Cambridge International Advanced Subsidiary and Advanced Level

9709 Mathematics November 2018

Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/11
Paper 11

Key messages

Where answers are given as in Question 5(i) and 7(ii) it is important that all steps to the solution are shown
clearly. When a question is unstructured, such as Question 9, candidates should be particularly careful in
explaining their methods.

When the sign change of the gradient is used to identify a turning point, the values of x that have been used
should be clearly identified, as should the values of the gradient given by them.

In questions where a numerical answer is required from use of calculus such as Question 7(iii) it is essential
that working is shown to justify correct solutions.

General comments

Again it needs to be noted that although no GCSE topics are directly examined basic algebraic, geometrical
and trigonometric techniques are required to answer some A-level questions.

Any study of the quadratic function should look closely at the conditions for the inverse function to exist.

Comments on specific questions

Question 1

The best answers to this question were produced by solving a quadratic in √ x and squaring the resulting
roots. Those who chose to isolate the √ x term and square the resulting equation could find the solutions
directly but there was more scope for making an algebraic error. Confusing the values of x and √ x still gained
candidates the first mark.

9
Answers: ,4
16

Question 2

Most candidates saw the need to form a single quadratic equation with a coefficient of x expressed in terms
of b . Some realised that the equal roots of this equation must be either x = −2 or x = 2 and used the sum of
the roots to find the possible critical values of b . The most frequently used method was to set the
discriminant equal to zero. The best answers went on to express the required ranges of b correctly.

Answer: b - −3, b . 5

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 3

Part (i) This question was well answered. Although it was acceptable for a to appear in the numerator and
denominator of the gradient for the first mark, it was expected that it should be cancelled and not appear in
the final equation.

Part (ii) Correct use of the distance formula was the route to most of the correct answers to this part. A few
candidates showed the points on a diagram and formed a right angle triangle with sides 3a and 4a before
using Pythagoras’ theorem to find a.

3x 2
Answers: (i) y = − , (ii)
4 3

Question 4

Part (i) Many completely correct solutions to this part were seen from candidates who knew and understood
how to use either version of the Sn formula for an arithmetic progression.

Part (ii) The formula for the sum to infinity of a geometric progression was used well with most attempts
using either the correct value of r or one in the required range for a sum to infinity to exist.

Answers: (i) –12160, (ii) 9

Question 5

Part (i) Candidates showed confidence in either expressing the left hand side of the identity as a single
fraction or multiplying through the identity by the two denominators. Most who reached this stage used the
trigonometric identity correctly to reach the given result. Many answers reflected an awareness that when a
result is given the working leading up to it must be shown clearly.

Part (ii) The given result from part (i) was used to form a quadratic equation in cos θ by every candidate
who attempted this part. Most realised that the quadratic formula or completing the square were required to
find the one acceptable value of cos θ . The best answers came from the candidates who realised valid
solutions existed in the first and fourth quadrants.

Answers: (ii) 78.6º and 218.4º

Question 6

Part (i) Setting the gradient equal to zero at x = 3 was seen from many candidates, most of whom went on
to find the non-zero value of a correctly.

Part (ii) This part was well answered by those candidates who realised reaching the solution involved a
routine integration and substitution to find the constant of integration.

Part (iii) Although the simplest route to identifying the turning point was to find the sign of the second
derivative at x = 3 , successful candidates also clearly showed the sign change of the gradient at either side
of x = 3 .

9x 2
Answers: (i) –3 (ii) y = − x 3 + − 4 (iii) maximum
2

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 7

Part (i) This was a straightforward question which was generally answered correctly.

Part (ii) The given result was most easily demonstrated by substitution of x = 5 into the equation of the
curve. Some chose the more complicated method of solving the equations of the line and the curve
simultaneously and showing that one of the results was x = 5 .

Part (iii) The most popular approach used by candidates was to find the area under the curve between the
required limits and subtract the area of the triangle, although the subtraction was not always seen. Some
candidates found the area by forming and using a single integral.

1 8
Answers: (i) (iii)
2 3

Question 8

Part (i) Correct interpretation of the diagram generally resulted in an acceptably presented correct vector.

Part (ii) The method for finding the magnitude of a vector was often seen. The best answers used the vector
magnitude correctly to form the required unit vector.

Part (iii) The candidates who had found the correct vectors in the previous two parts were usually able to
apply the scalar product formula correctly to find the required angle. It was acceptable to use an obtuse
angle to find the correct acute angle.

1
Answers: (i) ( −6i + 2k ) (ii) ( −6i − 3 j + 2k ) (iii) 25.4o
7

Question 9

The most popular and straightforward route to the solution involved finding the areas of the two sectors and
subtracting the sum of these from the area of the triangle OAB. The required trigonometrical skills were
evident in the answers of many candidates as was the use of the sector area formula. Whist some
candidates preferred to work in degrees, most used radians as the given angle in the question suggested.
The best answers were set out clearly with intermediate answers given to more than the three significant
figures required in the final answer.

Answer: 1.56 – 1.57

Question 10

Part (i)(a) The candidates who realised the gradient of the curve was required were usually able to carry out
the necessary differentiation. Many of them went on to find the gradient of the normal correctly and use it
with the coordinates of point A to find the normal equation.

Part (i)(b) This part depended on the answer from part (i) and most of the candidates who produced an
answer there were able to solve their equation with the equation of the curve to attempt to form a quadratic
and find the required coordinate of the point of intersection.

Part (ii) The correct detailed use of the chain rule was a feature of the better candidates’ answers. Some of
these realised that the rate of change of x was a decrease and were able to reach the correct answer.

x −1
Answers: (i)(a) y = (i)(b) (ii) 0.6
2 4

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 11

Part (a)(i) The provision of the function in completed square form enabled candidates who appreciated the
significance of f being a one to one function to state the required value of a without any calculation.

Part (a)(ii) Careful thought was required to deduce the range of f , ensuring f ( x ) = −1 was not included.
Many candidates were able to change the subject of the given function to find an expression for the inverse,
but few realised the range of the inverse demanded the use of the negative square root they had in their
expressions.

Part (b)(i) The candidates who kept the expression for g ( 2 x ) in factorised form usually went on to find
gg ( 2x ) in the required form.

Part (b)(ii) Nearly all completely correct attempts came from the expansion and simplification of the first
stage or final result in part (b)(i). Re-checking the products of negative and positive terms would have helped
eliminate some errors.

Answers: (a)(i) 3 (ii) y > −1, f −1 ( x ) = 3 − 1 + x (b)(i) (2 − 3) − 6(2 − 3) + 9

(ii) 16 − 96 + 192 − 144 + 36

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/12
Paper 12

Key messages

It is very important that candidates show all working clearly. For example, correctly labelling angles using 3
capital letters in the proof in Question 8 and not relying upon Examiners looking at the diagram given in the
question. Similarly, when a scalar product is calculated, the individual components should be shown
multiplied together, and when definite integrals are evaluated it is important that both of the limits can be
 63 36   23 36 
clearly seen to have been substituted into the integral. For example  − 6 ( 6) −  −  − 6 ( 2) −  .
 12 6   12 2
This is partly so that Examiners can clearly see that the integral function, available on some calculators, has
not been used.

General comments

The paper seemed to be generally very well received by the candidates and many good and excellent scripts
were seen. There were, however, a number of candidates who missed out a significant number of questions
and scored very poorly overall. The paper contained a number of reasonably straightforward questions,
giving all candidates the opportunity to show what they had learned and understood, but also contained
questions which provided more of a challenge, even for the strongest candidates. The vast majority of
candidates appeared to have sufficient time to complete the paper. The standard of presentation was
generally good, with candidates setting their work out in a clear readable fashion. Writing answers in pencil
and then over writing them in pen later should be discouraged as it makes the answers very difficult to read.
The question candidates found easiest was Question 3, while Question 2 proved to be the easiest question
for candidates.

Comments on specific questions

Question 1

This question proved to be a very accessible start to the paper, with a great many candidates demonstrating
a good knowledge of the binomial expansion. They were very often able to write down the relevant term and
1
evaluate it correctly. Weaker candidates were sometimes unsure which term would have a coefficient of 2
x
and a good number forgot to raise 3 to the power 4. Those who bracketed the term ( 3 x )4 were usually more
successful.

Answer: 840

Question 2

Most candidates demonstrated a very good understanding of indices and definite integration and full marks
were very commonly scored in this question. Generally sufficient working out was shown but it is important
that both of the limits can be clearly seen to have been substituted into the integral. A number of candidates
were unsure about the required indices and small number forgot to actually integrate the function.

26
Answer:
3

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 3

This question was slightly different to normal and many candidates struggled with it. In part (i) many tried to
use Pythagoras’ Theorem and failed to understand that the length PQ was vertical and that the difference in
the x co-ordinates was therefore 0. Almost 40% of candidates missed out part (ii) probably due to the
complex answers they had reached for part (i). Those candidates with the correct length PQ were generally
able to differentiate, set their value to 0 and solve, although many of these stopped when they had found the
value of t and did not find the maximum length of PQ.

16 3
Answer: (i) 4t – t3, (ii)
9
Question 4

In part (i) the vast majority of candidates were able to form the function fg(x) and go on to solve the given
equation correctly although many answers were given in degrees rather than the required radians. Part (ii)
proved to be more difficult, with many candidates failing to draw a curve which clearly turned at both ends.
Those who attempted to transform the graph of y = cosx to the given function were generally much more
successful than those who plotted points and joined them up.

Answer: (i) x = 2.46

Question 5

Many fully correct solutions to this series question were seen, but in part (i) weaker candidates did not form
correct equations from the information given in the question. Those who were successful in part (i) could
generally identify the fourth term in part (ii), but those who were unsuccessful in (i) could not make much
meaningful progress in (ii) and almost a third of candidates missed this part out completely.

Answers: (i) x = 8, y = 12 (ii) 16, 27

Question 6

The vast majority of candidates attempted part (i) of this question and they generally made some progress,
usually finding that BD was equal to 20 sin θ by using triangle BDC. Many realised then that, in triangle ABD,
9
angle DBA would also equal θ and concluded that BD would be equal to . The two expressions were
tanθ
then equated and the given result shown. Weaker candidates were often unable to make the second
connection correctly. Some did not obtain full marks because they did not find BD in terms of θ , even though
they showed the final given result. Although part (ii) was independent of part (i), nearly 20% of candidates
omitted it. Those who did attempt it were almost always successful in forming a quadratic in cos θ and
correctly solving it.

Answer: (ii) 36.9

Question 7

Candidates coped better with the information presented in this vector question than in some previous years.
The majority were able to correctly identify the required vectors and use the scalar product, as instructed, to
correctly find the required angle. A significant number, though, incorrectly rounded their final answer, with 31
and 40 quite common.

Answer: 31.0

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 8

There were a good number of fully correct solutions seen to this question. Strong candidates clearly
identified which angles they were finding in part (i) with isosceles triangle ABC often being split into two right-
angled triangles and angles CAB or ABC being correctly identified. Weaker candidates often did not clearly
identify which angles were being found. In part (ii) many strong candidates correctly found the required
perimeter of the shaded region but weaker candidates often did not identify which angles and arc lengths
were required, with the shaded area sometimes being found.

Answer: (ii) 24.2

Question 9

Many fully correct solutions were seen for this question. The vast majority of candidates understood the need
to complete the square in part (i), although some did not know what was required for parts (ii) and (iii). In
part (iv) many realised the need to use the completed square form and made some progress, but only the
strongest were able to correctly identify fully the inverse function.

x + 11
Answer: (i) 2( x − 3)2 − 11, (ii) f(x) ⩾ –11, (iii) 3, (iv) 3 −
2

Question 10

This question was quite straightforward for strong candidates and many completely correct solutions were
seen. Part (i) proved to be more difficult for weaker candidates, who often knew what was required, equating
and using the discriminant, but were unable to correctly carry out the algebraic manipulation needed. Some
made the discriminant > 0 and others could not go from k 2 < 144 to the correct final answer. Part (ii) was
more accessible for weaker candidates as less algebraic and more numerical manipulation was needed.
Those who were successful in (ii) were generally able to complete part (iii) but some forgot to use the
perpendicular gradient or did not use the mid-point.

Answers: (i) –12 < k < 12, (ii) (1, 14) and (4, 11) (iii) y = x + 10

Question 11

Parts (i) and (ii) of this final question were very well done, but many candidates struggled to find the required
area in part (iii). The standard rules for the differentiation and integration of this type of function were well
known, but weaker candidates sometimes forgot to multiply or divide by 4. In part (ii), again many candidates
knew to equate their differential to 0 and were able to solve the subsequent equation correctly, although
some forgot to find the y as well as the x co-ordinate of the stationary point. In part (iii) candidates generally
knew to use the integral from part (i) but often had the wrong limits, some using both of the co-ordinates
found in (ii). The line was even more problematic with some not including it at all and others using the
tangent at A. Those integrating the line and those using the area of a trapezium were equally likely to be
successful as were those who combined the line and curve together into one integral.

3
1
− (4 x + 1) 2
Answers: (i) 6(4 x + 1) 2 − 2, − x 2 + c, (ii) (2, 5) (iii) 1
2

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/13
Paper 13

Key message

In general, the presentation of work by candidates was satisfactory. However, Question 3(ii) required
candidates to find certain lengths and certain areas of regions in order to find the area of the required region.
Examiners saw numerous different methods and, unless candidates explicitly made it clear at each stage
what they were finding (e.g. BD or ∆BDC) it was sometimes difficult to see what each calculation referred to
and what method was being attempted. Candidates should lay out steps in their working clearly, showing
what they are finding at each stage

One of the messages contained in last June’s report was that an answer unsupported by correct working will
not gain credit. Whilst there were still isolated instances of essential working not being shown, it is pleasing
to be able to report an improvement in this aspect of presentation.

General comments

Many very good scripts were seen and candidates all seemed to have sufficient time to finish the paper. One
issue, that of accuracy of numerical answers, remains a source of many lost marks. Unless specified
otherwise, non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place in
the case of angles in degrees. In order to achieve this it is usually necessary to use more than 3 significant
figures in the working. This was particularly necessary in Question 3.

Comments on specific questions

Question 1

This question was done very well, with many candidates scoring full marks. The most efficient approach was
5
1  –2 
, terms in x2 and   were needed, together with a
to realise that in order to achieve a final term in
x3  x
7
binomial coefficient of C5. A common approach, however, was to write out all, or part, of the expansion and a
number of candidates stopped before reaching the required term, which was the sixth. A common mistake
was to lose the minus sign.

Answer: ‒672

Question 2

Most candidates started this question correctly by differentiating the given function and many went on to set
this to zero in order to find stationary points. After this, many candidates were influenced by the fact that at
x = ‒2, for example, the gradient was zero which they thought automatically meant that the function was
neither increasing nor decreasing. Although it was the right answer, it was for the wrong reason. Candidates
2
trying to follow this approach needed to show that at x = , the second derivative was positive, thereby
3
establishing there was a minimum at this point. Therefore gradients to the left of this point were negative and
gradients to the right were positive, so it was neither an increasing function nor a decreasing function. Some
candidates found the second derivative but did not appreciate what it showed them. An alternative approach
2
was to find the critical values of ‒2 and , and evaluate the gradient of any point between them. Finding
3

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

2
this to be negative, evaluating the gradient of any point to the right of x = , and finding it to be positive, was
3
sufficient to determine that f was neither an increasing function nor a decreasing function. A less frequently
2
seen approach was to find f(x) for three or four values of x in the domain of the function and spanning x =
3
in order to show at first a decreasing pattern, and then an increasing pattern, of values. It was not at all
uncommon to see scripts which started off in the right way, scoring the first one or two marks, but then
reached an incorrect conclusion. Not many used enough specific values of x substituted into f(x) or into f′(x)
to be able to make a proper decision and hence often stated ‘increasing’ or ‘decreasing’ after only one valid
substitution. The wording used by some candidates in their conclusion also requires comment. A function
cannot be an increasing function for part of the domain and a decreasing function for another part of the
domain. The word function needs to be omitted and the conclusion is then that it is neither an increasing
function nor a decreasing function. Examiners were looking for the word ‘Neither’ being used, or words to
that effect.

Answer: Neither

Question 3

Part (i) was well answered with almost all candidates scoring the one mark available for this part. Part (ii)
was notable for the number of different routes candidates found to the end result. The most straightforward
route was to find the area of the trapezium and to subtract the area of the sector. As reported in previous
reports, some candidates showed a preference for working in degrees even though calculations in radians
are less complicated and less subject to rounding errors. In addition to this, as reported in General
comments above, many candidates obtained an incorrect final answer because they did not work with more
than 3 significant figures in their calculations. However, many of these candidates employed a correct
method and were able to score four of the five available marks.

Answers: (i) 0.8 (ii) 1.69

Question 4

In part (i), the vast majority of candidates were able to find the gradient of the line AB and most were then
able to find the equation of BC. However, a significant proportion of candidates misinterpreted the question
and found the equation of the perpendicular line through A or through the mid-point of AB. When finding the
position of point C some candidates substituted x = 0 instead of y = 0. In part (ii), some candidates did not
spot that an answer correct to 4 significant figures was required. This question should have provided a good
source of marks to most candidates but, in practice, marks were lost through minor mistakes.

–4 x
Answers: (i) y = + 8 x = 6; (ii) 7.071
3

Question 5

This question was quite well done, with a few candidates dropping the final mark as they did not appreciate
that n had to be an integer. Candidates had to eliminate either a or n (most choosing a) leading to a
quadratic equation and this was a stage that had to be shown. A number of candidates, after stating the
initial two equations, went straight to the solution with no working shown. These candidates scored only the
first two marks.

Answer: n = 40 a = ‒23

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 6

The first step required in this unstructured question was to decide which two vectors to combine in a scalar
product in order to find the required angle. Most candidates correctly chose BO and BF or OB and FB. If one
of these vectors was reversed it gave rise to a negative solution which lost the final mark. A few candidates
used OF, which resulted in finding a different angle and it was usually only possible to award the first mark. A
significant number of candidates found BF incorrectly. These candidates thought that CG was a vertical edge
and therefore set CG equal to 7 k. This did not give a final answer which could be expressed as a fraction in
which the denominator was an integer. There was usually no indication that the candidate had gone back to
check whether an error had been made.

 28 
Answer: cos–1  
 45 

Question 7

In part (i), candidates were generally able to express the left hand side as a single fraction, despite brackets
often being omitted. Candidates were then able in the vast majority of cases to reach the required result.
Many candidates had more trouble with part (ii). The first difficulty for candidates was in realising they had to
set the numerator to zero. Some of the candidates who successfully negotiated the first obstacle and
reached a quadratic equation in sinθ were not then able to deal with the quadratic equation.

Answer: (ii) 38.2°

Question 8

Part (i) was very well done, with and the vast majority of candidates scoring full marks. Part (ii) proved to be
dy
more difficult. Some candidates made mistakes in deriving the two initial equations, e.g. setting =3
dx
instead of 0 when x = 2. Mistakes were often made in the simplification, or solving, of the equations.

1 3 1 2
Answers: (i) y = ax + bx − 4 x + 11 (ii) a = 3 b = ‒4
3 2

Question 9

In part (i), most candidates knew to find the discriminant. However, mistakes were fairly common. Some
candidates made mistakes when collecting like term together to express the 3–term quadratic equation.
Brackets were also not always used and this usually led to errors. Finally, b2 ‒ 4ac was not always applied
correctly. Careful candidates usually found their way to the correct quadratic expression, but many
candidates did not know how to progress from that point. For example, a significant number of candidates
1
introduced an inequality, arriving perhaps at 9k2 + 6k + 1 ⩾ 0 from which they then stated their answer, k ⩾ .
3
This was not what was required. Having obtained the correct quadratic expression some candidates simply
stated that it was greater than or equal to zero, which was not sufficient. Candidates needed to express the
quadratic expression as a perfect square e.g. (3k + 1)2, state that this was greater than or equal to zero and
conclude that, for all values of k, the line and curve meet. Many candidates scored the first three marks if
their arithmetic and algebraic manipulation were correct, but relatively few candidates scored the final mark.
In part (ii) it was permissible to refer to the work done in the previous part, but many candidates repeated the
work in order to derive the value of k required. If this value of k was correct, candidates were often able to
continue and score full marks by following one of several possible routes to the right answer.

