Practice Paper – Set 1

A Level Further Mathematics A

Y540/01 Pure Core 1

MARK SCHEME

Duration: 1 hour 30 minutes

MAXIMUM MARK 75

DRAFT

This document consists of 15 pages

Y540/01 Mark Scheme Practice Paper (March 2018)

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
SC Special case
^ Omission sign
MR Misread
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 *
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 question included the instruction: In this question you must show detailed reasoning.

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Y540/01 Mark Scheme Practice Paper (March 2018)

2. Subject-specific Marking Instructions for A Level Further Mathematics A

a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for
responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded.
For subsequent marking you must make it clear how you have arrived at the mark you have awarded.

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 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.

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
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
Mark for explaining a result or establishing a given result. 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.
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Y540/01 Mark Scheme Practice Paper (March 2018)

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.

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. This rule should be applied to each case. When a value is not given in the paper accept any answer that agrees with the correct
value to 2 s.f. Follow through should be used so that only one mark is lost for each distinct accuracy error, except for errors due to premature
approximation which should be penalised only once in the examination. There is no penalty for using a wrong value for g. E marks will be lost except when
results agree to the accuracy required in the question.

g Rules for replaced work: if a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners
should do as the candidate requests; if there are two or more attempts at a question which have not been crossed out, examiners should mark what
appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.

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 mark in the question. Marks designated as cao may be awarded as long as there are no other errors. E marks are
lost unless, by chance, the given results are established by equivalent working. ‘Fresh starts’ will not affect an earlier decision about a misread. 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, of course, that there
is nothing in the wording of the question specifying that analytical methods are required). 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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Y540/01 Mark Scheme Practice Paper (March 2018)

Y540 Practice paper 1

Question Answer Marks AO Guidance

1 DR
Let 24 + 10i =a + bi where a and b are real numbers. M1 3.1a

Then 24 + 10i = ( a + bi ) = a 2 − b 2 + 2abi

2
A1 1.1
⇒=a − b 24 and
2 2
= 2ab 10
M1 2.2a
5 25
⇒ b = ⇒ a 2 − 2 = 24
a a
⇒ a − 24a − 25 =
4 2
0 A1 1.1
⇒ ( a 2 + 1)( a 2 − 25 ) =0⇒a =±5, b =±1
⇒ ( 5 + i ) and − ( 5 + i ) A1 2.2a

OR
 θ
Square root of (r , θ ) is  r ,  M1
 2
A1
1 10
=
where r 26, = θ tan −= 22.62°
24 M1
⇒± ( ) (
26,11.31 = ± 26 cos11.31 + 26 sin11.31i ) A1

=± (5 + i) A1

[5]

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Y540/01 Mark Scheme Practice Paper (March 2018)

Question Answer Marks AO Guidance

2 (i) 1 a  4 2  4 + 3a 2 + 3a  M1 2.1
= AB =    
3 0 3 3  12 6 
 4 21 a  10 4a  A1 1.1
= BA =    
 3 33 0 12 3a 

Equate any element give a = 2 A1 1.1

[3]
2 (ii) Choose any two matrices that are not commutative M1 2.1 Choose
A1 2.1 Demonstrate BC
[2]
2 (iii) Det B = 6 gives area = 24 B1 1.1
[1]
2 (iv)  4 2  1   λ  M1 2.1
   =  
 3 3  m   λ m  A1 1.1
⇒ 4 + 2m= λ , 3 + 3m= λ m
⇒ 3 + 3m = ( 4 + 2m ) m
⇒ 2m 2 + m − 3 = 0 ⇒ ( m − 1)( 2m + 3) = 0 A1 1.1
3
⇒m=
1 and m =−
2
i.e. 2 y += =
3 x 0 and y x
A1 2.2a

[4]

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Y540/01 Mark Scheme Practice Paper (March 2018)

Question Answer Marks AO Guidance

3 True for k = 1, since 15-1= 0 which is divisible by 5 B1 2.1
=
Assume true for n k , then k 5 − k is divisble by 5 M1 2.1
Then ( k + 1) − ( k + 1)
5

