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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Level
MARK SCHEME for the October/November 2010 question paper
for the guidance of teachers
9231 FURTHER MATHEMATICS
9231/01
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 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 October/November 2010 question papers for most IGCSE,
GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level
syllabuses.
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Mark Scheme: Teachers’ version
GCE A LEVEL – October/November 2010
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Syllabus
Paper
9231
01
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.
© UCLES 2010
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Mark Scheme: Teachers’ version
GCE A LEVEL – October/November 2010
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Syllabus
Paper
9231
01
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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Mark Scheme: Teachers’ version
GCE A LEVEL – October/November 2010
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Syllabus
Paper
9231
01
2
1
1 2x
1 2x
 dy 
−2 x 2
−2 x 2
1 +   = 1 + ( (e − e )) = (e + e )
x
d
2
4
 
1
2
Length =
∫ (
0
=
2
(
[(
)
1 2x
1
e + e − 2 x dx = e 2 x − e − 2 x
2
4
) (
)
)]
1
2
0
11
1 
 −

2n n + 2
1 3
1
1 
−
−

2  2 N + 2 N + 1 
Limit = ¾
4
3
M1
integrate
A1
cao
[4]
M1A1
 1

1   1
1   1
1
 N − N + 2  +  N − 1 − N + 1  +  N − 2 − N  + ...
 
 

1 

SN = 


2  1 1  1 1 
 −  +  − 

 2 4   1 3 

=
expression simplified
2
1 1
1 0
e −1
−1
0
AG
e −e − e −e =
4
4
4e
nth term is
M1A1
M1
sum of terms
A1
after cancellation
B1√
[4]
[1]
4
1
1

 1
2 3
− 
Area =  x 2 − x 2 dx =  x 2 − 2 x 2  = 8 / 3



 3
1

1
∫
B1
4
1
y= 2
∫
4
1

1  x2
1
( x − 2 + )dx 2  2 − 2 x + ln x 

1
x
=
A
A
M1
M1
A1
Final answer:
3
9
3
3
3
3
etc (ACF)
 ln 2 +  or
 ln 4 +  or ln 2 +
8
32
8
4
16 
2
© UCLES 2010
1 2
y dx
use of 2
A
integrate
correct
A1
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∫
[5]
Page 5
4
5
Mark Scheme: Teachers’ version
GCE A LEVEL – October/November 2010
www.studyguide.pk
Syllabus
Paper
9231
n = 0:
71 + 53 = 132 which is divisible by 44
Assume 72k + 1 + 5k + 3 is divisible by 44
Consider 72(k + 1) + 1 + 5(k + 1) + 3 = 7272k + 1 + 5.5k + 3
= 49(72k + 1 + 5k + 3) – 44.5k + 3
which is divisible by 44
B1
B1
M1
M1
A1
(k + 1) th term
in appropriate form
convincing argument
Alternative solution for final three marks:
Consider (72k + 3 + 5k + 4) – (72k + 1 + 5k + 3)
= 48(72k + 1 + 5k + 3) – 44.5k + 3
which is divisible by 44
M1
M1
A1
in appropriate form
convincing argument
∫
In + 2 = [–(1–x)n + 2 cosx] – (n + 2)(1 − x)
n +1
cos xdx
M1A1
= (1 + (n + 2)) + (n + 2)[((1 − x) n +1 sin x) + (1 − x) n sin xdx]
∫
M1
In + 2 = 1 – (n + 1)(n + 2) In AG
I6 = 1 – 5 × 6I4; I4 = 1 – 4 × 3I2; I2 = 1 – 1 × 2I0
A1
M1
01
[5]
integrate by parts again
[4]
1
∫
B1
I 0 = sin xdx = 1 − cos1
0
6
I6 = 1 – 30(1 – 12(1 – 2I0)) = 0.0177
M1A1
OR
I0 = 1 – cos1
I2 = 2cos1 – 1
I4 = 13 – 24cos1
I6 = 0.0177
Accept decimal versions
B1
M1
A1
A1
2
−1
α
1

