Cambridge International Advanced Subsidiary and Advanced Level

9709 Mathematics March 2016

Principal Examiner Report for Teachers

MATHEMATICS
Paper 9709/12
Pure Mathematics

Key message

In question 10, the phrase: ‘showing all necessary working, find by calculation’ was deliberately inserted so
that candidates who simply obtained the area by using the integration function on their calculators with no
method shown, received no credit. Many candidates showed their working out which meant that wrong final
answers could still receive credit for correct working. All candidates would benefit from following this very
important advice as similar questions in the future will very probably have the same pattern of wording.

General comments

The paper proved to be a difficult challenge for many candidates, particularly questions 7 and 9b. The
standard of presentation was generally good with candidates setting their work out in a clear readable
fashion with very few candidates dividing the page into two vertically. Most candidates appeared to have
sufficient time to complete the paper.

Comments on specific questions

Question 1

This question was generally well answered by most candidates. Some candidates struggled with minus signs
and either made both terms positive or both terms negative. Some candidates seemed unfamiliar with the
technique required for part (ii) and ended up with only one term in x 5 or equated their two terms rather than
adding and equating to 0.

Answer: (i) 80 and –32 (ii) 2/5

Question 2

A large majority of candidates realised the need to integrate but many weaker candidates used the equation
of a line formula and received no credit. Some candidates forgot to use + c after integrating and so made no
further progress. The number of minus signs involved in substituting –1 also proved challenging for some but
many fully correct answers were seen.

Answer: y = x 3 + x −2 + 3

Question 3

The majority of candidates gained full marks on this question however some weaker candidates struggled to
interpret the given information and were unable to form the required equations.

Answer: 93

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

Question 4

There was a mixed response to part (a) of this question. Many candidates knew how to solve the equation
but did not give an exact answer as requested in the question. Others were unsure how to start or gave a
positive rather than a negative answer. Part (b) proved more challenging than expected with many
candidates unable to factorise the equation. Attempting to square each term was a common mistake of
weaker candidates. Stronger candidates often lost marks due to cancelling and losing one of the solutions or
giving answers in degrees rather than radians.

− 3 π π
Answer: (a) (b) ,
6 3 2

Question 5

Many candidates scored full marks on this question but weaker candidates were sometimes confused by the
information presented and seemed unsure of what was expected from them. A sketch may well have helped
and candidates should perhaps be encouraged to do one of these for co-ordinate geometry questions. In
part (ii) some candidates correctly found the point X but did not continue to find the distance BX.

1
Answer: (i) y − 3 = ( x − 7 ) (ii) 7.33
2

Question 6

The majority of candidates scored very well on this question although weaker ones struggled in part (i) as
they were unaware of the formulae for the volume and surface area of a cylinder. In part (ii) the vast majority
of candidates realised the need to differentiate and set to zero and were able to cope well with the negative
power. Some weaker candidates were unsure how to deal with π when differentiating and a greater number
were unsure how to confirm that the flask was most efficient at their value of r. Many found the value of A
rather than confirming that the area was a minimum.

Answer: (ii) 5.4

Question 7

This question proved to be one of the more difficult ones on the paper. In part (i) the expected method was
3
to find the length of CA and then deduce that CP was of CA. Many candidates struggled with this and
5
some weaker ones used a circular argument with the answers from part (ii). Part (ii) was similarly found
difficult and although most candidates were very familiar with the method for the scalar product in part (iii)
quite a few used OP with BP instead of using CP.

Answer: (ii) 2.4i+1.2k, 2.4i–2.4j+1.2k (iii) 70.5°

Question 8

Many almost completely correct answers were given to this question but weaker candidates often made no
progress in either part. In part (i) many of them failed to substitute –2 and –3 into the original expression and
so received no credit. The vast majority of candidates however realised the need to do this and then to solve
the equations obtained. A number of sign errors occurred and where candidates then solved these incorrect
simultaneous equations directly from a calculator then no method marks could be awarded. In part (ii) a
significant number of candidates did not realise the need to complete the square first in order to find the
inverse function and so made no progress. A number of sign errors occurred but those who did complete the
square were generally successful, although only the very best candidates realised that due to the given
domain the – rather than the + option was required after square rooting.
3 5 1  13  −13
Answer: (i) a = −2 or , b = 3 or (ii) −  x +  , x≥
2 4 2  4  4

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

Question 9

Part (a) was generally well done but part (b) proved to be a significant challenge even for many good
candidates. In part (i) the vast majority were able to use the angles in a triangle adding up to 180 degrees or
2π radians to prove the given answer. In part (ii) a surprisingly large number of candidates failed to use the
1
formula ab sin C for the area of a triangle. A number of candidates who did find the required area then
2
1
cancelled the with the 2 in sin2α . In part (b) many candidates realised the link with part (a) but were
2
unable to determine the required values of r and α and so were unable to make much progress. Those who
found the area of the triangle ABC and then took away the 3 equal segments were generally successful.

