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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Level
MARK SCHEME for the May/June 2010 question paper
for the guidance of teachers
9231 FURTHER MATHEMATICS
9231/13
Paper 13, 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 must be read in conjunction with the question papers and the report on the
examination.
•
CIE will not enter into discussions or correspondence in connection with these mark schemes.
CIE is publishing the mark schemes for the May/June 2010 question papers for most IGCSE, GCE
Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.
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Page 2
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
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.
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Page 3
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
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.
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1
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
Syllabus
9231
Paper
13
Relevant working from
 0 −3 0 


A – 5I =  1 − 3 1  or the equivalent in equations
 − 1 3 − 1


M1
1
 
to obtain an eigenvector of the form  0 
 −1
 
A1
[2]
An eigenvalue of A + A2 is 5 + 25 = 30
Corresponding eigenvector, as above
B1
B1
[2]
No penalty for not hence methods.
 − 1
 
Accept λ = 6 with  − 1
1
 
3
 
 
and λ = 20 with  1 
 − 1
 
Accept linear scaling of eigenvectors.
2
2 sin α
∑
N
n=1
sin (2nα ) = cos α − cos[(2 N + 1)α ]
M1A1
⇒ displayed result (AG)
M1A1
[4]
cos(2 N + 1)π / 3 oscillates finitely as n → ∞ ⇒
Require α =
3
∑
∞
n=1
sin (2nπ / 3) does not converge (CWO)
π
π
1
, ‘oscillate’ or values of cos(2 N + 1) given as
or –1
3
3
2
H k : x k > 2 for some k
(
B1
[1]
B1
)
x k +1 − 2 = 2 x k2 − 8 / (2 x k + 3)
M1A1
H k ⇒ 2 x k2 − 8 > 0 ⇒ x k +1 > 2 ⇒ H k +1
A1
x1 = 3 > 2 ⇒ H 1 is true
B1 CWO
Completion of the induction argument
A1
[6]
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Page 5
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
Syllabus
9231
Paper
13
Alternatively for lines 2 and 3:
x k +1 = x k + 12 −
3 12
(2 x k
M1A1
+ 3)
H k ⇒ 2 x k + 3 > 7 ⇒ H k +1
OR xk +1 = xk +
A1
xk − 2
(2 xk + 3)
M1A1
xk > 2 ⇒ xk +1 > 2
OR xk +1 − xk =
A1
xk − 2
(2 xk + 3)
M1A1
xk > 2 ⇒ xk +1 > xk > 2
A1
Minimum conclusion is ‘Hence true for n [ 1’.
4
dx / dt = t cos t , dy / dt = t sin t (both)
(dx / dt )2 + (dy / dt )2
B1
=t
(
B1
)
S = 2π ∫ π0 / 2 t sin t − t 2 cos t dt (AEF)
M1A1
∫ t sin t dt = −t cos t + sin t
B1
∫ t 2 cos t dt = t 2 sin t + 2t cos t − 2 sin t
B1
[
S = (π / 2 ) 12 − π 2
]
Accept forms such as 6π − π 3 / 2 , etc.
A1
[7]
OR for lines 4 and 5
[
{(
) ∫
2π (− t cos t ) + cos t dt − t 2 sin t − 2t sin t dt
∫
[
= 2π − t cos t + sin t − t 2 sin t − 2t cos t + 2 sin t
]
}]
π /2
0
π /2
0
(LNR)
(LNR)
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M1
A1
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Page 6
5
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
Syllabus
9231
Paper
13
(c + is )5 = c5 + 5c 4 (is ) + 10c3 (− s 2 ) + 10c 2 (is )3 + 5c(is )4 + (is )5
M1
θ = 5c 4 s − 10c 2 s 3 + s 5
M1
(
)
(
2
)
= 5s 1 − s 2 − 10s 3 1 − s 2 + s 5
M1
= 16 s 5 − 20 s 3 + 5s (AG)
