Version 1.0
General Certificate of Education (A-level)
January 2012
Mathematics
MPC3
(Specification 6360)
Pure Core 3
Final
Mark Scheme
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 events which all examiners 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 examiner
understands and applies it in the same correct way. As preparation for standardisation each
examiner 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,
examiners encounter unusual answers which have not been raised 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 students’ 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.
Further copies of this Mark Scheme are available from: aqa.org.uk
Copyright © 2012 AQA and its licensors. All rights reserved.
Copyright
AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material
from this booklet for their own internal use, with the following important exception: AQA cannot give permission to
schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre.
Set and published by the Assessment and Qualifications Alliance.
The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 3644723) and a registered
charity (registered charity number 1073334).
Registered address: AQA, Devas Street, Manchester M15 6EX.
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.
MPC3
Q
Solution
x
0
1(a)
y
1
1
2
1
1 12
2
2 12
3
Marks
4
all 7 x values correct (and no extra)
(PI by 7 correct y values)
B1
5 or more correct y values, exact
 12 1 
 4 , 4 ...  or evaluated (in table or in


formula)
16
32
64
f (x) =
4 x + 2 x – 8 or
]
correct substitution of their 7 y-values into
Simpson’s rule
M1
A1
4
CAO
g (x) =
8 – 2x – 4x
change of sign
∴ 1.2 < α < 1.3
(f(x) must be defined and all working
correct)
(
(
B1
8
=
f ( 1.2 ) –=
0.3 or g ( 1.2 ) 0.3 

=
f ( 1.3 ) 0.7
=
or g ( 1.3 ) – 0.7 
AWRT ±0.3 and ± 0.7
condone f (1.2) < 0 , f (1.3) > 0 if f is
defined
(ii)
Comments
2
A = 1 × 1 [ 65 + 4 × 42 + 2 × 20
3 2
= 91 or 45.5 or 273
2
6
(b)(i)
Total
x2 = ) 1.243
M1
attempt at evaluating f(1.2) and f(1.3)
alternative method
1.2
4=
5.3, 8 − 2 × 1.2
= 5.6 

1.3
4= 6.1, 8 − 2 × 1.3
= 5.4 
A1
2
at 1.2 LHS < RHS
at 1.3 LHS > RHS
∴ 1.2 < α < 1.3
B1
x3 = )1.232
B1
Total
M1
2
8
these values only
A1
MPC3 (cont)
Q
2(a)
(b)(i)
Solution
(c)(i)
(ii)
Total
Comments
f (1) = 21
f (16) = 1
M1
1 ≤ f ( x) ≤ 21
A1
63
4x – 1
x = 63
4y –1
x (4 y – 1) = 63 or better
M1
reverse x, y
M1
one correct step
sight of 1 and 21
2
allow f(x) replaced by f, y
y=
–1
f=
( x ) 1  63 + 1 
4 x

(ii)
Marks
Either order
OE
1  63 + 1  =

 1
4 x

63 + 1 =4, or better
x
( x =) 21
( fg (x) = )
A1
3
condone y =
A1
2
one correct step from their (b)(i) = 1,
or x = f(1)
note: 21 scores 2/2
B1
1
M1
63
4x2 – 1
63 = 1
4 x2 – 1
4 x 2 – 1 = 63 or better
M1
A1
A1
x = 16 OE
x = − 4 ONLY
2
Total
one correct step from their (c)(i) = 1
eg (2x + 8) (2x – 8) = 0, or x = ± 4
3
11
MPC3 (cont)
Q
3(a)
(b)
Solution
 dy 
=  12 x 2 – 6

 dx 
Marks
Total
B1
1
Comments
do not ISW
3
2 x 2 – 1 dx
∫2 4 x3 – 6 x + 1
(
(3)
)
=  1 ln 4 x3 – 6 x + 1 
6
 (2)
(
(
k ln 4 x 3 – 6 x + 1 , k is a constant
A1
k=1
6
m1
correct substitution in F(3) – F(2).
condone poor use or lack of brackets.
A1F
k ln 91 – k ln 21
only follow through on their k
)
= 1 ln 4 × 33 – 6 × 3 + 1
6
− 1 ln 4 × 23 – 6 × 2 + 1
6
= 1 ln 91 – 1 ln 21
6
6
1
91
 1 13 
= =
ln
or 
ln 
6
21
 6 3
(
)
)
M1
A1
5
or if using the substitution
u = 4 x3 − 6 x + 1
du
M1
∫=k ∫ u
A1
= 1 ln u
6
then, either change limits to 21 and 91 m1
then A1F A1as scheme or changing back
to ‘x’, then m1 A1F A1 as scheme
Total
4(a)
6
sec 2 θ − 1 =
...
B1
sec 2 θ + 3secθ – 10 (=
0)
M1
0
( secθ + 5)( secθ – 2 ) =
m1
secθ = – 5, 2
A1
correct use of sec 2 θ = 1 + tan 2θ
quadratic expression in sec θ with all
terms on one side
attempt at factors of their quadratic,
( secθ ± 5)( secθ ± 2 ) ,
or correct use of quadratic formula
1 1

 cos θ = – , 
5 2

60° , 300° ,101.5° , 258.5°
(b)
(AWRT)
B1
B1
4 x – 10° = 60° , 101 ⋅ 5° , 258 ⋅ 5° ,300°
M1
4 x = 70° , 111 ⋅ 5° , 268 ⋅ 5° ,310°
A1F
x =⋅
17 5° , 27 ⋅ 9° ,67.1° ,77 ⋅ 5°
(AWRT)
Total
A1
6
3 correct, ignore answers outside interval
all correct, no extras in interval
3
4 x – 10 = any of their (60),
all their answers from (a), BUT must have
scored B1
CAO, ignore answers outside interval
9
MPC3 (cont)
Q
Solution
5(a) stretch
SF 4
in y-direction
Marks
Total
Comments
I
II
III
M1A1
I + (II or III)
either order
translate
E1
e
 
