Examiners’ Report
January 2011
IGCSE
GCE
GCE Mathematics (6663-6689)
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January 2011
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2
Core Mathematics Unit C1
Specification 6663
Introduction
The paper seemed to be appropriate to the entry, differentiating between strong and poor
candidates. There were some excellent solutions and most candidates were able to make
reasonable attempts at the questions. Lack of time did not appear to be an issue. Poor basic
arithmetic was a common deficiency and it was disappointing to see errors of the type:
5d = 8 so d=5/8 or even 3(1) = 4 and –4  (–1)2 = +4, etc.
Some candidates had difficulty dealing with fractions in any form. Algebraic errors indicated
a failure to see the difference between an equation with 2 sides (where both sides may be
multiplied by the same number to eliminate a fraction) and an expression, which one cannot
simply ‘double up’ to conveniently lose the denominator.
Graphs were easier to see this session and few had used a light pencil making it difficult to
read on ePEN. Candidates should be advised that black ink is best and pale blue or pencil can
appear illegible on scanned scripts. Also if two graphs are drawn then it can be difficult to
ascertain which is the correct one. Candidates should be encouraged to re-start if they have
drawn incorrectly rather than try to resurrect their answer.
Report on individual questions
Question 1
Overall this question was done poorly, with very few candidates scoring full marks. There
were, however, many correct answers to part (a). Candidates were usually able to deal with
either the negative part of the power or the fractional part, but some had problems in dealing
with both. There were also some who did not understand the significance of the negative sign
in front of the 14 and the fact that it implied a reciprocal. Some assumed it implied a negative
final answer, e.g. –2.
Part (b) was successfully completed by very few candidates, with the vast majority of errors
being caused by the failure to raise 2 to the power 4. The power of the x inside the bracket
was also often incorrectly calculated. It was common for candidates to multiply by x before
raising to the power and so ending up with either 2x3 or 16x3. Others added the powers – 14
and 4. Even when the bracket was correctly expanded, the extra x was often omitted or not
combined with the other term.
Question 2
This question was attempted by all and was well answered by most candidates. The
integration was generally recognised and usually carried out correctly with many candidates
scoring full marks with two or three lines of working. Only very few tried to differentiate.
Virtually all knew that they had to increase the power by 1 and then divide by the new power.
Usually if mistakes occurred it was when simplifying. The third term presented the most
3
challenge, highlighting weaknesses in dealing with fractions and reciprocals: many had
difficulty when trying to simplify terms involving fractions such as 4/(4/3). The third term
4
3
was often left as 4x . The constant of integration was missed in only a minority of cases
resulting in the loss of the final mark.
Question 3
This was generally well done with most candidates correctly multiplying both numerator and
denominator by the same correct expression. (√3 + 1) was the expected choice but a
surprising number used (–√3 – 1) instead. They then usually obtained 2 (or –2) in the
denominator and most candidates were able to expand the numerator to obtain 4 terms.
Some expanded 5 – 23 (3 + 1) instead of (5 – 23) (3 + 1), simplifying the question and
not earning the method mark. Some had difficulty dealing with the simplification of 2√3  √3.
A number of candidates lost the final mark by unwisely multiplying through by 2 or by failing
to express their answer as two separate terms.
Question 4
On the whole, this was a high scoring question, with most candidates understanding the
notation and 45% obtaining full marks. Almost all candidates earned the first mark for 6 – c
or 3  2 – c, given as their answer in part (a). A correct expression for the third term was seen
regularly, occasionally followed by incorrect simplification to 18 – 2c or even 18 – c.
Candidates who attempted to use the formula for the sum of an Arithmetic Progression lost
the final three marks. A few candidates simply equated the expression for the third term to
zero and solved to find c, ignoring or not understanding the summation.
Question 5
In part (a) there were many well drawn correct graphs with the new asymptotes clearly
labelled. Where asymptotes were correct the most common error lay in the position of the left
hand branch of the curve, which was either drawn through the origin or crossed the negative
axes. Most candidates recognised a translation and all manner of one unit translations,
including movement in both x and y directions at once, were seen.
The first mark in part (b) was gained by many for marking the required point on the x-axis.
A number of candidates stopped at this point. Others tried substituting x = 0 into the original
equation. Better candidates obtained the y-intercept by evaluating f(–1) and usually scored full
marks (with only a few leaving their answer as “ 13 ” without indicating anywhere that this was
the y coordinate of the intercept). Those that attempted to find an algebraic expression for
f(x – 1) often scored the first M1, but a number of these did not make sensible use of it (i.e.
did not substitute x = 0) and so did not score the second M1. M0M1A0 was reasonably
common, often awarded for using x = 0 in f(x) – 1.
Some horrendous algebra was seen by those struggling to find the y intercept in this part and
even attempts to solve (x – 1) = x/(x – 2) were tried in some cases.
4
Question 6
There were many excellent, well presented solutions with 55% gaining full marks. The
majority gained full marks for (a) using the Sn formula. The formula was not always stated
and candidates should take care to show sufficient method in ‘show that’ questions.
A minority worked from first principles, writing out all of the terms and adding them but did
not get the credit if they missed out terms.
Common incorrect equations seen in (b) were: 6a + 15 d = 17 (from finding the sum of
6 terms), a + 16d = 17 and a + 6d = 17. In some cases, 17 and (a + 5d) were seen but not
equated.
In part (c), the elimination method was favoured, but some careless arithmetical errors were
8
made. Sometimes
was changed to an incorrect decimal, e.g. 1.4, which meant that their
5
value for a was incorrect (if they found d first). There were several algebraic mistakes in
5
part (c), such as 5d = 8 then d = .
8
A few candidates omitted to calculate a second variable.
Question 7
This question was done well by most candidates and the actual process of integration was well
practised. There were a large number of completely correct responses. Some candidates
however did not realise that the constant of integration had to be found and stopped after
integrating. They lost the final two marks. For those who continued the majority of errors
arose because they incorrectly evaluated their expression with x = –1. This was due to the
minus sign, which had to be cubed and squared. Some who did substitute correctly failed to
realise that their expression in c needed setting equal to zero and so they made a false
conclusion leading to c = –9.
Question 8
In part (a) most candidates appreciated the need to use b2 – 4ac and the majority of these
stated that b2 – 4ac >0 is necessary for two real roots. Some candidates however only
included the inequality in the final line of the answer. They should be aware that a full method
is needed in a question where the answer is given. The algebraic processing in solutions was
usually correct but common errors were squaring the bracket to give k2 + 9 and incorrect
multiplication by –4.
In part (b), the critical values of –3 and 1 were generally found by factorisation but many
candidates struggled to give the correct region; others used poor notation 1 < k < –3.
Candidates who gave their final answer in terms of x lost the final accuracy mark.
5
Question 9
Part (a) was done well by the majority of candidates. Most were able to obtain k = 5 after the
substitution of the coordinates for A.
To find the gradient in part (b) most candidates realised that they needed to rearrange the
equation of the line into the form y = mx + c, and the vast majority were able to do this
accurately, with only a few getting mixed up with signs. Unsurprisingly, those candidates that
attempted differentiation on the given equation without first rearranging to y = mx + c were
generally unsuccessful in determining the gradient. A significant number of candidates found
a second point on the line and used the two points to find the gradient. Many candidates gave
their answer as 1.5 which sometimes caused them problems when finding the negative
3x
reciprocal in part (c). Common incorrect gradients were 3 and
.
2
Part (c) was done quite well. Most candidates were able to write down an expression for the
negative reciprocal. It was pleasing to see so many of them writing down the correct form, i.e.
1
before attempting to work it out. The most common error here was the ‘half remembered’
m
2
3
negative reciprocal leading to
or – . Many of the candidates who failed to obtain the
3
2
correct gradient in part (b) were able to score the majority of the marks here. Most candidates
were able to use the negative reciprocal gradient to write down an expression for equation of
L2. Methods of approach were roughly equally divided between those using y – y1 = m(x – x1)
and y = mx + c. Those using the former method were generally more successful in scoring the
first accuracy mark. Only the better candidates were able to simplify their equation into the
correct form.
In part (d), many candidates were able to substitute y = 0 into their equation to find the
coordinates of B. By far the most common mistake (from about 20% of the candidates) was to
substitute x = 0 into their equation. The next most common error here was to substitute y = 0
correctly but then not being able to solve their equation for x.
In part (e), it was pleasing to see so many candidates able to make a good attempt at finding
the distance between the points A and B. Many drew diagrams and many quoted the formula.
Relatively few candidates this session got mixed up when determining the differences in the
x values and the differences in the y values. However, candidates should still be advised to
draw a diagram or to quote the formula before attempting to work out the differences. The
correct answer of √52 was frequently seen with 38% of candidates scoring full marks on this
question.
Question 10
For part (a)(i) the majority of candidates drew a curve which was recognisably of a cubic
form, although the occasional straight line and other non-cubic curves were seen. Very few
candidates did not label the points where the curves crossed the axes, but it was quite common
to see the curve passing through (–3, 0), (–2, 0) and the origin.
6
The most common error was to draw a “positive” cubic curve, not appreciating that the
equation of the curve was of the form y = –x3 + ....... ; even having made this error, however,
many candidates were still able to gain three marks for this curve.
Most candidates seem to know that the equation in part (a)(ii) represents a rectangular
hyperbola, and the majority placed the branches in the correct quadrants, although it was not
uncommon to see them placed in the first and third quadrants, and occasionally in the first and
second. Although the curves were sympathetically marked, it should be said that some of the
sketches of the hyperbola were quite poor, some looking as though they had asymptotes at
x = –2 and x = +2, and some needing examiners to have quite an imagination to see the axes
as asymptotes.
In part (b), only candidates who had correctly positioned graphs were able to gain both marks
in this part; some, but by no means all of this group, clearly had a good understanding of what
was being tested here and gained both marks. Candidates with an incorrect sketch were still
able to gain the first mark, if their answer was compatible with their sketch, and supported
with an acceptable reason. A disappointingly large number of candidates, however, did not
seem to appreciate how their graphs could be used to provide the number of real roots, often
giving the number of intersections with the x-axis. Some candidates did not refer to their
sketch at all and often did quite a bit of work trying to find the actual roots.
Question 11
As a last question this enabled good candidates to demonstrate an understanding of the
techniques of gradients, applying problem solving and logical skills to achieve the final
equation. 26% achieved full marks in this question. There were very few blank scripts or
evidence of candidates who did not have time to complete the question. Usually if candidates
did have difficulty, it was because they had made a mistake in answering the early part of the
question. In part (a) most candidates were able to differentiate the equation correctly, although
there were some problems with coefficients. Most mistakes occurred when differentiating 8x
with candidates being unable to rewrite it as 8x1 prior to differentiation, or losing the term
completely on differentiation. This term also caused candidates problems in the subsequent
substitution of numbers which resulted in many strange results. Again, as in question 2, an
inability to deal with fractions was seen.
In part (b), the usual approach was to substitute (4, 8) into the equation and show
that –8 = –8. Cases where candidates substituted x = 4 mistakenly into their gradient instead
of the equation of the curve C were frequent, although sometimes corrected. Substituting into
fractional items proved to be too much for some candidates and consequently elementary
mistakes were made. Simplification of the third term to 72 caused the most problems (many
getting 54).
In part (c) there was again the occasional mistake of substitution into the wrong expression.
Those candidates who correctly found the gradient of the curve, at the point P, usually went
on and found the equation of the normal without any trouble.
Arithmetic was often poor and it was common to see 24  27  12   52 and other numerical
slips. However even those candidates who had made an error initially then attempted to find a
7
perpendicular gradient and went on to use it successfully in finding the equation of their
normal. Very few used the gradient of the tangent in error. Where candidates used y = mx +c
the calculations for c were often numerically incorrect and followed long, complex (often
messy) workings.
