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A Level Further Mathematics Sample Assessment Materials DRAFT Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) First teaching from September 2017 First certification from 2019 This draſt qualification has not yet been accredited by Ofqual. It is published to enable teachers to have early sight of our proposed approach to Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0). Further changes may be required and no assurance can be given at this time that the proposed qualification will be made available in its current form, or that it will be accredited in time for first teaching in September 2017 and first award in 2019. DRAFT S u b j e c t t o O f q u a l a c c r e d i t a t i o n

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Page 1: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

A Level Further Mathematics

Sample Assessment Materials DRAFTPearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0)First teaching from September 2017First certification from 2019

This draft qualification has not yet been accredited by Ofqual. It is published to enable teachers to have early sight of our proposed approach to Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0). Further changes may be required and no assurance can be given at this time that the proposed qualification will be made available in its current form, or that it will be accredited in time for first teaching in September 2017 and first award in 2019.

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Page 2: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

Edexcel, BTEC and LCCI qualifications Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding body offering academic and vocational qualifications that are globally recognised and benchmarked. For further information, please visit our qualification websites at www.edexcel.com, www.btec.co.uk or www.lcci.org.uk. Alternatively, you can get in touch with us using the details on our contact us page at qualifications.pearson.com/contactus

About Pearson Pearson is the world's leading learning company, with 40,000 employees in more than 70 countries working to help people of all ages to make measurable progress in their lives through learning. We put the learner at the centre of everything we do, because wherever learning flourishes, so do people. Find out more about how we can help you and your learners at qualifications.pearson.com References to third party material made in this sample assessment materials are made in good faith. Pearson does not endorse, approve or accept responsibility for the content of materials, which may be subject to change, or any opinions expressed therein. (Material may include textbooks, journals, magazines and other publications and websites.) 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 2016

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Page 3: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

Contents

Introduction 1

General marking guidance 3

Paper 1 – sample question paper and mark scheme 5

Paper 2 – sample question paper and mark scheme 31

Paper 3A – sample question paper and mark scheme 57

Paper 4A – sample question paper and mark scheme 83

Paper 3B/4B – sample question paper and mark scheme 121

Paper 4C – sample question paper and mark scheme 155

Paper 3C/4D – sample question paper and mark scheme 183

Paper 4E – sample question paper and mark scheme 217

Paper 3D/4F – sample question paper, answer book, mark scheme 251

Paper 4G – sample question paper, answer book, mark scheme 285

Mathematical formulae and statistical tables

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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) – Draft 1.0 – June 2016 © Pearson Education Limited 2016

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Page 6: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample assessment materials (SAMs) – Draft 1.0 – June 2016 © Pearson Education Limited 2016

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Page 7: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

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.

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

or d… The second mark is dependent on gaining the first mark

Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample assessment materials (SAMs) – Draft 1.0 – June 2016 © Pearson Education Limited 2016

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3. 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 manifestly absurd answers 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. If a candidate makes more than one attempt at any question:

If all but one attempt is crossed out, mark the attempt which is NOT crossed out.

If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt.

6. Ignore wrong working or incorrect statements following a correct answer.

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Pearson Edexcel Level 3 GCE

Further Mathematics Advanced Paper 1: Further 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 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.

1. The roots of the equation

3 28 2 8 3 2 0x x x

are , and

(a) Write down the values of , and (1)

(b) Hence find the value of

(i) 1 1 1

(ii) 2 2 2

(iii) 2 2 2

(7)

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Question 1 continued

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(Total for Question 1 is 8 marks)

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2.

2 4

1 1 1

1 2 1

a

M

where a is a constant.

(a) Find the value of a for which the matrix M is singular. (2)

Given that M is non-singular,

(b) find 1M in terms of a (4)

(c) Solve the equations

2 4 1

1

2 2

x ay z

x y z

x y z

giving your answer in terms of a (2)

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Question 2 continued

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(Total for Question 2 is 8 marks)

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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)

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Page 16: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

4. 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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Question 4 continued

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(Total for Question 4 is 8 marks)

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Page 18: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

5. (i) Prove by induction that for all positive integers n,

3 1 2 1f 2 3 5n nn

is divisible by 17 (6)

(ii) Prove by induction that for all positive integers n,

2 9 3 1 9

1 4 3 1

nn n

n n

(6)

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Question 5 continued

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(Total for Question 5 is 12 marks)

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6.

Figure 1

Figure 1 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

1cos sin 2 , 1 sin 0 2

2x y

(a) Show that a Cartesian equation of the curve C is

2 4 32x y y

(4)

(b) Hence, using the model, find the maximum number of pendants that can be made from a bar of gold which has a volume of 500 cm3.

(6)

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Question 6 continued

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(Total for Question 6 is 10 marks)

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7. The line l1 has equation 2 4 6

4 2 1

x y z

and the plane Π has equation

2 6x y z

(a) Find the acute angle between l1 and Π, giving your answer in degrees to one decimal place.

(4)

(b) Find the coordinates of the point of intersection of l1 with Π. (3)

The line l2 is the reflection of the line l1 in the plane Π.

(c) Find a vector equation of the line l2 (5)

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Question 7 continued

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(Total for Question 7 is 12 marks)

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8. A complex number z is represented by the point P in the Argand diagram. Given that z – 3i = 3

(a) sketch the locus of P. (2)

(b) Find the complex number z which satisfies both z – 3i = 3 and arg (z – 3i) = 43 .

(3)

Given that there are no complex numbers z for which both z – 3i = 3 and arg (z – 6 – 3i) = θ when ,

(c) find the value of α. (3)

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Question 8 continued

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Total for Question 8 is 8 marks)

TOTAL FOR PAPER IS 75 MARKS

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Paper 1: Further Pure Mathematics 1 Mark Scheme

Question Scheme Marks

1(a) 8, 28, 32 B1

(1) (b)(i) 1 1 1

M1

7

8 A1

(2) (b)(ii) 22 2 2 2 M1

28 2 28 8 A1

(2)

(b)(iii) 2 2 2 2 2 4 2 M1

2 4 8 A1

32 2 28 4 8 8 128 A1

(3)

(8 marks)

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Question Scheme

Marks

2(a) 2 1 2 1 1 4 2 1 0a M M1

5a A1

(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 1adj M M

M M1

1

3 8 41

2 2 62 10

1 4 2

a a

aa a

M

2 correct rows or columns

A1

All correct

A1

(4) (c) 3 8 4 1

12 2 6 1 ...

2 101 4 2 2

a a

aa a

M1

13 12 3 9, ,

2 10 2 10 2 10

a a

a a a

A1

(2)

(8 marks)

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Question Scheme Marks

3 3 2iz is also a root B1

3 2i 3 2i ...z z M1

2 6 13z z A1

4 3 2 2 26 65 6 13 5 ...z az z bz z z z cz c

M1

2c A1

2 2 5 0 ...z z z M1

1 2iz A1

B1 3 ± 2i

Plotted correctly

B1 ̶ 1 ± 2i Plotted

correctly

(9 marks)

Re

Im

(3, 2)

(3, -2)

(-1, 2)

(-1, -2)

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Question Scheme

Marks

4(a) 3 6

4 2 18 8 24

2 12

M1

2 2 2

18 8 24 5

3 4 2d

M1

29 A1

(3) (b) 1 2 1 1

3 1 ... and 3 1 ...

1 5 1 2

M1

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) (c) 1 3

3 4 3 12 2

1 2

M1

2 2 2 22 21 3 1 3 4 2 cos 11 A1

o52 A1

(3)

(8 marks)

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Question Scheme

Marks

5(i) 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

Assume true for n = k so 3 1 2 12 3 5k k is divisible by 17 M1

3 4 2 3 3 1 2 1f 1 f 2 3 5 2 3 5k k k kk k M1

3 1 2 1 2 17 2 7 3 5 17 3 5k k k

2 17f 17 3 5 kk A1

2 1f 1 8f 17 3 5 kk k A1

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

(6) (ii)

When n = 1, lhs = 2 9

1 4

, rhs =

3 1 1 9 1 2 9

1 3 1 1 1 4

So the statement is true for n = 1

B1

Assume true for n = k so 2 9 3 1 9

1 4 3 1

kk k

k k

M1

12 9 3 1 9 2 9

1 4 3 1 1 4

kk k

k k

M1

6 2 9 27 9 36

2 3 1 9 12 4

k k k k

k k k k

A1

3 1 1 9 1

1 3 1 1

k k

k k

A1

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

(6)

(12 marks)

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Question Scheme

Marks

6(a) cos sin cos cosx y M1

sin 1y M1

2

21 1

xy

y

M1

2 4 32 *x y y A1*

(4) (b) 2 4 3d 2 dV x y y y y M1

5 4

5 2

y y

A1

5 4 5 40 0 2 2

5 2 5 2

M1

3 31.6 cm 5.03cm A1

500

1.6 M1

= 99 pendants A1

(6)

(10 marks)

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Question Scheme

Marks

7(a) 4 1

2 2 4 4 1

1 1

M1

2 2 2 2 2 24 2 1 1 2 1 cos 9 M1

o 990 arccos

21 6

M1

o53.3 A1

(4) (b) 2 4 2 4 2 6 6 ... M1

2 Required point is 2 2 4 , 4 2 2 , 6 2 1 M1

10, 0, 4 A1

(3) (c) 2 2 4 2 6 6 ...t t t t M1

3so reflection of 2,4, 6 is 2 6 1 ,4 6 2 , 6 6 1t M1

8, 8, 0 A1

10 8 2

0 8 8

4 0 4

M1

10 1

0 4 or equivalent

4 2

k

r A1

(5)

(12 marks)

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Question Scheme

Marks

8(a) Im 3 Re

M1 Circle

A1 Correct circle

(2) (b) 22 3 9, 3x y y x M1

22 3 9y M1

3 33 i

2 2z

A1

(3) (c) 1

Uses arcsin2

M1

1arcsin

2

M1

5

6

A1

(3)

(8 marks)

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Pearson Edexcel Level 3 GCE

Further Mathematics Advanced Paper 2: Further 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.

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Answer ALL questions. Write your answers in the spaces provided.

1.

2

2f

9

xx

x

(a) Find f dx x

(5)

(b) Hence find the mean value of f(x) over the interval [0, 3], giving your answer in the form ln 2p q where p and q are rational numbers to be found.

(3)

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Question 1 continued

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(Total for Question 1 is 8 marks)

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2.

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 2 continued

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(Total for Question 2 is 9 marks)

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3. (a) Express

1

1 3r r in partial fractions.

(1)

(b) Hence prove, by the method of differences, 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. (4)

(c) Find the exact value of

30

10

1

1 3r

r r

(2)

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Question 3 continued

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(Total for Question 3 is 7 marks)

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Page 42: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

4. In an experiment, Rob attaches a small weight to one end of an elastic spring. The other end of the spring is attached to a point P and the weight hangs vertically below P. Rob holds the weight at a fixed point A and releases it from rest. As he releases the weight, Rob starts a stopwatch. The weight is then made to oscillate in a vertical plane.

The vertical displacement, x cm, of the particle from A at time t seconds is modelled by the differential equation

2

2

d d3 4 5 cos , 0

dt d

x xx t t

t

Solve the differential equation to find the vertical distance of the weight from A, 9 seconds after the weight is released.

(12)

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Question 4 continued

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(Total for Question 4 is 12 marks)

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5. A complex number z has modulus 1 and argument θ.

(a) Show that

zn + 1nz

= 2cos nθ, n

(2)

(b) Show that

cos4 θ = 1

8(cos 4θ + 4cos 2θ + 3)

(5)

(c) Hence find all the solutions of cos 4θ + 4cos 2θ + 2 = 2cos2 θ

in the interval 0 θ < 2π.

(5)

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Question 5 continued

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(Total for Question 5 is 12 marks)

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6.

sin sinhy x x

(a) Show that 4

4d 4d

y yx

(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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Question 6 continued

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(Total for Question 6 is 10 marks)

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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

dd

0.2 0.4d

ff r

tr

f rt

(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 limitations of the model. (7)

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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 17 marks)

TOTAL FOR PAPER IS 75 MARKS

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Paper 2: Further Pure Mathematics 2 Mark Scheme

Question Scheme

Marks

1(a) 2 2 2

2 2f

9 9 9

x xx

x x x

B1

22

d ln 99

xx k x c

x

M1

2

2d arctan

9 3

xx k c

x M1

22

1d ln 9

9 2

xx x c

x

or

2

2 2d arctan

9 3 3

xx c

x

A1

22

2 1 2d ln 9 arctan

9 2 3 3

x xx x c

x

A1

(5) (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

xx x x

M1

Mean value = 1 1

ln 23 0 2 6

M1

1 1ln 2

6 18

1 1,

6 18p q

A1

(3)

(8 marks)

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Question Scheme Marks

2 94 cos 2 ...

