UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary and Advanced Level

MARK SCHEME for the June 2005 question paper

9709 MATHEMATICS

8719 HIGHER MATHEMATICS

9709/03, 8719/03 Paper 3, maximum raw mark 75

This mark scheme is published as an aid to teachers and students, to indicate the requirements of

the examination. It shows the basis on which Examiners were initially instructed to award marks. It

does not indicate the details of the discussions that took place at an Examiners’ meeting before

marking began. Any substantial changes to the mark scheme that arose from these discussions will

be recorded in the published Report on the Examination.

All Examiners are instructed that alternative correct answers and unexpected approaches in

candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills

demonstrated.

Mark schemes must be read in conjunction with the question papers and the Report on the

Examination.

• CIE will not enter into discussion or correspondence in connection with these mark schemes.

CIE is publishing the mark schemes for the June 2005 question papers for most IGCSE and GCE

Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.
Grade thresholds for Syllabus 9709/8719 (Mathematics and Higher Mathematics) in the

June 2005 examination.

maximum minimum mark required for grade:

mark
A B E
available

Component 3 75 61 55 27

The thresholds (minimum marks) for Grades C and D are normally set by dividing the mark

range between the B and the E thresholds into three. For example, if the difference

between the B and the E threshold is 24 marks, the C threshold is set 8 marks below the B

threshold and the D threshold is set another 8 marks down. If dividing the interval by three

results in a fraction of a mark, then the threshold is normally rounded down.

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.
• 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.

June 2005

GCE AS AND A LEVEL

MARK SCHEME

MAXIMUM MARK: 75

SYLLABUS/COMPONENT: 9709/03, 8719/03

MATHEMATICS AND HIGHER MATHEMATICS

PAPER 3
 Page 1 Mark Scheme Syllabus Paper

A AND AS LEVEL – JUNE 2005 9709/8719 3

Obtain correct unsimplified version of the x or x term

2 3
1 EITHER: or x M1

State correct first two terms 1 – 2x A1

Obtain next two terms 6 x

2 3
− 20 x A1 + A1

[The M mark is not earned by versions with unexpanded binomial

− 
1

coefficients, e.g.   .]
2

 
 2 

OR: Differentiate expression and evaluate f(0) and f’(0),

3

where f ′(x) = k (1 + 4 x )
−
M1
2

State correct first two terms 1 – 2x A1

Obtain next two terms 6 x

2 3
− 20 x A1 + A1 4

2 (i) Show or imply correct decimal ordinates 1.2755…, 1, 0.8223… B1

Use correct formula, or equivalent, with h = 0.6 and three ordinates M1

Obtain correct answer 1.23 with no errors seen A1 3

[SR: if the area is calculated with one interval, or three or more, give D1

for a correct answer.]

(ii) Give an adequate justification, e.g. one trapezium over-estimates area

and the other under-estimates, or errors cancel out B1 1

3 (i) Use quadratic formula, or the method of completing the square, or the

substitution z = x + iy to find a root, using i

2
= -1 M1

Obtain a root, e.g. 2 + i A1

Obtain the other root –2 + i A1 3

[Roots given as ± 2 + i earn A1 + A1.]

(ii) Obtain modulus 5 (or 2.24) of both roots B1√

Obtain argument of 2 + i as 26.6° or 0.464 radians

(allow ±1 in final figure) B1√

Obtain argument of –2 + i as 153.4° or 2.68 radians

(allow ±1 in final figure) B1√ 3

[SR: in applying the follow through to the roots obtained in (i), if both

roots are real or pure imaginary, the mark for the moduli is not available

and only B1√ is given if both arguments are correct; also if one of the

two roots is real or pure imaginary and the other is neither then B1√ is

given if both moduli are correct and B1√ if both arguments are correct.]

(iii) Show both roots on an Argand diagram in relatively correct positions B1√ 1

[This follow through is only available if at least one of the two roots is

of the form x + iy where xy ≠ 0.]

