Cambridge International Advanced Subsidiary and Advanced Level

9709 Mathematics November 2024

Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/11
Pure Mathematics 1 (11)

Key messages

The question paper contains a statement in the rubric on the f ront cover that ‘no marks will be given f or
unsupported answers from a calculator.’ This means that clear working must be shown to justif y solutions,
particularly in questions involving quadratic equations or trigonometric equations. In the case of quadratic
equations, f or example, it would be necessary to show f actorisation, use of the quadratic f ormula or
completing the square, as stated in the syllabus. Using calculators to solve equations and writing down only
the solution is not sufficient for certain marks to be awarded. It is also insufficient to quote only the f ormula:
candidates need to show values substituted into it. When f actorising, candidates should ensure that the
f actors always expand to give the coef f icients of the quadratic equation.

General comments

Some very good responses were seen but the paper proved very challenging f or many candidates. In AS
and A-Level Mathematics papers, the knowledge and use of basic algebraic and trigonometric methods from
IGCSE or O-Level is expected, as stated in the syllabus.

Comments on specific questions

Question 1
2
Many f ully correct answers to this question were seen. Successf ul candidates obtained 6(kx)2  
2
,
x
5
simplified and equated to 150, giving 24k 2 = 150 and then k = . Successful candidates then identif ied the
2
2 5
correct term 4(kx)3   and substituted k = to f ind the correct coef f icient of 125. Some candidates
 
x 2
simplified (kx) to kx and thus were unable to achieve full marks. This type of mistake in simplif ying (ax)n to
2 2

ax n is seen each year in this paper and candidates should be caref ul to avoid such mistakes.

Question 2

Several candidates achieved full marks on this question by differentiating to obtain 2x + ax –2 or equivalent,
equating this derivative to zero, substituting x = –3 and f inding the value of a.

a
Candidates who wrote x2 – as x 2 – ax -1 before differentiating were usually more successful than those who
x
attempted dif f erentiating without rearrangement.

Question 3

Most candidates earned the first mark for finding the area of sector OBC. Successf ul candidates then went
1 2 1 2 209
on to obtain × 152 ×  – × x 2 ×  =  or equivalent, and hence the length OA or OD.
2 5 2 5 5
Candidates who f ound the correct value of OA or OD generally went on to f ind the correct perimeter
required. Candidates should note that where a question asks for an answer in terms of π, f ull marks cannot
be gained f or decimal answers.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 4

Successful candidates often substituted for y in the first equation and simplified to obtain 10x2 + 3kx – 40 = 0
bef ore using b2 – 4ac to show that 9k 2 + 1600  0.

Some candidates attempted to solve 10x 2 + 3kx – 40 = 0 whilst others attempted to solve 9k2 + 1600 = 0, not
appreciating the signif icance of the discriminant in this context .

Question 5

(a) This proved to be a challenging question f or most candidates. Stronger responses rearranged
11 9
4x – 3 x + 1 = to f ind 4x – 3 x – = 0 which was then solved as a quadratic in x .
2 2
Candidates were expected to show their method for solving the quadratic in x as emphasised in
the Key messages.

dy
(b) Candidates were f amiliar with how to find the equation of a curve given
and a point on the line,
dx
and many candidates gained f ull marks on this question. Successf ul candidates integrated to
3
obtain y = 2x2 – 2x 2 + x + c, substituted x = 4 and y = 11 to f ind the value of c and hence stated
3
y = 2x 2 – 2x 2 + x – 9 .

Question 6

( x2 – x1 ) + ( y 2 – y1 ) to find
2 2
(a) Successful responses found the centres of the circles and then used
the distance between the centres. Many candidates were able to f ind the centre of the circle
(x – 9)2 + (y + 4)2 – 64 = 0. Candidates needed to rearrange x 2 + y 2 + 6x – 10y + 18 = 0 to f ind the
centre of this circle and some responses showed algebraic or sign errors when doing so.

(b) Stronger responses stated R = 4 and R = 8 and then used their answer f rom part (a) to f ind the
least and greatest distances required. Some candidates stated R = 4 and R = 8 but were unable to
f ind the greatest and least distances because they had not f ound the distances between the
centres in part (a).

Question 7

1 4
− dy − 7
(a) 3
Successful candidates differentiated 12(2x + 1) to obtain = –8(2x + 1) 3 , substituted x =
dx 2
7
to f ind the gradient and f ound the equation of the tangent by substituting x = , y = 6 into
2
y = mx + c or (y – y 1) = m(x – x 1).
1
Some candidates dif f erentiated 12(2x + 1) 3 and were unable to gain all the marks. Many
candidates were able to differentiate correctly and the method for finding the equation of a straight
line given the gradient and a point on the line was well understood.

(b) Many candidates were familiar with how to find the area under a curve using integration. As in part
1
(a), some candidates wrote y = 12(2x + 1) 3 and integrated this, so they were unable to gain f ull
marks.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 8

(a) This proved to be a challenging question for most candidates. Many responses were able to gain
sin 2 β
one mark f or substituting sin2 = a2 or cos 2 = 1 – a2. Successful candidates used tan2 =
cos 2 β
a2
and cos  = – 1 − a2 to obtain + 3a 1 − a 2 .
1− a 2

(b) Many candidates were f amiliar with how to approach questions of this type and used
cos 2 = 1 – sin2 to obtain sin2 + 4sin + 1 = 0. Solving this equation leads to sin = 3 – 2 and
hence θ = –15.5°; some candidates then instead used θ = +15.5° and hence did not find the correct
angles of 195.5° and 344.5°.

Question 9

(a) Stronger responses differentiated the given expression to obtain 5 + 12x – 9x 2, f actorised this
1 5 1 5
quadratic to obtain the critical values of − and and hence concluded x  − , x  .
3 3 3 3

Some responses did not show a method f or solving the quadratic and were theref ore unable to
gain f ull marks.

(b) Stronger responses equated their dif f erentiated expression f rom part (a) to 9, simplif ied to
28
9x 2 – 12x + 4 = 0, solved this quadratic and hence found k = . Whilst many responses gained
9
f ull marks on this question, many others were unsure about how to proceed.

Question 10

(a) Stronger responses stated that the first three terms of the geometric progression are 5 + d, 5 + 4d,
5 + 10d, f ormed the equation (5 + 4d)2 = (5 + d) (5 + 10d) or equivalent, simplif ied to a three-term
quadratic and solved it to obtain d = 2.5.

Candidates who stated 5 + d = a, 5 + 4d = ar, 5 + 10d = ar2 were generally unable to make any
signif icant f urther progress.

(b) Successful candidates found the sum of the arithmetic progression and the sum of the geometric
progression and subtracted the two values. Most candidates were familiar with f inding the sum of
an arithmetic progression. Some candidates used the correct f ormula f or f inding the sum of a
geometric progression but used 5 rather than 7.5 as the f irst term.

Question 11
2
15  3  15 15
(a) Stronger responses obtained – 2  x −  and stated the range as y  or f (x)  .
12  12  12 12
Candidates who attempted to complete the square sometimes made sign errors or had diff iculty in
dealing with fractional values. Candidates who completed the square correctly were often unable to
give a correct range f or the f unction.

 3
− 2
(b) Successful candidates stated that the reflection is in the x-axis and that the translation is   or
 15 
 
 2 
equivalent. Many responses stated the correct reflection but finding the correct translation proved
to be more challenging.

(c) Many candidates correctly stated that g is a one-to-one f unction because each y–value is
associated with a single x-value, or equivalent. Some candidates sketched the graph of g(x) = 3 +
6x + 2x 2 f or x  R rather than x  0.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

(d) This question proved to be challenging for most candidates. Candidates who attempted to f ind the
inverse f unction sometimes made sign errors or had dif f iculty in dealing with f ractional values.
Candidates were awarded one mark for drawing the line y = x and many candidates were able to
gain this mark.

© 2024
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/12
Pure Mathematics 1 (12)

Key messages

Candidates would benefit from spending time understanding the structure of exam papers. If inf ormation is
given in the introduction to the question, then it is true and relevant f or the whole question. In Question 2,
values f or the first term and common difference were given in the introduction and so were valid f or both
parts (a) and (b). Some candidates did not realise this and so were unable to complete part (b). However, if
inf ormation is given in part (a) then it is only true and relevant for that part of the question. The phrase ‘it is
given instead’ is of ten used to emphasise that this relates to a subsequent part of a question. Many
candidates failed to appreciate this distinction in Question 9 and incorrectly used those values found in part
(a) in part (b). This meant that no marks could be scored in part (b). Similarly, many candidates appeared
unsure of the required method in part (b)(ii) of Question 8. The f act that it was labelled ‘(b)(ii)’ rather than
‘(c)’ indicated that there was a link with part (b)(i). Those who used the result f ound in part (b)(i) were very
of ten able to complete part (b)(ii).

Previous reports have highlighted the fact that on the front of the question paper, in the list of instructions,
there is a statement ‘You must show all necessary working clearly; no marks will be given f or unsupported
answers f rom a calculator.’ Although most candidates now realise that they must do this to score f ull marks,
there is still a significant minority who continue to omit necessary working. To score full marks for the solution
of quadratic equations, for example, it would be necessary to show factorisation, use of the quadratic formula
or completing the square, as stated in the syllabus. Use of calculators to solve equations and writing down
the solutions is not sufficient. Neither is it suf f icient to quote only the f ormula: candidates need to show
values substituted into this. Some candidates appeared to be inventing f actors af ter having used their
calculators. Factorised quadratics must always produce the coefficients of the original quadratic when they
are expanded.

General comments

The paper was generally found to be reasonably accessible f or most candidates , although the f irst f ew
questions proved more challenging than usual. Many very good scripts were seen, and candidates generally
seemed to have suf f icient time to f inish the paper.

Presentation of work was mostly good, although some answers still seem to be written in pencil and then
overwritten with ink. This practice often produces unclear responses and makes it difficult to mark accurately.
Consequently, appropriate marks may not be awarded. Centres should strongly advise candidates to make
their responses clear, utilising the Additional page if more answer space is needed .

Comments on specific questions

Question 1

This question proved to be a challenging start to the paper for many candidates, especially part (b). Part (a)
was reasonably well answered with stronger candidates often scoring all three marks, but weaker candidates
of ten unable to score any. A common mistake was to think that b was 0.5 rather than 2. Part (b) was omitted
by many candidates and others tried to solve the given equations rather than using the given graph to work
out the number of points of intersection. Candidates may benef it f rom more practice on the relationship
between graphs and the number of solutions to an equation.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 2

Part (a) was very well answered, with most candidates able to score both marks. Part (b) proved more
challenging. Weaker responses were sometimes unable to state either the sum of the f irst k terms, or the
sum of the first 2k terms. As mentioned in the Key messages, some candidates did not realise that the
values given for a and d in the introduction were valid for part (b). Another common error was to multiply the
wrong side of their equation by 10.

Question 3

Some candidates seemed unaware of the required approach for this question and a clear understanding of
the required reasoning in part (b) was rarely seen. Candidates would benefit from time spent understanding
the f irst principles of differentiation. In part (a), many candidates knew that the required y-coordinate of B
would be 2(2 + h)2 − 3 and were able to write down the required gradient, but most were then unable to
simplify it correctly. In part (b), many responses did not use part (a) as instructed, but instead dif f erentiated
the original f unction and substituted x = 2 .

Question 4

This question proved to be reasonably straightforward with a good number of responses able to score f ull
marks. For both parts, some candidates wrote out the full expansion rather than f ocussing on the required
terms. Some weaker responses did not show understanding of the phrase ‘independent of x’, and answers of
135x and 1485x were seen f ollowing otherwise correct working. Some candidates were unable to f ind a
1
term in 3 in part (b) and were theref ore unable to make meaningf ul progress.
x

Question 5

Part (a)(i) was the best answered part of the whole paper, with 90 per cent of responses stating the correct
−1
value. Candidates need to be aware, though, that was not accepted if given as their f inal answer.
−3
Conversely, part (a)(ii) had one of the lowest success rates on this paper. Many candidates drew y = − x as
the mirror line rather than y = x. In part (a)(iii), most responses found the inverse correctly, but most did not
state its domain. Some candidates used y or f −1 rather than x to describe the domain. Part (b) was generally
 1
well answered, although some responses simply found gf   rather than using this value to form and solve
4
an equation.

Question 6

Most candidates seemed confident using the arc length and area formulae to f orm the initial equations f or
the perimeter and area, but many then had problems simplif ying them. The main problems resulted f rom
having to square 2r in the expression f or the area and also having to use 2θ as one of the angles.
Candidates were often able to form an equation in one variable and produce answers, although omission of
the necessary working meant that some responses were sometimes unable to score f ull marks.

Question 7

This question was generally well answered, especially part (a). Many correct answers were seen, although
some candidates did not seem to understand the word ‘vertex’ or it’s connection to the completed square
f orm. In part (b), most responses were able to equate the line and the curve and therefore f ind the required
points of intersection. Many candidates also realised that integration was required and were able to complete
this successfully. However, some responses only considered the curve and not the line, and others indicated
that the area under the line would be a triangle rather than a trapezium.

It should be noted that responses which omitted crucial working, possibly as this was done in an equation
solver and an integral evaluator on calculators, were unable to score any marks.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 8

Most candidates attempted part (a) but often struggled to find the correct a and b values. Candidates would
benef it from more time practicing the required techniques with both algebraic and numerical values. Part (b)
1
was usually completed correctly with most candidates realising that the gradient of the tangent was − and
2
theref ore the gradient of the normal would be 2. Part (b)(ii) was missed out by a third of candidates. As
mentioned in the Key messages, the fact that it was labelled ‘(b)(ii)’ rather than ‘(c)’ indicated that there was
a link with part b(i). Those who used this link were, usually, able to f ind p and then q correctly.

Question 9

Overall, this was the least successfully answered question on this examination paper. Part (b) was missed
out by almost 40 per cent of candidates. In part (a), many candidates equated the line and the curve and
attempted to use the discriminant, even though one point of intersection was given and a second one
required. Some realised their error and re-started, but of ten substituted only the given x-value into their
combined equation and so had one equation with 2 unknowns. In contrast, those who substituted both co -
ordinates into the given curve were able to solve it correctly, and a good number of f ully correct answers
were seen. The main error made by candidates in part (b) was to assume that the values calculated in part
(a) were still valid in part (b). Candidates need to understand that the phrase ‘it is given instead’ means that
this is not the case and only the information given in the introduction, bef ore part (a), is true f or the whole
question.

Question 10

Parts (a) and (b) of this question were generally well answered, although part (c) proved very challenging
and was only completed correctly by a small proportion of candidates. Some responses mixed up parts (a)
and (b) and did the integration needed f or part (b) in part (a). Some responses showed unnecessary
dif f erentiation in part (a) rather than using the given gradient f unction. In part (b), many were able to
integrate correctly, although some did not divide the f irst integral by 2 and others omitted the +c and
consequently could not be awarded full credit. In part (c), many candidates appeared confused by the given
equation and very few realised that because there were no solutions to it, there were therefore no stationary
points on the given curve. Because the gradient was negative in part (a), this then meant that it was a
decreasing f unction.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/13
Pure Mathematics 1 (13)

Key messages

Candidates would benefit from spending time understanding the structure of exam papers. If inf ormation is
given in the introduction to the question, then it is true and relevant for the whole question. If it is given in a
part of a question, it is only relevant to that part. In Question 7(a), a value for θ was given for use in that part
only and so was not valid for part (b). Some candidates did not realise this and, as a result, gained very f ew
marks in part (b). The phrase ‘[it is] given instead’ is often used to emphasise this in a subsequent part of a
question. This contrasts with Question 10 where the data required to solve both parts is given in the
introduction, so can be used in both of parts (a) and (b). Additionally, the result from part (a) can, and of ten
should, be used in part (b).