1 2
Answer: (a)(ii) r 2 α − r sin2α (b) 16 3 − 8π
2

Question 10

This question was generally very well done with many candidates scoring full marks. In part (i) a number of
candidates failed to realise the significance of the words ‘write down’ and spent a significant amount of time
expanding and attempting to solve the subsequent equation. In part (ii) the vast majority of candidates
realised the need to differentiate and find the equation of the tangent although some weaker candidates then
set the differential equal to 0 rather than substituting in 3.As was indicated in the key message above, the
phrase: ‘showing all necessary working’ was deliberately inserted in part (iii) so that candidates who simply
obtained the area by using the integration function on their calculators with no method shown, received no
credit. Most candidates realised this and attempted to show their working although some simply wrote down
the answer from their calculator and showed either no working or completely incorrect working. A number of
candidates used the wrong limits after integrating the curve correctly and then took more time than was
necessary integrating the equation of the line rather than simply finding the area of the triangle.

1 5 8
Answer: (i) (ii) (iii)
3 3 9

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

MATHEMATICS
Paper 9709/22
Pure Mathematics

Key messages

Candidates should always ensure that they are working to the correct level of accuracy and giving their
answers to the correct level of accuracy as specified on the front of the examination paper, unless told
otherwise. They should also ensure that they have answered the questions fully.

General comments

The number of entries for this first time of sitting was relatively small, hence making it difficult to comment
fully on the questions and candidate performance. For some questions, an expectation of what was
necessary to answer the question fully has been given. There did not appear to be an issue with the length
of the paper and the time given to answer the questions.

Comments on specific questions

Question 1

Most candidates made use of algebraic long division as was intended. Problems occurred when attempting
to deal with the lack of a term in x , which subsequently lead to errors in the quotient. Correct use of the
Remainder Theorem to obtain a remainder was given credit.

Answer: Quotient 2x 2 − x + 2 , Remainder 6

Question 2

Most candidates attempted to solve the problem by squaring both sides of the given inequality, with
subsequent work leading to the correct critical values. Obtaining the correct range for x was seldom done
correctly with many candidates giving their inequality signs in the ‘incorrect’ direction. Centres should
encourage candidates to check their solutions using a couple of numerical values. A graphical method,
inspection or use of 2 equations were also acceptable methods which could have been used.

2
Answer: x < −8 , x >
3

Question 3

Whilst most candidates were able to deal with the simplification of 2ln x to ln x 2 , very few recognised the
need to use the addition rule for 2 logarithms and were unable then to obtain the required 3-term quadratic
equation. It was expected that candidates would reject the negative solution to this quadratic equation in
order to obtain the final accuracy mark.

3
Answer: x = k
2

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

Question 4

(i) Most candidates do this type of question well, but they must ensure that they are working to the
correct number of decimal places as specified in the question and to also give their final answer as
a rounded answer to 3 decimal places, not a truncated answer. Unfortunately the final answer of
1.515 was too common.

(ii) Very few correct solutions were seen, with candidates mistakenly using their answer to part (i)
1 2
rather than finding an exact solution to the equation x = x + 4 x −3 .
2

Answer: (i) 1.516, (ii) 5

8 or 80.2

Question 5

A correct integration was given by most candidates, together with the correct application of the limits and
equating to 65. Few candidates were able to re-arrange the resulting equation correctly, with problems
involving logarithms being the cause of most incorrect solutions. Some candidates did not give their final
answer to the required level of accuracy as specified in the question.

Answer: 1.097

Question 6

(i) It was important that candidates recognise the need to differentiate the given equation as a product
in order to proceed. Quite a few correct solutions were seen with candidates correctly substituting
x = 0 into their gradient function to subsequently obtain the correct equation for the tangent.

(ii) Usually done correctly by those candidates who attempted this part, many candidates found the x
coordinate as required but omitted to find the corresponding y coordinate; thus the prompt for
candidates to ensure that they have completed a question completely before moving on.

Answer: (i) y = 6 x , (ii) ( 0.554,1.543 )

Question 7

(i) Most candidates recognised the need to differentiate the given equation implicitly, however some
dy
failed to deal with 24 correctly whilst other introduced a spurious into their equation. Many
dx
correct derivatives were seen, but candidates were unable to justify why the gradient of the curve
was never positive.

(ii) Many candidates were able to obtain the correct relationship between x and y , with many
obtaining the solution ( 2,2) . No attempts to gain the second solution were seen. Candidates
should be alert to the fact that equations of this type often have more than one solution.

dy 6x 2
Answer: (i) = − 2 , (ii) ( 2,2 ) and ( 2.88, −2.88 )
dx 3y

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

Question 8

(i) Very well done by the great majority of candidates, recognising the need to obtain both of the terms
on the left hand side of the identity in terms of sine and cosine first.

(ii) (a) Provided candidates realised the need to use the double angle formula most were able to obtain
the given expression in a form that would enable them to find the least possible value. However,
once in this form, candidates did not recognise that the least value of the trigonometric term was
zero and thus the least value of the expression was the value of the constant term obtained.
1
(b) Candidates were expected to obtain the integrand sec 2 2 x , having made use of the identity in
2
part (i).

1 1
Answer: (ii)(a) 3, (b) 3−
4 4

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

MATHEMATICS
Paper 9709/32
Pure Mathematics

General comments

In general the presentation of the work was acceptable and most candidates attempted all the questions. It
was a concern to see that many candidates had not taken on board the comments in the 2014 and 2015
reports. Namely that when attempting a question it is essential that sufficient working is shown to indicate
how they arrive at their answer, whether they are working towards a given answer or an answer that is not
given. For example question 9(ii), clearly showing the substitution of their limits, or question 2, clearly
displaying the substitution of their values in the quadratic equation solution formula.