2 M1

6

A1

2 21 14 cos 2 d 16 8 cos 2 cos 2 d

2 2 M1

2 1 1cos 2 cos 4

2 2 M1

21 1 sin 416 8cos2 cos 2 d 16 4sin 2

2 2 8 2

A1

1 33 3Usinglimits 0 and their : 2 3 0

6 2 12 16

M1

Area of triangle = 1 1 81 1 3cos sin

2 2 4 2 2r r M1

Area of R = 33 33 3 81 3

24 32 32

M1

33 3 3 33 3,

24 2 24 2p q

A1

(9 marks)

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Question Scheme

Marks

3(a)

1 1 1

1 3 2 1 2 3r r r r

B1

(1) (b)

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

1 1 1 1

4 6 2 2 2 3n n

A1

5 2 3 6 3 6 2

12 2 3

n n n n

n n

M1

5 13

12 2 3

n n

n n

A1

(4) (c)

30

10

1f 30 f 9

1 3r

r r

M1

30 5 30 13 9 5 9 13

12 32 33 12 11 12

119

2112 A1

(2)

(7 marks)

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Question Scheme

Marks

4 2 13 4 1 0 1,

3m m m M1

13e e ttx A B A1

cos sinx a t b t B1 2

2

d dsin cos , cos sin

d d

x xa t b t a t b t

t t M1

4 2 5, 2 4 0 ..., ...b a b a a b M1

1sin cos

2x t t A1

13

1e e sin cos

2ttx A B t t A1

10, 0

2t x A B M1

13

d 1 1 10, cos sin 0 1

d 3 2 3ttx

x Ae Be t t A Bt

M1

5 3,

4 4A B A1

13

5 3 1e e sin cos

4 4 2ttx t t A1

9 0.830t x cm A1

(12 marks)

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Question Scheme

Marks

5(a) cos i sin cos i sinn nz z n n n n M1

2cos *n A1

(2) (b) 41 416 cosz z B1

41 4 2 2 44 6 4z z z z z z M1

4 4 2 24 6z z z z A1

2cos4 4 2cos2 6 M1

cos4 θ = 1

8(cos 4θ + 4cos 2θ + 3)* A1

(5) (c) cos 4θ + 4cos 2θ + 2 = 2cos2 θ 4 28 cos 2 cos 1 M1

2 2 24cos 1 2cos 1 0 cos ... M1

2 1 1cos cos

2 2 A1

7,

4 4

A1

3 5,

4 4

A1

(5)

(12 marks)

Pearson Edexcel Level 3 Advanced GCE in Further Mathematics – Sample assessment materials (SAMs) – Draft 1.0 – June 2016 © Pearson Education Limited 2016

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Question Scheme

Marks

6(a) dsin cosh cos sinh

d

yx x x x

x M1

2

2

dcos cosh sin sinh cos cosh sin sinh

d2 cos cosh

yx x x x x x x x

xx x

M1

3

3

d2 cos sinh 2 sin cosh

d

yx x x x

x A1

4

4

d4 sinh sin 4 *

d

yx x y

x A1*

(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

Uses 2 3

0 0 0 0 ...2 ! 3!

x xy y xy y y with their values M1

2 6 10

2 8 322! 6! 10!

x x x A1

6 102

90 113400

x xx A1

(4) (c)

4 2

12 4 ,

4 2 !

nn x

n

B1, B1

(2)

(10 marks)

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Question Scheme

Marks

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

2

d d d10 2 0.2 0.4 10 2

d d d

f f ff f

t t t

M1

2

2

d d0.6 0.1 0 *

d d

f ff

t t A1*

(3) (b) 2

2 0.6 0.6 4 0.10.6 0.1 0

2m m m

M1

0.3 0.1im A1

e cos sintf A t B t M1

0.3e cos0.1 sin0.1tf A t B t A1

(4) (c) 0.3 0.3d

0.3e cos 0.1 sin 0.1 0.1e cos 0.1 sin 0.1d

t tfA t B t B t A t

t

M1

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

0.3e cos0.1 sin0.1tr A B t B A t A1

(3) (d)(i) 0, 6 6t f A M1

0, 20 14t r B M1

0.3e 20cos0.1 8sin0.1 0tr t t M1

tan 0.1 2.5t A1

2019 A1

(d)(ii) 3750 foxes B1

(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.

The model may only be valid for a short period of time.

B1

(7)

(17 marks)

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Pearson Edexcel Level 3 GCE

Further Mathematics Advanced Paper 3: Further Mathematics Option 1 Option 3A: Further Pure Mathematics 3 Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes

Paper Reference(s)

9FM0/3A

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.

1. Use Simpson’s Rule with 6 intervals to estimate

4 3

11 dx x

(5)

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Question 1 continued

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(Total for Question 1 is 5 marks)

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2.

2

2

d, 3 when 1

d

y y xy x

x x xy

Use the approximation 1 0

0

d

d

y yy

x h

with a step-length of 0.1 to estimate the value

of y when x = 1.1 and value of y when x = 1.2 (6)

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Question 2 continued

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(Total for Question 2 is 6 marks)

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3. A vibrating spring C has an external force acting on it. At time t seconds the displacement of the centre of the spring from a fixed point O is x micrometres. The motion of C is modelled by the differential equation

22 2 4

2

d d2 (2 ) (I)

d d

x xt t t x t

t t

(a) Show that the transformation x t transforms equation (I) into the equation

2

2

d(II)

dt

t

(5)

(b) Hence find the general equation for the displacement of C from O at time t seconds. (7)

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Question 3 continued

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(Total for Question 3 is 12 marks)

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4. (i) Find the Taylor Series form of 3 24 3 6 2x x x in ascending powers of (x – 2) (4)

(ii) Find a series solution in ascending powers of x, as far as the term in x5, of the differential equation

2

2

d d2 0

d d

y yx y

x x

where 0y and d

1d

y

x at 0x

(9)

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Question 4 continued

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(Total for Question 4 is 13 marks)

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5. The normal to the parabola 2 4y ax at the point 2 , 2P ap ap passes through the

parabola again at the point Q.

OP is perpendicular to OQ where O is the origin.

Show that 2 2p (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 has equation 2 3 3 0x y

The point E lies on the line AC and the point F lies on the line AD

Given that contains the point B, the point E and the point F

(c) find the value of k such that AE k AC

(3)

Given that 1

9AF AD

(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. (i) Given that

3ekxy x

use Leibnitz theorem to show that

3 3 3 2 2de ( 3 3 ( 1) ( 1)( 2))

d

nn kx

n

yk k x nk x n n kx n n n

x

(4)

(ii) Given that

cos(ln ) sin(ln )y x x

show that

2 12 2

2 1

d d d(2 1) ( 1) 0

d d d

n n n

n n n

y y yx n x n

x x x

(7)

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Question 7 continued

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(Total for Question 7 is 11 marks)

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8. P and Q are two distinct points on the ellipse described by the equation 2 24 4x y

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

Find an equation of the line l in terms of r and s. (8)

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Question 8 continued

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(Total for Question 8 is 8 marks)

TOTAL FOR PAPER IS 75 MARKS

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Paper 3: Further Mathematics Option 1 Option 3A: Further Pure Mathematics 3 Mark Scheme

Question Scheme

Marks

1 Step 0.5 B1

0y 1y 2y 3y 4y 5y 6y

x 1 1.5 2 2.5 3 3.5 4 y 2 4.375

3 16.625

28 43.875

65 M1

0 1 2 3 4 5 64 2 4 2 4 "77.23"y y y y y y y M1

4 3

1

0.51 d "77.23"

3x x M1

12.9 A1

(5)

(5 marks)

Question Scheme

Marks

2 2

0 0 20

d 3 11, 3, 2

d 1 3

yx y

x

B1

1 00

d3 0.1 2

d

yy y h

x

M1

3.2 A1 2

21

d 3.2 1.1

d 1.1 1.1 3.2

y

x

M1

9.141.932...

4.73 A1

2 11

d3.2 0.1 1.932 3.39

d

yy y h

x

to 3 sf A1

(6)

(6 marks)

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Question Scheme

Marks

3(a) d d

d d

x tv

x vv t

t t

Attempt Product Rule

M1

2 2

2 2

d d d d

d d d d

x v v vt

t t t t

Differentiate and attempt Product Rule

M1 A1

2 2 4their 2nd derivative 2 their 1st derivative (2 )t t t x t M1

23 3 4

2

d

d

vt t v t

t

2

2

d

dtt

*

A1

(5)

(b) 2 21 0 1 M1

cos sinA t B t A1

Particular Integral kt l B1 2

2

d d, 0

d d

v vk

t t M1

0 1, 0kt l t k l M1

Solution: cos sinA t B t t A1

Displacement of C from O is given by

cos sinx t A t B t t B1

(7)

(12 marks)

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Question Scheme Marks

4(i) (i) 2f '( ) 12 6 6, f '(2) 54x x x f ''( ) 24 6, f ''(2) 54x x

M1

f '''( ) 24x A1

2 3f ''(2) f '''(2)f ( ) f (2) f '(2)( 2) ( 2) ( 2)

2 6x x x x M1

2 3f ( ) 34 54( 2) 27( 2) 4( 2)x x x x A1

(4)

(ii) 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

0, 0, ' 1 ''(0) 0x y y y from equation B1

'' 2 ' ''' 2 '' 2 ' 'y xy y y xy y y M1

'''(0) 2 0 '' 1 1y y A1

''' 2 '' ' '''' 2 ''' 2 '' ''y xy y y xy y y

''''(0) 2 0 1 3 0 0y A1

'''' 2 ''' 2 '' '' ''''' 2 '''' 5 '''y xy y y y xy y M1

0, ''' 1, '''' 0 '''''(0) 5x y y y A1

Series solution: 3 51 1, ...

6 24y x x x A1, A1

(9)

(13 marks)

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Question Scheme Marks

5 2 d4 2 4

d

yy ax y a

x M1

d 2

d

y a

x y Gradient of normal is

2

yp

a

A1

Equation of normal: 22 ( )y ap p x ap

M1

Normal passes through 2( , 2 )Q aq aq so 2 32 2aq apq ap ap M1

Grad OP × Grad OQ = 1 2 2

2 21

ap aq

ap aq M1

4q

p

A1

Sub for q

32

4 162 2a ap ap ap

p p

M1

4 22 8 0p p

2 2( 2)( 4) 0p p M1

2 2p * A1

(9 marks)

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Question Scheme

Marks

6(a)

1 1 2 1 3

1 1 1 2 1

1 2 1 1 2

AB AC M1

3 0 3

1 1 . 1 1

2 0 2

r. M1

3 2 1x y z A1

(3)

(b) Volume of Tetrahedron =

1.( )

6n AD

M1

3 10 11

1 . 5 26

2 5 1

M1

= 1 8

( 27 3 8)6 3

A1

(3)

(c) kAE AC so E is (1 , 2 ,1 2 )k k k M1

E lies on plane so 2(1 ) 3(2 ) 3 0k k M1

1

5k A1

(3)

(d) Vol ABEF= 1 1 1 1

. .6 6 5 9

AB AE AF AB AC AD M1

1 1. .

45 6 AB AC AD and hence result. A1

(2)

(14 marks)

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Question Scheme

Marks

7(i) 3ekxy x so 2 3 4

3 22 3 4

d d d dand 3 and 6 and 6 and 0

d d d d

u u u uu x x x

x x x x

M1

1 2

1 21 2

d d d = e and e and e and e and ...

d d d

n n nkx n kx n kx n kx

n n n

v v vv k k k

x x x

M1

3 2 1 2 3d ( 1) ( 1)( 2)

e 3 e 6 e 6 e ( 0)d 2 3!

nn kx n kx n kx n kx

n

y n n n n nx k n x k xk k

x

A1

So 3 3 3 2 2de 3 3 ( 1) ( 1)( 2)

d

nn kx

n

yk k x nk x n n kx n n n

x A1

(4)

(ii) d 1 1sin(ln ). cos(ln )

d

yx x

x x x M1

2

2

d d 1 1cos(ln ) sin(ln )

d d

y yx x x

x x x x M1, A1

22

2

d d0

d d

y yx x y

x x A1

2 1 12

2 1 1

d d ( 1) d d d d2 2 0

d d 2 d d d d

n n n n n n

n n n n n n

y y n n y y y yx n x x n

x x x x x x

M1,

M1 2 1

2 22 1

d d d(2 1) ( 1) 0

d d d

n n n

n n n

y y yx n x n

x x x

A1

(7)

(11 marks)

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Question Scheme

Marks

8 2 2 d4 4 2 8 0

d

yx y x y

x M1

Equation of tangent at 1 1( , )P x y is

11 1

1

( ) ( )4

xy y x x

y

M1

2 21 1 1 14 4 4xx yy x y A1

And at 2 2( , )Q x y : 2 24 4xx yy

Intersect at ( , )r s

1 14 4rx sy and 2 24 4rx sy B1

Attempting substitution

M1

2 1

2 1 4

y y r

x x s

A1

Equation of PQ is 1 1( )4

ry y x x

s

M1

1 14 4 4sy rx sy rx A1

(8 marks)

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Pearson Edexcel Level 3 GCE

Further Mathematics Advanced Paper 4: Further Mathematics Option 2 Option 4A: Further Pure Mathematics 4 Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes

Paper Reference(s)

9FM0/4A

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.

