Mark Scheme (Results)
January 2022
Pearson Edexcel International GCSE
In Further Pure Mathematics (4PM1)
Paper 2R
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January 2022
Question Paper Log Number P66307A
Publications Code 4PM1_02R_2201_MS
All the material in this publication is copyright
© Pearson Education Ltd 2022
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 penalised for omissions.
Examiners should mark according to the mark scheme - not according to
their perception of where the grade boundaries may lie.
All the marks on the mark scheme are designed to be awarded. Examiners
should always award full marks if deserved, i.e. if the answer matches the
mark scheme. Examiners should also be prepared to award zero marks if
the candidate’s response is not worthy of credit according to the mark
scheme.
Where some judgement is required, mark schemes will provide the
principles by which marks will be awarded and exemplification/indicative
content will not be exhaustive.
When examiners are in doubt regarding the application of the mark
scheme to a candidate’s response, a senior examiner must be consulted
before a mark is given.
Crossed out work should be marked unless the candidate has replaced it
with an alternative response.
•
Types of mark
o M marks: method marks
o A marks: accuracy marks
o B marks: unconditional accuracy marks (independent of M marks)
•
Abbreviations
o cao – correct answer only
o ft – follow through
o isw – ignore subsequent working
o SC - special case
o oe – or equivalent (and appropriate)
o dep – dependent
o indep – independent
o awrt – answer which rounds to
o eeoo – each error or omission
•
No working
If no working is shown then correct answers normally score full marks
If no working is shown then incorrect (even though nearly correct) answers score
no marks.
•
With working
If the final answer is wrong, always check the working in the body of the script (and
on any diagrams), and award any marks appropriate from the mark scheme.
If it is clear from the working that the “correct” answer has been obtained from
incorrect working, award 0 marks.
If a candidate misreads a number from the question. Eg. Uses 252 instead of 255;
method marks may be awarded provided the question has not been simplified.
Examiners should send any instance of a suspected misread to review.
If there is a choice of methods shown, then award the lowest mark, unless the
answer on the answer line makes clear the method that has been used.
If there is no answer achieved then check the working for any marks appropriate
from the mark scheme.
•
Ignoring subsequent work
It is appropriate to ignore subsequent work when the additional work does not
change the answer in a way that is inappropriate for the question: eg. Incorrect
cancelling of a fraction that would otherwise be correct.
It is not appropriate to ignore subsequent work when the additional work
essentially makes the answer incorrect eg algebra.
Transcription errors occur when candidates present a correct answer in working,
and write it incorrectly on the answer line; mark the correct answer.
•
Parts of questions
Unless allowed by the mark scheme, the marks allocated to one part of the
question CANNOT be awarded to another.
General Principles for Further Pure Mathematics Marking
(but note that specific mark schemes may sometimes override these general principles)
Method mark for solving a 3 term quadratic equation:
1. Factorisation:
(x
2
( ax
+ bx + c ) = ( x + p )( x + q ) , where pq = c
2
leading to x = ....
+ bx + c ) = ( mx + p )( nx + q ) where pq = c and mn = a
leading to x = ....
2. Formula:
Attempt to use the correct formula (shown explicitly or implied by working) with values for a, b
and c, leading to x = ....
3. Completing the square:
x 2 + bx + c = 0 : ( x  b2 )  q  c = 0,
2
leading to x = ....
q0
Method marks for differentiation and integration:
1. Differentiation
Power of at least one term decreased by 1.
(x
n
→ x n −1 )
2. Integration:
Power of at least one term increased by 1.
(x
n
→ x n +1 )
Use of a formula:
Generally, the method mark is gained by either
quoting a correct formula and attempting to use it, even if there are mistakes in
the substitution of values
or, where the formula is not quoted, the method mark can be gained by implication
from the substitution of correct values and then proceeding to a solution.
Answers without working:
The rubric states "Without sufficient working, correct answers may be awarded no marks".
General policy is that if it could be done "in your head" detailed working would not be
required. (Mark schemes may override this eg in a case of "prove or show...."
Exact answers:
When a question demands an exact answer, all the working must also be exact. Once a
candidate loses exactness by resorting to decimals the exactness cannot be regained.
Rounding answers (where accuracy is specified in the question)
Penalise only once per question for failing to round as instructed - ie giving more
digits in the answers. Answers with fewer digits are automatically incorrect, but
the isw rule may allow the mark to be awarded before the final answer is given.
Question
number
1 (a)
(b)
Scheme
x3
Part
(a)
(b)
1
and x = 5
2
1
 x5
2
3 x 5
M1
B1ft
B1ft
For solving the given 3TQ by any method.
See General Guidance.
For the correct values; x =
1
and x = 5
2
1
 x5
2
ft their solutions to the 3TQ.
For 3  x  5
For
ft from parts (a) and (b)
A1
B1ft
(3)
B1ft
(1)
Total 5 marks
Mark Notes
For x  3
B1
A1
(c)
B1
(1)
M1
(2 x − 1)( x − 5)
Critical values are x =
(c)
Marks
Question
number
2 (a)
Scheme
Gradient =
Marks
4 +1 1
=
3+ 7 2
M1
1
( x + 7)
2
x − 2y + 5 = 0
y +1 =
(b)
k=
Part
(a)
A1
(3)
AB = 102 + 52 = 5 5
AC =
(c)
M1
( −7 − −3) + ( 7 − −1)
2
2
=4 5
5
4
M1
A1
(2)
7− p
−3 − 3
12 = 7 − p
p = −5
−2 =
M1
dM1
A1
(3)
Total 8 marks
Mark Notes
M1 For finding the gradient of AB
M1
A1
(b)
For a fully correct method for finding the equation of a straight line.
If y = mx + c is used, then they must find a value for c for the award of this mark.
For x − 2 y + 5 = 0 in the required form.
Accept the terms in any order provided they are all on one side of the equation with
the other = 0
e.g., −2 y + x + 5 = 0 or 2 y − x − 5 = 0 etc
For using Pythagoras to find both AB and AC
M1
AB =
( 7 − −3) + ( 4 − −1)
A1
For k =
5
4
(c)
M1
dM1
A1
2
2
= 5 5 and AC = 42 + 82 = 4 5
Obtains an equation using the perpendicular of their gradient from (a) and the point
(3, p)
−2 =
7− p
−3 − 3
For a linear equation in p
12 = 7 − p
For p = −5
ALT
M1
dM1
A1
Finds the equation of the perpendicular using their gradient form part (a)
y − 7 = −2 ( x + 3)  y = −2 x + 1
Substitutes x = 3 into their equation of the perpendicular to find a value for y
For p = −5
Question
number
3 (a)
(b)
Part
(a)
(b)
Scheme
dy
5e2 x
= 2 e2 x 5 x − 3 +
dx
2 5x − 3
dy 3x 2 cos 3x − x 3 (−3sin 3x)
=
dx
(cos 3x) 2
Marks
M1 A1
A1
(3)
M1 A1
A1
(3)
Total 6 marks
Mark Notes
M1 For use of the product rule. Sum of two terms (either way round)
There must be an attempt to differentiate both terms.
See below.
1
1
1
−
( 5 x − 3) 2   k  ( 5 x − 3) 2 k  0
2
2x
2x
e  le
l0
A1
For either term correct
5e2 x
2x
2e
5 x − 3 or
2 5x − 3
A1
For the correct derivative.
dy
5e2 x
Or
= 2e2 x 5 x − 3 +
dx
2 5x − 3
M1
1
1
dy
5
−
= 2e2 x ( 5 x − 3) 2 + e2 x ( 5 x − 3) 2 oe
dx
2
For an attempt at the use of the quotient rule.
-
There must be an acceptable attempt to differentiate both terms
x3  3x 2
cos ( 3x )  −m sin 3x m  0
-
A1
A1
The denominator must be squared.
- The terms in the numerator must be subtracted in either order.
For one term correct
3x2 cos3x or x3 (−3sin 3x)
For the fully correct derivative.
Question
number
4 (a)
Scheme
Marks
3
2
2
2
( +  ) − 2 =  +  2
 =
 +  = −2
( −2 )
2
B1
M1
3
− 2  = 4 − 3 =1 *
2
M1
A1cso
(4)
ALT
2
2
4
2
2
2
2
(b)
3
0
and
4
2
2
2
6
2
3
2
4
3
0
{B1}
0
{M1}
{M1}
{A1cso}
(4)
1*
 4 +  4 = ( 2 +  2 )( 2 +  2 ) − 2 2  2
M1
2
7
3
1 1 − 2   = −
2
2
M1 A1
(3)
ALT
2 2
2
4
(c)
4
1
1
9
2
4
2
2
2
4
4
4
2
9
4
{M1}
{M1}
{A1}
(3)
7
2
4
 3  81
Product of the roots:   =   =
 2  16
7
(Sum of the roots:  4 +  4 = − )
2
2
16 x + 56 x + 81 = 0
4
4
M1
M1 A1ft
(3)
Total 10 marks
Part
(a)
Mark Notes
3
2
B1
For  +  = −2 and  =
M1
For ( +  ) − 2 =  2 +  2
2
This must be correct
For substituting their sum and product into their expansion for  2 +  2
M1
3
2
( −2 ) − 2  
2
A1cso For obtaining the given expression  2 +  2 = 1
ALT
For 2 2 4
3 0 and
2 2 4
3 0
B1
M1
For 2
M1
For
2
2
2
2
4
6
0
3
A1cso For obtaining the given expression  2 +  2 = 1
(b)
M1
For  4 +  4 = ( 2 +  2 )( 2 +  2 ) − 2 2  2
This must be correct
For substituting  2 +  2 = 1 and their product from (a) into their expansion
M1
for  4 +  4
3
 +  = 1 1 − 2  
2
4
A1
For the correct value  4 +  4 = −
ALT
M1
For 1
4
M1
For 1
4
A1
For  4 +  4 = −
(d)
M1
M1
A1
2
4
2
2
4
2
2
7
2
4
9
4
7
2
  3  4 81 
For product of the roots:    =   = 
  2  16 


