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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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GCE Advanced Level
MARK SCHEME for the October/November 2013 series
9231 FURTHER MATHEMATICS
9231/12
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 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 – October/November 2013
Syllabus
9231
Paper
12
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 – October/November 2013
Syllabus
9231
Paper
12
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
Commentary
1 (i)
Uses area formula.
Mark Scheme
GCE A LEVEL – October/November 2013
Solution
Area =
1
2
π
2θ
2
π 4e dθ
6
∫
Paper
12
Marks
π
2
π
6
[ ]
= e 2θ
Syllabus
9231
Part Total
Mark
M1
π
Obtains result.
(ii)
Uses arc length
formula.
= eπ − e 3 ( = 20.3)
Arc length =
Finds S2.
∫
4e 2θ + 4e 2θ dθ = 2
π
2
π
6
[ ]
2 eθ
=2
=2
2
π
2 θ
π e dθ
6
∫
π
 π

2 e 2 − e 6 


(= 8.83)
M1A1
3
α 2 + β 2 + γ 2 = (α + β + γ ) − 2(αβ + βγ + γα )
= 0 − 2(− p ) = 2 p (AG)
B1
1
M1A1
2
(ii)
Finds S3.
α 3 + β 3 + γ 3 = p ∑ α + 3q = 0 + 3q = 3q (AG)
(iii)
Finds S5.
α5 + β5 +γ 5 = p
∑α
3
+q
∑α
2
M1
= p.3q + q.2 p = 5 pq
3
⇒ 6 ∑ α = 30pq = 5 ∑ α ∑ α
2
A1
(AG)
A1
Writes first four
sums.
S1…S4 ~ 3, 10, 21, 36
Deduces first four
terms, conjectures
and justifies result.
u1…u4 ~ 3, 7, 11, 15 ⇒ ur = 4r ‒ 1
n
since S n = {6 + 4(n − 1)} = 2n 2 + n as given.
2
B1B1
Or ur = S r − S r −1 = 2r 2 + r − 2(r − 1) − (r − 1)
= 4r – 1
B1B1
B1
∑ n2 n+1 (4r − 1) = 4.
2n(2n + 1)

 n(n + 1)
− 2 n −  4.
− n
2
2


(
)
= 8n 2 + 2 n − 2 n 2 + n = 6 n 2 + n
Or Sum of AP =
3
[6]
B1
2
Obtains required
sum.
[5]
2
5
3
2
A1
Obtains result.
2 (i)
π
2
π
6
A1
n
(4n + 3 + 8n − 1) = 6n 2 + n
2
© Cambridge International Examinations 2013
B1
4
M1A1
A1
3
[7]
Page 5
Qu
No
Mark Scheme
GCE A LEVEL – October/November 2013
Commentary
1
4
Integrates by parts.
1


I n  x n (1 + 2 x )2  −
0

= 3−n
∫
Marks
(1 + 2 x )dx
1
(1 + 2 x )2
M1A1
1 nx n −1
0
1
dx − 2n
x n −1
0
(1 + 2 x )2
Paper
12
Solution
∫
1
Syllabus
9231
∫
1
xn
0
(1 + 2 x )2
1
Part Total
Mark
dx
Obtains result.
⇒ (2n + 1)I n = 3 − nI n −1 (AG)
Alternatively:
1
1
1
d  n

−
n −1
n
 x (1 + 2 x )2  = (1 + 2 x )2 .nx + x (1 + 2 x ) 2
dx 

A1
1
n(1 + 2 x )x n −1


+ In
⇒  x n (1 + 2 x )2  =
1 + 2x

0
3
(M1)
1
⇒ (2n + 1)I n = 3 − nI n −1
Finds I0 (or I1).
Uses Red. Form.
Finds I2 and I3.
5 (i)
Differentiates once,
twice
and three times.
(ii)
[
I0 = 1 + 2x
(
]
1
0
(AG)
(A1)
= 3 −1
B1
)
1
3
3 2
2
⇒ I2 =
− ⇒ I3 =
3 +1
5 15
35
3I1 = 3 −
(A1)
3 − 1 ⇒ I1 =
(
)
M1
(AG)
y′ = 2(1 + x )ln (1 + x ) + (1 + x )
y′′ = 2ln (1 + x ) + 3
2
y′′′ =
1+ x
(Allow B1 if constant term in previous line incorrect.)
d 3 y (− 1) .2.0!
2
=
=
⇒ H 3 is true.
3
1+ x
1+ x
dx
k −1.
d k y (− 1) .2.(k − 3)!
Hk : k =
for some k .
dx
(1 + x )k − 2
A1A1
4
[7]
B1
B1
B1
3
2
Proves base case.
States inductive
hypothesis.
Differentiates
Proves inductive
step and
states conclusion.
d k +1 y
k −1
− ( k −1)
= (− 1) .2(k − 3)!(− 1)(k − 2 )(1 + x )
k +1
dx
k
(
− 1) .2.(k − 2)!
=
⇒ H k +1 is true
(1 + x )k −1
Hence by PMI Hn is true for all integers [=3
© Cambridge International Examinations 2013
B1
B1
M1
A1
A1
5
[8]
Page 6
Qu
No
Mark Scheme
GCE A LEVEL – October/November 2013
Commentary
6
Reduces to echelon
form.
Obtains rank
Solution
Evaluates matrix
product.
Finds general
solution of
equations.
1 − 3 −1 2 
1 − 3 −1 2 




