INTERNATIONAL ADVANCED LEVEL
EDEXCEL INTERNATIONAL GCSE
MATHEMATICS/
FURTHER
MATHEMATICS/
ECONOMICS
PURE MATHEMATICS
SPECIFICATION
Edexcel International GCSE in Economics (9-1) (4ET0)
SAMPLE ASSESSMENT
MATERIALS
First examination June
Pearson Edexcel International Advanced Subsidiary in Mathematics (XMA01)
Pearson Edexcel International Advanced Subsidiary in Further Mathematics (XFM01)
Pearson Edexcel International Advanced Subsidiary in Pure Mathematics (XPM01)
Pearson Edexcel International Advanced Level in Mathematics (YMA01)
Pearson Edexcel International Advanced Level in Further Mathematics (YFM01)
Pearson Edexcel International Advanced Level in Pure Mathematics (YPM01)
First teaching September 2018
First examination from January 2019
First certification from August 2019 (International
Advanced Subsidiary) and August 2020
(International Advanced Level)
Issue 3
Edexcel, BTEC and LCCI qualifications
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body offering academic and vocational qualifications that are globally recognised and
benchmarked. For further information, please visit our qualification website at
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Acknowledgements
These sample assessment materials have been produced by Pearson on the basis of
consultation with teachers, examiners, consultants and other interested parties. Pearson
would like to thank all those who contributed their time and expertise to the development.
References to third party material made in these 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 going to publication.
ISBN 978 1 4469 4982 5
All the material in this publication is copyright
© Pearson Education Limited 2018
Summary of Pearson Edexcel International Advanced
Subsidiary/Advanced Level in Mathematics, Further
Mathematics and Pure Mathematics Sample Assessment
Materials Issue 3 changes
Summary of changes made between previous issue and this
current issue
An alternative solution for Statistics 3, Question 5 has been inserted
in the mark scheme.
Page numbers
548
If you need further information on these changes or what they mean, contact us via our
website at: qualifications.pearson.com/en/support/contact-us.html.
Contents
Introduction
1
General marking guidance
3
Unit P1 – sample question paper and mark scheme
5
Unit P2 – sample question scheme and mark scheme
45
Unit P3 – sample question paper and mark scheme
87
Unit P4 – sample question paper and mark scheme
137
Unit FP1 – sample question paper and mark scheme
181
Unit FP2 – sample question paper and mark scheme
235
Unit FP3 – sample question paper and mark scheme
277
Unit M1 – sample question paper and mark scheme
323
Unit M2 – sample question paper and mark scheme
357
Unit M3 – sample question paper and mark scheme
399
Unit S1 – sample question paper and mark scheme
433
Unit S2 – sample question paper and mark scheme
473
Unit S3 – sample question paper and mark scheme
513
Unit D1 – sample question paper and mark scheme
557
Introduction
The Pearson Edexcel International Advanced Subsidiary in Mathematics, Further
Mathematics and Pure Mathematics and the Pearson Edexcel International Advanced
Level in Mathematics, Further Mathematics and Pure Mathematics are part of a suite of
International Advanced Level qualifications offered by Pearson.
These sample assessment materials have been developed to support these qualifications
and will be used as the benchmark to develop the assessment students will take.
For units P1, P2, P3, P4 and D1, the sample assessment materials have been formed
using questions from different past papers from legacy qualifications, together with some
new questions. For units FP1-FP3, M1-M3 and S1-S3, the sample assessment materials
have been formed using whole past question papers from legacy qualifications.
The booklet ‘Mathematical Formulae and Statistical Tables’ will be provided for use with
these assessments and can be downloaded from our website, qualifications.pearson.com.
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
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Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 2 – March 2018 © Pearson Education Limited 2018
General marking guidance
All candidates must receive the same treatment. Examiners must mark the last
candidate in exactly the same way as they mark the first.
Mark schemes should be applied positively. Candidates must be rewarded for what
they have shown they can do rather than be penalised for omissions.
Examiners should mark according to the mark scheme – not according to their
perception of where the grade boundaries may lie.
All the marks on the mark scheme are designed to be awarded. Examiners should
always award full marks if deserved, i.e. if the answer matches the mark scheme.
Examiners should also be prepared to award zero marks if the candidate’s response is
not worthy of credit according to the mark scheme.
Where some judgement is required, mark schemes will provide the principles by
which marks will be awarded and exemplification/indicative content will not be
exhaustive. However different examples of responses will be provided at
standardisation.
When examiners are in doubt regarding the application of the mark scheme to a
candidate’s response, a senior examiner must be consulted before a mark is given.
Crossed-out work should be marked unless the candidate has replaced it with an
alternative response.
Specific guidance for mathematics
1. These mark schemes use the following types of marks:
M marks: Method marks are awarded for ‘knowing a method and attempting to
apply it’, unless otherwise indicated.
A marks: Accuracy marks can only be awarded if the relevant method (M) marks
have been earned.
B marks are unconditional accuracy marks (independent of M marks)
Marks should not be subdivided.
2. Abbreviations
These are some of the traditional marking abbreviations that may appear in the mark
schemes.
bod
benefit of doubt
SC:
special case
ft
follow through
o.e.
this symbol is used for
correct ft
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
cao
correct answer only
cso
correct solution only.
There must be no errors in
this part of the question to
obtain this mark
isw
awrt answers which round to
ignore subsequent working
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Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
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or d… The second mark is
dependent on gaining the
first mark
3. All M marks are follow through.
All A marks are ‘correct answer only’ (cao.), unless shown, for example, as A1 ft to
indicate that previous wrong working is to be followed through. After a misread
however, the subsequent A marks affected are treated as A ft, but answers that don’t
logically make sense e.g. if an answer given for a probability is >1 or <0, should
never be awarded A marks.
4. For misreading which does not alter the character of a question or materially simplify
it, deduct two from any A or B marks gained, in that part of the question affected.
5. Where a candidate has made multiple responses and indicates which response they
wish to submit, examiners should mark this response. If there are several attempts at
a question which have not been crossed out, examiners should mark the final answer
which is the answer that is the most complete.
6. Ignore wrong working or incorrect statements following a correct answer.
7. Mark schemes will firstly show the solution judged to be the most common response
expected from candidates. Where appropriate, alternative answers are provided in the
notes. If examiners are not sure if an answer is acceptable, they will check the mark
scheme to see if an alternative answer is given for the method used. If no such
alternative answer is provided but deemed to be valid, examiners must escalate the
response to a senior examiner to review.
4
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Please check the examination details below before entering your candidate information
Candidate surname
Pearson Edexcel
Other names
Centre Number
Candidate Number
International
Advanced Level
Sample Assessment Materials for first teaching September 2018
(Time: 1 hour 30 minutes)
Paper Reference WMA11/01
Mathematics
International Advanced Subsidiary/Advanced Level
Pure Mathematics P1
You must have:
Mathematical Formulae and Statistical Tables, calculator
Total Marks
Candidates may use any calculator permitted by Pearson regulations.
Calculators must not have the facility for symbolic algebra manipulation,
differentiation and integration, or have retrievable mathematical
formulae stored in them.
Instructions
black ink or ball-point pen.
•• Use
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
in the boxes at the top of this page with your name, centre number and
• Fill
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.
Inexact
answers
should
be given to three significant figures unless otherwise
• stated.
Information
booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
•• AThere
are 10 questions in this question paper. The total mark for this paper is 75.
The
marks
each question are shown in brackets
• – use this asfora guide
as to how much time to spend on each question.
Advice
each question carefully before you start to answer it.
•• Read
Try to answer every question.
your answers if you have time at the end.
•• Check
If you change your mind about an answer, cross it out and put your new answer
and any working underneath.
Turn over
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Answer ALL questions. Write your answers in the spaces provided.
Given that y = 4x 3 –
5
x2
, x ≠ 0, find in their simplest form
dy
,
dx
(a)
(b)
(3)
∫ y dx
(3)
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1.
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Question 1 continued
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Q1
(Total for Question 1 is 6 marks)
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2.
(a) Given that 3–1.5 = a 3 find the exact value of a
1
( 2 x 2 )3
4 x2
(3)
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(b) Simplify fully
(2)
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4
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Q2
(Total for Question 2 is 5 marks)
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Solve the simultaneous equations
y 2 + 5x 2 + 2x = 0
(6)
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y + 4x + 1 = 0
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6
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Q3
(Total for Question 3 is 6 marks)
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4.
The straight line with equation y = 4x + c, where c is a constant, is a tangent to the curve
with equation y = 2x 2 + 8x + 3
(5)
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Calculate the value of c
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Q4
(Total for Question 4 is 5 marks)
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5.
(b) On your sketch, show, by shading, the region R defined by the inequalities
y < x + 2 and y > x 2 – x – 6
(c) Hence, or otherwise, find the set of values of x for which x 2 – 2x – 8 < 0
(1)
(3)
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(a) On the same axes, sketch the graphs of y = x + 2 and y = x 2 – x – 6 showing the
coordinates of all points at which each graph crosses the coordinate axes.
(4)
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Q5
(Total for Question 5 is 8 marks)
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6.
y
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C
6
O
x
(3, –1)
Figure 1
Figure 1 shows a sketch of the curve C with equation y = f(x)
The curve C passes through the origin and through (6, 0)
The curve C has a minimum at the point (3, –1)
(a) y = f(2x)
(3)
(b) y = f(x + p), where p is a constant and 0 < p < 3
(4)
On each diagram show the coordinates of any points where the curve intersects the
x‑axis and of any minimum or maximum points.
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On separate diagrams, sketch the curve with equation
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Question 6 continued
Q6
(Total for Question 6 is 7 marks)
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A curve with equation y = f(x) passes through the point (4, 25)
1
−
3
f ′(x) = x2 – 10 x 2 + 1,
8
x>0
find f(x), simplifying each term.
(5)
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Given that
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Q7
(Total for Question 7 is 5 marks)
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8.
y
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l2
B
C
O
x
l1
Figure 2
The line l2 passes through the origin O and is perpendicular to l1
(a) Find an equation for the line l2
(4)
The line l2 intersects the line l1 at the point C. Line l1 crosses the y‑axis at the point B as
shown in Figure 2.
(b) Find the area of triangle OBC. Give your answer in the form
integers to be found.
a
, where a and b are
b
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The line l1, shown in Figure 2 has equation 2x + 3y = 26
(6)
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Q8
(Total for Question 8 is 10 marks)
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y
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A
O
x
B
C
Figure 3
A sketch of part of the curve C with equation
18
,
x
x>0
is shown in Figure 3.
Point A lies on C and has x coordinate equal to 2
(a) Show that the equation of the normal to C at A is y = –2x + 7.
(6)
The normal to C at A meets C again at the point B, as shown in Figure 3.
(b) Use algebra to find the coordinates of B.
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y = 20 – 4x –
(5)
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*S59754A02126*
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Q9
(Total for Question 9 is 11 marks)
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*S59754A02226*
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10.
X
Z
m α
4c
6 cm
W
Y
9 cm
Figure 4
The triangle XYZ in Figure 4 has XY = 6 cm,
= 9 cm, ZX = 4 cm and angle ZXY = α.
The point W lies on the line XY.
The circular arc ZW, in Figure 4, is a major arc of the circle with centre X and radius 4 cm.
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(a) Show that, to 3 significant figures, α = 2.22 radians.
(b) Find the area, in cm2, of the major sector XZWX.
(2)
(3)
The region, shown shaded in Figure 4, is to be used as a design for a logo.
Calculate
(c) the area of the logo
(d) the perimeter of the logo.
(3)
(4)
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*S59754A02326*
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*S59754A02526*
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Q10
(Total for Question 10 is 12 marks)
TOTAL FOR PAPER IS 75 MARKS
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30
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Pure Mathematics P1 Mark scheme
Question
Scheme
1(a)
y = 4x3 −
Marks
5
x2
xn x n 1
e.g. sight of x2 or x−3 or
3 × 4x2 or −5 × −2x−3 (o.e.)
12x2 +
M1
1
x3
10
or 12x2 + 10x−3
3
x
(Ignore + c for this mark)
A1
A1
all on one line and no + c
(3)
(b)
x n x n1
e.g. sight of x4 or x−1 or
1
x1
M1
Do not award for integrating their answer to part (a)
x4
x 1
4
or −5 ×
4
1
For fully correct and simplified answer with + c all on one line. Allow
1
Allow x4 + 5 × + c
x
4
4
Allow 1x for x
A1
A1
(3)
(6 marks)
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31
Question
2(a)
Scheme
3−1.5 =
1
3 3
Marks
3
3
3
9
M1
so a
1
9
A1
(2)
Alternative
31.5
3 a 3 a 0.5 31.50.5
3
1.5
a 32
(b)
1
9
3
3
12
3 2
2
x
2
x
M1
A1
3
One correct power either 23 or x 2 .
3
8x 2
1
2x 2 or
2
4x
M1
dM1
A1
2
x
(3)
(5 marks)
Notes:
(a)
M1:
Scored for a full attempt to write 31.5 in the form a 3 or, as an alternative, makes a the
subject and attempts to combine the powers of 3
A1:
For a
1
9
Note: A correct answer with no working scores full marks
(b)
3
M1:
1
For an attempt to expand 2x 2 Scored for one correct power either 23 or x .
3
2
12 12 12
2 x 2 x 2 x on its own is not sufficient for this mark.
dM1: For dividing their coefficients of x and subtracting their powers of x. Dependent upon the previous
M1
A1:
32
Correct answer 2x
12
or
2
x
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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Question
3
Scheme
y = − 4x − 1
(−4x − 1)2 + 5x2 + 2x = 0
21x2 + 10x + 1 = 0
(7x+1)(3x+1) = 0 x
17 , 13
3 1
y=– ,
7 3
Marks
Attempts to makes y the subject of the
linear equation and substitutes into
the other equation.
M1
Correct 3 term quadratic
A1
dM1: Solves a 3 term quadratic by
the usual rules
1
1
7
3
A1: (x = ) ,
M1: Substitutes to find at least one y
value
A1: y =
3
7
1
,
dM1A1
M1 A1
3
(6)
Alternative
1
1
x
y
4
4
2
1
1
1
1
y 5 y 2 y
0
4
4
4
4
Attempts to makes x the subject of the
linear equation and substitutes into
the other equation.
2
21 2 1
3
y y
0
16
8
16
(21y 2 2 y 3
0)
Correct 3 term quadratic
A1
Solves a 3 term quadratic
3 1
(7 y 3)(3 y 1) 0 ( y
)
,
7 3
1 1
x
,
7 3
M1
y
3
7
,
1
3
dM1
A1
Substitutes to find at least one x
value.
M1
x= ,
A1
1
1
7
3
(6)
(6 marks)
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33
Question
4
Scheme
Marks
Sets 2x2 + 8x + 3 = 4x + c and collects x terms together
2
Obtains 2x + 4x + 3 − c = 0 o.e.
M1
A1
States that b2 – 4ac = 0
dM1
2
4 – 4×2×(3 − c) = 0 and so c =
c = 1 cso
dM1
A1
(5)
Alternative 1A
Sets derivative " 4x + 8 " = 4 x =
M1
x = −1
A1
2
Substitute x = −1 in y = 2x + 8x + 3 ( y = −3)
dM1
Substitute x = −1 and y = −3 in y = 4x + c or into (y + 3)=4(x + 1) and
expand
dM1
c = 1 or writing y = 4x + 1 cso
A1
(5)
Alternative 1B
Sets derivative "4 x 8" 4 x ,
M1
x = −1
2
Substitute x = -1 in 2 x 8x 3
A1
4x c
Attempts to find value of c
c = 1 or writing y = 4x + 1 cso
dM1
dM1
A1
(5)
Alternative 2
Sets 2 x2 8x 3 4 x c and collects x terms together
Obtains 2 x2 4 x 3 c
0 or equivalent
States that b2 4ac
0
2
4 4 2 (3 c)
0 and so c =
c = 1 cso
M1
A1
dM1
dM1
A1
(5)
Alternative 3
Sets 2 x2 8x 3 4 x c and collects x terms together
Obtains 2 x2 4 x 3 c
0 or equivalent
Uses 2( x 1)2 2 3 c
0 or equivalent
Writes -2 +3 – c = 0
So c = 1 cso
M1
A1
dM1
dM1
A1
(5)
(5 marks)
34
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Question 4 continued
Notes:
Method 1A
dy
4 . They must reach x =... (Just differentiating is M0 A0).
dx
dy
A1: x = –1 (If this follows
4 x + 8, then give M1 A1 by implication).
dx
dM1: (Depends on previous M mark) Substitutes their x = -1 into f(x) or into “their f(x) from (b)”
to find y.
dM1: (Depends on both previous M marks) Substitutes their x = -1 and their y = -3 values into y =
4x + c to find c or uses equation of line is (y + “3”) = 4(x + “1”) and rearranges to y = mx+c
A1: c = 1 or allow for y = 4x + 1 cso.
M1:
Attempts to solve their
Method 1B
M1A1: Exactly as in Method 1A above.
dM1: (Depends on previous M mark) Substitutes their x = -1 into 2 x2 8x 3
dM1: Attempts to find value of c then A1 as before.
4x c
Method 2
M1: Sets 2 x2 8x 3 4 x c and tries to collect x terms together.
A1: Collects terms e.g. 2 x2 4 x 3 c
0 or 2 x2 4 x 3 c
0 or 2 x2 4 x 3
c or even
2
2 x 4 x c 3 . Allow “=0” to be missing on RHS.
2
0 (Allow 2(x+1)2 – k + 3 – c = 0)
dM1: Then use completion of square 2( x 1) 2 3 c
where k is non zero. It is enough to give the correct or almost correct (with k) completion of
the square.
dM1: -2 + 3 - c = 0 AND leading to a solution for c (Allow -1 + 3 - c = 0) (x = –1 has been used)
A1: c = 1 cso
Method 3
M1: Sets 2 x2 8x 3 4 x c and tries to collect x terms together. May be implied by
2 x2 8x 3 4 x c on one side.
2
0 or 2 x2 4 x 3 c
0 or 2 x2 4 x 3
c even
Collects terms e.g. 2 x 4 x 3 c
2
2 x 4 x c 3 . Allow “=0” to be missing on RHS.
dM1: Then use completion of square 2(x+1)2 – k + 3 – c = 0 (Allow 2(x+1)2 –k + 3 – c = 0)
where k is non zero. It is enough to give the correct or almost correct (with k) completion of
the square.
dM1: -2 + 3 - c = 0 AND leading to a solution for c (Allow -1 + 3 - c = 0) (x = -1 has been used)
A1: c = 1 cso
A1:
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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35
Question
Marks
5(a)
(b)
Straight line, positive gradient positive
intercept
B1
Curve ‘U’ shape anywhere
B1
Correct y intercepts 2, 6
B1
Correct x-intercepts of 2 and 3
with intersection shown at (−2, 0)
B1
(4)
B1
Finite region between line and curve shaded
(1)
(c)
2
2
(x – x − 6 < x + 2 ) x – 2x − 8 < 0
(x − 4)( x + 2) < 0 Line and curve intersect at x = 4 and x = −2
−2 < x < 4
M1 A1
A1
(3)
(8 marks)
Notes:
36
(a)
As scheme.
(b)
As scheme.
(c)
M1:
A1:
A1:
For a valid attempt to solve the equation x2 – 2x − 8 = 0
For x = 4 and x 2
−2 < x < 4
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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Question
6(a)
Scheme
Marks
y
Shape
through (0, 0)
B1
(3, 0)
B1
(1.5, 1)
B1
x
(3)
(b)
y
Shape
x
, not through (0, 0)
B1
Minimum in 4th quadrant
B1
(p, 0) and (6 p, 0)
B1
(3 p, 1)
B1
(4)
(7 marks)
Notes:
(a)
B1:
B1:
B1:
(b)
B1:
B1:
B1:
B1:
U shaped parabola through origin.
(3,0) stated or 3 labelled on x - axis (even (0,3) on x - axis).
(1.5, -1) or equivalent e.g. (3/2, -1) labelled or stated and matching minimum point on the
graph.
Is for any translated curve to left or right or up or down not through origin
Is for minimum in 4th quadrant and x intercepts to left and right of y axis
(i.e. correct position).
Coordinates stated or shown on x axis (Allow (0 – p, 0) instead of (-p, 0))
Coordinates stated.
Note: If values are taken for p, then it is possible to give M1A1B0B0 even if there are
several attempts. (In this case none of the curves should go through the origin for M1 and
all minima should be in fourth quadrant and all x intercepts need to be to left and right of
y axis for A1
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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37
Question
Scheme
Marks
7
1
3
f ( x) x 2 10 x 2 1 dx
8
1
M1
A1
A1
3 x3
x2
x n x n1 f ( x) 10 x (c)
1
8 3
2
Substitute x = 4, y = 25 25 = 8 – 40 + 4 + c
M1
c=
1
x3
f ( x)
20 x 2 x 53
8
A1
(5)
(5 marks)
Notes:
M1:
Attempt to integrate xn xn1
A1:
A1:
M1:
Term in x 3 or term in x correct, coefficient need not be simplified, no need for +x nor +c
ALL three terms correct, coefficients need not be simplified, no need for + c
For using x = 4, y = 25 in their f(x) to form a linear equation in c and attempt to find c
A1:
38
1
2
1
x3
20 x 2 x 53 cao (all coefficients and powers must be simplified to give this
8
answer- do not need a left hand side and if there is one it may be f(x) or y). Need full
expression with 53. These marks need to be scored in part (a).
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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Question
Scheme
Marks
8(a)
2x + 3y = 26 3y = 26 ± 2x and attempt to find m from y = mx + c
M1
( y
2
26 2
x ) so gradient =
3
3 3
Gradient of perpendicular =
1
their gradient
Line goes through (0, 0) so y
A1
3
(= )
2
M1
3
x
2
A1
(4)
(b)
Solves their y
3
x with their 2x + 3y = 26 to form equation in x or in y
2
M1
Solves their equation in x or in y to obtain x = or y =
x = 4 or any equivalent e.g.
dM1
156
or y = 6 o.a.e
39
B= (0,
A1
26
) used or stated in (b)
3
Area =
26
3
x=4
B1
1
"26"
"4"
2
3
52
(o.e. with integer numerator
3
dM1
A1
and denominator)
(6)
(10 marks)
Notes:
(a)
M1:
Complete method for finding gradient. (This may be implied by later correct answers.) e.g.
Rearranges 2 x 3 y 26 y mx c so m =
Or finds coordinates of two points on line and finds gradient e.g.
(13, 0) and (1,8) so m
A1:
80
1 13
2
2
x
States or implies that gradient
3 condone
3 if they continue correctly. Ignore errors
in constant term in straight line equation.
M1:
A1:
1
Uses m1×m2= –1 to find the gradient of l2. This can be implied by the use of their gradient
3
3
y
x 0 Also accept 2y = 3x, y= 39 x or even
x or 2y – 3x = 0 Allow
26
2
2
3
y 0 ( x 0) and isw.
2
y
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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39
Question 8 notes continued
(b)
M1:
Eliminates variable between their y
3
x and their (possibly rearranged) 2 x 3 y
26 to
2
form an equation in x or y. (They may have made errors in their rearrangement).
dM1: (Depends on previous M mark) Attempts to solve their equation to find the value of x or y
A1: x = 4 or equivalent or y = 6 or equivalent
B1:
y coordinate of B is
26
26
(stated or implied) - isw if written as (
, 0).
3
3
Must be used or stated in (b)
dM1: (Depends on previous M mark) Complete method to find area of triangle OBC (using their
values of x and/or y at point C and their
A1:
Cao
26
)
3
104
1352
52
or
or
o.e
3
6
78
Alternative 1
Uses the area of a triangle formula ½×OB ×(x coordinate of C)
Alternative methods: Several Methods are shown below. The only mark which differs from
Alternative 1 is the last M mark and its use in each case is described below:
Alternative 2
In 8(b) using
1
BC OC
2
dM1: Uses the area of a triangle formula ½×BC ×OC Also finds OC (=√52 ) and BC= (
4
13 )
3
Alternative 3
In 8(b) using
104 00
2 0 6 263 0
dM1: States the area of a triangle formula
104 00
or equivalent with their values
2 0 6 263 0
Alternative 4
In 8(b) using area of triangle OBX – area of triangle OCX where X is point (13, 0)
dM1: Uses the correct subtraction
1
26 1
13 " " 13 "6"
2
3 2
Alternative 5
In 8(b) using area = ½ (6 × 4) + ½ (4 × 8/3) drawing a line from C parallel to the x axis and
dividing triangle into two right angled triangles
dM1: For correct method area = ½ (“6” × “ 4”) + ½ (“4” × [“26/3”-“6”])
Method 6 Uses calculus
4
4
26 2 x 3x
26
x 2 3x 2
dM1: " " dx = x
3
3
2
3
4 0
3
0
40
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
9(a)
Scheme
Substitutes x = 2 into y 20 4 2
Marks
18
and gets 3
2
B1
dy
18
4 2
dx
x
Substitute x 2
M1 A1
dy 1
then finds negative reciprocal ( 2 )
dx 2
dM1
States or uses y – 3 = –2(x – 2) or y = –2x + c with their (2, 3)
ddM1
to deduce that y = –2x + 7
A1*
(6)
(b)
18
2 x 7 and simplify to give 2x2 – 13x + 18 = 0
x
18
7 y
y
20 4
to give y2 − y – 6 = 0
7 y
2
2
Put 20 4 x
Or put
(2x − 9)(x − 2) = 0 so x =
or
(y – 3) (y + 2) = 0
M1 A1
so y =
dM1
9
, 2
2
A1 A1
(5)
(11 marks)
Notes:
(a)
B1:
M1:
Substitutes x = 2 into expression for y and gets 3 cao (must be in part (a) and must use
curve equation – not line equation). This must be seen to be substituted.
For an attempt to differentiate the negative power with x−1 to x−2.
A1:
Correct expression for
dy
18
4 2
dx
x
dM1: Dependent on first M1 substitutes x = 2 into their derivative to obtain a numerical gradient
1
2
and find negative reciprocal or states that −2× 1
Alternative 1
dM1: Dependent on first M1. Finds equation of line using changed gradient (not their
1
1
but
2
2
2 or −2) e.g. y – ″3″= – ″2″(x –2) or y= ″−2″ x + c and use of (2, ″3″) to find c =
A1*: cso. This is a given answer y= −2x + 7 obtained with no errors seen and equation should be
stated.
Alternative 2 – checking given answer
dM1: Uses given equation of line and checks that (2, 3) lies on the line.
A1*: cso. This is a given answer y = −2x + 7 so statement that normal and line have the same
gradient and pass through the same point must be stated.
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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41
Question 9 notes continued
(b)
M1:
Equate the two given expressions, collect terms and simplify to a 3TQ. There may be sign
errors when collecting terms but putting for example 20x − 4x2 − 18= −2x + 7 is M0 here.
A1: Correct 3TQ = 0 (need = 0 for A mark) 2x2 −13x + 18 = 0
dM1: Attempt to solve an appropriate quadratic by factorisation, use of formula, or completion of
the square (see general instructions).
9
o.e or y = −2 (allow second answers for this mark so ignore x = 2 or y = 3)
2
A1:
x
A1:
Correct solutions only so both x
9
9
, y = −2 or , 2
2
2
If x = 2, y = 3 is included as an answer and point B is not identified then last mark is A0.
Answer only – with no working – send to review. The question stated ‘use algebra’.
42
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
10(a)
Scheme
92 42 62 2 4 6cos cos .....
42 62 92
cos
2 46
Marks
Correct use of cosine rule
leading to a value for cos
M1
29
0.604..
48
= 2.22
A1
cso
(2)
Alternative
XY 2 42 62 2 4 6cos 2.22 XY 2 .. Correct use of cosine rule
leading to a value for XY2
XY = 9.00....
M1
A1
(2)
(b)
2 2.22(
4.06366......)
1
2
42 "4.06"
32.5
2 2.22 or 2 2.2 or awrt
4.06 (May be implied)
B1
Correct method for major
sector area. Allow 2.22
for the major sector angle.
