2 grams of pollutant in every litre of
water.
It is assumed that the pollutant instantly dissolves throughout the pond upon entry.
Given that there are x grams of the pollutant in the pond after t days,
(a) show that the situation can be modelled by the differential equation,
d 450
d 200
x x
t t
(4)
(b) Hence find the number of grams of pollutant in the pond after 8 days.
(5)
(c) Explain how the model could be refined.
(1)
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Question 5 continued
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(Total for Question 5 is 10 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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6.
2
2f
9
xx
x
(a) Show that
2f d ln 9 arctan3
xx x A x B c
where c is an arbitrary constant and A and B are constants to be found.
(4)
(b) Hence show that the mean value of f(x) over the interval [0, 3] is
1 1ln 2
6 18
(3)
(c) Use the answer to part (b) to find the mean value, over the interval [0, 3], of
f(x) + lnk
where k is a positive constant, giving your answer in the form 1
ln6
p q , where p and q are
constants and q is in terms of k.
(2)
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Question 6 continued
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(Total for Question 6 is 9 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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7.
Figure 2
Figure 2 shows the image of a gold pendant which has height 2 cm. The pendant is
modelled by a solid of revolution of a curve C about the y-axis. The curve C has
parametric equations
1
cos sin 2 , 1 sin 0 22
x y
(a) Show that a Cartesian equation of the curve C is
2 4 32x y y
(4)
(b) Hence, using the model, find, in cm3, the volume of the pendant.
(4)
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Question 7 continued
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Question 7 continued
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Question 7 continued
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(Total for Question 7 is 8 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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8. The line l1 has equation 2 4 6
4 2 1
x y z
The plane Π has equation x – 2y + z = 6
The line l2 is the reflection of the line l1 in the plane Π.
Find a vector equation of the line l2
(7)
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Question 8 continued
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Question 8 continued
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question 8 continued
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(Total for Question 8 is 7 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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9. 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,
2
2
d d4 200cos , 0
d d
x xm x t t
t t
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 30 000N.
Taking the value of g to be 10 ms-2 and assuming the capsule is at its maximum
permissible weight,
(a) (i) explain why the value of m is 3
(ii) show that a particular solution to the differential equation is
40sin 20cosx t t
(iii) hence find the general solution of the differential equation.
(8)
(b) 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.
(4)
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question 9 continued
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Question 9 continued
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Question 9 continued
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(Total for Question 9 is 12 marks)
TOTAL FOR PAPER IS 75 MARKS
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Paper 1: Core Pure Mathematics 1 Mark Scheme
Question Scheme Marks AOs
1
1
..., ...1 3 1 3
A BA B
r r r r
M1 3.1a
1
1
1 3
1 1 1 1 1 1 1 1...
2 2 2 4 2 3 2 5 2 2 2 2 1 2 3
n
rr r
n n n n
M1 2.1
1 1 1 1
4 6 2 2 2 3n n
A1 2.2a
5 2 3 6 3 6 2
12 2 3
n n n n
n n
M1 1.1b
5 13
12 2 3
n n
n n
A1 1.1b
(5)
Alternative by Induction:
2 21 1 11 , 2
8 12 3 4 8 15 12 4 5
a ba bn n
18, 2 23 ..., ...a b a b a b
M1 3.1a
Assume true for n = k so
1
5 131
1 3 12 2 3
k
r
k k
r r k k
1
1
5 131 1
1 3 12 2 3 2 4
k
r
k k
r r k k k k
M1 2.1
5 13 5 13 4 12 31
12 2 3 2 4 12 2 3 4
k k k k k k
k k k k k k k
A1 2.2a
3 2 1 2 5 185 33 52 12 36
12 2 3 4 12 2 3 4
k k kk k k k
k k k k k k
M1 1.1b
1 5 1 13
12 1 2 1 3
k k
k k
So true for n = k + 1
So
1
5 131
1 3 12 2 3
n
r
n n
r r n n
A1 1.1b
(5)
(5 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Notes:
(Main Scheme)
M1: Valid attempt at partial fractions
M1: Starts the process of differences to identify the relevant fractions at the start and end
A1: Correct fractions that do not cancel
M1: Attempt common denominator
A1: Correct answer
(Alternative by Induction)
M1: Uses n = 1 and n = 2 to identify values for a and b
M1: Starts the induction process by adding the (k + 1)th term to the sum of k terms
A1: Correct single fraction
M1: Attempt to factorise the numerator
A1: Correct answer and conclusion
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme
Marks AOs
2 When n = 1, 3 1 2 12 3 5 16 375 391n n
391 17 23 so the statement is true for n = 1 B1 2.2a
Assume true for n = k so 3 1 2 12 3 5k k is divisible by 17 M1 2.4
3 4 2 3 3 1 2 1f 1 f 2 3 5 2 3 5k k k kk k M1 2.1
3 1 2 1 2 17 2 7 3 5 17 3 5k k k
2 17f 17 3 5 kk A1 1.1b
2 1f 1 8f 17 3 5 kk k A1 1.1b
If the statement is true for n = k then it has been shown true for n
= k + 1 and as it is true for n = 1, the statement is true for all
positive integers n.
A1 2.4
(6)
(6 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
3 3 2iz is also a root B1 1.2
3 2i 3 2i ...z z
or
Sum of roots = 6, Product of roots = 13 ...
M1 3.1a
2 6 13z z A1 1.1b
4 3 2 2 26 65 6 13 5 ...z az z bz z z z cz c
M1 3.1a
2 2 5 0z z A1 1.1b
2 2 5 0 ...z z z M1 1.1a
1 2iz A1 1.1b
B1
3 ± 2i
Plotted
correctly
1.1b
B1ft
1 ± 2i
Plotted
correctly
1.1b
(9 marks)
Notes:
B1: Identifies the complex conjugate as another root
M1: Uses the conjugate pair and a correct method to find a quadratic factor
A1: Correct quadratic
M1: Uses the given quartic and their quadratic to identify the value of c
A1: Correct 3TQ
M1: Solves their second quadratic
A1: Correct second conjugate pair
B1: First conjugate pair plotted correctly and labelled
B1ft: Second conjugate pair plotted correctly and labelled (Follow through their second conjugate pair)
Re
Im
(3, 2)
(3, -2)
(-1, 2)
(-1, -2)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
4 94 cos 2 ...
2 M1 3.1a
6
A1 1.1b
2 21 1
4 cos 2 d 16 8cos 2 cos 2 d2 2
M1 3.1a
2 1 1 1 1 1cos 2 cos 4 16 8cos 2 cos 4 d
2 2 2 2 2A
M1 3.1a
1 sin 416 4sin 2
2 8 2
A1 1.1b
1 33 3
Using limits 0 and their : 2 3 06 2 12 16
M1 1.1b
Area of triangle = 1 1 81 1 3
cos sin2 2 4 2 2
r r M1 3.1a
Area of R = 33 33 3 81 3
24 32 32
M1 1.1b
11 3 3 11 3,
8 2 8 2p q
A1 1.1b
(9 marks)
Notes:
M1: Realises the angle for A is required and attempts to find it.
A1: Correct angle
M1: Uses a correct area formula and squares r to achieve a 3TQ integrand in cos 2θ
M1: Use of the correct double angle identity on the integrand to achieve a suitable form for integration
A1: Correct integration
M1: Correct use of limits
M1: Identifies the need to subtract the area of a triangle and so finds the area of the triangle
M1: Complete method for the area of R
A1: Correct final answer
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
5(a) Pond contains 1000 + 5t litres after t days M1 3.3
If x is the amount of pollutant in the pond after t days
Rate of pollutant out = 201000 5
x
t
g per day
M1 3.3
Rate of pollutant in = 25 2 g = 50g per day B1 2.2a
d 450 *
d 200
x x
t t
A1* 1.1b
(4)
(b)
4
d 4 4 4200e 200 200 50 200 d
ttI t x t t t
M1 3.1b
4 5
200 10 200x t t c A1 1.1b
120, 0 3.2 10x t c M1 3.4
12
4
3.2 108 10 200 8
200 8t x
M1 1.1b
= 370g A1 2.2b
(5)
(c) Examples
The model should take into account the fact that the
pollutant does not dissolve throughout the pond upon
entry
The rate of leaking could be made to vary with the volume
of water in the pond
B1 3.5c
(1)
(10 marks)
Notes:
(a)
M1: Forms an expression of the form 1000 + kt for the volume of water in the pond at time t
M1: Expresses the amount of pollutant out in terms of x and t
B1: Correct interpretation for pollutant entering the pond
A1*: Puts all the components together to form the correct differential equation
(b)
M1: Uses the model to find the integrating factor and attempts solution of their differential equation
A1: Correct solution
M1: Interprets the initial conditions to find the constant of integration
M1: Uses their solution to the problem to find the amount of pollutant after 8 days
A1: Correct number of grams
(c)
B1: Suggests a suitable refinement to the model
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
6(a) 2 2 2
2 2f
9 9 9
x xx
x x x
B1 3.1a
2
2d ln 9
9
xx k x c
x
M1 1.1b
32
2d arctan
9xx k c
x
M1 1.1b
2
2
2 1 2d ln 9 arctan
9 2 3 3
x xx x c
x
A1 1.1b
(4)
(b)
3 3
2
00
1 2f d ln 9 arctan
2 3 3
1 2 3 1 2ln18 arctan ln 9 arctan 0
2 3 3 2 3
1 18 2 3ln arctan
2 9 3 3
xx x x
M1 1.1b
Mean value = 1 1
ln 23 0 2 6
M1 2.1
1 1ln 2 *
6 18 A1* 2.2a
(3)
(c) 1 1ln 2 ln
6 18k M1 2.2a
61 1ln 2
6 18k A1 1.1b
(2)
(9 marks)
Notes:
(a)
B1: Splits the fraction into two correct separate expressions
M1: Recognises the required form for the first integration
M1: Recognises the required form for the second integration
A1: Both expressions integrated correctly and added together with constant of integration included
(b)
M1: Uses limits correctly and combines logarithmic terms
M1: Correctly applies the method for the mean value for their integration
A1*: Correct work leading to the given answer
(c)
M1: Realises that the effect of the transformation is to increase the mean value by lnk
A1: Combines ln’s correctly to obtain the correct expression
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
7(a) cos sin cos cosx y M1 2.1
sin 1y M1 2.1
2
21 1
xy
y
M1 2.1
2 4 32 *x y y A1* 1.1b
(4)
(b)
2 4 3d 2 dV x y y y y M1 3.4
5 4
5 2
y y
A1 1.1b
5 4 5 4
0 0 2 2
5 2 5 2
M1 3.4
31.6 cm or awrt 5.03 3cm A1 1.1b
(4)
(8 marks)
Notes:
(a)
M1: Obtains x in terms of y and cos θ
M1: Obtains an equation connecting y and sin θ
M1: Uses Pythagoras to obtain an equation in x and y only
A1*: Obtains printed answer
(b)
M1: Uses the correct volume of revolution formula with the given expression
A1: Correct integration
M1: Correct use of correct limits
A1: Correct volume
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
8
2 4 2 4 2 6 6 ... M1 1.1b
2 Required point is 2 2 4 , 4 2 2 , 6 2 1
10, 0, 4 A1 1.1b
2 2 4 2 6 6 ...t t t t M1 3.1a
3soreflection of 2,4, 6 is 2 6 1 ,4 6 2 , 6 6 1t M1 3.1a
8, 8, 0 A1 1.1b
10 8 2
0 8 8
4 0 4
M1 3.1a
10 1
0 4 or equivalent
4 2
k
r e.g.
10 1
0 4
4 2
r 0 A1 2.5
(7)
(7 marks)
Notes:
M1: Substitutes the parametric equation of the line into the equation of the plane and solves for λ
A1: Obtains the correct coordinates of the intersection of the line and the plane
M1: Substitutes the parametric form of the line perpendicular to the plane passing through (2, 4, 6) into
the equation of the plane to find t
M1: Find the reflection of (2, 4, 6) in the plane
A1: Correct coordinates
M1: Determines the direction of l by subtracting the appropriate vectors
A1: Correct vector equation using the correct notation.
(8,8, 0)
(2, 4,6)
(10, 0,4)
π
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
9(a)(i) Weight = massg
300003000m
g
But mass is in thousands of kg, so m = 3
M1 3.3
(ii) 2
2
d d40cos 20sin , 40sin 20cos
d d
x xt t t t
t t M1 1.1b
3 40sin 20cos 4 40cos 20sin
40sin 20cos ...
t t t t
t t
M1 1.1b
200cos t so PI is 40sin 20cosx t t A1* 2.1
or
Let cos sinx a t b t
2
2
d dsin cos , cos sin
d d
x xa t b t a t b t
t t
M1 1.1b
4 2 200, 2 4 0 ..., ...b a b a a b M1 2.1
40sin 20cosx t t A1* 1.1b
(iii) 2 13 4 1 0 1,
3 M1 1.1b
13e ettx A B
A1 1.1b
x PI CF M1 1.1b
13e e 40sin 20costtx A B t t
A1 1.1b
(8)
(b) 0, 0 20t x A B M1 3.4
13
d 10, 40cos 20sin 0
d 3
140
3
ttxx Ae Be t t
t
A B
M1 3.4
1350e 30e 40sin 20costtx t t
A1 1.1b
9 33mt x A1 3.4
(4)
(12 marks)
Notes:
(a)(i)
M1: Correct explanation that in the model, m = 3
(ii)
M1: Differentiates the given PI twice
M1: Substitutes into the given differential equation
A1*: Reaches 200cost and makes a conclusion
Or
M1: Uses the correct form for the PI and differentiates twice
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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M1: Substitutes into the given differential equation and attempts to solve
A1*: Correct PI
(iii)
M1: Uses the model to form and solve the auxiliary equation
A1: Correct complementary function
M1: Uses the correct notation for the general solution by combining PI and CF
A1: Correct General Solution for the model
(b)
M1: Uses the initial conditions of the model, t = 0 at x = 0, to form an equation in A and B
M1: Uses dx/dt = 0 at x = 0 in the model to form an equation in A and B
A1: Correct PS
A1: Obtains 33m using the assumptions made in the model.
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Pearson Edexcel Level 3 GCE
Further Mathematics Advanced Paper 2: Core Pure Mathematics 2
Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes
Paper Reference(s)
9FM0/02
You must have: Mathematical Formulae and Statistical Tables, calculator
Candidates may use any calculator permitted by Pearson regulations. Calculators must not have the facility for algebraic manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them.
Instructions
Use black ink or ball-point pen.
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
Fill in the boxes at the top of this page with your name, centre number and candidate number.
Answer all questions and ensure that your answers to parts of questions are clearly labelled.
Answer the questions in the spaces provided – there may be more space than you need.
You should show sufficient working to make your methods clear. Answers without working may not gain full credit.
Answers should be given to three significant figures unless otherwise stated. Information
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
There are 7 questions in this question paper. The total mark for this paper is 75.
The marks for each question are shown in brackets – use this as a guide as to how much time to spend on each question.
Advice
Read each question carefully before you start to answer it.
Try to answer every question.
Check your answers if you have time at the end.
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Answer ALL questions. Write your answers in the spaces provided.
1. The roots of the equation
3 28 28 32 0x x x
are , and
Without solving the equation, find the value of
(i) 1 1 1
(ii) 2 2 2
(iii) 2 2 2
(8)
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Question 1 continued
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(Total for Question 1 is 8 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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2. The plane Π1 has vector equation
. 3 4 2 5 r i j k
(a) Find the perpendicular distance from the point (6, 2, 12) to the plane Π1
(3)
The plane Π2 has vector equation
2 5 2 r = i j k i j k
where λ and µ are scalar parameters.
(b) Show that the vector 3 i j k is perpendicular to Π2
(2)
(c) Show that the acute angle between Π1 and Π2 is 52o to the nearest degree.
(3)
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question 2 continued
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(Total for Question 2 is 8 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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3. (i)
2 4
1 1 1
1 2 1
a
M
where a is a constant.
(a) For which values of a does the matrix M have an inverse?
(2)
Given that M is non-singular,
(b) find 1
M in terms of a
(4)
(ii) Prove by induction that for all positive integers n,
3 03 0
6 1 3 3 1 1
nn
n
(6)
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question 3 continued
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(Total for Question 3 is 12 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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4. A complex number z has modulus 1 and argument θ.
(a) Show that
zn + 1nz
= 2cos nθ, n
(2)
(b) Hence, show that
cos4 θ = 1
8(cos 4θ + 4cos 2θ + 3)
(5)
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question 4 continued
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(Total for Question 4 is 7 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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5.
sin sinhy x x
(a) Show that 4
4
d4
d
yy
x
(4)
(b) Hence find the first three non-zero terms of the Maclaurin series for y, giving each
coefficient in its simplest form.
(4)
(c) Find an expression for the nth non-zero term of the Maclaurin series for y.
(2)
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question 5 continued
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(Total for Question 5 is 10 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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6. (a) (i) Show on an Argand diagram the locus of points given by the values of z satisfying
4 3i 5z
Taking the initial line as the positive real axis with the pole at the origin and given that
, , where 43
arctan ,
(ii) show that this locus of points can be represented by the polar curve with equation
8cos 6sinr
(6)
The set of points A is defined by
: 0 arg : 4 3i 53
A z z z z
(b) (i) Show, by shading on your Argand diagram, the set of points A.
(ii) Find the exact area of the region defined by A, giving your answer in simplest form.
(7)
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question 6 continued
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(Total for Question 6 is 13 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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7. 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
d0.2 0.1
d
d0.2 0.4
d
ff r
t
rf r
t
(a) Show that 2
2
d d0.6 0.1 0
d d
f ff
t t
(3)
(b) Find a general solution for the number of foxes on the island at time t years.
(4)
(c) Hence find a general solution for the number of rabbits on the island at time t years.
(3)
At the start of the year 2000 there were 6 foxes and 20 rabbits on the island.
(d) (i) According to this model, in which year are the rabbits predicted to die out?
(ii) According to this model, how many foxes will be on the island when the rabbits
die out?
(iii) Use your answers to parts (i) and (ii) to comment on the model.
(7)
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question 7 continued
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question 7 continued
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question 7 continued
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Total for Question 7 is 17 marks)
TOTAL FOR PAPER IS 75 MARKS
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Paper 2: Core Pure Mathematics 2 Mark Scheme
Question Scheme Marks AOs
1(i) 8, 28, 32 B1 3.1a
1 1 1
M1 1.1b
7
8 A1ft 1.1b
(3)
(ii) 2 2 2 2 2 4 2 M1 1.1b
2 4 8 A1 1.1b
32 2 28 4 8 8 128 A1 1.1b
(3)
Alternative for part (ii)
3 2
2 8 2 28 2 32 0x x x M1 1.1b
... 8 ... 32 ... 56 32 128 A1 1.1b
2 2 2 128 A1 1.1b
(3)
(iii) 22 2 2 2 M1 3.1a
28 2 28 8 A1ft 1.1b
(2)
(8 marks)
Notes
(i)
B1: Identifies the correct values for all 3 expressions (can score anywhere)
M1: Uses a correct identity
A1ft: Correct value (follow through their 8, 28 and 32)
(ii)
M1: Attempts to expand
A1: Correct expansion
A1: Correct value
Alternative:
M1: Substitutes x – 2 for x in the given cubic
A1: Calculates the correct constant term
A1: Changes sign and so obtains the correct value
(iii)
M1: Establishes the correct identity
A1ft: Correct value (follow through their 8, 28 and 32)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme
Marks AOs
2(a) 3 6
4 2 18 8 24
2 12
M1 3.1a
2 2 2
18 8 24 5
3 4 2d
M1 1.1b
29 A1 1.1b
(3)
(b) 1 2 1 1
3 1 ... and 3 1 ...
1 5 1 2
M1 2.1
1 2 1 1
3 1 0 and 3 1 0
1 5 1 2
3 i j k is perpendicular to Π2
A1 2.2a
(2)
(c) 1 3
3 4 3 12 2
1 2
M1 1.1b
2 2 2 22 2
2 2 2 22 2
1 3 1 3 4 2 cos 11
11cos
1 3 1 3 4 2
M1 2.1
So angle between planes o52 * A1* 2.4
(3)
(8 marks)
Notes
(a)
M1: Realises the need to and so attempts the scalar product between the normal and the position
vector
M1: Correct method for the perpendicular distance
A1: Correct distance
(b)
M1: Recognises the need to calculate the scalar product between the given vector and both
direction vectors
A1: Obtains zero both times and makes a conclusion
(c)
M1: Calculates the scalar product between the two normal vectors
M1: Applies the scalar product formula with their 11 to find a value for cos θ
A1*: Identifies the correct angle by linking the angle between the normal and the angle between
the planes.
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
3(i)(a) 2 1 2 1 1 4 2 1 0 ...a a M M1 2.3
The matrix M has an inverse when 5a A1 1.1b
(2)
(b)
3 2 1
Minors : 8 2 4
4 6 2
a a
a a
or
3 2 1
Cofactors : 8 2 4
4 6 2
a a
a a
B1 1.1b
1 1adj M M
M M1 1.1b
1
3 8 41
2 2 62 10
1 4 2
a a
aa a
M
2 correct rows or columns.
Follow through their detM. A1ft 1.1b
All correct.
Follow through their detM. A1ft 1.1b
(4)
(ii) When n = 1, lhs =
3 0
6 1
, rhs =
1
1
3 0 3 0
6 13 3 1 1
So the statement is true for n = 1
B1 2.2a
Assume true for n = k so
3 03 0
6 1 3 3 1 1
kk
k
M1 2.4
1 3 03 0 3 0
6 1 6 13 3 1 1
kk
k
M1 2.1
3 3 0
3 3 3 1 6 1
k
k
A1 1.1b
1
1
3 0
3 3 1 1
k
k
A1 1.1b
If the statement is true for n = k then it has been shown true for n = k
+ 1 and as it is true for n = 1, the statement is true for all positive
integers n.
A1 2.4
(6)
(12 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Notes:
(i)(a)
M1: Attempts determinant, equates to zero and attempts to solve for a in order to establish the restriction for
a
A1: Provides the correct condition for a if M has an inverse
(i)(b)
B1: A correct matrix of minors or cofactors
M1: For a complete method for the inverse
A1ft: Two correct rows following through their determinant
A1ft: Fully correct inverse following through their determinant
(ii)
B1: Shows the statement is true for n = 1
M1: Assumes the statement is true for n = k
M1: Attempts to multiply the correct matrices
A1: Correct matrix in terms of k
A1: Correct matrix in terms of k + 1
A1: Correct complete conclusion
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme
Marks AOs
4(a) cos isin cos isinn nz z n n n n M1 2.1
2cos *n A1* 1.1b
(2)
(b) 4
1 416cosz z B1 2.1
4
1 4 2 2 44 6 4z z z z z z M1 2.1
4 4 2 24 6z z z z A1 1.1b
2cos4 4 2cos2 6 M1 2.1
cos4 θ = 1
8(cos 4θ + 4cos 2θ + 3)* A1* 1.1b
(5)
(7 marks)
Notes
(a)
M1: Identifies the correct form for zn and z-n and adds to progress to the printed answer
A1*: Achieves printed answer with no errors
(b)
B1: Begins the argument by using the correct index with the result from part (a)
M1: Realises the need to find the expansion of 4
1z z
A1: Terms correctly combined
M1: Links the expansion with the result in part (a)
A1*: Achieves printed answer with no errors
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme
Marks AOs
5(a) dsin cosh cos sinh
d
yx x x x
x M1 1.1a
2
2
dcos cosh sin sinh cos cosh sin sinh
d
2cos cosh
yx x x x x x x x
x
x x
M1 1.1b
3
3
d2cos sinh 2sin cosh
d
yx x x x
x M1 1.1b
4
4
d4sinh sin 4 *
d
yx x y
x A1* 2.1
(4)
(b) 2 6 10
2 6 10
0 0 0
d d d2, 8, 32
d d d
y y y
x x x
B1 3.1a
Uses 2 3
0 0 0 0 ...2! 3!
x xy y xy y y with their values M1 1.1b
2 6 10
2 8 322! 6! 10!
x x x A1 1.1b
6 102
90 113400
x xx A1 1.1b
(4)
(c)
4 21
2 44 2 !
nn x
n
M1 A1 3.1a
2.2a
(2)
(10 marks)
Notes
(a)
M1: Realises the need to use the product rule and attempts first derivative
M1: Realises the need to use a second application of the product rule and attempts the second
derivative
M1: Correct method for the third derivative
A1*: Obtains the correct 4th derivative and links this back to y
(b)
B1: Makes the connection with part (a) to establish the general pattern of derivatives and finds
the correct non-zero values
M1: Correct attempt at Maclaurin series with their values
A1: Correct expression un-simplified
A1: Correct expression and simplified
(c)
M1: Generalising, dealing with signs, powers and factorials.
A1: Correct expression.
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
6(a)(i)
M1 1.1b
A1 1.1b
(a)(ii) 2 2
4 3i 5 i 4 3i 5 4 3 ...z x y x y M1 2.1
2 2
4 3 25 or anycorrect formx y A1 1.1b
2 2
2 2 2 2
2
cos 4 sin 3 25
cos 8 cos 16 sin 6 sin 9 25
8 cos 6 sin 0
r r
r r r r
r r r
M1 2.1
8cos 6sin *r A1* 2.2a
(6)
(b)(i)
B1 1.1b
B1ft 1.1b
(b)(ii)
22
2 2
1 1d 8cos 6sin d
2 2
164cos 96sin cos 36sin d
2
A r
M1 3.1a
1
32 cos 2 1 96sin cos 18 1 cos 2 d2
M1 1.1b
1
14cos 2 50 48sin 2 d2
A1 1.1b
3
0
1 1 7 3 507sin 2 50 24cos 2 12 24
2 2 2 3
M1 2.1
7 3 2518
4 3
A1 1.1b
(7)
Im
Re
Im
Re
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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A
Alternative:
Candidates may take a geometric approach
E.g. by finding sector + 2 triangles
Angle ACB = 2
3
so area sector ACB = 21 2
52 3
Area of triangle OCB 1
8 32
M1 3.1a
Sector area ACB + triangle area OCB = 25
3
+ 12 A1 1.1b
Area of triangle OAC:
Angle ACO = 2 2 2
12 5 5 82 cos
3 2 5 5
so area OAC = 2 11 4 7
5 sin cos2 3 25
M1 1.1b
1 125 4 7 4 7sin cos cos cos sin cos
2 3 25 3 25
225 7 3 1 7 7 3
1 62 50 2 25 4
Total area = 25 1 7 3
8 3 63 2 4
M1 2.1
7 3 2518
4 3
A1 1.1b
(13 marks)
C
B O
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Notes
(a)(i)
M1: Draws a circle which passes through the origin
A1: Fully correct diagram.
(a)(ii)
M1: Uses z = x + iy in the given equation and uses modulus to find equation in x and y only
A1: Correct equation in terms of x and y in any form – may be in terms of r and θ
M1: Introduces polar form, expands and uses 2 2cos sin 1 leading to a polar equation
A1*: Deduces the given equation (ignore any reference to r = 0 which gives a point on the curve)
(b)(i)
B1: Correct pair of rays added to their diagram
B1ft: Area between their pair of rays and inside their circle from (a) shaded, as long as there is an
intersection.
(b)(ii)
M1: Selects an appropriate method by linking the diagram to the polar curve in (a), evidenced by use
of the polar area formula
M1: Uses double angle identities
A1: Correct integral
M1: Integrates and applies limits
A1: Correct area
(b)(ii) Alternative:
M1: Selects an appropriate method by finding angle ACB and area of sector ACB and finds area of
triangle OCB to make progress towards finding the required area
A1: Correct combined area of sector ACB + triangle OCB
M1: Starts the process of finding the area of triangle OAC by calculating angle ACO and attempts
area of triangle OAC
M1: Uses the addition formula to find the exact area of triangle OAC and employs a full correct
method to find the area of the shaded region
A1: Correct area
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
7(a) 2
2
d d d d10 2 10 2
d d d d
f r f fr f
t t t t M1 2.1
2
2
d d d10 2 0.2 0.4 10 2
d d d
f f ff f
t t t
M1 2.1
2
2
d d0.6 0.1 0*
d d
f ff
t t A1* 1.1b
(3)
(b) 22 0.6 0.6 4 0.1
0.6 0.1 02
m m m
M1 3.4
0.3 0.1im A1 1.1b
e cos sintf A t B t M1 3.4
0.3e cos0.1 sin 0.1tf A t B t A1 1.1b
(4)
(c) 0.3 0.3d
0.3e cos0.1 sin 0.1 0.1e cos0.1 sin 0.1d
t tfA t B t B t A t
t M1 3.4
0.3 0.3
d10 2
d
e 3 cos0.1 3 sin 0.1 2e cos0.1 sin 0.1t t
fr f
t
A B t B A t A t B t
M1 3.4
0.3e cos0.1 sin 0.1tr A B t B A t A1 1.1b
(3)
(d)(i) 0, 6 6t f A M1 3.1b
0, 20 14t r B M1 3.3
0.3e 20cos0.1 8sin 0.1 0tr t t M1 3.1b
tan0.1 2.5t A1 1.1b
2019 A1 3.2a
(d)(ii) 3750 foxes B1 3.4
(d)(iii) E.g.
The model predicts a large number of foxes are on the island
when the rabbits have died out and this may not be sensible.
B1 3.5a
(7)
(17 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Notes
(a)
M1: Attempts to differentiate the first equation with respect to t
M1: Proceeds to the printed answer by substituting into the second equation
A1*: Achieves the printed answer with no errors
(b)
M1: Uses the model to form and solve the auxiliary equation
A1: Correct values for m
M1: Uses the model to form the CF
A1: Correct CF
(c)
M1: Differentiates the expression for the number of foxes
M1: Uses this result to find an expression for the number of rabbits
A1: Correct equation
(d)(i)
M1: Realises the need to use the initial conditions in the model for the number of foxes
M1: Realises the need to use the initial conditions in the model for the number of rabbits to find
both unknown constants
M1: Obtains an expression for r in terms of t and sets = 0
A1: Rearranges and obtains a correct value for tan
A1: Identifies the correct year
(d)(ii)
B1: Correct number of foxes
(d)(iii)
B1: Makes a suitable comment on the outcome of the model
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Pearson Edexcel Level 3 GCE
Further Mathematics Advanced Paper 3: Further Pure Mathematics 1
Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes
Paper Reference(s)
9FM0/3A
You must have: Mathematical Formulae Calculator
Candidates may use any calculator permitted by Pearson regulations. Calculators must not have the facility for algebraic manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them.
Instructions
Use black ink or ball-point pen.
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
Fill in the boxes at the top of this page with your name, centrenumber and candidate number.
Answer all questions and ensure that your answers to parts ofquestions are clearly labelled.
Answer the questions in the spaces provided– there may be more space than you need.
You should show sufficient working to make your methods clear.Answers without working may not gain full credit.
Inexact answers should be given to three significant figures unlessotherwise stated.