2
For use of x − ( their sum ) x + ( their product ) =  0
4
For 16 x 2 + 56 x + 81 = 0
Must include = 0
4
Question
number
5 (a)
(b)
Scheme
 S = 
12
96
=
3 5
1−
8
M1 A1
(2)
5
3
ar = 12  
8
2
2  3  35 36
=
= 13
215
2
5
(c)
Marks
3
8
M1
M1
A1cso
(3)
*
n −1
e.g. un = 12  
M1
n −1
3
log 2 un = log 2 12 + log 2  
8
log 2 un = log 2 12 + (n − 1) log 2 3 − log 2 8
log 2 un = log 2 3 + 2log 2 2 + (n − 1)[log 2 3 − 3log 2 2]
log 2 un = n log 2 3 − 3n + 5 *
M1
M1
M1
A1 cso
(5)
Total 10 marks
Part
(a)
Mark Notes
M1
A1
(b)
M1
M1
(c)
a
1− r
96
For the correct value
5
For the correct use of ar n −1
For expressing 12 22  3 and 8
For the correct use of
(
)
( 2 ) as powers of 2
3
A1cso For obtaining the given expression with no errors seen.
For the correct use of ar n −1 and the given values of a and r to
M1
M1
M1
n −1
3
give un = 12  
8
For taking logs [base 2] of both sides and applying the addition law.
3
For applying the power law and subtraction laws to log 2  
8
n−1
n −1
3
log 2   = (n − 1)  log 2 3 − log 2 8
8
M1 For obtaining log 2 12 = log 2 3 + 2log 2 2
A1cso For obtaining the given equation with no errors seen.
ALT
For the correct use of ar n −1 and the given values of a and r to
M1
M1
M1
M1
A1
cso
n −1
3
give un = 12  
8
For rearranging the equation to obtain:
n −1
3n −1
3n
3
U n = 12   = 22  3  3( n −1) = 3n −5
2
2
8
For taking logs of both sides and applying the addition (subtraction) law.
log 2 U n = log 2 3n − log 2 ( 3n − 5 )
For applying the power law to obtain:
log 2 U n = n log 2 3 − log 2 ( 3n − 5 )
For obtaining the given equation with no errors seen.
Question
number
6 (a)
Scheme
dy
2
=
dx
x
When x = 9
M1
dy 2
=
dx 3
A1
2
y − 12 = ( x − 9)
3
When y = 0 −12 =
x = −9 So (−9, 0)
(b)
M1
2
( x − 9)
3
Gradient of Normal: −
When y = 0 −12 = −
x = 17 So (17, 0)
1
12  26 = 156
2
M1
A1
(5)
3
2
3
y − 12 = − ( x − 9)
2
(c)
Marks
M1
M1
3
( x − 9)
2
M1
A1
(4)
M1 A1
(2)
Total 11 marks
Part
(a)
(b)
(c)
Mark Notes
1
M1
−
dy
= 2x 2
For an attempt to differentiate y wrt x. Accept
dx
1
−
dy
= kx 2 where k is a constant  0
Allow as a minimum
dx
A1
dy 2
For
=
dx 3
M1 For a complete method to find the equation of a straight line.
If they use y = mx + c then they must reach a value for c for this mark.
dy
Accept the equation of the line with their value of
using the given
dx
coordinates.
M1 For substitution of y = 0 to find a value for x
A1
For (−9, 0)
M1
3
For gradient of Normal: −
2
Which is the negative reciprocal of their gradient obtained in (a)
M1 For a complete method to find the equation of a straight line with a gradient
3
of − (ft their gradient of normal)
2
If they use y = mx + c then they must reach a value for c for this mark.
M1 For substitution of y = 0 to find a value for x
A1
For (17, 0)
M1 For any correct method for finding the area of a triangle.
e.g.,
1
A = 12  ( '17 '− '− 9 ' ) = 156
2
OR
A=
A1
1  9 −9 17 9  1
= ( 0 + 0 + 204 ) − ( 0 + 0 − 108 )  = 156
2 12 0 0 12  2 
For 156
Question
number
7 (a)
Scheme
3 + (1 − cos 2  )
= 3cos 
cos  − 2
4 − cos2  = 3cos2  − 6cos 
(2cos  + 1)(2cos  − 4) = 0
( cos  = 2 does not exist so) cos  = −
(b)
Marks
M1
M1
M1
1
2
A1 cso
(4)
*
1
2
3x = 120 , 240 , 480
cos3x = −
M1
A1
A1 A1
(4)
Total 8 marks
x = 40 , 80 , 160
Part
(a)
Mark Notes
M1 For the correct use of sin 2  + cos 2  = 1
M1
For multiplying both sides by cos  − 2 and expanding brackets
For solving their 3TQ using any method.
M1
4 cos 2  − 6 cos  − 4 = 0  2 ( 2 cos  − 1)( cos  − 2 ) = 0
See General Guidance.
A1cso
ALT
M1
M1
M1
A1cso
(b)
M1
For obtaining the given equation: cos  = −
Must reject cos  = 2
1
2
For the correct use of sin 2  + cos 2  = 1
( 2 − cos  )( 2 + cos  ) = 3cos 
4 − cos 2 
= 3cos  
Factorises LHS
cos  − 2
cos  − 2
Cancels through by cos  − 2 and solves their linear equation in terms of
cos 
− ( 2 + cos  ) = 3cos   cos  = ...
For obtaining the given equation: cos  = −
For cos3x = −
1
2
1
2
For 3x = 120 or any other correct angle, e.g. even 3x = −120
A1
Allow an angle in radians for this mark. E.g. 3x =
A1
For one from x = 40 , 80 , 160
A1
2
3
For all angles correct with no additional angles within range. x = 40 , 80 , 160
Ignore any angles out of range.
Penalise any extra angles within range by deducting the final A mark.
Question
number
8 (a)
(b)
Part
(a)
(b)
Scheme
4 1.5 = 6
 r2 = 6
6
r=

