Version
General Certificate of Education (A-level)
January 2013
Mathematics
MFP3
(Specification 6360)
Further Pure 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
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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.
MFP3 - AQA GCE Mark Scheme 2013 January series
Q
Solution
1(a) y(3.2) = y(3) + 0.2 2  3  5
Marks
M1
= 5 + 0.2 × 11
A1
= 5.66332… = 5.6633 to 4dp
A1
(b) y(3.4) = y(3) + 2(0.2){f [3.2, y(3.2)]}
M1
Total
3
A1F
…. = 5 + 2(0.2) 2  3.2  5.6633...
Comments
Condone >4dp
Ft on cand’s answer to (a)
(= 5 + (0.4) 12.0633... )
= 6.389 to 3dp
A1
Total
3
6
2
(a) e3x = 1 + 3x + 4.5x2
(b)
B1
1  2 x 3 / 2 = 1  3x  15 x 2
CAO Must be 6.389
Ignore higher powers beyond x2
throughout this question
1
M1
1  2 x 3 / 2 =1±3x+kx2 or 1+kx±7.5x2 OE
A1
1−3x+7.5x2 OE (simplified PI)
M1
Product of c’s two expansions with an
attempt to multiply out to find x2 term
2
e3x 1  2 x 
3 / 2
=
(1 + 3x + 4.5x2)(1−3x+7.5x2)
x2 term(s): 7.5x2 − 9x2 + 4.5x2 = 3x2 .
A1
Total
4
5
MFP3 - AQA GCE Mark Scheme 2013 January series
Q
Solution
Marks
Total
Comments
M1
3 PI: y  kx 2 e x
PI
y ' PI  2kxe x  kx 2 e x
y '' PI  2ke x  4kxe x  kx 2 e x
m1
Product rule used in finding
both derivatives
m1
Subst. into DE
2k = 6 ; k = 3; y PI  3x 2 e x
A1
CSO
(GS: y =) ex(Ax+B) + 3 x 2 e x
B1F
5
E1
1
2ke x  4kxe x  kx 2 e x − 4kxe x  2kx 2 e x + kx 2 e x =6e
x
Total
4(a) Integrand is not defined at x = 0
5

…… =

1
0

lim
a0

1
a
A1
x 4 ln x dx }
lim a 5
1
a5
–
[ ln a  ]
25
25
a0 5
lim
But
a0
involving the ‘original’ ln x
A1
x5
x5
(+ c)
ln x 
5
25
x 4 ln x dx = {
= 
M1
x5
x5  1 
ln x 
  dx
5
5  x
OE
...= kx 5 ln x   f x  , with f(x) not
(b)
x 4 ln x dx =
ex(Ax+B) + kx 2 e x , ft c’s k .
Limit 0 replaced by a limiting
process and F(1)−F(a) OE
M1
a 5 ln a = 0
E1
Accept
lim
x0
x k ln x = 0 for
any k>0
So

1
0
x 4 ln x dx = 
1
25
A1
Total
6
7
Dep on M and A marks all scored
MFP3 - AQA GCE Mark Scheme 2013 January series
Q
Solution
5
(a)
Marks
Comments
dy sec x

y  tan x
dx tan x
IF is exp (

sec 2 x
dx)
tan x
M1
= eln(tanx) = tan x
(b)
Total
2
tan x
A1
2
AG Be convinced
dy
 (sec 2 x) y  tan 2 x
dx
d
 y tan x  tan2x
dx
M1

y tanx = tan 2 x dx

y tanx = tan x − x (+c)

4
c2
 tan

4

4


4
LHS as differential of y×IF PI
A1
 y tanx = (sec 2 x  1) dx
3 tan
and with integration attempted
c
so y tanx = tanx − x+ 2 
y = 1 +(2 − x 

4
m1
Using tan2x = ± sec2x ± 1 PI
or other valid methods to integrate tan2x
A1
Correct integration of tan2x; condone
absence of +c.
m1
Boundary condition used in attempt to
find value of c

4
) cotx
Total
A1
6
8
ACF
MFP3 - AQA GCE Mark Scheme 2013 January series
Q
Solution
6(a)(i) y = ln e 3 x cos x = ln e3x + ln cos x = 3x + ln cos x
dy
1
3
 ( sin x)
dx
cos x
dy
 3  tan x
dx

(ii)

d2 y
d3 y
2
;


 2 sec x(sec x tan x)
sec
x
dx 2
dx 3
d4 y
 4 sec x(sec x tan x) tan x  2 sec 4 x
4
dx


y″(0)= −1;
1 2 0 3  2 4
x  x 
x ...
2!
3!
4!
1
1
= 3x  x 2  x 4
2
12
(d)(i)
{ln(1+px)} = px 

1 2 2
p x
2
A1
Comments
Chain rule for derivative of ln cosx
3
CSO AG
M1 for d/dx{ [f(x)]2 }= 2f(x)f′(x)
3
ACF
M1
Mac. Thm with attempt to
evaluate at least two derivatives
at x=0
A1F
At least 3 of 5 terms correctly
obtained. Ft one miscopy in (a)
A1
B1
3
CSO AG Be convinced
1
accept (px)2 for p2x2;
ignore higher powers;