A1 OEW
[4]
x = sin θ ⇒ sin 5θ = −1 / 2
M1
Roots are sin qπ where q = 7/30, 11/30, 31/30, 35/30, 43/30
A3
A1 for 1 root: + A1 for 2 further roots: + A1 for completion
CWO
CWO
[4]
Alternative answers
q=
11 23 35 47 59
,
,
,
,
30 30 30 30 30
or
q=
6
7 19 31 43 55
,
,
,
,
30 30 30 30 30
(i) One asymptote is x = –1
y = x – 4 – 3/(x + 1)
Require y = x + non-zero constant.
Other asymptote is y = x – 4
B1
M1
A1
[3]
Alternatively for last two marks:
OR x + k ≈ x 2 − 3 x − 7 / (x + 1) for large x ⇒ x 2 + (k + 1)x + k ≈ x 2 − 3 x − 7 for large x
⇒ k + 1 = −3 ⇒ k = −4 ⇒ other asymptote is y = x – 4
(
(
)
)
OR x 2 − 3 x − 7 / ( x + 1) = mx + c ⇒ (m − 1)x 2 + (m + c + 3)x + c + 7 = 0
Put m − 1 = 0, m + c + 3 = 0 to obtain other asymptote is y = x – 4
(ii) Obtains any correct result for dy/dx
[
]
M1
A1
M1
⇒ K ⇒ dy / dx = (x + 1) + 3 / (x + 1) = 1 + 3 / (x + 1)
No comment required.
2
M1
A1
2
2
(iii) Axes and asymptotes correctly placed
Right-hand branch
Left-hand branch
A1
[2]
B1
B1
B1
Deduct 1 overall for bad form(s) at infinity
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Page 7
7
(i)
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
(
)
(
2
)
Syllabus
9231
2
dx / dt = 2t − 2t 3 e − t , dy / dt = 1 − 2t 2 e −t (both, AEF)
Obtains displayed result (AG)
(− 1 + t
2
(
)
− 2t 4 e t
)
M1A1A1
M1
2
3
4t 3 1 − t 2
Two terms in numerator must be combined.
8
B1
M1A1
[3]
(ii) Any correct result for d (dy / dx ) / dt in terms of t (*)
M1 – Quotient Rule A1A1 for terms in numerator
d 2 y / dx 2 = (∗) × dt / dx expressed in terms of t
Simplify to, e.g.,
Paper
13
Complementary function is Ae − x + Be −4 x
A1
[5]
M1A1
Particular integral of form P sin 3x + Q cos 3 x , so that
− 5 P + 15Q = 10 and 15 P − 5Q = −20
M1
⇒ P = −7 / 5, Q = −1 / 5
A1
General solution is y = Ae − x + Be −4 x − 1.4 sin 3 x − 0.2 cos 3 x
A1
[5]
Ae − x + Be −4 x → 0 as x → +∞ ,
M1
tan φ =
1
7
M1
R = 2 = 1.41, φ = π + arctan(1 / 7 ) = 3.28
B1A1
Accept R = 2 , but must be positive.
[4]
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Page 8
9
[
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
Syllabus
9231
]
D s n +1c = − s n + 2 + (n + 1)s n c 2 , where D ≡ d / dx, s = sin x, c = cos x
Paper
13
M1
= K = −(n + 2 )s n + 2 + (n + 1)s n
A1
Integrates w.r.t. x over the range [0, π / 2] to obtain 0 = −(n + 2 )I n + 2 + (n + 1)I n
M1
Result (AG)
A1
[4]
[
]
π /2
OR I n = − cs n−1 0
+ (n − 1)
∫
π /2
0
c 2 s n−2 dx
M1
⇒ (n − 1)I n − 2 − (n − 1)I n (n [=2)
M1A1
⇒ I n = [(n − 1) / n]I n − 2 ⇒ I n + 2 = [(n + 1) / (n + 2 )]I n for n [ 0
A1
OR Starts with I n + 2 and relates to I n directly: mark as above
OR I n =
∫
π /2
0
sin n+2 θ cosec 2θ dθ
[
= − sin n+2 θ cot θ
= 0 + (n + 2 )
∫
π /2
0
]
π /2
0
+
π /2
∫ (n + 2)sin
(
0
n+1
θ cos θ cot θdθ (LNR)
)
sin n θ 1 − sin 2 θ dθ (LR)
M1
= (n + 2 )I n − (n + 2 )I n + 2
⇒ I n+ 2 =
M1
A1
(n + 1) I
(n + 2) n
A1
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Page 9
y=
∫
∫
1
2
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
π / 2m
0
π / 2m
0
Syllabus
9231
sin 8 mx dx
M1 (LR)
sin 4 mx dx
let u = mx
y=
∫
∫
1
2
M1
π /2
0
π /2
0
sin 8 u du
π
2
I6 =
5 3 1 π
× × ×
6 4 2 2
1
2
A1
sin 4 u du
I0 =
y=
Paper
13
I2 =
1 π
×
2 4
I4 =
I8 =
3 1 π  3π 
× × =

