Cambridge International A Level

MATHEMATICS 9709/33
Paper 3 Pure Mathematics 3 May/June 2024
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 International will not enter into discussions about these mark schemes.

Cambridge International is publishing the mark schemes for the May/June 2024 series for most
Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some
Cambridge O Level components.

m
co
e.
at
-m
am
x
.e
w
w
w

This document consists of 19 printed pages.

© Cambridge University Press & Assessment 2024 [Turn over

9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Generic Marking Principles

These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the
mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these marking principles.

GENERIC MARKING PRINCIPLE 1:

Marks must be awarded in line with:

 the specific content of the mark scheme or the generic level descriptors for the question
 the specific skills defined in the mark scheme or in the generic level descriptors for the question
 the standard of response required by a candidate as exemplified by the standardisation scripts.

GENERIC MARKING PRINCIPLE 2:

Marks awarded are always whole marks (not half marks, or other fractions).

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:

 marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the
syllabus and mark scheme, referring to your Team Leader as appropriate
 marks are awarded when candidates clearly demonstrate what they know and can do
 marks are not deducted for errors
 marks are not deducted for omissions
 answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as
indicated by the mark scheme. The meaning, however, should be unambiguous.

GENERIC MARKING PRINCIPLE 4:

Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

GENERIC MARKING PRINCIPLE 5:

Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited
according to the quality of the candidate responses seen).

m
co
GENERIC MARKING PRINCIPLE 6:

e.
at
Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in

-m
mind.

am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 2 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Mathematics-Specific Marking Principles

1 Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required
then no marks will be awarded for a scale drawing.

2 Unless specified in the question, non-integer answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the
degree of accuracy is not affected.

3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points.

4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).

5 Where a candidate has misread a number or sign in the question and used that value consistently throughout, provided that number does not alter the
difficulty or the method required, award all marks earned and deduct just 1 A or B mark for the misread.

6 Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.

m
co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 3 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Mark Scheme Notes

The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons
outside the scope of these notes.

Types of mark

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.

DM or DB 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 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.

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 or B marks are given for correct work only (not for results obtained from incorrect working) unless follow through is allowed (see abbreviation FT above).
 For a numerical answer, allow the A or B mark if the answer is correct to 3 significant figures or would be correct to 3 significant figures if rounded (1
decimal place for angles in degrees).
 The total number of marks available for each question is shown at the bottom of the Marks column.
 Wrong or missing units in an answer should not result in loss of marks unless the guidance indicates otherwise.
 Square brackets [ ] around text or numbers show extra information not needed for the mark to be awarded.

m
co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 4 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Abbreviations

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

ISW Ignore Subsequent Working

SOI Seen Or Implied

SC Special Case (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)

WWW Without Wrong Working

AWRT Answer Which Rounds To

m
co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 5 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

1 Use law of the logarithm of product or quotient on each side *B1 Allow logs to any base, as well as decimals, throughout.
ln 83 + ln 8−6x and ln 4 + ln 5−2x.
83
Allow for ln and ln 86x – ln 52x.
4
(3 − 6x) ln 8 and ln 4 + ln 5−2x gains next DB1 as well.

Use law of logarithm of a power involving x on ONE side, DB1 SC If *B0 DB0, then allow B1 (1/4) for a correct logarithm law
e.g. ln 83 + (−)6x ln 8 or (3 − 6x) ln 8 or  9  18x  ln 2 seen anywhere.
or ln 4 − 2x ln 5

Obtain a correct linear equation in x, e.g. B1 If in decimals, allow small errors in 2nd and 3rd dp.
 3  6 x  ln 8   9  18 x  ln2  ln 4  2 x ln 5
Obtain answer x = 0.524 B1 3dp required.
No working scores 0/4 marks.
After *B1 DB1 to correct answer with no more log working seen,
then SC B1 for x = 0.524. Maximum 3/4 possible.

