Examiners’ Report

Principal Examiner Feedback

Summer 2014 (R)

GCE Mathematics

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Core Mathematics C1 (R) (6663)

Introduction

Mostly students answered this paper well with the modal mark being full marks on every question. Weaker students made slips and errors in arithmetic, in basic algebra and in copying down information but there were also some very good students who had been well prepared for the examination and who avoided these errors.

Report on Individual Questions

Question 1

This question was generally very well answered. Most gave the answer in the form given in the main scheme. Some students only took out the x term then stopped and others lost two marks as their first line was with wrong signs. It was very unusual to see anyone misunderstanding the instruction to factorise and continuing their answer by ‘solving’ to x = …. This was a good opening question.

Question 2

This question was generally done well. In part (a) the common mistakes were not evaluating 93 or 36 or incorrectly evaluating 93. Some evaluated 813 first, then struggled to find the square root on this non calculator examination.

In part (b) some stated incorrectly that or that .

Others stated that or . Another common error was not squaring the 4 at

the start of the bracket and only dealing with the x term.

Question 3

This question was generally very well done. Part (a) was usually correct.

In part (b) there were some fairly common errors. A minority used the formula for the sum of an AP. Some put a3 equal to 66 and another group of students did not put their sum equal to 66. There was also some weak algebra solving the linear equation in k.

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Question 4

Part (a) was usually correct. Any errors were usually in dealing with the second term,

expressing as a multiple of an index and then differentiating.

In part (b) the main difficulty was in the integration of . There were a few but not many,

who integrated their answer to part (a).

Question 5

This question was well answered. Some however found it challenging to get beyond

or . Other errors were, for example, replacing 20 by and then by

, or replacing 10 by .

Some rationalised the denominator creating extra work instead of simplifying it to 2. Others thought they could remove the roots by squaring, which creates a second, invalid, solution.

Question 6

Most students did fairly well on this question. The most common mistake was putting their inequality in part (b) as their answer to part (c) and not trying to combine their two inequalities.

In part (a) the most common error was to give a wrong expression for perimeter (usually neglecting one or two of the sides). Those who gave a correct expression usually completed part (a) successfully.

In part (b) a wrong expression for area was less common - most divided up the inside region to obtain their expression. The resulting quadratic was usually solved correctly and most students chose the inside region, as required. A minority put so did not have an inequality or put and obtained the wrong inequalities. Credit was given to those giving the answer as and also to those who realised that was a length and so gave the answer as .

In part (c) some students made no attempt to combine their answers to part (a) and part (b).

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Question 7

This proved more challenging than the earlier questions.

In part (a) students were usually able to get the first three marks by a variety of methods, although a few transposed the x and y coordinates when substituting, or mixed them up. Signs were an issue for some, particularly when finding the gradient.

Having got the correct equation, some made no attempt to change it to the correct form. Others made arithmetic/algebraic errors. Those who did have all three terms on one side sometimes ignored the need for ‘integers’ or ‘= 0’.

In part (b) the most common method used was finding the equation of MN, then substituting x = 16. Those using Pythagoras were often successful.

In part (c) it was usual to see their final answer as the coordinates of K, rather than just the y coordinate as requested. (This was not penalised). Some realised that they just needed to add 6 to p (although they did it in a variety of different ways). Quite a few correctly solved the simultaneous equations generated by the line equations for KL and KN. A handful used vectors. Those who tried to use Pythagoras were usually unsuccessful, not realising that it would generate two solutions, so were confused if they managed to reduce it to a quadratic equation. A significant minority assumed wrongly that x = 7. An interesting method came from those who realised that, as it is a rectangle, the diagonals LN and KM bisect each other, hence the midpoints are the same.

Question 8

Most students integrated the two terms correctly, though a few could not deal correctly with

. Those who gave it as to the power usually had no problem integrating and

dividing by the fraction . A minority missed the constant hence losing the last three marks.

Some students made arithmetic mistakes in working out the constant. A very small minority tried to differentiate instead of using integration as the reverse of differentiation.

Question 9

This question was a reasonable discriminator.

In part (a) the quadratic and linear graphs were generally well drawn. Marks were lost due to

the omission of co-ordinates particularly the .

For part (b) students were asked to determine a value for k for which the given line was a tangent to the given curve. There were several possible methods of solution. The method

using was the most popular approach. Those who began correctly by this method putting

the gradient expression for the curve equal to the gradient of the line, usually completed it to find x, then y, then k. Many who attempted instead to set the curve expression equal to the

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line expression obtained a quadratic but proceeded no further. Of those who continued with this method, use of the condition for equal roots, putting the discriminant equal to zero usually was more successful than completion of square methods.

Question 10

Part (a) required students to show a printed result. It was extremely rare to have brackets missing. The formula was not usually quoted, but students should be advised that including the formula would make their method clearer.

In part (b) some students when copying made errors such as wrong signs or 13 changing to 3. The time for Yin was usually correct, but not always simplified correctly. A few students used A instead of A – 13, as their first term. Most students equated the two times and solved to find d. A few treated their times as simultaneous equations usually coupled with ‘= 0’ and obtained d = 3 after incorrect assumptions.

For part (c) the formula for sum was not usually quoted but students were able to use it with n = 14 and usually got the first M1. However, depending on how they had set this up initially, many had not taken into account that for Xin the difference between the terms is actually d + 1, and so were unable to gain any more marks. There were sometimes further arithmetic errors. Several students did not use the sum formula instead putting Xin’s time for day 14 = 784.

Question 11

This was a well answered question .

In part (a) most students understood the method and used the curve equation to find the value for y, then differentiated to find an expression for the gradient of the curve. They found a numerical gradient at x = 2, then used the negative reciprocal to obtain a numerical gradient for the normal. A few students found y = 3 by an incorrect method, using the line equation

which they were trying to find, hence producing a circular argument. Differentiation of

was a challenge for some, and others made errors calculating the numerical gradients. The printed answer gave them an opportunity to check for errors.

Part (b) was particularly well answered, and most showed good algebraic skills on this question. Very few students attempted the quadratic in y, mostly using the most concise method of solution, involving x. Some students forgot to find the second coordinate.

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Core Mathematics C2 (R) (6664)

Introduction

This paper proved a good test of students’ knowledge and students’ understanding of Core 2 material. There were plenty of accessible marks available for students who were competent in topics such as binomial expansions, integration, geometric series, trigonometric equations and differentiation. Therefore, a typical E grade student had enough opportunity to gain marks across the majority of questions. At the other end of the scale, there was sufficient material, particularly in later questions to stretch and challenge the most able students.

Trigonometry in general proved to be an area of weakness for a significant number of students. Some responses showed that students were either unfamiliar or uncomfortable when working in radians and often converted to degrees, e.g. qus.5 and 7(ii). Question 10(d) involving the trigonometry of a right angled triangle also caused significant problems, possibly due to not using a suitable diagram.

Report on Individual Questions

Question 1

This question was generally well answered and responses showed that students could work confidently with binomial expansions. Although the majority of responses gained full marks the error of not squaring the denominator in the term when expanding the bracket was seen occasionally, leading to an incorrect expansion of .

Since the bracket did not contain a negative term, sign errors were all but eliminated, increasing the likelihood of maximum marks.

Question 2

Part (a) was well answered by the majority, though a few seemed unaware of the sum to infinity formula. Those with a correct solution used appropriate algebra rather than verification.

Students had few difficulties with part (b) with only a handful using instead of the

correct .

Part (c) was well done by most although premature rounding cost a significant number the final accuracy mark, with “3” given as the final answer. A small number found S29 instead of S30.

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Question 3

This question was well done by most students. Most errors seen were either bracketing problems or issues finding the value of h. Many who did use an incorrect h often divided by 5 not 4 in finding the width of the strips. As the trapezia were of width two, the multiplying factor outside the bracket was 1. This meant that it was not realistically possible to identify genuine bracketing errors so that expressions such as

or

were condoned and it was assumed that students were interpreting the trapezium rule correctly.

