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General Certificate of Education
Mathematics 6360
MFP2 Further Pure 2
Mark Scheme
2006 examination - January series
Mark schemes are prepared by the Principal Examiner and considered, together with the relevant
questions, by a panel of subject teachers. This mark scheme includes any amendments made at
the standardisation meeting attended by all examiners and is the scheme which was used by them
in this examination. The standardisation meeting ensures that the mark scheme covers the
candidates’ responses to questions and that every examiner understands and applies it in the
same correct way. As preparation for the standardisation meeting each examiner analyses a
number of candidates’ scripts: alternative answers not already covered by the mark scheme are
discussed at the meeting and legislated for. If, after this meeting, examiners encounter unusual
answers which have not been discussed at the meeting they are required to refer these to the
Principal Examiner.
It must be stressed that a mark scheme is a working document, in many cases further developed
and expanded on the basis of candidates’ reactions to a particular paper. Assumptions about
future mark schemes on the basis of one year’s document should be avoided; whilst the guiding
principles of assessment remain constant, details will change, depending on the content of a
particular examination paper.
Copyright © 2006 AQA and its licensors. All rights reserved.
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MFP2 – AQA GCE Mark Scheme, 2006 January series
Key To Mark Scheme And Abbreviations Used In Marking
M
mark is for method
m or dM
A
B
E
mark is dependent on one or more M marks and is for method
mark is dependent on M or m marks and is for accuracy
mark is independent of M or m marks and is for method and accuracy
mark is for explanation
or ft or F
CAO
CSO
AWFW
AWRT
ACF
AG
SC
OE
A2,1
–x EE
NMS
PI
SCA
follow through from previous
incorrect result
correct answer only
correct solution only
anything which falls within
anything which rounds to
any correct form
answer given
special case
or equivalent
2 or 1 (or 0) accuracy marks
deduct x marks for each error
no method shown
possibly implied
substantially correct approach
MC
MR
RA
FW
ISW
FIW
BOD
WR
FB
NOS
G
c
sf
dp
mis-copy
mis-read
required accuracy
further work
ignore subsequent work
from incorrect work
given benefit of doubt
work replaced by candidate
formulae book
not on scheme
graph
candidate
significant figure(s)
decimal place(s)
No Method Shown
Where the question specifically requires a particular method to be used, we must usually see evidence of
use of this method for any marks to be awarded. However, there are situations in some units where part
marks would be appropriate, particularly when similar techniques are involved. Your Principal Examiner
will alert you to these and details will be provided on the mark scheme.
Where the answer can be reasonably obtained without showing working and it is very unlikely that the
correct answer can be obtained by using an incorrect method, we must award full marks. However, the
obvious penalty to candidates showing no working is that incorrect answers, however close, earn no
marks.
Where a question asks the candidate to state or write down a result, no method need be shown for full
marks.
Where the permitted calculator has functions which reasonably allow the solution of the question directly,
the correct answer without working earns full marks, unless it is given to less than the degree of accuracy
accepted in the mark scheme, when it gains no marks.
Otherwise we require evidence of a correct method for any marks to be awarded.
2
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AQA GCE Mark Scheme, 2006 January series – MFP2
MFP2
Q
Solution
1(a)
1
r
=
(b)
−
2
2
1
( r + 1)
( r + 1) − r
2
r 2 ( r + 1)
=
2
Marks
r
A1
( r + 1)2
3
2
1 ×2
2
5
22 × 32
7
2
3 ×4
2
1
=
2
1
=
22
1
3
2n + 1
n 2 ( n + 1)
2
2
=
−
1
32
1
M1A1
42
1
2
−
( n + 1)2
M1
1
A1F