When f ormulating methods, candidates would be well advised to consider the marks available for a question
or question part. In Question 2, some candidates chose to expand tan 2x leading to many lines of working
of ten without a final result. With only two marks available they could have realised that a better method must
be available and then looked for this method. Similarly, in Question 6, those candidates who used the nth
term instead of the sum to n terms might have realised they obtained answers f ar too easily f or a question
worth f ive marks and then reconsidered their approach. Previous reports have highlighted that on the front of
the question paper, in the list of instructions, there is a statement ‘You must show all necessary working
clearly; no marks will be given for unsupported answers from a calculator.’ Although this message has been
taken on board by most candidates, there is still a significant minority for whom it has not. For the solution of
quadratic equations, it is necessary to show factorisation, use of the quadratic f ormula or completing the
square, as stated in the syllabus. Using calculators to solve equations and writing down the solutions is not
suf ficient. Neither is it sufficient only to quote the formula: candidates need to show values substituted into it.
Factors stated must always produce the coef f icients of the quadratic when expanded.

General comments

The paper was accessible to nearly all candidates and many excellent scripts were seen. The f inal three
questions were nearly always attempted showing candidates had suf f icient time in which to complete the
paper.

Some candidates made multiple attempts at questions. When no indication is made as to which attempt the
candidate wants to be their final attempt, the last attempt is marked. If this is not the response candidates
wish to be their f inal answer, they must make it clear which attempt they want to be marked.

Comments on specific questions

Question 1

For nearly all candidates, this was a very good f irst question. The nth term f ormula f or an arithmetic
progression was well understood and used effectively to find the 30th term. The relatively f ew errors which
were reported were usually caused by misreads or sign errors in the algebraic manipulation.

Question 2

Most answers showed that candidates appreciated the requirement to give an exact answer, and most
candidates were able to use calculators or basic skills to f ind the only answer in the given range.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 3

(a) Use of the general term of the binomial expansion produced many correct answers to this part.
Apart f rom some observed sign errors and the omission of brackets, candidates’ responses were
strong f or this question part.

(b) The results f rom part (a) were used effectively to produce a simple quartic equation. Those who
f ound both solutions correctly nearly always selected the correct positive value.

Question 4

The solution of this equation by f orming a quadratic equation in sin2 θ was usually seen. Although many
1
correct solutions for all four angles were seen, the negative solution of sin2 θ = was f requently omitted
2
resulting in only two solutions appearing. No method marks were awarded to candidates who obtained the
correct answers but did not include any working; as emphasised in the Key Messages, marks cannot be
given f or unsupported answers. When the range of the solutions is given in degrees it is expected that the
solutions will be given in degrees not radians.

Question 5

(a) The name of each of three possible transf ormation was of ten given correctly. Examiners
overlooked poor spelling if the candidates’ answers were phonetically correct. The stretch was the
transf ormation most often described correctly and the translation leas t of ten. Some candidates
f ound and described the correct transf ormations but did not present them in the correct order.
Better responses gave the description of the translation in the most economical way , i.e. as a
vector. Many different sets of correct answers were possible but only two of these sets were helpful
in answering part (b).

(b) The ef f ect of the stretch on the function was well understood but the ref lection’s ef f ect and the
translation’s effects were only described correctly in a minority of answers. Some candidates used
the incorrect relationship pf ( − x − q ) = − pf ( x + q ), losing a correct expression in the process.

Question 6

This question was challenging f or some candidates who did not make any progress with this question.
1 − r 4 17 1 − r 7 17 10(1 − r 4 ) 10(1 − r 8 )
Wrong starting points included: = , = or 16 = and 17 = .
1 − r 8 16 1 − r 3 16 1− r 1− r

Candidates who set up a correct equation mostly progressed accurately to f orm, and then solve, a correct
three term quadratic in r 4 , or to factorise and divide to form a simple equation in r 4 . Those who arrived at
1
r4 = did not always get both solutions resulting in only one correct sum to inf inity, while others thought
16
1
the solution was r = (  ) . Some responses incorrectly stated that a sum to inf inity could be f ound using
4
r = ±1.

Question 7

(a) Most responses involved finding the areas of two triangles, two semicircles and the major sector.
Some candidates found the area of the large circle, subtracted the areas of two segments and
added the areas of the two semicircles. A higher proportion of correct answers were seen f rom
candidates who worked in radians rather than degrees. The requirement to f ind the radius of the
semicircles was not always appreciated and some responses incorrectly assumed the diameters of
the semicircles were equal to the radius of the large circle. The f ormula f or sector area was well
known, as was the need to use an angle at the centre of 2π − 2θ.

(b) This part proved to be more difficult even for some candidates who gained f ull marks in part (a).
Using the area of a semicircle was problematic for some, and a significant number did not realise
that the value of θ was now changed from that in part (a). The strong hint that the perimeter could

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

be expressed exactly was of ten missed and numerical values of π were sometimes used. The
f ormula for arc length was nearly always seen correctly expressed using radians and occasionally
degrees.

Question 8

(a) This was the most successfully answered question part on this paper with completion of the square
f or this type of f unction proving to be well understood by nearly all candidates.

(b) With so many correct answers to part (a) it was anticipated that the required value of k would be
f ound from most of these. This proved to not always be the case, with k given as x or use of the
value of b rather than a of ten seen.

(c) Change of subject of a formula was used to good effect by most candidates to obtain a correct form
of the inverse f unction.

(d) The quickest method using x = f −1f −1 ( 29 ) was rarely seen, with most candidates attempting to
express ff ( x ) in algebraic form and equating their result to 29. This produced more opportunities
f or algebraic errors and the resulting quartic equation often proved too diff icult to solve. Although
some candidates found a correct solution, they did not realise that this method gives an additional
incorrect solution of x = 1 and this was seldom discounted.

Question 9

(a) Most candidates knew to equate the equations and obtained a correct cubic in x. Some responses
not worthy of full credit either omitted showing the factorisation or omitted the solution x = 0. Some
went on to f ind the values of y unnecessarily.

(b) The required integration was done well with many responses showing full working and gaining f ull
marks. There seemed to be an even proportion of candidates subtracting the results of two
integrations and candidates subtracting the equations before integration. Some responses did not
show the substitution of limits and, as such, could not be awarded f ull credit. A small number of
responses obtained the correct answer with no working shown and theref ore gained no marks.

Question 10

(a) Two main methods were used, and some very good answers were seen. Most who f ound the
gradient and mid-point of AB and used these to find the perpendicular bisector of AB reached the
required equation. Those who equated the distances of A and B f rom a centre (a, b ) of the circle
tended to make more algebraic errors, and some did not show their result, a = 2b + 8, to be
equivalent to the given equation.

(b) Several successful methods were used to f ind the two possible centres. Those who f ound the
centres usually used a general f orm of the circle equation ef f ectively to obtain both answers
correctly. Most correct answers came from substitution of the equation given in part (a) into the
distance equation of A or B f rom the centre of the circle. However, clever use of vectors and
geometry also produced correct answers. A significant number of scripts saw candidates repeating
part (a) to f ind the equation of the perpendicular bisector of AB even though they were entitled to
use the given result in this part.

Question 11

(a) The vast majority of candidates f ound the correct f irst and second dif f erential.

(b) This part of the question proved more challenging, with many candidates who set up the correct
1
−
equation x 2 − 8 x = 0 unable to solve this correctly. Some of those candidates squared each term
individually and arrived at the correct value for x from incorrect algebra. Other candidates did not
deal correctly with the negative fractional index, resulting in an incorrect solution. Some candidates

© 2024
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers
3
3
1  1 2
correctly arrived at x = but incorrectly stated the solution was   . A common error was to
2
8 8
only f ind the x-coordinate of the stationary point. Candidates usually knew that the nature of the
stationary point could be determined by finding the value of the second derivative, which many of
the candidates did accurately f or their x-coordinate. Occasionally, a candidate thought that
‘increasing’ or ‘decreasing’ was required rather than ‘maximum’ or ‘minimum’ when determining the
nature of the stationary point.

(c) Responses to this part of the question were mixed. Stronger responses attempted to f ind both the
gradients and the y-coordinates, then used these to f ind the equation of the tangents bef ore
equating and using x = 0.6 to find the value of k. It was not uncommon for there to be mistakes in
the working which often occurred when finding the y-coordinate at x = 0.25. This resulted in one
incorrect equation and theref ore an incorrect value f or k. Occasionally, having f ound the two
equations correctly, some then made errors when simplifying. Some candidates correctly found the
two y-coordinates but then attempted to use this in an equation involving the change in y/change in
x.

© 2024
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/21
Pure Mathematics 2 (21)

Key messages

Candidates need to ensure that they have fully met the demands of the question by reading it in f ull bef ore
attempting a solution. Some candidates still do not appreciate the meaning of the word ‘exact’ in the context
of an answer. The implication is that no calculator use is required. Greater care is needed in the
simplification of some algebraic expressions, with simple sign slips leading to an unnecessary loss of marks.

General comments

A wide range of responses was seen. It was evident that some candidates had very little confidence with the
topic on iterative methods for the solution of equations. Most candidates appeared to have been otherwise
prepared well for the examination. There were no timing issues, and most candidates had sufficient space for
their solutions.

Comments on specific questions

Question 1

(a) Most candidates attempted to take logarithms as required. Some candidates wrote down the result
3 k
y= x+ but did not identify the gradient from this equation, as required . This highlights
2ln a 2ln a
the need f or candidates to ensure that they have met the demand of the question.

(b) Most candidates f ound the gradient of the straight line, making use of the given coordinates.
Stronger responses then equated this to the given result f rom part (a) and the value of a was
f ound. Correct use of the equation of the straight line initially f ound in part (a) yielded the correct
value of k. Some candidates chose to use the straight line equation from part (a) and f ormed two
simultaneous equations using the given coordinates. This was less common, but equally
successful. Very few candidates attempted to use the original equation and re-write the coordinates
accordingly.

Question 2

The method of squaring to obtain a three-term quadratic equation was the most common method used.
Fewer candidates used the method of dealing with two appropriate linear equations. Most candidates
4 10 10
obtained the critical values and − ; very few candidates discounted the value of − . Candidates should
5 3 3
be reminded to check the values around the critical values to see if the inequality is f ully satisf ied.
Alternatively, a small sketch of the lines y = x − 7 and y = 4 x − 3, considering the relative gradients, would
have clarif ied that there was only one point of intersection and helped with the f inal inequality.

Question 3

(a) Few candidates made use of the chain rule, and those that did of ten made errors with the
coefficients of the terms involved. Most chose to rewrite the f unction using an identity bef ore

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 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

x
2 tan  
 x  1 − cos x x  2 .
attempting dif f erentiation, f or example tan2   = or tan2   = 1 −
 
2 1 + cos x  2  tan x
Subsequent use of the quotient rule and substitution could then be made to f ind the exact value
required.

x
(b) Few candidates made any progress unless they realised that tan2   needed to be written as
2  
x
sec 2   − 1 bef ore integration could take place. Correct answers were seldom seen.
2

Question 4

(a) Most candidates were able to find the correct value of a using the factor theorem. Few errors were
seen.

(b) Most candidates were able to find the quadratic factor needed to go on and fully factorise p( x ). This
was of ten done either by algebraic long division or inspection. Some candidates thought that this
was suf ficient and did not attempt any further factorisation. There was evidence some candidates
made use of their calculator to solve the equation p( x ) = 0 and ‘work backwards’ f rom the
2
 3
solutions, as evidenced by answers such as ( x + 2 )  x − . This result is not the factorised f orm
 2 
of p( x ) and as such was not worthy of f ull credit.

(c) Few candidates made any attempt at this part of the question, likely not recognising the connection
with part (b). Of those that did make progress, most missed the negative solution.

Question 5

(a) The integration was performed well by the candidates who attempted this question part. There
were occasional errors made in the application of the rules of logarithms, but candidates who
applied the rules correctly usually went on to show the given result with suf f icient detail.

(b) There were a large number of blank responses for this question, indicating that this is a topic with
which many candidates are not confident. For the candidates who did attempt a solution, most
work was completely correct, including sufficient iterations stated to the required level of accuracy.

Question 6

(a) Most candidates made a reasonable attempt at dif f erentiation. The quotient rule was f requently
dx dy
used to find , but errors in simplification were quite common. Most were also able to f ind
dt dt
dy
correctly, and hence an expression for as required. It was expected that an attempt be made
dx
dx
to simplify the numerator of to a single term, and that there were no ‘fractions within f ractions’
dt
in the f inal answer.

(b) Most candidates realised that they needed to find the value of t when x = 0 and obtained a correct
value. Unf ortunately, errors in simplification in part (a) of ten prevented candidates obtaining f ull
marks. There was also some evidence of calculator use even though an exact answer was
required; candidates are reminded to reread the question once they have completed their work to
ensure they have met the stated requirements.

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 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 7

(a) Many candidates used correct trigonometric expansions and substitutions f or the required
trigonometric ratios. Use of the double angle formula f or sin 2θ was usually applied together with
use of the identity cos2 θ + sin2 θ = 1, leading to the required result. There were the occasional sign
errors, but most candidates showed suf f icient detail and were accurate in their solutions.

(b) Many candidates successfully made the connection with the result from part (a). Some candidates
solved an equation in sin 2α rather than the correct sin 4α. Whilst some candidates did obtain a
correct solution of 6.9o , very few continued and f ound the second solution in the stated range.

© 2024
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/22
Pure Mathematics 2 (22)

Key messages

Candidates should be aware of the required level of accuracy in their final answers, as specified by the rubric
of the paper or in the question itself. It should be noted that calculations prior to giving a f inal answer should
be conducted with greater accuracy than the level required in the answer itself. It is essential that candidates
ensure that they have fully met the demands of the question. When they are asked to show a certain result,
they need to show suf f icient detail.

General comments

A wide range of responses was seen. It was evident that some candidates had very little confidence with the
topic on iterative methods for the solution of equations. Most candidates appeared to have been otherwise
prepared well for the examination. There were no timing issues, and most candidates had sufficient space for
their solutions.

Comments on specific questions

Question 1

Most candidates were able to obtain a mark for correctly applying logarithms and using the power law. Many
7ln6
candidates appeared to be unable to rearrange the equation 8y ln5 = 7 x ln6 correctly and obtain y = x,
8ln5
ln 6 6
with some candidates mistakenly rewriting as ln . Some candidates expressed the coef f icients in the
ln 5 5
equation 8y ln5 = 7 x ln6 in decimal form, which was acceptable provided a sufficient level of accuracy was
used. However, many candidates were unable to obtain the f inal accuracy mark due to premature
approximation.

Question 2

(a) A large number of candidates were unable to gain marks here as they did not identif y a suitable
method of solution. It was intended that candidates make use of the chain rule. Of those that did
use the chain rule, errors in coefficients were common. Another valid method of solution was to
make use of the double angle f ormula and write 4sin2 3x as 2 − 2cos 6x and then dif f erentiate.
Candidates using this approach were often successful and were then able to make use of this initial
work in part (b).

(b) It was essential that candidates make use of the double angle f ormula to rewrite the integrand as
2 − 2cos 6x bef ore attempting integration. Few candidates made use of this method, and f ew
correct solutions were seen.

Question 3

It was evident that some candidates were not conf ident with implicit dif f erentiation and, as a result, were
unable to gain any marks in this question. For those candidates that were familiar with the topic, many were
−x 2 dy
unable to differentiate the term 6e y as a product but were able to continue to obtain an expression for
dx
and hence f ind a value f or the gradient.

© 2024
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 4

Many candidates were unable to produce a correct sketch of the graph of y = 1 + e , although most
2x
(a)
candidates were f amiliar with the f orm of the graph of the modulus of a f unction.

(b) The stated equation gave many candidates a good hint so that they considered the equation
1 + e2 x = − x + 4. A correct result usually followed although there were instances of the incorrect use
of logarithms. Candidates who chose to consider a squaring method were seldom successf ul ,
being unable to deal with the resulting algebraic terms successf ully.

