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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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Cambridge International Advanced Level
MARK SCHEME for the May/June 2015 series
9231 FURTHER MATHEMATICS
9231/11
Paper 1 (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 2015 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 – May/June 2015
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 2015
Page 3
Mark Scheme
Cambridge International A Level – May/June 2015
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
PA –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.
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 2015
Page 4
Mark Scheme
Cambridge International A Level – May/June 2015
Qn &
Part
Solution
3 ×
1
Syllabus
9231
13 × 14 × 27
13 × 14
− 5×
+13 = 2015
6
2
Paper
11
Marks
M1A1
2
 9 × 10 
 2  −10 = 2015 (Award M1 for subtracting 9 or 10 here.)


2 − 3 4
3 − 1 0  = 0 ⇒ 7 k = − 28 ⇒ k = − 4


1 2 k 
Add 1st and 3rd ⇒ 3x – y =2 ( Same as 2nd) (OE)
Set x = t (for example) (OE)
7
5
⇒ y = 3t − 2 , z = t −
(OE) – many forms
4
4
2
3
a1 > 5 (given) ⇒ H1 is true.
Assume Hk is true for some positive integer k, i.e. ak = 5 + δ, where δ > 0.
2
2
4a + 25
4a + 25 − 25ak (4ak − 5)(ak − 5)
ak +1 − 5 = k
−5= k
=
> 0 , ⇒ ak +1 > 5
5ak
5ak
5ak
Or
4
5
4
δ δ2
ak +1 = (5 + δ ) +
, = 4 + δ + (1 − +
− ...) for 0 < δ < 5
5
5+δ
5
5 25
3
= 5 + δ + 0 (δ 2 ) [ ak+1 > 5, ( δ [ 5 is trivial).
5
Hk ⇒ Hk +1 and H1 is true, hence by mathematical induction, the result is true for all===
n ∈ Z + =(N.B. The minimum requirement is ‘true for all positive integers’.)
5
1
a k +1 − a k =
− ak
ak 5
1
5
< 1 and ak > 1 ⇒ ak +1 − ak < 0 ⇒ ak +1 < ak
5
ak
M1A1
(4)
Total: 4
M1A1
(2)
M1
M1
A1A1
(4)
Total: 6
B1
B1
M1A1
(M1)
(A1)
A1
(5)
M1
A1
(2)
Total: 7
(i)
α + β + γ = 7 , αβ + βγ + γα = 2 , αβγ = 3 (Stated or implied by working.)
1
1
1
=
=
2
(αβ )( βγ )(γα ) (αβγ )
9
(ii)
1
1
1
α + β +γ
7
+
+
=
=
αβ
βγ
γα
αβγ
3
M1A1
(iii)
1 1
1 1
1  αβ + βγ + γα  2
1
1
1
=
+
+
 + + =
=
2
2
αβγ  α β γ  αβγ 
αβγ
α βγ αβ γ αβγ
 9
M1A1
(6)
4
2
⇒ x3 −
7 2 2
1
x + x − = 0 ⇒ 9 x 3 − 21x 2 + 2 x −1 = 0
3
9
9
© Cambridge International Examinations 2015
B1
B1
M1A1√
(2)
Total: 8
Page 5
Mark Scheme
Cambridge International A Level – May/June 2015
Qn &
Part
5
Syllabus
9231
Solution
1
1
1
sin θ = ⇒ θ = π ; Intersection at
2
2
3
Marks
 1 1 
1

, π  (Accept 1.05 for π .)
3
 2 3 
C1: Circle centre at pole and radius 1/√2
C2: Curve, approx. correct orientation, from (0,0) to (1,π).
Completely correct correct shape.
M1A1
M1A1
(4)
Total: 9
dy
 dy

