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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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Cambridge International Advanced Level
MARK SCHEME for the October/November 2014 series
9231 MATHEMATICS
9231/11
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 October/November 2014 series for
most Cambridge IGCSE®, Cambridge International A and AS Level components and some
Cambridge O Level components.
® IGCSE is the registered trademark of Cambridge International Examinations.
Page 2
Mark Scheme
Cambridge International A Level – October/November 2014
Syllabus
9231
Paper
11
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 2014
Page 3
Mark Scheme
Cambridge International A Level – October/November 2014
Syllabus
9231
Paper
11
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 2014
Page 4
Mark Scheme
Cambridge International A Level – October/November 2014
Qn &
Part
1
Solution
(
1
25
1
27
−
)+ (
1
27
∑
= 15 −
∑
= 15
n
u
r =13 k
2
∞
u
r =13 k
1
29
−
)+ ... + (
2
1
2 n −1
=
∫
)
y& = et sin t + et cos t
(
2
2
)
2t
2et dt
0
[ 2e ]
t
1π
2
0
M1A1
M1A1
B1
M1
(
)
1π
= 2 e2 −1
( = 5.39)
2! – S1 = 1 , 3! – S2 = 1 , 4! – S3 = 1 , 5! – S4 = 1 (Two correct B1, all four correct B2)
Sn = (n +1)! – 1
2! – 1 = 2 – 1 = 1 ⇒ H1 is true.
Hk: Sk = (k +1)! – 1
(k +1)! – 1 + (k + 1) × (k + 1)!
= (k + 1)!(1 + k +1) – 1
= ([k +1] + 1)! – 1 Hence Hk ⇒ Hk+1
So result holds for all positive integers (by PMI).
4
Marks
M1A1
(4)
x& + y& = 2e cos t + sin t = 2e
s=
1
2 n +1
Paper
11
B1
(1)
[5]
2t
1
π
2
−
1
2 n+1
x& = e t cos t − e t sin t
2
3
Syllabus
9231
Vertical asymptote is x = 1
−1
y = 2 x + 3 + 2( x − 1) ⇒ y = 2 x + 3 is the oblique asymptote.
A1A1
(6)
[6]
B2,1,0
(2)
B1
(1)
B1
B1
M1
A1
(4)
[7]
B1
M1A1
(3)
2 x 2 + (1 − y )x + ( y − 1) = 0 has real roots
⇔ (1 − y ) − 8( y − 1) > 0
2
⇔ y 2 − 10 y + 9 > 0
⇔ ( y − 1)( y − 9 ) > 0
Hence ( y − 1)( y − 9 ) < 0 ⇒ no real roots.
i.e. 1 < y < 9 ⇒ no points on C. (AG) Thumbnail sketch, or similar, required.
© Cambridge International Examinations 2014
M1A1
M1
A1
(4)
[7]
Page 5
5
Mark Scheme
Cambridge International A Level – October/November 2014
1 −1
Paper
11
2
1 −1 − 3 = 0
1 −1 7
⇒ 5 a + 5 = 0 ⇒ a = −1
M1
A1A1
(3)
1 − 1 2 | 4 
 1 −1 2 | 4




1 − 1 − 3 | − 5  → ... →  0 0 5 | 9  , or by elimination methods,
1 − 1 7 | 13 
0 0 0 | 0




x – y + 2z = 4
5z = 9
 x   0.4   1 
     
⇒  y  =  0  + t 1 
 z   1.8   0 
     
(OE)
M1A1
(4)
B1
(1)
[8]
(c + is )5 = c5 + 5c 4is + 10c3 (is )2 + 10c 2 (is )3 + 5(is )4 + (is )5
5
3 2
4
)
(
cos 5θ = c − 10c s + 5cs
2
= c 1 − s 2 − 10 s 2 1 − s 2 + 5s 4 