M1
Awrt 32.5
A1
(3)
Alternative – Circle Minor – sector
42
1
2
32.5
42 42 2.22
(c)
32.5
Area of triangle =
Correct expression for circle
area
Correct method for
circle - minor sector area
Awrt 32.5
B1
M1
A1
(3)
Correct expression for the area of
triangle XYZ (allow 2.2 or awrt
2.22)
B1
So area required = “9.56” + “32.5”
Their Triangle XYZ + part (b) or
correct attempt at major sector
(Not triangle ZXW)
M1
Area of logo = 42.1 cm2 or 42.0 cm2
Awrt 42.1 or 42.0 (or just 42)
A1
1
4 6 sin 2.22 9.56
2
(3)
(d)
Arc length
4 4.06
16.24
M1: 4 their 2 2.22
or
or circumference – minor arc
M1
A1ft
8 4 2.22
A1: Correct ft expression
Perimeter = ZY + WY + Arc Length
9 + 2 + Any Arc
M1
Perimeter of logo = 27.2 or 27.3
Awrt 27.2 or awrt 27.3
A1
(4)
(12 marks)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Please check the examination details below before entering your candidate information
Candidate surname
Pearson Edexcel
Other names
Centre Number
Candidate Number
International
Advanced Level
Sample Assessment Materials for first teaching September 2018
(Time: 1 hour 30 minutes)
Paper Reference WMA12/01
Mathematics
International Advanced Subsidiary/Advanced Level
Pure Mathematics P2
You must have:
Mathematical Formulae and Statistical Tables, calculator
Total Marks
Candidates may use any calculator permitted by Pearson regulations.
Calculators must not have the facility for symbolic algebra manipulation,
differentiation and integration, or have retrievable mathematical
formulae stored in them.
Instructions
black ink or ball-point pen.
•• Use
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
in the boxes at the top of this page with your name, centre number and
• Fill
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.
Inexact
answers
should
be given to three significant figures unless otherwise
• stated.
Information
booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
•• AThere
are 9 questions in this question paper. The total mark for this paper is 75.
The
marks
each question are shown in brackets
• – use this asfora guide
as to how much time to spend on each question.
Advice
each question carefully before you start to answer it.
•• Read
Try to answer every question.
your answers if you have time at the end.
•• Check
If you change your mind about an answer, cross it out and put your new answer
and any working underneath.
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Answer ALL questions. Write your answers in the spaces provided.
f(x) = x4 + x3 + 2x2 + ax + b,
where a and b are constants.
When f(x) is divided by (x – 1), the remainder is 7
(a) Show that a + b = 3
(2)
When f(x) is divided by (x + 2), the remainder is –8
(b) Find the value of a and the value of b
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1.
(5)
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Q1
(Total for Question 1 is 7 marks)
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2.
(a) Find the value of S∞
(2)
The sum to N terms of the series is SN
(b) Find, to 1 decimal place, the value of S12
(c) Find the smallest value of N, for which S∞ – SN < 0.5
(2)
(4)
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7
The first term of a geometric series is 20 and the common ratio is . The sum to infinity
8
of the series is S∞
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Question 2 continued
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Q2
(Total for Question 2 is 8 marks)
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3.
y = (3x + x)
x
0
0.25
y
1
1.251
0.5
0.75
1
2
(2)
(b) Use the trapezium rule with all the values of y from your table to find an approximation
for the value of
∫
1
0
(3x + x) dx
You must show clearly how you obtained your answer.
(4)
1
0
(3x + x) dx
(1)
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(c) Explain how the trapezium rule could be used to obtain a more accurate estimate for
the value of
∫
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(a) Complete the table below, giving the values of y to 3 decimal places.
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Q3
(Total for Question 3 is 7 marks)
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4.
Given n ∈ , prove, by exhaustion, that n2 + 2 is not divisible by 4.
(4)
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Q4
(Total for Question 4 is 4 marks)
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5.
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An arithmetic series has first term a and common difference d.
(a) Prove that the sum of the first n terms of the series is
1
2
n[2a + (n – 1)d]
(4)
A company, which is making 200 mobile phones each week, plans to increase its
production.
The number of mobile phones produced is to be increased by 20 each week from 200 in
week 1 to 220 in week 2, to 240 in week 3 and so on, until it is producing 600 in week N.
(b) Find the value of N
(2)
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The company then plans to continue to make 600 mobile phones each week.
(c) Find the total number of mobile phones that will be made in the first 52 weeks starting
from and including week 1.
(5)
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Q5
(Total for Question 5 is 11 marks)
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(i) Find the exact value of x for which
log2(2x) = log2(5x + 4) – 3
(4)
(ii) Given that
loga y + 3loga2 = 5
express y in terms of a. Give your answer in its simplest form.
(3)
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Q6
(Total for Question 6 is 7 marks)
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y
P
r
C
Q
O
x
x = 13
Figure 1
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The circle with equation
x2 + y2 – 20x – 16y + 139 = 0
had centre C and radius r.
(a) Find the coordinates of C.
(b) Show that r = 5
(2)
(2)
The line with equation x = 13 crosses the circle at the points P and Q as shown in Figure 1.
(c) Find the y coordinate of P and the y coordinate of Q.
(3)
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A tangent to the circle from O touches the circle at point X.
(d) Find, in surd form, the length OX.
(3)
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Q7
(Total for Question 7 is 10 marks)
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y
y = x3
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8.
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B
y = 10x – x2 – 8
R
A
x
O
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Figure 2
Figure 2 shows a sketch of part of the curves C1 and C2 with equations
C1: y = 10x – x2 – 8
x>0
C2: y = x3
x>0
The curves C1 and C2 intersect at the points A and B.
(a) Verify that the point A has coordinates (1, 1)
(b) Use algebra to find the coordinates of the point B
(1)
(6)
The finite region R is bounded by C1 and C2
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(c) Use calculus to find the exact area of R
(5)
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Q8
(Total for Question 8 is 12 marks)
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9.
(i) Solve, for 0 θ < π, the equation
sin 3θ −
3 cos3θ = 0
giving your answers in terms of π
(3)
(ii) Given that
4sin2x + cos x = 4 − k,
0k3
(a) find cos x in terms of k
(b) When k = 3, find the values of x in the range 0 x < 360°
(3)
(3)
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Q9
(Total for Question 9 is 9 marks)
TOTAL FOR PAPER IS 75 MARKS
30
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Pure Mathematics P2 Mark scheme
Question
Scheme
1(a)
f ( x) x4 x3 2 x2 ax b
Marks
Attempting f(1) or f(−1)
(b)
M1
f(1) = 1 + 1 + 2 + a + b = 7 or 4 + a + b = 7 a + b = 3
(as required) AG
A1*
cso
Attempting f(−2) or f(2)
(2)
M1
f(−2) = 16 – 8 + 8 – 2a + b = –8 { – 2a + b = –24}
A1
Solving both equations simultaneously to get as far as a =… or b =…
dM1
Any one of a = 9 or b = −6
A1
Both a = 9 and b = −6
A1
(5)
(7marks)
Notes:
(a)
M1:
A1:
For attempting either f (1) or f (1).
For applying f (1) , setting the result equal to 7, and manipulating this correctly to give
the result given on the paper as a b
3. Note that the answer is given in part (a).
Alternative
M1: For long division by ( x 1) to give a remainder in a and b which is independent of x.
A1:
Or Remainder b a 4 7 leading to the correct result of a b
3 (answer given).
(b)
M1:
Attempting either f ( 2) or f (2).
A1:
correct underlined equation in a and b; e.g. 16 8 8 2a b 8 or equivalent,
e.g. 2a b 24.
dM1: An attempt to eliminate one variable from 2 linear simultaneous equations in a and b .
Note that this mark is dependent upon the award of the first method mark.
A1:
Any one of a 9 or b 6 .
A1:
Both a 9 and b 6 and a correct solution only.
Alternative
M1: For long division by ( x 2) to give a remainder in a and b which is independent of x.
A1:
For Remainder b 2(a 8) 8 2a b 24 .
Then dM1A1A1 are applied in the same way as before.
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75
Question
2(a)
Scheme
S
20
; = 160
1 78
Marks
Use of a correct S formula
M1
160
A1
(2)
(b)
S12
20 1 ( )
; 127.77324...
1 78
= 127.8 (1 dp)
7 12
8
M1: Use of a correct S n formula
with n = 12 (condone missing
brackets around
7
)
8
M1 A1
A1: awrt 127.8
(2)
(c)
20(1 ( 78 ) N )
160
0.5
1 78
N
N
7
7
0.5
160 0.5 or
8
160
8
7
0.5
N log log
8
160
N
0.5
log 160
43.19823...
7
cso
log 8
Applies S N (GP only) with a 20 ,
r 78 and “uses” 0.5 and their S
at any point in their working.
M1
Attempt to isolate 160 78 or
dM1
N
78
N
Uses the law of logarithms to obtain an
equation or an inequality of the form
0.5
7
N log log
8
their S
or
0.5
N log 0.875
their S
M1
N 44 (Allow N 44 but no N 44
A1 cso
N
44
An incorrect inequality statement at any stage in a candidate’s working loses
the final mark. Some candidates do not realise that the direction of the
inequality is reversed in the final line of their solution. BUT it is possible to
gain full marks for using =, as long as no incorrect working seen.
(4)
Alternative: Trial & Improvement Method in (c):
Attempts 160 S N or S N with at least one value for N > 40
M1
Attempts 160 S N or S N with N = 43 or N = 44
For evidence of examining 160 S N or S N for both N 43 and N 44 with
both values correct to 2 DP
Eg:
160 S43 awrt 0.51 and
160 S44 awrt 0.45 or
dM1
N 44
A1 cso
M1
S43 awrt159.49and
S44 awrt159.55
Answer of N = 44 only with no working scores no marks
(4)
(8 marks)
76
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
3(a)
Scheme
Marks
x
0
0.25
0.5
0.75
1
y
1
1.251
1.494
1.741
2
B1 B1
(2)
(b)
1
×0.25, {(1 + 2) +2(1.251 + 1.494 + 1.741)} o.e.
2
B1 M1
A1ft
= 1.4965
A1
(4)
(c)
Gives any valid reason including
Decrease the width of the strips
Use more trapezia
Increase the number of strips
Do not accept use more decimal places
B1
(1)
(7 marks)
Notes:
(a)
B1:
B1:
For 1.494
For 1.741 (1.740 is B0). Wrong accuracy e.g. 1.49, 1.74 is B1B0
(b)
Need ½ of 0.25 or 0.125 o.e.
M1: Requires first bracket to contain first plus last values and second bracket to include no
additional values from the three in the table. If the only mistake is to omit one value
from second bracket this may be regarded as a slip and M mark can be allowed ( An
extra repeated term forfeits the M mark however) x values: M0 if values used in brackets are
x values instead of y values
A1ft: Follows their answers to part (a) and is for {correct expression}
A1: Accept 1.4965, 1.497, or 1.50 only after correct work. (No follow through except one
special case below following 1.740 in table).
B1:
Separate trapezia may be used: B1 for 0.125, M1 for
1
2
h(a b) used 3 or 4 times (and
A1ft if it is all correct) e.g. 0.125(11.251) 0.125(1.251+1.494) 0.125(1.741 2) is
M1 A0 equivalent to missing one term in { } in main scheme.
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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77
Question
4
Scheme
Marks
A solution based around a table of results
n
n2
n2 2
1
1
3
Odd
2
4
6
Even
3
9
11
Odd
4
16
18
Even
5
25
27
Odd
6
36
38
Even
When n is odd, n 2 is odd (odd odd = odd) so n2 2 is also odd
M1
So for all odd numbers n, n2 2 is also odd and so cannot be divisible by 4
(as all numbers in the 4 times table are even)
A1
When n is even, n 2 is even and a multiple of 4, so n2 2 cannot be a
multiple of 4
M1
Fully correct and exhaustive proof. Award for both of the cases above plus a
final statement ''So for all n, n2 2 cannot be divisible by 4''
A1*
(4)
Alternative - (algebraic) proof
n2 2
If n is even, n 2k , so
4
2k
2
2 4k 2 2
1
k2
4
4
2
2
If n is odd,
n 2k 1, so
n2 2 2k 1 2
4k 2 4k 3
3
k2 k
4
4
4
4
For a partial explanation stating that
3
1
either of k 2 or k 2 k are not a whole numbers.
4
2
with some valid reason stating why this means that n2 +2 is not a
multiple of 4.
Full proof with no errors or omissions. This must include
The conjecture
Correct notation and algebra for both even and odd numbers
M1
M1
A1
A1*
A full explanation stating why, for all n, n2 2 is not divisible
by 4
(4)
(4 marks)
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Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
Scheme
Marks
(S =)a + (a + d) + …+[a+(n − 1)d]
B1: requires at least 3 terms,
must include first and last
terms, an adjacent term and
dots!
B1
(S =)[a+(n − 1)d] + ……+ a
M1: for reversing series (dots
needed)
M1
5(a)
2S = [2a+(n − 1)d] + ……+ [2a+(n − 1)d]
and be a genuine attempt.
Either line is sufficient.
Dependent on 1st M1.
dM1
(NB −Allow first 3 marks for
use of l for last term but as
given for final mark )
2S = n[2a+(n − 1)d]
S=
dM1: for adding, must have 2S
n
[2a+(n − 1)d] cso
2
A1
(4)
(b)
600 = 200 + (N – 1)20 N
...
N = 21
Use of 600 with a correct formula in an
attempt to find N.
M1
cso
A1
(2)
(c)
Look for an AP first:
S=
21
2
(2 × 200 + 20 × 20) or
21
2
S=
20
2
(200 + 600)
(2 × 200 + 19 × 20) or
20
2
(200 + 580)
(= 8400 or 7800)
M1: Use of correct sum formula with
their integer n = N or N – 1 from part
(b) where 3 < N < 52 and a = 200 and d
= 20.
M1A1
M1: Use of correct sum formula with
their integer n = N or N – 1 from part
(b) where 3 < N < 52 and a = 200 and d
= 20.
Then for the constant terms:
600 × (52 − " N " ) (= 18600)
So total is 27000
M1: 600 k where k is an integer and 3
< k < 52
A1: A correct un-simplified follow
through expression with their k
consistent with n so that
n + k = 52
cao
M1
A1ft
A1
There are no marks in (c) for just finding S52
(5)
(11 marks)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
79
Question
6(i)
Scheme
2x
3
5x 4
5x 4
5x 4
or log 2
3 or log 2 x 4
2
x
log 2
2x
5x 4
Marks
5x 4
3
=2
2x
−3
or
2
M1
M1
5x 4
4
=2
x
or
16x = 5x + 4 x = (depends on Ms and must be this equation or equiv)
x=
4
or exact recurring decimal 0.36 after correct work
11
dM1
A1 cso
Alternative
log2 (2 x
) 3 log 2 (5x 4)
So log2 (2 x) log2 (8) log2 (5x 4) earns 2nd M1 (3 replaced by log 2 8 )
log2 (5x 4) earns 1st M1 (addition law of logs)
Then log
2 (16 x)
2nd M1
Then final M1 A1 as before
dM1A1
1st M1
(4)
(ii)
M1
3
loga y loga 2
5
log a 8 y 5
1
y a5
8
Applies product law of logarithms
cso
1
y a5
8
dM1
cso
A1
(3)
(7 marks)
Notes:
(i)
M1:
Applying the subtraction or addition law of logarithms correctly to make two log terms
into one log term .
1
M1: For RHS of either 2 3 , 2 3 , 2 4 or log 2 , log 2 8 or log 2 16 i.e. using connection
8
between log base 2 and 2 to a power. This may follow an error. Use of 3 2 is M0
dM1: Obtains correct linear equation in x. usually the one in the scheme and attempts x =
A1: cso. Answer of 4/11 with no suspect log work preceding this.
(ii)
M1:
Applies power law of logarithms to replace 3loga 2 by loga 23 or loga 8
dM1: (Should not be following M0) Uses addition law of logs to give loga 23 y 5 or
loga 8 y 5
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Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
7(a)
Scheme
Marks
(x ± 10)2 and (y ± 8)2
Obtain
M1
(10, 8)
A1
(2)
(b)
2
2
2
See (x ± 10) + (y ± 8) = 25 (= r )
or
(r =) " 100 " + " 64 " − 139
M1
2
r = 5*
A1
(2)
(c)
Substitute x = 13 into the equation of circle and solve quadratic to give y =
M1
e.g. x = 13 (13 − 10)2 + (y − 8)2 = 25 ( y − 8)2 = 16
so y = 4 or 12
A1 A1
N.B. This can be attempted via a 3, 4, 5 triangle so spotting this and
achieving one value for y is M1 A1. Both values scores M1 A1 A1
(d)
(3)
M1
2
2
OC = 10 8 164
M1 A1
2
Length of tangent = 164 5 139
(3)
(10 marks)
Notes:
(a)
M1:
A1:
Obtains (x ± 10)2 and (y ± 8)2 May be implied by one correct coordinate
(10, 8) Answer only scores both marks.
Alternative: Method 2: From x 2 y 2 2 gx 2 fy c 0 centre is ( g , f )
M1:
Obtains 10, 8
A1:
Centre is ( g , f ) , and so centre is (10, 8).
(b)
M1:
For a correct method leading to r = …, or r 2
r 2 form to identify r=
Allow ''100" "64" 139 or an attempt at using ( x 10)2 ( y 8)2
A1*: r = 5 This is a printed answer, so a correct method must be seen.
Alternative:
(b)
M1:
Attempts to use
g 2 f 2 c or (r 2 )"100" "64" 139
A1*: r = 5 following a correct method.
(c)
M1:
A1:
A1:
Substitutes x 13 into either form of the circle equation, forms and solves the quadratic
equation in y
Either y = 4 or 12
Both y = 4 and 12
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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Question 7 notes continued
(d)
M1:
Uses Pythagoras' Theorem to find length OC using their 10,8
M1:
Uses Pythagoras' Theorem to find OX. Look for
A1:
82
OC 2 r 2
139 only
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
8(a)
Scheme
Marks
Substitutes x 1 in C1: y = 10x – x2 – 8 = 10 – 1 – 8 = 1
and in C2: y = x3 = 13 = 1 (1, 1) lies on both curves.
B1
(1)
(b)
2
3
10x – x – 8 = x
3
B1
2
x + x – 10x + 8 = 0
(x – 1)(x2 + 2x – 8) = 0
M1 A1
x=2
(x − 1)(x + 4)(x − 2) = 0
M1 A1
(2, 8)
A1
(6)
(c)
10x x 8 x dx
2
3
M1
x3
x4
5x 8x
3
4
2
8
3
M1 A1
1
3
1
4
Using limits 2 and 1: 20 16 4 5 8
M1
11
12
A1
(5)
(12 marks)
Notes:
(a)
B1:
(b)
B1:
M1:
A1:
M1:
A1:
A1:
Substitutes x =nto both y 10 x x 2 8 and y x3 AND achieves y =1 in both.
Sets equations equal to each other and proceeds to x3 + x2 – 10x + 8 = 0
Divides by (x −1) to form a quadratic factor. Allow any suitable algebraic method including
division or inspection.
Correct quadratic factor (x2 + 2x – 8)
For factorising of their quadratic factor.
Achieves x= 2
Coordinates of B = (2, 8)
(c)
10x x 8 x dx
2
3
M1:
For knowing that the area of R =
M1:
This may also be scored for finding separate areas and subtracting.
For raising the power of x seen in at least three terms.
A1:
Correct integration. It may be left un-simplified. That is allow
10 x 2
for 5x 2
2
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Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
83
Question 8 notes continued
84
M1:
For using the limits ''2'' and 1 in their integrated expression. If separate areas have been
attempted, ''2'' and 1 must be used in both integrated expressions.
A1:
For
11
or exact equivalent.
12
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
Scheme
9(i)
Way 1
Divides by cos 3 to give
tan 3 3 so (3 )
Marks
3
1
3
or sin 3
cos 3
2
2
so (3 )
Adds π or 2π to previous value of angle( to give
So
4
9
,
9
,
2
3
4
7
)
or
3
3
M1
7
(all three, no extra in range)
9
2
4(1 – cos x) + cos x = 4 – k
(ii)(a)
M1
Way 2
Or Squares both sides, uses
cos2 3 sin 2 3
1 , obtains
A1
(3)
M1
2
Applies sin x = 1 – cos x
2
Attempts to solve 4 cos x − cos x – k = 0, to give cos x =
cos x
1 1 16k
8
dM1
A1
1
1 k
cos x
8
64 4
or
or other correct equivalent
(3)
(b)
cos
x
M1
1 49
3
1 and (see the note below if errors are made)
8
4
Obtains two solutions from 0 , 139 , 221
dM1
(0 or 2.42 or 3.86 in radians)
x = 0 and 139 and 221
(allow awrt 139 and 221) must be in degrees
A1
(3)
(9 marks)
Notes:
(i)
M1: Obtains
9
3
. Allow x
3
or even
3
. Need not see working here. May be implied by
in final answer (allow (3 ) 1.05 or 0.349 as decimals or (3 ) 60 or 20 as
1
3
M1: Adding or 2 to a previous value however obtained. It is not dependent on the previous
4
7
or
mark. (May be implied by final answer of
). This mark may also be given for
degrees for this mark). Do not allow tan 3 3 nor tan 3
9
9
answers as decimals [4.19 or 7.33], or degrees (240 or 420).
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
85
Question 9 notes continued
A1:
Need all three correct answers in terms of and no extras in range.
NB:
20 ,80 , 140 earns M1M1A0 and 0.349, 1.40 and 2.44 earns M1M1A0
(ii)(a)
M1: Applies sin 2 x 1 cos2 x (allow even if brackets are missing e.g. 4 × 1 − cos2 x).
This must be awarded in (ii) (a) for an expression with k not after k = 3 is substituted.
dM1: Uses formula or completion of square to obtain cos x = expression in k
(Factorisation attempt is M0)
A1: cao - award for their final simplified expression
(ii)(b)
M1: Either attempts to substitute k = 3 into their answer to obtain two values for cos x
Or restarts with k = 3 to find two values for cos x (They cannot earn marks in ii(a) for
this). In both cases they need to have applied sin 2 x 1 cos2 x (brackets may be missing)
and correct method for solving their quadratic (usual rules – see notes) The
values for
cos x may be >1 or < -1.
dM1: Obtains two correct values for x
A1: Obtains all three correct values in degrees (allow awrt 139 and 221) including 0.
Ignore excess answers outside range (including 360 degrees) Lose this mark for excess
answers in the range or radian answers.
86
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Please check the examination details below before entering your candidate information
Candidate surname
Pearson Edexcel
Other names
Centre Number
Candidate Number
International
Advanced Level
Sample Assessment Materials for first teaching September 2018
(Time: 1 hour 30 minutes)
Paper Reference WMA13/01
Mathematics
International Advanced Level
Pure Mathematics P3
You must have:
Mathematical Formulae and Statistical Tables, calculator
Total Marks
Candidates may use any calculator permitted by Pearson regulations.
Calculators must not have the facility for symbolic algebra manipulation,
differentiation and integration, or have retrievable mathematical
formulae stored in them.
Instructions
black ink or ball-point pen.
•• Use
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
in the boxes at the top of this page with your name, centre number and
• Fill
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.
Inexact
answers
should
be given to three significant figures unless otherwise
• stated.
Information
booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
•• AThere
are 10 questions in this question paper. The total mark for this paper is 75.
The
marks
each question are shown in brackets
• – use this asfora guide
as to how much time to spend on each question.
Advice
each question carefully before you start to answer it.
•• Read
Try to answer every question.
your answers if you have time at the end.
•• Check
If you change your mind about an answer, cross it out and put your new answer
and any working underneath.
Turn over
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Answer ALL questions. Write your answers in the spaces provided.
Express
6x + 4
2
9x − 4
−
2
3x + 1
as a single fraction in its simplest form.
(4)
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Q1
(Total for Question 1 is 4 marks)
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2.
f(x) = x 3 + 3
2
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+ 4x – 12
x=
The equation x 3 + 3
2
4(3 − x)
(3 + x)
x ≠ –3
(3)
+ 4x – 12 = 0 has a single root which is between 1 and 2
(b) Use the iteration formula
xn + 1 =
4(3 − xn )
(3 + x )
n
n0
with x0 = 1 to find, to 2 decimal places, the value of x1, x2 and x3
(3)
(2)
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The root of f(x) = 0 is α.
(c) By choosing a suitable interval, prove that α = 1.272 to 3 decimal places.
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(a) Show that the equation f(x) = 0 can be written as
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Q2
(Total for Question 2 is 8 marks)
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3.
y
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y = f(x)
O
x
Figure 1
Figure 1 shows a sketch of part of the graph y = f(x) where
f(x) = 2½3 – x ½ + 5
x0
f(x) =
1
x + 30
2
(3)
Given that the equation f(x) = k, where k is a constant, has two distinct roots,
(b) state the set of possible values for k.
(2)
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(a) Solve the equation
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8
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Q3
(Total for Question 3 is 5 marks)
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(i) Find
5
1
dx
(2 x − 1)
writing your answer in its simplest form.
(4)
(ii) Use integration to find the exact value of
∫
π
2
0
sin 2x + sec
1
1
x tan x dx
3
3
(3)
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∫
13
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Q4
(Total for Question 4 is 7 marks)
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Given that
show that
( x − 1) 2
x≠1
dy
k
=
, where k is a constant to be found.
d x ( x − 1)3
(6)
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y=
5 x 2 − 10 x + 9
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Q5
(Total for Question 5 is 6 marks)
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The functions f and g are defined by
x∈
g : x ln x
x>0
(a) State the range of f.
(b) Find fg(x), giving your answer in its simplest form.
(c) Find the exact value of x for which f(2x + 3) = 6
(2)
(4)
(3)
(e) On the same axes sketch the curves with equation y = f(x) and y = f –1(x), giving the
coordinates of all the points where the curves cross the axes.
(4)
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(d) Find f –1 stating its domain.
(1)
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f : x ex + 2
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Q6
(Total for Question 6 is 14 marks)
*S59757A01935*
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7.
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The point P lies on the curve with equation
π
Given that P has (x, y) coordinates p, , where p is a constant,
2
(a) find the exact value of p
(1)
The tangent to the curve at P cuts the y-axis at the point A.
(b) Use calculus to find the coordinates of A.
(6)
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x = (4y – sin 2y)2
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*S59757A02135*
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Q7
(Total for Question 7 is 7 marks)
*S59757A02335*
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8.
In a controlled experiment, the number of microbes, N, present in a culture T days after
the start of the experiment were counted.
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where a and b are constants
N = aT b
(a) Show that this relationship can be expressed in the form
log10 N = m log10 T + c
giving m and c in terms of the constants a and/or b.
(2)
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N and T are expected to satisfy a relationship of the form
log10 N
5.0 –
4.5 –
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4.0 –
3.5 –
3.0 –
2.5 –
2.0 –
1.5 –
1.0 –
–
–
–
–
0.4
0.6
0.8
1.0
1.2
–
–
–
0.2
1.4 log10 T
Figure 2
Figure 2 shows the line of best fit for values of log10 N plotted against values of log10 T
(b) Use the information provided to estimate the number of microbes present in the
culture 3 days after the start of the experiment.
(4)
(c) With reference to the model, interpret the value of the constant a.
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(1)
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0–
0
–
0.5 –
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*S59757A02535*
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Question 8 continued
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Q8
(Total for Question 8 is 7 marks)
*S59757A02735*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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9.
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(a) Prove that
cos A + sin A
cos A − sin A
A≠
(2n + 1)π
4
n∈
(5)
(b) Hence solve, for 0 θ < 2π
sec 2θ + tan 2θ =
1
2
Give your answers to 3 decimal places.
(4)
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sec 2A + tan 2A ≡
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Question 9 continued
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Q9
(Total for Question 9 is 9 marks)
*S59757A03135*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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Turn over
10. The amount of an antibiotic in the bloodstream, from a given dose, is modelled by the
formula
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where x is the amount of the antibiotic in the bloodstream in milligrams, D is the dose
given in milligrams and t is the time in hours after the antibiotic has been given.
