A-LEVEL
Mathematics
Further Pure 2 – MFP2
Mark scheme
6360
June 2015
Version/Stage: 1.0 Final
Mark schemes are prepared by the Lead Assessment Writer and considered, together with the
relevant questions, by a panel of subject teachers. This mark scheme includes any amendments
made at the standardisation events which all associates participate in and is the scheme which was
used by them in this examination. The standardisation process ensures that the mark scheme covers
the students’ responses to questions and that every associate understands and applies it in the same
correct way. As preparation for standardisation each associate analyses a number of students’
scripts: alternative answers not already covered by the mark scheme are discussed and legislated for.
If, after the standardisation process, associates encounter unusual answers which have not been
raised they are required to refer these to the Lead Assessment Writer.
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expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark
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assessment remain constant, details will change, depending on the content of a particular
examination paper.
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MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015
Key to mark scheme abbreviations
M
m or dM
A
B
E
or ft or F
CAO
CSO
AWFW
AWRT
ACF
AG
SC
OE
A2,1
–x EE
NMS
PI
SCA
c
sf
dp
mark is for method
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
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
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.
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.
3 of 11
MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015
Q1
Solution
(a)
Mark
r + 1= A( r + 2) + B or
A( r + 2)
B
=
1
+
r +1
r +1
either A = 1 or B = −1
(b)
Total
OE with factorials removed
M1
correctly obtained
A1
1
1
1
=
−
( r + 2) r ! ( r + 1)!
( r + 2)!
A1
1 1
1 1
+ − +…
−
2! 3! 3! 4!
M1
3
1
1
−
2
( n + 2)!
A1
Total
allow if seen in part (b)
use of their result from part (a) at least
twice
1
1
−
( n + 1)!
( n + 2)!
Sum =
Comment
2
must simplify 2!
and must have scored at least M1 A1 in
part (a)
5
(a) Alternative Method Substituting two values of r to obtain two correct equations in A and B with factorials
evaluated correctly
1
B
1 A B
; r=1 ⇒ = +
earns M1 then A1, A1 as in main scheme
r = 0 ⇒ = A+
2
2
3 2 6
1
1
−
earns 3 marks.
( r + 1)!
( r + 2)!
However, using an incorrect expression resulting from poor algebra such as 1 = A( r + 2)!+ B ( r + 1)! with
candidate often fluking A = 1, B = –1 scores M0 ie zero marks which you should denote as FIW
These candidates can then score a maximum of M1 in part (b).
NMS
(b) ISW for incorrect simplification after correct answer seen
4 of 11
MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015
Q2
Solution
(a)
Mark
Total
Comment
y
1
O
x
–1
(b)
Graph roughly correct through O
M1
Correct behaviour as x → ±∞ & grad at O
A1
Asymptotes have equations y = 1 & y = −1
B1
sech x =
2
e x − e− x
; tanh x = x
−x
e +e
e + e− x
condone infinite gradient at O for M1
3
both correct ACF or correct squares of
these expressions seen
B1
x
22 + ( e x − e )
( sech 2 x + tanh 2 x =
)
2
( e x + e− x )
must state equations
−x 2
2
x
=
sech 2 x + tanh
(c)
e 2 x + 2 + e −2 x
= 1
e 2 x + 2 + e −2 x
A1
6(1 − tanh 2 x ) =
4 + tanh x
6 tanh x + tanh x − 2 ( =
0)
2
2
3
1 1+ k 
tanh x = k ⇒ x = ln 

2 1− k 
1
1 1
,
x = ln 3
x = ln
2
2 5
tanh x =
1
,
2
attempt to combine their squared terms
with correct single denominator
M1
tanh x = −
3
AG valid proof convincingly shown to
equal 1 including LHS seen
B1
M1
correct use of identity from part (b)
A1
obtained from correct quadratic
forming quadratic in tanh x
A1F
A1
5
Total
FT a value of k provided k < 1
both solutions correct and no others
any equivalent form involving ln
11
(a) Actual asymptotes need not be shown, but if asymptotes are drawn then curve should not cross them for A1.
Gradient should not be infinite at O for A1.
(b) Condone trailing equal signs up to final line provided “ sech 2 x + tanh 2 x = ” is seen on previous line for A1
Denominator may be e 4 x + 4e 2 x + 6 + e 4 x + 4e −2 x + e −4 x etc for M1 and A1
Accept sech x + tanh=
x
2
2
(e
(e
+ e− x )
=
1 for A1
2
x
+ e− x )
x
2
x
−x
1
 1
sinh 2 x  1 + ( 2 (e − e ) )
Alternative : 
+
=