(
(
M1
A1
Planes form a prism, or words to that effect.
6
Syllabus
9231
B1
M1A1
)
M1
) (
)
= c 1 − 2s 2 + s 4 − 10s 2 + 10s 4 + 5s 4 = c 16s 4 − 12s 2 + 1
(
(AG)
)
A1
(5)
n
Alternative: using z + z −1 = 2 cos nθ
(2 cosθ )5 = 2 cos 5θ + 10 cos 3θ + 20 cosθ
(M1)
5
⇒ 16 cos θ = cos 5θ + 5 cos 3θ + 10 cos θ
But cos 3θ = 4 cos 3 θ − 3 cos θ (Can be quoted, if known.)
(
4
2
⇒ cos 5θ = cosθ 16 cos θ − 20 cos θ + 5
)
(
)
Uses cos 2 θ = 1 − sin 2 θ to obtain cos 5θ = c 16s 4 − 12s 2 + 1
(B1)
(A1)
(AG)
cos 5θ = 0 ⇒ θ = 101 π , 103 π , 12 π , 107 π , 109 π
12 ± 144 − 64 3 ± 5
=
32
8
2 1
2 3
Since sin (10 π ) < sin (10 π ), sin 2 (101 π ) = 3−8 5 (Or by any other valid method.)
(AG) Justification required for final mark.
s2 =
© Cambridge International Examinations 2014
(M1A1)
B1
M1A1
A1
(4)
[9]
Page 6
7
Mark Scheme
Cambridge International A Level – October/November 2014
[
I n = (1 − x ) e x
n
] + ∫ n(1 − x )
1
1
0
n −1
0
Syllabus
9231
e x dx
M1A1
= 0 – 1 + nIn–1 = nIn–1 – 1
I0 =
∫ e dx = [e ]
1
x 1
0
x
A1
(3)
= e −1
B1
I 4 = 4 I 3 − 1 = 12 I 2 − 5 = 24 I1 − 17 = 24 I 0 − 41
= 24e − 65
Or for the last 3 marks:
I1 = e – 2, I2 = 2e – 5
I3 = 6e – 16 and I4 = 24e – 65
M1A1
A1
0
(M1A1)
(A1)
(4)
B1
M1
Draws appropriate sketch, or explains clearly. (Must be seen.)
0 < Area below graph < 1 ⇒ 0 < 24e − 65 < 1
Hence
8
Paper
11
65
24
<e<
66
24
⇒ 65
< e < 114
24
A1
(3)
[10]
Circle sketched
Cardioid – correct location and orientation – correct indentation near pole.
(a, π2 ) and (a, 32π ) (B1 for reverse, or a = a(1 − cosθ ) ⇒ θ = π2 , 32π
=
2
1
2 πa
+
= 12 πa 2 +
π
2
0
∫ a (1 − cos θ ) dθ
∫ (1 − 2 cos θ + cos θ ) dθ
Area = 12 πa 2 + 2 × 12
π
a 2
0
π
a 2 2 32
0
2
∫ (
2
2
seen.)
(Half circle + Area of sector)
2
− 2 cos θ + 12 cos 2θ ) dθ
= 12 πa 2 + a 2 [32θ − 2 sin θ + 14 sin 2θ ] 02
π
= 12 πa 2 + a 2 ( 34 π − 2 ) = ( 54 π − 2 ) a 2
B1
B1B1
(3)
B1B1
(2)
B1M1
A1
(Use of double angle formula.)
M1
(Integration)
M1
(AG)
A1
(6)
[11]
© Cambridge International Examinations 2014
Page 7
9
Mark Scheme
Cambridge International A Level – October/November 2014
dv
dy
d 2v
dy
d2 y
= y+x ⇒ 2 =2 +x 2
dx
dx
dx
dx
dx
2
d y
dy
x 2 + (2 x + 2) + (2 − 3x ) y = 10e 2 x
dx
dx
2
dy
d y
dy 
d 2v
dv

⇒ 2 + x 2 + 2 y + x  − 3 xy = 10e 2 x ⇒ 2 + 2 − 3v = 10e 2 x
dx
dx 
dx
dx
dx

v = xy ⇒
Syllabus
9231
Paper
11
B1B1
M1A1
(4)
m 2 + 2m − 3 = 0 ⇒ (m + 3)(m − 1) = 0 ⇒ m = −3, 1
M1
CF: Ae −3 x + Be x
A1
v = ke 2 x ⇒ v′ = 2 ke 2 x ⇒ v′′ = 4 ke 2 x
⇒ 4 ke 2 x + 4 ke 2 x − 3ke 2 x = 10e 2 x ⇒ 5k = 10 ⇒ k = 2 ⇒ PI : 2e 2 x
GS: v = Ae −3 x + Be x + 2e 2 x ⇒ y =
(
1
A e −3 x + B e x + 2 e 2 x
x
)
© Cambridge International Examinations 2014
M1
M1A1
M1A1
(7)
[11]
Page 8
Mark Scheme
Cambridge International A Level – October/November 2014
i
10
(i)
j
 − 3  1 
   