4 2 2  16 
B1
7 5 3 1 π  105π 
× × × × =

8 6 4 2 2  768 
I 8 35
(or 0.365)
=
I 4 96
B1
B1
[6]
OR for last 3 marks
I 8 35
=
(oe)
I 4 48
∴y =
10 (i)
(ii)
M1A1
35
96
A1
x = 1 / y ⇒ 2 y 4 − 4 y 3 − cy 2 − y − 1 = 0
∑α
2
∑α
−2
= 1 − 2c
M1A1
= 4+c
(M1 is for use of
(iii) S =
M1A1
[2]
A1
∑α
2
=
(∑α )
2
∑ (α − α ) = ∑α + ∑α
−1 2
2
−2
−2
∑αβ
in either part.)
− 8 = −c − 3
A1ft is for adding answers to (ii) correctly and subtracting 8.
[3]
M1A1√
[2]
(iv) c = −3 ⇒ S = 0 so that if all roots are real then α = ±1
and similarly for β , γ , δ
M1A1 CWO
This is impossible since e.g., αβγδ = −2, or any other contradiction
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11 (i)
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
Syllabus
9231
dr / dθ = − a / (1 + θ )
2
Paper
13
M1
Since a > 0, then dr / dθ < 0, ∀ points of C
OR a(> 0 ) constant and as θ (> 0 ) increases, 1 + θ increases
A1
[2]
M1
a
decreases
1+θ
A1
(ii) y = a sin θ / (1 + θ )
M1
∴
dy / dθ = 0 ⇒ K ⇒ (1 + θ ) cos θ − sin θ = 0
A1
⇒ tan θ = 1 + θ (AG)
A1
tan θ − 1 − θ = −0.135 when θ = 1.1 : tan θ − 1 − θ = +0.372 when θ = 1.2
B1
[4]
OR equivalent argument for B1 such as:
tan (1.1) ≈ 1.96 < 2.1, tan (1.2 ) ≈ 2.37 > 2.2
B1
or f (θ ) = tan θ − (1 + θ ) ⇒ f (1.1) = −0.14 f (1.2 ) = 0.37
(iii) Sketch:
Approximately correct shape and placement for 0 Y θ Y π/2 and passing through
(a, 0) and (0.4a, π / 2), approximately, indicated in some way
Maximum in interval (π / 4, π / 2 )
(
(iv) A = a 2 / 2
(
)[
)∫
π /2
0
B1
[3]
(1 + θ )−2 dθ
M1
]
−1 π / 2
0
= a 2 / 2 − (1 + θ )
B1B1
A1
= K = πa 2 / 2(π + 2 )
A1
Do not accept double minus signs or fractions in the denominator for the final mark.
[3]
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Page 11
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
Syllabus
9231
Paper
13
12 EITHER
(i) (i + j) × (j + 2k) = 2i – 2j + k
M1A1
PQ = (4i + j + 3k).(2i – 2j + k)/3 = 9/3 = 3
M1A1
[4]
(ii) (4 + λ)/2 = (1 + λ – µ)/(–2) = (3 – 2µ)/1 (AEF)
OR (4 + λ) + (1 + λ – µ) = 0, (1 + λ – µ + 2(3 – 2µ) = 0, both
M1A1
⇒K⇒ µ =1
M1A1
Position vector of Q is –i + j + k
A1
[5]
Parts (i) and (ii) together:
 4+λ 


