Version 1.0
General Certificate of Education (A-level)
January 2012
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.
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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.
Q
Solution
 2 − 1
1(a) y(1.1) = y(1) + 0.1 

 4 + 1
= 2 + 0.02 = 2.02
Marks
Total
M1A1
A1
(b) y(1.2) = y(1) + 2(0.1){f[1.1, y(1.1)]}
3
M1
 2.02 − 1.1 
= 2 + 2(0.1) 
2

 2.02 + 1.1 
= 2.035518… = 2.036 to 3dp
A1F
A1
Total
ft on c’s answer to (a)
3
6
M1

x
2  1
 4 + x − 2   4 + O( x )   4 + O( x) 
=


=
2
2
 x + x
  x + x   1 + x 

 

m1
Division by x stage before taking the
limit
 4 + x − 2 1
lim 
=
2
x →0
 x+x  4
A1
 1 x

x 2

4 + x = 21 +  = 2 1 +   + O( x 2 )
4

 24

Total
3
CAO Must be 2.036
Attempt to use binomial theorem OE
The notation O(xn) can be replaced by a
term of the form kxn
1
2
Comments
3
CSO NMS 0/3
3
m + 2m + 10 = 0
m = −1 ± 3i
M1
A1
PI
Complementary function is
(y =) e − x ( A cos 3 x + B sin 3 x)
A1F
OE Ft on incorrect complex value of m
Particular integral: try y = kex
k + 2k + 10k = 26 ⇒ k = 2
M1
A1
(GS y =) e − x ( A cos 3 x + B sin 3 x) + 2ex
B1F
c’s CF+ c’s non-zero PI but must have 2
arb consts
x = 0, y = 5 ⇒ 5 = A + 2 so A = 3
B1F
ft c’s k ie A = 5 − k, k ≠ 0
dy
=
dx
e − x (−3 A sin 3 x + 3B cos3 x − A cos3 x − B sin 3 x) + 2e x
M1
Attempt to differentiate c’s GS
(ie CF + PI)
11 = 3B − A + 2
( B = 4)
y = e − x (3 cos 3x + 4 sin 3x) + 2ex
A1
A1
2
Total
10
10
CSO
Q
Solution
4(a) IF is exp (
Marks
2
∫ x dx)
= e2lnx
= x2
[ ]
d
yx 2 = x 2 ln x
dx
⇒ yx2 =
=
∫
(ln x)
d  x3

