w
w
ap
eP
m
e
tr
.X
w
CAMBRIDGE INTERNATIONAL EXAMINATIONS
s
er
om
.c
GCE Advanced Level
MARK SCHEME for the May/June 2013 series
9231 FURTHER MATHEMATICS
9231/13
Paper 1, maximum raw mark 100
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of
the examination. It shows the basis on which Examiners were instructed to award marks. It does not
indicate the details of the discussions that took place at an Examiners’ meeting before marking began,
which would have considered the acceptability of alternative answers.
Mark schemes should be read in conjunction with the question paper and the Principal Examiner
Report for Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE
Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
Page 2
Mark Scheme
GCE A LEVEL – May/June 2013
Syllabus
9231
Paper
13
Mark Scheme Notes
Marks are of the following three types:
M
Method mark, awarded for a valid method applied to the problem. Method marks are
not lost for numerical errors, algebraic slips or errors in units. However, it is not
usually sufficient for a candidate just to indicate an intention of using some method or
just to quote a formula; the formula or idea must be applied to the specific problem in
hand, e.g. by substituting the relevant quantities into the formula. Correct application
of a formula without the formula being quoted obviously earns the M mark and in some
cases an M mark can be implied from a correct answer.
A
Accuracy mark, awarded for a correct answer or intermediate step correctly obtained.
Accuracy marks cannot be given unless the associated method mark is earned (or
implied).
B
Mark for a correct result or statement independent of method marks.
•
When a part of a question has two or more "method" steps, the M marks are generally
independent unless the scheme specifically says otherwise; and similarly when there are
several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a
particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme.
When two or more steps are run together by the candidate, the earlier marks are implied and
full credit is given.
•
The symbol √ implies that the A or B mark indicated is allowed for work correctly following on
from previously incorrect results. Otherwise, A or B marks are given for correct work only.
A and B marks are not given for fortuitously "correct" answers or results obtained from
incorrect working.
•
Note:
B2 or A2 means that the candidate can earn 2 or 0.
B2/1/0 means that the candidate can earn anything from 0 to 2.
The marks indicated in the scheme may not be subdivided. If there is genuine doubt
whether a candidate has earned a mark, allow the candidate the benefit of the doubt.
Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong
working following a correct form of answer is ignored.
•
Wrong or missing units in an answer should not lead to the loss of a mark unless the
scheme specifically indicates otherwise.
•
For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f.,
or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated
above, an A or B mark is not given if a correct numerical answer arises fortuitously from
incorrect working. For Mechanics questions, allow A or B marks for correct answers which
arise from taking g equal to 9.8 or 9.81 instead of 10.
© Cambridge International Examinations 2013
Page 3
Mark Scheme
GCE A LEVEL – May/June 2013
Syllabus
9231
Paper
13
The following abbreviations may be used in a mark scheme or used on the scripts:
AEF
Any Equivalent Form (of answer is equally acceptable)
AG
Answer Given on the question paper (so extra checking is needed to ensure that
the detailed working leading to the result is valid)
BOD
Benefit of Doubt (allowed when the validity of a solution may not be absolutely
clear)
CAO
Correct Answer Only (emphasising that no "follow through" from a previous error
is allowed)
CWO
Correct Working Only – often written by a ‘fortuitous' answer
ISW
Ignore Subsequent Working
MR
Misread
PA
Premature Approximation (resulting in basically correct work that is insufficiently
accurate)
SOS
See Other Solution (the candidate makes a better attempt at the same question)
SR
Special Ruling (detailing the mark to be given for a specific wrong solution, or a
case where some standard marking practice is to be varied in the light of a
particular circumstance)
Penalties
MR –1
A penalty of MR –1 is deducted from A or B marks when the data of a question or
part question are genuinely misread and the object and difficulty of the question
remain unaltered. In this case all A and B marks then become "follow through √"
marks. MR is not applied when the candidate misreads his own figures – this is
regarded as an error in accuracy. An MR–2 penalty may be applied in particular
cases if agreed at the coordination meeting.
PA –1
This is deducted from A or B marks in the case of premature approximation. The
PA –1 penalty is usually discussed at the meeting.
© Cambridge International Examinations 2013
Page 4
Qu
No
1
Mark Scheme
GCE A LEVEL – May/June 2013
Commentary
Simplifies.
Syllabus
9231
Solution
Marks
f(r + 1) – f(r) = r ( r + 1)!−(r − 1)r!
= r!(r 2 + r − r + 1) = r!(r 2 + 1)
Uses difference
method.
Obtains result.
n
∑
1
∴
M1
= n(n + 1)!−0 = n(n + 1)!
A1
∑
A1
= 2n(2n + 1)!− n(n + 1)!
Part
Mark
Total
M1
A1
= f(2) – f(1) + f(3) – f(2) +…f(n + 1) – f(n)
2n
Paper
13
5
[5]
n +1
2n
(Or directly using
∑ = f (2n + 1) − f (n + 1) from the
n +1
method of differences.)
2
1
Makes substitution.
Squares.
y 2 − 4 y + 3y 2 − 2 = 0
4
2
2
⇒ 9 y = 4 + y + 16 y − 4 y + 16 y − 8 y
3
M1
M1
(N.B. Must see both terms in y2.)
Obtains result.
⇒ y 4 − 8 y 3 + 12 y 2 + 7 y + 4 = 0 (AG)
S 2 = 0 2 − 2 × (−4) = 8
S 8 = 8S 6 − 12 S 4 − 7 S 2 − 16
⇒ S 8 = 8S 6 − 12 S 4 − 56 − 16 = 8S 6 − 12 S 4 − 72 (AG)
A1
3
B1
M1
A1
3
[6]
7
[7]
Alternatively – for final two marks.
S2 = 8 , S3 = –9 , S4 = 40 , S5 = –60 , S6 =203 , S7 = –378
S8 = 1072 (generated by substitution of roots in equations
and summing.)
Then 8S 6 − 12 S 4 − 72 = 1624 − 480 − 72 = 1072 = S 8
M1 requires a complete method, A1 if all correct.
3
Differentiates once.
Rearranges.
Shows true for n = 1 .
States inductive
hypothesis.
(May be seen by
implication.)
(
)
d x
e sin x = e x sin x + e x cos x
dx
 1

