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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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Cambridge International Advanced Level
MARK SCHEME for the October/November 2014 series
9709 MATHEMATICS
9709/33
Paper 3, maximum raw mark 75
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
9709
Paper
33
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
9709
Paper
33
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
1
Mark Scheme
Cambridge International A Level – October/November 2014
Either
2
2
2
2
M1
Obtain value 6 from graph, inspection or solving linear equation
4
Obtain value − similarly
5
4
State − < x < 6
5
B1
Use correct product rule or correct chain rule to differentiate y
dy
dy dθ
Use
= dx
dx dθ
− 4 cos θ sin 2 θ + 2 cos 3 θ
or equivalent
sec 2 θ
dy
in terms of cos θ
Express
dx
Confirm given answer 6 cos 5 θ − 4 cos 3 θ legitimately
(i) Either
Or
B1
Solve a three-term quadratic or two linear equations 5 x − 26 x − 24 < 0
4
Obtain − and 6
5
4
State − < x < 6
5
Obtain
3
Paper
33
State or imply non-modular inequality ( 3x − 1) < ( 2 x + 5 ) or corresponding
quadratic equation or pair of linear equations 3 x − 1 = ± (2 x + 5)
Or
Syllabus
9709
Equate p(–1) or p(–2) to zero or divide by (x + 1) or (x + 2 ) and
equate constant remainder to zero.
Obtain two equations a − b = 6 and 4a − 2b = 34 or equivalents
Solve pair of equations for a or b
Obtain a = 11 and b = 5
State or imply third factor is 4 x − 1
Carry out complete expansion of (x + 1)(x + 2 )(4 x − 1) or
(x + 1)(x + 2 )(Cx + D )
Obtain a = 11
Obtain b = 5
(ii) Use division or equivalent and obtaining linear remainder
Obtain quotient 4 x + a , following their value of a
Indicate remainder x − 13
© Cambridge International Examinations 2014
A1
A1
B2
B1
[4]
M1
M*1
A1
DM*1
A1
[5]
M*1
A1
DM*1
A1
B1
M1
A1
A1
[4]
M1
A1
A1
[3]
Page 5
4
Mark Scheme
Cambridge International A Level – October/November 2014
(i) Either
Use cos ( A ± B ) correctly at least once
State correct complete expansion
Confirm given answer cos θ with explicit use of cos 60o =
A1
Use correct cos A + cos B formula
 2θ   − 120 
State correct result e.g. 2 cos  cos

 2   2 
M1
A1
(
)
Confirm given answer cos θ with explicit use of cos ± 60o =
1
2
cos 2 x
=3
cos x
Obtain equation 2 cos 2 x − 3 cos x − 1 = 0
Solve a three-term quadratic equation for cos x
1
Obtain 3 − 17 or exact equivalent and, finally, no other
4
(ii) State or imply
(
5
)
A1
B1
M1
A1
B1
Carry out multiplication by
M1
4−i
4−i
7 23
Obtain final answer − + i or equivalent
17 17
[3]
B1
(i) State or imply iw = −3 + 5i
(ii) Multiply w by z to obtain 17 + 17i
3
1
State arg w = tan −1 or arg z = tan −1
5
4
State arg wz = arg w + arg z
3
1 1
Confirm given result tan −1 + tan −1 = π legitimately
5
4 4
6
Paper
33
M1
A1
1
2
SR: “correct” answer from sign errors in both expansions is B1 only
Or
Syllabus
9709
A1
[4]
[3]
B1
B1
M1
A1
[4]
(i) State or imply correct ordinates 1, 0.94259…, 0.79719…, 0.62000…
Use correct formula or equivalent with h = 0.1 and four y values
Obtain 0.255 with no errors seen
B1
M1
A1
[3]
(ii) Obtain or imply a = −6
Obtain x 4 term including correct attempt at coefficient
Obtain or imply b = 27
27 5
x , following their values of a and b
Either Integrate to obtain x − 2 x 3 +
5
Obtain 0.259
Or
Use correct trapezium rule with at least 3 ordinates
Obtain 0.259 (from 4)
B1
M1
A1
© Cambridge International Examinations 2014
B1
B1
M1
A1
[5]
Page 6
7
Mark Scheme
Cambridge International A Level – October/November 2014
(i) State at least two of the equations 1 + λ = a + µ , 4 = 2 + 2 µ , − 2 + 3λ = −2 + 3aµ
Solve for λ or for µ
Obtain λ = a (or λ = a + µ − 1 ) and µ = 1
Confirm values satisfy third equation
Paper
33
B1
M1
A1
A1
[4]
(ii) State or imply point of intersection is (a + 1, 4, 3a − 2 )
B1
Use correct method for the modulus of the position vector and equate to 9, following their
point of intersection
M*1
2
a −a−6 =0
Solve a three-term quadratic equation in a
DM*1
Obtain –2 and 3
A1
[4]
(
8
Syllabus
9709
)
(i) Sensibly separate variables and attempt integration of at least one side
1
2
Obtain 2 y = ... or equivalent
M1
A1
1
1
1
Correct integration by parts of x sin x as far as ax cos x ± b cos xdx
3
3
3
1
1
Obtain − 3x cos x + 3 cos xdx or equivalent
3
3
1
1
Obtain − 3x cos x + 9 sin x or equivalent
3
3
∫
∫
M1
A1
A1
2
1
9
1
 3

Obtain y =  − x cos x + sin x + c  or equivalent
10
3
10
3


(ii) Use x = 0 and y = 100 to find constant
Substitute 25 and calculate value of y
Obtain 203
9
(i) Sketch increasing curve with correct curvature passing through origin, for x ≥ 0
Recognisable sketch of y = 40 − x 3 , with equation stated, for x > 0
Indicate in some way the one intersection, dependent on both curves being roughly
correct and both existing for some x < 0
(ii) Consider signs of x 3 + ln (x + 1) − 40 at 3 and 4 or equivalent or compare values of
relevant expressions for x = 3 and x = 4
Complete argument correctly with correct calculations (–11.6 and 25.6)
(iii) Use the iterative formula correctly at least once
Obtain final answer 3.377
Show sufficient iterations to justify accuracy to 3 d.p. or show sign change in interval
(3.3765, 3.3775)
(iv) Attempt value of ln (x + 1)
Obtain 1.48
© Cambridge International Examinations 2014
A1
[6]
M*1
DM*1
A1
[3]
B1
B1
B1
[3]
M1
A1
[2]
M1
A1
A1
[3]
M1
A1
[2]
Page 7
Mark Scheme
Cambridge International A Level – October/November 2014
du
= ex
dx
Substitute throughout for x and dx
u
du or equivalent (ignoring limits so far)
Obtain
2
u + 3u + 2
10 State or imply
∫
A
B
+
, following their integrand
u + 2 u +1
Carry out a correct process to find at least one constant for their integrand
2
1
−
Obtain correct
u + 2 u +1
State or imply partial fractions of form
Integrate to obtain a ln (u + 2 ) + b ln (u + 1)
Obtain 2 ln (u + 2 ) − ln (u + 1) or equivalent, follow their A and B
Apply appropriate limits and use at least one logarithm property correctly
8
Obtain given answer ln legitimately
5
SR for integrand
Syllabus
9709
Paper
33
B1
M1
A1
B1
M1
A1
M1
A1
M1
A1 [10]
u2
u (u + 1)(u + 2 )
A
B
C
+
+
u u +1 u + 2
Carry out a correct process to find at least one constant
2
1
−
Obtain correct
u + 2 u +1
…complete as above.
State or imply partial fractions of form
© Cambridge International Examinations 2014
(B1)
(M1)
(A1)