( n + 1)2
Total
2(a)
p = −4
(α + β + γ )
4
6
B1
2
= ∑ α 2 + 2 ∑ αβ
M1
16 = 20 + 2 ∑ αβ
∑ αβ = −2
q = −2
(b)
A1 for at least 3 lines
1
Clear cancellation
1−
AG
22
−
n
2
1
−
1
=
Comments
M1
2r + 1
2
Total
2
A1
A1F
A1F
3 − i is a root
B1
Third root is −2
B1F
αβγ = ( 3 + i )( 3 − i )( −2 )
M1
= −20
r = +20
A1F
A1F
Alternative to (b)
Substitute 3 + i into equation
5
Real αβγ
5
Real r
M1
B1
2
( 3 + i ) = 8 + 6i
3
( 3 + i ) = 18 + 26i
B1
r = 20
A2,1,0
Total
Provided r is real
10
3
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MFP2 – AQA GCE Mark Scheme, 2006 January series
MFP2 (cont)
Q
Solution
Marks
Total
M1A1
2
M1A1
2
Comments
2
3(a)
(b)
(c)
1 + i (1 + i )
=
=i
1 − i 1 − i2
z2 =
1 3
+ = 1 = z1
4 4
r =1
1
1
θ = π, π
2
3
B1
AG
PI
B1B1
3
Deduct 1 mark if extra solutions
B2,1F
2
Positions of the 3 points relative to each
other, must be approximately correct
(d)
(e)
Arg ( z1 + z2 ) =
5
tan π =
12
1+
5
π
12
1
3
2
1
2
=2+ 3
B1
Clearly shown
M1
Allow if BO earned
A1
Total
4
3
12
AG must earn BO for this
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AQA GCE Mark Scheme, 2006 January series – MFP2
MFP2 (cont)
Q
Solution
4(a) Assume result true for n = k
Marks
Total
Comments
k
∑ ( r + 1)2r −1 = k 2k
r =1
k +1
∑ ( r + 1) 2r −1 = k 2k + ( k + 2 ) 2k
M1A1
r =1
= 2k ( k + k + 2 )
m1
= 2k ( 2k + 2 )
= 2k +1 ( k + 1)
2 × 20 = 2 = 1 × 21
B1
Pk ⇒ Pk +1 and P1 is true
E1
n =1
(b)
A1
2n
n
r =1
r =1
∑ ( r + 1)2r −1 − ∑ ( r + 1) 2r −1
M1
= 2n 22 n − n 2n
A1
(
)
= n 2n+1 − 1 2n
A1
Total
6
Sensible attempt at the difference between
2 series
3
9
5
Provided previous 5 marks earned
AG
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MFP2 – AQA GCE Mark Scheme, 2006 January series
MFP2 (cont)
Q
Marks
Solution
Total
Comments
5(a)
B1
Circle
B1
Correct centre
B1
(b)
3
Touching both axes
z max = OK
M1
Accept
= 42 + 42 + 4
A1F
Follow through circle in incorrect position
=4
(
)
2 +1
A1F
(c) Correct position of z , ie L
3
42 + 42 + 4 as a method
AG
M1
1 ⎞
⎛
a = − ⎜ 4 − 4cos π ⎟
6 ⎠
⎝
(
=− 4−2 3
)
A1F
1
b = 4 + 4sin π = 6
6
A1F
Total
6
Follow through circle in incorrect position
3
9
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AQA GCE Mark Scheme, 2006 January series – MFP2
MFP2 (cont)
Q
6(a)(i)
Solution
z+
z2 +
1
z2
(b)
1
= eiθ + e –iθ
z
M1
A1
2
AG
2
OE
3
AG
= cos 2θ + i sin 2θ
z2 − z + 2 −
+ cos ( −2θ ) + i sin ( −2θ )
M1
A1
1 1
+
z z2
= 2 cos 2θ − 2 cos θ + 2
M1
Use of cos 2θ = 2cos 2 θ − 1
m1
= 4cos 2 θ − 2cos θ
A1
z+
Comments
Or z +
cos ( −θ ) + i sin ( −θ )
= 2cos 2θ
(iii)
Total
1
= cos θ + i sin θ +
z
= 2cos θ
(ii)
Marks
1
=0
z
z = ±i
M1A1
Alternative:
1
z + =1
z
z2 − z + 1 = 0
M1A1
z = ±i
1± i 3
2
Accept solution to (b) if done otherwise
A1F
Alternative
1
1
If θ = + π θ =
π
2
3
M1
z=
5
M1
A1
B1
Total
12
7
θ =± π
M1
A1
cos θ =
z=e
A1
1+ 3 i
2
Or any correct z values of θ
Any 2 correct answers
One correct answer only
z =i z=
1
2
cos θ = 0
1
2
± 1 πi
3
1
3
M1
)
A1 A1
θ =± π
=
1
1± i 3
2
(
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MFP2 – AQA GCE Mark Scheme, 2006 January series
MFP2 (cont)
Q
7(a)(i)
θ
⎛e −e
2⎜
⎜
2
⎝
=
(ii)
−θ
Solution
⎞ ⎛ e + e −θ ⎞
⎟⎟ ⎜⎜
⎟⎟
2
⎠⎝
⎠
e 2θ − e −2θ
= sinh 2θ
2
⎛ eθ − e −θ
⎜⎜
2
⎝
2
⎞
⎛ eθ + e −θ
⎟⎟ + ⎜⎜
2
⎠
⎝
⎞
⎟⎟
⎠
Total
Comments
M1A1
2
AG
3
AG
2
M1
e 2θ − 2 + e − 2θ + e 2θ + 2 + e −2θ
4
= cosh 2θ
=
(b)(i)
Marks
θ
A1
A1
x! = 3cosh 2 θ sinh θ ′′
M1A1
Allow M1 for reasonable attempt at
differentiation, but M0 for putting in
terms of ekθ or sinh 3 θ unless real
progress made towards x! 2 + y! 2
y! = 3sinh 2 θ cosh θ
A1
Allow this M1 if not squared out, must be
clear sum in question is x! 2 + y! 2
x! 2 + y! 2 = 9cosh 4 θ sinh 2 θ
M1
+9 sinh 4 θ cosh 2 θ
2
2
(
2
2
= 9 sinh θ cosh θ cosh θ + sinh θ
AG
)
A1
Accept
1
∫0
9
4
sinh 2 2θ cosh 2θ dθ
but limits must appear somewhere
9
= sinh 2 2θ cosh 2θ
4
(ii)
S=
∫
13
0
2
A1
sinh 2θ cosh 2θ d θ
6
AG
6
AG
M1
u = cosh 2θ
d u = 2 sinh 2θ d θ
1
3 2
3 2 3
I=
u du = × u2
4
4 3
M1A1
∫
A1F
1
3⎫
⎧⎪ 1
⎪
S = ⎨ ( cosh 2θ ) 2 ⎬
2
⎩⎪
⎭⎪0
A1F
3
1⎧
⎫
⎨( cosh 2 ) 2 − 1⎬
2⎩
⎭
A1
=
Total
TOTAL
8
17
75