1
cos x 
= 2e x 
sin x +
2
 2

1 

= 2e x sin x + π  ⇒ H1 true .
4 

Hk :
dk
(e x sin x) =
k
x
d
Differentiates.
d k +1
=
dx k +1
Rearranges.
=
( 2)
( 2) e
k
x
1 

sin  x + kπ 
4 

( 2 )  sin x + 14 kπ e
k +1
B1
M1
A1
B1
1 

+ e x cos x + kπ  
4 

M1
 1
1  1
1 


sin  x + kπ  +
cos x + kπ  
e x 
4 
4 
2


 2
A1
k

x
© Cambridge International Examinations 2013
Page 5
Shows H k ⇒ H k +1
and states
conclusion.
4
Mark Scheme
GCE A LEVEL – May/June 2013
=
( 2)
k +1
1


e x sin x + (k + 1)π 
4


⇒ H k +1 true.
Syllabus
9231
A1
∴ ,by PMI, true for all positive integers n. (CWO)
3 ଶ ᇱ 3 ଶ ᇱ 6 0
(B1 for 1st term and = 0, but allow recovery)
At (1,–2) 12 ᇱ 3 ᇱ 12 0
ସ
(AG)
⇒ 9 ᇱ 12 ⇒ ᇱ B1B1
Differentiates again.
(One mark for
each pair of terms.)
3 ଶ ᇳ 6 ᇱ ଶ 6 ᇱ 3 ଶ ᇳ 6 ᇱ 6 0
At (1,–2)
B1 for each pair of terms. 3rd mark includes = 0, but
allow recovery.
B1B1
B1
Substitutes values.
12y″ –12 ×
Differentiates.
Paper
13
B1
ଷ
Obtains result.
⇒ 9 ᇳ 5
Finds I1.
16
I1 =
∫
1
0
I 2 n +1 =
xe
∫
− x2
1
0
– –8+3y″ + 6×
9
ଶ଴
ଷ
⇒ ᇳ  e − x2
dx =  2
M1
(Allow – 0.741)
A1
8
M1A1
2
I3 =
1
∫ 2nx
2
2 n −1
0
e -x
dx
2
M1A1
A1
3
M1
1 1
1 1 1
− −
= −
2 2e 2e 2 e
Obtains I 5 , or some
intermediate result,
correctly.
Obtains I 7 .
5
1 1 1
= 1−
I 5 = 2 −  −
2e
 2 e  2e
A1
5 1
8