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 the numbers 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)

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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)

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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)

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Page 96: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

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 2a 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)

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Page 100: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

5. An engineering student makes a miniature arch as part of the design for a theme park. The arch stands on a table.

The cross section of this arch is modelled by the curve with equation

12 cosh 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.

(a) Sketch the curve with the given equation. (1)

(b) Find the length of this curve, giving your answer in the form 22

1k a

a

, stating

the value of the constant k. (5)

Given that the length of the curved cross section of the arch is 2 m,

(c) find the height of the arch, according to this model. (5)

(d) Explain which dimension of the arch is given by the value of 2lna. (1)

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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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Page 104: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

6. The curve C has equation

6 2 6z z z

(a) Find a Cartesian equation for C (2)

(b) Sketch the curve C on an Argand diagram. (2)

The line l has equation * * 0az a z , where a and z and where a* and z* are the complex conjugates of a and z respectively.

Given that the line l is a tangent to C and that arg a = θ

(c) find the possible values of tan (5)

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Question 6 continued

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Page 106: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

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. 2

0sin dn

nI x x

, n 0

(a) Prove that, for n 2,

2( 1)n nnI n I (4)

(b)

Figure 1

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 0.7m and width of π m.

Figure 1 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 white, as shown in Figure 1.

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) Find, using a characteristic polynomial method, an exact expression for the number of ways in which she could climb up to the top of the staircase.

(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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Paper 4: Further Mathematics Option 2 Option 4A: Further Pure Mathematics 4 Mark Scheme

Question Scheme Marks

1(i) 602 3 161 119 M1

161 119 42, 119 2 42 35 A1

42 35 7, 35 5 7, hcf 7 A1

(3)

(ii) Number of codes under old system = 5 4 4 3 2( 480) B1

Number of codes under new system = 4 3 7 6 5( 2520) B1

Subtracts first answer from second M1

Increase in number of codes is 2040 A1

(4)

(7 marks)

Question Scheme

Marks

2(a) Let z = x + i M1

2 2( i) ( 1) 2 iw x x x A1

Let w = u + iv, then 2( 1) and 2u x v x M1

2 4( 1)v u , which therefore represents a parabola A1

(4)

(b)

M1

A1

(2)

(6 marks)

M1: Sketches a parabola with symmetry about the real axis A1: Accurate sketch

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Question Scheme

Marks

3(a) Finds the characteristic equation 2(2 ) (4 ) (4 ) 0 M1

so 2(4 )( 4 3) 0 so 4 A1

Solves quadratic to give M1

1 and 3 A1

(4) (b) Uses a correct method to find an eigenvector M1

Obtains a vector parallel to one of

0

0

1

or

1

1

1

or

3

3

1

A1

Obtains two correct vectors A1

Obtains all three correct A1

(4) (c) Uses their three vectors to form a matrix M1

0 1 3

0 1 3

1 1 1

or other correct answer with columns in a different order.

A1

(2)

(10 marks)

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Question Scheme Marks

4(i) 2 2If we assume ; as thenab ba a b ba ab a b M1

1 1 1 2 1So a abb a a bb M1

So e a A1

But this is a contradiction, as the elements e and a are distinct so ab ba A1

(4)

(ii)(a) 2 has order 4 and 4 has order 2 M1

7, 8 and 13 have order 4 A1

11 and 14 have order 2 and 1 has order 1 A1

(3)

(ii)(b) Finds the subgroup {1, 2, 4, 8} or the subgroup {1, 7, 4, 13} M1

Finds both and refers to them as cyclic groups, or gives generator 2 and generator 7

A1

Finds {1, 4, 11, 14} B1

States each element has order 2 or refers to it as Klein Group B1

(4)

(ii)(c) J has an element of order 8, or J is a cyclic group or other valid reason

M1

They are not isomorphic

A1

(2)

(13 marks)

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Question Scheme

Marks

5(a)

B1

(1)

(b) d

sinh 2d

yx

x

B1

so s = 21 sinh 2 dx x M1

s cosh 2 dx x A1

= ln12 ln

sinh 2a

ax

or ln0sinh 2

ax M1

= 2ln -2ln12sinh 2 ln [e e ]a aa = 21

2 2

1a

a

(so k = ½) A1

(5)(c) 21

2 2

1a

a

= 2 so 4 24 1 0a a M1

2 2 5a and a = 2.06 (approx.) M1

a = 2.06 A1

When x = lna, y = 0 so 12 cosh 2 lnA a M1

Height = A – 0.5 = awrt 0.62m or 5 1

2

A1

(5)

(d) The width of the base B1

(1)

(12 marks)

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Question Scheme

Marks

6(a) 2 2 2 2( 6) 4[( 6) ]x y x y M1

2 2 20 36 0x y x A1

(2) (b)

M1

A1

(2) (c)

Let a = c+id and a* = c-id then (c+id)(x-iy)+(c-id)(x+iy)=0

M1

So c

y xd

A1

The gradients of the tangents (from geometry) are 4

3

B1

So 4

3

c

d and

3

4

d

c M1

So 3

tan4

A1

(5)

(9 marks)

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Question Scheme

Marks

7(a) 2 1

0sin sin dn

nI x x x

M1

221 2 2

0 0cos sin ( ) cos ( 1)sin dn nx x x n x x

A1

Obtains 2 2 2

00 ( ) (1 sin )( 1)sin dnx n x x

M1

So 2( 1) ( 1)n n nI n I n I and hence 2( 1)n nnI n I * A1

(4) (b)

uses 2

( 1)n n

nI I

n

to give 10 8

9

10I I or 2 0

1

2I I M1

So 10 0

9 7 5 3 1

10 8 6 4 2I I

M1

0 2I

B1

Required area is 2 102( )I I or 12 1048 ( )I I M1

263 652 m

4 512 256

A1

(5)

(9 marks)

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Question Scheme

Marks

8(a) 1 1u as there is only one way to go up one step B1

2 2u as there are two ways: one step then one step or two steps B1 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

(3)(b) sequence begins 1, 2, 3, 5, 8, 13, 21, 34,… so 34 ways of

climbing 8 steps B1

(1)(c) To find general term use 2

1 2 gives 1n n nu u u M1

This has roots 1 5

2

M1

So general form is 1 5 1 5

2 2

n n

A B

A1

Uses initial conditions to find A and B reaching M1 401 401

1 1 5 1 5

2 25

A1

(5)

(9 marks)

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Pearson Edexcel Level 3 GCE

Further Mathematics Advanced Paper 3: Further Mathematics Option 1 Option 3B: Further Statistics 1 Paper 4: Further Mathematics Option 2 Option 4B: 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. 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) Find the probability density function of Y. (2)

(c) Find the mode of Y, giving a reason for your answer. (2)

(d) Using algebraic integration, show that the Var (Y) is 0.9375 (4)

(e) Describe the skewness of the distribution of Y, Give a reason for your answer (1)

(f) Find P(Y > 3). (2)

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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 14 marks)

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3. 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

(a) Calculate Spearman’s rank correlation coefficient between b and s. (3)

(b) Stating your hypotheses clearly, test whether or not the data provides support for the researcher’s claim. Use a 1% level of significance.

(4)

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 3 continued

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(Total for Question 3 is 9 marks)

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Page 134: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

4. (a) Write down the conditions under which the Poisson distribution may be used as an approximation to the binomial distribution.

(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.

(b) Find the mean and variance of the number of wrongly connected calls a day. (7)

(c) 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)

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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 11 marks)

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Page 138: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

5. 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 1.3 1.2 1.15 1.25 1.15

(a) Find Sxx, Smm and Sxm for these data. (3)

(b) Find the equation of the regression line of m on x in the form m = a + bx as a model for these results.

(2)

(c) Show that the residual sum of squares (RSS) is 0.064 to 3 decimal places. (2)

(d) Calculate the residual values. (2)

(e) Write down the outlier. (1)

(f) (i) Ignoring the outlier, produce another model.

(ii) Use this model to estimate the average weight at birth if x = 15.

(iii) Comment, giving a reason, on the reliability of your estimate (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 14 marks)

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Page 142: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

6. 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 calculated 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 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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Page 146: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

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) 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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Paper 3: Further Mathematics Option 1 Option 3B: Further Statistics 1 Mark Scheme

Paper 4: Further Mathematics Option 2 Option 4B Further Statistics 1 Mark Scheme

Question Scheme Marks

Q1 Ho : = 5 (= 2.5) H1 : > 5 (> 2.5) B1

X ~ Po (2.5) B1

7 1 ( ) 6 P X P X M1

= 1 − 0.9858 = 0.0142 CR X 6

A1

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

(5 marks)

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Question Scheme

Marks

Q2(a) P(A > 3) =

2

5 B1

32 8

5 125

M1 A1

(3)

(b) 230 5

125f ( )

0 otherwise

yy

y

M1 A1

(2)

(c) Mode = 5 B1

Or reason based on df( )

0d

y

y

B1

(2)

(d) 35

0

3E( ) d

125

yY y

54

0

3=

500

y

M1

15

= 4

A1

25 4

0

3 15Var( ) d

125 4

yY y

M1

= 0.9375 A1

(4)

(e) From a sketch or mode > mean therefore it has negative skew B1

(1)

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Question Scheme

Marks

(f) P( 3) 1 F(3)Y

271

125 M1

98

125 A1

(2)

(14 marks)

Question Scheme

Marks

Q3(a) Distance rank

1 2 3 4 5 6 7

Depth rank

1 2 4 3 6 7 5

d 0 0 1 1 1 1 2 2d 0 0 1 1 1 1 4

2 8d M1

6 81

7 48sr

M1

= awrt 0.857 or 6

7oe A1

(3)

(b) 0 1H : 0, H : 0 B1

Critical value at 1% level is 0.8929 B1

sr < 0.8929 so not significant evidence to reject 0H , M1

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

A1

(4)

(c) The mean of the tied ranks is given to each then use PMCC

B1 B1

(2)

(9 marks)

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Question Scheme

Marks

Q4(a) [If X ~ B(n,p) and]

n is large, B1

p is small, B1

then X can be approximated by Po(np)

(2)

(b) P(X 1) = 1 – P(X = 0) B1

1 – P(X = 0) = 0.049

P(X = 0) = 0.951 B1

5 0.951x M1

0.99x

p = 0.01 A1

X ~B(1000, 0.01) M1

Mean = np = 10 A1

Variance = np(1 – p) = 9.9 A1

(7)

(c) X ~ Po(10)

P(X > 6) = 1 – P (X 6) M1

= 1 – 0.1301

= 0.870 A1

(2)

(11 marks)

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Question Scheme

Marks

Q5(a) Sxx = 82.5

M1

Smm = 0.12756 A1

Sxm = –2.29 A1

(3)

(b) b =

S

Sxm

xx

= – 0.0277576 M1

a = m bx = 1.278 + 0.0277576 8.5 = 1.5139

m = 1.5139 – 0.02775x A1

(2)

(c) RSS =

22.29

0.1275682.5

M1

= 0.06399 A1

(2)

(d) 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

(2)

(e) The point (5, 1.2) is an outlier B1

(1)

(f)(i) a m bx =1.28667 +0.03765 8.88889 = 1.6213 M1

m =1.6213 – 0.03765x A1

(ii) m =1.6213 – 0.03765 15

= 1.056 or awrt 1.06 B1

(ii) Not reliable since it involves extrapolation B1

(4)

(14 marks)

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Question Scheme

Marks

Q6(a) Expected value for 2 = 2 18150 ( 2) 150 0.05 0.95P X M1

= 28.3015… A1

Expected value for 4 or more = 150 – (53.8 + 56.6 + 28.3 + 8.9) = 2.4

A1

H0: Bin(20, 0.05) is a suitable model H1: Bin(20, 0.05) is not a suitable model

B1

Combining last two groups

3 Observed frequency 19 Expected frequency 11.3

M1

B1

Critical value, χ2 (0.05) = 7.815 B1

Test statistic = 2 243 53.8 62 56.6

...53.8 56.6

M1

= 8.117 A1

In critical region, sufficient evidence to reject H0, accept H1

Significant evidence at 5% level to reject the manager’s model A1

(10)

(b) = 4 – 2 = 2

4 classes due to pooling

B1

2 restrictions (equal total and mean/proportion) B1

(2)

(c) H0: Binomial distribution is a good model H1: Binomial distribution is not a good model

B1

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

(2)

(14 marks)

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Question Scheme

Marks

Q7(a)

Sxx = 22 1010

10

ss

M1

2658.9 = 22 100100

10

ss

M1

2658.9 = 100 Sss

Sss = 26.589 * A1

(3)

(b) 10

164 10( 9) id M1

10

164 10 900id

10

196.4id A1

Sdd = 2"96.4"

1081.74 -10

M1

= 152.444

r = 0.935 A1

(4)

(c) Linear correlation is significant but scatter diagram suggests a non-linear relationship

B1

(1)

(8 marks)

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Pearson Edexcel Level 3 GCE

Further Mathematics Advanced Paper 4: Further Mathematics Option 2 Option 4C: Further Statistics 2 Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes

Paper Reference(s)

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. 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. 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 2̂

A 95% confidence interval for was found to be ( 1.128, 2.232 )

(a) Show that 2̂ = 1.79 (correct to 3 significant figures) (4)

(b) Obtain a 95% confidence interval for 2 (4)

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Question 1 continued

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(Total for Question 1 is 8 marks)

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2. 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 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 13 marks)

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3. 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 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 13 marks)

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4. The probability of Richard winning a prize in a game at the fair is 0.15

Richard plays a number of games.