2
2
2
 cosh x cosh x 
( 12 (e x + e− x ) )
and then A1 for
2
scores B1 M1
1 2 x 1 −2 x 1
e + e +
4=
2 1 , (all like terms combined in any order).
+ tanh 2 x 4
sech 2 x =
1 2 x 1 1 −2 x
e + + e
4
2 4
5 of 11
MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015
Q3
(a)
Solution
Mark
dx
1
= 1− 2
dt
t
B1
dy 2
=
dt
t
  d x 2  d y 2
  +   =
  dt   dt 

(b)


( 2π ) ( 2ln t )  t −


t (2t ) − (t 2 + 1)
t2
ACF
squaring and adding their expressions and
attempting to multiply out
2
A1
1
dt
2 

1
OE eg
M1
∫ ( 2 ln t ) 1 + t
2
Comment
B1

2 1 4
 1− 2 + 4 + 2

t t t

1

= 1 + 2 
 t 
2 1
1+ 2 + 4
t t
2π
Total
1
2  1

−
 t −  ( dt ) 
t
t  t

∫
2 

 1 
2π ( 2ln t )  t −  −  2t +  
t 
 t 

= 2π (3ln 2 − 5 + 4)
= π (6ln 2 − 2)
4
AG be convinced
B1
must have 2π , limits and dt
M1
integration by parts - clear attempt to
1
integrate 1 + 2 and differentiate 2ln t
t
A1
correct (may omit limits, 2 π and dt)
A1
correct including 2 π (no limits required)
A1
5
Total
9
(b) May have two separate integrals and attempt both using integration by parts for M1
{
(
∫
)}
Must see (2π ) 2t ln t − ∫ 2 ( dt ) − 2t −1 ln t − 2t −2 ( dt ) (may omit limits, 2 π and dt) for first A1
and 2π ( 2t ln t − 2t ) − ( 2t −1 ln t + 2t −1 ) for second A1
Condone poor use of brackets if later recovered.
6 of 11
MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015
Q4
(a)
Solution
Mark
f ( k + 1)= 24 k +7 + 33k + 4
M1
7
16 × 24 k +3
convincingly showing 2 4 k +=
f ( k + 1) − 16f ( k )
= (81 − 16 × 3) × 33k
E1
= 33 × 33k
(b)
A1
f (1) = 209 therefore f (1) is a multiple of 11
Total
Comment
must see 16 = 2 4 OE
3
B1
19 etc
f (1)= 209= 11 × 19 or 209 ÷ 11 =
therefore true when n=1
f ( k +=
1) 16f ( k ) + 33 × 33k
M1
= 11M + 11N = 11( M + N )
Therefore f ( k + 1) is a multiple of 11
attempt at f (k + 1) =
... using their result
from part (a)
where M and N are integers
A1
Assume f ( k ) is a multiple of 11 (*)
Since f(1) is multiple of 11 then f(2), f(3),…
are multiples of 11 by induction
(or is a multiple of 11 for all integers n ≥ 1 )
Total
E1
4
must earn previous 3 marks and have (*)
before E1 can be awarded
7
(a) It is possible to score M1 E0 A1
(b) Withhold E1 for conclusion such as “a multiple of 11 for all n ≥ 1 ” or poor notation, etc
7 of 11
MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015
Q5
Solution
(a)
Im(z)
2
Mark
Total
Comment
Ignore the line OP drawn in full or circles
drawn as part of construction for locus L.
A Re(z)
O
B
–4
(b)(i)
+P
Straight line
Through midpoint of OP, P correct
Perpendicular to OP, P correct
M1
A1
A1
( x − 2) 2 + ( y + 4) 2 = x 2 + y 2
M1
2y − x +5 =
0
A(5,0) & B (0, −2.5)
5 5
5 5
C  , −  ⇒ complex num = − i
2 4
2 4
A1
A1
(ii) either
α=
A1
5 5
5 5
− i or k =
2 4
4
M1
5 5
5 5
+ i =
2 4
4
A1
z−
Total
P represents 2 – 4i
3
may have 5 + 0i and 0 – 2.5i
4
allow statement with correct value for
centre or radius of circle
2
must have exact surd form
9
(a) Withhold the final A1 (if 3 marks earned) if the straight line does not go beyond the Re(z) axis and
negative Im(z) axis.
The two A1 marks can be awarded independently.
1
(b)(i) Alternative 1: grad OP = –2 ⇒ grad L =
( x − 1) OE A1 then A1, A1 as per scheme
0.5 M1 ; y + 2=
2
Alternative 2: substituting z= x (or a) then z= iy ( or ib) into given locus equation
Both ( x − 2) 2 + 4 2 =
0 and 4 + 8 y + 16 =
x 2 and 22 + ( y + 4) 2 =
y 2 M1; 4 − 4 x + 16 =
0 OE for A1
then A1, A1 as per scheme.
8 of 11
MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015
Q6
Solution
(a)
Mark
1
( x − 2) (4 − 2 x )
2
5 + 4x − x +
5 + 4 x − x2
9 × 13
(+)
2
 x−2
1− 