Presentation in this question varied from some excellent easily followed solutions to some
with little coherence.
8
Core Mathematics C2
Specification 6664
Introduction
This paper proved to be very accessible to many of the candidature and there was little
evidence of candidates being short of time. The paper afforded a typical E grade candidate
plenty of opportunity to gain marks across the majority of questions.
The standard of algebra seen was good, although a number of candidates made basic sign or
manipulation errors. The design of the question booklet continues to help candidates to
present their solutions well and an overwhelming majority of them were able to give their
solutions to all questions in the spaces provided.
In question 2 there were a significant number of candidates who worked in degrees and
converted their final answers to radians. Some candidates, however, worked completely in
degrees.
In summary, questions 1, 3, 4, 6(a), 6(b), 7(a) and 10 were a good source of marks for the
average candidate, mainly testing standard ideas and techniques; and questions 5, 7(b), 9(d)
and especially question 8 were discriminating questions at the higher grades.
Report on individual questions
Question 1
This question was accessible to the nearly all the candidates with the majority of them
attempting to use the remainder theorem. In part (a), many candidates substituted x  1 into
the f ( x) expression and were able to achieve the required a  b  3. A few candidates,
however, substituted x   1 into f ( x).
In part (b), most candidates attempted to find f ( 2) and applied f (2)   8 to give
16  8  8  2a  b   8 . A significant minority of candidates incorrectly simplified (2) 4
as 16 and a few candidates incorrectly set their f ( 2) equal to 8 or even 0.
Poor manipulation was also a common feature, with some candidates simplifying
16  8  8  2a  b   8 incorrectly to give  2a  b  8. The need to solve the two
equations simultaneously was clearly understood and generally correctly applied, although
those candidates who had made sign or manipulation slips earlier were unable to access the
final two marks for finding both a and b.
A small minority of candidates attempted to use a method of long division in parts (a) and (b).
The majority of these candidates usually failed to achieve a remainder in a and b which was
independent of x. Some able candidates, however, handled long division with confidence and
gained full marks in both parts of the question.
9
Question 2
This question was well answered with a considerable number of candidates gaining full
marks. It was rare to see a solution assuming that the triangle was right-angled, although there
were a few candidates who did not proceed beyond using right-angled trigonometric ratios.
In part (a), the majority of candidates were able to correctly state or apply the correct cosine
rule formula. In rearranging to make cos C the subject a significant minority of candidates
incorrectly deduced that cos C = 141 . A negative sign leading to an obtuse angle appeared to
upset these candidates. The more usual error, however, was to use the formula to calculate
one of the other two angles. This was often in spite of a diagram with correctly assigned
values being drawn by candidates, thus indicating a lack of understanding of how the labeling
of edges and angles on a diagram relates to the application of the cosine rule formula.
Although the question clearly stated that the answer should be given in radians, it was not
unusual to see an otherwise completely correct solution losing just one mark due to candidates
giving the answer to part (a) in degrees. It was also fairly common to see evidence of
candidates preferring to have their calculator mode in degrees, by evaluating their answer in
degrees and then converting their answer to radians.
Part (b) was a good source of marks, with most candidates showing competence in
1
using ab sin C correctly. Of those candidates who “really” found angle A or B in part (a),
2
1
most assumed it was angle C and applied (7)(8) sin(their C ) , thus gaining 2 out of the
2
possible 3 marks available. A few candidates correctly found the height of the triangle and
1
applied (base)(height) to give the correct answer.
2
Question 3
The vast majority of candidates found this question to be accessible and problems seen were
usually concerning signs.
In part (a), a majority candidates were able to write down both ar  750 and ar 4   6 and
proceed to correctly find the value of r. A minority of candidates displayed poor algebraic
6
6
skills, giving incorrect results such as ar 3  
or r 4  r  
or r 3  756 .
750
750
A significant minority of candidates were unhappy with a negative value for r 3 and thus r and
this problem with signs would then persist in parts (b) and (c). A very small number of
candidates confused geometric series with arithmetic series.
In part (b), most candidates were able to substitute their value for r into a correct equation that
they had written down in part (a) in order to find the first term of the series.
In part (c), many candidates were able to write down the correct formula for S . Some
1
candidates who had correctly found r as  , incorrectly interpreted the condition of r  1 to
5
10
1
. Some candidates believed that a sum to infinity
5
can only be positive and so arrived at an incorrect answer of 3125. Some candidates who had
earlier found a value of r whose modulus was not less than 1, were happy with substituting
this into the correct sum to infinity formula, and did not then deduce or were aware that their
value for r found in part (a) must then be incorrect.
mean that their r in part (c) should then be
Question 4
This question was very well attempted by the majority of candidates. It was rare to see errors
in part (a). In part (b), most candidates expanded correctly and went on to integrate
successfully, gaining the first four marks, although a few candidates differentiated instead of
integrating. Some candidates could not cope with the negative result and tried a range of
100
ingenious tricks to create a positive result. A common error was to take 
to be positive
3
8
and then subtract . This incorrect use of limits meant some candidates lost the final two
3
marks. There were a significant number of errors in evaluating the definite integral.
Disappointing calculator use and inability to deal with a negative lower limit meant that a
significant minority of candidates lost the final accuracy mark. Some candidates used 1 as
their lower limit instead of 1, and lost the final two marks for part (b). A few candidates
correctly dealt with a negative result by reversing their limits whilst others multiplied their
expression by 1 before integration to end up with a “positive area”.
Question 5
A significant number of candidates failed to answer part (a) correctly, due to the unfamiliarity
n
with the formula for   . Common incorrect answers for b included 1, 4, 10, 36! or 91390.
r
In part (b), most candidates were able to write down the binomial expansion of (1 + x)n.
Although a minority of candidates picked out wrong terms, most commonly terms in x3 and
36
q
x 4 rather than x 4 and x 5 , the majority of candidates were able to to give
. Other
as
5
p
p
q
common errors included finding
or giving
as 7.2x, which is not independent of x.
q
p
11
Question 6
In part (a), the vast majority of candidates correctly evaluated both y-values to 2 decimal
places, although a significant minority of candidates lost marks due to incorrect rounding or
truncating, with the most common error being either writing 0.3 or 0.29 instead of 0.30.
In part (b), some candidates incorrectly used the formula h 
ba
, with n  5 instead of
n
1
. Many candidates, however, were able to
5
look at the given table and deduce the value of h. The correct structure of the trapezium rule
inside the brackets was usually evident, although as usual there were the inevitable ‘invisible
brackets’ and bracketing errors.
n  4 to give the width of each trapezium as
In part (c), most candidates identified the correct triangle and correctly found its area.
A significant number of candidates did not realise that the ‘height’ of the triangle was given in
the table and re-calculated it. A small minority of candidates found the equation of the straight
line segment between (2, 0) and (3, 0.2) and used integration to find the area of the triangle.
Candidates should be encouraged to look at the available marks for a question before
embarking on such a long and complicated method. Almost all candidates who correctly
found the area of the triangle applied the correct method of subtracting this from their answer
to part (b). Some candidates used elaborate incorrect methods for finding the area of the
triangle and so gained no credit in part (c).
Those candidates who gave the incorrect answer in part (b) could gain full credit in part (c) if
they correctly applied their part (b) answer – 0.1.
Question 7
Most candidates were able to score both marks in part (a). Most candidates proceeded by
replacing 1  sin 2 x for cos 2 x . A few candidates, however, made algebraic errors or slips in
rearranging the equation correctly into the result given.
The need to use the alternative form was understood in part (b) and most candidates made a
valid attempt at factorisation, with correct factors being seen much more frequently than
incorrect ones. Some candidates correctly wrote (4sin x  3)(sin x  1)  0 and solved this
3
incorrectly to give one of their solutions as sin x  . Of those candidates achieving the
4
correct two values for sin x many only gave two correct solutions, usually 228.6 and 270 or
311.4 and 270. Sometimes extra incorrect solutions were given, usually 131.4 and/or 90. A
small number of candidates found (270  their  ) rather than (180 +  ) and (360 –  ).
Some candidates incorrectly stated that sin x – 1had no solutions and a few gave their answers
to the nearest degree. A significant number of candidates used a sketch of sin x to help them
to correctly identify their answers.
12
Question 8
Many good sketches were seen in part (a), with a significant number of candidates
constructing a table of x and y-values in order to help them sketch the correct curve. Some
candidates had little idea of the shape of the curve, whilst others omitted this part completely
and a significant number failed to show the curve for x  0. For x  0, some candidates
believed the curve levelled off to give y  1, whilst others showed the curve cutting through
the x-axis. Many candidates were able to state the correct y-intercept of (0 , 1) , but a few
believed the intercept occurred at (0, 7 ).
Responses to part (b) varied considerably with a number of more able candidates unable to
produce work worthy of any credit. A significant number of candidates incorrectly took logs
of each term to give the incorrect result of 2 x log 7  x log 28  log 3  0. Some candidates
provided many attempts at this part with many of them failing to appreciate that 7 2 x is
equivalent to (7 x ) 2 and so they were not able to spot the quadratic equation in 7 x . Those
candidates who wrote down the correct quadratic equation of y 2  4 y  3  0 proceeded to
gain full marks with ease, but sometimes final answers were left as 3 and 1. Some candidates
wrote down incorrect quadratic equations such as 7 y 2  4 y  3  0 or 7 y 2  28 y  3  0.
Notation was confusing at times, especially where the substitution x  7 x appeared.
Question 9
This question was answered more successfully by candidates than similar ones in the past. It
was pleasing to see that a significant number of candidates used diagrams to help them to
answer this question.
In part (a), most candidates were able to verify that (3, 6) was the centre of the circle, usually
by finding the midpoint of A and B, although other acceptable methods were seen.
In part (b), most candidates were able to write down an expression for the radius of the circle
(or the square of the radius). A significant number of candidates found the length of the
diameter AB and halved their result to find the radius correctly. Most candidates were also
familiar with the form of the equation of a circle, although some weaker candidates gave
equations of straight lines. The most common error in this part was confusion between the
diameter and radius of a circle leading to the incorrect result of ( x  3)2  ( y  6)2  50.
In part (c), the majority of those candidates who had found a correct equation in part (b) were
able to substitute both x  10 and y  7 into the left-hand side of their circle equation and
show that this gave a result of 50. Other candidates successfully substituted x  10
(or y  7 ) into the circle equation, solved the resulting quadratic and showed that one
resulting y (or x ) value was correct. Those candidates who gave an incorrect answer in part
(b) were usually unable to gain any credit in part (c).
13
In part (d), many candidates knew the method for finding the equation of the tangent at
(10, 7). Typical mistakes here included candidates finding the gradient of the radius AB or
finding a line parallel to the radius or finding a line through the centre of the circle. A few
candidates attempted to find the gradient of the line by differentiating their circle equation.
This method was rarely successful, as most candidates were not able to apply the method of
implicit differentiation correctly.
Question 10
In part (a), most candidates expanded V to obtain a cubic equation of the correct form and
then differentiated this to give the correct result. Occasional slips, usually with signs,
appeared as did the loss of a term when squaring (5  x)2 . A few candidates attempted to use
the product rule but most of them made slips.
In part (b), nearly all candidates were able to put their answer from part (a) equal to 0 and
5
many candidates obtained x 
with most of them realising that x  5 was outside the
3
range. Unfortunately a significant number of candidates did not substitute their x-value into an
expression for V in order to find the maximum volume. A significant minority of candidates
d 2V
 0.
tried to find the value of x which satisfied
dx 2
d 2V
.
dx 2
dV
The final mark was often lost, however, due to candidates differentiating an incorrect
or
dx
equating their second differential to zero or failing to evaluate the second differential, and
then stating that this was negative which meant that the volume found in part (b) was
maximum.