= k 5 + 5k 4 + 10k 3 + 10k 2 + 5k + 1 − k − 1
= ( 5k 4
+ 10k 3 + 10k 2 + 5k ) + k 5 − k
= 5(k 4 +2k 3 + 2k 2 + k )+k 5 − k A1 2.5
So if divisible by 5 for n= k then also for n= k + 1
True for n = 1, so true generally A1 2.2a
[4]

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Y540/01 Mark Scheme Practice Paper (March 2018)

Question Answer Marks AO Guidance

4 (i) Find normal by vector product M1 1.1a
1  2   7 
      A1 1.1
n =  2  ×  −1 =  −1 
 1   3   −5 
     
Taking b = (7, 1, 6) and a = (2, 6, −2)
b − a = (5, −5, 8) M1 2.1
Using distance formula:
57
( b − a ) . n = −5  .  −1  =0
A1 1.1

 8   −5 
  
So distance = 0 so they intersect.
M1
A1
Alternative:
Any point on l1 is ( 7 + 2λ ,1 − λ , 6 + 3λ ) M1
Any point on l2 is ( 2 + µ , 6 + 2 µ , −2 + µ ) A1
Equate any two: ⇒ λ =−3, µ =−1
and the point is (1, 4, −3)
[4]
4 (ii) Use of normal vector B1 1.1
7 x − y − 5z =c
Through a or b M1 1.1
⇒c= 18
A1 1.1
⇒ 7 x − y − 5z =
18
[3]

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Y540/01 Mark Scheme Practice Paper (March 2018)

Question Answer Marks AO Guidance

5 5
2
5
2
1 1
= I ∫3 4 x 2 − 12 x + 13 dx
= ∫
3 (2 x − 3) 2 + 4
dx M1 2.1 Completing the square
2 2

Substitute 2 x − 3 =2sinh θ M1 3.1a Substitution

=⇒ dx cosh θ dθ , (2
= x − 3) 2 + 4 4 cosh 2 θ
3 5
=
When x = , θ 0. When
= x =
, sinh θ 1 A1 1.1 Correct reduction (ignore limits)
2 2
sinh −1 1
1 1 sinh −1 1
=⇒I ∫0 2 cosh θ
= cosh θ d θ [θ ] A1 2.2a Correct integral (ignore limits)
2 0 M1 1.1 Limits
=
1
2
=
sinh −1
1
1
2
ln 1 + 2 ( ) A1 1.1

OR
5
2
5
2 u= x − 32
1 1
3
∫ =
4( x − 32 ) 2 − 9 + 13
dx 1
2∫3 ( x − 3 )2 + 1 dx
= du = dx
2
2 2
=x 5
2
=,u 1
1
1 ln(u + u 2 + 1 =
∫ dx= ln(1 + 2) = =
1 1 1 3
x ,u 0
2
0 u +1
2 2
  2 2

[6]

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Y540/01 Mark Scheme Practice Paper (March 2018)

Question Answer Marks AO Guidance

6 (i) =θ A cos 2t + B sin 2t B1 1.2
[1]
6 (ii) =θ A cos 2t + B sin 2t M1 3.3
dθ
⇒ =
−2 A sin 2t + 2 B cos 2t A1 3.4
dt
dθ
=0 when t =0 ⇒ B =0 A1 1.1
dt
θ =θ 0 when t =0 ⇒ A =θ 0
⇒θ =
θ 0 cos 2t A1 3.4
[4]
6 (iii) Angle has to be small otherwise the model is invalid. B1 3.5b
[1]

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Y540/01 Mark Scheme Practice Paper (March 2018)