dA
= 2 r
dr
dr dr dA  1

=

=
 1.5
dt dA dt  2 r

1
=
1.5
6
2

= 0.173
Marks
B1
M1
A1
(3)
M1
M1
M1
A1
(4)
Total 7 marks
Mark Notes
For ( 4  1.5 = ) 6
B1
M1
For  r 2 = 6 an attempt to rearrange to make r the subject
A1
For r =
M1
For
M1
M1
A1
6

dA
= 2 r This must be correct.
dr
dr dr dA
For application of
=

dt dA dt
dr
For substituting their r into
dt
For awrt 0.173
Question
number
9 (a)
Scheme
cos  =
3
B1
(1)
13
(b)
h = 17 2 − ( 92 + 122 ) = 8
(c)
Let M be the midpoint of BC
h
8 2
tan  =
=
= *
OM 12 3
Let N be the midpoint of EM
(d)
Marks
M1 M1
A1
(3)
M1 A1
(2)
EM = 82 + 122 = 4 13
M1
 3 
− 2 (12 ) 2 13 
  NO = 2 13
 13 
Hence triangle ONM is isosceles
180 − 2  33.7 = 112.6
(
NO = 122 + 2 13
)
2
(
)
M1
M1 A1
(4)
Total 10 marks
Part
(a)
(b)
(c)
(d)
Mark Notes
3
B1
For cos  =
M1
For use of Pythagoras to OC find e.g.
M1
For use of Pythagoras to find h e.g.
A1
For 8
M1
For tan  =
A1
cso
M1
M1
M1
A1
13
(9
2
+ 122 ) oe
17 2 − ( 92 + 122 )
h
8
(condone the omission of the degree sign)
=
OM 12
For obtaining the given result
For use of Pythagoras to find EM e.g. 82 + 122
For use of the cosine rule to find NO e.g.
 3 
− 2 (12 ) 2 13 

