Cambridge International A Level

MATHEMATICS 9709/33
Paper 3 Pure Mathematics 3 October/November 2023
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 October/November 2023 series for most
Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level
components.

This document consists of 21 printed pages.

© UCLES 2023 [Turn over

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
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 descriptors 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).

GENERIC MARKING PRINCIPLE 6:

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
mind.

© UCLES 2023 Page 2 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
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.

© UCLES 2023 Page 3 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
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.

© UCLES 2023 Page 4 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
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

© UCLES 2023 Page 5 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

1 State or imply non-modular inequality −0.5  2 x+1 − 2  0.5 , can be in two B1 −0.25  2 x − 1  0.25 , can be in two separate statements,
separate statements,
( )
2
or 2 x − 1  0.252 or corresponding pair of linear
( )
2
or 2 x+1 − 2  0.52
equations 0.25 = 2x – 1 and −0.25 = 2x – 1 or quadratic
– 2 and − 0.5 = 2 –2
( )
x+1 x+1
or corresponding pair of linear equations 0.5 = 2 2
equation 2 x − 1 = 0.252 .
( )
2
or quadratic equation 2 x+1 − 2 = 0.52
Incorrect inequality mark recoverable by correct final
answer or x < 0.32 and x > –0.42 .

Use correct method for solving an equation or inequality of the form M1 Reach (x + 1)ln2 = lna or equivalent, do not need to reach
2 x +1 = a or 2 x = b where a, b > 0 x=…

Obtain critical values x = 0.322 and –0.415 A1 ln 2.5 ln1.5

or awrt x = 0.32 and –0.42 e.g. − 1 and −1 .
ln 2 ln 2
or exact equivalents

State final answer –0.415 < x < 0.322 or (–0.415, 0.322) A1 Need 3 significant figures.
Need combined result, not x < 0.32 and x >–0.42 .
Must be strict inequalities. No working, 0/4.

Alternative method for Question 1

Use correct method for solving an equation or inequality of the form M1 May see 2x+1 = 1.5 and 2x+1 = 2.5 .
2 x +1 = a or 2 x = b where a, b > 0 Reach (x + 1)ln2 = lna or equivalent, don’t need to reach
x=…

Obtain one critical value, e.g. 0.322 A1 ln 2.5

or awrt x = 0.32 e.g. −1 .
ln 2
or exact equivalent

Obtain the other critical value e.g. –0.415 or awrt x = –0.42 or exact A1 ln1.5
equivalent e.g. −1 .
ln 2

© UCLES 2023 Page 6 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

1 State final answer –0.415 < x < 0.322 or (–0.415, 0.322) A1 Need 3 significant figures.
Need combined result, not x < 0.32 and x > – 0.42 .
Must be strict inequalities. No working, 0/4.

Question Answer Marks Guidance

2 Show a circle centre (2, 0) B1

Show the relevant part of a circle with radius 1 B1 FT FT centre not at the origin even if centre at 1 – 2i.
Must clearly go through (1, 0) or (3, 0) (oe for FT mark).

Show the point representing 1 – 2i B1 Can be implied by correct perpendicular bisector

Show the perpendicular bisector of the line joining 1 – 2i and the origin. B1 FT FT on the position of 1 – 2i.
Perpendicular to OP by eye and at midpoint of OP by eye sufficient.
Must reach midpoint of OP and if extended will cut BE.

© UCLES 2023 Page 7 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

2 Shade the correct region. Dependent on all previous marks, except in case B1
3 below, and the perpendicular must cut axes between CF and BE, but not
actually through C or F and not through B or E
Scale can be implied by dashes

1 Scale only on y-axis and 2OA = OC B1, B1FT, B1, B1FT, B1

2 Scale only on x-axis and 2OB = OE B1, B1FT, B1, B1FT, B1

3 No scale on either axis, but 2OA = OC B0, B1FT, B0, B1FT, B1

then 2OB = OE

Question Answer Marks Guidance

3 2(−2)3 + a(−2)2 +b(−2) + 6 = −38 M1 Substitute x = –2 and equate the result to –38
Allow errors or divide by x + 2 to obtain quadratic quotient, and equate
2 x 2 + ( a − 4 ) x + b − 2a + 8 constant remainder to –38.
x+2 3
2 x + ax 2 + bx +6
3 2
2x + 4x
(a – 4)x2 + bx
(a – 4)x2 + (2a – 8)x
(b – 2a + 8)x + 6
(b – 2a+8)x + 2b – 4a + 16
4a – 2b –10
© UCLES 2023 Page 8 of 21
9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