I 7 = 31 −  −
= 3−
e
 2e  2e
A1
Reduces to echelon
form
−1   − 2
3
− 2 5

 
−4 −2   0
1
 0
 6 − 14 − 13 1  ~  0

 
 1
− 2 − 11  0
1

 − 2 5 3 −1   − 2 5

 
 0 1 − 4 − 2  0 1
~
~
0 0 0
0   0 0

 
 0 0 27 − 9   0 0

 
(N.B. Allow matrix
with a row of
zeros – not in
[8]
1

1 1
(AG)
 = −
 0 2 2e
1
 1
= −  − [0] + nI 2 n−1 = nI 2 n−1 −
(AG)
2e
 2e 
Obtains reduction
formula.
6
– 12 =0
2
1
Attempts to use
reduction formula at
least once.
ଶ଻
3
x 2 n +1e − x dx
 2n e -x2 
= − x
 +
2 

0
Integrates by parts.
ଶ଴
–4
−1 
3

1 −4 −2 
1 −4 −2 

7 − 1 − 23 
5
−1 

− 4 − 2
(α ≠ 0)
3 −1 

0
0 
M1A1
3
© Cambridge International Examinations 2013
A1
3
[8]
Page 6
echelon form.)
Solves set of
equations.
Obtains basis.
Mark Scheme
GCE A LEVEL – May/June 2013
y − 4 z − 2t = 0
3z − t = 0
 25 
 
10
⇒ K1  
(OE)
 1 
 3 


If α = 0
6
ctd
Paper
13
− 2 x + 5 y + 3z − t = 0
Solves equations in
second case.
Obtains basis.
Syllabus
9231
⇒ K2
− 2 x + 5 y + 3z − t = 0
y − 4 z − 2t = 0
 23   9 
5
0
   
0
10
 8   4 
  ,   (OE) e.g. ൮ 2 ൲ or ൮ 9 ൲
 2   0 
–4
–23
 0   2 


M1
A1
M1
A1A1
8
[8]
Other Methods
Working from the
start with equations
Use of transpose
matrices
Sets up both sets of equations
M1A1
Solves in the case α ≠ 0
States K1. correctly
M1A1
A1
Solves in the case α = 0
States K2. correctly
M1A1
A1
Uses row operations to reduce transpose matrices to
echelon form.
When α ≠ 0
− 2

 0
T
M ~ 
0

 0

⇒
K1
α ≠0
− 2

 0
T
M ~ 
0

 0

r1
1





5r1 + 2r2
2 2 7


by 


5r1 + 2r3 − 4r4
0 0 45



 50r + 20r + 2r + 6r 
0 0 0 
2
3
4
 1
M1A1
 50  
 25 
  
 
 20  
 10 
   or  
 2  
 1 


 6  
 3 




M1A1
r1
0 6 0





5
2
r
r
+
2 2 0


2
1
by 

23r1 + 8r2 + 2r3 
0 0 0



 − 9r − 4r − 2r 
0 0 0 
1
2
4


M1A1
0 6
© Cambridge International Examinations 2013
[8]
Page 7
Mark Scheme
GCE A LEVEL – May/June 2013
 23   9 
   