Candidates should realise that where they have made an error, as often happened in question 2, then it is
even more essential to show the details of the solution of their quadratic equation and not just two incorrect
answers from their calculator. The latter approach will result in the method mark for the solution to this
equation being withheld. Once again the reason for this decision is that the examination is a mathematics
examination, not a calculator examination.

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

Comments on specific questions

Question 1

The solution to this question required, (i) application of the laws of logarithms for product/quotient and power,
(ii) reduction of a logarithmic equation to an equation free of logarithms by taking the inverse, (iii) solution of
a two term quadratic to obtain two roots then (iv) the selection of only the exact positive root, based on
knowing that the logarithmic function is undefined for negative values. Candidates who only produced the
positive solution were awarded full marks, however it would have been better to have seen the negative root
and for this to have been rejected, rather than wonder whether the candidate had failed to realise that there
was another solution to the quadratic equation.
2
Answers: x =
3

Question 2

This question required the use of the correct tan(A±B) formulae. These were usually correctly applied, but
occasionally the signs were incorrect, for example both positive or both negative. Some of the candidates
then made errors in removing the denominators to produce a correct un-simplified horizontal equation. Such
an error meant that only the method mark was still available to score. Unfortunately, as mentioned in the
general comments, some candidates even failed to gain this mark by solving the incorrect quadratic equation
on their calculator.

Answers: θ = 28.7° and 165.4°

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

Question 3

This proved to be a question where candidates could score most of the marks.

(i) It was not sufficient to state f(1) < 0 and f(2) > 0, the actual values were required. However, even
f(1) = − 5 and f(2) = 8, did not score both marks since it was necessary to complete the argument
with a statement that there was a change of sign, and hence a root of the equation between 1 and
2.

(ii) Some candidates opted to work from the rearranged equation to reach the given equation and this
was perfectly acceptable.

(iii) Far too many candidates rounded incorrectly at the end and gave the final answer as x = 1.77.

Answers: (ii) 1.78

Question 4

1
(i) Various successful approaches were seen, for example substituting x = − and equating to zero;
2
dividing by (2x +1) and equating remainder to zero; or showing that factorisation produced the
specific value of a, in fact the latter approach also produced the solution to (ii)(a).

(ii) (a) Since the value of a was virtually always correct candidates had little difficulty in factorising p(x).

(b)
Most candidates obtained the correct answer, but often failed to justify this and so lost marks. The
solution required much more than just stating that the quadratic factor was always positive. Some
candidates correctly showed that the quadratic factor had a constant sign, either by using the
discriminant or by actually finding the complex roots of the quadratic equation. However, this was
still insufficient since it is essential to show that the quadratic term is actually positive. This could
have been achieved by evaluating the quadratic factor at any value of x or by considering the
coefficient of the x2 term.
1
Answers: (i) a = 3 (ii)(a) 2x2 – x +2 (ii)(b) x > −
2

Question 5

(i) dx/dθ was usually correct, but sometimes √3 was omitted. Again with a given answer all the
detailed working was required, for example substitution to obtain 1 + tan2θ, followed by its
replacement by sec2θ and the appropriate cancellation of terms, in order to gain full marks.

(ii) Most candidates used the double angle formula, but sign errors were common, as they were in the
integration process. Too often integration had either sinθ instead of sin2θ or the omission of 2 from
the denominator.

1 3
Answers: 3π+
12 8

Question 6

(i) Some candidates ignored the differentiation of any term involving y. However those candidates that
realised that implicit differentiation, together with the product rule, was required were usually
successful apart from the occasional sign error. Other good solutions were by candidates obtaining
an expression for siny before commencing their differentiation.

(ii) Provided candidates had only made a small sign error in (i) they were able to solve for lnx and gain
dy
the first two method marks. However some candidates believed that setting to zero required
dx
both the numerator AND the denominator to be set to zero.

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 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

dy 1 + ln x 1
Answers: (i) = (ii) x =
dx (2 + ln x )2 e

Question 7

(i) Separation of variables was required in order to make any progress with this question. However,
the candidates who managed to undertake this often experienced difficulty with the integration of
1
. Common to see either this converted to ex or e−x incorrectly integrated as e−x. Although the
ex
integration of xex was usually correct.

(ii) As in the final part of Q6, unless the result of the previous section was virtually completely correct
then it was difficult to score this mark.

Answers: (ii) y = − ln(ex(1 – x))

Question 8

(i) Most candidates successfully showed that the vectors 2i + j + 3k and 2i – j – k were perpendicular,
but omitted to test that no point on the line lies in the plane.

(ii) Few candidates made much progress with this part, incorrectly using either i + 2j –k or 5i + 3j + k
in trying to establish the direction of the line m. Only candidates who opted to determine the vector
product of 2i + j +3k and 2i – j – k were really successful.

Answers: (ii) r = 5i +3j + k + µ(i + 4j +2k)

Question 9

(i) Many fully correct answers, but also lots of arithmetical errors. Too many candidates wasted time in
first opting to find the rational fraction of f(x) – A. This un-necessary working usually lead to more
arithmetical errors.