3 Obtain a correct evaluated equation, e.g. –16 + 4a – 2b + 6 = –38 or 4a – A1

2b = −28

1
3
1
2
1
M1 1 19
2   + a   +b   + 6 =
19 Substitute x = and equate the result to
2 2
 
2  
2  
2 2
Allow errors
or divide by 2x – 1 to obtain quadratic quotient, and
a +1 b a 1
x2 + x+ + + 19
2 2 4 4 equate constant remainder to .
2x −1 3 2
2 x + ax 2 + bx + 6
2x – x
3 2

(a + 1)x2 + bx
( a + 1) x 2 −  +  x
a 1
 2 2

 a 1
b + +  x + 6
 2 2
 a 1 b a 1
b + +  x −  + + 
 2 2 2 4 4
b a 1
6+ + +
2 4 4

1 a b 19 a b 13 A1
Obtain a correct evaluated equation, e.g. + + +6= or + =
4 4 2 2 4 2 4

Obtain a = –3 and b = 8 A1 ISW

© UCLES 2023 Page 9 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

4
2 ( −2 )2 − 4 ( 3 + i )( 3 − i ) M1 Use quadratic formula to solve for w

2 (3 + i )

Use i2 = –1 in (3 + i)(3 − i) M1

2 + 6i 2 − 6i A1 Must be simplified to this form.

Obtain one of the answers w = or w =
6 + 2i 6 + 2i

Show intention to multiply numerator and denominator by conjugate of M1 Independent of previous M marks but must be of the same
their denominator. a
form, e.g. .
b + ci

3 4 A1 2 + 6i
Obtain final answers + i and – i SC Both correct final answers from w = and
5 5 6 + 2i
Accept 0.6 + 0.8i and 0 – i 2 − 6i
w= seen, no evidence of conjugate, then SC B1
6 + 2i
for both.
3 4
Allow x = , y = or x = 0, y = –1.
5 5
3 + 4i
A0 for .
5

Alternative method for Question 4

Multiply the equation by 3 – i M1

Use i2 = –1 in (3 + i)(3 − i) M1

Obtain 10w2 – 2(3 – i)w + (3 – i)2 = 0 or equivalent A1

Use quadratic formula or factorise to solve for w M1

© UCLES 2023 Page 10 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

4 3 4 A1 SC Both correct final answers from

Obtain final answers + i and – i 10w2 – 2(3 – i)w + (3 – i)2 = 0 with no working then SC
5 5
Accept 0.6 + 0.8i and 0 – i B1 for both.
3 4
Allow x = , y = or x = 0, y = –1.
5 5
3 + 4i
A0 for .
5

Alternative method for Question 4

Substitute w = x + iy and form equations for real and imaginary parts M1

Use i2 = –1 in (x + iy)2 M1

Obtain 3(x2 – y2) – 2xy – 2x + 3 = 0 and x2 – y2 + 6xy – 2y – 1 = 0 A1 OE

Form quartic equation in x only or y only using the correct substitution M1 (3 − x )

and solve for x or y Use correct y = to attempt to form and solve
10 x − 3
50x4 – 60x3 + 63x2 – 27x = 0
x (5x – 3)(10x2 − 6x + 9) = 0.
3(1 + y )
Use correct x = to attempt to form and solve
10 y + 1
100y4 + 40y3 – 66y2 – 14y – 8 = 0
(y + 1)(5y – 4)(20y2 + 4y + 2) = 0.

3 4 A1 SC Both correct final answers from

Obtain final answers + i and – i 3(x2 – y2) – 2xy – 2x + 3 = 0 and x2 – y2 + 6xy – 2y – 1 = 0
5 5
Accept 0.6 + 0.8i and 0 – i with no working then SC B1 for both.
3 4
Allow x = , y = or x = 0, y = –1.
5 5
3 + 4i
A0 for .
5

© UCLES 2023 Page 11 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

5 Use correct product or quotient rule M1 Need attempt at both derivatives condone errors in chain
rule.
In quotient rule allow BOD in formula if ± 2x seen unless
clear that incorrect formula has been used.
If omit denominator or forget to square or complete
reversal of signs then M0 A0 M1 A1 A1 A1.

