Cambridge International Examinations

Cambridge International Advanced Level

MATHEMATICS 9709/33
Paper 3 May/June 2017
MARK SCHEME
Maximum Mark: 75

Published

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 2017 series for most Cambridge IGCSE®,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

® IGCSE is a registered trademark.

This document consists of 9 printed pages.

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9709/33 Cambridge International A Level – Mark Scheme May/June 2017
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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 FT 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.

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The following abbreviations may be used in a mark scheme or used on the scripts:

AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent

AG Answer Given on the question paper (so extra checking is needed to ensure that the
detailed working leading to the result is valid)

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

SOI Seen or implied

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.

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Question Answer Marks

1 Express the LHS in terms of either cos x and sin x or in terms of tan x B1

Use Pythagoras M1

Obtain the given answer A1

Total: 3

2 EITHER: (M1
State a correct unsimplified version of the x or x 2 term in the expansion of
(1 + 23 x )
−3
or ( 3 + 2 x )
−3

 −3 
[Symbolic binomial coefficients, e.g.   , are not sufficient for M1.]
 2

State correct first term 1 B1

27

Obtain term − 272 x A1

Obtain term 8
x2 A1)
81

OR: (M1
Differentiate expression and evaluate f ( 0 ) and f ' ( 0 ) , where f ' ( x ) = k ( 3 + 2 x )
−4

State correct first term 1 B1

27

Obtain term − 272 x A1

Obtain term 8
x2 A1)
81

Total: 4

3 Rearrange as 3u 2 + 4u − 4 = 0 , or 3e2 x + 4e x − 4 = 0 , or equivalent B1

Solve a 3-term quadratic for e x or for u M1

Obtain e x = 2
or u = 23 A1
3

Obtain answer x = –0.405 and no other A1

Total: 4

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Question Answer Marks

4 *M1
∫
Integrate by parts and reach aθ cos 12 θ + b cos 12 θ dθ

Complete integration and obtain indefinite integral −2θ cos 12 θ + 4sin 12 θ A1

Substitute limits correctly, having integrated twice DM1

Obtain final answer ( 4 − π ) / 2 , or exact equivalent A1

Total: 4

5(i) Use the chain rule M1

Obtain correct derivative in any form A1

Use correct trigonometry to express derivative in terms of tan x M1

dy 4 tan x A1
Obtain =− , or equivalent
dx 4 + tan 2 x

Total: 4

5(ii) Equate derivative to –1 and solve a 3–term quadratic for tan x M1

Obtain answer x=1.11 and no other in the given interval A1

Total: 2

6(i) Calculate the value of a relevant expression or expressions at x = 2.5 and at another M1
relevant value, e.g. x = 3

Complete the argument correctly with correct calculated values A1

Total: 2

6(ii) State a suitable equation, e.g. x = π + tan −1 (1 / (1 − x ) ) without suffices B1

Rearrange this as cot x = 1 − x , or commence working vice versa B1

Total: 2

6(iii) Use the iterative formula correctly at least once M1

Obtain final answer 2.576 only A1

Show sufficient iterations to 5 d.p. to justify 2.576 to 3 d.p., or show there is a sign A1
change in the interval (2.5755, 2.5765)

Total: 3

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Question Answer Marks

7(i) Use correct quotient rule or product rule M1

Obtain correct derivative in any form A1

Equate derivative to zero and solve for x M1

Obtain x = 2 A1

Total: 4

7(ii) State or imply ordinates 1.6487…, 1.3591…, 1.4938… B1

Use correct formula, or equivalent, with h = 1 and three ordinates M1

Obtain answer 2.93 only A1

Total: 3

7(iii) Explain why the estimate would be less than E B1

Total: 1

8(i) Justify the given differential equation B1

Total: 1

8(ii) Separate variables correctly and attempt to integrate one side B1

Obtain term kt, or equivalent B1

Obtain term − ln ( 50 − x ) , or equivalent B1

Evaluate a constant, or use limits x = 0, t = 0 in a solution containing terms M1*
a ln ( 50 − x ) and bt