C + Dx
(ii) Unless candidates had found the constant D in (i) to be zero their integration of the term
x2 + 2
proved difficult. However, this was only penalised in the final accuracy mark. The expression A +
B
was nearly always integrated correctly however the coefficient of the ln(x2 + 2) term was often
x
incorrect. To obtain the final mark with the answer given it was necessary to see the clear
substitution of the limits and ALL the SUBSEQUENT logarithm work. Omission of steps in this
logarithm work, for example −4ln2 +2ln6 – 2ln3 to answer given is NOT acceptable.

Answers: (i) A = 3, B = − 4, C = 4, D = 0

Question 10

(a) Few candidates made any progress with this question after they had substituted for z and z*, since
they did not realise that they had to take real and imaginary parts and solve the resulting equations.
Of the few candidates who did know what to do, often they then had arithmetical errors in their
solution.

(b) (i) The circle inequality was often correct, but again far too many candidates had either the incorrect
centre or a radius which was not equal to 1. The Im z ⩾ 3 statement was usually incorrectly
represented by a circle of radius 3 centred at the origin.

(ii) Only those with the correct diagram in (b)(i) could really attempt this section.

1 2
Answers: (a) z = − i (b)(ii) 0.588 radians or 33.7°
3 3

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

MATHEMATICS
Paper 9709/42
Mechanics

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.

Some candidates lost marks due to not giving answers to 3sf as requested. Marks were also lost due to
prematurely approximating within their calculations leading to the final answer. This was particularly
noticeable in questions 2, 3, 4 and 5 where exact trigonometric values were given and hence there was no
need to evaluate the angles. If angles were evaluated in these cases this lead immediately to a loss of
accuracy.

Students should also be reminded that if an answer is required to 3sf, as is the rubric on this paper for non-
exact answers, then their working should be performed to at least 4sf.

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 it is impossible to achieve the correct given answer unless
this value is used.

Comments on specific questions

Question 1

Since it is not given in the question that acceleration is constant, the best approach here was to use the
Work/Energy principle. Most candidates correctly evaluated the gain in kinetic energy and the work done
against the resistive force. The total work done by the cyclist is the sum of these two values. However, some
candidates found the difference between the values and hence lost the final mark.

For those candidates who assumed constant acceleration, it was necessary to find this acceleration which
most performed successfully and then apply Newton’s second law to find the driving force. Again several
candidates made sign errors when applying this method. In order to find the work done this force is then
multiplied by the distance travelled to give the result. Although it was not given that the acceleration was
constant, candidates applying this method were given the benefit and allowed to score full marks in this
case. However, it is important to read the questions carefully as it may be that only an energy method could
be used in some similar cases.

Answer: Total work done by the cyclist = 5940 J (to 3sf)

Question 2

(i) As the speed of the car is given in the question as constant, then the total force acting on the car is
zero and so the driving force is exactly balanced by the resistive force and hence driving force = F
= 1350 N. Candidates then had to apply the equation P = Fv in order to find the rate at which the
engine was working. Most candidates performed well on this part, although some lost a mark by
not giving the answer in the required form. Again it is well worth reading the question carefully.

Answer: The rate at which the engine of the car is working = 43.2 kW

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

(ii) In this part the car travels up a hill. The sine of the angle is given and this can be used without
finding the angle itself. The car is still travelling at constant speed but now the driving force, F, is
balanced by the resistive force plus the component of the weight of the car. Some candidates only
included one of these forces here. Some also lost marks by forgetting to write 76.5 kW as 76500 W
when using v = P/F to find the speed in ms–1

Answer: The constant speed up the hill = 30 ms–1

Question 3

(i) The majority of candidates scored well on this question. Most resolved the forces horizontally (two
of the forces having horizontal components) and vertically (all three forces having a vertical
component). A few candidates resolved along the directions of the 30 N and 40 N force which was
a perfectly correct method. In order to find the magnitude of the resultant in either case the square
root of the sum of the squares of the two components was needed. An inverse trigonometric
calculation was needed to find the required direction. Care had to be taken if the directions of the
30 N and 40 N forces were used rather than the horizontal and vertical. Many candidates correctly
found the answer to both parts.

Answers: Magnitude of the resultant = 31.6 N Direction of the resultant = 18.4° with the positive x-axis

(ii) In this part all that was needed was to note that the net force in the vertical has to be zero and
since the vertical component had been found by most candidates as 10 N in part (i) then by
replacing the 50 N force by one of 40 N would produce the required result.

Answer: P = 40

Question 4

(i) Most candidates resolved horizontally to find that F = 5cosα. It was then necessary to resolve
vertically to find R from the equation R + 5sinα = 8. Although most candidates correctly found F
many assumed that R was merely mg and so wrongly used R = 8. The required coefficient of
friction can be then be found using F = µR.

Answer: µ = 0.8

(ii) In this part many candidates did not realise that it was necessary to recalculate the normal reaction
due to the new force of 10 N that was now acting. Some also misread the question and added an
extra 10 N to the force and continued with the question using a force of 15 N. In order to find the
acceleration it was necessary to use Newton’s second law applied to the particle using the value of
µ found in part (i) with the new reaction found in part (ii) to determine the friction term.