(c) It is evident that some candidates were unprepared to answer a question on this topic, as
evidenced by the lack of responses to this question part. For the candidates who did attempt a
solution, most work was completely correct, with sufficient iterations done to the required level of
accuracy.

Question 5

(a) Most candidates were able to gain marks in this question part, showing an understanding of both
the f actor and remainder theorems. Sign errors and errors in simplification were usually the reason
candidates did not obtain the accuracy marks.

(b) Most candidates attempted algebraic lo ng division and, provided part (a) was correct, usually
obtained a correct quadratic factor. Some candidates simply stopped at this point, stating that the
resulting quadratic factor had no solutions when equated to zero. Candidates should be aware that
the demand of the question required them to show that there was only one real root to the equation
p ( x ) = 0. To obtain f ull marks, it was essential that candidates show that the discriminant of
2x 2 − 3 x + 4 was −23, (or use an equivalent valid method) and conclude appropriately, also stating
the only real root of the equation.

(c) This question part was completed with varying levels of success. Well prepared candidates were
f amiliar with the approach needed and many obtained a critical value of 14.5o. Fewer candidates
gave the correct answer of −14.5o. Many candidates were unable to make any valid attempt at
solution.

Question 6

(a) It was pleasing to see many correct solutions gaining full marks. However, it was evident that some
candidates were unaware of the trapezium rule and were unable to proceed.

(b) Some candidates did not read the question caref ully and considered the given area as being
2 27
calculated f rom both 0 3 5 x 2 + 7 dx and 
2
dx, when only the latter was needed. Many
0 2x + 5

2 27
responses were able to gain credit f or a valid attempt at  dx, but incorrect answers of
0 2x + 5

9
27ln were common.
5

(c) It was essential that candidates show the deductive process that they used, that is their answer to
part (b) – their answer to part (a), rather than just write down an unsupported f igure. Candidates
should be reminded of the significance of the command word ‘deduce’ and the expectations in their
solutions associated with this.

(d) For most candidates, the answer was an incorrect ‘over-estimate’. Of the candidates who stated
’under-estimate’, f ew were able to justif y their choice correctly and with suf f icient detail.

© 2024
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 7

(a) Many candidates were able to expand 4 sin θ sin ( θ + 60o ) correctly and obtain

2sin2 θ + 2 3 cos θ sin θ. While many candidates went on to make use of the f act that
2 3 cos θ sin θ = 3 sin2θ, most did not apply the appropriate double angle f ormula to 2sin2 θ. No
f urther progress could be made until this was completed. As a result, very f ew f ully correct
solutions were seen.

(b) Very f ew correct solutions were seen. Many candidates, having no f inal response to part (a), did
not attempt this part. Other candidates who attempted to make use of their incorrect result f rom
part (a) usually did not deal with the double angle correctly. Some candidates did not appreciate
the meaning of the word ‘hence’ and attempted to solve the given equation by other incorrect
means.

© 2024
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/23
Pure Mathematics 2 (23)

Key messages

Candidates need to ensure that they have fully met the demands of the question by reading it in f ull bef ore
attempting a solution. Some candidates still do not appreciate the meaning of the word ‘exact’ in the context
of an answer. The implication is that no calculator use is required. Greater care is needed in the
simplification of some algebraic expressions, with simple sign slips leading to an unnecessary loss of marks.

General comments

A wide range of responses was seen. It was evident that some candidates had very little confidence with the
topic on iterative methods for the solution of equations. Most candidates appeared to have been otherwise
prepared well for the examination. There were no timing issues, and most candidates had sufficient space for
their solutions.

Comments on specific questions

Question 1

(a) Most candidates attempted to take logarithms as required. Some candidates wrote down the result
3 k
y= x+ but did not identify the gradient from this equation, as required . This highlights
2ln a 2ln a
the need f or candidates to ensure that they have met the demand of the question.

(b) Most candidates f ound the gradient of the straight line, making use of the given coordinates.
Stronger responses then equated this to the given result f rom part (a) and the value of a was
f ound. Correct use of the equation of the straight line initially f ound in part (a) yielded the correct
value of k. Some candidates chose to use the straight line equation from part (a) and f ormed two
simultaneous equations using the given coordinates. This was less common, but equally
successful. Very few candidates attempted to use the original equation and re-write the coordinates
accordingly.

Question 2

The method of squaring to obtain a three-term quadratic equation was the most common method used.
Fewer candidates used the method of dealing with two appropriate linear equations. Most candidates
4 10 10
obtained the critical values and − ; very few candidates discounted the value of − . Candidates should
5 3 3
be reminded to check the values around the critical values to see if the inequality is f ully satisf ied.
Alternatively, a small sketch of the lines y = x − 7 and y = 4 x − 3, considering the relative gradients, would
have clarif ied that there was only one point of intersection and helped with the f inal inequality.

Question 3

(a) Few candidates made use of the chain rule, and those that did of ten made errors with the
coefficients of the terms involved. Most chose to rewrite the f unction using an identity bef ore

© 2024
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

x
2 tan  
 x  1 − cos x x  2 .
attempting dif f erentiation, f or example tan2   = or tan2   = 1 −
 
2 1 + cos x  2  tan x
Subsequent use of the quotient rule and substitution could then be made to f ind the exact value
required.

x
(b) Few candidates made any progress unless they realised that tan2   needed to be written as
2  
x
sec 2   − 1 bef ore integration could take place. Correct answers were seldom seen.
2

Question 4

(a) Most candidates were able to find the correct value of a using the factor theorem. Few errors were
seen.

(b) Most candidates were able to find the quadratic factor needed to go on and fully factorise p( x ). This
was of ten done either by algebraic long division or inspection. Some candidates thought that this
was suf ficient and did not attempt any further factorisation. There was evidence some candidates
made use of their calculator to solve the equation p( x ) = 0 and ‘work backwards’ f rom the
2
 3
solutions, as evidenced by answers such as ( x + 2 )  x − . This result is not the factorised f orm
 2 
of p( x ) and as such was not worthy of f ull credit.

(c) Few candidates made any attempt at this part of the question, likely not recognising the connection
with part (b). Of those that did make progress, most missed the negative solution.

Question 5

(a) The integration was performed well by the candidates who attempted this question part. There
were occasional errors made in the application of the rules of logarithms, but candidates who
applied the rules correctly usually went on to show the given result with suf f icient detail.

(b) There were a large number of blank responses for this question, indicating that this is a topic with
which many candidates are not confident. For the candidates who did attempt a solution, most
work was completely correct, including sufficient iterations stated to the required level of accuracy.

Question 6

(a) Most candidates made a reasonable attempt at dif f erentiation. The quotient rule was f requently
dx dy
used to find , but errors in simplification were quite common. Most were also able to f ind
dt dt
dy
correctly, and hence an expression for as required. It was expected that an attempt be made
dx
dx
to simplify the numerator of to a single term, and that there were no ‘fractions within f ractions’
dt
in the f inal answer.

(b) Most candidates realised that they needed to find the value of t when x = 0 and obtained a correct
value. Unf ortunately, errors in simplification in part (a) of ten prevented candidates obtaining f ull
marks. There was also some evidence of calculator use even though an exact answer was
required; candidates are reminded to reread the question once they have completed their work to
ensure they have met the stated requirements.

© 2024
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 7

(a) Many candidates used correct trigonometric expansions and substitutions f or the required
trigonometric ratios. Use of the double angle formula f or sin 2θ was usually applied together with
use of the identity cos2 θ + sin2 θ = 1, leading to the required result. There were the occasional sign
errors, but most candidates showed suf f icient detail and were accurate in their solutions.

(b) Many candidates successfully made the connection with the result from part (a). Some candidates
solved an equation in sin 2α rather than the correct sin 4α. Whilst some candidates did obtain a
correct solution of 6.9o , very few continued and f ound the second solution in the stated range.

© 2024
 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/31
Pure Mathematics 3 (31)

Key messages

Candidates should:
• take time to think about the most ef f icient method to use
• think about standard results and how these could be relevant
• check that their solution answers the question.

General comments

There was only a small entry for this paper. Some of the candidates demonstrated a good understanding of
the topics examined, but many scored low marks, even on the more accessible topics.

There were f ewer blank responses compared to previous series, with more candidates showing knowledge
of most of the specif ication.

The candidates scored well on numerical methods (Questions 5(b) and 5(c)), partial f ractions
(Question 7(a)) and solving the dif f erential equation (Question 10(b)). In some places, they adopted
methods that made the solution unnecessarily complicated, such as using long division rather than the factor
and remainder theorems (Question 1) and not using the most obvious trigonometrical identity (Question
4(a)). Particular areas candidates can improve on include sketching (Question 5(a)), integration of
trigonometric functions (Question 6(b)), the vector equation of a straight line (Question 9) and f orming a
dif f erential equation (Question 10(a)).

Comments on specific questions

Question 1

Those candidates who were f amiliar with the factor theorem and the remainder theorem were quick to f orm
the pair of simultaneous equations required and had f ew dif f iculties in solving them. In contrast, the
responses which attempted to use algebraic long division often found this challenging and rarely reached the
point where they had the necessary remainders to f orm the required simultaneous equations. It is not
incorrect to use long division, but the method is more complicated than necessary f or this task.

Question 2

The majority of candidates recognised the need to use integration by parts. Most obtained an expression of
the correct form but with the correct coefficients rarely seen. This was of ten due to errors in dif f erentiating
ln3 x. The limits were used correctly, but due to errors in integrating and in simplif ying the log terms, f ully
correct responses were rare.

Question 3

Responses which recognised the need to use implicit differentiation usually obtained the correct derivative
f or the right-hand side of the equation. The lef t-hand side was more challenging, but the stronger candidates
of ten completed this correctly. Some candidates then rearranged their derivative to obtain an expression of

© 2024
 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

dy
the f orm = ..., which was not necessary and made the solution more complicated than needed.
dx
dy
Substituting x = 1, y = 0 into the unsimplif ied derivative gives a simple equation in .
dx

Question 4

(a) There are two simple approaches to this question. One is to f actorise the lef t-hand side of the
identity using a difference of two squares and the other is to use the substitution sec 2 θ = 1 + tan2 θ.
The second was the more popular option, but both lead to the solution. Some candidates
attempted to produce an expression in sinθ and cos θ, which usually resulted in a complicated
expression which did not lead to the required f orm.

(b) A small number of candidates used the result f rom part (a) to f orm and solve a quadratic in
tan2 2α. To obtain the final answer, candidates needed to consider both solutions of the equation
and to remember that their equation gives values of 2α, so there is a f urther step necessary to
obtain the values f or α. Fully correct solutions were rare.

Question 5

(a) This question demonstrated that many candidates need to improve upon the quality of their
sketches. Some candidates showed an understanding of the exponential curve and s ome gave a
good representation of y = ln (1 + x ) . Other candidates must remember to show both curves on the
same sketch and remembered to indicate that the point of intersection represent s a solution of the
equation.

(b) The majority of candidates were aware of the ‘sign change’ method, and many used this effectively.

(c) Those candidates who used an initial value of 8 f ound that the iteration converged very quickly.
There were several f ully correct responses. Other responses did not work to the required level of
accuracy or omitted the f inal conclusion.

Question 6

(a) Most candidates knew that they needed to differentiate to locate M. Some dif f erentiated the given
equation as a product, which was usually the more successf ul method, and some expanded the
bracket before differentiating. The question instructs candidates that they are looking for a value of
1 3
x in the interval π  x  π, but few candidates noticed this, and f ully correct responses were
2 4
rare.

1 − cos 4 x
(b) A f ew candidates recognised that they needed to rewrite sin2 2x as , but it was more
2
common to see an incorrect solution involving sin3 2x or cos3 2x.

Question 7

(a) The majority of candidates used the correct f orm f or the partial f ractions and there were several
f ully correct answers. Incorrect answers were usually due to arithmetic errors.

Many candidates obtained the correct value for the coefficient of x 3 in the expansion of (1 + 2x ) .
−1
(b)

The expansion of ( 2 + x 2 ) was more challenging, with several responses incorrectly dealing with
−1

the 2 at the start of the bracket. Some candidates gave the expansion up to and including the term
in x 3 . This involved additional work, but so long as the coefficient of x 3 was correct the other terms
were ignored.

© 2024
 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 8

1 1
(a) The candidates who understood that they needed to write as and to multiply the top and
z 1 + yi
the bottom by 1 − yi usually obtained the correct answer.

(b) Demonstrating the given answer requires the correct answer f rom part (a). However, two of the
three marks were f or correct use of the results f rom part (a). These marks depended on correct
algebra, but also on having an answer to part (a).

(c) Most candidates drew the correct vertical line and understood that the second locus was a circle.
Responses which were not awarded f ull credit were normally due to the circle being located
incorrectly or having the incorrect radius.

(d) This part of the question required candidates to draw together the previous parts of the question.
This was a novel question, but there were many correct responses seen.

Question 9

(a) This was a straightf orward question about determining a vector line equation f rom basic
inf ormation. Most responses included some of the key processes in how to form the equation, but
many overlooked the requirement that the equation should commence r = …

(b) There were a f ew correct solutions, but many candidates did not appear to be conf ident working
with vectors. Candidates should be aware that it is necessary to form simultaneous equations using
the components of the line equation f or questions such as this.

(c) Those candidates who knew how to use the scalar product were able to f orm an equation in a.
Some solutions were incorrect as they did not use the directions of the lines, and some due to
algebraic or numerical errors.

Question 10

(a) There were only a f ew correct solutions seen for this part of the question. Several candidates did
not f orm a correct equation using the information about proportionality, and several did not attempt
to f orm an equation using related rates of change.

(b) There was little work to do to separate the variables in this dif f erential equation, and several
candidates had a correct strategy f or solving the equation. The question does ask f or an
expression for t, so candidates who did not form an equation with t as the subject were unable to
gain f ull credit.

© 2024
 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/32
Pure Mathematics 3 (32)

Key messages

Candidates should:
• ensure their responses meet all the demands of the question
• take care in the details of their algebra and in the accuracy of their arithmetic calculations
• ensure that, if the question asks f or a particular method, this request is f ollowed
• note that diagrams can be helpf ul, especially in vector questions
• not replace a solution by writing over it, as the result may be illegible to the examiner. This applies
equally to pencil working overwritten in ink.

General comments

This paper proved to be challenging for many candidates, but also was very accessible f or many others.
While there were many candidates scoring fewer than 10 marks, the number of candidates scoring 60 marks
or more was higher than in previous series.

The paper was a mixture of familiar questions: Question 1 (binomial expansion), Question 3 (square roots
of a complex number), Question 6 (linear graph), Question 7 (trigonometric identities) and Question 8
(parametric form) should all have been recognised and were accessible. Question 4 (index notation) and
Question 5 (complex numbers) proved to be particularly dif f icult. There was no obvious pattern to which
questions the candidates found difficult – some weaker candidates scored close to full marks in Question 11
and some otherwise strong candidates made limited progress with it.

Incorrect use of mathematical notation continues to be a problem, with brackets f requently omitted, and
many responses treated terms like tan and ln as algebraic objects. There were many very basic algebraic
4e2 x (e2 x − 3e x + 2) − 2e2 x (2e2 x − 3e x ) 4e2 x − 2e2 x (2e2 x − 3e x )
errors: in particular, the incorrect cancelling =
(e 2x
− 3e x + 2 )
2
(e 2x
− 3e x + 2 )
was common. In Question 3 and Question 7(b), several candidates were unable to obtain f ull marks f rom
only considering positive square roots of a number.