A first dose of 15 mg of the antibiotic is given.
(a) Use the model to find the amount of the antibiotic in the bloodstream 4 hours after the
dose is given. Give your answer in mg to 3 decimal places.
(2)
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x = De–0.2t
A second dose of 15 mg is given 5 hours after the first dose has been given. Using the
same model for the second dose,
(b) show that the total amount of the antibiotic in the bloodstream 2 hours after the second
dose is given is 13.754 mg to 3 decimal places.
(2)
b
(c) Show that T = a ln b + , where a and b are integers to be determined.
e
(4)
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No more doses of the antibiotic are given. At time T hours after the second dose is given,
the total amount of the antibiotic in the bloodstream is 7.5 mg.
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Q10
(Total for Question 10 is 8 marks)
TOTAL FOR PAPER IS 75 MARKS
*S59757A03535*
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Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Pure Mathematics P3 Mark scheme
Question
1
Scheme
Marks
9x2 – 4 = (3x − 2) (3x + 2 )
at any stage
Eliminating the common factor of (3x + 2)
2 (3x 2)
(3x 2) (3x 2)
B1
at any stage
2
3x 2
M1
Use of a common denominator
2(3x 2)(3x 1)
2(9 x 2 4)
2(3x 1)
2(3x 2)
or
2
2
(9 x 4)(3x 1) (9 x 4)(3x 1)
(3x 2)(3x 1) (3x 1)(3x 2)
M1
6
6
or 2
(3x 2)(3x 1)
9 x 3x 2
A1
(4)
(4 marks)
Notes:
B1:
B1:
M1:
For factorising 9 x2 4 (3x 2)(3x 2) using difference of two squares. It can be
awarded at any stage of the answer but it must be scored on E pen as the first mark.
For eliminating/cancelling out a factor of (3x+2) at any stage of the answer.
For combining two fractions to form a single fraction with a common denominator. Allow
slips on the numerator but at least one must have been adapted. Condone invisible brackets.
Accept two separate fractions with the same denominator as shown in the mark scheme.
Amongst possible (incorrect) options scoring method marks are
2(3x 2)
2(9 x 2 4)
. Only one numerator adapted, separate fractions
(9 x 2 4)(3x 1) (9 x 2 4)(3x 1)
2 3x 1 2 3x 2
Invisible brackets, single fraction.
(3x 2)(3x 1)
A1:
6
(3x 2)(3x 1)
This is not a given answer so you can allow recovery from ‘invisible’ brackets.
Alternative
2(3x 2)
2
2(3x 2)(3x 1) 2(9 x 2 4)
18x 12
2
2
2
(9 x 4) (3x 1)
(9 x 4)(3x 1)
(9 x 4)(3x 1)
has scored 0,0,1,0 so far
6(3x 2)
is now 1,1,1,0
(3x 2)(3x 2)(3x 1)
6
and now 1,1,1,1
(3x 2)(3x 1)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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123
Question
Scheme
2(a)
x3 + 3x2 + 4x – 12 = 0 x3 + 3x2 = 12 – 4x
Marks
x2 (x + 3) = 12 – 4x
M1
dM1
12 4 x
x2
( x 3)
x
(b)
x1
4(3 x)
( x 3)
A1*
(3)
M1 A1
4(3 1)
1.41
(3 1)
A1
awrt x2 1.20
x3 1.31
(3)
(c)
()0.00827... f (1.2715)
0.00821....
Attempts f (1.2725)
M1
Values correct with reason (change of sign with f(x) continuous) and
conclusion ( α = 1.272)
A1
(2)
(8 marks)
Notes:
(a)
12 4 x
Moves from f(x) = 0, which may be implied by subsequent working, to x2 ( x 3)
by separating terms and factorising in either order. No need to factorise rhs for this mark.
dM1: Divides by ‘(x+3)’ term to make x2 the subject, then takes square root. No need for rhs to
be factorised at this stage.
A1*: CSO. This is a given solution. Do not allow sloppy algebra or notation with root on just
numerator for instance. The 12−4x needs to have been factorised.
M1:
(b)
M1:
An attempt to substitute x0 1 into the iterative formula to calculate x1 . This can be awarded for
the sight of
A1:
4(3 1) 8
,
, 2 and even 1.4
(3 1)
4
x1 1.41 . The subscript is not important. Mark as the first value found,
2 is A0
x2 awrt
1.20 x3 awrt 1.31 . Mark as the second and third values found. Condone 1.2
A1:
for x2
(c)
M1:
A1:
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Calculates f(1.2715) and f(1.2725), or the tighter interval with at least 1 correct to 1 sig fig
rounded or truncated. Accept f(1.2715) = -0.008 1sf rounded or truncated. Also accept
f(1.2715) = -0.01 2dp. Accept f(1.2725) = (+) 0.008 1sf rounded or truncated. Also accept
f(1.2725) = (+)0.01 2dp
Both values correct (see above),
A valid reason; Accept change of sign, or >0 <0, or f(1.2715) ×f(1.2725)<0
And a (minimal) conclusion; Accept hence root or α=1.272 or QED or
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
3(a)
Scheme
Uses −2(3 − x) + 5 =
Marks
M1
1
x + 30
2
M1
Attempts to solve by multiplying out bracket, collect terms etc.
3
x = 31
2
x
A1
62
only
3
(3)
(b)
Makes the connection that there must be two intersections.
Implied by either end point k 5 or k 11
M1
A1
5 < k 11
(2)
(5 marks)
Notes:
(a)
1
x 30 can be found by solving
2
M1:
Deduces that the solution to f (
x)
M1:
1
x 30
2
Correct method used to solve their equation. Multiplies out bracket/ collects like terms.
62
only. Do not allow 20.6
x
3
2(3 x) 5
A1:
(b)
M1:
A1:
Deduces that two distinct roots occurs when y k intersects y f ( x) in two places. This
may be implied by the sight of either end point. Score for sight of either k 5 or k 11
Correct solution only k : k , 5 k 11
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Question
4(i)
Scheme
13
5
M1 A1
1
1
dx
ln(2 x 1)
2
2 x 1
Marks
dM1
1
1
1
1 25
dx ln 25 ln 9 ln
2
2
2 9
2 x 1
A1
5
ln
3
(4)
(ii)
M1
1
Integrates to give cos2 x sec x c where 0, 0
3
1
1
2 cos 2 x 3sec 3 x c
1
π
1 π 1
Substitutes limits of 0
cos 2 3sec cos 0 3sec 0 and 𝜋𝜋 and subtracts the
2
3 2 2
2
2
dM1
correct way around
A1
2 32
(3)
(7 marks)
Notes:
(i)
1
dx k ln(2 x 1) where k is a constant.
2 x 1
M1:
For
A1:
Correct integration
1
1
dx
ln(2 x 1)
2
2 x 1
dM1: Scored for substituting in the limits, subtracting and using correctly at least one log law.
25
You may see the subtraction law k ln 25 k ln 9
k ln or the index law
9
1
1
ln 25 ln 9
ln 5 ln 3
2
2
A1:
(ii)
M1:
5
cao ln
3
1
Integrates to a form cos2 x sec x c where 0, 0
3
𝜋𝜋
dM1: Dependent upon the previous M1. It is scored for substituting limits of 0 and 2 and
subtracting the correct way around.
A1:
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cao 2 3 2
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
Scheme
5
y
Marks
5 x 2 10 x 9
( x 1)2
Differentiates numerator to 10x −10 and denominator to 2(x − 1) o.e.
B1
Uses the quotient rule
M1 A1
2
2
dy ( x 1) 10 x 10 5 x 10 x 9 2 x 1
dx
( x 1)4
M1
Takes out a common factor from the numerator and cancels
2
dy x 1 ( x 1) 10 x 10 5 x 10 x 9 2
dx
( x 1) 4 3
M1
Simplifies the numerator by multiplying and collecting terms
10 x 2 20 x 10 10 x 2 20 x 18
dy
dx
( x 1)3
A1
dy
8
dx ( x 1)3
(6)
(6 marks)
Notes:
B1:
See scheme.
M1:
dy ( x 1)
Uses the quotient rule to reach a form
dx
2
Ax B 5 x2 10 x 9 Cx D
( x 1)4
o.e.
Alternatively uses the product rule to reach a for
dy
3
( x 1)2 Ax B 5 x 2 10 x 9 C x 1
dx
A1:
Fully correct
dy
If the product rule is used
dx
dy
3
( x 1)2 10 x 10 5 x 2 10 x 9 2 x 1
dx
dy
g( x)
. See scheme when using the
dx ( x 1)3
M1:
This is for using a correct method to reach a form
M1:
quotient rule. If the product rule is used it is for combining the terms using a common
denominator.
Scored for simplifying the numerator (By multiplying out and collecting terms).
A1:
dy
8
dx ( x 1)3
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
127
Question
Scheme
Marks
6(a)
f(x) > 2
B1
(1)
(b)
fg( x) eln x 2 , x 2
M1 A1
(2)
(c)
e
2 x3
26e
2 x3
M1 A1
4
2x 3
ln 4
x
ln 4 3
3
or ln 2
2
2
M1 A1
(4)
(d)
Let y e x 2 y 2 e x ln( y 2) x
M1
A1
B1ft
f 1 (
x) ln( x 2), x > 2
(3)
(e)
Shape for f(x)
B1
(0, 3)
B1
Shape for f -1(x)
B1
(3, 0)
B1
(4)
(14 marks)
Notes:
(a)
B1:
Range of f(x)>2. Accept y>2, (2,∞), f>2, as well as ‘range is the set of numbers bigger
than 2’ but don’t accept x >2
(b)
M1:
For applying the correct order of operations. Look for eln x 2 . Note that ln e x 2 is M0
A1:
Simplifies eln x 2 to x 2 . Just the answer is acceptable for both marks.
(c)
M1:
6 and proceeds to e2 x3 ...
Starts with e2 x3 2
A1:
M1:
e2 x3 4
ln.. and proceeds to x=….
Takes ln’s both sides, 2 x 3
128
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question 6 notes continued
A1:
ln 4 3
3
Remember to isw any incorrect working after a correct
oe. eg ln2 2
2
answer.
x
(d)
M1:
x e y 2 and attempts to change the subject. All ln work must be
Starts with y e x 2 or
correct. The 2 must be dealt with first. Eg. y e x 2 ln y x ln 2 x ln y ln 2 is
M0.
A1:
f 1 ( x)
ln( x 2) or y= ln( x 2) or y= ln x 2 There must be some form of bracket.
B1ft: Either x >2, or follow through on their answer to part (a), provided that it wasn’t y
Do not accept y>2 or f-1(x)>2.
(e)
B1:
B1:
B1:
B1:
Shape for y=ex. The graph should only lie in quadrants 1 and 2. It should start out with a
gradient that is approx. 0 above the x axis in quadrant 2 and increase in gradient as it moves
into quadrant 1. You should not see a minimum point on the graph.
(0, 3) lies on the curve. Accept 3 written on the y axis as long as the point lies on the curve.
Shape for y= ln x. The graph should only lie in quadrants 4 and 1. It should start out with
gradient that is approx. infinite to the right of the y axis in quadrant 4 and decrease in
gradient as it moves into quadrant 1. You should not see a maximum point. Also with
hold this mark if it intersects y=ex.
(3, 0) lies on the curve. Accept 3 written on the x axis as long as the point lies on the curve.
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
129
Question
Scheme
Marks
p = 4π2 or (2π)2
7(a)
B1
(1)
(b)
x = (4y – sin 2y)2
Sub y
π
2
dx
dy
24π (= 75.4) OR
Equation of tangent y
Using y
π
2
M1 A1
dx
= 2(4y – sin 2y) (4 – 2cos 2y)
dy
π
2
dy
dx
1
(= 0.013)
24π
M1
1
x 4π 2
24π
1
x 4π 2 with x
24π
M1
0
π
3
y
cso
M1 A1
(6)
Alternative I for first two marks
2
x 4 y sin 2 y x0.5 4 y sin 2 y
0.5 x 0.5
dx
dy
M1A1
4 2cos 2 y
Alternative II for first two marks
x
16 y 2 8 y sin 2 y sin 2 2 y
1 32 y
dy
dy
dy
dy
8sin 2 y
16 y cos 2 y
4sin 2 y cos 2 y
dx
dx
dx
dx
M1A1
Or 1dx
32 y dy 8sin 2 y dy 16 y cos 2 y dy 4sin 2 y cos 2 y dy
(7 marks)
Notes:
(a)
B1: B1:
(b)
M1:
p
2
4π 2 or exact equivalent 2π . Also allow x
4π 2
Uses the chain rule of differentiation to get a form
A(4 y sin 2 y) B C cos 2 y , A, B, C 0 on the right hand side.
Alternatively attempts to expand and then differentiate using product rule and chain rule to
dx
a form x 16 y 2 8 y sin 2 y sin 2 2 y Py Q sin 2 y Ry cos 2 y S sin 2 y cos 2 y P, Q, R, S 0
dy
A second method is to take the square root first. To score the method look for a
differentiated expression of the form Px0.5... 4 Q cos 2 y
A third method is to multiply out and use implicit differentiation. Look for the correct
terms, condoning errors on just the constants.
130
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question 7 notes continued
A1:
dx
dy
1
2 4 y sin 2 y 4 2cos 2 y or
with both sides
dy
dx 2 4 y sin 2 y 4 2cos 2 y
correct. The lhs may be seen elsewhere if clearly linked to the rhs. In the alternative
dx
32 y 8sin 2 y 16 y cos 2 y 4sin 2 y cos 2 y
dy
M1: M1:
Sub y
into their dx or inverted dx . Evidence could be minimal, eg y dx ...
2
dy
dy
2
dy
2
It is not dependent upon the previous M1 but it must be a changed
x 4 y sin 2 y
M1:
M1:
2
Allow for y
Allow for
Even allow for
π
1
2
their numerical dx
y
π
2
y
dy
their numerical dx
π
2
2
x their 4π 2
1
their numerical dx
x
dy
x
their 4π 2
p
dy
It is possible to score this by stating the equation y
M1:
Score for a correct method for finding the equation of the tangent at ' 4 ', .
1
x
24π
c as long as ' 4 ',
2
used in a subsequent line.
y mx c and stating the value of 'c'
Score for writing their equation in the form
or setting x 0 in their y
π
2
1
x
24π
4π 2 and solving for
is
2
y.
Alternatively using the gradient of the line segment AP = gradient of tangent.
y
1
y .. Such a method scores the previous M mark as well.
Look for 2 2
24
4
At this stage all of the constants must be numerical. It is not dependent and it is possible to
score this using the ''incorrect'' gradient.
A1:
cso y
π
. You do not have to see 0,
3
3
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
131
Question
8(a)
Scheme
Marks
N
aT b log10 N
log10 a log10 T b
M1
log10 N log10 a b log10 T so m = b and c log10 a
A1
(2)
(b)
(c)
Uses the graph to find either a or b a 10intercept or b = gradient
M1
Uses the graph to find both a and b a 10intercept and b = gradient
M1
b
Uses T 3 in N aT with their a and b
M1
Number of microbes 800
A1
States that 'a' is the number of microbes 1 day after the start of the experiment.
(4)
B1
(1)
(7 marks)
Notes:
(a)
M1:
Takes log10' s of both sides and attempts to use the addition law. Condone log log10 for
this mark.
A1:
log10 a b log10 T and states m = b and c log10 a
Proceeds correctly to log
10 N
(b)
M1:
M1:
Way One: Main scheme
For attempting to use the graph to find either a or b using a 10intercept or b = gradient. This
may be implied by a 101.75 to 1.85 or b = 2.27 to 2.33
For attempting to use the graph to find BOTH a and b (See previous M1)
Uses T 3 in N aT b with their a and b
Number of microbes 800
Way Two: Alternative using line of best fit techniques.
For log10 3 0.48 and using the graph to find log10 N
M1:
For using the graph to find log10 N (FYI log10 N 2.9 )
M1:
For log10 N k N 10k
A1:
Number of microbes 800
(c)
B1:
See scheme.
M1:
M1:
A1:
132
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
9(a)
Scheme
sec 2 A
B1
sin 2 A
cos 2 A
1
cos 2 A
tan 2 A
Marks
1 sin 2 A
cos 2 A
M1
1 2 sin A cos A
cos2 A sin 2 A
M1
cos2 A sin 2 A 2 sin A cos A
cos2 A sin 2 A
(cos A sin A)(cos A sin A)
(cos A sin A)(cos A sin A)
M1
cos A sin A
cos A sin A
A1*
(5)
(b)
sec 2θ
tan 2θ
1
2
cos θ sin θ
cos θ sin θ
1
2
2cos θ + 2sin θ = cos θ – sin θ
tan θ
M1 A1
1
3
dM1
A1
θ = awrt 2.820, 5.961
(4)
(9 marks)
Notes:
(a)
B1:
A correct identity for sec 2 A
1
or tan 2 A
cos 2 A
sin 2 A
.
cos 2 A
It need not be in the proof and it could be implied by the sight of sec 2 A
M1:
1
cos A
2
sin 2 A
For setting their expression as a single fraction. The denominator must be correct for their
fractions and at least two terms on the numerator.
This is usually scored for 1 cos 2 A tan 2 A or 1 sin 2 A
M1:
cos 2 A
cos 2 A
For getting an expression in just sin A and cos A by using the double angle identities
sin 2 A 2sin A cos A and cos 2 A cos2 A sin 2 A , 2 cos2 A 1 or 1 2sin 2 A .
Alternatively for getting an expression in just sin A and cos A by using the double angle
identities sin 2 A
For example
2sin A cos A and tan 2 A
1
cos2 A sin 2 A
2 tan A
with tan A
1 tan 2 A
sin A
.
cos A
2 sin A
cos A is B1M0M1 so far
2
sin
A
1
cos2 A
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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133
Question 9 notes continued
M1:
In the main scheme it is for replacing 1 by cos2 A sin2 A and factorising both numerator and
denominator.
A1*: Cancelling to produce given answer with no errors. Allow a consistent use of another
variable such as θ , but mixing up variables will lose the A1*.
(b)
M1:
A1:
For using part (a), cross multiplying, dividing by cosθ to reach tan θ
Condone tan 2θ k for this mark only.
tan θ
k
1
3
dM1: Scored for tan θ k leading to at least one value (with 1 dp accuracy) for between 0
and 2π . You may have to use a calculator to check. Allow answers in degrees for this mark.
θ awrt 2.820, 5.961with no extra solutions within the range. Condone 2.82 for 2.820.
A1:
You may condone different/ mixed variables in part (b)
134
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
10(a)
Scheme
Marks
x = 15e−0.2×4 = 6.740 (mg)
Subs D 15 and t = 4
M1 A1
(2)
(b)
(c)
M1
15e−0.2×7 + 15e−0.2×2 = 13.754 (mg)
−0.2×T
15e
−0.2×(T+5)
+ 15e
A1*
(2)
M1
= 7.5
15e−0.2×T + 15e−0.2×Te−1 = 7.5
7.5
15e0.2T (1+e1 )
7.5 e0.2T
15 1+e1
dM1
7.5
2 2
T
5ln
5ln
1
e
15 1 e
A1 A1
(4)
(8 marks)
Notes:
(a)
M1:
A1:
(b)
M1:
Attempts to substitute both D 15 and t =4 in x De
15e
0.8
0.24
0.2 t
. It can be implied by sight of
, 15e
or awrt 6.7. Condone slips on the power. Eg you may see -0.02
Cao. 6.740 (mg) Note that 6.74 (mg) is A0
Attempt to find the sum of two expressions with D =15 in both terms with t values of 2
and 7. Evidence would be 15e0.27 15e0.22 or similar expressions such as
15e1 15 e0.22 . Award for the sight of the two numbers awrt 3.70 and awrt 10.05,
followed by their total awrt 13.75. Alternatively finds the amount after 5 hours,
15e−1 = awrt 5.52 adds the second dose = 15 to get a total of awrt 20.52 then multiplies this
by e0.4 to get awrt 13.75. Sight of 5.52+15=20.52 13.75 is fine.
A1*:
Cso so both the expression 15e
0.27
15e0.22 and 13.754( mg ) are required
Alternatively both the expression 15e0.25 15 e0.22 and 13.754 (mg) are required.
Sight of just the numbers is not enough for the A1*
(c)
M1:
dM1:
Attempts to write down a correct equation involving T or t. Accept with or without correct
bracketing Eg. accept 15e0.2T 15e0.2(T 5)
7.5 or similar equations
1
0.2T
7.5
15e 15 e
Attempts to solve their equation, dependent upon the previous mark, by proceeding to
n
e0.2T .. An attempt should involve an attempt at the index law x m
x m x n and
0.2T
0.2T
taking out a factor of e
Also score for candidates who make e
the subject using
the same criteria.
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
135
Question 10 notes continued
A1:
A1:
7.5
15 1 e1
Any correct form of the answer, for example, 5ln
2
Cso.
T 5ln 2 Condone t appearing for T throughout this question.
e
(c)
Alternative 1
1st Mark (Method): 15e0.2T awrt 5.52e0.2T
7.5 e0.2T awrt 0.37
7.5
2nd Mark (Accuracy): T=-5ln awrt 0.37 or awrt 5.03 or T=-5ln
awrt
20.52
Alternative 2
7.5
1st Mark (Method ): 13.754e0.2T
5ln
7.5 T
or equivalent such as 3.03
13.754
7.5
2nd Mark (Accuracy): 3.03 + 2 = 5.03 Allow 5ln
2
13.754
Alternative 3 (by trial and improvement)
0.25
15e0.210
7.55 or 15e0.25.1 15e0.210.1
7.40 or any value
1st Mark (Method): 15e
between.
2nd Mark (Accuracy): Answer T =5.03.
136
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Please check the examination details below before entering your candidate information
Candidate surname
Pearson Edexcel
Other names
Centre Number
Candidate Number
International
Advanced Level
Sample Assessment Materials for first teaching September 2018
(Time: 1 hour 30 minutes)
Paper Reference WMA14/01
Mathematics
International Advanced Level
Pure Mathematics P4
You must have:
Mathematical Formulae and Statistical Tables, calculator
Total Marks
Candidates may use any calculator permitted by Pearson regulations.
Calculators must not have the facility for symbolic algebra manipulation,
differentiation and integration, or have retrievable mathematical
formulae stored in them.
Instructions
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•• Use
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in the boxes at the top of this page with your name, centre number and
• Fill
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.
Inexact
answers
should
be given to three significant figures unless otherwise
• stated.
Information
booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
•• AThere
are 9 questions in this question paper. The total mark for this paper is 75.
The
marks
each question are shown in brackets
• – use this asfora guide
as to how much time to spend on each question.
Advice
each question carefully before you start to answer it.
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Leave
blank
Answer ALL questions. Write your answers in the spaces provided.
Use the binomial series to find the expansion of
1
(2 + 5 x)
3
½x½ <
2
5
in ascending powers of x, up to and including the term in x 3
Give each coefficient as a fraction in its simplest form.
(6)
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Q1
(Total for Question 1 is 6 marks)
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2.
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A curve C has the equation
(a) Find
dy
in terms of x and y.
dx
(5)
(b) Find an equation of the tangent to C at the point (3, –2), giving your answer in the
form ax + by + c = 0, where a, b and c are integers.
(2)
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x 3 + 2xy – x – y 3 – 20 = 0
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Q2
(Total for Question 2 is 7 marks)
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3.
f(x) =
1
=
B
C
A
+
+
x (3 x − 1) (3 x − 1) 2
(a) Find the values of the constants A, B and C
(4)
(b) (i) Hence find ∫ f(x) dx
2
(ii) Find ∫1 f(x) dx, giving your answer in the form a + ln b, where a and b are
constants.
(6)
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x(3 x − 1)
2
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Q3
(Total for Question 3 is 10 marks)
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4.
y
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C
O
x
Figure 1
x=
(a) Show that
3 sin 2t
y = 4 cos2 t
0tπ
dy
= k 3 tan 2t, where k is a constant to be found.
dx
(b) Find an equation of the tangent to C at the point where t =
(5)
π
3
Give your answer in the form y = ax + b, where a and b are constants.
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Figure 1 shows a sketch of the curve C with parametric equations
(4)
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*S59758A01130*
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Q4
(Total for Question 4 is 9 marks)
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5.
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y
R
A
O
x
Figure 2
1
x
Figure 2 shows a sketch of part of the curve with equation y = 4x – xe 2 , x 0
(a) Find, in terms of ln 2, the x coordinate of the point A.
1
(b) Find
(2)
x
∫ xe 2 d x
(3)
The finite region R, shown shaded in Figure 2, is bounded by the x‑axis and the curve with
1
x
equation y = 4x – xe 2 , x 0
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The curve meets the x‑axis at the origin O and cuts the x‑axis at the point A.
(c) Find, by integration, the exact value for the area of R.
Give your answer in terms of ln 2
(3)
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*S59758A01530*
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Q5
(Total for Question 5 is 8 marks)
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6.
Prove by contradiction that, if a, b are positive real numbers, then a + b 2 ab
(4)
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*S59758A01930*
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Q6
(Total for Question 6 is 4 marks)
20
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7.
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y
C
O
x
Figure 3
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Figure 3 shows a sketch of the curve C with parametric equations
π
x = 4 cos t +
6
y = 2 sin t
0 t 2π
(a) Show that
x + y = 2 3 cos t
(3)
(b) Show that a cartesian equation of C is
(x + y)2 + ay 2 = b
where a and b are integers to be found.
(2)
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*S59758A02130*
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Q7
(Total for Question 7 is 5 marks)
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8.
Water is being heated in a kettle. At time t seconds, the temperature of the water is θ °C.
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The rate of increase of the temperature of the water at time t is modelled by the differential
equation
dθ
= λ(120 – θ )
dt
θ 100
where λ is a positive constant.
Given that θ = 20 when t = 0
(a) solve this differential equation to show that
θ = 120 – 100e –λt
(8)
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When the temperature of the water reaches 100 °C, the kettle switches off.
(b) Given that λ = 0.01, find the time, to the nearest second, when the kettle switches off.
(3)
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*S59758A02330*
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*S59758A02530*
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Q8
(Total for Question 8 is 11 marks)
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9.
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With respect to a fixed origin O, the line l1 is given by the equation
8
−5
r = 1 + μ 4
−3
3
where μ is a scalar parameter.
The point A lies on l1 where μ = 1
(a) Find the coordinates of A.
(1)
1
The point P has position vector 5
2
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The line l2 passes through the point P and is parallel to the line l1
(b) Write down a vector equation for the line l2
(2)
(c) Find the exact value of the distance AP.
Give your answer in the form k 2 , where k is a constant to be found.
(2)
The acute angle between AP and l2 is θ
(d) Find the value of cos θ
(3)
A point E lies on the line l2
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Given that AP = PE,
(e) find the area of triangle APE,
(f) find the coordinates of the two possible positions of E.
(2)
(5)
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Q9
(Total for Question 9 is 15 marks)
TOTAL FOR PAPER IS 75 MARKS
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Pure Mathematics P4 Mark scheme
Question
1
Scheme
Marks
M1
1
−3
(2 + 5x)
3
2 5x
(2) 3 1
5x
2
3
1 5x
1
8
2
B1
3
M1 A1
( 3)( 4)
( 3)( 4)(5)
1
1 ( 3)(k x)
( k x) 2
(k x)3 ...
2!
3!
8
2
3
( 3)( 4)(5) 5 x
1
5 x ( 3)( 4) 5 x
1 ( 3)
...
2!
3!
8
2
2
2
1
15
75 2 625 3
1
x
x
x ...
8
2
2
4
1
[ 1 – 7.5x + 37.5x2 −156.25 x3 + …]
8
1 15
75 2 625 3
x ;
x
x ...
8 16
16
32
or
A1 A1
1 15
11
17
x ; 4 x 2 19 x3 ...
8 16
16
32
(6)
Notes:
M1:
B1:
M1:
1
Mark can be implied by a constant term of (2) 3 or .