In part (c), most candidates knew an appropriate method with almost all opting to find
14
Core Mathematics Unit C3
Specification 6665
Introduction
This paper was accessible for almost all candidates with no real evidence of candidates failing
to finish.
There was an increase in the number of candidates relying on graphical calculators for
calculus/trigonometry type questions. It must be noted that the rubric on the front of the paper
states that answers without working may not score full marks. This was applied in questions
on this paper.
There was a lack of bracketing in many questions, especially in 2, 5 and 7. This potentially
could lead to the loss of many marks
Report on individual questions
Question 1
Question 1 was a familiar one to most candidates. It was generally well done by the majority
of candidates although part (b) and finding answers in the range 0 to 2π in part (c) did
discriminate.
In part (a) most candidates were able to find R and to make a worthwhile attempt at α usually
via the tangent ratio. Degrees were occasionally used despite the range being given in radians.
Some candidates were undecided and gave both degrees and radians, sometimes continuing
with this throughout the question.
Part (b) was frequently incorrect with +25 as common as the correct answer of 25. Another
common answer was 1 and more surprisingly 0. A less common error was to identify the
value of x for which the maximum/minimum would occur.
The majority of candidates attempted part (c) and realised the need to use the form found in
part (a). There were therefore some very good solutions, in many of which the only error was
to omit the second correct answer. Candidates should remember to derive additional values
from their principal value before rearranging their equation. Not many gave all three values of
1.16, 5.12 and 7.44 for (x + 1.287). Rounding errors were common with 3.83 and 6.15 popular
answers.
15
Question 2
In part (a) there were many fully correct and well-presented answers. Most candidates were
able to combine the two fractions, although some used unnecessarily complex denominators.
A number of responses included errors made when cancelling the 2 from the numerator and
denominator. A number of responses failed to simplify their answer and lost the final mark.
A few candidates, having correctly given the fraction with a common denominator on the first
line, cancelled one or more of the brackets leaving both numerator and denominator as linear
expressions. A common thread running through the paper for numerous candidates, not just
the weaker ones, was the lack of consistent bracketing. Candidates need to be aware that this
could lead to the potential loss of many marks.
Part (b), most candidates realised that they could use their answer from part (a) and many
were able to successfully demonstrate the proof. A few candidates started from scratch and
often went on to gain full marks.
The method used in part (c) was equally split between those deciding to use the chain rule and
those using the quotient rule. For the quotient rule, a number differentiated 3 incorrectly,
usually as 1. Having found the expression for dy/dx , a number of candidates multiplied out
the denominator in terms of x first, before substituting in x = 2. A surprising number reached
the correct fractional answer for dy/dx but, on substituting in x = 2, gave an answer of 6/25
(stating that 2  2  1 = 5). A small number of candidates decided to differentiate their
incorrect answer for (b). Candidates should be advised that this should be avoided at all costs,
especially where an answer is given in the paper.
Question 3
This question was attempted by most candidates but was not answered that well. Only the best
candidates produce fully correct solutions. Most understood that cos 2x should be replaced,
although 1  sin2x was sometimes seen instead of 1 – 2 sin2 x. The use of brackets was
careless in some cases. Many used cos2 x  sin2 x first, which was acceptable as long as the
cos2 x was replaced subsequently by 1  sin2 x. The majority of good candidates did arrive at
the correct quadratic equation but solving it was one of the least successful parts of this paper.
Many candidates could not believe that it would not factorise! Often several attempts were
made before moving on, some just gave up. Those using the quadratic formula did not always
quote it correctly. Others made errors in the substitution. A minority used ‘completing the
square’ to solve, though with mixed success. A few more successfully used equation solver on
their calculators. If they got this far, one or two angles were found, but many didn’t find all
four. They either rejected the negative value or only gave one answer for it within the range.
16
Question 4
The first 4 marks in this question were accessible to almost all candidates. The final 4 marks
however were far more demanding and were only gained by the best. In part (a) the majority
of candidates gained full marks, some able to write down the correct answer with a minimum
of working.
In part (b) most candidates followed the desired route of substituting t = 5 and θ = 55 into the
equation, then rearranging to make 5k the subject. Unfortunately the given answer was then
just written down without any adequate reason. Candidates needed to recognise that e5k
1
could be replaced by 5 k , or show that ln 12 = ln 2 , or even simply state it.
e
The final part (c) was poorly done, with a significant number of candidates failing to
recognise rate of change as differentiation. Many simply found the temperature when t = 10.
Others found the difference between the temperatures at t = 10 and t = 0, subtracted and
divided by 10. A number of candidates realised that they needed to differentiate to find the
rate of decrease in temperature but unfortunately put t = 10 into their expression before
attempting to differentiate, not appreciating that their expression was now constant. Another
common mistake in the derivative was finding expressions of the form Atekt.
Question 5
Question 5 proved to be a useful source of marks for all candidates. Grade A candidates
scored almost all marks and E grade candidates picked up at least 5 marks. Part (a) proved a
positive start to the question for nearly all candidates with most writing down both correct
x-coordinates although a few did struggle in solving ln x = 1. “Write down” should have been
a hint that no real calculation was required.
In part (b) apart from a few who confused the notation with the inverse function most realised
the need to use the product rule and proceeded correctly. The majority of good candidates
scored full marks in this part. Candidates should still be advised to quote formulae before they
are used.
While some candidates in part (c) mistakenly used f(x) instead of f′(x), most successfully
substituted both 3.5 and 3.6 into their derivative and knew to look for a sign change. There
were a multitude of wordings applied to the significance of this – “hence root” being the most
common  and candidates would be advised to read the text of the question in order to set
their conclusion in the right context. There were, however, some excellent answers where
candidate clearly understood the question and in some cases added diagrams to illustrate their
point.
A significant number of candidates omitted part (d) altogether or tried a variation on (c).
Those who realised that they had to equate the derivative to zero usually gained full marks.
Trying to work backwards from the answer rarely proved a good idea with candidates unsure
of how far they needed to go with their solution. The neatest solutions often resulted in
continuing from a simplified version of the derivative they had found earlier.
17
Part (e), this as with (a) proved a good source of marks even where there was little gained in
other parts of the question. Often just the correct values appeared and it was good to note that
virtually all complied with the question and gave all 3 answers to 3 decimal places.
Question 6
In part (a) it is worth noting that a number of candidates were weak on notation with a
significant number finding the derivative f ′(x) rather than the inverse function f1(x). Those
who tried to find the inverse were generally successful, although a worryingly large minority
found the algebraic manipulation beyond them.
Many candidates gave the correct answer in part (b) which could be easily found from the
graph. A few used domain notation rather than the range.
Parts (c) and (d) were often not attempted. Of those that did, many failed to see that g(2) and
g(8) could be read from the graph, and instead worked out the two linear equations for the
function g. This could lead to the correct solution but rarely did they were not always
correctly applied. A popular incorrect solution involved finding g(2) correctly, but then
simply squaring to get gg(2) = g(2) × g(2) = 0. Part (d) was generally more successful as the
function f(x) was given.
In part (e) accurate sketch graphs were usually seen in (i), with candidates generally familiar
with the idea of a modulus. Incorrect or missing co-ordinates lead to the loss of some marks.
There was less success in (ii) with the sketch of the inverse function. Many were able to
remember to reflect in the line y = x but there were many incorrect attempts, again with
missing or incorrect co-ordinates
In part (f) a substantial majority realised that the domain of the inverse was the same as the
range of the original function, but there was again some confusion about which variable ‘x’ or
‘y’ should be used.
Question 7
In part (a) most candidates knew that the quotient rule should be used. It would again be wise
to quote this formula. Only high achieving candidates produced full solutions to part (a), with
common mistakes including differentiating (3 + sin 2x) to give either (2 cos x) or (cos 2x), and
similarly with (2 + cos 2x). Predictably, the final accuracy mark was frequently lost through
candidates not being explicit enough in their working to demonstrate the given result –
jumping from (2 cos2 (2x) + 2 sin2 2x) to 2 was common. This was a given solution and hence
there was an expectation that the result should be shown. This could be achieved by writing
(2 cos2 2x + 2 sin2 2x) as 2(cos2 2x + sin2 2x) = 2 × 1 = 2.
Some candidates struggled in part (b) to give a correct value for either y or m. A very common
incorrect result was obtained by using the calculator set in degrees to work out these values.
Occasionally, a “perpendicular method” was used to replace a gradient of 2 with 0.5. Most
candidates arriving at an answer understood that exact answers for a and b were required.
18
Question 8
Part (a) question has been set before and most attempts proceeded in the correct manner by
either using the chain rule or quotient rule. Again this was a ‘show that’ question and
candidates are expected to demonstrate that the answer is true and not simply write it down.
The successful candidates in part (b) used the result in part (a) to simply write down the
answer. Some unfortunately went back to first principles wasting valuable time. Marks were
lost by candidates who wrote the solution as sec 2y tan 2y, sec 2x tan 2x or indeed the LHS
dy
as
.
dx
In part (c) most candidates recognised the need to invert their answer for (b) reaching
dy
dx
= 1 / . Many also replaced sec 2y by x often stopping at that point. The candidates who
dx
dy
continued to find an expression for tan 2y in terms of x generally obtained the correct final
result. This could be described as a grade A type question and it certainly did discriminate
between very good candidates.
19
Core Mathematics C4
Specification 6666
Introduction
The majority of candidates who took this paper found it straightforward and many correct
solutions were seen to all questions on the paper. The quality of algebraic manipulation was
generally good and the work seen in the question on the binomial theorem was particularly
impressive. The standard of presentation was good and almost all candidates now recognise
that, when a question asks for an exact answer, a decimal approximation is not acceptable.
Problems did arise, however, with candidates giving exact answers to questions, presumably
derived from calculators with functions that gave such answers, without any supporting
working. The rubric on the front of the paper advises candidates that they “should show
sufficient working to make your methods clear to the examiner”. When, for example, a
questions states, as question 2 does on this paper, “Use differentiation to find the value of …”
then, if the process of differentiation is not shown, the conditions of the question have not
been complied with and little or no credit can be awarded.
Report on individual questions
Question 1
This proved a good starting question and full marks were common. A few candidates
differentiated the expression, using the product rule. However, the great majority realised that
integration by parts was necessary and such errors as were seen usually arose from integrating
sin 2x and cos 2x incorrectly. Both errors of sign and multiplying (rather than dividing) by 2
were not uncommon. Most knew how to use the limits and complete the question. In some

cases the numerically correct answer, , was obtained after incorrect working. In these cases,
4
the final accuracy mark was not awarded.
Question 2
d x
 a   a x ln a usually found this
dx
question straightforward. Those who did not, tried a number of methods and these were
t 1
frequently incorrect. Errors seen included 16t  0.5  ,  16  0.5  ln t and 8t ln t . Nearly all
Those who knew, and often quoted, a formula of the form
dI
but a significant number of candidates failed to give
dt
their answer in the form ln a , as required by the question, leaving their answer in the
form n ln a .
candidates substituted t  3 into their
20
Question 3
Partial fractions are well understood and part (a) was usually fully correct. The majority used
substitution to find the constants and comparing constants was rare, as was the use of the
cover up rule. One error that was seen from time to time was 5 A  5  A  5 . Part (b) was
3
also well done and the common error 
dx  3ln  3x  2  was seen less often than in

 3x  2
some recent examinations.