1
0
−1
1
− 2α  →  0
0 − 3
0
4
− 2 − 2α 
0

0
1
−
3
−
2



Dim = 4 ⇒ α ≠ 1 AG
a + 2b – c = 0
2a + 3b – c = 0
2a + b + 2c = 0
b – 3c = 0
2
−1
0
0
−1
1
1
0
α 
− 2α 
4α − 2 
6α − 6 
[4]
(use of RF)
cao
M1A1
A1
[3]
Show a = b = c = 0
M1
Linearly independent and dim R(T) not 4: basis
A1
[2]
a + 2b – c = p
2a + 3b – c = 1 Attempt to find a, b, c in terms of q or p
2a + b + 2c = 1
b – 3c = q
6p + q = 3
M1A1
A1
[3]
Alternative solution:
Use row operations as in (i)
p


 1− 2p 

Final column 
 4 p − 2 
 6 p + q − 3
6p + q = 3
attempt to solve
M1
A1
A1
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Mark Scheme: Teachers’ version
GCE A LEVEL – October/November 2010
1− y
1
∴x =
x +1
y
Gives 6y3 – 7y2 + 3y – 1 = 0 AG
n = 1: given expression = sum of roots = 7/6
y=
n = 2:
1
∑ (α + 1)
2

= 

∑
2
1 
 −2
(α + 1) 
M1
13
∑ "αβ " = 36
From cubic in y,
3
13  7 
 1 
6 
 − 7. + 3  − 3 = 0
36  6 
 α +1
∑
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Syllabus
Paper
9231
01
use in given cubic equation
A1
B1
[2]
B1
[2]
M1
3
 1 

 = 73 / 216
 α +1
 ( β + 1)(γ + 1)(α + 1 


LHS =
3


(α + 1)


∑
A1
∑
[2]
M1
−1
73
1
=  ×
216
6
= 73/36 AG
8
(i)
M1
recognise product of roots
A1
1 + sin θ = 3 sin θ ⇒ sin θ =
1
2
[3]
M1
3 π 
 3 5π 
 ,  and  , 
2 6
2 6 
A1
(both)
[2]
B1
B1
B1
circle
cardioid behaviour at origin
cardioid closed and symmetry
[3]
(ii)
(iii) Subtract integrands
π
2×
1
2
M1
2
∫ (3 − 4 cos 2θ − 2 sinθ )dθ
π
M1
6
π
= [3θ − 2 sin 2θ + 2 cos θ ]π 2
M1A1
6
=π
AG
Alternative:
Area inside C1:
π
π
2
1
9
1

2×
9 sin 2 θ dθ = θ − sin 2θ 
2π
2
2

∫
6
=
A1
2
M1
π
6
9  π
3 
+
2  3
4 
A1
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Mark Scheme: Teachers’ version
GCE A LEVEL – October/November 2010
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Syllabus
Paper
9231
01
Area inside C2:
π
1 2
1
2×
1 + 2 sin θ + (1 − cos 2θ )dθ
2π
2
∫
6
π
1
 3θ
 2
=  − 2 cos θ − sin 2θ 
4
2
π 6
M1
π 9 3 

=  +
8 
2
Subtraction
Required area = π AG
9
(A1
M1
A1
(3 − λ )[(2 − λ )(3 − λ ) − 1] + 1(−(3 − λ )) = 0
(3 − λ )(λ − 1)(λ − 4) = 0
λ = 1, 3, 4
3 − λ
 −1
 0

if not earned earlier)
M1
M1
A1
[5]
characteristic equation
factorise
0  x   0 
− 1  y  =  0 
3 − λ  z   0 
−1
2−λ
−1
Solve for λ = 1: (1, 2, 1)
Solve for λ = 3: (1, 0, –1)
Solve for λ = 4: (1, –1, 1)
M1A1
A1
A1
1
M = 2
1