1  2 1
Answer: (ii) k = – ,  ,– 
3 3 9

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 10

While the method for finding volume of revolution was familiar to most candidates, the challenge was in
2
 – 
1
2
finding  ( 3 x – 1) 3  . Candidates did not seem confident finding the resultant power and powers such as ,
  3
5 1
and were seen frequently. Integrating first and then trying to square the result was a route some took,
3 9
whilst others integrated y or, having found y2, forgot to integrate. The actual process of integrating was safer
ground and was mostly done effectively. Some final answers were given without showing the working for
substituting the limits (for which marks were lost) and 1 was seen occasionally as the lower limit. In contrast
to part (i), candidates seemed more comfortable with the process of differentiation in part (ii) and many fully
correct solutions were seen. Using the gradient of the tangent instead of the normal or assuming the y
coordinate of A was zero were errors that were occasionally seen.

 1 5
Answers: (i) 4π (ii) y =   x +
 2 3

Question 11

The familiar request in part (i) was done very well, with most candidates scoring full marks. Part (ii) required
a little more thought and a variety of wrong attempts were seen. The question asks for the value of k, so
answers left in terms of x did not score. In part (iii), most candidates demonstrated that they knew the
process for finding the inverse function, but many lost a mark by not choosing the negative square root and
some candidates got the domain wrong, often by using the wrong inequality. In part (iv), it tended to be the
stronger candidates only who recognised the need to use x + 3 ⩽ 1 and went on to score full marks. Most
candidates were successful in finding fg(x), although most employed a more laboured method in order to find
f(x + 3) rather than using the form found for f(x) in part (i).

1
Answers: (i) 2 ( x − 3 ) − 7 (ii) Largest value of k is 3 (iii) f −1 ( x ) = 3 − ( x + 7) x ⩾ – 7
2

2
(iv) largest p is ‒2, fg(x) = 2x2 ‒ 7

© 2018
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/21
Paper 21

Key messages

It is essential that candidates give their answers to the required level of accuracy of 3 significant figures,
which is stated in the rubric on the front of the examination paper. If a different level of accuracy is required,
it will be stated within the question itself. This means that working must be done to a greater accuracy. When
an exact answer is required, a calculator must not be used in the simplification and evaluation involved in the
solution.

General comments

The cohort taking this examination was relatively small, so where it is difficult to make generalisations, the
intended method of solution will be indicated.

Some candidates did not attempt a large number of questions, or gave working which demonstrated a
complete lack of understanding of what was required in some of the questions. Basic mathematical errors
were common.

Some candidates did demonstrate a good knowledge of the syllabus and were able to apply the techniques
they had used both appropriately and correctly. It was clear from these candidates that there were no issues
with timing.

Comments on specific questions

Question 1

(i) Most candidates chose to square each side of the equation and attempt to solve the resulting 2
term quadratic equation. Some candidates showed poor algebraic skills when attempting to square
and simplify, with many candidates failing to identify the solution x = 0. Better success was had by
those candidates who chose to produce 2 separate linear equations.

(ii) It was intended that candidates make use of their solutions to the first part of the question. This
was the reason the word ‘Hence’ was used. Candidates should recognise the use of this word.
Very few candidates did so and as a result, correct solutions were few. It was intended that x in
2
part (i) was equivalent to 3 y in part (ii) and so a solution to the equations 3 y = , making use of
3
logarithms, was required. It was essential that the answer was given to the correct level of
accuracy as some candidates gave an answer of −0.37.

2
Answer: (i) 0, (ii) −0.369
3

© 2018
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 2

In spite of the hint given in the question that the integral involved a logarithm, some candidates did not
integrate correctly to obtain an integral of the form k ln ( 2 x + 1) . It was pleasing to see that many candidates
did obtain this form and applied the limits correctly to their integral. However, many did not go on to use the
power law and subtraction law for logarithms, choosing instead to resort to the use of a calculator. Checking
that ln125 = 4.83 or similar, does not ‘show’ that the integral is equal to ln125. It is essential that each step
of working is given in a ‘Show that..’ question and the use of calculators is not usually required in a question
ln15
of this type. Candidates must also take care when using the subtraction law as ln15 − ln3 ≠ .
ln3

Question 3

There were few correct responses to this question as many candidates were unsure of which trigonometric
identities to use. There were 2 different initial approaches that could be used. The first one was to write
1
sec 2 θ as and then make use of cos2 θ = 1 − sin2 θ which then led to a quadratic equation in sin θ.
cos2 θ
sin θ
The second, but longer approach was to sec 2 θ = 1 + tan2 θ and then make use of tan θ = and the
cos θ
identity cos2 θ = 1 − sin2 θ to obtain the same quadratic equation in sin θ. However, many candidates made
the incorrect assumption that if cosec 2θ = 1 + cot 2 θ then cosecθ = 1 + cot θ. The resulting quadratic equation
in sin θ does not factorise so use of the quadratic formula or equivalent was necessary.

Answer: 57.9o , 122.1o

Question 4

(i) It was intended that x = −2 be substituted into the given expression. Many candidates did so, but
immediately stated that the expression was equal to zero. As the candidates were required to show
that x + 2 is a factor, it is essential that each term is evaluated and shown. Many candidates did
not do this and lost the accuracy mark as a result. It is also good practice to conclude questions of
the type with words to the effect that ‘ the remainder is zero, hence x + 2 is a factor’. Algebraic
long division was acceptable provided a remainder of zero was reached and a conclusion made, as
was synthetic division. Forming an identity was also acceptable, but a remainder was needed and
this had to be shown to be equal to zero and a conclusion made for both marks. Candidates should
recognise that a question of this type is straightforward and little work is needed to get full marks.
Algebraic long division, synthetic division and forming an identity are time consuming and also
prone to errors.

(ii) It was intended that candidates make use of algebraic long division and obtain a cubic quotient
which could be equated to zero and then re-arranged. Those candidates that had used this method
in part (i) were then able to make use of their quotient from part (i) immediately. It was evident that
some candidates were unable to produce the steps required. A good approach to questions with
several parts is to check if a previous part can be used, if a candidate is having a problem in how to
proceed.

(iii) Many candidates demonstrated high levels of competence with the iterative process, making good
use of their calculators and the answer function to cut down time consuming calculations. It was
necessary that each iteration be shown to the required level of accuracy and that there were
enough iterations to deduce the final answer, which also had to be given to the required level of
accuracy.

Answer: 2.256

© 2018
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 5

dy
(i) Many candidates were able to obtain a correct expression for but seldom used it to obtain the
dx
equation of the tangent as required. It is essential that candidates check that they have met the
dx 1
demands of the question. Common errors included writing = rather than the correct
dt t + 1
dx 1
= 1+ and also failing to recognise that y = 3te2t needed to be differentiated as a product.
dt t +1
Most candidates did not attempt to find the value of t at the origin.
dy
(ii) Most candidates attempted to equate their to zero, but needed to make a valid attempt to solve
dx
for t before being able to obtain a method mark. Very few correct solutions were seen.

3x
Answer: (i) y = (ii) x = −1.19, y = −0.55
2

Question 6

(i) The use of the trapezium rule continues to cause problems for candidates. Common errors
included having too many x values, which implies that the interval width is incorrect, calculators
being in the incorrect mode, incorrect evaluations and just using the x values rather than the y
values. As a result there were few correct solutions.

(ii) Many candidates were able to obtain the correct integrand for the required volume. However many
did not recognise the need to make use of an appropriate double angle formula before attempting
to integrate. It should also be noted that an exact answer was required and even though
candidates were requested not to use a calculator, many did.

5π 2
Answer: (i) 4.84 (ii)
2

Question 7

(i) Most candidates were able to gain at least one mark in this question, with many completely correct
responses seen. A common error was to obtain cos 2 x − 3 sin2 x.

(ii) Many candidates knew the steps that they had to take in the solution of this part. However, errors
meant that full marks were rarely obtained. Some candidates obtained an incorrect value for α as
1
they had written tan α = rather than tan α = 3. Some worked in degrees and did not convert
3
back to radians. Questions involving calculus need to have any work done in terms of radians
unless otherwise specified. Many candidates obtained the first solution in the range, but were
dy
unable to find the second solution. Of those candidates whose expression for was incorrect,
dx
many were able to gain method marks for correct steps being used.

Answer: (i) 2cos 2 x − 6 sin2 x (ii) 1.979, 3.055

© 2018
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/22
Paper 22

Key messages

It is essential that candidates give their answers to the required level of accuracy of 3 significant figures,
which is stated in the rubric on the front of the examination paper. If a different level of accuracy is required,
it will be stated within the question itself. This means that working must be done to a greater accuracy. When
an exact answer is required, a calculator must not be used in the simplification and evaluation involved in the
solution.

General comments

Some candidates did not attempt a large number of questions, or gave working which demonstrated a
complete lack of understanding of what was required in some of the questions. Basic mathematical errors
were common.

Some candidates did demonstrate a good knowledge of the syllabus and were able to apply the techniques
they had used both appropriately and correctly. It was clear from these candidates that there were no issues
with timing.

Comments on specific questions

Question 1

It was essential that candidates obtained the critical values using either a quadratic equation or a pair of
linear equations or inequalities. By far the most popular method of obtaining these critical values was to form
a quadratic equation by squaring each side of the given inequality and equating. However, many candidates
squared the left hand side of the inequality and forgot about squaring the coefficient of x 2 on the right hand
side which resulted in the incorrect quadratic equation 7 x 2 − 15 x + 25 = 0 rather than the correct quadratic
equation 5 x 2 − 15 x + 25 = 0.

Candidates making use of either 2 linear inequalities or 2 linear equations to find the critical values were
usually more successful.

Answer: 1 < x < 5

Question 2

Very few candidates were able to make much progress with this question. It was intended that a quadratic
equation in 3 x be formed, the solutions of which could then be used to solve for x. Quite a few candidates
recognised that 9 x = 32 x , but were unable to progress from there is a correct manner.

Most candidates attempted to take logarithms of each term in the given equation. This was a completely
incorrect approach which could not be awarded any marks. It is essential that candidates read the question
carefully. In this case, the first demand was to find value of 3 x and then use logarithms to find the value of
x.

Answer: 15, 2.465

© 2018
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 3

Many candidates were able to attempt the differentiation required to find the x – coordinate of the stationary
point. Most were able to obtain the form p cos 2 x − q sec 2 2 x , with many obtaining the correct values of
p and q . Some candidates forgot to take into account the double angle when they were differentiating and
dy
gave an incorrect statement of = 5cos 2 x − 3 sec 2 2 x. Rearranging the trigonometric result when
dx
dy 3
= 0 to obtain cos3 2 x = was problematic for some who either tried to use the identity
dx 5
1
tan2 2 x + 1 = sec 2 2 x ( this would then involve a lot more work) or mistakenly thought that sec 2 x =
.
sin2 x
There were however, candidates who completed this question with well thought out and well-presented
correct solutions.

Answer: 0.284

Question 4

Most candidates realised that implicit differentiation was needed and differentiated each term of the given
equation, with varying degrees of success. Some candidates did not recognise the need to differentiate
3 ye2 x as a product and others did not differentiate the right hand side of the equation, resulting in the
derivatives of the left hand side of the equation being equated to 10 rather than zero. It is far easier to
substitute values for x and y into an equation which has not been re-arranged and simplified, than to re-
arrange the equation and them make substitutions of values. In this case, most candidates attempted to find
dy
an expression for and then substitute in x = 0 and y = 2 . However, mistakes in simplification prior to
dx
substitution often led to incorrect answers. For those candidates who did not realise that implicit
differentiation was needed, there was no opportunity for gaining any marks.

16
Answer: −
7

Question 5

(i) Candidates with algebraic skills gave good answers to this part of the questions It is essential that
each step of a candidate’s working is shown in a question of this type as the final result is given.

(ii) There were 2 different approaches that could be taken. The first approach, which was attempted by
the majority of those who attempted this part, was to make use of the equation obtained in part (i).
Some candidates substituted both values into both sides of the equation but made no explanation as
to what this was showing. The more successful candidates chose to use the equation in part (i) and
1
( )
form a new function of the type f ( x ) = x − ln 1.6 x 2 + 4 or equivalent and use substitution of both
2
given values to show a sign change.

The second approach was to make use of the original equation y = 5e2 x − 8 x 2 − 20 and substitute
the given values to again obtain a sign change. Whichever approach had been used, it was expected
that the candidate would make an appropriate conclusion after their calculations.

(iii) It was expected that a value between 0.75 and 0.85 be used as a starting point for the iterative
process. It was evident that many candidates were not making the correct use of their calculator as
evidenced by the incorrect results obtained after substitution into a correct formula. Some candidates
did not give enough iterations. In most cases at least 9 or 10 were needed and it was essential that
each of these was shown to the correct level of accuracy. Many gave their final answer as either
0.8095 or 0.8096, not the level of accuracy required in the question. Some candidates chose
incorrectly to ‘round down’ to 0.80955.

© 2018
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

(iv) Of those candidates who attempted this part, most were able to gain marks for a correct
differentiation and attempt to substitute in their value from part (iii) to obtain a value for the gradient
of the curve at the given point.

Answer: (iii) 0.80956 (iv) 37.5

Question 6

(a) Most candidates recognised that the integral involved a logarithm, the clue to this being in the given
answer. As the answer was given, it was important that each step of the solution be shown and the
actual integral was expected. There were some errors involving the numerical coefficient, with 12
being a common error. The correct use of the square bracket notation was needed and then a clear
use of both the subtraction law involving logarithms and the power law involving logarithms. These
could be used in either order. Many candidates did not obtain the mark for stating that 4ln 4 = ln 44 .

(b) As an exact result was asked for, it was expected that no use of calculators be made. It was
necessary to make use of the double angle formulae 2 sin2 x = 1 − cos 2 x and
tan2 2 x = sec 2 2 x − 1. If either of these was not used, candidates were unable to make much
progress. There were occasional slips in coefficients and signs. Few correct exact answers were
seen as many candidates were unable to obtain the correct integral at the start.

π 3
Answer: (b) −
2 2

Question 7

3
(i) It was expected that candidates substitute x = −into the given expression and obtain a result of
2
zero. The question demanded a result to be shown and those candidates that wrote
3 2
 3  3  3
8  −  + 4  −  − 10  −  + 3 = 0 with no evaluation of each of the separate terms were unable
 2  2  2
to gain the accuracy mark. It is important to show each step in a solution. Algebraic long division
was also acceptable provided each step in the process was shown clearly and correctly. An
appropriate conclusion was also expected for the accuracy mark, whichever method was being
used.

(ii) Many candidates were able to make use of the correct double angle formula and appropriate
simplification to obtain the required result. There were occasional slips in manipulation and
simplification.

(iii) It was necessary for candidates to make the connection between this part of the question and parts
(i) and (ii). Those candidates who had completed algebraic long division in part (i) were at an
advantage over those that had not as they were more readily able to obtain a factorised form of the
equation in part (ii). There was only one solution for cos θ . For those candidates who did not make
the connection, often no marks were awarded. Some of the stronger candidates were able to gain
full marks for this part by making the correct connections between each of the parts. Candidates
should be encouraged to look for ‘similar’ expressions or equations in longer questions of this type.

Answer: (iii) 60o , 300o

© 2018
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/23
Paper 23

Key messages

It is essential that candidates give their answers to the required level of accuracy of 3 significant figures,
which is stated in the rubric on the front of the examination paper. If a different level of accuracy is required,
it will be stated within the question itself. This means that working must be done to a greater accuracy. When
an exact answer is required, a calculator must not be used in the simplification and evaluation involved in the
solution.

General comments

The cohort taking this examination was relatively small, so where it is difficult to make generalisations, the
intended method of solution will be indicated.

Some candidates did not attempt a large number of questions, or gave working which demonstrated a
complete lack of understanding of what was required in some of the questions. Basic mathematical errors
were common.

Some candidates did demonstrate a good knowledge of the syllabus and were able to apply the techniques
they had used both appropriately and correctly. It was clear from these candidates that there were no issues
with timing.

Comments on specific questions

Question 1

(i) Most candidates chose to square each side of the equation and attempt to solve the resulting 2
term quadratic equation. Some candidates showed poor algebraic skills when attempting to square
and simplify, with many candidates failing to identify the solution x = 0. Better success was had by
those candidates who chose to produce 2 separate linear equations.

(ii) It was intended that candidates make use of their solutions to the first part of the question. This
was the reason the word ‘Hence’ was used. Candidates should recognise the use of this word.
Very few candidates did so and as a result, correct solutions were few. It was intended that x in
2
part (i) was equivalent to 3 y in part (ii) and so a solution to the equations 3 y = , making use of
3
logarithms, was required. It was essential that the answer was given to the correct level of
accuracy as some candidates gave an answer of −0.37.

2
Answer: (i) 0, (ii) −0.369
3

© 2018
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 2

In spite of the hint given in the question that the integral involved a logarithm, some candidates did not
integrate correctly to obtain an integral of the form k ln ( 2 x + 1) . It was pleasing to see that many candidates
did obtain this form and applied the limits correctly to their integral. However, many did not go on to use the
power law and subtraction law for logarithms, choosing instead to resort to the use of a calculator. Checking
that ln125 = 4.83 or similar, does not ‘show’ that the integral is equal to ln125. It is essential that each step
of working is given in a ‘Show that..’ question and the use of calculators is not usually required in a question
ln15
of this type. Candidates must also take care when using the subtraction law as ln15 − ln3 ≠ .
ln3

Question 3

There were few correct responses to this question as many candidates were unsure of which trigonometric
identities to use. There were 2 different initial approaches that could be used. The first one was to write
1
sec 2 θ as and then make use of cos2 θ = 1 − sin2 θ which then led to a quadratic equation in sin θ.
cos2 θ
sin θ
The second, but longer approach was to sec 2 θ = 1 + tan2 θ and then make use of tan θ = and the
cos θ
identity cos2 θ = 1 − sin2 θ to obtain the same quadratic equation in sin θ. However, many candidates made
the incorrect assumption that if cosec 2θ = 1 + cot 2 θ then cosecθ = 1 + cot θ. The resulting quadratic equation
in sin θ does not factorise so use of the quadratic formula or equivalent was necessary.

Answer: 57.9o , 122.1o

Question 4

(i) It was intended that x = −2 be substituted into the given expression. Many candidates did so, but
immediately stated that the expression was equal to zero. As the candidates were required to show
that x + 2 is a factor, it is essential that each term is evaluated and shown. Many candidates did
not do this and lost the accuracy mark as a result. It is also good practice to conclude questions of
the type with words to the effect that ‘ the remainder is zero, hence x + 2 is a factor’. Algebraic
long division was acceptable provided a remainder of zero was reached and a conclusion made, as
was synthetic division. Forming an identity was also acceptable, but a remainder was needed and
this had to be shown to be equal to zero and a conclusion made for both marks. Candidates should
recognise that a question of this type is straightforward and little work is needed to get full marks.
Algebraic long division, synthetic division and forming an identity are time consuming and also
prone to errors.

(ii) It was intended that candidates make use of algebraic long division and obtain a cubic quotient
which could be equated to zero and then re-arranged. Those candidates that had used this method
in part (i) were then able to make use of their quotient from part (i) immediately. It was evident that
some candidates were unable to produce the steps required. A good approach to questions with
several parts is to check if a previous part can be used, if a candidate is having a problem in how to
proceed.

(iii) Many candidates demonstrated high levels of competence with the iterative process, making good
use of their calculators and the answer function to cut down time consuming calculations. It was
necessary that each iteration be shown to the required level of accuracy and that there were
enough iterations to deduce the final answer, which also had to be given to the required level of
accuracy.

Answer: 2.256

© 2018
 Cambridge International Advanced Subsidiary Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 5

dy
(i) Many candidates were able to obtain a correct expression for but seldom used it to obtain the
dx
equation of the tangent as required. It is essential that candidates check that they have met the
dx 1
demands of the question. Common errors included writing = rather than the correct
dt t + 1
dx 1
= 1+ and also failing to recognise that y = 3te2t needed to be differentiated as a product.
dt t +1
Most candidates did not attempt to find the value of t at the origin.
dy
(ii) Most candidates attempted to equate their to zero, but needed to make a valid attempt to solve
dx
for t before being able to obtain a method mark. Very few correct solutions were seen.