Part (c) proved more difficult. Many could not separate the variables correctly and some did
not even realise that this was necessary. Some kept the 5 with the y and this caused problems
in applying the result of part (b) correctly. Those who established the appropriate method
usually included a constant of integration and were able to obtain an equation to find its value.
Making y the subject of the formula proved difficult and moving from an expression of the
form ln y  ln  f  x    ln k to y  f  x   k was a common error.
Question 4
This question was well done and full marks were common. Part (a) was almost always correct
and such errors as were seen were errors of arithmetic. Even at this level 2  3  1 is seen
from time to time. For (b), the majority knew the form of a straight line. A few got the vectors
the wrong way round or produced an answer of the form r  a  b . The commonest error
 1
 3 


 
was to give an answer of the form  3     5  , not recognising that, as the question asks
 2
 3 
 
 
for an equation, this expression must be preceded by r  ... .
Part (c) was more demanding and many were unable to choose vectors in the appropriate
directions. Almost all knew that they had to form an equation by equating a scalar product to
zero. However, often one of the position vectors was used; finding the scalar product of OC
with AB being a common choice. Some formed the scalar product of AB with the vector
equation of AC obtaining equations involving parameters as well as p. Almost all knew the
appropriate method for part (d).
Question 5
2
 3 
as 22 1  x  nearly always
 2 
expanded correctly and those who could not were usually able to gain 2 or 3 marks by
showing that they knew how to expand binomial expressions. The distribution of marks for
part (b) was bimodal; the majority of candidates obtaining either 0 or 5 marks. Those who
knew how to proceed were able to obtain two linear equations by comparing coefficients and
solve them for a and b. Apart from occasional algebraic slips, these candidates usually
obtained full marks. Those who did not know the appropriate method often gave up very
quickly and, wisely, went on to the next question. Those who were able to solve part (b)
almost always completed the question.
Part (a) was well done. Those who could write  2  3x 
21
2
Question 6
Although there were many correct solutions to part (a), a surprising number of candidates
dy
dy
dx
made mistakes in establishing
from their
and their
. Both 2 and 2t 3 were seen and,
dx
dt
dt
in many cases, the method used was not clearly shown and this resulted in the loss of both the
dy
method and the accuracy marks. If
was correctly found, the majority were able to
dx
complete this part correctly. A significant number, however, failed to read the question and
gave the equation of the tangent rather than that of the normal. Part (b) was well done and
nearly all could eliminate the parameter. Quite a number of candidates thought that (ex)2 was
2
e x and this often caused a major loss of marks in part (c).
In part (c), the majority of candidates knew the volume formula but the attempts at integration
were of a very variable quality. Many used the lead given in part (b) but the resulting squaring
out of the brackets was often incorrect. Examples of errors seen are
e
2x
 e
e 
2x 2
 e4 x
2
and
 2   e4 x  4 or e 4 x  4 . There were also attempts at direct integration, for example
2x
2
 2
2
e
dx 
2x
 2
3
3
. Those who used parameters often made similar mistakes and
dx
was omitted. The choice of limits also gave some difficulty; those who
dt
integrated using the variable x using the t limits and vice versa. Despite these frequent
mistakes, there were many completely correct solutions.
sometimes the
Question 7
Parts (a) and (b) were usually fully correct and few lost marks through failing to work to the
accuracy specified in the question. The trapezium rule is well known and the only error
commonly seen was obtaining an incorrect width of an individual trapezium.
Part (c) proved more demanding. Many got off to a bad start. The substitution was
dx
2
 2  u  4  could be
deliberately given in the form x   u  4   1 so that the essential
du
dx 1
1
  x  1 2 and
found easily. Many, however, rearranged the substitution and obtained
du 2
the resulting algebraic manipulations often proved beyond candidates. Those who did
complete the substitution often failed to complete the definite integral. An unexpected
difficulty was that a number of candidates failed, in the context, to simplify 4   u  4 to u.
22
2u  8
8
reduced to 2  and embarked upon complicated solutions
u
u
using integration by parts which, although theoretically possible, were rarely completed. The
choice of limits also gave some difficulties. The convention is used that a surd is taken as the
positive square root. If this were not the case the expression given at the head of the question
would be ambiguous. Some however produce limits resulting from negative square roots, 3
and 2, as well as the correct 5 and 6, and did not know which to choose. As in question 6,
there was also some confusion in choosing the limits, some choosing the x limits when the u
limits were appropriate and vice versa. Despite these difficulties, nearly 34% of the
candidates gained full marks for this question.
Many did not see that
23
Further Pure Mathematics Unit FP1
Specification 6667
Introduction
The questions on the whole were well answered with many fully correct answers. Candidates
found the paper very accessible and standard methods were well known and accurately
applied.
The standard of presentation was generally good with solutions showing logical steps making
the work easy to follow. The questions that proved most challenging were question 6,
question 8, question 9 and question 10 (b).
Report on individual questions
Question 1
This question was well done by the vast majority of candidates. In part (a) virtually all
obtained the method mark but a few lost the accuracy mark for either having −9i2 or failing to
use i2 = −1 when simplifying. There were a surprising number of numerical errors. In part (b)
anyone who gave an answer was using the correct multiplier and most expanded and
simplified correctly. A few lost the final accuracy mark for leaving their answer as a single
fraction.
Question 2
Part (a) was well answered with most candidates gaining full marks. A few lost accuracy
marks for sign errors but most had used the matrix multiplication method correctly. In part
(b) there were many candidates correctly identifying it as a reflection. A lot of these
candidates went on to give the correct line of reflection but some did give the wrong axis.
Many candidates also mentioned a centre at this point which was erroneous but did not lose
any credit. Candidates who failed to identify it as a reflection generally thought that it was a
rotation. Part (c) was also generally answered correctly with the original matrix as the most
common of the incorrect responses.
Question 3
Many candidates gained full marks for this question. In part (a) some candidates lost marks
for incorrect signs and struggled with interpolation. Those with good diagrams tended to
produce the best responses and a lot of variety was seen in the correct methods. In part (b)
there were rarely any errors and in part (c) many were able to use Newton-Raphson method
correctly to gain full marks.
24
Question 4
In part (a) virtually all candidates wrote down the correct conjugate. In part (b) a variety of
methods were used with varying degrees of success. Candidates using expansion of brackets
were usually able to gain full marks provided they set out the terms carefully. Of those who
used other methods, some candidates found only the product of the roots and so gained no
marks here. Of those that also found the sum of the roots correctly, a significant number then
gave the p value as 4 and so lost the final accuracy mark. Methods involving simultaneous
equations were also quite common and had varying degrees of success.
Question 5
In part (a) many candidates were able to expand the brackets and use the standard formulae
correctly. The most successful candidates then began factorising immediately and were
usually able to gain full marks. Candidates who expanded all of the brackets found it more
difficult to complete the proof. Some correct polynomial division was used and some
candidates also expanded the final expression to meet in the middle. Some candidates
attempted to use proof by induction here which typically gained no credit. In part (b)
candidates generally gained full marks and if not, the method mark was usually awarded and
then a numerical error made. A small number of candidates used S50 – S20 which lost the
method mark.
Question 6
Marks in parts (a), (b) and (c) were usually gained without difficulty. Unfortunately part (d)
proved more problematic but most did still complete it correctly. In part (e) most candidates
also gained the marks using the area of a trapezium. Candidates splitting it into a rectangle
and two triangles had often had an incorrect answer in part (d) and so were only able to gain
the method mark here.
Question 7
Part (a) was almost always gained although some candidates seemed to think it had to be
plotted on fully accurate scaled axes. In part (b) most candidates were able to achieve at least
the method mark and many got full marks. In part (c) the few candidates who used the
w = r cos θ + ir sin θ method often gave correct solutions here. Many candidates were also
correct using the modulus and argument but there were a lot of incorrect signs used, as many
failed to refer to which quadrant contained w when deciding whether to use the positive or
negative solutions from their quadratic. In part (d) those candidates who expanded before
finding the modulus were only able to gain the method mark since they were using an
incorrect w. Many however recovered by using the product of the moduli and gained full
marks here. Candidates with part (c) fully correct usually had no problem in part (d) either.
25
Question 8
In part (a) most candidates gave correct solutions but some did give 14 but then still used 14 as
1
in part (b) thus not following through their own determinant and displaying a
ad  bc
confusion regarding the terminology. Most candidates did however give the fully correct
inverse in part (b).
In part (c) most candidates also gained full marks with a significant number of those that had
an incorrect determinant still achieving the follow through mark to gain full marks here.
Part (d) was generally well done. Candidates who had been fully correct in part (a) to part (c)
were usually also correct here although a few lost the final mark for failing to give
coordinates.
A significant majority of those who had an incorrect inverse still gained the first two marks
here for correct processing and a follow through mark. Some candidates however were
unable to identify the correct method and some were just multiplying by A.
Question 9
This proved to be the most demanding question on the paper. Solutions often evaluated u2
rather than u1 and so lost the first mark. They were often unaware that they had not in fact
found u1 though since they had used k = 1 in the expression for uk + 1 + and so were able to
then state the four necessary elements of the method confidently and gain the final accuracy
mark. There was the usual difficulty with getting the logic right, as candidates struggled to
state the precise wording, probably due to a superficial understanding of the method.
However, the most striking thing was the very large number of candidates who thought it
necessary to show it was true for n = 2 which did not gain any credit.
Question 10
dy
using a variety
dx
of successful methods. A small number omitted to show their method for gradient and as the
equation was given in the question, the initial marks were lost.
Part (a) was usually done successfully with few candidates failing to find
Part (b) proved more problematic with many candidates failing to use the most efficient
method. There were some slips with signs too which meant that the final coordinates were
wrong in quite a few cases. Some candidates however did achieve full marks here.
26
Mechanics Unit M1
Specification 6677
General
The paper seemed to be of a suitable length for the vast majority with very few candidates
unable to attempt all the questions. The final part of the last question was very discriminating
and so it wasn’t always clear whether the weaker candidates had run out of time or run out of
ideas. Overall the paper seemed to be very accessible and there were several very
straightforward questions. The first question proved to be a very easy starter for most and
question 5 also provided an opportunity for many to score highly. Other good sources of
marks were questions 4(a) and 6(a). The questions which caused difficulties were 4(c), 6(c)
and 7(c).
Overall, candidates who used large and clearly labelled diagrams and who employed clear and
concise methods were the most successful. It should also be emphasised that candidates
should “show sufficient working to make your methods clear to the Examiner” as stated on
the front page and usually correct answers only, with no working, will gain no marks. In
calculations the numerical value of g which should be used is 9.8, as advised on the front of
the question paper. Final answers should then be given to 2 (or 3) significant figures – more
accurate answers will be penalised. If a candidate runs out of space in which to give his/her
answer than he/she is advised to use a supplementary sheet – if a centre is reluctant to supply
extra paper then it is crucial for the candidate to say whereabouts in the script the extra
working is going to be done.
Report on individual questions
Question 1
This provided a very straightforward start for most candidates. In part (a), nearly all used an
appropriate conservation of linear momentum equation and sign errors due to incorrect
interpretation of directions were comparatively rare. Occasionally attempts to equate impulses
were seen; these were acceptable provided it was recognised that they were in opposite
directions. The majority used impulse = change in momentum in part (b), generally with
appropriate signs, although 3(3 – 2) was sometimes seen. The marks could be achieved by
considering either particle, but one of them depended on a correct value of m from part (a).
Most remembered to give the magnitude as (+)15 for the final mark.
