CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level

MARK SCHEME for the May/June 2013 series

9280 MATHEMATICS (US)

9280/21 Paper 2, maximum raw mark 50

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.
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GCE AS LEVEL – May/June 2013 9280 21

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

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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

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1 Either ( ) 2
State or imply non-modular equation 2 x − 7 = 12, or corresponding pair of equations M1
Obtain 2x = 8 and 2x = 6 A1
State answer 3 B1
Use logarithmic method to solve an equation of the form 2x = k, where k > 0 M1
State answer 2.58 A1

Or State or imply one value for 2x, e.g. 8, by solving an equation or by inspection B1
State answer 3 B1
State second value for 2x B1
Use logarithmic method to solve an equation of the form 2x = k, where k > 0 M1
State answer 2.58 A1 [5]

2 Use 2 ln x = ln(x2) M1
Use law for addition or subtraction of logarithms M1
Obtain correct quadratic equation in x A1
Make reasonable solution attempt at a 3-term quadratic DM1
(dependent on previous M marks)
3
State x = and no other solutions A1 [5]
5

3 (i) Either
Use sin 2x = 2sin x cos x to convert integrand to k sin2 2x M1
Use cos 4x = 1 – 2 sin2 2x M1
1 1
State correct expression − cos 4 x or equivalent A1
2 2
Or
1 − cos 2 x 1 − cos 2 x
Use cos 2 x = and/or x = to obtain an equation in cos 2 x only M1
2 2
1 + cos 4 x
Use cos 2 2 x = M1
2
1 1
State correct expression − cos 4 x or equivalent A1 [3]
2 2

3 3
(ii) State correct integral x − sin 4 x , or equivalent B1
2 8
Attempt to substitute limits, using exact values M1
Obtain given answer correctly A1 [3]

3
4 (i) Substitute x = − , equate to zero M1
2
Substitute x = −1 and equate to 8 M1
Obtain a correct equation in any form A1
Solve a relevant pair of equations for a or for b M1
Obtain a = 2 and b = −6 A1 [5]

© Cambridge International Examinations 2013

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2
(ii) Attempt either division by 2x + 3 and reach a partial quotient of x + kx , use of an identity or
observation M1
2
Obtain quotient x − 4 x + 3
Obtain linear factors x – 1 and x – 3 A1
[Condone omission of repetition that 2x + 3 is a factor.] A1
[If linear factors x – 1, x − 3 obtained by remainder theorem or inspection, award B2 + B1.] [3]

5 (i) Use product rule to differentiate y M1

Obtain correct derivative in any form A1
dy dy dx
Use = ÷ M1
dx dt dt
Obtain given answer correctly A1 [4]

(ii) Substitute t = 0 in dy and both parametric equations B1

dx
dy
Obtain = 2 and coordinates (1, 0) B1
dx
Form equation of the normal at their point, using negative reciprocal of their dy M1
dx
1 1
State correct equation of normal y = − x + or equivalent A1 [4]
2 2

6 (i) Make a recognisable sketch of a relevant graph, e.g. y = cot x or y = 4x – 2 B1

Sketch a second relevant graph and justify the given statement B1 [2]

(ii) Consider sign of 4x – 2 – cot x at x = 0.7 and x = 0.9, or equivalent M1

Complete the argument correctly with appropriate calculations A1 [2]

1+ 2 tan x
(iii) Show that given equation is equivalent to x = , or vice versa B1 [1]
4 tan x

(iv) Use the iterative formula correctly at least once M1

Obtain final answer 0.76 A1

Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change
in the interval (0.755, 0.765) B1 [3]

© Cambridge International Examinations 2013

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7 (i) State R = 29 B1
Use trig formula to find α M1
Obtain α = 21.800 with no errors seen A1 [3]
4
(
(ii) Carry out evaluation of sin −1   ≈ 47.97 0) M1
R
Carry out correct method for one correct answer M1
Obtain one correct answer e.g. 13.1o A1
Carry out correct method for a further answer M1
Obtain remaining 3 answers 55.10, 193.10, 235.10 and no others in the range A1 [5]

(iii) Greatest value of 10 sin 2θ + 4 cos 2θ = 2√29 M1

1
A1 [2]
116

© Cambridge International Examinations 2013