GCE

Further Mathematics B (MEI)

Y420/01: Core Pure

Advanced GCE

Mark Scheme for Autumn 2021

Oxford Cambridge and RSA Examinations

OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of
qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications
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needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is
invested back into the establishment to help towards the development of qualifications and
support, which keep pace with the changing needs of today’s society.

This mark scheme is published as an aid to teachers and students, to indicate the requirements
of the examination. It shows the basis on which marks were awarded by examiners. It does not
indicate the details of the discussions which took place at an examiners’ meeting before marking
commenced.

All examiners are instructed that alternative correct answers and unexpected approaches in
candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills
demonstrated.

Mark schemes should be read in conjunction with the published question papers and the report
on the examination.

© OCR 2021

Oxford Cambridge and RSA Examinations

Y420/01 Mark Scheme October 2021

Text Instructions

1. Annotations and abbreviations

Annotation in scoris Meaning

✓and 
BOD Benefit of doubt
FT Follow through
ISW Ignore subsequent working
M0, M1 Method mark awarded 0, 1
A0, A1 Accuracy mark awarded 0, 1
B0, B1 Independent mark awarded 0, 1
E Explanation mark 1
SC Special case
^ Omission sign
MR Misread
BP Blank page
Highlighting

Other abbreviations in Meaning

mark scheme
E1 Mark for explaining a result or establishing a given result
dep* Mark dependent on a previous mark, indicated by *. The * may be omitted if only previous M mark.
cao Correct answer only
oe Or equivalent
rot Rounded or truncated
soi Seen or implied
www Without wrong working
AG Answer given
awrt Anything which rounds to
BC By Calculator
DR This indicates that the instruction In this question you must show detailed reasoning appears in the question.

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Y420/01 Mark Scheme October 2021

2. Subject-specific Marking Instructions for AS/A Level Further Mathematics B (MEI)

a a Annotations must be used during your marking. For a response awarded zero (or full) marks a single appropriate annotation (cross, tick, M0 or ^)
is sufficient, but not required.

For responses that are not awarded either 0 or full marks, you must make it clear how you have arrived at the mark you have awarded and all responses
must have enough annotation for a reviewer to decide if the mark awarded is correct without having to mark it independently.

It is vital that you annotate standardisation scripts fully to show how the marks have been awarded.

Award NR (No Response)

- if there is nothing written at all in the answer space and no attempt elsewhere in the script
- OR if there is a comment which does not in any way relate to the question (e.g. ‘can’t do’, ‘don’t know’)
- OR if there is a mark (e.g. a dash, a question mark, a picture) which isn’t an attempt at the question.
Note: Award 0 marks only for an attempt that earns no credit (including copying out the question).

If a candidate uses the answer space for one question to answer another, for example using the space for 8(b) to answer 8(a), then give benefit of doubt
unless it is ambiguous for which part it is intended.

b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking
incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and
answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar
must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect
method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, escalate the question
to your Team Leader who will decide on a course of action with the Principal Examiner.
If you are in any doubt whatsoever you should contact your Team Leader.

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Y420/01 Mark Scheme October 2021

c The following types of marks are available.

M
A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost
for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method
or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula.
In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words “Determine” or “Show
that”, or some other indication that the method must be given explicitly.

A
Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark
is earned (or implied). Therefore M0 A1 cannot ever be awarded.

B
Mark for a correct result or statement independent of Method marks.

E
A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown
result.

Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes
this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct
answer as part of a wrong argument.

d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and
similarly where there are several B marks allocated. (The notation ‘dep*’ is used to indicate that a particular mark is dependent on an earlier, asterisked,
mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is
worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the
earlier marks are implied and full credit must be given.

e The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B
marks are given for correct work only – differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from
incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally
acceptable. In such cases, what is acceptable will be detailed in the mark scheme. If this is not the case, please escalate the question to your Team Leader
who will decide on a course of action with the Principal Examiner.
Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such
cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it
easier to mark follow through questions candidate-by-candidate rather than question-by-question.