PQ = 1 + λ − µ 
 3 − 2µ 


B1
1
 
PQ. 1  = 0 ⇒ 5 + 2λ − µ = 0
 0
 
M1A1
0
 
PQ. 1  = 0 ⇒ 7 + λ − 5µ = 0
 2
 
A1
λ = –2, µ = 1
M1A1
 2 
 
PQ. − 2  ⇒ PQ = 3
 1 
 
M1A1
 − 1
 
OQ =  1 
1
 
A1
[9]
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Page 12
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
Syllabus
9231
(iii) (i + j) × (4i + j + 3k) = 3i – 3j – 3k
Paper
13
M1A1
p2 = [(i − j − k )(
. 4i + k )]/ 3
M1
=K= 3
A1
[4]
for final 2 marks
π :x− y−z =0
 −1−1−1
perpendicular distance = 
 2 2 2
 1 +1 +1

= 3


or any other method.
M1 is for a complete strategy and A1 for
3.
12 OR
1

3
(i) (a) 
1

3

1
5
7
1


9 17 25 
0
→K→ 

7 7 11
0




6 16 23 
0
1 5 7

6 2 4
0 0 0

0 0 0 
OR Establishes equivalent result for MT
OR Establishes 2 linearly independent relations for rows or columns
⇒ r (M ) = 2 ⇒ dim(R ) = 2
 1 
 
 3 
(b) Basis for RT is  ,
 1 
 3 

M1A1
A1
 1 
 
 9 
 7 
 
 6 
 
M1A1
[5]
 1 
1  1 


   
 − 15 
 3  9 
= p  + q 
(ii) Let 

1
7
− 17


   
 −6 
 3  6 


   
Solves any 2 of:
p + q = 1, 3p + 9q = –15, p + 7q = –17, 3p + 6q = –6 to obtain p = 4, q = –3
Checks consistency with other 2 equations
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A1
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Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
Syllabus
9231
 4   1 
  

 − 3   − 15 
(iii) M  = 
OR Works with equations (M written out fully.) (AG)
0
− 17 

  
 0   −6 
  

Paper
13
B1 OEW
e1 and e2 basis vectors for T ⇒ Me1 = 0, Me 2 = 0
M1
 4 
 1 
 1 
 




 − 15 
 − 3
 − 15 
+ λ0 + µ0 = 
, ∀λ , µ
M  + λMe1 + µMe 2 = 
0
− 17 
− 17 
 




 0 
 −6 
 −6 
 




A1
[3]
(iv) Need –3 + λ + 2µ = 0
M1
 37 
 
 0 
λ =1⇒ µ =1⇒ x =  
−3
 
 − 3
 
A1
[2]
Accept a parametric answer
 46   − 3 
   
 0   0 
e.g.   + t  
−9
2
   
 0   − 1
   
Augmented Matrix Method
1

3
(i) (a) 
1

3

1 5 7
9 17 25
7 7 11
6 16 23
1  1
 
− 15   0
~
− 17   0
 
− 6   0
1 5 7
6 2 4
6 2 4
3 1 2
1  1
 
− 18   0
~
− 18   0
 
− 9   0
1
3
0
0
5
1
0
0
7
2
0
0
⇒ r (M ) = dim(R ) = 2
 1 
 
 3 
(b) Basis for R is  ,
 1 
 3 

1 

− 9
0 

0 
M1A1
A1
 1 
 
 9 
 7 
 
 6 
 
M1A1
[5]
© UCLES 2010
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Page 14
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2010
Syllabus
9231
(ii) and (iii)
x + y + 5 z + 7t = 1
3 y + z + 2t = −9
Paper
13
M1A1
 x   46 
 14 
 3 
   
 
 
 y  0 
 1 
 0 
⇒   =   + λ  + µ 
z
−9
−3
−2
   
 
 
t  0 
 0 
 1 
   
 
 
A1
 x  4 
   
 y   − 3
λ = −3, µ = 0 ⇒   =  
z
0
   
t  0 
   
B1
 4   1 
  

 − 3   − 15 
∴ M  = 
0
− 17 
  

 0   −6 

  
Me1 = Me2 = 0
M1
 4 
  1 
 
 

  − 3
  − 15 
∴ M   + λe1 + µe 2  = 

 0 
  − 17 
 0 
  −6 

 
 
A1
 3 
 x   46 
 
   
 0 
 y  0 
(iv) λ = 0   =   + µ  
−2
z
−9
 
   
 1 
t  0 
 
   
M1A1
[8]
© UCLES 2010
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