Comments on specific questions

Question 1

There were a good number of fully correct solutions seen. The most successful method was to start by taking
1
a f actor of 9 2 from the bracket. Some candidates then omitted multiplication by 3 af ter they had expanded
the bracket. The incorrect statement ( 9 − 3 x ) 2 = 9 (1 − 3 x ) 2 was common and there were many sign errors
1 1

seen. Some candidates who attempted to use the general form for the expansion of ( a + b ) got no f urther
n

1 1
than the initial substitution, being unable to give a value f or 2C1 or 2C2 .

Question 2

(a) The quality of the sketches was lower than in previous series. Many curves seen were notably
dif ferent to the curves expected. Some candidates were unclear about the horizontal scale they
were using, often conf using cot 2x with cot x or cot 21 x. Curves of the correct shape sometimes

© 2024
 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

crossed the asymptote at 1

2
π or appeared to be heading f or an asymptote much less than 21 π. A
small number of candidates took an alternative approach and attempted to sketch tan 2x and
cos x.

(b) The most successf ul method was to start with the iterative f ormula and show that, at the root,
cot 2 x = sec x. Candidates who worked in the reverse direction f requently stopped at
x = 21 tan−1 ( cos x ) and did not state the iterative formula. Quite a f ew candidates thought that part
(b) was an invitation to find the root and embarked on the iterative process, reaching the value
0.3747. This does not answer the question and candidates should ensure that, as highlighted in the
Key messages, their responses meet the demands of the question.

Question 3

The majority of candidates were familiar with the method required in this question. There were some errors in
squaring and sign errors seen, particularly when comparing the imaginary parts. Responses which got as f ar
as comparing both real and imaginary parts usually obtained the quartic equation correctly. The f inal mark
could often not be awarded due to candidates not considering both the positive and negative square roots or
an incorrect combination of signs.

A significant minority of candidates could not get started with the question. The most common errors were to
start by squaring 6 − 8i or to f ind the expansion of ( x + iy ) .
4

Question 4

Many candidates scored no marks for this question. In the majority of cases, this was due to starting with the
incorrect use of logarithms; use of the incorrect formula log ( a + b ) = log a + log b was the most common f irst
step.

Those candidates who started by using 5x +2 = 5x  52 usually obtained the correct answer. There were only a
f ew responses which did not give the answer to the required level of accuracy.

Question 5

(a) Very f ew candidates were sufficiently familiar with the properties of a number written in the f orm
r ( cos θ + i sin θ ) and the associated form reiθ . Only a small minority of candidates spotted that this
was a very straightforward question merely requiring subtraction of two arguments, albeit with one
of them negative. There were many blank responses. Some responses showed an attempt to
multiply top and bottom of the fraction by the conjugate of the denominator, which was fine, but the
significance of the result and how this related to what they had been asked to f ind was of ten not
f ully appreciated.

(b) Several candidates were aware that u* is a ref lection of u, but they often omitted which line it was
ref lected in or gave an incorrect line. The line y = x was a common incorrect answer. Few
candidates seemed aware that the argument of u* is simply the negative of the argument of u.
Although both marks here were accessible to a candidate aware of the basic properties of a
number and its conjugate, there were many blank responses seen.

Question 6

Candidates who started by forming a pair of simultaneous equations in a and b of ten became conf used
between y and ln y . Equations involving ln 2.24 and ln 8.27 were common. Candidates who started by finding
the gradient of the line usually equated this to ln b and went on to f ind b correctly. Finding the value of a
proved to be more challenging. Some responses starting with a correct equation lost the sign and obtained
ln a = 1.20... . Some who had ln a = −1.20... thought that the answer f or a would be −e1.20 . There was some
conf usion between signif icant f igures and decimal places.

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 Cambridge International Advanced Level
9709 Mathematics November 2024
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Question 7

(a) Most candidates recognised that they need to start by using the f ormula f or tan2x, which was
usually stated correctly. Some candidates overlooked that they need to double this, and some
doubled both the numerator and the denominator. Careful manipulation of the left-hand side usually
resulted in the correct expression. Common errors included tan3 x  tan2 x = tan6 x and
( )
− tan x 1 − tan2 x = − tan x − tan3 x. Both of these were of ten f ollowed by f orced algebraic
manipulation of incorrect working to obtain the given answer.

(b) Most candidates recognised that they needed to use the result f rom part (a) to obtain
4tan4 2θ − 2tan2 2θ − 3 = 0. From here, several responses went on to obtain the correct solutions.
The common errors were to overlook the negative square root of 3, to replace 2θ with θ and never
halve the solutions to obtain the required answers and to state tan2 2θ = −1  tan2θ = −1 = 1.
Several candidates obtained f ortuitous answers by rewriting 4 tan4 2θ − 2tan2 2θ − 3 = 0 as
( )
tan 2θ 4 tan 2θ − 2 = 3 and concluding that tan2 2θ = 3. There were some decimal answers seen
2 2

despite the question asking f or exact values.

Question 8

(a) The basic method of differentiating each of x and y with respect to t and then using the chain rule
was widely understood. In differentiating y, the (–)2 was of ten omitted. In dif f erentiating x, the 2t
was of ten replaced by t or x. Several candidates did not obtain a derivative of the correct f orm,
of ten settling on 2sec 2 2x. Despite these errors, most candidates claimed to have obtained the
given answer. Very few candidates with errors in their working attempted to identify these and thus
could not be awarded f ull credit.

(b) There were several f ully correct solutions. A f ew responses showed errors in simplif ying their
answers to the required form. Some candidates substituted the given value of t correctly but then
f ound the equation of the tangent rather than the normal. Some candidates were not aware of the
values of tan 41 π and cos 41 π, and some responses stated decimal values which were very different
f rom the true values of these.

Question 9

Three common errors were seen in this question: candidates paid insufficient attention to the directions of
the vectors (they would find BA in place of AB and there were many errors in copying vectors), minus signs
were f requently lost in the middle of working and many candidates did not appear to be aware that the
vertices of a polygon are named sequentially (several worked with ABDC in place of ABCD).

(a) There were several f ully correct solutions. Many errors occurred between stating a correct equation
in OD and stating the answer. There were several errors due to using AB = OA + OB. A signif icant
 20 
 
minority believed that AB =  1 4  .
 −3  1
 

(b) Most responses showed an attempt to solve this using the intersection of two lines. There were
several f ully correct solutions, but several went wrong due to lost signs and other numerical slips.
The most common error was to use the sides AD and BC, rather than the diagonals AC and BD. A
f ew responses even tried to f ind the point of intersection of AB and DC. Very f ew responses
showed a diagram of the trapezium, which might have helped candidates find a correct method. A
small number of candidates solved the problem by using similar triangles.

(c) This question asks for the use of a scalar product, so otherwise correct solutions using the cosine
rule were unable to be awarded any credit. The majority of candidates understood the correct
process for using the scalar product and many fully correct solutions were seen. Some errors were
due to use of the incorrect vectors and some were due to slips in the arithmetic. Candidates should
be aware that in order to f ind the angle at B, they need to use BA and BC or AB and CB.

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 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 10

(a) Few candidates gave a correct solution to this part. For many responses, the only mark which
could be awarded was for the derivative of V with respect to r. Correctly dealing with the derivative
of V with respect to t proved the main dif f iculty f or many, with many attempts looking at the
dV dV
components = 40 π and = 0.8 πr separately, but not combining them. Some responses
dt dt
showed a misunderstanding of the request and tried to work f rom the given answer.

(b) There were several clear and concise solutions seen. Some solutions started correctly but did not
go f ar enough to earn full credit. Several quotients involved r2 and r3. Some solutions were untidy,
with multiple crossings out, largely because many candidates appeared not to have a structured
approach to algebraic division. A small number of candidates used the remainder theorem to obtain
the remainder.

(c) The majority of candidates separated the variables correctly. Those candidates with a correct
solution in part (b) were then able to solve the differential equation quite quickly. Some candidates
who had not been successf ul in part (b) completed the division correctly here and solved the
equation. Responses which did not have a correct f orm f or the quotient and remainder were not
able to make progress beyond the first two marks. The question asked for an expression f or t, so
responses which did not get as far as an equation with t as the subject could not score the f inal
mark.

(d) The responses with a correct equation in t usually scored this mark.

Question 11

(a) The marks f or the correct use of the quotient rule were of ten scored but there were many
responses which did not use this formula correctly. The denominator was often not squared or was
missing entirely. The omission of brackets was a common issue, causing errors in the later work.

Most responses did score the mark for setting their derivative equal to zero and many then, despite
errors, still managed to retain an equation that allowed them to obtain an exact value f or x.

Several candidates tried to make the problem look simpler by substituting for e x. What followed was
dif ferentiation with respect to ex, and this was rarely accompanied by use of the chain rule to obtain
f ’(x).

Some candidates started by trying to apply partial fractions to split the given f raction. The majority
of these attempts took no account of the fact that the degree of the numerator equals that of the
denominator. A few correct solutions using partial f ractions were seen, but they involved much
more work than necessary to obtain the derivative.

(b) There were some concise and fully correct solutions to this question. However, many candidates
did not work through the substitution process correctly, so they did not appreciate that the partial
f ractions required were straightforward, as was the integration that f ollowed. Many tried to apply
partial f ractions before they attempted the substitution. This could have worked if they had then
completed the substitution correctly and used partial f ractions a second time. Many candidates
worked on a more complicated fraction than necessary, and they did not choose an appropriate
f orm of partial f ractions f or their f raction. This resulted in a lot of unproductive work.

© 2024
 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/33
Pure Mathematics 3 (33)

Key messages

Candidates need to:

• know what is required in their response when attempting a question with a given answer – Questions
7(a) and 10(a)
• know what is required when sketching on an Argand diagram – Questions 1(a) and 1(b)
• know how to apply the chain rule in dif f erentiation − Question 11(a)
• think in both trigonometrical relation and complex number questions about which method best suits that
problem, as numerous dif f erent approaches are of ten possible − Questions 4 and 5.

As mentioned above, candidates need to show working to justify their result in complex number questions, so
reliance on a calculator is not advised here.

General comments

The standard of work on this paper was very variable, with some of the short initial questions proving f ar
more difficult than expected. Furthermore, many candidates were still not showing sufficient working in their
responses, f or example Questions 7(a) and 10(a). In Question 7(a), some candidates went f rom an
expression in cos t and sin t to the final given answer in sin 2t with no mention of what double angle f ormula
they were using. Such large jumps are very likely to lead to errors and, even if that is not the case, working
needs to be seen to gain the marks, especially in questions with the answer stated. Whilst dif f erent
candidates will always display different levels of working, a good gauge of what is deemed adequate can be
f ound in past question papers and their respective mark schemes.

Comments on specific questions

Questions 1

(a) (b) Many candidates struggled with this question. The common partial attempts were complete circles
of radii 2 and 4, although in some cases neither radius was stated or indicated . Candidates of ten
1 1
shaded the region between 0 and π, and between 0 and π. This question had nothing to do
4 2
with regions and was simply about points satisf ying two conditions: one condition being on the
modulus and the other on the argument defined over an interval. The solution to part (a) should
1
have been an arc of the circle centred at the origin and of radius 2 between 0 and π and nothing
4
1
else, f or example a line drawn from the centre of the circle to the circle itself at π. Likewise, the
4
solution to part (b) should have been an arc of a circle centred at the origin and of radius 4
1
between 0 and π.
2

Question 2

(a) Candidates adopted one of two dif f erent approaches, one f ar more successf ul than the other.
Those who assumed convergence of the iterative f ormula, say to the limit x, had little trouble
showing that this limit was the solution to the equation f (x) = 0. However, those candidates who

© 2024
 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

started from x being the solution of f(x) = 0, were usually not successf ul in recovering the given
iterative f ormula.

(b) Most candidates scored full marks. Of the few responses which did not obtain the f inal mark, this
was caused by not showing convergence, not working to 4 decimal places or not quoting the
converged root to 2 decimal places.

Question 3

(a) This was another question that saw many candidates score full marks. However, it was common to
see P = 3 being used instead of ln P = 3.

(b) Many candidates became confused in doubling their value of P, taking P = 3 and 2P = 6. In f act,
the value of P initially is e3 and 2e3 when doubled. Although any value of P above e3 was
acceptable, values below this were not as they arose f rom a negative time.

Question 4

Many candidates struggled with this question, as highlighted in the Key messages. The obvious approach is
to cross multiply, then either solve the linear equation for z and subsequently rationalise the denominator, or
to introduce z = x + iy and consider real and imaginary parts separately. Far too many candidates showed a
lack of understanding of complex numbers by taking real and imaginary parts of an equation involving z as
opposed to one in x and y. The most common approach saw candidates rationalising z + 3i in the
denominator on the left-hand side of the equation, resulting in z2 + 9 or a similar such non-linear expression
involving x2, y 2 and xy terms. Having done this, few responses were able to make much progress, even af ter
many attempts and pages of working.

Question 5

(a) This identity proof generated several different approaches. The most common successful approach
was using the difference of two squares, together with the double angle f ormula f or cosine, allied
with recognition that 4sin2 θ cos2 θ = sin2 2θ. However, candidates who directly expressed
cos4 θ − sin4 θ as cos 2θ could not receive f ull credit as the “show that” demand in the question
requires complete working. A f ew of the common errors observed were that
1
cos4 θ − sin4 θ = cos2 2θ and including/excluding f actors of and 2 in the sin 2θ identity.
2
Candidates who chose to work from the right-hand side often ended up with a mixture of dif f erent
sign errors, thus preventing much further credit. However, most candidates gained at least one
mark on this question for a single correct use of a double angle f ormula. Better responses were
able to show the proof in just a f ew lines of working, but f or a signif icant number their written
answers were not structured or clear.

(b) The majority of candidates recognised the relevance of part (a) in part (b). A minority did not
realise the disguised quadratic was in cos 2α not cos α, and thus did not solve f or the required
variable. Of those who solved successfully for one solution, the vast majority f ound both without
issue.

Question 6

(a) The majority of candidates were unable to gain credit for this question. Many lef t their answers in
terms of position vectors rather than coordinates, which was explicitly required in the question.
Some candidates unnecessarily solved simultaneous equations to f ind the point of intersection,
even though both lines shared the same position vector. Candidates should pay close attention to
the terminology used in the question: a ‘state’ question should not involve complex calculations.

(b) This part was done well, with candidates recognising that they needed to use the scalar product to
f ind the cosine of the angle. Some did not obtain the f inal mark by giving their answer in an
8
unsimplified form and leaving it as, f or example, . Some candidates used the position
5 30
vector for P and therefore gained no marks. A minority of candidates used the cosine rule to f ind

© 2024
 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

the required cosine value. The most common errors seen were misreads of the direction vectors or
arithmetic errors in the dot product.

(c) Most candidates earned the first mark, which was awarded for the two lengths required to f ind the
area of the triangle. However, many struggled to secure the next two marks because they were
unable to appreciate that the phrase ‘exact area’ required avoiding the use of a decimal
approximation for cos θ. Instead, they proceeded to calculate the area of the triangle using this
approximated angle, leading to incorrect results. The simplest approach, which was rarely seen,
was to use the identity sin2 θ = 1 − cos2 θ to find the value of sinθ and then proceed to find the area
1
by using ab sin C or draw a ratio triangle to evaluate sin θ. Some candidates used the vector
2
product to compute the area and some even used the dot product to find the perpendicular height
1
and used × base × height. Many who did this did not manage to obtain full marks, usually due to
2
a careless mistake or not working in exact values.

Question 7

(a) Most candidates had the correct approach and differentiated at least one of x and y correctly with
1
respect to t. Common errors included incorrectly taking the derivative of cot t to be . It was
cos2 t
also common for candidates with incorrect derivatives to manipulate their expressions and claim to
have reached the given answer. Weaker responses of ten presented derivatives with various
−1
combinations of x, t and θ. Of those that got down to , a signif icant proportion then
6cos2 t sin2 t
just stated the given answer without any justification of the change in numerator or denominator.
Candidates should be reminded that ‘show that...’ questions require f ull working with correct
notation to arrive at the conclusion as stated in the question paper.