−1 −1
( ) −1 −1
2 2 2 2
6 x(1 − x 2 )e3 x + 2 xe3 x A1 If 6 x 1 − x 2 e3 x + 2 xe3 x = 0 from the start, with no
Obtain correct derivative in any form, e.g.
(1 − x ) 2 2
wrong formula seen, award M1A1.

Equate derivative (or its numerator) to zero and solve for x M1 6x – 6x3 + 2x = 0 and solve. Allow for just one x value.
Allow if from solution of 3 term quadratic equation, but if
they get x = 0 the x must factorise out

Obtain the point (0,e −1 ) or exact equivalent A1 2 3

Or for all three x coordinates found 0,  oe and no
3
extras but if this is the case then one pair of correct
coordinates A1 and both other pairs of correct coordinates
A1.
Accept, e.g. x = 0, y = e–1
ISW for last 3 marks.

2 3  A1 Allow √(4/3).
Obtain the point  , −3e3  or exact equivalent
 3 

 2 3  A1
Obtain the point  − , −3e3  or exact equivalent
 3 

© UCLES 2023 Page 12 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

6(a) Use correct Pythagoras cot2 = cosec2 – 1 or cot2 = 1/sin2 – 1 M1 If consistent omission of brackets, e.g. (sinθ)2 written as
or cot² = cos² /sin² and then cos² = 1 – sin² , sinθ2 then SC B1 in place of M1A1.
together with double angle formula cos2 = 1 – 2sin2 ,
to obtain an equation in sin 𝜃 or sin 𝜃 and cosec2

Obtain a correct equation in sin 𝜃 in any form A1 e.g. 1/sin2 − 1 + 2(1 – 2sin2 ) = 4
or
1 – sin²
sin²
( )
+ 2 1 – 2sin 2 = 4 .

If
cos²
sin²
( )
+ 2 1 – 2sin 2 = 4 then

( )
e.g. 1 − sin² + 2 1 – 2sin 2 sin² = 4 .
2
(missing sin on right) allow M1A1A0.

Reduce to the given answer of 4sin 4  + 3sin 2  − 1 = 0 correctly A1 AG Must follow from a horizontal equation (no
denominators).
If s = sin 𝜃 used and defined, allow all marks. If not
defined, award M1A1A0.

© UCLES 2023 Page 13 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

6(b) Solve the given quadratic to obtain a value for 𝜃 M1 (4sin2 − 1)(sin2 + 1) = 0 and solve for 𝜃.

Incorrect sign in solution of quadratic seen,

e.g. (4sin2 − 1)(sin2 – 1) = 0 then M0 A0 A0 but if
only see (4sin2 − 1) = 0 and nothing incorrect seen allow
3/3.

Obtain answer, e.g. 𝜃 = 30° A1 /6 award A0

Obtain three further answers, e.g. 𝜃 = 150°, 210° and 330° and no others A1 Ignore any answers outside interval.
in the interval 5/6 7/6 11/6 award A1.

Question Answer Marks Guidance

7(a)
State or imply 2 y
dy
as the derivative of y2
B1 Allow for 3x2dx + 2ydy or Fx = 3x 2 + 6 x and Fy = 2 y + 3 .
dx

dy M1 dy dy
Equate derivative of LHS to zero and solve for 3x2 + 2 y + 6x + 3 =0
dx dx dx
dy F
or 3x2dx + 2 ydy + 6xdx + 3dy = 0 or = − x need
dx Fy
evidence from B1 mark or formula must be seen. Allow
errors.

Obtain the given answer A1 dy 3x 2 + 6 x −3x 2 − 6 x

AG =− not .
dx 2y + 3 2y + 3
dy dy
Must factorise with e.g. 3x2 + 6x + (2 y + 3) = 0
dx dx
or 3x2dx + 6xdx + dy ( 2 y + 3) = 0.

3
© UCLES 2023 Page 14 of 21
9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

7(b) Equate numerator to zero and solve for x *M1 Allow for just one x value.

Obtain x = 0 and x = –2 only A1

Substitute their x, [x = 0 or x = –2] in curve equation to obtain quadratic DM1 y2 + 3y – 4 = 0 or y2 + 3y = 0.

equation in y equal to 0

Obtain y = 1 and y = –4 [when x = 0] A1

Obtain y = 0 and y = –3 [when x = –2] A1 ISW If forget x = 0 then max 3/5.