Information
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
There are 8 questions in this question paper. The total mark for thispaper is 75.
The marks for each question are shown in brackets– use this as a guide as to how much time to spend on each question.
Advice
Read each question carefully before you start to answer it.
Try to answer every question.
Check your answers if you have time at the end.
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Answer ALL questions. Write your answers in the spaces provided.
1. Use Simpson’s Rule with 6 intervals to estimate
(5)
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Question 1 continued
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(Total for Question 1 is 5 marks)
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2. Given k is a constant and that
y = x3 ek x
use Leibnitz theorem to show that
dn y
dxn= k n-3ek x k3x3 + 3nk 2x2 + 3n(n -1)k x + n(n -1)(n - 2)( )
(4)
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Question 2 continued
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(Total for Question 2 is 4 marks)
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3. A vibrating spring, fixed at one end, has an external force acting on it such that the
centre of the spring moves in a straight line. At time t seconds, the displacement
of the centre C of the spring from a fixed point O is x micrometres.
The displacement of C from O is modelled by the differential equation
t2 d 2x
dt2- 2t
dx
dt+ (2 + t2 )x = t4 (I)
(a) Show that the transformation x = t v transforms equation (I) into the equation
d 2v
dt2+ v = t (II)
(5)
(b) Hence find the general equation for the displacement of C from O at time t seconds.
(7)
(c) (i) State what happens to the displacement of C from O as t becomes large.
(ii) Comment on the model with reference to this long term behaviour.
(2)
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Question 3 continued
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(Total for Question 3 is 14 marks)
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4.
d 2 y
dx2- 2x
dy
dx+ y = 0 (I)
(a) Show that
d 5 y
dx5= ax
d 4 y
dx4+ b
d3y
dx3
where a and b are integers to be found.
(4)
(b) Hence find a series solution, in ascending powers of x, as far as the term in x5,
of the differential equation (I) where y = 0 and d
1d
y
x at x = 0
(5)
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Question 4 continued
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(Total for Question 4 is 9 marks)
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5. The normal to the parabola y2 = 4ax at the point
P ap2 , 2ap( ) passes through the
parabola again at the point Q aq2 , 2aq( ).
The line OP is perpendicular to the line OQ, where O is the origin.
Prove that p2 = 2
(9)
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Question 5 continued
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(Total for Question 5 is 9 marks)
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6. A tetrahedron has vertices A(1, 2, 1), B(0, 1, 0), C (2, 1, 3) and
D(10, 5, 5).
Find
(a) a Cartesian equation of the plane ABC.
(3)
(b) the volume of the tetrahedron ABCD.
(3)
The plane P has equation 2x - 3y + 3 = 0
The point E lies on the line AC and the point F lies on the line AD.
Given that P contains the point B, the point E and the point F,
(c) find the value of k such that
(3)
Given that
(d) show that the volume of the tetrahedron ABCD is 45 times the volume of the
tetrahedron ABEF.
(2)
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Question 6 continued
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(Total for Question 6 is 11 marks)
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7. P and Q are two distinct points on the ellipse described by the equation x2 + 4y2 = 4
The line l passes through the point P and the point Q.
The tangent to the ellipse at P and the tangent to the ellipse at Q intersect at the
point (r , s).
Show that an equation of the line l is
4s y + r x = 4
(8)
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Question 7 continued
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(Total for Question 7 is 8 marks)
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8.
Figure 1
Figure 1 shows the graph of the function h(x) with equation
h( ) 45 15sin 21sin 25cos [0, 40]2 2
x xx x x
(a) Show that
dh
dx=
t2 - 6t - 17( ) 9t2 + 4t - 3( )2 1 + t2( )
2
where tan .4
xt
(6)
Figure 2
Figure 2 shows a graph of predicted tide heights, in metres, for Portland harbour from
08:00 on the 3rd January 2017 to the end of the 4th January 20171.
The graph of k h(x), where k is a constant and x is the number of hours after 08:00 on 3rd
of January, can be used to model the predicted tide heights, in metres, for this period of
time.
(b) (i) Suggest a value of k that could be used for the graph of h( )k x to form a suitable
model.
(ii) Why may such a model be suitable to predict the times when the tide heights are
at their peaks, but not to predict the heights of these peaks?
(3)
(c) Use Figure 2 and the result of part (a) to estimate, to the nearest minute, the time of
the highest tide height on the 4th January 2017.
(6) 1Data taken on 29th December 2016 from http://www.ukho.gov.uk/easytide/EasyTide
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Question 8 continued
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Question 8 continued
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(Total for Question 8 is 15 marks)
TOTAL FOR PAPER IS 75 MARKS
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AL Paper 3: Further Pure Mathematics 1 Mark Scheme
Question Scheme Marks AOs
1. Step 0.5 B1 1.1b
1 1.5 2 2.5 3 3.5 4 3
M1 1.1b
0 1 2 3 4 5 64 2 4 2 4 "77.23"y y y y y y y M1 1.1b
1 x3 dx1
4
0.5
3 "77.23" M1 1.1b
12.9 A1 1.1b
(5) (5 marks)
Notes B1 Use of step length 0.5 M1 Attempt to find y values with at least 2 correct. M1 Use of formula" y
0 4y
1 2y
2 4y
3 2y
4 4y
5 y
6" with correct coefficients
A1 0.5
3 their 77.23
A1 awrt 12.9
Question Scheme Marks AOs
2. y x3ekx so u x3 and
2 3 4
22 3 4
d d d d3 and 6 and 6 and 0
d d d d
u u u ux x
x x x x
M1 1.1b
1 21 2
1 2
d d de and e and e and e
d d d(and...)
n n nkx n kx n kx n kx
n n n
v v vv k k k
x x x
M1 2.1
( 1) ( 1)( 2)3 2 1 2 3
2 3!
de 3 e 6 e 6 e
d
nn n n n nn kx n kx n kx n kx
n
yx k n x k xk k
x
and remaining terms disappear. M1 2.1
So dn y
dxn k n3ekx k3x3 3nk 2x2 3n(n1)kx n(n1)(n 2) * A1* 1.1b
(4) (4 marks)
Notes M1 Differentiate 3u x three times. M1 Use u ekx and establish the form of the derivatives, with at least the three shown. M1 Uses correct formula, with 2 and 3! (or 6) and with terms shown to disappear after the
fourth term. A1* Correct solution leading to the given answer stated. No errors seen.
0y 1y 2y 3y 4y 5y 6y
xy 2 4.375 16.625 28 43.875 65
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Question Scheme Marks AOs
3. (a) Use of x tv to give
dx
dt v t
dv
dt M1 1.1b
Hence
2 2
2 2
d d d d
d d d d
x v v vt
t t t t
M1 2.1 A1 1.1b Uses
t2 their 2ndderivative 2t their 1stderivative (2 t2 )x t4
and simplifies LHS.
M1 2.1
t3 d2v
dt2 t3v t4 leading to
2
2
d*
d
vv t
t A1* 1.1b
(5)
(b) Solve 2 1 0 to give 2 1 M1 1.1b
Acos t Bsin t A1ft 1.1b Particular Integral is kt l B1 2.2a
2
2
d dand 0 and solve 0 to give 1, 0
d d
v vk kt l t k l
t t M1 1.1b
Solution: Acos t Bsin t t A1 1.1b
Displacement of C from O is given by ...x tv M1 3.4
cos sinx t A t B t t A1 2.2a
(7) (c)(i) For large t, the displacement gets very large (and positive). B1 3.2a (ii) Model suggests midpoint of spring moving relative to fixed point
has large displacement when t is large, which is unrealistic. The spring may reach elastic limit / will break.
B1 3.5a
(2) (14 marks)
Notes 3. (a) M1 Uses product rule to obtain first derivative. M1 Continues to differentiate again, with product rule and chain rule as appropriate, in order
to establish the second derivative. A1 Correct second derivative. Accept equivalent expressions. M1 Shows clearly the substitution into the given equation in order to form the new equation
and gathers like terms. A1* Fully correct solution leading to the given answer. (b) Accept variations on symbols for constants throughout. M1 Form and solve a quadratic Auxiliary Equation. A1ft Correct form of the Complementary Function for their solutions to the AE. B1 Deduces the correct form for the Particular Integral (note v mt2 kt l is fine). M1 Differentiates their Particular Integral and substitutes their derivatives into the equation to
find the constants (m = 0 if used). A1 Correct general solution for equation (II). M1 Links the solution to equation (II) to the solution of the model equation correctly to find
the displacement equation. A1 Deduces the correct general solution for the displacement.
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Notes continued
3. (c)(i)
B1 States that for large t the displacement is large o.e. Accept e.g. as t , x. (c)(ii) B1 Reflect on the context of the original problem. Accept ‘model unrealistic’ / ‘spring will
break’.
Question Scheme Marks AOs
4. (a) '' 2 ' ''' 2 '' 2 ' 'y xy y y xy y y M1 1.1b A1 1.1b
''' 2 '' ' '''' 2 ''' 2 '' ''y xy y y xy y y M1 2.1 y '''' 2xy ''' 3y '' y ''''' 2xy '''' 5y ''' A1 2.1 (4)
(b) 0, 0, ' 1 ''(0) 0x y y y from equation B1 2.2a
y '''(0) 2 0 y ''(0)1 1; y ''''(0) 2 01 3 0 0; x 0, y '''(0) 1, y ''''(0) 0 y '''''(0) 5
M1 1.1b A1 1.1b
2 3 4 5''(0) '''(0) ''''(0) '''''(0)(0) '(0) ...
2 6 24 120
y y y yy y y x x x x x M1 2.5
Series solution: y x
1
6x3
1
24x5 ... A1ft 1.1b
(5) (9 marks)
Notes 4. (a) M1 Attempts to differentiate equation with use of the product rule. A1 cao. Accept if terms all on one side. M1 Continues the process of differentiating to progress towards the goal. Terms may be kept
on one side, but an expression in the fourth derivative should be obtained. A1 Completes the process to reach the fifth derivative and rearranges to the correct form to
obtain the correct answer by correct solution only. (b) B1 Deduces the correct value for '' (0)y from the information in the question. M1 Finds the values of the derivatives at the given point. A1 All correct M1 Correct mathematical language required with given denominators. Can be in factorial
form. A1ft Correct series, must start y = …. Follow through the values of their derivatives at 0.
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Question Scheme Marks AOs
5. 2 d4 2 4
d
yy ax y a
x M1 2.1
d 2
d
y a
x y
Gradient of normal is
2
yp
a
A1 1.1b
Equation of normal is : 22 ( )y ap p x ap M1 1.1b
Normal passes through 2( ,2 )Q aq aq so 2 32 2aq apq ap ap M1 3.1a
Grad OP Grad OQ = 12 2
2 21
ap aq
ap aq M1 2.1
4
qp
A1 1.1b
3
2
4 162 2a ap ap ap
p p
p4 2 p2 8 0 M1 2.1
( p2 2)( p2 4) 0 p2 ... M1 1.1b
Hence (as p2 4 0), p2 2* A1* 1.1b
(9)
5. ALT 1
First three marks as above and then as follows. M1 2.1 A1 1.1b M1 1.1b
Solves y2 4ax and their normal simultaneously to find, in terms
of a and p, either 22
44Q
ax ap a
p
or 4
2Q
ay ap
p
M1 3.1a
Finds the second coordinate of Q in terms of a and p M1 1.1b
Both 22
44Q
ax ap a
p and
42Q
ay ap
p A1 1.1b
Grad OP Grad OQ = 12
4
2 2 4
221
4
ap
ap
apap
ap ap a
M1 2.1
Simplifies expression and solves: 4 p2 8 p4 4 p2 4
p4 4 0 ( p2 2)( p2 2) 0 p2 ...M1 2.1
Hence (as p2 2 0), p2 2* A1* 1.1b
(9)
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Question Scheme Marks AOs
5. ALT 2
First three marks as above and then as follows. M1 2.1 A1 1.1b M1 1.1b
Solves y2 4ax and their normal simultaneously to find, in terms
of a and p, either 22
44Q
ax ap a
p
or 4
2Q
ay ap
p
M1 3.1a
Forms a relationship between p and q from their first coordinate:
either 2 2
2Qy a p q pp p
or xQ a p
2
p
2
q p 2
p
M1 2.1
2
q pp
(if x coordinate used the correct root must be clearly
identified before this mark is awarded). A1 1.1b
Grad OP Grad OQ = 1 2 2
2 21
ap aq
ap aq
4.q
p
M1 2.1
Sets 2 4
q pp p
and solves to give p2 ... M1 1.1b
2 4
Hence as = gives no solution ,q pp p
p2 2 (only)* A1* 1.1b
(9) (9 marks)
Notes M1 Begins proof by differentiating and using the perpendicularity condition at point P in
order to find the equation of the normal. A1 Correct gradient of normal, p only.
M1 Use of 1 1( )y y m x x . Accept use of y mx c and then substitute to find c. M1 Substitute coordinates of Q into their equation to find an equation relating p and q. M1 Use of 1 2 1m m with OP and OQ to form a second equation relating p and q.
A1 4
qp
only.
M1 Solves the simultaneous equations and cancels a from their results to obtain a quadratic equation in p2 only.
M1 Attempts to solve their quadratic in p2. Usual rules. A1* Correct solution leading to given answer stated. No errors seen.
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Notes continued
5. ALT 1 M1A1M1 As main scheme. M1 Solves y2 4ax and their normal simultaneously to find one of the coordinates for
Q in terms of a and p as shown. M1 Finds the second coordinate of Q in terms of a and p. A1 Both coordinates correct in terms of a and p. M1
Use of m1m
2 1 with OP and OQ. i.e.
2
their21
theirQ
Q
yap
ap x with coordinates of
P and their expressions for xQ
and yQ
.
M1 Cancels the a’s, simplifies to a quadratic in p2 and solves the quadratic. Usual rules. A1* Correct solution leading to the given answer stated. No errors seen. ALT 2 M1A1M1 As main scheme. M1 Solves y2 4ax and their normal simultaneously to find one of the coordinates for
Q in terms of a and p as shown. M1 Uses their coordinate to form a relationship between p and q. Allow
2q p
p
for this mark. A1 For
2.q p
p If the x coordinate was used to find q then consideration of the
negative root is needed for this mark. Allow for 2
.q pp
M1 Use of 1 2 1m m with OP and OQ to form a second equation relating p and q only.
M1 Equates expressions for q and attempts to solve to give p2 .... A1* Correct solution leading to the given answer stated. No errors seen. If x coordinate
used, invalid solution must be rejected.
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Question Scheme Marks AOs
6. (a) AB AC
1
1
1
1
12
211 2
11
31
2
M1 1.1b
r.31
2
0
1
0
.
3
1
2
1 M1 1.1b
Hence 3x y 2z 1 A1 1.1b (3)
(b) Volume of Tetrahedron = 1
.( )6
n AD M1 3.1a
1
6
31
2
.105
5
1
2
1
M1 1.1b
1
6(27 3 8)
8
3 A1 1.1b
(3)
(c) kAE AC so E is (1 k , 2 k , 1 2k) M1 3.1a E lies on plane so 2(1 k) 3(2 k) 3 0, leading to k ... M1 3.1a
Hence
k
1
5 A1 1.1b
(3)
(d) Volume ABEF = 1
6AB AE .AF
1
6AB
1
5AC
.1
9AD M1 3.1a
1 1
45 6
AB AC . AD and hence result. * A1* 2.2a
(2) (11 marks)
Notes 6. (a) M1 Attempting a suitable cross product. Accept use of unit vectors. M1 Complete method that would lead to finding the Cartesian equation of plane. A1 Accept any equivalent form. (b) M1 Identifies suitable vectors and attempts to substitute into a correct formula. Accept use of
unit vectors. M1 Correct form of scalar triple product using their n from part (a).
A1 8
3 or exact equivalent form.
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Notes continued
6. (c) M1 Uses that E is on AC in order to find an expression for E. M1 Uses that E is in the plane to form and solve an expression in k.
A1 1
5 o.e. only.
(b) M1 Uses formula for volume of tetrahedron and substitutes for AE and AF
A1* Deduces result: Use of
1
6AB AC .AD is required and no errors seen in solution.
Question Scheme Marks AOs
7. x2 4y2 4 2x 8y
dy
dx 0
dy
dx ... M1 3.1a
Equation of tangent at 1 1( , )P x y is ( y y1)
x1
4y1
(x x1) M1 3.1a
xx1 4yy
1 x
12 4y
12 4 and at Q (x
2, y
2) : xx
2 4yy
2 4 A1 2.2a
Intersect at (r , s) gives rx1 4sy
1 4 and rx
2 4sy
2 4 B1 2.1
Uses their previous results to find the gradient of the line l. M1 3.1a
y
2 y
1
x2 x
1
r
4s A1 1.1b
Equation of l is y y
1 r
4s(x x
1) M1 2.1
4sy rx 4sy1 rx
1 4* A1* 2.2a
(8) (8 marks)
Notes
M1 Attempts to solve the problem by using differentiation to obtain an expression for dy
dx.
M1 Realise the need to form a general equation of the tangent at 1 1( , ).x y May use alternative
variables. A1 Deduces 2 2
1 14 4x y to obtain a correct equation and deduces a correct second equation.
B1 Uses (r, s) in both equations to form the two given equations or exact equivalents.
M1 Uses their previous results to find the gradient of the line l.
A1
r
4s
M1 Formulates the line l with their
r
4s. Use of 1 1( )y y m x x or y mx c with their
gradient and an attempt to find c. A1* Correct solution leading to 1 14 4sy rx sy rx with deduction that this equals 4 as
1 1( , )x y is on the ellipse. No errors seen.
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Question Scheme Marks AOs
8. h( ) 45 15sin 21sin 25cos2 2
x xx x
(a)
dh
dx
21 2515cos cos sin
2 2 2 2
x xx
M1 1.1b
dh
dx
2
2 2
1 2... .. ..
1 1
t t
t t
M1 1.1a
E.g.
dh
dx
22
2
1.. 2 1 ... or
1
t
t
dh
dx
2
2
2
2
21
1.. ...
21
1
tt
tt
M1 3.1a
E.g.
dh
dx
22 2
2 2 2
1 21 1 25 215 2 1
1 2 1 2 1
t t t
t t t
A1 1.1b
…2 2 2 2 2 2 2
2 2
15[4(1 ) 2(1 ) ] 21(1 )(1 ) 50 (1 )
2(1 )
t t t t t t
t
M1 2.1
…4 3 2 2 2
2 2 2 2
9 50 180 50 51 ( 6 17)(9 4 3)*
2(1 ) 2(1 )
t t t t t t t t
t t
A1* 2.1
(6)
(a) ALT 1
2
2 2
2 1h( ) ... 21 25
1 1
t tx
t t
M1 1.1a
2
2 2
2 1... 15 2 ...
1 1
t t
t t
2
2
2
22
1or ... 15 ...
21
1
tt
tt
M1 2.1
2 2 4
2 2
15(4 (1 )) 42 (1 ) 25(1 )h( ) 45
(1 )
t t t t tx
t
M1 1.1b
4 3 4 3 2
2 2 2 2
25 18 102 25 20 18 90 102 70h( ) 45 or
(1 ) (1 )
t t t t t t tx
t t
A1 1.1b
dh
dx
dh
dt
dt
dx
('u ')(1 t2 )2 ('u ')(4t(1 t2 ))
(1 t2 )4 M1 3.1a
4 3 2 2 2
2 2 2 2
9 50 180 50 51 ( 6 17)(9 4 3)... *
2(1 ) 2(1 )
t t t t t t t t
t t
A1* 2.1
(6)
(b)(i) Accept any value between 1
40 0.025and
1
60 0.167inclusive. B1 3.3
(ii) Suitable for times since the graphs both oscillate bi-modally with about the same periodicity.
B1 3.4
Not suitable for predicting heights since the heights of the peaks vary over time, but the graph of h (x) has fixed peak height.
B1 3.5b
(3)
1
4(1 t2 )
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Question Scheme Marks AOs
8. (c) Solves at least one of the quadratics
M1 1.1b
Finds corresponding x values, for at least one value
of t from the factor.M1 1.1b
One correct value for these x. A1 1.1b
Maximum peak height occurs at smallest positive value of x, from first graph, but the third of these peaks needed, so is the required time.
M1 3.4
corresponds to 26 hours and 39 minutes
(nearest minute) after 08:00 on 3rd January. (Allow if a different greatest peak height used.)
M1 3.4
Time of greatest tide height is approximately 10:39 (am) (also allow 10:38 or 10:40)
A1 3.2a
(6) (15 marks)
Notes 8. (a) M1 Differentiates h(x).
M1 Applies t-substitution to both terms with their coefficients.
M1 Forms a correct expression in t for the cos x term, using double angle formula and t-substitution, or double ‘t’-substitution.
A1 Fully correct expression in t for dh
dx.
M1 Gets all terms over the correct common factor. Numerators must be appropriate for their terms.
A1* Achieves the correct answer via expression with correct quartic numerator before factorisation.
t 6 36 4117
2 3 26
or t 4 16 4 9 (3)
182 31
9x 4tan1(t)
29 4 3t t e.g. x awrt 2.797 or 9.770,1.510
t 1.509...8 26.642...26.642....x
x
2
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Notes Continued
8. (a) ALT 1
M1 Applies t-substitution to both terms.
M1 Forms a correct expression in t for the sin x term, using double angle formula and t-substitution, or double ‘t’-substitution.
M1 Gets all terms in t over the correct common factor. Numerators must be appropriate for their terms. May include the constant term too.
A1 Fully correct expression in t for . M1 Differentiates, using both chain rule and quotient rule with their ‘u’. A1* Achieves the correct answer via expression with correct quartic numerator before
factorisation. NOTE The individual terms may be differentiated before putting over a common denominator.
In this case score the third M for differentiating with chain rule and quotient rule, then return to the original scheme.
(b) (i) B1 Any value between
1
40 (e.g. taking h(0) as reference point) or
1
60 (taking lower peaks
as reference). NB: Taking high peak as reference gives 1
50.
(b)(ii) B1 Should mention both the bimodal nature and periodicity for the actual data match the
graph of h. B1 Mentions that the heights of peaks vary in each oscillation. (c) M1 Solves (at least) one of the quadratic equations in the numerator. M1 Must be attempting to solve the quadratic factor from which the solution comes
and using to find a corresponding value for x.
A1 At least one correct x value from solving the requisite quadratic: awrt any of – 2.797, 1.510, 9.770, 14.076, 22.336, 26.642, 34.902 or 39.208.
M1 Uses graph of h to pick out their x = 26.642 as the time corresponding to the third of the higher peaks, which is the highest of the peaks on 4th January on the tide height graph. As per scheme or allow if all times listed and correct one picked.
M1 Finds the time for one of the values of t corresponding to the highest peaks. E.g. 1.5096...~ 09:31 (3rd January) or 14.076... ~ 22:05 (3rd January) or 26.642…~ 10:39 (4th January) or 39.208…~ 23:13 (4th January). (Only follow through on use of the smallest positive t solution 4k .)
A1 Time of greatest tide height on 4th January is approximately 10:39. Also allow 10:38 or 10:40.
x
2
h( )x
2(9 4 3)t t tan4
xt
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Pearson Edexcel Level 3 GCE
Further Mathematics Advanced Paper 4: Further Mathematics Option 2 Option 4A: Further Pure Mathematics 2
Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes
Paper Reference(s)
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Answer ALL questions. Write your answers in the spaces provided.
1. (i) Use the Euclidean algorithm to find the highest common factor of 602 and 161.
Show each step of the algorithm.
(3)
(ii) The digits which can be used in a security code are 1, 2, 3, 4, 5, 6, 7, 8 and 9
Originally the code used consisted of two distinct odd digits, followed by three
distinct even digits.
To enable more codes to be generated, a new system is devised. This uses two
distinct even digits, followed by any three other distinct digits. No digits are
repeated.
Find the increase in the number of possible codes which results from using the new
system.
(4)
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Question 1 continued
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(Total for Question 1 is 7 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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2. A transformation from the z-plane to the w-plane is given by
2w z
(a) Show that the line with equation Im(z) = 1 in the z-plane is mapped to a parabola in
the w-plane, giving an equation for this parabola.
(4)
(b) Sketch the parabola on an Argand diagram.
(2)
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Question 2 continued
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(Total for Question 2 is 6 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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3. The matrix M is given by
M =
2 1 0
1 2 0
1 0 4
(a) Show that 4 is an eigenvalue of M, and find the other two eigenvalues.
(4)
(b) For each of the eigenvalues find a corresponding eigenvector.
(4)
(c) Find a matrix P such that 1P MP is a diagonal matrix.
(2)
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Question 3 continued
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Question 3 continued
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Question 3 continued
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(Total for Question 3 is 10 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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4. (i) A group G contains distinct elements a, b and e where e is the identity element and
the group operation is multiplication.
Given 2
a b ba , prove ab ba
(4)
(ii) The set H = {1, 2, 4, 7, 8, 11, 13, 14} forms a group under the operation of
multiplication modulo 15
(a) Find the order of each element of H.
(3)
(b) Find three subgroups of H each of order 4, and describe each of these subgroups.
(4)
The elements of another group J are the matrices
4 4
4 4
cos sin
sin cos
k k
k k
where k =1, 2, 3, 4, 5, 6, 7, 8 and the group operation is matrix multiplication.
(c) Determine whether H and J are isomorphic, giving a reason for your answer.
(2)
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Question 4 continued
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Question 4 continued
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Question 4 continued
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(Total for Question 4 is 13 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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5.
Figure 1
An engineering student makes a miniature arch as part of the design for a piece of
coursework.
The cross-section of this arch is modelled by the curve with equation
12cosh 2y A x , ln lna x a
where a >1 and A is a positive constant. The curve begins and ends on the x-axis, as
shown in Figure 1.
(a) Show that the length of this curve is 2
2
1k a
a
, stating the value of the constant k.
(5)
The length of the curved cross-section of the miniature arch is required to be 2 m long.
(b) Find the height of the arch, according to this model, giving your answer to
2 significant figures.
(4)
(c) Find also the width of the base of the arch giving your answer to 2 significant
figures.
(1)
(d) Give the equation of another curve that could be used as a suitable model for the
cross-section of an arch, with approximately the same height and width as you found
using the first model.
(You do not need to consider the arc length of your curve)
(2)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question 5 continued
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Question 5 continued
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Question 5 continued
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(Total for Question 5 is 12 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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6. A curve has equation
6 2 6z z z
(a) Show that the curve is a circle with equation 2 2 20 36 0x y x
(2)
(b) Sketch the curve on an Argand diagram.
(2)
The line l has equation * * 0az a z , where a and z
Given that the line l is a tangent to the curve and that arg a = θ
(c) find the possible values of tan
(5)
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Question 6 continued
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Question 6 continued
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Question 6 continued
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(Total for Question 6 is 9 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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7.
2
0sin dn
nI x x
, n 0
(a) Prove that, for n 2,
2( 1)n nnI n I
(4)
(b)
Figure 2
A designer is asked to produce a poster to completely cover the curved surface area of a
solid cylinder which has diameter 1m and height 0.7 m.
He uses a large sheet of paper with height 0.7 m and width of π m.
Figure 2 shows the first stage of the design, where the poster is divided into two sections
by a curve.
The curve is given by the equation
2 10sin (4 ) sin (4 )y x x
relative to axes taken along the bottom and left hand edge of the paper.
The region of the poster below the curve is shaded and the region above the curve
remains unshaded, as shown in Figure 2.
Find the exact area of the poster which is shaded.
(5)
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Question 7 continued
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Question 7 continued
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Question 7 continued
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(Total for Question 7 is 9 marks)
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8. A staircase has n steps. A tourist moves from the bottom (step zero) to the top (step n).
At each move up the staircase she can go up either one step or two steps, and her overall
climb up the staircase is a combination of such moves.
If nu is the number of ways that the tourist can climb up a staircase with n steps,
(a) explain why nu satisfies the recurrence relation
1 2n n nu u u , with 1 1u and 2 2u
(3)
(b) Find the number of ways in which she can climb up a staircase when there are eight
steps.
(1)
A staircase at a certain tourist attraction has 400 steps.
(c) Show that the number of ways in which she could climb up to the top of this staircase
is given by
401 401
1 1 5 1 5
2 25
(5)
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Question 8 continued
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Question 8 continued
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Question 8 continued
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(Total for Question 8 is 9 marks)
TOTAL FOR PAPER IS 75 MARKS
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Paper 4: Further Pure Mathematics 2 Mark Scheme
Question Scheme Marks AOs
1(i) 602 3 161 119 M1 1.1b
161 119 42, 119 2 42 35 M1 1.1b
42 35 7, 35 5 7, hcf 7 A1 1.1b
(3)
(ii) Number of codes under old system = 5 4 4 3 2 ( 480) B1 3.1b
Number of codes under new system = 4 3 7 6 5 ( 2520) B1 3.1b
Subtracts first answer from second M1 1.1b
Increase in number of codes is 2040 A1 1.1b
(4)
(7 marks)
Notes:
(i)
M1: Attempts Euclid’s algorithm – (there may be an arithmetic slip finding 119)
M1: Uses Euclid’s algorithm a further two times with 161 and “their 119” and then with “their 119”
and “their 42”
A1: This should be accurate with all the steps shown
(ii)
B1: Correctly interprets the problem and uses the five odd digits and four even digits to form a correct
product
B1: Interprets the new situation using the four even digits, then the seven digits which have not been
used, to form a correct product
M1: Subtracts one answer from the other
A1: Correct answer
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
2(a) Let z = x + i M1 2.1
2 2( i) ( 1) 2 iw x x x A1 1.1b
Let w = u + iv, then 2
( 1) and 2u x v x M1 2.1
2
4( 1)v u , which therefore represents a parabola A1ft 2.2a
(4)
(b)
M1
1.1b
A1 1.1b
(2)
(6 marks)
Notes:
(a)
M1: Translates the information that Im(z) = 1 into a cartesian form; e.g. z = x + i
A1: Obtains a correct expression for w
M1: Separates the real and imaginary parts and equates to u and v respectively
A1ft: Obtains a quadratic equation and states that their quadratic equation represents a parabola
(b)
M1: Sketches a parabola with symmetry about the real axis
A1: Accurate sketch
M1: Sketches a
parabola with
symmetry about
the real axis
A1: Accurate
sketch
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Question Scheme Marks AOs
3(a) Finds the characteristic equation
2(2 ) (4 ) (4 ) 0 M1 2.1
so 2
(4 )( 4 3) 0 so 4 * A1* 2.2a
Solves quadratic equation to give M1 1.1b
1 and 3 A1 1.1b
(4)
(b) Uses a correct method to find an eigenvector M1 1.1b
Obtains a vector parallel to one of
0
0
1
or
1
1
1
or
3
3
1
A1 1.1b
Obtains two correct vectors A1 1.1b
Obtains all three correct vectors A1 1.1b
(4)
(c) Uses their three vectors to form a matrix M1 1.2
0 1 3
0 1 3
1 1 1
or other correct answer with
columns in a different order.