Question 4

For part (a) the majority could obtain the correct value for a by solving f(2) = 0. There were very few students who chose long division or comparison of coefficients.

Many students could at least make a start in part (b) and used inspection or long division to establish the quadratic factor –4x2 + 9. Interestingly, many students stopped at (x – 2)(–4x2 + 9) for the factorised form of f(x), presumably not spotting the difference of two squares. Of those who did attempt to factorise –4x2 + 9, a significant number of students chose to change the sign and obtained (2x + 3)(2x – 3) without subsequently compensating for the change of sign.

The method in part (c) was well known and most chose to evaluate , with a few students

opting for long division. The scheme allowed for a follow through accuracy mark for those with an incorrect value for a in part (a).

Question 5

Part (a) was extremely well done. Most problems occurred because students were not comfortable using radians and changed 2.1 radians to degrees before making an attempt at the arc length DEA.

In part (b), to find the width and height of triangle BCD many students resorted to using the sine rule instead of basic trigonometry. This then caused problems for some who wrote down

equations such as and then proceeded to work in radians, including

using 90 degrees as a radian measure. There were some cases where students did not appreciate what was meant by the perimeter and included BD in their total. There were also a significant number of cases where students rounded prematurely which meant that the final A mark was lost.

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Question 6

The most common error in this question occurred when students integrated the given expression, substituted the limits 2 and –4 to give 10.5 and then stopped. The most popular correct strategy was to find the area of the enclosing rectangle and then to subtract their 10.5. With this approach, some students found different values after substituting x = –4 and x = 2 into the curve and so did not have a rectangle, although it was treated as one. Others had difficulty evaluating their integrated expression with the required limits.

A significant number of students chose a different strategy and attempted the area by subtracting the curve from the line first and then integrating. This approach was met with varying degrees of success. Those who worked this method carefully often produced the correct answer but there were a surprising number of sign errors and it was not uncommon to

see interpreted as .

Question 7

Responses to part (i) were varied. The majority of students could at least reach

following a fairly straightforward rearrangement. Occasionally just one value was found for theta but the most common mistake came from premature rounding with 9.8 and 80.3 seen often. Some students obtained the first value correctly (9.7) but then subtracted this value from 180 degrees with 9.7 and 170.3 resulting.

In part (ii) the majority of students recognised the need to apply the appropriate trigonometric identity, , although some incorrect identities were seen including

. Those who did obtain a quadratic in sometimes made errors when rearranging or made mistakes when solving the quadratic. Commonly students dealt with the constant but then ‘lost’ the negative to give leading incorrectly to

.

A large number of students chose to work in degrees and although some converted back into radians at the end, most lost the first A1 mark by leaving their answer in degrees. The final B1 mark was also occasionally lost when students gave only one value from or cancelled their quadratic, losing one factor altogether.

Question 8

Very few errors were seen in part (i). The majority took logs base 10 and divided but some took logs base 5 to give the correct answer directly.

Success in part (ii) was very varied. Those with a clear understanding of the properties of logs could make significant progress although the resulting quadratic in √x confused many.

The most common error was from those whose understanding of logs was weak, wrote as . Some credit was given for any evidence of understanding of

either the power law or addition/subtraction laws and some students could gain at least one or

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two marks. Solving the quadratic involving √x was challenging for many and of those who chose to square x + 15, sometimes produced x2 + 225. More successful students substituted y = √x to help with the factorising and solving of the equation.

Question 9

In part (a) most students were able to attempt an equation with three areas and these were often correct in the un-simplified form. Students regularly used Pythagoras to find the height of the triangle in finding the area and this led to a complicated expression that was sometimes simplified incorrectly. Students chose this approach more regularly than the area formula in terms of sine. For the semi-circle, not squaring the denominator when removing the bracket led to an incorrect simplified expression. Some students gave a final answer with a subtraction inside the bracket instead of an addition.

Part (b) saw the first B1 lost with an incorrect term for the perimeter of the semi-circle for some. Students were using 2πr for the circumference but then often they used x as the radius. Most gained the M1 for the substitution. There were errors in the manipulation of the expression to reach the given equation but the most common was in expanding the bracket to

reach a positive term.

In part (c) most students made a good attempt at differentiation and gained the M1 for at least one term correct (usually the x term). Many students found solving the equation difficult and errors in manipulation often led to an incorrect value for x. A large proportion of students failed to use their value of x to find the minimum value for P.

Many correct responses were seen in part (d) and students usually differentiated again successfully. Almost all substituted their value for x from part (c) but then some failed to consider the sign and/or give a conclusion.

Question 10

Part (a) was generally well done although there were some errors seen.

In part (b) many students were familiar with the equation for a circle but difficulties often occurred with finding the radius. Methods were often muddled and students did not make it clear if they were finding the radius or the diameter.

For part (c) and part (d) a clear labelled diagram was of great benefit, but seen only rarely. A fair proportion of students were successful in part (c) but far fewer scored well in part (d). It was here in particular that the clear diagram came into its own. As it was, many students found the wrong angle with a common substitute for .

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Mathematics C3 (R) (6665)

Introduction

This paper was found to be accessible by most students. It contained a mixture of straightforward questions that tested the student’s ability to perform routine tasks, as well as some more challenging and unstructured questions that tested the most able students. Most students were able to apply their knowledge on questions 1, 2, 3, 4(ii) and 7(i). Timing did not seem to be a problem as most students seemed to finish the paper. Questions 4(i), 6 and 7 required a deeper level of understanding. Overall the level of algebra was pleasing, although there are many examples of students not using brackets correctly. Points that could be addressed in future exams are the lack of explanation given by some students in questions involving proof. When an answer is given it is important to show all stages of the calculation. It is also useful to quote a formula before using it. Examples of this are when using the product and quotient rule (qu.4) and using trigonometrical identities (qus.4(i) and 3).

Report on Individual Questions

Question 1

The vast majority could obtain the fully simplified answer accurately and efficiently. However there were a few students who failed to factorise the 4x2 – 9 term, so could not complete their solution, and some who used more terms in the denominator than necessary, resulting in quadratic or cubic numerators. Some of these did proceed to a complete solution, but most made errors in the process. A frequent transcription error was 2 instead of 3 in the numerator of the first term. Only occasionally did students use partial fractions on the final term.

Question 2

Part (a) was generally well done, with accurate differentiation and manipulation of the given expression. Some students failed to recognise that, as a proof, there was an expectation to set

.

Part (b) the graph of y = x3 was often correct, although some students drew a straight line, and others showed a cubic with two turning points. The exponential graph was much less successful, with some students showing a reflection of the correct graph. Some very pleasing and accurate sketches were seen, but a correct asymptote was rare – often given as x = –2 instead of y = –2.

Those students who had graphs with one point of intersection generally answered part (c) correctly, although there were also references to graphs crossing axes, and to one-to-one mappings.

Part (d) was nearly always successful and mostly given to the accuracy asked for by the question.

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Part (e), most students obtained the x coordinate, although many did a lot of extra work, either by continuing the iteration or by looking for a change of sign of the function. However many either forgot to find a y coordinate, or substituted into an incorrect function finding an incorrect y value.

Question 3

Part (i)(a) This question was generally tackled well, with most students knowing the trigonometrical identities for . A surprisingly frequent error in part (a) was the omission of cosec x from the original equation, but those who started with the correct form could usually follow an appropriate process, with some making minor arithmetical/algebraic errors.

Part (i)(b) Nearly all could solve their quadratic, and most gave two values for x. There were very few answers given in degrees.

Part (ii) was usually successfully attempted, although some students who worked from both sides of the required identity did not give an adequate conclusion to complete their work. One

of the most common errors was to obtain , as they had difficulty with the in the

denominator.

Question 4

Part (i) Many students wrote down correctly. Some did it implicitly, and a few used

and quotient rule. The most common errors seen were and

.

Most then went on to find by inverting their result, often not until after expressing in

terms of x.

A number used a triangle with ratios as expressions of ‘x’ to reach their final expression. Of course some just manipulated the given answer, but there were many completely correct solutions. It was important that students use the identity and not

in proving their result.