2x − 6  x
+ y  + 50 y
=0
dx
 dx

M1A1
A1
A1
(4)
B1B1
At (3,1) 6 − 18 y ' − 6 + 50 y ' = 0 ⇒ y ' = 0 (AG)
2 − 6( x y" + y ' + y ' ) + 50 [( y ' ) 2 + y y" ] = 0
1
At (3,1) 2 − 18 y" + 50 y" = 0 ⇒ y" = −
16
⇒ maximum. (All previous marks required for final mark, i.e. CSO)
7
In =
π
2
0
∫
π
x n sin x dx = [− x n cos x ]02 +
= 0 +  nx
π 
= n 
2
π
2
0
n −1
π
2
0
∫
π
2
0
n x n −1 cos x dx
sin x  −
n−1
π
2
0
∫
M1A1
A1
(5)
Total: 9
(LNR)
M1A1
n(n − 1) x n − 2 sin x dx (LR)
M1A1
− n(n − 1) I n−2
(AG)
π
2
0
I 0 = ∫ x sin x dx = [ − cos x ] = 1
n
M1A1
(2)
B1
B1
B1
(3)
1
1 1 1π
1
× π × − ∫ 3 sin θ dθ
6
2 2 0
2
1
1
1  1π 1
3
π × − 2 cos θ  3 = π +
=
−1
12
2
2 0
12
2
6
Paper
11
I 2 = π − 2 , using reduction formula.
3
1
π 
I 4 = 4 ×   −12 (π − 2) = π 3 −12π + 24
2
2
(If I2 found by integration M1A1 (if correct) then M1A1 for I4 from reduction
formula.)
© Cambridge International Examinations 2015
A1
(5)
B1
M1A1
A1
(4)
Total: 9
Page 6
8
Mark Scheme
Cambridge International A Level – May/June 2015
∑
n
z 2 r −1 =
r =1
(
Syllabus
9231
)
z 1 − [ z 2 ]n z − z 2 n +1
=
1 − z2
1− z2
M1A1
1 − z 2n
1 − (cos 2 nθ + i sin 2nθ )
1 − cos 2nθ − i sin 2nθ
=
=
−1
− 2 i sin θ
z − z cosθ − i sin θ − (cos θ + i sin θ )
Equating imaginary parts:
=
∑ r =1 sin (2r −1)θ =
n
Paper
11
2sin 2 nθ sin 2 nθ
=
2 sin θ
sin θ
(AG)
M1A1
A1
M1A1
(7)
Differentiating:
∑
n
r =1
sin ( 2r − 1) cos ( 2r −1)θ = 2n sin nθ cos nθ cosec θ − sin 2 nθ cosec θ cot θ
Putting θ =
∑
⇒
n
r =1
M1A1
π
:
2n
π
π
π 
π 
π  π 
2 π
 = n sin cos cosec   − sin cosec   cot   dM1
2
2
2
 2n 
 2n 
 2n   2n 
(2 r − 1) cos (2 r − 1) 
∑
n
r =1
 ( 2r −1)π 
π 
π 
( 2r −1) cos 
= − cosec   cot   . (AG)

 2n 
 2n 
 2n 
A1
(4)
Total: 11
1
9
x& = 4 + 3t 2
1
y& = 4 − 3t 2
B1
x& 2 + y& 2 = 32 + 18t
B1
4
1
3
1

s = ∫ 04 (32 + 18t ) 2 dt =  (32 + 18t ) 2 
 27
0
=
∫
32
0
ydx = ∫
4
0
M1A1
3
3

1 
104 2 − 32 2  = 32.6


27 

4
dx
y dt = ∫ t 4 − 2 t
0
dt
(
(CAO)
)( 4 + 3 t ) dt = ∫
4
0
M1A1
(6)
3

2
2
16t + 4t − 6t  dt


M1
4
 2 8 5

= 8t + t 2 − 2t 3 
5

 0
MV =
( = 51.2)
∫ ba y dx
51.2
= =
= 1.6 (CAO)
32
b−a
© Cambridge International Examinations 2015
M1A1
M1A1
(5)
Total: 11
Page 7
10
Mark Scheme
Cambridge International A Level – May/June 2015
 2 2 − 3  1   − 3

   
 2 2 3   − 1 =  3  ⇒ eigenvalue is – 3.
 − 3 3 3   1   − 3

   
i j k 7
 
−2 2 − 3 =  7  ~
−3 3 − 1  0 
i j k
 3
 
−4 2 − 3 =  −3 
 −6 
−3 3 − 3
 
λ =4:
λ =6:
 1 1 1


P = −1 1 −1 
 1 0 − 2


1 
 
(Or by equations)
1 
0
 
 1
 
~  − 1
 −2 
 
M1A1
A1
(3)
B1√B1√
(2)
M1;A1
B1
B = QAQ –1 = QPAP–1Q–1; = QPA (QP) –1
(CAO)
Eigenvalues are –3, 4 and 6 (same as those for A).
11 (e)
Paper
11
M1A1
(2)
 − 3 0 0