A1 1.1b
(2)
(10 marks)
Notes:
(a)
M1: Attempts to find the characteristic equation (there may be one slip)
A1*: Deduces that 4 is a solution by the method shown or by checking that 4 satisfies the
characteristic equation
M1: Solves their quadratic equation
A1: Obtains the two correct answers as shown above
(b)
M1: Uses a correct method to find an eigenvector
A1: Obtains one correct vector (may be a multiple of the given vectors)
A1: Obtains two correct vectors (may be multiples of the given vectors)
A1: Obtains all three correct vectors (may be multiples of the given vectors)
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Notes: (continued)
(c)
M1: Forms a matrix with their vectors as columns
A1:
0 1 3
0 1 3
1 1 1
or
1 0 3
1 0 3
1 1 1
or
3 1 0
3 1 0
1 1 1
or other correct alternative
Question Scheme Marks AOs
4(i) 2 2If we assume ; as thenab ba a b ba ab a b M1 2.1
1 1 1 2 1So a abb a a bb M1 2.1
So e a A1 2.2a
But this is a contradiction, as the elements e and a are distinct so
ab ba A1 2.4
(4)
(ii)(a) 2 has order 4 and 4 has order 2 M1 1.1b
7, 8 and 13 have order 4 A1 1.1b
11 and 14 have order 2 and 1 has order 1 A1 1.1b
(3)
(ii)(b) Finds the subgroup {1, 2, 4, 8} or the subgroup {1, 7, 4, 13} M1 1.1b
Finds both and refers to them as cyclic groups, or gives generator 2 and
generator 7 A1 2.4
Finds {1, 4, 11, 14} B1 2.2a
States each element has order 2 or refers to it as Klein Group B1 2.5
(4)
(ii)(c) J has an element of order 8, (H does not) or J is a cyclic group (H is
not) or other valid reason
M1 2.4
They are not isomorphic
A1 2.2a
(2)
(13 marks)
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Notes:
(i)
M1: Proof begins with assumption that 2and deduces that this impliesab ba ab a b
M1: A correct proof with working shown follows, and may be done in two stages
A1: Concludes that assumption implies that e a
A1: Explains clearly that this is a contradiction, as the elements e and a are distinct so ab ba
(ii)(a)
M1: Obtains two correct orders (usually the two in the scheme)
A1: Finds another three correctly
A1: Finds the final three so that all eight are correct
(ii)(b)
M1: Finds one of the cyclic subgroups
A1: Finds both subgroups and explains that they are cyclic groups, or gives generators 2 and 7
B1: Finds the non cyclic group
B1: Uses correct terms that each element has order 2 or refers to it as Klein Group
(ii)(c)
M1: Clearly explains how J differs from H
A1: Correct deduction
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Question Scheme Marks AOs
5(a)
dsinh 2
d
yx
x
B1 2.1
so S = 21 sinh 2 dx x M1 2.1
s cosh 2 dx x A1 1.1b
= ln
12 lnsinh 2
a
ax
or
ln
0sinh 2
ax M1 2.1
= 2ln -2ln1
2sinh 2ln [e e ]a a
a = 21
2 2
1a
a
(so k = ½) A1 1.1b
(5)
(b) 21
2 2
1a
a
= 2 so
4 24 1 0a a M1 1.1b
2 2 5a (and a = 2.06 (approx.) ) M1 1.1b
When x = lna, y = 0 so 1
2cosh 2lnA a M1 3.4
Height = A – 0.5 = awrt 0.62m A1 1.1b
(4)
(c) The width of the base = 2lna = 1.4m B1 3.4
(1)
(d) A parabola of the form 20.62 1.19y x , or other symmetric curve
with its equation e.g. 0.62cos(2.2x) M1A1
3.3
3.3
(2)
(12 marks)
Notes:
(a) B1: Starts explanation by finding the correct derivative
M1: Uses their derivative in the formula for arc length
A1: Uses suitable identity to simplify the integrand and to obtain the expression in scheme
M1: Integrates and uses appropriate limits to find the required arc length
A1: Uses the definition of sinh to complete the proof and identifies the value for k
(b)
M1: Uses the formula obtained from the model and the length of the arch to create a quartic equation
M1: Continues to use this model to obtain a quadratic and to obtain values for a
M1: Attempts to find a value for A in order to find h
A1: Finds a value for the height correct to 2sf (or accept exact answer)
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Notes: (continued)
(c) B1: Finds width to 2 sf i.e. 1.4m
(d)
M1: Chooses or describes an even function with maximum point on the y axis
A1: Gives suitable equation passing through (0, 0.62) and (0.7, 0) and (- 0.7, 0)
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Question Scheme Marks AOs
6(a) 2 2 2 2( 6) 4[( 6) ]x y x y M1 2.1
2 2
20 36 0x y x which is the equation of a circle* A1* 2.2a
(2)
(b)
M1 1.1b
A1 1.1b
(2)
(c)
Let a = c + id and a* = c − id then (c + id)(x − iy) + (c − id)(x + iy) = 0
M1 3.1a
Soc
y xd
A1 1.1b
The gradients of the tangents (from geometry) are 4
3
B1 3.1a
So 4
3
c
d and
3
4
d
c M1 3.1a
So 3
tan4
A1 1.1b
(5)
(9 marks)
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Notes:
(a)
M1: Obtains an equation in terms of x and y using the given information
A1*: Expands and simplifies the algebra, collecting terms and obtains a circle equation correctly,
deducing that this is a circle
(b)
M1: Draws a circle with centre at (10, 0)
A1: (Radius is 8) so circle does not cross the y axis
(c)
M1: Attempts to convert line equation into a cartesian form
A1: Obtains a simplified line equation
B1: Uses geometry to deduce the gradients of the tangents
M1: Understands the connection between arg a and the gradient of the tangents and uses this
connection
A1: Correct answers
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Question Scheme Marks AOs
7(a) 2 1
0sin sin dn
nI x x x
M1 2.1
221 2 2
0 0cos sin ( ) cos ( 1)sin dn nx x x n x x
A1 1.1b
Obtains 2 2 2
00 ( ) (1 sin )( 1)sin dnx n x x
M1 1.1b
So 2( 1) ( 1)
n n nI n I n I
and hence 2
( 1)n n
nI n I
* A1* 2.1
(4)
(b) uses
2
( 1)n n
nI I
n
to give
10 8
9
10I I or
2 0
1
2I I M1 3.1b
So 10 0
9 7 5 3 1
10 8 6 4 2I I
M1 2.1
0
2I
B1 1.1b
Required area is 2 102( )I I or
1
2 1048 ( )I I M1 3.1b
263 65
2 m4 512 256
A1 1.1b
(5)
(9 marks)
Notes:
(a) M1: Splits the integrand into the product shown and begins process of integration by parts (there may
be sign errors)
A1: Correct work
M1: Uses limits on the first term and expresses cos2 term in terms of sin2
A1*: Completes the proof collecting nI terms correctly with all stages shown
(b)
M1: Attempts to find 10I and/or 2
I
M1: Finds 10I in terms of 0
I
B1: Finds 0I correctly
M1: States the expression needed to find the required area
A1: Completes the calculation to give this exact answer
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Question Scheme Marks AOs
8(a) 1 1u as there is only one way to go up one step B1 2.4
2 2u as there are two ways: one step then one step or two steps B1 2.4
If first move is one step then can climb the other (n1) steps in 1nu
ways
If first move is two steps can climb the other (n2) steps in 2nu
ways
So 1 2n n nu u u
B1 2.4
(3)
(b) Sequence begins 1, 2, 3, 5, 8, 13, 21, 34,… so 34 ways of climbing 8
steps B1 1.1b
(1)
(c) To find general term use
2
1 2gives 1
n n nu u u
M1 2.1
This has roots 1 5
2
A1 1.1b
So general form is 1 5 1 5
2 2
n n
A B
M1 2.2a
Uses initial conditions to find A and B reaching two equations in A and B M1 1.1b
Obtains 1 5
2 5A
and 1 5
2 5B
and so when n = 400
obtains
401 401
1 1 5 1 5
2 25
*
A1* 1.1b
(5)
(9 marks)
Notes:
(a)
B1: Need to see explanation for 1 1u
B1: Need to see explanation for 2 2u with the two ways spelled out
B1: Need to see the first move can be one step or can be two steps and clear explanation of the
Iterative expression as in the scheme
(b)
B1: The answer is enough for this mark
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Notes: (continued)
(c)
M1: Obtains this characteristic equation
A1: Solves quadratic – giving exact answers
M1: Obtains a general form
M1: Use initial conditions to obtains two equations which should be 1 5 1 5 2A B o.e.
and 3 5 3 5 4A B but allow slips here.
A1*: Must see exact correct values for A and B and conclusion given for n = 400
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Pearson Edexcel Level 3 GCE
Further Mathematics Advanced
Further Mathematics Option 1 Paper 3: Further Statistics 1
Further Mathematics Option 2 Paper 4: Further Statistics 1
Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes
Paper Reference(s)
9FM0/3B 9FM0/4B
You must have: Mathematical Formulae and Statistical Tables, calculator
Candidates may use any calculator permitted by Pearson regulations. Calculators must not have the facility for algebraic manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them.
Instructions
Use black ink or ball-point pen.
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
Fill in the boxes at the top of this page with your name, centre number and candidate number.
Answer all questions.
Answer the questions in the spaces provided – there may be more space than you need.
You should show sufficient working to make your methods clear. Answers without working may not gain full credit.
Answers should be given to three significant figures unless otherwise stated. Information
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
There are 7 questions in this question paper. The total mark for this paper is 75.
The marks for each question are shown in brackets – use this as a guide as to how much time to spend on each question.
Advice
Read each question carefully before you start to answer it.
Try to answer every question.
Check your answers if you have time at the end.
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Answer ALL questions. Write your answers in the spaces provided.
1. Bacteria are randomly distributed in a river at a rate of 5 per litre of water. A new factory
opens and a scientist claims it is polluting the river with bacteria. He takes a sample of
0.5 litres of water from the river near the factory and finds that it contains 7 bacteria.
Stating your hypotheses clearly test, at the 5% level of significance, whether there is
evidence that the level of pollution has increased.
(5)
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Question 1 continued
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(Total for Question 1 is 5 marks)
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2. A call centre routes incoming telephone calls to agents who have specialist knowledge to
deal with the call. The probability of a caller, chosen at random, being connected to the
wrong agent is p.
The probability of at least 1 call in 5 consecutive calls being connected to the wrong agent
is 0.049
The call centre receives 1000 calls each day.
(a) Find the mean and variance of the number of wrongly connected calls a day.
(7)
(b) Use a Poisson approximation to find, to 3 decimal places, the probability that more than
6 calls each day are connected to the wrong agent.
(2)
(c) Explain why the apporoximation used in part (b) is valid.
(2)
The probability that more than 6 calls each day are connected to the wrong agent using the
binomial distribution is 0.8711 to 4 decimal places.
(d) Comment on the accuracy of your answer in part (b)
(1)
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Question 2 continued
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Question 2 continued
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Question 2 continued
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(Total for Question 2 is 12 marks)
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3. Bags of £1 coins are paid into a bank. Each bag contains 20 coins.
The bank manager believes that 5% of the £1 coins paid into the bank are fakes. He
decides to use the distribution X ~ B(20, 0.05) to model the random variable X, the
number of fake £1 coins in each bag.
The bank manager checks a random sample of 150 bags of £1 coins and records the
number of fake coins found in each bag. His results are summarised in Table 1. He then
calculates some of the expected frequencies, correct to 1 decimal place.
Number of fake coins in each bag 0 1 2 3 4 or more
Observed frequency 43 62 26 13 6
Expected frequency 53.8 56.6
8.9
Table 1
(a) Carry out a hypothesis test, at the 5% significance level, to see if the data supports
the bank manager’s statistical model. State your hypotheses clearly.
(10)
The assistant manager thinks that a binomial distribution is a good model but
suggests that the proportion of fake coins is higher than 5%. She calculates the actual
proportion of fake coins in the sample and uses this value to carry out a new
hypothesis test on the data. Her expected frequencies are shown in Table 2.
Number of fake coins in each bag 0 1 2 3 4 or more
Observed frequency 43 62 26 13 6
Expected frequency 44.5 55.7 33.2 12.5 4.1
Table 2
(b) Explain why there are 2 degrees of freedom in this case.
(2)
(c) Given that she obtains a χ2 test statistic of 2.67, test the assistant manager’s
hypothesis that the binomial distribution is a good model for the number of fake
coins in each bag. Use a 5% level of significance and state your hypotheses clearly.
(2)
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Question 3 continued
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Question 3 continued
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Question 3 continued
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(Total for Question 3 is 14 marks)
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4. A random sample of 100 observations is taken from a Poisson distribution with mean 2.3
Estimate the probability that the mean of the sample is greater than 2.5
(4)
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(Total for Question 4 is 4 marks)
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5. The probability of Richard winning a prize in a game at the fair is 0.15
Richard plays a number of games.
(a) Find the probability of Richard winning his second prize on his 8th game,
(2)
(b) State two assumptions that have to be made, for the model used in part (a) to be valid.
(2)
Mary plays the same game, but has a different probability of winning a prize. She plays
until she has won r prizes. The random variable G represents the total number of games
Mary plays.
(c) Given that the mean and standard deviation of G are 18 and 6 respectively, determine
whether Richard or Mary has the greater probability of winning a prize in a game.
(4)
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Question 5 continued
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(Total for Question 5 is 8 marks)
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6. The probability generating function of the discrete random variable X is given by
GX (t) = 2
23 2k t t
(a) Show that k = 1
36
(2)
(b) Find P(X = 3)
(2)
(c) Show that Var(X) = 29
18
(8)
(d) Find the probability generating function of 2X + 1
(2)
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Question 6 continued
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Question 6 continued
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Question 6 continued
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(Total for Question 6 is 14 marks)
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7. Sam and Tessa are testing a spinner to see if the probability, p, of it landing on red is less
than 1
5. They both use a 10% significance level.
Sam decides to spin the spinner 20 times and record the number of times it lands on red.
(a) Find the critical region for Sam’s test.
(2)
(b) Write down the size of Sam’s test.
(1)
Tessa decides to spin the spinner until it lands on red and she records the number of
spins.
(c) Find the critical region for Tessa’s test.
(6)
(d) Find the size of Tessa’s test.
(1)
(e) (i) Show that the power function for Sam’s test is given by
19
1 1 19p p
(ii) Find the power function for Tessa’s test.
(4)
(f) With reference to parts (b), (d) and (e), state, giving your reasons, whether you
would recommend Sam’s test or Tessa’s test when p = 0.15
(4)
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Question 7 continued
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Question 7 continued
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Question 7 continued
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(Total for Question 7 is 18 marks)
TOTAL FOR PAPER IS 75 MARKS
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AL Further Statistics 1
Question Scheme Marks AOs
Q1 Ho : = 5 (= 2.5) H1 : > 5 (> 2.5) B1 2.5
X ~ Po (2.5) B1 3.3
Method 1 Method 2
P 7 1 P )6 ( X X
= 1 − 0.9858
P(X 5) = 0.1088
P(X 6) = 0.042 M1 1.1b
= 0.0142 CR X 6 A1 1.1b
0.0142 < 0.05 7 6 or 7 is in critical region or 7 is significant
Reject H0. There is evidence at the 5% significance level that the
level of pollution has increased.
or
There is evidence to support the scientists claim is justified
A1cso 2.2b
(5 marks)
Notes
B1: Both hypotheses correct using or and 5 or 2.5
B1: Realising that the model Po(2.5) is to be used. This may be stated or used.
M1: Using or writing 1 P ( )6 X or ( ) 1 P < 7 X
a correct CR or P(X 5) = awrt 0.109 and P(X 6) = awrt 0.042
A1: awrt 0.0142 or CR X 6 or X > 5.
A1: A fully correct solution and drawing a correct inference in context
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Question Scheme
Marks AOs
Q2(a) P(X 1) = 1 – P(X = 0)
1 – P(X = 0) = 0.049 B1 3.1b
P(X = 0) = 0.951 B1 1.1b
5 0.951x
0.99x M1 3.1b
p = 0.01 A1 1.1b
X ~B(1000, 0.01) M1 3.3
Mean = np = 10 A1ft 1.1b
Variance = np(1 – p) = 9.9 A1ft 1.1b
(7)
(b) X ~ Po(“10”) then require: P(X > 6) = 1 – P (X 6) M1 3.4
= 1 – 0.1301
= 0.870 A1 1.1b
(2)
(c) The approximation is valid as : the number of calls is large B1 2.4
the probability of connecting to the wrong agent is small B1 2.4
(2)
(d) The answer is accurate to 2 decimal place B1 3.2b
(1)
(12 marks)
Notes
(a) B1: Realising that the P(at least 1 call ) = 1 – P(X = 0)
B1: Calculating P(X = 0) = 0.951
M1: Forming the equation 5 "their 0.951"x may be implied by p = 0.01
A1: 0.01 only
M1: Realising the need to use the model B(1000, 0.01) This may be stated or used
A1: mean =10 or ft their p but only if 0 < p < 1
A1: Var = 9.9 or ft their p but only if 0 < p < 1
(b) M1: Using the model Po(“their 10”) ( this may be written or used) and 1 – P (X 6)
A1: awrt 0.870 Award M1 A1 for awrt 0.870 with no incorrect working
(c) B1:Explaining why approximation is valid - need the context of number and calls
B1: need the context connecting, wrong agent
(d) B1: Evaluating the accuracy of their answer in (b). Allow 2 significant figures
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Question Scheme
Marks AOs
Q3(a) Expected value for 2 =150 P( 2)X M1 3.4
= 28.3015… A1 1.1b
Expected value for 4 or more = 150 – (53.8 + 56.6 + 28.3 + 8.9)
= 2.4 A1ft 1.1b
H0: Bin(20, 0.05) is a suitable model
H1: Bin(20, 0.05) is not a suitable model B1 2.5
Combining last two groups
3
Observed frequency 19
Expected frequency 11.3
M1 2.1
B1 1.1b
Critical value, χ2 (0.05) = 7.815 B1 1.1a
Test statistic =
2 243 53.8 62 56.6
...53.8 56.6
M1 1.1b
= 8.117 A1 1.1b
In critical region, sufficient evidence to reject H0, accept H1
Significant evidence at 5% level to reject the manager’s model A1 3.5a
(10)
(b) = 4 – 2 = 2
4 classes due to pooling B1 2.4
2 restrictions (equal total and mean/proportion) B1 2.4
(2)
(c) H0: Binomial distribution is a good model
H1: Binomial distribution is not a good model B1 3.4
Critical value, χ2 (0.05) = 5.991
Test statistic is not in critical region, insufficient evidence to
reject H0
There is evidence that the Binomial distribution is a good model.
B1 3.5a
(2)
(14 marks)
Notes
(a) M1: Using the binomial model 2 18150 (1 )p p may be implied by 28.3
A1: awrt 28.3
A1: awrt 2.4 or ft their “28.3”
B1: Both hypotheses correct using the correct notation or written out in full.
M1: For recognising the need to combine groups
B1: Number of degrees of freedom = 3 may be implied by a correct CV
B1: awrt 7.82
M1: Attempting to find
2
i i
i
O E
E or
2
i
i
ON
E may be implied by awrt 8.12
A1: awrt 8.12
A1: Evaluating the outcome of a model by drawing a correct inference in context
(b) B1: Explaining why there are 4 classes
B1: Explanation of why 2 is subtracted
(c) B1: Correct hypotheses for the refined model
B1: The CV awrt 5.99 and drawing the correct inference for the refined model
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Question Scheme
Marks AOs
Q4. Po(2.3) n = 100 = 2.3 2 2.3
CLT 2.3
2.3,100
NX
M1
A1
3.1a
1.1b
2.5 2.3( 2.5)
0.023P PX Z
M1 3.4
= P(Z > 1.318..)
= 0.09632… A1 1.1b
(4)
(4 marks)
M1: For realising the need to use the CLT to set X normal with correct mean.
May be implied by using the correct normal distribution.
A1: For fully correct normal stated or used
M1: Use of the normal model to find ( 2.5)P X . Can be awarded for 2.5 2.3
0.023
or
awrt 1.32
A1: awrt 0.0963
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Question Scheme
Marks AOs
Q5(a)
627
0.15 0.851
M1 3.3
= 0.05940… = awrt 0.0594 A1 1.1b
(2)
(b) The model is only valid if:
the games (trials) are independent B1 3.5b
the probability of winning a prize, 0.15, is constant for each
game B1 3.5b
(2)
(c) 2
2
118 and 6
r pr
p p
M1
A1
3.1b
1.1b
Solving: 2p = 1 – p M1 1.1b
p = 1
3 (> 0.15) so Mary has the greater chance of winning a
prize
A1 3.2a
(4)
(8 marks)
Notes
5(a) M1: For selecting an appropriate model negative binomial or B(7, 0.15) with an
extra success in 8th trial e.g.
67
0.15 0.85 0.151
Allow
670.85 0.15 0.85
1
may be implied by awrt
0.0594
A1: awrt 0.0594
(b) B1: Stating the first assumption that games are independent
B1: Stating the second assumption that the probability remains constant
(c) M1: Forming an equation for the mean or for the standard deviation.
A1: Both equations correct
M1: Solving the 2 equations leading to 2p = 1 – p
A1: For p = 1
3followed by a correct deduction
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Question Scheme Marks AOs
Q6(a) G (1) 1X gives M1 2.1
26 1k so k = 1
36 * A1*cso 1.1b
(2)
(b) P(X = 3) = coefficient of t3 so 3G ( ) ... 4 ...X t k t M1 1.1b
[ P(X = 3) =] 1
9 A1 1.1b
(2)
(c) 2G ( ) 2 3 2 1 4X t k t t t M1 2.1
E(X) = G (1) 2 3 1 2 1 4X k M1 1.1b
= 5
3 A1 1.1b
22G ( ) 2 3 2 4 1 4X t k t t t
M1
A1
2.1
1.1b
2G (1) 2 [6 4 5 ]X k 49
18
M1 1.1b
Var(X) = 2 49 5 25
G (1) G (1) G (1)18 3 9
X X X M1 2.1
= 29
18* A1*cso 1.1b
(8)
(d)
22
2 2
2 1G ( ) 3 236
X
tt t t [ t or sub t2 for t] M1 3.1a
= 2
2 4
2 1G ( ) 3 236
X
tt t t A1 1.1b
(2)
(14 marks)
Notes
6(a) M1: Stating G (1) 1X
A1*cso: Fully correct proof with no errors
(b) M1: Attempting to find the coefficient of t3. May be implied by obtaining
1
9 or awrt
0.11
A1: 1
9, allow awrt 0.111
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Notes (continued)
(c) M1: Attempting to find G ( )X t . Allow Chain rule or multiplying out the brackets
and differentiating
M1: Substituting t = 1 into G ( )X t
A1: 5
3, allow awrt 1.67
M1: Attempting to find G ( )X t
A1: 222 3 2 4 1 4k t t t
or
2(48 24 26)k t t o.e.
A1: 22 [6 4 5 ]k o.e.
M1: Using 2
G (1) G (1) G (1)X X X to find the Variance
A1*cso: 29
18
(d) M1:Realising the need to t or sub t2 for t
A1: 2
2 43 236
tt t , or 2 4 6 89 6 13 4 4
36
tt t t t o.e.
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Question Scheme
Marks AOs
Q7(a) X ~ B(20, 0.2) and seek c such that P(X c) < 0.10 M1 3.3
[P(X 1) = 0.0692] CR is X 1 A1 1.1b
(2)
(b) Size = 0.0692 B1ft 1.2
(1)
(c) Y = no. of spins until red obtained so Y~ Geo(0.2) M1 3.3
1
p so if p < 0.2 then mean is larger so seek d so that
P(Y d) < 0.10
M1 2.4
1P( ) 0.8 dY d M1 3.4
1 log(0.1)0.8 0.10 1
log(0.8)
d d M1 1.1b
d > 11.3.. A1 1.1b
CR is Y 12 A1 2.2b
(6)
(d) Size = [ 110.8 = 0.085899…] = 0.0859 B1 1.1b
(1)
(e)(i) Power = P(reject H0 when it is false) = P(X 1 | X ~B(20, p)) M1 2.1
= 20 19
1 20 1p p p M1 1.1b
= 19
1 1 19p p * A1*cso 1.1b
(ii) Power = 11
1 p B1 1.1b
(4)
(f) Sam’s test has smaller P(Type I error) (or size) so is better B1 2.2a
Power of Sam’s test = 0.1755… B1 1.1b
Power of Tessa’s test = 110.85 = 0.1673… B1 1.1b
So for p = 0.15 Sam’s test is recommended B1 2.2b
(4)
(18 marks)
Notes
7(a) M1: Realising the need to use the model Using B(20,0.2) with method for finding
the CR or implied by a correct CR
A1: X 1 or X < 2
(b) B1: awrt 0.0692
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Notes (continued)
(c) M1: Realising that the model Geo(0.2)is needed. This may be written or used
M1: Realising the key step that they need to find P(Y d) < 0.10
M1: Using the model 10.8 d
M1: Using the model 10.8 0.10d and finding a method to solve leading to a
value/range of values for d
A1: For d > 11.3..
A1: For Y 12 or Y > 11 (a correct inference)
(d) B1ft: awrt 0.0692. ft their answer to part (c)
(e)(i) M1: Using B(20, p) and realizing they need to find P(X 1) o.e. This may be used
or written
M1: Using P(X = 0) + P(X = 1)
A1*cso: Fully correct proof ( no errors)
(ii) B1: For 11
1 p
(f) B1: Making a deduction about the tests using the answers to part(b) and (d)
B1: awrt 0.0176
B1: awrt 0.167
B1: A correct inference about which test is recommended
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Pearson Edexcel Level 3 GCE
Further Mathematics Advanced
Further Mathematics Option 2 Paper 4: Further Statistics 2
Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes
Paper Reference(s)
9FM0/4E
You must have: Mathematical Formulae and Statistical Tables, calculator
Candidates may use any calculator permitted by Pearson regulations. Calculators must not have the facility for algebraic manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them.
Instructions
Use black ink or ball-point pen.
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
Fill in the boxes at the top of this page with your name, centre number and candidate number.
Answer all questions.
Answer the questions in the spaces provided – there may be more space than you need.
You should show sufficient working to make your methods clear. Answers without working may not gain full credit.
Answers should be given correct to 3 significant figures unless otherwise stated. Information
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
There are 6 questions in this question paper. The total mark for this paper is 75.
The marks for each question are shown in brackets – use this as a guide as to how much time to spend on each question.
Advice
Read each question carefully before you start to answer it.
Try to answer every question.
Check your answers if you have time at the end.
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Answer ALL questions. Write your answers in the spaces provided.
1. The three independent random variables A, B and C each have a continuous uniform
distribution over the interval [0, 5].
(a) Find the probability that A, B and C are all greater than 3
(3)
The random variable Y represents the maximum value of A, B and C.
The cumulative distribution function of Y is
F(y) = 3
0 0
0 5125
1 5
y
yy
y
(b) Using algebraic integration, show that Var (Y) = 0.9375
(4)
(c) Find the mode of Y, giving a reason for your answer.
(2)
(d) Describe the skewness of the distribution of Y. Give a reason for your answer.
(1)
(e) Find the value of k such that P(k < Y < 2k) = 0.189
(3)
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Question 1 continued
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Question 1 continued
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Question 1 continued
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(Total for Question 1 is 13 marks)
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2. A researcher claims that, at a river bend, the water gradually gets deeper as the distance
from the inner bank increases. He measures the distance from the inner bank, b cm, and
the depth of a river, s cm, at 7 positions. The results are shown in the table below.
Position A B C D E F G
Distance from
inner bank b cm 100 200 300 400 500 600 700
Depth s cm 60 75 85 76 110 120 104
The Spearman’s rank correlation coefficient between b and s is 6
7
(a) Stating your hypotheses clearly, test whether or not the data provides support for the
researcher’s claim. Use a 1% level of significance.
(4)
(b) Without re-calculating the correlation coefficient, explain how the Spearman’s rank
correlation coefficient would change if
(i) the depth for G is 109 instead of 104
(ii) an extra value H with distance from the inner bank of 800 cm and depth 130 cm
is included.
(3)
The researcher decided to collect extra data and found that there were now many tied ranks.
(c) Describe how you would find the correlation with many tied ranks.
(2)
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Question 2 continued
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(Total for Question 2 is 9 marks)
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3. A nutritionist studied the levels of cholesterol, X mg/cm3, of male students at a large
college. She assumed that X was distributed N(, 2) and examined a random sample of
25 male students. Using this sample she obtained unbiased estimates of and 2 as
and 2
A 95% confidence interval for was found to be ( 1.128, 2.232 )
(a) Show that 2 = 1.79 (correct to 3 significant figures)
(4)
(b) Obtain a 95% confidence interval for 2
(3)
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Question 3 continued
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(Total for Question 3 is 7 marks)
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4. The times, x seconds, taken by the competitors in the 100 m freestyle events at a school
swimming gala are recorded. The following statistics are obtained from the data.
No. of competitors Sample mean x
2x
Girls 8 83.1 55 746
Boys 7 88.9 56 130
Following the gala, a mother claims that girls are faster swimmers than boys. Assuming
that the times taken by the competitors are two independent random samples from
normal distributions,
(a) test, at the 10% level of significance, whether or not the variances of the two
distributions are the same. State your hypotheses clearly.
(7)
(b) Stating your hypotheses clearly, test the mother’s claim. Use a 5% level of
significance.
(6)
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Question 4 continued
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Question 4 continued
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Question 4 continued
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(Total for Question 4 is 13 marks)
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5. Scaffolding poles come in two sizes, long and short. The length L of a long pole has the
normal distribution N(19.6, 0.62). The length S of a short pole has the normal distribution
N(4.8, 0.32). The random variables L and S are independent.
A long pole and a short pole are selected at random.
(a) Find the probability that the length of the long pole is more than 4 times the length
of the short pole. Show your working clearly.
(6)
Four short poles are selected at random and placed end to end in a row. The random
variable T represents the length of the row.
(b) Find the distribution of T.
(3)
(c) Find P(L – T < 0.2)
(4)
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Question 5 continued
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Question 5 continued
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Question 5 continued
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(Total for Question 5 is 13 marks)
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6. A random sample of 10 female pigs was taken. The number of piglets, x, born to each
female pig and their average weight at birth, m kg, was recorded. The results were as
follows:
Number of piglets, x 4 5 6 7 8 9 10 11 12 13
Average weight at
birth, m kg
1.50 1.20 1.40 1.40 1.23 1.30 1.20 1.15 1.25 1.15
(You may use Sxx = 82.5 and Smm = 0.12756 and Sxm = 2.29 )
(a) Find the equation of the regression line of m on x in the form m = a + bx as a model
for these results.
(2)
(b) Show that the residual sum of squares (RSS) is 0.064 to 3 decimal places.