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Y420/01 Mark Scheme October 2021

f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in
SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be
assumed to be the unspecified unit.)
We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme
explicitly says so.
• When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
• When a value is not given in the paper accept any answer that agrees with the correct value to 2 s.f. unless a different level of accuracy has
been asked for in the question, or the mark scheme specifies an acceptable range.
NB for Specification A the rubric specifies 3 s.f. as standard, so this statement reads “3 s.f”
Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of 9.80, 9.81 or 10 for g should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8
which is given in the rubric.

g Rules for replaced work and multiple attempts:

• If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
• If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less
complete.
• if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks
appropriately.

h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the
scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units.
This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors. If a
candidate corrects the misread in a later part, do not continue to follow through. E marks are lost unless, by chance, the given results are established by
equivalent working. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.

i If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers provided that there is nothing in
the wording of the question specifying that analytical methods are required such as the bold “In this question you must show detailed reasoning”, or the
command words “Show” and “Determine. Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong
answers with no supporting method score zero. If in doubt, consult your Team Leader.

j If in any case the scheme operates with considerable unfairness consult your Team Leader.

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

1 (a) 1 A B
= +
(2r − 1)(2r + 1) 2r − 1 2r + 1
 1 = A(2r + 1) + B(2r − 1) M1 1.1a or cover-up method
r = 12  A = 12 A1 1.1
r = − 12  B = − 12 A1 1.1
[3]
1 (b)
( )
n
1 1 n 1 1
 (2r − 1)(2r + 1) 2  2r − 1 − 2r + 1
= M1* 1.1 SC
r =1 r =1 B3 For fully correct
=
1
2 (
1 1 1
1 − + − + ... +
3 3 5
1
−
1
+
1
−
1
2n − 3 2n −1 2n −1 2n + 1 ) M1de
p*
1.2 showing cancellation clearly summation with A = -
½ and B = ½
1
= 1−
2 (
1
2n + 1 ) A1
1.1

n 1.1
=
2n + 1 A1
[4]

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

2 DR
dy 1 B1 1.1 𝑘
= 6 2 √1−4𝑥 2
dx 1 − 4 x2 Chain rule (x 2)
M1 1.1
𝑑𝑦
A1 1.1 Fully correct
dy 1 𝑑𝑥
when x = 14 , = 6 2 =8 3 A1 1.1
dx 3 [4]
4
Alternative solution
DR
𝑦 𝑑𝑥 𝜋
sin
= 2𝑥 For = 𝑘 cos
6 M1 𝑑𝑦 6
1
𝑑𝑥 1 𝑦 M1 Using chain rule (× )
= cos 6
𝑑𝑦 12 6
𝑑𝑦 12 1
= 𝑦 and 𝑥 = , 𝑦 = 𝜋
𝑑𝑥 cos 4
6 A1 dy 1
Or = 6 2
𝑑𝑦 dx 1 − 4 x2
→ = 8√3
𝑑𝑥 A1

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

3 (a) DR
z1 = 8 B1 1.1
3
arg( z1 ) = E1 1.1 Must see some reasoning for arg(z)
4
[2]
3 (b) DR
z1 8
= = 2 B1 1.1
z2 2

arg ( )
z1 3  7
z2
= − =
4 6 12
M1
A1
1.1
1.1
arg ( zz ) = arg( z ) − arg( z ) used
1
2
1 2

z
so 1 = 2 cos
z2
7
12(+ isin
7
12 ) B1
[4]
1.1

4 DR
1
1  1
mean =  dx B1 1.1
1 − (−1) −1 1 + 4 x2
1 2 1 1 M1 for rearranging
or u = 2 x  
1
1 1 M1 1.1  du
=  dx 2 −2 1 + u 2 2 denominator correctly
8 −1 4 + x 2
1
into appropriate form
1 1 1 2 or for karctan2x
=  2arctan 2 x  A1 1.1 = arctan u 
8 −1 −2
4
1 A1 1.1
= arctan 2  0.554
2 [4]

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

5 (a) 1
ln(1 + 2 x)  2 x − (2 x) 2 = 2 x − 2 x 2 B1 1.1
2 [1]
5 (b) ln(1.2)  0.18 B1ft 1.1
0.18 − ln(1.2) M1 1.1
% error = 100 
ln(1.2)
= (−)1.27% A1 1.1
[3]
5 (c)
−1  2 x  1  − 12  x  12 M1 1.1 substitute x = 1 in quadratic
the Maclaurin series is not convergent for this A1 2.3
approximation when x = 1 [2]

6
(12 −22) ( xy ) = ( 2xx +− 22 yy ) M1 1.1

substituting y = mx and 2 x − 2 y = m( x + 2 y ) M1 2.1 Could see mx instead of y in the initial

2 x − 2mx = m( x + 2mx) A1 1.1 matrix multiplication for this mark
 2m2 + 3m − 2 = 0
 m = −2, 12 A1 2.2a
[4]