© University of Cambridge International Examinations 2005

 Page 2 Mark Scheme Syllabus Paper

A AND AS LEVEL – JUNE 2005 9709/8719 3

dx
State or imply dx = sec θ dθ
2 2
4 (i) or = sec θ B1
dθ

Substitute for x and dx throughout the integral M1

Obtain integral in terms of θ in any correct form A1

Reduce to the given form correctly A1 4

State integral
1
(ii) sin 2θ B1
2

Use limits θ = 0 and θ = π correctly in integral of the form k sin 2θ

1
M1
4

Obtain answer ½ or 0.5 A1 3

5 (i)

Attempt division by x − x + 3 reaching a partial quotient x

2 2
EITHER: + x B1

Complete division and equate constant remainder to zero M1

Obtain answer a = −6 A1

Commence inspection and reach unknown factor of x

2
OR: + x + c B1

Obtain 3c = a and an equation in c M1

Obtain answer a = −6 A1

State or obtain factor x

2
+ x - 2 B1

State or obtain factors x + 2 and x – 1 B1 + B1 6

State that x
2
(ii) + x – 2 = 0, has two (real) roots B1

Show that x
2
− x + 3 = 0, has no (real) roots B1 2

6 (i)

EITHER: Express cos 4θ in terms of cos 2θ and/or sin 2θ B1

Use double angle formulae to express LHS in terms of cos θ

(and maybe sin θ) M1

Obtain any correct expression in terms of cos θ alone A1

Reduce correctly to the given form A1

OR: Use double angle formula to express RHS in terms of cos 2θ M1

Express cos 2θ in terms of cos 4θ

2
. B1

Obtain any correct expression in terms of cos 4θ and cos2θ A1

Reduce correctly to the given form A1 4

(ii) Using the identity, carry out method for calculating one root M1

Obtain answer 27.2° (or 0.475 radians) or 27.3° (or 0.476 radians) A1

Obtain a second answer, e.g. 332.8° (or 5.81 radians) A1√

Obtain remaining answers, e.g. 152.8° and 207.2°

(or 2.67 and 3.62 radians) and no others in range A1√ 4

© University of Cambridge International Examinations 2005

 Page 3 Mark Scheme Syllabus Paper

A AND AS LEVEL – JUNE 2005 9709/8719 3

7 (i) Make recognisable sketch of a relevant graph over the given range,

e.g. y = cosec x B1

Sketch the other relevant graph, e.g. y = ½ x + 1, and justify the given

statement B1 2

(ii) Consider sign of cosec x − ½ x − 1 at x = 0.5 and x = 1, or equivalent M1

Complete the argument correctly with appropriate calculations A1 2

Rearrange cosec x = x + 1 in the given form, or vice versa

1
(iii) B1 1
2

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

Obtain final answer x = 0.80 A1

Show sufficient iterations to at least 3 d.p. to justify its accuracy to 2 d.p.,

or show there is a sign change in the interval (0.795, 0.805) A1 3

8 (i) Attempt to express integrand in partial fractions,

A B
e.g. obtain A or B in + M1
y 4 − y

1 1 1
Obtain ( + ) , or equivalent A1
4 y 4 − y

Integrate and obtain ln ( 4 − y ) , or equivalent

1 1
ln y − A1√ + A1√ 4
4 4

A B
(ii) Separate variables correctly, integrate + and obtain further
y 4 − y

term x, or equivalent M1*

Use y = 1 and x = 0 to evaluate a constant, or as limits M1(dep*)

Obtain answer in any correct form A1

Obtain final answer y = 4 /(3 e + 1) , or equivalent

−4 x
A1 4

(iii) State that y approaches 4 as x becomes very large B1 1

9 (i) Use quotient or product rule M1

Obtain derivative in any correct form A1

Equate derivative to zero and solve for x or x

2
M1

Obtain x = 1 correctly A1 4

[Differentiating ( x + 1)y = x using the product rule can also earn the
2

first M1A1.]

[SR: if the quotient rule is misused, with a ‘reversed’ numerator or v

instead of v² in the denominator, award M0A0 but allow the

following M1A1.]

Obtain indefinite integral of the form k ln ( x

2
(ii) + 1) , where k = ½, 1 or 2 M1*

Use limits x = 0 and x = p correctly, or equivalent M1(dep*)

Obtain answer ½ ln(p

2
+1) A1 3

[Also accept –ln cos θ or ln cos θ , where x = tan θ , for the first M1*.]