3x
Answer: (i) y = (ii) x = −1.19, y = −0.55
2

Question 6

(i) The use of the trapezium rule continues to cause problems for candidates. Common errors
included having too many x values, which implies that the interval width is incorrect, calculators
being in the incorrect mode, incorrect evaluations and just using the x values rather than the y
values. As a result there were few correct solutions.

(ii) Many candidates were able to obtain the correct integrand for the required volume. However many
did not recognise the need to make use of an appropriate double angle formula before attempting
to integrate. It should also be noted that an exact answer was required and even though
candidates were requested not to use a calculator, many did.

5π 2
Answer: (i) 4.84 (ii)
2

Question 7

(i) Most candidates were able to gain at least one mark in this question, with many completely correct
responses seen. A common error was to obtain cos 2 x − 3 sin2 x.

(ii) Many candidates knew the steps that they had to take in the solution of this part. However, errors
meant that full marks were rarely obtained. Some candidates obtained an incorrect value for α as
1
they had written tan α = rather than tan α = 3. Some worked in degrees and did not convert
3
back to radians. Questions involving calculus need to have any work done in terms of radians
unless otherwise specified. Many candidates obtained the first solution in the range, but were
dy
unable to find the second solution. Of those candidates whose expression for was incorrect,
dx
many were able to gain method marks for correct steps being used.

Answer: (i) 2cos 2 x − 6 sin2 x (ii) 1.979, 3.055

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/31
Paper 31

Key messages

Candidates need to know:

• that they cannot take logs of negative numbers
• exactly what is meant by a tangent parallel to the x-axis or parallel to the y-axis
• how to sketch a cubic graph
• how to use the chain rule for differentiation of a trig function such as sin3(4x)
• that a 3-term partial fraction with two linear terms and one linear term squared for the denominators
should have only three unknown coefficients. Introducing Dx + E in the numerator above the squared
linear term is over-specifying the system.

General comments

The standard of work on this paper varied considerably, although all questions were accessible to the
stronger candidates. A significant number of candidates found certain questions extremely difficult.
Candidates are reminded that working carefully through past papers and mark schemes will give them a
good idea of what to expect in the examination.

Questions or parts of questions that were generally done well were Question 2 (solution of exponential
equation), Question 3(ii) (iterative convergence), (iii) (iterative formula to find root) Question 4(i)
(parametric equations), Question 5 (solution of differential equation), Question 6(ii) (solution of trig
equation), Question 9(i) (partial fractions) and Question 10(ii) (angle between planes). Questions that were
done less well were Question 3(i) (sketching graphs), Question 4(ii) (finding where tangent to curve is
parallel to y-axis), Question 6(ii) (solving trig equation), Question 7(i), (ii) (establishing stationary point on
curve given by a trig function and determining the exact area under this curve), Question 9(ii) (integration of
partial fractions) and Question 10(iii) (finding position of point on a line a given distance from a plane).

In general the presentation of the work was good, though there were some rather untidy scripts. Candidates
should bear in mind that scripts will be scanned for marking and they should use a black pen, reasonable
sized lettering and symbols, and present their work clearly. Candidates should avoid using ink that is
absorbed into the paper and then appears on the reverse side as this can make it difficult to read the pages
when scanned.

It was pleasing to see that candidates are aware of the need to show sufficient working in their solutions.
Previous reports mentioned this in the context of solving a quadratic equation and substituting limits into an
integral. In recent years additional rubric has been added to individual questions, for example ‘show all your
working.’ This is to ensure that candidates can demonstrate understanding of the methods even when their
calculator is capable of finding the answer. For example, modulus of a vector, scalar product of two vectors,
product or division of two complex numbers, modulus or argument of complex number. No credit is given to
answers written directly from the calculator, without working. This will be even more important when the
modified syllabus commences, since it has a new rubric ‘no marks will be given for unsupported answers
from a calculator’. Very strong candidates who can do some of these tasks mentally should also be
encouraged to show the steps in their solution in order to gain credit.

Where answers are given after the comments on individual questions, it should be understood that the form
given is not necessarily the only ‘correct answer’.

Comments on specific questions

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 1

This question proved to be challenging for candidates. The omission of the parameter a was usually the
source of the problem. Many candidates failed to square the 2 outside the modulus sign or poor algebra
prevented them from arriving at the correct equation. Although many candidates took the quadratic equation
approach, this question is more easily solved by considering the pair of linear equations or linear inequalities.
Many candidates who successfully obtained a correct quadratic equation ran into problems because they did
not have the correct power of a where required. Factorising was generally more successful. Some
x
candidates decided to set a to unity, i.e. to solve for , but they did not reintroduce a after solving their
a
a 5a
equation. Those who did obtain x = − and usually made the correct choice of region at the very end.
5 3

a 5a
Answer: − <x<
5 3

Question 2

This question was usually well done, however there were still many candidates who did not reach an
b
equation of the form ae2x = b, aex = be−x or ae x = x with the correct values of both a and b. Many arithmetic
e
and algebraic errors were seen, as well as multiplication of indices to give an exponential term to the power
x2. Most candidates knew they then needed to take lns but some did not show this step despite the additional
instruction to show all necessary working. Too many candidates had a correct equation, but with a and b
negative. Then, before moving to an equation with positive terms, they attempted to take lns. Candidates
should be aware that they cannot take lns of negative numbers

Answer: 0.46

Question 3

(i) This part proved challenging more many candidate, for various reasons. The graph of y = x3 should
be sketched for at least −2 ⩽ x ⩽ 2 to show its shape. Stopping at x = −1 does not highlight how the
graph behaves for negative x. Similarly the graph of y = 3 – x should be sketched for at least
–1 ⩽ x⩽ 4. However, leaving the graphs in this form is still insufficient, and adding the comment
‘there is exactly one real root’ is not enough. The question required a clear indication of the single
point of intersection with a dot or some other mark, together with a comment that there is only one
point of intersection so only one root. This final statement should be clearly made without it being
left to the examiner to interpret.

(ii) Convergence, starting with either form, was usually well done. As it is a proof there should be no
jumps in the working; it is better to show too many lines than too few.

(iii) Most candidates found the root to the required accuracy but occasionally candidates left their
answer to 2 d.p. or 4 d.p.

Answer: (iii) 1.213

Question 4

dx dy
(i) Most candidates knew what was required and were successful with either or . However
dθ dθ
many did not apply the chain rule correctly to either sin 2θ or cos 2θ. Although the question didn’t
ask for the expression in a particular form, most candidates tried to simplify it into an expression in
just sinθ and cosθ. This often introduced errors and if candidates continued to use this form into (ii)
little progress was made in that section. Fortunately most candidates returned to their initial
expression that was hopefully free of errors.

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

(ii) This is where candidates really ran into trouble. Most candidates believed that this question must
dy
be about finding where = 0 , 1 or ∞, with about half using 0, 10% using 1 and the remainder
dx
using ∞. If candidates cannot reason this mentally they should draw a small sketch to help
themselves, as a wrong guess meant zero out of four immediately. Most of the candidates making
the correct choice scored three or four marks, since some stopped after finding the value of θ.
Some candidates factorised their quadratic incorrectly or had a sign error in their double angle
formula.

( 2sinθ + 2sin2θ )  3 3 1
Answers: (i) − (ii)  , 

( 2cosθ + 2cos 2θ )  2 2

Question 5

This proved a high scoring question as most candidates could separate variables correctly and obtain the ln
2
y term. The integral in x proved more difficult, since instead of considering and x, some candidates
x
resorted to integrating by parts and increased the possibility of sign errors. A considerable number of
candidates left the expression as ln y instead of obtaining an expression for y as requested in the question.
Candidates who left the term exp(2ln x) in their final answer were penalised in their final mark.

1 1 
Answer: y = x 2 exp  − x 2 
2 2 

Question 6

(i) Many candidates found it difficult to obtain the given expression, taking far more steps than were
required. Fortunately most did achieve it, although several had sign and coefficient errors in their
cos x and sin x terms. Done correctly, R cos α and R sin α should have been 3 and 1
respectively, and the working for R and α should not have had any negative signs present.
However, a large number of candidates simply quoted the expressions R = (a 2
+ b2 ) and

b
α = tan−1   , with (−1)2 seen in the expression for R. Whilst examiners allowed this error in the
a
 
evaluation of R = 2 they were less generous if the sign error was present in finding α. There is a
similar issue arising in Question 9(i) when determining the partial fractions. In the present case the
use of the formula for R, which is perfectly valid, has covered up the incorrect mathematics. A few
candidates made the error of writing cos α = 3 and sin α = 1 (omitting R), leading incorrectly to the
1
correct expression tan α = and α = 30°.
3

(ii) A few candidates did not spot the link between (i) and (ii) and as a result were unable to make any
 2
progress. However, most candidates readily found the values of sin−1   and hence both
 R 
required solutions. Some incorrectly believed that there was only one angle arising from
 2
sin−1 
 R  and that the other angle should come from 180° minus their correctly found angle.
 

Answer: x = 75° and 165°

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 7

(i) This was a question where candidates usually scored either just a single method mark, for using
the product rule, or full marks. Correct solutions required candidates to apply the chain rule
accurately twice. It was still possible to proceed to a solution if the errors in the chain rule were
restricted to the coefficients. Other errors such as incorrect powers of trig functions meant that it
was impossible to reduce the equation to one that could be solved.

(ii) Nearly all candidates struggled with this question despite being given the substitution required. This
was evidenced by the amount of working most produced, with little progress made. Common errors
were poor substitution of du = cos x dx, inability to express cos2 x correctly in terms of u, substituting
π
limits 0 and for u or substituting limits 0 and 1 for x. In fact very few candidates reached the
2
correct integrand in terms of u, together with the correct limits.

2
Answers: (i) x = 0.685 (ii)
3

Question 8

(a) The rubric demanded that candidates show detailed working in order to gain marks. Most
candidates divided out the complex numbers then converted to polar form, as opposed to
converting to polar form and then dividing out. When multiplying the numerator and the
denominator by the conjugate of the denominator, candidates often had errors either in the
denominator, e.g. 1 + (2i)2 instead of 1 – (2i)2, or arithmetic errors when collecting the real or
imaginary parts in the numerator. Basic errors such as these meant that none of the accuracy
marks could be gained. However, many candidates who performed the division correctly then left
their modulus in the exact form instead of correct to 3 significant figures. Few candidates managed
to obtain the correct argument, the usual answer being θ = −1.05, something that candidates
arrived at from their calculators without any reference to the position of the complex number
−4 7
+ i in the Argand diagram. A quick sketch would have helped them.
5 5

(b) Many candidates scored full marks, although some restricted their solution to the Argand diagram
part only, with no attempt at the least value of z . Several candidates finished with their circles in
the wrong quadrant, whilst others did not indicate that the radius of the circle was unity.

Answers: (a) 1.61ei2.09 (b) 13 − 1

Question 9

(i) This question was extremely well done by most candidates. However, a few candidates introduced
a constant term into their partial fractions, something only necessary when the powers of the
numerator and denominator are the same. A few others incorrectly used a linear term as opposed
to a constant term for the numerator of the term with the denominator (3 + 2x)2 in the 3-term partial
fractions. This form will see two of the constants determined uniquely but only a combination of the
other two as unique. Candidates need to check very carefully when they remove the denominator
in order to find their constants, since slips are possible, e.g. multiplication by a factor from the
denominator that is of the incorrect power. Any error at this point leads to errors in the three
equations obtained by equating the coefficients of the powers of x and incorrect constants.
However, if the candidate opts instead to construct their equations by using the roots of the factors
of the denominator then in some cases the incorrect terms will vanish and the correct values of the
constants will be obtained. Unfortunately, these cannot be given credit since the equations were in
error and it is only by the incorrect terms dropping out that correct answers have been obtained.
This feature is like that arising in Question 6(i) where the correct solution only resulted from the
squaring of an incorrect negative sign.

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

(ii) Integrating the partial fractions proved testing, as many candidates had the incorrect coefficients for
both their ln terms. Hence it was possible to see some candidates who had the correct forms for all
three integrals gaining no marks due to mistakes in determining the appropriate coefficients. In
obtaining the given answer, it is necessary for candidates to ensure that there are no gaps in their
working; again too much detail is better than not enough.

1 1 3 1 2x
Answers: (i) − + or −
( 2 − x ) ( 3 + 2 x ) ( 3 + 2 x )2 ( 2 − x ) ( 3 + 2 x )2

Question 10

(i) All details of the working using the scalar product were required. Most candidates were able to
show the scalar product was zero, but omitted to verify that one point of l does not lie in the plane.
If, instead, the candidate substituted coordinates of a general point of l in the equation of the plane,
it was necessary to establish and state there is no point that satisfies this equation. Again few
candidates made the concluding statement.

(ii) Candidates usually showed the full details for their scalar product and moduli hence even with the
presence of errors it was possible for them to gain both method marks.

(iii) Those candidates who knew the formula for the distance of a point from a plane usually quickly
scored full marks, whilst most others struggled. There were a few candidates who didn’t appear to
know this formula yet made a reasonable attempt by finding where the normal vector from a point
on the line met the plane and then setting the distance between these points equal to 2. Others
appeared to think they had solved the problem when they found where the line met the plane at (5,
3, 3) and stated this as their answer. In fact they still had to find the distance between this point and
2
a point on the line and set this equal to the length , where α is the acute angle between the
( cos α )
direction of the line and the normal vector to the plane.

Answers: (ii) 74.5° or 1.30 radians (iii) 7i + 5j + 7k and 3i + j − k

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/32
Paper 32

General comments

The response to this paper was very varied. There were some very strong candidates who produced clear,
accurate solutions, and at the other extreme there were some scripts that were very difficult to follow,
sometimes because one solution had been overwritten with another, and sometimes because there was no
discernible method being followed.

Most candidates offered solutions to all ten questions, and very few candidates needed to use additional
answer booklets. In general the candidates were familiar with the methods examined, but there were many
examples of major errors in cancelling fractions, errors in basic arithmetic and the use of incorrect rules such
as ln ( a + b ) = ln a + ln b .

Candidates scored well on Question 1 (modular inequality), Question 2 (solving equations using
trigonometry formulae), Question 5 (root of equation and iterative process) and Question 8 (partial fractions
and binomial expansion). Parts of some questions were challenging to candidates, most notably Question
3ii (evaluate definite integral) Question 4 (recognition of the quadratic equation in e x ), Question 6 (variable
separable differential equation), Question 7ii (calculus and trigonometry), Question 9b (finding the
argument of a complex number) and Question 10i and iii (vectors).

Key messages for candidates

● Read the questions very carefully.

● If the question asks for full and exact working then the method must be clear at each stage, and
calculator approximations will not be accepted.
● If the answer is given in the question then full working is expected.
● Never underestimate the need to practise basic arithmetic and algebraic processes – many candidates
lose marks through elementary errors.
● Use brackets correctly when evaluating complex calculations on the calculator, and when writing down
formulae.
● Never write one solution on top of another (even if you think that you have erased the original) – when
the work is scanned it is often illegible.

Where numerical and other answers are given after the comments on individual questions that follow it
should be understood that alternative forms are often acceptable and that the form given is not necessarily
the only ‘correct answer’.

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Comments on specific questions

Question 1

This question gave many candidates a confident start to the paper. A few drew diagrams or rewrote the
modular inequality as two linear inequalities, but the majority started by squaring both sides. A large number
of candidates reached the correct critical values. Some candidates gave a decimal approximation to the
negative root, which resulted in an inaccurate final answer. The most common errors were in squaring the
brackets and in simplifying to obtain a quadratic equation/inequality. Some candidates did not square the 3.
1 7
The incorrect answer − > x > was a common error. Similarly, some candidates thought that it was
7 5
1 7
possible for “ x < − and x > “.
7 5
1 7
Answer: x < − , x >
7 5

Question 2

The candidates recognised that the first step required was to expand sin (θ − 30°) . This was often done
correctly, although several candidates made a sign error, or used the expansion of cos (θ − 30° ) . Some
candidates scored no marks at all because they started with sin (θ − 30°) = sin θ − sin30° .

In simplifying the equation, some candidates overlooked the "+ cos θ " , but the majority went on to form an
equation in tan θ. Some candidates were unsure of the properties of tan θ and gave an additional incorrect
answer of 156.2°.

The alternative method of solving cos θ + ( )

3 − 4 sin θ = 0 by first expressing the left hand side in the form
R cos (θ + α ) , or equivalent, was seen from a small number of candidates, and was usually completed
correctly.

Answer: θ = 23.8°

Question 3

(i) Most candidates did attempt to use integration by parts, and the majority obtained an answer with
terms of the correct form, although several had difficulty with the coefficients, making sign errors
and placing the 2 in the numerator. Some candidates made errors in integrating x −3 , obtaining
answers involving x 4 or x −4 .

(ii) This was the first stage in the paper where a significant number of candidates offered no response.
The first mark was available to any candidate who had an answer of the correct form in part (i). As
the answer is given in the question, the candidates were expected to explain how they moved from
an expression involving ln 2 to one involving ln 4 .

ln x 1
Answer: (i) − 2
− 2
2x 4x

Question 4

Those candidates who recognised that the given equation simplified to a quadratic equation in e x usually
went on to obtain a value for e x . Some candidates worked on an incorrect quadratic equation, usually
( )
because they had 4 e x + 1 = 4e x + 1 .

The question asks candidates to show “all necessary working”, so some indication of the method for
obtaining the final answer was expected. Some candidates gave their final answer to 3 significant figures
rather than 3 decimal places, and some attempted to obtain a solution for x from a negative value of e x .

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

A significant minority of candidates made no progress at all because they attempted to apply incorrect rules
of logarithms to the original fraction, or to e x + e − x = 4e x + 4 , in order to remove the exponentials.

Answer: x = −1.536

Question 5

(i) Most candidates made a correct start to this question by attempting to use the product rule for
d 1
differentiation. The error ln ( 8 − x ) = was common, and very few candidates traced their
dx 8−x
error when they could not obtain the given answer. Processing errors meant that several
candidates with the correct derivative did not reach the given answer.

(ii) There were many correct solutions to this question, but also evidence of many candidates not
really understanding the difference between looking for a root of f ( x ) = 0 and looking for a root of
8
f ( x ) = x. There were several instances of candidates using f ( x ) = 8 − , obtaining
ln ( 8 − x )
f ( 2.9) = 3.09 and f ( 3.1) = 2.97 and concluding that there was no root in the interval because there
was no sign change.

(iii) The majority of candidates applied the process for using the iterative formula correctly. Some
candidates did not work to the accuracy required by the question, but many correct answers were
seen. Some solutions were longer than necessary; given that the root is known to lie in the interval
( 2.9, 3.1) candidates should have been working from a starting point in that interval, but this was
not always the case.

Answer: (iii) 3.02

Question 6

This should have been a straight forward question about a variable separable differential equation. The first
key step was to express the statement of proportionality as an equation. Some candidates did this, but the
constant was often implied to be equal to 1, or an incorrect value was used. The incorrect constant usually
came from using the gradient of the chord joining the two given points.

Most candidates achieved the next step, separation of the variables, correctly.

1 1
The majority of candidates completed ∫ x dx correctly, but ∫y 2
dy proved to be more difficult. Many

candidates gave the answer as a multiple of ln y 2 .

1 1
Those candidates who completed the integration correctly usually reached the equivalent of = 1 − ln x .
y 2
Some candidates stopped at that point, but most went on to rearrange the equation to express y in terms of
x. This process generated several algebraic errors.

2
Answer: y =
2 − ln x

Question 7

(i) The candidates recognised the need to start by differentiating the equation of the curve. Most
candidates used the quotient rule, although some preferred the product rule. Several candidates
made errors in quoting the rule they chose to use, and some used only the numerator of the
quotient rule without giving any justification for this. There were many errors in moving from the
derivative to an equation in sin x. Several candidates used “invisible brackets”, writing down terms
like 2 + sin x. − 3 sin x that were accepted if they were then multiplied out correctly at the next

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

stage. However, the majority confirmed that they had not intended to use brackets at all. Some
candidates multiplied out the terms in the numerator and the denominator and then “cancelled” the
terms in sin x and sin2 x . Many of the candidates who did reach 6 sin x + 3 = 0 did not obtain a
negative value for sin x , and those who did have a negative answer often ignored the minus sign.
Although the question asks for the exact values of the coordinates, many candidates gave only
decimal approximations.