© UCLES 2023 Page 15 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

8 Separate variables correctly and reach asec2 3y or be−4x B1 Condone missing integral signs or dy and dx, but allow if
recognisable integrals follow. Not for 1/cos2 3y and 1/e4x.

1 B1 Can recover the previous B1 if de–4x seen here.

Obtain term − e −4 x
4

Obtain only a term of the form a tan 3 y M1 Can recover the first B1 if a tan 3 y seen here.

1 A1
Obtain term tan 3 y
3

Use x = 2, y = 0 to evaluate a constant or as limits in a solution containing M1 May see tan by and e 4 x here.
terms of the form a tan by and ce4 x

Obtain correct answer in any form A1 1 1 1

e.g. tan3y = − e−4x + e−8
3 4 4
1 1 −4x
or tan3y = − e + 8.39  10–5
3 4

1 3 3  A1 ISW
Obtain final answer y = tan −1  e −8 − e −4 x 
  1  3 
3 4 4 OE e.g. y = tan −1  2.52 10−4 − e−4 x 
3  4 

© UCLES 2023 Page 16 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

9(a) Ax + B C B1 If incorrect partial fractions e.g. A = 0 or Ax2 + B then

State or imply the form + M1, A1 A0 for correct C. Only allow single A1 even if
2 + 3x 2
2− x
other coefficients correct.
B1 recoverable by a correct form end statement.

Use a correct method for finding a coefficient M1 e.g. (Ax + B)(2 – x) + C(2 + 3x2)
= (3C −A)x2 + (2A −B)x + (2B + 2C)
= 17x2 – 7x + 16.

Obtain one of A = –2, B = 3 and C = 5 A1 If error present in above still allow A1 for C.

Obtain a second value A1

Obtain the third value A1 Extra term in partial fractions, D/(2 + 3x2), that is 4
unknowns A, B, C and D then B0 unless recover at end,
e.g. by setting B or D = 0. If B or D set to any value other
than 0 and all coefficients correctly found to their new
values then allow all A marks, but still B0 for partial
fraction expression unless B + D combined. Hence A1 for
each coefficient, but nothing for coefficient set to specific
value.
Another case of extra term in partial fraction expression,
namely + F, mark as above but need F = 0 to recover B1.

© UCLES 2023 Page 17 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

9(b) Use a correct method to find the first two terms of the expansion M1 Symbolic coefficients are not sufficient for the M1.
(2 − x)−1 = 2−1 + (−1) 2−2(−x) + [(−1)(−2)2−3(−x)2/2!],
−1 −1
 3x 2  3x2  x  −x 
1 +  =1– or 1 −  = 1 –  
 2  2  2  2 

Ax + B  3x 2  A1 FT Obtain correct un-simplified expansions up to the term in

1 + ( −1)  A = –2 B = 3 x 3 of each partial fraction.
2  2 
  − x  ( −1)( −2 )  − x  
2

1 + ( −1)  +   
C  2  2  2  
C=5
2 
 + ( −1)( −2 )( −3)  − x   
3

   
6  2 

3 − 2 x  3x 2  5  x x
2
x 
3 A1 FT Unsimplified (2 − x)−1 expanded correctly, error in
=  1 −  +  1 + + +  simplifying before their C is involved in the expression,
2  2  2  2 4 8  allow A1FT when their C is introduced.
3 5  5  9 5 3 5  The FT is on A, B, C.
=  +  +  −1 +  x +  − +  x 2 +  +  x3
2 2  4  4 8  2 16 

 3x 2 
−1
 3x 2  M1 Allow either ±2 or ±2−1 outside bracket or missing.
Multiply expansion of 1 +   must reach 1  by Ax + B, Allow one error in actual multiplication to acquire the 4
 2   2  terms [all terms needed].
where AB ≠ 0, up to the term in x 3 . Allow if used Cx + D (Ax + B Ignore errors in higher powers.
miscopied).

© UCLES 2023 Page 18 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

9(b) 1 13 29 3 A1 Maclaurin’s Series:

Obtain final answer 4 + x − x2 + x , or equivalent f(0) = 4 B1 f’(0) = 1/4 B1.
4 8 16
f”(0) = –13/4 and f’”(0) = 87/8 B1.
[If final answer has been multiplied throughout, e.g. by 16 then A0 at the 13 2 87 3
x x
end] 1
4+ x− 4 + 8 or equivalent M1 A1.
4 2 6
13 2 87 3
x x
1
If 1 + x − 4 + 8 then M0 A0 unless their f(0)
4 2 6
actually is 1.