Equate to 1 and convert equation to the form p

2
(iii) + 1 = exp(1/k ) M1

Obtain answer p = 2.53 A1 2

© University of Cambridge International Examinations 2005

 Page 4 Mark Scheme Syllabus Paper

A AND AS LEVEL – JUNE 2005 9709/8719 3

10 (i) State or imply a direction vector for AB is –i + 2j + 2k , or equivalent B1

EITHER: State equation of AB is r = 2i + 2j + k +t(−i + 2j + 2k) , or equivalent B1√

Equate at least two pairs of components of AB and l and solve for

s or for t M1

Obtain correct answer for s or for t, e.g. s = 0 or t = −2; s = −

5 1
or t = −
3 3

or s = 5 or t = 3 A1

Verify that all three pairs of equations are not satisfied and that the

lines fail to intersect A1

x − 2 y − 2 z −1
OR: State a Cartesian equation for AB, e.g. = = , and for l,
−1 2 2

x − 4 y + 2 z − 2
e.g. = = B1√
1 2 1

Solve a pair of equations, e.g. in x and y, for one unknown M1

Obtain one unknown, e.g. x = 4 or y = −2 A1

Obtain corresponding remaining values, e.g. of z, and show lines

do not intersect A1

OR: Form a relevant triple scalar product,

e.g. (2i – 4j + k).((−i + 2j + 2k)×(i + 2j + k)) B1√

Attempt to use correct method of evaluation M1

Obtain at least two correct simplified terms of the three terms of the

complete expansion of the triple product or of the corresponding

determinant A1

Obtain correct non-zero value, e.g. – 20, and state that the lines

do not intersect A1 5

(ii)

EITHER: Obtain a vector parallel to the plane and not parallel to l, e.g. 2i –4j + k B1

Use scalar product to obtain an equation in a, b and c, e.g. a + 2b + c = 0 B1

Form a second relevant equation, e.g. 2a – 4b + c = 0 and solve for

one ratio, e.g. a : b M1

Obtain final answer a : b : c = 6 : 1 : −8 A1

Use coordinates of a relevant point and values of a, b and c in

general equation and find d M1

Obtain answer 6x + y – 8z = 6, or equivalent A1

OR: Obtain a vector parallel to the plane and not parallel to l, e.g. 2i –4j + k B1

Obtain a second relevant vector parallel to the plane and attempt to

calculate their vector product, e.g. (i + 2j + k)×( 2i – 4j + k) M1

Obtain two correct components of the product A1

Obtain correct answer, e.g. 6i + j – 8k A1

Substitute coordinates of a relevant point in 6x + y – 8z = d, or equivalent,

to find d M1

Obtain answer 6x + y – 8z = 6, or equivalent A1

OR: Obtain a vector parallel to the plane and not parallel to l, e.g. 2i – 4j + k B1

Obtain a second relevant vector parallel to the plane and correctly form

a 2-parameter equation for the plane,

e.g. r = 2i +2j + k + λ(2i – 4j +k) +µ(i + 2j + k) M1

State 3 correct equations in x, y, z, λ and µ A1

Eliminate λ and µ M1

© University of Cambridge International Examinations 2005

Page 5 Mark Scheme Syllabus Paper

A AND AS LEVEL – JUNE 2005 9709/8719 3

Obtain equation in any correct form A1

Obtain answer 6x + y – 8z = 6, or equivalent A1

OR: Using the coordinates of A and two points on l, state three simultaneous

equations in a, b, c and d, e.g. 2a + 2b + c = d, 4a – 2b + 2c = d

and 5a + 3c = d B1

Solve and find one ratio, e.g. a:b M1

State one correct ratio A1

Obtain a ratio of three unknowns, e.g. a:b:c = 6:1:−8, or equivalent A1

Either use coordinates of a relevant point and found ratio to find fourth

unknown, e.g. d, or find the ratio of all four unknowns M1

Obtain answer 6x + y – 8z = 6, or equivalent A1 6

© University of Cambridge International Examinations 2005