(ii) Many candidates did recognise the form of the integral correctly. The common errors were to
attempt to use integration by parts, or to integrate the numerator and denominator separately.
Candidates with correct integration often reached a correct final answer. Candidates with errors in
their working who found that sin a > 1 did not seem to realise that this indicated an error in their
working. A common error in the final stages of the solution was the incorrect statement
ln ( 2 + sin x ) = ln2 + ln ( sin x ) .

π
Answer: (i) x = − , y = 3 (ii) a = 0.913
6

Question 8

(i) This question provided a welcome source of marks for many candidates, the great majority of
whom selected a correct form for their partial fractions. Most did attempt to split the original into
three parts, although several opted for the two term alternative. Most errors were caused by slips in
the arithmetic rather than by the use of incorrect methods.

The expansion of (1 − 2x ) was often correct, although there were errors in simplifying ( −2 x ) .
−1 2
(ii)
The expansion of ( 2 − x ) and ( 2 − x ) was more challenging, with several attempts to expand
−1 −2

−1
 x
2  1 −  or even 2 (1 − x ) . Correct expansions were not always combined with the correct
−1

 2 
coefficients, and the correct final answer was not common.

1 3 2 9
Answers: (i) + − , (ii) 2 + x + 4 x 2
1 − 2x 2 − x ( 2 − x ) 2
4

Question 9

(a) (i) Most candidates understood exactly what they needed to do to divide by a complex number, and
many obtained the correct answer.

(ii) The majority of candidates understood how to obtain the values of r and θ , with many reaching a
correct value for r. Only a minority of candidates obtained the correct value for θ ; despite the
π
frequent use of sketch diagrams, the most common answer was θ = − .
4

(b) The majority of candidates recognised z − 3i = 1 as a circle, and most drew a circle of the correct
size in the correct place. Although the question asked for a “sketch” of an Argand diagram,
candidates were expected to use even scales on their axes. The scales on the two axes should be
identical, but unequal scales were acceptable so long as the shape drawn looked like an ellipse.

Many candidates did not attempt to find the greatest value of argz . Those who did offer a solution
usually overlooked the fact that it is the angle between the tangent and the radius that is a right
1 1
angle; many attempts involved tan−1 in place of sin−1 .
3 3

3
Answers: (i) −2 + 2i , (ii) r = 2 2, θ = π , (iii) 1.91
4

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 10

(i) Although there have been questions in the past using the form r.n = a.n , many candidates did not
recognise this use of (r − a ) .n = 0. Many simply ignored the −i − 2 j , and some worked with a
vector ri − j − 2k.

(ii) The majority of candidates were familiar with the process for finding the angle between two vectors,
although several did not work with the direction vector of the line and the normal to the plane. Many
of the candidates who were working with the correct vectors found the angle between the line and
the normal to the plane and did not then go on to find the angle between the line and the plane.

(iii) Several candidates offered no attempt at this part of the question. Those candidates who identified
that they needed to find a vector perpendicular to both the direction of the line and to the normal
vector of the plane usually had a correct method to achieve this. Provided they had an answer to
part (i) they could complete the task. The final mark was only given for answers expressed in the
correct form.

Answer: (i) 2i + 3 j − 4k , (ii) 14.3° , (iii) r = 2i + 3 j − 4k + µ ( 3i − 2 j − 7k )

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/33
Paper 33

Key messages

Candidates need to know:

• that they cannot take logs of negative numbers
• exactly what is meant by a tangent parallel to the x-axis or parallel to the y-axis
• how to sketch a cubic graph
• how to use the chain rule for differentiation of a trig function such as sin3(4x)
• that a 3-term partial fraction with two linear terms and one linear term squared for the denominators
should have only three unknown coefficients. Introducing Dx + E in the numerator above the squared
linear term is over-specifying the system.

General comments

The standard of work on this paper varied considerably, although all questions were accessible to the
stronger candidates. A significant number of candidates found certain questions extremely difficult.
Candidates are reminded that working carefully through past papers and mark schemes will give them a
good idea of what to expect in the examination.

Questions or parts of questions that were generally done well were Question 2 (solution of exponential
equation), Question 3(ii) (iterative convergence), (iii) (iterative formula to find root) Question 4(i)
(parametric equations), Question 5 (solution of differential equation), Question 6(ii) (solution of trig
equation), Question 9(i) (partial fractions) and Question 10(ii) (angle between planes). Questions that were
done less well were Question 3(i) (sketching graphs), Question 4(ii) (finding where tangent to curve is
parallel to y-axis), Question 6(ii) (solving trig equation), Question 7(i), (ii) (establishing stationary point on
curve given by a trig function and determining the exact area under this curve), Question 9(ii) (integration of
partial fractions) and Question 10(iii) (finding position of point on a line a given distance from a plane).

In general the presentation of the work was good, though there were some rather untidy scripts. Candidates
should bear in mind that scripts will be scanned for marking and they should use a black pen, reasonable
sized lettering and symbols, and present their work clearly. Candidates should avoid using ink that is
absorbed into the paper and then appears on the reverse side as this can make it difficult to read the pages
when scanned.

It was pleasing to see that candidates are aware of the need to show sufficient working in their solutions.
Previous reports mentioned this in the context of solving a quadratic equation and substituting limits into an
integral. In recent years additional rubric has been added to individual questions, for example ‘show all your
working.’ This is to ensure that candidates can demonstrate understanding of the methods even when their
calculator is capable of finding the answer. For example, modulus of a vector, scalar product of two vectors,
product or division of two complex numbers, modulus or argument of complex number. No credit is given to
answers written directly from the calculator, without working. This will be even more important when the
modified syllabus commences, since it has a new rubric ‘no marks will be given for unsupported answers
from a calculator’. Very strong candidates who can do some of these tasks mentally should also be
encouraged to show the steps in their solution in order to gain credit.

Where answers are given after the comments on individual questions, it should be understood that the form
given is not necessarily the only ‘correct answer’.

Comments on specific questions

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 1

This question proved to be challenging for candidates. The omission of the parameter a was usually the
source of the problem. Many candidates failed to square the 2 outside the modulus sign or poor algebra
prevented them from arriving at the correct equation. Although many candidates took the quadratic equation
approach, this question is more easily solved by considering the pair of linear equations or linear inequalities.
Many candidates who successfully obtained a correct quadratic equation ran into problems because they did
not have the correct power of a where required. Factorising was generally more successful. Some
x
candidates decided to set a to unity, i.e. to solve for , but they did not reintroduce a after solving their
a
a 5a
equation. Those who did obtain x = − and usually made the correct choice of region at the very end.
5 3

a 5a
Answer: − <x<
5 3

Question 2

This question was usually well done, however there were still many candidates who did not reach an
b
equation of the form ae2x = b, aex = be−x or ae x = x with the correct values of both a and b. Many arithmetic
e
and algebraic errors were seen, as well as multiplication of indices to give an exponential term to the power
x2. Most candidates knew they then needed to take lns but some did not show this step despite the additional
instruction to show all necessary working. Too many candidates had a correct equation, but with a and b
negative. Then, before moving to an equation with positive terms, they attempted to take lns. Candidates
should be aware that they cannot take lns of negative numbers

Answer: 0.46

Question 3

(i) This part proved challenging more many candidate, for various reasons. The graph of y = x3 should
be sketched for at least −2 ⩽ x ⩽ 2 to show its shape. Stopping at x = −1 does not highlight how the
graph behaves for negative x. Similarly the graph of y = 3 – x should be sketched for at least
–1 ⩽ x⩽ 4. However, leaving the graphs in this form is still insufficient, and adding the comment
‘there is exactly one real root’ is not enough. The question required a clear indication of the single
point of intersection with a dot or some other mark, together with a comment that there is only one
point of intersection so only one root. This final statement should be clearly made without it being
left to the examiner to interpret.

(ii) Convergence, starting with either form, was usually well done. As it is a proof there should be no
jumps in the working; it is better to show too many lines than too few.

(iii) Most candidates found the root to the required accuracy but occasionally candidates left their
answer to 2 d.p. or 4 d.p.

Answer: (iii) 1.213

Question 4

dx dy
(i) Most candidates knew what was required and were successful with either or . However
dθ dθ
many did not apply the chain rule correctly to either sin 2θ or cos 2θ. Although the question didn’t
ask for the expression in a particular form, most candidates tried to simplify it into an expression in
just sinθ and cosθ. This often introduced errors and if candidates continued to use this form into (ii)
little progress was made in that section. Fortunately most candidates returned to their initial
expression that was hopefully free of errors.

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

(ii) This is where candidates really ran into trouble. Most candidates believed that this question must
dy
be about finding where = 0 , 1 or ∞, with about half using 0, 10% using 1 and the remainder
dx
using ∞. If candidates cannot reason this mentally they should draw a small sketch to help
themselves, as a wrong guess meant zero out of four immediately. Most of the candidates making
the correct choice scored three or four marks, since some stopped after finding the value of θ.
Some candidates factorised their quadratic incorrectly or had a sign error in their double angle
formula.

( 2sinθ + 2sin2θ )  3 3 1
Answers: (i) − (ii)  , 

( 2cosθ + 2cos 2θ )  2 2

Question 5

This proved a high scoring question as most candidates could separate variables correctly and obtain the ln
2
y term. The integral in x proved more difficult, since instead of considering and x, some candidates
x
resorted to integrating by parts and increased the possibility of sign errors. A considerable number of
candidates left the expression as ln y instead of obtaining an expression for y as requested in the question.
Candidates who left the term exp(2ln x) in their final answer were penalised in their final mark.

1 1 
Answer: y = x 2 exp  − x 2 
2 2 

Question 6

(i) Many candidates found it difficult to obtain the given expression, taking far more steps than were
required. Fortunately most did achieve it, although several had sign and coefficient errors in their
cos x and sin x terms. Done correctly, R cos α and R sin α should have been 3 and 1
respectively, and the working for R and α should not have had any negative signs present.
However, a large number of candidates simply quoted the expressions R = (a 2
+ b2 ) and

b
α = tan−1   , with (−1)2 seen in the expression for R. Whilst examiners allowed this error in the
a
 
evaluation of R = 2 they were less generous if the sign error was present in finding α. There is a
similar issue arising in Question 9(i) when determining the partial fractions. In the present case the
use of the formula for R, which is perfectly valid, has covered up the incorrect mathematics. A few
candidates made the error of writing cos α = 3 and sin α = 1 (omitting R), leading incorrectly to the
1
correct expression tan α = and α = 30°.
3

(ii) A few candidates did not spot the link between (i) and (ii) and as a result were unable to make any
 2
progress. However, most candidates readily found the values of sin−1   and hence both
 R 
required solutions. Some incorrectly believed that there was only one angle arising from
 2
sin−1 
 R  and that the other angle should come from 180° minus their correctly found angle.
 

Answer: x = 75° and 165°

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 7

(i) This was a question where candidates usually scored either just a single method mark, for using
the product rule, or full marks. Correct solutions required candidates to apply the chain rule
accurately twice. It was still possible to proceed to a solution if the errors in the chain rule were
restricted to the coefficients. Other errors such as incorrect powers of trig functions meant that it
was impossible to reduce the equation to one that could be solved.

(ii) Nearly all candidates struggled with this question despite being given the substitution required. This
was evidenced by the amount of working most produced, with little progress made. Common errors
were poor substitution of du = cos x dx, inability to express cos2 x correctly in terms of u, substituting
π
limits 0 and for u or substituting limits 0 and 1 for x. In fact very few candidates reached the
2
correct integrand in terms of u, together with the correct limits.

2
Answers: (i) x = 0.685 (ii)
3

Question 8

(a) The rubric demanded that candidates show detailed working in order to gain marks. Most
candidates divided out the complex numbers then converted to polar form, as opposed to
converting to polar form and then dividing out. When multiplying the numerator and the
denominator by the conjugate of the denominator, candidates often had errors either in the
denominator, e.g. 1 + (2i)2 instead of 1 – (2i)2, or arithmetic errors when collecting the real or
imaginary parts in the numerator. Basic errors such as these meant that none of the accuracy
marks could be gained. However, many candidates who performed the division correctly then left
their modulus in the exact form instead of correct to 3 significant figures. Few candidates managed
to obtain the correct argument, the usual answer being θ = −1.05, something that candidates
arrived at from their calculators without any reference to the position of the complex number
−4 7
+ i in the Argand diagram. A quick sketch would have helped them.
5 5

(b) Many candidates scored full marks, although some restricted their solution to the Argand diagram
part only, with no attempt at the least value of z . Several candidates finished with their circles in
the wrong quadrant, whilst others did not indicate that the radius of the circle was unity.

Answers: (a) 1.61ei2.09 (b) 13 − 1

Question 9

(i) This question was extremely well done by most candidates. However, a few candidates introduced
a constant term into their partial fractions, something only necessary when the powers of the
numerator and denominator are the same. A few others incorrectly used a linear term as opposed
to a constant term for the numerator of the term with the denominator (3 + 2x)2 in the 3-term partial
fractions. This form will see two of the constants determined uniquely but only a combination of the
other two as unique. Candidates need to check very carefully when they remove the denominator
in order to find their constants, since slips are possible, e.g. multiplication by a factor from the
denominator that is of the incorrect power. Any error at this point leads to errors in the three
equations obtained by equating the coefficients of the powers of x and incorrect constants.
However, if the candidate opts instead to construct their equations by using the roots of the factors
of the denominator then in some cases the incorrect terms will vanish and the correct values of the
constants will be obtained. Unfortunately, these cannot be given credit since the equations were in
error and it is only by the incorrect terms dropping out that correct answers have been obtained.
This feature is like that arising in Question 6(i) where the correct solution only resulted from the
squaring of an incorrect negative sign.

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

(ii) Integrating the partial fractions proved testing, as many candidates had the incorrect coefficients for
both their ln terms. Hence it was possible to see some candidates who had the correct forms for all
three integrals gaining no marks due to mistakes in determining the appropriate coefficients. In
obtaining the given answer, it is necessary for candidates to ensure that there are no gaps in their
working; again too much detail is better than not enough.

1 1 3 1 2x
Answers: (i) − + or −
( 2 − x ) ( 3 + 2 x ) ( 3 + 2 x )2 ( 2 − x ) ( 3 + 2 x )2

Question 10

(i) All details of the working using the scalar product were required. Most candidates were able to
show the scalar product was zero, but omitted to verify that one point of l does not lie in the plane.
If, instead, the candidate substituted coordinates of a general point of l in the equation of the plane,
it was necessary to establish and state there is no point that satisfies this equation. Again few
candidates made the concluding statement.

(ii) Candidates usually showed the full details for their scalar product and moduli hence even with the
presence of errors it was possible for them to gain both method marks.

(iii) Those candidates who knew the formula for the distance of a point from a plane usually quickly
scored full marks, whilst most others struggled. There were a few candidates who didn’t appear to
know this formula yet made a reasonable attempt by finding where the normal vector from a point
on the line met the plane and then setting the distance between these points equal to 2. Others
appeared to think they had solved the problem when they found where the line met the plane at (5,
3, 3) and stated this as their answer. In fact they still had to find the distance between this point and
2
a point on the line and set this equal to the length , where α is the acute angle between the
( cos α )
direction of the line and the normal vector to the plane.

Answers: (ii) 74.5° or 1.30 radians (iii) 7i + 5j + 7k and 3i + j − k

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/41
Paper 41

Key messages

● Non-exact numerical answers are required correct to three significant figures as stated on the question
paper. Candidates should be reminded that if an answer is required to 3 sf then their working should be
performed to at least 4 sf. This was often seen to affect answers to Questions 1, 2, 3, 4, 5 and 6 in this
paper, and led to particular problems in Question 5(ii)
● When answering questions such as Question 7(i), 7(ii) and 7(iii) on this paper, where the acceleration
is given as a function of time, the equations of constant acceleration cannot be used.
● In questions such as Question 4, where the motion of connected particles takes place, or Question 6,
involving motion on an inclined plane, it is always advisable to draw a force diagram. In order to give the
answer more clarity, candidates should also state the origin of their equations, e.g. applying Newton’s
law to particle A along the plane.

General comments

There were some candidates who produced very good answers on this paper but also a significant number
who scored low marks. Overall a wide range of performance was seen.

The examination allowed candidates at all levels to show what they knew, whilst differentiating well between
the stronger candidates. Question 1 was the question candidates found easiest whilst Question 6 proved to
be the one candidates found most challenging.

One of the rubrics on this paper is to take g = 10 and virtually all candidates followed this instruction. In some
cases it is impossible to achieve a correct given answer unless this value has been used.

Comments on specific questions

Question 1

Most candidates made a good attempt at this question. It was first necessary to use the given information to
determine the acceleration of the particle. As all of the given forces were constant, the constant acceleration
equations could be used. Once the acceleration was found, Newton’s second law of motion could be used to
enable the required value of F to be found.

Answer: F = 1.58

Question 2

(i) As the train in this question was travelling at constant speed there was no net force acting on it.
This means that the driving force provided by the engine is exactly balanced by the resistance
force. Using the given values of the power and the speed, and applying Resistance force = Driving
P
force = , the resistance force could be found. Some candidates made the error of thinking that
v
the driving force was directly related to the weight of the train.

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

(ii) In this part of the question the train moved up a hill but maintained its speed of 85 m s–1 and so
again the forces were balanced. Newton’s second law had to be applied to the train with
acceleration = 0. As the train was going up a hill there was a force acting due to the component of
the weight of the train acting against the direction of motion. This force takes the form
1
490 000 g sin θ = 490 000 g × . The three forces, namely the driving force, the weight
200
component and the resistance, were balanced. Application of Newton’s second law takes the form
P 1
– 48 000 – 490 000 g × = 0. Some candidates forgot that the resistance force was still
85 200
acting. Others did not include the weight component. Some also forgot to include the factor of g in
the weight component term.

Answers: (i) 48 000 N (Answer given) (ii) 6.16 MW (to 3 sf)

Question 3

Many candidates made the incorrect assumption that the driving force provided by the van’s engine was
constant. This is not given in the question and so cannot be assumed. In cases such as this, it is necessary
to use work and energy principles rather than Newton’s second law. There are several stages to go through
in order to solve this problem. Firstly there is a loss of potential energy (PE) as the van descends the hill and
this could be evaluated as PE loss = 2500 × g × 400 sin 4. Secondly there is a gain in kinetic energy (KE) as
1
the speed increases and this is given by KE gain = × 2500 × (302 – 202). Thirdly the work done against the
2
constant resistance force is given by WD against resistance = 600 × 400. Finally these three elements had to
be combined using the work-energy principle in order to find the required work done by the van’s engine.
This takes the form: WD by van’s engine + PE lost = KE gain + WD against resistance. Substituting the
values found earlier enabled the required work done to be found. Many candidates made the wrong
assumption that the driving force acting on the van was constant and proceeded to use constant acceleration
equations and lost some marks in doing so. Some candidates who used the work-energy principle did not
include all of the terms, while others used incorrect signs within their equation.

Answer: 167 000 J

Question 4

(i) In this question involving a simple pulley it was a good idea to draw a diagram showing the
situation and the forces acting on each particle. The question gave the distance travelled by the
particles before particle A reached the ground as 0.8 m, and the speed of particle A as it reached
the ground as 0.6 m s-1. This enabled the acceleration of the particles to be found by using the
constant acceleration equations with u = 0, s = 0.8 and v = 0.6 and this gave a = 0.225. When
applying Newton’s second law to particle B in the form T – 3 = 0.3 a where a = 0.225, the tension T
in the string could be found. Most candidates found the acceleration but some involved the
unknown mass m at this stage rather than applying the equations to the 0.3 kg particle.

(ii) With the value of T determined, the equation of motion for the particle of mass m kg in the form
mg – T = ma could be used with the values of a and T found in part (i). This enabled the value of m
to be determined. Since the question stated that particle A reached the ground, it was clear that
m > 0.3 and so the equations of motion took the form T – 3 = 0.3a and mg – T= ma. Some
candidates took the signs to be opposite to those stated here and found the tension and mass to
be negative.

Answers: (i) 3.07 N (to 3 sf) (ii) m = 0.314 (to 3 sf)

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 5

(i) Most candidates made a good attempt at this question. It is important when solving this type of
problem for candidates to state the directions in which they are resolving. The majority of
candidates resolved forces horizontally and vertically and found the components of the system in
each of these directions. The majority of candidates also made good attempts to find the resultant
force, generally using Pythagoras. Very few candidates correctly identified the direction of the
force. Although an angle was found in most cases, this angle had to be correctly described and the
most straightforward method was by drawing a diagram.