0
B1
4
accept ‘e in positive x-direction’
(b)
e
(c)(i)
4ln ( x – e
A1
curve touches x-axis at 1 + e (or 3.7 or
better), and labelled (ignore scale)
3
correct curvature, including at their 1+ e,
approx. asymptote at x = e
=4
=
4ln ( x – e ) 4=
4ln ( x – e ) – 4
or better
(x =) 2e
mod graph, in 2 connected sections, both
in the first quadrant, touching x-axis
A1
1+e
)
M1
do not ISW
( x= ) e + e−1 or ( x = ) e + 1e
do not ISW
M1
must see 2 equations, condone omission
of brackets
A1
accept values of AWRT 5.42, 5.43, 5.44
A1
3
accept values of AWRT 3.08, 3.09
if M0 then x = 2e with or without working
scores SC1
(ii)
x ≥ 2e
e<x≤e+
B1
1
e
B2
accept values of AWRT 5.42, 5.43, 5.44
3
accept values of AWRT 2.72, 3.08, 3.09
if B2 not earned, then SC1for any of
1
1
1
e ≤ x ≤ e+ , e < x < e+ , e ≤ x < e+
e
e
e
Total
13
MPC3 (cont)
Q
Marks
Solution
6(a)
Total
 dx =  ( sin θ × 0 ) – 1 × cos θ


sin 2 θ
 dθ 
Comments
± sin θ × k ± 1 × cos θ
sin 2 θ
where k = 0 or 1
quotient rule
M1
must see the ‘0’ either in the quotient or in
A1
=
(b)
–
cos θ
sin 2 θ
or =
−
eg
cos θ
sin θ sin θ
du
= 0 etc
dθ
or equivalent
CSO, AG must see one of the previous
expressions
= – cosecθ cot θ
A1
x = cosecθ
dx
= – cosec θ cot θ
dθ
Replacing
B1
OE, eg dx = – cosecθ cot θ dθ
B1
at any stage of solution
M1
all in terms of θ , and including their
attempt at dx, but condone omission of dθ
A1
fully correct and must include dθ
(at some stage in solution)
A1
OE eg ∫ − sin θ ( dθ )
( cosec θ – 1)
by cot 2 θ , or better
2
∫=∫
– cosecθ cot θ
cosec 2θ
(
cosec 2θ – 1
)
dθ
–cosecθ cot θ
( dθ )
cosec 2θ cot θ
–1
=∫
( dθ )
cosecθ
= cos θ
x 2,=
θ 0.524 AWRT
=
3
∫
x
=
A1
B1
2, θ 0.785 AWRT
=
correct change of limits

1 
or ( ± ) cos θ =
( ± )  1 – 2  
x  
 
0.8660 – 0.7071
c's F ( 0.52 ) – F ( 0.79 )
substitution into ± cos θ only
m1
 3 1 
or 
–

2 
 2
= 0.159
A1
Total
9
12
2
OE
2
MPC3 (cont)
Q
Solution
 dy
=
7(a)
 dx
–

 p e
–

 ⇒ e
e
1x
4
1x
4
x 2 + qxe
Marks
– 1x
4
A1
Comments
p, q constants
1
p = – and q = 2
4

 – 1 x2 + 2 x  =

 0
 4


– 1x
– 1x
4
≠0
 − 14 x 
2
0
 e
 ax + bx =


x = 0, 8
(
(b)(i)
Total
M1
)
E1
or e 4 = 0 impossible OE
(may be seen later)
m1
or e
=
x 0,=
y 0
A1
A1
=
x 8,=
y 64e –2
B1
2
∫x e
– 1x
4
− 1x
4
x ( ax + b ) =
0
– 8
dv e – 14 x
2
dx =
u x=
dx
– 1x
du 2=
=
x v ke 4
dx
7
M1
condone y = 82 e 4 etc
ignore further numerical evaluation
where k is a constant
A1
k= – 4
1
2 – 4x
− ∫ −4e
– 4x e
– 1x
4
× 2 x(dx), or better
A1F
correct substitution of their terms
m1
both differentiation and integration must
be correct
– 1x
dv ne 4
=
u mx
=
dx
– 1x
du
= m=
v – 4ne 4
dx
1
2 – 4x
∫ =– 4 x e
–

=  – 4 x2e

1x
4
–

+ 8  – 4 xe

– 32 xe
– 1x
4
1x
4
+ ∫ 4e
– 128e
– 1x
4
– 1x
4

dx 

(4)


(0)
Al
m1
= –e
–1
[
64 + 256 ] – [ –128
]
= 128 – 320
e
(4)
(ii)
v=π
1
2 – 4x
∫ 9x e
(0)
correct substitution and attempt at
(dep
on M1
only)
A1
( dx )
– 1x
subtraction in ax 2 e 4 + bxe
(may be in 3 stages)
7
M1
= 9π  128 – 320 
e 

A1F
Total
TOTAL
– 1x
4
+ ce
or 128 − 320e −1
ignore further numerical evaluation
condone omission of brackets, limits
2
16
75
9π × (their exact b(i))
– 1x
4