8
1
1
as candidate’s constant term in their binomial expansion.
2 3 or outside brackets or
8
8
Expands (...+ k x ) , k = a value ≠ 1 to give any 2 terms out of 4 terms simplified or un−3
( 3)( 4)
( 3)( 4)(5)
( k x) 2
(k x)3 or
2!
3!
( 3)( 4)
(
3)(
4)
(
3)(
4)(
5)
1 ...
(k x) 2 or
( k x) 2
(k x)3 are fine for M1.
2!
2!
3!
( 3)( 4)
( 3)( 4)(5)
A correct simplified or un-simplified 1 ( 3)(k x)
( k x) 2
( k x)3
2!
3!
simplified, Eg: 1 ( 3)(k x) or
A1:
( )
expansion with consistent k x . Note that k x must be consistent and k = a value ≠ 1.
the RHS, not necessarily the LHS) in a candidate’s expansion.
(on
1 15
x (simplified) or also allow 0.125 0.9375x.
8 16
A1:
For
A1:
Accept only
11
17
75 2 625 3
x
x or 4 x 2 − 19 x 3 or 4.6875x2 19.53125x3
16
32
16
32
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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167
Question
Scheme
Marks
2(a)
x3 + 2xy – x – y3 – 20 = 0
dy
dy
2
2 dy
3x 2 y 2 x 1 3 y 0
dx
dx
dx
3x 2 2 y 1 2 x 3 y 2
dy
3x 2 2 y 1
dx
3 y2 2x
or
ddyx 0
M1 A1
B1
dM1
1 3x 2 2 y
2x 3y2
cso
A1
(5)
(b)
At P(3, −2), m T
dy
dx
2
3(3) 2(2) 1
;
3(2)2 2(3)
11
22
or
3
6
11
"11 "
and either T: y
2) (3) c
c ... ,
2
x 3 or (
3
3
T: 11x – 3y – 39 = 0
or
K(11x – 3y – 39) = 0 cso
M1
A1
(2)
(7 marks)
Notes:
(a)
M1:
Differentiates implicitly to include either 2 y
dx
).
dy
dx
x3 3x 2
and
dy
dx
dx
dx
or x3 kx 2
or x
dy
dy
dy
(Ignore
A1:
B1:
2 xy 2 y
x y 3 20 0
dx
3 y 2 0
dy
dx
2x
dy
dM1: Dependent on the first method mark being awarded. An attempt to factorise out all the
terms in
A1:
For
dx
dx
as long as there are at least two terms in
.
dy
dy
1 2 y 3x 2
3x 2 2 y 1
or
equivalent.
Eg:
2x 3y2
3y2 2x
(b)
M1:
Some attempt to substitute both x 3 and y 2 into their
y to find mT and
y 2
either applies
A1:
168
dy
which contains both x and
dx
their mT x 3 , where mT is a numerical value.
( 2) their mT (3) c , where mT is a numerical value.
or finds c by solving
Accept any integer multiple of 11x 3 y 39
0 or 11x 39 3 y
0 or
11x 3 y 39
0 , where their tangent equation is equal to 0.
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
Scheme
Marks
1 = A(3x – 1)2 + Bx(3x – 1) + Cx
3(a)
B1
x 0 (1 = A)
M1
C C = 3 any two constants correct coefficients of x2
1
x 13
1
3
0 = 9A + 3B
B = − 3 all three constants correct
A1
A1
(4)
(b)(i)
1
3
3
dx
x 3x 1 3x 12
3
3
1
ln x ln 3x 1
3x 1
3
1 3
1
ln x ln 3x 1
3x 1
M1
A1ft
A1ft
C
C
(3)
(b)(ii)
2
1
2
1
f x d
x ln x ln 3x 1
3x 1 1
1
1
ln 2 ln 5 ln1 ln 2
5
2
ln
M1
2 2
...
5
M1
3
4
ln
10
5
A1
(3)
(10 marks)
Notes:
(a)
B1:
M1:
A1:
A1:
1 A(3x 1)2 Bx 3x 1 Cx at any stage. This will usually be at the
Obtaining
beginning of the solution but, if the cover-up rule is used, it could appear later.
A complete method of finding any one of the three constants. If either A 1 or C 3 is given
without working or, at least, without incorrect working, allow this M1 – use of the cover-up
rule is acceptable. In principle, an alternative method is equating coefficients (or substituting
three values other than 0 and 13 ), obtaining a sufficient set of equations and solving for any
one of the three constants.
Any two of A, B and C correct. These will usually, but not always, be A and C.
All three of A, B and C correct. If all three constants are correct and the answers do not
clearly conflict with any working, allow all 4 marks (including the B1) bod. There are a
number of possible ways of finding B but, as long as the M has been gained, you need not
consider the method used.
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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169
Question 3 notes continued
(b)(ii)
M1: Dependent upon the M mark in (b). Substituting in the correct limits and subtracting, not
necessarily the right way round. There must be evidence that both 1 and 2 have been used
but errors in substitution do not lose the mark.
M1: Dependent upon both previous Ms. Applies the addition and/or subtraction rules of logs to
obtain a single logarithm. Either the addition or the subtraction rule of logs must be used
correctly at least once to gain this mark and this must be seen in the attempt at (b)(ii).
A1: The correct answer in the form specified. Accept equivalent fractions including exact
decimals for a and or b.
3 .
Accept ln 54 10
3 ln 5
10 4
170
is not acceptable.
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
Scheme
4(a)
Marks
dx
2 3 cos 2t
dt
B1
dy
−8 cos t sin t
dt
M1 A1
d y 8cos t sin t
dx 2 3 cos 2t
M1
4sin 2t
2 3 cos 2t
dy
2
3 tan 2t
dx
3
2
k
3
A1
(5)
(b)
When t
3
x
3
, y 1 can be implied
2
B1
2
2
3 tan
m
2
3
3
M1
3
y
1 2 x
2
dM1
y = 2x − 2
A1
(4)
(9 marks)
Notes:
(a)
dx
dt
B1:
The correct
M1:
dy
k cos t sin t or k sin 2t , where k is a non-zero constant. Allow k 1
dt
A1:
dy
8cos t sin t or 4sin 2t or equivalent. In this question, it is possible to get a correct
dt
M1:
answer after incorrect working, e.g. 2cos 2t 2 4sin 2t . This should lose this mark and
the next A but ignore in part (b).
dx
dt
dy
dy
divided by their
, or their
multiplied by their
. The answer must be a
Their
dt
dx
dt
dt
function of t only.
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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171
Question 4 notes continued
A1:
The correct answer in the form specified. They don’t have to explicitly state k = 23 but
there must be evidence that the constant is 23 . Accept equivalent fractions.
(b)
B1:
That when t 3 , x 32 and y 1 . Exact numerical values are required but the values can
be implied, for example by a correct final answer, and can occur anywhere in the question.
M1:
Substituting t 3 into their
dy
dx
. Trigonometric terms, e.g. tan 23 need not be evaluated.
dM1: Dependent on the previous M. Finding an equation of a tangent with their point and their
numerical value of the gradient of the tangent, not the normal. Expressions like tan 23
must be evaluated. The equation must be linear. Using y y m x x . They should get
A1:
172
y
x and y the right way round. Alternatively writing
their point, the right way round, to find c.
cao. The correct answer in the form specified.
their m x c
and using
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
Scheme
Marks
5(a)
y 4x x e 2 , x 0
1x
1x
1x
y
0 4 x x e 2 0 x(4 e 2 ) 0
1x
e
1x
2
4 xA 4ln 2
Attempts to solve e 2 4 giving x = ...
in terms of ln where 0
M1
4ln2 cao (Ignore x = 0 )
A1
(2)
(b)
1
dx
x e
1x
2
2x e
1
x
2
2e
1
x
2
1
x
x
M1
x
A1
xe 2 e 2 dx , 0, 0
dx
1
x
1
2 x e 2 2e 2 dx , with or without dx
1
1
x
= 2xe 2 − 4e 2
x
{+ c}
A1
(3)
4x dx 2x
(c)
2
4ln 2
0
(4 x x e ) dx
1x
2
1
1
2
x
x
2
2 x 2 x e 4e 2
B1
4ln 2 or ln16 or their limits
0
1
1
1
1
(4ln 2)
(4ln 2)
(0)
(0)
2
2(4ln 2)2 2(4ln 2)e 2
4e 2
2(0) 2(0)e 2 4e 2
M1
= (32(ln 2)2 – 32(ln 2) + 16) − (4)
A1
= 32(ln 2)2 – 32(ln 2) + 12
(3)
(8 marks)
Notes:
(a)
M1:
A1:
1x
Attempts to solve e 2 4 giving x = ... in terms of ln where 0
4ln 2 cao stated in part (a) only (Ignore x = 0 )
(b)
M1:
1
x
1
x
Integration by parts is applied in the form xe 2 e 2 dx , where α > 0, β > 0 .
(must be in this form) with or without dx
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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173
Question 5 notes continued
1
x
1
x
A1:
2 x e 2 2e 2 dx or equivalent, with or without dx . Can be un-simplified.
A1:
2 x e 2 4e 2 or equivalent with or without + c. Can be un-simplified.
1
x
1
x
(c)
4 x2
oe
2
B1:
4 x 2 x 2 or
M1:
Complete method of applying limits of their x A and 0 to all terms of an expression of the
A1:
form ± Ax 2 ± Bx e 2 ± Ce 2 . (Where A 0, B 0 and C 0 ) and subtracting the correct way
round.
A correct three term exact quadratic expression in ln 2 . For example allow for A1
1
x
1
x
32(ln2)2 − 32(ln2) + 12
8(2ln2)2 − 8(4ln2) + 12
2(4ln2)2 − 32(ln2) + 12
2(4ln 2)2 − 2(4ln 2)e 2
1
(4ln2)
+ 12
1
Note that the constant term of 12 needs to be combined from 4e 2
(4ln 2)
1
4e 2
(0)
o.e.
12
Also allow 32ln2(ln2 − 1) + 12 or 32ln 2 ln 2 1
for A1.
32ln 2
Allow 32(ln2 2) − 32(ln2) + 12 for the final A1.
174
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
6
Scheme
Marks
Assumption: there exists positive real numbers a, b such that
B1
a b 2 ab
Method 1
a+b−2
(
a
b
ab
<0
)2 < 0
Method 2
(a + b)2 = (2
2
ab )
a2 + 2ab + b2 < 4ab
a2 − 2ab + b2 < 0
A complete method for
creating
M1A1
(f(a,b))2 < 0
(a − b)2 < 0
This is a contradiction, therefore
If a, b are positive real numbers, then a + b 2 ab
A1
(4)
(4 marks)
Notes:
B1:
As this is proof by contradiction, the candidate is required to start their proof by assuming
that the contrary. That is ''if a, b are positive real numbers, then a b 2 ab '' is true.
Accept, as a minimum, there exists a and b such that a b 2 ab
M1:
For starting with a b 2 ab and proceeding to either ( a b )2 < 0 or (a − b)2 < 0
A1:
All algebra is required to be correct. Do not accept, for instance, (a + b)2 = 2 ab 2 even
when followed by correct lines.
A fully correct proof by contradiction. It must include a statement that (a − b)2 < 0 is a
A1:
contradiction so if a, b are positive real numbers, then a + b 2 ab
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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175
Question
Scheme
7(a)
x 4cos t ,
6
Marks
y = 2sin t
x 4 cos t cos sin t sin
6
6
M1
x y 4 cos t cos sin t sin + 2sin t
So,
6
6
Adds their expanded x
(which is in terms of t)
to 2sin t
dM1
3
1
4
cos
t
sin
t
+ 2sin t
2
2
2 3 cos t *
A1*
cso
(3)
(b)
x y
2 3
2
2
Applies cos t + sin t = 1 to
achieve an equation
containing only x’s and y’s.
2
2
y
1
2
M1
2
( x y)2
y
1
12
4
(x + y)2 + 3y2 = 12
(x + y)2 + 3y2 = 12
A1
{a = 3, b = 12}
(2)
( x y)2 12 3 y 2
Applies cos2 t sin 2 t
1 to
achieve an equation
containing only x’s and y’s.
M1
( x y )2 3 y 2
12
( x y)2 3 y 2
12
A1
Alternative
( x y)2 12cos2 t 12(1 sin 2 t ) 12 12sin 2 t
(2)
(5 marks)
Notes:
(a)
M1:
cos t cos t cos sin t sin or cos t
6
6
6
6
3
1
2 cos t 2 sin t
dM1: Adds their expanded x (which is in terms of t) to 2sin t .
A1*: Evidence of cos and sin evaluated and the proof is correct with no errors.
6
6
(b)
M1:
Applies cos2 t sin 2 t
1 to achieve an equation containing only x’s and y’s.
A1:
leading ( x y)2 3 y 2
12
176
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
8(a)
Scheme
Marks
d
= λ(120 – θ), θ 100
dt
1
d
120
B1
dt
For integrating lhs M1 A1
For integrating rhs M1 A1
–ln(120 – θ); = λt + c
M1A1;
M1A1
{t = 0, θ = 20 } –ln (100) = λ(0) + c
−ln (120 – θ) = λt − ln 100
−λt = ln (120 – θ) − ln 100
M1
−λt ln 120
e t
100
120
100
dddM1
100 e–λt = 120 − θ
A1*
leading to θ = 120 − 100e–λt
(8)
(b)
{λ = 0.01, θ = 100
100
e–0.01t = 120 −100
}
t
M1
120 100
0.01t
ln
100
t
100 = 120 – 100 e–0.01t
1
120 100
ln
0.01 100
Uses correct order of operations by
moving from 100
120 100e 0.01t
dM1
to give t ... and t A ln B , where
B0
1
1
ln 100ln 5
0.01 5
t = 160.94379… 161 (s) (nearest second) awrt 161
A1
(3)
(11
marks)
Notes:
(a)
B1M1A1M1A1: Mark as in the scheme.
M1: Substitutes t 0 AND 20 in an integrated equation leading to
t
ln f
dddM1: Uses a fully correct method to eliminate their logarithms and writes down an
equation containing their evaluated constant of integration.
A1*: Correct answer with no errors. This is a given answer
(b)
M1:
M1:
A1:
Substitutes λ = 0.01, θ = 100 into given equation
See scheme
Awrt 161 seconds.
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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177
Question
Scheme
Marks
9 (a)
A(3, 5, 0)
B1
(1)
(b)
a d or a d , a + td a 0, d 0
1
5
l2 :
r 5 4
2
3
with either a i 5j 2k or d
5 i 4 j 3k ,
M1
or a multiple of 5 i 4 j 3k
Correct vector equation using r or l or l2 =
A1
(2)
(c)
1 3
AP OP OA 5 5
2 0
2
0
2
Full method for finding AP
AP ( 2)2 (0)2 ( 2)2 8 2 2
M1
A1
2 2
(2)
(d)
Realisation that the dot
product is required
2
5
2 5
and
So AP 0
d 2 4 0 . 4
2
3
2 3
cos
AP . d 2
AP d 2
M1
between
and ± Kd2 or ± Kd1
dM1
2 5
0 . 4
2 3
( 2) 2 (0) 2 ( 2) 2 . ( 5) 2 (4) 2 (3) 2
{cos θ } = ± (10 + 0 + 6)
8 . 50
=
4
5
A1 cso
(3)
(e)
Area
APE
1
(their 2 2)2 sin
2
= 2.4
M1
A1
(2)
178
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
9(f)
Scheme
Marks
PE
(5 )i (4 ) j (3 )k and PE
their 2 2 from part (c)
{PE2 =} (−5λ)2 + (4λ)2 + (3λ)2 = (their 2
2
)2
4
2
2
8 2
50
25
5
l2
:r
OE
M1
2
2
or
5
5
A1
Either
dependent on the
previous M mark
Substitutes at least one
of their values of into
1
5
2
5 5 4
2
3
3
3
17
or 3.4 , OE
5
0.8
4
5
This mark can be
implied.
dM1
l2 .
1
1
33 or 6.6
5
3.2
16
5
At least one set of
coordinates are correct.
A1
Both sets of coordinates
are correct.
A1
(5)
(15 marks)
Notes:
(a)
B1:
(b)
A1:
3
3
Allow A(3, 5, 0) or 3i 5j or 3i +5j + 0k or 5 or benefit of the doubt 5
0
0
Correct vector equation using r or l or l2 = or Line 2 =
1
5
r 5 4
i.e. Writing
2
3
æ 1 ö
or r = ç 5 ÷ + λ d , where d is a multiple of
ç
÷
çè 2 ÷ø
5
4.
3
Note: Allow the use of parameters µ or t instead of λ .
(c)
M1:
Finds the difference between
AP
and their
and applies Pythagoras to the result to find
æ 2 ö
Note: Allow M1A1 for ç 0 ÷ leading to AP = (2)2 + (0)2 + ( 2)2 = 8 = 2 2 .
ç
÷
çè 2 ÷ø
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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179
Question 9 notes continued
(d)
M1:
Realisation that the dot product is required between
dM1: Full method to find cos (dependent upon the previous M),
4
A1:
cos or exact equivalent
5
(e)
1
1
awrt 2.40
M1 A1: For (2 2)2 sin(36.869... ) or (2 2) 2 sin(180 36.869... );
2
2
Candidates must use their θ from part (d) or apply a correct method of finding their
4
3
sin from their cos
5
5
(f)
M1:
Allow special case 1st M1 for λ = 2.5 from comparing lengths or from no working.
for
( 5 )2 (4 )2 (3 )2
(their 2 2)
1st M0 for ( 5 )2 (4 )2 (3 )2
(their 2 2) or equivalent.
1st M1 for
So
d1
their AP "2 2 "
2
2
(5) (4) (3)
2
and 1st A1 for
2 2
5 2
5
5
1
2
2
4 "vector"
4 is M1A1
5 2 3
5 2 3
dM1: In part (f) can be implied for at least 2 (out of 6) correct x, y, z ordinates from their
values of λ .
180
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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Please check the examination details below before entering your candidate information
Candidate surname
Pearson Edexcel
Other names
Centre Number
Candidate Number
International
Advanced Level
Sample Assessment Materials for first teaching September 2018
(Time: 1 hour 30 minutes)
Paper Reference WFM01/01
Mathematics
International Advanced Subsidiary/Advanced Level
Further Pure Mathematics FP1
You must have:
Mathematical Formulae and Statistical Tables, calculator
Total Marks
Candidates may use any calculator permitted by Pearson regulations.
Calculators must not have the facility for symbolic algebra manipulation,
differentiation and integration, or have retrievable mathematical
formulae stored in them.
Instructions
black ink or ball-point pen.
•• Use
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
in the boxes at the top of this page with your name, centre number and
• Fill
candidate number.
Answer all questions and ensure that your answers to parts of questions are
• clearly
labelled.
the questions in the spaces provided
• Answer
– there may be more space than you need.
should show sufficient working to make your methods clear.
• You
Answers without working may not gain full credit.
answers should be given to three significant figures unless
• Inexact
otherwise stated.
Information
booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
•• AThere
are 10 questions in this question paper. The total mark for this paper is 75.
The
marks
each question are shown in brackets
• – use this asfora guide
as to how much time to spend on each question.
Advice
each question carefully before you start to answer it.
•• Read
Try to answer every question.
your answers if you have time at the end.
•• Check
If you change your mind about an answer, cross it out and put your new answer
and any working underneath.
Turn over
S59759A
©2018 Pearson Education Ltd.
1/1/1/1/1/1/
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181
Leave
blank
Answer ALL questions. Write your answers in the spaces provided.
Use the standard results for
∑r
and for
∑ r (r
r =1
2
− 3) =
∑r
3
to show that, for all positive integers n,
r =1
r =1
n
n
n
(n + a)(n + b)(n + c)
4
where a, b and c are integers to be found.
(4)
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*S59759A0232*
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Q1
(Total for Question 1 is 4 marks)
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2.
A parabola P has cartesian equation y 2 = 28x. The point S is the focus of the parabola P.
(1)
Points A and B lie on the parabola P. The line AB is parallel to the directrix of P and cuts
the x-axis at the midpoint of OS, where O is the origin.
(b) Find the exact area of triangle ABS.
(4)
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(a) Write down the coordinates of the point S.
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Q2
(Total for Question 2 is 5 marks)
*S59759A0532*
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3.
3
− 1,
x
x<0
The only real root, α, of the equation f(x) = 0 lies in the interval [−2, −1].
(a) Taking −1.5 as a first approximation to α, apply the Newton-Raphson procedure once
to f(x) to find a second approximation to α, giving your answer to 2 decimal places.
(5)
(b) Show that your answer to part (a) gives α correct to 2 decimal places.
(2)
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f(x) = x2 +
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Q3
(Total for Question 3 is 7 marks)
*S59759A0732*
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Given that
3
, where k is a constant
k + 2
(a) show that det(A) > 0 for all real values of k,
(b) find A−1 in terms of k.
(3)
(2)
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k
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Q4
(Total for Question 4 is 5 marks)
*S59759A0932*
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2z + z* =
Find z, giving your answer in the form a + bi, where a and b are real constants. You must
show all your working.
(5)
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3 + 4i
7+i
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Q5
(Total for Question 5 is 5 marks)
*S59759A01132*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
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6.
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The rectangular hyperbola H has equation xy = 25
The point A on H has parameter t =
(1)
1
2
(b) Show that the normal to H at the point A has equation
8 y − 2 x − 75 = 0
This normal at A meets H again at the point B.
(c) Find the coordinates of B.
(5)
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5
(a) Verify that, for t ¹ 0, the point P 5t , is a general point on H.
t
(4)
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*S59759A01332*
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Question 6 continued
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Q6
(Total for Question 6 is 10 marks)
*S59759A01532*
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7.
−
12
13
5
13
(a) Describe fully the single geometrical transformation U represented by the matrix P.
(3)
The transformation V, represented by the 2×2 matrix Q, is a reflection in the line with
equation y = x
(b) Write down the matrix Q.
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5
13
P=
12
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(1)
Given that the transformation V followed by the transformation U is the transformation T,
which is represented by the matrix R,
(c) find the matrix R.
(d) Show that there is a value of k for which the transformation T maps each point on the
straight line y = kx onto itself, and state the value of k.
(4)
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*S59759A01732*
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Question 7 continued
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Q7
(Total for Question 7 is 10 marks)
*S59759A01932*
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f(z) = z 4 + 6 z 3 + 76 z 2 + az + b
Given that −3 + 8i is a complex root of the equation f(z) = 0
(a) write down another complex root of this equation.
(b) Hence, or otherwise, find the other roots of the equation f(z) = 0
(c) Show on a single Argand diagram all four roots of the equation f(z) = 0
(1)
(6)
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where a and b are real constants.
(2)
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*S59759A02132*
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Q8
(Total for Question 8 is 9 marks)
*S59759A02332*
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The quadratic equation
has roots α and β.
Without solving the quadratic equation,
(a) find the exact value of
(i) α2 + β 2
(ii) α3 + β 3
(5)
(b) Find a quadratic equation which has roots (α 2 + β) and ( β 2 + α), giving your
answer in the form ax2 + bx + c = 0, where a, b and c are integers.
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2x2 + 4x −3 = 0
(4)
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*S59759A02532*
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Q9
(Total for Question 9 is 9 marks)
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10. (i) A sequence of positive numbers is defined by
un+1 = 3un+ 2,
n 1
Prove by induction that, for n Î +,
un = 2 × (3) − 1
n
(5)
(ii) Prove by induction that, for n Î +,
n
∑
r =1
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u1 = 5
4r
(3 + 2n)
= 3−
r
3
3n
(6)
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*S59759A02932*
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*S59759A03132*
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Q10
(Total for Question 10 is 11 marks)
TOTAL FOR PAPER IS 75 MARKS
32
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Further Pure Mathematics FP1 Mark scheme
Question
Scheme
1
n
r (r
r 1
2
n
3)
r
Marks
3
n
r
3
r 1
r 1
1 2
1
n (n 1)2 3 n(n 1)
4
2
Attempts to expand r (r 2 3) and
attempts to substitute at least one
correct standard formula into their
resulting expression.
M1
Correct expression (or equivalent)
A1
dM1
1
n(n 1) n(n 1) 6
4
dependent on the previous M
mark
Attempt to factorise at least n(n 1)
having attempted to substitute both
the standard formulae
1
n(n 1) n2 n 6
4
{this step does not have to be
written]
1
n(n 1)(n 3)(n 2)
4
Correct completion with no errors
A1 cso
(4)
(4 marks)
Notes:
Applying eg.
n 1,
n 2,
n 3 to the printed equation without applying the standard formulae to give
a 1, b 3, c 2 or another combination of these numbers is M0A0M0A0.
Alternative Method:
Obtains
n
r (r
2
3)
r 1
1
1
n(n 1) n(n 1) 6 n(n a)(n b)(n c)
4
4
1
1
(1)(2)(1 b)(1 c) and n 2 0 (2)(3)(2 b)(2 c)
4
4
leading to either b
2, c
3 or b 3, c 2
So a 1.
n 1 2
dM1: dependent on the previous M mark.
Substitutes in values of n and solves to find b ... and c ...
A1:
Finds a 1, b 3, c 2 or another combination of these numbers.
Using only a method of “proof by induction” scores 0 marks unless there is use of the
standard formulae when the first M1 may be scored.
1 4 1 3 5 2 3
1
n n n n or n(n3 2n2 5n 6)
4
2
4
2
4
1 4
1
or
(n 2n3 5n2 6n) n(n 1)(n 3)(n 2) , from no incorrect working.
4
4
1
1
n(n 1)( x 3)( x 2) unless recovered.
Give final A0 for eg. n(n 1) n2 n 6
4
4
Allow final dM1A1 for
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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Question
2(a)
Scheme
Marks
P : y 2 28x or P(7t 2 ,14t )
Accept (7,0) or
x 7,
y 0 or
7 marked on the x-axis in a
sketch
2
( y 4ax a 7) S (7,0)
B1
(1)
(b)
{A and B have x coordinate}
Divides their x coordinate from
(a) by 2
and
substitutes this into the
parabola equation and takes the
sqaure root to find y ...
7
2
7
So y 2 28 y 2 98 y ...
2
or
(2(7) 3.5)2 (3.5)2
y
(10.5)2 (3.5)2
or
7t 2 3.5 t
0.5 y 2(7) 0.5
or applies
2
y
"7" "7"
2("7") 2 2
2
M1
or solves
7t 2 3.5 and finds
y 2(7)"their t "
At least one correct exact
value of y. Can be unsimplified or simplified.
y ()7 2
A1
7
A, B have coordinates , 7 2 and
2
7
, 7 2
2
Area triangle ABS =
1
7
2(7 2 )
2
2
1
2 7
3.5
0
3.5
7 2 7 2
49
2
2
7
dependent on the previous M
mark
A full method for finding
the area of triangle ABS.
dM1
0
Correct exact answer.
A1
(4)
(5 marks)
214
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Question 2 continued
Notes:
(a)
You can give B1 for part (a) for correct relevant work seen in either part (a) or part (b).
(b)
1st M1: Allow a slip when candidates find the x coordinate of their midpoint as long as
0 their midpoint their a
Give 1st M0 if a candidate finds and uses y 98
1st A1: Allow any exact value of either 7 2 , 7 2 , 98 , 98 , 14 0.5 , awrt 9.9 or awrt 9.9
2nd dM1: Either
1
2 their "7 2 " their xmidpoint
2
7
or
Condone area triangle ABS = 7 2 , i.e.
2
2nd A1: Allow exact answers such as
or
1
2 their "7 2 " their "7" xmidpoint
2
their 7
49
49
2,
, 24.5 2 ,
2
2
7
1
98 but do not allow 3.5 2 98
2
2
"
their "7"
2"
2
4802
,
2
4802
1
, 3.5 2 ,49
4
2
seen by itself.
Give final A0 for finding 34.64823228… without reference to a correct exact value.