(b) Most candidates recognised the connection between part (a) and part (b). The majority were
successfully able to derive the required gradient of the normal, and the co -ordinates needed to
establish the normal. The most common errors seen were forgetting to take the negative reciprocal
f or the normal gradient and not putting the answer in the required form. Candidates who used the
gradient of the tangent were unable to gain any marks f or this part.

Question 8

(a) Many candidates spent a considerable amount of time and ef f ort on this question only to f ail to
score any marks. One relatively quick and ef f icient approach is substituting key values af ter
equating numerators. Some candidates seem to want to avoid this method, intent instead on
employing the method of equating coefficients. Whilst perfectly acceptable, this approach suf f ers
f rom the fact that if a candidate gets one coef f icient wrong then the other one f ollows likewise.
Other approaches do not have this weakness.

The presence of a did confuse some candidates and some changed the 7a2, substituting an x or a
value f or a. Others made an error with the calculation of (a − 2x), despite having stated x = −3a. To
conf irm their success, candidates should check their answer quickly by recombining their two
terms. One or two candidates miscopied in going f rom part (a) to part (b).

(b) Many responses did not correctly deal with the a’s and the need to get the bracket into the f orm
(1 +/−..). Some did the reciprocal of the a but not of the 3, and some forgot to bring in the original
2a and a af ter they had performed the expansions. Quite a f ew errors were seen with the sign of
the 'x term in the expansion of (a − 2x) −1. However, many good solutions were observed and f or
those who stayed accurate throughout, the process was manageable. Several responses received
partial credit having got the expansions correct but incorrectly combined their expansions. It was
common to see 215 instead of 217 or to have lost some a’s in the denominators.

(c) The vast majority of candidates found this part challenging and sometimes omitted this altogether.
Very f ew correct responses were seen.

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 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 9

(a) Large numbers of candidates reached the correct answers and correctly identified the quotient and
remainder. However, not all really understood the significance of this part and did not connect it
properly with part (b).

(b) This part was generally well answered. Where responses started off with the form found in part (a),
the integration was generally correct and simplified to the given answer. Some responses did not
receive f ull credit due to insuf f icient detail, and those who started with the wrong setup usually
scored nothing other than the f irst mark.

Question 10

(a) Few f ully correct responses were seen here, with many candidates f inding even the f irst mark
challenging to earn. Candidates need to concentrate more on the information given and correctly
identify what each part of the question is telling them and to what variables it relates. Many were
too driven by the final answer to identify correctly what they needed. Many responses did not state
any chain rule equations but produced the answer by putting numbers together; responses scoring
f ull marks were not widely seen.

(b) Candidates who recognised a differential equation and separated variables correctly did extremely
well and almost always obtained the required answer, dealing well with both the –3 coefficient of h
and the ln term. It was also very rare to see the constant omitted. Responses which did not
separate variables correctly could not score any marks.

Question 11

(a) Many candidates completed the differentiation involving the chain rule accurately and obtained the
correct form. A good number of these who got through to the required quadratic were able to solve
it correctly. At this stage, some chose the correct answer f or a, some of f ered two answers f or a,
some chose just 1.35 and some stated 1.35 + π. Obtaining only 4.93 would score the f inal mark.

(b) Six marks were available here f or a relatively straightforward integration by substitution, with the
substitution given in the question. If candidates followed the method and structure f or solving this
sort of question they usually scored very well. Many dealt deftly and clearly with the negative and
the inverted limits. Some candidates did not obtain the final mark, usually f rom stating a negative
area.

© 2024
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/41
Mechanics (41)

Key messages

• When answering questions involving any system of forces, a well annotated f orce diagram could help
candidates to make sure that they include all relevant terms when f orming either an equilibrium situation
or a Newton’s Law equation. Such a diagram would have been particularly usef ul here in Questions 3
and 6.
• Non-exact numerical answers are required correct to three significant figures (or correct to one decimal
place f or angles in degrees) as stated on the question paper. Candidates would be advised to carry out
all working to at least four significant f igures if a f inal answer is required to three signif icant f igures.
• In questions such as Question 8 in this paper, where acceleration is given as a f unction of time, then
calculus must be used, and it is not possible to apply the equations of constant acceleration.

General comments

The requests were well answered by many candidates. Candidates at all levels were able to show their
knowledge of the subject. Questions 1, 2(a), 3, 4(a) and 7(a)(i) were f ound to be the easiest questions whilst
Questions 4(c), 5 and 6 proved to be the most challenging .

7
In Questions 6 and 7(b), the angles were given exactly as sin θ = and sin−1 ( 0.15 ) respectively. There is
25
no need to evaluate the angle in these cases and in doing so can often lead to inexact answers (and so any
approximation of the angle can lead to a loss of accuracy).

One of the rubric points on the front cover of the question paper was to take g = 10 and it was noted that
almost all candidates f ollowed this instruction.

Comments on specific questions

Question 1

This question was answered well with most candidates setting up the equations of motion f or both particles
separately and then solving simultaneously to f ind the required magnitude of the acceleration and the
tension in the string. The most common error was to either give a negative acceleration or to set up two
equations that were not consistent with each other (e.g. having the same sign f or T in both equations).

Question 2

(a) Almost all candidates correctly used the principle of conservation of energy to find the speed of the
particle at B.

(b) Approaches to this part were almost equally split between those who continued to use an energy
approach, and those who used Newton’s Second Law and the equations of constant acceleration.
Generally, both groups were equally successful with the only common error being sign errors in
either the application of the work-energy principle or Newton’s Second Law.

© 2024
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 3

This was the best answered question on the entire paper with almost all candidates correctly resolving
horizontally and vertically to f ind the values of P and θ.

Question 4

(a) Almost all candidates sketched the correct velocity-time graph f or the bus’s journey f rom A to B.

(b) It is important that candidates read the question carefully as many in this part correctly calculated
the time when the bus was either accelerating or decelerating, but very few correctly added these
two expressions together and subtracted f rom the given value of 240 to obtain the required
expression, in terms of a, f or the length of time that the bus was travelling at constant speed.

(c) Responses to this part were varied with several candidates leaving this part blank or not
appreciating that to work out the distance from A to B then the area below the velocity-time graph
needed to be found. Those candidates who had correctly calculated the required expression in part
(b) were usually successful in this part. Finally, those candidates who could recall and correctly
apply the f ormula f or the area of a trapezium were generally more successf ul than those who
needed to split this area into a rectangle and a triangle.

Question 5

(a) Most candidates were successf ul in obtaining at least one mark in this part f or applying the
1
constant acceleration formula s = ut + at 2 at least once, but only the most able could set up a
2
correct equation for when the two particles collided e.g. 100 (T − 1) − 5 (T − 1) = 80T − 5T 2 and then
2

solve this to obtain the given result that T = 3.5. The most common errors were to either use the
same expression f or the time in both equations or to use T + 1 instead of T − 1.

(b) Surprisingly this part was left blank by a good number of candidates with many not realising that all
1
that was required was to use s = ut + at 2 again with the given value of T (or T – 1) and u = 80 (or
2
100).

(c) Very f ew candidates scored full marks here and many left this part blank. Not all appreciated that
there were three parts to finding the required time. The f irst part was to f ind the velocity of each
particle at the point of collision (which was best done using v = u + at ) , then apply conservation of
linear momentum to find the speed of the combined particle after impact, and then use a complete
method to f ind the required time (the most ef f icient way of doing this was to again use
1
s = ut + at 2 ).
2

Question 6

This was a question in which many candidates struggled, and it was clear that many were unsure where to
begin with this type of extended response. Examiners commented that it was very unclear at times what
candidates were doing and what mechanical principles were being applied. It would be extremely beneficial if
candidates stated which principle (e.g. resolving parallel to the plane) they were applying at each stage of
their working.

One common error was to assume that the normal contact f orce of the plane on the particle was a
component of the weight only when in fact it was a combination of this component and a component of the
horizontal force P. A second common error was to have the frictional force acting in the wrong direction; as
the least possible value of P was required this meant that the f rictional f orce would be acting up (and not
down) the plane.

Finally, candidates are reminded that in questions such as this that it is advisable to be working with exact
trigonometric expressions until the end of the problem and to only use their calculators at this f inal stage to
work out the numerical value of P f rom an exact expression.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 7

(a) (i) This part was done extremely well with almost all candidates correctly showing that k = 40 .

(ii) This part was also well done by candidates with many applying P = Dv , and then correctly
applying Newton’s 2nd Law to f ind the required acceleration. The most common errors were to
either misread the fact that speed had changed to 45 (f rom 48 in the previous part) or to not include
the resistive f orce in their application of Newton’s 2 nd Law.

(b) In this part many candidates appreciated the need to apply both P = Dv and Newton’s 2nd Law
with a = 0, but a number f ailed to include the correct weight component as 1200  g  0.15 and
some included the acceleration f rom part (a)(ii). Many candidates who correctly obtained a
quadratic equation in v went on to solve it correctly.

Question 8

The responses to this final question were mixed; Examiners reported seeing several perfect responses (that
scored all 7 marks) to some that were either blank or little in the way of progress was made. Most candidates
appreciated the need to use calculus (rather than the constant acceleration f ormulae) so many scored at
least 2 of the 7 marks f or correctly integrating the given expression f or the acceleration twice to obtain
corresponding expressions for the velocity and displacement of the particle (it was surprising how many
candidates though believed incorrectly that this expression needed dif f erentiating too). It was relatively
common (and pleasing) to see candidates correctly interacting with the fact that the velocity at t = 0 was 4.2
to correctly calculate the f irst constant of integration and realising too that they needed to solve their
expression f or the velocity set equal to zero to f ind the two required times when the particle was at
instantaneous rest. However, it was only the most able candidates who recognised that they needed to
integrate their expression f or v between these two values of t to f ind the required distance.

© 2024
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/42
Mechanics (42)

Key messages

• In previous reports, it has been mentioned that when answering questions involving any system of
f orces, a well annotated force diagram could help candidates to include all relevant terms when forming
either an equilibrium situation or a Newton’s Law equation. Force diagrams are now being seen more
of ten than in previous examination series.

• In questions such as Question 6 on this paper, where acceleration is given as a f unction of time,
calculus must be used, and it is not possible to apply the equations of constant acceleration. When
integrating, a constant of integration must be considered as this constant may not necessarily be zero.

• Non-exact numerical answers are required correct to three significant f igures or angles correct to one
decimal place as stated on the front of the question paper. Candidates are strongly advised to carry out
all working to at least four significant f igures if a f inal answer is required to three signif icant f igures.

General comments

The requests were well answered by many candidates. Candidates at all levels were able to show their
knowledge of the subject. Questions 2, 3(a) and 7(a) were f ound to be the most accessible questions whilst
Questions 4, 5, and 7(b) proved to be the most challenging.

In Question 7(a), the angle θ was given exactly as sin θ = 0.6 . It was pleasing to see that many candidates
used this exact value rather than evaluate the angle which would have led to a loss of accuracy.

One of the rubric points on the front cover of the question paper was to take g = 10 and it was noted that
almost all candidates f ollowed this instruction.

Comments on specific questions

Question 1

(a) A well answered question by a significant majority of candidates by using the fact that the gradient
of the line between t = 4 and t = 10 is equivalent to deceleration. The most common error seen
5
was f inding a negative gradient of the line and equate it to m s –2 leading to V = −10 , when the
3
diagram has V as a positive value and described in the text as a speed.

(b) Another well answered question by using the area between the graph and the t -axis is equivalent
to distance. The main error seen was to use a value of –3 for the distance moved between t = 10
and t = T , using displacement rather than distance. A minority of candidates wrongly assumed that
the time accelerating and decelerating between t = 10 and t = T were equal.

Question 2

Given that the resistance force is not described as being constant, the only valid approach to answering the
request is to use energy. Most candidates who used energy gained some credit for a kinetic energy term or a
potential energy term. When forming a work energy equation, errors were seen in signs with the f our terms,

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 Cambridge International Advanced Subsidiary and Advanced Level
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or occasionally a term was omitted. Those who assumed that the resistance f orce was constant, gained
some credit but not f ull marks.

Question 3

(a) This was a well answered question by most candidates.

(b) This question was only correctly answered by a minority of candidates. The two main errors seen
was to omit the weight component when forming an equation, or for the cyclist to be accelerating
even though the request was f or the steady speed the cyclist could maintain.

Question 4

This question proved to be challenging for most candidates. Dif f iculty arose f rom candidates labelling the
tensions in the three distinct strings with the same label, usually T , when these tensions were dif f erent.

(a) This is essentially a three-force problem with forces acting at the point A . Once this is realised,
there are a variety of approaches that could be used. The simplest of which is a triangle of forces to
get an isosceles right-angled triangle so that the tension in AB is equal to the weight of the 0.2 kg
particle. The most common approach was to resolve in two directions, usually horizontally and
vertically, to get two equations involving TAB and TAP .

(b) This is also a three-force problem with forces acting at the point B . For those who attempted this
question, the most common approach was to resolve in two directions to form two equations in TBP
and θ and solve simultaneously.

Question 5

This question proved to be the most challenging f or the majority of candidates .

(a) The approach to this question is to get expressions in terms of t f or the distance that P and Q
1 1
move, usually sP = 2t − gt 2 and sQ = gt 2 , and then use the fact that the particles are 1 m apart
2 2
so that sP + sQ = 1 leads to the particles meeting after 0.5 seconds. This time then can be used in
the equation v = u + at to find speeds of 3 m s –1 and 5 m s –1. Candidates who did get a correct
time, invariably gave a negative value f or at least one of the speeds.

(b) The f irst part of this question is to use conservation of momentum, but there was a great deal of
conf usion about the directions that P and Q are travelling at the time they meet. In f act, the
particles are travelling in the same direction. It was quite common to see that the speed of P found
f rom using the conservation of momentum, to be given as the f inal answer. Once this speed is
f ound, a further calculation using constant acceleration is needed, but f irstly the distance that the
particles are above the ground at collision is required.

Question 6

This topic is usually a good source of marks. However, a signif icant number of candidates assumed the
constant of integration to be zero with a resulting loss of marks.

(a) This was usually successfully answered for the candidates who included a constant of integration.
3
The most common incorrect answer seen was v = −t 2 leading to a velocity of –1 m s –1.

(b) A constant of integration was seen here but was either not evaluated, or used t = 0 to evaluate it,
even though t = 0 is not within the domain of this stage of the motion.

(c) Those who obtained correct expressions for the velocities in each stage of the motion answered
this part well, integrating and applying the correct limits correctly.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 7

(a) This was a straightforward pulley question. Most candidates correctly wrote down the two Newton’s
second Law equations, sometimes leading to a negative acceleration which then failed to score the
associated mark unless the magnitude, as requested, was stated. One error seen was to omit the
f riction in the equation for particle A. Another error was to use the coefficient of friction as though it
was f riction. Overall, most candidates scored well on this question.

(b) This question proved challenging for most candidates. The speed of A (or B ) at the time when B
reaches the ground needed to be calculated first. As B remains on the ground, A is now moving
with the string slack, so a new acceleration needs to be calculated so that the distance that B
moves subsequently can be calculated and added to the 0.25 m. Often, candidates assumed that
the acceleration remained the same once the string was slack.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/43
Mechanics (43)

Key messages

• Non-exact numerical answers are required correct to three significant figures as stated on the question
paper. Candidates would be advised to carry out all working to at least four significant f igures if a f inal
answer is required to three signif icant f igures.

• In questions where the total work done over a period of time is given, candidates should be advised that
a method involving acceleration is not appropriate and an energy method should be used.

• Candidates should be advised that the wording at the start of a question, then applies to the whole of
that question, and not just the f irst part of the question.

General comments

• The requests were well answered by many candidates. Candidates at all levels were able to show their
knowledge of the subject. The f irst five questions together with Question 6(b) and 7(a) were f ound to be
the most accessible questions, whilst Questions 6(a), 6(c), 6(d) and 7(b) proved to be the most
challenging.