Question 2
In part (a) relatively few were able to show that u = 0.9 exactly. It must be stressed that when
an answer is given, the method used must be clear and fully correct. Many candidates fudged
the signs in their methods, failing to appreciate that u was a speed and therefore positive, or
else used an inexact method. The second part elicited many correct responses for finding the
height reached above the point of projection. However, the answer was not always given to
the required 2 or 3 significant figures, consistent with the use of g = 9.8 and answers to one or
four significant figures were penalised. There were many possible approaches for finding the
required height in part (c); sign errors were fairly common and some found the total height
reached by the ball. Others found the correct value but added or subtracted another distance to
27
produce their final answer showing a lack of real understanding of the situation. Nevertheless,
there were some entirely correct systematic solutions seen.
Question 3
The first part was done well, with the most common error being to give Rc as 117.6 N which
was penalised for being over-accurate. The question required two equations and those who
used a vertical resolution were almost always successful whereas those who used two
moments equations often made errors. The same was true in part (b), but candidates often
made errors when expressing the distances used in terms of AD. The omission of g was
penalised in the first part but not in part (b), provided it was consistent, where it was not
needed to obtain a fully correct solution. A few used the same values for the reactions in
part (b) as those found in the first part and received little credit.
Question 4
In part (a) almost all candidates found the magnitude of the velocity to give the speed
correctly. Most derived the acceleration, in the second part, by subtracting the velocities and
dividing by the time appropriately (often by setting up a v = u + at equation first); however, a
significant minority continue to lack confidence in dealing with vectors and vector notation.
Most realised they then had to multiply by the mass to find the force, but sometimes the
answer given was the same as the acceleration. The final part proved to be a good
discriminator. Some failed to set up an equation in terms of t for the velocity, using their
acceleration and the initial velocity; various combinations of terms were often seen. Many
realised the j-component had to be zero, but some equated the i-component to zero whilst
others equated the two components. Nevertheless, there were a significant number of
candidates who correctly deduced the answer, often with very little working.
Question 5
The vast majority of candidates achieved full marks for the speed-time graph in part (a) and
for equating the area under the graph to the given distance in order to find v, in part (c).
Occasionally ‘v’ was left off the axis, or ‘40, 50’ labelled instead of ‘60, 70’ which also led to
errors in part (c). The acceleration-time graph in the second part provided a greater challenge
and some non-horizontal lines were seen. Those who had a graph with the correct basic shape
were penalised if they included vertical lines on their sketch, although dotted lines were
acceptable.
28
Question 6
Most candidates were able to resolve perpendicular to the plane, in part (a), to obtain a correct
expression for the reaction. The given answer was exact so evidence of rounding, such as
using a rounded value for the angle, was penalised. In the second part, some did not realise
that they needed to find the new reaction and so lost a number of marks. Many did complete
the resolutions correctly although occasionally the friction was acting up the slope instead of
down. Attempts at part (c) tended to be less successful, with the weight component often
omitted or else limiting friction was used. Those who found a correct numerical value for the
magnitude of the frictional force (by resolving parallel to the plane) did not always deduce the
correct direction.
Question 7
Despite the lack of structure in part (a) most candidates knew the methods required. Many
gained the marks for resolving perpendicular to the slope and for using F = 23 R. The equation
of motion for the 7 kg mass was also often correct, but a common error was to replace T by 7g
when resolving parallel to the plane. Sometimes a term, either friction or the weight
component, was omitted from this equation. Although some candidates failed to complete
successfully all the substitution and rearranging required to find the acceleration, there were a
number of entirely correct solutions. An appropriate constant acceleration formula was
generally used for the velocity in part (b), although an incorrect answer from part (a) led to
loss of the accuracy mark. In the final part, many did not manage to find the new acceleration
by a valid method; some used the value from part (a) or quoted ‘9.8’ without any justification,
whilst others realised that a new value was required but omitted a term from the equation of
motion. Amongst those who resolved the two forces, a number had friction acting in the
wrong direction. There were a small number of entirely correct solutions seen.
29
Mechanics Unit M2
Specification 6678
Introduction
This paper proved to be accessible to most candidates, who commonly offered responses to all
eight questions. Much of the work was of a high standard, with many clearly presented and
concise solutions, particularly in questions 1, 2, 3 and 8. Some candidates were not so
confident with the motion of a projectile described in vector form, and the lack of direction
towards an appropriate method in question 7 made this more challenging. Candidates do need
to be reminded to read the questions carefully, and to ensure that they have answered
appropriately - many found the change in kinetic energy in question 2, rather than the kinetic
energy after the impulse. Similarly, many candidates lost a mark because they did not give the
final answer in question 5 to the nearest degree. Although it has been commented on in
previous reports, candidates are still losing marks through giving final answers to an
inappropriate level of accuracy following the use of 9.8 m s2 as an approximate value for g.
The general level of arithmetic and algebraic skills was good, with fewer candidates
sacrificing marks through careless errors. The given answers in some questions allowed
candidates to identify and correct errors. As usual, the best work was accompanied by clearly
labelled diagrams - all candidates should be encouraged to use diagrams to present and
summarise information whenever possible.
Report on individual questions
Question 1
This proved to be a friendly starter for most candidates. Part (a) most candidates obtained the
correct answer, although there was often no clear statement that the driving force must be
equal to the resistance. In part (b) there were many completely correct solutions. The most
common error was to omit the resistance when writing down the equation of motion, often
when a candidate had not drawn a diagram of forces.
Question 2
This was well answered by the majority of candidates, many of whom gained full marks.
Most equated impulse to change in momentum, but the subtraction was sometimes done the
wrong way round. A few candidates made errors due to the poor use of brackets. Most
candidates went on to find the kinetic energy correctly. Calculation of the magnitude of the
velocity was often correct, but there were arithmetical errors and slips such as squaring a
component twice. Some candidates did not appear to understand that kinetic energy is a scalar
and gave their answer in terms of i and j. √29 and 3 were common incorrect answers.
30
Question 3
The vast majority of candidates knew that integration was required for parts (a) and (b) and
they performed this competently with only a small minority omitting the constants of
integration. A small number did try to use suvat inappropriately, and one or two differentiated
instead of integrating. Part (c) most candidates knew that they needed to put v = 0 and most of
these recognised the equation as a quadratic in t 2 and factorised or sometimes completed the
square to obtain values of 2 and 4 for t 2. The final mark was occasionally lost by a failure to
reject negative values of t. Candidates who did not recognise the quartic as a quadratic in t 2
sometimes went to considerable lengths to use the factor theorem and/or trial and
improvement to find factors of the quartic, but they rarely reached the correct final answer.
Another common error was to rearrange the equation as t 4  6t 2 = 8 and attempt to set
factors of the left hand side equal to factors of 8.
Question 4
In part (a), some candidates interpreted the question as requiring just the work done against
friction. Another frequent mistake was finding the correct frictional and gravitational forces
but then failing to multiply by the distance. Some candidates double counted by including
both the increase in gravitational potential energy and the work done against the weight of the
box. The final answer was often given as 8481 J, which is inappropriate following the use of
an approximate value for g.
In part (b) the solution was often correct. Some candidates using the work-energy principle
did make errors through double counting, and sometimes made a sign error by attempting to
use their answer from (a). The alternative method of using F = ma and suvat was usually
successful provided the candidate did not omit the friction.
Question 5
In part (a) there were many entirely correct solutions to this question, with candidates
employing a number of different strategies to split this shape into standard components. Most
commonly this involved expressing it as the difference between two triangles, or splitting it
into two rectangles and two triangles. Using just two triangles tended to produce the most
concise and accurate solutions, although there was some confusion over the positions of the
centres of mass of the triangles. Some candidates did not realise that they could work in terms
of the horizontal and vertical distances from these vertices at B and E and went to
considerable lengths to calculate the heights of the triangles measured from these vertices and
then to use trigonometry. Candidates who divided the shape into four or more pieces
frequently made errors in calculating the areas of these pieces or in locating their centres of
mass. Another cause of errors was to double count a region, or even to leave it out entirely.
A small number of candidates treated this as a structure made of rods rather than as a uniform
lamina. A surprising number of candidates did not use the symmetry of the lamina to find the
second distance, with many reworking a moments equation to get the same answer – or in
some cases a different answer. Part (b) was very well answered by most candidates; the
required angle was usually identified correctly and candidates could gain two marks for work
clearly following from incorrect values in part (a). A common incorrect answer was 45, from
candidates who did not consider the geometry of the situation or use a diagram to help. Many
candidates did not round their final answer to the nearest degree.
31
Question 6
Solutions in part (a) often lacked a clear method. Candidates should be reminded of the need
for detail when deriving a given answer. Candidates showed a poor knowledge of vector
analysis and little understanding of the use of a displacement vector with a position vector.
There were plenty of fudges to include 10j, only rarely was r = r0 + s used. Many candidates
considered the horizontal and vertical components separately. The horizontal component was
easily found but the candidates found it difficult to justify the 10 in the vertical. Many,
incorrectly, attempted to equate the vertical displacement to 10 without any reference to initial
conditions. The best solutions used integration, with the 10 being found by using the initial
conditions to find the constant of integration.
The best solutions in part (b) were where candidates equated the j component of their position
vector to 0 and solved the resulting quadratic equation. Many started again and found the
vertical displacement equation from scratch leaving a greater scope for error. A common error
was to equate the j component from (a) to 10, failing to realise that the 10 was already
included in the equation. As usual, there were a few unnecessarily long methods involving
calculation of the time to reach the maximum height and then the time from there to the
ground. Some candidates lost the final mark due to ‘over accurate’ answers following the use
of a decimal approximation for g.
In part (c) some candidates clearly differentiated the result from part (a), and others derived
the velocity from the initial information. There was evidence of confusion on some candidates
who found the speed or velocity at a particular time, rather than a general expression for the
velocity. Part (d) surprisingly, many candidates had difficulty here , commonly equating their
j component to +3 rather than 3, often despite having a correct diagram. Others did not
connect “45° below the horizontal” with equal horizontal and vertical components of velocity.
In part (e) many candidates had success here despite earlier problems, with most finding the
modulus of a vector of the form 3i + nj. Candidates should be encouraged to read all parts of
questions as later parts do not always rely on success in earlier ones.
Question 7
Although there were many fully correct responses to this question, the unstructured nature of
the problem did present difficulties for some candidates. A clearly labelled diagram showing
all the forces was essential. Some candidates were unsure of the direction of action of the
normal reactions at A and at C. Some gave them both the same name, and appeared to believe
that they were equal in magnitude. Others omitted at least one of the normal reactions. Some
candidates used horizontal and vertical components for the force at C but were usually unable
to connect them later in their solution. Many candidates recognised the need to take moments
and to resolve but errors were often made in doing so. The most straightforward approach of
taking moments about A and then resolving vertically and horizontally was often seen. Many
candidates took moments about C or tried to resolve parallel to and perpendicular to the rod,
but this frequently resulted in a missing term. There were a small number of more imaginative
solutions involving moments about points not on the plank. Although there was evidence of
confusion between sine and cosine when resolving forces, incorrect solutions usually involved
an attempt to resolve when it was not necessary, or a failure to do so when it was required.
Equally, it was common to find distances missing from a moments equation. Despite having
32
correct equations many candidates could not combine them to find friction and reaction forces
correctly. Some candidates did demonstrate that they were finding the least possible value
of μ, but many did not address this point and used F = μR throughout. A few candidates
misread the question, using 100 g as the weight.
Question 8
Candidates found this impact question more straight forward than some in recent years. In
part (a) many candidates derived the given answer correctly. A few made sign errors in
equating the change in KE to energy lost and a small number were not able to complete this
part because they could not find the speed immediately after the impact. In part (b), a lack of
clear diagrams sometimes led to sign errors and confusion over the direction of motion of the
particles after the collision, but many candidates gained full marks in this question. Most
candidates formed a correct equation for the conservation of momentum, and there were
fewer errors this time in applying Newton’s Experimental Law. There were arithmetic and
algebraic errors in solving the simultaneous equations, but the standard of justification of the
second collision was good.