 8   4 
  ,  
 2   0 
 0   2 


⇒ K2
7
y ′ = λe − x − λxe − x , y ′′ = −2λe − x + λxe − x
Finds complementary
function
(m + 1)(m + 4) = 0 ⇒ m = −1 or − 4
and hence general
solution.
8
y ′′ + 5 y ′ + 4 y = 3λe − x = 6e − x ⇒ λ = 2
y = Ae
C.F:
y = Ae
G.S.:
+ Be
−4 x
−x
−x
⇒ y ′ = − Ae − 4 Be + 2e − 2 xe
y (0) ⇒ A + B = 2 , y′(0) ⇒ 2 − A − 4 B = 3
⇒ A = 3 and B = –1
⇒ y = 3e − x − e −4 x + 2 xe − x
Differentiates and
attempts
ds
to find
.
dt
Integrates to find arc
length.
x& = 3t , y& = 3t 2 ⇒
s=
∫
0
x=
∫
∫
0
xydx
6
Uses correct
formulae for
y-coordinate.
Finds value by
integration.
3
= 2
Eliminating t.
∫
∫ t dt =
∫ t dt
0
2
6
[10]
t2 3
.t .3tdt
2
M1
A1
2
0
M1
A1
3
∫
2
0
4
t 3 .3tdt
3 1 7 
t
2  7  0
=
2
4
∫
∫
3 5
30
× ×4 =
(Or 4.29)
2 7
7
∫
∫
0
M1
A1
M1
0
2
2
∫ t dt =
∫ t dt
0
2
6
M1A1
(Allow 10.2)
1 5 
0
5 t 
0

1 6 2
1 2 6
y dx
.t .3tdt
0
0
y= 2 6
= 2 2
ydx
t 3 .3tdt
1
= 2
A1
2
2
0
Alternative layout:
ydx
=
M1
2
⇒ s = 5 5 −1
6
ds
= 9t 2 + 9t 4 = 3t 1 + t 2
dt
3


dt = (1 + t 2 ) 2 

 0
1
3t (1 + t 2 ) 2
2
0
Finds value by
integration.
4
B1√
−x
Differentiates G.S.
Uses initial
conditions to find
constants, and obtain
particular solution.
Uses correct
formulae for
x-coordinate.
B1B1
M1A1
[8]
M1
A1
A1
−4 x
+ Be −4 x + 2 xe − x
−x
Paper
13
M1A1
Differentiates twice
and substitutes to
find value of λ.
−x
Syllabus
9231
7
4
1 1 8 
t
2  8  0
2
1 5 
5 t 
0

=
1 5
5
(Or 2.5)
× ×8 =
2 8
2
1
ydx (B1) xydx (B1) 2 y2 dx (B1) (in terms of t.)
Then award M1A1 for each of x and y .
1
Area (B1) xydx (B1) y2 dx (B1)
2
Then award M1A1 for each of x and y.
© Cambridge International Examinations 2013
M1
A1
7
[11]
Page 8
9
Proves initial result.
States eigenvalues.
Finds one
eigenvector.
All correct.
States eigenvalues of
B.
Finds eigenvectors of
B.
(N.B. Same as A’s is
M0)
10
Uses identity.
Mark Scheme
GCE A LEVEL – May/June 2013
Syllabus
9231
Ae = λe
BMe = MAM-1Me = (MAIe)
= MAe = Mλe = λMe
(CWO)
(Me ≠ 0 since M non-singular ⇒ λ is an eigenvalue.)
B1
M1
A1
Eigenvalues are: –1 , 1 , 2
i j k  6
1
 
 
0 2 1 =  0 ~  0
λ = –1
 0
0 2 4  0 
 
i
j k 8
1
 
 
− 2 2 1 = 8 ~ 1
λ=1
 0
0 2 4  0 
 
i
j k 9
 3
 
 
− 3 2 1 = 12  ~  4 
λ=2
1
0 − 1 4  3 
 
B1
A1
1
1
1
4
4
1
4
1
1
2
4
√2
= cos π – 2θ+ π +
3
= cos 2θ– π +
4
1
√2
Closed loop through origin, in correct position.
Obtains line of
symmetry.
For line of symmetry 2θ– π=0⇒θ= π.
Integrates correctly.
Substitutes limits.
Obtains given
answer.
π
3
3
4
8
3
3
(AG)
3
π
4
1
3
1
3
1
3
16
= –
3
1
16
1
3
4
8
2
1
1
2√2
1
16
4
3
3
θ 4π
4
2 0
1
3
M1
A1
A1
N.B Method marks are dependent in final part.
If factor missing throughout – award M’s (Max 3)
3
3
dM1
(AG)
8
3
dM1A1
sin 2θ– π + + + π – – = π+1
M1
A1
B1B1
= 0 cos 4θ– π +√2 cos 2θ– π +1 dθ
2
2
2
4
= sin 4θ– π +
[11]
B1
A= 04 cos2 2θ– π +√2 cos 2θ– π + dθ
2
4
4
2
1
4
M1
Sketches graph.
Uses area of sector
formula.
Rearranges.
M1A1
A1
2sin θ cos θ– π = sin 2θ – π + sin π
3
4
B1
 1  1  4
     