2 
2 − 3
 4 − 10 0
0 1
1 − 1 3 − 4 → K → 0 0
0
0 




 5 − 12 1
0 0
1 
0
0 


r(M) = 4 ‒ 2 = 2
Part Total
Mark
M1A1
A1
 − 5   7 
   
 − 2   3 
Basis is :   ,   (OE) e.g.
 1   0 
 0   1 


 1  2
   
 − 2   16 
M  = 
−3
10
   
 − 4   22 
   
 1 
0
 
 
 0 
1
 − 3  or  7 
 
 
 2 
 5
 
 
 1 
 
 − 2
x −  ∈ K
−3
 
 − 4
 
M1
A1
A1
6
B1
M1
 1 
 
 − 2
x =   + λe1 + µe 2
−3
 
 − 4
 
7
Paper
12
Marks
x − 3 y − z + 2t = 0

y + 2 z − 3t = 0
⇒ t = µ , z = λ , y = 3µ − 2λ , x = 7 µ − 5λ
and basis for null
space.
Syllabus
9231
(AG)
A1
3
Ae = λe
⇒ A2e = AAe = A λe = λ Ae = λ2e ⇒ result.
(e ≠ 0 ⇒ λ2 is an eigenvalue of A2.)
B1
M1A1
3
Obtains
eigenvalues of B.
(1 − λ )(λ − 4)(λ + 2) = 0
M1A1
A1A1
4
Obtains
eigenvalues of
related matrix.
14 e + 2 × 12 e + 3e = 6e ⇒ 6 is an eigenvalue.
Proves first result.
⇒ λ = ‒2, 1, 4
9
M1A1
(− 2) e + 2 × (− 2) e + 3e = 27e ⇒ 27 is an eigenvalue.
4
2
4 4 e + 2 × 4 2 e + 3e = 291e ⇒ 291 is an eigenvalue.
© Cambridge International Examinations 2013
A1
3
10
Page 7
Qu
No
Mark Scheme
GCE A LEVEL – October/November 2013
Commentary
8
Solution
i
Finds normal to Π1 .
j
1 0
1 = i + 3j − k
1 −1 − 2
Equation of Π1 :
Finds angle
between normals,
using scalar
product.
cosθ =
Finds point
common to both
planes.
States vector
equation.
Marks
Part Total
Mark
M1A1
3 12
2 − 3 −1
11 6
2
=
⇒ θ = 75.7° or 1.32 rad.
66
i
Paper
12
k
Finds Cartesian
equation.
Finds direction of
line of intersection,
using vector
product.
Syllabus
9231
j
A1
3
M1
A1
2
k
1 3 − 1 = 2i − 3 j − 7 k
2 −1 1
M1A1
Point on both planes is e.g. (6,2,0)
r = 6i + 2 j + t (2i − 3 j − 7k )
(OE)
© Cambridge International Examinations 2013
M1A1
A1
5
[10]
Page 8
9
Obtains area of
surface of
revolution.
Mark Scheme
GCE A LEVEL – October/November 2013
x& = 2t
Syllabus
9231
y& = 1 − t 2
,
2
(
 ds 
⇒   = 4t 2 + 1 − 2t 2 + t 4 = 1 + t 2
 dt 
1
1 
2π ∫ y ds = 2π ∫  t − t 3  1 + t 2 dt
0
3 
(
)
2
M1A1
)
M1A1
1
t2 t4 t6 
2
1 
= 2π  t + t 3 − t 5  dt = 2π  + − 
0
3
3 
 2 6 18  0
11
= π or 3.84
9
∫
Finds coordinates
of centroid, using
relevant formulae.
1
dx
dt =
dt
∫
y
∫
dx
xy dt =
dt
∫
M1
A1
∫
B1
1
 t5 2 t7 
2 6
32
 2t − t  dt = 2 − .  =
0
3 
 5 3 7  0 105
1
∫
6
1
 t3 2 t5 
2 4
8
2
 2t − t  dt = 2 −
 =
0
3 
 3 3 5  0 15
1
1 2 dx
dt =
y
2
dt
∫
Paper
12
4
M1A1
1
t 4 t 6 t8 
2 5 1 7
11
 t − t + t  dt =  − +  =
0
3
9 
 4 9 72  0 72
1
3
M1A1
Centroid is
 4 55 
 ,