(2)
(c) Calculate the residual values.
(2)
(d) Write down the outlier.
(1)
(e) (i) Comment on the validity of ignoring this outlier .
(ii) Ignoring the outlier, produce another model.
(iii) Use this model to estimate the average weight at birth if x = 15
(iv) Comment, giving a reason, on the reliability of your estimate.
(5)
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Question 6 continued
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Question 6 continued
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Question 6 continued
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(Total for Question 6 is 12 marks)
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7. Over a period of time, researchers took 10 blood samples from one patient with a blood
disease. For each sample, they measured the levels of serum magnesium, s mg/dl, in the
blood and the corresponding level of the disease protein, d mg/dl. One of the researchers
coded the data for each sample using x = 10s and 10( 9) y d but spilt ink over his
work.
The following summary statistics and unfinished scatter diagram are the only remaining
information.
2 1081.74d Sds = 59.524
and
64y Sxx = 2658.9
(a) Use the formula for Sxx to show that Sss = 26.589
(3)
(b) Find the value of the product moment correlation coefficient between s and d.
(4)
(c) With reference to the unfinished scatter diagram, comment on your result in part (b).
(1)
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Question 7 continued
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Question 7 continued
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Question 7 continued
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(Total for Question 7 is 8 marks)
TOTAL FOR PAPER IS 75 MARKS
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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AL Further Statistics 2
Question Scheme Marks AOs
Q1(a) P(A > 3) =
2
5 B1 1.1b
32 8
5 125
M1
A1
1.1a
1.1b
(3)
(b)
23f ( )
125
yy M1 2.1
35
0
3E( ) d
125
yY y
54
0
3=
500
y
15
= 4
M1 1.1b
25 4
0
3 15Var( ) d
125 4
yY y
M1 1.1b
= 0.9375* A1*cso 1.1b
(4)
(c) Mode = 5 B1 1.2
Or reason based on
( )
0df
d
y
y
B1
2.4
(2)
(d) From a sketch or mode > mean therefore it has negative skew B1ft 2.4
(1)
(e)
3 320.189
125 125
k k M1 3.1a
37
0.189125
k A1 1.1b
k = 1.5 A1 1.1b
(3)
(13 marks)
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Notes
1(a) B1:
2
5 o.e. may be implied by a correct answer
M1:
32
"their "5
may be implied by a correct answer
A1: 8
125 o.e.
(b) M1: realising that firstly need to find pdf f(y) and attempt to differentiate F(y)
M1: Continuing the argument with an attempt to integrate " ( )"y y their f
1n ny y
M1: Integrating 2 " ( )"y y their f - [“their E(Y)”]2 1n ny y
A1*: Complete correct solution no errors.
(c) B1: 5 only
B1: Explain their reason by either an accurate sketch or ( )
0df
d
y
y therefore an
increasing function oe
(d) B1ft: Explaining the reason for their answer. Follow through their part(b) or mean
from(d) and mode from(c). A correct sketch of “their f(y)” – may be seen anywhere
in question or ft their mean and mode plus a correct conclusion
NB: Watch for gaming. A student who writes both negative skew with a reason and
positive skew with a reason. Please send these to your Team Leader
(e) M1: Attempting to translate the problem into an equation using 2k and k. Allow if
the brackets are missing e.g.3 32
125 125
k k . No need for the 0.189
A1: A correct equation in any form
A1: a correct answer only.
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Question Scheme Marks AOs
Q2(a) 0 1H : 0,H : 0 B1 2.5
Critical value at 1% level is 0.8929 B1 1.1b
sr < 0.8929 so not significant evidence to reject 0H , M1 2.1
The researcher's claim is not correct (at 1% level).
or insufficient evidence for researcher’s claim
or there is insufficient evidence that water gets deeper further from
inner bank.
or no (positive) correlation between depth of water and distance from
inner bank
A1ft
2.2b
(4)
(b)(i) The ranks will remain the same therefore there will be no change to
the spearman’s rank correlation coefficient
B1 2.4
(ii) Spearman’s rank correlation coefficient will increase since B1 2.2a
The ranks are the same for both distance and depth therefore
d = 0 however, n has increased or the new position follows the
pattern that large b is assosciated with large s and so rs will increase
B1 2.4
(3)
(c) The mean of the tied ranks is given to each… B1 2.4
… then use PMCC B1 2.4
(2)
(9 marks)
Notes
(a) B1: Both hypotheses correct written using the notation
B1: awrt 0.893
M1: Drawing a correct inference using their answer to part(a) and their CV
A1ft: Drawing a correct inference in context using their answer to part(a) and their
CV
(b)(i) B1: Stating no change and an explanation including ranks remain unchanged oe
and no change oe
(b)(ii) B1: Interpreted the outcome of adding a point as increased oe
B1: Explaining why. Need to mention the ranks are the same for both oe and n has
increased oe
(c) B1: Explaining that The mean of the values for the tied ranks is given to both values
B1: Explaining that the PMCC must be used
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Question Scheme Marks AOs
Q3(a) 95% CI for uses t value of 2.064 B1 3.3
ˆ 1
"2.064" 2.232 1.128225
or
ˆ1
2.232 1.128 "2.064" 2.2322 25
(oe)
M1 2.1
2.76
ˆ"2.064"
or 1.3372… M1 1.1b
2ˆ 1.788... [=1.79 (3sf)] * A1*cso 1.1b
(4)
(b) 12.401, <
2
79.124
<, 39.364
B1
M1
1.1b
1.1a
1.09 < 2 < 3.46 A1 1.1b
(3)
(7 marks)
Notes
(a) B1: Realising that the t-distribution must be used as a model and finding the correct
value awrt 2.06
M1: Using the correct formula with a t-value, ˆ 1
" value" 2.232 1.128225
t
or ˆ1
2.232 1.128 " value" 2.2322 25
t
or ˆ1
2.232 1.128 " value" 1.1282 25
t
M1: Rearranging one of these formula accurately to find a value of
A1cso*: A correct solution only using awrt 1.79
(b) B1: awrt 12.4 or 39.4 May be implied by a correct confidence interval
M1: 2
79.124
May be implied by a correct confidence interval
A1: awrt 1.09 and awrt 3.46
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Question Scheme Marks AOs
Q4(a) H0: 2
G = 2
B , H1: 2
G 2
B , B1 2.5
2
Bs = 61 (56130 – 7 88.92) =
6
53.807= 134.6
M1
A1
2.1
1.1b
2
Gs = 71 (55746 – 8 83.12) =
7
12.501= 71.58 A1 1.1b
2
2
G
B
s
s= 1.880… M1 3.4
critical value F6, 7 = 3.87 B1 1.1b
not significant, variances can be treated as the same A1 ft 2.2b
(7)
(b) H0: B = G , H1: B > G B1 2.5
pooled estimate of variance s2 = 13
58.7176.1346 = 100.6653… M1 3.1b
test statistic t =
81
71
1.839.88
s = awrt 1.12
M1
A1
1.1b
1.1b
critical value t13(5%) = 1.771 B1 1.1b
Insufficient evidence to support mother’s claim A1 ft 2.2b
(6)
(13 marks)
Notes
4(a) B1: Both hypotheses correct using the notation2 . Allow rather than
2 .
M1: Using a correct Method for either 2
Bs or 2
Gs May be implied by a correct value
A1: awrt 135
A1: awrt 71.6
M1: Using the F-distribution as the model eg2
2
G
B
s
s
B1: awrt 3.87
A1ft: Drawing a correct inference following through their CV and value for 2
2
G
B
s
s
(b) B1: Both hypotheses correct using the notation .
M1: For realising the need to find the pooled estimate for the test require from a
correct interpretation of the question.
M1: Correct method for test statistic t = 1 17 8
88.9 83.1
""their s
May be implied by a
correct awrt 1.12
A1: awrt 1.12
B1: awrt 1.77
A1ft Drawing a correct inference following through their CV and test statistic
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Question Scheme Marks AOs
Q5(a) Let X = L – 4S then E(X) = 19.6 – 4 4.8 M1 2.3
= 0.4 A1 1.1b
Var(X) = 2 2 2Var( ) 4 Var( ) 0.6 16 0.3L S M1 2.1
= 1.8 A1 1.1b
P(X > 0) = [P(Z > 0 0.4
0.298...1.8
…)] M1 2.1
= 0.617202.. awrt 0.617 A1 1.1b
(6)
(b) 1 2 3 4T S S S S (May be implied by 0.36) M1 3.3
T ~ N(19.2, 0.36) E(T) = 19.2 B1 1.1b
Var(T) = 0.36 or 20.6 A1 1.1b
(3)
(c) Let Y = L – T E(Y ) = E(L) – E(T) = [ 0.4] M1 3.3
Var(Y) = Var(L) + Var(T) = [ 0.72] M1 1.1b
Require P( – 0.2 < Y < 0.2) M1 3.1a
= 0.16708… awrt 0.167 A1 1.1b
(4)
(13 marks)
Notes
(a) M1: Selecting and using an appropriate model i.e – 4 L S . May be implied by
0.4
A1: 0.4 oe
M1: For realising the need to use Var(L) + 42Var(S). Allow use of 0.6 for Var(L)
instead of 20.6 and/or 0.3 for Var(S) instead of
20.3 may be implied by 1.8
A1: 1.8 only
M1: For realising P(X > 0) is required and an attempt to find it e.g.
0 0.4
"their Var( )"X
but do not allow a negative Var(X)
A1: awrt 0.617
(b) M1: Selecting and using an appropriate model ie 1 2 3 4S S S S : may be implied
by 0.36
B1: 19.2 only
A1: 0.36
(c) M1: Setting up and using the model Y = L – T. May be implied by
–E Y E L E T
M1: Using Var(Y) = Var(L) + Var(T)
M1: Dealing with the modulus and realising they need to find P( – 0.2 < Y < 0.2)
A1: awrt 0.167
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Question Scheme Marks AOs
Q6(a) – 0.02775
S76
Sxm
xx
b
M1 3.3
[a = m bx = 1.278 + 0.0277576× 8.5 = 1.5139]
m = 1.5139 – 0.02775…x A1 1.1b
(2)
(b) RSS =
2
2.290.12756
82.5
M1 1.1b
= 0.06399* A1* 1.1b
(2)
(c)
x m m = a + bx
4 1.50 1.4029 +0.0971
5 1.20 1.3752 – 0.1752
6 1.40 1.3474 +0.0526
7 1.40 1.3196 +0.0804
8 1.23 1.2919 – 0.0619
9 1.30 1.2641 +0.0359
10 1.20 1.2364 – 0.0364
11 1.15 1.2086 – 0.0586
12 1.25 1.1808 +0.0692
13 1.15 1.1531 – 0.0031
M1
A1
3.4
1.1b
(2)
(d) The point (5, 1.2) is an outlier B1ft 2.2b
(1)
(e)(i)
It is a valid piece of data so should be used
or It does not follow the pattern according to the residuals so may
contain an error making the result invalid so should be removed
B1 2.4
(ii)
a m bx =1.28667 +0.03765 × 8.88889 = 1.6213 M1 3.3
m =1.6213 – 0.03765x A1 1.1b
(iii)
m =1.6213 – 0.03765 15
= 1.056 or awrt 1.06 B1ft 3.4
(iv) The model is only reliable if the values are limited to those in the
given range so probably not reliable B1 3.5b
(5)
(12 marks)
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Notes
6(a) M1: Realsing the need to use b =
S
S
xm
xx
and a = m bx
A1: m = awrt 1.51) – (awrt 0.0278) x. Award M1A1 for correct equation
(b) M1: Using
2
xm
mm
xx
SS
S
A1*: awrt 0.064
(c) M1: Using the model in part (a) i.e. m – (“1.5139” – “0.02775”x) implied by a
correct value
A1: All correct.
Award M1A1 for a list of correct residuals
(d) B1: Inferring from the residuals that the outlier is (5, 1.2) ft their residuals.
(e)(i) B1: Explaining why the outlier should be removed or not.
(ii) M1: Removing the outlier and refining the model by finding a new regression line.
A1: ( 1.62) ( 0.0377)awrt awrt m x
(iii) B1ft: using their model in e(i) with x = 15. awrt 1.06 or ft their e(ii)
(iv) B1: Realising the limitations of the model by stating it is not reliable and giving the
reason why ie extrapolation/out of range o.e.
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Question Scheme Marks AOs
Q7(a)
Sxx =
2
2 1010
10
ss
M1 2.1
2658.9 =
2
2 100100
10
ss
M1 1.1b
2658.9 = 100 Sss
Sss = 26.589 * A1*cso 1.1b
(3)
(b) 10
1
64 10( 9) id M1 3.1a
10
1
64 10 900id
10
1
96.4id A1 1.1b
Sdd =
2"96.4"
1081.74 -10
M1 1.1b
= 152.444
r = 0.935 A1ft 1.1b
(4)
(c) Linear correlation is significant but scatter diagram suggests a
non-linear relationship between the level of serum magnesium,
and the level of the disease protein
B1 3.5a
(1)
(8 marks)
Notes
(a)
M1: Attempting to use
2
2
10xx
xS x
with x = 10s
M1: Substituting in 2658.9 and dealing with the 10 correctly
A1*: cso A complete solution with no errors leading to 26.589 only
(b) M1: Realising that either
10
1
64 10( 9) id or 10
1
64 10 900id o.e. must be used.
May be implied by seeing 96.4
A1: 96.4 only
M1: Attempting to use
2
2
10dd
dS d
may be implied by 0.935
A1ft: awrt 0.935 ft “their 96.4”
(c) B1: A correct comment comparing their value of r and the scatter diagram in context
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Pearson Edexcel Level 3 GCE
Further Mathematics Advanced Paper 3: Further Mathematics Option 1 Option 3C: Further Mechanics 1
Paper 4: Further Mathematics Option 2 Option 4C: Further Mechanics 1
Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes
Paper Reference(s)
9FM0/3C 9FM0/4C
You must have: Mathematical Formulae and Statistical Tables, calculator
Candidates may use any calculator permitted by Pearson regulations. Calculators must not have the facility for algebraic manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them.
Instructions
Use black ink or ball-point pen.
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
Fill in the boxes at the top of this page with your name, centre number and candidate number.
Answer all questions and ensure that your answers to parts of questions are clearly labelled.
Answer the questions in the spaces provided – there may be more space than you need.
You should show sufficient working to make your methods clear. Answers without working may not gain full credit.
Answers should be given to three significant figures unless otherwise stated. Information
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
There are 8 questions in this question paper. The total mark for this paper is 75.
The marks for each question are shown in brackets – use this as a guide as to how much time to spend on each question.
Advice
Read each question carefully before you start to answer it.
Try to answer every question.
Check your answers if you have time at the end.
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Answer ALL questions. Write your answers in the spaces provided.
Unless otherwise indicated, whenever a numerical value of g is required, take g = 9.8 m s-2 and
give your answer to either 2 significant figures or 3 significant figures.
1. A particle P of mass 0.5 kg is moving with velocity 4 i j m s-1 when it receives
an impulse 2 i j N s.
Show that the kinetic energy gained by P as a result of the impulse is 12 J.
(6)
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Question 1 continued
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(Total for Question 1 is 6 marks)
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2. A parcel of mass 5 kg is projected with speed 8 m s-1 up a line of greatest slope of a
fixed rough inclined ramp.
The ramp is inclined at angle to the horizontal, where 1
sin7
The parcel is projected from the point A on the ramp and comes to instantaneous
rest at the point B on the ramp, where 14AB m.
The coefficient of friction between the parcel and the ramp is .
In a model of the parcel's motion, the parcel is treated as a particle.
(a) Use the work-energy principle to find the value of .
(5)
(b) Suggest one way in which the model could be refined to make it more realistic.
(1)
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Question 2 continued
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(Total for Question 2 is 6 marks)
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3. A particle of mass m kg lies on a smooth horizontal surface.
Initially the particle is at rest at a point O between two fixed parallel vertical walls.
The point O is equidistant from the two walls and the walls are 4 m apart.
At time t = 0 the particle is projected from O with speed u m s-1 in a direction
perpendicular to the walls.
The coefficient of restitution between the particle and each wall is 3
4
The magnitude of the impulse on the particle due to the first impact with a wall is mu N s.
(a) Find the value of .
(3)
The particle returns to O, having bounced off each wall once, at time 7t seconds.
(b) Find the value of u.
(5)
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Question 3 continued
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(Total for Question 3 is 8 marks)
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4.
Figure 1
Figure 1 represents the plan view of part of a horizontal floor, where AB and BC
are perpendicular vertical walls.
The floor and the walls are modelled as smooth.
A ball is projected along the floor towards AB with speed u m s-1 on a path at an
angle of 60 to AB. The ball hits AB and then hits BC.
The ball is modelled as a particle.
The coefficient of restitution between the ball and wall AB is 1
3
The coefficient of restitution between the ball and wall BC is 2
5
(a) Show that, using this model, the final kinetic energy of the ball is 35% of the
initial kinetic energy of the ball.
(8)
(b) In reality the floor and the walls may not be smooth. What effect will the model
have had on the calculation of the percentage of kinetic energy remaining?
(1)
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60°
C
BA
u
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Question 4 continued
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Question 4 continued
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Question 4 continued
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(Total for Question 4 is 9 marks)
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5. A car of mass 600 kg is moving along a straight horizontal road.
At the instant when the speed of the car is v m s-1, the resistance to the motion of the car is
modelled as a force of magnitude 200 2v N.
The engine of the car is working at a constant rate of 12 kW.
(a) Find the acceleration of the car at the instant when 20v
(4)
Later on the car is moving up a straight road inclined at an angle to the horizontal, where
1sin
14
At the instant when the speed of the car is v m s-1, the resistance to the motion of the car from
non-gravitational forces is modelled as a force of magnitude 200 2v N.
The engine is again working at a constant rate of 12 kW.
At the instant when the car has speed w m s-1, the car is decelerating at 0.05 m s-2.
(b) Find the value of w.
(5)
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Question 5 continued
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(Total for Question 5 is 9 marks)
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6. [In this question i and j are perpendicular unit vectors in a horizontal plane.]
A smooth uniform sphere A has mass 2m kg and another smooth uniform sphere B,
with the same radius as A, has mass 3m kg.
The spheres are moving on a smooth horizontal plane when they collide obliquely.
Immediately before the collision the velocity of A is 3 3i j m s-1 and the velocity
of B is 5 2 i j m s-1.
At the instant of collision, the line joining the centres of the spheres is parallel to i.
The coefficient of restitution between the spheres is 1
4
(a) Find the velocity of B immediately after the collision.
(7)
(b) Find, to the nearest degree, the size of the angle through which the direction of
motion of B is deflected as a result of the collision.
(2)
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Question 6 continued
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Question 6 continued
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(Total for Question 6 is 9 marks)
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7. A particle P of mass m is attached to one end of a light elastic string of natural length a and
modulus of elasticity 3mg.
The other end of the string is attached to a fixed point O on a ceiling.
The particle hangs freely in equilibrium at a distance d vertically below O.
(a) Show that 4
3d a .
(3)
The point A is vertically below O such that 2OA a .
The particle is held at rest at A, then released and first comes to instantaneous rest at the point B.
(b) Find, in terms of g, the acceleration of P immediately after it is released from rest.
(3)
(c) Find, in terms of g and a, the maximum speed attained by P as it moves from A to B.
(5)
(d) Find, in terms of a, the distance OB.
(3)
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Question 7 continued
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Question 7 continued
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Question 7 continued
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(Total for Question 7 is 14 marks)
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8. A particles P of mass 2m and a particle Q of mass 5m are moving along the same
straight line on a smooth horizontal plane.
They are moving in opposite directions towards each other and collide directly.
Immediately before the collision the speed of P is 2u and the speed of Q is u.
The direction of motion of Q is reversed by the collision.
The coefficient of restitution between P and Q is e.
(a) Find the range of possible values of e.
(8)
Given that 1
3e
(b) show that the kinetic energy lost in the collision is 240
7
mu.
(5)
(c) Without doing any further calculation, state how the amount of kinetic energy
lost in the collision would change if 1
3e
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Question 8 continued
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Question 8 continued
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Question 8 continued
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(Total for Question 8 is 14 marks)
TOTAL FOR PAPER IS 75 MARKS
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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AL Further Mechanics 1 Mark Scheme
Question Scheme Marks AOs
1
Use Impulse-momentum principle M1 2.1
2 0.5 0.5 4 i j v i j A1 1.1b
1 1
4 , 82 2
v i j v i j (m s-1) A1 1.1b
Use of KE = 2 21 1
2 2m mv u M1 2.1
1
0.5 64 1 16 12
A1 1.1b
1
48 124
(J) * A1* 1.1b
(6)
(6 marks)
Notes:
M1: Difference of terms & dimensionally correct
A1: Correct unsimplified equation
A1: C.A.O.
M1: Must be a difference of two terms.
Must be dimensionally correct.
A1: Correct unsimplified equation
A1*: Complete justification of given answer
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
2(a)
4 3
5 cos 5 48.497...7
R g g
M1 3.4
Force due to friction = 5 cosg M1 3.4
Work-Energy equation M1 3.4
1
5 64 5 9.8 14sin 142
R A1 1.1b
0.0913 or 0.091 A1 1.1b
(5)
(b) Appropriate refinement B1 3.5c
(1)
(6 marks)
Notes:
(a)
M1: Condone sin/cos confusion
M1: Use of their R
M1: Must be using work-energy. Requires all terms.
Condone sin/cos confusion, sign errors and their R
A1: Correct in and R .
A1: Accept 0.0913 or 0.091
(b)
B1: e.g.
- Do not model the parcel as a particle and therefore take air resistance into account.
- Take into account the dimensions/uniformity of the parcel.
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
3(a) Use NEL to find the speed of particle after the first impact
3
4eu u
B1 3.4
Impulse = 3
4mu mv mu mu mu
M1 3.1b
7
4 A1 1.1b
(3)
(b) Use NEL to find the speed of the particle after the second impact
3 3 9
4 4 16u u
B1 3.4
Use of s vt to find total time M1 3.1b
2 4 2 2 16 32
73 9 3 9
4 16
u u u uu u
A1 1.1b
Solve for u: 63 18 48 32u M1 1.1b
98 14
1.563 9
u A1 1.1b
(5)
(8 marks)
Notes:
(a)
B1: Using Newton's experimental law as a model to find the
speed after the first impact
M1: Must be a difference of two terms, taking account of the
change in direction of motion.
A1: cao
(b)
B1: Using NEL as a model to find the speed after the second
impact.
M1: Needs to be used for at least one stage of the journey
A1: or equivalent
M1: Solve their linear equation for u
A1: Accept 1.56 or better
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
4(a) Complete strategy to find the kinetic energy after the second impact M1 3.1b
Parallel to AB after collision: cos60u M1 3.1b
Perpendicular to AB after collision: 1
sin 603
u M1 3.4
Components of velocity after first impact: , 2 2
u u A1 1.1b
Parallel to BC after collision: 2
u
1sin 60
3u
M1 3.1b
Perpendicular to BC after collision: 2 1
5 2 10
uu
2
cos605
u
M1 3.4
Components of velocity after second impact: , 2 10
u u A1 1.1b
Final KE =
2 2 21 7
2 4 10 2 20
u u mum
Fraction of initial KE
2
2
7
2 20
2
mu
mu
7
35%20
* A1* 2.2a
(8)
(b) The answer is too large - rough surface means resistance so final
speed will be lower. B1 3.5a
(1)
(9 marks)
Notes:
(a)
M1: Use of CLM parallel to the wall. Condone sin/cos confusion
M1: Use NEL as a model to find the speed perpendicular to the wall. Condone sin/cos confusion
A1: Both components correct with trig substituted (seen or implied)
M1: Use of CLM parallel to the wall. Condone sin/cos confusion
M1: Use NEL as a model to find the speed perpendicular to the wall. Condone sin/cos confusion
A1: Both components correct with trig substituted (seen or implied)
M1: Correct expression for total KE using their components after 2nd collision.
A1*: Obtain given answer with sufficient working to justify it
(b)
Clear explanation of how the modelling assumption has affected the outcome
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
5(a)
Use of P Fv :
12000
20F B1 3.3
Equation of motion: 200 2 600F v a M1 3.4
600 240 600a A1ft 1.1b
360 600a , 0.6a (m s-2) A1 1.1b
(4)
(b) Equation of motion M1 3.3
12000
200 2 600 sin 600 0.05w gw
A1
A1
1.1b
1.1b
3 term quadratic and solve: 22 590 12000 0w w M1 1.1b
2590 590 96000
19.14
w
(m s-1) A1 1.1b
(5)
(9 marks)
Notes:
(a)
B1: 600 or equivalent
M1: Use the model to form the equation of motion.
Must include all terms.Condone sign errors.
A1ft: Correct for their F
A1: cao.
(b)
M1: Use the model to form the equation of motion.
All terms needed.
Condone sign errors and sin/cos confusion.
A1: All correct A1A1
One error A1A0
M1: Dependent on the preceding M1. Use the equation of motion to form a 3-term quadratic in w only
A1: Accept 19. Do not accept more than 3 s.f.
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Question Scheme Marks AOs
6(a)
Overall strategy to find Av M1 3.1a
Velocity of A perpendicular to loc after collision 3 j (m s-1) B1 3.4
CLM parallel to loc. M1 3.1a
2 3 3 5 3 2m m mw mv 9 3 2w v A1 1.1b
Correct use of impact law M1 3.1a
1
3 54
v w 2 A1 1.1b
Solve for w 3 2 9
2 2 4
w v
v w
2B v i j (m s-1), A1ft 1.1b
(7)
(b) ( 5 2 2cos
29 5
i j).( i j) M1 3.1a
41.63... 42 (nearest degree) A1 1.1b
Alternative method: 1 1 2tan 2 tan 41.63... 42
5
(nearest
degree)
(2)
(9 marks)
Notes:
(a)
M1: Correct overall strategy to form sufficient equations and solve for Av
B1: Use the model to find the component of Av perpendicular to the line of centres.
M1: Use CLM to form equation in v and w. Need all 4 terms, dimensionally correct
A1: Correct unsimplified
M1: Must be used the right way round
A1: Correct unsimplified
A1ft: vB correct. Follow their 2j
(b)
M1: Complete method for finding the required angle. Follow their vB
A1: cao
wivi
3j 2j
-5i+2j3i+3j
B(3m)A(2m)
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Question Scheme Marks AOs
7(a)
In equilibrium no resultant vertical force. M1 2.1
3mgx
mga
A1 1.1b
3
ax ,
4
3d a * A1* 2.2a
(3)
(b) Equation of motion. M1 3.1a
3mga
mg mxa
A1 1.1b
2x g A1 1.1b
(3)
(c) Max speed at equilibrium position B1 3.1a
Work energy & use of 2
EPE2
x
a
. M1 3.1a
2
22
33 1 23
2 2 2 3
amg
mga amv mg
a a
A1
A1
1.1b
1.1b
21 3 1 2 2
2 2 6 3 3v ga ga
,
4
3
gav A1 1.1b
(5)
(d) At max ht. KE = 0. EPE lost = GPE gained M1 3.1a
23
2
mgamgh
a A1 1.1b
2
aOB A1 1.1b
(3)
(14 marks)
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Question 7 continued
Notes:
(a)
M1: Use x
Ta
to form equation for equilibrium
A1: Correct unsimplified equation
A1*: Requires sufficient working to justify given answer
plus a 'statement' that the required result has been achieved.
(b)
M1: Use x
Ta
to form equation of motion.
Need all 3 terms. Condone sign errors
A1: Correct unsimplified equation
A1: cao
(c)
B1: Seen or implied
M1: Form work-energy equation. All 4 terms needed.
Condone sign errors
A1: Correct unsimplified equation A1A1
One error in the equation A1A0
A1: cao
(d)
M1: Form energy equation
A1: Correct unsimplified equation
A1: cao
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Question Scheme Marks AOs
8(a)
Complete overall strategy to find v M1 3.1a
Use of CLM M1 3.1a
2 2 5 5 2m u m u m v m w ,
5 2u v w A1 1.1b
Use of Impact law M1 3.1a
2v w e u u A1 1.1b
Solve for v: 5 2
6 2 2
u v w
eu v w
7 6 1v u e 6 17
uv e
A1 1.1b
Direction of Q reversed: 0v M1 3.4
1
16
e A1 1.1b
(8)
(b) 1
3e
7
uv ,
6
7
uw B1 2.1
Equation for KE lost M1 2.1
2 22 21 36 1
2 4 52 49 2 49
u um u m u
A1
A1
1.1b
1.1b
2
21 72 5 408 5
2 49 49 7
mumu
* A1* 2.2a
(5)
(c) Increase e more elastic less energy lost B1 2.2a
(1)
(14 marks)
w v
u2u
Q
5m
P
2m
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Question 8 continued
Notes:
(a)
M1: Complete strategy to form sufficient equations in v and w and solve for v.
M1: Use CLM to form equation in v and w.
Needs all 4 terms & dimensionally correct
A1: Correct unsimplified equation
M1: Use NEL as a model to form a second equation in v and w. Must be used the right way round
A1: Correct unsimplified equation
A1: for v or 7v correct
M1: Use the model to form a correct inequality for their v
A1: Both limits required
(b)
B1: or equivalent statements
M1: terms of correct structure combined correctly
A1: Fully correct unsimplified A1A1
One error on unsimplified expression A1A0
A1*: cso. plus a 'statement' that the required result has been achieved
(c)
B1: "less energy lost" or equivalent
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Pearson Edexcel Level 3 GCE
Further Mathematics Advanced
Further Mathematics Option 2 Paper 4: Further Mechanics 2
Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes
Paper Reference(s)
9FM0/4F
You must have: Mathematical Formulae and Statistical Tables, calculator
Candidates may use any calculator permitted by Pearson regulations. Calculators must not have the facility for algebraic manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them.
Instructions
Use black ink or ball-point pen.
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
Fill in the boxes at the top of this page with your name, centre number and candidate number.
Answer all questions and ensure that your answers to parts of questions are clearly labelled.
Answer the questions in the spaces provided – there may be more space than you need.
You should show sufficient working to make your methods clear. Answers without working may not gain full credit.
Answers should be given to three significant figures unless otherwise stated. Information
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
There are 7 questions in this question paper. The total mark for this paper is 75.
The marks for each question are shown in brackets – use this as a guide as to how much time to spend on each question.
Advice
Read each question carefully before you start to answer it.
Try to answer every question.
Check your answers if you have time at the end
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Answer ALL questions. Write your answers in the spaces provided.
Unless otherwise indicated, whenever a numerical value of g is required, take 2
9.8 m sg and give your answer to either 2 significant figures or 3 significant figures.
1. A flag pole is 15 m long.
The flag pole is non-uniform so that, at a distance x metres from its base, the mass per
unit length of the flag pole, m kg m-1 is given by the formula 10 125
xm
.
The flag pole is modelled as a rod.
(a) Show that the mass of the flag pole is 105 kg.