(ii) Candidates performed well on this part, realising that the component of the 25 N force to the left
had to balance the component of the F N force to the right and that this gave an equation which
enabled F to be determined. The final part of this question involved combining the vertical
components of the three forces and is a very good example of making sure that full accuracy is
maintained throughout the calculations. Although the individual components of the forces were
large, the final answer when combining the three forces was very small and so any premature
approximation made during the calculation led to an inaccurate final result. This affected a
significant number of candidates. All calculations should be kept to at least 4 sf when performing
these calculations.

Answers: (i) 12.8 N; Direction: angle of 37.7° (to 3 sf) below the negative x-axis
(ii) F = 28.3 (to 3 sf) The magnitude of the new force is 0.667 N (to 3 sf) in a direction vertically
upwards

Question 6

(i) Most candidates found this question difficult to answer. When the result is given in the question, as
it is here, candidates must be particularly careful when setting out their responses. There were
several different approaches which could be taken to this question. One method was to use the
1
constant acceleration equation s = ut + at2 applied to the section PQ and then to section PR.
2
This produced two simultaneous equations involving the variables initial speed, u, and the
acceleration, a. Solving these two equations gave the required value of u and confirmed the given
value of the deceleration. Many candidates applied the equation to the sections PQ and QR but
wrongly assumed that the initial speed was the same in both cases, leading to incorrect solutions.
Some candidates took the given value of the deceleration and used it to find u but this did not
provide a proof for the value of the deceleration.

(ii) In this part of the question a force diagram was particularly helpful. The motion was given to be on
an inclined plane. Even candidates who could not prove the result from part one could now assume
the given value of the deceleration and use it when applying Newton’s second law. The normal
reaction on the particle could be shown to be R = mg cos 3 and Newton’s second law takes the
2
form –mg sin 3 – µR = m × – . This equation could be solved for the required coefficient of friction,
3
µ. Most candidates attempted this but often with incorrect signs in their equation.

2 23
Answers: (i) a = – m s–2 (Answer given) u =
3 15
(ii) µ = 0.0144 (to 3 sf)

Question 7

(i) This question gave the acceleration as a function of time, t. In cases such as this it is not possible
to use the constant acceleration equations and calculus techniques must be used. In order to find
an expression for velocity, the given expression for a had to be integrated and the given condition
that the particle starts from rest used. This expression for the velocity had to be set to zero and the
resulting equation solved in order to answer the question. Note that the answer had to be given as
an exact fraction. Some candidates did not do this and hence lost marks.

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

(ii) In this part the expression found in part (i) for velocity had to be evaluated at t = 10. The velocity-
time graph could be drawn using the information that v = 0 at t = 0, the value of v found at t = 10 and
20
the fact that from part (i) it is known that the velocity is again zero at t = . Also the expression for
3
v is a quadratic function of t which means that the graph is the shape of a parabola. Using all of this
information the v-t graph could be drawn.

(iii) Integration of the expression for velocity was required here to find the distance travelled. However,
20
from the v-t graph it could be seen that the part of the region from t = 0 to t = consisted of a
3
20
curve above the t-axis while that from t = to t = 10 lay below the axis. In order to find the total
3
distance travelled over the first ten seconds these two areas had to be evaluated separately,
remembering that when evaluating areas below the t-axis the answer will be negative

20 2
Answers: (i) t = =6
3 3
20
(ii) The v-t graph is an inverted parabola, passing through the points (0,0), ( , 0) and (10,–27)
3
(iii) 80 m

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/42
Paper 42

Key messages

● Non-exact numerical answers are required correct to three significant figures as stated on the question
paper and cases where this was not adhered to were seen in Question 1, Question 2, Question 6 and
Question 7. Candidates would be advised to carry out all working to at least 4sf if a final answer is
required to 3sf.
● When answering questions involving an inclined plane, a force diagram could help candidates to include
all relevant terms when forming a Newton’s Law equation or a work/energy equation. This was
particularly noticeable here in Question 6 parts (ii) and (iii).
● In questions such as Question 5 in this paper, where acceleration is given as a function of time, then
calculus must be used and it is not possible to apply the equations of constant acceleration.

General comments

The paper was generally well done by many candidates although as usual a wide range of marks was seen.

The presentation of the work was good in most cases and as the papers are now scanned, it is important to
write answers clearly using black pen.

In Question 4 the tangent of an angle which is used in the question was given. In questions such as this it is
not necessary to calculate the angle itself as the sine and cosine of the angle which are required in order to
achieve the question could be evaluated exactly. However, many candidates often proceeded to find the
relevant angle to 1 decimal place and immediately lost accuracy and in some cases lost marks.

The examination allowed candidates at all levels to show what they knew, whilst differentiating well between
even the stronger candidates. Question 2 was found to be the easiest question whilst Question 7(ii) proved
to be the most challenging for almost all candidates.

One of the rubrics on this paper is to take g = 10 and it has been noted that virtually all candidates are now
following this instruction. In fact in some cases such as question 4(i) it is impossible to achieve a correct
given answer unless this value is used.

Comments on specific questions

Question 1

In this question it is vital that a force diagram is drawn so that the directions of the tension forces acting on
the ring can be seen. Some candidates wrongly assumed that the tension forces in the two parts of the string
were different but as it is a continuous string the forces must be the same. There are two possible
approaches to this question. One method is to resolve forces horizontally and vertically. Alternatively the
resolution of forces could be made along the direction of the 2.5 N force and along the direction
perpendicular to this force. The horizontal resolution does not involve the mass, m, of the ring and
immediately gives the value of the tension, T, in the string. Vertical resolution immediately gives the value of
the mass, m, of the ring. If forces are resolved along and perpendicular to the 2.5 N force, this produces two
simple simultaneous equations for T and m which can readily be solved. It is possible to solve the problem
using Lami’s theorem but two of the forces must be combined as it is currently a 4–force system. Some good
solutions were seen to this question and the major problem was where candidates assumed that two
different tensions were acting in the string
2
Answers: The tension in the string is 1.25 N m = = 0.177 (to 3sf)
8

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 2

Almost all candidates made a good attempt at this question. Firstly the normal reaction must be found in the
form R = 5g cos 6. Since the block moves with constant speed there is no net force acting on the block. The
equilibrium equation can be written as T = 5g sin 6 + µR where T is the required tension in the rope. On
substituting the given value for µ and the expression for R this gives the required value of the tension in the
rope. Errors that were seen included misreading 6º as 60º and having the wrong sign on the µR term.

Answer: The tension in the rope is given by T = 20.1 N (to 3 sf)

Question 3

(i) This was a very straightforward use of the given velocity-time graph and candidates had to use the
fact that the gradient of the graph represents the acceleration. Some candidates appeared to mix
up the signs of acceleration and deceleration since the gradient of the graph over this period is
negative. This means that if acceleration is stated then it is negative whereas if deceleration is
quoted then it is positive.

Answer: The acceleration of the particle over the first 2 seconds is –1 ms–2

(ii) Again most candidates made a good attempt at this question. It is necessary to use the graph
either between t = 4 and t = 10 or between t = 6 and t = 10, since the acceleration is constant over
those periods. The required result can be achieved either by use of the constant acceleration
V (V + 2)
formulae or simply by use of similar triangles in the velocity-time graph such as = .
4 6

Answer: V = 4

(iii) Some candidates had difficulty with this part of the question, particularly in understanding how to
use the one third factor that was referred to in the question. It was necessary to evaluate the area
beneath the graph between the times t = 0 and t = 6, since this represents the distance that the
particle has travelled, which is one third of the distance AB. Multiplying this area by 3 gives the
required distance AB. At t = 6 the particle reaches A and its direction of travel reverses. It then
moves from A to B, with the area under the triangle between t = 6 and t = T representing the
distance AB. The area of this triangle is then equated to three times the distance travelled in the
1
first 6 seconds, leading to an equation for T. One error seen was to mix up the factors of 3 and
3
when setting up the equation for T. Some candidates lost marks because although they found the
correct value of T, they did not fully answer the question by specifically stating the distance AB.

Answers: Distance AB = 24 m T = 18

Question 4

(i) This is a case of a given answer and so care must be taken to show the full working of this
problem. It is also a case where the tangent of the angle α is given and so throughout the question
there is no requirement for the specific angle to be found. The sine and cosine of this angle can be
3 4
evaluated exactly as sin α = and cos α = . In addition, if candidates did not use g = 10 in this
5 5
problem then they would not reach the given answer. As it is given that P remains stationary, the
system is in equilibrium but with particle P on the point of moving down the plane. As particle Q is
stationary, the tension in the string T = 0.7g = 7. The normal reaction at P is given by R = 0.4g cos
α = 3.2 and the friction force F = µR = 1.6 The weight component acting down the plane is given by
0.4g sin α = 2.4 and hence the equilibrium condition can simply be stated as X + 2.4 = 7 + 1.6 and
hence the given result follows. Many candidates performed well on this part, with the main loss of
marks being due to rounding errors when using the angle as 36.9 degrees. Some candidates
misread the direction of the force X, some taking it vertically downwards and others assuming that
it was acting in a direction perpendicular to the plane. Some candidates introduced acceleration
into their calculations even though the question stated that the particles were stationary.

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Answer: X = 6.2 (answer given)

(ii) In this part of the question, since X has a given value which is smaller than in 4(i) with X = 0.8, the
particle P must begin to move up the plane and particle Q must move vertically downwards. It is
necessary to write down the equations of motion for both P and Q or to consider the system of the
two particles. If T is the tension in the string and a is the magnitude of the acceleration of the
particles, then the equation for the motion of P takes the form T – µR – 0.4g sin α – 0.8 = 0.4a,
while the equation for particle Q is 0.7g – T = 0.7a. By substituting the given values of µ and α, the
tension can be eliminated and the value of the required acceleration can be found. Some common
errors seen were to forget to include some of the terms in the equation of motion for P or , where
the system equation was used, often the forces acting on the system were equated to 0.4a
whereas they should have been equated to (0.4 + 0.7)a

Answer: The acceleration of P is 2 ms–2

Question 5

(i) In this question the acceleration, a, is given as a function of t and so calculus must be used. The
question asks for the time at which maximum velocity is reached and this happens when a = 0.
Many candidates found difficulty manipulating the fractional power which occurs in the expression
for a when attempting to solve this equation. Some candidates unnecessarily found an expression
for the velocity before considering the equation a = 0. The majority of problems encountered by
candidates in this question related to algebraic errors in the manipulation of the equation.

Answer: T = 4

(ii) In order to find when the acceleration is a maximum, one approach is to differentiate the given
expression for a and to set this to zero and solve to find the required value of t. This was the
method used by most candidates, although some realised that the given expression for a is a
quadratic in √t and so the method for finding the vertex of a quadratic was also seen. Once the time
at which maximum acceleration occurs has been found, the expression for a must be integrated in
order to produce an expression for the velocity. Care is needed here since the velocity at t = 0 is
given as v = 1 ms–1 and this must be used to find the constant of integration. The value of t at which
the maximum acceleration takes place can then be substituted into the expression for v in order to
complete the answer. Most candidates found the correct expression for v by using integration.
Some forgot to use the given condition to determine the constant of integration. Many did not
differentiate a in order to find the time at which maximum acceleration occurred but instead used
the value of t found in 5(i).

Answer: The velocity of the particle when the acceleration is maximum is 1.5 ms–1

Question 6

(i) In this problem the car is moving at a constant speed and so there is no net force acting on it. This
means that the driving force exactly balances the resistance force. In order to find the required
power of the car’s engine, use must be made of the formula P = Fv where the driving force F is the
resistance force 350 N and v is the given speed. Almost all candidates successfully solved this
problem.

Answer: The power of the car’s engine is 5250 W

(ii) In this part of the question the car is accelerating under the effect of three forces, namely, the
driving force, the component of the weight of the car and the resistance force. The equation of
motion for this situation can be written as P/15 + 1200g sin 1 – 350 = 1200 × 0.12 and by
rearranging this equation, the required power P could be found. Some errors seen included either
ignoring the weight component or taking the wrong sign for the weight component term, and
forgetting to include the resistance force. However, most candidates performed well on this part.

Answer: The new power of the car’s engine is 4270 W (to 3 sf)

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

(iii) There are two distinct methods which can be applied to solve this problem, either being acceptable.
As the only forces acting are constant, namely the weight component and the resistance, Newton’s
second law can be applied in the form 1200g sin 1 – 350 = 1200a to determine the deceleration of
the car as a = –0.117. Once this has been found, the constant acceleration equations can be
applied with the values u = 20, v = 18 and a = –0.117. Use can be made of the equation v2 = u2 + 2
as and this can be solved for the distance, s, travelled. Alternatively the problem can be solved
using work-energy methods. The PE loss can be shown to be 1200g × s sin 1, the KE loss is ½ ×
1200 × (202 – 182) and the work done against the resistance force is 350s. In this situation the work
energy principle can be written as PE loss + KE loss = WD against friction. Both of these methods
were seen. Errors that occurred when using Newton’s laws included missing terms, a driving force
still included in the equations and the factor of g missing in the weight term. When using the energy
method some candidates used the correct terms but with wrong signs and others mixed up units by
including the resistance force 350 N in the energy equation rather than the work done of 350s J.

Answer: The distance travelled = 324 m (to 3 sf)

Question 7

(i) This question specifically requests an energy approach and so any candidate who did not use this
method was penalised. There are several approaches to the problem. One method is to find the
speed of the particle as it reaches the surface of the liquid. This can be done by equating the loss
in PE to the gain in KE. Alternatively the initial height of the particle above the liquid surface can be
found. Once either of these values has been found, energy principles can be used either for the
motion from the liquid’s surface to the bottom of the tank or for the whole motion from the instant
when the particle is released until it reaches the bottom of the tank. In both methods PE is lost, KE
is gained and work is done against the resistance force while the particle is in the tank. The work-
energy principle can be used in either method to find the speed at which the particle reaches the
bottom of the tank. Some candidates found the acceleration in the tank by assuming a constant
resistance force. This assumption was incorrect and also did not use an energy method and so
scored few marks. Some candidates who used energy methods missed out some of the terms
within the energy equation or combined the terms incorrectly.

Answer: The speed of the particle as it reaches the bottom of the tank is 9 ms–1

(ii) This question proved to be by far the hardest on the paper for the majority of candidates. This was
caused in part by a misunderstanding of what was happening in the problem. Once the particle
rebounded from the bottom of the tank, almost all candidates assumed that it would come to rest
within the liquid. However, once the acceleration in the liquid was found, it could be shown that the
particle was still moving after it had risen 1.25 m and reached the surface of the liquid. The majority
of candidates only scored the marks available for finding the acceleration in the liquid by using
–3g – 1.8 = 0.3a and finding a = –16. From this point, the most common incorrect approach was to
7
use the equation v = u + at with u = 7, v = 0 and a = –16, which gave an answer of t = . A correct
16
approach, once the acceleration in the liquid is found as a = –16, is to find the speed at the liquid
surface using the equation v2 = u2 + 2 as with u = 7, a = –16 and s = 1.25. This gives the speed of
the particle at the liquid surface as 3 ms–1. Using v = u + at in the liquid with u = 7, v = 3 and a = –
16 leads to a time of 0.25 seconds which the particle takes to reach the surface of the liquid. The
particle leaves the liquid with a speed of 3 ms–1 and slows down to rest under gravity. Using the
equation v = u + at with u = 3, v = 0 and a = –g = –10 leads to 0 = 3 – 10t and solving this shows
that the additional time taken for the particle to come to rest once it has left the liquid is 0.3
seconds. Adding the time spent in the liquid and the time taken to come to rest outside the liquid
gives the final answer. There are several alternative methods which could be used including energy
based methods.

Answer: The value of t is t = 0.25 + 0.3 = 0.55 s

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/43
Paper 43

Key messages

Candidates are reminded to maintain sufficient accuracy in their working to achieve 3 significant figure
accuracy in their final answers e.g. Question 1 and Question 6 parts (ii) and (iii).

When forming an equation of motion or an energy equation, candidates are reminded to check that all
relevant forces or all energies have been considered e.g. Question 3 and Question 6 parts (ii) and (iii).

General comments

As usual much of the work seen was of a very high standard. The paper provided the opportunity for
candidates of all levels to demonstrate what they knew. The easiest questions were found to be Question 4
part (i), Question 5 part (i) and Question 6 part (i). The least well answered were Question 3 part (ii),
Question 6 part (iii) and Question 7 part (iii).

Comments on specific questions

Question 1

This was a straightforward question for candidates who realised that the tension in the one string was the
same throughout. However, many resolved with different tensions for parts AR and BR of the string and thus
formed two equations in three unknowns. Some solutions were complicated by using angles, other than the
given 70o and 45o, which were not always calculated correctly. A few attempted to resolve along and
perpendicular to either AR or BR which had more scope for error than the usual choice to resolve
horizontally and vertically. The value of P was sometimes given as 0.44, correct to two, rather than the
required three, significant figures.

Answer: Tension in string = 1.21 N P = 0.443

Question 2

Whilst there were many correct solutions seen, the main difficulty for candidates was in determining the
normal reaction. Some used R = mg – 50 sin 20o rather than R = mg + 50 sin 20o leading to a mass of
17.4 kg. Others used R = mg, omitting to consider the vertical component of the 50 N force and leading to a
mass of 15.7 kg. A few obtained the equation 50 cos 20o = 0.3 × 50 sin 20o + mg either following R = 50 sin 20o
or possibly from omitting brackets.

Answer: 14.0 kg

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 3

The majority of candidates appropriately attempted an energy solution for both parts of this question and
thus avoided a solution which depended on a constant resistance.

(i) In equating the change in kinetic energy to the work done against the resistance, it was common to
see 21 × 1.2(v2 – 7.52) = 25 rather than 21 × 1.2(7.52 – v2) = 25. The resulting 9.90 ms–1 suggested
an increase in velocity as a result of the resisting force. This example provides a reminder that
some errors may be avoided by checking that an answer is suitable for the given situation. A few
candidates ignored the initial kinetic energy and mistakenly solved 21 × 1.2v2 = 25.

Answer: 3.82 ms–1

(ii) In the expected four term work/energy equation the usual errors were: to omit a term such as the
original kinetic energy; to duplicate a term e.g. potential energy included also as work done by the
component of weight down the plane; or to make a sign error, usually resulting in a negative
distance. Some calculated ‘h’ the vertical distance without involving sin 30o to obtain the required
distance AB.

Answer: 6.64 m

Question 4

(i) Most candidates sketched a correct graph and confirmed the distance 96 m either by calculating
the area of the trapezium or by totalling the distances for the three stages of motion. It was
expected that the v–t sketch would show three straight lines joining the four points (0,0), (5,6),
(17,6) and (20,0) with the key values 5, 17 and 6 seen on the relevant axes. A few graphs showed
the acceleration and/or the deceleration with a curve rather than a straight line. Occasionally a
value was missing or incorrect such as 1.2 ms–1 shown as the maximum velocity.

Answer: 96 m (AG)

(ii) This part was more challenging. A variety of methods were attempted, the most successful being to
use the area property to calculate the maximum velocity before applying v = u + at. The cyclist’s
journey was sometimes misinterpreted to include a section of constant speed either in place of the
acceleration or the deceleration, and sometimes showing a journey time of 30 s rather than an end
time of 30 s. Candidates who attempted a ‘suvat’ solution often produced erroneous equations
such as 96 = 21 a(20)2 or 96 = 21 a(10)2.

Answer: 0.96 ms–2

Question 5

(i) This was a routine pulley problem which was solved confidently by most candidates. Any errors
were usually due to an incorrect sign in setting up or in solving simultaneous equations, or due to
an incorrect system equation such as 0.5g – 0.3g = 0.5a. A very few candidates oversimplified and
used a = g as the acceleration.

Answer: a = 2.5 ms–2 h = 0.45

(ii) Candidates often applied an appropriate ‘suvat’ formula to find the required velocity of particle P
using the acceleration found in part (i). The majority of candidates understood that the acceleration
of P changed to g ms–2 when Q reached the ground and calculated the time taken to reach a
maximum height, usually either by using v = u – gt or by using a longer two stage method of finding
the maximum height and then applying s = ut – 21 gt2. Candidates often misunderstood or misread
the total time required, with common final answers of 0.3 s (the total time with the string slack) or
0.75 s (the total time for P in upward motion).