Part (ii) was well completed. Students know the product rule really well but all should be advised to write it down first before using it. The derivative of ln 2x was also often correct,

although was frequently seen. The substitution of was often correct, but a considerable

number of students made algebraic errors in manipulating the resulting expression.

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Part (iii) Most students differentiated correctly, but couldn’t simplify the numerator to gain the final mark. Most used the quotient rule although those who used the product rule found the expression easier to simplify. There were a lot of basic algebraic errors shown in the attempt to reach an expression of the right form. Again it must be stressed that writing down the quotient rule formula would have resulted in more marks for students making slips in applying the method.

Question 5

Nearly all students drew a V-shaped graph in part (a), and most were in the right position with coordinates correctly labelled. Generally marks were lost for not labelling one of the intersections.

In part (b) and part (c) most students retained inequalities throughout their working, frequently having problems when the x term was negative. Many only used one inequality and hence only achieved half of the required region. Those who had sketched a graph were most likely to pick the correct regions in part (b) some clearly changed a correct inequality because they did not understand which region was required. Those who found the critical values by squaring often failed to consider whether they had introduced extra solutions. Only careful and thoughtful students scored all 4 marks here.

In part (c) it was particularly noticeable that many students reversed the sign of the wrong

expression, so obtained an incorrect inequality. Many students did not realise that .

And meant that was excluded from the solution set.

Question 6

Most students could find the inverse function correctly, although a few differentiated. However a lot of students gave an incorrect range for f(x), and often an incorrect domain for f –1.

Many students misunderstood part (c) and used ln (4x2) etc. They usually proceeded with correct log and exponential work, but obtained an incorrect answer. Very few students who started this part correctly could simplify to obtain .

Part (d) and part (e) were moderately successful, although many students failed to square both the 2 and the when simplifying the expression. Many then lost marks in part (e) because of incorrect solutions to part (d). A few managed to restart from and solve correctly.

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Question 7

The majority of students could complete part (a), with only a few giving answers in degrees.

A very common error in part (b) was to forget the intersection of the graph with the y-axis; many found at least one correct intersection with the x-axis.

Part (c) was probably the least successful part of the question. Although a good number of students could write down the maximum and minimum easily, some of those who had a correct value for the maximum then gave either –18.5 or 12 as the minimum. There were quite a number of students who had no idea how to tackle this part, often using values of H when t = 0 and t = 52.

Part (d) many students successfully reached one or more correct values for but then

made calculator errors in reaching their value for t. Many students showed very little working at this stage, so it was sometimes unclear how much of the work was accurate.

There were still a number of students who did not realise that part (a) should be used in solving part (c) and part (d).

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Core Mathematics C4 (R) (6666)

Introduction

This paper proved to be a good test of students’ knowledge and understanding of the specification. There was plenty of opportunity in the first 6 questions and Q8 for grade E students to demonstrate their skills. At the other end of the scale, there were some testing questions involving vectors, differential equations and parametric equations that allowed the paper to discriminate well across the higher ability levels.

Examiners were impressed with quality of students’ presentation and how, in many cases, solutions were methodical and easy for examiners to follow.

The standard of algebra was usually good, although a number of students made basic sign or manipulation errors in questions 1(b), 2(c), 3(b), 4(b), 6(c), 7(b) and 8(c). In summary, questions 3(b), 6, 7 and 8 were discriminating at the higher grades. Question 8 proved to be the most discriminating question on the paper, with only about 7% of the candidature able to gain all 12 marks.

Report on individual questions

Question 1

This question was well answered with about 60% of students gaining full marks.

Part (a), most students started by manipulating to give , although a

few incorrectly wrote . The majority were able to use a correct method for

expanding a binomial expression of the form (1 + ax)n. A variety of incorrect values of a such

as or and n such as , –1 or –2 were seen at this stage. The majority of

students expanded to give , but some forgot to multiply this

by to give the answer to part (a). Sign errors, bracketing errors and simplification errors

were also seen this part.

Part (b), most students multiplied (3 + x) by their binomial expansion from part (a). A small minority, however, attempted to divide (3 + x) by their binomial expansion. A surprising number of students attempted to expand (3 + x) by writing it in the form k(1 + ax)n. Other students omitted the brackets around 3 + x although they progressed as if “invisible” brackets were there.

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Question 2

Question 2 was well answered, with about 42% of students gaining full marks and about 61% of students gaining at least 8 of the 9 marks available.

Part (a), most students applied the trapezium rule correctly in order to find the approximate area for R. The most common errors were using an incorrect strip width of 0.4 or not rounding their final answer to 2 decimal places.

Part (b), examiners saw a number of acceptable reasons to explain how the trapezium rule can be used to give a more accurate approximation. These included increase the number of strips, make h smaller or increase the number of x and/or y values used. Incorrect reasons included use more decimal places, use smaller values of x and/or y or use definite integration.

Part (c), the majority of students employed a method of integration by parts with u = 2 – x

and , although a minority multiplied out (2 – x)e2x to give prior to

integrating. Common mistakes included sign errors when integrating or evaluating their final answer; or integrating e2x to give either 2e2x, e2x or ke4x.

Question 3

This question discriminated well, with about 36% of students gaining full marks and about 48% of students gaining at least 9 of the 10 marks available.

Part (a), many students were able to differentiate correctly, factorise out , and rearrange

their equation to arrive at a correct expression for the gradient function such as

. A significant minority, however, did not simplify this expression as

required by the question. A minority did not apply the product rule correctly when differentiating –4xy, whilst a small number left the constant term of 10 on the right hand side of their differentiated equation.

Part (b) proved a test for the higher ability students. Most recognised that the numerator of

their answer to part (a) had to be set to zero and obtained x = 2y – 5 or , but then a

minority gave up at this point. Whilst most substituted their x = 2y – 5 (or equivalent) into a significant minority who had problems with the resulting

algebra and found difficulty in reaching a correct or equivalent. Those who progressed this far were usually able to solve the quadratic equation to give both correct values of y. A minority of successful students applied the alternative method (as detailed in

the mark scheme), of finding both values of x, followed by using to find both

values of y.

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Question 4

In this question the majority of students were able to score full marks in part (a) with part (b) offering good discrimination for the more able students. About 39% of the candidature were able to score all 10 marks.

Part (a), the majority of students were able to split up in the correct form of

, although a significant number missed the factor to give the incorrect

form of . Many students were successful in either substituting values and/or

equating coefficients in order to find their constants.

Part (b), the majority of students were able to write down a correct expression for the volume formed, although a number of students omitted π and applied . A number of weaker students did not make the link with part (a) and made no further progress.

The majority of students attempted to integrate their partial fraction, although a few tried to integrate either the square root or the square of their partial fraction. Most were able to

integrate both and correctly with a few integrating to give . The most

common error was to integrate to give . Most students applied the

limits of 4 and 1 correctly, but a significant minority struggled to apply the laws of logarithms to manipulate their answer into the form .

Question 5

This question was well answered, with about 54% of students gaining full marks and about 77% of students gaining at least 4 of the 6 marks available.

Part (a), many students wrote down and used the Chain Rule correctly to set up

an equation for . They applied 3 divided by their and substituted r = 4 to find a value

for . Common errors in this part included applying substituting r = 3 into

their , giving their final answer as , or incorrectly rounding their answer to

give either 0.015 or 0.01.

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Part (b), the majority of students applied and substituted r = 4 to give the

correct answer for . Some applied divided by their whilst others made no

attempt at this part.

Question 6

This question discriminated well across students of all abilities, with about 22% of students gaining full marks and about 65% of students gaining at least 7 of the 10 marks available.

Part (a), the overwhelming majority of students found the correct answer of p = 5, with a few incorrectly stating p = –5.

Part (b), the majority of students applied the simplest method of equating the i components of l1 and l2 leading to μ = –2. They then substituted this value into the equation for l2 to give the coordinates of C. There were a significant minority, however, who did not attempt to show that that l1 and l2 intersected.