33
Mechanics Unit M3
Specification 6679
Introduction
The overall standard seemed very high, even though the paper was not unusually
straightforward. Algebraic manipulation was good and there were a great many wellexpressed, concise solutions from well-prepared candidates. There were relatively few blank
pages or other worthless attempts. The length seemed about right; there were no signs of haste
towards the end but neither was there evidence of spare time being used to do questions twice
although there were some blank responses to 7(c) from a few candidates. It was not possible
to decide whether these blanks were due to lack of time or an inability to realise where the
maximum and minimum tensions occurred.
Some candidates would be well-advised to make more use of the generous space allowed for
their solutions. Solutions written in small or poor hand-writing on consecutive lines of the
answer booklet can be difficult for examiners to follow and are not easily checked for errors
by the candidates either. Well-spaced working tends to contain far fewer mistakes.
Solutions to “show that” questions did not always contain sufficient working to convince the
examiner that the candidate would have arrived at the correct answer had it not been given in
the question. In some cases it was obvious that the candidate had failed to arrive at the given
answer and had either simply introduced extra terms or had tried to correct work but the
correction was incomplete.
Report on individual questions
Question 1
As intended, this was a straightforward opening question for most candidates. The majority
dv
knew that the v
form of the acceleration was needed and proceeded to a correct integration
dx
dv
and then obtained a correct value for the constant. Some candidates integrated
, obtaining
dx
dv
was seen, resulting in a zero score
v  ... , leading to an incorrect solution. Occasionally
dt
on this question. Virtually all candidates set v  0 and managed to solve the resulting
quadratic although some made hard work of solving even the correct quadratic.
34
Question 2
Part (a) was usually successful. Very few candidates failed to get the masses correct (a very
small number considered the volumes of the shapes) but some managed to interchange the
distances. Most candidates took moments about the centre of the common face, but some
successfully took moments about either end of the toy.
Very poor diagrams did not help with the solution of part (b). Basic geometry/trigonometry let
many candidates down as they took the slant edge to be 2r and/or put the 30 angle in the
wrong place. Most attempts tried to set up a moments equation about B but few of these led to
a correct result. The best candidates realised that the forces were parallel and so taking
moments about O or the centre of mass of the combined toy allowed them to use
measurements along the axis and thereby avoid complicated trigonometry. These candidates
used surds confidently and quickly arrived at a correct result. The candidates who found the
centre of mass of the combined shape and then used that, in conjunction with 30 to work out
the value for  also achieved a higher than average success rate.
Question 3
Part (a) was almost always right, the only occasional errors occurring in integrating e 2 x .
1
A few failed to realise that (e x ) 2 was e 2 x and similar small numbers missed the
when
2
integrating. Some of these “corrected” their error by quoting the volume formula as
1
 y 2 dx and so lost everything. The method for (b) was also well known but there were

2
quite frequent mistakes. Most errors occurred in the integration by parts, often as a result of
trying to substitute the limits and simplify without writing all the detailed steps. There were
inevitably some incorrect formulae quoted, the most common of which was for the xcoordinate of the centre of mass of an area. Some started with a mixture of formulae such as
 xy dx but all such errors were in the minority. A few candidates left answers in terms of e
2
 y dx
and one or two gave answers to 4 significant figures. A small minority appeared not to have
covered the necessary C4 integration as they had no idea of how to integrate by parts.
Question 4
The specification for this unit states that “proof that a particle moves with simple harmonic
motion in a given situation may be required”, i.e. showing that x   2 x . It was clear from
many of the attempts seen for part (a) of this question that not all candidates are aware of this.
While many did complete the necessary differentiation and final substitution, although not
always correctly, some used v and a instead of x and x . Full marks could have been gained
had a been replaced with x for the final statement, but many failed to do this. Part (b) was
usually correct, although some who had incorrect final results in (a) used an incorrect  to
obtain the period. Most used v  a and obtained the correct result in part (c). Part (d) also
rarely produced problems, the most frequent error being to add the two times instead of
subtracting. A few candidates had their calculators in degree mode here.
35
Question 5
This question was answered well by most candidates. Only in rare cases did candidates fail to
recognise that the angle was 45. Similarly, in only a few cases, some candidates considered
the tensions to be equal. In general, candidates were able to find the equation of motion,
vertical resolution and value for r correctly. However, some computational errors were made
collecting terms which led to a loss of accuracy. In most cases, those who gained the marks
for part (a) understood that the tension had to be greater than zero in the lower string and so
were able to answer part (b) correctly as well. In some instances, candidates confused
themselves by comparing the two tensions instead.
The candidates who simplified their equations before trying to eliminate a tension achieved a
better rate of success at getting the correct tensions  many quickly spotted the simplified
version of the horizontal resolution obtained by cancelling sin (or cos) 45. It was
disappointing to see candidates get all of the mechanics in place and then fall down on
processing. A small number of candidates tried to resolve along the strings but failed to
resolve the acceleration. Some did not read the question and left final answers in terms of r
instead of l.
Question 6
Part (a) was answered quite well although in some cases excessively convoluted work was
done to produce the correct value of k rather than obtaining the tension using Hooke’s Law
and separately from resolving vertically and then equating. A very small number treated the
problem as a single string of length 2l, and when they did, they often mixed up the extension
for the double string with the natural length of the single one.
In part (b), some candidates had difficulty finding the new initial extension – a few continued
1
to use the l they had found in part (a). A few candidates used the single string approach but
4
too many using two strings failed to double the EPE they had found. Many candidates failed
to realise that the mass of the ball was 3m resulting in errors in PGE and KE; some candidates
even had 3m for the mass in one energy term and m for the mass in the other. It was rare to
see a candidate attempting to solve this part of the question by any method other than using
energy considerations. Those who did quickly ran into problems. A solution by SHM requires
SHM to be proved, not just stated as a fact, and use of the equation of motion should include a
variable acceleration. Either approach requires much work!
Question 7
Part (a) caused few problems but having the answer given in the question helped a few with
convenient changes in signs. Apart from a few who used v 2  u 2  2as , it was done very
successfully, if not always convincingly. Many candidates made things more complicated
than needed by using various reference lines for the potential energy rather than simply
finding the height fallen.
36
Part (b) was generally answered well but again, having a given answer helped some
candidates choose a correct value for cos . Some tried T > 0 as the required condition even
though they were given that it was a rod. There were some very well explained energy
solutions (initial energy > PE needed to reach the top), but the majority of correct solutions
were via the main method on the scheme.
Part (c) was a good discriminator. There were a lot of good solutions, and some outstandingly
brief ones but also a great many that lost track of what they were trying to do. The most
efficient solutions found a general expression for T in the  position shown in the diagram
and then just substituted 1 and –1 for cos . This bypassed a lot of the algebraic manipulation
which might go wrong. The majority of correct solutions, though, used the main method on
the scheme. The positions of maximum and minimum T were well known, and correct, or
almost correct, equations for these positions were written by a majority. There were some
very weak attempts, completed very quickly, which assumed the same v in both equations
for T. Because this answer was also given, there were the inevitable optimistic fudges from
some candidates.
37
Statistics Unit S1
Specification 6683
Introduction
The paper appeared to be of about the right standard and length. Each of the questions were
accessible to all candidates but the longer questions 6, 7 and 8 proved more challenging
towards the end. Question 3(c) also proved quite discriminating as many candidates failed to
realise that the lower quartile was the point of focus rather than the upper quartile.
There was some good and accurate use of the formulae in the formula booklet and simple
applications of the normal distribution are now handled confidently by most candidates.
Report on individual questions
Question 1
This proved to be a friendly starter for most candidates with many scoring all 5 marks in part
(a) and (b). Most errors here were arithmetic such as writing
= 596.666... rather than
569.666... or accuracy problems in part (b) where an answer of 0.57 was often seen, rather
than the 3sf accuracy that we look for. A minority still have difficulty in using the printed
2
 4027 
formulae and Sll  327754.5  
 or
 50 
seen.
 l   w instead of  lw in
Slw were sometimes
Part (c) caused problems for many candidates who simply wrote “positive correlation” but did
not interpret this statement in the context by mentioning that longer salmon usually weigh
more. Some candidates tended to “overstate” their conclusion by implying that as a salmon
grows it gets longer (not strictly true in this instance as the study was of 50 different salmon
not one salmon at 50 different time intervals ) and others referred to a proportionate
relationship such as “for every cm increase in length the salmon weighs 0.572 kg more”.
Whilst such indiscretions were overlooked for the single mark on this occasion, these
examples should provide useful points of discussion for teachers with future cohorts of
candidates.
Question 2
Again the majority of the candidates encountered few problems here and many scored 3 or 4
marks, although a number in part (a) found both means and then simply added them and
divided by 2 rather than taking the weighted average. Most chose to find the sum of all 28
records and then simply divided this by 28. A few candidates misinterpreted the 84.6 as the
mean for the 21 days despite what should have been an obvious discrepancy with Keith’s
data.
In part (b) many realised that the changes would have no effect on the mean but sometimes
they failed to give any numerical values to support this claim. There were many excellent
answers though showing clearly that 9.4 + 0.5 = 4.9 + 5.0 or some other suitable calculation.
38
Question 3
There were many fully correct box plots in part (a) that gained full marks but also a number of
almost correct plots with no supporting working that scored very few marks. Those who did
show their calculations for determining the outliers usually got the 24.5 value correct but a
large number had 3.5 rather than – 3.5 as their lower limit. Some candidates drew two upper
whiskers one ending at 20 (the next highest non-outlier data point) and one ending at 24.5 (the
outlier limit). A correct answer should only have one whisker and it appears that some
candidates had been copying the practice on the mark scheme of illustrating both alternatives
for the benefit of the examiners. It should be remembered that the published mark schemes
are not model solutions and they should not be offered to candidates as such.
In part (b) most candidates stated that Q2  Q1  Q3  Q2 or something equivalent though some
then claimed that the skewness was positive but this was usually answered well.
Part (c) caught many candidates out with a large number agreeing with the company’s claim
because Q3 = £14 000 which is greater than £10 000. Those who did appreciate that the lower
quartile was the significant figure to be considering often gave excellent answers but the
success rate here was lower than expected.
Question 4
Part (a) was answered well by many but a number still have difficulties in determining which
variable corresponds to y and which to x and then using the formulae given in the booklet.
The question clearly stated that the regression line of p on v was required and this was further
emphasised by giving the form as p = a + bv but despite this a number calculated 1.688/1.168
for b and some used a = 4.42 – 3.32b. There was the usual crop of accuracy errors with
candidates failing to work accurately enough to give their final coefficients to the usual 3sf (or
better) accuracy. A significant number of candidates simply substituted 85 into their equation
and received no marks for part (b) but a good many did appreciate the need to find the value
for v and often went on to obtain an answer rounding to 4.3 as well.
Question 5
The use of interpolation, which was expected in part (a), is improving and many candidates
made a reasonable attempt. Some failed to use correct class limits and had a width of 9 rather
than 10 but the correct answer was often seen.
Despite the values for  t and  t 2 being given in the question a number of candidates
chose to estimate these values from the table and subsequently obtained an incorrect estimate
for the mean but the mark scheme did permit them to obtain the remaining marks. The
question did not ask for estimates of the mean and standard deviation and so using the midpoints was inappropriate in this question. The usual crop of errors arose when calculating the
standard deviation with many failing to divide the 69378 by 32. Candidates can usually
calculate S xx correctly using the given formula and they may find it helpful to simply
remember that standard deviation is simply
S xx
but few seem to use this approach.
n
39
In part (c) most compared the mean and median and correctly concluded that the skewness
was positive. Some used a formula which, apart from the extra work, was fine but those who
used a (median – mean) formula rarely gave the correct conclusion. Some candidates went to
great lengths to calculate the quartiles and conducted a quartile test for skewness which, if
correct, led to a conclusion of negative skewness. This was allowed but was clearly not the
intended approach for 2 marks.