1

3x
 x 2 {ln e cos x  ln(1  px)} =


1 
1 2
1 2 2

4
3  
 2 3 x  x  O( x )   px  p x  O( x ) 
2
2

 
x 
For
(ii)
M1
A1
ln e 3 x cos x = 0  3 x 
(c)
Total
B1; M1
(b) Maclaurin’s Thm:
x 4 (iv )
x2
x2
y(0)+x y′(0)+
y″(0)+
y′′′(0) +
y (0)
4!
2!
3!
y(0) = ln1 = 0; y′(0) =3;
y′′′(0) = 0; y (iv ) (0) = −2
Marks
B1
 1  e 3 x cos x 
 to exist, p = 3
 ln
x  0  x 2  1  px 
lim
lim
 3 p

1 p2
…… =
(
)


 O( x)

2
2
x0  x

1
2
Value of limit   
p2
= 4.
2
M1
Law of logs and expansions used;
A1
p=3 convincingly found
m1
Divide throughout by x2 before
taking limit. (m1 can be awarded
before or after the A1 above)
A1
Total
4
14
Must be convincingly obtained
MFP3 - AQA GCE Mark Scheme 2013 January series
Q
7(a)
Solution
Solving
Total
Auxl. Eqn. m2 −6m + 10 = 0
(m − 3)2 + 1= 0
M1
m=3±i
A1
CF (yCF =) e3t(A cos t + B sin t)
M1
OE Condone x for t here; ft c’s
2 non-real values for ‘m’.
For PI try (yPI =) ke2t
M1
Condone x for t here
1
4k  12k  10k  1  k 
2
A1
1 2t
e
2
B1F
PI Completing sq or using
quadratic formula to find m.
6
M1
OE
dy
dy
 2x
dx
dt
A1
OE
d2 y
d2 y
dy
4

2
t
2
2
dt
dx
dt
A1
3
3
1
 d2 y
dy 
dy
2
2
+ 40t y  4t 2 e 2t
t 4t 2  2  − (12t  1)2t
dt
dt 
 dt
M1
1
2
4t
3
d2 y
2{
2
dt
Product rule OE used dep on
previous M1 being awarded at
some stage
5
CSO
A.G.
Subst. using (b) into given DE to
eliminate all x
3
dy
− 6
+ 10 y}  4t 2 e 2t
dt
3
2
t≠0 so divide by 4t gives
2
Chain rule
d
dt d
( f (t ) ) 
( f (t ) ) OE
dx
dx dt
d
dx d
eg
( g( x) )
( g( x) ) 
dt
dt dx
M1
m1
(d)
CF +PI with 2 arb. constants and
both CF and PI functions of t
only
dy
d t dy
=
dx dx dt
d 2 y d  dy 
dt d  dy  dy
=
 2 x  = (2x)
  +2
2
dx  dt 
dx dt  dt 
dt
dx
2
dy
d y
= (2x)(2x)
+2
2
dt
dt
(c)
Comments
d y
dy
6
 10 y  e 2t (*)
2
dt
dt
GS of (*) is (yGS =) e3t(A cos t + B sin t) +
(b)
Marks
2
d2 y
dy
6
 10 y  e 2t (*)
2
dt
dt
y = e 3 x ( A cos x 2  B sin x 2 ) 
1 2 x2
e
2
Total
A1
2
CSO
B1
1
OE Must include y=
14
A.G.
MFP3 - AQA GCE Mark Scheme 2013 January series
Q
Solution
8(a)(i)
r  sin
2
3
(2 
1
3
9 3 3

=
cos ) =

2
3
2
4
4
(ii) x = ON = (3√3)/8
Marks
Total
M1; A1
2
Polar eqn of PN is r cos = ON
3 3
r=
sec
8
M1
(iii) Area Δ ONP = 0.5 × rN × rP × sin (/3)
M1
=
(b)(i)
A1
1 3 3 3 3
3
27 3



=
2
8
4
2
128
 sin
n
 cos  d =
u
n
A1
(+c)
AG Be convinced
OE
With correct or ft from
(a)(i) (ii), values for rP and rN .
2
M1
du
sin n 1 
n 1
=
A1
2
Comments
Be convinced
PI
2
(ii) Area of shaded region bounded by line OP
1
and arc OP =
2
1
2


2
3
1


sin 2  2  cos   d
2


2


2
(1  cos 4 ) d +
3
1
4

2
4 sin 2  cos 2  cos  d
2
B1
Correct limits
M1
2 sin 2 2  1  cos 4
B1
sin22 cos = 4sin2 cos2 cos
A1
Correct integration of
0.5(1−cos4)
3


Use of


 sin 4  2
=  
+
8  
2
(sin 2   sin 4  ) cos  d
3
m1
Writing 2nd integrand in a suitable
form to be able to use (b)(i) OE
PI
A1
Last two terms OE
3

 sin 4 sin  sin   2
=  



8
3
5 
2
3
3
=

12

1 2
r d
2
M1
5
21 3 2

160 15
A1
Total
TOTAL
8
16
75
CSO