Alternative Method for Question 1

Use laws of indices to get to a = b ±2x or c ± x in a correct form so now (B2) (83/4) and (5/83)−2x or (52/86)−x opposite sides or
only ONE log power law required (4/83) and (83/5)–2x or (86/52)–x opposite sides

83 83 (B1) −2x ln (5/83) or 2x ln (83/5) or x ln (86/52) or – x ln (52/86).

Obtain a correct linear equation in x, e.g. ln  2 x ln SC: If B0 then allow B1 (1/4) for a correct term seen anywhere.
4 5
If in decimals, allow small errors in 2nd and 3rd dp.

Obtain answer x = 0.524 (B1) 3dp required.

No working scores 0/4 marks.
From the first line to correct answer with no log working seen, then

m
B2 and SC B1 for x = 0.524. Maximum 3/4 possible.

co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 6 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

1 Alternative Method 2 for Question 1

Use laws of indices to get to any correct form with indices (*B1) Allow 27 – 18x and 5–2x on opposite sides
combined so now TWO log power laws are required or 29 – 18x and 22 – 4.64x on opposite sides.

Use law of logarithm of a power involving x on ONE side, (DB1) e.g. (7 – 18x) ln 2 or  9  18 x  ln2 or –2x ln 5 or (2 – 4.64x) ln 2
e.g. (7 – 18x) ln 2 = ln 5–2x or ln 27 – 18x = − 2x ln 5 or … SC: If *B0 DB0 then allow B1 (1/4) for a correct term seen
anywhere.
Allow 7 – 18x ln 2 or 9  18 x ln2 E.g. any term in *B1 shown above.

Obtain a correct linear equation in x, (B1) If in decimals, allow small errors in 2nd and 3rd dp.
e.g. (7 – 18x) ln 2 = − 2x ln 5 or (9 – 18x) ln 2 = (2 – 4.64x) ln 2

Obtain answer x = 0.524 (B1) 3dp required.

No working scores 0/4 marks.
From the first line to correct answer with no log working seen, then
*B1 and SC B1 for x = 0.524. Maximum 2/4 possible.

m
co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 7 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

2 Use correct product rule M1 ae2xsin2x + e2xbcos2x. Need a or b = 2.

cos 2x may be 1 – 2 sin2x or … Allow M1 if only error is ex instead of e2x in one of terms, then
maximum 1/5.

Obtain correct derivative 2e 2 x sin2 x  2e 2 x cos2 x  

A1 OE, e.g. 4e 2 x sinx cos x  2e 2 x cos 2 x  sin 2 x .

Equate derivative of the form ae2xsin2x + e2xbcos2x to 0 and solve M1 Obtain 2x = tan−1(− their b/their a) OE.
for 2x or x using a correct method Allow one slip in rearranging.
Note may have substituted for sin2x and/or cos2x Allow degrees.
Variety of other methods available, such as solving quadratic
equation in sin x or tan x e.g. tan² x – 2tan x – 1 = 0 leading to x
= tan-1(1 + √2).

Obtain x = 83 π only or exact equivalent A1 CWO

67.5° gets A0.
π
Ignore any answers outside interval 0 ⩽ x ⩽ .
2

1 π
3 A1 CWO, ISW.
Obtain y = 2e 4 only or exact simplified equivalent  3 3π 
2 Not sin  πe 4  .
4 
 
π
Ignore any answers using x outside interval 0 ⩽ x ⩽ .
2

m
co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 8 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

3 Square x + iy obtaining three terms when simplified and equate real M1 Having used i2 = −1.
and imaginary parts to 24 and −7 respectively

Obtain equations x2 – y2 = 24 and 2xy = –7 A1 Allow 2xyi = –7i.