Part (c), the majority of students applied the scalar product formula in order to find the angle ACB. The majority achieved the correct answer by applying the scalar product formula between and (or and ). Some students applied the scalar product formula between the direction vectors of l1 and l2 which gave an obtuse angle. Only a handful, however, manipulated this angle to give the correct answer of 27.7°. Comparatively, few chose to use the cosine rule.

Part (d), the majority of students used , with and and achieved

the correct answer of 14.7. Some unsuccessful students applied a and b as the length of their direction vectors.

Question 7

This question discriminated well across the higher ability students, with about 24% of students gaining full marks and about 44% of students gaining at least 8 of the 10 marks available.

Those students who separated the variables correctly in part (a), were usually able to integrate at least one side of their equation correctly. Common errors at this stage included integrating

to give and omitting a constant of integration “ ”, whilst a

number of students struggled to integrate . Some students did not show sufficient

steps in order to progress from to . Other students struggled to remove logarithms correctly and gave an equation such as

which was then sometimes manipulated to the answer given on the question paper.

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Part (b), many students wrote down two equations each containing A and k and attempted to solve them simultaneously. Algebraic manipulation and dealing with exponentials caused problems for a significant minority of students.

Those who were successful in finding the exact values of A and k, usually achieved the correct answer of 4400 fish in part (c).

Question 8

This question discriminated well across students of all abilities, with about 7% of students gaining full marks and about 55% of students gaining at least 6 of the 12 marks available.

Part (a), the majority of students were able to apply a full method of setting y = 1 in order to find t and substituting t into in order to find k. Only a minority of students

found and applied to give the correct answer of The majority used

values of t such as and leading to incorrect answers of and

respectively.

Part (b), the majority of students were able to apply the process of parametric differentiation

followed by substitution of their t into their . Occasional sign errors were seen in the

differentiation of both x and y and a number of students obtained for . Only a

minority of students used to obtain a correct answer of Most used

to give a final answer of , when it was clear from the diagram that the gradient of

tangent to the curve at A must be negative.

Part (c), many students achieved (or equivalent) after setting their

. The majority, however, were not able to find a correct strategy for solving their

trigonometric equation. Some students who squared (or equivalent) to give .

The most popular method was to square both sides of , and apply the identity

sin2 t + cos2 t ≡ 1 to achieve a quadratic equation in either cos t or sin t. Some students squared both sides of 4sin t – 4cos t = –1, applied the identities sin2 t + cos2 t ≡ 1 and

to achieve Few students correctly rewrote

as (or equivalent). The majority of students who

used a correct strategy usually achieved the correct answer of .

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Further Pure Mathematics FP1 (R) (6667)

Introduction

The standard of work was high with a lot of well organised and clear solutions. Diagrams would have helped students make better progress in question 7 and greater attention to accuracy in arithmetical calculations was often needed with matrices. Students found question 8 and question 9 challenging.

Report on individual questions

Question 1

A large majority of students made a reasonable start to the paper and were able to write down 1 – 2i. Some students then found it difficult to form a quadratic factor. Those that did form a quadratic found the linear factor 2z + 1. After this good work a proportion of students did not

go on to state that .

Question 2

Part (a) students often gained full marks with the use of degrees rather than radians being the most common error. Some students gained most of the marks but then omitted to round their final answer to the required degree of accuracy. Some students were confused in their approach to linear interpolation and made a sign error without quoting a correct formula so they gained no credit.

Part (c) the majority of students made some progress towards the correct interval by finding the first two values, but inaccurate calculations meant they then made incorrect decisions for subsequent values. It was clear that this method was less well known than the earlier parts of this question.

Question 3

Incomplete descriptions of the correct transformation meant marks were lost in part (a). The most common error was to miss the origin and hence lose the first mark. The majority of students knew the correct matrix for the enlargement part (b) with the occasional swapping of elements of omission of correct signs.

Part (c) ‘14’ was seen very often and then an attempt was made with the determinant. A number of students then did not know how to proceed and lost the final marks for this part.

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Question 4

Almost all students knew they were required to multiply through by the conjugate of the denominator in part (a), but some lost out accuracy to sign errors or not collecting terms correctly. There were a number of students who did not present the solution in the required form.

Part (b) there were some confused attempts with the argument of a complex number, but most students realised that the real and imaginary parts had to be equal and made some progress. If accuracy was lost part (a) then this impacted on the accuracy marks awarded here too.

Question 5

The first three marks were regularly awarded to students with the expansion of brackets and the use of the summation formulae carried out accurately. However a number of students failed to spot the obvious factors and decided to multiply out to a cubic or quartic. This was recovered by some with detailed working from a cubic, but other simply quoted the given answer and moved on which did not gain full marks. The majority of students produced fully correct solutions in part (b) and 1619910 was seen very often. A minority struggled as they did not see the link to part (a), while other lost accuracy as they subtracted the sum of ten terms rather than nine.

Question 6

This was expected to be a good source of marks for all students, but unfortunately accuracy was an issue for some students. Part (a) sums, differences and / or products of matrices often had inaccurate elements which lost at least two marks.

Part (b) a number of students did not take the importance of the order of matrix multiplication into account and a large group could not find an inverse of a 2 × 2 matrix accurately. Those students who decided to use simultaneous equations were less likely to produce an accurate answer due to accuracy errors.

Question 7

Part (a) was well rehearsed and convincing and full solutions leading to the given answer were not unusual. However, a large number of students did not write down the correct answer

Part (b) with sign errors on the right hand side of the equation being very common. This impacted on the accuracy PART (c) as students often attempted to solve their two equations simultaneously rather than using at the intersection of the two straight lines.

Students who had a correct answer part (c) typically went on to gain all the marks part (d). Those who were successful usually drew a sketch which helped with the formulation of the area. Students who did not draw a sketch usually struggled without any visual clues and usually missed out the factor of 2 in their areas. Some weaker attempts showed good exam technique and did well to salvage a mark by writing down and labelling the coordinates of the focus.

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Question 8

This question proved to be challenging for some students and a lot of incomplete or very confused methods were seen. Typically, if a student realised they needed to substitute their coordinates of the intersection into the formula for the line, they usually found the coordinates, although sometimes these were presented with sign errors.

Question 9

This proved to be a demanding induction question; especially given students usually finding the topic a challenge.

Part (a) the better attempts made good progress, but some students were confused over the method of adding the (K + 1)th term. However once they did so correctly then they almost always completed satisfactorily.

Part (b) was a challenge for a large number of students and many did not show the statement to be true for both n = 1 and n = 2. The majority of students could not make any progress with U(k+2) or were unable to handle the powers of 4 and 2 convincingly. If the students had done the earlier work correctly then they usually produced a correct final statement. Those with incorrect final statements usually had made significant errors earlier in the question.

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Further Pure Mathematics FP2 (R) (6668)

Introduction

This was a paper with some straightforward questions and some more challenging ones and thus every student was able to show what they had learnt. There was little evidence of students running out of time before completing all they could do.

Sometimes the presentation of the work is poor, with equations straddling lines or very small handwriting with lots of scribbled out work. Poor presentation can lead to a student miscopying their own work or making other errors and so achieving a lower score. It is good practice to quote formulae before substitution. When an error is made on substitution the examiner needs to be sure that the correct formula is being used before the method mark can be awarded.

If a student 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 student to say whereabouts in the script the extra working is going to be done.

Report on Individual Questions

Question 1

The vast majority of students confidently separated the given function into partial fractions and demonstrated good understanding of the method of differences. Some failed to appreciate the need to divide by 2 for the required summation. Good explanation and presentation of the method was generally in evidence.

Question 2

The majority of students used a valid method to obtain the critical values in this problem (most common was to multiply through by x2 and subtract 2x) and it was pleasingly rare to see students simply multiplying through by x. However, using the critical values to identify the regions for which the inequality held proved more problematic as some students did not appear to know how to interpret the values they had obtained. It was common to see 1 < x < 6 and/or 2 < x < 11/3 as answers.

Question 3

Most students displayed a good knowledge of first order linear differential equations and the need to establish an integrating factor. Integration of e4x was generally correct. The most common error in part (a) was the failure to multiply the constant of integration by cos2x but complete solutions were in pleasing abundance.