1 − 2 − 3 =  3  ~  − 1
 − 3  1 
−2 1
3
   
Syllabus
9231
Paper
11
k
M1A1
BA = 6i + 4 j − 6k
Shortest distance
B1
 6  1 
  
 4 . −1
 −6   1 
  
12 +12 +12
=
4
3
( = 2.31)
M1A1
(5)
Alternative
 − 6 − 2λ − µ   1 

  
 − 4 + λ + 2µ  .  − 2  = 0 ⇒ −13λ − 14µ = 16
 6 + 3λ + 3µ   − 3 

  
(M1)
 − 6 − 2λ − µ   − 2 

  
 − 4 + λ + 2µ  .  1  = 0 ⇒ 14λ + 113µ = −26
 6 + 3λ + 3µ   3 

  
(A1)
⇒ λ = − 529 , µ = 389
Shortest distance =
i
(ii)
j
(M1A1)
4
3
.
(= 2.31)
 − 4  4
   
1 −1 1 =  − 5 ~  5
− 2 1 3  − 1   1 
k
M1A1
Cartesian equation of Π: 4 x + 5 y + z = −12 − 5 + 2 = −15
(iii)
(A1)
 6   4
   
 4  .5
 − 6  1
   
4 2 + 52 + 12
Distance of A from Π:
=
38
42
M1A1
(4)
M1A1
( = 5.86)
© Cambridge International Examinations 2014
A1
(3)
[12]
Page 9
11E (i)
Mark Scheme
Cambridge International A Level – October/November 2014
Syllabus
9231
α + β + γ + δ = −4
Paper
11
B1
(1)
(ii)
α 2 + β 2 + γ 2 + δ 2 = (− 4) − 2 × 2 = 12
(iii)
1 1 1 1 − (− 4)
+ + + =
=4
α β γ δ
1
(iv)
α
β
γ
δ
α 2 + β 2 + γ 2 + δ 2 12
+
+
+
=
=
= 12
βγδ αγδ αβδ αβγ
αβγδ
1
2
M1A1
(2)
M1A1
(2)
M1A1
(2)
y = x +1⇒ x = y −1
( y − 1)4 + 4( y − 1)3 = y 4 − 6 y 2 + 8 y − 3
2
2( y − 1) − 4( y − 1) + 1 = 2 y 2 − 8 y + 7
(ii)
A1
⇒ x 4 + 4 x3 + 2 x 2 − 4 x + 1 = y 4 − 4 y 2 + 4 = 0
A1
(y
A1
2
)
2
− 2 = 0 ⇒ y = ± 2 (twice).
⇒ x = ± 2 − 1 (twice).
11O (i)
M1A1
(Some indication of four roots for final mark.)
Ae = λe; since A is non-singular Ae ≠ 0 ⇒ λ ≠ 0 (e ≠ 0) .
Ae = λe ⇒ A–1Ae = A–1λe
1
⇒ e = λA–1e ⇒ A −1e = e
λ
Eigenvalues of A are –2 , –1 , 3
M1A1
(7)
[14]
M1A1
(2)
M1
A1
(2)
(1 mark for any one, 2 marks for all three.)
1  2  − 6
     
Corresponding eigenvectors are:  0  ,  1  ,  25  (M1A1 for one, A1 for each other.)
 0   0   20 
     
B2,1,0
M1A1
A1 A1
(N.B. May come from using eigenvalues of A + 3I.)
1 2 − 6


P =  0 1 25 
 0 0 20 


Eigenvalues of A + 3I are 1 , 2 , 6 (Award B3 if obtained from A + 3I.)
Eigenvalues of (A + 3I)–1 are 1, 12 , 16
B1
B1
1 0 0


D =  0 12 0 
0 0 1 
6

B1
(10)
[14]
(CAO)
© Cambridge International Examinations 2014
B1