• In Question 3 the angle was given exactly. There is no need to evaluate the angle in situations like this,
and it is better not to do so, as it can lead to a loss of accuracy. A signif icant number of candidates
showed all of their working in terms of sin α , rather than 0.08. If the final answer given by a candidate
was incorrect then it was not clear whether they had used the correct value of sin α and so partial
marks could sometimes not be awarded.

Comments on specific questions

Question 1

This question was well answered with many candidates gaining all four marks. Candidates had to f orm an
energy equation relating the work done by the athlete to the work done against f riction and the change in
kinetic energy. The most common mistake was to think that the athlete did work at a constant rate.
Candidates who thought this, divided the work done by the athlete by the distance to get the f orce (which
they thought was constant) to use in Newton’s second law. Although this method produced the right answer
it was not correct as the work done by the athlete was given rather than the force exerted by the athlete, and
so the method could only gain 2 marks.

Question 2

This f our-force problem was again generally well answered with many candidates gaining at least five marks
out of six. Many candidates did not gain the final mark, either getting the direction wrong (of ten giving the
direction of the resultant force) or stating that the angle was 4 degrees or 86 degrees but not specif ying in
which direction it was. Some gave a general indication such as north-west without specifying an angle. A few
candidates had not resolved correctly and had sine rather than cosine or vice versa. Some candidates only
gave the horizontal and vertical components and went no f urther, and some missed one of the f orces
(usually the 24 Newton f orce).

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 Cambridge International Advanced Subsidiary and Advanced Level
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Question 3

(a) Many candidates answered this question well, but more errors were made than in previous
questions. The most common error was to omit the f act that the car is traveling up a slope and
simply multiply the resistance f orce by the speed (f or which no marks were available). Some
candidates thought that the angle was sin 0.08 , instead of sin−1 0.08.

(b) This question proved more challenging. Candidates had to use the new power to f ind the
acceleration of the car. A considerable number of candidates omitted either the resistance force or
the f act that the car was travelling up a slope or sometimes both in their answer. Such candidates
could still score a mark for the follow through of their power f rom part (a). It should be noted that
the wording at the start of the question before part (a) applies to the whole question, not just part
(a). A large number of candidates rounded the answer to two significant f igures, perhaps thinking
that the zero in f ront of the decimal point was a significant f igure. A f ew candidates made a sign
error in resolving up the slope and others again used an angle of sin 0.08.

Question 4

In this question candidates had to use conservation of momentum to find the new velocity or velocities and
then f ind the greater change in kinetic energy. Although most f ound two velocities, some only f ound the
lower one as they presumably realised that this would produce the greater loss in kinetic energy. Most
candidates did find both possible velocities and then went on to f ind the loss in kinetic energy f or both,
usually correctly. Some candidates found the loss for each particle, but did not combine them , and some
f ound both losses but then did not state which was the greater and so could not gain the f inal mark. A f ew
candidates made errors in using conservation of momentum and a f ew found the velocity of 3 ms –1, but then
thought that the other option was that the particles coalesced, which was incorrect.

Question 5

(a) Candidates had to resolve f orces to f ind the normal reaction R , and then relate the horizontal
component of the tension to μR . Most candidates found no difficulty in doing this. A f ew made a
sign error in R , but more candidates thought that R was equal to the weight, missing out the
component of the tension perpendicular to the plane. These candidates could still get a mark if they
equated T cos30 to their friction. A few tried to include acceleration, which they could not do since
the particle was at rest.

(b) This part was rather similar to part (a), but this time there was acceleration involved. Candidates
of ten correctly used Newton’s second law to find the required tension and those who thought that
R was equal to the weight could still get a mark f or the equation T cos30 − F = 12  0.2 , with F
replaced by 0.5  120 .

Question 6

(a) In this part, candidates had to integrate the acceleration a = 0.6t to f ind the velocity, and then
substitute t = 4 . Many tried to use constant acceleration equations instead to show that the
velocity was 4.8 ms –1, which was an incorrect method. Some candidates did integrate correctly but
then simply stated that when t = 4 was substituted, the value obtained was 4.8, but did not
actually show the substitution and so did not get the mark.

(b) Candidates had to draw a velocity time graph for this part of the question. Although some produced
a f ully correct graph, many others produced a graph where the f irst section was straight, so
producing a trapezium. This was the case even with some candidates who had a correct part (a).
Some candidates who had used constant acceleration equations, produced a f ully correct graph. A
f ew candidates thought that the particle started slowing down at t = 11 rather than t = 15 , and
some had an incorrect maximum speed or no f igures on their diagram at all.

(c) In this part, almost all candidates f ound the gradient of the velocity time graph correctly. Many
wrongly f ound the constant term to be 4.8 rather than 19.2, by using t = 0 rather than t = 15.
Candidates who substituted the two points (15, 4.8) and (20, 0) almost always f ound the equation
correctly.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

(d) Candidates had to find the total distance travelled by the particle in this part. Those who had a
trapezium in part (b) f ound the area of this trapezium (of ten as two triangles and one rectangle)
and so their final answer (of 74.4 m) was incorrect. Those candidates who had a correct graph in
part (b) usually f ound the area correctly.

Question 7

(a) This question involved finding the tension in a string joining two particles over a pulley, and the time
taken f or one of the particles to reach the plane. Most candidates answered this well and produced
a f ully correct response. A small minority of candidates made an error in one or more of the
Newton’s second law equations.

(b) This question was the most challenging question in the paper. Many candidates correctly found the
speed of the particles when the particle B reached the plane. However, the remaining stages in
f inding the overall time for which the particle is at least 3.25 m above the plane proved to be more
challenging. There were two popular methods of f inding this time. The most popular was to use
1
s = ut + at 2 with s = 0.25 , v = 10 and a = −10 to find the two times at which the particle was
2
3.25 m above the plane, and then subtract them. Several candidates who used this method had
a = 10 or had a different value of s , of ten 2.25 or 3.25. Many candidates who tried to use this
method did not subtract the two times but gave one or other as the total time. The other method
was to f ind the speed of the particle when its height was 3.25 and then use v = u + at with u = 5 ,
v = 0 and a = −10 to f ind the time to the highest point, and then double it. Most candidates who
used this method came to the correct answer. There were also a number of ‘hybrid’ methods,
which sometimes came to the correct answer. Those who tried to use an energy method rarely had
any success at all. A considerable number of candidates had a f ully correct method but due to
premature approximation did not f ind the f inal answer to three signif icant f igures.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/51
Probability and Statistics 1 (51)

Key messages

Candidates need to be aware of the necessity for clear communication in their solutions. Answers should be
supported by the relevant calculations linked to the appropriate stage in their workings. Where dif f erent
scenarios are selected to fulfil the demands of a question these should be clearly identified; an organised list
or table is of ten helpf ul.

Where a statistical diagram is required, accuracy and clarity are vital. Scales should be chosen appropriately
and the axes labelled f ully.

Candidates should state only non-exact answers correct to three significant figures, exact answers should be
stated exactly. There is no requirement f or f ractions to be converted to decimals. Intermediate working
values correct to at least four significant figures should be used throughout in order to justif y a f inal answer
correct to three signif icant f igures.

Candidates should be aware that numerical accuracy is expected, and the efficient use of a calculator should
enable the initial calculation to be stated and then fully evaluated as the next stage of the process. Some
simple multiplication and addition errors were noted.

General comments

Most candidates used the response space effectively. Where there is more than one attempt at a question,
the solution to be presented for marking should be identified clearly. If extra space is required, the additional
page should be used in the f irst instance.

Good solutions were clearly organised and explained. Helpf ul lists of scenarios, tables of results and
diagrams of the Normal distribution of ten supported accurate answers.

Many candidates were able to tackle Questions 3 and 5 while Questions 6 and 7 were more challenging.
Suf ficient time seems to have been available for most candidates to attempt all of the paper however some
topic areas of the syllabus did not appear to have been prepared well f or.

Comments on specific questions

Question 1

Most candidates recognised that the geometric approximation was appropriate. However, inconsistent
interpretation of the success criteria limited overall success.

(a) Candidates who used the longer P(X = 0,1,2,3,4,5,6,7) approach tended to be more successf ul
than those who attempted to use the P(X < 8) f ormula as this was of ten interpreted as P(X  8)
leading to an incorrect f inal probability.

(b) Good solutions considered the second 6 as being f ixed on the 8 th throw and then identif ied the
possibilities f or the placement of the f irst 6. A simple ‘diagram’ with the f irst 6 placed in the 7
dif ferent positions was often helpful. Some candidates did not include the probability of the second
6 in their calculation, but the most common error was failing to use ‘× 7C1’ in the solution. Over 10
per cent of candidates submitted no attempt f or this part.

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 Cambridge International Advanced Subsidiary and Advanced Level
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Question 2

Many candidates found this question challenging, with the requirement to establish the value of constant k
bef ore completing the probability distribution table.

(a) Good solutions included clear workings to show the possible values of X substituted into the
probability formula and created an equation in k which was then used to complete the probability
distribution table. Weaker solutions of ten omitted 0, or having obtained a value f or k did not
generate a numerical probability distribution table. Some more able candidates stated a value for k
without the supporting work that is expected at this level. The values in some probability distribution
tables did not sum to 1, which should be a standard check that candidates are aware of .

(b) Where there was an appropriate probability distribution table present, candidates were usually
successful in f inding E(X), but f ound the variance f ormula more challenge, with less numerical
accuracy or f ailing to use (E(X))2. The techniques required are quite standard, and candidates
could be encouraged to develop a scaffolded approach to answering this topic. Because 2(a) was
of ten incomplete, over 15 per cent of candidates did not attempt this part.

Question 3

(a) Some excellent cumulative frequence graphs were noted. These included plotting clearly at the
upper boundary of the classes using a ‘x’, joining the points with a f reehand curve, labelling both
axes appropriately, including the units on the time axis. Weaker candidates used a ruler to join the
points, which is inappropriate at this level, or plotted the values at the mid-point of the range. A f ew
candidates attempted to draw a histogram.

(b) Many candidates were able to use their graphs ef f ectively to f ind the median, and with less
accuracy the interquartile range. As the question states that the graph must be used, there should
be a clear indication on the graph where values are being obtained, with the best solutions using a
ruler to draw horizontal/vertical lines to obtain the readings. Some candidates did show some
marking on the graph, but many did not communicate their working process. An unexpected error
was using 35 minutes on the time axis to obtain the median, rather than f inding the time of the
‘middle’ value f or the appropriate terms.

(c) Candidates who used a table format to collate the required data were often more successful. Some
candidates used the space near the original data table before 3(a) to find both the class midpoints
and the class f requencies, but were not always caref ul in transf erring their values to f urther
calculations. Many good attempts were noted, with clear working to show how the estimated mean
f ormula was being used. A surprising number of candidates made slight slips in either calculating
the f requencies or the mid-points for just a single class. A small number of candidates presented
their f inal answer as an improper f raction, which is not appropriate in this context.

Question 4

The best solutions included a tree diagram in 4(a) to clarify the information presented, and then use this to
support their work throughout the question.

(a) The best solutions used a tree diagram to identify the four possible ways of obtaining marbles of
the same colour, stating clearly the scenarios when evaluating the individual probabilities bef ore
summing to present the answer as a proper fraction. Some candidates did convert the fraction to a
decimal which is not required. Weaker solutions f ailed to identif y the possible scenarios to
communicate their thinking. A major misconception was not including the probability of obtaining
the initial head or tail in the scenario, even when the marble probabilities clearly identif ied which
had been obtained.

(b) There were many good attempts at calculating the conditional probability seen, including f rom
candidates who had not used the ‘coin probability’ in 3(a). Some accuracy errors in the calculation
f rom the correct expression were seen. Weaker solutions again omitted the ‘coin probabilities’.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 5

The use of the Binomial distribution leading to a Normal approximation is a common application. A large
number of candidates did not interpret the success criteria accurately. A signif icant number of scripts had
little or no attempt at part (b).

(a) This was a standard application of the normal distribution. Most candidates were able to f orm at
least one of the normal standardisation formulas correctly, although some introduced a continuity
correction even though mass is a continuous variable. Good solutions then used a simple sketch of
the normal distribution to identif y the required probability area to f orm the correct calculation.
Weaker solutions often simply found the difference between the two probabilities obtained. It was
unexpected that some candidates assumed that the required probability area was symmetrical.

(b) Good solutions identified that the Normal distribution was an appropriate approximation in this
context. The best answers started with clear, unsimplified and appropriately identified calculations
f or the mean and variance which were then substituted into the standardisation f ormula. Many
realised that a continuity correction was required as the variable was discrete. Candidates are
reminded that the use of the standardisation formula must be seen to gain full marks. The inclusion
of a simple sketch was of ten seen and used to clarif y the required probability area. Weaker
solutions omitted the continuity correction or simply gave the probability for their z-value, not finding
the required area.

Question 6

Many candidates found this question challenging, with 10 per cent not making any attempt. It is possible that
candidates did not read the entire question, as 6(b) was a f airly standard question f or the Binomial
distribution, but was omitted by over 40 per cent of candidates.

(a) Most solutions included an attempt to form at least one standardisation f ormula equated to a z-
value. Candidates should be aware that 90 per cent is a critical value, and as such the value stated
in the tables provided must be used. Weaker solutions equated with either the original probabilities,
or used the tables incorrectly and treated the given probabilities as z-values and f ound the linked
probability. Most candidates who formed the two simultaneous equations were able to solve them
accurately.

(b) The best solutions interpreted the success criteria accurately, stated the individual terms within the
calculation and evaluated accurately. A common error was to include obtaining 8 in the required
probability. A small number of candidates did not use the probabilities given in the question. Most
candidates used the 1 – P(8,9,10) approach, but accurate solutions using the alternative, but
longer, P(0,1,2,3,4,5,6,7) were also noted. Candidates should be aware that they should work to at
least f our significant figures if a three significant figure answer is anticipated. There did appear to
be some confusion between four significant figures and four decimal places within the calculations
presented.

Question 7

Whilst many candidates were able to tackle 7(a), the f ollowing parts provided more of a challenge.

(a) This question was completed correctly by most candidates. Whilst almost all realised that the
number of ways of arranging 9 letters was 8! because the Ts were together, a small number did not
allow f or the repeated Es and Ls. The weakest attempts did not apply the Ts condition and simply
divided by an additional 2! As the Ts were repeated.

(b) A good number of candidates started by providing a simple diagram showing a T at each end of a
row of 9 letters. Most successful solutions then excluded the Es and considered the number of
arrangements that the remaining five letters could make and then the number of ways that the Es
could be inserted so that they were not together. More conf ident candidates used 6C2 within their
calculation. Candidates who attempted to use a subtraction approach were less successful, f ailing
to identify an appropriate ‘total arrangements’ term initially, often using either their answer to 7(a) or
9!. Again, candidates who used a simple diagram to represent the criteria were of ten more
successf ul.

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 Cambridge International Advanced Subsidiary and Advanced Level
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Principal Examiner Report f or Teachers

(c) Most candidates found this question demanding. Accurate answers were often characterised by a
clear and systematic approach to identifying the possible options, calculating the total number of
selections without restrictions that could be made and then calculated the percentage accurately.
Common errors were the omission of TEE, the inclusion of TTEE or not including the number of
ways that the T and Es could be selected initially. A significant number of candidates found a value
f or the number of selections but failed to make any attempt at finding the required percentage, or
used an incorrect divisor, typically their answer to 7(a) or 9!. Few complete and correct solutions
were seen.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/52
Probability and Statistics 1 (52)

Key messages

Candidates should be aware of the need to communicate their method clearly. Simply stating values of ten
does not provide sufficient evidence of the calculation undertaken, especially when there are errors earlier in
the solution. The use of algebra to communicate processes is anticipated at this level and enables
candidates to review their method effectively and is an essential tool when showing given statements are
true. When errors are corrected, candidates would be well advised to cross through and replace the term. It
is extremely dif f icult to interpret accurately terms that are overwritten.