Question Answer Marks AO Guidance

( e − e ) , sinh 3 x = (e − e )
7 (i) 1 x −x 1 3 x −3 x M1 1.1a B1 for one error
sinh x =
2 2
A1 1.1
sinh 3 x = ( e x − e − x ) = ( e3 x − 3e x + 3e − x − e −3 x )
1 3 1
8 8
1e −e3x −3 x
3e − 3e − x  1
x
3 A1 2.1
= −  =sinh 3 x − sinh x
4 2 2  4 4
AG
⇒ 4sinh 3 x =sinh 3 x − 3sinh x
[3]
7 (ii) Substitute u = sinhx M1 3.1a
=
4sinh 3
x sinh 3x − 3sinh x
A1 1.1
⇒ 16sinh 3 x + 12sinh x =
4sinh 3x
3 A1 1.1
u= sinh x ⇒ 4sinh 3x = 3 ⇒ sinh 3x =
4
3 9 
⇒ 3 x= ln  + 1 + = ln 2 M1 1.1
4 16 
 13 − 
1 1
ln 2
= 23
1
Use of e 3
 2 − 2 3

1 1   
=
x ln 2 ⇒=
u sinh  ln 2=

A1 3.2a
3 3  2
[5]

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Y540/01 Mark Scheme Practice Paper (March 2018)

Question Answer Marks Guidance

8 (i)   x3    b2 x2 
f ( x) =(1 + a sin x ) ebx =1 + a  x −   1 + bx +  B1 1.2 Sin expansion
  6  2 
B1 1.2 Exponential expansion
b2 x2 2b 
2
=1 + bx + + ax + abx =1 + (a + b) x + x  + ab 
2
M1 1.1a Multiply out and compare
2  2  coefficients
3 2
≈ 1+ 2x + x
2
b2 3
⇒a=
+ b 2, += ab
2 2
A1 1.1 Both equations
⇒ b + 2b(2 − b) = 3 ⇒ b 2 − 4b + 3 = 0
2
M1 1.1 Solve
⇒ b = 1,3 ⇒ a = 1, − 1
⇒ (−1,3) A1 1.1 As a<0
[6]
8 (ii) There is no restriction on x because there is no restriction on the B1 2.3
maclaurin's expansion for sinx and ex.
[1]

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Y540/01 Mark Scheme Practice Paper (March 2018)

Question Answer Marks Guidance

−k (1 + Q 2 ) ⇒ ∫
9 dQ dQ B1 3.3
= −k ∫ dt
= 3.1b
dt 1 + Q2 M1
⇒ tan −1 Q =c − kt A1 1.1

t= 0, Q= 100 ⇒ c= tan −1 100 A1 3.4

t= 50 ⇒ tan −1 50 =
100, Q = tan −1 100 − 100k A1 3.4

=⇒k
1
100
( tan −1 100 − tan −1 50 ) ≈ 0.0001 M1 1.1a

When t = 400, Q = 20 A1 3.4

20 g A1 3.2b Must have unit

[8]

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Y540/01 Mark Scheme Practice Paper (March 2018)

Question Answer Marks AO Guidance

10 (i) (a) r B1 1.1
sec θ =
x
r
⇒ r = 2− B1 3.1a
x
2x B1 3.1a
⇒r=
x +1
2
 2x 
y =r −x ⇒ y =
2 2 2
 −x
2 2
A1 2.1 AG
 x +1
[4]
2
10 (i) (b) Because the curve is y = ..... and so x = p gives y = ±q B1 2.4
[1]
10 (i) (c) a=−1 B1 2.2a Accept x=− 1
[1]
10 (ii) V = π ∫ y 2 dx M1 1.1a Use of formula

  2 x 2    2x + 2  4  M1 3.1a any equivalent form that can be

= π ∫ − 2
 = π ∫
 − 2
− + dx
  x + 1   2 
x d x 4 x 4 integrated
  x + 2 x + 1  ( + )
2

   x 1  A1 1.1

 x3 4  M1 3.1a Attempt to integrate

π 4 x − − 4 ln( x + 1) 2 −
( x + 1) 
A1 1.1
 3
Limits are 0 and 1 B1 3.1a
 1   M1 1.1
⇒=V π   4 − − 4 ln 4 − 2  − ( 0 − 0 − 0 − 4 ) 
 3  
 17  A1 1.1 4ln4 and 8ln2 are equivalent
= π  − 4 ln 4 
 3 
[8]

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Y540/01 Mark Scheme Practice Paper (March 2018)

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