dx  3




x3
x2
ln x −
dx
3
3
∫
x3
x3
ln x −
+A
3
9
x
x
{ y = ln x − + Ax − 2 }
3
9
Total
Comments
M1
and with integration attempted
A1
A1
PI
M1
LHS; PI
M1
Attempt integration by parts in correct
direction to integrate xp lnx
A1
RHS
yx2 =
A1
7
(b) Now, as x → 0, xk lnx → 0
E1
Must be stated explicitly for a value of
k>0
As x → 0, y → 0 ⇒ A = 0
B1
Const of int = 0 must be convincing
yx2 =
x3
x3
ln x −
3
9
When x = 1, y = −
1
9
B1F
Total
3
10
ft on one slip but must have made a
realistic attempt to find A
Q
Solution
5(a) The interval of integration is infinite
Marks
E1
(b)
=
u x 2 e −4 x + 3 ⇒ du = (2xe–4x – 4x2e–4x) dx
M1
1 2 x(1 − 2 x)e −4 x
∫ 2 × x 2e−4 x + 3 dx
1
1
= ×
du
2
u
1
1
= ln u + c = ln ( x 2 e −4 x + 3) {+ c}
2
2
A1
Total
1
Comments
OE
du/dx or ‘better’
x(1 − 2 x)
∫ x 2 + 3e4 x dx =
∫
(c) I =
∫
∞
1
2
a
= lim
a→∞
=
3
OE Condone missing c. Accept later
substitution back if explicit
x(1 − 2 x)
dx
x 2 + 3e 4 x
= lim ∫ 1
a →∞
A1
2
x(1 − 2 x)
dx
x 2 + 3e 4 x
(
)
(
)
M1
1
e −2
{ln a 2 e −4 a + 3 – ln(
+ 3)}
2
4
M1
Uses part (b) and F(a) – F(1/2)
E1
Stated explicitly (could be in general
form)
1
1
e −2
ln{ lim a 2 e −4 a + 3 } – ln(
+ 3)
a→∞
2
2
4
(
Now lim a 2 e −4 a
a→∞
I=
)
=0
1
1
e −2
ln 3 – ln(
+ 3)
2
2
4
A1
Total
4
8
CSO ACF
Q
Solution
1
(−2 sin 2 x)
6(a) y = ln cos 2x ⇒ y′(x) =
cos 2 x
Marks
M1
A1
Total
Comments
Chain rule
y′′ (x) = − 4sec22x
m1
λ sec22x OE
y′′′ (x) = − 8sec 2x (2sec 2x tan 2x)
M1
K sec22x tan2x OE
{y′′′ (x) = −16tan 2x (sec2 2x)}
y′′′′(x) = −16[2sec2 2x(sec2 2x) +
tan 2x(2sec 2x (2sec 2x tan 2x))]
M1
A1
(b) y(0) = 0, y′(0) = 0, y′′(0) = −4, y′′′(0) = 0,
y′′′′(0) = −32
x2
x4
ln cos 2x ≈ 0 + 0 +
(−4) + 0 +
(−32)
2
4!
4
≈ − 2x 2 − x 4
3
B1F
(c) ln (sec2 2x) = − 2ln (cos 2x)
8
≈ 4x 2 + x 4
3
M1
Product rule OE
ACF
ft c’s derivatives
M1
A1
A1
Total
6
3
CSO throughout parts (a) and (b) AG
PI
2
11
Q
Solution
7(a) u = xy
dy
du
= y+x
dx
dx
2
dy
d2 y
d u dy
=
+
(
)
+
x
dx
dx
dx 2
dx 2
Marks
Total
Comments
M1
A1
Product rule OE
OE
A1
OE
d2 y
dy
+ 2(3 x + 1)
+ 3 y (3 x + 2) = 18 x
2
dx
dx
dy
d2 y
dy
+ y ) + 9 xy = 18 x
( x 2 + 2 ) + 6( x
dx
dx
dx
x
(b)
d 2u
du
+6
+ 9u = 18 x
2
dx
dx
A1
d 2u
du
+6
+ 9u = 18 x
2
dx
dx
CF: Aux eqn m 2 + 6m + 9 = 0
(m + 3) 2 = 0 so m = −3
CF: (u =) e−3x (Ax + B)
M1
PI
A1
A1F
PI
M1
PI. Must be more than just stated
PI: Try (u =) px + q
0 + 6p + 9(px + q) = 18x
9p = 18, 6p + 9q = 0
12
p=2; q=−
9
4
u = e −3 x ( Ax + B ) + 2 x −
3
xy = e −3 x ( Ax + B) + 2 x −
4
CSO AG Be convinced
m1
A1
Both
B1F
c’s CF + c’s PI but must have 2 constants,
also must be in the form u = f(x)
4
3
1
4
y = {e −3 x ( Ax + B ) + 2 x − }
x
3
A1
Total
8
12
Q
Solution
Marks
Comments
π
1 2
2
d
θ
or
r
∫0 r dθ
2∫
1
(3 + 2cos θ ) 2 dθ
2∫
M1
Use of
1
(9 + 12cos θ + 4cos 2 θ ) dθ
∫
20
B1
B1
Correct expn of [3 + 2cosθ ]2
Correct limits
M1
Attempt to write cos 2 θ in terms of
cos 2θ
A1F
Correct integration ft wrong coefficients
8(a) Area =
2π
=
Total
2π
= ∫ (4.5 + 6cos θ + (1 + cos 2θ )) dθ
0
2π
1


=  4.5θ + 6sin θ + θ + sin 2θ 
2

0
= 11π
(b)(i)
A1
x 2 + y 2 − 8 x + 16 = 16
M1
r 2 − 8r cos θ + 16 = 16 ⇒ r = 8 cos θ
A1
At intersection, 8 cos θ = 3 + 2 cos θ
3
⇒ cos θ =
6
 π
Points  4,  and
 3
π

AB = 2 ×  4sin 
3

= 4 3
 5π 
 4,

 3 
Length of arc AOB of circle = 4 ×
Perimeter of segment AOB =
2π
3
2π
3
8π
+4 3
3
Total
CSO
Use of any two of x = r cosθ,
y = r sinθ , x 2 + y 2 = r 2
M1
Equating rs or equating cosθ s with a
further step to solve eqn.
(OE eg 4r = 12 + r ⇒ 4r − r = 12)
A1
OE
M1
Valid method to find AB, ft c’s r and θ
values
A1
(ii) Let M=centre of circle then ∠ AMB =
6
6
OE surd
π
3
B1
Accept equiv eg ∠ AMO =
M1
Use of arc = 4 × ( ∠ AMB in rads)
A1
3
15
Alternative to (b)(i):
Writing r= 3 + 2cos θ in cartesian form (M1A1)
Finding cartesian coordinates of points A and B ie (2, ± 2 2) (M1A1)
Finding length AB (M1A1)
TOTAL
75