9(c) 2 2 2 B1 Or exact equivalent.

State answer |x| < or − <x< clear conclusion required Strict inequality.
3 3 3

Question Answer Marks Guidance

10(a) Use the product rule correctly on y = x cos 2x M1 dx/dx cos 2x + x d/dx(cos 2x) attempted.

Obtain the correct derivative in any form A1 e.g. cos 2x – 2x sin 2x.
If cos 2x + x–2sin 2x, not recovered, max M1A0A1FTA0
but can recover for full marks by seeing correct
substitution.

π dy π A1FT dy π
Obtain y = − and = −1 when x = FT their with x = substituted.
2 dx 2 dx 2

Obtain answer x + y = 0 A1 π
OE CWO Need to see y and dy/dx at x = .
2

© UCLES 2023 Page 19 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

10(b) Integrate by parts and reach ax sin 2 x + b  sin 2 xdx *M1

1 1 A1 OE
Obtain x sin 2 x −  sin 2 xdx
2 2

1 1 A1 OE
Complete integration and obtain x sin 2 x + cos 2 x
2 4

π DM1 1 π 2π 1 2π 1
Use limits of x = 0 and x = in the correct order, having integrated twice If correct,   sin + cos − cos 0
4 2 4 4 4 4 4
to obtain ax sin 2x + ccos 2x 1 π 2π 1
or   sin − cos 0 .
2 4 4 4
Max one substitution error.

π 1 A1 π−2
Obtain answer − or exact simplified two term equivalent ISW Accept .
8 4 8
1 1
Accept x sin 2 x + cos 2 x then final answer.
2 4

Question Answer Marks Guidance

11(a) Use correct process for modulus on direction vector of l, e.g. M1 SOI Allow −12.
(−1)2 + 12 + 22 (− ) 2 +  2 + ( 2 ) .
2
Allow

1 A1 OE Allow coordinates as row or column, but not row or

 ( −i + j + 2k ) column with i, j and k included.
6

© UCLES 2023 Page 20 of 21

9709/33 Cambridge International A Level – Mark Scheme October/November 2023
PUBLISHED
Question Answer Marks Guidance

11(b) Use a correct method to form an equation for line m M1 Allow even if all signs of point incorrect, namely use
+2i − 2j + k or –3i + j – k.

Obtain r = –2i + 2j – k + μ1(–5i + 3j – 2k) A1 OE, e.g. r = 3i – j + k + μ2(−5i + 3j − 2k)

Must have r = …

11(c) Justify lines are not parallel B1 ( −5, 3, −2) ≠ d (− 1, 1, 2) or ( −5, 3, −2)x(− 1, 1, 2) ≠ 0.
Can find angle (105°, 74.6°, 1.84c or 1.3(0)c) instead but if
incorrect B0 and A0 at end.
Accept direction vectors don’t have common factor
but not direction vectors are not equal or direction vectors
are different or μ ≠ λ or scalar product ≠ 0.
Not the line equations are not multiples of each other.

Express l or m in component form B1

e.g. (–2 – 5μ1, 2 + 3μ1, –1 – 2μ1) or (3 – 5μ2, − 1 + 3μ2, 1 – 2μ2) or
(1 – λ, –2 + λ, –3 + 2λ)

Equate two pairs of components of general points on l and their m M1

and solve simultaneously for λ or for μ

11 1 A1
Obtain correct answer for λ or μ, e.g. λ = , μ1 =
2 2

Determine that all three equations are not satisfied and the lines fail to A1
1 λ μ1 2 λ μ2
intersect and conclude the lines are skew.
Conclusion needs to follow correct working ij 11/2 1/2 8 ≠ –2 ij 11/2 3/2 8 ≠ –2

ik 4/3 –1/3 –2/3 ≠ 1 ik 4/3 2/3 –2/3 ≠ 1

jk 7/4 –3/4 –3/4≠7/4 jk 7/4 1/4 –3/4≠7/4

Dependent on 4 previous marks gained.

5
© UCLES 2023 Page 21 of 21