Question 6
Most candidates showed us clearly that they were using the sum of the probabilities to reach
10k = 1 from which they showed that k = 0.1 and the calculation of E(X) was usually correct
too. In part (c) some confusion between E( X 2 ) and Var(X) and also over what to square
(some choosing the probabilities rather than the values of x) caused difficulties and a few
simply squared their answer to part (b). Part (d) was a fairly standard request and most knew
that Var(2 – 5X) = 25 Var(X) and were sometimes able to recover from errors made in part (c)
but then many solutions ground to a halt.
Many candidates did not recognise the 3 cases required in part (e) and were therefore unable
to emulate this approach in part (f) but it was encouraging to see that some of these candidates
did realise what was required for part (g) and often gained both marks here.
Question 7
Many candidates were able to complete the tree diagram correctly but common errors were to
have probabilities of 5/9 and 4/9 on the top two and bottom two branches.
Part (b) was often answered correctly and even those with incorrect tree diagrams could
achieve 2 or even all 3 marks here. Few candidates explained which four probabilities they
were using to answer part (c) and the examiners were often left trying to deduce this from
their tree diagram. Simply writing P(RRR) + P(RYR) + P(YRR) + P(YYR) would have earned
them the first mark and made their solution much clearer. Some candidates failed to
appreciate that a sum of 4 products of 3 probabilities from their tree diagram was required and
16 5 5 4
a popular “fiddle” was to calculate      using the probabilities from the third
49 9 9 9
branches and magically dividing by 4 to reach the printed answer. Part (d) was answered
poorly with most attempts assuming that A and B were independent events (they were but this
was never justified). Part (e) though was usually answered well with candidates clearly using
the addition rule and the given answers.
In the final part many candidates failed to identify the conditional probability and those who
did often did not clearly state what their ratio of probabilities represented: once again a
statement such as (" P "( RRR)) / (" P "( RRR)  " P "(YYY )) with some suitable probabilities
attempted would have secured the first mark.
Question 8
40
There were some good responses to this question and even some of those who had struggled
with parts of questions 6 and 7 were able to pick up a good score here.
Part (a) was answered well and it was good to see diagrams being used to assist the
candidates. Part (b) still causes problems for many candidates. They can usually standardise
but then far too often equate this expression to 0.01 or 0.99. Those who did use a z value often
used the “small” table and found the 2.3263 value but the minus sign was often missed and
the final answer was therefore incorrect. Most seemed to try part (c) and the standardising was
usually correct and suitable z values were often seen (use of 2.32 or 2.33 and 1.28 were
acceptable here) but, when suitable equations were formed, a minus sign was often missing
from the second equation. Solving their two linear equations was carried out quite well but
full marks were only secured by those who worked carefully and accurately throughout this
part.
41
Statistics Unit S2
Specification 6684
Introduction
Candidates would appear to have had enough time to complete this paper. There were few
questions where no attempt had been made to produce an answer.
The level of work was generally very good. Many candidates were well prepared for the
examination and candidates’ answers were well presented.
Report on individual questions
Question 1
This question proved to be a very good start to the paper for a large majority of candidates. In
part (a) the assumptions were written “in context”. However, there are still too many scripts
where there was no mention of context at all. Many simply wrote a list of reasons as to why a
Poisson distribution should be used rather than stating the assumptions that had been made.
In parts (b) and (c) fully correct solutions were seen in a large majority of candidates.
However, there were a small number who failed to spot the change in the value of the
parameter n from part (b) to part (c).
In part (d) most candidates provided a clear and accurate solution. However, there were a few
candidates who used a Normal approximation, which is clearly inappropriate in this situation
since “n is large and p is small” applies in this case and therefore indicates the use of a
Poisson approximation. A small but significant number of candidates used a Binomial
distribution despite the question requesting that a suitable approximation be used. A common
error from those using a Poisson approximation was the use of P(X ≤ 4) instead of P(X ≤ 5).
Question 2
The overall response to this question was disappointing. A common incorrect ‘alternative
hypothesis’ of p < 0.2 was frequently seen implying that ‘not guessing’ is an inferior strategy
to ‘guessing’. Other common errors included the use of p = 0.4 instead of p = 0.2 and using
P(X = 4) or P(X ≤ 4) instead of P(X  4). There were also problems with candidates’
conclusions. It was fairly common for candidates to provide complete and correct responses
to the entire question except the final contextual conclusion. Their correct statement “do not
reject the null hypothesis” was often followed by an incorrect comment in context such as:
“So she was not guessing” or “So reject the teacher’s claim”.
42
Question 3
This question was a good source of marks for many candidates. In parts (a) and (b) the
candidates who chose to use the formula generally did so successfully. The few who
attempted integration to obtain a solution did so with variable success. In part (c) a few
2
candidates who attempted to use the formula Var ( X )  EX 2   E( X )  were unable to
correctly rearrange it to obtain EX 2   Var ( X )  E( X )  .
2
The most common error was to get EX 2   Var ( X )  E( X )  .
2
A small number of candidates used an alternative method, starting from ‘first principles’:
3
 x3 
1
27  1 7
E X   x 2 dx    

 . Most of these candidates were successful,
4
12
12
12
3


1
1
although a final answer of 13/6 was obtained by a few candidates who failed to deal
successfully with the two negatives. Part (e) was very well done by a large majority of
candidates: a clear and concise method was provided together with fully detailed working
leading to the correct answer.
 
3
2
Question 4
Many candidates gained full marks in this question. In particular, it is to be noted that most
candidates had few problems with either the hypotheses or the conclusion. A sizeable
minority of candidates used > instead of < in H1. The most common error was to use
P ( X  3) instead of P ( X  3) . There were also a number of candidates who failed to place
their conclusion in context.
Question 5
The majority of candidates were able to attempt all parts of this question. However, part (a),
proved to be a challenge to many. It was common to find candidates verifying rather than
showing that y = 4 – 8x, by substituting in either value in each pair of co-ordinates to get the
other or showing that

0.5
0
(4  8x) dx  1 and then stating ‘ f ( x)  4  8 x for 0  x  0.5 ’. For
a minority finding the gradient of the line also proved challenging, with a variety of methods
4
 8 using a diagram as an aid, with only the more observant
seen. Often seen was m 
0.5
candidates adding a note to explain why it must be negative. In some cases, candidates gave
exemplary responses to the first part of part (a) but did not then proceed to specify f(x) fully,
hence losing two marks. Part (b) was generally well answered by the majority of candidates.
In part (c) the most common errors were to find F(0.5) or to solve f(x) = 0.5. The common
incorrect modes given in part (d) were 4 or 0.5.
The majority of candidates were able to follow through their answers to part (c) and (d) to
give the correct direction and reason for the skewness of X. A few candidates also calculated
the mean, usually correctly, which was unnecessary.
43
Question 6
This question was answered well by a high proportion of candidates reflecting a good
understanding of the Poisson distribution and also the use of the normal approximation to
Poisson with many gaining full marks. In part (a) the main error was using Po(150).
In part (b) a minority of candidates failed to use a context when stating the conditions for any
Poisson distribution or, if in context, failed to use words that implied “cars arrive” or “rate of
arrival”. For part (c)(i) the most common error seen was a rounded answer of 0.082.
When finding the probability in part (c)(ii), a small minority of candidates calculated
P(X > 3) = 1 – P(X ≤ 2) or found P(X ≤ 3).
In part (d), the most common error was for candidates to write P(X > m) = 1– P(X ≤ m – 1)
and, having successfully shown that P(X ≤ 15│X~Po(10)) = 0.93150, then write m – 1 = 15 so
m = 16. The majority of candidates used a normal approximation successfully in part (e) and
gained full marks.
Question 7
Many competent and exemplary responses were seen here showing that candidates were well
prepared for this type of question and a high percentage gained full marks on parts (a) to (c).
In part (a) the majority of candidates realised that they had to find F(X) with only a small
minority neglecting to put the integral = 1.
9
Common errors in part (b) included writing E( X )   xf ( x)dx and then finding
0
9
 f ( x)dx
0
or
when multiplying xf(x) making the very basic error of omitting to multiply the second term
by x. In part (c) the most common error when using
9
 k (81x  x
5
3
)dx was to use a lower limit
of 6 rather than 5. A small minority of candidates who used P(X > 5) = 1  P(X ≤ 5) found
P(X ≤ 5) then forgot to find 1  P(X ≤ 5).
Part (d) was perhaps the most challenging part of a question in the paper. There were many
exemplary responses but also a high proportion of incorrect attempts at using the binomial.
The most common error was to swap the p and 1 – p over. Candidates who used a ‘common
sense’ approach and listed the possibilities were generally successful.
44
Decision Mathematics Unit D1
Specification 6689
General
This paper proved accessible to the candidates. All questions contained marks available to the
E grade candidate and there seemed to be sufficient material to challenge the A grade
candidates also. In general candidates seemed well prepared for the examination and, with the
possible exception of question 2(c), set out their work in a clear and efficient way.
Candidates are reminded that they should not use methods of presentation that depend on
colour, but are advised to complete diagrams in (dark) pencil. This remains a particular
problem in the questions on matchings (question 4 on this paper).
Candidates are also reminded that this is a ‘methods’ paper. They need to make their method
clear, ‘spotting’ the correct answer, with no working, rarely gains credit.
Some very poor handwriting was seen, not only making it difficult for the examiners to
decipher, but with a number of candidates misreading their own writing.
Report on individual questions
Question 1
This proved a good starter and was well answered by many candidates with around 35%
getting full marks and over half the candidates getting 7 or 8 marks out of 8. Nearly all
candidates illustrated an understanding of Dijkstra’s algorithm. A common error was to have
a value of 18 at F from labelling G and F in the wrong order, giving an incorrect shortest path
of 22. Some candidates did not work systematically, leading to a jumble of working values,
particularly at F. Some candidates were over-zealous in crossing out working values, the
examiners need to be able to read these to award marks. Those who had a final route length of
21 were able to state a correct route and able to use their answer to (b) to help with part (c),
most candidates were able to use the fact that C lay on the shortest path from A to H to
deduce the shortest path from C to H.
Question 2
Around 50% of the candidates gained full marks on this question showing good preparation.
Most candidates scored full marks in (a), although a few did not state an integer value. Most
candidates were able gain some credit in (b) although many did not place 10 in bin 1. It was
good to see many correct bubble sorts in (c). Some candidates did not perform a final pass or
make a statement that indicated that they had done so. A significant number of candidates
wasted time in showing each exchange, or even each comparison, when they only needed to
show the result of each pass. Some candidates misread their own handwriting and ended with
different numbers to the ones they started with. Most candidates got at least the first two
marks in (d) with a small number getting the 14 and 17 in bins 1 and 2 the wrong way round.
45
Question 3
Around 25% of the candidates scored full marks on this question. In part (a) a number of
candidates only stated the arcs they were including in their tree and did not state the arcs that
they rejected, as they rejected them. Some candidates only referred to the length of the arc
rather than by its end vertices, this makes it difficult for the examiners to determine which arc
is being considered. Part (b) was usually completed correctly, although some candidates
referred to arcs that they had rejected. Some candidates wasted time drawing a table to run
Prim, and then showed their working on the table. If they listed the arcs in order they gained
full credit, but some only listed the nodes in order. Many candidates knew that Kruskal
needed to be used, though some were unclear about how they would modify it. Incorrect
answers included Prim, Dijkstra, Route Inspection and Hamiltonian cycle.