Answer: Velocity of P when Q reaches floor = 1.5 ms–1 Total time = 0.9 s

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 6

Whilst part (i) was found to be one of the easiest questions, part (iii) differentiated between candidates and
included the highest proportion of non-attempts.

(i) The application of P = Fv and Newton’s Second Law to this situation was usually carried out
successfully, although the resistance was sometimes given as 1800 + 3200 × 0.2 (= 2240 N)
instead of 1800 – 3200 × 0.2 (= 1160 N).

Answer: 1160 N

(ii) Whilst most candidates attempted to resolve forces along the incline, some omitted to include
either the component of weight down the hill or the resistance. Others appeared to be unaware of
the constant speed and equated to ‘ma’ as in part (i). 60 000 W was sometimes given as the
answer, correct to two, instead of three, significant figures.

Answer: 59900 W

(iii) As in part (ii), some candidates oversimplified the situation by forming an equation of motion with
two instead of three terms. Some overlooked the component of weight down the hill and solved
±1160 = 3200a. The application of v2 = u2 + 2 as sometimes suggested an unawareness that the
van was decelerating up the hill to rest with an unexplained negative value found for s.

The alternative work/energy approach was regularly seen. The common errors with this method
were: to omit one term; to include an extra term (e.g. potential energy duplicated as work done by
the component of weight down the hill); to use 1160 instead of 1160s or to make a sign error. 720
m was sometimes seen as the answer, either following premature approximation in the working or
from a final approximation correct to two significant figures.

Answer: 721 m

Question 7

(i) Part (i) was frequently well answered with accurate integration of a(t) and substitution to find the
maximum velocity. Many found the velocity when t = 5 as expected, but a common error was to
evaluate v(9), assuming that the maximum velocity occurred for the greatest value of t.

Answer: 83.3 ms–1

(ii) Whilst a majority of candidates recognised that a further integration was needed, a significant
number of candidates applied s = 21 (u + v)t often using v = 83.3 and t = 5 from part (i) to obtain a
total distance of 208 m. Others used integration but with an upper limit of 9 instead of 5. Errors in
t3 t4 t4
integration usually involved the cubic term e.g. ∫ dt = instead of .
3 4 12

Answer: 260 m

(iii) A complete solution for this part depended on a consideration of the limits or the constant of
integration. Many candidates ignored the constant of integration (C) or assumed C=0 without
considering both functions when t = 9 to obtain C. Some found v(9) = –18 and stated erroneously C
= –18, leading to a velocity of –48 ms–1. The integration of −3 caused a few problems, mainly
1 3
–
3t
2
3t 2
with the coefficient but also with the index e.g. − or .
2 1.5

Answer: –30 ms–1

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/51
Paper 51

General comments

The paper proved to be harder than the one set last November.

The presentation of a few candidates was rather untidy and difficult to read.

Candidates should be reminded that an answer should be given to 3 significant figures unless otherwise
stated in the question. This means that they should work to at least 4 significant figures. Very few candidates
gave answers to 2 significant figures.

Most candidates now use g = 10 as requested on the question paper.

Occasionally candidates used an incorrect formula. A formula booklet is provided so it is recommended that,
when using a formula, they should refer to the booklet.

The questions candidates found easier proved to be 1, 3(i), 4(i), 5(i)

The questions candidates found harder proved to be 2(i), 2(ii), 6(ii), 7(i) and 7(ii)

Comments on specific questions

Question 1

This question was generally well done by candidates. The vertical velocity, v, of the ball at B was 20 ms–1.
Because it was the second time, v = –20 had to be used.

Answer: 4.6(0)

Question 2

(i) Many candidates found this part of the question quite difficult. If θ is the angle between the
0.2 x
horizontal and the axis of the cone, then cosθ = and so tanθ = , where x is the required
0.3 0.3
distance.

(ii) This part of the question proved to be too difficult for many candidates. It was necessary to realise
that the centre of mass would be at the centre of the base of the hemisphere. To solve this part of
the question, it was necessary to take moments about the point A. The resulting equation would be
 π 0.32 h   h   2π × 0.23   3 × 0.2 
 3  ×  4  =  3   8  where h was the height of the cone.

Answers: (i) 0.335 (ii) 0.231

Question 3

(i) This part of the question was generally well done. Candidates needed to find the extension at the
equilibrium position, which is where the greatest speed would occur. From here a 3 term energy
equation could be set up.

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

(ii) Many candidates scored well on this part of the question. A 2 term energy equation was required,
and from this a three term quadratic equation could be found.

Answers: (i) 3.32 ms–1 (ii) 0.932 m

Question 4

(i) This part of the question was usually well done.

(ii) Many candidates made a good attempt to solve this part of the question. The first step was to
equate the acceleration to zero. The next step was to integrate. This led to an equation in v and x.
When the correct values were substituted, the required value of v could be calculated.

vdv
Answers: (i) = 32 − 40 x − 48 x 2 (ii) 1.5
dx

Question 5

(i) Many candidates made a good attempt at this part of the question. The value of T could be
calculated by using Newton’s Second Law horizontally for P. The value of R was then found by
resolving vertically for P.

(ii) This part of the question was generally well done. Candidates needed to resolve vertically for P
and then to use Newton’s Second Law horizontally.

Answers: (i) T = 0.703 R = 0.578 (ii) 3.54 rads–1

Question 6

(i) This part of the question was well done by many candidates. The required distances could be
found by taking moments about BC and AB.

(ii) Very few candidates scored well on this part of the question. Because the prism is going to topple
about D, candidates were required to take moments about D. If W is the weight then 2cos45 ×
(0.7– 0.32) = 2cos45 × (0.3 – 0.276) + W(0.3 – 0.276) is the required moment equation. This
proved far too difficult for many candidates to set up.

Answers: (i) 0.32 m (ii) 21(.0) N

Question 7

(i) Many candidates were able to express x and y in terms of t. Unfortunately very few candidates
realised that x = y. With this information, an equation in t could be found and then solved to give the
required answer.

(ii) This part of the question proved to be too difficult for many of the candidates. The total height
1
above the ground at time t is 24sin60t – gt 2 and the height to the plane is 24cos60t, so the
2
1 dh
required height, h, was 24sin60t – gt 2 –24cos60t. By differentiating to find and equating to
2 dt
zero, the required time could be found and hence the greatest height.

1
Answers: (i) x = 24cos60t, y = 24sin60t – gt 2 t = 1.76 s (ii) 3.86 m
2

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/52
Paper 52

General comments

The paper proved to be much harder than the one set last November.

The presentation of a few candidates was rather untidy and difficult to read.

Candidates should be reminded that an answer should be given to 3 significant figures unless otherwise
stated in the question. This means that they should work to at least 4 significant figures. Very few candidates
gave answers to 2 significant figures.

Most candidates now use g = 10 as requested on the question paper.

Occasionally candidates used an incorrect formula. A formula booklet is provided so it is recommended that,
when using a formula, candidates should refer to the booklet.

The questions candidates found easier proved to be 1, 3(i), and 4(ii)

The questions candidates found harder proved to be 2(ii), 3(iii), 6(i), 6(ii) and 7(i)

Comments on specific questions

Question 1

This question was generally well done by candidates. A number of candidates reached the final answer
using circuitous methods. The quickest and most efficient method was to say initial horizontal velocity = the
final horizontal velocity i.e. vcos20 = 38 cos30, where v is the required speed, and solve for v.

Answer: 35(.0) ms–1

Question 2

(i) This part of the question proved to be rather difficult due to the complex expression resulting from
the moment equation. Candidates were required to take moments about the vertex of the cone.

(ii) Only a few candidates successfully solved this part of the question. Candidates were required to
take moments about a point on the circumference of the base of the cone. The resulting equation is
kW cos30 × 0.3 + kW sin30 × 0.8 = 0.3 W

Answers: (i) 0.525 m (ii) 0.455

Question 3

(i) This part of the question was generally well done.

(ii) A number of candidates were unable to integrate the –5e–x term. The result should have been 5e–x.
After the integration it was necessary to use the correct limits of 0.5 and 0.

(iii) Only a handful of candidates were able to solve this part of the question. The answer was found by
λ ( x – 0.5 )
using 8 x – 4 = T = with x = 0 used.
0.5

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

dv
Answers: (i) v = 20 x − 10 − 5e − x (ii) 5.2(0) (iii) 4 N
dx

Question 4

(i) Some candidates did not follow the instruction to express x and y in terms of t. These candidates
simply used the trajectory equation quoted in the formula booklet.

(ii) This part of the question was generally well done. Candidates needed to substitute the two given
points into the equation found in part (i) and then solve the two equations to find a.

(iii) This part of the question was generally well done.

5x 2
Answers: (i) y = − (ii) a = 4 (iii) Height = 100 m
V

Question 5

(i) Many candidates realised that it was necessary to set up a 4 term energy equation. Sign errors
occurred on a number of occasions.

(ii) Candidates needed to realise that the greatest speed occurred at the equilibrium position. The first
λx
step was to use T = with T = 0.7 g. Having found x, a 5 term energy then had to be set up.
L

Answers: (i) 0.424 m (ii) 2.06 ms–1

Question 6

(i) Very few candidates were able to solve this part of the question. Quite often the first mistake was in
finding the centre of mass of the quadrilateral OAB, simply because the wrong formula was used.
Candidates needed to take moments about O. The resulting equation was rather complicated and
proved to extremely difficult to manipulate.

(ii) Only a few candidates were able to give a satisfactory answer to this part of the question.

Answers: (i) x = 1.01 r (ii) Within quadrant as the square will be smaller than the rectangle

Question 7

(i) Most candidates realised they had to set up two equations in the two tensions. The two equations
were set up by resolving vertically and by using Newton’s Second Law horizontally. Errors occurred
when trying to solve the two equations.

(ii) This part of the question proved to be very difficult for many candidates. Candidates were required
to resolve vertically and to use Newton’s Second Law horizontally before using F = µR.

Answers: (i) 1.66 N (ii) 0.556

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/53
Paper 53

General comments

The paper proved to be harder than the one set last November.

The presentation of a few candidates was rather untidy and difficult to read.

Candidates should be reminded that an answer should be given to 3 significant figures unless otherwise
stated in the question. This means that they should work to at least 4 significant figures. Very few candidates
gave answers to 2 significant figures.

Most candidates now use g = 10 as requested on the question paper.

Occasionally candidates used an incorrect formula. A formula booklet is provided so it is recommended that,
when using a formula, they should refer to the booklet.

The questions candidates found easier proved to be 1, 3(i), 4(i), 5(i)

The questions candidates found harder proved to be 2(i), 2(ii), 6(ii), 7(i) and 7(ii)

Comments on specific questions

Question 1

This question was generally well done by candidates. The vertical velocity, v, of the ball at B was 20 ms–1.
Because it was the second time, v = –20 had to be used.

Answer: 4.6(0)

Question 2

(i) Many candidates found this part of the question quite difficult. If θ is the angle between the
0.2 x
horizontal and the axis of the cone, then cosθ = and so tanθ = , where x is the required
0.3 0.3
distance.

(ii) This part of the question proved to be too difficult for many candidates. It was necessary to realise
that the centre of mass would be at the centre of the base of the hemisphere. To solve this part of
the question, it was necessary to take moments about the point A. The resulting equation would be
 π 0.32 h   h   2π × 0.23   3 × 0.2 
 3  ×  4  =  3   8  where h was the height of the cone.

Answers: (i) 0.335 (ii) 0.231

Question 3

(i) This part of the question was generally well done. Candidates needed to find the extension at the
equilibrium position, which is where the greatest speed would occur. From here a 3 term energy
equation could be set up.

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

(ii) Many candidates scored well on this part of the question. A 2 term energy equation was required,
and from this a three term quadratic equation could be found.

Answers: (i) 3.32 ms–1 (ii) 0.932 m

Question 4

(i) This part of the question was usually well done.

(ii) Many candidates made a good attempt to solve this part of the question. The first step was to
equate the acceleration to zero. The next step was to integrate. This led to an equation in v and x.
When the correct values were substituted, the required value of v could be calculated.

vdv
Answers: (i) = 32 − 40 x − 48 x 2 (ii) 1.5
dx

Question 5

(i) Many candidates made a good attempt at this part of the question. The value of T could be
calculated by using Newton’s Second Law horizontally for P. The value of R was then found by
resolving vertically for P.

(ii) This part of the question was generally well done. Candidates needed to resolve vertically for P
and then to use Newton’s Second Law horizontally.

Answers: (i) T = 0.703 R = 0.578 (ii) 3.54 rads–1

Question 6

(i) This part of the question was well done by many candidates. The required distances could be
found by taking moments about BC and AB.

(ii) Very few candidates scored well on this part of the question. Because the prism is going to topple
about D, candidates were required to take moments about D. If W is the weight then 2cos45 ×
(0.7– 0.32) = 2cos45 × (0.3 – 0.276) + W(0.3 – 0.276) is the required moment equation. This
proved far too difficult for many candidates to set up.

Answers: (i) 0.32 m (ii) 21(.0) N

Question 7

(i) Many candidates were able to express x and y in terms of t. Unfortunately very few candidates
realised that x = y. With this information, an equation in t could be found and then solved to give the
required answer.

(ii) This part of the question proved to be too difficult for many of the candidates. The total height
1
above the ground at time t is 24sin60t – gt 2 and the height to the plane is 24cos60t, so the
2
1 dh
required height, h, was 24sin60t – gt 2 –24cos60t. By differentiating to find and equating to
2 dt
zero, the required time could be found and hence the greatest height.

1
Answers: (i) x = 24cos60t, y = 24sin60t – gt 2 t = 1.76 s (ii) 3.86 m
2

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/61
Paper 61

Key messages

Candidates should be aware of the need to work to at least 4 significant figures to achieve the required
degree of accuracy. Efficient use of a calculator is expected, but candidates should be encouraged to show
sufficient workings in all questions to communicate their reasoning.

General comments

Candidates would be well advised to read the question again after completing their solution to ensure they
have included all the relevant details.

Candidates who presented their work in a clear and ordered fashion often avoided careless mistakes.

A significant numbers of candidates did not perform well on this paper and had correspondingly low marks.

Comments on specific questions

Question 1

Many candidates did not recognise that this was a ‘selection’ question which required the use of
combinations to be answered successfully. Good solutions presented a logical working progression,
identifying the three groups separately with linked expressions before finding the product of their three
expressions. Most candidates used the question order of the groups, but this was not necessary to be
successful.

Answer: 1260

Question 2

This question was the best answered question on the paper.

(i) Many candidates produced an expression from the table and equated the sum of the probabilities
to 1. They then successfully solved the equation. A common error was to omit the 0.1 in the
expression for the sum of the probabilities.

(ii) Although E(X) was stated, a reasonable number of candidates recalculated from the initial data, not
always accurately. Good solutions provided an unsimplified expression for the variance and then
used the calculator efficiently to evaluate this expression. Weaker solutions failed to square the
mean in the variance formula.

Answer: (i) 0.15 (ii) 1.93

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 3

The best solutions included a representation of the context. For example, a list of all of the possible
scenarios or a diagram to illustrate the question restrictions.

(i) Many partial solutions were seen, where not all possible scenarios were identified. Many solutions
only used the boundary values stated within the question to obtain one value. Candidates needed
to have a good understanding of the meaning of ‘at least’ and ‘no more than’ to be successful.
Many solutions omitted the group containing four violinists and two cellists. As the value obtained
was exact, candidates should not round their answer to 3 significant figures.

(ii) Here, a simple diagram, showing the four violinists sitting together plus a separate cellist and
double bass player, helped clarify the restrictions. Strong responses demonstrated understanding
that this was an arrangement problem with two conditions: the arrangement of violins within their
group and the arrangement of the three different instruments. However, weaker responses were
seen which started to list the possible arrangements, but did not consider the different ways the
violins could sit together.

Answers: (i) 12 210 (ii) 144

Question 4

(a) This question required the use of the standardisation formula. Strong responses demonstrated both
correct interpretation of the notation and correct substitution to find the required standard deviation.
Many candidates stated the complement of the required answer. Drawing a simple diagram would
have helped to avoid this error. Candidates should be aware of how to use the normal tables
provided accurately.

(b) Candidates with a good understanding of the normal distribution performed well on this question.
On this unstructured question, good solutions demonstrated logical progress through the
information to produce two simultaneous equations and then calculate the mean and standard
deviation. The accurate use of the normal tables was essential to reach an acceptable final answer.
Weak solutions often solved the simultaneous equations formed by equating the standardisation
formula with the calculated probabilities. Setting out work in a structured way is advisable in order
to avoid the transfer errors which were seen in some candidates’ algebraic manipulation.

Answers: (i) 0.146 (ii) µ = 15.8, σ = 3.03

Question 5

Most candidates recognised that this was a context which would be modelled by the binomial distribution.

(i) Solutions generally contained at least one correct binomial term, although not always with the
probabilities stated in the question. Again, candidates needed to have a clear understanding of ‘at
most’ as a common error was to interpret this as ‘less than’ and omit the probability of 6 students
passing. Good solutions stated an appropriate expression to calculate and then evaluated
efficiently with the calculator, which reduced the possibility of working to insufficient accuracy.

(ii) Many candidates did not progress from part (i). Good solutions recognised that, as the data were
discrete, a continuity correction was required. These solutions went on to evaluate the mean and
variance accurately and often included a simple sketch of the normal curve to help identify the
required probability area. Weaker solutions omitted the continuity correction or found the
complement probability.

(iii) Few successful solutions were seen. Where candidates are asked to justify the use of the normal
distribution as an approximation for the binomial distributions, np and nq both need to be evaluated
and stated as fulfilling the required condition.

Answers: (ii) 0.262 (ii) 0.125 (iii) np = 160 nq = 40, both > 5 so normal approximation holds

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 6

(i) Although most candidates attempted the cumulative frequency graph, few complete solutions were
seen. Candidates needed to use a scale which allowed them to plot points accurately, to label axes
fully including units, and to draw a smooth curve. Most candidates plotted at the upper class
boundary, although some introduced an unnecessary continuity correction.

(ii) Very few solutions interpreted the question correctly. Almost all candidates stated their reading for
a cumulative frequency of 100, rather than (250–100) as required.

(iii) Again few fully correct answers were seen. The unstructured nature of the question required
candidates to work logically, applying their knowledge of mean and variance from grouped data.
Good solutions often copied and extended the original data table to include ‘frequency (f)’, ‘mid-
point (m)’, ‘m × f’ and ‘m2 × f’. They then used these values to calculate both the mean and the
standard deviation. In many solutions inaccurate midpoints were used throughout, which limited
progress. A number of candidates calculated the mean but did not attempt the standard deviation.
A few candidates did not find the square root of their variance to obtain the standard deviation.
Some candidates did not use the square of their mean when calculating the variance.

Answers: (ii) 42 (iii) µ = 39.9 σ = 23.2

Question 7

(i) Almost all candidates successfully used the data table to state the required probability. Some
candidates unsuccessfully attempted to simplify or convert their answer to a decimal. Candidates
should be aware that using fractions in probability generally leads to more accurate answers.

(ii) Although many candidates attempted this question, the use of P(B) × P(M) = P(B∩M) as the
required condition was limited. A number of solutions produced circular numerical arguments rather
than comparing data from the original table. The best solutions identified the probabilities required
and made an appropriate numerical comparison before reaching a conclusion. The weakest
solutions presented a logic argument without evidence.

(iii) The most successful solutions used the original data table to identify the groups that fulfilled the
conditional probability conditions. Where candidates attempted to use the alternative approach of
substituting probabilities into the standard conditional probability formula, some produced quite
complex and incorrect calculations.

(iv) Few fully successful solutions to this part were seen. Good solutions identified the scenarios that
fulfilled the question conditions, recognised that once a candidate had been selected they could not
be chosen again, and accounted for different orders, since order was not important. The most
common error seen was finding the products of probabilities with the same denominator, indicating
that replacement had occurred.

13 27
Answers: (i) (ii) Not independent (iii) (iv) 0.201
40 64

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/62
Paper 62

Key messages

In order to gain full credit, candidates should show sufficient method to justify their conclusions.