(3)
(b) Find the distance of the centre of mass of the flag pole from its base.
(4)
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Question 1 continued
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(Total for Question 1 is 7 marks)
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2.
A hollow right circular cone, of base diameter 4a and height 4a is fixed with its axis
vertical and vertex V downwards, as shown in Figure 1.
A particle of mass m moves in a horizontal circle with centre C on the rough inner
surface of the cone with constant angular speed .
The height of C above V is 3a.
The coefficient of friction between the particle and the inner surface of the cone is 1
4
Find, in terms of a and g, the greatest possible value of .
(8)
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Question 2 continued
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(Total for Question 2 is 8 marks)
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3.
A uniform solid cylinder has radius 2a and height h h a .
A solid hemisphere of radius a is removed from the cylinder to form the vessel V.
The plane face of the hemisphere coincides with the upper plane face of the cylinder.
The centre O of the hemisphere is also the centre of the upper plane face of the cylinder,
as shown in Figure 2.
(a) Show that the centre of mass of V is
2 23 8
8 6
h a
h a
from O.
(5)
The vessel V is placed on a rough plane which is inclined at an angle to the horizontal.
The lower plane circular face of V is in contact with the inclined plane.
Given that 5h a , the plane is sufficiently rough to prevent V from slipping and V is on
the point of toppling,
(b) find, to three significant figures, the size of the angle .
(4)
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Question 3 continued
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Question 3 continued
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(Total for Question 3 is 9 marks)
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4. A car of mass 500 kg moves along a straight horizontal road.
The engine of the car produces a constant driving force of 1800 N.
The car accelerates from rest from the fixed point O at time 0t and at time t seconds
the car is x metres from O, moving with speed v m s-1.
When the speed of the car is v m s-1, the resistance to the motion of the car has
magnitude 22v N.
At time T seconds, the car is at the point A, moving with speed 10 m s-1.
(a) Show that 25
ln 26
T
(6)
(b) Show that the distance from O to A is 9
125ln8
m.
(5)
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Question 4 continued
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(Total for Question 4 is 11 marks)
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5.
Figure 3
A shop sign is modelled as a uniform rectangular lamina ABCD with a semicircular
lamina removed.
The semicircle has radius a, 4BC a and 2 .CD a
The centre of the semicircle is at the point E on AD such that AE d , as shown in
Figure 3.
(a) Show that the centre of mass of the sign is
44
3 16
a
from AD.
(4)
The sign is suspended using vertical ropes attached to the sign at A and at B and
hangs in equilibrium with AB horizontal.
The weight of the sign is W and the ropes are modelled as light inextensible strings.
(b) Find, in terms of W and , the tension in the rope attached at B.
(2)
The rope attached at B breaks and the sign hangs freely in equilibrium suspended
from A, with AD at an angle to the downward vertical.
Given that11
tan18
(c) find d in terms of a and .
(6)
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Question 5 continued
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Question 5 continued
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Question 5 continued
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(Total for Question 5 is 12 marks)
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6. A small bead B of mass m is threaded on a circular hoop.
The hoop has centre O and radius a and is fixed in a vertical plane.
The bead is projected with speed 7
2ga from the lowest point of the hoop.
The hoop is modelled as being smooth.
When the angle between OB and the downward vertical is , the speed of B is v.
(a) Show that 2 3
2cos2
v ga
(3)
(b) Find the size of at the instant when the contact force between B and the hoop is
first zero.
(5)
(c) Give a reason why your answer to part (b) is not likely to be the actual value of .
(1)
(d) Find the magnitude and direction of the acceleration of B at the instant when B is
first at instantaneous rest.
(5)
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Question 6 continued
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Question 6 continued
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Question 6 continued
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(Total for Question 6 is 14marks)
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7. Two points A and B are 6 m apart on a smooth horizontal surface.
A light elastic string of natural length 2 m and modulus of elasticity 20 N, has one end
attached to the point A.
A second light elastic string of natural length 2 m and modulus of elasticity 50 N, has
one end attached to the point B.
A particle P of mass 3.5 kg is attached to the free end of each string.
The particle P is held at the point on AB which is 2 m from B and then released from
rest.
In the subsequent motion both strings remain taut.
(a) Show that P moves with simple harmonic motion about its equilibrium position.
(7)
(b) Find the maximum speed of P.
(2)
(c) Find the length of time within each oscillation for which P is closer to A than to B.
(5)
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Question 7 continued
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Question 7 continued
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Question 7 continued
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(Total for Question 7 is 14 marks)
TOTAL FOR PAPER IS 75 MARKS
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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AL Further Mechanics 2
Question Scheme Marks AOs
1(a)
Total mass = 15
010 1 d
25
xx
M1 2.1
15
2
0
105
xx
A1 1.1b
225
150 1055
(kg) * A1* 1.1b
(3)
(b) Taking moments about the base :
15
010 1 d
25
xx x
M1 3.4
15
2 3
0
25 675
15x x
A1 1.1b
105 675d M1 3.4
6.43d (m) 3
67
(m) A1 1.1b
(4)
(7 marks)
Notes:
(a)
M1: Use integration (usual rules)
A1: Correct integration
A1*: Use limits and show sufficient working to justify given answer.
(b)
M1: Use the model to find the moment about the base (usual rules for integration)
A1: Correct integration
M1: Use the model to complete the moments equation.
Require 105 and their 675 used correctly
A1: 6.43 or better
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
2
Complete overall strategy M1 3.1b
Resolve vertically M1 3.3
cos sinmg F R A1 1.1b
Horizontal equation of motion M1 3.3
2 cos sinmr R F A1 1.1b
Use of limiting friction since maximum M1 3.3
Substitute for trig ratios:
23 9
2 2
a
g
M1 1.1b
Maximum 3g
a A1 1.1b
(8 marks)
Notes:
M1: Overall strategy to form equation in only e.g.
consider vertical and horizontal motrion and limiting friction
M1: needs all 3 terms. Condone sign errors and sin/cos confusion
A1: correct unsimplified equation
M1: needs all 3 terms. Condone sign errors and sin/cos confusion
A1: correct unsimplified equation
M1: seen or implied
M1: substitute to achieve equation in , and a g only
A1: or equivalent exact form
F
mg
R
θ
C
V
3a
4a
4a
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Question Scheme Marks AOs
3(a)
mass c of m from O
cylinder 24 a h
2
h
hemisphere 32
3a
3
8a
V 2 324
3a h a d
Mass ratios B1 1.2
Correct distances B1 1.2
Moments about a diameter through O. M1 2.1
2 3 22 3 14 2 2
2 3 8 3
ha h a a a h a d
A1 1.1b
22
2 23 88
8 62
3
ah h a
da h a
h
* A1* 2.2a
(5)
(b)
5 2.573...h a d a B1 1.1b
About to topple so c of m above tipping point M1 2.2a
2
tan5 2.573
a
a a
A1ft 1.1b
39.5 or 0.689 rads. A1 1.1b
(4)
(9 marks)
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Notes:
(a)
B1: correct mass ratios
B1: correct distances
M1: All three terms & dimensionally correct.
Could use a parallel axis but final answer must be for the distance from O.
A1: Correct unsimplified equation
A1*: Deduce the given answer. Their working must make it clear how they reached
their answer.
(b)
B1: Distance of com from base
M1: Condone tan the wrong way up.
A1ft: Correct unsimplified expression for trig ratio for following their d
A1: 39.5 or 0.689 rads.
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Question Scheme Marks AOs
4(a)
Equation of motion 21800 2 500v a (when seen) B1 2.1
Select form for a: d
500d
v
t M1 2.5
2
2 1 1 1 1d d d
500 900 60 30 30t v v
v v v
M1 2.1
1 1
ln 30 ln 30 250 60 60
tv v C A1 1.1b
25 30 10 25
ln ln 26 30 10 6
T
* M1
A1*
2.1
2.2a
(6)
(b) Equation of motion: 2d500 1800 2
d
vv v
x M1 2.5
2
500d 1d
1800 2
vv x
v
M1 2.1
2125ln 1800 2 v x C A1 1.1b
Use boundary conditions: 125ln1600 125ln1800x M1 2.1
9
125ln8
x (m) * A1* 2.2a
(5)
(11 marks)
Notes:
(a)
B1: all three terms & dimensionally correct
M1: use of correct form for acceleration to give equation in v, t only
M1: Separate variables and integrate
A1: Condone missing C
M1: Use boundary conditions correctly
A1*: Show sufficient working to justify given answer and a 'statement' that the required
result has been achieved
(b)
M1: Correct form of acceleration in the equation of motion to give equation in v, x only
M1: Separate variables and integrate.
A1: Condone missing C
M1: Extract and use boundary conditions
A1*: Show sufficient working to justify given answer and a 'statement' that the required result
has been achieved
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
5(a)
Mass From AD
Rectangle 28a a
Semicircle 21
2a
4
3
a
Sign 2
82
a
h
Mass ratios B1 1.2
Moments about AD M1 2.1
2 2 2 3 3 31 4 2 228 8 8
2 2 3 3 3
aa h a a a a a a
A1 1.1b
22 448
3 2 3 16
ah a
* A1* 2.2a
(4)
(b) Moments about A
442
3 16
aaT W
M1 3.1b
22
2 3 16
hW WT
a
A1 1.1b
(2)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
(c)
Take moments about AB to find distance of com from AB M1 3.1b
2 2 21 1
8 2 82 2
a a a d a v
A1 1.1b
32
16
a dv
A1 1.1b
Correct trig for the given angle M1 3.1b
11 44tan
18 3 32
h a
v a d
A1ft 1.1b
( 24 32a a d , 8a d ) 8a
d
A1 1.1b
(6)
(12 marks)
D
C
B
A
α
h
v
Notes:
(a)
B1: correct mass ratios
M1: need all three terms, must be dimensionally correct
A1: Correct unsimplified equation
A1*: Show sufficient working to justify the given answer and a 'statement' that the required result
has been achieved.
(b)
M1: Could also take moments about B or about the c.o.m. and use A BT T W
A1: cso
(c)
M1: all terms and dimensionally correct
A1: Correct unsimplified equation
A1: or equivalent
M1: Condone tan the wrong way up.
A1: Equation in a and d; follow through on their v
A1: cao.
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
6(a)
Conservation of energy M1 2.1
21 1 71 cos
2 2 2mv mga m ga
A1 1.1b
2 3
2cos *2
v ga
A1* 2.2a
(3)
(b)
Resolve parallel to OB and use
2mv
a M1 3.1b
2
cosmv
R mga
A1 1.1b
Use 0R
2
cosv
ga
M1 3.1b
Solve for 3
cos 2cos2
g g
M1 1.1b
120 A1 1.1b
(5)
(c) Any appropriate comment
e.g. The hoop is unlikely to be smooth B1 3.5b
(1)
Rv
7
2ga
B
O
θ
mg
a
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Question Scheme Marks AOs
(d) At rest 0 v M1 3.1b
3
cos4
A1 1.1b
Acceleration is tangential M1 3.1b
Magnitude cos 90 6.48g m s-2 or 7
4g A1 1.1b
At 1 3
cos 90 48.64
to the downward
vertical
A1 1.1b
(5)
(14 marks)
Notes:
(a)
M1: All terms required. Must be dimensionally correct
A1: Correct unsimplified equation
A1*: Show sufficient working to justify the given answer and a 'statement' that the required
result has been achieved.
(b)
M1: Resolve parallel to OB
A1: correct equation
M1: Use 0R seen or implied
M1: Solve for
A1: Accept 2
3
(c)
B1: Any appropriate comment e.g.
- hoop may not be smooth;
- air resistance could affect the motion
(d)
M1: 0v seen or implied
A1: correct equation in
M1: correct direction for acceleration
A1: Accept 6.48, 6.5 or exact in g
A1: Accept 0.848 (radians)
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Question Scheme Marks AOs
7(a)
20
2A
eT ,
50 2
2B
eT
M1 3.1a
In equilibrium A BT T , 10 25 2e e M1 3.1a
10
35 50 , 7
e e A1 1.1b
Equation of motion for P when distance x from equilibrium
position towards B: M1 3.1a
50 2 20
3.5 2 2
B A
e x e xx T T
A1
A1
1.1b
1.1b
4 1050 20
7 7
2 2
x x
3.5 35 , 10x x x x and hence SHM about
the equilibrium position A1 3.2a
(7)
(b) Amplitude 10 4
27 7
B1 ft 2.2a
Use of max speed a M1 1.1b
4
10 1.817
(m s-1) A1 ft 1.1b
(3)
P
50 N20 N
6m
BA
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
7(c) Nearer to A than to B: 3
7x B1 3.1a
Solve for 10t : 3
cos 104
t ,
10 2.418.............t
M1 3.1a
Length of time: 2
2.418...10
M1 1.1b
0.457 (seconds) A1 1.1b
Alternative: 3.864 2.419
0.45710
Alternative:
4 3
sin 107 7
x t 10 0.8481 or 10 2.29353t t
1 20.2682, 0.72527t t
time 0.457 (seconds)
(4)
(14 marks)
Notes:
(a)
M1: Use of x
Ta
M1: Dependent on the preceding M1. Equate their tensions
A1: cao
M1: Condone sign error
A1: Correct unsimplified equation in e and x A1A1
Equation with one error A1A0
A1: Full working to justify conclusion that it is SHM about the
equilibrium position.
(b)
B1ft: Seen or implied. Follow their e
M1: correct method for max. speed
A1ft: 1.81 or better. Follow their ,a
(c)
B1: Seen or implied
M1: Use of cosx a wt
M1: Correct strategy for the required interval
A1: 0.457 or better
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Pearson Edexcel Level 3 GCE
Further Mathematics Advanced
Further Mathematics Option 1 Paper 3: Decision Mathematics 1
Further Mathematics Option 2 Paper 4: Decision Mathematics 1
Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes
Paper Reference(s)
9FM0/3D 9FM0/4D
You must have: Decision Mathematics Answer Book (enclosed), calculator
Instructions
Use black ink or ball-point pen.
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
Write your answers for this paper in the Decision Mathematics answer book provided.
Fill in the boxes at the top of the answer book with your name, centre number and candidate number.
Do not return the question paper with the answer book.
Answer all questions and ensure that your answers to parts of questions are clearly labelled.
Answer the questions in the spaces provided – there may be more space than you need.
You should show sufficient working to make your methods clear. Answers without working may not gain full credit.
Answers should be given to three significant figures unless otherwise stated. Information
There are 8 questions in this question paper. The total mark for this paper is 75.
The marks for each question are shown in brackets – use this as a guide as to how much time to spend on each question.
Advice
Read each question carefully before you start to answer it.
Try to answer every question.
Check your answers if you have time at the end.
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Answer ALL questions. Write your answers in the answer book provided.
1. A list of n numbers needs to be sorted into descending order starting at the left-hand end
of the list.
(a) Describe how to carry out the first pass of a bubble sort on the numbers in the list.
(2)
Bubble sort is a quadratic order algorithm.
A computer takes approximately 0.021 seconds to apply a bubble sort to a list of
2000 numbers.
(b) Estimate the time it would take the computer to apply a bubble sort to a list of 50 000
numbers. Make your method clear.
(2)
(Total for Question 1 is 4 marks)
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2.
Figure 1
(a) Define what is meant by a planar graph.
(2)
(b) Starting at A, find a Hamiltonian cycle for the graph in Figure 1.
(1)
Arc AG is added to Figure 1 to create the graph shown in Figure 2.
Figure 2
Taking ABCDEFGA as the Hamiltonian cycle,
(c) use the planarity algorithm to determine whether the graph shown in Figure 2 is
planar. You must make your working clear and justify your answer.
(4)
(Total for Question 2 is 7 marks)
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3.
(a) Explain clearly the difference between the classical travelling salesperson problem
and the practical travelling salesperson problem.
(2)
The table shows the least distances, in km, by road between seven towns, A, B, C, D,
E, F and G. The least distance between B and G is x km, where 25x
Preety needs to visit each town at least once, starting and finishing at A. She wishes
to minimise the total distance she travels.
(b) Starting by deleting B and all of its arcs, find a lower bound for Preety’s route.
(3)
Preety found the nearest neighbour routes from each of A and C. Given that the sum
of the lengths of these routes is 331 km,
(c) find ,x making your method clear.
(4)
(d) Write down the smallest interval that you can be confident contains the optimal
length of Preety’s route. Give your answer as an inequality.
(2)
(Total for Question 3 is 11 marks)
A B C D E F G
A - 17 24 16 21 18 41
B 17 - 35 25 30 31 x
C 24 35 - 28 20 35 32
D 16 25 28 - 29 19 45
E 21 30 20 29 - 22 35
F 18 31 35 19 22 - 37
G 41 x 32 45 35 37 -
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4.
Figure 3
The network in Figure 3 shows the roads linking a depot, D, and three collection points
A, B and C. The number on each arc represents the length, in miles, of the corresponding
road. The road from B to D is a one-way road, as indicated by the arrow.
(a) Explain clearly why Dijkstra’s algorithm cannot be used to find a route from D to A.
(1)
The initial distance and route tables for the network are given in the answer book.
(b) Use Floyd’s algorithm to find a table of least distances. You should show both the
distance table and the route table after each iteration.
(7)
(c) Explain how the final route table can be used to find the shortest route from D to B.
State this route.
(2)
There are items to collect at A, B and C. A van will leave D to make these
collections in any order and then return to D. A minimum route is required.
Using the final distance table and the Nearest Neighbour algorithm starting at D,
(d) find a minimum route and state its length.
(2)
Floyd’s algorithm and Dijkstra’s algorithm are applied to a network. Each will find
the shortest distance between vertices of the network.
(e) Describe how the results of these algorithms differ.
(2)
(Total for Question 4 is 14 marks)
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5. A garden centre makes hanging baskets to sell to its customers. Three types of hanging
basket are made, Sunshine, Drama and Peaceful. The plants used are categorised as
Impact, Flowering or Trailing.
Each Sunshine basket contains 2 Impact plants, 4 Flowering plants and 3 Trailing plants.
Each Drama basket contains 3 Impact plants, 2 Flowering plants and 4 Trailing plants.
Each Peaceful basket contains 1 Impact plant, 3 Flowering plants and 2 Trailing plants.
The garden centre can use at most 80 Impact plants, at most 140 Flowering plants and at
most 96 Trailing plants each day.
The profit on Sunshine, Drama and Peaceful baskets are £12, £20 and £16 respectively.
The garden centre wishes to maximise its profit.
Let x, y and z be the number of Sunshine, Drama and Peaceful baskets respectively,
produced each day.
(a) Formulate this situation as a linear programming problem, giving your constraints as
inequalities.
(5)
(b) State the further restriction that applies to the values of ,x y and z in this context.
(1)
The Simplex algorithm is used to solve this problem. After one iteration, the tableau
is
(c) State the variable that was increased in the first iteration. Justify your answer.
(2)
(d) Determine how many plants in total are being used after only one iteration of the
Simplex algorithm.
(1)
(e) Explain why for a second iteration of the Simplex algorithm the 2 in the z column is
the pivot value.
(2)
b.v. x y z r s t Value
r 1
4 0
1
2 1 0
3
4 8
s 5
2 0 2 0 1
1
2 92
y 3
4 1
1
2 0 0
1
4 24
P 3 0 6 0 0 5 480
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After a second iteration, the tableau is
(f) Use algebra to explain why this tableau is optimal.
(1)
(g) State the optimal number of each type of basket that should be made.
(1)
The manager of the garden centre is able to increase the number of Impact plants
available each day from 80 to 100. She wants to know if this would increase her
profit.
(h) Use your final tableau to determine the effect of this increase. (You should not carry
out any further calculations.)
(2)
(Total for Question 5 is 15 marks)
b.v. x y z r s t Value
r 3
8 0 0 1
1
4
7
8 31
z 5
4 0 1 0
1
2
1
4 46
y 1
8 1 0 0
1
4
3
8 1
P 21
2 0 0 0 3
7
2 756
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6.
Figure 4
A project is modelled by the activity network shown in Figure 4. The activities are
represented by the arcs. The number in brackets on each arc gives the time, in days, to
complete that activity. Each activity requires one worker. The project is to be completed
in the shortest possible time.
(a) Calculate the early time and the late time for each event, using Diagram 1 in the
answer book.
(3)
(b) On Grid 1 in the answer book, complete the cascade (Gantt) chart for this project.
(3)
(c) On Grid 2 in the answer book, draw a resource histogram to show the number of
workers required each day when each activity begins at its earliest time.
(3)
The supervisor of the project states that only three workers are required to complete
the project in the minimum time.
(d) Use Grid 2 to determine if the project can be completed in the minimum time by only
three workers. Give reasons for your answer.
(3)
(Total for Question 6 is 12 marks)
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7. A linear programming problem in ,x y and z is described as follows.
Maximise 3 2 2P x y z
subject to 2 2 25x y z
4 15x y
3x
(a) Explain why the Simplex algorithm cannot be used to solve this linear programming
problem. (1)
The big-M method is to be used to solve this linear programming problem.
(b) Define, in this context, what M represents. You must use correct mathematical
language in your answer.
(1)
The initial tableau for a big-M solution to the problem is shown below.
b.v. x y z 1s 2s
3s 1t Value
1s 2 2 1 1 0 0 0 25
2s 1 4 0 0 1 0 0 15
1t 1 0 0 0 0 1 1 3
P (3 + M) 2 2 0 0 M 0 3M
(c) Explain clearly how the equation represented in the b.v. 1t row was derived.
(1)
(d) Show how the equation represented in the b.v. P row was derived.
(2)
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The tableau obtained from the first iteration of the big-M method is shown below.
b.v. x y z 1s
2s 3s
1t Value
1s 0 2 1 1 0 2 2 19
2s 0 4 0 0 1 1 1 12
x 1 0 0 0 0 1 1 3
P 0 2 2 0 0 3 3 M 9
(e) Solve the linear programming problem, starting from this second tableau. You must
give a detailed explanation of your method by clearly stating the row operations
you use and
state the solution by deducing the final values of , , and . P x y z
(7)
(Total for Question 7 is 12 marks)
TOTAL FOR PAPER IS 75 MARKS
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Pearson Edexcel Level 3 GCE
Further Mathematics Advanced Further Mathematics Decision Mathematics 1
Sample assessment material for first teaching September 2017
Paper Reference(s)
9FM0/3D 9FM0/4D
Answer Book Do not return the question paper with the answer book.
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1.
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Question 1 continued
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(Total for Question 1 is 4 marks)
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2.
Figure 1
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Question 2 continued
Figure 2
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(Total for Question 2 is 7 marks)
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3.
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A B C D E F G
A - 17 24 16 21 18 41
B 17 - 35 25 30 31 x
C 24 35 - 28 20 35 32
D 16 25 28 - 29 19 45
E 21 30 20 29 - 22 35
F 18 31 35 19 22 - 37
G 41 x 32 45 35 37 -
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Question 3 continued
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(Total for Question 3 is 11 marks)
A B C D E F G
A - 17 24 16 21 18 41
B 17 - 35 25 30 31 x
C 24 35 - 28 20 35 32
D 16 25 28 - 29 19 45
E 21 30 20 29 - 22 35
F 18 31 35 19 22 - 37
G 41 x 32 45 35 37 -
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4.
Figure 3
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(b)
A B C D A B C D
A - 5 11 8 A A B C D
B 5 - 3 2 B A B C D
C 11 3 - 4 C A B C D
D 8 ∞ 4 - D A B C D
A B C D A B C D
A A
B B
C C
D D
A B C D A B C D
A A
B B
C C
D D
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Question 4 continued
A B C D A B C D
A A
B B
C C
D D
A B C D A B C D
A A
B B
C C
D D
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(Total for Question 4 is 14 marks)
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5.
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Question 5 continued
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(Total for Question 5 is 15 marks)
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6.
Diagram 1
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Question 6 continued
Grid 1
(There is a spare grid on the next page)
Grid 2
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Question 6 continued
Copy of Grid 1
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(Total for Question 6 is 12 marks)
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7.
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b.v. x y z 1s 2s 3s 1t Value
1s 0 2 1 1 0 2 2 19
2s 0 4 0 0 1 1 1 12
x 1 0 0 0 0 1 1 3
P 0 2 2 0 0 3 3 M 9
b.v. x y z 1s 2s 3s 1t Value Row Ops
P
b.v. x y z 1s 2s 3s 1t Value Row Ops
P
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Question 7 continued
b.v. x y z 1s 2s 3s 1t Value Row Ops
P
b.v. x y z 1s 2s 3s 1t Value Row Ops
P
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(Total for Question 7 is 12 marks)
TOTAL FOR PAPER IS 75 MARKS
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AL Decision Mathematics 1 Mark Scheme
Question Scheme Marks AOs
1(a)
In the first pass we compare the first value with the second
value and we swap these values if the second is larger than
the first.
B1 2.4
We then compare the value which is now second with the
third value and swap if the third is larger than the second.
We continue in this way until we reach the end of this list.
B1 2.4
(2)
(b) 250000
0.0212000
t
M1 1.1a
13.125t (seconds) A1 1.1b
(2)
(4 marks)
Notes:
(a)
B1: Comparing first value with second value, swap if second is larger (oe) – in their
reasoning it must be clear that the first value in the list is being compared with the second
value in the list and swapping if the second is larger than the first
B1: Compare second with third, (third with fourth), and so on until the end of the list –
must be clear in their reasoning that after the first comparison the second value in the
list is compared with the third value and so on until the end of the list
(b)
M1: Correct method seen – accept 25 for 50000/2000
A1: CAO
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Question Scheme Marks AOs
2(a)
A planar graph is a graph that can be drawn so that…….. B1 1.2
….. no arc meets another arc except at a vertex. B1 1.2
(2)
(b)
E.g. ABCDEGFA B1 1.1b
(1)
(c) Creates two lists of arcs, M1 2.1
e.g. BG AD
CG BD M1 1.1b
EG AE
CE AF A1 1.1b
Since no arc appears in both lists, the graph is planar
(or draws a planar version) A1 2.4
(4)
(7 marks)
Notes:
(a)
B1: A clear indication that a planar graph ‘can be drawn’ – allow this mark even if candidate
implies that arcs can cross each other
B1: CAO – no arc meets another arc except at a vertex – technical language must be correct
(b)
B1: Any correct Hamiltonian cycle (must start and finish at A) – must contain 8 vertices with
every vertex appearing only once (except A)
(c)
M1: Creates two list of arcs (with at least three arcs in each list) which contain no common
arcs
c2M1: Four arcs (in each list) and within each list there are no crossing arcs
c1A1: CAO
c2A1: Correct reasoning that no arc appears in both lists + so the graph is therefore planar
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Question Scheme Marks AOs
3(a) e.g. in the practical problem each vertex must be visited at least
once. In the classical problem each vertex must be visited just
once
B2,1,0 2.4
2.4
(2)
(b)
Prim’s algorithm on reduced network starting at A: AD, AF, AE,
CE, CG M1
1.1b
Lower bound = 107 + 17 + 25 = 149 (km) M1
A1
1.1b
1.1b
(3)
(c)
NNA from A: A – D – F – E – C – G – B – A = 126 x
NNA from C: C – E – A – D – F – B – G – C = 139 x
M1
A1
A1
1.1b
1.1b
1.1b
(126 ) (139 ) 331 33x x x A1 1.1b
(4)
(d) 149 optimal 159
M1
A1
2.2b
1.1b
(2)
(11 marks)
Notes:
(a)
B1: Understands the difference is connected to the number of times each vertex may be visited
(but maybe incorrectly attributed). Must be an attempt at a difference (so must refer to
both the classical and practical problems explicitly). Technical language (vertex/node)
must be correct. Need not imply each/every/all (oe) vertices for this first mark
B1: Correctly reasons which is classical and which is practical and correctly states the
difference. Must imply that each/every/all (oe) vertices are visited, so for example, ‘the
practical problem visits a vertex at least once while the classical visits a vertex only once’
is B1B0 (note that B0B1 is not possible in (a))
(b)
M1: Correctly applying Prim’s algorithm from node A for the first four arcs (or five nodes)
M1: Candidates weight of their RMST + 17 + 25 (the two smallest arcs incident to B)
A1: CAO (condone lack of units)
(c)
M1: Either one route, must return to A
A1: One correct route, must return to A and corresponding length correct (do not is in part
(c) if correct lengths seen but are then doubled)
A1: Both routes correct and their corresponding lengths correct
A1: CAO for x
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Notes: (continued)
(d)
M1: Their numbers correctly used, accept any inequalities or any indication of interval from
their 149 to their 159 (so 149 – 159 can score this mark). This mark is dependent on two
routes seen in (c), however, neither of the two totals need to be correct. Please note that
UB > LB for this mark
A1: CAO (no follow through on their values) including correct inequalities or equivalent set
notation (but condone 149 optimal 159)
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Question Scheme Marks AOs
4(a) Dijkstra’s algorithm cannot be applied to directed
networks (in this problem there is a directed arc from
B to D)
B1 3.5b
(1)
(b)
Initial tables
5 11 8
5 3 2
11 3 4
8 4
A B C D
A B C D
A B C D
A B C D
1st iteration
3 2
3 4
4
5 11 8
5
11
8 13
A B C D
A B C D
B D
A CA
A C
D
M1
A1
1.1b
1.1b
2nd iteration
4
8
5
5 3 2
3
1 43
8 7
8
A B
A B C D
B C D
A A C
B
B
D
B
M1
A1ft
1.1b
1.1b
3rd iteration
8
3
5 7
5 2
8
8 3
47
4
A B B B
A B C D
B B C D
A C C D
M1
A1ft
1.1b
1.1b
4th iteration
5 8
5 3
7
2
8 4
8 3 4
7
A B B B
A B C D
B B C D
A C C D
no changes therefore optimal
A1 1.1b
(7)
(c) Start at D (4th) row and read across to the B (2nd) column,
there is a C there
so the route starts DC. Look at the C row, B column and you
see B B1 2.4
The route is therefore DCB B1 2.2a
(2)
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Question Scheme Marks AOs
(d)
D – C –B – A – B – D M1 2.2a
Length 19 (miles) A1 1.1b
(2)
(e) Dijkstra’s algorithm finds the shortest distances from one
vertex to all the others. Floyd’s algorithm finds the shortest
distance between every pair of vertices.