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Question
Scheme
3(a)
f ( x) x 2
3
1 ,
x
Marks
x0
At one of either x2 Ax or
f ( x
) 2 x 3x
2
3
Bx 2
x
M1
Correct differentiation
A1
where A and B are non-zero constants.
Either f (1.5)
0.75 or
13
0.75 , f (1.5)
f (1.5)
3
13
or awrt 4.33 or a
f (1.5)
3
correct numerical expression for
either f (1.5) or f (1.5)
B1
Can be implied by later working
1.5
f (1.5)
f (1.5)
1.5
0.75
4.333333...
87
1.67307692... or
1.67
52
dependent on the previous
M mark
Valid attempt at NewtonRaphson using their values
of f (1.5) and f (1.5)
dM1
dependent on all 4
previous marks
1.67 on their first iteration
(Ignore any subsequent
iterations)
A1
cso
cao
Correct differentiation followed by a correct answer scores full marks in
(a)
Correct answer with no working scores no marks in (a)
(5)
(b)
Way 1
f (1.675)
0.01458022...
f (1.665)
0.0295768...
Sign change (positive, negative) (and f ( x)
is continuous) therefore (a root)
1.67 (2 dp)
Chooses a suitable interval for x,
which is within 0.005 of their
answer to (a) and at least one
attempt to evaluate f ( x).
M1
Both values correct awrt (or
truncated)
1 sf, sign change and conclusion.
A1 cso
(2)
216
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Question
3(b)
continued
Scheme
Marks
Way 2
Alt 1: Applying Newton-Raphson again Eg. Using
1.67, 1.673 or
87
52
0.007507185629...
1.671700115...
4.415692926...
1.67
1.673
0.005743106396...
1.671700019...
4.41783855...
87
0.006082942257...
1.67170036...
4.417893838...
52
So 1.67 (2 dp)
Evidence of
applying
NewtonRaphson for a
second time on
their answer to
part (a)
M1
1.67
A1
(2)
(7 marks)
Notes:
(a)
B1:
Incorrect differentiation followed by their estimate of with no evidence of applying the
NR formula is final dM0A0.
B1 can be given for a correct numerical expression for either f (1.5) or f (1.5)
Eg. either (1.5)2
3
3
are fine for B1.
1 or 2(1.5)
(1.5)
(1.5)2
Final -This mark can be implied by applying at least one correct value of either f (1.5) or
f (1.5)
dM1: in 1.5
f (1.5)
f (1.5)
. So just 1.5
with an incorrect answer and no other evidence
f (1.5)
f (1.5)
scores final dM0A0.
Give final dM0 for applying 1.5
f (1.5)
without first quoting the correct N-R formula.
f (1.5)
(b)
A1:
Way 1: correct solution only
Candidate needs to state both of their values for f ( x) to awrt (or truncated) 1 sf along with
a reason and conclusion. Reference to change of sign or eg. f (1.675) f (1.665) 0
or a diagram or < 0 and > 0 or one positive, one negative are sufficient reasons. There must
be a (minimal, not incorrect) conclusion, eg. 1.67, root (or or part (a)) is correct, QED and a
square are all acceptable. Ignore the presence or absence of any reference to continuity.
A minimal acceptable reason and conclusion is “change of sign, hence root”.
No explicit reference to 2 decimal places is required.
Stating “root is in between – 1.675 and – 1.665” without some reference to is not
sufficient for A1
Accept 0.015 as a correct evaluation of f (1.675)
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Question 3 notes continued
(b)
A1:
Way 2: correct solution only
Their conclusion in Way 2 needs to convey that they understand that 1.67 to 2 decimal
places. Eg. “therefore my answer to part (a) [which must be 1.67 ] is correct” is fine for A1.
1.67
f (1.67)
1.67(2 dp) is sufficient for M1A1 in part (b).
f (1.67)
The root of f ( x) 0 is 1.67169988... , so candidates can also choose x1 which is less than
lying
1.67169988... and choose x2 which is greater than 1.67169988... with both x1 and x2
in the interval 1.675, 1.665 and evaluate f ( x1 ) and f ( x2 ).
Helpful Table
x
218
f ( x)
1.675
0.014580224
1.674
0.010161305
1.673
0.005743106
1.672
0.001325627
1.671
0.003091136
1.670
0.007507186
1.669
0.011922523
1.668
0.016337151
1.667
0.020751072
1.666
0.025164288
1.665
0.029576802
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
4(a)
Scheme
k
Marks
3
2
A=
where k is a constant and let g(k ) k 2k 3
1 k 2
Correct det(A) , un-simplified
or simplified
B1
(k 1) 1 3
Attempts to complete the
square
[usual rules apply]
M1
(k 1)2 2 0
(k 1)2 2 and 0
det(A) k (k 2) 3 or
k 2 2k 3
Way 1
2
A1 cso
(3)
Way 2
det(A) k (k 2) 3 or
b
2
k 2 2k 3
Correct det(A) , un-simplified
or simplified
Applies “ b2 4ac ” to their
4ac 22 4(1)(3)
det( A)
B1
M1
All of
b2 4ac 8 0
some reference to k 2 2k 3 being
above the x-axis
so det(A) 0
Complete solution
A1 cso
(3)
Way 3
g(k)
det(A
)
k (k 2) 3 or
g(k ) 2k 2 0 k 1
gmin (1)2 2(1) 3
gmin 2, so det(A) 0
k 2 2k 3
Correct det(A) , un-simplified
or simplified
B1
Finds the value of k for which
g(k ) 0 and substitutes this
value of k into g(k )
M1
g min 2 and states det(A) 0
A1 cso
(3)
(b)
-1
A =
k 2 3
1
k
k 2k 3 1
2
1
their det( A) k 2 3
1
k
M1
Correct answer in terms of k
A1
(2)
(5 marks)
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Question 4 continued
Notes:
(a)
B1:
Also allow k (k 2) 3
Way 2: Proving b2 4ac 8 0 by itself could mean that det(A) 0 or det(A) 0 .
To gain the final A1 mark for Way 2, candidates need to show b2 4ac 8 0 and make
some reference to k 2 2k 3 being above the x-axis (eg. states that coefficient of k 2 is
positive or evaluates det(A) for any value of k to give a positive result or sketches a
quadratic curve that is above the x-axis) before then stating that det(A) 0.
Attempting to solve det(A) 0 by applying the quadratic formula or finding 1 2i is
enough to score the M1 mark for Way 2. To gain A1 these candidates need to make some
reference to k 2 2k 3 being above the x-axis (eg. states that coefficient of k 2 is positive or
evaluates det(A) for any value of k to give a positive result or sketches a quadratic curve that
is above the x-axis) before then stating that det(A) 0.
(b)
A1:
220
k 2 3
1
Allow either
2
k
(k 1) 2 1
k2
2
or k 2k 3
1
2
k 2k 3
3
k 2k 3
or equivalent.
k
k 2 2k 3
2
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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Question
5
Scheme
Marks
3 4i
2 z z
7i
Way 1
Left hand side 2(a ib) (a ib)
2z z 2(a ib) (a ib)
..........
..........
(3 4i) (7 i)
(7 i) (7 i)
25 25i
50
1 1
2 2
1
1
1 1
a ,b
or z i
6
2
6 2
So, 3a ib i
Can be implied by eg. 3a ib
Note: This can be seen anywhere in
their solution
Multiplies numerator and denominator
of the right hand side by 7 i or
7 i
Applies i 2 1 to and collects like
terms to give
25 25i
right hand side =
or equivalent
50
dependent on the previous B and M
marks
Equates either real parts or imaginary
parts to give at least one of a ... or
B1
M1
A1
ddM1
b ...
Either a
1
1
1 1
and b
or z
i
6
2
6 2
A1
(5)
Way 2
2z z 2(a ib) (a ib)
(3a ib)(7 i)
..........
21a 3ai 7bi b
..........
Left hand side 2(a ib) (a ib)
Can be implied by eg. 3a ib
Multiplies their (3a ib) by (7 i)
Applies i 2 1 to give
left hand side = 21a 3ai 7bi b
gives 21a
b 3 , 3a 7
b 4
dependent on the previous B and M
marks
Equates both real parts and imaginary
parts to give at least one of a ... or
1
1
1 1
a ,b
or z i
6
2
6 2
Either a
So, (21a b) (3a 7b) 3 4i
B1
M1
A1
ddM1
b ...
1
1
1 1
and b
or z
i
6
2
6 2
A1
(5)
(5 marks)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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221
Question 5 continued
Notes:
Some candidates may let z x iy and z x iy .
So apply the mark scheme with x a and y b.
For the final A1 mark, you can accept exact equivalents for a, b.
222
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Question
6(a)
Scheme
Marks
5
H : xy 25 , P 5t , is a general point on H
t
25 25 5
5
Either 5t 25 or
y
x
t
5t
t
x
or
25 25
5t
5
y
t
or
states
B1
c5
(1)
(b)
25
dy
25
25 x 2
2
y
25 x 1
x
dx
x
xy 25 x
dy
y 0
dx
dy dy dt
5 1
. 2
dx dt dx
t 5
1
5
dy
10
4
At A , t , x , y
2
2
dx
dy
k x 2
dx
where k is a numerical value
Correct use of product rule. The
sum of two terms, one of which is
correct.
dy
1
dt their dx
dt
Correct numerical gradient at A,
which is found using calculus.
Can be implied by later working
Applies mN
So, mN
1
4
1
, to find a
mT
numerical mN , where mT is found
from using calculus.
Can be implied by later working
1
5
y 10
x
4
2
15
75
1
75
10
y
x
c c
4 2
8
4
8
leading to 8 y 2 x 75
0 (*)
M1
A1
M1
Correct line method for a
normal where a numerical
mN mT is found from
using calculus.
Can be implied by later
working
M1
Correct solution only
A1
(5)
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223
Question
Scheme
6(c)
y
25
25
8 2 x 75 0 or
x
x
or x 5t , y
Substitutes y
25
x
or x
Marks
25
25
8 y 2 75 0
y
y
x
5
5
8 5t 2 75 0
t
t
M1
5
25
or x 5t and y into the printed equation
y
t
or their normal equation to obtain an equation in either x only, y only or t only
0 or 2t 2 15t 8
0 or
2 x2 75x 200
0 or 8 y 2 75 y 50
10t 2 75t 40
0
(2 x 5)( x 40) 0 x ... or ( y 10)(8 y 5) 0 y ... or
(2t 1)(t 8) 0 t ...
dM1
dependent on the previous M mark
Correct attempt of solving a 3TQ to find either
x ...,
y ... or
t ...
Finds at least one of either x 40 or y
5
B 40,
8
5
8
A1
Both correct coordinates (If coordinates
are not stated they can be paired together
as
x ...,
y ... )
A1
(4)
(10 marks)
Notes:
(a)
A conclusion is not required on this occasion in part (a).
B1:
c
Condone reference to c 5 (as xy c 2 and ct , are referred in the Formula book.)
t
(b)
dy dy dt
5 1
1
5
.
2
2 mN t 2 y 10 t 2 x scores only the first M1.
dx dt dx
5
2
t
t
When t
1
1
5
is substituted giving y 10 x the response then automatically gets A1(implied)
4
2
2
M1(implied) M1
224
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Question 6 notes continued
(c)
You can imply the final three marks (dM1A1A1) for either
5
25
8 2 x 75 0 40,
8
x
25
5
8 y 2 75 0 40,
8
y
5
5
8 5t 2 75 0 40,
8
t
with no intermediate working.
You can also imply the middle dM1A1 marks for either
25
8 2 x 75
0 x
40
x
25
5
8 y 2 75
0 y
8
y
5
5
8 5t 2 75 0 x 40 or y
8
t
with no intermediate working.
5
5
5
Writing x
40, y
followed by B 40, or B , 40 is final A0.
8
8
8
5
5
Ignore stating B , 10 in addition to B 40,
8
2
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Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
225
Question
7(a)
Scheme
Rotation
Marks
Rotation
B1
1 5
Either arctan 125 , tan 1 125 , sin 1 12
13 , cos 13 ,
67 degrees
(anticlockwise)
awrt 67 degrees, awrt 1.2, truncated 1.1
(anticlockwise), awrt 293 degrees clockwise or awrt
5.1 clockwise
The mark is dependent on at least one of the
previous B marks being awarded.
About (0, 0) or about O or about the origin
about (0, 0)
Note: Give 2nd B0 for 67 degrees clockwise
o.e.
(b)
0 1
1 0
Q
B1 o.e.
dB1
(3)
Correct matrix
B1
(1)
(c)
{R = PQ}
5
13
12
13
12
1
13 0
;=
5
0
13 1
12
13
5
13
5
13
12
13
Multiplies P by their Q in the
correct order and finds at least
one element
M1
Correct matrix
A1
(2)
(d)
Way 1
12
13
5
13
5
13
12
13
x x
Forming the equation "their matrix R "
kx kx
x x
kx kx
Allow x being replaced by any non-zero
number eg. 1.
Can be implied by at least one correct ft equations
below.
12
5kx
5
12kx
x
x or
x
kx k
...
13
13
13
13
Uses their matrix equation to
form an equation in k and
progresses to give
M1
k numerical value
dependent on only the previous M mark
So k 5
M1
k 5
A1 cao
Dependent on all previous marks being scored in this part. Either
12
5kx
5
12kx
x
x and
x
kx to give k 5
13
13
13
13
Solves both
Finds k 5 and checks that it is true for the other component
12
13
Confirms that
5
13
x x
5
13
12
13
5x 5x
A1 cso
(4)
226
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
7(d)
continued
Scheme
Marks
Way 2
Either
12
5
5
, sin 2
cos 2
or tan 2
13
13
12
Correct follow through
equation in
2 based on their matrix R
M1
Full method of finding 2 ,
then and applying tan
M1
1
12
tan arccos or
13
2
1
12
k tan arccos
13
2
tan awrt 78.7
or
A1
tan awrt 1.37 . Can be
implied.
k 5 by a correct solution
So k 5
only
A1
(4)
(10 marks)
Notes:
(a)
Condone “Turn” for the 1st B1 mark.
Penalise the first B1 mark for candidates giving a combination of transformations.
(c)
1
for eg. "their matrix R "
Allow 1 M1
k
st
k
1
or "their matrix R "
k
k2
k
k2
or "their matrix R " 1 1 or equivalent
k
1
y
k
1
(tan
)x :
sin 2
cos 2
sin 2 cos 2
12
13
5
13
5
13
12
13
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
227
Question
8(a)
Scheme
Marks
f ( z) z 4 6 z 3 76 z 2 az b , a, b are real constants. z1 3 8i is given.
3 8i
3 8i
B1
(1)
(b)
Attempt to expand ( z ( 3 8i))( z ( 3 8i))
or any valid method to establish a quadratic
factor
eg z 3 8i z 3 8i z 2 6 z 9 64
or sum of roots 6, product of roots 73
z 2 6 z 73
M1
to give z 2 (sum) z product
z 2 6 z 73
A1
Attempts to find the other quadratic factor.
2
2
f ( z) ( z 6 z 73)( z 3)
eg. using long division to get as far as z 2 ...
2
M1
2
or eg. f ( z) ( z 6 z 73)( z ...)
A1
z2 3
z
2
3 0 z
3i
dependent on only the previous M mark
Correct method of solving the 2nd quadratic
factor
dM1
A1
3 i and 3 i
(6)
(c)
Criteria
3 8i plotted correctly in
quadrants 2 and 3 with some
evidence of symmetry
Their other two complex
roots (which are found from
solving their 2nd quadratic in
(b)) are plotted correctly with
some evidence of symmetry
about the x-axis
Satisfies at least one of the two
criteria
B1 ft
Satisfies both criteria with some
indication of scale or coordinates
stated. All points (arrows) must
be in the correct positions
relative to each other.
B1 ft
(2)
(9 marks)
228
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question 8 continued
Notes:
(b)
Give 3rd M1 for z 2 k 0, k 0 at least one of either z k i or z k i
Give 3rd M0 for z 2 k 0, k 0 z ki
Give 3rd M0 for z 2 k 0, k 0 z k or z k
Candidates do not need to find
a 18,
b 219
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
229
Question
Scheme
9(a)
2 x2 4 x 3
0 has roots ,
(i)
4
3
or 2,
2
2
2 2 2 ......
2
( 2)2 2 32 7
(ii)
3 3 3 ......
3
or 2 2 ......
2 3 32 2 17
Both
Marks
4
3
and
.
2
2
This may be seen or implied
anywhere in this question.
Use of a correct identity for 2 2
(May be implied by their work)
7 from correct working
B1
M1
A1 cso
Use of an appropriate and correct
identity for 3 3
(May be implied by their work)
M1
3
or 2 7 32 17
17 from correct working
A1 cso
(5)
(b)
Sum = 2 2
2 2
7 ( 2) 5
Product = ( 2 )( 2 )
( )2 3 3
3 2
2
17 32 654
x2 5x 65
0
4
4 x2 20 x 65
0
Uses at least one of their 2 2 or
in an attempt to find a
numerical value for the sum of
( 2 ) and ( 2 )
M1
Expands ( 2 )( 2 ) and uses at
least one of their or 3 3 in an
attempt to find a numerical value for
the product of ( 2 ) and ( 2 )
Applies x2 (sum) x product (Can
be implied)
(“ = 0” not required)
Any integer multiple of
4 x2 20 x 65
0,
M1
M1
A1
including the “= 0”
(4)
230
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
9(b)
continued
Scheme
Marks
Alternative: Finding 2 and 2 explicitly
4 40
4 40
and so
,
4
4
5 3 10
5 3 10
2
, 2
2
2
Eg. Let
Uses x 2 x 2
5 3 10 5 3 10
x
x
2
2
M1
with exact numerical values. (May
expand first)
Attempts to
expand using
exact
5 3 10
5 3 10 5 3 10 numerical
2 5 3 10
x
x
x
2
2
2
2
values for
2 and
M1
2
x2 5x 65
0
4
2
4 x 20 x 65
0
Collect terms to give a 3TQ.
(“ = 0” not required)
M1
Any integer multiple of
4 x2 20 x 65
0,
A1
including the “= 0”
(4)
(9 marks)
Notes:
(a)
3
2 2
4 2 32
7 is M1A0 cso
2
3
4 40 4 40
Finding
but then
2,
by writing down or applying
,
2
4
4
1st A1:
2,
writing 2 2 2 4 3 7 and 3 3 3 8 9 17
scores B0M1A0M1A0 in part (a).
2
Applying
3
4 40 4 40
explicitly in part (a) will score B0M0A0M0A0
,
4
4
2
2
4 40
4 40
7
Eg: Give no credit for
4
4
3
3
4 40
4 40
17
or for
4
4
(b)
4 40 4 40
explicitly in part (b).
,
4
4
A correct method leading to a candidate stating a
4, b 20, c
65 without writing a
Candidates are allowed to apply
0 is final M1A0
final answer of 4 x2 20 x 65
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
231
Question
Scheme
Marks
10
u1 5, un1 3un 2, n 1 . Required to prove the result,
un
2 (3)n 1 , n
(i)
u
2(3) 1
5 or u1 6 1 5
1
n 1: u
2(3)
1 5
1
B1
(Assume the result is true for n k )
u
3 2(3)k 1 2
k 1
2(3)k1 1
uk 2(3)k 1 into uk
Substitutes
3uk 2
1
M1
dependent on the previous M mark
Expresses uk 1 in term of 3k 1
dM1
uk 1 2(3)k 1 1 by correct solution only
If the result is true for n k , then it is true for n k 1. As the result has been
shown to be true for n 1, then the result is true for all n
A1
A1 cso
(5)
Required to prove the result
n
4r
3
r
r 1
(ii)
(3 2n)
3
, n
3n
Shows or states both LHS
4
5 4
n 1: LHS , RHS 3
3
3 3
RHS
4
3
or states LHS
RHS
4
and
3
B1
4
3
(Assume the result is true for n k )
k 1
r 1
4r
(3 2k ) 4(k 1)
3
k 1
r
3
3k
3
3(3 2k ) 4(k 1)
3
k 1
3k 1
3
Adds the (k 1)th term to the sum of k
terms
dependent on the previous M mark
Makes 3k1 or (3)3k a common
denominator for their fractions.
Correct expression with common
denominator 3k1 or (3)3k for their
fractions.
M1
dM1
A1
3(3 2k ) 4(k 1)
3
3k 1
5 2k
3 k 1
3
(3 2(k 1))
(3 2(k 1))
3
by correct solution only
k 1
3
3k 1
If the result is true for n k , then it is true for n k 1. As the result has been
3
shown to be true for n 1, then the result is true for all n
A1
A1 cso
(6)
(11 marks)
232
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question 10 continued
Notes:
(i) & (ii)
Final A1 for parts (i) and (ii) is dependent on all previous marks being scored in that part.
It is gained by candidates conveying the ideas of all four underlined points either at the end of their
solution or as a narrative in their solution.
(i)
u1 5 by itself is not sufficient for the 1st B1 mark in part (i).
2(3) 1
5 or u1 6 1 5 is B0
u1 3 2 without stating u
1
(ii)
LHS RHS by itself is not sufficient for the 1st B1 mark in part (ii).
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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233
234
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Please check the examination details below before entering your candidate information
Candidate surname
Pearson Edexcel
Other names
Centre Number
Candidate Number
International
Advanced Level
Sample Assessment Materials for first teaching September 2018
(Time: 1 hour 30 minutes)
Paper Reference WFM02/01
Mathematics
International Advanced Subsidiary/Advanced Level
Further Pure Mathematics FP2
You must have:
Mathematical Formulae and Statistical Tables, calculator
Total Marks
Candidates may use any calculator permitted by Pearson regulations.
Calculators must not have the facility for symbolic algebra manipulation,
differentiation and integration, or have retrievable mathematical
formulae stored in them.
Instructions
black ink or ball-point pen.
•• Use
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
in the boxes at the top of this page with your name, centre number and
• Fill
candidate number.
Answer all questions and ensure that your answers to parts of questions are
• clearly
labelled.
the questions in the spaces provided
• Answer
– there may be more space than you need.
should show sufficient working to make your methods clear.
• You
Answers without working may not gain full credit.
answers should be given to three significant figures unless
• Inexact
otherwise stated.
Information
booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
•• AThere
are 8 questions in this question paper. The total mark for this paper is 75.
The
marks
each question are shown in brackets
• – use this asfora guide
as to how much time to spend on each question.
Advice
each question carefully before you start to answer it.
•• Read
Try to answer every question.
your answers if you have time at the end.
•• Check
If you change your mind about an answer, cross it out and put your new answer
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S59763A
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Answer ALL questions. Write your answers in the spaces provided.
Using algebra, find the set of values of x for which
x
2
<
x+2 x+5
(7)
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Q1
(Total for Question 1 is 7 marks)
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2.
(a) Express
1
in partial fractions.
(1)
(b) Hence show that
n
∑ ( r + 6 ) ( r + 8)
2
=
r =1
n ( an + b )
56 ( n + 7 ) ( n + 8)
where a and b are integers to be found.
(4)
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( r + 6 ) ( r + 8)
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Q2
(Total for Question 2 is 5 marks)
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(a) Show that the substitution z = y–2 transforms the differential equation
(I)
into the differential equation
2
dz
− 4 xz = −2 xe − x
dx
(II)
(b) Solve differential equation (II) to find z as a function of x.
(4)
(5)
(c) Hence find the general solution of differential equation (I), giving your answer in the
form y2 = f(x).
(1)
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2
dy
+ 2 xy = xe − x y 3
dx
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Q3
(Total for Question 3 is 10 marks)
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A transformation T from the z-plane to the w-plane is given by
z −1
, z ≠ −1
z +1
The line in the z-plane with equation y = 2x is mapped by T onto the curve C in the
w-plane.
(a) Show that C is a circle and find its centre and radius.
(7)
The region y < 2x in the z-plane is mapped by T onto the region R in the w-plane.
(b) Sketch circle C on an Argand diagram and shade and label region R.
(2)
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w=
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Q4
(Total for Question 4 is 9 marks)
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Given that y = cot x,
d2 y
= 2 cot x + 2 cot 3 x
d x2
(3)
(b) Hence show that
d3 y
= p cot 4 x + q cot 2 x + r
3
dx
where p, q and r are integers to be found.
(3)
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(a) show that
π
(c) Find the Taylor series expansion of cot x in ascending powers of x − up to and
3
3
π
including the term in x − .
3
(3)
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Q5
(Total for Question 5 is 9 marks)
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(a) Find the general solution of the differential equation
Given that y = 0 and
(I)
(8)
dy
= 1 when x = 0
dx
(b) find the particular solution of differential equation (I).
(5)
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d2 y
dy
−2
− 3 y = 2 sin x
2
dx
dx
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Q6
(Total for Question 6 is 13 marks)
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7.
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R
O
Initial line
Figure 1
Figure 1 shows the two curves given by the polar equations
r = 3 sinθ,
0θπ
r = 1 + cosθ,
0θπ
3 π
(a) Verify that the curves intersect at the point P with polar coordinates , .
2 3
The region R, bounded by the two curves, is shown shaded in Figure 1.
(
)
(b) Use calculus to find the exact area of R, giving your answer in the form a π − 3 ,
where a is a constant to be found.
(6)
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(2)
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Q7
(Total for Question 7 is 8 marks)
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(a) Show that
3
3
where k is a constant to be found.
(3)
Given that z = cosθ + isin θ, where θ is real,
(b) show that
1
= 2cosnθ
zn
1
n
(ii) z − n = 2isin nθ
z
n
(i) z +
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1
1
1
1
2
6
z + z − = z − 6 − k z − 2
z
z
z
z
(3)
(c) Hence show that
1
(3sin 2θ – sin 6θ)
32
(4)
(d) Find the exact value of
π
8
0
∫
cos3 θ sin3 θ dθ
(4)
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cos3 θ sin3 θ =
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Q8
(Total for Question 8 is 14 marks)
TOTAL FOR PAPER IS 75 MARKS
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Further Pure Mathematics FP2 Mark scheme
Question
Scheme
1
Marks
x
2
x2 x5
Seen anywhere in solution
Both correct B1B1; one correct
B1B0
Critical Values 2 and 5
B1
B1
x
2
0
x2 x5
x 2 3x 4
0
x 2 x 5
x 4 x 1 0
x 2 x 5
Attempt single fraction and factorise
numerator or use quad formula
M1
Critical values 4 and 1
Correct critical values May be seen
on a graph or number line.
A1
dM1: Attempt an interval inequality using one
of 2 or 5 with another cv
5 x 4, 2 x 1
5, 4 2,1
A1, A1: Correct intervals
Can be in set notation
One correct scores A1A0
Award on basis of the inequalities seen ignore any and/or between them
Set notation answers do not need the union
sign.
dM1
A1
A1
(7)
Alternative
Critical Values 2 and 5
Seen anywhere in solution
B1, B1
x
2
2
2
x x 5 x 2 2 x 2 x 5
x2 x5
x 5 x 2 x x 5 2 x 2 0
Multiply by x 5 x 2
x 5 x 2 x 1 x 4 0
and attempt to factorise a
quartic or use quad formula
Critical values 4 and 1
Correct critical values
dM1: Attempt an interval
inequality using one of 2 or
5 x 4, 2 x 1
5 with another cv
A1, A1: Correct intervals
5, 4 2,1
Can be in set notation
One correct scores A1A0
Any solutions with no algebra (eg sketch graph followed by critical values
with no working) scores max B1B1
2
2
M1
A1
dM
A1
A1
(7 marks)
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Question
Scheme
2(a)
1
Marks
r 6 r 8
1
1
oe
2 r 6 2 r 8
Correct partial fractions, any
equivalent form
B1
(1)
(b)
1 1 1 1 1 1 1
1
1
1
1
2 ...
2 7 9 8 10 9 11
n 5 n 7 n 6 n 8
Expands at least 3 terms at start and 2 at end (may be implied)
The partial fractions obtained in (a) can be used without multiplying by 2.