Candidates are reminded that only non-exact answers should be rounded to 3 significant figures.

Candidates need to consider whether their scale will enable their graphical solutions to be plotted to an
appropriate degree of accuracy.

General comments

The majority of candidates presented their solutions in a logical manner. However, there was sometimes a
lack of structure which made determining the final answer difficult for examiners. In Question 7, this was
possibly the cause for the number of solutions that did not provide the required information.

Many good solutions were seen for Questions 3 and 5. The context in Question 6 was found challenging by
many. Sufficient time seems to have been available for candidates to complete all the work they were able
to. However, a significant number of candidates made little or no attempt at the final parts of Question 7.

Candidates should read the question again once they have finished their solution to ensure that they have
provided the required information.

Comments on specific questions

Question 1

This was a standard permutations and combinations question, which was attempted by almost all candidates

(i) Good solutions simply stated the standard result to calculate the number of arrangements. Weaker
candidates did not divide by all the terms required to remove the effect of repeated letters. A few
candidates provided solutions which assumed that the repeated letters could be identified, so
gained little credit. There appeared to be some inefficient use of calculators, as a number of
solutions displayed the correct numerical formula, evaluated inaccurately.

(ii) A number of different approaches were seen to this question. The most efficient considered the
three probabilities for selecting the double letters and summed the results. A common error was to
treat the situation as if replacement occurred and not reduce the denominator. The alternative
efficient approach was to consider the number of selections for each of the double letter scenarios
and then convert the total into the required probability by dividing by the total number of possible
selections. Some candidates assumed that the letters were individually identifiable here. In this
approach, weaker solutions simply stated either the numerator or denominator value.

13
Answers: (i) 34 650 (ii)
55

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 2

The majority of candidates correctly interpreted the key given for the back-to-back stem-and-leaf data and
then used it appropriately throughout. Candidates did not always use a scale that would enable the box-and-
whisker plot to be drawn accurately using their values.

(i) The median was identified correctly by almost all candidates. Good solutions stated the candidate’s
values for the upper and lower quartiles before their calculation of the inter-quartile range. There
was less consistency in identifying the upper quartile, with the weakest solutions stating the
position of the piece of data, rather than its value. When finding the interquartile range, some
candidates incorrectly found the mean of their upper and lower quartile values or restated the
median value as the interquartile range.

(ii) Few good comparison box-and-whisker plots were seen overall. Many candidates plotted only one,
and failed to identify which data they were using. Good solutions had a linear scale that used the
graph grid well, with 2 cm = 0.01 seconds allowing for accurate representation. Good solutions also
labelled the axis both with the item being recorded and measurement units and identified the two
plots. The use of a ruler is expected at this level to draw straight lines.

A number of candidates used the notation for the mean when plotting their median value.

Some candidates used separate axes for each plot, which only gained full credit if they aligned
exactly, as otherwise they did not allow visual comparison of the data as required.

Answers: (i) median = 0.225 IQR = 0.021

Question 3

(i) This was a standard binomial distribution probability question, and almost all candidates used the
appropriate method. Candidates should be aware that ‘at least 3 times’ means that ‘3 times’ is
included in the solution set, as omission of this value was the most common error. Good solutions
stated the unsimplified expression and then used a calculator efficiently to evaluate the probability.
Some candidates who calculated each value did not work to sufficient accuracy, which resulted in
premature approximation affecting their final answer.

(ii) The use of a tree diagram assisted many candidates to identify the outcomes required to fulfil the
stated condition and link the different probabilities successfully. The most common approach was
to calculate the probabilities of being successful on both days, on Monday only and on Tuesday
only and sum the results. A few candidates identified that this was equivalent to 1 – P(unsuccessful
on both days). Common errors were to link the probabilities from the question inaccurately, or to
omit P(successful on Monday, unsuccessful on Tuesday).

Answers: (i) 0.896 (ii) 0.92

Question 4

Candidates were generally more successful with (i) and it was encouraging to more explanation of the
candidates’ logic than in previous papers.

(i) Many good solutions were seen, with almost all realising that each person could be identified.
86 400 was a common incorrect answer, where candidates did not allow the boy and girl blocks to
change places.

(ii) The common feature in many good solutions was a simple diagram which interpreted the
requirement for no boy to stand next to another boy. A number of solutions appeared not to allow a
boy to stand at the end of the row, and so eliminated two possible positions. A few candidates only
stated a non-exact answer in this question.

Answers: (i) 172 800 (ii) 1 814 400

© 2018
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 5

(i) The best solutions for this question were well structured, laying out the steps involved clearly. They
identified that Σx for the Junior members was required and presented a single expression to sum
the two age totals and divide by the total number of members. Weaker solutions were often poorly
structures and included information that was not required for the answer. Candidates used the
incorrect method of calculating the mean of the Senior members and then finding the average of
the two means. A number of candidates correctly found the total of all the ages, but only divided
this by the number of Junior members. At this level candidates should not accept final answers
which are unrealistic, such as ages over 100. Candidates should be aware that exact answers
should not be rounded in this context.

(ii) Many candidates answered this question with confidence, applying the variance formula
appropriately for the Juniors to calculate Σx2 and then using all the appropriate data to calculate the
variance for the whole group. A small number of candidates did not continue to finally state the
standard deviation. Common errors were either to use a 3 significant figure mean from (i), which led
to inaccurate answers, or to use an incorrect value from (i) accurately. Weaker candidates
calculated the standard deviation for each group of members and averaged their values.

Answers: (i) 34.25 (ii) 16.0

Question 6

There was evidence that some candidates did not read the question fully and assumed that the random
variable X was the difference between the spinners rather than the score on the red spinner minus the score
on the blue spinner.

(i) Almost all solutions contained a clear probability distribution table. Good solutions often included a
sample space table as well, which enabled candidates to complete the PDF without further
calculation. Many candidates showed clear calculation to obtain each of their required probabilities.
In a few solutions, 4 was incorrectly included as a value for the random variable.

A significant number of solutions only considered positive values, using the difference between the
spinners. A few candidates correctly calculated the probabilities of the positive random variable
values and omitted the negative values as if scores could not be negative.

(ii) Good solutions clearly presented unsimplified expressions using the information from their
probability distribution table, and calculated first the mean and then the variance. Inaccurate
transfer of probabilities was seen, often where presentation was not clear. A common error was not
squaring the mean within the variance formula.

(iii) Many candidates found this question challenging, and it was omitted by a significant number.
Candidates who had used a sample space table in (i) were often able to simply write down the
required conditional probability by identifying the outcomes fulfilling the conditions. The most
common method was to apply the standard conditional probability formula and calculate the
required values for the numerator and denominator. A very common incorrect solution was simply
stating their P(1) value.

23 1
Answers: (i) (ii) (iii)
12 3
–2 –1 0 1 2 3

1 2 3 3 2 1
12 12 12 12 12 12

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 7

This was a fairly standard normal distribution question. However, many candidates used the normal tables
inaccurately. Many solutions were poorly structured, and some candidates did not show understanding of the
purpose of standardisation.

(a) (i) The majority of candidates applied the normal standardisation formula correctly, without continuity
correction, and found the required probability. A few candidates did not take account of the
statement in the question that the variable had a normal distribution and used a binomial
distribution. Good solutions realised that the probability was not a final answer and continued to
work at 4 significant figure accuracy to calculate the expected number of days. The context required
an integer answer, and candidates needed to interpret their calculation rather than simply round.

(ii) Although a fairly standard question, good solutions presented the solution in a logical progression,
linking the standardisation to the z-value and manipulating the algebra accurately. Weaker
solutions often used an inaccurate z-value. A number of solutions equated with the stated
probability, or treated the probability as a z-value and found a new probability to use. The use of a
probability in this question could gain little credit.

(iii) This question acted as a discriminator and enabled candidates with a good understanding of the
principles of the normal distribution to work efficiently.

Good solutions often had a simple sketch of the normal curve with a visual interpretation of the
required conditions. The symmetry properties were identified, and there was an appreciation that
‘within 1.5 standard deviations of the mean’ provided the required z-value without further
calculation. This led to an expression for the required probability area which was then evaluated.
Weaker solutions calculated the mean ± 1.5 x standard deviation as stated in the question, then
used their expression to calculate the z-values which, with the extra stages, often became
inaccurate. These candidates were usually able to calculate the correct probability area from their
values. Candidates should be aware that probabilities cannot be greater than 1, so their final value
should meet this requirement.

(b) Many candidates found this question challenging, and no attempt was made by a significant
proportion. Good solutions recognised that this was a generalisation of the standardisation formula,
with 0 as the value to be standardised. Rearrangement of the question condition was performed
before substitution into the formula, which was followed by appropriate cancellation to produce the
z-value. A simple diagram often helped candidates appreciate the probability area that was
required.

Some candidates did not achieve the required accuracy, either rounding their decimal equivalent to
4
− to 3 significant figures or stating their probability as a 2 significant figures value. A number of
3
solutions introduced x as a variable to replace 0, which limited progress.

Answers: (a)(i) 286 or 287 (ii) 4.05 (iii) 0.866 (iv) 0.909

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/63
Paper 63

Key messages

A few questions in this paper could be solved in several ways and a number of candidates chose to do more
than one method to arrive at the final answer. Candidates need to be aware that we will only mark one
solution when more are presented – normally the ‘more complete’ solution. However, if more than one
‘complete solution’ is offered, we will mark the last solution.

Those who choose to use a calculator rather than tables when finding answers using the Binomial or Normal
distributions need to remember to write down the unsimplified binomial or standardisation expression before
writing the final answer. This advice is even more important when working on a question like Question 5i
where the answer is given.

General comments

Candidates need to remember that all non-exact numerical answers must be given correct to 3 significant
figures. This requires them to work with numbers to at least 4 significant figures before the final answer is
reached. Premature approximation was apparent in several questions, particularly the probabilities in
Question 2i, where some candidates chose to work in decimals rather than fractions and Question 5i,
where the answer was given to 3 significant figures and we required them to work with probabilities to 4
decimal places. Candidates also need to be aware that a decimal rounded to 3 decimal places which results
in a zero in the third decimal place cannot be treated as an exact decimal and truncated as was seen
frequently in Question 5ii.

Comments on specific questions

Question 1

This question was answered accurately by many candidates, with about the same number choosing to
subtract the number of ways with two women next to each other from the total number of ways (7! – 2 × 6!)
as finding the number of ways to position two women apart once five men have been arranged (5! X 6P2). Of
those who chose the first method, not all candidates remembered to multiply 6! by 2 i.e. they forgot that the
two women together could be ordered in two ways. Among those who chose the second method, 5!
frequently appeared on its own or multiplied by other factorials. A few chose a longer method, considering
the five different scenarios where the women are separated by one, two, three, four or five men and
summing the number of ways.

Answer: 3600

Question 2

(i) This question was well answered by most candidates. Just a few candidates appeared not to
understand what a probability distribution table was and produced a possibility table instead. Of
those who slipped up with the probability distribution, the most common error was to omit one of
the X values, usually 3.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Answer:

x –2 –1 0 1 2 3
P(X = x) 2 4 5 4 2 1
18 18 18 18 18 18

(ii) The strongest candidates appreciated that they should show their method for obtaining both the
mean and the variance, and presented clear unsimplified expressions for both before writing down
their final answer. We insisted on the probabilities from part i summing to 1 to justify the first
method mark and most candidates made sure that this was the case. The most common error was
to subtract the mean, rather than the mean squared, in the variance formula.

65
Answer: (1.81)
36

Question 3

(i) This question was well answered by most candidates despite the introduction of a third colour
possibility for the second ball. Most knew to label the branches with both the colour and the
assigned probability. The few who made mistakes either introduced 9 as the denominator for the
First Ball or, more commonly, introduced 7 as the denominator for the Second Ball. Some of those
using 7 still considered the outcomes with 8 balls, making their numerators sum to 8 and the
probabilities to more than 1.

Answer:

(ii) Most candidates realised that they needed to consider the two possibilities of both balls being red
or both being blue and knew to multiply the probabilities in each of these cases. Only a few
misunderstood the question and gave two separate answers instead of summing the products.
Even those with incorrect tree diagrams usually earned the method mark as long as the
probabilities in each branch summed to 1.

13
Answer: (0.406)
32

(iii) The majority of candidates were able to use the Conditional Probability formula correctly and find
the correct answer. Of those who found the question more challenging, some calculated the
probability of red then blue and presented it as their final answer, others found both required
probabilities but did not put them in a fraction and still others calculated two probabilities to put in a
fraction but one or both were incorrect.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

3
Answer: (0.429)
7

Question 4

(i) This question was successfully answered in a number of different ways. Most successful
candidates divided the number of ways of choosing 6 people from the 11 after one particular boy is
removed (11C6 = 462) by the total number of ways of choosing 7 people out of 12 (12C7 = 792).
Some took a lot longer to arrive at the numbers 462 and 792 by summing five products of
combinations for one or both. Very few used the probability method to find the probability that the
 11 10 5
particular boy was chosen by multiplying six fractions  × × …… ×  and subtracting the
 12 11 6
 5
answer   from 1.
 12 

The most commonly seen incorrect approaches were either to assume constant probability and use
a Binomial distribution or to forget what the question asked and give the number of ways that the
particular boy could be chosen in a group of seven rather than a probability.

7
The few candidates who offered as the answer with no working or justification for their answer
12
were not given the marks.

7
Answer: (0.583)
12

(ii) This question proved to be more challenging than part i and was approached in a variety of ways.
As in part i, a number of candidates incorrectly assumed constant probability and tried to use the
Binomial distribution. Of those who made a good attempt at the question, most chose to find the
number of ways 2, 3 or 4 girls could be chosen (with 5, 4 or 3 boys) using products of combinations
and then summing them. A simpler way was to subtract the number of combinations with 0 or 1 girl
(and 7 or 6 boys) from 12C7. A surprising number of candidates then presented the total (672) as
their answer and forgot to find the probability of this happening by dividing by the total number of
ways.

Those who attempted the Probability methods were generally less successful. The most common
slip when attempting ‘1 – Probability (0 or I girl)’ was to forget to multiply by 7 when finding the
probability of 1 girl. Only a handful of candidates who tried calculating the Probability (2, 3 or 4
girls) had any success, as candidates generally omitted the combination term (7C2, 7C3 and 7C4
respectively) in the three probability calculations.

28
Answer: (0.848)
33

Question 5

(i) Most candidates appreciated that in a ‘Show that’ question where the answer is given, all stages of
the working must be shown. The majority knew to standardise and they generally obtained the z-
5 5
values and − . At that point some did not know how to proceed and abandoned their working.
6 4
Others drew useful diagrams to demonstrate the area that they needed and explained their working
using phi notation, before they found the probabilities from the tables or a calculator. Only a few
‘fudged’ their numbers to arrive at the given answer using incorrect numbers. Some candidates
who used calculators rather than tables lost marks by jumping to the final answer too quickly.
However, a pleasing number of candidates presented detailed and careful solutions, appreciating
that they needed to write down probabilities to at least 4dp if their final answer was to be correct to
3dp.

Answer: 0.692 AG

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

(ii) Most candidates recognised this as a Binomial Distribution question and produced at least one
binomial term with n = 4 and p = 0.692. A few used an incorrect probability from part i, despite
having been given the correct answer. The words ‘at least two’ caused confusion, with some
candidates interpreting it as ‘more than 2’, but most appreciated that they either had to sum the
probabilities of 2, 3 or 4 medium apples or subtract the probabilities of 0 or 1 medium apple from 1.
Some forgot the combination element in a binomial term and a few did not use two probabilities
that summed to exactly 1 in each term. A surprising number did not realise that a zero as the final
digit in a non-exact decimal is still significant. 0.91 was frequently presented as the answer without
showing a longer answer before the rounding.

Answer: 0.910

Question 6

(i) This question was well answered. Most candidates knew to find the z-value corresponding to 0.96
and equate it to a standardised expression. A few used the tables the wrong way round or equated
the result of their standardisation to 0.96 or 0.04, and some tried to deal with the negative z-value
by subtracting 1. Those who drew a diagram generally dealt well with the negative value of the
standardised expression. However, a significant number of candidates either gave their final
answer as a negative value or crossed out the negative sign when they realised that a standard
deviation has to be positive.

Answer: 114

(ii) Most candidates recognised this as a Normal approximation from the Binomial distribution (300,
0.2) and correctly found the mean and variance to be 60 and 48. A pleasing number knew to apply
a continuity correction when standardising, although 70.5 was frequently seen instead of 69.5. A
more common mistake was to divide by 48 instead of the square root of 48, either because they
thought the variance was 48 squared or because they did not know to divide by the standard
deviation.

Answer: 0.915

(iii) Surprisingly few candidates understood what was being asked here. Some had quoted the
conditions in part ii and still did not recognise that it was those conditions being asked for in this
question. Of those who did understand the question, only a few realised that quoting the theory
was not enough and that they needed to evaluate np and nq as 60 and 240, and state that they are
both greater than 5. Many thought that it was the mean and variance that had to be greater than 5.

Answer: np = 60 > 5; nq = 240 > 5

Question 7

(i) Candidate who scored well on this question knew to read the question carefully and to make sure
that they followed the instructions. They needed to put the Anvils on the left and the Brecons on the
right in a back-to-back stem and leaf with both the diagram and the key having clear labels and the
‘leaves’ being lined up carefully to demonstrate the shape of the data. For three digit data like this,
the first two figures appear in the ‘stem’ and the ‘leaves’ are single digits. Many candidates omitted
the ‘1’ in the ‘stem’ but usually ‘recovered’ in the key. Marks were deducted for the presence of
commas, lack of labels, data the wrong way round, misalignment of the ‘leaves’, splitting the
‘leaves’ into two rows, units and labels omitted from the key, or the key being presented separately
for the Anvils and Brecons.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Answer:

Anvils Brecons
8 15
95 16 6
5320 17 01228
410 18 1233
6 19 2
Key: 5|16|6 means 165 cm
for Anvils and 166 cm for
Brecons

(ii) This question was well answered by most candidates. Most knew that the median from 11 items of
data is the 6th item and identified this to be 173. Just a few divided 11 by 2 and took the average of
the 5th and 6th items. Those who did not identify the median correctly usually made a similar
mistake with the quartiles, but gained a mark if they calculated the difference between them. A few
candidates found the quartiles but did not do the subtraction.

Answer: median = 173; IQ range = 12

(iii) This question was well answered by most candidates. They knew to add the three extra heights to
337221 to find Σx for 14 members, although a few tried to sum the 14 numbers from scratch.
Calculating the new Σx2 presented a greater challenge. Some examples of incorrect thinking were
3372212 and (19232 + 1662 + 1722 + 1822).

Most candidates used the correct variance formula and knew to write their unsimplified expression
for the variance in full before writing the final answer. Only a few forgot to subtract the mean
squared, or gave the variance as their final answer, rather than the standard deviation.

Answer: 9.19

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/71
Paper 71

Key messages

As stated on the front of the paper, non-exact numerical answers are required to 3 significant figures.
Sometimes this requires working to be carried out to more than 3 significant figures, as in Question 6(iii)
and in Question 7.

The conclusion to a significance test should be expressed in the context of the question, as in Question 2(ii)
and in Question 7(i).

General comments

Many candidates presented their solutions in a clear and logical manner.

When standardising within a normal distribution the appropriate form for the numerator was “ x – µ”.

In Question 4(ii) this was 59.5 – 61. In Question 5(i) this was 6 – 5 and in Question 5(ii) this was 0 – 0.16.

Comments on specific questions

Question 1

(i) In order to score full marks here it was necessary to provide the correct form for the interval and
the correct value of z (2.24). Many candidates found both of these and calculated correctly. Other
candidates gave an incorrect value for z, such as 1.96.

7.22
Other candidates did not use the variance correctly.
200

Also the answer for the confidence interval had to be given as an interval. Acceptable forms were
“175 to 177” and “(175, 177)”.

(ii) The necessary condition for the calculations to be valid was that a random sample was obtained.

Many candidates suggested different conditions which were not appropriate such as “the heights
must be normally distributed”, “n must be large”, “the data must be continuous” and “the standard
deviation must be the same”.

