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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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GCE Advanced Level
MARK SCHEME for the May/June 2013 series
9231 FURTHER 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 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
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 2013
Page 3
Mark Scheme
GCE A LEVEL – May/June 2013
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 2013
Page 4
Qu
No
1
Mark Scheme
GCE A LEVEL – May/June 2013
Commentary
Use of
Solution
1 2
r dθ
2
∫
A=
Use of double angle formula
and attempt to integrate.
2
3
Syllabus
9231
=
1
2
∫
2π
0
4(1 + 2 cos θ + cos 2 θ ) dθ
2π
∫ (3 + 4 cos θ + cos 2θ ) dθ
Paper
11
Marks
A1
Finds value.
= 6π (CWO) Accept 18.8
A1
Proves base case.
Pn: 5 2 n − 1 is divisible by 8.
5 2 − 1 = 24 = 3 × 8 ⇒ P1 is true
B1
B1
States inductive hypothesis.
Assume Pk is true: 5 2 k − 1 = 8λ for some k.
5 2 k + 2 − 1 = 25.5 2 k − 1 = 24.5 2 k + 5 2 k − 1
= 3 × 8.5 2 k + 8λ
M1
Proves inductive step.
∴ Pk ⇒ Pk+1
A1
States conclusion.
(Since P1 is true and Pk ⇒ Pk+1)∴ Pn is for every
positive integer n (by PMI).
A1
c=2
B1
( α + β = c − γ etc.) ⇒ y = c − x ⇒ x = c − y
Uses substitution
3
2
(2 − y ) − 2(2 − y ) − 3(2 − y ) + 4 = 0 …(their c)
to obtain required cubic
equation.
⇒ y3 − 4y2 + y + 2 = 0
Obtains equation whose roots
are reciprocals of those in
previous cubic equation.
Uses z = y −1 to obtain 2 z 3 + z 2 − 4 z + 1 = 0
Uses
∑α
2
=
(∑ α )
2
−2
∑ αβ
∑
[4]
0
2π
−b
.
a
4
M1
sin 2θ 

= 3θ + 4 sin θ +
2  0

∑α =
Total
M1
Integrates correctly.
Uses
Part
Mark
5
[5]
1
M1
M1
A1
3
M1A1
2
M1A1
2
2
1
1
1
=   − 2(−2) = 4
2
2
4
(α + β )
 
© Cambridge International Examinations 2013
[8]
Page 5
4
Mark Scheme
GCE A LEVEL – May/June 2013
Integrates by parts.
In =
∫
1
dx
0 (1 + x 2 ) n
1
5
Paper
11
1


x
=
+
2 n 
 (1 + x )  0
Rearranges.
Syllabus
9231
∫
∫
1
0
n (1 + x 2 ) − ( n +1) .2 x 2 dx
1
M1A1
M1A1
= 2 − n + 2n (1 + x 2 ) −( n +1) (1 + x 2 − 1) dx
0
Obtains result.
∴ 2nI n +1 = 2 − n + (2n − 1) I n . (AG)
Uses reduction formula to find
I2 .
2I 2 =
1 1
1 1
+ π ⇒ I2 = + π
2 4
4 8
Uses reduction formula to find
I3 .
4I 3 =
1 3 3
1 3
+ + π ⇒ I3 = + π
4 4 8
4 32
Finds partial fractions.
1
1 1
1 
= 
−
(`2r + 1)(2r + 3) 2  2r + 1 2r + 3 
N
1
(
2
r
+
1
)(
2r + 3)
r =1
A1
5
M1A1
A1
3
[8]
M1A1
∑
Expresses terms as differences.
Shows cancellation.
2N
Uses
∑
N +1
=
2N
N
∑ ∑
−
1
Applies result
and simplifies.
Deduces inequality.
1
=
1 1 1
1 1
1 
−