Corresponding eigenvectors are:  0  ,  1  ,  4 
 0  0 1
     
1
3
M1A1
Eigenvalues of B are –1 , 1 , 2
Uses sine-cosine
link.
Obtains result.
Paper
13
1
1
If 2× 08 r2 dθ , penultimate line is = π – – 2
8
8 2
© Cambridge International Examinations 2013
6
[12]
Page 9
11 E
Finds PQ.
Finds direction of
common perpendicular.
Mark Scheme
GCE A LEVEL – May/June 2013
 − 3 + µ − 3λ 


PQ =  − 6 µ − 2λ 
 12 − 2 µ + λ 


i
j
k
 − 10 
2


 
3 2 − 1 =  5  ~  − 1
4
1 − 6 − 2  − 20 
 
Syllabus
9231
Paper
13
M1A1
M1
(Or uses scalar product
of PQ with the
direction vector of each
line.)
Obtains two correct
equations. e.g.
Solves.
Finds p and q.
Finds common
perpendicular.
Finds PA (or QA or PB
or QB)
Uses triple scalar product
to find shortest distance.
Alternative for last 4
marks:
− 3 + µ − 3λ = 12 µ + 4λ
− 24 µ − 8λ = −12 + 2 µ − λ
M1A1
µ = 1 , λ = −2
 1+ µ   2
 4 + 3λ   − 2 

  

  
p =  7 + 2λ  =  3  q =  7 − 6 µ  =  1 
11 − 2 µ   9 
 −1− λ   1 

  

  
i
AB × PQ = − 1
2
A1
A1
j
0
k 4
 
4 = 12 
− 1 4  1 
 6 
2
3
–1
 
PA  4  QA 6 PB 4 QB 6  − 2
–10
10
2
 
 6  4
   
 4  ⋅ 12 
 − 2  1 
    = 24 + 48 − 2 = 70
= 5.52
16 + 144 + 1
161
161
Plane through e.g. P in direction PA :
4 12 29 0 . Award M1A1.
Then use of distance of point from line
70
. Award M1A1.
formula to get
161
© Cambridge International Examinations 2013
8
A1
M1A1
B1
M1A1
A1
6
[14]
Page 10
11 O
Mark Scheme
GCE A LEVEL – May/June 2013
Syllabus
9231
Draws Argand diagram.
Shows position of 3 cube roots on Argand diagram.
E.g. Uses De M’s
theorem.
1 = cos2kπ + isin 2kπ ; k = 0, 1, 2.
Gives a + ib form.
2π
2π
4π
4π
and ω 2 = cos
+ isin
+ isin
3
3
3
3
1
3
1
3
, ω2 = − −i
ω = − +i
2
2
2
2
⇒ ω = cos
B1
Paper
13
1
M1A1
A1
3
S.C. Award B1 for cube roots without/incorrect
working. Look out for algebraic forms from
z–1z2 +z+1=0 then squaring one to get the
other, which scores M1A1 A1.
(6 − ω )− 3ω (9ω
3
Expands determinant.
2
)
− 2ω 2 + 2ω 2 (3ω − 4ω )
Uses ω 3 = 1 etc.
= 5 – 21 – 2 = –18
Uses a + ib forms
correctly.
 1
3   3 1 
z = 4 3  − −
i  − 4
− i
 2 2   2 2 
= −2 3 − 6i − 2 3 + 2i
Simplifies.
Rearranges
Reverts to r, θ form.
One cube root.
Other two.
(
= −4 3 + i
)
 3 1 
= −8
+ i 
2
2 

1
1 
7
7 


= −8  cos π + isin π  = 8  cos π + isin π 
6
6 
6
6 


7
7 

Cube roots are: 2  cos π + isin π  ,
18
18 

19
19 
31
31 


2  cos π + isin π  , 2  cos π + isin π 
18
18 
18
18 


© Cambridge International Examinations 2013
M1A1
B1
3
M1A1
A1
M1A1
5
B1
B1
2
[14]