 7 192 
Or (0.571,0.286)
© Cambridge International Examinations 2013
A1
6
[12]
Page 9
Qu
No
10 (i)
(ii)
(iii)
11E
Mark Scheme
GCE A LEVEL – October/November 2013
Commentary
Syllabus
9231
Solution
Paper
12
Marks
Part Total
Mark
Vertical asymptote.
Oblique asymptote.
Asymptotes : x = –1
−1
y = px + 4 − p + ( p − 3)(x + 1) ⇒ y = px + 4 − p
B1
M1A1
3
Obtains value of p.
Sketches curve.
p = 4 ⇒ x-axis is a tangent
Correct location of turning points and asymptotes.
Each branch.
B1
B1
B1B1
4
Proves required result.
Sketches graph.
p = 1 ⇒ y = x + 3 − 2(x + 1) ⇒ y′ = 1 + 2( x + 1)
−1
(
Intersections on x-axis at − 2 ± 3 , 0
Each branch.
−2
)
([ 1)
2kπ
2kπ
+ i sin
, k = 0 , ± 1, ± 2 .
5
5
Obtains all fifth roots.
z = cos
Simplifies expression.
x 2 − 2 cos
Obtains factors.
2π
4π
 2


+ 1 x 2 − 2 cos
+ 1(x − 1)
 x − 2 cos
5
5



Solves quadratic in x3.
x3 =
Expresses them in
polar form.
or cos
7π
7π
13π
13π
± i sin
or cos
± i sin
3
3
3
3
x = cos
π
π
7π
7π
13π
13π
± i sin , cos
± i sin
, cos
± i sin
9
9
9
9
9
9
Finds factors.
2π
x +1
5
3
1
π
π
±i
= cos ± i sin
2
2
3
3
7π
π
 2
 2

+ 1
 x − 2 cos x + 1 x − 2 cos
9
9



13π
 2

+ 1
 x − 2 cos
9


(ACF)
© Cambridge International Examinations 2013
M1A1
B1
B1B1
5
B1B1
2
M1A1
2
M1A1
2
[12]
M1A1
A1
A1
M1A1
6
M1A1
2
[14]
Page 10
Mark Scheme
GCE A LEVEL – October/November 2013
Qu
No
Commentary
Solution
11O
Uses substitution
v = y 3 ⇒ v′ = 3 y 2
to obtain v–x
equation.
1 d 2v
dv
− 2 + 3v = 25e − 2 x
2
3 dx
dx
2
d v
dv
⇒ 2 − 6 + 9v = 75e − 2x
dx
dx
Syllabus
9231
Paper
12
Marks
Part Total
Mark
2
dy
d2 y
 dy 
⇒ v′′ = 6 y  + 3 y 2 2
dx
dx
 dx 
B1B1
M1
(AG)
A1
Finds CF.
m 2 − 6m + 9 = 0 ⇒ m = 3
v = Ae3 x + Bxe 3 x
M1
A1
Finds PI.
v = ke −2 x ⇒ v′ = −2ke −2 x ⇒ v′′ = 4ke −2 x
4 k + 12 k + 9k = 75k ⇒ k = 3
v = Ae 3 x + Bxe 3 x + 3e −2 x
M1
A1
A1
Uses initial
conditions to find
constants.
x = 0, y = 2,v =8⇒8 = A+3⇒ A = 5
B1
v′ = 15e 3 x + 3Bxe 3 x + Be 3 x − 6e −2 x
4
M1A1
x = 0 , y = 2 , y ′ = 1 ⇒ v′ = 12
12 = 15 + B − 6 ⇒ B = 3
A1
Writes solution of
y–x equation
y 3 = v = 5e3 x + 3 xe3 x + 3e −2 x
explicitly.
y = 5e 3 x + 3 xe 3 x + 3e − 2 x
{
1
3
}
© Cambridge International Examinations 2013
A1
10
[14]