There should be a clear understanding of how significant figures work for decimal values less than one. It is
important that candidates realised the need to work to at least four significant figures throughout to justif y a
three significant value. Many candidates rounded prematurely in normal approximation questions which
produced inaccurate values from the tables and lost accuracy in their solutions. It is an inefficient use of time
to convert an exact fractional value to an inexact decimal equivalent, there is no requirement for probabilities
to be stated as a decimal.

The interpretation of success criteria is an essential skill f or this component. Candidates would be well
advised to include this within their preparation.

General comments

Although many well-structured responses were seen, some candidates made it difficult to follow their thinking
within their solution by not using the response space in a clear manner. The best solutions of ten included
some simple notation to clarif y the process that was being used.

The use of simple sketches and diagrams can help to clarify both context and inf ormation provided. These
were of ten seen in successful solutions. It was encouraging that more candidates f ormed the back -to-back
stem-and-leaf diagram more accurately, although the omission of units was still common in the key provided.

Suf ficient time seems to have been available f or candidates to complete all the work they were able to,
although some candidates may not have managed their time effectively. It is good practice to look quickly
through the paper initially to identify the syllabus content for each question, as this should help candidates
manage their time more efficiently. A few candidates did not appear to have prepared well for some topics, in
particular when more than one technique was required within a solution. Many good solutions were seen f or
Questions 4 and 6. The context in Questions 1, 2 and 7 was f ound to be challenging f or many.

Comments on specific questions

Question 1

Although this probability question was quite standard, many candidates did not use the raw data accurately
when determining the required values.

number of students in Year 1 who played netball

(a) Good solutions used the data table and stated .
number of students who played netball
However, many solutions used either the total number of candidates in Year 1 or the total number
of candidates in the table as the denominator. A small number of solutions misinterpreted the
question and f ound P(N | X).

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

(b) Again, good solutions used the data table and stated
number of students in Year 1 who played netball
. However, many solutions used either the total
number of students in Year 1
number of candidates who played netball or the total number of candidates in the table as the
denominator. A small number of solutions misinterpreted the question and f ound P(X | N).

(c) The majority of candidates used the relationship P(X) × P(N) = P(X ∩ N) to determine
independence. The instruction ‘determine’ in the question requires candidates to justif y their
conclusion, so clear communication using correct notation is expected and numerical values must
be present to compare. In many solutions, some or all of these f eatures were omitted and so
gained no credit. There were a number of inaccurate probabilities stated f rom the data table and
candidates should be reminded to always be caref ul with simple arithmetic. The alternative
approach required comparing either 1(a) with P(X) or 1(b) with P(N), and where attempted this was
usually successful with the required relationship stated initially, e.g. P(N | X) = P(N), the values
substituted and the conclusion stated.

(d) Most candidates recognised that this part was a standard binomial approximation and used an
appropriate approach for the task. The best solutions stated the unsimplified calculation for P(0,1,2)
and used a calculator to evaluate accurately, providing an answer accurate to more than three
significant figures before rounding to three significant figures. Weaker solutions of ten included the
values of the individual terms bef ore summing, which allowed rounding errors to occur. A very
small number of candidates used the alternative approach and calculated 1 – P(3,4,5,6,7,8) but
these were of ten unsuccessf ul. As in previous years, a common error was to misinterpret the
success criteria and include P(3) in the solution. Candidates are well advised to practice
interpreting the common success criteria accurately as part of their preparation.

Question 2

Many candidates found the context of this question challenging, although it was encouraging to see more
diagrams showing the possible scenarios within solutions. This is good practice in ‘permutation and
combination’ questions.

(a) Good solutions stated the initial calculation and evaluated accurately. A very small number of
candidates did round their exact answer to three significant f igures, which is inappropriate in this
component when an exact numerical value has been found. Weaker solutions did not consider the
ef f ect of the repeated As and omitted dividing by 2!.

(b) Most candidates found this part challenging. The most successf ul approach was to identif y the
three possible scenarios which f ulf illed the success criteria, and then determine the number of
arrangements possible for each. As in 1(d), interpreting the success criteria inaccurately was a
common error with the omission of A ^ ^ A ^ ^ ^ ^ ^ as a possible scenario. Many solutions f ound
the number of arrangements possible with the As in their initial position, but did not continue to
identify the different places that the As could be positioned and still fulfil the success criteria. A f ew
candidates multiplied their answers by two, as the position of the As could be interchanged, but
should then have divided their value by two because of the repeated As.

Question 3

The context of this question was found challenging by many, with a common misconception being that the
outcomes must be no greater than 6. The successful solutions often generated an outcome space initially.

(a) Candidates who included an outcome space were of ten more successf ul in identif ying all the
possible outcomes f or the probability distribution table. Good solutions set up the probability
distribution table so that it could be used to support the calculations required in 3(b). A number of
tables had probabilities that did not sum to 1, or included odd numbers which should not have been
possible with the random variable conditions. The inclusion of 1, but no other odd variable was
unexpected.

(b) Over 15 per cent of candidates made no attempt at this part, even when a good attempt at 3(a)
was present. Candidates should be aware that as this is syllabus content, clear supporting working
is expected at all stages. Where errors were present in 3(a), only the unsimplified calculations are
able to gain credit. A small number of candidates did not use the variance f ormula accurately and

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 Cambridge International Advanced Subsidiary and Advanced Level
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subtracted E(X). It is good practice not to state f inal answers as improper f ractions in most
contexts. Good solutions of ten included supporting work f or E(X) and Var(X) as part of the
‘probability distribution table’ in 3(a). However, the final solution f or Var(X) must be stated in the
answer space f or 3(b).

Question 4

This normal distribution question was answered well by most candidates.

(a) This part was answered well by most candidates. The best solutions clearly stated the
standardisation formula with an appropriate evaluation. The probability was f ound by using the
provided tables accurately and the criteria interpreted correctly to f ind the appropriate probability
area. A clear calculation was stated, using a probability to at least four signif icant f igures, and the
answer interpreted to state an integer number of trees having a height less than 18.2 m. As in
previous years, some candidates did not attempt to f ind the expected number of trees, and it is
recommended that candidates do re-read the question to conf irm that they have completed the
task set before moving on. Candidates should be aware that they are interpreting the value f ound,
rather than simply approximating or stating to a given number of significant figures. A small number
of candidates introduced a continuity correction, which was not necessary as heights is a
continuous variable.

(b) The majority of candidates used an appropriate standardisation f ormula with a variable f or the
denominator. The majority of errors seen were equating this to either the original probability or
using the tables incorrectly and not f inding a z-value. The context was such that an improper
f raction was not appropriate f or the f inal answer.

Question 5

Although this was a quite a standard probability question, many candidates f ound it challenging, of ten
because the requirements were not considered in a logical manner.

(a) Many good solutions were seen, with the initial criteria applied accurately. A common
misconception was that the selection of the violinists, guitarists and pianists were separate
outcomes, and the values were added instead of multiplied. A few candidates multiplied the correct
expression by 3! or 6, which appears to be considering that the order the type of musician was
selected was significant. The solutions of some candidates suggested that the question had not
been read f ully, as 21C7 was presented as the answer, which is the number of ways that a group of
seven can be chosen f rom the class with no restrictions.

(b) Candidates who listed possible scenarios in a logical order were often more successful in this part.
Good solutions then clearly linked the calculations with the scenarios bef ore f inding the total.
Weaker solutions often omitted one or more possible scenario. Again, some candidates considered
the scenarios as a combination of separate outcomes and added their values.

Question 6

(a) Many excellent stem-and-leaf diagrams were seen. The best used a ruler to ensure that the stem
was vertical, which then enables the vertical alignment of the leaves to be achieved more easily.
Most candidates followed the instruction to place the Falcons on the lef t and very f ew diagrams
were seen with ‘punctuation’ used between terms. The most f requent error was to omit the units
f rom the key, which should be included at this level.

(b) Many f ully correct solutions were seen. Candidates should be aware that when more than one
value is required in a question, these should be clearly identified to obtain credit. Most candidates
correctly identif ied the median as the 8 th value, but the Upper Quartile and Lower Quartile are
expected to be the median of the two set of remaining values, so here the 12 th and 4th values.

(c) Many candidates found this part challenging, and the question was not attempted by a number of
them. The best solutions recognised that the data given needed to be combined into the mean and
variance f ormulae with a group size of 30 and presented clear calculations. Weaker solutions often
f ound the mean and variance of the groups separately and then either combined both values, or
f ound the mean average of the values. This is a common misconception which candidates can be
supported to avoid. Some poor arithmetic was noted in this question.

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 Cambridge International Advanced Subsidiary and Advanced Level
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Principal Examiner Report f or Teachers

Question 7

This question using the geometric and normal approximations was found challenging by many and a large
number of incomplete solutions were noted.

(a) Many good solutions to this style of question were seen. The best included a clear supporting
calculation. A few candidates used the complements of the given probabilities or omitted to include
the probability of scoring a goal on the 5th attempt. A misinterpretation of the success criteria was
calculating scoring 5 goals in total instead of scoring a goal on the 5th attempt.

(b) There were two common approaches to this question. The most successf ul recognised that
P(3,4,5,6,7) needed to be found and extended the work in 7(a) to obtain the result. Candidates who
attempted the more efficient process of calculating P(X  7) – P(X  2), often failed to interpret the
success criteria accurately and omitted P(3) or f ailed to evaluate the two required probabilities
consistently, frequently omitting ‘1 – ‘ f rom P(X  2). Rounding intermediate answer values was
noted in this question, which of ten resulted in an inaccurate f inal answer.

(c) Most candidates found this part challenging, and the question was not attempted by a surprising
number. Again, solutions which listed possible scenarios were of ten more successf ul. The best
solutions recognised in each scenario that the position of the first goal was variable and included
an appropriate multiplier in each scenario. Weaker solutions assumed that the first goal was scored
on the f irst attempt. A common incorrect approach was to find the probability of scoring a goal on
the f if th attempt and adding to the probability of scoring at least one goal in f our attempts.
An alternative approach seen from some better candidates was to use a binomial approximation
and calculating P(2,3,4,5) from 5 attempts, which was sometimes more ef f iciently f ound as 1 –
P(0,1).

(d) This standard normal approximation question was omitted by over 10 per cent of candidates.
Candidates who did attempt the question often scored well, as clear working was shown even if
their f inal answer was incorrect. Good solutions included clear calculations to f ind the mean and
variance, used a continuity correction when f orming both standardisation formulas and included a
simple sketch of the normal distribution to identify the required probability area. The interpretation
of the success criteria was of ten inconsistent, with of ten one continuity correction applied
incorrectly. A number of candidates, having formed the standardisation formulas assumed that the
required probability area was symmetrical about the mean and did not use both of their values.
Candidates should be aware that incorrect terminology will be penalised, f or example calculating
the variance and stating that it is the standard deviation.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/53
Probability and Statistics 1 (53)

Key messages

When answering complex questions candidates are reminded that they should make their working clear and
explain their approach to the reader. This was particularly true in Question 5c and Questions 6b, c and d.
In Question 5c they needed to explain what the numerator and denominator of their probability f raction
represented. In Questions 6b, c and d, where they were adding or subtracting different totals, they needed
to explain what those totals represented. This was particularly true in Question 6b where there was a variety
of acceptable approaches. No part marks can be awarded for incorrect final answers if the subtotals are not
explained.

Another issue in this paper was premature approximation. Premature approximation is penalised every time
and some candidates lost several marks across the paper. In Question 1b a rounded z-value of 1.56
produced an inaccurate f inal answer, as did a rounded value f or E(X) in the calculation of the standard
deviation in Question 4c and rounded values of the individual binomial terms in Question 5a. Candidates
need to understand that all numbers they work with should be correct to four or more significant figures if the
f inal answer is to be correct to three signif icant f igures.

General comments

Being able to use tables for the Normal Distribution is an essential requirement f or this syllabus and most
candidates appreciate that they need to show the complete standardisation formula with substituted values
f or the mean and standard deviation. Those who produced the final answer f rom a calculator in Questions
1b and 3 with no standardisation f ormula written out lost unnecessary marks.

The style of grouping of the data in Question 4 caused a variety of challenges. Candidates who listed the
calculated cumulative frequencies were far more likely to produce an accurate graph than those who plotted
the points without having written down the totals. Candidates need to plan their graphs caref ully. If they
recognise that the boundary values are not integers, they will be more likely to plot the points accurately if
they allocate thick graph lines to the boundary values. Those who aimed to plot points between graph lines
rarely managed to plot them all accurately. Candidates need to remember that a graph should tell a story
and if the axes are not fully labelled (‘cumulative f requency’ and ‘height in cm’ in this case), the graph is
meaningless.

Comments on specific questions

Question 1

(a) Most candidates knew to cube the probabilities of both owning an electric car (0.3) and not owning
an electric car (0.7), and the majority knew that the ‘either...or...’ meant they needed to add the two
cubed values. However, a signif icant number gave two separate values, not realising that they
needed to sum them.

(b) Candidates seemed very comf ortable with tackling this question. The question asked f or an
approximation and only the weaker candidates ignored this instruction and tried unsuccessf ully to
use a binomial calculation. Most candidates remembered the continuity correction, and almost all
took it in the correct direction. A few forfeited an easy mark by prematurely rounding the z-value
which resulted in a narrowly inaccurate f inal answer.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 2

(a) This question was answered well with the majority producing a clear table, of ten af ter having
investigated the outcomes in a sample space diagram. An unfortunate number misunderstood the
question and dealt with the sum rather than the product of the scores and a special case was made
f or them, allowing them one of the three available marks. Careful candidates often wisely included
a check that their probabilities added to 1. Most candidates knew and were able to apply the
f ormula for E(X). Those who had worked with sums rather than products in part (a) were eligible for
the f ollow through mark. The most common error seen was candidates dividing the correct
calculation by either 6 or 36.

Question 3

(a) This question was answered well with candidates knowing that we needed to see the f ull
standardisation f ormula with 170, 176 and 4.8 substituted correctly. Caref ul candidates of ten
supported their working with a diagram.

(b) This second part of the question proved to be more challenging. Strong candidates knew to
subtract their answer to part (a) f rom 1 and to add this to 60 per cent or 0.6 to find the probability of
an adult male having a height less than k cm (0.7056). They then used their tables to f ind the z-
(k − 176)
value corresponding to this probability and equated that z-value to to f ind k. However,
4.8
many solutions were very confused with z-values and probabilities being interchanged and added
together with little understanding.Those who did use a correct process of ten gave an incorrect or
truncated version of the required z-value (0.541).

Question 4

(a) The style of grouping of the heights in this question seemed to be unf amiliar to many candidates
and it was insisted on seeing points plotted above 9.5, 19.5 etc. (not 10, 20 …). Strong and caref ul
candidates clearly listed the cumulative frequencies rather than relying on their Examiner having to
read them off the graph, and they knew to plot the points above the upper boundaries of each
group with the initial point being at (9.5, 0) and to join the points to make a curve without the use of
a ruler. A number of graphs had points plotted above the mid-points or lower boundaries instead of
upper boundaries. Axis had to be labelled, and the horizontal axis had to start on or bef ore 9.5.
Those who chose to allocate 9.5, 19.5 etc. to the thick graph lines usually produced the most
accurate graphs.

(b) This question asked for the graph to be used, and we needed to see evidence of this, pref erably a
line drawn across f rom 45 on the cumulative f requency axis.

(c) The majority of candidates knew that they needed to find the mid-point of each group to estimate
the mean and standard deviation, with only a few using other measurements and most correctly
f ound an accurate estimate of the mean. Many were less familiar with the formula for the standard
deviation, and they often resorted to a calculator for the answer. The calculations were bulky, and it
was allowed for a shortened calculation, as long as the first, last and one of the middle terms in the
summation could be seen.