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Question 4

The majority of students knew what to do and differentiated correctly, setting their derivative to zero and proceeding to attempt to solve the equation. Some students lost marks through incorrect double angle formulae. It was fairly evenly split between those who found sin θ and those who found cos θ first. Some found cos 2θ directly. The last three marks were more problematic with many students resorting to arcsin or similar (not realising that they needed to be working with exact values) and many did not give the final answer as r = f(θ).

Question 5

This question was well answered and students seemed confident applying standard techniques to obtain the required Maclaurin expansion. Mistakes in part (a) tended to be seen

in incorrect attempts to differentiate and some students lost marks because having

differentiated perfectly, they then made no attempt to form an expression for as required

by the question.

Part (b) was accessible to almost all students who were able to gain most if not all marks available.

Question 6

This question was either attempted confidently or quite poorly. Most students chose the first method for part (a) and alternative 3 for part (b). Few mistakes were made once the student had decided which method to use although some only used the numerator of the realised expression to compare with v = –1 (for part (a)) or y = 0.5 (for part (b)). Some of the weaker students tried to realise the denominator of an expression still containing w or z. Presentation and poor handwriting were an issue with this question as it was difficult to distinguish sometimes between u and v, i and 1, 2 and z. It was very rare to see students approaching the problem using loci, which would have produced a much simpler solution than using Cartesian equations.

Question7

Part (a) was generally attempted with confidence and good appreciation of de Moivre’s Theorem. The binomial expansion was dealt with successfully and the manipulation of trigonometric functions caused few difficulties. The use of imaginary parts for the sin θ expansion was well presented although some students failed to illustrate the specific application of de Moivre’s theorem with the index of 5.

Part (b) most students realised the need to link the given equation part (a) by the simple trigonometric substitution but the multiple solutions required proved to be a real differentiator. Only the best students were able to provide five angles with different sine ratios.

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The majority of students again recognised the need to express the given expression in terms of sin 5θ when working part (c) but issues of sign in integration of the sine function and numerical errors in dealing with the multiple angle prevented many students from achieving the required result.

Question 8

Most students had some idea of how to tackle part (a) although many were let down by

getting the differentiation wrong to achieve usually by missing the factor. It

appeared from the solutions that many students had ‘differentiate’ rather than ‘differentiate with respect to’ in their heads and therefore the chain rule was not used. Quite a number of students tried to fudge the answer by just changing letters sometimes or adding in an extra

if one was lacking. Some students differentiated again and started with the second

equation, substituting to end up with the first one. As with question 6, presentation and handwriting were a real issue with here as it was difficult to distinguish between x and z and 2 (and even y sometimes) and many students overwrote mistakes. The organisation of some students’ work was poor and hard to follow which as this was a proof question made it difficult to mark.

The second part of the question was more successfully attempted with most students understanding the full method and getting the first 2 marks although some lost the 3rd by using x in their CF rather than z. There were quite a few students who used an expression of the form ax (not ax + b) for their particular integral and hence lost marks and others made arithmetic and sign errors in finding their constants. Although almost everyone was able to reverse the substitution for the final part, earlier mistakes cost them this final B mark.

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Further Pure Mathematics FP3 (R) (6669)

Introduction

This paper proved a good test of students’ knowledge and students’ understanding of FP3 material. There were plenty of easily accessible marks available for students who were competent in topics such as vector methods, integration, hyperbolic functions and differentiation. Therefore, a typical E grade student had enough opportunity to gain marks across the majority of questions. At the other end of the scale, there was sufficient material, particularly in later questions to stretch and challenge the most able students. Presentation was also good but there were some cases where students showed insufficient working, particularly in places where the answer was given in the question. Students should be aware that in these cases, sufficient working must be shown to fully justify the printed answer.

Report on Individual Questions

Question 1

The majority of students substituted the exponential forms of tanh x and sech x and went on to establish the correct quadratic in ex and solve it correctly to find the correct values of x.

A significant number of students multiplied through by cosh x and then substituted for exponentials and were equally successful. A minority of students opted to rearrange the given equation and square to obtain a quadratic in sinh x or tanh and then went on to introduce exponential functions and again were very successful.

Question 2

Part (a) almost all students could obtain the correct values for a, b and c although a small number struggled with the algebra.

Part (b) and part (c), students needed to identify the correct forms to be able to deal with the integration. Although the majority recognised the arctan form for part (b) and the arsinh form part (c), many students did not obtain the correct coefficients in one case and/or the other. Students are advised to consider carefully situations like these where the coefficient of x2 in the quadratic is something other than unity.

Question 3

Part (a) the majority of students used the chain rule successfully to establish the printed result. However, there were some cases where insufficient working was shown and some

students wrote followed by . Examiners would

expect some intermediate working to justify this given answer. Some students wrote the equation as and used implicit differentiation and were largely successful with this approach.

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For the arc length part (b), the majority could make a start by using the correct formula. Most then opted to use the identity although a surprising number of students used which would potentially lose them 3 marks. Those who reached an expression involving coth then often spotted the need for a natural logarithm

although those with coth 2x sometimes missed the .

Question 4

Students struggled to establish the reduction formula in part (a). Of those who made progress,

the majority took u as (3 – x2)n and as 1. The alternative of writing (3 – x2)n as

(3 – x2)n–1(3 – x2) and then multiplying out to obtain 2 integrals was seen occasionally. However, many students were unable to make a start.

Part (b) was accessible to students of all abilities and apart from a few slips, many students obtained the correct answer although some did not leave it in exact form.

Question 5

This coordinate geometry question proved to be a good source of marks for many students with part (d) discriminating at the top end.

Part (a) was a write down and students invariably gave the correct values for a and b.

Part (b) involved routine work to find the equation of a tangent. Although the work was often sound, a significant number of students did not show enough working to establish the printed

result. Some students correctly reached and then just wrote

down . Students are expected to show enough working to establish a given answer and this would be regarded as insufficient.

Part (c) was a straightforward demand to find the area of a triangle although a surprising

number of students missed off the .

In part (d), many students found the mid-point correctly although some got them the wrong way round. Most could then at least make a start establishing the cartesian equation of the mid-point but a significant number of students struggled to make y2 the subject and there were sometimes some basic algebraic misconceptions such as

.

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Question 6

Performance on this question was very variable. Those students familiar with this part of the specification usually obtained all 11 marks but a significant number of students could only manage the matrix algebra in part (b) and/or gained very few marks in part (a) and part (c).

Part (a) students usually wrote down a matrix of eigenvectors for P but not always of unit length. Often the matrix D was not attempted but those who did, usually knew it needed to have the eigenvalues on the leading diagonal and were usually consistent with their matrix P.

Part (b) was accessible to the majority of students and many showed clearly each step of their working to establish the given result.

Those students who had matrices for P and D could make further progress in part (c) and those with correct matrices often proceeded to obtain the correct matrix M.

Question 7

Part (a) nearly all students could establish the printed result for the surface area. Some students did, however, fail to show enough working. Students must be clear that a solution

along the lines of is not enough to score the 3 marks. It

was expected that students should at least quote the general formula for surface area and then substitute their derivative.

Part (b) the majority of students knew how to use the substitution but a few failed to replace the correctly and could make little progress. Those who did substitute correctly were

confused by the minus when correctly obtaining and sometimes crossed it

out in an attempt to establish the printed answer. The better students realised that the minus sign could be eliminated by swapping the limits.

Students were helped with the subsequent integration required and the majority could establish the result in part (c) by using the correct hyperbolic identity. Correct answers to part (d) were surprisingly rare and students often forgot to reintroduce π or omitted the 2 in the

. Students also sometimes just wrote down an incorrect answer and did not give themselves opportunity to score the method mark for substitution of limits.

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Question 8

Part (a), students used a variety of methods to find an equation for the line of intersection of the two planes. Perhaps the most common approach was to attempt the cartesian equation of the line by expressing one of x or y or z in terms of the other variables from the Cartesian equations of the two planes. This was met with varying degrees of success but largely the method was sound with some occasional algebraic slips. From the Cartesian equation, most could identify the position and direction of the line correctly. A significant number of students correctly used a vector product to find the direction of the line.