(a) Find

(i) the probability of Richard winning his second prize on his 8th game,

(ii) the expected number of games Richard will need to play in order to win 3 prizes. (3)

(b) State two assumptions that you have made in part (a). (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 4 continued

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(Total for Question 4 is 9 marks)

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5. The probability generating function of the discrete random variable X is given by

GX (t) = 223 2k t t

(a) Show that k = 1

36

(2)

(b) Find P(X = 3) (2)

(c) Calculate E(X) and Var(X) (8)

(d) Find the probability generating function of 2X + 1 (2)

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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 14 marks)

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6. 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) Write down the size of Tessa’s test. (1)

(e) (i) Show that the power function for Sam’s test is given by

191 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 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 18 marks)

TOTAL FOR PAPER IS 75 MARKS

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Paper 3: Further Mathematics Option 2 Option 4C: Further Statistics 2 Mark Scheme

Question Scheme

Marks

1(a) 95% CI for uses t value of 2.064 B1

ˆ 1"2.064" 2.232 1.128

225

or

ˆ12.232 1.128 "2.064" 2.232

2 25

(oe)

M1

2.76

ˆ"2.064"

or 1.3372… M1

2ˆ 1.788... [=1.79 (3sf)] (*) A1cso (4)

(b) 12.401, <

2

79.124

<, 39.364 B1 M1 B1

1.09 < 2 < 3.46 A1 (4)

(8 marks)

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Question Scheme

Marks

2(a) H0: 2G = 2

B , H1: 2G 2

B , B1

2Bs = 6

1 (56130 – 7 88.92) = 6

53.807= 134.6

M1

A1

2Gs = 7

1 (55746 – 8 83.12) = 7

12.501= 71.58 A1

2

2

G

B

s

s= 1.880… M1

critical value F6, 7 = 3.87 B1

not significant, variances are the same A1 ft

(7)

(b) H0: B = G , H1: B > G B1

pooled estimate of variance s2 = 13

58.7176.1346 = 100.6653… M1

test statistic t = 81

71

1.839.88

s = awrt 1.12 M1

A1

critical value t13(5%) = 1.771 B1

Insufficient evidence to support mother’s claim A1 ft

(6) (13 marks)

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Question Scheme

Marks

3(a) Let X = L – 4S then E(X) = 19.6 – 4 4.8 M1

= 0.4 A1

Var(X) = 2 2 2Var( ) 4 Var( ) 0.6 16 0.3L S M1

= 1.8 A1

P(X > 0) = [P(Z > 0 0.4

0.298...1.8

…)] M1

= 0.617202.. awrt 0.617 A1

(6) (b)

1 2 3 4T S S S S (May be implied by 0.36) M1

T ~ N(19.2, 0.36) E(T) = 19.2 B1

Var(T) = 0.36 or 20.6 A1

(3) (c) Let Y = L – T E(Y ) = E(L) – E(T) = [ 0.4] M1

Var(Y) = Var(L) + Var(T) = [ 0.72] M1

Require P( – 0.2 < Y < 0.2) M1

= 0.16708… awrt 0.167 A1

(4) (13 marks)

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Question Scheme

Marks

4(a)(i) 67

0.15 0.85 0.151

M1

= 0.05940… = awrt 0.0594 A1

(ii) ~ NegB( , ) E( )

rX r p X

p

3

0.15

20 B1

(3) (b) The games (trials) are independent B1

The probability of winning a prize, 0.15, is constant for each game

B1

(2) (c) 2

2

118 and 6

r pr

p p

M1

A1

Solving: 2p = 1 – p M1

p = 1

3 (> 0.15) so Mary has the greater chance of winning a

prize A1

(4) (9 marks)

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Question Scheme

Marks

5(a) G (1) 1X gives M1

26 1k so k = 1

36 (*) A1cso

(2) (b) P(X = 3) = coefficient of t3 so 3G ( ) ... 4 ...X t k t M1

[ P(X = 3) =] 1

9 A1

(2) (c) 2G ( ) 2 3 2 1 4X t k t t t M1

E(X) = G (1) 2 3 1 2 1 4X k M1

= 5

3 A1

22G ( ) 2 3 2 4 1 4X t k t t t M1 A1

2G (1) 2 [6 4 5 ]X k 49

18

M1

Var(X) = 2 49 5 25G (1) G (1) G (1)

18 3 9X X X M1

= 29

18 A1

(8) (d) 222 2

2 1G ( ) 3 236X

tt t t [ t or sub t2 for t] M1

= 22 42 1G ( ) 3 2

36X

tt t t A1

(2) (14 marks)

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Question Scheme

Marks

6(a) X ~ B(20, 0.2) and seek c such that P(X c) < 0.10 M1

[P(X 1) = 0.0692] CR is X 1 A1

(2)

(b) Size = 0.0692 B1

(1)

(c) Y = no. of spins until red obtained so Y~ Geo(0.2) M1

1

p so if p < 0.2 then mean is larger so seek d so that

P(Y d) < 0.10 M1

1P( ) 0.8 dY d M1

1 log(0.1)0.8 0.10 1

log(0.8)d d M1

d > 11.3.. A1

CR is Y 12 A1

(6) (d) Size = [ 110.8 = 0.085899…] = 0.0859 B1

(1) (e)(i) Power = P(reject H0 when it is false) = P(X 1 | X ~B(20, p)) M1

= 20 191 20 1p p p M1

= 191 1 19p p (*) A1cso

(ii) Power = 111 p B1

(4)

(f) Sam’s test has smaller P(Type I error) (or size) so is better B1

Power of Sam’s test = 0.1755… B1

Power of Tessa’s test = 110.85 = 0.1673… B1

So for p = 0.15 Sam’s test is recommended B1

(4)

(18 marks)

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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 4D: Further Mechanics 1 Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes

Paper Reference(s)

9FM0/3C 9FM0/4D

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 g = 9.8 m s-2 and give your answer to either 2 significant figures or 3 significant figures.

1. A particle of mass m kg lies on a smooth horizontal surface. Initially the particle is at rest on the surface 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 1 continued

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(Total for Question 1 is 8 marks)

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2. A particle P of mass 0.5 kg is moving with velocity (4i + j) m s-1 when it receives an impulse (2i – j) N s.

(a) Show that the kinetic energy gained by P as a result of the impulse is 12 J. (6)

(b) Find, to the nearest degree, the size of the angle through which the direction of motion of P is deflected as a result of the impulse.

(3)

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Question 2 continued

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(Total for Question 2 is 9 marks)

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3.

The non-uniform rod BC of mass m and length 5a is suspended from a fixed point A by two light strings, AB and AC. The rod rests horizontally in equilibrium with AB

at an angle to the rod, where 3

tan4

, and with AB perpendicular to AC, as

shown in Figure 1. The centre of mass of the rod is at the point G of the rod such that BG x .

(a) Show that 16

5x a .

(3)

The string AC is elastic, with natural length 5

2a and modulus of elasticity kmg,

where k is a constant.

(b) Find the value of k. (6)

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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 9 marks)

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4. A car of mass 600 kg is moving along a straight horizontal road. The resistance to the motion of the car when its speed is v m s-1, is modelled as a force of magnitude

200 2v N. The engine of the car is working at a constant rate of 12 kW. The car

is modelled as a particle.

(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, where1

sin14

. The resistance to the motion of the car from

non-gravitational forces when its speed is v m s-1 is modelled as a force of magnitude 200 2v N. The engine of the car 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 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.

Figure 2

A shop sign is modelled as a uniform rectangular lamina ABCD with a semi-circular lamina removed. The radius of the semicircle is 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 2.

(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 B, so that the sign 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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Page 204: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

6. A particle 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)

(b) Given that 1

3e , show that the kinetic energy lost in the collision is

240

7

mu.

(5)

(c) If 1

,3

e without doing any further calculation, state how your answer to

part (b) would change. (1)

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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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Page 208: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

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 2 .OA 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)

TOTAL FOR PAPER IS 75 MARKS

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Paper 3: Further Mathematics Option 1 Option 3C: Further Mechanics 1 Mark Scheme

Paper 4: Further Mathematics Option 2 Option 4D: Further Mechanics 1 Mark Scheme

Question Scheme

Marks

1(a) Speed of particle after the first impact

3

4eu u B1

Impulse = 3

4mv mu mu mu

M1

7

4 A1

(3)

(b) Speed of the particle after the second impact

3 3 9

4 4 16u u B1

Use of s vt to find total time M1

2 4 2 2 16 32

73 9 3 94 16

u u u uu u

A1

Solve for u: 63 18 48 32u M1

98 141.5

63 9u A1

(5)

(8 marks)

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Question Scheme

Marks

2(a)

Impulse momentum equation – difference of terms & dimensionally correct

M1

2 0.5 0.5 4 i j v i j A1

1 1

4 , 82 2

v i j v i j (m s-1) A1

Use of KE = 2 21 1

2 2m mv u M1

10.5 64 1 16 1

2 A1

1

48 124

(J) *Given Answer* A1

(6)

(b)

Use of trig or scalar product to find the correct angle M1

1 11 1tan tan

4 8 or 1 32 1

cos17 65

A1

21.16….= 21 to the nearest degree A1

(3)

(9 marks)

tan-11

8

tan--11

4

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Question Scheme

Marks

3(a) Moments about A G is vertically below A M1

Correct use of trig: 2cos cosBG BA BC

Or : Similar triangles: cosAB BG

BC AB ,

2 2

4

5

AB aBG

BC a

M1

16 16

525 5

a a *Given answer* A1

(3)

(b) Moments about B - both terms and dimensionally correct M1

16

45CT a mg a A1

Use of x

Ta

M1

0.5

2.5C

aT

a

A1

Solve for : 0.5

2.5C

aT

a

=

4

5mg M1

4mg , 4k A1

(6)

(9 marks)

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Question Scheme

Marks

4(a) Use of P Fv :

12000

20F B1

Equation of motion: 200 2 600F v a M1

600 240 600a A1

360 600a , 0.6a (m s-2) A1

(4)

(b) Equation of motion – all terms needed M1

12000200 2 600 sin 600 0.05w g

w A1

A1

3 term quadratic and solve: 22 590 12000 0w w M1

2590 590 96000

19.14

w

(m s-1) A1

(5)

(9 marks)

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Question Scheme

Marks

5(a)

Mass From AD Rectangle 28a a

Semicircle 21

2a

4

3

a

Sign 2 82

a

h

Mass ratios B1

Moments about AD – all three terms, dimensionally correct M1

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

22 448

3 2 3 16

ah a

*Given Answer* A1

(4)

(b) Moments about A (or equivalent complete method)

442

3 16

aaT W

M1

22

2 3 16

hW WT

a

A1

(2)

(c)

Moments about AB - all terms and dimensionally correct M1

2 2 21 18 2 8

2 2a a a d a v

A1

32

16

a dv

A1

Correct trig for the given angle M1

11 44tan

18 3 32

h a

v a d

A1

24 32a a d , 8a d , 8a

d

A1

(6)

(12 marks)

D

C

B

A

α

h

v

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Question Scheme Marks

6(a)

CLM – all 4 terms & dimensionally correct M1

2 2 5 5 2m u m u m v m w , 5 2u v w A1

Impact law – used the right way round M1

2v w e u u A1

Solve for v: 5 2

6 2 2

u v w

eu v w

M1

7 6 1v u e 6 17

uv e

A1

Use of correct inequality for their v: 0v M1

1

16

e A1

(8)

(b) 1

3e

7

uv ,

6

7

uw B1

KE lost – terms of correct structure combined correctly M1

2 2

2 21 36 12 4 5

2 49 2 49

u um u m u

A1 A1

2

21 72 5 408 5

2 49 49 7

mumu

*Given Answer* A1

(5)

(c) Increase e more elastic less energy lost B1

(1)

(14 marks)

w v

u2u

Q5m

P2m

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Question Scheme Marks

7(a) In equilibrium no resultant vertical force. Use

xT

a

M1

3mgx

mga

A1

3

ax ,

4

3d a *Given Answer* A1

(3)

(b) Equation of motion. Use

xT

a

M1

3mga

mg mxa

A1

2x g A1

(3)

(c) Max speed at equilibrium position B1

Work energy & use of 2

EPE2

x

a

. All 4 terms needed. M1

2

22

33 1 23

2 2 2 3

amg

mga amv mg

a a

A1 A1

21 3 1 2 2

2 2 6 3 3v ga ga

,

4

3

gav A1

(5)

(d) At max ht. KE = 0. EPE lost = GPE gained M1

23

2

mgamgh

a A1

3

2

ah ,

2

aOB A1

(3)

(14 marks)

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Pearson Edexcel Level 3 GCE

Further Mathematics Advanced Paper 4: Further Mathematics Option 2 Option 4E: Further Mechanics 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 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 g = 9.8 m s-2 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 modelled as a non-uniform rod 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 10 125

xm

.