Eliminate one variable by correct method and find a horizontal M1 All powers of x or y are positive and are in the numerator.
equation in the other

Obtain 4x4 – 96x2 – 49 = 0 or 4y4 + 96y2 – 49 = 0 A1

or 3-term equivalents

7 2 2 7 2 2 A1 7 2 2  7 2 2 
Obtain answers  i and   i or exact equivalents E.g.    i  , but not    i  or
2 2 2 2  2 2   2 2 
and no others
 7 2 2 
   i .
 2 2 
Allow coordinates or x =…, y =… paired correctly.
ISW converting to different form.
Must simplify 49.

m
co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 9 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

4 1 B1 Allow ln k + ln y = cx lne
State or imply that ln k + ln y = cx or ln y = cx + ln etc.
k

Carry out a completely correct method for finding ln k or c M1 Equations must have been formulated correctly.

Obtain value c = 0.80 A1 AWRT

Allow 0.8 for 0.80.
Not a fraction.
Accept in the equation ky = ecx.

Obtain value k = 6.5 A1 AWRT

Not a fraction.
Accept in the equation ky = ecx.

m
co
e.
at
-m
xam
.e
w
w
© Cambridge University Press & Assessment 2024 Page 10 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

5 B C B1
State or imply the form A  
x 1 2x  1

Use a correct method for finding a constant M1 Correct appropriate method.

Obtain one of A = 3, B = 2 and C = –3 A1

Obtain a second value A1

Obtain a third value A1

Alternative Method for Question 5

Divide numerator by denominator to reach A = 3 (M1) ax  b

May be implied by 3 [+] with a and b not both 0.
 x  1 2 x  1
x5 (A1)
Obtain 3 +
 x  1 2 x  1
D E (B1)
State or imply the form 
x 1 2x  1

Obtain one of D = 2 and E = − 3 (A1)

Obtain a second value (A1)

m
co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 11 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

6(a) Show a circle centre (4, 3) B1 Note full circle is not required but must show centre and include
Allow dashes for coordinates on axes relevant arc.

Show a circle with radius 2. B1FT FT centre not at the origin.

Can be implied by at least two of the points (2, 3), (6, 3), (4, 1) and
(4, 5) being correct

Point representing (2, 1) B1 Half-line or ‘correct’ full line extending into the third quadrant
implies point (2, 1).

Show a half-line at their (2, 1) at an angle of 13  , cutting top of B1FT FT the point (±2, ±1) or (±1, ±2).
circle between x = 3 and x = 5

Shade the correct region B1

Needs correct half-line or “correct” full line extending into the third
quadrant AND correct circle

6(b) Carry out a correct method for finding the greatest value of arg z in M1 E.g. sin−1(2/√(25)) + tan−1(3/4) or
the correct region in (a) sin−1(2/√(25)) + sin−1(3/5).
Or, e.g., substitute y = kx in circle equation, solve when
 6  21 
discriminant = 0, to get tan−1   .
 6 

Obtain answer 1.06, or 1.05 or 1.055 or 1.056 or 60.4° or 60.5° A1 The marks in (b) are available even if errors in (a).

m
No working seen scores 0/2 marks.

co
e.
at
2

-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 12 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

7(a) Show 8 × (–7)3 + 54 × (–7)2 – 17 × (–7) – 21 = 0 B1 No errors allowed.

This is sufficient if no errors seen.
[ – 2744 + 2646 +119 – 21 = 0] Correct division:
8x2 −2x −3 .
Or complete division of 8x3 + 54x2 – 17x – 21 by x + 7 to get x + 7 8x3 + 54x2 − 17x − 21
quotient 8x2 – 2x – 3 and remainder of 0 8x3 + 56x2 .
− 2x2 −17x
Or state (x + 7)(8x2 – 2x – 3) is sufficient − 2x2 −14x .
Factors must be stated again in (b) to collect marks there − 3x − 21
− 3x − 21

7(b) Commence division and reach partial quotient of the form 8x2 ± 2x M1 Condone no visible working.
or
8x3 + 54x2 – 17x – 21 = (x + 7)(Ax2 + Bx + C) and reach A = 8 and B
= ± 2 or C = –3