B1
B1
2.5
2.5
(2)
(14 marks)
Notes:
(a)
B1: CAO (must include mention of ‘directed’ network)
M1: No change in the first row and first column of both tables with at least one value in the
distance table reduced and one value in the route table changed
(b)
A1: CAO
M1: No change in the second row and second column of both tables with at least two values
in the distance table reduced and two values in the route table changed
A1ft: Correct second iteration follow through from the candidate’s first iteration
M1: No change in the third row and third column of both tables with at least one value in
the distance table reduced and one value in the route table changed
A1ft: Correct third iteration follow through from the candidate’s second iteration
A1: CSO (no change after the fourth iteration) – all previous marks must have been
awarded in this part
(c)
B1: Clear indication of how the final route table can be used to get from D to B (therefore
must mention the correct rows and columns in their reasoning)
B1: Completely correct argument + correct route (DCB)
(d)
M1: Deduce correctly their minimum route from their final distance table (dependent on all
M marks in (a)) must begin and end at D
A1: CAO (length of 19)
(e)
B1: CAO – must use correct language ‘one vertex to all other vertices’
B1: CAO – must use correct language ‘every pair of vertices’
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Question Scheme Marks AOs
5(a)
Maximise P = 12x + 20y + 16z B1 3.3
Subject to
2 3 80
4 2 3 140
3 4 2 96
, , 0
x y z
x y z
x y z
x y z
M1
A1
A1
B1
3.3
1.1b
1.1b
3.3
(5)
(b) The values must all be integers B1 3.3
(1)
(c)
Variable y entered the basic variable column…. M1 2.4
…so y was increased first A1 2.2a
(2)
(d) (80 140 96) (8 92) 216 plants B1 3.2a
(1)
(e)
The next pivot must come from a column which has a
negative value in the objective row so therefore the pivot
must come from column z.
M1 2.4
The pivot must be positive and the least of 92/2 = 46 and
24/0.5 = 48 so the pivot must be the 2 (from column z). A1 2.2a
(2)
(f) P + 10.5x + 3s + 3.5t = 756 so increasing x, s or t will
decrease profit B1 2.4
(1)
(g) Make 1 Drama basket and 46 Peaceful baskets B1 2.2a
(1)
(h) The slack variable, r, associated with this type of plant, is
currently at 31. Increasing the number of Impact plants by
a further 20 would have no effect.
M1
A1
3.1b
3.2a
(2)
(15 marks)
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Notes:
(a)
B1: Correct objective function/expression (accept in pence rather than pounds e.g. 1200x +
2000y + 1600z)
M1: Correct coefficients and correct right-hand side for at least one inequality – accept any
inequality or equals
A1: Two correct (non-trivial) inequalities
A1: All three non-trivial inequalities correct
B1: , , 0x y z
(b)
B1: CAO
(c)
M1: Correct reasoning that y has become a basic variable
A1: Correct deduction that y was therefore increased first
(d)
B1: CAO
(e)
M1: Correct reasoning given that the pivot value must come from column z
A1: Correctly deduce (from correctly stated calculations) that the pivot value is the 2 in
column z
(f)
B1: States correct objective function and mention of increasing x, s or t will decrease profit
(g)
B1: CAO – in context so not in terms of y and z
(h)
M1: Identifies the slack variable r and its current value of 31
A1: Correct interpretation that increasing the number of Impact plants would have no effect
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Question Scheme Marks AOs
6(a)
See diagram on next page. Top and bottom boxes
Top boxes correct
M1
A1
2.1
1.1b
Bottom boxes correct A1 1.1b
(3)
(b)
See diagram below
At least 8 activities + 4 floats with clear distinction
between activity and their corresponding float M1 2.5
Correct critical activities + 4 correct non-critical activities A1 1.1b
All 13 correct A1 1.1b
(3)
(c)
M1 1.1b
Bars correct to time = 13 A1 1.1b
Bars correct from 14 to 24 A1 1.1b
(3)
(d) Until time 4 only A and B can happen.
After time 4, there are 6 worker-days to cover, but only 4
worker-days available.
Hence the project cannot be completed by time 24 with
three workers.
B1
M1
A1
3.1a
2.4
2.2a
(3)
(12 marks)
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Notes:
(a)
M1: All top boxes and all bottom boxes completed. For the top boxes all values must be
increasing in the direction of the arrows for both the activities and the dummies. For
the bottom boxes all values must be decreasing in the opposite direction to the arrows
for both the activities and the dummies. While the values need not be correct each
value must be increasing or decreasing (as appropriate) in a logical and sequential
manner.
A1: CAO for top boxes
A1: CAO for bottom boxes
M1: At least 8 activities including 4 floats. Scheduling diagram scores M0 – clear
distinction must be shown between the notation used for an activity and its float
(b)
A1: Correct critical activities and 4 correct non-critical activities
A1: CAO (all 13 correct activities)
(c)
M1: Plausible histogram with no holes or overhangs (must go to at least 10 on the time axis)
A1: Histogram correct to time 13
A1: Histogram correct from time 14 to time 24
(d) B1: Considering an appropriate process to adjust Grid 2 so that no activity must be
completed by a 4th worker, for example, a correct argument that until time 4 only
activities A and B can happen (so no activity can use the spare worker before time 4)
M1: Uses their histogram to explain when the number of workers is greater or less than the
minimum number found in (b)
A1: Correctly deduces that the project cannot be completed by time 24 – this mark is
dependent on a correct histogram seen in (d)
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Diagram for Question 6(a)
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Question Scheme Marks AOs
7(a) Simplex can only work with constraints B1 3.5b
(1)
(b) M is an arbitrary large real number B1 2.5
(1)
(c) 3 13 3x x s t where
3s is a surplus variable and 1t is an
artificial variable B1 2.4
(1)
(d) Let 13 2 2P x y z Mt (where M is an arbitrary large
number)
33 2 2 (3 )P x y z M x s
3(3 ) 2 2 3M x y z Ms M
3(3 ) 2 2 3P M x y z Ms M
M1
A1
2.1
1.1b
(2)
(e)
b.v. x y z 1s 2s
3s
1t Value Row Ops
3s 0 1 ½ ½ 0 1 1 19/2 1 1(1/ 2)r R
2s 0 3 1/2 1/2 1 0 0 5/2 2 1R r
x 1 1 ½ ½ 0 0 0 25/2 3 1R r
P 0 1 1/2 3/2 0 0 M 75/2 4 13R r
M1
A1
A1
1.1b
1.1b
1.1b
b.v. x y z 1s 2s
3s 1t Value Row Ops
z 0 2 1 1 0 2 2 19 1 12r R
2s 0 4 0 0 1 1 1 12 2 1(1/ 2)R r
x 1 0 0 0 0 1 1 3 3 1(1/ 2)R r
P 0 2 0 2 0 1 1M 47 4 1(1/ 2)R r
M1
A1
B1
1.1b
1.1b
2.4
47, 3, 0, 19P x y z B1ft
1.1b
(7)
(12 marks)
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Notes:
(a)
B1: Correctly states the limitation of the Simplex model – Simplex involves iterations which
allow movement from one vertex in the feasible region to another vertex (in the feasible
region). If all constraints are of the form this means that the origin is always a feasible
solution and therefore can act as the initial starting point for the problem. However, the
constraint 3x means that the origin is not feasible and so the algorithm is unable to
begin.
(b)
B1: CAO including the correct mathematical language (must include ‘arbitrary’, ‘large’ and
‘real’)
(C)
(i)B1: Correctly states both the inequality 3x and the equation 3 1 3x s t together with an
explanation of the meaning behind the variables 3s and
1t
(ii)M1: 13 2 2P x y z Mt and substitutes their expression for
1t
(ii)A1: Correct mathematical argument including sufficient detail to allow the line of reasoning to
be followed to the correct conclusion – dependent on previous B mark in (c)
(d)
M1: Correct pivot located, attempt to divide row. If negative value used then no marks
A1: Pivot row correct (including change of b.v.) and row operations used at least once, one of
columns 1 1, , ,y z s t or Value correct
A1: CAO for values (ignore b.v. column and Row Ops)
M1: Pivot row consistent (following their previous table) including change of b.v. and row
operations used at least once, one of columns 1 3 1, , ,y s s t or Value correct
A1: CAO on final table (ignore Row Ops)
B1: The correct Row Operations explained either in terms of the ‘old’ or ‘new’ pivot rows
B1ft: Correctly states the final values of P, , and x y z from their correct corresponding rows of
the final table
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Pearson Edexcel Level 3 GCE
Further Mathematics Advanced Decision Mathematics 2
Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes
Paper Reference(s)
9FM0/4G
You must have: Decision Mathematics Answer Book (enclosed), calculator
Instructions
Use black ink or ball-point pen.
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
Write your answers for this paper in the Decision Mathematics answer book provided.
Fill in the boxes at the top of the answer book with your name, centre number and candidate number.
Do not return the question paper with the answer book.
Answer all questions and ensure that your answers to parts of questions are clearly labelled.
Answer the questions in the spaces provided – there may be more space than you need.
You should show sufficient working to make your methods clear. Answers without working may not gain full credit.
Answers should be given to three significant figures unless otherwise stated. Information
There are 8 questions in this question paper. The total mark for this paper is 75.
The marks for each question are shown in brackets – use this as a guide as to how much time to spend on each question.
Advice
Read each question carefully before you start to answer it.
Try to answer every question.
Check your answers if you have time at the end.
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Answer ALL questions. Write your answers in the answer book provided.
1. (a) Find the general solution of the recurrence relation
2 1 , 1 n n nu u u n
(3)
Given that 1 1u and 2 1u
(b) find the particular solution of the recurrence relation.
(3)
(Total for Question 1 is 6 marks)
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2.
A company has three factories, A, B and C. It supplies mattresses to three shops, D, E and
F. The table shows the transportation cost, in pounds, of moving one mattress from each
factory to each shop. It also shows the number of mattresses available at each factory and
the number of mattresses required at each shop. A minimum cost solution is required.
(a) Use the north-west corner method to obtain an initial solution.
(1)
(b) Show how the transportation algorithm is used to solve this problem.
You must state, at each appropriate step, the
shadow costs,
improvement indices,
route,
entering cell and exiting cell,
and explain clearly how you know that your final solution is optimal.
(11)
(Total for Question 2 is 12 marks)
D E F Available
A 15 19 9 25
B 11 18 10 55
C 11 12 18 20
Required 38 24 38
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3. Four workers, A, B, C and D, are to be assigned to four tasks, P, Q, R and S.
Each worker must be assigned to at most one task and each task must be done by just one
worker.
The amount, in pounds, that each worker would earn while assigned to each task is
shown in the table below.
The Hungarian algorithm is to be used to find the maximum total amount which may be
earned by the four workers.
(a) Explain how the table should be modified.
(1)
(b) Reducing rows first, use the Hungarian algorithm to obtain an allocation which
maximises the total earnings, stating how each table was formed.
(7)
(c) Formulate the problem as a linear programming problem. You must define your
decision variables and make your objective function and constraints clear.
(5)
(Total for Question 3 is 13 marks)
P Q R S
A 32 32 33 35
B 28 35 31 37
C 35 29 33 36
D 36 30 36 33
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4. A game uses a standard pack of 52 playing cards.
A player gives 5 tokens to play and then picks a card. If they pick a 2, 3, 4, 5 or 6 then
they gain 15 tokens. If any other card is picked they lose.
If they lose, the card is replaced and they can choose to pick again for another 5 tokens.
This time if they pick either an ace or a king they gain 40 tokens. If any other card is picked
they lose.
Daniel is deciding whether to play this game.
(a) Draw a decision tree to model Daniel’s possible decisions and the possible outcomes.
(6)
(b) Calculate Daniel’s optimal EMV and state the optimal strategy indicated by the
decision tree.
(2)
(Total for Question 4 is 8 marks)
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5.
A two person zero-sum game is represented by the pay-off matrix for player A given
above.
(a) Explain, with justification, how this matrix may be reduced to a 3 3 matrix.
(2)
(b) Find the play-safe strategy for each player and verify that there is no stable solution
to this game.
(4)
The game is formulated as a linear programming problem for player A.
The objective is to maximise ,P V where V is the value of the game to player A.
One of the constraints is that 1 2 3 1,p p p where 1 2 3, ,p p p are the probabilities
that player A plays 1, 2, 3 respectively.
(c) Formulate the remaining constraints for this problem. Write these constraints as
inequalities.
(3)
The Simplex algorithm is used to solve the linear programming problem. The
solution obtained is 1 2 3
3 40, ,
7 7p p p
(d) Calculate the value of the game to player A.
(3)
(Total for Question 5 is 12 marks)
B plays 1 B plays 2 B plays 3 B plays 4
A plays 1 4 2 3 2
A plays 2 3 1 2 0
A plays 3 1 2 0 3
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6.
Figure 1
Figure 1 shows a capacitated, directed network. The number on each arc ( , )x y represents
the lower ( )x capacity and upper ( )y capacity of that arc.
(a) Calculate the value of the cut 1C and cut 2C
(2)
(b) Explain why the flow through the network must be at least 12 and at most 16
(1)
(c) Explain why arcs DG, AG, EG and FG must all be at their lower capacities.
(1)
(d) Determine a maximum flow pattern for this network and draw it on Diagram 1 in the
answer book. You do not need to use the labelling procedure.
(2)
(e) (i) State the value of the maximum flow through the network.
(ii) Explain why the value of the maximum flow is equal to the value of the
minimum flow through the network.
(3)
Node E becomes blocked and no flow can pass through it. To maintain the maximum
flow through the network the upper capacity of exactly one arc is increased.
(f) Explain how it is possible to maintain the maximum flow found in (d).
(3)
(Total for Question 6 is 12 marks)
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7. A company assembles boats.
They can assemble up to five boats in any one month, but if they assemble more than three
they will have to hire additional space at a cost of £800 per month.
The company can store up to two boats at a cost of £350 each per month.
The overhead costs are £1500 in any month in which work is done.
Boats are delivered at the end of each month. There are no boats in stock at the beginning
of January and there must be none in stock at the end of May.
The order book for boats is
Month January February March April May
Number ordered 3 2 6 3 4
Use dynamic programming to determine the production schedule which minimises the
costs to the company. Show your working in the table provided in the answer book and
state the minimum production cost.
(Total for Question 7 is 12 marks)
TOTAL FOR PAPER IS 75 MARKS
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Pearson Edexcel Level 3 GCE
Further Mathematics Advanced Decision Mathematics 2
Sample assessment material for first teaching September 2017
Paper Reference(s)
9FM0/4G
Answer Book Do not return the question paper with the answer book.
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1. .....................................................................................................................................................................
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Question 1 continued
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(Total for Question 1 is 6 marks)
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2.
D E F Available
A 15 19 9 25
B 11 18 10 55
C 11 12 18 20
Required 38 24 38
(a)
D E F
A
B
C
(b)
D E F
A
B
C
D E F
A
B
C
D E F
A
B
C
D E F
A
B
C
D E F
A
B
C
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Question 2 continued
D E F
A
B
C
D E F
A
B
C
D E F
A
B
C
D E F
A
B
C
D E F
A
B
C
(Total for Question 2 is 12 marks)
D E F
A
B
C
D E F
A
B
C
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3.
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P Q R S
A 32 32 33 35
B 28 35 31 37
C 35 29 33 36
D 36 30 36 33
P Q R S
A
B
C
D
P Q R S
A
B
C
D
P Q R S
A
B
C
D
P Q R S
A
B
C
D
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Question 3 continued
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(Total for Question 3 is 13 marks)
P Q R S
A
B
C
D
P Q R S
A
B
C
D
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4.
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Question 4 continued
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(Total for Question 4 is 8 marks)
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5.
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B plays 1 B plays 2 B plays 3 B plays 4
A plays 1 4 2 3 2
A plays 2 3 1 2 0
A plays 3 1 2 0 3
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Question 5 continued
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(Total for Question 5 is 12 marks)
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6.
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Question 6 continued
Diagram 1
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(Total for Question 6 is 12 marks)
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7.
Stage State Action Dest Value
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Month January February March April May
Number
assembled
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Question 7 continued
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(Total for Question 7 is 12 marks)
TOTAL FOR PAPER IS 75 MARKS
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AL Decision Mathematics 2 Mark Scheme
Question Scheme Marks AOs
1(a) Auxiliary equation: 2 1 0 and attempt to solve M1 1.1b
1 5
2
1 5 1 5
2 2
n n
nu A B
, where A and
B are arbitrary constants
M1
A1
1.1b
2.2a
(3)
(b) Use given conditions to obtain two equations in A and B M1 1.1b
Attempt to solve to obtain an A and B M1 1.1b
1 1 5 1 5
2 25
n n
nu
A1
1.1b
(3)
(6 marks)
Notes:
(a)
M1: writes down correct auxiliary equation and attempts to solve using either the formula or
completing the square
M1: writes down the general solution in the form 1 2( )n n
nu A B using their roots 1 2,
- dependent on the first M mark
A1: CAO – both lhs and rhs correct including defining A and B as (arbitrary) constants
(b)
M1: uses the correct initial conditions to write down two equations in A and B – for reference
these equations are 1 5 1 5 2A B and 2 2
1 5 1 5 4A B
M1: Attempts to solve these two equations (using a correct method but condone sign slips) to
achieve a value for A and B
A1: CAO
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Question Scheme Marks AOs
2(a)
D E F Available
A 25 25
B 13 24 18 55
C 20 20
Required 38 24 38
B1 1.1b
(1)
(b) Shadow
costs
15 22 14
D E F
0 A X -3 -5
-4 B X X X
4 C -8 -14 X
M1
A1
2.1
1.1b
D E F
A
B
24
18
C 20
Entering CE, exiting CF
D E F
A 25
B
13 4 38
C 20
M1
A1
2.1
1.1b
Shadow
costs
15 22 14
D E F
0 A X -3 -5
-4 B X X X
-10 C 6 X 14
M1
A1
1.1b
1.1b
D E F
A 25
B
13 38
C
Entering AF, exiting AD
D E F
A 25
B
38 4 13
C 20
M1
A1
1.1b
2.2a
Shadow
costs
10 17 9
D E F
0 A 5 2 X
1 B X X X
-5 C 6 X 14
No negative IIs so optimal solution of £1085
M1
A1
A1
2.1
1.1b
2.4
(11)
(12 marks)
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Notes:
B1: CAO
M1: Finding all 6 shadow costs and the 4 improvement indices for the correct 4 entries –
candidates must clearly identify these two sets of results
A1: Shadow costs and II CAO
M1: A valid route, their most negative II chosen, only one empty square used, ’s balance
A1: CAO
M1: Finding all 6 shadow costs and the 4 improvement indices for the correct 4 entries
A1: Shadow costs and II CAO
M1: A valid route, their most negative II chosen, only one empty square used, ’s balance
A1: CAO – including the deduction of all entering and exiting cells
M1: Finding all 6 shadow costs and the 4 improvement indices for the correct 4 entries – this
mark is depedent on all previous M marks which will therefore indicate a correct
mathematical argument leading from the initial solution to the confirmation of the optimal
solution
A1: Shadow costs and II CAO
A1: CSO including the correct reasoning that the solution is optimal because there are no
negative IIs
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Question Scheme Marks AOs
3(a) Subtract each entry from a constant (eg 40) B1 2.4
(1)
(b)
e.g.
P Q R S
A 5 5 4 2
B 9 2 6 0
C 2 8 4 1
D 1 7 1 4
B1
1.1b
Reducing row A by 2, no reduction for row B, reduce row C by 1
and row D by 1. No reduction of columns P, R and S, reduce
column Q by 2.
P Q R S
A 3 3 2 0
B 9 2 6 0
C 1 7 3 0
D 0 6 0 3
then
P Q R S
A 3 1 2 0
B 9 0 6 0
C 1 5 3 0
D 0 4 0 3
B1
M1
A1
2.4
2.1
1.1b
Three lines required to cover the zeros hence solution is not
optimal – augment by 1 B1 2.4
P Q R S
A 2 1 1 0
B 8 0 5 0
C 0 5 2 0
D 0 5 0 4
M1 2.1
A – S, B – Q, C – P, D – R A1 2.2a
(7)
(c) 1 if worker does task
0 otherwise ij
i jx
B1 3.3
Where A,B,C,Di and P,Q,R,Sj B1 3.3
e.g. Minimise AP AQ AR AS BP BQ BR5 5 4 2 9 2 6x x x x x x x
CP CQ CR CS DP DQ DR DS2 8 4 7 4x x x x x x x x B1 3.3
Subject to:
P Q R S1, 1, 1, 1i i i ix x x x M1 3.3
A B C D1, 1, 1, 1j j j jx x x x A1 3.3
(5)
(13 marks)
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Notes
(a)
B1: valid statement regarding converting a max. problem to a min. problem
(b)
B1: CAO
B1: Correct statements regarding row and column reduction
M1: Simplifying the initial matrix by reducing rows and then columns
A1: CAO
B1: Correct statements regarding both max. number of lines to cover zeros and augmentation
M1: Develop an improved solution – need to see one double covered +e; one uncovered –e;
and one single covered unchanged. 3 lines needed to 4 lines needed (so getting to the
optimal table)
A1: CSO on final table (so must have scored all previous marks in this part ) + deduction of
the correct allocation
(c)
B1: possible values of ijx defined
B1: definine the set of values for i and j
B1: Correct objective function and either ‘minimise’ or ‘maximise’ (dependent on if problem
is defined in terms of original values or modified values)
M1: at least four equations, unit coefficient and equal to 1
A1: CAO (all eight equations)
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Question Scheme Marks AOs
4(a)
M1
A1
M1
A1
M1
A1
3.3
1.1b
3.4
1.1b
3.4
1.1b
(6)
(b) EMV is 1.48 (tokens) per game (correct to 3 sf)
Analysis: Play the game and if the player doesn’t pick a 2 – 6 on
the first go then they should pick again
B1
B1
3.4
3.2a
(2)
(8 marks)
Notes:
(a)
M1: Tree diagram with at least three end pay-offs, two decision nodes and two chance nodes
A1: Correct structure of tree diagram with each arc labelled correctly (including probabilities)
M1: At least three end-pay offs consistent with their stated probabilities; all five attempted
A1: CAO for end-pay offs
M1: End chance node follow through their end pay-offs and other chance/decision nodes
completed
A1: CAO for decision and chance nodes including double lines through inferior options
(b)
B1: Correct EMV
B1: Correct analysis
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Question Scheme Marks AOs
5(a) Column 2 dominates column 4 B1 2.5
Because 2 2,0 1 and 3 2 B1 2.4
(2)
(b) Row minima: -2, -1, -1 max is -1
Column maxima: 4, 2, 3 min is 2
M1
A1
1.1b
1.1b
Play safe is A plays 2 or 3 and B plays 2 A1 1.1b
Row maximin (-1) Column minimax (2) so not stable A1 2.4
(4)
(c) e.g.
4 2 3 6 0 5
3 1 2 5 1 4
1 2 0 1 4 2
1 2 3
2 3
1 2 3
Subject to 6 5 0
4 0
5 4 2 0
V p p p
V p p
V p p p
B1
B1
B1
1.1b
3.3
3.3
(3)
(d) Substitute p values to obtain 19 19 20 19
, ,7 7 7 7
V V M1 3.4
Value of the game to player A 19 5
27 7
M1
A1
1.1b
1.1b
(3)
(12 marks)
Notes:
(a)
B1: Correct statement – must include the word ‘dominate’
B1: Correct inequalities – must be clear that all three inequalities must hold
(b)
M1: Attempt at row minima and column maxima – condone one error
A1: Correct max(row min) and min(col max)
A1: Correct play safe for both players
A1: Correct reasoning that the game is not stable (accept 1 2 + statement)
(c)
B1: Correct augmentation to make all entries non-negative
B1: At least one (of the three) equations or inequalities correct in V, 1 2 3, ,p p p (with all ip terms
in the constraint equations having correct signs)
B1: CAO - all three constraints correct involving V and ip expressed as inequalities
(d)
M1: Substitute p values to obtain three values for V
M1: Their least value of V minus their augmented value
A1: CAO for the value of the game to player A
Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
357
Question Scheme Marks AOs
6(a)
1 3 3 3 5 7 21C
2 8 5 3 6 3 19C
B1
B1
1.1b
1.1b
(2)
(b) e.g. the minimum flow out of the source S is at least 5 + 3 + 4 =
12 and the maximum flow into the sink T is 6 + 10 = 16
B1
2.4
(1)
(c) The minimum flow into G is 1 + 1 + 1 + 3 = 6 but the maximum
flow out of G is 6 therefore the arcs into G must be at their lower
capacities
B1
2.4
(1)
(d)
M1
A1
3.1a
1.1b
(2)
(e)
Maximum flow is 15 B1 1.1b
The minimum flow out of the source is 12 but the flow out of C
is at least 3 + 4 = 7 B1 2.4
Therefore the minimum flow through the network is 5 + 3 + 3 +
4 = 15 which is equal to the maximum flow B1 2.2a
(3)
(f) Increase the upper capacity of arc BF to at least 9 and therefore
increase the flow in this arc to 9 B1 2.1
Therefore increase the flow in FH and HT to 10 B1 2.4
The flow in GT decreases to 5 and all other arcs are unchanged B1 2.2a
(3)
(12 marks)
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Notes:
(a)
B1: correct capacity for 1C
B1: correct capacity for 2C
(b)
B1: correct statement regarding the min. flow out of the sink and max. flow into the sink
(c)
B1: correct statement regarding the flow into node G
(d)
M1: consistent flow pattern ( 12) throughout the network - so the flow into each node must
equal the flow out of each node (and this flow must be greater than or equal to 12 but not
necessarily the maximum flow of 15) - one number only on each arc
A1: CAO
(e)
B1: CAO (for max. flow)
B1: Consideration of both the min. flow from the source and the flow through node C
B1: Completely correct argument that the max. flow = min. flow
(f) B1: Correct argument regarding increasing the upper capacity of arc BF and hence the flow in
that arc
B1: Correct reasoning regarding increasing the flow in arcs FH and HT
B1: Correct deduction that the flow in GT decreases to 5 and conclude that all other arcs are
unchanged
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Question Scheme Marks AOs
7
Stage Stat
e
Actio
n
Dest Value
May 2 2 0 700 + 1500 = 2200*
(4) 1 3 0 350 + 1500 = 1850*
0 4 0 1500 + 800 = 2300*
April 2 1 0 700 + 1500 + 2300 = 4500
(3) 2 1 700 + 1500 + 1850 = 4050*
3 2 700 + 1500 + 2200 = 4400
1 2 0 350 + 1500 + 2300 = 4150
3 1 350 + 1500 + 1850 = 3700*
4 2 350 + 1500 + 800 +2200 = 4850
0 3 0 1500 + 2300 = 3800*
4 1 1500 + 800 + 1850 = 4150
5 2 1500 + 800 + 2200 = 4500
March 2 4 0 700 + 1500 +800 + 3800 = 6800
(6) 5 1 700 + 1500 + 800 + 3700 = 6700*
1 5 0 350 + 1500 + 800 + 3800 = 6450*
Feb 2 1 1 700 + 1500 + 6450 = 8650*
(2) 2 2 700 + 1500 + 6700 = 8900
1 2 1 350 + 1500 + 6450 = 8300*
3 2 350 + 1500 + 6700 = 8550
0 3 1 1500 + 6450 = 7950*
4 2 1500 + 800 + 6700 = 9000
Jan 0 3 0 1500 + 7950 = 9450*
(3) 4 1 1500 + 800 + 8300 = 10600
5 2 1500 + 800 + 8650 = 10950
M1
A1
M1
A1
M1
A1ft
M1
A1ft
M1
A1
3.1b
1.1b
3.1b
1.1b
1.1b
1.1b
1.1b
1.1b
1.1b
1.1b
Month January February March April May
Number
made 3 3 5 3 4
B1 1.1b
Minimum production cost: £9450 B1 1.1b
(12 marks)
Notes:
All M marks – must bring optimal result from previous stage into calculations so for the
second stage (April) if none of their 2200, 1850 or 2300 (the optimal results from May) are
used then M0. Ignore extra rows. Condone and credit rows that have been crossed out if
they can still be read. Must have right ‘ingredients’ (storage costs, additional space costs,
overhead cost) at least once per stage. Must have values in two of the three colums (State,
Action, Dest). If no working seen then the number stated in the Value column must be
correct to imply the correct method has been used
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1M1: First stage (May) completed. At least 3 rows, ‘something’ in each cell (but see M mark
guidance above) including the correct structure (e.g. no value greater than 5 in the action
column) in each of the first four columns
1A1: CAO for first stage.
2M1: Second stage (April) completed. At least 9 rows, something in each cell (see M mark
guidance above) including the correct structure for the fifth (Value) column (e.g. bringing
forward values from the previous stage)
2A1: CAO for second stage. No extra rows
3M1: Third stage (March) completed. At least 3 rows, something in each cell (see M mark
guidance above)
3A1ft: CAO on the ft for third stage. No extra rows
4M1: Fourth stage (February) completed. At least 6 rows, something in each cell (see M mark
guidance above)
4A1ft: CAO on the ft for fourth stage. No extra rows
5M1: Fifth stage (January) completed. At least 3 rows, something in each cell (see M mark
guidance above)
5A1: CAO for the fifth stage. No extra rows
1B1: CAO – but must have scored all previous M marks
2B1: CAO – condone lack of units - but must have scored all previous M marks
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Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample Assessment Materials (SAMs) – Issue 1 – Accredited pre-publication – May 2017 © Pearson Education Limited 2017
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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE
Mathematics and
Further Mathematics
Mathematical formulae and statistical tables
For first certification from June 2018 for:
Advanced Subsidiary GCE in Mathematics (8MA0)
Advanced GCE in Mathematics (9MA0)
Advanced Subsidiary GCE in Further Mathematics (8FM0)
For first certification from June 2019 for:
Advanced GCE in Further Mathematics (9FM0)
This copy is the property of Pearson. It is not to be removed from the
examination room or marked in any way.
Contents
1 Introduction 1
2 AS Level in Mathematics 2
Pure Mathematics 2
Statistics 2
Mechanics 3
3 A Level in Mathematics 4
Pure Mathematics 4
Statistics 6
Mechanics 7
4 AS Level in Further Mathematics 8
Pure Mathematics 8
Statistics 12
Mechanics 14
5 A Level in Further Mathematics 15
Pure Mathematics 15
Statistics 21
Mechanics 25
6 Statistical Tables 26
Binomial Cumulative Distribution Function 26
Percentage Points Of The Normal Distribution 31
Poisson Cumulative Distribution Function 32
Percentage Points of the 2 Distribution 33
Critical Values for Correlation Coefficients 34
Random Numbers 35
Percentage Points of Student’s t Distribution 36
Percentage Points of the F Distribution 37
Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 1
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
1 Introduction The formulae in this booklet have been arranged by qualification. Students sitting AS or A Level
Further Mathematics papers may be required to use the formulae that were introduced in AS or
A Level Mathematics papers.
It may also be the case that students sitting Mechanics and Statistics papers will need to use
formulae introduced in the appropriate Pure Mathematics papers for the qualification they are
sitting.
2 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
2 AS Level in Mathematics
Pure Mathematics
Mensuration
Surface area of sphere = 4 r 2
Area of curved surface of cone = r slant height
Binomial series
1 2 2
2( )
1 n n n r r nn n
n na b a b a b b
r
na b a
(n ℕ)
where C( )
n
r
n n!
r r ! n r !