(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. A model car of mass m moves on the rough inner surface of the cone in a horizontal circle with centre C. The car moves with constant angular speed . The height of C above V is 3a, as shown in Figure 1. The coefficient

of friction between the car and the inner surface of the cone is 1

4. The car is modelled as

a particle.

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 circular 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 ϕ to the horizontal. The lower plane circular face of V is in contact with the inclined plane. It is given that

5h a and the plane is sufficiently rough to prevent V from slipping. If V is on the point of toppling,

(b) find, in degrees to three significant figures, the size of angle ϕ. (4)

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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 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 starts from rest at the fixed point O at time t = 0 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 newtons. At time T seconds, the car is at the point A, moving with speed 10 m s-1.

Show that

(a) 25

ln 26

T

(6)

(b) 9

125ln8

OA metres.

(5)

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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 11 marks)

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5. 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

(i) the velocity of A immediately after the collision,

(ii) the velocity of B immediately after the collision. (7)

(b) Find the impulse received by A in the collision. (2)

(c) 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.

(3)

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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 thin smooth 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. When the angle between OB and the

downward vertical is , the speed of B is v.

(a) Show that 2 3

2cos2

v ga

(4)

(b) Find the size of at the instant when the contact force between B and the hoop is first zero.

(5)

(c) 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 14 marks)

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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. Another 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 on the surface at the point on AB which is 2 m from B and then released from rest. In the subsequent motion neither string goes slack.

(a) Show that P moves with simple harmonic motion about its equilibrium position. (7)

(b) Find the maximum speed of P. (3)

(c) Find the length of time within one oscillation for which P is closer to A than to B. (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 14 marks)

TOTAL FOR PAPER IS 75 MARKS

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Paper 4: Further Mathematics Option 2 Option 4E: Further Mechanics 2 Mark Scheme

Question Scheme

Marks

1(a)

Total mass = 15

010 1 d

25

xx

M1

152

0

105

xx

A1

225150 105

5 (kg) *Given Answer* A1

(3)

(b)

15

010 1 d

25

xx x M1

15

2 3

0

25 675

15x x

A1

Taking moments about the base end: 105 675d M1

6.43d (m) 36

7 (m) A1

(4)

(7 marks)

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Question Scheme

Marks

2

Resolve vertically (needs all 3 terms) M1

cos sinmg F R A1

Horizontal equation of motion (all 3 terms) M1

2 23cos sin

2mr R F ma

A1

Use of F R M1

Eliminate R M1

Substitute for trig ratios: 23 9

2 2

a

g

M1

3g

a A1

(8 marks)

F

mg

R

θ

C

V

3a

4a

4a

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Question Scheme

Marks

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

Correct distances B1

Moments about a diameter through O. All three terms & dimensionally correct

M1

2 3 22 3 14 2 2

2 3 8 3

ha h a a a h a d

A1

22 2 23 88

8 623

ah h a

da h ah

*Given answer* A1

(5)

(b)

5 2.573...h a d a B1

About to topple so c of m above tipping point M1

2tan

5 2.573

a

a a

A1

39.5 A1

(4)

(9 marks)

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Question Scheme

Marks

4(a)

Equation of motion – all three terms & dimensionally correct 21800 2 500v a (when seen)

B1

Select form for a: d500

d

v

t M1

Separate and prepare to integrate:

2

2 1 1 1 1d d d

500 900 60 30 30t v v

v v v

M1

1 1ln 30 ln 30

250 60 60

tv v C A1

Use limits 25 30 10 25

ln ln26 30 10 6

T

*Given answer* M1 A1

(6)

(b) Equation of motion: 2d500 1800 2

d

vv v

x M1

Separate and integrate: 2

500d 1d

1800 2

vv x

v

M1

2125ln 1800 2v x C A1

Use limits 0 & 10: 125ln1600 125ln1800x M1

9125 ln

8x (m) *Given answer* A1

(5)

(11 marks)

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Question Scheme

Marks

5(a)

CLM parallel to loc – all 4 terms dimensionally correct. M1

2 3 3 5 3 2m m mw mv 9 3 2w v A1

Correct use of impact law M1

13 5

4v w 2 A1

Solve for v or w 3 2 9

2 2 4

w v

v w

M1

3 3A v i j , 2B v i j A1 A1

(7)

(b) Impulse = m mv u : 2 3 3 2 3 3m m i j i j M1

12m i (Ns) A1

(2)

(c) 5 2 i j 2 i j correct use of scalar product M1

5 4

cos29 5

A1

41.63... 42 (nearest degree) A1

Alternative method: 1 1 2tan 2 tan 41.63... 42

5

(nearest degree)

(3)

(12 marks)

wjvi

3j 2j

-5i+2j3i+3j

B(3m)A(2m)

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Question Scheme

Marks

6(a)

Conservation of energy – all terms required M1

21 1 71 cos

2 2 2mv mga m ga

A1

A1

2 7 3 3 2 1 cos 2 cos 2cos

2 2 2v ga ga ga ga ga

*Given Answer*

A1

(4)

(b) Resolve parallel to OB and use

2mv

a M1

2

cosmv

R mga

A1

Use 0R 2

cosv

ga

M1

Solve for 3

cos 2cos2

g g

M1

1

cos , 1202

A1

(5)

Rv

7

2ga

B

O

θ

mg

a

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Question Scheme

Marks

(c) At rest 0 v M1

3 cos =-

4 A1

Acceleration is tangential M1

Magnitude cos 90 6.48g (6.5) m s-2 or 7

4g A1

At 1 3cos 90 48.6

4

to the downward

vertical

A1

(5)

(14 marks)

Question Scheme

Marks

7(a)

Use of xT

a

: 20

2A

eT ,

50 2

2B

eT

M1

In equilibrium A BT T , 10 25 2e e M1

1035 50,

7e e A1

Equation of motion for P when distance x from equilibrium towards B:

M1

50 2 20

3.5 2 2B A

e x e xx T T

A1 A1

4 1050 20

7 7

2 2

x x

3.5 35 , 10x x x x and hence SHM A1

(7)

(b) Amplitude 10 42

7 7 B1

Use of max speed a M1

410 1.81

7 (m s-1) A1

(3)

P

50 N20 N

6m

BA

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Question Scheme

Marks

(c) Nearer to A than to B: 3

7x B1

Solve for 10t : 3cos 10

4t ,

10 2.418.............t

M1

Correct strategy for the required interval: 22.418...

10 M1

0.457 (seconds) A1

Alternative: 3.864 2.419

0.45710

Alternative:

4 3sin 10

7 7x t 10 0.8481 or 10 2.29353t t

1 20.2682, 0.72527t t

time 0.457 (seconds)

(4)

(14 marks)

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Pearson Edexcel Level 3 GCE

Further Mathematics Advanced Paper 3: Further Mathematics Option 1 Option 3D: Decision Mathematics 1 Paper 4: Further Mathematics Option 2 Option 4F: Decision Mathematics 1 Sample assessment material for first teaching September 2017 Time: 1 hour 30 minutes

Paper Reference(s)

9FM0/3D 9FM0/4F

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. 24 23 29 15 16 20 18 25

The numbers above are the scores for eight pupils in an examination.

(a) Show how a bubble sort is used to sort this list of numbers into ascending order. You need only give the state of the list after each pass.

(3)

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 5 marks)

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2. (a) Explain why it is not possible to draw a graph with exactly 5 nodes with orders 1, 3, 4, 4 and 5

(1)

A connected graph has exactly 5 nodes and contains 18 arcs. The orders of the 5 nodes are 2 1, 3 , 2, 4 and .x x x x

(b) (i) Calculate .x

(ii) State whether the graph is Eulerian, semi-Eulerian or neither. You must justify your answer.

(5)

(c) Draw a graph which satisfies all of the following conditions:

The graph has exactly 5 nodes.

The nodes have orders 2, 2, 4, 4 and 4

The graph is not Eulerian. (2)

(Total for Question 2 is 8 marks)

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3.

[The total weight of the network is 189]

Figure 1 represents a network of pipes in a building. The number on each arc is the length, in metres, of the corresponding pipe.

(a) Use Dijkstra’s algorithm to find the shortest path from A to F. State the path and its length.

(5)

On a particular day, Gabriel needs to check each pipe. A route of minimum length, which traverses each pipe at least once and which starts and finishes at A, needs to be found.

(b) Use an appropriate algorithm to find the pipes that will need to be traversed twice. You must make your method and working clear.

(4)

(c) Write down a possible route and state its length. (2)

A new pipe, BG, is added to the network. A route of minimum length that traverses each pipe, including BG, needs to be found. The route must start and finish at A. Given that this new pipe increases the length of the route found in (c) by twice the length of the new pipe BG,

(d) calculate the length of the new pipe BG. You must show your working. (3)

(Total for Question 3 is 14 marks)

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4.

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.

(a) Starting by deleting B, and all of its arcs, find a lower bound for Preety’s route. (4)

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,

(b) find ,x making your method clear. (4)

(c) 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 4 is 10 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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5. Jonathan makes two types of information pack for an event, Standard and Value.

Each Standard pack contains 25 posters and 500 flyers.

Each Value pack contains 15 posters and 800 flyers.

He must use at least 150 000 flyers.

Between 35% and 65% of the packs must be Standard packs.

Posters cost 20p each and flyers cost 4p each.

Jonathan wishes to minimise his costs.

Let x and y represent the number of Standard packs and Value packs produced respectively.

Formulate this as a linear programming problem, stating the objective and listing the constraints as simplified inequalities with integer coefficients.

You should not attempt to solve the problem.

(Total for Question 5 is 6 marks)

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6.

The network shown in Figure 2 shows the roads linking four towns, A, B, C and D. The number on each arc represents the distance, in miles, of the corresponding road. The initial distance and route tables for this network are as follows A B C D A B C D

A - 10 13 A A B C D B 10 - 26 8 B A B C D C 13 26 - 4 C A B C D D 8 4 - D A B C D

Use Floyd’s algorithm to find the complete table of shortest distances, show the distance and route tables after each iteration.

(Total for Question 6 is 8 marks)

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7.

The table above shows the activities required for the completion of a building project. For each activity, the table shows the time taken in days to complete the activity, the immediately preceding activities and the number of workers required. The project is to be completed in the shortest possible time.

(a) Draw the activity network described in the table, using activity on arc. Your activity network must contain only the minimum number of dummies.

(4)

(b) Show that the project can be completed in 21 days, identifying the critical activities. You must show your working.

(3)

(c) Draw a resource histogram to show the number of workers required each day when each activity begins at its earliest possible start time.

(3)

(d) Explain how it is possible to complete the project in the minimum project completion time of 21 days when only four workers are available.

(2)

(Total for Question 7 is 12 marks)

Activity Time taken (days)

Immediately preceding activities

Number of workers

A 5 - 1 B 7 - 2 C 3 - 2 D 4 A, B 1 E 4 D 2 F 2 B 1 G 4 B 1 H 5 C, G 2 I 10 C, G 1

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8. 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

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

(a) Using the information given, show how the equations represented in the final two rows of the above tableau were derived.

(3)

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

(b) Solve the linear programming problem, starting from this tableau. Make your method clear by stating the row operations you use and state the final values of

, , and . P x y z (9)

(Total for Question 8 is 12 marks)

TOTAL FOR PAPER IS 75 MARKS

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Pearson Edexcel Level 3 GCE

Further Mathematics Advanced Paper 3: Further Mathematics Option 1 Option 3D: Decision Mathematics 1 Paper 4: Further Mathematics Option 2 Option 4F: Decision Mathematics 1 Sample assessment material for first teaching September 2017

Paper Reference(s)

9FM0/3D 9FM0/4F

Total Marks

Answer Book Do not return the question paper with the answer book.

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1. 24 23 29 15 16 20 18 25

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Question 1 continued

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(Total for Question 1 is 5 marks)

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2. .....................................................................................................................................................................

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Question 2 continued

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(Total for Question 2 is 8 marks)

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3.

Shortest path: ........................................................................................................................................................................................................................

Length of shortest path: ........................................................................................................................................................................................................

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Question 3 continued

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(Total for Question 3 is 14 marks)

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4. 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 4 continued

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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(Total for Question 4 is 10 marks)

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5. ...............................................................................................................................................................................................................................

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(Total for Question 5 is 6 marks)

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6.