D =  0 4 0
 0 0 6


 − 2 15 − 17 


QP =  − 1 5 − 7 
 0 3 −3 


Syllabus
9231
(CAO)
B1
 2  15  17 
     
Eigenvectors are: 1  ,  5  ,  7  .
0  3   3 
     
B1
(5)
Total: 12
dv
1 dy
=− 2
y dx
dx
B1
2
d 2v
1 d 2 y  dy  d 2 v
1 d 2 y 2  dy 
=
−
+
=
−
+  
 
y 2 dx 2  dx  dy 2
y 2 dx 2 y 3  dx 
dx 2
2
M1A1
2
2  dy 
1 d 2 y 2 dy 5
d 2v
dv
2
+ = 17 + 6 x − 5 x ~ 2 + 2
+ 5v = 17 + 6 x − 5 x 2 A1
 − 2 2− 2
3 
dx
dx
y  dx  y dx
y dx y
(4)
(AG)
m 2 + 2 m + 5 = 0 ⇒ m = − 1 ± 2i ⇒ CF : e − x ( A cos 2 x + B sin 2 x )
M1A1
PI : v = px 2 + qx + r ⇒ v ' = 2 px + q ⇒ v" = 2 p
Equate coefficients: 5 p = − 5 4 p + 5q = 6 2 p + 2q + 5r = 17
p = −1 , q = 2 , r = 3
M1
M1
A1
GS : v = e − x ( A cos 2 x + B sin 2 x ) + 3 + 2 x − x 2
When x = 0
y=
dy
dv
1
⇒ v = 2 and
= −1 ⇒
=4
2
dx
dx
© Cambridge International Examinations 2015
A1
Page 8
Mark Scheme
Cambridge International A Level – May/June 2015
Syllabus
9231
⇒ 2 = A + 3⇒ A = −1
v' = e
−x
( −2 A sin 2 x + 2 B cos 2 x ) − e
B1
−x
( A cos 2 x + B sin 2 x )
1
⇒ 4 = 2 B + 1+ 2 ⇒ 2 B = 1 ⇒ B =
2
−1
A1
(10)
Total: 14
5
3+λ
 8+λ 










p =  2 − 2λ  , q =  3 + 2µ  ⇒ QP =  − 1 − 2λ − 2µ 
 3 
 − 14 − 3µ 
 17 + 3µ 






(ii) (a)
M1A1
3+λ
 1  

  

 − 2  .  − 1 − 2λ − 2µ  = 0 ⇒ 5λ + 4µ = − 5
 0   17 + 3µ 
  

M1A1
3+λ
 0 

  

 2  .  − 1 − 2λ − 2µ  = 0 ⇒ − 4λ − 13µ = 53
 − 3   17 + 3µ 
  

A1
Solving :
λ = 3, µ = – 5
M1A1
Whence:
 11 
 5
 
 
p =  − 4 , q =  − 7 .
 3
 1 
 
 
A1
(8)
3 
− 3
 
 
AP =  − 6  , AQ =  − 9 
 0 
 − 2
 
 
i j k
 12 


3 − 6 0 =  6  (CAO) ;
 −45 
−3 − 9 − 2


(b)
M1
A1
 1


⇒ y = e − x  sin 2 x − cos 2 x  + 3 + 2 x − x 2 

 2

110 (i)
Paper
11
 −3 
uuur 
1 1

AB =  1  ⇒ Volume = ×
3 2
 −17 


B1
Area =
1
2205
2
(=23.5)
 − 3   12 



 1  . 6 
 −17   −45 

 = 735 (= 122.5)
2205 
6
2205
© Cambridge International Examinations 2015
M1A1
A1
M1A1
(6)
Total: 14
Page 9
Mark Scheme
Cambridge International A Level – May/June 2015
Syllabus
9231
Paper
11
Alternative: for marks 3,4,5,6 and 7 in part (i)
i j
1 −2
0
k
6
 
0 = 3
 2
2 −3
 
6k − λ = 3
3k + 2λ + 2 = − 1
2k − 3µ = 17
Solving; k = 1 , λ = 3 and µ = – 5 (If k missing, or assumed to be 1, deduct 1
mark.)
© Cambridge International Examinations 2015
(M1A1)
(A1)
(M1A1)