 −  + ... + 
23 5
2  2N + 1 2N + 3 
=
1
1
(AG)
−
6 2( 2 N + 3)
2N
.
1
1
 1
M1A1
A1
1

∑ =  6 − 2(4 N + 3)  −  6 − 2(2 N + 3) 
5
M1
N +1
=
1 1
1 
−


2  2N + 3 4N + 3 
A1
=
N
(2 N + 3)(4 N + 3)
M1
<
N
1
(AG)
=
2 N .4 N 8 N
A1
© Cambridge International Examinations 2013
4
[9]
Page 6
6
7
Mark Scheme
GCE A LEVEL – May/June 2013
Syllabus
9231
Paper
11
Shows e is an eigenvector of
A
and gives eigenvalue.
Ae = 2e ⇒ e is an eigenvector with eigenvalue 2.
M1A1
Finds characteristic equation.
λ3 − 2λ 2 − λ + 2 = 0
M1A1
Factorises.
⇒ (λ − 2)(λ2 − 1) = 0
A1
States other eigenvalues.
Other eigenvalues are –1 and 1.
A1
Repeats for B.
Be = 3e ⇒ e is an eigenvector with eigenvalue 3
B1
States result for AB.
ABe = A.3e = 3Ae = 3.2e = 6e
AB has eigenvector e with eigenvalue 6
Expands and groups.
Use of z − z −1 and z + z −1 .
Correctly.
(z − z ) = (z
Obtains result.
(2i sin θ )6 = 2 cos 6θ − 12 cos 4θ + 30 cos 2θ − 20
−1 6
6
) (
)
(
M1A1
)
+ z −6 − 6 z 4 + z − 4 + 15 z 2 + z − 2 − 20
1
(10 − 15 cos 2θ + 6 cos 4θ − cos 6θ )
32
(Allow p = 10 , q = –15 , r = 6 , s = –1)
sin 6 θ =
Integrates correctly.
Inserts limits and evaluates.
π
 5θ 15 sin 2θ 3 sin 4θ sin 6θ  4
 16 − 64 + 64 − 192 

0
5π 15
1
5π 11
−
+
=
−
64 64 192 64 48
(SC: 1f power of 2 consistently wrong ¾ for 2nd part.)
© Cambridge International Examinations 2013
2
4
3
[9]
M1A1
M1
A1A1
A1
6
M1A1
M1A1
4
[10]
Page 7
8
Reduces matrices to
echelon form.
Finds basis for each null
space.
Mark Scheme
GCE A LEVEL – May/June 2013
1

3
5
4

1

2
4
6

3
−2
− 4 17
− 9 20
− 7 16
0
−2
0
−1
1
−7
− 10 0
5
1 − 2


33 
0
1
→ ... → 
36 
0
0


29 
0
0
−3 
1 − 2


0 
0
1
→ ... → 
−9 
0
0


− 14 
0
0
Syllabus
9231
3
4
1
0
0
0
1
0
 1 
x − 2 y + 3 z + 5t = 0  1 
y + 4 z + 9t = 0  ⇒  
  2 
z + 2t = 0
 
 − 1
 1 
x − 2 y − 3t = 0  2 
y + 2t = 0  ⇒  
z + t = 0   1 
 
 − 1
5

9
2
0 
− 3

2 
1 
0 
Paper
11
M1
A1
A1
M1A1
A1
Writes x1 and x2
appropriately.
1
1
1
1
 
 
 
 
2
1
2
2



and x 2 =
x1 =
+λ
+ µ 
 2
1
 3
 3
 − 1
 − 1
 4
 4
 
 
 