Answer: The acceleration of P is 8 ms–2

Question 5

(i) The most straightforward approach to finding the acceleration in this part of the question is to apply
Newton’s second law to the system of car and trailer. This does not involve the tension in the cable
which joins them. The alternative approach is to write down the equation of motion for the car and
for the trailer separately and solve the resulting simultaneous equations in T, the tension in the
cable, and a, the acceleration of the system. In order to find T, one of these equations has to be
used. Many candidates were confused as to which forces were acting on each part of the system.
A clear force diagram showing the forces acting on the car and another for the trailer would help
greatly in problems such as these.

1
Answers: The acceleration of the system a = = 0.125 ms–2. The tension in the cable T = 1050 N
8

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

(ii) In this part again the best method was to apply Newton’s second law to the system as only the new
acceleration of the system was needed. Many candidates assumed that the acceleration was
unchanged when the driving force was removed whereas it takes a constant negative value until
the system comes to rest. Once the acceleration was found one of the constant acceleration
equations could be used to find the required time. Most candidates used the correct method to find
the time taken but many used an incorrect value for acceleration.

80
Answer: The time before the system comes to rest is = 26.7 seconds
3

Question 6

(i) This question was well done by many candidates. Horizontal resolution of forces acting on A is
needed and vertical resolution of forces acting on B. Most candidates correctly resolved at B but a
few wrongly included the weight component when resolving horizontally at A. Most correctly found
the acceleration of the system and then used s = ut + ½at 2 with u = 0 and s = 2.5 to find the
required time.

1
Answer: The time taken for A to reach the pulley is 10 = 1.58 seconds
2

(ii) In this part the frictional force acting on A had to be found from the given value of the coefficient of
friction. The equation of motion for B was unchanged. Most candidates found the friction force
correctly and applied it to the equation at A. An error that was often seen was to wrongly use the
same tension as was found in part (i) rather than eliminate the new tension and find the
acceleration a. Once a was found most candidates correctly chose to use the constant acceleration
formula v2 = u2 + 2as with u = 0 and s = 2.5 to find the required speed.

Answer: The speed of A immediately before it reaches the pulley is 6 =2.45 ms–1

Question 7

(i) This part was correctly found by many who realised that they had to use the fact that the velocity is
continuous either at t = 4 or at t = 14 to find the value of k.

Answer: k = 40

(ii) Candidates needed to produce a v–t graph showing the three parts of the motion correctly and
many made a good attempt at this. The first part of the motion from t = 0 to t = 4 caused most
problems. It should be shown as a quadratic curve starting at the origin, passing through the value
of v = –5 at t = 1, crossing the t-axis at t = 2 and taking a value of v = 40 at t = 4. Many wrongly
showed this as a straight line from the origin to (4,40) or showed it as a series of straight lines
rather than a curve and hence lost the mark. The second part of the motion was represented by a
horizontal line from (4,40) to (14,40) and was correctly shown by most candidates. The final part
was a straight line with negative slope from (14,40) to (20,28). Some wrongly showed this part as a
line which returned to the t-axis at t = 20. Most candidates scored 2 out of 3 on this part. When
drawing a sketch of a v-t graph or similar, the axes should be well annotated at all of the key points
in order to score full marks.

Answers: A quadratic curve from (0,0) passing through (1,–5), (2,0) and (4,40)
A straight line from (4,40) to (14,40)
A straight line from (14,40) to (20,28)

(iii) This part could be answered directly from the v-t graph by spotting that the acceleration is positive
only from t = 1 up to t = 4. Alternatively the expression for v could be differentiated to find the
acceleration and in the first region this gives a = 10t – 10 which shows that a is zero at t = 1.

Answer: 1 < t ⩽ 4

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

(iv) Almost all candidates found difficulty with this question. Since the question asked for the total
distance travelled, great care had to be taken over the treatment of the first 4 seconds of motion.
Most candidates integrated the quadratic expression from t = 0 to t = 4 which gives the
displacement at t = 4 not the distance travelled due to the fact that between t = 0 and t = 2 the
integral is negative. Between t = 4 and t = 20 the distance travelled can be evaluated directly from
2 4
the area under the graph and most candidates found this part correctly. If A = ∫ 0
v dt , B = ∫ 2
v dt

and C is the area under the v–t graph from t = 4 to t = 20 then the total distance travelled is given
by – A + B + C

Answer: The total distance travelled by P in the interval 0 ⩽ t ⩽ 20 is 644 m

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

MATHEMATICS
Paper 9709/52
Mechanics

General comments

This paper cannot be compared with a previous one as this is the first time a March paper has been set. If
compared to last November's paper or last June's paper then the standard was very similar.

Most candidates are now using g = 10 as instructed on the front page of the question paper.

The easier questions were 1, 2(ii), 4, 5(i) and 6(i).

The harder questions were 2(i), 3, 5(ii), 5(iii) and 7.

The hardest question proved to be number 7.

Comments on specific questions

Question 1

Many candidates used tanθ = y/x instead of tanθ = v y / v x , where θ is the angle of projection.