1 1 1 1
Fractions may be etc These comments apply to both M1 and
2 7 2 9
A1
1 1
1
1
7 8 n 7 n 8
15 n 7 n 8 56 2n 15
56 n 7 n 8
n 15n 113
56 n 7 n 8
M1
Identifies the terms that do not
cancel
A1
Attempt common denominator
Must have multiplied the
fractions from (a) by 2 now
M1
A1 cso
(4)
(5 marks)
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Question
Scheme
3(a)
2
dy
xe x y 3
2 xy
dx
z y 2 y z
12
M1:
dy
3 dz
12 z 2
dx
dx
12 z
dy
3 dz
kz 2
dx
dx
M1 A1
A1: Correct differentiation
2 3
dz 2 x
xe x z 2
1
dx z 2
32
Marks
2
dz
4 xz
2 xe x *
dx
Substitutes for dy/dx
Correct completion to printed
answer with no errors seen
M1
A1 cso
(4)
Alternative 1
M1:
dz
2 y 3 oe
dy
12 y 3
dz
ky 3
dy
M1 A1
A1: Correct differentiation
2
dz
xe x y 3
2 xy
dx
2
dz
4 xz
2 xe x *
dx
Substitutes for dy/dx
M1
Correct completion to printed
answer with no errors seen
A1
Alternative 2
dz
dy
ky 3
inc chain rule
dx
dx
A1: Correct differentiation
M1:
dz
dy
2 y 3
dx
dx
12 y 3
2
dz
xe x y 3
2 xy
dx
2
dz
4 xz
2 xe x *
dx
(b)
I e
e
4 x dx
2
ze2 x
2 xe
Substitutes for dy/dx
M1
Correct completion to printed
answer with no errors seen
A1
M1: I e
2 x2
A1: e2 x
3 x2
1 3 x2
e c
3
2
1 2
z ce2 x e x
3
dx
M1 A1
4 x dx
M1 A1
2
z I 2 xe x I dx
2
dx
xe
qx2
peqx c
2
Or equivalent
dM1
M1
A1
(5)
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Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
269
Question
3(c)
Scheme
2
1
1 2
1
ce2 x e x y 2
2
2
1 2
y
3
ce2 x e x
3
Marks
1
y
(b)
2
3e x
3 x2
1 ke
2
B1ft
(1)
(10 marks)
270
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
Scheme
4(a)
w
z 1
z 1
z 1
wz w z 1 z ...
z 1
w
Marks
Attempt to make z the subject
M1
w 1
1 w
Correct expression in terms of
w
A1
u iv 1 1 u iv
1 u iv 1 u iv
Introduces “u + iv” and
multiplies top and bottom by
the complex conjugate of the
bottom
M1
y
2 x 2v
2u 2 2v2 2
Uses real and imaginary parts
and
y = 2x to obtain an equation
connecting “u” and “v” Can
have the 2 on the wrong side.
M1
u 2 v 12 14
1
Processes their equation to a
form that is recognisable as a
circle
ie coefficients of u2 and v 2 are
the same and no uv terms
M1
z
x
u 2 v 2 1
2v
, y
....
....
2
Centre (0, 12 ), radius
5
2
A1: Correct centre (allow -½i)
A1: Correct radius
A1,A1
(7)
Special Case:
w
x
x iy 1
x iy 1
2
x 1 2 xi x 1 2 xi
x 1 2 xi x 1 2 xi
1 4 x 2 2 xi x 1 x 1
x 1
2
4 x2
(b)
M1: rationalise the
denominator, may have 2x or y
A1: Correct result in terms of x
only. Must have rational
denominator shown, but no
other simplification needed
B1ft: Their circle correctly
positioned provided their
equation does give a circle
B1ft B1
R
B1: Completely correct sketch
and shading
(2)
(9 marks)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
271
Question
Scheme
5(a)
Marks
y cot x
dy
cosec2 x
dx
M1: Differentiates using the chain
rule or product/quotient rule
A1: Correct derivative
A1: Correct completion to printed
answer
1 cot 2 x
cosec2 x or
cos2 x sin 2 x
1
must be used
Full working must be shown
d2 y
2cosec x cosec x cot x
dx 2
2cosec2 x
cot x 2cot x 2cot 3 x *
M1A1
A1cso*
(3)
Alternative
cos x
dy sin 2 x cos 2 x
1
y
2
2
sin x
dx
sin x
sin x
2
d y
2sin 3 x cos x ...
2
dx
Correct completion to printed answer see above
M1A1
A1
(3)
(b)
d3 y
2cosec2 x 6cot 2 x cosec2 x
3
dx
2 1 cot 2 x 6cot 2 x 1 cot 2 x
6cot 4 x 8cot 2 x 2
Correct third derivative
B1
cosec2 x
Uses 1 cot 2 x
M1
cso
A1
(3)
(c)
4
8
16
1
,f
f
,f
,f
3 3 3 3
3
3 3
3
3
M1: Attempts all 4 values at
No working need be shown
3
2
M1
3
1 4
4 8
y x
x x
3 3 3
3 9
3
3 3
M1: Correct application of Taylor using their values. Must be up to and
3
including x
3
A1: Correct expression Must start y = .... or cot x
f(x) allowed provided defined here or above as f x cot x or y
M1A1
Decimal equivalents allowed (min 3 sf apart from 0.77), 0.578, 1.33, 0.770,
(0.7698.., so accept 0.77) 0.889
(3)
(9 marks)
272
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
Scheme
6(a)
d2 y
dy
2 3y
2sin x
2
dx
dx
Marks
AE: m2 2m 3
0
m2 2m 3 0 m ... 1,3
y
Ae3x Be x
PI:
y
p sin x q cos x
Forms Auxiliary Equation and
attempts to solve (usual rules)
M1
Cao
A1
Correct form for PI
B1
y
p cos x q sin x
y
p sin x q cos x
p sin x q cos x 2 p cos x q sin x 3 p sin x 3q cos x
2sin x
M1
Differentiates twice and substitutes
2q 4 p 2, 4q 2 p 0
Correct equations
2
1
p
, q
5
5
A1A1 both correct
A1A0 one correct
y
A1
A1 A1
1
2
cos x sin x
5
5
1
2
y Ae3 x Be x cos x sin x
5
5
Follow through their p and q and
their CF
B1ft
(8)
(b)
1
2
y 3 Ae3 x Be x sin x cos x
5
5
1
2
0 A B , 1 3A B
5
5
Differentiates their GS
M1
M1: Uses the given conditions to
give two equations in A and B
M1 A1
A1: Correct equations
A
3
1
, B
10
2
Solves for A and B Both correct
y
3 3x 1 x 1
2
e e cos x sin x
10
2
5
5
Sub their values of A and B in
their GS
A1
A1ft
(5)
(13 marks)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
273
Question
Scheme
7(a)
Marks
3
3 sin
3 2
Attempt to verify
coordinates in at least one of
the polar equations
M1
3
r 1 cos
Coordinates verified in both
curves
(Coordinate brackets not
needed)
A1
3
r
3
3
2
(2)
Alternative
Equate rs:
3 sin 1 cos and verify (by substitution) that
solution or solve by using t tan
2
3
is a
or writing
1
sin
6 2
3
1
1
sin cos
2
2
2
M1
3
Squaring the original equation allowed as is known to be between 0 and
Use
in either equation to obtain r
3
3
2
A1
(2)
(b)
1
2
( 3 sin ) d ,
1
(1 cos ) 2 d
2
1
2
2
1
3sin 2 d ,
2
1
2
Correct formula used on at
least one curve (1/2 may
appear later)
Integrals may be separate or
added or subtracted.
M1
(1 2cos cos 2 )d
3
1
1
1 cos2 d , (1 2cos 1 cos2 )d
2
2
2
1 1
Attempt to use sin 2 or cos2 cos 2 on either integral
2 2
M1
Not dependent 1/2 may be missing
3
1
3
sin 2 ,
4
2
0
1 3
1
2sin sin 2
2 2
4
3
Correct integration (ignore limits) A1A1 or A1A0
Correct use of limits for both
1 3
3
3
3 integrals
R
0 3 Integrals must be added.
43 4
8
2 2 2
Dep on both previous M
marks
3
Cao
3
No equivalents allowed
4
A1, A1
ddM1
A1
(6)
(8 marks)
274
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
8(a)
Scheme
3
3
1
1 2 1
z z z 2
z
z
z
z 6 3z 2
z6
Marks
3
M1: Attempt to expand
3
z 6
2
z
A1: Correct expansion
1
1
3 z 2 2
6
z
z
Correct answer with no errors
seen
M1A1
A1
(3)
Alternative
3
3
1
3 1
1
3 1
3
3
z z 3z 3 , z z 3z 3
z
z z
z
z z
M1A1
M1: Attempt to expand both cubic brackets A1: Correct expansions
z6
1
1
3 z 2 2
6
z
z
Correct answer with no
errors
A1
(3)
(b)(i)(ii)
z n cos n i sin n
z n
cos n i sin n
cos n sin n
but must be different from their z
n
1
1
z n n 2cos n *,
z n n 2i sin n *
z
z
Correct application of de
Moivre
B1
Attempt z-n
M1
z n cos n i sin n
must be seen
A1*
(3)
(c)
3
3
1
1
3
3
2 cos 2i sin
z z
z
z
z6
1
1
3 z2
2i sin 6 6i sin 2
6
z
z2
64i sin3 cos3
2i sin 6 6i sin 2
3
cos
sin 3
B1
Follow through their k in
place of 3
B1ft
Equating right hand sides and
M1
simplifying 23 2i (B mark
needed for each side to gain
M mark)
3
1
3sin 2 sin 6 *
32
A1cso
(4)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
275
Question
Scheme
8(d)
8
0
3
cos
sin 3 d
0
8
Marks
1
3sin 2 sin 6 d
32
M1:
p cos 2 q cos 6
1 3
1
8
cos 2 cos 6
32 2
6
0
1
3
1 3 1 1 4 5 2
32 2 2 6 2 2 6 32 3
6
A1: Correct
integration
Differentiation
scores M0A0
M1 A1
dM1: Correct use
of limits – lower
limit to have nonzero result.
Dep on previous M dM1 A1
mark
A1: Cao (oe) but
must be exact
(4)
(14 marks)
276
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Please check the examination details below before entering your candidate information
Candidate surname
Pearson Edexcel
Other names
Centre Number
Candidate Number
International
Advanced Level
Sample Assessment Materials for first teaching September 2018
(Time: 1 hour 30 minutes)
Paper Reference WFM03/01
Mathematics
International Advanced Subsidiary/Advanced Level
Further Pure Mathematics FP3
You must have:
Mathematical Formulae and Statistical Tables, calculator
Total Marks
Candidates may use any calculator permitted by Pearson regulations.
Calculators must not have the facility for symbolic algebra manipulation,
differentiation and integration, or have retrievable mathematical
formulae stored in them.
Instructions
black ink or ball-point pen.
•• Use
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.
the questions in the spaces provided
• Answer
– there may be more space than you need.
should show sufficient working to make your methods clear. Answers
• You
without working may not gain full credit.
answers should be given to three significant figures unless otherwise
• Inexact
stated.
Information
booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
•• AThere
are 8 questions in this question paper. The total mark for this paper is 75.
The
marks
each question are shown in brackets
• – use this asfora guide
as to how much time to spend on each question.
Advice
each question carefully before you start to answer it.
•• Read
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your answers if you have time at the end.
•• Check
If you change your mind about an answer, cross it out and put your new answer
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Answer ALL questions. Write your answers in the spaces provided.
The curve C has equation
y = 9 cosh x + 3 sinh x + 7x
Use differentiation to find the exact x coordinate of the stationary point of C, giving your
answer as a natural logarithm.
(6)
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Q1
(Total for Question 1 is 6 marks)
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An ellipse has equation
The point P lies on the ellipse and has coordinates (5cos θ, 2sin θ), 0 < θ <
π
2
The line L is a normal to the ellipse at the point P.
(a) Show that an equation for L is
5x sin θ – 2y cos θ = 21sin θ cos θ
(5)
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x2
y2
+
=1
25
4
Given that the line L crosses the y-axis at the point Q and that M is the midpoint of PQ,
(b) find the exact area of triangle OPM, where O is the origin, giving your answer as a
multiple of sin 2θ
(6)
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Q2
(Total for Question 2 is 11 marks)
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Without using a calculator, find
(b)
∫
∫
4
1
dx, giving your answer as a multiple of π,
−2 x + 4 x + 13
2
−1
1
4 x 2 − 12 x + 34
(5)
(
)
dx, giving your answer in the form p ln q + r 2 ,
where p, q and r are rational numbers to be found.
(7)
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(a)
1
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Q3
(Total for Question 3 is 12 marks)
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(a) Find M–1 in terms of k.
(5)
Hence, given that k = 0
(b) find the matrix N such that
3 5 6
MN = 4 −1 1
2 −3
3
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1 k 0
M = −1 1 1 , where k is a constant
1 k 3
(4)
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Q4
(Total for Question 4 is 9 marks)
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5.
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Given that y = artanh(cos x)
dy
= −cosec x
dx
(2)
(b) Hence find the exact value of
∫
π
6
cos x artanh(cos x) dx
0
(
)
giving your answer in the form a ln b + c 3 + dπ, where a, b, c and d are rational
numbers to be found.
(5)
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(a) show that
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Q5
(Total for Question 5 is 7 marks)
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6.
The coordinates of the points A, B and C relative to a fixed origin O are (1, 2, 3),
(–1, 3, 4) and (2, 1, 6) respectively. The plane Π contains the points A, B and C.
(5)
The point D has coordinates (k, 4, 14) where k is a positive constant.
Given that the volume of the tetrahedron ABCD is 6 cubic units,
(b) find the value of k.
(4)
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(a) Find a cartesian equation of the plane Π.
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Q6
(Total for Question 6 is 9 marks)
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7.
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The curve C has parametric equations
y = 4t 3 ,
0 t 1
The curve C is rotated through 2π radians about the x-axis. The area of the curved surface
generated is S.
(a) Show that
∫
1
1
S = k π t 5 (t 2 + 1) 2 d t
0
where k is a constant to be found.
(4)
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x = 3t 4 ,
(b) Use the substitution u2 = t2 + 1 to find the value of S, giving your answer in the form
pπ 11 2 − 4 where p is a rational number to be found.
(7)
(
)
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Q7
(Total for Question 7 is 11 marks)
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8.
0
tanh 2 n x dx,
n0
(a) Show that, for n 1
I n = I n −1
1 3
−
2n − 1 5
2 n −1
(5)
(b) Hence show that
∫
ln 2
0
tanh 4 x dx = p + ln 2
where p is a rational number to be found.
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∫
In =
ln 2
(5)
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Q8
(Total for Question 8 is 10 marks)
TOTAL FOR PAPER IS 75 MARKS
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Further Pure Mathematics FP3 Mark scheme
Question
Scheme
Marks
1
y 9cosh x 3sinh x 7 x
dy
9sinh x 3cosh x 7
dx
Correct derivative
B1
e
9
Replaces sinhx and coshx by
the correct exponential
forms
M1
x
e x
2
3 e + e 7 0
x
x
2
Note that the first 2 marks can score the other way round:
e + e 3 e
y 9
x
M1:
x
2
x
e e
dy
B1:
9
dx
2
2x
x
x
e x
2
7x
3 e + e 7
x
x
2
M1: Obtains a quadratic in
ex
x
12e 14e 6
0 oe
3e
x
M1 A1
A1: Correct quadratic
Solves their quadratic as far
as ex = ...
1 2e x 3 0 e x ...
cso (Allow –ln3) e x
1
x ln
3
need not be seen. Extra
answers, award A0
3
2
M1
A1
Alternative
dy
9sinh x 3cosh x 7
dx
Correct derivative
B1
9sinh x 3cosh x 7 81sinh 2 x 9cosh 2 x 42cosh x 49
Squares and attempts
quadratic in coshx
72cosh 2 x 42cosh x 130
0
5
3
3cosh x 512cosh x 13 0 cosh x
M1: Solves quadratic
A1: Correct value
M1
M1 A1
2
5
5
x ln 1
3
3
Use of ln form of arcosh
M1
1
x ln
3
cso (Allow – ln3)
A1
NB: Ignore any attempts to find the y coordinate
(6 marks)
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Question
2(a)
Scheme
Marks
x2 y 2
1, P(5cos , 2sin )
25 4
dx
dy
5sin ,
2 cos
d
d
or
Correct derivatives or correct
implicit differentiation
B1
Divides their derivatives
correctly or substitutes and
rearranges
M1
Correct perpendicular gradient
rule
M1
5sin
y 2sin
x 5cos
2cos
Correct straight line method (any
complete method) Must use
their gradient of the normal.
M1
5x sin 2 y cos
21sin cos *
cso
A1*
2 x 2 y dy
0
25 4 dx
dy 2cos
dx 5sin
MN
5sin
2cos
(5)
(b)
At Q, x = 0 y
21
sin
2
0 5cos 2sin 212 sin
,
M is
2
2
5
17
cos , sin
4
2
21
cos
5
Area OPM = OLP
+OLM
1 21
1 21
17
. cos .2sin . cos . sin
2 5
2 5
4
105
sin 2
16
B1
Correct mid-point method for at
least one coordinate
Can be implied by a correct x
coordinate
L cuts x-axis at
M1
B1
M1: Correct triangle area
method using their coordinates
M1 A1
A1: Correct expression
Or 6.5625sin 2 must be
positive
A1
(6)
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Question
2(b)
continued
Scheme
Marks
Alternative 1: Using Area OPM
See above for B1M1
B1 M1
1 0 5cos 52 cos
OPM
Area
2 0 2sin 174 sin
0
0
1
85
0 5sin cos 0 0 sin cos 0
2
4
105
sin cos
4
105
sin 2
16
M1: Correct determinant
with their coords, with 2 or
0
3 points. should be at
0
both or neither end.
A1: Correct determinant
(There are more
complicated determinants
using the 3 points.)
A1
M1 A1
A1
A1
(6)
Alternative 2: Using Area OPQ
At Q, x = 0 y
21
sin
2
1 5cos
Area OPQ
2 2sin
1 105
sin cos
2 2
0
212 sin
Can be implied by the
following line
OQ is base, x coord of P is
height
M1 A1
A1
105
sin 2
8
Area OPM =
B1
1
Area OPQ
2
105
sin 2
16
M1
A1
(6)
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Question
2(b)
continued
Scheme
Marks
Alternative 3
At Q, x = 0 y
21
sin
2
0 5cos 2sin 212 sin
M is
,
2
2
B1
5
17
cos , sin
4
2
OP
4sin 2 25cos 2
4 21cos 2
5
2sin 17
cos
sin
5cos 4
d 2
4sin 2
1
25cos 2
105
sin 2
16
B1
21
sin
4
4 21cos 2
25cos 2
21
sin
1
4
4 21cos 2
Area
2
2
4 21cos
25cos 2
M1
M1 A1
A1
(6)
(11 marks)
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Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
3(a)
Scheme
Marks
x 2 4 x 13 x 2 9
2
1
x2
dx arctan
3
3
x 2 9
1
2
1
1
1
x 2
arctan1 arctan 0
3 arctan
3 2 3
12
B1
M1: karctan f x .
A1: Correct expression
M1 A1
Correct use of limits
arctan0 need not be shown
M1
cao
A1
(5)
Alternative
Sub x 2
3tan t
x 2 4 x 13 x 2 9
2
B1
dx
3sec2 t
dt
x
2, tan t
0, t
0; x
1, tan t
1, t
4
M1 sub and integrate inc use of
2
3sec t
1
1
tan 2 1
sec2
d
t
d
t
t
A1 Correct expression Ignore
9 tan 2 t 9
3
3
limits
Either change limits and
substitute
4
.
12
Or reverse substitution and
0
substitute original imits
12
cao
M1 A1
M1
A1
(5)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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313
Question
3(b)
Scheme
M1: 4 x p q, p, q 0
2
2
3
4 x 12 x 34 4 x 25
2
2
2
3
A1: 4 x 25
2
or 2x 3 25
2
Marks
x 3
1
1
1
1
dx
dx
arsinh 5 2
2
2
2
2
2
4 x 32 25
x 32 254
M1 A1
M1 A1
M1: karsinh f x . A1: Correct expression
4
1
x 32
1
ar sinh 1 ar sinh 1
arsinh 5
2 1 2
2
1
ln 1 2 ln 1 2
2
1
ln 3 2 2 or ln 1 2
2
Correct use of limits
M1
Uses the logarithmic
form of arsinh
M1
cao
A1
(7)
Alternative: Second M1 A1
Sub
2 x 3 u or
2x 3 5sinh u
arsinh1
arsinh 1
5
1
u
5cosh udu arsinh
5 5
2
25sinh 2u 25
1
5
1
u
5 2 u 2 25 du 2 arsinh 5
5
5
1
M1 A1
(12 marks)
314
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
4(a)
Scheme
Marks
1 k 0
M 1 1 1
1 k 3
Correct determinant in any
form
M 3 k k 3 1 3k 3
1 1 1
3 k
M k 1 k or minors 3k
0 1 3
k
3 k 4 k 1
or cofactors 3k 3
0
k
1 1 k
T
4 k 1
3
0
1 1 k
B1
B1
M1: Identifiable full attempt
at inverse including
reciprocal of determinant.
Could be indicated by at
least 6 correct elements.
k
3 k 3k
1
M
4
3
1
3 3k
k 1 0 1 k
A1ft: Two rows or two
columns correct (follow
through their determinant
but not incorrect entries in
the matrices used)
1
M1
A1ft
A1ft
A1ft: Fully correct inverse
(follow through as before)
NB: If every element is the negative of the correct element, allow M1A1A0
(5)
(b)
3 5 6
3 5 6
1
MN 4 1 1 N M 4 1 1
3 2 3
3 2 3
3 0 0 3 5 6
1
N
4 3 1 4 1 1
3
3 2 3
1
0
1
3 5 6
7 5 10
0 1 3
Correct statement
M1: Multiplies the given
matrix by their M 1 in the
correct order Must include
1
the " "
3
A2: Correct matrix
( 1 each error). If left with
1
outside the matrix
3
award A0
B1
M1
A(2, 1,
0)
(4)
(9 marks)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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315
Question
Scheme
5(a)
y artanh(cos x)
dy
1
sin x
dx 1 cos 2 x
sin x
1
cosec x
2
sin x sin x
*
Marks
Correct use of the chain rule
M1
A1: Correct completion with no
errors
A1
(2)
Alternative 1
tanh y
cos x sech 2 y
dy
sin x
dx
dy sin x
sin x
2
dx sech y
1 cos2 x
sin x
1
cosec x
2
sin x sin x
*
Correct differentiation to obtain a
function of x
M1
A1: Correct completion with no
errors
A1
(2)
Alternative 2
artanh(cos x)
1
2
1 cos x
1 cos x
ln
dy 1 1 cos x sin x 1 cos x sin x 1 cos x
2
dx 2 1 cos x
1 cos x
2sin x
cosec x
2 1 cos 2 x
*
Correct differentiation
to obtain a function of
x
M1
A1: Correct
completion with no
errors
A1
(2)
(b)
artanh cos x dx sin x artanh cos x sin x cosecx dx
cos x
M1 A1
M1: Parts in the correct direction A1: Correct expression
3
1
x x 06
artanh
sin x artanh cos
0
2
2 6
M1: Correct use of limits on either part (provided both parts are integrated).
Lower limit need not be shown
1 1
ln
4 1
3
2
3
2
6
1
1
ln 7 4 3 or ln 2 3
4
6
2
6
The last 2 M marks may be gained in
reverse order.
M1
Use of the logarithmic form of
artanh
M1
Cao (oe)
A1
(5)
(7 marks)
316
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
6(a)
Scheme
Marks
2
1
3
AB
1 , AC
1 , BC
2
1
3
2
Two correct vectors in Π
Can be negatives of those shown
i
2
M1: Attempt cross product of two
vectors lying in Π (At least one
no. to be correct.)
j k 4
1 1
7
1 1 3 1
B1
M1 A1
A1: Correct normal vector
4 1
7 2 4 14 3
1 3
Attempt scalar product with their
normal and a point in the plane
4x 7 y z
21
Cao (oe)
dM1
A1
(5)
Alternative 1
a 2b 3c
d
a 3b 4c d
Correct equations
B1
2a b 6c
d
4
1
1
a
d, b
d, c
d
21
3
21
M1: Solve for a, b and c in terms
of d
M1 A1
A1: Correct equations
d 21 a ..., b ..., c ...
Obtains values for a, b, c and d
M1
4x 7 y z
21
Cao (oe)
A1
(5)
Alternative 2: Using r a sb tc where b and c are vectors in
Two correct vectors in the plane
x
Eg r y
z
See main scheme
1 2 1
2 s 1 t 1
3 1 3
x 1 2s t
y 2 s t
B1
M1
Deduce 3 correct equations
A1
z 3 s 3t
4x 7 y z
21
M1: Eliminate s, t
A1: Cao
M1 A1
(5)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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317
Question
6(b)
Scheme
AD AB AC
Marks
Attempt suitable triple product
M1
M1: Set
1
(their triple product) = 6
6
A1: Correct equation
dM1
A1
Cao (oe)
A1
4 k 1
7 2 4k 4 14 11
1 11
1
4k 21
6
6
k
15
4
(4)
Alternative
Area ABC
1
1
6 11
= AB AC
2
2
D to Π is
4k 28 14 21
16 49 1
1
4k 28 14 21
6 11
6
6
16 49 1
Attempt area ABC and distance
between DandΠ
1
(their area x their
3
distance) = 6
M1: Set
M1
dM1
A1
A1: Correct equation
k
15
4
Cao (oe)
A1
(4)
(9 marks)
318
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
Scheme
Marks
7(a)
x 3
t 4 , y 4t 3
dx
dy
12
t3,
12t 2
dt
dt
S
2
Correct derivatives
1
2
dx 2 dy 2
y
dt
dt dt
2
4t 3
B1
12t 12t
3
2
2
2
dt
M1
1
2 4t 3 144t 6 144t 4 2 dt
M1: Substitutes their derivatives into a correct formula (2 not required)
S
2 4t 3 144t 4
S 96
1
0
1
2
t
2
1
2
1 dt
1
2
t 5 t 2 1 dt
Attempt to factor out at least t4 numerical factor may be left
M1
Correct completion
A1
(4)
(b)
u 2 t 2 1 2u
du
dt
2t or 2u 2t
dt
du
t 0 u 1, t 1 u 2
S
96
u
t 5 u du
t
S
96
u
Correct differentiation
B1
Correct limits
Alternative:
Reverse the substitution later.
(Treat as M1 in this case and
award later when work seen)
B1
M1: Complete substitution
2
1 u 2 du
2
A1: Correct integral in terms of u.
Ignore limits, need not be
simplified
u 7 2u 5 u 3
S 96
u 2u u
du 96
5
3
7
M1: Expands and attempts to integrate
6
4
2
dM1
3
2
2 7 2 2 5
u 7 2u 5 u 3
2 1 2 1
S 96
96
7 5 3
7
5
3
7
5
3
1
M1: Correct use of their changed limits (both to be changed)
Alternative: If sub reversed, substitute the original limits
192
64
S
11 2 4
Cao eg
105
35
M1 A1
ddM1
A1
(7)
(11 marks)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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319
Question
Scheme
8(a)
In
Marks
ln 2
tanh 2 n x dx,
n0
0
tanh 2n x tanh 2 n1 x tanh 2 x
B1
tanh 2n x
tanh 2 n1 x 1 sech 2 x
In
ln2
tanh
2 n 1
x dx
0
ln2
M1
tanh 2 n 1 x sech 2 x dx
0
M1: Correctly substitutes for In-1 and
obtains
ln 2
1
tanh 2 n 1 x
I n I n 1
2n 1
0
tanh
2 n 1
x sech 2 x dx k tanh 2n1 x
M1 A1
A1: Correct expression
I n 1
1 3
2n 1 5
2 n 1
Correct completion with no errors
*
A1*
(5)
Alternative
I n I n 1
ln 2
0
ln 2
0
ln 2
0
tanh
2n
x tanh 2 n 1 x dx
tanh 2 n 1 x tanh 2 x 1 dx
B1
2
tanh 2 n 1 x sech
x dx
ln 2
0
ln 2
1
tanh 2 n 1 x
I n I n 1
2
1
n
0
tanh 2 n 1 x sech 2 x dx
M1
M1:
Obtains
tanh 2 n 1 x sech 2 x dx k tanh 2 n 1 x
M1 A1
A1: Correct expression
1 3
I n 1
2n 1 5
2 n 1
*
Correct completion with no errors
A1*
(5)
320
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
8(b)
Scheme
I 0 ln 2
13
I 2 I1
3 5
The integration must be seen.