Answers: (i) 175 to 177 (ii) random sample

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 2

(i) Many candidates stated the correct hypotheses in terms of p. Other candidates incorrectly used
1 1
µ = instead of p = , omitted p = altogether, or tried µ = 50.
3 3

(ii) Some candidates proceeded correctly by comparing the given probability (0.0084) with the
significance level (1%) and then making the correct conclusion about p from this significant result.
1
Other candidates incorrectly compared 0.0084 with or wrote general phrases without using the
3
values.

(iii) The largest number of children was 150. Other values such as 50, 36 and 114 were suggested by
some candidates.

1 1
Answers: (i) H0: p = H1: p < (ii) 0.0084 < 0.01 There is evidence that p has decreased
3 3
(iii) 150

Question 3

Many candidates correctly found the value of n to be 600, or at least reached the value 600.25, though the
2.52
final answer did need to be the whole number value. It was necessary to use the variance ( ) in a correct
n
form, to use the correct value for z (1.96), and to rearrange the equation accurately. Some candidates tried
2.5
an incorrect form for the variance, such as or an incorrect value for z, such as 2.24. Some candidates
n
made algebraic errors, such as reaching n = 24.5 but then taking the square root instead of squaring.

Answer: 600

Question 4

(i) Many candidates found the combined Poisson distribution with parameter 6.1 and used the
relevant terms to obtain the correct answer 0.857. Some candidates used λ = 0.61. Other
candidates added an extra Poisson term (P(4)) or omitted the first term (P(0)). Most candidates
used P(⩾4) = 1 – P(⩽3). A few candidates attempted to use a Binomial distribution or a normal
distribution. These were not appropriate.

(ii) As the Poisson distribution for the total number of drops in 100 cm3 of air had a parameter of 61
which was large (> 15), the appropriate approximating distribution was the normal distribution N(61,
61). Many candidates realised this and used the necessary continuity correction (59.5) to find the
probability. Some candidates found λ = 61, but then used an incorrect variance such as 23.79 (from
npq). Other candidates omitted the continuity correction factor and used 60, or used an incorrect
value such as 60.5. A diagram can assist in deciding on the value of this continuity correction factor
and also in choosing the correct area (the tail) for the probability.

Answers: (i) 0.857 (ii) 0.424

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 5

(i) Many candidates found the distribution for the total time (T1 + T2) correctly as N(5.0, 0.41),
standardised correctly and found the probability. Some candidates inserted a continuity correction
factor. This was not appropriate. Other candidates incorrectly used 0.4 + 0.5 = 0.9 as the variance.
Other candidates found 0.41 but used it as the standard deviation. The larger area was required
here for the probability.

(ii) Some candidates used the appropriate new variable “Y” = T2 – 1.2 × T1 and found the probability
that Y > 0. Some candidates found the correct distribution N(0.16, 0.4804) for this.

Other candidates incorrectly multiplied T2 by 1.2 or made errors when trying to find the variance.
These included multiplying by 1.2 instead of 1.22 or subtracting instead of adding the separate
variances (common errors for the variance were 0.0196 and 0.442).

A diagram can be helpful for identifying the correct area and probability for Y = T2 – 1.2 × T1 > 0.

This is especially so when the value 0 is not the mean.

Answers: (i) 0.941 (ii) 0.591

Question 6

(i) Many candidates correctly integrated kx–1 between the limits 2 and 6 and equated the result to 1.
The combination of the ln 6 and ln 2 was carried out to obtain the given answer for k. Some
candidates used incorrect limits such as 0 and 6. Other candidates wrongly combined the ln terms
6
to ln .
ln 2

(ii) Many candidates correctly integrated x.kx–1 and applied the limits 2 and 6 to the result of the
4
integration (kx). The result was which then gave 3.64 to the required accuracy. Some
ln 3
candidates multiplied the x and the x–1 incorrectly. Other candidates omitted the x or tried to
integrate the x and the x–1 separately. This was usually unsuccessful as the approach did not
become a correct form of “integration by parts”.

(iii) Many candidates attempted to find the median and then the probability. The median required the
integration of kx–1 for the limits 2 to m (or m to 6) and the use of 0.5. Some candidates followed this
route correctly and found the median to be 3.46(4) (the exact value was 12 ). Other candidates
used incorrect limits such as 0 to m or used limits 2 to 3.64 and incorrectly equated the result to m.
The required probability was found by using the limits 12 to E(X). To obtain an accurate answer
the earlier working needed to be carried out to more than 3 sig. figs.

Some candidates used the method listed on the left hand side of the mark scheme. First they found
P(2 < X < E(X)). Then they noted that P(2 < X < m) = 0.5 so subtracted this 0.5 from 0.545(08) to
obtain the probability.

A few candidates found the median, but then found the variance and attempted to use a normal
distribution. The second part of this was not valid.

1
Answers: (i) k = (ii) 3.64 (iii) 0.045 (2 sig. figs.)
ln 3

Question 7

(i) Many candidates followed a logical route through this significance test. The first stage required the
hypotheses and the unbiased estimate of the variance of the masses. Some candidates stated the
hypotheses incorrectly in terms of x or “the mean”. Some candidates found only the sample biased

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

variance. The second stage required the standardisation of the sample mean (49.8667) using the
46.962
normal distribution N(51, ). Some candidates omitted the 150.
150

The third stage required the comparison and conclusion. The comparison could involve the values
of z (–2.025 < – 1.96) or the probabilities (0.0215 < 0.025) or the critical value. It was necessary to
show the comparison very clearly. Some candidates used inconsistent signs, while others
compared incorrect probability tails. The conclusion should have been given in the context of the
question. Some candidates wrote contradictory statements.

(ii) It was necessary to find the critical value for the rejection (or acceptance) region for the normal
6.8562
distribution N(51, ) as defined by H0. As this critical value was very close to the sample
150
mean value, it was necessary to provide this value to several sig. figs. (49.903). To find the Type II
error it was necessary to work with this critical value in the new distribution with population mean
mass 49 kg. Some candidates successfully followed this procedure. Some candidates reversed the
means and some candidates omitted the 150. Other candidates used +1.96 instead of –1.96.
Some of these errors created difficulties in selecting the relevant probability tail. Some candidates
omitted the first stage and attempted only a standardisation for the second stage.

A range of answers was allowed to score full marks here.

Answers: (i) there was evidence that the population mean mass of sacks was less than 51 kg
(ii) 0.0534

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/72
Paper 72

Key messages

There were places on this paper where answers indicated that candidates needed to read the question more
carefully (see below).

Candidates need to know how to round answers to 3.s.f. Candidates need to be aware of the difference
between rounding to a specified number of significant figures, and a specified number of decimal places.

In a ‘show that’ question, all steps in the calculation need to be shown (see comments below). If essential
stages of the working out are missing, full marks will not be obtained.

All of the steps required for a Hypothesis test should be fully presented (see comments below on Question
5(ii) and 6(i). It is important that the conclusion to the test is fully justified.

General comments

On this paper, candidates were largely able to demonstrate and apply their knowledge in the situations
presented. Script from across the complete attainment range were seen. In general, candidates scored well
on Questions 1, 2(i), 4(i) and 7(i), whilst Questions 5(ii), 6 and 7(ii) proved particularly challenging.

Most candidates kept to the required level of accuracy, though if a question requires the final answer to be to
a specified level of accuracy (e.g. to 1 decimal place as in Question 2(i) this instruction must be adhered to.

Timing did not appear to be a problem for candidates.

Detailed comments on individual questions follow. Whilst the comments indicate particular errors and
misconceptions, it should be noted that there were also some good and complete answers.

Comments on specific questions

Question 1

In general, this was a well answered question. Most candidates correctly chose to calculate P(2,3,4) with
λ = 2.3. There were only a few cases where P(5) was incorrectly included or P(2) was omitted from the
calculation. Weaker candidates separated the inequalities and attempted to calculate separate probabilities
and then subsequently combine. This approach was seldom successful.

Answer: 0.585

Question 2

Many candidates correctly found the required confidence interval in part (i). Some candidates did not read
the question correctly and did not give the end-points of their interval correct to 1 decimal place, as
requested. Hence they could not gain the final answer mark. Other errors included use of an incorrect z
value and confusion between standard deviation and variance. It is important that a confidence interval is
written as an interval and not as two separate values.

Part (ii) proved challenging as candidates were not precise enough in their responses. It had to be clear that
the candidate knew that it was the population that was not stated to be normal (therefore the CLT was

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

required). Many candidates did not specify this clearly, indicating a possible confusion between the random
sample and the population. Some candidates did not give a clear yes or no answer, and some did not state a
reason.

Answers: (i) 329.4 to 330.8 (ii) Yes, because volume of all cans (i.e. the parent population) was not
stated to be normal

Question 3

Candidates were mostly aware of how to approach this question, though errors were made in calculating the
variance. Calculation of the mean and the standardisation calculation were usually very well attempted
despite the often incorrect value used for the variance. A small number of candidates found the probability of
less than 1310 rather than greater than.

Answer: 0.108

Question 4

In part (i) candidates were required to show that the value of k was
( a + 1) . In questions of this kind it is
a
important that candidates show all relevant working. There were occasions here where full marks were not
awarded due to lack of essential working. There were also occasions when candidates incorrectly reached
the required answer, offering a solution containing an error or errors (sign errors were particularly noted
here). For full marks to be awarded, no errors should be seen. Other candidates did not evaluate the lower
limit of the integration, or incorrectly assumed it was zero.

An answer in the context of the question was required in part (ii). Questions of this type are not always well
answered. Many candidates merely said that ‘a’ was the maximum value of the pdf. This statement on its
own was not sufficient to gain the marks as no context was given. The maximum time for the runners to
finish would have given the required context.

Part (iii) was not always well answered. Candidates did not always use the information given in the correct
way. An attempt to integrate f(x) between 0 and 0.5, and then equate to 0.75, was required. Some
candidates indicated a lack of understanding by interchanging the 0.5 and 0.75, while some candidates
equated their integral to 1, and many candidates used incorrect limits.

Answers: (ii) Maximum time allowed by the model for the runners to finish (iii) 0.8

Question 5

Part (i) produced a mixed response. There was confusion in finding the values required from the given table
(Σfx, Σfx2 and Σf were required, but these were not always correctly found). There was also confusion
between the two alternative formulae for the unbiased estimate for the population variance. Candidates
would be advised to consistently use just one of these formulae to avoid this confusion.

Part (ii) proved to be challenging. Candidates should be aware of all the steps required to carry out the test.
The null and alternative Hypotheses should be stated; the calculation of test statistic clearly shown, and a
comparison made between the test statistic and the critical value in order to justify the conclusion.
Candidates often gave no Hypotheses or incorrect ones. Calculation of the test statistic was not always done
well. Even the candidates who realised they needed to use the values found in (i) often did not use √70. A
significant number of candidates did not show the comparison clearly; often diagrams were seen that
referred to ‘RR’ or ‘AR’ without defining what was meant by this. A clear inequality statement or a diagram
with both figures for comparison clearly marked was required. (The comparison here could either be between
z values or between areas). The conclusion to the test should be non-definite and in context. Some
candidates drew a correct conclusion but then made a contradictory statement.

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

In part (iii) many candidates realised that they needed to check whether Ho had or had not been rejected. A
few candidates did not make the reason for their choice clear.

Answers: (i) 0.858 (ii) No evidence that mean number of courts in S is less than in N
(iii) Type II because Ho was not rejected

Question 6

As in Question (5), all steps for the Hypothesis test need to be clearly followed. Some candidates did not
state their Hypotheses, or gave incorrect ones. Use of N(9,7.65), or equivalent, was not always seen and
candidates again did not show their comparison clearly, meaning that their conclusion was not justified. As
an alternative method, some candidates correctly used Bin(60,0.15) and were able to successfully carry out
the test, though errors here included calculating P(<6) rather than P(⩽6). Conclusions to the test should
again be non-definite and in context and should not contain contradictory statements.

6
In part (ii) the main error noted was using 0.15 rather than ( = 0.1) . The calculation required to find α from
60
the z value, using the symmetry of the confidence interval, was not always successfully done. The question
specified that the answer should be given to the nearest integer. This was not always done, illustrating the
need for candidates to read the question carefully.

Answers: (i) No evidence that the train is late less often (ii) 80

Question 7

Most candidates used the correct value of λ (5.6) and successfully found the required probability in part (i).

Part (ii), however, was not well attempted, with many candidates not appreciating that the question required
a conditional probability. Thus, there were few candidates who attempted the required calculation
P ( X = 2 and Y = 1)
of .
P ( X + Y = 3)

2.1
Part (iii) was reasonably well attempted with the use of N(2.1, ), or equivalent, often seen. Common
100
errors included the omission of 100, and confusion between different methods. A continuity correction, if
included, was often incorrect.

Answers: (i) 0.108 (ii) 0.264 (iii) 0.245

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

MATHEMATICS

Paper 9709/73
Paper 73

Key messages

As stated on the front of the paper, non-exact numerical answers are required to 3 significant figures.
Sometimes this requires working to be carried out to more than 3 significant figures, as in Question 6(iii)
and in Question 7.

The conclusion to a significance test should be expressed in the context of the question, as in Question 2(ii)
and in Question 7(i).

General comments

Many candidates presented their solutions in a clear and logical manner.

When standardising within a normal distribution the appropriate form for the numerator was “ x – µ”.

In Question 4(ii) this was 59.5 – 61. In Question 5(i) this was 6 – 5 and in Question 5(ii) this was 0 – 0.16.

Comments on specific questions

Question 1

(i) In order to score full marks here it was necessary to provide the correct form for the interval and
the correct value of z (2.24). Many candidates found both of these and calculated correctly. Other
candidates gave an incorrect value for z, such as 1.96.

7.22
Other candidates did not use the variance correctly.
200

Also the answer for the confidence interval had to be given as an interval. Acceptable forms were
“175 to 177” and “(175, 177)”.

(ii) The necessary condition for the calculations to be valid was that a random sample was obtained.

Many candidates suggested different conditions which were not appropriate such as “the heights
must be normally distributed”, “n must be large”, “the data must be continuous” and “the standard
deviation must be the same”.

Answers: (i) 175 to 177 (ii) random sample

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 2

(i) Many candidates stated the correct hypotheses in terms of p. Other candidates incorrectly used
1 1
µ = instead of p = , omitted p = altogether, or tried µ = 50.
3 3

(ii) Some candidates proceeded correctly by comparing the given probability (0.0084) with the
significance level (1%) and then making the correct conclusion about p from this significant result.
1
Other candidates incorrectly compared 0.0084 with or wrote general phrases without using the
3
values.

(iii) The largest number of children was 150. Other values such as 50, 36 and 114 were suggested by
some candidates.

1 1
Answers: (i) H0: p = H1: p < (ii) 0.0084 < 0.01 There is evidence that p has decreased
3 3
(iii) 150

Question 3

Many candidates correctly found the value of n to be 600, or at least reached the value 600.25, though the
2.52
final answer did need to be the whole number value. It was necessary to use the variance ( ) in a correct
n
form, to use the correct value for z (1.96), and to rearrange the equation accurately. Some candidates tried
2.5
an incorrect form for the variance, such as or an incorrect value for z, such as 2.24. Some candidates
n
made algebraic errors, such as reaching n = 24.5 but then taking the square root instead of squaring.

Answer: 600

Question 4

(i) Many candidates found the combined Poisson distribution with parameter 6.1 and used the
relevant terms to obtain the correct answer 0.857. Some candidates used λ = 0.61. Other
candidates added an extra Poisson term (P(4)) or omitted the first term (P(0)). Most candidates
used P(⩾4) = 1 – P(⩽3). A few candidates attempted to use a Binomial distribution or a normal
distribution. These were not appropriate.

(ii) As the Poisson distribution for the total number of drops in 100 cm3 of air had a parameter of 61
which was large (> 15), the appropriate approximating distribution was the normal distribution N(61,
61). Many candidates realised this and used the necessary continuity correction (59.5) to find the
probability. Some candidates found λ = 61, but then used an incorrect variance such as 23.79 (from
npq). Other candidates omitted the continuity correction factor and used 60, or used an incorrect
value such as 60.5. A diagram can assist in deciding on the value of this continuity correction factor
and also in choosing the correct area (the tail) for the probability.

Answers: (i) 0.857 (ii) 0.424

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

Question 5

(i) Many candidates found the distribution for the total time (T1 + T2) correctly as N(5.0, 0.41),
standardised correctly and found the probability. Some candidates inserted a continuity correction
factor. This was not appropriate. Other candidates incorrectly used 0.4 + 0.5 = 0.9 as the variance.
Other candidates found 0.41 but used it as the standard deviation. The larger area was required
here for the probability.

(ii) Some candidates used the appropriate new variable “Y” = T2 – 1.2 × T1 and found the probability
that Y > 0. Some candidates found the correct distribution N(0.16, 0.4804) for this.

Other candidates incorrectly multiplied T2 by 1.2 or made errors when trying to find the variance.
These included multiplying by 1.2 instead of 1.22 or subtracting instead of adding the separate
variances (common errors for the variance were 0.0196 and 0.442).

A diagram can be helpful for identifying the correct area and probability for Y = T2 – 1.2 × T1 > 0.

This is especially so when the value 0 is not the mean.

Answers: (i) 0.941 (ii) 0.591

Question 6

(i) Many candidates correctly integrated kx–1 between the limits 2 and 6 and equated the result to 1.
The combination of the ln 6 and ln 2 was carried out to obtain the given answer for k. Some
candidates used incorrect limits such as 0 and 6. Other candidates wrongly combined the ln terms
6
to ln .
ln 2

(ii) Many candidates correctly integrated x.kx–1 and applied the limits 2 and 6 to the result of the
4
integration (kx). The result was which then gave 3.64 to the required accuracy. Some
ln 3
candidates multiplied the x and the x–1 incorrectly. Other candidates omitted the x or tried to
integrate the x and the x–1 separately. This was usually unsuccessful as the approach did not
become a correct form of “integration by parts”.

(iii) Many candidates attempted to find the median and then the probability. The median required the
integration of kx–1 for the limits 2 to m (or m to 6) and the use of 0.5. Some candidates followed this
route correctly and found the median to be 3.46(4) (the exact value was 12 ). Other candidates
used incorrect limits such as 0 to m or used limits 2 to 3.64 and incorrectly equated the result to m.
The required probability was found by using the limits 12 to E(X). To obtain an accurate answer
the earlier working needed to be carried out to more than 3 sig. figs.

Some candidates used the method listed on the left hand side of the mark scheme. First they found
P(2 < X < E(X)). Then they noted that P(2 < X < m) = 0.5 so subtracted this 0.5 from 0.545(08) to
obtain the probability.

A few candidates found the median, but then found the variance and attempted to use a normal
distribution. The second part of this was not valid.

1
Answers: (i) k = (ii) 3.64 (iii) 0.045 (2 sig. figs.)
ln 3

Question 7

(i) Many candidates followed a logical route through this significance test. The first stage required the
hypotheses and the unbiased estimate of the variance of the masses. Some candidates stated the
hypotheses incorrectly in terms of x or “the mean”. Some candidates found only the sample biased

© 2018
 Cambridge International Advanced Level
9709 Mathematics November 2018
Principal Examiner Report for Teachers

variance. The second stage required the standardisation of the sample mean (49.8667) using the
46.962
normal distribution N(51, ). Some candidates omitted the 150.
150

The third stage required the comparison and conclusion. The comparison could involve the values
of z (–2.025 < – 1.96) or the probabilities (0.0215 < 0.025) or the critical value. It was necessary to
show the comparison very clearly. Some candidates used inconsistent signs, while others
compared incorrect probability tails. The conclusion should have been given in the context of the
question. Some candidates wrote contradictory statements.

(ii) It was necessary to find the critical value for the rejection (or acceptance) region for the normal
6.8562
distribution N(51, ) as defined by H0. As this critical value was very close to the sample
150
mean value, it was necessary to provide this value to several sig. figs. (49.903). To find the Type II
error it was necessary to work with this critical value in the new distribution with population mean
mass 49 kg. Some candidates successfully followed this procedure. Some candidates reversed the
means and some candidates omitted the 150. Other candidates used +1.96 instead of –1.96.
Some of these errors created difficulties in selecting the relevant probability tail. Some candidates
omitted the first stage and attempted only a standardisation for the second stage.

A range of answers was allowed to score full marks here.

Answers: (i) there was evidence that the population mean mass of sacks was less than 51 kg
(ii) 0.0534

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