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

7 1
r 1+ 2
 2r −1 = 1 = 4 − 20
so true for n = 1 B1 2.1
r =1
k
r k +2
Assume true for n = k so  2r −1 = 4 − 2k −1 M1 2.1
r =1
k +1
r k +2 k +1
 2r −1 = 4 − 2k −1 + 2k
M1 2.1
r =1
2k + 4 − k − 1 M1 1.1
= 4−
2k
k +1+ 2 A1 2.2a
= 4− so true for n = k + 1
2k
So true for n = 1 and if true for n = k then true for
n=k+1
 true for all n A1 2.2a Dependent on all previous marks
[6] awarded

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

8 (a) 1 9
  (− )    =− M1 2.1
 4
3
𝛼 = (±) A1 2.2a
2
1
Sum of roots =  + =1 M1 2.1

  2 −  +1 = 0
1 i 3 1 1 i 3 A1 2.1
 = , =
2  2
3 1 i 3 A1 2.2a
so roots are  and
2 2
Alternative solution

(
( x −  )( x +  )( x −  ) x −
1
 ) M1

(
= x4 +  +
1
 ) (
x3 + (1 −  2 ) x 2 −  2  +)1

x − 2 A1

9 3 A1
2 =  = 
4 2
1 M1
 + = 1  2 −  +1 = 0

1 i 3 1 1 i 3
 = , =
2  2
3 1 i 3 A1
so roots are  and
2 2
[5]

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

8 (b)  
Sum of pairs =  (− ) +  + −  − + 1
  M1 2.1 Unsimplified
5
= 1 −  2 = −  p = −5 A1 2.2a
4

(
Sum of triples = − 2  +
1
 ) 9
= − 1 = −
4
9
4
M1 2.1 simplified

so q = 9 A1 2.2a
Alternative solution

( )( ) (
x−
3
2
x+
3
2
x−
2 )(
1+ i 3
x−
2 )
1− i 3 M1

( )
9
= x 2 − ( x 2 − x + 1)
4
A1

5 9 9 Award for either 𝑥 2 𝑜𝑟 𝑥 coefficient

= x 4 − x3 − x 2 + x − A1
4 4 4 correct
so p = −5 and q = 9 A1
[4]

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

9 (a) SC B1 For unlabelled
M1 1.1 ( −−21 10)(00 1 1 0
0 1 1
= )(
0 −1 −1 0
0 −2 −1 1 ) diagram with no
working

A′ (−1, −2), B′ (−1, −1),

A1 1.1 C′ (0, 1) plotted correctly

[2]

9 (b) (i) det M = −11 − 0  (−2) = −1 B1 1.1

[1]
9 (b) (ii) area is preserved B1 1.1
orientation is reversed B1 1.1
[2]
9 (c) (i) Reflection in y axis B1 3.1a
then shear M1 3.1a
invariant line y-axis, mapping (−1, 0) to (−1, −2) A1 1.1 oe, e.g. mapping (1, 0) to (1, 2)
Alternatively
Shear M1
invariant line y-axis, mapping (1, 0) to (1, −2) A1
then reflection in y-axis B1
[3]
9 (c) (ii)
Reflection: ( −01 10) B1 1.1 Or shear ( −12 10) first
Shear: (
2 1)
1 0 B1 1.1  −1 0 
then reflection  
 0 1

(12 10)(−01 10) = (−−21 10) B1 2.2a ( )(

−1 0 1 0
0 1 −2 1
= )(
−1 0
−2 1 ) Must match the order
described in ci for final
[3] mark

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

10 (a)

B1 1.1 z=1
B1 1.1 other two roots forming correct
1
equilateral triangle
[2]

10 (b) (i) 1+ 3 i = 2 B1 1.1


arg (1 + 3 i ) = B1 1.1
3
i
z = 3 2e 9 B1ft 2.5 if roots correct but in cis form,
7i withhold one mark only
3 2e 9 B1ft 1.1
− 5i9
3 2e B1ft 1.1
[5]
10 (b) (ii) Rotated through 20 (oe) B1 3.1a
Enlarged by 3 2 B1 3.1a
[2]
10 (b) (iii)
(1 − (e ) )
−5i 2i 3
3 2e 9 3
−5i i 7i
3 2e 9 + 3 2e 9 + 3 2e 9 = M1 3.1a sum of GP formula May use cos 𝜃 +
2i
1− e 3 𝑖 sin 𝜃 but must take
−5i out factor for M1
3 2e 9 (1 − e2i ) A1 2.2a
= 2i =0
1− e 3 [2]
10 (b) (iv) Imaginary part of sum of roots is zero

 sin + sin
9
7
9
+ sin −
5
9
=0 ( ) M1 2.1

 sin 20 + sin140 = − sin(−100) = sin100 A1 2.2a AG

[2]