B1
Applies the reduction formula once
M1
3
1
1 3 1 3
I2
I0
1 5 3 5
I 2 ln 2
Marks
M1: Second application of the
reduction formula
3
M1A1
A1: Correct expression
84
125
cao
A1
Special Case: If I4 is found award B1 for Io or I1 and M1M0A0A0
(5)
Alternative
I1
ln 2
2
tanh
x dx
0
I
1
ln 2
0
1 sech x dx
x tanh x0
ln 2
13
I 2 I1
3 5
Correct integration
B1
Applies the reduction formula once
M1
3
3
I1
ln 2 tanh ln 2
ln 2
5
3 13
I 2 ln 2
5 3 5
ln 2
2
M1: Uses limits
M1A1
A1: Correct expression
3
84
125
A1
(5)
(10 marks)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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Please check the examination details below before entering your candidate information
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Pearson Edexcel
Centre Number
Candidate Number
International
Advanced Level
Sample Assessment Materials for first teaching September 2018
(Time: 1 hour 30 minutes)
Paper Reference WME01/01
Mathematics
International Advanced Subsidiary/Advanced Level
Mechanics M1
You must have:
Mathematical Formulae and Statistical Tables, calculator
Total Marks
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should show sufficient working to make your methods clear. Answers
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each question are shown in brackets
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Answer ALL questions. Write your answers in the spaces provided.
1.
A car is moving along a straight horizontal road with constant acceleration
a m s–2 (a > 0). At time t = 0 the car passes the point P moving with speed u m s–1.
In the next 4 s, the car travels 76 m and then in the following 6 s it travels a further 219 m.
Find
(i) the value of u,
(ii) the value of a.
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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.
(7)
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Q1
(Total for Question 1 is 7 marks)
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2.
(a) Find the value of k.
(4)
(b) Find, in terms of m and u only, the magnitude of the impulse exerted on Q by P in the
collision.
(2)
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Two particles P and Q are moving in opposite directions along the same horizontal straight
line. Particle P has mass m and particle Q has mass km. The particles collide directly.
Immediately before the collision, the speed of P is u and the speed of Q is 2u. As a result
of the collision, the direction of motion of each particle is reversed and the speed of each
particle is halved.
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Q2
(Total for Question 2 is 6 marks)
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3.
(a) Find the value of h.
(2)
Immediately after the impact the blocks move downwards together with the same speed
and both come to rest after sinking a vertical distance of 12 cm into the ground. Assuming
that the resistance offered by the ground has constant magnitude R newtons,
(b) find the value of R.
(8)
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A block A of mass 9 kg is released from rest from a point P which is a height h metres
above horizontal soft ground. The block falls and strikes another block B of mass 1.5 kg
which is on the ground vertically below P. The speed of A immediately before it strikes
B is 7 m s–1. The blocks are modelled as particles.
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Q3
(Total for Question 3 is 10 marks)
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4.
0.6m C
B
4m
Figure 1
A diving board AB consists of a wooden plank of length 4m and mass 30kg. The plank
is held at rest in a horizontal position by two supports at the points A and C, where
AC = 0.6 m, as shown in Figure 1. The force on the plank at A acts vertically downwards
and the force on the plank at C acts vertically upwards.
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A
A diver of mass 50 kg is standing on the board at the end B. The diver is modelled as a
particle and the plank is modelled as a uniform rod. The plank is in equilibrium.
(a) Find
(i) the magnitude of the force acting on the plank at A,
(6)
The support at A will break if subjected to a force whose magnitude is greater than 5000 N.
(b) Find, in kg, the greatest integer mass of a diver who can stand on the board at B
without breaking the support at A.
(3)
(c) Explain how you have used the fact that the diver is modelled as a particle.
(1)
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(ii) the magnitude of the force acting on the plank at C.
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Q4
(Total for Question 4 is 10 marks)
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Two forces, F1 and F2, act on a particle A.
Given that the resultant of F1 and F2 is parallel to (i + 2j),
(a) show that 2p – q + 7 = 0
(5)
Given that q = 11 and that the mass of A is 2 kg, and that F1 and F2 are the only forces
acting on A,
(b) find the magnitude of the acceleration of A.
(5)
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F1 = (2i – 3j) N and F2 = (pi + qj) N, where p and q are constants.
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Q5
(Total for Question 5 is 10 marks)
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6.
P
Figure 2
Two cars, A and B, move on parallel straight horizontal tracks. Initially A and B are
both at rest with A at the point P and B at the point Q, as shown in Figure 2. At time
t = 0 seconds, A starts to move with constant acceleration a m s–2 for 3.5 s, reaching a
speed of 14 m s–1. Car A then moves with constant speed 14 m s–1.
(a) Find the value of a.
(2)
(b) On a diagram, sketch, on the same axes, a speed-time graph for the motion of A
for the interval 0 t T and a speed-time graph for the motion of B for the interval
0 t T.
(3)
(d) Find the distance of car B from the point Q when B overtakes A.
(8)
(1)
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Car B also starts to move at time t = 0 seconds, in the same direction as car A. Car B
moves with a constant acceleration of 3 m s–2. At time t = T seconds, B overtakes A. At
this instant A is moving with constant speed.
(c) Find the value of T.
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Q
(e) On a new diagram, sketch, on the same axes, an acceleration-time graph for the
motion of A for the interval 0 t T and an acceleration-time graph for the motion
of B for the interval 0 t T.
(3)
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Q6
(Total for Question 6 is 17 marks)
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7.
β
α
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P (4 kg)
Q (m kg)
Figure 3
1
The first plane is rough and the coefficient of friction between P and the plane is . The
4
second plane is smooth. The system is in limiting equilibrium.
Given that P is on the point of slipping down the first plane,
(a) find the value of m,
(b) find the magnitude of the force exerted on the pulley by the string,
(10)
(4)
(1)
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(c) find the direction of the force exerted on the pulley by the string.
342
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A particle P of mass 4 kg is attached to one end of a light inextensible string. A particle
Q of mass m kg is attached to the other end of the string. The string passes over a
small smooth pulley which is fixed at a point on the intersection of two fixed inclined
planes. The string lies in a vertical plane that contains a line of greatest slope of each of
the two inclined planes. The first plane is inclined to the horizontal at an angle α, where
3
and the second plane is inclined to the horizontal at an angle β, where
tan α =
4
4
tan β = . Particle P is on the first plane and particle Q is on the second plane with the
3
string taut, as shown in Figure 3.
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Q7
(Total for Question 7 is 15 marks)
TOTAL FOR PAPER IS 75 MARKS
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Mechanics M1 Mark scheme
Question
1
Scheme
76
76
Marks
1
Use of s ut at 2 for
2
t 4,
s 76 and u 0 (use
of u 0 is M0)
or
4u 12 a.42
1
u u 4a 4
2
(38
2u 4a)
Correctly substituted
equation
295
10u 12 a.102
1
u u 10a 10
2
1
or 295 u 10a 10 a 100
2
or 295
19 2a 6
A1
1
Use of s ut at 2 for
2
t 10,
s 295
M1
1
s ut at 2 for
or
2
t 6,
s 219, u u
Correctly substituted
equation
(59
2u 10a)
or 219
M1
1
a 62
2
1
a 62
2
1
or 219 u 4a 6 a 62
2
1
u 4a u 10 6
or 219
2
1
or 219 (u 10a) 6 a 36
2
Solve simultaneous for u or for a.
This marks is not available if they have assumed a value for u or a in the
preceding work - it is dependent on the first 2 M marks.
u 12
or 219
38 u 6
A1
DM1
A1
a 3.5
A1
(7)
Alternative
Find the speed at
t 2,
t 7
76
19
4
219
t 7,
v7 36.5
6
t 2, v
2
Both values correct
Averages with no links to times is
M0
M1
A1
36.5 19 5a a 3.5
Use of v u 5a with their u,v
Correct a
M1
A1
19 u 2a
Complete method for finding u
Correct equation in u
DM1
A1
u 19 7 12
A1
(7)
(7 marks)
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347
Question
Scheme
2(a)
mu 2kmu
12 mu kmu
or
1
m u u km u 2u
2
Use of CLM
or Equal and opposite impulses
Need all 4 terms dimensionally
correct.
Masses and speeds must be paired
correctly
Condone sign errors
Condone factor of g throughout.
Marks
M1
Unsimplified equation with at most one error
A1
Correct unsimplified equation
A1
k 12
From correct working only
A1
(4)
Impulse on P or impulse on Q.
Mass must be used with the correct
speeds
(b)
For P : I m( 12 u u )
For Q : I km(u 2u )
3mu
2
1
e.g. km u is M0
2
If working on Q, allow equation
using
their k.
Terms must be dimensionally
correct.
Use of g is M0
Only
From correct working only
M1
A1
(2)
(6 marks)
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Question
3(a)
Scheme
72 2 x 9.8h
Marks
2
Use of v
u 2 2as with
u 0,
v 7
M1
or alternative complete method to
find h
h 2.5
Condone h 2.5 in the working
but the final answer must be
positive.
A1
(2)
(b)
9 x 7 = 10.5 u
Use CLM to find the speed of the
blocks after the impact. Condone
additional factor of g throughout.
M1
u 6
2
0
62 2a x 0.12
A1
2
Use of v
u 2 2as with
u 6,
v 0
Allow for their u and v 0
M1
Allow for
u 7,
v 0
Accept alternative suvat method to
form an equation in a.
Condone use of 12 for 0.12
Correctly substituted equation in a
with
u 6,
s 0.12
implied by a 150
() 10.5g R
10.5 x (-a)
A1
Use of F ma with their a g .
M1
Must have all 3 terms and 10.5
Condone sign error(s)
() 10.5g R
10.5 x (-150)
Unsimplified equation with a
substituted and at most one error
(their a with the wrong sign is 1
error)
A1
Correct unsimplified equation with a
substituted
A1
R 1680 or 1700
A1
(8)
Alternative for the last 6 marks:
Energy equation ( needs all three
1
10.5 62 10.5 9.8 0.12 R 0.12terms)
2
M2
-1 each error
A1A1A0 for 1 error, A1A0A0 for 2
errors
R 1680 or 1700
A3
A1
(10 marks)
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349
Question
Scheme
4(a)
Marks
R
A
1.4 m
0.6 m C
S
G
2m
50 g
30 g
M ( A) (30g x 2) + (50g x 4) = 0.6 S
M C
0.6 R 1.4 30g 3.4 50g
M G
2 R 1.4 S 2 50g
M B
4 R 2 30 g
3.4 S
() R 30 g 50 g
S
S
R 784
B
Moments equation. Requires all
terms and dimensionally correct.
Condone sign errors.
Allow M1 if g missing
M1
Correct unsimplified equation
A1
Resolve vertically. Requires all 4
terms. Condone sign errors
M1
Correct equation (with R or their R)
A1
NB: The second M1A1 can also be earned for a second moments equation
R = 3460 or 3500 or
1060 g (N)
3
Not 353.3g
S 4250 or 4200 or
1300 g (N)
3
Not 433.3g
One force correct
Both forces correct
If both forces are given as decimal
multiples of g mark this as an
accuracy penalty A0A1
A1
A1
(6)
(b)
Use R 5000 and complete
method to form an equation in M
or weight. Needs all terms present
M (C) (30g x 1.4) + (Mg x 3.4) = 0.6 x 5000
and dimensionally correct.
Condone sign errors.
Accept inequality.
Use of R and S from (a) is M0
M1
Correct equation in M (not weight)
implied by M 77.68
A1
77.7 is A0 even is the penalty for
over-specified answers has already
been applied
A1
M 77 kg
(3)
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Question
4(c)
Scheme
The weight of the diver acts at a point.
Marks
Accept “the mass of the diver is at
a point”.
B1
(1)
(10 marks)
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351
Question
5(a)
Scheme
(2i 3j) ( pi qj) ( p 2)i (q 3) j
p2 1
q 3 2
or
p2
n
for n 1
q 3
2n
4 2 p 3 q
Marks
Resultant force = F1+F2 in
the form ai bj
M1
Use parallel vector to form
a scalar equation in p and q.
M1
Correct equation (accept
any equivalent form)
A1
Dependent on no errors
seen in comparing the
vectors.
Rearrange to obtain given
answer.
At least one stage of
working between the
fraction and the given
answer
DM1
Given Answer
2p q 7
0
A1
(5)
5(b)
q 11 p 2
B1
R 4i 8j
2 p i 8j for their p
M1
4i 8j 2a (a 2i 4 j)
Use of F ma
M1
a
Correct method for a
22 42
Dependent on the preceding
M1
20 = 4.5 or 4.47 or better (m s2 )
DM1
A1
2 5
(5)
Alternative for the last two M marks:
F 16 64
80 2 a
80
M1
Correct method for F
DM1
Use of F m a
Dependent on the preceding
M1
(5)
(10 marks)
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Question
Scheme
Use of suvat to form an equation
in a
v u at 14 3.5a
6(a)
Marks
a4
M1
A1
(2)
(b)
v
B
14
A
3.5
T
t
Graph for A or B
B1
Second graph correct and both
graphs extending beyond the
point of intersection
B1
Values 3.5, 14, T shown on axes,
with T not at the point of
intersection. Accept labels with
delineators.
B1
NB: 2 separate diagrams scores max B1B0B1
(c)
1
2
T .3T ,
(T T 3.5)
.14
2
Find distance for A or B in terms
of T only.
Correct area formulae: must see
1
in area formula and be adding
2
in trapezium
M1
One distance correct
A1
Both distances correct
A1
1
2
(T T 3.5)
.14
2
T .3T 12 4 3.5 14 T 3.5
Equate distances and simplify to
a 3 term quadratic in T in the
form aT 2 bT c
0
M1
3T 2 28T 49
0
Correct quadratic
A1
Solve 3 term quadratic for T
M1
Correct solution(s) - can be
implied if only ever see
T 7 from correct work.
A1
1
2
T .3T
2
(3T 7)(T 7)
0
T 73 or 7
but T 3.5, T 7
(d)
(3)
73.5 m
A1
From correct work only. B0 if
extra answers.
(8)
B1
(1)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
353
Question
Scheme
6(e)
Marks
(A) Condone missing 4
4
(B) Condone graph going
beyond T 7
(A)
(B)
3
(A)
O
B1
3.5
B1
Must go beyond 3.5. Condone
no 3.
(A) Condone graph going
beyond T 7
Must go beyond 3.5. B0 if see a
solid vertical line.
Sometimes very difficult to see.
If you think it is there, give the
mark.
B1
(3)
Condone separate diagrams.
Alternative for (c) for candidates with a
sketch like this:
B1B1B0 on the graph and then
max 5/8 for (c) if they do not
solve for the T in the question.
v
B
3T
14
Treat as a special case.
B1
B1
B0
A
3.5
1
2
t
T
3 T 3.5
2
1
4 3.52 14T
2
2
12T 28T 49
0
(2T 7)(6T 7)
0
T 72 or
7
6
Total time 7
Use diagram to find area
M1
One distance correct
A1
Both distances correct
A1
Simplify to a 3 term quadratic in
T
M1
Correct quadratic
A1
Complete method to solve for the
T in the question
M1
Correct solution(s) - can be
implied if only ever see
Total 7
A1
A1
(8)
(17 marks)
354
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
7(a)
Scheme
Marks
F 0.25R
sin
3
5
sin
4
5
B1
or cos
4
5
or cos
3
5
R 4 g cos
(31.36)
T F
4 g sin
23.52
T 7.84
T 15.68
Use of correct trig ratios for or
B1
Normal reaction on P
Condone trig confusion (using )
M1
Correct equation
A1
Equation of motion for P. Requires all 3
terms.
Condone consistent trig confusion
Condone an acceleration not equated to 0 :
T F 4 g sin
4a
M1
Correct equation
A1
T mg sin
Equation of motion for Q
Condone trig confusion
Condone an acceleration not equated to 0:
T mg sin
ma
M1
T 7.84m
Correct equation
A1
Solve for m
Dependent on the 3 preceding M marks
Not available if their equations used a 0
m2
DM1
A1
NB Condone a whole system equation 4 g sin F
mg sin followed by
m 2 for 6/6
M2 for an equation with all 3 terms. Condon trig confusion. Condone an
acceleration 0
A2 (-1 each error) for a correct equation:
(10)
7(b)
F
T 2 T 2 or 2T cos 45 or
T
cos 45
Complete method for finding F in
terms of T
Accept
Rh
2
Rv
2
Correct expression in T
A1
Substitute their T into a correct expression. Dependent on the previous M
mark
F 2
8g
= 22 or 22.2 N
5
M1
Watch out - resolving vertically is not a
correct method and gives 21.9 N.
DM1
A1
(4)
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Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
355
Question
7(c)
Scheme
Along the angle bisector at the
pulley
Marks
Or equivalent - accept angle + arrow
shown on diagram.
( 8.1 to downward vertical)
Do not accept a bearing
(1)
(15 marks)
356
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Please check the examination details below before entering your candidate information
Candidate surname
Pearson Edexcel
Other names
Centre Number
Candidate Number
International
Advanced Level
Sample Assessment Materials for first teaching September 2018
(Time: 1 hour 30 minutes)
Paper Reference WME02/01
Mathematics
International Advanced Subsidiary/Advanced Level
Mechanics M2
You must have:
Mathematical Formulae and Statistical Tables, calculator
Total Marks
Candidates may use any calculator permitted by Pearson regulations.
Calculators must not have the facility for symbolic algebra manipulation,
differentiation and integration, or have retrievable mathematical
formulae stored in them.
Instructions
black ink or ball-point pen.
•• Use
If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).
in the boxes at the top of this page with your name, centre number and
• Fill
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.
Inexact
answers
should
be given to three significant figures unless otherwise
• stated.
Information
booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
•• AThere
are 7 questions in this question paper. The total mark for this paper is 75.
The
marks
each question are shown in brackets
• – use this asfora guide
as to how much time to spend on each question.
Advice
each question carefully before you start to answer it.
•• Read
Try to answer every question.
your answers if you have time at the end.
•• Check
If you change your mind about an answer, cross it out and put your new answer
and any working underneath.
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Answer ALL questions. Write your answers in the spaces provided.
1.
A car of mass 900 kg is travelling up a straight road inclined at an angle θ to the horizontal,
where sin θ =
1.
The car is travelling at a constant speed of 14 m s–1 and the resistance
25
to motion from non-gravitational forces has a constant magnitude of 800 N. The car
takes 10 seconds to travel from A to B, where A and B are two points on the road.
(a) Find the work done by the engine of the car as the car travels from A to B.
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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.
(4)
When the car is at B and travelling at a speed of 14ms–1 the rate of working of the engine
of the car is suddenly increased to P kW, resulting in an initial acceleration of the car
of 0.7 m s–2. The resistance to motion from non-gravitational forces still has a constant
magnitude of 800 N.
(4)
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(b) Find the value of P.
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Q1
(Total for Question 1 is 8 marks)
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2.
(a) Show that e <
Given that e =
7
12
(7)
1
4
(b) find the magnitude of the impulse exerted on Q in the collision.
(3)
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A particle P of mass 0.7 kg is moving in a straight line on a smooth horizontal surface.
The particle P collides with a particle Q of mass 1.2 kg which is at rest on the surface.
Immediately before the collision the speed of P is 6ms–1. Immediately after the collision
both particles are moving in the same direction. The coefficient of restitution between the
particles is e.
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6
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Q2
(Total for Question 2 is 10 marks)
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3.
At time t seconds (t 0) a particle P has velocity vm s–1, where
When t = 0 the particle P is at the origin O. At time T seconds, P is at the point A and
v = λ(i + j), where λ is a constant.
Find
(a) the value of T,
(3)
(b) the acceleration of P as it passes through the point A,
(c) the distance OA.
(3)
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v = (6t2 + 6t)i + (3t2 + 24)j
(5)
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Q3
(Total for Question 3 is 11 marks)
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4.
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B
P (2 kg)
Q (4 kg)
2.5 m
A
2.5 m
α
Figure 1
Two particles P and Q, of mass 2 kg and 4 kg respectively, are connected by a light
inextensible string. Initially P is held at rest at the point A on a rough fixed plane inclined
3
at α to the horizontal ground, where sin α = . The string passes over a small smooth
5
(a) Find the work done against friction as P moves from A to B.
(b) Find the total potential energy lost by the system as P moves from A to B.
(c) Find, using the work-energy principle, the speed of P as it passes through B.
(4)
(3)
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pulley fixed at the top of the plane. The particle Q hangs freely below the pulley and
2.5 m above the ground, as shown in Figure 1. The part of the string from P to the pulley
lies along a line of greatest slope of the plane. The system is released from rest with the
string taut. At the instant when Q hits the ground, P is at the point B on the plane. The
1
coefficient of friction between P and the plane is .
4
(4)
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(Total for Question 4 is 11 marks)
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5.
A
a
B
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3a
C
D
a
F
3a
E
Figure 2
Given that tan θ =
4
, find the value of k.
7
(10)
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The uniform lamina ABCDEF, shown in Figure 2, consists of two identical rectangles with
sides of length a and 3a. The mass of the lamina is M. A particle of mass kM is attached
to the lamina at E. The lamina, with the attached particle, is freely suspended from A and
hangs in equilibrium with AF at an angle θ to the downward vertical.
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Q5
(Total for Question 5 is 10 marks)
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6.
F
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B
C
b
2a
θ
A
Figure 3
(a) Show that the magnitude of F is
5mga
cos θ
b
(4)
The force exerted on the rod by the hinge at A is R, which acts upwards at an angle f
above the horizontal, where f > θ.
(b) Find
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A uniform rod AB, of mass 3m and length 2a, is freely hinged at A to a fixed point on
horizontal ground. A particle of mass m is attached to the rod at the end B. The system
is held in equilibrium by a force F acting at the point C, where AC = b. The rod makes
an acute angle θ with the ground, as shown in Figure 3. The line of action of F is
perpendicular to the rod and in the same vertical plane as the rod.
(i) the component of R parallel to the rod, in terms of m, g and θ,
(ii) the component of R perpendicular to the rod, in terms of a, b, m, g and θ.
(5)
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(c) Hence, or otherwise, find the range of possible values of b, giving your answer in
terms of a.
(2)
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*S59764A02128*
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Q6
(Total for Question 6 is 11 marks)
*S59764A02328*
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7.
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A
45°
6 m s–1
u m s–1
θ°
O
Figure 4
At time t = 0, a particle P of mass 0.7 kg is projected with speed u m s–1 from a fixed
point O at an angle θ ° to the horizontal. The particle moves freely under gravity. At time
t = 2 seconds, P passes through the point A with speed 6 m s–1 and is moving downwards
at 45° to the horizontal, as shown in Figure 4.
(a) the value of θ,
(6)
(b) the kinetic energy of P as it reaches the highest point of its path.
(3)
For an interval of T seconds, the speed, v m s–1, of P is such that v 6
(c) Find the value of T.
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Find
(5)
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Q7
(Total for Question 7 is 14 marks)
TOTAL FOR PAPER IS 75 MARKS
28
384
*S59764A02828*
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Mechanics M2 Mark scheme
Question
1(a)
Scheme
Resolving parallel to the plane
Marks
Condone trig confusion
D 900 g sin 800
M1
A1
900
g 800
1152.8 (N)
25
Work done :
Their D distance
1152.8 14 10
161392
161 kJ (160)
Independent.
For use of 14 x 10 x their D
M1
Accept 161000 (J), 160000
(J). Ignore incorrect units.
A1
(4)
Alternative using energy
Work
done 900 gd sin 800d
Allow with incorrect d
Use of d 14 10
Independent – allow in an
incorrect expression
161392
161 kJ (160)
M1A1
M1
A1
(4)
Equation of motion
All terms required.
Condone trig confusion and
sign errors.
Allow with 900a
M1
D 900 g sin 800 900 0.7
Correct unsimplified with
a 0.7 used
Accept with their 1152.8
arising from a 2 term
expression in (a)
A1
Independent
Treat missing 1000 as
misread, so allow for
14 theirD
P
1000 P
Allow for
(or
)
14
14
in their equation of motion
M1
cao
A1
1(b)
900 0.7 )
( D 1152.8
D 1782.8 (N)
Use of P Fv
P 14
their D
1000
P 25.0 (25)
(4)
(8 marks)
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385
Question
Scheme
2(a)
Marks
6 m s-1
0.7 kg
1.2 kg
w
v
CLM: 0.7 6 0.7 v 1.2w
Requires all terms &
dimensionally correct
M1
42
Correct unsimplified
A1
Used the right way round
Condone sign errors
M1
7v 12w
Impact:
wv
6e
A1
19v
Equation in e and v only: 42 72e
Dependent on the two
previous M marks
Use direction to form an inequality:
Independent.
Applied correctly for their v
M1
*Given answer*
A1
42 72e 0
e
7
12
DM1
(7)
Impulse on Q: I w 1.2
2(b)
Solve for w : w v 6e
I 1.2
42 72
19
1
4 6 1
4
42 5 63
3.32 (N s)
19 4 19
Accept unsimplified with e
substituted.
Have to be using w in part
(b)
105
w 2.763..... seen or
38
implied
3.3 or better
M1
B1
A1
(3)
Alternative
Impulse on Q = impulse on P
386
0.7 v 6
Accept negative here
M1
1
42 4 72
0.7
6
19
Substitute for e in their v
24
v 1.263.... seen or
19
implied
Accept negative here.
B1
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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Question
2(b)
continued
Scheme
63
19
Marks
Final answer must be
positive.
3.3 or better
A1
(3)
(10 marks)
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387
Question
3(a)
Scheme
v i j :
Use
Form an equation in t, T or
2 108 2592
0
6T 2 6T 3T 2 24
Solve for T
Marks
3T 2 6T 24
0,
0, T 2
T 4T 2
M1
Simplify to quadratic in t, T or
and solve.
M1
T 2 only
A1
If they score M1 and then state T 2 allow 3/3
If they guess T 2 and show that it works then allow 3/3.
If all we see is T 2 with no equation then 0/3 for (a) but full marks are
available for (b) and (c).
(3)
3(b)
Differentiate: a 12t 6 i 6tj
-2
30i 12 j (m s )
Majority of powers going down
Need to be considering both
components
M1
Correct in t or T
A1
Cao
A1
(3)
3(c)
Integrate :
r 2t 3 3t 2 ( A) i t 3 24t ( B) j
Clear evidence of integration.
Need to be considering both
components.
Do not need to see the constant(s).
-1 each error
M1
A2
If the integration is seen in part (a) it scores no marks at that stage, but if the
result is used in part (c) then the M1A2 is available in part (c)
OA
28i 56 j
Use their T
28
5 62.6 (m)
Distance
63 or better ,
Dependent on previous M1
Use of Pythagoras on their
OA
3920
DM1
A1
NB: Incorrect T can score 2/3 in (b) and 4/5 in (c)
(5)
(11 marks)
388
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Question
4(a)
Scheme
Marks
Resolve perpendicular to the plane:
R 2 g cos
B1
1
4 2g
Use F R : F 2 g
4
5 5
1
with and their R (3.92)
4
M1
2.5 F
Work done: WD
For their F
dM1
2g
9.8 (J)
5
If a candidate has found the total work
done but you can see the correct
terms/processes for finding the work done
against friction, give B1M1DM1A0 (3/4)
2.5
Accept g
A1
(4)
4(b)
Change in PE :
4 g 2.5 2 g 2.5sin
= 4 g 2.5 2 g 1.5
PE lost
7
g 68.6 (J)
Requires one gaining and one
losing
Condone trig confusion
M1
(correct unsimplified)
A1
or 69 (J) Accept 7g
A1
(3)
KE gained + WD = loss in GPE
The question requires the use
of work-energy. Alternative
methods score 0/4.