Logarithms and exponentials
loglog
log
ba
b
xx
a
lnex a xa
Differentiation
First Principles
0
f ( ) f ( )f ( ) = lim
h
x h xx
h
Statistics
Probability
P(A) = 1 – P(A)
Standard deviation
Standard deviation = (Variance)
Interquartile range = IQR = Q3 – Q1
For a set of n values 1 2, ,... ,...i nx x x x
22 2 ( )
S ( ) ixx i i
xx x x
n
Standard deviation =
2
2or
S xx
xx
n n
Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 3
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
Statistical tables
The following statistical tables are required for A Level Mathematics:
Binomial Cumulative Distribution Function (see page 25)
Random Numbers (see page 34)
Mechanics
Kinematics
For motion in a straight line with constant acceleration:
v = u + at
s = ut + ½ at2
s = vt - ½ at2
v2 = u2 + 2as
s = ½ (u + v)t
4 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
3 A Level in Mathematics
Pure Mathematics
Mensuration
Surface area of sphere = 4 r 2
Area of curved surface of cone = r slant height
Arithmetic series
Sn = 1
2n(a + l) =
1
2n[2a + (n 1)d]
Binomial series
1 2 2
21 n n n r r nn n
n na b a b a b b
r
n( a b ) a
(n ℕ)
where C( )
n
r
n n!
r r ! n r !
2( 1) ( 1) ( 1)(1 ) 1 ( 1 )
1 2 1 2
n rn n n n n rx nx x x x , n
r
Logarithms and exponentials
loglog
log
ba
b
xx
a
lnx a xae
Geometric series
Sn = (1 )
1
na r
r
S =1
a
r for r < 1
Numerical integration
The trapezium rule:
b
a
xy d 21 h{(y0 + yn) + 2(y1 + y2 + ... + yn – 1)}, where
b ah
n
Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 5
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
Trigonometric identities
sin ( ) sin cos cos sinA B A B A B
cos( ) cos cos sin sinA B A B A B
1
2
tan tantan ( ) ( ( ) )
1 tan tan
A BA B A B k
A B
sin sin 2sin cos2 2
A B A BA B
sin sin 2cos sin2 2
A B A BA B
cos cos 2cos cos2 2
A B A BA B
cos cos 2sin sin2 2
A B A BA B
Differentiation
First Principles
0
f ( ) f ( )f ( ) = lim
h
x h xx
h
f(x) f (x)
tan kx k sec2 kx
seckx kseckx tankx
cotkx – kcosec2kx
cosec kx – kcosec kx cot kx
f( )
g( )
x
x
2
f ( ) g( ) f( ) g ( )
(g( ))
x x x x
x
6 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
Integration (+ constant)
f(x) f( ) d
x x
sec2 kx k
1 tan kx
tankx k
1ln seckx
cot kx k
1ln sinkx
coseckx 12
1 1ln cosec cot , ln tan( )kx kx kx
k k
seckx 1 12 4
1 1ln sec tan , ln tan( )kx kx kx
k k
xx
uvuvx
x
vu d
d
dd
d
d
Numerical solution of equations
The Newton-Raphson iteration for solving 0)f( x : )(f
)f(1
n
n
nnx
xxx
Statistics
Probability
P(A) = 1 – P(A)
P( ) P( ) P( ) P( )A B A B A B
P( ) P( )P( )A B A B | A
)P()|P()P()|P(
)P()|P()|P(
AABAAB
AABBA
For independent events A and B,
P(BA) = P(B), P(AB) = P(A),
P(A B) = P(A) P(B)
Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 7
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
Standard deviation
Standard deviation = (Variance)
Interquartile range = IQR = Q3 – Q1
For a set of n values 1 2, ,... ,...i nx x x x
n
xxxxS i
iixx
2
22)(
)(
Standard deviation =
2
2xxxS
xn n
or Discrete distributions
Distribution of X P(X = x) Mean Variance
Binomial ),B( pn (1 )x n xnp p
x
np )1( pnp
Sampling distributions
For a random sample of n observations from 2N( , )
~ N(0, 1)/
X
n
Statistical tables
The following statistical tables are required for A Level Mathematics:
Binomial Cumulative Distribution Function (see page 25)
Percentage Points Of The Normal Distribution (30)
Critical Values for Correlation Coefficients: Product Moment Coefficient (see page 33)
Random Numbers (see page 34)
Mechanics
Kinematics
For motion in a straight line with constant acceleration:
v = u + at
s = ut + ½ at2
s = vt - ½ at2
v2 = u2 + 2as
s = ½ (u + v)t
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4 AS Level in Further Mathematics
Students sitting a AS Level Further Mathematics paper may also require those formulae listed for
A Level Mathematics in Section 3.
Pure Mathematics
Summations
)12)(1(6
1
1
2
nnnrn
r
22
4
1
1
3 )1(
nnrn
r
Matrix transformations
Anticlockwise rotation through about O:
cos sin
sincos
Reflection in the line xy )(tan :
2cos2sin
2sin 2cos
Area of a sector
A = d
2
1 2r (polar coordinates)
Complex numbers
{ (cos i sin )} (cos i sin )n n
r r n n
The roots of 1nz are given by
2 i
ek
nz
, for 1 , ,2 ,1 ,0 nk
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Maclaurin’s and Taylor’s Series
)0(f!
)0(f!2
)0(f)0f()f( )(2
rr
r
xxxx
xr
xxxx
rx allfor
!
!21)exp(e
2
2 31ln (1 ) ( 1) ( 1 1)
2 3
rrx x x
x x xr
xr
xxxxx
rr allfor
)!12()1(
!5!3sin
1253
xr
xxxx
rr allfor
)!2()1(
!4!21cos
242
3 5 2 1
arctan ( 1) ( 1 1)3 5 2 1
rrx x x
x x xr
Vectors
Vector product:
1221
3113
2332
321
321ˆ sin
baba
baba
baba
bbb
aaa
kji
nbaba
)()()(
321
321
321
bac.acb.cba.
ccc
bbb
aaa
If A is the point with position vector kjia 321 aaa and the direction vector b is given by
kjib 321 bbb , then the straight line through A with direction vector b has cartesian equation
)( 3
3
2
2
1
1
b
az
b
ay
b
ax
The plane through A with normal vector kjin 321 nnn has cartesian equation
1 2 3 0 where n x n y n z d d a.n
The plane through non-collinear points A, B and C has vector equation
cbaacabar )1()()(
The plane through the point with position vector a and parallel to b and c has equation
cbar ts
The perpendicular distance of ) , ,( from 1 2 3 0n x n y n z d is 1 2 3
2 2 2
1 2 3
n n n d
n n n
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Hyperbolic functions
1sinhcosh 22 xx
xxx coshsinh22sinh
xxx 22 sinhcosh2cosh
)1( 1lnarcosh }{ 2 xxxx
}{ 1lnarsinh 2 xxx
12
1artanh ln ( 1)
1
xx x
x
Differentiation
f(x) f(x)
xarcsin 21
1
x
xarccos 21
1
x
xarctan 21
1
x
xsinh xcosh
xcosh xsinh
xtanh x2sech
xarsinh 21
1
x
xarcosh 1
1
2 x
artanh x 21
1
x
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Integration (+ constant; 0a where relevant)
f(x)
xx d)f(
xsinh xcosh
xcosh xsinh
xtanh xcoshln
22
1
xa
)( arcsin axa
x
22
1
xa
a
x
aarctan
1
22
1
ax
)( ln,arcosh }{ 22 axaxxa
x
22
1
xa
}{ 22ln,arsinh axxa
x
22
1
xa )( artanh
1ln
2
1ax
a
x
axa
xa
a
22
1
ax
ax
ax
a
ln
2
1
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Statistics
Discrete distributions
For a discrete random variable X taking values ix with probabilities P(X = xi)
Expectation (mean): E(X) = = ix P(X = ix )
Variance: Var(X) = 2 = ( ix – )2 P(X = ix ) = 2
ix P(X = ix ) – 2
Discrete distributions
Standard discrete distributions:
Distribution of X P(X = x) Mean Variance
Binomial B(n, p) 1n xx
np p
x
np np(1 – p)
Poisson Po ( ) e!
x
x
Continuous distributions
For a continuous random variable X having probability density function f
Expectation (mean): xxxX d)f()E(
Variance: 2222 d)f(d)f()()Var( xxxxxxX
For a function )g( X : xxxX d)f()g())E(g(
Cumulative distribution function:
0
0 0F( ) P( ) f ( ) d
x
x X x t t
Standard continuous distribution:
Distribution of X P.D.F. Mean Variance
Normal ) ,N( 2
2
21
e2
1
x
2
Uniform (Rectangular) on [a, b] ab
1
1
2( )a b
2
121 )( ab
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Correlation and regression
For a set of n pairs of values ) ,(ii
yx
22 2 ( )
S ( ) ixx i i
xx x x
n
22 2 ( )
S ( ) iyy i i
yy y y
n
( )( )S ( )( ) i i
xy i i i i
x yx x y y x y
n
The product moment correlation coefficient is:
2 2 2 22 2
( )( )S ( )( )
S S ( ) ( ) ( ) ( )
{ }{ }
i ii i
xy i i
xx yy i i i ii i
x yx y
x x y y nrx x y y x y
x yn n
The regression coefficient of y on x is 2
S ( )( )
S ( )
xy i i
xx i
x x y yb
x x
Least squares regression line of y on x is bxay where xbya
Residual Sum of Squares (RSS) =
2
2S
S S 1S
xy
yy yy
xx
r
Spearman’s rank correlation coefficient is
2
2
61
( 1)s
d
n nr
Non-parametric tests
Goodness-of-fit test and contingency tables: 2
2
~)(
i
ii
E
EO
Statistical tables
The following statistical tables are required for AS Level Further Mathematics:
Binomial Cumulative Distribution Function (see page 25)
Poisson Cumulative Distribution Function (see page 31)
Percentage Points of the 2 Distribution (see page 32)
Critical Values for Correlation Coefficients: Product Moment Coefficient and Spearman’s Coefficient
(see page 33)
Random Numbers (see page 34)
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Mechanics
Centres of mass
For uniform bodies:
Triangular lamina: 23
along median from vertex
Circular arc, radius r, angle at centre 2 :
sinr from centre
Sector of circle, radius r, angle at centre 2 : 2 sin
3
r
from centre
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5 A Level in Further Mathematics
Students sitting a A Level Further Mathematics paper may also require those formulae listed for
A Level Mathematics in Section 3.
Pure Mathematics
Summations
)12)(1(6
1
1
2
nnnrn
r
22
4
1
1
3 )1(
nnrn
r
Matrix transformations
Anticlockwise rotation through about O:
cos sin
sincos
Reflection in the line xy )(tan :
2cos2sin
2sin 2cos
Area of a sector
A = d
2
1 2r (polar coordinates)
Complex numbers
{ (cos i sin )} (cos i sin )n n
r r n n
The roots of 1nz are given by
2 i
ek
nz
, for 1 , ,2 ,1 ,0 nk
16 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics
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Maclaurin’s and Taylor’s Series
)0(f!
)0(f!2
)0(f)0f()f( )(2
rr
r
xxxx
xr
xxxx
rx allfor
!
!21)exp(e
2
2 31ln (1 ) ( 1) ( 1 1)
2 3
rrx x x
x x xr
xr
xxxxx
rr allfor
)!12()1(
!5!3sin
1253
xr
xxxx
rr allfor
)!2()1(
!4!21cos
242
3 5 2 1
arctan ( 1) ( 1 1)3 5 2 1
rrx x x
x x xr
Vectors
Vector product:
1221
3113
2332
321
321ˆ sin
baba
baba
baba
bbb
aaa
kji
nbaba
)()()(
321
321
321
bac.acb.cba.
ccc
bbb
aaa
If A is the point with position vector kjia 321 aaa and the direction vector b is given by
kjib 321 bbb , then the straight line through A with direction vector b has cartesian equation
)( 3
3
2
2
1
1
b
az
b
ay
b
ax
The plane through A with normal vector kjin 321 nnn has cartesian equation
1 2 3 0 where n x n y n z d d a.n
The plane through non-collinear points A, B and C has vector equation
cbaacabar )1()()(
The plane through the point with position vector a and parallel to b and c has equation
cbar ts The perpendicular distance of ) , ,( from 1 2 3 0n x n y n z d is
1 2 3
2 2 2
1 2 3
n n n d
n n n
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Hyperbolic functions
1sinhcosh 22 xx
xxx coshsinh22sinh
xxx 22 sinhcosh2cosh
)1( 1lnarcosh }{ 2 xxxx
}{ 1lnarsinh 2 xxx
12
1artanh ln ( 1)
1
xx x
x
Conics
Ellipse Parabola Hyperbola
Rectangular
Hyperbola
Standard
Form 1
2
2
2
2
b
y
a
x axy 42 1
2
2
2
2
b
y
a
x
2cxy
Parametric
Form )sin ,cos( ba )2 ,( 2 atat
(a sec , b tan )
(a cosh , b sinh )
t
cct,
Eccentricity 1e
)1( 222 eab 1e
1e
)1( 222 eab e = 2
Foci )0 ,( ae )0 ,(a )0 ,( ae ( 2c , 2c )
Directrices e
ax ax
e
ax x + y = 2c
Asymptotes none none b
y
a
x 0 ,0 yx
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Differentiation
f(x) f(x)
xarcsin 21
1
x
xarccos 21
1
x
xarctan 21
1
x
xsinh xcosh
xcosh xsinh
xtanh x2sech
xarsinh 21
1
x
xarcosh 1
1
2 x
artanh x 21
1
x
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Integration (+ constant; 0a where relevant)
f(x)
xx d)f(
xsinh xcosh
xcosh xsinh
xtanh xcoshln
22
1
xa
)( arcsin axa
x
22
1
xa
a
x
aarctan
1
22
1
ax
)( ln,arcosh }{ 22 axaxxa
x
22
1
xa
}{ 22ln,arsinh axxa
x
22
1
xa )( artanh
1ln
2
1ax
a
x
axa
xa
a
22
1
ax
ax
ax
a
ln
2
1
Arc length
xx
ys d
d
d1
2
(cartesian coordinates)
tt
y
t
xs d
d
d
d
d22
(parametric form)
2
2 drd
ds r
(polar form)
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Surface area of revolution
Sx =
2d
2 1 dd
yy x
x
(cartesian coordinates)
Sx =
2 2d d
2 dd d
x yy t
t t
(parametric form)
Sx =
2
2 dr2 sin d
dr r
(polar form)
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Statistics
Discrete distributions
For a discrete random variable X taking values ix with probabilities P(X = xi)
Expectation (mean): E(X) = = xi P(X = xi)
Variance: Var(X) = 2 = (xi – )2 P(X = xi) = 2
ix P(X = xi) – 2
For a function )g( X : E(g(X)) = g(xi) P(X = xi)
The probability generating function of X is G ( ) EX
Xt t and
2
E( ) G (1) and Var( ) G (1) G (1) G (1)X X X X
X X
For Z = X + Y, where X and Y are independent: G ( ) G ( ) G ( )Z X Y
t t t
Discrete distributions
Standard discrete distributions:
Distribution of X P(X = x) Mean Variance P.G.F.
Binomial B(n, p) 1n xx
np p
x
np np(1 – p) 1
np pt
Poisson Po ( ) e!
x
x
1
et
Geometric Geo(p)
on 1, 2, …
11
xp p
1
p
2
1 p
p
1 (1 )
pt
p t
Negative binomial
on r, r + 1, …
1(1 )
1
r x rx
p pr
r
p
2
(1 )r p
p
1 (1 )
r
pt
p t
Continuous distributions
For a continuous random variable X having probability density function f
Expectation (mean): xxxX d)f()E(
Variance: 2222 d)f(d)f()()Var( xxxxxxX
For a function )g( X : xxxX d)f()g())E(g(
Cumulative distribution function:
0
0 0F( ) P( ) f ( ) d
x
x X x t t
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Standard continuous distribution:
Distribution of X P.D.F. Mean Variance
Normal ) ,N( 2
2
21
e2
1
x
2
Uniform (Rectangular) on [a, b] ab
1
1
2( )a b
2
121 )( ab
Correlation and regression
For a set of n pairs of values ) ,(ii
yx
22 2 ( )
S ( ) ixx i i
xx x x
n
22 2 ( )
S ( ) iyy i i
yy y y
n
( )( )S ( )( ) i i
xy i i i i
x yx x y y x y
n
The product moment correlation coefficient is
2 2 2 22 2
( )( )S ( )( )
S S ( ) ( ) ( ) ( )
{ }{ }
i ii i
xy i i
xx yy i i i ii i
x yx y
x x y y nrx x y y x y
x yn n
The regression coefficient of y on x is 2
S ( )( )
S ( )
xy i i
xx i
x x y yb
x x
Least squares regression line of y on x is bxay where xbya
Residual Sum of Squares (RSS) =
2
2S
S S 1S
xy
yy yy
xx
r
Spearman’s rank correlation coefficient is
2
2
61
( 1)s
d
n nr
Expectation algebra
For independent random variables X and Y
E( ) E( ) E( )XY X Y , 2 2Var( ) Var( ) Var( )aX bY a X b Y
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Sampling distributions
(i) Tests for mean when is known
For a random sample n
XXX , , ,21 of n independent observations from a distribution having
mean and variance 2
:
X is an unbiased estimator of , with n
X2
)Var(
2S is an unbiased estimator of 2 , where
1
)( 2
2
n
XXS i
For a random sample of n observations from 2
N( , ) , ~ N(0, 1)/
X
n
For a random sample of xn observations from ) ,N( 2
xx and, independently, a random
sample of yn observations from ) ,N( 2
yy , )1 ,0N(~
)()(
22
y
y
x
x
yx
nn
YX
(ii) Tests for variance and mean when is not known
For a random sample of n observations from ) ,N( 2 :
2
12
2
~)1(
n
Sn
1~
/
n
tnS
X (also valid in matched-pairs situations)
For a random sample of xn observations from ) ,N( 2
xx and, independently, a random
sample of y
n observations from ) ,N( 2
yy
1 ,122
22
~/
/ ynxn
yy
xx FS
S
If 222
yx (unknown) then
2
2
( ) ( )~
1 1
x y
n nx y
p
x y
X Yt
Sn n
where
2 2
2( 1) ( 1)
2
x x y y
p
x y
n S n SS
n n
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Non-parametric tests
Goodness-of-fit test and contingency tables: 2
2
~)(
i
ii
E
EOStatistical tables
The following statistical tables are required for A Level Further Mathematics:
Binomial Cumulative Distribution Function (see page 25)
Percentage Points Of The Normal Distribution (see page 30)
Poisson Cumulative Distribution Function (see page 31)
Percentage Points of the 2 Distribution (see page 32)
Critical Values for Correlation Coefficients: Product Moment Coefficient and Spearman’s Coefficient
(see page 33)
Random Numbers (see page 34)
Percentage Points of Student’s t Distribution (see page 35)
Percentage Points of the F Distribution (see page 36
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Mechanics
Centres of mass
For uniform bodies:
Triangular lamina: 32
along median from vertex
Circular arc, radius r, angle at centre 2 :
sinr from centre
Sector of circle, radius r, angle at centre 2 : 2 sin
3
r
from centre
Solid hemisphere, radius r: r8
3 from centre
Hemispherical shell, radius r: r21
from centre
Solid cone or pyramid of height h: 14
h above the base on the line from centre of base to vertex
Conical shell of height h: 13
h above the base on the line from centre of base to vertex
Motion in a circle
Transverse velocity: rv
Transverse acceleration: rv
Radial acceleration: r
vr
22
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6 Statistical Tables
Binomial Cumulative Distribution Function
The tabulated value is P(X x), where X has a binomial distribution with index n and parameter p.
p = 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
n = 5, x = 0 0.7738 0.5905 0.4437 0.3277 0.2373 0.1681 0.1160 0.0778 0.0503 0.0312
1 0.9774 0.9185 0.8352 0.7373 0.6328 0.5282 0.4284 0.3370 0.2562 0.1875
2 0.9988 0.9914 0.9734 0.9421 0.8965 0.8369 0.7648 0.6826 0.5931 0.5000
3 1.0000 0.9995 0.9978 0.9933 0.9844 0.9692 0.9460 0.9130 0.8688 0.8125
4 1.0000 1.0000 0.9999 0.9997 0.9990 0.9976 0.9947 0.9898 0.9815 0.9688
n = 6, x = 0 0.7351 0.5314 0.3771 0.2621 0.1780 0.1176 0.0754 0.0467 0.0277 0.0156
1 0.9672 0.8857 0.7765 0.6554 0.5339 0.4202 0.3191 0.2333 0.1636 0.1094
2 0.9978 0.9842 0.9527 0.9011 0.8306 0.7443 0.6471 0.5443 0.4415 0.3438
3 0.9999 0.9987 0.9941 0.9830 0.9624 0.9295 0.8826 0.8208 0.7447 0.6563
4 1.0000 0.9999 0.9996 0.9984 0.9954 0.9891 0.9777 0.9590 0.9308 0.8906
5 1.0000 1.0000 1.0000 0.9999 0.9998 0.9993 0.9982 0.9959 0.9917 0.9844
n = 7, x = 0 0.6983 0.4783 0.3206 0.2097 0.1335 0.0824 0.0490 0.0280 0.0152 0.0078 1 0.9556 0.8503 0.7166 0.5767 0.4449 0.3294 0.2338 0.1586 0.1024 0.0625
2 0.9962 0.9743 0.9262 0.8520 0.7564 0.6471 0.5323 0.4199 0.3164 0.2266
3 0.9998 0.9973 0.9879 0.9667 0.9294 0.8740 0.8002 0.7102 0.6083 0.5000
4 1.0000 0.9998 0.9988 0.9953 0.9871 0.9712 0.9444 0.9037 0.8471 0.7734
5 1.0000 1.0000 0.9999 0.9996 0.9987 0.9962 0.9910 0.9812 0.9643 0.9375
6 1.0000 1.0000 1.0000 1.0000 0.9999 0.9998 0.9994 0.9984 0.9963 0.9922
n = 8, x = 0 0.6634 0.4305 0.2725 0.1678 0.1001 0.0576 0.0319 0.0168 0.0084 0.0039 1 0.9428 0.8131 0.6572 0.5033 0.3671 0.2553 0.1691 0.1064 0.0632 0.0352
2 0.9942 0.9619 0.8948 0.7969 0.6785 0.5518 0.4278 0.3154 0.2201 0.1445
3 0.9996 0.9950 0.9786 0.9437 0.8862 0.8059 0.7064 0.5941 0.4770 0.3633
4 1.0000 0.9996 0.9971 0.9896 0.9727 0.9420 0.8939 0.8263 0.7396 0.6367
5 1.0000 1.0000 0.9998 0.9988 0.9958 0.9887 0.9747 0.9502 0.9115 0.8555
6 1.0000 1.0000 1.0000 0.9999 0.9996 0.9987 0.9964 0.9915 0.9819 0.9648
7 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9998 0.9993 0.9983 0.9961
n = 9, x = 0 0.6302 0.3874 0.2316 0.1342 0.0751 0.0404 0.0207 0.0101 0.0046 0.0020 1 0.9288 0.7748 0.5995 0.4362 0.3003 0.1960 0.1211 0.0705 0.0385 0.0195
2 0.9916 0.9470 0.8591 0.7382 0.6007 0.4628 0.3373 0.2318 0.1495 0.0898
3 0.9994 0.9917 0.9661 0.9144 0.8343 0.7297 0.6089 0.4826 0.3614 0.2539
4 1.0000 0.9991 0.9944 0.9804 0.9511 0.9012 0.8283 0.7334 0.6214 0.5000
5 1.0000 0.9999 0.9994 0.9969 0.9900 0.9747 0.9464 0.9006 0.8342 0.7461
6 1.0000 1.0000 1.0000 0.9997 0.9987 0.9957 0.9888 0.9750 0.9502 0.9102
7 1.0000 1.0000 1.0000 1.0000 0.9999 0.9996 0.9986 0.9962 0.9909 0.9805
8 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9992 0.9980
n = 10, x = 0 0.5987 0.3487 0.1969 0.1074 0.0563 0.0282 0.0135 0.0060 0.0025 0.0010 1 0.9139 0.7361 0.5443 0.3758 0.2440 0.1493 0.0860 0.0464 0.0233 0.0107
2 0.9885 0.9298 0.8202 0.6778 0.5256 0.3828 0.2616 0.1673 0.0996 0.0547
3 0.9990 0.9872 0.9500 0.8791 0.7759 0.6496 0.5138 0.3823 0.2660 0.1719
4 0.9999 0.9984 0.9901 0.9672 0.9219 0.8497 0.7515 0.6331 0.5044 0.3770
5 1.0000 0.9999 0.9986 0.9936 0.9803 0.9527 0.9051 0.8338 0.7384 0.6230
6 1.0000 1.0000 0.9999 0.9991 0.9965 0.9894 0.9740 0.9452 0.8980 0.8281
7 1.0000 1.0000 1.0000 0.9999 0.9996 0.9984 0.9952 0.9877 0.9726 0.9453
8 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9995 0.9983 0.9955 0.9893
9 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9990
Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 27
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
p = 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
n = 12, x = 0 0.5404 0.2824 0.1422 0.0687 0.0317 0.0138 0.0057 0.0022 0.0008 0.0002 1 0.8816 0.6590 0.4435 0.2749 0.1584 0.0850 0.0424 0.0196 0.0083 0.0032
2 0.9804 0.8891 0.7358 0.5583 0.3907 0.2528 0.1513 0.0834 0.0421 0.0193
3 0.9978 0.9744 0.9078 0.7946 0.6488 0.4925 0.3467 0.2253 0.1345 0.0730
4 0.9998 0.9957 0.9761 0.9274 0.8424 0.7237 0.5833 0.4382 0.3044 0.1938
5 1.0000 0.9995 0.9954 0.9806 0.9456 0.8822 0.7873 0.6652 0.5269 0.3872
6 1.0000 0.9999 0.9993 0.9961 0.9857 0.9614 0.9154 0.8418 0.7393 0.6128
7 1.0000 1.0000 0.9999 0.9994 0.9972 0.9905 0.9745 0.9427 0.8883 0.8062
8 1.0000 1.0000 1.0000 0.9999 0.9996 0.9983 0.9944 0.9847 0.9644 0.9270
9 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9992 0.9972 0.9921 0.9807
10 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9989 0.9968
11 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9998
n = 15, x = 0 0.4633 0.2059 0.0874 0.0352 0.0134 0.0047 0.0016 0.0005 0.0001 0.0000 1 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353 0.0142 0.0052 0.0017 0.0005
2 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268 0.0617 0.0271 0.0107 0.0037
3 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969 0.1727 0.0905 0.0424 0.0176
4 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155 0.3519 0.2173 0.1204 0.0592
5 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216 0.5643 0.4032 0.2608 0.1509
6 1.0000 0.9997 0.9964 0.9819 0.9434 0.8689 0.7548 0.6098 0.4522 0.3036
7 1.0000 1.0000 0.9994 0.9958 0.9827 0.9500 0.8868 0.7869 0.6535 0.5000
8 1.0000 1.0000 0.9999 0.9992 0.9958 0.9848 0.9578 0.9050 0.8182 0.6964
9 1.0000 1.0000 1.0000 0.9999 0.9992 0.9963 0.9876 0.9662 0.9231 0.8491
10 1.0000 1.0000 1.0000 1.0000 0.9999 0.9993 0.9972 0.9907 0.9745 0.9408
11 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9995 0.9981 0.9937 0.9824
12 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9989 0.9963
13 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9995
14 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
n = 20, x = 0 0.3585 0.1216 0.0388 0.0115 0.0032 0.0008 0.0002 0.0000 0.0000 0.0000 1 0.7358 0.3917 0.1756 0.0692 0.0243 0.0076 0.0021 0.0005 0.0001 0.0000
2 0.9245 0.6769 0.4049 0.2061 0.0913 0.0355 0.0121 0.0036 0.0009 0.0002
3 0.9841 0.8670 0.6477 0.4114 0.2252 0.1071 0.0444 0.0160 0.0049 0.0013
4 0.9974 0.9568 0.8298 0.6296 0.4148 0.2375 0.1182 0.0510 0.0189 0.0059
5 0.9997 0.9887 0.9327 0.8042 0.6172 0.4164 0.2454 0.1256 0.0553 0.0207
6 1.0000 0.9976 0.9781 0.9133 0.7858 0.6080 0.4166 0.2500 0.1299 0.0577
7 1.0000 0.9996 0.9941 0.9679 0.8982 0.7723 0.6010 0.4159 0.2520 0.1316
8 1.0000 0.9999 0.9987 0.9900 0.9591 0.8867 0.7624 0.5956 0.4143 0.2517
9 1.0000 1.0000 0.9998 0.9974 0.9861 0.9520 0.8782 0.7553 0.5914 0.4119
10 1.0000 1.0000 1.0000 0.9994 0.9961 0.9829 0.9468 0.8725 0.7507 0.5881
11 1.0000 1.0000 1.0000 0.9999 0.9991 0.9949 0.9804 0.9435 0.8692 0.7483
12 1.0000 1.0000 1.0000 1.0000 0.9998 0.9987 0.9940 0.9790 0.9420 0.8684
13 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9985 0.9935 0.9786 0.9423
14 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9984 0.9936 0.9793
15 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9985 0.9941
16 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9987
17 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998
18 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
28 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
p = 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
n = 25, x = 0 0.2774 0.0718 0.0172 0.0038 0.0008 0.0001 0.0000 0.0000 0.0000 0.0000 1 0.6424 0.2712 0.0931 0.0274 0.0070 0.0016 0.0003 0.0001 0.0000 0.0000
2 0.8729 0.5371 0.2537 0.0982 0.0321 0.0090 0.0021 0.0004 0.0001 0.0000
3 0.9659 0.7636 0.4711 0.2340 0.0962 0.0332 0.0097 0.0024 0.0005 0.0001
4 0.9928 0.9020 0.6821 0.4207 0.2137 0.0905 0.0320 0.0095 0.0023 0.0005
5 0.9988 0.9666 0.8385 0.6167 0.3783 0.1935 0.0826 0.0294 0.0086 0.0020
6 0.9998 0.9905 0.9305 0.7800 0.5611 0.3407 0.1734 0.0736 0.0258 0.0073