If V is the speed of projection and θ the angle of projection, then 2 equations can be found by resolving
horizontally and vertically. These equations are Vcosθ = 15 and Vsinθ = 20.
By squaring and adding V = 25 and by dividing tanθ = 20/15, θ = 53.1.

Answer: Initial speed = 25 m s −1 , Angle of projection = 53.1°

Question 2

(i) Few candidates were able to do this part of the question. It was necessary to take moments about
the point of contact of the hemi-sphere with the plane. The problem was finding the perpendicular
distance of P from this point. This distance was 0.8 – 0.8cosθ.

(ii) This part was quite well done. It was done by applying F = µR, where µ = coefficient of friction.

Answers: (i) P = 8.75 (ii) Coefficient of Friction = 0.146

Question 3

This question proved to be quite difficult.

Let V be the speed of the particle at the 2 points when the particle makes an angle of 45° with the
horizontally. The horizontal speed is unchanged and so Vcos45 = 9cos60. This leads to V = 9 2 /2. Using
the vertical motion between the 2 points leads to –Vsin45 = Vsin45 – gt.
This results in t = 0.9. Finally by using horizontal motion s = 9cos60 × 0.9 = 4.05 m.

Answer: Distance = 4.05 m

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

Question 4

This question was generally well done.

(i) Candidates were required to take moments about both BC and BAG.

(ii) Most candidates gave the correct answer of 45°.

(iii) A few candidates found the angle with the horizontal instead of the vertical.

Answers: (i) h = 0.775, v = 0.775 (ii) 45° (iii) 66.7°

Question 5

(i) Most candidates scored both available marks.

(ii) Only a few candidates were able to solve this part of the question. A 5 term energy equation was
needed in order solve this part. Too often either 1 or 2 terms were omitted.
If d is the distance below the equilibrium position, when the speed is 3.5 m s −1 , then the equation is
0.6 × 4.52 /2 + 24 × 0 .22 /(2 × 0.8) + 0.6gd = 0.6 × 3.52 /2 + 24(d + 0.2 )2 /(2 × 0.8). This leads to d =
0.4 and so AP = 0.8 + 0.2 + 0.4 = 1.4 m.

(iii) Only a few candidates scored any marks on this part of the question. This time a 4 term energy
equation was required, as follows:
24 × 0.22 /(2 × 0.8) + 0.6 × 4.52 /2 = 0.6 v 2 /2 + 0.6g × 0.5, where v is the required speed.

Answers: (i) Extension = 0.2 m (ii) AP = 1.4 m (iii) Speed = 3.5 m s −1

Question 6

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

(ii) Most candidates attempted to integrate to find v. Some candidates arrived at v = f(x) and not v 2 =
f(x).

(iii) The value of x, where P comes to rest, was attempted by equating v to zero.

Answers: (i) 2vdv/dx = 10 –( 3 ) x 2 (ii) Maximum speed of P = 4(.00) m s −1 (iii) x = 4.16

Question 7

This question proved to be the most difficult question on the paper.

(i) Too many candidates mixed up the trigonometric ratios when attempting to resolve vertically. The
result should be Rcos60 + Tcos30 = 0.2g.

(ii) An attempt to use Newton's Second Law horizontally was often used. Unfortunately the radius used
was 0.6 instead of 0.6sin60.

(iii)a Candidates realised that they needed to solve the 2 simultaneous equations already found.
Unfortunately the 2 equations were often incorrect.

(iii)b Only a few candidates realised that the greatest value of T was when R was zero. An attempt at
the speed was seen but the wrong radius was again used.

Answers: (i) R + T 3 = 4 (ii) T – R 3 = 0.12 ω 2 3 (iii)a R = 0.6

(iii)b Maximum T =2.31 and speed = 1.73 m s −1

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

MATHEMATICS
Paper 9709/62
Probability and
Statistics

Key messages

To do well in this paper, candidates must work with 4 significant figures or more in order to achieve the
accuracy required.

Candidates should be encouraged to show all workings. In the event of a mistake being made, credit can be
given for the method. A wrong answer with no workings scores no marks. A number of candidates did not
show sufficient working to make their approach clear and were unable to gain full credit.

When drawing graphs, candidates should use sensible scales that enable accurate readings to be achieved
and label axes including units.

General comments

It was pleasing that many of the candidates who took this paper had a good knowledge of the syllabus. The
paper allowed candidates to demonstrate their knowledge of basic skills.

Comments on specific questions

Question 1

The use of the coded mean appeared to cause some difficulty for candidates.

(i) The majority of candidates who attempted the question recognised that all that was required was to
multiply the mean by the number of values. A number chose to calculate the value of the constant
a initially and then solved Σ(x – a) = 362

(ii) Good solutions expanded Σ(x – a) as Σx – Σa, substituting their answer from (i) into the equation
before solving. Weaker solutions failed to replace Σa with 10a, and thus solved an incorrect
equation. An alternative approach used was to consider a single term x – a, which was then
equated with 36.2.

Answer: (i) 862 (ii) 50

Question 2

This was a standard ‘probability without replacement’ question. Good solutions often included a tree diagram
to help identify required outcomes, with the best recognising that there were only 2 options necessary (white
or non-white roses). The alternative approach of considering combinations of roses was successfully used by
many. A common error was to ignore choosing the third rose as there were only 2 white roses available.
There requirements of the probability distribution table were understood, but an attempt at the probabilities
was necessary before credit was available. Solutions were seen where candidates used the fact that the total
probability was 1 to calculate the third probability.

