Edexcel Maths A Level Pure Maths 2 MS Nov 2021

Edexcel Maths A Level Pure Maths 2
MS Nov 2021

, Mark Scheme (Results)



November 2021


Pearson Edexcel GCE
In Mathematics (9MA0)
Paper 02 Pure Mathematics 2




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October 2021
Question Paper Log Number P68732A
Publications Code 9MA0_02_2111_MS
All the material in this publication is copyright
© Pearson Education Ltd 2021

,General Marking Guidance




 All candidates must receive the same treatment. Examiners
must mark the first candidate in exactly the same way as they
mark the last.
 Mark schemes should be applied positively. Candidates must
be rewarded for what they have shown they can do rather than
penalised for omissions.
 Examiners should mark according to the mark scheme not
according to their perception of where the grade boundaries
may lie.
 There is no ceiling on achievement. All marks on the mark
scheme should be used appropriately.
 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 may be limited.
 When examiners are in doubt regarding the application of the
mark scheme to a candidate’s response, the team leader must
be consulted.
 Crossed out work should be marked UNLESS the candidate has
replaced it with an alternative response.

, EDEXCEL GCE MATHEMATICS

General Instructions for Marking



1. The total number of marks for the paper is 100.


2. The Edexcel Mathematics 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.


3. Abbreviations


These are some of the traditional marking abbreviations that will appear in the
mark schemes.

 bod – benefit of doubt
 ft – follow through
 the symbol will be used for correct ft
 cao – correct answer only
 cso - correct solution only. There must be no errors in this part of the
question to obtain this mark
 isw – ignore subsequent working
 awrt – answers which round to
 SC: special case
 oe – or equivalent (and appropriate)
 dep – dependent
 indep – independent
 dp decimal places
 sf significant figures
 🞸 The answer is printed on the paper
 The second mark is dependent on gaining the first mark


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,
alternatives 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.




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with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom

,Question Scheme Marks AOs

1(a) 16   211   d  24  d  ... M1 1.1b
d  0.4 A1 1.1b
Answer only scores both marks.
(2)
n2a   n 1 d  S
1 1
(b) S    5002 16  499 "0.4" M1 1.1b
n 500
2 2
 57900 A1 1.1b
Answer only scores both marks
(2)
na  l
1
(b) Alternative using S n
2
1
l  16   500 1"0.4"  215.6  S   500 16 "215.6" M1 1.1b
500
2
 57900 A1 1.1b
(4 marks)
Notes
(a)
M1: Correct strategy to find the common difference – must be a correct method using a = 16, and
n = 21 and the 24. The method may be implied by their working.
If the AP term formula is quoted it must be correct, so use of e.g. un  a  nd scores M0
A1: Correct value. Accept equivalents e.g. 8 , 4 , 2 etc.
20 10 5
(b)
M1: Attempts to use a correct sum formula with a = 16, n = 500 and their numerical d from
part (a)
If a formula is quoted it must be correct (it is in the formula book)
A1: Correct value
Alternative:
M1: Correct method for the 500th term and then uses S n 1 na  l with their l
2
A1: Correct value

Note that some candidates are showing implied use of un  a  nd by showing the following:
24 16 8 1  8
(a) d  21  21 (b) S 500  2  500  2 16  499  21 55523.80952...
 
This scores (a) M0A0 (b) M1A0

,Question Scheme Marks AOs

2(a) y 7 B1 2.5
(1)
3 0.52
(b) f  1.8  7  2 1.82  0.52  gf  1.8  g 0.52   ... M1 1.1b
5 0.52 1
39
gf 1.8  0.975 oe e.g. A1 1.1b
40
(2)
3x
(c) y  5xy  y  3x  x 5 y  3  y M1 1.1b
5x 1
g  x  
1 x
A1 2.2a
5x  3
(2)
(5 marks)
Notes
(a)
B1: Correct range. Allow f (x) or f for y. Allow e.g. y  ℝ : y 7 ,    y 7, , 7
(b)
M1: Full method to find f (1.8) and substitutes the result into g to obtain a value.
Also allow for an attempt to substitute x = 1.8 into an attempt at gf (x).
3  7  2x 2  
3 7  2 1.8   2


gf  x    ...
5  7  2  1.8   1
E.g.
5  7  2x 2  1 2



A1: Correct value
(c)
M1: Correct attempt to cross multiply, followed by an attempt to factorise out x from an xy term
and an x term.
If they swap x and y at the start then it will be for an attempt to cross multiply followed by an
attempt to factorise out y from an xy term and a y term.
x 1 3
A1: Correct expression. Allow equivalent correct expressions e.g. , 
3  5x 5 25x 15
Ignore any domain if given.

,Question Scheme Marks AOs

12 y  5
log 12 y  5  log 1 3y   2  log
3
2
3 3 3 B1
1 3y
M1 on 1.1b
or e.g. EPEN
2  log3 9
12 y  5 12 y  5 2
log 2  3  9  27 y  12 y  5  y  ...
3
1 3y 1 3y
or e.g. M1 2.1
log 12 y  5  log 3 1 3y   12 y  5  32 1 3y   y  ...
2
3 3

4
y A1 1.1b
39
(3)
(3 marks)
Notes
B1(M1 on EPEN): Applies at least one addition or subtraction law of logs correctly.
Can also be awarded for using 2  log3 9 . This may be implied by e.g.
log3 ...  2  ...  9
M1: A rigorous argument with no incorrect working to remove the log or logs correctly and
obtain a correct equation in any form and solve for y.
A1: Correct exact value. Allow equivalent fractions.


Guidance on how to mark particular cases:


log 3 12 y  5 
log 3 12 y  5 log3 1 3y  2  2
log 3 1 3y 
12 y  5 4
  32  9  27 y  12 y  5  y 
1 3y 39

B1M0A0



log 3 12 y  5  12 y  5
log 3 12 y  5 log3  1 3y  2   2  log 3 2
log 3 1 3y  1 3y
12 y  5 4
  32  9  27 y  12 y  5  y 
1 3y 39

B1M0A0



12 y  5 4
log3 12 y  5  log 3 1 3y  2   3 2  9  27 y  12 y  5  y 
1 3y 39
B1M1A1

, Question Scheme Marks AOs

4 Examples:
2
    2
4 sin  4   , 3cos2   3  1 
2 2  2 


3cos2   31 sin2    31 2 
M1 1.1a


cos 2 1 3 4 2 
3cos   3
2
 1 1
2 2 2 
Examples:
2
     2
4 sin  3cos2   4    3  1 
2 2  2 

  
4 sin  3cos2   4  31 sin2    2  31  2  dM1 1.1b
 2 
2  

 2  cos 2 1   3  4 2 
4 sin  3cos   4 sin 3  4    1 1
2 2 2  2 2 2 

 2  31  2  ...  3  2  3 2 A1 2.1
(3)
(3 marks)
Notes
M1: Attempts to use at least one correct approximation within the given expression.
  2
or e.g. sin   if they write cos  as 1 sin  or e.g.
2 2
Either sin  or cos  1
2 2 2
 2 
2
1 cos 2
cos 2  1 (condone missing brackets) if they write cos2  as .
2 2
Allow sign slips only with any identities used but the appropriate approximations must be
applied.
dM1: Attempts to use correct approximations with the given expression to obtain an expression
in terms of θ only. Depends on the first method mark.
A1: Correct terms following correct work. Allow the terms in any order and ignore any extra
terms if given correct or incorrect.