7 1.0000 0.9977 0.9745 0.8909 0.7265 0.5118 0.3061 0.1536 0.0639 0.0216
8 1.0000 0.9995 0.9920 0.9532 0.8506 0.6769 0.4668 0.2735 0.1340 0.0539
9 1.0000 0.9999 0.9979 0.9827 0.9287 0.8106 0.6303 0.4246 0.2424 0.1148
10 1.0000 1.0000 0.9995 0.9944 0.9703 0.9022 0.7712 0.5858 0.3843 0.2122
11 1.0000 1.0000 0.9999 0.9985 0.9893 0.9558 0.8746 0.7323 0.5426 0.3450
12 1.0000 1.0000 1.0000 0.9996 0.9966 0.9825 0.9396 0.8462 0.6937 0.5000
13 1.0000 1.0000 1.0000 0.9999 0.9991 0.9940 0.9745 0.9222 0.8173 0.6550
14 1.0000 1.0000 1.0000 1.0000 0.9998 0.9982 0.9907 0.9656 0.9040 0.7878
15 1.0000 1.0000 1.0000 1.0000 1.0000 0.9995 0.9971 0.9868 0.9560 0.8852
16 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9992 0.9957 0.9826 0.9461
17 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9988 0.9942 0.9784
18 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9984 0.9927
19 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9996 0.9980
20 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9995
21 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999
22 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
n = 30, x = 0 0.2146 0.0424 0.0076 0.0012 0.0002 0.0000 0.0000 0.0000 0.0000 0.0000 1 0.5535 0.1837 0.0480 0.0105 0.0020 0.0003 0.0000 0.0000 0.0000 0.0000
2 0.8122 0.4114 0.1514 0.0442 0.0106 0.0021 0.0003 0.0000 0.0000 0.0000
3 0.9392 0.6474 0.3217 0.1227 0.0374 0.0093 0.0019 0.0003 0.0000 0.0000
4 0.9844 0.8245 0.5245 0.2552 0.0979 0.0302 0.0075 0.0015 0.0002 0.0000
5 0.9967 0.9268 0.7106 0.4275 0.2026 0.0766 0.0233 0.0057 0.0011 0.0002
6 0.9994 0.9742 0.8474 0.6070 0.3481 0.1595 0.0586 0.0172 0.0040 0.0007
7 0.9999 0.9922 0.9302 0.7608 0.5143 0.2814 0.1238 0.0435 0.0121 0.0026
8 1.0000 0.9980 0.9722 0.8713 0.6736 0.4315 0.2247 0.0940 0.0312 0.0081
9 1.0000 0.9995 0.9903 0.9389 0.8034 0.5888 0.3575 0.1763 0.0694 0.0214
10 1.0000 0.9999 0.9971 0.9744 0.8943 0.7304 0.5078 0.2915 0.1350 0.0494
11 1.0000 1.0000 0.9992 0.9905 0.9493 0.8407 0.6548 0.4311 0.2327 0.1002
12 1.0000 1.0000 0.9998 0.9969 0.9784 0.9155 0.7802 0.5785 0.3592 0.1808
13 1.0000 1.0000 1.0000 0.9991 0.9918 0.9599 0.8737 0.7145 0.5025 0.2923
14 1.0000 1.0000 1.0000 0.9998 0.9973 0.9831 0.9348 0.8246 0.6448 0.4278
15 1.0000 1.0000 1.0000 0.9999 0.9992 0.9936 0.9699 0.9029 0.7691 0.5722
16 1.0000 1.0000 1.0000 1.0000 0.9998 0.9979 0.9876 0.9519 0.8644 0.7077
17 1.0000 1.0000 1.0000 1.0000 0.9999 0.9994 0.9955 0.9788 0.9286 0.8192
18 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9986 0.9917 0.9666 0.8998
19 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.9971 0.9862 0.9506
20 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9991 0.9950 0.9786
21 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9984 0.9919
22 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.9974
23 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9993
24 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998
25 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 29
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
p = 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
n = 40, x = 0 0.1285 0.0148 0.0015 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
1 0.3991 0.0805 0.0121 0.0015 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000
2 0.6767 0.2228 0.0486 0.0079 0.0010 0.0001 0.0000 0.0000 0.0000 0.0000
3 0.8619 0.4231 0.1302 0.0285 0.0047 0.0006 0.0001 0.0000 0.0000 0.0000
4 0.9520 0.6290 0.2633 0.0759 0.0160 0.0026 0.0003 0.0000 0.0000 0.0000
5 0.9861 0.7937 0.4325 0.1613 0.0433 0.0086 0.0013 0.0001 0.0000 0.0000
6 0.9966 0.9005 0.6067 0.2859 0.0962 0.0238 0.0044 0.0006 0.0001 0.0000
7 0.9993 0.9581 0.7559 0.4371 0.1820 0.0553 0.0124 0.0021 0.0002 0.0000
8 0.9999 0.9845 0.8646 0.5931 0.2998 0.1110 0.0303 0.0061 0.0009 0.0001
9 1.0000 0.9949 0.9328 0.7318 0.4395 0.1959 0.0644 0.0156 0.0027 0.0003
10 1.0000 0.9985 0.9701 0.8392 0.5839 0.3087 0.1215 0.0352 0.0074 0.0011
11 1.0000 0.9996 0.9880 0.9125 0.7151 0.4406 0.2053 0.0709 0.0179 0.0032
12 1.0000 0.9999 0.9957 0.9568 0.8209 0.5772 0.3143 0.1285 0.0386 0.0083
13 1.0000 1.0000 0.9986 0.9806 0.8968 0.7032 0.4408 0.2112 0.0751 0.0192
14 1.0000 1.0000 0.9996 0.9921 0.9456 0.8074 0.5721 0.3174 0.1326 0.0403
15 1.0000 1.0000 0.9999 0.9971 0.9738 0.8849 0.6946 0.4402 0.2142 0.0769
16 1.0000 1.0000 1.0000 0.9990 0.9884 0.9367 0.7978 0.5681 0.3185 0.1341
17 1.0000 1.0000 1.0000 0.9997 0.9953 0.9680 0.8761 0.6885 0.4391 0.2148
18 1.0000 1.0000 1.0000 0.9999 0.9983 0.9852 0.9301 0.7911 0.5651 0.3179
19 1.0000 1.0000 1.0000 1.0000 0.9994 0.9937 0.9637 0.8702 0.6844 0.4373
20 1.0000 1.0000 1.0000 1.0000 0.9998 0.9976 0.9827 0.9256 0.7870 0.5627
21 1.0000 1.0000 1.0000 1.0000 1.0000 0.9991 0.9925 0.9608 0.8669 0.6821
22 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9970 0.9811 0.9233 0.7852
23 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9989 0.9917 0.9595 0.8659
24 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.9966 0.9804 0.9231
25 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9988 0.9914 0.9597
26 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.9966 0.9808
27 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9988 0.9917
28 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.9968
29 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9989
30 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997
31 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999
32 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
30 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
p = 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
n = 50, x = 0 0.0769 0.0052 0.0003 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
1 0.2794 0.0338 0.0029 0.0002 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
2 0.5405 0.1117 0.0142 0.0013 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000
3 0.7604 0.2503 0.0460 0.0057 0.0005 0.0000 0.0000 0.0000 0.0000 0.0000
4 0.8964 0.4312 0.1121 0.0185 0.0021 0.0002 0.0000 0.0000 0.0000 0.0000
5 0.9622 0.6161 0.2194 0.0480 0.0070 0.0007 0.0001 0.0000 0.0000 0.0000
6 0.9882 0.7702 0.3613 0.1034 0.0194 0.0025 0.0002 0.0000 0.0000 0.0000
7 0.9968 0.8779 0.5188 0.1904 0.0453 0.0073 0.0008 0.0001 0.0000 0.0000
8 0.9992 0.9421 0.6681 0.3073 0.0916 0.0183 0.0025 0.0002 0.0000 0.0000
9 0.9998 0.9755 0.7911 0.4437 0.1637 0.0402 0.0067 0.0008 0.0001 0.0000
10 1.0000 0.9906 0.8801 0.5836 0.2622 0.0789 0.0160 0.0022 0.0002 0.0000
11 1.0000 0.9968 0.9372 0.7107 0.3816 0.1390 0.0342 0.0057 0.0006 0.0000
12 1.0000 0.9990 0.9699 0.8139 0.5110 0.2229 0.0661 0.0133 0.0018 0.0002
13 1.0000 0.9997 0.9868 0.8894 0.6370 0.3279 0.1163 0.0280 0.0045 0.0005
14 1.0000 0.9999 0.9947 0.9393 0.7481 0.4468 0.1878 0.0540 0.0104 0.0013
15 1.0000 1.0000 0.9981 0.9692 0.8369 0.5692 0.2801 0.0955 0.0220 0.0033
16 1.0000 1.0000 0.9993 0.9856 0.9017 0.6839 0.3889 0.1561 0.0427 0.0077
17 1.0000 1.0000 0.9998 0.9937 0.9449 0.7822 0.5060 0.2369 0.0765 0.0164
18 1.0000 1.0000 0.9999 0.9975 0.9713 0.8594 0.6216 0.3356 0.1273 0.0325
19 1.0000 1.0000 1.0000 0.9991 0.9861 0.9152 0.7264 0.4465 0.1974 0.0595
20 1.0000 1.0000 1.0000 0.9997 0.9937 0.9522 0.8139 0.5610 0.2862 0.1013
21 1.0000 1.0000 1.0000 0.9999 0.9974 0.9749 0.8813 0.6701 0.3900 0.1611
22 1.0000 1.0000 1.0000 1.0000 0.9990 0.9877 0.9290 0.7660 0.5019 0.2399
23 1.0000 1.0000 1.0000 1.0000 0.9996 0.9944 0.9604 0.8438 0.6134 0.3359
24 1.0000 1.0000 1.0000 1.0000 0.9999 0.9976 0.9793 0.9022 0.7160 0.4439
25 1.0000 1.0000 1.0000 1.0000 1.0000 0.9991 0.9900 0.9427 0.8034 0.5561
26 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9955 0.9686 0.8721 0.6641
27 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9981 0.9840 0.9220 0.7601
28 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9993 0.9924 0.9556 0.8389
29 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9966 0.9765 0.8987
30 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9986 0.9884 0.9405
31 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9995 0.9947 0.9675
32 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9978 0.9836
33 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9991 0.9923
34 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9967
35 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9987
36 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9995
37 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998
38 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 31
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
Percentage Points Of The Normal Distribution
The values z in the table are those which a random variable Z N(0, 1) exceeds with probability p;
that is, P(Z > z) = 1 (z) = p.
p z p z
0.5000 0.0000 0.0500 1.6449
0.4000 0.2533 0.0250 1.9600
0.3000 0.5244 0.0100 2.3263
0.2000 0.8416 0.0050 2.5758
0.1500 1.0364 0.0010 3.0902
0.1000 1.2816 0.0005 3.2905
32 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
Poisson Cumulative Distribution Function
The tabulated value is P(X x), where X has a Poisson distribution with parameter .
= 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
x = 0 0.6065 0.3679 0.2231 0.1353 0.0821 0.0498 0.0302 0.0183 0.0111 0.0067
1 0.9098 0.7358 0.5578 0.4060 0.2873 0.1991 0.1359 0.0916 0.0611 0.0404
2 0.9856 0.9197 0.8088 0.6767 0.5438 0.4232 0.3208 0.2381 0.1736 0.1247
3 0.9982 0.9810 0.9344 0.8571 0.7576 0.6472 0.5366 0.4335 0.3423 0.2650
4 0.9998 0.9963 0.9814 0.9473 0.8912 0.8153 0.7254 0.6288 0.5321 0.4405
5 1.0000 0.9994 0.9955 0.9834 0.9580 0.9161 0.8576 0.7851 0.7029 0.6160
6 1.0000 0.9999 0.9991 0.9955 0.9858 0.9665 0.9347 0.8893 0.8311 0.7622
7 1.0000 1.0000 0.9998 0.9989 0.9958 0.9881 0.9733 0.9489 0.9134 0.8666
8 1.0000 1.0000 1.0000 0.9998 0.9989 0.9962 0.9901 0.9786 0.9597 0.9319
9 1.0000 1.0000 1.0000 1.0000 0.9997 0.9989 0.9967 0.9919 0.9829 0.9682
10 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9990 0.9972 0.9933 0.9863
11 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9991 0.9976 0.9945
12 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9992 0.9980
13 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9993
14 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9998
15 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999
16 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
17 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
18 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
19 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
= 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0
x = 0 0.0041 0.0025 0.0015 0.0009 0.0006 0.0003 0.0002 0.0001 0.0001 0.0000
1 0.0266 0.0174 0.0113 0.0073 0.0047 0.0030 0.0019 0.0012 0.0008 0.0005
2 0.0884 0.0620 0.0430 0.0296 0.0203 0.0138 0.0093 0.0062 0.0042 0.0028
3 0.2017 0.1512 0.1118 0.0818 0.0591 0.0424 0.0301 0.0212 0.0149 0.0103
4 0.3575 0.2851 0.2237 0.1730 0.1321 0.0996 0.0744 0.0550 0.0403 0.0293
5 0.5289 0.4457 0.3690 0.3007 0.2414 0.1912 0.1496 0.1157 0.0885 0.0671
6 0.6860 0.6063 0.5265 0.4497 0.3782 0.3134 0.2562 0.2068 0.1649 0.1301
7 0.8095 0.7440 0.6728 0.5987 0.5246 0.4530 0.3856 0.3239 0.2687 0.2202
8 0.8944 0.8472 0.7916 0.7291 0.6620 0.5925 0.5231 0.4557 0.3918 0.3328
9 0.9462 0.9161 0.8774 0.8305 0.7764 0.7166 0.6530 0.5874 0.5218 0.4579
10 0.9747 0.9574 0.9332 0.9015 0.8622 0.8159 0.7634 0.7060 0.6453 0.5830
11 0.9890 0.9799 0.9661 0.9467 0.9208 0.8881 0.8487 0.8030 0.7520 0.6968
12 0.9955 0.9912 0.9840 0.9730 0.9573 0.9362 0.9091 0.8758 0.8364 0.7916
13 0.9983 0.9964 0.9929 0.9872 0.9784 0.9658 0.9486 0.9261 0.8981 0.8645
14 0.9994 0.9986 0.9970 0.9943 0.9897 0.9827 0.9726 0.9585 0.9400 0.9165
15 0.9998 0.9995 0.9988 0.9976 0.9954 0.9918 0.9862 0.9780 0.9665 0.9513
16 0.9999 0.9998 0.9996 0.9990 0.9980 0.9963 0.9934 0.9889 0.9823 0.9730
17 1.0000 0.9999 0.9998 0.9996 0.9992 0.9984 0.9970 0.9947 0.9911 0.9857
18 1.0000 1.0000 0.9999 0.9999 0.9997 0.9993 0.9987 0.9976 0.9957 0.9928
19 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9995 0.9989 0.9980 0.9965
20 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9998 0.9996 0.9991 0.9984
21 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9998 0.9996 0.9993
22 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9999 0.9997
Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 33
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
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Percentage Points of the 2 Distribution
The values in the table are those which a random variable with the 2 distribution on degrees of
freedom exceeds with the probability shown.
0.995 0.990 0.975 0.950 0.900 0.100 0.050 0.025 0.010 0.005
1 0.000 0.000 0.001 0.004 0.016 2.705 3.841 5.024 6.635 7.879
2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.597
3 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838
4 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.860
5 0.412 0.554 0.831 1.145 1.610 9.236 11.070 12.832 15.086 16.750
6 0.676 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.812 18.548
7 0.989 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.475 20.278
8 1.344 1.646 2.180 2.733 3.490 13.362 15.507 17.535 20.090 21.955
9 1.735 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666 23.589
10 2.156 2.558 3.247 3.940 4.865 15.987 18.307 20.483 23.209 25.188
11 2.603 3.053 3.816 4.575 5.580 17.275 19.675 21.920 24.725 26.757
12 3.074 3.571 4.404 5.226 6.304 18.549 21.026 23.337 26.217 28.300
13 3.565 4.107 5.009 5.892 7.042 19.812 22.362 24.736 27.688 29.819
14 4.075 4.660 5.629 6.571 7.790 21.064 23.685 26.119 29.141 31.319
15 4.601 5.229 6.262 7.261 8.547 22.307 24.996 27.488 30.578 32.801
16 5.142 5.812 6.908 7.962 9.312 23.542 26.296 28.845 32.000 34.267
17 5.697 6.408 7.564 8.672 10.085 24.769 27.587 30.191 33.409 35.718
18 6.265 7.015 8.231 9.390 10.865 25.989 28.869 31.526 34.805 37.156
19 6.844 7.633 8.907 10.117 11.651 27.204 30.144 32.852 36.191 38.582
20 7.434 8.260 9.591 10.851 12.443 28.412 31.410 34.170 37.566 39.997
21 8.034 8.897 10.283 11.591 13.240 29.615 32.671 35.479 38.932 41.401
22 8.643 9.542 10.982 12.338 14.042 30.813 33.924 36.781 40.289 42.796
23 9.260 10.196 11.689 13.091 14.848 32.007 35.172 38.076 41.638 44.181
24 9.886 10.856 12.401 13.848 15.659 33.196 36.415 39.364 42.980 45.558
25 10.520 11.524 13.120 14.611 16.473 34.382 37.652 40.646 44.314 46.928
26 11.160 12.198 13.844 15.379 17.292 35.563 38.885 41.923 45.642 48.290
27 11.808 12.879 14.573 16.151 18.114 36.741 40.113 43.194 46.963 49.645
28 12.461 13.565 15.308 16.928 18.939 37.916 41.337 44.461 48.278 50.993
29 13.121 14.256 16.047 17.708 19.768 39.088 42.557 45.722 49.588 52.336
30 13.787 14.953 16.791 18.493 20.599 40.256 43.773 46.979 50.892 53.672
34 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
Critical Values for Correlation Coefficients
These tables concern tests of the hypothesis that a population correlation coefficient is 0. The values
in the tables are the minimum values which need to be reached by a sample correlation coefficient in order to be significant at the level shown, on a one-tailed test.
Product Moment Coefficient Sample
Size, n
Spearman’s Coefficient
0.10 0.05 Level
0.025 0.01 0.005 0.05
Level
0.025 0.01
0.8000 0.9000 0.9500 0.9800 0.9900 4 1.0000 - -
0.6870 0.8054 0.8783 0.9343 0.9587 5
0.9000 1.0000 1.0000
0.6084 0.7293 0.8114 0.8822 0.9172 6 0.8286 0.8857 0.9429
0.5509 0.6694 0.7545 0.8329 0.8745 7 0.7143 0.7857 0.8929
0.5067 0.6215 0.7067 0.7887 0.8343 8 0.6429 0.7381 0.8333
0.4716 0.5822 0.6664 0.7498 0.7977 9 0.6000 0.7000 0.7833
0.4428 0.5494 0.6319 0.7155 0.7646 10
0.5636 0.6485 0.7455
0.4187 0.5214 0.6021 0.6851 0.7348 11 0.5364 0.6182 0.7091
0.3981 0.4973 0.5760 0.6581 0.7079 12 0.5035 0.5874 0.6783
0.3802 0.4762 0.5529 0.6339 0.6835 13 0.4835 0.5604 0.6484
0.3646 0.4575 0.5324 0.6120 0.6614 14 0.4637 0.5385 0.6264
0.3507 0.4409 0.5140 0.5923 0.6411 15
0.4464 0.5214 0.6036
0.3383 0.4259 0.4973 0.5742 0.6226 16 0.4294 0.5029 0.5824
0.3271 0.4124 0.4821 0.5577 0.6055 17 0.4142 0.4877 0.5662
0.3170 0.4000 0.4683 0.5425 0.5897 18 0.4014 0.4716 0.5501
0.3077 0.3887 0.4555 0.5285 0.5751 19 0.3912 0.4596 0.5351
0.2992 0.3783 0.4438 0.5155 0.5614 20
0.3805 0.4466 0.5218
0.2914 0.3687 0.4329 0.5034 0.5487 21 0.3701 0.4364 0.5091
0.2841 0.3598 0.4227 0.4921 0.5368 22 0.3608 0.4252 0.4975
0.2774 0.3515 0.4133 0.4815 0.5256 23 0.3528 0.4160 0.4862
0.2711 0.3438 0.4044 0.4716 0.5151 24 0.3443 0.4070 0.4757
0.2653 0.3365 0.3961 0.4622 0.5052 25
0.3369 0.3977 0.4662
0.2598 0.3297 0.3882 0.4534 0.4958 26 0.3306 0.3901 0.4571
0.2546 0.3233 0.3809 0.4451 0.4869 27 0.3242 0.3828 0.4487
0.2497 0.3172 0.3739 0.4372 0.4785 28 0.3180 0.3755 0.4401
0.2451 0.3115 0.3673 0.4297 0.4705 29 0.3118 0.3685 0.4325
0.2407 0.3061 0.3610 0.4226 0.4629 30
0.3063 0.3624 0.4251
0.2070 0.2638 0.3120 0.3665 0.4026 40 0.2640 0.3128 0.3681
0.1843 0.2353 0.2787 0.3281 0.3610 50 0.2353 0.2791 0.3293
0.1678 0.2144 0.2542 0.2997 0.3301 60 0.2144 0.2545 0.3005
0.1550 0.1982 0.2352 0.2776 0.3060 70 0.1982 0.2354 0.2782
0.1448 0.1852 0.2199 0.2597 0.2864 80
0.1852 0.2201 0.2602
0.1364 0.1745 0.2072 0.2449 0.2702 90 0.1745 0.2074 0.2453
0.1292 0.1654 0.1966 0.2324 0.2565 100 0.1654 0.1967 0.2327
Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 35
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
Random Numbers
86 13 84 10 07 30 39 05 97 96 88 07 37 26 04 89 13 48 19 20
60 78 48 12 99 47 09 46 91 33 17 21 03 94 79 00 08 50 40 16
78 48 06 37 82 26 01 06 64 65 94 41 17 26 74 66 61 93 24 97
80 56 90 79 66 94 18 40 97 79 93 20 41 51 25 04 20 71 76 04
99 09 39 25 66 31 70 56 30 15 52 17 87 55 31 11 10 68 98 23
56 32 32 72 91 65 97 36 56 61 12 79 95 17 57 16 53 58 96 36
66 02 49 93 97 44 99 15 56 86 80 57 11 78 40 23 58 40 86 14
31 77 53 94 05 93 56 14 71 23 60 46 05 33 23 72 93 10 81 23
98 79 72 43 14 76 54 77 66 29 84 09 88 56 75 86 41 67 04 42
50 97 92 15 10 01 57 01 87 33 73 17 70 18 40 21 24 20 66 62
90 51 94 50 12 48 88 95 09 34 09 30 22 27 25 56 40 76 01 59
31 99 52 24 13 43 27 88 11 39 41 65 00 84 13 06 31 79 74 97
22 96 23 34 46 12 67 11 48 06 99 24 14 83 78 37 65 73 39 47
06 84 55 41 27 06 74 59 14 29 20 14 45 75 31 16 05 41 22 96
08 64 89 30 25 25 71 35 33 31 04 56 12 67 03 74 07 16 49 32
86 87 62 43 15 11 76 49 79 13 78 80 93 89 09 57 07 14 40 74
94 44 97 13 77 04 35 02 12 76 60 91 93 40 81 06 85 85 72 84
63 25 55 14 66 47 99 90 02 90 83 43 16 01 19 69 11 78 87 16
11 22 83 98 15 21 18 57 53 42 91 91 26 52 89 13 86 00 47 61
01 70 10 83 94 71 13 67 11 12 36 54 53 32 90 43 79 01 95 15
36 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
Percentage Points of Student’s t Distribution
The values in the table are those which a random variable with Student’s t distribution on degrees
of freedom exceeds with the probability shown.
0.10 0.05 0.025 0.01 0.005
1 3.078 6.314 12.706 31.821 63.657
2 1.886 2.920 4.303 6.965 9.925
3 1.638 2.353 3.182 4.541 5.841
4 1.533 2.132 2.776 3.747 4.604
5 1.476 2.015 2.571 3.365 4.032
6 1.440 1.943 2.447 3.143 3.707
7 1.415 1.895 2.365 2.998 3.499
8 1.397 1.860 2.306 2.896 3.355
9 1.383 1.833 2.262 2.821 3.250
10 1.372 1.812 2.228 2.764 3.169
11 1.363 1.796 2.201 2.718 3.106
12 1.356 1.782 2.179 2.681 3.055
13 1.350 1.771 2.160 2.650 3.012
14 1.345 1.761 2.145 2.624 2.977
15 1.341 1.753 2.131 2.602 2.947
16 1.337 1.746 2.120 2.583 2.921
17 1.333 1.740 2.110 2.567 2.898
18 1.330 1.734 2.101 2.552 2.878
19 1.328 1.729 2.093 2.539 2.861
20 1.325 1.725 2.086 2.528 2.845
21 1.323 1.721 2.080 2.518 2.831
22 1.321 1.717 2.074 2.508 2.819
23 1.319 1.714 2.069 2.500 2.807
24 1.318 1.711 2.064 2.492 2.797
25 1.316 1.708 2.060 2.485 2.787
26 1.315 1.706 2.056 2.479 2.779
27 1.314 1.703 2.052 2.473 2.771
28 1.313 1.701 2.048 2.467 2.763
29 1.311 1.699 2.045 2.462 2.756
30 1.310 1.697 2.042 2.457 2.750
32 1.309 1.694 2.037 2.449 2.738
34 1.307 1.691 2.032 2.441 2.728
36 1.306 1.688 2.028 2.435 2.719
38 1.304 1.686 2.024 2.429 2.712
40 1.303 1.684 2.021 2.423 2.704
45 1.301 1.679 2.014 2.412 2.690
50 1.299 1.676 2.009 2.403 2.678
55 1.297 1.673 2.004 2.396 2.668
60 1.296 1.671 2.000 2.390 2.660
70 1.294 1.667 1.994 2.381 2.648
80 1.292 1.664 1.990 2.374 2.639
90 1.291 1.662 1.987 2.369 2.632
100 1.290 1.660 1.984 2.364 2.626
110 1.289 1.659 1.982 2.361 2.621
120 1.289 1.658 1.980 2.358 2.617
Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 37
Mathematical Formulae and Statistical Tables – Issue 1 – Accredited pre-publication - April 2017
© Pearson Education Limited 2017
Percentage Points of the F Distribution
The values in the table are those which a random variable with the F distribution on 1 and 2
degrees of freedom exceeds with probability 0.05 or 0.01.
Probability
1
2
1 2 3 4 5 6 8 10 12 24
0.05
1 161.4 199.5 215.7 224.6 230.2 234.0 238.9 241.9 243.9 249.1 254.3
2 18.51 19.00 19.16 19.25 19.30 19.33 19.37 19.40 19.41 19.46 19.50
3 10.13 9.55 9.28 9.12 9.01 8.94 8.85 8.79 8.74 8.64 8.53
4 7.71 6.94 6.59 6.39 6.26 6.16 6.04 5.96 5.91 5.77 5.63
5 6.61 5.79 5.41 5.19 5.05 4.95 4.82 4.74 4.68 4.53 4.37
6 5.99 5.14 4.76 4.53 4.39 4.28 4.15 4.06 4.00 3.84 3.67
7 5.59 4.74 4.35 4.12 3.97 3.87 3.73 3.64 3.57 3.41 3.23
8 5.32 4.46 4.07 3.84 3.69 3.58 3.44 3.35 3.28 3.12 2.93
9 5.12 4.26 3.86 3.63 3.48 3.37 3.23 3.14 3.07 2.90 2.71
10 4.96 4.10 3.71 3.48 3.33 3.22 3.07 2.98 2.91 2.74 2.54
11 4.84 3.98 3.59 3.36 3.20 3.09 2.95 2.85 2.79 2.61 2.40
12 4.75 3.89 3.49 3.26 3.11 3.00 2.85 2.75 2.69 2.51 2.30
14 4.60 3.74 3.34 3.11 2.96 2.85 2.70 2.60 2.53 2.35 2.13
16 4.49 3.63 3.24 3.01 2.85 2.74 2.59 2.49 2.42 2.24 2.01
18 4.41 3.55 3.16 2.93 2.77 2.66 2.51 2.41 2.34 2.15 1.92
20 4.35 3.49 3.10 2.87 2.71 2.60 2.45 2.35 2.28 2.08 1.84
25 4.24 3.39 2.99 2.76 2.60 2.49 2.34 2.24 2.16 1.96 1.71
30 4.17 3.32 2.92 2.69 2.53 2.42 2.27 2.16 2.09 1.89 1.62
40 4.08 3.23 2.84 2.61 2.45 2.34 2.18 2.08 2.00 1.79 1.51
60 4.00 3.15 2.76 2.53 2.37 2.25 2.10 1.99 1.92 1.70 1.39
120 3.92 3.07 2.68 2.45 2.29 2.18 2.02 1.91 1.83 1.61 1.25
3.84 3.00 2.60 2.37 2.21 2.10 1.94 1.83 1.75 1.52 1.00
0.01
1 4052. 5000. 5403. 5625. 5764. 5859. 5982. 6056. 6106. 6235. 6366.
2 98.50 99.00 99.17 99.25 99.30 99.33 99.37 99.40 99.42 99.46 99.50
3 34.12 30.82 29.46 28.71 28.24 27.91 27.49 27.23 27.05 26.60 26.13
4 21.20 18.00 16.69 15.98 15.52 15.21 14.80 14.55 14.37 13.93 13.45
5 16.26 13.27 12.06 11.39 10.97 10.67 10.29 10.05 9.89 9.47 9.02
6 13.70 10.90 9.78 9.15 8.75 8.47 8.10 7.87 7.72 7.31 6.88
7 12.20 9.55 8.45 7.85 7.46 7.19 6.84 6.62 6.47 6.07 5.65
8 11.30 8.65 7.59 7.01 6.63 6.37 6.03 5.81 5.67 5.28 4.86
9 10.60 8.02 6.99 6.42 6.06 5.80 5.47 5.26 5.11 4.73 4.31
10 10.00 7.56 6.55 5.99 5.64 5.39 5.06 4.85 4.17 4.33 3.91
11 9.65 7.21 6.22 5.67 5.32 5.07 4.74 4.54 4.40 4.02 3.60
12 9.33 6.93 5.95 5.41 5.06 4.82 4.50 4.30 4.16 3.78 3.36
14 8.86 6.51 5.56 5.04 4.70 4.46 4.14 3.94 3.80 3.43 3.00
16 8.53 6.23 5.29 4.77 4.44 4.20 3.89 3.69 3.55 3.18 2.75
18 8.29 6.01 5.09 4.58 4.25 4.01 3.71 3.51 3.37 3.00 2.57
20 8.10 5.85 4.94 4.43 4.10 3.87 3.56 3.37 3.23 2.86 2.42
25 7.77 5.57 4.68 4.18 3.86 3.63 3.32 3.13 2.99 2.62 2.17
30 7.56 5.39 4.51 4.02 3.70 3.47 3.17 2.98 2.84 2.47 2.01
40 7.31 5.18 4.31 3.83 3.51 3.29 2.99 2.80 2.66 2.29 1.80
60 7.08 4.98 4.13 3.65 3.34 3.12 2.82 2.63 2.50 2.12 1.60
120 6.85 4.79 3.95 3.48 3.17 2.96 2.66 2.47 2.34 1.95 1.38
6.63 4.61 3.78 3.32 3.02 2.80 2.51 2.32 2.18 1.79 1.00
If an upper percentage point of the F distribution on 1 and 2 degrees of freedom is f , then the corresponding
lower percentage point of the F distribution on 2 and 1 degrees of freedom is 1/ f.
38 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics
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© Pearson Education Limited 2017