Answer: P(0) = 7/15, P(1) = 7/15, P(2) = 1/15

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

Question 3

Candidates should be aware that not all dice have 6 sides, as a number of solutions were presented which
ignored the 7 and 8. It was unfortunate that a few candidates did not show any workings in this question as
they stated incorrect probabilities which may have come from undertaking processes which would have
gained credit.

(i) The best solutions drew an outcome space and identified the outcomes that fulfilled the
requirement of R. Many solutions listed just the 5 different pairs of values that were possible.
Weaker solutions did not recognise that the values could be reversed, and so did not double the
probability.

(ii) The best solutions used the outcome space constructed for (i) to identify the necessary terms.
There was little confusion in understanding the meaning of ‘product’ although the value was not
always calculated accurately. A number of solutions were seen where inaccurate counting of
correctly marked terms occurred.

(iii) Most candidates recognised that they needed to consider P(R∩S) to determine independence.
Good solutions stated the requirements for independence, used the previous parts within their
arguments, stating clearly each calculation and reaching a final conclusion. A number of circular
arguments were presented, where P(R) x P(S) was calculated and then stated as equal to P(R∩S).
Errors made in (i) were often repeated in this part, but appropriate comparisons were valid.
Candidates should be aware that they are expected to state a conclusion from their investigation,
rather than just calculating the values. A few candidates considered whether R and S were
exclusive, and provided the contra evidence.

Answers: (i) 10/64 (ii) 28/64 (iii) not independent

Question 4

(i) The majority of candidates calculated the frequency correctly. A surprising error was stating the
class width here.

(ii) The best solutions presented a table which allowed for the calculation of both the class width and
the class frequencies. These were then used to calculate the frequency density within the table.
However, some simple arithmetical errors were identified at this stage. Weaker solutions used a
continuity correction, which was unnecessary as time is a continuous variable. Good solutions then
used scales to ensure that all values necessary to construct were on the grid lines, which resulted
in accurate histograms. Common errors were failing to state the units on the horizontal axis, or to
label as class width. Weaker solutions often drew either a cumulative frequency graph or a bar
chart for either the frequency or cumulative frequency. A number of histograms were seen drawn
without a ruler, which is not acceptable at this level.

(iii) The best solutions continued using the table constructed in (ii) to calculate the mid-value, and
hence the estimate of the mean. Where a continuity correction had been applied in (ii), there was a
resulting accuracy error at this stage. The weakest solutions used the mid-value and the
cumulative frequency within their calculations.

Answers: (i) 32 (iii) 34.7

Question 5

(i) This question attempted by almost all candidates. The best solutions used a tree diagram, which
clarified the possible outcomes and ensured the appropriate probability was identified. More able
candidates often simplified the question by considering only camping and non-camping holidays.
Weaker solutions did not recognise that this was a conditional probability question, and simple
stated P(A∩C). A surprising number of simple arithmetical errors were identified within solutions.

(ii) Where attempted, candidates had a good understanding of the process required. The best
solutions stated clearly the required power equation, and then either stated answers calculated
from Trial and Improvement or used the properties of logarithms to solve. A number of candidates

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

penalised themselves by not copying the values carefully from the question, using 0.02 was the
most frequent error.

Answers: (i) 0.168 (ii) 15

Question 6

The context of this question was accessible by all but the weakest candidates. A common error was to use
the values for males and females inconsistently through the question.

(i) The majority of candidates recognised that this was an ‘arrangement’ question and used
permutations correctly, although a few considered 5P5. A number of candidates rounded their exact
answer to 3 significant figures. Candidates can be advised that more accurate exact answers are
acceptable.

(ii) The anticipated method was to consider the number of ways each position could be chosen, and
then recognising that the values should be multiplied to determine the number of lists. Good
candidates often used permutations for the male and female groups before multiplying. A number
of candidates considered the combinations for the males and females and then the number of ways
these could be arranged.

(iii) The best solutions identified the different gender combinations that met the stated requirement, and
then calculated the separate possibilities before summing. Almost all candidates recognised the
key word ‘selections’ and used combinations appropriately. Some solutions were difficult to follow
because the work was not presented in a logical way. A significant minority did not include the
group of 5 males and 0 females within their final answer.

(iv) The best solutions used the same process as in (iii) for the 3 remaining places. Common errors
were not reducing the number of possible applicants once Mr and Mrs Blake had been chosen, or
doing so inconsistently.

Answers: (i) 360 360 (ii) 5400 (iii) 501 (iv) 58

Question 7

(i) Many good solutions contained a sketch of the normal distribution to clarify the information
contained within the question. Candidates should be aware that the tables provided include critical
values for the normal distribution, which was the anticipated source for the z-value. Most solutions
used the standardisation formula correctly, with accurate algebraic manipulation to achieve an
answer. A few solutions appear to have used a continuity correction, which was not required as
time is a continuous variable. There was some confusion as to the area that was being considered,
which the use of a sketch could have assisted with.