Question 4
Around 48% of the candidates gained full marks in this question. In part (a) many gave the
correct name, although with a wide variation in spelling. Most candidates were able to find a
correct alternating path, but some did not make their path clear, or show the change status step
or the improved matching. Some candidates chose to show their improved matching on one of
the diagrams but some failed to make their matching clear.
Question 5
Around 25% of the candidates were able to secure full marks on this question. Most
candidates correctly paired up the odd nodes and found the shortest routes between them.
A number of candidates stated incorrect totals and a few merely listed arcs but made no
attempt to pair them. Some candidates only showed two pairings. Some omitted the route and
others did not find a correct one. In part (b) most candidates attempted to add their least total
to 31.6. Part (c) proved discriminating, though a number of perfect solutions were seen. Some
candidates did not state clearly that FI (FHI) was the least route. A significant number stated
that the repeated pairing needed to include D.
Question 6
22% of the candidates gained full marks in this question and around 80% gained full marks.
Many candidates struggled to write down the constraints correctly, getting either the equation
or the inequality incorrect. Common errors were; interchanging the coefficients or x and y;
omitting the constant in the second inequality; reversing the inequality. Many candidates had
a significant amount of working in (a), those that did answer correctly showed the most
succinct working. Part (b) was often completed very well. Examiners were pleased to see
more rulers in use. A number of candidates did not label the feasible region R. In (c) the
majority of candidates drew a correct profit line although some drew a line with reciprocal
gradient. Many candidates did not draw a line at all and used point testing losing marks.
A small number of candidates found the maximum point rather than the minimum point.
46
Question 7
This question gave rise to a good spread of marks. Many candidates completed the precedence
table, The most common errors were omitting E and F for I and omitting G and H for K
and L. In (b) many candidates gave a clear and correct explanation for the 6-7 dummy. Some
made reference to G and H but a number made no reference to I and J. The explanation for the
8-9 dummy was poorer. All but a few candidates were able to attempt the completion of the
boxes in the diagram. Errors tended to be at events 5, 7 and 8. In (d) many candidates omitted
at least one critical activity, or included B as a critical activity. Most candidates showed their
calculation for part (e) correctly. Part (f) was usually completed correctly, although some
candidates tried to find a lower bound by drawing a scheduling diagram.
47
Grade Boundaries:
January 2011 GCE Mathematics Examinations
The tables below give the lowest raw marks for the award of the stated uniform marks
(UMS).
Module
80
70
60
50
40
6663 Core Mathematics C1
62
53
44
35
27
6664 Core Mathematics C2
65
57
50
43
36
6665 Core Mathematics C3
61
53
45
38
31
6666 Core Mathematics C4
69
61
53
46
39
6667 Further Pure Mathematics FP1
64
57
50
43
36
6677 Mechanics M1
64
57
50
43
37
6678 Mechanics M2
62
55
48
41
35
6679 Mechanics M3
63
56
50
44
38
6683 Statistics S1
59
52
45
38
31
6684 Statistics S2
69
62
55
48
42
6689 Decision Maths D1
66
60
54
48
42
48
Pass rate statistics:
January 2011 GCE Mathematics Examinations
The percentage of candidates obtaining at least the given number of uniform marks (UMS) at
the time of grading are given below (the final figures may vary slightly from these).
80
70
60
50
40
6663 Core Mathematics C1
37.1
55.5
67.9
78.0
85.7
6664 Core Mathematics C2
39.0
60.3
73.4
82.6
89.0
6665 Core Mathematics C3
27.3
46.0
62.9
75.4
84.7
6666 Core Mathematics C4
32.9
52.3
67.2
75.9
83.8
6667 Further Pure Mathematics FP1
61.1
76.0
86.3
91.8
95.7
6677 Mechanics M1
32.7
53.3
68.4
78.7
84.6
6678 Mechanics M2
43.2
62.7
76.1
85.7
90.8
6679 Mechanics M3
41.1
57.0
68.6
77.5
85.5
6683 Statistics S1
28.1
43.7
59.4
74.1
86.0
6684 Statistics S2
37.7
62.7
77.4
85.7
90.2
6689 Decision Maths D1
28.9
49.0
65.6
78.5
86.4
Module
49
The Uniform Mark Score (UMS):
All unit results are given in the manner of a Uniform Mark Score (UMS).
It was decided by QCA that the reporting of results on modular
examinations should be standardised among all Awarding Bodies and all
subjects, since as modular syllabuses proliferated there were more and
more scoring systems for varying numbers of modules. All Advanced
GCEs are now out of a total of 600 UMS. AS is out of 300 UMS.
Different units in some subjects carry different weightings, but each
Edexcel AS/Advanced GCE Mathematics combination comprises six
equally weighted units with a maximum of 100 UMS possible on each
unit.
The example on the right shows the UMS score for each total mark on a
hypothetical 6663 Core Mathematics C1 unit. The grade A boundary is
set at 57 (out of 75), grade B at 50 and grade E at 29 and these
correspond to 80, 70 and 40 UMS respectively. The notional grade
boundaries are shown in bold type.
Further weighting-UMS correspondences can be found, if you can get
hold of it, in a publication called Guidance notes for Modular Syllabuses
(4th Edition October 1995), although they're not very difficult to work
out (and are given in the table below). Of course, the boundaries vary
from unit to unit and from year to year. Consequently the same
percentage score in different modules may lead to different UMS scores.
JCQ, under prompting from QCA have agreed that the stretching of
UMS between the grade A boundary and the maximum mark means it is
harder for candidates to score uniform marks once their raw mark is
above the grade A threshold. An adjustment has been made so that the
maximum UMS mark is achieved by scoring twice the mark range
between the A and B grade above that of the grade A boundary.
Core Mathematics
Module C1 (6663)
(rm = raw mark)
rm
UMS
rm
UMS
75
74
73
72
71
70
69
68
67
66
65
64
63
62
61
60
59
58
57
56
55
54
53
52
51
50
49
48
47
46
45
44
43
42
41
40
39
38
37
36
100
100
100
100
100
99
97
96
94
93
91
90
89
87
86
84
83
81
80 A
79
77
76
74
73
71
70 B
69
67
66
64
63
61
60 C
59
57
56
54
53
51
50 D
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
49
47
46
44
43
41
40 E
39
37
36
34
33
32
30
29
28
26
25
23
22
21
19
18
17
15
14
12
11
10
8
7
6
4
3
1
Thus the example for 6663 Core Mathematics C1 has candidates
achieving a raw mark of 71 or more scoring a maximum 100 UMS
(i.e. 57 + 2(57 – 50) = 71).
The change to the UMS in June 2001 was communicated to centres at the time results were
issued with the following statement:
“The UMS has been very effective in giving centres a clear picture of progress during the
course. In conjunction with the regulatory bodies, the awarding bodies have considered ways
in which it can be made even more effective. It has been agreed therefore that, with effect
from the outcomes of the 1998 Summer module tests, the system for converting raw marks to
UMS scores for high raw marks should be slightly modified. The modification ensures that
the ‘rate of exchange’ between raw marks and uniform marks is closer at all grades.”
50
UMS Ready Reckoner – GCE Mathematics January 2011
A
B
C
UMS
C1
C2
C3
C4
FP1
100
99
98
97
96
95
94
93
92
91
90
89
88
87
86
85
84
83
82
81
80
79
78
77
76
75
74
73
72
71
70
69
68
67
66
65
64
63
62
61
60
59
58
57
56
55
75
75
75
74
75
75
74
73
74
73
72
71
70
69
72
66
65
62
61
60
59
58
57
56
55
54
53
52
51
50
49
48
47
46
45
44
43
42
41
40
73
69
68
67
68
66
65
64
63
65
64
63
58
57
56
55
66
65
64
63
58
57
56
54
53
52
51
62
61
60
59
53
52
50
49
48
47
58
57
56
55
51
50
49
46
45
44
54
53
52
55
54
48
47
43
42
51
50
75
57
56
67
66
60
59
60
59
58
57
59
58
55
54
56
64
58
57
56
58
57
56
56
55
54
57
56
53
52
51
63
62
61
70
62
61
60
59
75
61
60
68
67
66
69
68
67
D1
60
59
69
65
62
61
60
59
S2
62
61
72
66
S1
62
61
71
70
71
M3
64
63
73
72
69
67
64
63
74
71
70
68
67
71
70
M2
75 74–75 75 73–75
72
74
73
74
72
73
71
73
70
72
71
72
69
71
70
71
68
70
70
69
67
68
69
66
69
65
67
68
68
66
67
64
63
67
65
66
66
62
64
65
61
65
63
64
60
64
62
63
59
63
61
62
58
74
73
72
M1
74
74
73
73
72
71
72
70
71
69
68
70
67
69
68
66
65
64
63
55
54
55
54
53
52
53
52
53
52
51
50
51
50
49
51
50
49
49
48
47
55
54
50
49
60
59
48
47
58
57
46
45
44
56
55
54
53
48
47
48
47
46
45
52
51
50
49
48
47
51
62
61
60
59
58
57
43
42
53
52
56
55
54
53
52
51
D
E
N
UMS
C1
C2
C3
C4
FP1
M1
M2
54
53
52
51
50
49
48
47
46
45
44
43
42
41
40
39
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
39
38
37
36
35
34
33
46
45
41
40
49
48
46
45
46
45
44
43
44
43
42
39
38
37
47
46
45
44
43
42
44
43
42
41
44
42
41
40
39
43
42
40
38
41
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
41
40
36
35
44
43
41
40
39
38
34
33
42
41
39
38
37
36
35
32
31
30
40
39
38
37
36
35
34
33
29
28
37
36
34
33
32
31
27
26
35
34
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
25
24
23
22
33
32
31
30
29
28
27
26
24–25
23
22
21
20
19
18
17
16
15
14
13
12
11
10
8–9
7
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
14
13
12
11
10
9
8
7
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
M3
46
45
39
38
37
36
40
39
37
35
38
36
35
34
33
37
36
34
32
35
33
32
31
30
34
33
S1
S2
41
40
51
50
39
38
37
49
48
48
47
46
47
46
45
45
44
43
44
43
42
42
41
40
41
40
39
39
38
37
38
37
36
35
34
33
32
31
30
29
28
27
26
D1
50
49
25
31
29
32
24
36
36
30
28
31
23
35
35
29
27
30
22 33–34 33–34
28
26
29
32
32
27
25
28
21
31
31
26
24
27
20
30
30
25
23
26
19
29
29
24
22 24–25 18 27–28 27–28
23
21
23
26
26
22
20
22
17
25
25
21
19
21
16
24
24
20
18
20
15
23
23
19
17
19
14 21–22 21–22
18
16
18
20
20
17
15
17
13
19
19
15 15–16
16
12
18
18
14
14
14
15
11
17
17
13
13
13
14
10 15–16 15–16
12
12
12
13
14
14
11
11
11
12
9
13
13
10
10
10
11
8
12
12
9
9
9
10
7
11
11
8
8
8
8–9
6 9–10 9–10
7
7
7
7
8
8
52
UMS
C1
C2
C3
C4
FP1
M1
M2
M3
S1
S2
D1
6
5
4
3
2
1
0
4
3
6
5
4
3
2
1
0
5
4
3
2
6
5
4
3
2
1
0
6
5
4
3
2
1
0
6
5
4
3
2
1
0
6
5
4
3
2
1
0
6
5
4
3
2
1
0
5
4
3
2
7
6
5
3–4
2
1
0
7
6
5
3–4
2
1
0
2
1
0
1
0
Maximum mark for each unit is 75
53
1
0