A B C D A B C D A - 10 13 A A B C D B 10 - 26 8 B A B C D C 13 26 - 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

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

(Total for Question 6 is 8 marks)

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7. (a) and (b)

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Question 7 continued

(c)

(d)

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(Total for Question 7 is 12 marks)

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8. (a)

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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 8 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 8 is 12 marks)

TOTAL FOR PAPER IS 75 MARKS

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Paper 3: Further Mathematics Option 1

Option 3D: Decision Mathematics 1 Mark Scheme

Paper 4: Further Mathematics Option 2

Option 4F: Decision Mathematics 1 Mark Scheme

Question Scheme

Marks

1(a)

24 23 29 15 16 20 18 25 23 24 15 16 20 18 25 29 23 15 16 20 18 24 25 29 15 16 20 18 23 24 25 29 15 16 18 20 23 24 25 29 15 16 18 20 23 24 25 29

M1

A1

A1

(3)

(b) 250000

0.0212000

t

M1

13.125t (seconds) A1

(2)

(5 marks)

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Question Scheme

Marks

2(a)

E.g. a graph cannot contain an odd number of odd nodes

E.g. number of arcs 1 3 4 4 5

8.52

B1

(1)

(b)(i)

(2 1) 3 ( 2) 4 2(18)x x x x M1

5x A1

(2)

(b)(ii) The order of the nodes are 9, 15, 3, 4, 5 M1

Therefore the graph is neither Eulerian nor semi-Eulerian as there are more than two odd nodes

A1

A1

(3)

(c)

M1

A1

(2)

(8 marks)

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Question Scheme

Marks

3(a)

M1

A1

A1

Path: ABECDGF A1

Length: 55 (metres) A1

(5)

(b) AB + DG = 13 + 11 = 24 ←

M1

A(BEC)D + B(ECD)G = 34 + 32 = 66

A1

A(BECD)G + B(EC)D = 45 + 21 = 66

A1

Repeat arcs: AB, DG A1

(4)

(c) Route (e.g.): ABACBECFGDGCDA B1

Length: 189 + 24 = 213 (metres) B1

(2)

(d) 189 34 213 2x x M1 A1

10x A1

(3)

(14 marks)

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Question Scheme

Marks

4(a)

Prim’s algorithm on reduced network starting at A: AD, AF, AE, CE, CG

M1 A1

Lower bound = 107 + 17 + 25 = 149 (km) M1 A1

(4)

(b) 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

(126 ) (139 ) 331 33x x x A1

(4)

(c) 149 optimal 159 M1 A1

(2)

(10 marks)

Question Scheme

Marks

5 Minimise ( ) 25 35C x y M1 A1

Subject to: (500 800 150 000 ) 5 8 1500x y x y B1

7 13

20 20x y x x y M1

Which simplifies to 7 13y x and 13 7y x , 0x y

A1 A1

(6 marks)

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Question Scheme

Marks

6

1st iteration:

A B C D A - 10 13 B 10 - 23 8 C 13 23 - 4 D 8 4 -

A B C D A A B C D B A B A D C A A C D D A B C D

M1

A1

2nd iteration:

A B C D A - 10 13 18 B 10 - 23 8 C 13 23 - 4 D 18 8 4 -

A B C D A A B C B B A B A D C A A C D D B B C D

M1

A1

3rd iteration:

A B C D A - 10 13 17 B 10 - 23 8 C 13 23 - 4 D 17 8 4 -

A B C D A A B C C B A B A D C A A C D D C B C D

M1

A1

4th iteration:

A B C D A - 10 13 17 B 10 - 12 8 C 13 12 - 4 D 17 8 4 -

A B C D A A B C C B A B D D C A D C D D C B C D

M1

A1

(8 marks)

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Question Scheme

Marks

7(a)(b)

Critical activities: B, G, I

M1 A1 A1 A1

(4)

M1 A1 A1

(3)

(c)

M1

A1

A1

(3)

(d) E.g. delay the start of activity A until time 3 then delay either activity E or H until the other activity is complete

M1 A1

(2)

(12 marks)

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Question Scheme

Marks

8(a)

3 13 3x x s t where 3s is a surplus variable and 1t is an

artificial variable B1

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

(3)

(b)

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

M1

A1

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

B1

M1

A1

47, 3, 0, 19P x y z M1 A1

(9)

(12 marks)

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Pearson Edexcel Level 3 GCE

Further Mathematics Advanced Paper 4: Further Mathematics Option 2 Option 4G: 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. (4)

(Total for Question 1 is 7 marks)

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2. Six workers, A, B, C, D, E and F, are to be assigned to five tasks, P, Q, R, S and T.

Each worker can be assigned to at most one task and each task must be done by just one worker.

The time, in minutes, that each worker takes to complete each task is shown in the table below.

Reducing rows first, use the Hungarian algorithm to obtain an allocation which minimises the total time. You must make your method clear and show the table after each stage.

(Total for Question 2 is 9 marks)

P Q R S T A 32 32 35 34 33 B 28 35 31 37 40 C 35 29 33 36 35 D 36 30 34 33 35 E 30 31 29 37 36 F 29 28 32 31 34

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3. In two-dimensional space, lines divide a plane into a number of different regions.

It is known that

One line divides a plane into 2 regions, as illustrated in Figure 1

Two lines divide a plane into a maximum of 4 regions, as illustrated in Figure 2

Three lines divide a plane into a maximum of 7 regions, as illustrated in Figure 3

Four lines divide a plane into a maximum of 11 regions, as illustrated in Figure 4

(a) Find the maximum number of regions when five, six and seven lines divide a plane. (1)

(b) Find, in terms of ,nu the recurrence relation for 1,nu the maximum number of regions

when a plane is divided by ( 1)n lines where 1.n (1)

(c) Solve the recurrence relation for nu and hence determine the maximum number of

regions created when 200 lines divide a plane. (3)

(Total for Question 3 is 5 marks)

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4.

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.

Using the north-west corner method to obtain an initial solution, show how the transportation algorithm is used to solve this problem. You must state the shadow costs, improvement indices, route, entering cell and exiting cell at each appropriate step.

(Total for Question 4 is 12 marks)

D E F AvailableA 15 19 9 25 B 11 18 10 55 C 11 12 18 20

Required 38 24 38

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5. 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 5 is 8 marks)

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6.

A two person zero-sum game is represented by the pay-off matrix for player A given above.

(a) Find the play-safe strategy for each player and verify that there is no stable solution to this game.

(4)

(b) Given that 1 2 3, ,p p p are the probabilities that A plays 1, 2, 3 respectively, formulate

the game as a linear programming problem for player A. You should write the constraints as inequalities.

(5)

The Simplex algorithm is used to solve the linear programming problem. The solution

obtained is 1 2 3

3 40, ,

7 7p p p

(c) Calculate the value of the game to player A. (3)

(Total for Question 6 is 12 marks)

B plays 1 B plays 2 B plays 3

A plays 1 4 2 3

A plays 2 3 1 2

A plays 3 1 2 0

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7.

Figure 5

Figure 5 shows a capacitated, directed network. The number on each arc ( , )x y represents the lower ( )x and upper ( )y capacities of that arc.

(a) Calculate the capacities of the cuts 1C and 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)

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.

(e) Explain how it is possible to maintain the maximum flow found in (d). (3)

(Total for Question 7 is 9 marks)

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8. 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 8 is 13 marks)

TOTAL FOR PAPER IS 75 MARKS

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Pearson Edexcel Level 3 GCE

Further Mathematics Advanced Paper 4: Further Mathematics Option 2 Option 4G: Decision Mathematics 2 Sample assessment material for first teaching September 2017

Paper Reference(s)

9FM0/4G Total Marks

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 7 marks)

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2. P Q R S T

A 32 32 35 34 33 B 28 35 31 37 40 C 35 29 33 36 35 D 36 30 34 33 35 E 30 31 29 37 36 F 29 28 32 31 34

P Q R S T A B C D E F

P Q R S T A B C D E F

P Q R S T A B C D E F

P Q R S T A B C D E F

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(Total for Question 2 is 9 marks)

P Q R S T A B C D E F

P Q R S T A B C D E F

P Q R S T A B C D E F

P Q R S T A B C D E F

P Q R S T A B C D E F

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3. .....................................................................................................................................................................

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Question 3 continued

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(Total for Question 3 is 5 marks)

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4. D E F Available

A 15 19 9 25 B 11 18 10 55 C 11 12 18 20

Required 38 24 38

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

D E F A B C

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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

D E F A B C

(Total for Question 4 is 12 marks)

D E F A B C

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5.

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Question 5 continued

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(Total for Question 5 is 8 marks)

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6.

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B plays 1 B plays 2 B plays 3

A plays 1 4 2 3

A plays 2 3 1 2

A plays 3 1 2 0

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Question 6 continued

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(Total for Question 6 is 12 marks)

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7. ......................................................................................................................................................................

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Question 7 continued

Diagram 1

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(Total for Question 7 is 9 marks)

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8. Stage State Action Dest Value

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Month January February March April May Number assembled

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Question 8 continued

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(Total for Question 8 is 13 marks)

TOTAL FOR PAPER IS 75 MARKS

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Paper 4: Further Mathematics Option 2 Option 4G: Decision Mathematics 2 Mark Scheme

Question Scheme

1(a) Auxiliary equation: 2 1 0 and attempt to solve

1 5

2

1 5 1 5

2 2

n n

nu A B

(b) Use given conditions to obtain two equations in A and B

Obtain 1 5 1 5 2A B and

2 2

1 5 1 5 4A B

Attempt to solve to obtain an A and B

1 1 5 1 5

2 25

n n

nu

(7 marks)

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Question Scheme

Marks

2

32 32 35 34 33 40

28 35 31 37 40 40

35 29 33 36 35 40

36 30 34 33 35 40

30 31 29 37 36 40

29 28 32 31 34 40

P Q R S T X

A

B

C

D

E

F

B1

Reducing rows and then columns

0 0 3 2 1 8

0 7 3 9 12 12

6 0 4 7 6 11

6 0 4 3 5 10

1 2 0 8 7 11

1 0 4 3 6 12

P Q R S T X

A

B

C

D

E

F

then

0 0 3 0 0 0

0 7 3 7 11 4

6 0 4 5 5 3

6 0 4 1 4 2

1 2 0 6 6 3

1 0 4 1 5 4

P Q R S T X

A

B

C

D

E

F

M1

A1

e.g. augment by 1 then augment by 1

1 1 3 0 0 0

0 7 2 6 10 3

6 0 3 4 4 2

6 0 3 0 4 1

2 3 0 6 6 3

1 0 3 0 4 3

P Q R S T X

A

B

C

D

E

F

followed by

2 2 3 1 0 0

0 7 1 6 9 2

6 0 2 4 3 1

6 0 2 0 3 0

3 4 0 7 6 3

1 0 2 0 3 2

P Q R S T X

A

B

C

D

E

F

M1

A1

M1

A1

A1

A – T, B – P, C – Q, (D – ), E – R, F – S A1

(9 marks)

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Question Scheme

Marks

3(a)

16, 22, 29 B1

(1)

(b) 1 1n nu u n B1

(1)

(c) As 1 ( )n nu u p n 2nu n n and attempt to solve

with 1,2,3n M1

11 1

2nu n n

20 101 (regions)

A1

A1

(3)

(5 marks)

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Question Scheme

Marks

4 D E F Available A 25 25 B 13 24 18 55 C 20 20 Required 38 24 38

B1

Shadowcosts

15 22 14

D E F 0 A X -3 -5 -4 B X X X 4 C -8 -14 X

M1

A1

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

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

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

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

(12 marks)

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Question Scheme

Marks

5(a)

M1

A1

M1

A1

M1

A1

(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

(2)

(8 marks)

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Question Scheme

Marks

6(a)

Row minima: -2, -1, -1 max is -1 Column maxima: 4, 2, 3 min is 2

M1 A1

Play safe is A plays 2 or 3 and B plays 2 A1

Row maximin (-1) Column minimax (2) so not stable A1

(4)

(b) e.g. 4 2 3 6 0 5

3 1 2 5 1 4

1 2 0 1 4 2

Maximise P V where V is the value of the (augmented) game

1 2 3

2 3

1 2 3

1 2 3

1 2 3

Subject to 6 5 0

4 0

5 4 2 0

1

, , 0

V p p p

V p p

V p p p

p p p

p p p

B1

B1

M1 A1 A1

(5)

(c) Substitute p values to obtain

19 19 20 19, ,

7 7 7 7V V M1

Value of the game to player A 19 5

27 7

M1 A1

(3)

(12 marks)

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Question Scheme

Marks

7(a)

1 3 3 3 5 7 21C

2 8 5 3 6 3 19C B1 B1

(2)

(b) 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

(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

(1)

(d)

M1

A1

(2)

(e) Increase the upper capacity of arc BF to at least 9 and therefore increase the flow in this arc to 9

B1

Increase the flow in FH and HT to 10 B1

The flow in GT decreases to 5 and all other arcs are unchanged B1

(3)

(9 marks)

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Question Scheme Marks

8 Stage State Action 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 A1

M1 A1

M1 A1

M1 A1

Month January February March April May Number made

3 3 5 3 4B1

Minimum production cost: £9450 B1

(13 marks)

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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 2018 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.

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Contents

1  Introduction 1 

Pure Mathematics 2 Statistics 2 

3  A Level in Mathematics 3 

Pure Mathematics 3 Statistics 5 

4  AS Level in Further Mathematics 7 

Pure Mathematics 7 Statistics 11 Mechanics 13 

5  A Level in Further Mathematics 14 

Pure Mathematics 14 Statistics 20 Mechanics 24 

6  Statistical Tables 25 

Binomial Cumulative Distribution Function 25 Poisson Cumulative Distribution Function 30 Percentage Points of the 2 Distribution 31 Critical Values for Correlation Coefficients 32 Random Numbers 33 Percentage Points of Student’s t Distribution 34 Percentage Points of the F Distribution 35 

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 1Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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.