(ii) The best solutions correctly interpreted the question as requiring all 4 cars to have taken less than
2 hours to fit. Weaker solutions considered P(>2) and then used this to calculated P(<2). A number
of candidates found these values correctly but then failed to complete the question. It was
unfortunate that many candidates were unable to gain full credit as they had not worked to a
sufficient degree of accuracy in their calculations. Candidates are reminded that to achieve 3
significant figures, then at least 4 significant figures is required throughout and that they need to
use a more accurate value for the mean than the acceptable answer in (i).

(iii) Good solutions developed the method from part (ii). The general standardisation formula was
stated, and then appropriate algebraic values substituted. Weaker solutions displayed poor
algebraic manipulation, with either the µ being replaced inconsistently by 3σ in the numerator, or
the µ / 3 in the denominator not being simplified accurately.

Answers: (i) 1.48 (ii) 0.758 (iii) 0.885

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

MATHEMATICS
Paper 9709/72
Probability and
Statistics

General comments

In general, candidates scored well on question 6 whilst questions 1 and 2 proved more demanding.
Candidates were largely able to demonstrate and apply their knowledge in the situations presented, though
interpretations of some scenarios proved difficult for some. There was a complete range of scripts from good
ones to poor ones.

Most candidates kept to the required level of accuracy, though as is often the case, there were situations
where candidates lost marks for giving final answers to less than three significant figure accuracy, this was
particularly seen on question 6(ii) where errors could have been made because of confusion between 3 s.f.
and 3 d.p.

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

This was not a well attempted question. Candidates who successfully found E(X) and E(Y) usually went on to
successfully find E(X+Y). However, Var(X) and Var(Y) were not always correctly found, and even candidates
who did have the correct answers of 25/9 and 5 did not always combine them successfully. A common error
was to leave the answer as the variance; the question required the standard deviation.

Answer: 13.3 2.79

Question 2

Candidates were required to set up their Null and Alternative Hypotheses; this was omitted by some
candidates. The probability of 18, 19 or 20 hits should then have been calculated using the parameters in the
given Binomial situation. A valid comparison was then required (which candidates must clearly show) before
the relevant conclusion can be drawn. Errors on this question included attempting invalid approximations, or
using incorrect parameters, or merely calculating P(18). It should be noted that the final conclusion drawn
should not be written as a ‘definite’ statement but merely as ‘probable’ or that ‘evidence indicates’ etc.

There is no evidence at the 1% level that she has improved

Question 3

Again, Null and Alternative Hypotheses should have been set up at the start of this question. Many
candidates successfully found the probability of a Type I error, but common mistakes were to omit √25 when
standardising. In part (ii), many candidates successfully explained why a Type II error was not possible.

This question was attempted more confidently than has been previously found with questions on Type I and
Type II errors.

Answer: 0.328
Ho is rejected but Type II error can only be made if Ho is not rejected

© 2016
 Cambridge International Advanced Subsidiary and Advanced Level
9709 Mathematics March 2016
Principal Examiner Report for Teachers

Question 4

This was, generally, a well attempted question. The main error seen was in calculating the variance of X–2Y;
the calculation required was 0.22 + 4 x 0.12 and not 0.22 + 2 x 0.12. Some candidates incorrectly attempted
2X–Y.

Most candidates successfully standardised with their values for the mean and variance, and chose the
appropriate probability.

Answer: 0.362

Question 5

This was a question on which many candidates scored well. Most candidates successfully found unbiased
estimates of the mean and variance (though a few candidates found the biased estimate of the variance, and
therefore could not gain full marks). Other errors included use of an incorrect ‘z’ value and it was important
that the final answer was written as an interval and not as two separate answers.

In part (ii) candidates should have explained that ‘100’ was within the confidence interval before drawing
their conclusion. Many candidates realised that a random sample was required to avoid any bias.

Answer: 95.8 to 103

100 lies within the CI, so ‘yes’
To avoid bias or to enable statistical inference

Question 6

Candidates demonstrated good knowledge on this question.

In part (i) most candidates used the correct value of λ to find the required probability. Some candidates
incorrectly interpreted ‘fewer than 3 times’ and consequently their calculation had an extra unrequired term.

In part (ii) (a), candidates mostly applied the correct approximating distribution (though not all were able to
justify this in part (b)). Common errors included an incorrect, or no, continuity correction when standardising.

Many candidates found the correct λ in part (iii) but, on occasion’s, premature approximation led to an
incorrect value of λ for some candidates. Similarly, as in part (i), a misinterpretation of ‘at least 4’ caused
errors to be made in this calculation as well.

Answer: 0.481
0.0513
λ>15
0.228

Question 7

Many candidates used integration for some parts of this question, rather than using the symmetry of the
given diagram. These candidates were often able to reach the required answers, but gave themselves time
penalties by using longer methods.

Part (a) was well attempted, though many candidates used an incomplete formula for the variance omitting
to subtract X 2 to reach the required answer. Few candidates were able to ‘write down’ E(X) and, as
mentioned above, attempted a full calculation to reach 0.5. Similarly part (iii) was mostly done by integration.

Part (b) was done well by the better candidates. Weaker candidates confused the method for finding the
median, and often, incorrectly, equated an integral to 2 rather than using 2 as a limit and equating the
integral to 0.5.

Answer: 0.45
0.5
7/27
2√2

© 2016