Obtain quotient 8x2 – 2x – 3 with no errors seen A1 Division can terminate with 0 or −3x – 21 stated once or twice.
Stating (x + 7)(8x2 – 2x – 3) is sufficient The working of division and finding quotient may be seen in (a)
but results required here to collect marks.

m
co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 13 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

7(c)  1  M1 (x + 7) (8x2 − 2x – 3) = (x + 7)(4x – 3)(2x + 1) = 0

Solve quadratic from (b) to obtain a value for 𝜃 = cos 1  
 2  2  4  96 1 3
x = cos    and .
3 16 2 4
or cos 1  
4

Obtain one answer, e.g. 𝜃 = 120° A1

Obtain three further answers, e.g. 𝜃 = 240°, 41.4° and 318.6° A1 Accept more accurate answers.
(condone 319°) and no others in the interval Answers in radians, maximum 2/3.

Question Answer Marks Guidance

8(a) State R = 12 or exact equivalent B1 ISW

Use trig formula to find α M1  3  3 −1  1 

Allow   30 or tan 1   or cos  
−1
 or sin   
 3   2   2
 3
Allow M1 if – tan 1   etc.
 3 
NB: If cos = 3 and sin = 3 seen then M0 A0.

1 A1  3
Obtain α = π CWO, so A0 if from tan 1    .
6
 3 

m
co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 14 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

8(b) Express integral in the form A  sec 2  2 x  ... dx or B1FT FT α from (a).
A  sec 2  2 x  ... dx

Integrate and reach B tan  2 x  ... or B tan  2 x  ... B1FT FT α from (a).
Where B = A or 2A or 0.5A.

1 B1FT OE
Obtain tan  2 x  ... FT α from (a).
8
1 1 1
Allow as  .
8 4 2
Coefficient must be correct.

Use limits of x  0 and x  121 π in the correct order in expression of M1 Allow with tan still present.
FT α from (a).
 
form B tan  2 x  ... so B tan   ...  −B tan ... 3 1  1 
 6  SC: B1 OE after tan  2 x  π  with no working.
12 8  6 
  
or B tan   ...  −B tan  ...
 6 

1 1 1 A1 1 1 1  3 1 
Obtain answer 3 or or
8  3 
12 ( 3 – )= needs simplifying.
4 3 48 8 3
or single term exact equivalent

5 1
Note: allow all marks in (b) even if α = π found by an incorrect
6
method in (a).

m
co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 15 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

9(a) dV k dV 1 B1
Obtain     or   
dt t dt kt

dV B1
Obtain  20 x  3x 2
dx

Correct use of chain rule involving k M1 dV dV dx

Use =  .
dt dx dt
dV dV
Expressions for and must be seen to get M1.
dt dx

dx k A1 If this expression is first seen with numerical values, allow A1

Obtain   or equivalent,
dt 
t 20 x  3x 2  when their value of k is substituted back into the general
expression.

1 1 dx 20 A1 dx 1
Use t  , x  and   to obtain given answer which  AG
10 2 dt 37 
dt 2t 20 x  3 x 2 
must be stated
20 k k
Need to at least see  = if
dx 20 37 1 3 t
 needed to score final A1 10  
dt 37 10  4
20 k k
or  = if  in working for correct k.
37 1 3 t
10  
10  4
dx 20
 seen anywhere, then A0.
dt 37

m
co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 16 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

9(b) Separate variables correctly & integrate at least one side correctly B1

Obtain terms 10x2 – x3 B1 May see –10x2 + x3 if negative sign moved across
or e.g. 20x2 – 2x3 if 2 moved across.
20 x 2 3x3
Allow  .
2 3

Obtain term ln t with ‘correct’ coefficient from their separation of B1FT FT sign and position of 2 from their separation but B0 if error from
a later manipulation.
variables, for example a ln t for .
t

1 1 M1 Allow numerical and sign errors and decimals.