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2 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further MathematicsMathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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 )

where C( )

nr

n n!

r r ! n r !

Logarithms and exponentials

loglog

logb

ab

xx

a

lne x a xa

Differentiation

First Principles

0

f ( ) f ( )f ( ) = lim

x

x x xx

x

Statistics

Standard deviation

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 Sxx

xx

n n

Statistical tables

The following statistical tables are required for A Level Mathematics:

Binomial Cumulative Distribution Function (see page 25)

Random Numbers (see page 33) 

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 3Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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 )

where C( )

nr

n n!

r r ! n r !

2( 1) ( 1) ( 1)(1 ) 1 ( 1 )

1 2 1 2n rn n n n n r

x nx x x x , nr

Logarithms and exponentials

loglog

logb

ab

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

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4 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further MathematicsMathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

Trigonometric identities

sin ( ) sin cos cos sinA B A B A B

cos ( ) cos cos sin sinA B A B A B

12

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

x

x x xx

x

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

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 5 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

Integration (+ constant)

f(x) f( ) d

x x

sec2 kx k

1 tan kx

tankx k

1ln sec kx

cot kx k

1ln sin kx

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

nnn x

xxx

Statistics

Probability

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

Standard deviation

For a set of n values 1 2, ,... ,...i nx x x x

n

xxxxS i

iixx

222 )(

)(

Standard deviation = 2

2xxxS

xn n

or

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6 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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)

Critical Values for Correlation Coefficients: Product Moment Coefficient (see page 32)

Random Numbers (see page 33)

 

 

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 7 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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(61

1

2

nnnrn

r

 

2241

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 nr r n n  

The roots of 1nz are given by 2 i

ek

nz

, for 1 , ,2 ,1 ,0 nk

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8 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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 21 2 3

n n n d

n n n

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 9 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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

xarccos    21

1

x  

xarctan    21

1

xsinh       xcosh  

xcosh      xsinh  

xtanh       x2sech  

xarsinh    21

1

xarcosh    1

12 x

 

artanh x  21

1

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10 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

Integration (+ constant; 0a where relevant)

f(x)

xx d)f(

xsinh       xcosh  

xcosh      xsinh  

xtanh       xcoshln  

22

1

xa     )( arcsin ax

a

x

 

22

1

xa    

a

x

aarctan

22

1

ax     )( ln,arcosh }{ 22 axaxx

a

x

 

22

1

xa     }{ 22ln,arsinh axx

a

x

 

22

1

xa     )( artanh

1ln

2

1ax

a

x

axa

xa

a

 

22

1

ax    

ax

ax

a

ln2

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 11 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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 ) = 2i

x P(X = ix ) – 2

Discrete distributions

Standard discrete distributions:

Distribution of X P(X = x) Mean Variance

Binomial B(n, p) 1n xxn

p px

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 ( ) dx

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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12 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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 ix y

x x y y x yn

The product moment correlation coefficient is:

2 2 2 22 2

( )( )S ( )( )

S S ( ) ( ) ( ) ( )

{ }{ }

i ii ixy i i

xx yy i i i ii i

x yx yx x y y nr

x x y y x yx y

n 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 yyxx

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:  22

~)(

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 30)

Percentage Points of the 2 Distribution (see page 31)

Critical Values for Correlation Coefficients: Product Moment Coefficient and Spearman’s Coefficient (see page 32)

Random Numbers (see page 33)

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 13 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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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14 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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(61

1

2

nnnrn

r

 

2241

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 nr r n n

The roots of 1nz are given by 2 i

ek

nz

, for 1 , ,2 ,1 ,0 nk

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 15 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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 21 2 3

n n n d

n n n

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Page 344: A Level Further Mathematics - Weebly · 2020. 2. 22. · A Level Further Mathematics ... Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding

16 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 17 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

Differentiation

f(x) f(x)

xarcsin    21

1

xarccos    21

1

x  

xarctan    21

1

xsinh       xcosh  

xcosh      xsinh  

xtanh       x2sech  

xarsinh    21

1

xarcosh    1

12 x

 

artanh x    21

1

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18 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

Integration (+ constant; 0a where relevant)

f(x)

xx d)f(

xsinh       xcosh  

xcosh      xsinh  

xtanh       xcoshln  

22

1

xa     )( arcsin ax

a

x

 

22

1

xa    

a

x

aarctan

22

1

ax     )( ln,arcosh }{ 22 axaxx

a

x

 

22

1

xa     }{ 22ln,arsinh axx

a

x

 

22

1

xa     )( artanh

1ln

2

1ax

a

x

axa

xa

a

 

22

1

ax    

ax

ax

a

ln2

Arc length

xx

ys d

d

d1

2

      (cartesian coordinates)

tt

y

t

xs d

d

d

d

d22

    (parametric form) 

22 dr

dd

s r

     (polar form) 

DRAFT

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 19 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

Surface area of revolution

Sx = 2

d2 1 d

d

yy x

x

    (cartesian coordinates)

Sx = 2 2

d d2 d

d d

x yy t

t t

   (parametric form) 

Sx = 2

2 dr2 sin d

dr r

  (polar form) 

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20 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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) = 2i

x 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 ( ) E X

X t t and

2E( ) G (1) and Var( ) G (1) G (1) G (1)X X X XX X

For Z = X + Y, where X and Y are independent: G ( ) G ( ) G ( )Z X Yt t t

Discrete distributions

Standard discrete distributions: 

Distribution of X P(X = x) Mean Variance P.G.F.

Binomial B(n, p) 1n xxn

p px

np np(1 – p) 1

np pt

Poisson Po ( ) e!

x

x 1e t

Geometric Geo(p) on 1, 2, … 1

1x

p p

1

p

2

1 p

p

1 (1 )

pt

p t

Negative binomial on r, r + 1, …

1(1 )

1r x rx

p pr

r

p

2

(1 )r p

p

1 (1 )

rpt

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 ( ) dx

x X x t t

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 21 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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 ix y

x x y y x yn

The product moment correlation coefficient is

2 2 2 22 2

( )( )S ( )( )

S S ( ) ( ) ( ) ( )

{ }{ }

i ii ixy i i

xx yy i i i ii i

x yx yx x y y nr

x x y y x yx y

n 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 yyxx

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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22 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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

)( 22

n

XXS i

For a random sample of n observations from 2N( , ) , ~ N(0, 1)/

X

n

For a random sample of xn observations from ) ,N( 2xx

and, independently, a random

sample of yn observations from ) ,N( 2yy

, )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 : 

212

2

~)1(

n

Sn

 

1~

/

n

tnS

X       (also valid in matched‐pairs situations) 

For a random sample of  xn  observations from  ) ,N( 2xx

 and, independently, a random sample 

of y

n  observations from  ) ,N( 2yy

 

1 ,122

22

~/

/ ynxn

yy

xx FS

S

 

If 222

yx (unknown) then  

2

2

( ) ( )~

1 1

x yn nx y

px y

X Yt

Sn n

    where   

2 22 ( 1) ( 1)

2

x x y yp

x y

n S n SS

n n

 

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 23 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

Non-parametric tests 

Goodness-of-fit test and contingency tables: 22

~)(

i

ii

E

EO

Statistical tables

The following statistical tables are required for A Level Further Mathematics:

Binomial Cumulative Distribution Function (see page 25)

Poisson Cumulative Distribution Function (see page 30)

Percentage Points of the 2 Distribution (see page 31)

Critical Values for Correlation Coefficients: Product Moment Coefficient and Spearman’s Coefficient (see page 32)

Random Numbers (see page 33)

Percentage Points of Student’s t Distribution (see page 34)

Percentage Points of the F Distribution (see page 35)

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24 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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: r83 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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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 25 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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 0.50n = 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 0.18752 0.9988 0.9914 0.9734 0.9421 0.8965 0.8369 0.7648 0.6826 0.5931 0.50003 1.0000 0.9995 0.9978 0.9933 0.9844 0.9692 0.9460 0.9130 0.8688 0.81254 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 0.01561 0.9672 0.8857 0.7765 0.6554 0.5339 0.4202 0.3191 0.2333 0.1636 0.10942 0.9978 0.9842 0.9527 0.9011 0.8306 0.7443 0.6471 0.5443 0.4415 0.34383 0.9999 0.9987 0.9941 0.9830 0.9624 0.9295 0.8826 0.8208 0.7447 0.65634 1.0000 0.9999 0.9996 0.9984 0.9954 0.9891 0.9777 0.9590 0.9308 0.89065 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 0.00781 0.9556 0.8503 0.7166 0.5767 0.4449 0.3294 0.2338 0.1586 0.1024 0.06252 0.9962 0.9743 0.9262 0.8520 0.7564 0.6471 0.5323 0.4199 0.3164 0.22663 0.9998 0.9973 0.9879 0.9667 0.9294 0.8740 0.8002 0.7102 0.6083 0.50004 1.0000 0.9998 0.9988 0.9953 0.9871 0.9712 0.9444 0.9037 0.8471 0.77345 1.0000 1.0000 0.9999 0.9996 0.9987 0.9962 0.9910 0.9812 0.9643 0.93756 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 0.00391 0.9428 0.8131 0.6572 0.5033 0.3671 0.2553 0.1691 0.1064 0.0632 0.03522 0.9942 0.9619 0.8948 0.7969 0.6785 0.5518 0.4278 0.3154 0.2201 0.14453 0.9996 0.9950 0.9786 0.9437 0.8862 0.8059 0.7064 0.5941 0.4770 0.36334 1.0000 0.9996 0.9971 0.9896 0.9727 0.9420 0.8939 0.8263 0.7396 0.63675 1.0000 1.0000 0.9998 0.9988 0.9958 0.9887 0.9747 0.9502 0.9115 0.85556 1.0000 1.0000 1.0000 0.9999 0.9996 0.9987 0.9964 0.9915 0.9819 0.96487 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 0.00201 0.9288 0.7748 0.5995 0.4362 0.3003 0.1960 0.1211 0.0705 0.0385 0.01952 0.9916 0.9470 0.8591 0.7382 0.6007 0.4628 0.3373 0.2318 0.1495 0.08983 0.9994 0.9917 0.9661 0.9144 0.8343 0.7297 0.6089 0.4826 0.3614 0.25394 1.0000 0.9991 0.9944 0.9804 0.9511 0.9012 0.8283 0.7334 0.6214 0.50005 1.0000 0.9999 0.9994 0.9969 0.9900 0.9747 0.9464 0.9006 0.8342 0.74616 1.0000 1.0000 1.0000 0.9997 0.9987 0.9957 0.9888 0.9750 0.9502 0.91027 1.0000 1.0000 1.0000 1.0000 0.9999 0.9996 0.9986 0.9962 0.9909 0.98058 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 0.00101 0.9139 0.7361 0.5443 0.3758 0.2440 0.1493 0.0860 0.0464 0.0233 0.01072 0.9885 0.9298 0.8202 0.6778 0.5256 0.3828 0.2616 0.1673 0.0996 0.05473 0.9990 0.9872 0.9500 0.8791 0.7759 0.6496 0.5138 0.3823 0.2660 0.17194 0.9999 0.9984 0.9901 0.9672 0.9219 0.8497 0.7515 0.6331 0.5044 0.37705 1.0000 0.9999 0.9986 0.9936 0.9803 0.9527 0.9051 0.8338 0.7384 0.62306 1.0000 1.0000 0.9999 0.9991 0.9965 0.9894 0.9740 0.9452 0.8980 0.82817 1.0000 1.0000 1.0000 0.9999 0.9996 0.9984 0.9952 0.9877 0.9726 0.94538 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9995 0.9983 0.9955 0.98939 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9990

 

   