Requires all terms but condone
sign errors (must be
considering both particles)
M1
1
1
4v 2 2v 2 their (a)
their (b)
2
2
Correct unsimplified. -1 each
error
A2
4(c)
3v 2 6 g
v
2 g 4.43 (m s-1)
or 4.4. Accept
2g
A1
(4)
Alternative
Equations of motion for each
Equations of motion for each particle leading particle leading to
12 g
12 g
T 23.52 followed by a W-E
to
T 23.52 followed
5
5
equation for P:
by a W-E equation for Q:
1
2
2.5T 2v 2 g 2.5sin a M1A2
1
4v 2 2.5T 4 g 2.5
2
2
v
2 g 4.43 (m s-1)
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A1
389
Question
4(c)
continued
Scheme
Marks
Use of 36.9 gives correct answers to 3 sf
Use of 37 gives correct answers to 2 sf and more than this is not
justified, so A0 if they give 3 sf in this case.
(11 marks)
390
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
5
Scheme
Moments about vertical axis (AF):
Mg 1
Mg
a
1.5a 3akMg Mg 1 k x
2 2
2
Marks
Requires all terms and
dimensionally correct but
condone g missing
-1 each error
Accept with M and/or g not
seen.
M1
A2
1 3k
a
x
1 k
Moments about horizontal axis (AB or FE):
Requires all terms and
dimensionally correct but
condone g missing
-1 each error.
Mg
Mg
Accept with M and/or g not
1.5a
3.5a 4akMg Mg 1 k y
seen.
2
2
Do not penalise repeated
errors.
M1
A2
2.5 4k
a
y
1 k
Working with axes through
F gives
1 3k
x
a and
1 k
1.5
y
a
1 k
SR: A candidate working with a mixture of
mass and mass ratio can score 4/6
M1A0A0M1A2
Use of tan with their distances from AF &
AB
tan
M 3kM 4
2.5M 4kM 7
Must be considering the
whole system. Allow for
inverted ratio.
M1
or exact equivalent
A1
4
and solve for k :
7
7M 21kM 10M 16kM
Equate their tan to
k
3
5
M1
cso
A1
(10)
Alternative for the people who start by considering only the L shape.
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391
Question
Scheme
5
continued
5
2
x a and y a or
3
a
2
Combine with the particle
Marks
M1 (for either) requires
all terms and
dimensionally correct
but condone g /M
missing. A1 for each
correct.
M1A2
M1 (for both) requires
all terms and
dimensionally correct
but condone g missing.
A1 for each correct.
M1A2
See over for a more geometrical approach
A
B
θ
Candidate starts by
finding centre of mass
3
at a, a relative to F
2
(or equivalent), M1A2
scored
θ
d1
a
2
a
D
C
a
F
θ
E
d2
Use of tan with their
distances for finding d1
or d2.
θ
5
2
A1
Obtain length of a side
in a triangle containing
d1
a
θ
d1
a
392
M1
θ
5
3
a tan a a
2
7
Correct for their centre
of mass
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Question
Scheme
5
continued
θ
Marks
3
d1 a cos
7
Correct for their centre of
mass
A1
Use of tan to find second
distance
5
3a 4a tan
a
7
M1
4a
d2
3a
θ
5
a cos
7
A1
d2
θ
Moments about A: Md1 kMd2
M1
3
5
3
a cos k a cos k
7
7
5
A1
(10)
(10 marks)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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393
Question
6(a)
Scheme
Taking moments about A:
bF 3mga cos mg 2a cos
bF 5mga cos
5mga
F
cos
b
Marks
Requires all terms - condone trig
confusion and sign errors
M1
-1 each error
A2
*Given answer*
A1
(4)
6(b)
Component of R parallel to AB:
R cos
Requires all terms - condone trig
confusion
M1
3mg sin mg sin 4mg sin
Correct unsimplified
A1
Component of R perpendicular to
AB:
Requires all terms - condone consistent
trig confusion and sign errors
M1
4mg cos
R sin F
Correct unsimplified
A1
2aR sin 3mga cos
F 2a b
Alternatives for: M ( B)
M1A1
bR sin 2a b mg cos
M (C )
3mg b a cos
5mga
cos
b
ISW for incorrect work after correct
components seen
R sin
4mg cos
Correct with F substituted.
A1
(5)
Alternative
X F
sin
5mga
cos sin
b
Allow with F. Requires all
terms - condone trig confusion
F substituted
Y
4mg F cos
4mg
M1
A1
5mga
cos 2
b
Allow with F. Requires all
terms - condone trig confusion
and sign errors.
M1
Correct unsimplified
A1
Correct substituted
A1
(5)
6(c)
Use of R sin 0
Solve for b in terms of a:
5a
5
4
, 2a b a
b
4
M1
2a not required CSO
A1
(2)
Special case:
Misread of directions in (b)
394
NB: This MR can score full
marks
(2)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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Question
6(c)
continued
Scheme
Marks
Alternative
For , tan tan
5a
4 cos 2
Y
b
tan
tan
X 5a cos sin
b
4
5a
5a
cos2 sin 2
b
b
4
5a
5
cos 2 sin 2 b a
b
4
M1
cso
A1
(2)
(11 marks)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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395
Question
7(a)
Scheme
Equate horizontal components of speeds:
u cos
6cos
45 3 2
4.24....
Use suvat for vertical speeds:
u sin 2 g 6sin 45
u sin
2g 3 2
Divide to find tan :
2 g 6sin 45
tan
6cos 45
M1
Correct unsimplified
A1
Condone sign errors
M1
Correct unsimplified
A1
Dependent on previous 2 Ms.
Follow their components.
u 15.93....
2g 3 2
3.61..
3 2
74.6 75
7(b)
Marks
At max height, speed
u
cos ( 3 2 (m s-1) )
2
1
KE 0.7 3 2 (J)
2
6.3 (J)
DM1
A1
(6)
B1
Correct for their v at the top,
v0
M1
accept awrt 6.30. CSO
A1
(3)
7(c)
When P is moving upwards at 6 m s
-1
Use suvat to find first time v 6
A1
u sin gt
3 2
2 g 3 2 gt
3 2
t
2g 6 2
1.13....
g
T
2 1.13
0.87
M1
Solve for t
M1
Sensitive to premature
approximation. Allow 1.14.
A1
CAO accept awrt 0.87
A1
(5)
Alternative
6sin 45 0 gt
find time from top to A:
12 2
2 0.87
T 2
t
g
Correct strategy
Correct unsimplified
M1A1
M1
A1
A1
(5)
396
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
7(c)
continued
Scheme
Marks
Alternative
: u sin gt (their u, )
Time to top
M1
t 1.567....
A1
T 2 2 1.567...
M1A1
= 0.87
A1
(5)
Alternative
Vertical speed at A = - (vertical speed at
B)
=
36 3 2
2
3 2
u at for A B
Use v
Or use the 45° angle
M1
A1
Correct use for their values
M1
3 2 3 2 gT
A1
T 0.87
See below for alt 7d
A1
(5)
Alternative 7d
2
v
3 2
2
u sin gt 36
2
Form expression for v 2 .
Inequality not needed at
this stage
M1
Correct inequality for v 2
.
A1
18 u sin gt 18
M1
u sin 18
u sin 18
t
g
g
A1
u sin 18 u sin 18 2 18
T
0.866
g
g
g
A1
(5)
(14 marks)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
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Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Please check the examination details below before entering your candidate information
Candidate surname
Pearson Edexcel
Other names
Centre Number
Candidate Number
International
Advanced Level
Sample Assessment Materials for first teaching September 2018
(Time: 1 hour 30 minutes)
Paper Reference WME03/01
Mathematics
International Advanced Subsidiary/Advanced Level
Mechanics M3
You must have:
Mathematical Formulae and Statistical Tables, calculator
Total Marks
Candidates may use any calculator permitted by Pearson regulations.
Calculators must not have the facility for symbolic algebra manipulation,
differentiation and integration, or have retrievable mathematical
formulae stored in them.
Instructions
black ink or ball-point pen.
•• Use
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.
the questions in the spaces provided
• Answer
– there may be more space than you need.
should show sufficient working to make your methods clear. Answers
• You
without working may not gain full credit.
answers should be given to three significant figures unless otherwise
• Inexact
stated.
Information
booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
•• AThere
are 6 questions in this question paper. The total mark for this paper is 75.
The
marks
each question are shown in brackets
• – use this asfora guide
as to how much time to spend on each question.
Advice
each question carefully before you start to answer it.
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your answers if you have time at the end.
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S59765A
©2018 Pearson Education Ltd.
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Answer ALL questions. Write your answers in the spaces provided.
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1.
r
1
r
2
P
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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.
Figure 1
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A hemispherical bowl, of internal radius r, is fixed with its circular rim upwards and
horizontal. A particle P of mass m moves on the smooth inner surface of the bowl. The
particle moves with constant angular speed in a horizontal circle. The centre of the circle
1
is at a distance r vertically below the centre of the bowl, as shown in Figure 1.
2
The time taken by P to complete one revolution of its circular path is T.
2r .
Show that T = π
g
(8)
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Q1
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2.
towards the centre of the Earth. The force has magnitude K , where K is a constant.
x2
2
(a) Show that K = mgR
(2)
When S is at a distance 3R from the centre of the Earth, the speed of S is V. Assuming
that air resistance can be ignored,
(b) find, in terms of g, R and V, the speed of S as it hits the surface of the Earth.
(7)
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A spacecraft S of mass m moves in a straight line towards the centre of the Earth. The
Earth is modelled as a sphere of radius R and S is modelled as a particle. When S is at
a distance x, x R, from the centre of the Earth, the force exerted by the Earth on S is directed
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Q2
(Total for Question 2 is 9 marks)
*S59765A0724*
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At time t = 0, a particle P is at the origin O, moving with speed 8 m s–1 in the
positive x direction. At time t seconds, t 0, the acceleration of P has
2
m s–2 and is directed towards O.
1
(a) Show that, at time t seconds, the velocity of P is 16 – 4(t + 4) 2 ms–1
(b) Find the distance of P from O when P comes to instantaneous rest.
(5)
(7)
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–1
magnitude 2(t + 4)
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Q3
(Total for Question 3 is 12 marks)
*S59765A01124*
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4.
a
A
u
Figure 2
A particle of mass 3m is attached to one end of a light inextensible string of length a. The
other end of the string is attached to a fixed point O. The particle is held at the point A,
where OA is horizontal and OA = a. The particle is projected vertically downwards from
A with speed u, as shown in Figure 2. The particle moves in complete vertical circles.
(a) Show that u2 3ag.
(7)
(5)
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Given that the greatest tension in the string is three times the least tension in the string,
(b) show that u2 = 6ag.
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O
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5.
5m
O
B
Figure 3
Two fixed points A and B are 5 m apart on a smooth horizontal floor. A particle P of
mass 0.5 kg is attached to one end of a light elastic string, of natural length 2 m and
modulus of elasticity 20 N. The other end of the string is attached to A. A second
light elastic string, of natural length 1.2 m and modulus of elasticity 15 N, has one end
attached to P and the other end attached to B.
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P (0.5 kg)
A
Initially P rests in equilibrium at the point O, as shown in Figure 3.
(a) Show that AO = 3 m.
(4)
(b) Show that, while both strings are taut, P moves with simple harmonic motion.
(c) Find the speed of P at the instant when the string PB becomes slack.
(4)
(4)
The particle first comes to instantaneous rest at the point D.
(d) Find the distance DB.
(5)
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The particle is now pulled towards A and released from rest at the point C, where ACB is
a straight line and OC = 1 m.
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6.
y = x2 + 3
R
O
x
2
Figure 4
The shaded region R is bounded by part of the curve with equation y = x2 + 3, the x-axis,
the y-axis and the line with equation x = 2, as shown in Figure 4. The unit of length on
each axis is one centimetre. The region R is rotated through 2π radians about the x-axis
to form a uniform solid S.
Using algebraic integration,
(a) show that the volume of S is
202
π cm3 ,
5
(b) show that, to 2 decimal places, the centre of mass of S is 1.30 cm from O.
(5)
7cm
V
2 cm
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(4)
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y
6 cm
Figure 5
(c) Find the distance from V to the centre of mass of T.
(5)
The point A lies on the circumference of the base of the cone. The solid T is suspended
from A and hangs freely in equilibrium.
(d) Find the size of the angle between VA and the vertical.
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A uniform right circular solid cone, of base radius 7 cm and height 6 cm, is joined to S to
form a solid T. The base of the cone coincides with the larger plane face of S, as shown
in Figure 5. The vertex of the cone is V.
The mass per unit volume of S is twice the mass per unit volume of the cone.
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*S59765A02324*
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Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
23
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Q6
(Total for Question 6 is 17 marks)
TOTAL FOR PAPER IS 75 MARKS
24
422
*S59765A02424*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
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___________________________________________________________________________
Mechanics M3 Mark scheme
Question
1
Scheme
Marks
(30o or for the first 3 lines)
M1
A1
R sin 30 mg
R
cos30 m r cos30
2
M1
A1
A1
2
R
g
mr r sin 30
DM1
2g
r
2
r
2r
Time
=
2
2g
g
A1
*
A1
cso
(8)
Alternative:
Resolve perpendicular to the reaction:
mg cos30 m rad 2 cos 60
M2
A1
(LHS)
A1
(RHS)
mr cos30 2 cos 60
A1
Obtain
M1
A1
Correct time
A1
(8)
(8 marks)
Notes:
Resolving vertically 30o or
Correct equation 30o or
Attempting an equation of motion along the radius, acceleration in either form 30o or
Allow with r for radius.
A1: LHS correct 30o or
A1: RHS correct, 30o or but not r for radius.
DM1: Obtaining an expression for 2 or for v2 and the length of the path 30o or Dependent
on both previous M marks.
A1: Correct expression for Must have the numerical value for the trig function now.
A1cso: Deducing the GIVEN answer.
M1:
A1:
M1:
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
423
Question
Scheme
2(a)
F
x R F mg
K mgR 2
Marks
K
x2
K
mg
R2
M1
*
A1
(2)
(b)
mgR 2
dv
mv
2
x
dx
M1
R2
g 2 dx v dv
x
g
R2
1
v2
x
2
dM1
A1ft
c
x 3R, v V g
R2
1
V2 c
3R
2
M1
Rg 1 2
c
V
3 2
A1
1 2
Rg 1 2
R2
V g
x R v
2
3 2
R
M1
2
v
V2
v
V2
4 Rg
3
4 Rg
3
A1
cso
(7)
(9 marks)
Notes:
(a)
M1:
A1:
Setting F mg and x R
Deducing the GIVEN answer
(b)
M1:
Attempting an equation of motion with acceleration in the form v
dv
. The minus sign may
dx
be missing.
dM1: Attempting the integration.
A1ft: Correct integration, follow through on a missing minus sign from line 1, constant of
integration may be missing.
M1: Substituting
x 3
R, v V to obtain an equation for c
A1: Correct expression for c.
M1: Substituting x R and their expression for c.
A1: Correct expression for v, any equivalent form.
424
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
3(a)
Scheme
Marks
1
dv
2 t 4 2
dt
M1
1
v
2 t 4 2 dt
1
v
4 t 4 2
dM1
A1
c
t 0, v 8 c 16
1
v 16 4 t 4 2
M1
A1
cso
(m s-1) *
(5)
(b)
v 0 16
4 t 4
1
2
M1
16
t4 t
12
A1
1
x 4 4 t 4 2 dt
3
2
x 4 4t t 4 2
3
M1
A1
d
2 32 64
t 0, x 0 d 4 4
3
3
A1
oe
3
64
2
2
t 12 x 4 4 12 16 2 42 (m)
3
3
3
dM1
A1
oe eg 43 or better
(7)
(12 marks)
Notes:
(a)
dv
; minus may be omitted.
dt
M1:
Attempting an expression for the acceleration in the form
DM1:
A1:
M1:
A1:
Attempting the integration
Correct integration, constant of integration may be omitted (no ft)
Using the initial conditions to obtain a value for the constant of integration
cso. Substitute the value of c and obtain the final GIVEN answer
(b)
M1:
A1:
M1:
A1:
Setting the given expression for v equal to 0
Solving to get t 12
dx
Setting v
and attempting the integration wrt t. At least one term must clearly be
dt
integrated.
Correct integration, constant may be omitted.
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
425
Question 3 notes continued
M1:
Substituting
t 0,
x 0 and obtaining the correct value of d. Any equivalent number, inc
decimals.
dM1: Substituting their value for t and obtaining a value for the required distance. Dependent on the
second M mark.
A1: Correct final answer, any equivalent form.
426
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
4(a)
Scheme
Energy to top:
NL2 at top:
T 3m
Marks
M1
A1
1
1
3m u 2 3mv 2
3mga
2
2
M1
A1
v2
T 3mg
3m
a
dM1
u2
6mg 3mg
a
T
0
u2
3ag
u2
a
M1
3g
A1
cso
(7)
(b)
Tension at bottom:
1
1
3m V 2 3mu 2
3mga
2
2
M1
V2
Tmax 3mg
3m
a
M1
Tmax 3mg 6mg 3m
Tmin 3m
A1
u2
a
u2
9mg
a
dM1
u2
u2
9mg 3m 3 3m 9mg
a
a
u 2 6ag
*
A1
cso
(5)
(12 marks)
Notes:
(a)
M1:
A1:
M1:
Attempting an energy equation, can be to a general point for this mark. Mass can be missing
2
u 2 2as scores M0
but use of v
Correct equation from A to the top.
Attempting an equation of motion along the radius at the top, acceleration in either form.
v2
r
2
dM1: Eliminate v to obtain an expression for T dependent on both previous M marks.
M1: Use T 0 at top to obtain an inequality connecting a, g and u
A1: Re-arrange to obtain the GIVEN answer.
A1:
Correct equation, acceleration in form
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
427
Question 4 notes continued
(b)
M1:
M1:
A1:
dM1:
A1:
428
Attempting an energy equation to the bottom, maybe from A or from the top.
Attempting an equation of motion along the radius at the bottom.
Correct expression for the max tension.
Forming an equation connecting their tension at the top with their tension at the bottom. If
the 3 is multiplying the wrong tension this mark can still be gained. Dependent on both
previous M marks.
cso. Obtaining the GIVEN answer.
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
5(a)
Scheme
T
Marks
20e 15 1.8 e
2
1.2
M1A1
10e 1.2
15 1.8 e
e 1
A1
AO 3 m
*
A1cso
(4)
(b)
0.5 x
20 1 x 15 0.8 x
2
1.2
x 45x
(c)
M1
A1
A1
A1
cso
SHM
String becomes slack when x
0.8
(4)
B1
(allow wo sign due to symmetry)
v2 2 a2 x2
v2
45 1 0.82
M1
A1 ft
16.2
v 4.024... m s-1 (4.0 or better)
A1ft
(4)
(d)
1 20 y 2 1 20 1.82 1
0.5 16.2
2
2
2
2
2
M1
A1
A1 ft
ft on v
20 y 2 64.8
16.2
A1
y 2 4.05
y 2.012....
A1ft
Distance DB
5 4.012...
0.988... m (accept 0.99 or better)
Alternative
0.5a = −10(1.8 + x)
v
dv
36 10 x
dx
(36 10 x)dx
v dv
M1
A1
v2
36 x 5x 2 c
2
x = 0, v =
A1
9 5
c = 8.1
5
M1
A1
Then v = 0 etc
(5)
(17 marks)
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
429
Question 5 continued
Notes:
(a)
M1:
A1:
A1:
A1:
Attempting to obtain and equate the tensions in the two parts of the string.
Correct equation, extension in AP or BP can be used or use OA as the unknown.
Obtaining the correct extension in either string (ext in BP =0.8 m) or another useful
distance.
cso. Obtaining the correct GIVEN answer.
(b)
M1:
Forming an equation of motion at a general point. There must be a difference of tensions,
both with the variable. May have m instead of 0.5 Accel can be a.
A1 A1: Deduct 1 for each error, m or 0.5 allowed, acceleration to be x now.
A1: cso Correct equation in the required form, with a concluding statement; m or 0.5 allowed.
Question 5 notes continued
(c)
B1:
M1:
For x 0.8 Need not be shown explicitly.
Using
v 2 2 a 2 x 2 with their (numerical) and their x
A1ft: Equation with correct numbers ft their
A1ft: Correct value for v 2sf or better or exact.
(d)
M1:
A1:
Attempting an energy equation with 2 EPE terms and a KE term.
2 correct terms may have 1.8 x instead of y.
A1ft: Completely correct equation, follow through their v from (c)
A1: Correct value for distance travelled after PB became slack. x 0.21
A1ft: Complete to the distance DB. Follow through their distance travelled after PB became slack.
430
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Question
6(a)
Scheme
2
Marks
M1
2
Vol = x 2 3 dx
0
x 4 6 x 2 9 dx
2
0
dM1
A1
2
1
x5 2 x3 9 x
5
0
202
cm3
5
A1
*
(4)
(b)
M1
x x 2 3 d
x x5 6 x3 9 x dx
2
2
2
0
0
2
3
9
1
x6 x4 x 2
2
2 0
6
A1
158
3
(Or by chain rule or substitution)
158 5
,
1.3036...
1.30 cm
C of m
3 202
A1
M1
A1
(5)
(c)
Mass ratio
Dist from V
2
202
5
6.7
1
72 6
3
4.5
B1
404
98
5
x
B1
M1
A1 ft
404
404
6.7 98 4.5
98 x
5
5
A1
404
6.7 98 4.5
5
x
5.494....
5.5 cm Accept 5.49 or better
404
98
5
(d)
tan
(5)
M1
6 x 0.5058...
7
7
6
M1
A1
0.5058...
36.468... 36 or better
7
tan 1 tan 1
7
(3)
(17 marks)
Notes:
(a)
M1:
Using y 2dx with the equation of the curve, no limits needed
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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431
Question 6 notes continued
dM1: Integrating their expression for the volume.
A1: Correct integration inc limits now.
A1: Substituting the limits to obtain the GIVEN answer.
(b)
M1:
Using xy 2dx with the equation of the curve, no limits needed, can be omitted.
A1:
A1:
Correct integration, including limits; no substitution needed for this mark.
Correct substitution of limits.
xy 2 dx
with their xy 2dx . must be seen in both numerator and
M1:
Use of
A1:
denominator or in neither.
cso. Correct answer. Must be 1.30
(c)
B1:
B1:
M1:
A1ft:
A1:
Correct mass ratio.
Correct distances, from V or any other point, provided consistent.
Attempting a moments equation.
Correct equation, follow through their distances and mass ratio.
Correct distance from V
(d)
M1:
M1:
A1:
Attempting the tan of an appropriate angle, numbers either way up.
Attempting to obtain the required angle.
Correct final answer 2sf or more.
432
y dx
2
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
Please check the examination details below before entering your candidate information
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Pearson Edexcel
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Centre Number
Candidate Number
International
Advanced Level
Sample Assessment Materials for first teaching September 2018
(Time: 1 hour 30 minutes)
Paper Reference WST01/01
Mathematics
International Advanced Subsidiary/Advanced Level
Statistics S1
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Mathematical Formulae and Statistical Tables, calculator
Total Marks
Candidates may use any calculator permitted by Pearson regulations.
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Instructions
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in the boxes at the top of this page with your name, centre number and
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Answer all questions and ensure that your answers to parts of questions are
• clearly
labelled.
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should
show sufficient working to make your methods clear. Answers
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Inexact
answers
should
be given to three significant figures unless otherwise
• stated.
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booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
•• AThere
are 7 questions in this question paper. The total mark for this paper is 75.
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each question are shown in brackets
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©2018 Pearson Education Ltd.
1/1/1/1/1/1/
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blank
Answer ALL questions. Write your answers in the spaces provided.
The percentage oil content, p, and the weight, w milligrams, of each of 10 randomly
selected sunflower seeds were recorded. These data are summarised below.
∑ w2 = 41252
∑ wp = 27557.8
∑ w = 640
∑ p = 431
Spp= 2.72
(a) Find the value of Sww and the value of Swp
(b) Calculate the product moment correlation coefficient between p and w
(c) Give an interpretation of your product moment correlation coefficient.
(3)
(2)
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1.
(1)
The equation of the regression line of p on w is given in the form p = a + bw
(d) Find the equation of the regression line of p on w
(2)
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(e) Hence estimate the percentage oil content of a sunflower seed which weighs
60 milligrams.
(4)
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Q1
(Total for Question 1 is 12 marks)
*S59767A0528*
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2.
The time taken to complete a puzzle, in minutes, is recorded for each person in a club. The
times are summarised in a grouped frequency distribution and represented by a histogram.
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Find the number of people in the club.
(3)
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One of the class intervals has a frequency of 20 and is shown by a bar of width 1.5 cm and
height 12 cm on the histogram. The total area under the histogram is 94.5 cm2
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(Total for Question 2 is 3 marks)
*S59767A0728*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
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3.
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The discrete random variable X has probability distribution
1
5
x = 1, 2, 3, 4, 5
(a) Write down the name given to this distribution.
(1)
Find
(b) P(X = 4)
(1)
(c) F(3)
(1)
(d) P(3X – 3 > X + 4)
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P( X = x) =
(2)
(e) Write down E(X )
(f) Find E(X 2)
(2)
(g) Hence find Var (X )
(2)
Given that E(aX – 3) = 11.4
(h) find Var (aX – 3)
(4)
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(1)
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*S59767A0928*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
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*S59767A01028*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
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Q3
(Total for Question 3 is 14 marks)
*S59767A01128*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
11
443
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4.
A researcher recorded the time, t minutes, spent using a mobile phone during a particular
afternoon, for each child in a club.
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Coded Time (v)
Frequency ( f )
Coded Time Midpoint (m)
0v<5
20
2.5
5 v < 10
24
a
10 v < 15
16
12.5
15 v < 20
14
17.5
20 v < 30
6
b
(You may use ∑
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t−5
The researcher coded the data using v =
and the results are summarised in the table
10
below.
fm = 825 and ∑ fm2 = 12 012.5)
(b) Calculate an estimate of the mean of v.
(c) Calculate an estimate of the standard deviation of v.
(d) Use linear interpolation to estimate the median of v.
(e) Hence describe the skewness of the distribution. Give a reason for your answer.
(1)
(1)
(2)
(2)
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(a) Write down the value of a and the value of b.
(2)
(f) Calculate estimates of the mean and the standard deviation of the time spent using a
mobile phone during the afternoon by the children in this club.
(4)
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*S59767A01228*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
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*S59767A01328*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
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*S59767A01428*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
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Q4
(Total for Question 4 is 12 marks)
*S59767A01528*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
15
447
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5.
A biased tetrahedral die has faces numbered 0, 1, 2 and 3. The die is rolled and the number
face down on the die, X, is recorded. The probability distribution of X is
0
1
2
3
P(X = x)
1
6
1
6
1
6
1
2
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x
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If X = 3 then the final score is 3
If X ≠ 3 then the die is rolled again and the final score is the sum of the two numbers.
The random variable T is the final score.
(a) Find P(T = 2)
(2)
(b) Find P(T = 3)
(3)
(3)
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(c) Given that the die is rolled twice, find the probability that the final score is 3
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*S59767A01628*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
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*S59767A01728*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
17
449
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Question 5 continued
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*S59767A01828*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
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Q5
(Total for Question 5 is 8 marks)
*S59767A01928*
Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018
19
451
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6.
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Three events A, B and C are such that
2
5
P(C) =
1
2
P(A ∪ B) =
5
8
Given that A and C are mutually exclusive find
(a) P(A ∪ C )
(1)
Given that A and B are independent
(b) show that P(B) =
3
8
(c) Find P(A | B)
(4)
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P(A) =
(1)
Given that P(C′ ∩ B′) = 0.3
(5)
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