Obtain solution − ln ( 50 − x ) = kt − ln 50 , or equivalent A1

Use x = 25, t = 10 to determine k DM1

Obtain correct solution in any form, e.g. ln 50 − ln ( 50 − x ) = 101 ( ln 2 ) t A1

Obtain answer x = 50 (1 − exp ( −0.0693t ) ) , or equivalent A1

Total: 8

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Question Answer Marks

9(i) A B C B1
State or imply the form + 2+
x x 3x + 2

Use a relevant method to determine a constant M1

Obtain one of the values A = 3, B = –2, C = –6 A1

Obtain a second value A1

Obtain the third value A1

Ax + B C
[Mark the form + using same pattern of marks.]
x 2
3x + 2

Total: 5

9(ii) 2 B3 FT
Integrate and obtain terms 3ln x = − 2ln ( 3x + 2 )
x
[The FT is on A, B and C]

3x − 2
Note: Candidates who integrate the partial fraction by parts should obtain
x2
2
3ln x + − 3 or equivalent
x

Use limits correctly, having integrated all the partial fractions, in a solution M1
containing terms a ln x + bx + c ln ( 3x + 2 )

Obtain the given answer following full and exact working A1

Total: 5

10(i) Carry out a correct method for finding a vector equation for AB M1

Obtain r = i − 2 j + 2k + λ ( 2i + 3j − k ) , or equivalent A1

Equate two pairs of components of general points on AB and l and solve for λ or for M1
µ

Obtain correct answer for λ or µ, e.g. λ = 75 or µ = 73 A1

Obtain m = 3 A1

Total: 5

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Question Answer Marks

10(ii) EITHER: (B1

Use scalar product to obtain an equation in a, b and c, e.g. a − 2b − 4c = 0

Form a second relevant equation, e.g. 2a + 3b − c = 0 and solve for one ratio, e.g. a M1
:b

Obtain final answer a : b : c = 14 : – 7 : 7 A1

Use coordinates of a relevant point and values of a, b and c and find d M1

Obtain answer 14 x − 7 y + 7 z = 42 , or equivalent A1)

OR 1:
Attempt to calculate the vector product of relevant vectors, e.g. (M1
( i − 2 j − 4k ) × ( 2i + 3j − k )
Obtain two correct components A1

Obtain correct answer, e.g. 14i − 7 j + 7k A1

Substitute coordinates of a relevant point in 14 x − 7 y + 7 z = d , or equivalent, and M1

find d

Obtain answer 14 x − 7 y + 7 z = 42 , or equivalent A1)

OR 2:
Using a relevant point and relevant vectors, form a 2–parameter equation for the (M1
plane

State a correct equation, e.g. r = i − 2 j + 2k + s ( i − 2 j − 4k ) + t ( 2i + 3j − k ) A1

State 3 correct equations in x, y, z, s and t A1

Eliminate s and t M1

Obtain answer 2x – y + z = 6, or equivalent A1)

OR 3: (M1
Using a relevant point and relevant vectors, form a determinant equation for the
plane

x −1 y + 2 z −1 A1
State a correct equation, e.g. 1 −2 −4 =0
2 3 −1

Attempt to expand the determinant M1

Obtain or imply two correct cofactors A1

Obtain answer 14 x − 7 y + 7 z = 42 , or equivalent A1)

Total: 5

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Question Answer Marks

11(a) Solve for z or for w M1

Use i 2 = −1 M1

i 2+i A1
Obtain w = or z =
2−i 2−i

Multiply numerator and denominator by the conjugate of the denominator M1

Obtain w = − 15 + 52 i A1

Obtain z = 53 + 54 i A1

Total: 6

11(b) EITHER: (B1

(
Find ±  2 + 2 − 2 3 i 
  )
Multiply by 2i (or –2i) M1*

Add result to v DM1

Obtain answer 4 3 − 1 + 6i A1)

OR: (M1
z −v
State = ki , or equivalent
v−u

State k = 2 A1

Substitute and solve for z even if i omitted M1

Obtain answer 4 3 − 1 + 6i A1)

Total: 4

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