 
B1√
Finds difference
 λ−µ 


λ − 2µ 

⇒ λ − 2 µ = 5 and
⇒ x1 − x 2 =
 2λ − µ 
− λ + µ 


2λ − µ = 7
⇒ λ = 3 and µ = −1
M1
6
A1
and solves.
x 1 − x 2 = ( 4 5 7 − 4 ) T ⇒ p = 4 and q = –4
© Cambridge International Examinations 2013
A1
4
[10]
Page 8
9
Mark Scheme
GCE A LEVEL – May/June 2013
Finds complementary
function.
4m 2 + 4m + 1 = 0 ⇒ ( 2m + 1) 2 = 0 ⇒ m = −
C.F.: x = Ae
Finds particular integral.
Adds for general solution.
Uses initial conditions to
find
constants.
Gives particular solution.
States limit.
−
t
2
+ Bt e
−
t
2
Syllabus
9231
1
2
−
t
2
+ Bte
−
t
2
+
2 − 2t
e
3
5
5
2
⇒ = A+ ⇒ A =1
3
3
3
t
t
t
−
−
1 −
1
4
x& = − e 2 + Be 2 − Bte 2 − e − 2t
2
2
3
7
7
1
4
x& (0) = ⇒ = − + B − ⇒ B = 3
6
6
2
3
x ( 0) =
x=e
−
t
2
+ 3te
−
t
2
+
M1
A1
P.I.: x = ke − 2t ⇒ x& = −2ke − 2t ⇒ &x& = 4ke − 2t
2
2
16 k − 8k + k = 6 ⇒ k = ⇒ x = e − 2t
3
3
G.S.: x = Ae
Paper
11
2 − 2t
e
3
lim x = 0
t →∞
© Cambridge International Examinations 2013
M1
A1
A1
B1
M1
A1
A1
9
B1
1
[10]
Page 9
10
Mark Scheme
GCE A LEVEL – May/June 2013
Syllabus
9231
Paper
11
States asymptotes.
Vertical: x = 1 and Horizontal: y = 2
B1B1
Obtains quadratic form in x.
yx 2 − 2 yx + y = 2 x 2 − 3 x − 2
M1A1
2
⇒ ( y − 2) x 2 − (2 y − 3) x + ( y + 2) = 0
Uses B 2 − 4 AC ≥ 0 for real
roots.
Finds condition for y ′ = 0 .
For real x (2 y − 3) 2 − 4( y − 2)( y + 2) ≥ 0
25
.
⇒ 12 y ≤ 25 ⇒ y ≤
12
M1
y′ = 0 ⇒
M1
2
A1
4
2
( x − 2 x + 1)(4 x − 3) − (2 x − 3x − 2)(2 x − 2) = 0
Solves
⇒ x 2 − 8 x + 7 = 0 ⇒ ( x − 7)( x − 1) = 0
⇒ x = 7 , (since x = 1 is vertical asymptote).
A1
Obtains stationary point.
 25 
Stationary point is  7 , 
 12 
A1
3
Sketch showing:
Axes and asymptotes
(-0.5,0) , (2,0) , (0,–2) and (4,2)
Left hand branch.
Right hand branch.
B1
B1
B1
B1
4
© Cambridge International Examinations 2013
[13]
Page 10
11 E
Mark Scheme
GCE A LEVEL – May/June 2013
Syllabus
9231
Paper
11
Differentiation.
y = 2 sec x ⇒ y ′ = 2 sec x tan x
M1
Use of sec 2 x = 1 + tan 2 x .
1 + ( y ′) = 1 + 4 sec 2 x(sec 2 x − 1)
M1
2
= 4 sec 4 x − 4 sec 2 x + 1
(
)
= 2 sec 2 x − 1
Substitute in arc length
formula.
s=
∫
π
4
0
2
A1
A1
( 2 sec 2 x − 1) dx
π
4
0
Integrate.
= [2 tan x − x ]
M1
Substitute limits.
1 

= 2 − π 
4 

A1
(i) Use surface area formula
Obtain correct form.
(ii) Differentiates
S = 2π
= 4π
∫
π
4
0
π
4
0
2 sec x( 2 sec 2 x − 1) dx
∫ (2 sec
3
)
x − sec x dx (AG)
d
(secxtanx ) = secxtan 2 x + sec 3 x
dx
= secx sec 2 x − 1 + sec 3 x
(
4
2
M1A1
A1
3
M1A1
)
and obtains printed result.
= 2sec 3 x − secx (AG)
Uses result to deduce
surface area.
A1
3
S = 4π [sec x tan x ] 40
M1
2
= 4π 2
A1
π
© Cambridge International Examinations 2013
[14]
Page 11
11 O Vector perpendicular to Π1
Mark Scheme
GCE A LEVEL – May/June 2013
i
2
1
j
0
−3
Syllabus
9231
k  3 
 
1 =  − 1
1  − 6 
Paper
11
M1A1
Obtains cartesian equation
3 × 2 + 1 × (−1) + (−2) × (−6) = 17
⇒ 3 x − y − 6 z = 17
Obtains area of triangle
ABC.
1 2
1
3 + 12 + 6 2 =
46 (=3.39)
2
2
M1 A1
Obtains length of
perpendicular
from D to triangle ABC.
9 − 6 − 12 − 17
M1A1
1
× Base
3
area × Height.
3 2 + 12 + 6 2
=
26
1 1
26
13
×
46 ×
=
3 2
3
46
Either
or triple scalar product
method.
1  3 
1     26 13
Or e.g.  5  ⋅  − 1  =
=
6   
6
3
 4  − 6
Obtains normal to ABD.
i
1
2
k  5 


4 = 7 
1  − 10 
Uses scalar product
2
2
3 +1 + 6
to find angle between
normals
and hence angle between
Π1
and Π2.
⇒ cos θ =
2
4
46
Uses
j
5
0
M1
A1
68
46 174
6
M1A1
 3   5 
  

5 + 7 + 10 cos θ =  − 1  ⋅  7 
 − 6   − 10 
  

2
M1A1
2
M1
2
⇒ θ = 40.5º
A1
4
[14]
© Cambridge International Examinations 2013