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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
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GCE Advanced Level
MARK SCHEME for the May/June 2011 question paper
for the guidance of teachers
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 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 2011 question papers for most IGCSE, GCE
Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.
Page 2
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2011
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.
© University of Cambridge International Examinations 2011
Page 3
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2011
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.
© University of Cambridge International Examinations 2011
Page 4
Qu
No
1
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2011
Commentary
Solution
1
1 1
1 
= 
−

(2r + 1)(2r + 3) 2  2r + 1 2r + 3 
Expresses all terms as
differences.
−
S n =   −  +  − .....

2   3 5   5 7   2n + 1 2n + 3  
1  1
=
S∞ =
1 1
1
Paper
11
Marks
Any method including
cover-up rule.
Finds sum.
2
Syllabus
9231

1
B1
1

M1A1
1
1
(acf)
−
6 2(2n + 3)
A1
4
1
(B0M1A1√ A0A1√ if signs reversed.)
6
A1
1
Sum of roots.
β + βk + βk 2
= −p
k
B1
Sum of products in pairs.
β2
+ kβ 2 + β 2 = q
k
B1
Factorises.
 k 2 + k +1
 k 2 + k +1
 = − p and β 2 
=q
⇒ β 



k
k




M1
⇒β =−
q
p
(AG)
β 3 = −r
Product of roots.
⇒−
3 (i)
q3
= − r ⇒ rp 3 = q 3 (AG)
3
p
A1
B1
Reduces matrix to echelon
form.
Uses rank = dimension –
nullity
r(A) = 4 – 2 = 2
A1
x + 3 y − 2 z + 4t = 0
z −t = 0
B1
Finds basis vectors.
 − 2   − 3 
   
 0   1 
Basis is   ,   (OE)
 1   0 
 1   0 


© University of Cambridge International Examinations 2011
[5]
4
B1
3 −2 4 
 1
1 3 − 2 4 




 5 15 − 9 19 
 0 0 1 − 1
− 2 − 6 3 − 7 . → 0 0 0
0




 3
0 0 0
9 − 5 11 
0 


(ii) Obtains a set of equations.
Part Total
Mark
2
[6]
M1A1
M1A1
3
3
[6]
Page 5
Qu
No
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2011
Commentary
4 (i) Establishes initial result.
Syllabus
9231
Solution
f(n) + f(n + 1) = 3 3n + 6 n −1 + 3 3n + 3 + 6 n
= 33n (1 + 27) + 6 n −1 (1 + 6)
Paper
11
Marks
M1
A1
= 28(33n ) + 7(6 n −1 ) (AG)
(ii) States inductive
hypothesis.
Proves base case.
Shows Pk ⇒ Pk+1.
States conclusion.
5
Right-hand loop
Left-hand loop
Deduct 1 mark for extra
loops (r < 0).
Uses A =
1 2
r dθ
2
∫
B1
33 + 6 0 = 28 = 4 × 7 ⇒ H1 is true
f(k + 1) + f(k) = f(k +1) + 7λ = 28(33k ) + 7(6 k −1 )
= 7µ
⇒ f (k + 1) = 7( µ − λ ) ∴ Hk ⇒ Hk+1
(Hence by the principle of mathematical
induction Hn is) true for all positive integers n.
B1
A=
1
4cos 2θ dθ dθ
2
∫
Uses double angle
formula.
= (cos 4θ + 1) dθ
Integrates.
= [sin 4θ + θ ]
∫
Inserts any appropriate
limits which legitimately
give the result.
2
Hk: f(k) = 7λ
Position and through pole and (2, 0).
Position and through pole and (2, π).
π
4
π
−
4
e.g. [sin 4θ + θ ]
=
Part Total
Mark
M1
A1
4
B1B1
B1B1
[6]
4
M1
(LNR)
M1
(LNR)
A1
π
2
A1
© University of Cambridge International Examinations 2011
4
[8]
Page 6
Qu
No
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2011
Commentary
6 (i) Uses vector product to find
vector perpendicular to
both lines.
Finds BA and its scalar
product with unit
perpendicular vector.
(ii)
Syllabus
9231
Solution
i
j
n= 4 3
Paper
11
Marks
Part Total
Mark
k
0 = –3i + 4j – 12j
M1A1
4 0 −1
BA = i + 10j – 11j
− 3 × 1 + 4 × 10 + 12 × 11 169
perp.dist. =
=
= 13
169
3 2 + 4 2 + 12 2
= 13 (AG) (No penalty for sign errors
made in n, which lead to correct result.)
 8 + 4λ 


p =  8 + 3λ  q =
 −7 


M1A1
4
 7 + 4µ 


 −2 
 4−µ 


 − 1 − 4λ + 4 µ 
 −3 




PQ =  − 10 − 3λ  = t  4 
 11 − µ

 − 12 




B1
M1A1
⇒ t = −1 λ = −2 µ = −1
M1
p = 2j – 7k
q = 3i – 2j + 5k
(Award B1B1B1 if t assumed to be ±1 i.e. 3/5)
A1
5
[9]
6(ii) Alternative Solution:
Finds two parameter
representation for PQ
 8 + 4λ 


p =  8 + 3λ  q =
 −7 


 7 + 4µ 


 −2 
 4−µ 


Uses scalar product
between PQ and direction
vector of at least one line
and equates to zero
 − 1 − 4λ + 4 µ 