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

11 (a) (i) u .v =  1 + 1 2 + (−3)  (−2)
=  +8 B1 1.1
[1]
11 (a) (ii)  1  4 
u  v =  1    2  =  2 − 3  oe i, j, k form B1 1.1a For one of i,j,k correct
      B1 1.1 For all i,j,k correct
 −3   −2   2 − 1 
[2]
11 (b) (i) angle between 2i + j − 3k and i + 2 j − 2k M1 3.1a
taking  = 2,
10
cos = M1 1.1
14 9
 = 27.0 or 0.472 rad A1 1.1
[3]
11 (b) (ii) direction vectors are 3i + j − 3k and i + 2 j − 2k , so take
 =3 M1 3.1a

with  = 3, u  v = 4i + 3j + 5k M1
A1
3−0 
1   
4
20 [3] 1.1
distance = 3 . 0−4  = =2 2 1.1
50     5 2
 5   2 − (−2) 

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

12

Let DOA =  , COA = 

z + w is represented by C
arg( z + w) =  +  B1 3.1a
BOC =  (diagonal of rhombus bisects BOA) M1 2.1
arg z + arg w =  + ( + 2 ) M1 2.1 finds arg z + arg w in terms of , 
= 2( +  )
so arg( z + w) = 12 (arg z + arg w) A1 2.2a AG
[4]

13 d2y dy
2
+ 2 − 3 y = 2e x
dx dx
AE:  2 + 2 − 3 = 0   = −3, 1 M1 2.1

CF: y = Ae −3 x + Be x A1 2.1
PI: y = Cxe x M1 2.1
dy
= C (e x + xe x ) A1 1.1
dx
d2y
= C (2e x + xe x )
dx 2 A1 1.1
 C (2e x + xe x ) + 2C (e x + xe x ) − 3Cxe x = 2e x
 4C = 2  C = 12 M1 2.1

GS: y = A e−3 x + B e x + 12 x e x
A1 2.2a
[7]

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

14 (a) Let cos + 2sin  = R cos( −  )
= R cos cos + R sin sin 
R cos  = 1, R sin  = 2
3.1a
M1A1
R= 5 M1
tan  = 2 A1 1.1
 = 1.107 1.1
or arctan(2)
1.1
so r = 5a cos( − 1.107) M1
r is maximum when cos( − 1.107) = 1
  = 1.107 rad 3.1a
A1 or arctan(2)
polar coordinates are  5a, 1.107  A1
1.1
1.1
Alternative solution
dr
= a(− sin  + 2cos  ) M1A1
d
dr M1
r is maximum when =0
d
 −2cos + sin = 0
 tan = 2   = 1.107 M1
when  = 1.107, r = 5a or 2.236a A1B1
polar coordinates are [2.236a, 1.107] B1 2.24 or better
[7]
14 (b) (i) r 2 = ar cos + 2ar sin  M1 3.1a attempt to find cartesian eqn
M1 3.1a multiplying by r
 x 2 + y 2 = ax + 2ay A1 1.1
 ( x − 12 a)2 + ( y − a)2 = 54 a 2
M1 2.1 completing the square
This is the cartesian equation of a circle A1 2.2a
radius 12 5a A1 3.2a
[6]

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

14 (b) (ii) centre  12 5a, 1.107  B1 1.1
[1]

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Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

15 −4 k 7
1 −2 5 = −4  ( −17) − k  ( −9) + 7  7 M1 3.1a finding det of matrix of coeffs
2 3 1
or B2 for use of linear
= 117 + 9k M1 1.1 setting det = 0 dependency to find k
so det = 0 when k = −13 A1 1.1 k = −13
−4 x − 13 y + 7 z = 4 (1)
x − 2 y + 5z = l (2)
2 x + 3 y + z = 2 (3) or
(1) + 2  (3) : −7 y + 9 z = 8 M1 1.1 finding eqn in 2 variables
M2 for 2(2) − 3(3)
(3) − 2  (2) : 7 y − 9 z = 2 − 2l M1 1.1 finding 2nd eqn in 2 variables
so 2l − 2 = 8  l = 5 A1 1.1 l=5
[6]