A significant number of candidates used the approximated value of 38.9 for E(X) in their calculation
of the standard deviation and produced an inaccurate f inal answer, usually 10.7. They need to
understand that, if the final answer is to be correct to three signif icant f igures, all input numbers
must be correct to at least f our signif icant f igures.

Question 5

(a) This was a f amiliar style of binomial question and was tackled conf idently by the majority of
candidates. A few candidates confused it with a geometric distribution and gave a response that
lacked binomial coefficients. The wording of ‘no more than’ was generally well interpreted with only
a f ew omitting the probability of either 2 or zero. Lack of necessary working was rarely seen with
candidates appreciating that they need to show all the binomial terms in f ull. Just a f ew lost
accuracy and a mark through premature rounding bef ore they had added the three terms.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

(b) This was another well answered question with the majority of correct responses using the method
of subtracting the probability of none of the first 6 chocolates being in red f oil f rom 1. However, a
signif icant number chose the longer method of summing six geometric terms. Those who
attempted the method of subtracting two probabilities from 1 were generally less successf ul, of ten
including an extra incorrect probability of 0.25 in their calculation. Whilst this question could be
answered with use of a binomial distribution within it, this was not the correct general approach and
those who tried this method were generally unsuccessf ul

(c) This was a dif ficult question to interpret and there were a number of errors seen as a result. The
most common was in assuming independence between the second chocolate being the first one in
gold foil and the f if th one being the f irst unwrapped. Many recognised the correct conditional
probability calculation and strong candidates clearly explained what they were doing. A large
number calculated the denominator correctly, the probability of the 5 th chocolate being the f irst
unwrapped. Finding the numerator, the intersection of the second chocolate being the f irst gold
wrapped and the 5th being the first unwrapped, was more taxing. Those who attempted to f ind the
single f ive factor product were generally more successf ul than those who considered the f our
scenarios that satisf ied the conditions and needed to sum f our f ive f actor products. A f ew
candidates spotted the shortcut in the question, realising the selection was based on the f irst two
choices alone. The f irst one had to be red wrapped and the second gold wrapped given that they
were both wrapped.

Question 6

(a) This part of the question was answered well by most candidates with only a few not taking account
of the repeats of the Ps and Ss and dividing by 2!2!.

(b) Candidates found this part of the question more challenging. The most successf ul candidates
subtracted the number of ways with Ps at either end or Ss at either end f rom the total number of
ways calculated in the previous part of the question. They needed to remember that there would
still be a repeated letter in the middle seven. Those who tried to consider all the situations when
there was not a repeat made the question much more complex than it needed to be. They often did
not make it clear what they were trying to do which made awarding marks difficult. Candidates who
tried to deal with all the five non-repeated letters separately, of ten gave up part way through the
question.

(c) Strong candidates added the number of ways the letters could be arranged with the two Ps
between the Ss (6!) to the number of ways they could be arranged with the two Ps outside the Ss
(5! X 5P2). Less successful candidates split the problem into more parts and usually looked either
at the positioning of the Ss or the positioning of the Ps. This made the problem signif icantly more
challenging and few of these achieved full marks. Candidates need to appreciate that they need to
make their method clear to their reader or marker. Unless there is a clear explanation f or working,
credit cannot be awarded.

(d) This part of the question proved to be more accessible than the previous two parts and most
candidates made a good attempt. Strong candidates recognised that if the Ps were in the group of
f ive it meant that the Ss were in the group of f our and vice versa. This meant they only had two
situations to consider rather than four. If they did not appreciate this, they tended to produce a
numerator that was double the correct value. A small number of candidates attempted a probability
approach and were rarely successf ul. A f ew others attempted to use Permutations rather than
Combinations where the denominator of their probability fraction was either 90720 or 9!. This made
the question unnecessarily complex by considering arrangements rather than selections and was
rarely successf ul.

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 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/61
Probability and Statistics 2 (61)

There were too f ew candidates f or a meaningf ul report to be produced.

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 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/62
Probability and Statistics 2 (62)

Key messages

• In all questions, suf f icient method must be shown to justif y answers.

• When the answer is given in the question, candidates need to show convincingly their f ull and clear
working leading to the answer.
• Candidates should be encouraged to read questions caref ully and extract the relevant inf ormation.
• Candidates should be encouraged to reread the question when they have completed their solution to
ensure that all the requirements have been f ulf illed (f or example answering ‘in context’ if required).
• Candidates need to leave their f inal non-exact answers to at least three significant figures unless stated
otherwise. This means they need to keep four significant figures or more in the working leading to the
answer.
• All working should be done in the correct question space of the answer booklet. If a candidate does not
have enough space to complete their answer, they should either use the additional page, or ask f or an
extra sheet of paper, and label it clearly with the question number.
• Arithmetical errors, especially sign errors are very common. Candidates should get into the habit of
checking their work caref ully.
• Mathematical notation and vocabulary should be used appropriately.
• Conclusions to Hypothesis tests should be written in context and with a level of uncertainty in the
language used.

General comments

Candidates, in general, were able to apply their knowledge in the situations presented. Questions that were
done well were Question 1 and Question 5b, whilst Question 2 and 3b and were f ound to be more
demanding. On the whole candidates showed the required amount of working.

Presentation was generally good, and there did not appear to be any issues with timing.

There were some very good scripts, but equally there were scripts where candidates appeared to be
unprepared f or the demands of the paper.

The f ollowing comments, on individual questions, reflect some common errors, but there were also some
complete and f ully correct solutions.

Comments on specific questions

Question 1

This question was well attempted with most candidates using the correct Poisson distribution with mean 4.5.
Calculation of 1 – P(0,1,2,3) was required; common errors seen were the inclusion of P(4) in this expression,
use of an incorrect value for λ and occasional errors in calculation. Most candidates gave evidence of their
method of calculation and showed the f ull Poisson expression used.

Question 2

(a) Many candidates did not know how to find the value of µ, not realising that the mean was the mid
value of 245 and 263. A common error was to find 245 + 263 and divide by 50 instead of 2. A lack
of understanding of conf idence intervals was evident in many cases.

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 Cambridge International Advanced Level
9709 Mathematics November 2024
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(b) Again, this was not particularly well attempted, though some candidates were able to find a correct
expression for the width of the confidence interval and some used an appropriate expression using
their mean f ound in part (a). Errors included an incorrect z-value in their expression. Some
candidates did not use information from the confidence interval and attempted to use the standard
deviation f ormulae.

(c) Only a small number of candidates gave a fully correct explanation that the given f act would not
make a difference to their calculation in part (b), by recognising that n was large (implying that the
Central Limit Theorem had been used) or by stating the Central Limit Theorem explicitly.

Question 3

(a) Many candidates gave correct hypotheses, but errors were made in finding the required probability
of X  23 using B(25,0.8). Some candidates f ound X > 23 and occasionally X = 23. A valid
comparison with 0.1 was required followed by a conclusion which was in context and was not a
def inite statement; many candidates used phrases such as ‘we can conclude that …’ without using
the necessary level of uncertainty in the language used. Some candidates attempted to use a
Normal approximation despite clear instructions in the question to use the Binomial distribution.

(b) This part was f ound challenging with many candidates stating that 30 was large enough f or a
normal distribution to be used rather than appreciating that the initial model was only based on 25
employees so using it f or 30 employees would not be suitable.

Question 4

(a) This part was reasonably well attempted. Common errors included omission of √ 140 when
standardising, using continuity corrections and not keeping to the required level of accuracy.

(b) Again, many candidates knew how to approach this question and set up a standardising equation,
though some used 0.986 rather than ϕ–1(0.986), or made sign mistakes in their equation and others
did not keep to the required level of accuracy.

Question 5

(a) The interpretation of ‘more than 2 but not more than 5’ proved difficult for some candidates, with a
particularly common error being to calculate P(3,4) rather than P(3,4,5). Most candidates correctly
used Po(3.7).

(b) Many candidates used Po(6.3) and reached the correct answer of 0.126. Use of a Normal
approximation was occasionally seen, but on the whole candidates did well on this question.

(c) This part was f ound more challenging. Candidates who f ound the approximating distributions
N(37,37) and N(26,26) usually went on to find N(–11,63) and the correct probability, though errors
f inding the variance were made. There was confusion between 10 sweets and 10 bags, highlighting
the importance of reading the question carefully. Full marks could be gained f or a correct method
with or without the appropriate continuity correction.

Question 6

(a) Many candidates did not answer in context and merely said that a and b were the limits of X, or
similar, without ref erence to the time taken.

(b) This part was reasonably well attempted. Errors included a sign error when integrating, and
reversed limits of integration. On the whole suf f icient working was shown.

(c) Some candidates assumed that b was 2 rather than showing it, and others used the correct method
but were unable to rearrange their equation in b correctly.

(d) There was a mixed response here, and again sign errors when integrating and incorrect limits were
noted.

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 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

Question 7

(a) Candidates did not always read the question carefully and many defined a Type I error here rather
than give its probability. Others gave an answer of 98 per cent or less than 2 per cent, and some
omitted this part.

(b) Marks were gained by many for finding unbiased estimates for the population mean and variance.
The biased variance was occasionally used and there was also some conf usion seen between
dif ferent formulae for the unbiased estimate of the population variance. Some candidates stated
correct hypotheses, but others omitted to include them or did not use µ. Errors when standardising
included omission of √ 100 and conf usion between the variance and standard deviation. The
comparison with 2.054 (or equivalent) was not always clearly stated, and 2.54 was often seen. The
conclusion should be in context and not a def inite statement; again, there were candidates who
used phrases such as ‘we can conclude that …’ without using the necessary level of uncertainty in
the language used.

(c) Many candidates omitted this part, and many defined a Type II error and did not comment on the
claim.

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 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

MATHEMATICS

Paper 9709/63
Probability and Statistics 2 (63)

Key messages

It is expected that candidates work to sufficient accuracy in the paper. This may require working to more than
three significant figures in order to achieve a final answer which is accurate to three significant f igures. This
can be seen in Question 3(b) and in Question 5(b).

General comments

Clear writing of words and numbers was desirable. Many candidates gave clear presentations of their
method and calculation steps. However, it was difficult to distinguish between certain letters and numbers in
the work of some candidates. For example, in Question 4 the ‘a’ and the ‘9’ could look very similar thus
making marking dif f icult.

Comments on specific questions

Question 1

(a) To f ind the 96 per cent conf idence interval it was necessary to f ind the appropriate value of z,
namely 2.054 or 2.055. Many candidates f ound this value. Some candidates used an incorrect
value such as 1.751 or 2.54. It was essential that √150 was included in the calculation and that the
answer was given as an interval.

(b) The explanation to this part required both an answer (number or equivalent) and a reason. The
distribution used in the calculation related to the mean height.

Question 2

To compare the total mass of 3 small bags to the mass of 1 large bag it was necessary to consider a new
variable of the f orm 3S – L with a normal distribution N(–3, 2.1) and to consider 3S – L > 0. Af ter
standardising the ‘0’ it was necessary to select the small probability. A sketch could help in deciding on
whether the small or large probability was appropriate. Use of S1 + S2 + S3 instead of 3S would be of help in
f inding the correct mean and variance.

Question 3

(a) Many candidates found the unbiased estimates correctly. Answers were accepted as a decimal or
as a f raction. Some candidates used an incorrect f ormula f or the variance.

(b) It was necessary to change f rom the given probability 0.234 to 0.766 and then f ind the
corresponding z value 0.726. Many candidates omitted at least one of these steps or made errors
in trying to find the new values. Having found this z-value it could be equated to the standardised
f orm for x in the distribution of means of samples. This required the use of √75. Some candidates
omitted this. Some accuracy could be lost if the values f or mean and variance f ound in part (a)
were prematurely rounded. An inaccurate answer of 3.22 was of ten seen.

Question 4

(a) As many candidates demonstrated, the property to use here was that the area under the curve of
f (x) was equal to 1. So, integration of f(x) between the limits 2 and 3 was required. The integration

© 2024
 Cambridge International Advanced Level
9709 Mathematics November 2024
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steps, the substitution of the limits and the simplif ication of the terms then needed to be clearly
shown.

(b) As the answer was also given in this part of the question, the various steps needed to be clearly
shown. The method required the correct integration of x f (x) between the limits 2 and 3. This
18
involved ln(x) and or equivalent and the clear substitution of the limits.
x

Question 5

(a) As the scientist wished to test whether the true value of µ was different to that given in the article, a
two-tailed test was suitable. To answer the question fully it was necessary to give both the type of
test and a brief reason.

(b) In order to carry out the significance test it was necessary to state the hypotheses, standardise the
sample mean, compare the z-value with the critical value and state the conclusion. The
standardisation required the use of the normal distribution of means of samples of size 50 and
theref ore the use of √50. The critical z-value f or this 1 per cent two-tailed test was –2.576. The
main alternative comparison could be between the probabilities 0.0099 and 0.005 or equivalent.
The conclusion needed to be stated in context, not def inite and with no contradictions. Many
candidates showed all these steps correctly. Some candidates used an incorrect critical value such
as 2.326 or an incorrect probability such as 1 per cent. Other candidates gave a def inite
conclusion, omitting a suitable phrase such as ‘there is insuf f icient evidence that’.

Question 6

(a) For the number of customers arriving at desk A in a 15-minute period the new Poisson parameter
2.7 was required. For more than two customers the method was to find the probability 1 – P(0, 1,
2). It was necessary to write down the relevant terms.

(b) For the number of customers arriving at desks A and B in a 5-minute period, the new Poisson
parameters 0.9 and 1.05 were required and hence 1.95 for the total number. It was necessary to
write down the relevant terms f or P(0, 1, 2, 3).

(c) For customers arriving at desk B in a t-minute period the Poisson distribution with parameter 0.21t
could be used. Then f or a 90 per cent probability for the inspector the inequality 1 – e–0.21t  0.90
needed to be solved. This required re-arrangement and the taking of ln(t). To demonstrate that
suf ficient accuracy was being used the value 2.3026 was required here and then 10.96 (or 10.97)
and then 11 minutes. Working with inequalities was also given marks. Some candidates showed
these steps successfully. Other candidates attempted to take ln(t) of inappropriate expressions or
did not show suf f icient accuracy.

Question 7

(a) In the past, the value of the Poisson parameter was 3.3 for 1 year and hence f or 2 years was 6.6.
For a Type I error the probability of rejecting Ho was required. Thus, calculation of the sums of the
early terms of the Poisson distribution were required. P(0, 1, 2) = 0.0399676 and P(0, 1, 2, 3) =
0.105. The f irst of these was less than 0.05 whilst the second of these was greater than 0.05.
Hence the critical range was X  2 and the probability of a Type I error was 0.0400 (to 3 sf). These
various terms, expressions and values were required to score f ull marks.

(b) To complete the test the hypotheses needed to be stated. This might have been done in part (a).
The calculation work for the probabilities might have also been done in part (a) and the results
quoted here. The results gave the comparison 0.04 < 0.05 f or the 5 per cent test and then the
appropriate conclusion.

(c) A type II error would occur when X > 2 with the Poisson parameter 1.2 for the 2 years. To f ind this
probability it was necessary to find 1 – P(0, 1, 2). Some candidates used 0.6. Other candidates
omitted the ‘1–’.

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 Cambridge International Advanced Level
9709 Mathematics November 2024
Principal Examiner Report f or Teachers

(d) For the 30-year period the Poisson parameter would be 18. As 18 > 15 the appropriate
approximating distribution was the normal distribution N(18, 18). This change required a continuity
correction (10.5) and standardisation and selection of the relevant probability area. Many
candidates answered this correctly. Some candidates omitted the correction f actor or used an
incorrect factor such as 9.5. Other candidates chose the incorrect area. A sketch could help with
the choice.

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