PQ =  − 10 − 3λ 
 11 − µ



B1
16 µ − 25λ = 34
M1A1
17 µ − 16λ = 15
Solves simultaneously.
µ = −1
Obtains position vectors
for P and Q
p = 2j – 7k
(i) Obtains length of PQ.
A1
λ = −2
M1A1
q = 3i – 2j + 5k
32 + (−4) 2 + 12 2 = 13
(AG)
© University of Cambridge International Examinations 2011
A1
7
M1A1
2
[9]
Page 7
Qu
No
7
8
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2011
Commentary
Syllabus
9231
Solution
Paper
11
Marks
Differentiates twice.
v = y 3 ⇒ v ′ = 3 y 2 y ′ ⇒ v′′ = 3 y 2 y ′′ + 6 y ( y ′) 2
Substitutes.
1
2
v ′′ − 2 y ( y ′) 2 + 2 y ( y ′) 2 + v ′ − 5v = 8e − x
3
3
Obtains v-x equation.
⇒
Solves AQE
m 2 + 2m − 15 = 0 ⇒ m = − 5 , 3
Finds CF.
CF:
Ae −5 x + Be 3 x
A1
Differentiates form for PI.
PI:
v = ke- x ⇒ v′ = − ke − x ⇒ v′′ = ke − x
M1
Substitutes.
ke − x − 2ke − x − 15ke − x = 24e − x
M1
Obtains PI
⇒ −16ke − x = 24e − x ⇒ k = −
3
2
A1
Obtains GS
GS:
Obtains y in terms of x
(from a complete,
if incorrect, GS).
3

3
y =  Ae −5 x + Be 3 x − e − x 
2


Finds characteristic
equation and solves.
Det(A – λI) = 0 ⇒ λ3 − 4λ2 − 11λ + 30 = 0
⇒ λ = –3, 2, 5 (Any one)
(All three)
M1A1
A1
A1
Uses vector product
(or equations) to find
corresponding
eigenvectors.
i
j
k  0   0
   
λ = −3 e1 = 7 − 1 1 =  20  = t  1 
− 1 3 − 3  20   1 
M1A1
Forms P from eigenvectors
and D with fifth powers of
eigenvalues on leading
diagonal.
Note: Columns of P and D
can be permuted, but must
match.
d 2v
dv
+ 2 − 15v = 24e- x
2
dx
dx
(AG)
3
v = Ae −5 x + Be 3 x − e − x
2
Part Total
Mark
B1B1
M1
A1
4
M1
A1
1
A1√
7
i
j
k
 5  1
   
λ = 2 e2 = 2 − 1 1 =  5  = t  1 
− 1 − 2 − 3  − 5   − 1
A1
i
j
k
 8   2
   
λ = 5 e3 = − 1 − 1 1 =  − 4  = t  − 1
− 1 − 5 − 3  4   1 
A1
8
B1√
M1
A1√
3
2
0 1


P =  1 1 − 1 D =
1 −1 1 


0 
 − 243 0


32
0 
 0
 0
0 3125 

© University of Cambridge International Examinations 2011
[11]
[11]
Page 8
Qu
No
9
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2011
Commentary
Uses formulae for coords
of centroid.
Syllabus
9231
Solution
4
∫
x=
1
4
∫
1
5
1
y= 2
x 2 dx
3
x 2 dx
∫
∫
4
1
4
Numerators.
2 7 
254
 x2  =
7
 7 1
Denominator.
2 5 
62
 x2  =
5
 5 1
Obtains values. (Accept
rational values.)
x=
Differentiates y wrt x and
squares.
y = x 2 ⇒ y′ =
Subs in arc length formula
(LR).
s=
4
1
Paper
11
Marks
Part Total
Mark
x 3 dx
M1M1
3
x 2 dx
4
255
1 4 
8 x  = 8
1

A1A1
4
B1
635
(= 2.93)
217
3
∫
28
5
1+
y=
1275
(= 2.57)
496
3
9x
x ⇒ ( y ′) 2 =
2
4
M1A1
7
B1
9x
dx
4
B1
28
Integrates.
3