 13 
For 180 − 2  33.7
For 112.6
(
122 + 2 13
)
2
(
)
Question
number
10 (a)
(b)
Scheme
Marks
sin x + 1 = cos x + 1  tan x = 1
 5
x= ,
4 4
Area R1 :
M1
A1 A1
(3)
5
4
5
4
4
4
 ( sin x + 1) − ( cos x + 1) dx =  (sin x − cos x ) dx
M1
5
 − cos x − sin x4
M1
4
 2
2 
2
2
+
−

 −  −
=2 2
2   2
2 
 2
Area R2 :
M1 A1
3
2
5
4
5
4
 ( cos x + 1) dx +  ( sin x + 1) dx
5
M1
3
sin x + x4 +  − cos x + x52
A1
4

  3  2 5  
2 5 
1
+
+
 −
 −   +  − 
  = − 2 + 
4 
4  
2
 2
  2  2
area of R1 : area of R2 = 2 :
2
4
1 oe
M1 A1
A1
(9)
Total 12 marks
Part
(a)
Mark Notes
M1 For tan x = 1

For x =
[Allow 45º]
A1
4
5
For x =
[Allow 225º]
A1
4
(b)
'
5
'
4
( sin x + 1) − ( cos x + 1) dx
M1
For stating A =  
M1
For attempting to integrate their expression for the area.
This must be correct for this mark.
 (sin x − cos x ) dx = − cos x − sin x
M1
A1
For substitution of correct limits
For 2 2
' '[
4
For attempting
M1
5
4

f ( x)dx +  g( x)dx with the correct limits.
3
 ( cos x + 1) dx + 52 ( sin x + 1) dx
4
M1
M1
A1
A1
For attempting to integrate, which must be correct for this mark.
For substitution of their limits correctly.
1
For − 2 + 
2
For the simplified ratio 2 :
2
4
1
Question Scheme
→
11(a)
AB = −a + b
→
NB = −
OR
→
NB = −
→
2
AC = −a + b
3
Marks
→
5
AM = −a + b
6
B2
1 → →
1
2 
 a 2b 
AC + AB = −  −a + b  + ( −a + b ) =  − + 
2
2
3 
 2 3
M1
1 → → 1
2  1
 a 2b 
AC + AB =  −a + b  + b =  − + 
2
2
3  3
 2 3 
→
Two statements for OP using two parameters
→ →
→
5  
5 

OP = OA +  AM = a +   −a + b  = a (1 −  ) + b 
6  
6 

→ →
→
 a 2b     
 2  
OP = OB −  NB = b −   − +
 = a   + b  1 −

3  
 2 3   2

Equates coefficients

5
2
 1−  =
 = 1−
2
6
3
Solves S. E.
2
2
 =
=
3
3
→
2
5  1
5
 OP = a +  −a + b  = a + b
3
6  3
9
M1
M1
M1
M1
M1A1
[9]
(b)
Any two from
→ 1 1
CP = a − b
3
9
→ 3 →
CQ =
CP
OR
2
with a conclusion
→ 1
1
PQ = a − b
6
18
→
→
PQ = 2 CP
OR
→ 1 1
CQ = a − b
2
6
→
→
CQ = 3 PQ
B2
M1
A1
[4}
Total 13 marks
Part
(a)
Mark Notes
→
→
2
AC = −a + b
3
B1
For at least one from AB = −a + b
B1
All three correct.
M1
For a correct vector statement for NB
→
5
AM = −a + b
6
→
→
There are several paths to OP . Those below are examples. For the award of each of
the two M marks, there must be two different paths using two distinct parameters.
→
For either vector statement for OP using a parameter
M1
→ →
→
5  
5 

e.g. OP = OA +  AM = a +   −a + b  = a (1 −  ) + b 
6  
6 

→
M1
M1
For a second vector statements for OP using a different parameter
→ →
→
 a 2b     
 2  
e.g. OP = OB −  NB = b −   − +
 = a   + b  1 −

3  
 2 3   2

For equating coefficients of a and b
For solving the simultaneous equations
This mark is dependent on the previous 3 M marks
dM1
M1
2

 = 3
OR
2
= 
3
→
→
→
→
→
or OP = OA +  AM
A1
→
For substituting their value of  or  into their OP = OB −  NB
→ 1
5
9
For the correct vector OP = a + b
3
→
→
SC – If CQ is assumed to be a straight line and is used to find used OP , award all the
(b)
marks as above but withhold the final A mark.
→ 1 1
→ 1
→ 1 1
1
For any two vectors from CP = a − b or PQ = a − b or CQ = a − b
B2
3
9
6
18
2
6
(B1 for one of the above)
M1
A1cso
→ 3 →
For CQ =
2
→
→
→
→
CP oe or PQ = 2 CP oe or CQ = 3 PQ
For a correct conclusion with supporting working
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