Part (b) students with a vector equation from part (a) could proceed correctly and substituted in to the third plane to identify the intersection of the 3 planes. Some students chose to solve 3 simultaneous equations to find the required point and were, in many cases, successful.

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Mechanics M1 (R) (6677)

Introduction

The vast majority of students seemed to find the paper to be of a suitable length, with no evidence of students running out of time. Students found some aspects of the paper challenging, in particular questions 2(a), 6(c) and 7(c). However, there were some parts of all questions which were accessible to the majority. The questions on equilibrium and v – t graphs were generally well understood and full marks for these questions were commonly seen. Generally, students who used large and clearly labelled diagrams and who employed clear, systematic and concise methods were the most successful.

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, including fractions.

If there is a printed answer to show then students need to ensure that they show sufficient detail in their working to warrant being awarded all of the marks available.

In all cases, as stated on the front of the question paper, students should show sufficient working to make their methods clear to the Examiner.

Report on Individual Questions

Question 1

This question proved to be an easy starter for the vast majority of students. Most chose to resolve parallel and perpendicular to the slope and achieved the correct answers. Because g was not involved, there was no upper limit to the accuracy of the answers but we were expecting at least 2 significant figures.

Question 2

Part (a), most scored the first two marks for adding the two vectors to find the resultant but then many just equated their resultant to 2i + j and obtained q = 3 instead of using a ratio. Some subtracted the vectors to get the resultant rather than added. In the second part, some found a scalar quantity for the acceleration and then continued just using scalars. A sizeable minority obtained a velocity vector but then forgot to use Pythagoras to find the speed.

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Question 3

Part (a) most v – t graphs were well constructed although the second horizontal line was sometimes missed leading to a number of lost marks in part (c) and part (d). The second part was answered well by the vast majority and in part (c) there were a few sign errors but the times were generally found successfully. In the final part occasionally halves were missed off areas of triangles calculations but generally fewer errors were made by using rectangles and triangles rather than trapezia to find the required total area.

Question 4

Part (a) was generally well done and most were able to eventually find the given answer in the second part. However, the final part was much more challenging and confusion between t and T caused many of the problems.

Question 5

Part (a) was mostly well done although some omitted the m from the ‘ma’ term in their equations of motion which nonetheless still led to the given answer for T and examiners had to be alert to ensure that unwarranted marks weren’t awarded.

The second part was only really successfully done by stronger students as again the algebra made it less accessible. Many used g not 0.2g and added 1.5 m instead of doubling their distance. Part (c), m was used in the impulse-momentum equation instead of 3m in the majority of cases.

Question 6

The first part was answered very well by the vast majority but in part (b) most marks lost were due to students assuming that TC = 2TB from part (a). The final part was poorly done by most and a lot of students used an = sign in their working and substituted the inequality at the end which lost marks.

Question 7

Part (a) there were a few errors with angles and some sin/cos confusion but generally this was well done. Most marks lost were due to over-accuracy of the answer after use of g = 9.8.

There were some sign errors in the second part but most students were able to make a good attempt at finding a value for . Part (c) most were able to find the component of the weight down the plane but many lost the rest of marks by using the original value of the friction. Also a good number of answer marks were lost due to the use of a rounded value of μ which gave a value of 4.66 N instead of 4.70 N for the limiting friction force.

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Mechanics M2 (R) (6678)

Introduction

Most questions on this paper proved to be accessible to all candidates, and a lot of good quality work was seen. The best work was clearly set out, with clear methods and accompanied by fully labelled diagrams. All candidates were able to demonstrate their skills in a number of familiar questions, but more demanding aspects such as qu.3 (the rod in limiting equilibrium), qu.4(b) (the suspended lamina), qu.6(c) (perpendicular motion) and qu.7 (b) (the range of values for e) presented a challenge for those at the top end.

Candidates need to be reminded to read the rubric and the questions carefully. In all cases, where a value for g is substituted, the value should be 9.8 m s–2. The use of 9.81 will be penalised as an accuracy error. The rubric on the paper gives candidates a very clear reminder about the accuracy expected after the use of 9.8, but many candidates lost marks for giving too many significant figures in their final answers. If the question asks for the magnitude of a quantity, then a positive answer will be expected.

Report on Individual Questions

Question 1

Most candidates were well prepared for a question on power, and gave confident answers. The most common error was an over-specified answer at the end of part (a). Many candidates left their answer as 882.5, which is inappropriate after the use of 9.8 as an approximate value for g.

Question 2

Part (a) Most candidates started with a correct impulse-momentum equation to find the velocity of the ball before the impact, but many did not go no to find the corresponding speed.

Part (b) Candidates with a clear diagram usually found the correct angle, and it was pleasing to see a number of candidates with enough knowledge of vectors to use the scalar product (although this method was not expected). Some candidates found the angle between a velocity and the impulse, and some found the angle between a velocity and a fixed direction (usually the unit vector i).

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Question 3

Some good responses to this question were seen, but many candidates had difficulty in working with a non-uniform rod. It was common to see attempts to use moments in part (a), which only required resolving. Candidates needed to name the forces for themselves, and this led to some confusion, often involving the same name being used for the friction between the rod and the ground, and for the friction between the rod and the wall. A significant minority of candidates had the friction at B preventing the rod from slipping up the wall.

Those candidates who formed a correct moments equation in part (b) usually went on to find the correct value for the distance AG. Although they were looking for a distance from A, several candidates chose to find this by taking moments about B.

Question 4

Part (a) Most candidates gave a correct answer to this part of the question. Errors were usually due to having the area of the triangle incorrect, or to adding the square to the triangle rather than removing it from the triangle.

Part (b) Many candidates made a correct start to this part of the question, and they often got as far as finding the distance of O from an axis through A or through B, but they found it more complicated to find the distance from DC. Some candidates did not recognise that they needed to work back from the 25° at some stage in their method.

Question 5

Many candidates demonstrated a good understanding of the work-energy principle and gave confident solutions to this question. Although only one part of the question specifically asked for use of the work-energy principle, it was common to see candidates working the whole question by this method rather than resorting to the equation of motion of P in the latter parts of the question. Although there were candidates “double counting” by considering both the change in gravitational potential energy of P and the work done against the weight of P, this happened less often than in previous examinations. It was disappointing to find a number of candidates working through the method correctly but then losing accuracy marks through the use of 9.81 as an approximate value for g or giving over-specified final answers. The rubric on the paper is very clear about what is expected, and candidates need to take note of this.

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Question 6

Part (a) Almost all candidates found the correct answer for the vertical distance from A to the maximum height, but several did not go on to find the greatest height above the ground.

Part (b) Many candidates understood the method required to find the distance OB, but there

were often sign slips in using , resulting in an incorrect quadratic equation in t.

Candidates should be aware that if they start with an incorrect quadratic in t, and simply state answers without demonstrating any method for solving the equation then they will not gain any credit for their solution.

Part (c) The final part of this question resulted in many incorrect assumptions. There were a good number of concise and correct solutions, usually using similar triangles or differentiation of the equation for the trajectory. However, many candidates incorrectly assumed that reversing the vertical component of the velocity would give a perpendicular direction of motion.

Question 7

Part (a) Many candidates earned full marks for correct work in this part of the question, although there were some slips in forming the equations and some inconsistent use of signs, usually from candidates with poor diagrams or no diagram at all.

Part (b) Although there have been similar questions to this in the past, some candidates struggled to find a way of introducing an inequality in e. Most of the marks in this part of the question depend on using the information about the direction of motion of P after the collision. The initial inequality does need to match the candidate's velocity for P, whether or not they have already taken account of the change in direction. For the final mark they also need to state the maximum value for e.

Part (c) The work on kinetic energy was usually very good, and often resulted in a correct answer for k.

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Mechanics M3 (R) (6679)

Introduction

Students seemed to find this paper difficult. Several questions were designed to encourage students to think about the problem in hand rather than to follow patterns learned from working through past papers. This is what Mathematics should be about but many students find this beyond them.