28
3
2
x
8
9
1





or  (4 + 9 x ) 2 
=
1 + 
 27 
4  
 27
5

 5
Obtains printed result.
=
84 − 7 3
16 3 − 7 3
or
= 139
27
27
(AG)
© University of Cambridge International Examinations 2011
M1A1
A1
5
[12]
Page 9
Qu
No
10
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2011
Syllabus
9231
Commentary
Solution
Puts cos n x = cos n −1 x. cos x
and integrates by parts.
I n = cos n−1 x sin x 02 + (n − 1) ∫ 2 cos n−2 x sin 2 x dx
Uses cos 2 x = 1 − sin 2 x
and obtains reduction
formula.
I n= (n − 1)
Uses mean value formula.
⇒ In =
Finds I0.
Uses reduction formula to
find I4 and I6.
∫
π
2
0
0
(cos n −2 x − cos n x) dx
n −1
I n−2 ( n ≥ 2 ) (AG)
n
=
π
2
0
π
2
0
Marks
∫
b
a
π
2
0
a cos 3 t.3a sin 2 t cos tdt
= 3a
∫
= 3a
∫
I0 =
π
or I2 = π/4.
2
a
cos 4 t sin 2 tdt
A1
4
M1A1
A1
4
(cos 4 t − cos 6 t )dt (AG)
3 1 π
3
5 3
5
. . = π , I 6 = . .π = π
4 2 2 16
6 16
32
5
1
(e.g. I 6 = I 4 ⇒ ( I 4 − I 6 ) = I 4 )
6
6
Substitutes in integral to
find mean value.
M1
M1
I4 =
(Or equivalent.)
M1A1
ydx
b−a
∫
Part Total
Mark
π
]
Mean value =
Transforms variable to t.
Uses sin 2 t = 1 − cos 2 t to
write in a form where
reduction formula can be
used.
π
[
Paper
11
Mean value =
3a
π
32
© University of Cambridge International Examinations 2011
B1
M1A1
A1
4
[12]
Page 10
Qu
No
11
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2011
Commentary
EITHER
Uses de Moivre’s theorem
and
binomial theorem.
Solution
(cosθ + i sin θ ) 3 = cos 3θ + i sin 3θ
and
cos 3 θ + 3i cos 2 θ sin θ − 3 cosθ sin 2 θ − i sin 3 θ
(If line 1 missing, or no
reference made to cos3θ
and sin3θ being real and
imaginary parts.)
Award B0M1A1M1M1A0 i.e.4/6
Equates real and imaginary
parts.
sin 3θ = 3 cos 2 θ sin θ − sin 3 θ
cos 3θ = cos 3 θ − 3 cosθ sin 2 θ
3 cos 2 θ sin θ − sin 3 θ
tan 3θ =
cos 3 θ − 3 cosθ sin 2 θ
3 tan θ − tan 3 θ
∴ tan 3θ =
(*) (AG)
1 − 3 tan 2 θ
Uses tan A =
sin A
cos A
Syllabus
9231
π
5π
9π
3π
=
,
,
12
12
12
4
Put tan 3θ = 1 in (*) ⇒ t 3 − 3t 2 − 3t + 1 = 0
States roots of tan 3θ = 1
between 0 and π.
Obtains cubic equation and
solves.
Evaluates each root.
Roots are tan
π
5π
3π
, tan
, tan
12
12
4
(t + 1)(t 2 − 4t + 1) = 0
Paper
11
Marks
Part Total
Mark
B1
M1A1
M1
M1
A1
B1
(one)
B1
(all)
6
2
M1
A1
(one)
A1
(all)
3
M1
⇒ t = −1 , 2 ± 3
3π
π
= −1
tan = 2 − 3
4
12
5π
tan
= 2+ 3
12
tan
© University of Cambridge International Examinations 2011
A1
(one)
A1
(all)
3
[14]
Page 11
Qu
No
11
Mark Scheme: Teachers’ version
GCE A LEVEL – May/June 2011
Commentary
Syllabus
9231
Solution
Paper
11
Marks
Part Total
Mark
OR
(i) Differentiates y wrt x.
dy ( x + 3)(2 x + λ ) − ( x 2 + λx − 6λ2 )
=
dx
( x + 3) 2
x 2 + 6 x + 3λ + 6λ2
=
( x + 3) 2
Uses discriminant and
factorises.
Obtains result.
(No equality.)
(ii) Uses division to obtain the
form for recognition of
oblique asymptote.
(iii) For 0 < λ < 1
(iv) For λ > 3
dy
= 0 has distinct roots if
dx
36 − 12λ − 24λ2 > 0 ⇒ 3 − λ − 2λ2 > 0
(3 + 2λ )(1 − λ ) > 0
3
⇒ − < λ <1
(AG)
2
M1
A1
M1
A1
A1
5
x 2 + λx − 6λ2
9 − 3λ − 6λ2
≡ x + λ −3+
x+3
x+3
(Tolerate error on remainder term.)
⇒ asymptotes are: x = −3
and y = x + λ − 3 .
M1
B1
A1
3
Axes and asymptotes.
Upper branch with minimum below x-axis.
Lower branch with maximum.
B1
B1
B1
3
Axes and asymptotes.
(n.b. oblique asymptote has positive y intercept.)
Left-hand branch.
Right-hand branch.
B1
Deduct 1 mark overall for
wrong forms
At infinity.
© University of Cambridge International Examinations 2011
B1
B1
3
[14]