 3 
5 + 4 x − x2
Total
Comment
M1
product rule ( condone one error)
A1
correct unsimplified
2
dy 
2
 =  2 5 + 4x − x
 dx 
1
k
A1cso

−1  x − 2  
2
( x − 2) 5 + 4 x − x + 9sin 

 3 

 3 27
1
1
+ 9sin −1 

" their " k  2 4
2
=
or
9
3
3+ π
8
4
32 − ( x − 2 )
2
correct unsimplified
5
k=2
M1
ft “their” k
m1
correct sub of limits (simplified at least
this far)
A1 cso
Total
9
last two terms above combined correctly
A1
5 + 4 x − x2
(b)
B1
3
must have earned 5 marks in part(a) to be
awarded this mark
8
2
(a) Second A1 ; may combine all three terms correctly and obtain 10 + 8 x − 2 x
5 + 4 x − x2
9 of 11
MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015
Q7
(a)
(b)(i)
Solution
Mark
αβ + βγ + γα = 0
4
αβγ = −
27
Total
B1
B1
0 ; αβ 2 = −
αβ + αβ + β 2 =
4
27
2
May use γ instead of β throughout (b)(i)
B1
Clear attempt to eliminate either α or β
from “their” equations
correct
M1
α3 = −
1
27
or
β3 =
8
27
A1
1
2
or β =
3
3
1
2
2
α = − , β= , γ =
3
3
3
either α = −
(ii)
k

−
1=
 ∑α =
27

A1

⇒ k =
−27

α 2 = −2i
(c)(i)
α =−2 − 2i
3
(ii)
27( −2 − 2i) − 2ik + 4 =0
1
1
+1⇒ z =
z
y −1
27
12
−
+4=
0
3
( y − 1) ( y − 1) 2
27 − 12( y − 1) + 4( y − 1)3 =
0
3
2
27 − 12 y + 12 + 4( y − 3 y + 3 y − 1) =
0
3
2
4 y − 12 y + 35 =
0
∑α ' β '=
3+
αβ + βγ + γα
3+
∑α ' =
αβγ
all 3 roots clearly stated
B1
1
or substituting correct root into equation
B1
B1
2
correctly substituting “their ” α 2 = −2i
and “their ” α 3 =−2 − 2i
2
B1
may use any letter instead of y
M1
sub their z into cubic equation
A1
A1
A1
removing denominators correctly
correct and (y–1)3 expanded correctly
5
3
=
(B1)
2(αβ + βγ + γα ) + α + β + γ
sum of new roots =3
M1 for either of the other two formulae
correct in terms of αβγ , αβ + βγ + γα and
α + β +γ
(M1)
αβγ
(A1)
=0
∏=
5
A1
y=
Alternative:
A1
M1
k=
−27 + 25i
(d)
Comment
αβ + βγ + γα + 1 + α + β + γ
αβγ
−35
=
4
1+
(A1)
4 y 3 − 12 y 2 + 35 =
0
Total
(A1)
(5)
may use any letter instead of y
17
10 of 11
MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015
Q8
Solution
(c)
Comment
So ω is a root of z 5 = 1
B1
1
must have conclusion plus verification that
ω5 = 1
ω2 , ω3 , ω4 .
B1
1
OE powers mod 5 ( must not include 1)
1
or clear statement that sum of roots (of
z5 − 1 =
0 ) is zero
5
(ii)
(ii)
Total
( ω = ) cos 2π + isin 2π = 1
(a)(i)
(b)(i)
Mark
1+ ω+ ω 2 + ω3=
+ ω4
1 − ω5
= 0
1− ω
B1
2
1 
1

 ω+  +  ω+  − 1
ω 
ω

1
1
= ω 2 + 2 + 2 + ω+ − 1
ω
ω
1+ ω+ ω 2 + ω3 + ω 4
=
=0
ω2
2π
2π
1
= cos
− isin
5
5
ω
1
2π
⇒ ω+ = 2cos
ω
5
1

Solving quadratic  ω+
ω

correct expansion
M1
A1
2
AG correctly shown to = 0
do not allow simply multiplying by ω 2
M1
−1 ± 5
2
2π
Rejecting negative root since cos
>0
5
2π
5 −1
Hence cos
=
5
4

=

A1
SC1 if result merely stated
M1
must see both values
must see this line for final A1
A1
4
It is possible to score SC1 M1 A1
Total
9
(b)(ii) May replace 1 by ω3 and 1 by ω 4 and/or 1 by ω5 in valid proof.
ω2
ω
1 1
Alternative: 1+ ω+ ω 2 + ω3 + ω 4 = 0 ⇒ 2 + +1+ω+ω 2 =
0 M1
ω
ω
2
2
1 
1
1
1



 ω+  − 2+  ω+  + 1 =0 ⇒  ω+  +  ω+  − 1 =0 A1
ω 
ω
ω
ω



11 of 11