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26 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

p = 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50n = 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 0.00322 0.9804 0.8891 0.7358 0.5583 0.3907 0.2528 0.1513 0.0834 0.0421 0.01933 0.9978 0.9744 0.9078 0.7946 0.6488 0.4925 0.3467 0.2253 0.1345 0.07304 0.9998 0.9957 0.9761 0.9274 0.8424 0.7237 0.5833 0.4382 0.3044 0.19385 1.0000 0.9995 0.9954 0.9806 0.9456 0.8822 0.7873 0.6652 0.5269 0.38726 1.0000 0.9999 0.9993 0.9961 0.9857 0.9614 0.9154 0.8418 0.7393 0.61287 1.0000 1.0000 0.9999 0.9994 0.9972 0.9905 0.9745 0.9427 0.8883 0.80628 1.0000 1.0000 1.0000 0.9999 0.9996 0.9983 0.9944 0.9847 0.9644 0.92709 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 0.996811 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 0.00001 0.8290 0.5490 0.3186 0.1671 0.0802 0.0353 0.0142 0.0052 0.0017 0.00052 0.9638 0.8159 0.6042 0.3980 0.2361 0.1268 0.0617 0.0271 0.0107 0.00373 0.9945 0.9444 0.8227 0.6482 0.4613 0.2969 0.1727 0.0905 0.0424 0.01764 0.9994 0.9873 0.9383 0.8358 0.6865 0.5155 0.3519 0.2173 0.1204 0.05925 0.9999 0.9978 0.9832 0.9389 0.8516 0.7216 0.5643 0.4032 0.2608 0.15096 1.0000 0.9997 0.9964 0.9819 0.9434 0.8689 0.7548 0.6098 0.4522 0.30367 1.0000 1.0000 0.9994 0.9958 0.9827 0.9500 0.8868 0.7869 0.6535 0.50008 1.0000 1.0000 0.9999 0.9992 0.9958 0.9848 0.9578 0.9050 0.8182 0.69649 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 0.940811 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9995 0.9981 0.9937 0.982412 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9997 0.9989 0.996313 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.999514 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 0.00001 0.7358 0.3917 0.1756 0.0692 0.0243 0.0076 0.0021 0.0005 0.0001 0.00002 0.9245 0.6769 0.4049 0.2061 0.0913 0.0355 0.0121 0.0036 0.0009 0.00023 0.9841 0.8670 0.6477 0.4114 0.2252 0.1071 0.0444 0.0160 0.0049 0.00134 0.9974 0.9568 0.8298 0.6296 0.4148 0.2375 0.1182 0.0510 0.0189 0.00595 0.9997 0.9887 0.9327 0.8042 0.6172 0.4164 0.2454 0.1256 0.0553 0.02076 1.0000 0.9976 0.9781 0.9133 0.7858 0.6080 0.4166 0.2500 0.1299 0.05777 1.0000 0.9996 0.9941 0.9679 0.8982 0.7723 0.6010 0.4159 0.2520 0.13168 1.0000 0.9999 0.9987 0.9900 0.9591 0.8867 0.7624 0.5956 0.4143 0.25179 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 0.588111 1.0000 1.0000 1.0000 0.9999 0.9991 0.9949 0.9804 0.9435 0.8692 0.748312 1.0000 1.0000 1.0000 1.0000 0.9998 0.9987 0.9940 0.9790 0.9420 0.868413 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9985 0.9935 0.9786 0.942314 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9984 0.9936 0.979315 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9985 0.994116 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.998717 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999818 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 27 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

p = 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50n = 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 0.00002 0.8729 0.5371 0.2537 0.0982 0.0321 0.0090 0.0021 0.0004 0.0001 0.00003 0.9659 0.7636 0.4711 0.2340 0.0962 0.0332 0.0097 0.0024 0.0005 0.00014 0.9928 0.9020 0.6821 0.4207 0.2137 0.0905 0.0320 0.0095 0.0023 0.00055 0.9988 0.9666 0.8385 0.6167 0.3783 0.1935 0.0826 0.0294 0.0086 0.00206 0.9998 0.9905 0.9305 0.7800 0.5611 0.3407 0.1734 0.0736 0.0258 0.00737 1.0000 0.9977 0.9745 0.8909 0.7265 0.5118 0.3061 0.1536 0.0639 0.02168 1.0000 0.9995 0.9920 0.9532 0.8506 0.6769 0.4668 0.2735 0.1340 0.05399 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 0.212211 1.0000 1.0000 0.9999 0.9985 0.9893 0.9558 0.8746 0.7323 0.5426 0.345012 1.0000 1.0000 1.0000 0.9996 0.9966 0.9825 0.9396 0.8462 0.6937 0.500013 1.0000 1.0000 1.0000 0.9999 0.9991 0.9940 0.9745 0.9222 0.8173 0.655014 1.0000 1.0000 1.0000 1.0000 0.9998 0.9982 0.9907 0.9656 0.9040 0.787815 1.0000 1.0000 1.0000 1.0000 1.0000 0.9995 0.9971 0.9868 0.9560 0.885216 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9992 0.9957 0.9826 0.946117 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9988 0.9942 0.978418 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9984 0.992719 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9996 0.998020 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.999521 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999922 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 0.00001 0.5535 0.1837 0.0480 0.0105 0.0020 0.0003 0.0000 0.0000 0.0000 0.00002 0.8122 0.4114 0.1514 0.0442 0.0106 0.0021 0.0003 0.0000 0.0000 0.00003 0.9392 0.6474 0.3217 0.1227 0.0374 0.0093 0.0019 0.0003 0.0000 0.00004 0.9844 0.8245 0.5245 0.2552 0.0979 0.0302 0.0075 0.0015 0.0002 0.00005 0.9967 0.9268 0.7106 0.4275 0.2026 0.0766 0.0233 0.0057 0.0011 0.00026 0.9994 0.9742 0.8474 0.6070 0.3481 0.1595 0.0586 0.0172 0.0040 0.00077 0.9999 0.9922 0.9302 0.7608 0.5143 0.2814 0.1238 0.0435 0.0121 0.00268 1.0000 0.9980 0.9722 0.8713 0.6736 0.4315 0.2247 0.0940 0.0312 0.00819 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 0.049411 1.0000 1.0000 0.9992 0.9905 0.9493 0.8407 0.6548 0.4311 0.2327 0.100212 1.0000 1.0000 0.9998 0.9969 0.9784 0.9155 0.7802 0.5785 0.3592 0.180813 1.0000 1.0000 1.0000 0.9991 0.9918 0.9599 0.8737 0.7145 0.5025 0.292314 1.0000 1.0000 1.0000 0.9998 0.9973 0.9831 0.9348 0.8246 0.6448 0.427815 1.0000 1.0000 1.0000 0.9999 0.9992 0.9936 0.9699 0.9029 0.7691 0.572216 1.0000 1.0000 1.0000 1.0000 0.9998 0.9979 0.9876 0.9519 0.8644 0.707717 1.0000 1.0000 1.0000 1.0000 0.9999 0.9994 0.9955 0.9788 0.9286 0.819218 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9986 0.9917 0.9666 0.899819 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.9971 0.9862 0.950620 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9991 0.9950 0.978621 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9984 0.991922 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.997423 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.999324 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999825 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

 

   

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28 Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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 0.00002 0.6767 0.2228 0.0486 0.0079 0.0010 0.0001 0.0000 0.0000 0.0000 0.00003 0.8619 0.4231 0.1302 0.0285 0.0047 0.0006 0.0001 0.0000 0.0000 0.00004 0.9520 0.6290 0.2633 0.0759 0.0160 0.0026 0.0003 0.0000 0.0000 0.00005 0.9861 0.7937 0.4325 0.1613 0.0433 0.0086 0.0013 0.0001 0.0000 0.00006 0.9966 0.9005 0.6067 0.2859 0.0962 0.0238 0.0044 0.0006 0.0001 0.00007 0.9993 0.9581 0.7559 0.4371 0.1820 0.0553 0.0124 0.0021 0.0002 0.00008 0.9999 0.9845 0.8646 0.5931 0.2998 0.1110 0.0303 0.0061 0.0009 0.00019 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 0.001111 1.0000 0.9996 0.9880 0.9125 0.7151 0.4406 0.2053 0.0709 0.0179 0.003212 1.0000 0.9999 0.9957 0.9568 0.8209 0.5772 0.3143 0.1285 0.0386 0.008313 1.0000 1.0000 0.9986 0.9806 0.8968 0.7032 0.4408 0.2112 0.0751 0.019214 1.0000 1.0000 0.9996 0.9921 0.9456 0.8074 0.5721 0.3174 0.1326 0.040315 1.0000 1.0000 0.9999 0.9971 0.9738 0.8849 0.6946 0.4402 0.2142 0.076916 1.0000 1.0000 1.0000 0.9990 0.9884 0.9367 0.7978 0.5681 0.3185 0.134117 1.0000 1.0000 1.0000 0.9997 0.9953 0.9680 0.8761 0.6885 0.4391 0.214818 1.0000 1.0000 1.0000 0.9999 0.9983 0.9852 0.9301 0.7911 0.5651 0.317919 1.0000 1.0000 1.0000 1.0000 0.9994 0.9937 0.9637 0.8702 0.6844 0.437320 1.0000 1.0000 1.0000 1.0000 0.9998 0.9976 0.9827 0.9256 0.7870 0.562721 1.0000 1.0000 1.0000 1.0000 1.0000 0.9991 0.9925 0.9608 0.8669 0.682122 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9970 0.9811 0.9233 0.785223 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9989 0.9917 0.9595 0.865924 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.9966 0.9804 0.923125 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9988 0.9914 0.959726 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.9966 0.980827 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9988 0.991728 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9996 0.996829 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.998930 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999731 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999932 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

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Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics 29 Mathematical Formulae and Statistical Tables – Draft 1.0 – June 2016 - © Pearson Education Limited 2016

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 0.00002 0.5405 0.1117 0.0142 0.0013 0.0001 0.0000 0.0000 0.0000 0.0000 0.00003 0.7604 0.2503 0.0460 0.0057 0.0005 0.0000 0.0000 0.0000 0.0000 0.00004 0.8964 0.4312 0.1121 0.0185 0.0021 0.0002 0.0000 0.0000 0.0000 0.00005 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 0.00007 0.9968 0.8779 0.5188 0.1904 0.0453 0.0073 0.0008 0.0001 0.0000 0.00008 0.9992 0.9421 0.6681 0.3073 0.0916 0.0183 0.0025 0.0002 0.0000 0.00009 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 0.000012 1.0000 0.9990 0.9699 0.8139 0.5110 0.2229 0.0661 0.0133 0.0018 0.000213 1.0000 0.9997 0.9868 0.8894 0.6370 0.3279 0.1163 0.0280 0.0045 0.000514 1.0000 0.9999 0.9947 0.9393 0.7481 0.4468 0.1878 0.0540 0.0104 0.001315 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 0.007717 1.0000 1.0000 0.9998 0.9937 0.9449 0.7822 0.5060 0.2369 0.0765 0.016418 1.0000 1.0000 0.9999 0.9975 0.9713 0.8594 0.6216 0.3356 0.1273 0.032519 1.0000 1.0000 1.0000 0.9991 0.9861 0.9152 0.7264 0.4465 0.1974 0.059520 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 0.161122 1.0000 1.0000 1.0000 1.0000 0.9990 0.9877 0.9290 0.7660 0.5019 0.239923 1.0000 1.0000 1.0000 1.0000 0.9996 0.9944 0.9604 0.8438 0.6134 0.335924 1.0000 1.0000 1.0000 1.0000 0.9999 0.9976 0.9793 0.9022 0.7160 0.443925 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 0.664127 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9981 0.9840 0.9220 0.760128 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9993 0.9924 0.9556 0.838929 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.9966 0.9765 0.898730 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 0.967532 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9998 0.9978 0.983633 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9999 0.9991 0.992334 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9997 0.996735 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 0.999537 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.999838 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

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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

   

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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 0.0051 0.000 0.000 0.001 0.004 0.016 2.705 3.841 5.024 6.635 7.8792 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.5973 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.8384 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.8605 0.412 0.554 0.831 1.145 1.610 9.236 11.070 12.832 15.086 16.7506 0.676 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.812 18.5487 0.989 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.475 20.2788 1.344 1.646 2.180 2.733 3.490 13.362 15.507 17.535 20.090 21.9559 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 25.18811 2.603 3.053 3.816 4.575 5.580 17.275 19.675 21.920 24.725 26.75712 3.074 3.571 4.404 5.226 6.304 18.549 21.026 23.337 26.217 28.30013 3.565 4.107 5.009 5.892 7.042 19.812 22.362 24.736 27.688 29.81914 4.075 4.660 5.629 6.571 7.790 21.064 23.685 26.119 29.141 31.31915 4.601 5.229 6.262 7.261 8.547 22.307 24.996 27.488 30.578 32.80116 5.142 5.812 6.908 7.962 9.312 23.542 26.296 28.845 32.000 34.26717 5.697 6.408 7.564 8.672 10.085 24.769 27.587 30.191 33.409 35.71818 6.265 7.015 8.231 9.390 10.865 25.989 28.869 31.526 34.805 37.15619 6.844 7.633 8.907 10.117 11.651 27.204 30.144 32.852 36.191 38.58220 7.434 8.260 9.591 10.851 12.443 28.412 31.410 34.170 37.566 39.99721 8.034 8.897 10.283 11.591 13.240 29.615 32.671 35.479 38.932 41.40122 8.643 9.542 10.982 12.338 14.042 30.813 33.924 36.781 40.289 42.79623 9.260 10.196 11.689 13.091 14.848 32.007 35.172 38.076 41.638 44.18124 9.886 10.856 12.401 13.848 15.659 33.196 36.415 39.364 42.980 45.55825 10.520 11.524 13.120 14.611 16.473 34.382 37.652 40.646 44.314 46.92826 11.160 12.198 13.844 15.379 17.292 35.563 38.885 41.923 45.642 48.29027 11.808 12.879 14.573 16.151 18.114 36.741 40.113 43.194 46.963 49.64528 12.461 13.565 15.308 16.928 18.939 37.916 41.337 44.461 48.278 50.99329 13.121 14.256 16.047 17.708 19.768 39.088 42.557 45.722 49.588 52.33630 13.787 14.953 16.791 18.493 20.599 40.256 43.773 46.979 50.892 53.672

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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 Level

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

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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

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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

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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.

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