Use t  , x  to evaluate a constant or as limits in a solution Allow if exponentiate before substitution, even if exponentiation
10 2
containing terms of the form x2, x3 and ln t (or ln 2t) done incorrectly, allow for c or ec.

Obtain correct answer in any form, for example A1 ln 2t 19 ln 0.2

10 x 2  x3    
ln t 19 ln 0.1 2 8 2
10 x 2  x3    
2 8 2 ln t
or 10 x 2  x3    2.5  0.125  1.15
2
Allow 1.14 to 1.16 for 1.15 and allow 2.44 to 2.46 for 2.45

2 x3  20 x 2 
19 A1 ISW
Obtain answer t  101 e 4 or equivalent Need t =………
19
2 x3  19
0.1 e 4 1 4 2 x3  20 x2
E.g. 19
, 2
, e e .
20 x 2  2 x3 
4 10e20 x 10
e
3
 20 x 2  2.45
Allow decimals, allow 2.44 to 2.46 for 2.45, e.g. e2 x .

m
1
ln

co
A0 if e 10 present in final answer.

e.
at
-m
6

xam
.e
w
w
© Cambridge University Press & Assessment 2024 Page 17 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

10(a) Carry out correct process for evaluating the scalar product of *M1  3   1   3   1 
direction vectors        
 4  .  2  or  4  . 2  
 a   2   a   2  
   
3(–1) + 4(2) + 2a or –3 + 8 + 2a or 5 + 2a.
Allow one slip in unsimplified form.

Using the correct process for the moduli, divide the scalar product *M1 *M1 marks independent of each other, so *M0 *M1 for failure to
2 use both direction vectors, but must be using scalar product and
by the product of the moduli and equate to  , same 2 vectors throughout.
2
2 2
or equate the scalar product to the product of the moduli and 
2 Allow or − throughout question.
2 2 2

5  2a 2 A1 5  2a 2
State a correct equation in any form, e.g.      OE
3 25  a 2 2 9  16  a 1 4  4
2 2
Allow unsimplified as in guidance 2
E.g. 5 + 2a =    9  16  a 2 1  4  4
2
If moduli initially correct but later has errors, award A1 when using
2 2 2
or  or − .
2 2 2

Form a quadratic equation in a with 3 or more terms all on one side DM1 Must square (5 + 2a) to get 3 terms and must remove square roots
and solve for a. from both terms on other side.
DM1 depends on BOTH *M1 9
25 + 20a + 4a2 = (25 + a2)
2
a2 − 40a + 175 = 0 hence (a – 5)(a – 35) = 0.

10(a) Obtain a = 5 and a = 35 A2 A1 for each, working not needed if quadratic correct.

m
co
6

e.
at
-m
xam
.e
w
w
© Cambridge University Press & Assessment 2024 Page 18 of 19

w
9709/33 Cambridge International A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks Guidance

10(b) Express general point of at least one line correctly in component B1 Often the third point on the line occurs after M1 A1 is gained.
form,
 1  3   –3 – µ 
   
i.e.  1  4  or  –1  2 µ 
 2 a  a   4  2µ 
   

Equate at least two pairs of corresponding components and solve M1 If solve for a first, they must have a complete method to eliminate
for λ or µ or a both λ and µ.
If using a to solve for λ or for µ, a must have been found from a
valid method.

Obtain λ = –1 or µ = –1 A1

Obtain a = 2 A1

Obtain position vector of the point of intersection is –2i – 3j + 2k A1 Accept coordinates, row or column, but not (–2i,– 3j,+ 2k) or
Two different answers for point of intersection scores A0 even if  2i 
one is correct  
 –3j  but ISW after correct form seen.
 2k 
 

m
co
e.
at
-m
am
x
.e
w
w
© Cambridge University Press & Assessment 2024 Page 19 of 19