B1√
1
0
−1
1
− 1
1 
[7]
eigenvectors as columns
0
(except  0  )
0
 
−1
D=  0
0

10
0
1
0
0
0
8 
5
M1A1√ ft on eigenvalues
3 2
cos 5θ = c − 10c s + 5cs
4
2 3
sin 5θ = 5c s − 10c s + s
5
tan 5θ =
3
t − 10t + 5t
2
1 − 10t + 5t
tan 5θ = 0 ⇒ θ =
4
4
M1A1
5
[3]
use of de Moivre for (c + is)5
A1
AG
M1A1
nπ
5
intermediate step needed
[5]
justify values of n
[2]
M1
nπ
for n = 1, 2, 3, 4
5
π
2π
Roots ± tan , ± tan
5
5
Product of these roots = 5
π
2π
tan tan
= 5
5
5
Solutions tan
A1
B1
M1
A1
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Mark Scheme: Teachers’ version
GCE A LEVEL – October/November 2010
11 z′ = y + xy′
z′′ = 2y′ + xy′′
Obtain result
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Syllabus
Paper
9231
01
B1
B1
B1
Auxiliary equation: m2 + 4 = 0 : m = ±2i
CF: A cos 2 x + B sin 2 x
PI: z = ax2 + bx + c
Differentiate twice and substitute
a = 2, b = 0, c = 3
2
GS: z = A cos 2 x + B sin 2 x + 2 x + 3
[3]
M1
A1
M1
A1
A1√
their CF + their PI
2
1
π
π : (z = 0) gives A =
+3
2
2
z′ = –2Asin2x + 2Bcos2x + 4x
π
3π
: (z′ = –π) gives B =
y′ = –2, x =
2
2
2


1  π
3π
2
y = 
+ 3  cos 2 x +
sin 2 x + 2 x + 3 


x   2
2


y = 0, x =
B1
M1
A1
A1
[9]
(i) y′ = 0 ⇒ (x2 – 2x + λ)(2x + 2λ) – (x2 + 2λx)(2x – 2) = 0
M1
2
2
⇒ ... ⇒ (λ + 1)x – λx – λ = 0
A1
Hence at most 2 values of x and at most 2 stationary points A1
[3]
12 EITHER
(ii) For 2 real distinct roots, λ2 > 4(λ + 1)(– λ2)
5
2
λ (5 + 4λ ) > 0 ∴ λ > − AG
4
M1
use of discriminant
A1
(iii) Vert. asymptotes when x2 – 2x + λ = 0
b2 – 4ac > 0 ⇒ 4 – 4 λ > 0
For two vert. asymp. λ < 1
M1
(iv) (a) y = 0 ⇒ x2 + 2λx = 0
⇒ x = 0 or –2λ
λ
(b) y = 1: x =
2λ + 2
M1
A1
[2]
A1
[2]
(both)
B1
[3]
(v) (a) λ < –2: no stat points: 2 vert. asymp
B1
B1
3 branches
completely correct shape
B1
B1
max, min, horiz asymp
correct shape
(b) λ < 2: 2 stats points: no vert. asymp
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Mark Scheme: Teachers’ version
GCE A LEVEL – October/November 2010
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Syllabus
Paper
9231
01
OR
Normal to plane: (2, 3, 4) × (–1, 0, 1) = (3, –6, 3)
r.(1, –2, 1) = d and point (2, 1, 4)
d = 4 x – 2y + z = 4
Alternative:
x = 2 + 2λ – µ
y = 1 + 2λ
x + z = 6 + 6λ
z = 4 + 4λ + µ
∴ x + z = 6 + 2( y − 1)
∴x − 2y + z = 4
}
M1A1
M1
A1
substitute point into plane eqn
[4]
M1A1
M1
A1
x – 4y + 5z = 12
x – 2y + z = 4 Solve by eliminating one variable
Use parameter and express all 3 variables in terms of it
e.g. x = 3t – 4, y = 2t – 4, z = t
r = (–4, –4, 0) + t (3, 2, 1)
M1
M1
A1
or equivalent
[3]
Alternative:
 1   1   3
Direction of line =  − 2  ×  − 4  = t  2 
 1   5  1
     
 − 4  2
Find any point on line e.g.  − 4  ,  0  etc.
 0   2
   
 − 4  3
∴ r =  − 4  + t 2 
 0  1
   
M1A1
B1
Line l: r = (a, 2a + 1, –3) + α(3c, –3, c)
Plane: x – 2y + z = 4
Distance A to plane:
a − 2(2a + 1) − 3 − 4
6
=
15
M1
6
3a + 9 =15
a=2
sin θ =
M1
A1
3c + 6 + c
2
correct use of modulus sign
M1A1
2
6 9c + 9 + c
4c + 6
2
∴
=
2
6
6 9 + 10c
2
6c – 12c = 0: c = 2
(Penalise only once for negative values.)
M1
solve for c
A1
[7]
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