19
Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

Alternative method

16 (a) e2u + e−2u

LHS = B1 2.1
2

( )
2
eu − e−u B1 2.1
R HS = 1 + 2
2

(
e − 2 + e−2u
) B1 2.1
2u
= 1+ 2
4

(
e − 2 + e−2u e2u + e−2u
)
2u
= 1+ = = LHS B1 2.2a
2 2 [4]
16 (b) dx
Let x = 2sinh u  = 2cosh u M1 3.1a
du
2 arsinh1
 x2  4sinh 2 u A1 1.1
 dx =  2cosh u du
0 4 + x2 0 4 + 4sinh u 2

arsinh1
 4sinh 2 u
= 2cosh u du M1 2.1 1 + sinh 2 u = cosh 2 u used
0 2cosh u
arsinh1 A1 2.1
= 4sinh 2 u du
0
arsinh1
= (2cosh 2u − 2) du M1 2.1 cosh 2u = 1 + 2sinh 2 u used
0

= sinh 2u − 2u 0
arsinh1
A1 2.1
soi
arsinh1 = ln (1 + 2 ) B1 2.1

(
 12 e2ln(1+ 2)
− e−2ln(1+ 2)
) − 2ln (1 + 2) M1 2.1
−2
or sinh 2u = 2sinh u cosh u M1
= 12 ( eln(1+ 2) 2
) − ( eln(1+ 2 ) )  − 2ln (1 + 2 )
= 2sinh u 1 + sinh 2 u M1
 
= 12 (1 + 2 ) − (1 + 2 )  − 2ln (1 + 2 )
2 −2
M1 2.1 =2 2

20
Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

−1 1 1− 2
(1 + 2) = = = 2 −1
1 + 2 (1 + 2 )(1 − 2 )

giving = 12 ( 3 + 2 2 ) − ( 3 − 2 2 ) − 2ln (1 + 2 )
= 2 2 − 2ln (1 + 2 ) A1 2.2a AG
Alternative for last 6 marks
sinh u = 12 (eu − e−u ) used
arsinh1
= (eu − e−u )2 du M1
0
arsinh1
= (e2u − 2 + e−2u ) du
0
arsinh1
=  12 e2u − 2u − 12 e−2u  A1
0
arsinh1 = ln (1 + 2 ) B1
e2ln(1+ 2)
= (1 + 2 )
2
M1
−2ln (1+ 2 ) −2
= (1 + 2 ) = ( 2 − 1)
2
e M1
3+ 2 2 3− 2 2
giving − − 2ln (1 + 2 )
2 2
= 2 2 − 2ln (1 + 2 ) A1 AG
[10]

21
Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

17 (a) (i) at litres of chemical in, bt litres of mixture out M1 3.1b
amount of liquid in container = 1 + (a − b)t litres
x
 proportion of chemical = A1 2.4 AG
1 + (a − b)t
[2]
17 (a) (ii) rate of chemical in = a litres/hr
bx
rate of chemical out = litres/hr
1 + (a − b)t
dx bx
 =a− M1 3.3
dt 1 + (a − b)t
dx bx
 + =a A1 3.3 AG
dt 1 + (a − b)t
[2]
17 (b) (i) dx
+ ax = a
dt

1
 dx =  a dt M1 1.1 separating variables or IF eat
 1− x
 − ln(1 − x) = at + c A1 1.1 x eat = eat + c
when t = 0, x = 0  c = 0 B1 3.3 c = −1
 1 − x = e−at
x = 1 − e−at A1 3.4
[4]
17 (b) (ii) 1 = 1 − e− a
2 M1 3.3
a = ln 2= 0.693
rate of inflow = 0.693 l/min A1 3.4
[2]

22
Y420/01 Mark Scheme October 2021

Question Answer Marks AOs Guidance

17 (c) (i) 1 1
when t = the container has no liquid left B1 3.5b oe e.g. volume negative when t 
a [1] a
17 (c) (ii) dx 2ax
+ =a B1 1.1
dt 1 − at
IF = e 1−at
2a dt
M1 3.1a
= e −2ln(1− at ) = (1 − at ) −2 A1 1.1

( x(1 − at )−2 ) = a(1 − at )−2

d
 M1 1.1
dt
 x(1 − at )−2 = (1 − at ) −1 + c A1 1.1
when t = 0, x = 0  c = −1 M1 3.4
 x = (1 − at ) − (1 − at ) 2 = at (1 − at ) A1 1.1
dx 1
= a − 2a 2t = 0 when t = M1 3.4 or completing the square
dt 2a
x= 2−4 = 4
1 1 1

so maximum amount of chemical is 0.25 l A1 3.2a

[9]

23
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