While many good students write beautifully presented, well explained answers, the standard of presentation of others is appalling. Handwriting is often so bad that they misread their own figures, mistakes are often corrected by scrawling illegibly over the original and the impression is given that the only thing which needs to be clear is the final answer. Even if they work like this at school, it is almost unbelievable that they will risk examiners not being able to follow their working in a potentially life-defining examination.

Those who work entirely in formulae until the final line of a calculation should be reminded how risky this is; if something goes wrong they could leave very little which is worth any marks. Values need to be substituted throughout the working. Also, surds are generally acceptable in any form. It is not necessary to waste time realising denominators unless they have to reach a given answer or give their answer in a pre-set form.

Report on Individual Questions

Question 1

Being given an expression for caused problems for many students who did not know how

to start part (a). Some of those who integrated forgot to include a constant of integration and often little progress was made as students did not realise that they could find a by using

and then obtain the magnitude of the force by using F = ma.

Part (b) required the integration of v with respect to x and those who had not integrated in part (a) were unable to start here too. There was a mixture of definite and indefinite integration and those who had found a correct expression for v in part (a) were usually successful here.

Question 2

This was a “conical pendulum” question and, as such, should have been predicted by most students. However, the inclusion of a normal reaction from the surface of the cone as a second non-gravitational force acting on the particle caused many students problems. They were accustomed to dealing with two tensions but this was different. The angles they needed to use when resolving were found by some simple geometry but many seemed to be confused.

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Part (b) brought another set of problems. Many seemed to think that the tension in the string had to be greater than (or equal to) zero for the particle to remain in contact with the cone. This was a necessary condition of the motion regardless of any contact with the cone. Those who were unsuccessful in part (a) were unlikely to obtain a correct expression for the reaction. It was disappointing to see so many students deal badly with inequalities and many simply wrote an answer using the one given in the question even though it clearly was not obtained from the working shown.

Question 3

Many students did not recognise part (a) as a work-energy problem and tried to find an acceleration which they could use, ignoring the fact that the acceleration was in fact variable; others mixed forces and energy terms in their equation. This is a "show that" question and so students must show every step of their working. Examiners can only read what the candidate

writes on the page, not the candidate's mind. Thus the equation

is not a suitable starting point as it will be interpreted as a forces equation (with an incorrect Hooke's Law) and not as a work-energy equation with a distance cancelled. The candidate may have intended the later but the work does not show that.

Part (b), as the particle has already slid down the plane and come to rest, any further motion will be up the plane. Students therefore needed to show that the tension in the string at the point where the particle stopped was less than or equal to the sum of the component of the weight down the plane and the maximum possible friction force. Some failed to realise that finding the tension was a simple application of Hooke's Law. Once again the answer was in the question and almost always appeared at the end of students' work regardless of whether it could be obtained from that work.

Question 4

Part (a) was a standard vertical circle question and most students could obtain valid energy and Newton's law equations. Setting the reaction equal to zero and solving usually gave the required expression for V. Strangely, although V was the required quantity here, many solved their equations for and then used their value to obtain V.

Part (b) considered the motion of the particle after it left the circular path and was moving as a projectile. As always with this type of question many students could not correctly identify the direction of motion of the particle at the instant when it left the surface of the sphere and as a consequence the following work contained a sine/cosine interchange. Students frequently failed to realise that use of the horizontal distance travelled to the wall would give them an expression for the time of flight. Those who obtained the time of flight could usually use it to obtain the vertical distance travelled. It was rare to see a candidate who managed this forget

to subtract this from to obtain the distance AX.

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Question 5

The majority of students found part (a) straightforward and managed to legitimately obtain the distance as given.

Part (b) was found much more taxing. The cut-out cylinder was offset and consequently the distance of the centre of mass of S from the axis of the original cylinder was needed before any work involving the given angle could be done. As many students omitted to find this distance they scored zero for this part of the question. Those who found this distance correctly rarely encountered any problem finding r in terms of h although occasionally the tangent ratio was used upside down.

Question 6

Most students could use Hooke's Law to obtain the equilibrium extension required in part (a).

The proof of the simple harmonic motion required in part (b) was, as usual, found difficult. Many students persist in using α for the acceleration instead of and so restrict themselves to a maximum of 3 of the available 5 marks, while others fail to give a concluding statement and so lose the last mark even if their work is fully correct. Those who achieved an equation from which ω could be deduced were usually able to obtain the amplitude as a multiple of l in part (c).

In part (d) it was not uncommon for those who had a correct ω and amplitude to be confused about the distance and trigonometric function to be used to obtain the time. Perhaps thinking about the motion in relation to the sine curve would have helped. Answers were almost always given using radians. If a numerical value was substituted for g the answer was usually given to 2 or 3 significant figures. However, students should be aware that without a numerical value for l there is little point in substituting for g (or π) and an exact answer is preferable.

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Mechanics M4 (R) (6680)

Introduction

The students demonstrated a good understanding of all sections of the specification. They found this paper accessible, with the majority of them offering responses to all parts of all questions.

Much of the work was clearly set out and of a high standard. In many instances, taking the time to draw a clear diagram was the key to a successful outcome. It is apparent that some students are using their calculators for basic processes such as solving equations. They need to be aware of the risk that they take by showing no working – one small slip in deriving the equation can cause them to lose all subsequent marks if they have not demonstrated a clear method in their working.

Report on Individual Questions

Question 1

All students understood that they needed to start by considering the motion parallel and perpendicular to the plane. In forming the equation for the impulse, some solutions did not take account of the change in direction of the motion perpendicular to the plane due to the impact. Using the initial equations to form an expression for I in terms of m, u and e proved to be quite challenging, with most students making some progress but only a few reaching the correct conclusion.

Question 2

There were several fully correct solutions to this question. The majority of errors were due to premature rounding in the course of the work, or slips, but a few students missed the essential starting point of a correct vector triangle with v perpendicular to the relative velocity in part (a), and did not have a correct vector triangle in part (b).

Question 3

Although they were given no guidance on how to start, the students showed a good understanding of the topic by forming the correct differential equation in v and x and attempting to solve it. Although the integration is relatively straightforward, the direct approach did require the student to recognise that they were starting with a top-heavy fraction in v, which needed to be split before integrating. This was where the majority of errors occurred.

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Question 4

Those students who started with a clear diagram showing the velocities of both spheres before and after the collision worked through part (a) with few problems other than algebraic slips. They all understood that if was at rest before the collision then it would move along the line of centres.

Part (b) required the students to recognise that the largest value of occurs when θ has the largest possible value. The majority of students made little progress with this part.

Question 5

A small number of students tackled part (b) of this question first, and then went back to work on part (a). In fact part (a) was relatively straightforward, and most students worked through it correctly.

In part (b) the method was well understood, but there were some slips in the algebra and the arithmetic, and at the very end some students gave no evidence in support of their conclusion.

It is not sufficient to state without showing an expression for the derivative which

takes account of the constant factor 4mgl, and clearly satisfies .

Question 6

In part (a) most students were able to use the information given to derive the required differential equation in x and t.

In part (b) apart from a small number of slips in the algebra and arithmetic, the students demonstrated a good understanding of how to solve the differential equation, and most reached a correct conclusion.

In part (c) with only a small number of exceptions, the students understood that the greatest value of occurs when and gave a correct solution, often in exact form.

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Mechanics M5 (R) (6681)

Introduction

All of the students seemed to find the paper to be of a suitable length, with no evidence of students running out of time. The majority of the students were very well prepared.

The paper discriminated well at all levels including the top end where there were some impressive, fully correct solutions seen to all questions. Generally, students who used large and clearly labelled diagrams and who employed clear, systematic and concise methods were the most successful.

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, including fractions.

If there is a printed answer to show then students need to ensure that they show sufficient detail in their working to warrant being awarded all of the marks available.

In all cases, as stated on the front of the question paper, students should show sufficient working to make their methods clear to the Examiner.

If a student runs out of space in which to give their answer than they are advised to use a supplementary sheet – if extra paper is unavailable then it is crucial for the student to say whereabouts in the script the extra working is going to be done.

Report on Individual Questions

Question 1

The most popular way to approach this question was using a scalar product and the work-energy principle and it proved to be an easy starter for the majority. Other approaches were unsuccessful.

Question 2

Most candidates applied the conservation of energy principle correctly and were then able to convert the angular speed into a linear speed. There were some problems with the distances involved in the energy equation.

Question 3

This question was successfully answered by the majority of students, with only a few demonstrating that they knew little about the concept of conservation of angular momentum and/or how to apply Newton’s Law of Restitution to such a scenario.

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Question 4

This question was answered completely correctly by the majority of the students. However, a few were unable to make much progress and struggled to obtain a correct expression for the moment of inertia of an elemental disc.

Question 5

This question proved to be a good source of marks for the majority of students. The integrating factor method was well known and correctly applied to find the general solution of the differential equation. The initial conditions were then used to find the required particular solution. Sometimes the trigonometry involved defeated a minority of the students.

Question 6

This question was extremely well answered by all of students. The first two marks in part (a) were scored by all. In the second part, the methods were well known but there were a few errors made in evaluating the vector products required and this led to a final incorrect answer for G.

Question 7

The printed answer given part (a) was easily obtained by most students but a minority did not really know where to start. In the second part, those who were unsuccessful in part (a) were able to go on and obtain credit for rearranging and solving the given differential equation. However, the integral involved defeated some candidates.

Question 8

This was a challenging question and provided discrimination at the top end. The first part was answered very well by some but for others, it proved to be a very challenging. For those who were unsuccessful in part (a), the second part provided a potential opportunity to go on and obtain some marks but this was not always taken advantage of.

The same also applied to part (c) and the final part was answered correctly by only the most able students.

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Statistics S1 (R) (6683)

Introduction

Apart from a very small group of students who did not seem to have been prepared for this paper the standard of the entry was high with many scoring full marks or nearly full marks on most questions. Some of the questions requiring explanations or interpretation, such as qu.3 and qu.5, were answered less well but the standard calculations and manipulation of probabilities were generally handled very confidently.

Report on individual questions

Question 1

This proved to be a straightforward start to the paper. Most students could clearly show that p = 0.2 and the majority went on to find E(X) correctly. Many students though did not know what to do for F(0) and responses to part (c) were poor.

Part (d) had a mixed response, some could simplify the inequality to reach P(X > 1) but were then unable to link this with the correct probabilities from the table but plenty of students were able to secure full marks here. Most students knew the formula for Var(aX + 3) but some failed to realise that a = +2 and just gave the answer 2.

Question 2

Some students could not recall the name of the distribution for X with several students calling it the discrete distribution rather than the discrete uniform distribution. The probabilities in part (b) were usually tackled well with only the very weak students giving answers of 1 and 9 rather than 0.1 and 0.9.

Part (c)(i) caused problems for a few and some stumbled upon the answer of zero after much dubious calculation but part (ii) was answered more successfully.

Question 3

Part (a) was, as usual, answered very well by almost all the students but some missed the “show that” part of the demand and only scored 4 of the 5 marks here.

Parts (b) and (c) were answered well too and most knew how to tackle part (d) but some forgot to multiply their answer by 100 and gave an answer of £37 rather than £ 3700.

A similar problem arose in part (e) with few explaining that the cost increases by £82.40 for each additional employee.

Most knew the answer to part (f)(i) but part (ii) was less routine and only the more astute students realised that the value of b would be halved.

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Question 4

This question was answered very well with over 55% of students scoring full marks.

Part (a) could be answered quite readily by using the formula in the formula booklet and the given values for P(B) and and then it should have been an easy step to complete the Venn diagram in part (b). However some forgot to subtract when completing their diagram but the follow through marks meant that some credit could still be given for the later parts.

Part (c) and part (d) were usually answered well but in part (e) some tried to use a formula (which was often incorrect) rather than their Venn diagram and made the question much more complicated.

Question 5

Most students secured the mark in part (a) although a few gave the boundaries for the whole interval.

Part (b) caused problems. Many students are not confident with the calculations for the height of bars on a histogram and in this case some could not deduce which bar was the second tallest.

Part (c) some found the correct fraction of the upper or lower group but often they failed to do this for both groups.

Most knew the correct technique to tackle part (d) but errors often occurred with the lower class boundaries or when rounding.

Part (e) was answered well with most using their calculated quartiles to justify their answer although a few calculated the mean and compared this to the median.

Question 6

A number of students failed to label their diagram correctly in part (a) and a few gave an answer of 25% for part (b) but there was plenty of fully correct answers here.

There were many correct answers to part (c) too but although many used the value of 1.6449 for the equation based on 28, they only used 0.84 (rather than 0.8416) for the other equation.

Part (d) was a relatively straightforward application of the normal distribution and even those with incorrect answers from part (c) could secure the method marks here.

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Question 7

Most students scored all the marks for drawing the tree diagram and could then use this to answer part (b).

In part (c) the conditional probability discriminated more with some failing to obtain the correct denominator.

Part (d) caused difficulties for a number of students: some did not realise they needed a probability of the form p(1 – p) where p was their denominator from part (c) and those who did secure this first mark often failed to multiply by 2 to account for both possible cases.

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Statistics S2 (R) (6684)

Introduction

On the whole this paper was well answered. Most questions on the paper seemed accessible to the candidates however time seemed to be a problem as question 7 was often not completed.

Generally the work was quite well presented. Often candidates knew the statistics needed to answer the questions but were let down by poor algebraic skills.

Report on Individual Questions

Question 1

This question was done well with the majority of candidates gaining at least 4 marks. A minority of candidates failed to give a correct contextual interpretation despite rejecting H0. Occasionally hypotheses were incorrectly written and a few candidates attempted to use the Poisson distribution.

Question 2

Part (a) was well done with a number of candidates justifying in some detail each answer they gave. A few candidates tried to ‘hedge their bets’ by writing ‘yes yes yes’ and ‘no no no’.

The majority of candidates answered part (b) correctly however part (c) proved to be more difficult. The most common error was that candidates thought that the first and third counter had the number 2 on rather than just the 3rd. The majority of candidates who interpreted the question correctly and then went on to gain the right answer found P . Those who tried listing the four probabilities usually only listed three.

Question 3

Parts (a) and (b) were well answered although in part (a) a few candidates wrote down Poisson but forgot to give the value of .

Since part (c) is a ‘show that’ question candidates were required to write down all the steps needed to reach 0.777. Many candidates went straight from 1 – P(Y = 0) to 0.7769 To gain full marks they were required to write down the figures between these two stages (1 – 0.2231).

Most candidates identified the Binomial with n = 6 and p = 0.777 in part (d) but few could progress any further. Many who did try either omitted 6C4 or used (0.777)2 (1 – 0.777)4.

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Question 4

Many candidates gained full marks in part (a). The most frequent error was to draw a straight line from (1, 3k) to (0, 0) rather than the horizontal line y = 3k. Others did not draw a curve but a series of short straight lines.

Part (b) proved to be quite challenging with few candidates getting the mode = 2. Many candidates spent some time trying to work the answer out algebraically with different degrees of success. The question asked them to “Write down” which indicates that the answer should be able to be gained without any working. In this case the diagram in part (a) shows the mode.

The majority of candidates were able to use the mean and their mode to comment on the skewness in part (c).

In part (d) the most common error was to forget about the area for 0 ≤ x < 1 or forget to equate the area to 1. Integration when attempted was often done correctly.

Part(e) was well answered although a few candidates did copious calculations rather than realise the area for 0 ≤ x < 1 was 0.25.

Few candidates were able to make much progress with part (e). A minority of candidates recognised that P(1 < X < 2) was the same as P(2 < X < 3) and used this fact to gain the answer. The majority used integration to work out the unknown areas.

Question 5

This question was generally well answered by the majority of candidates. The most common errors were to forget to write down the hypotheses in part (a) and in part (c) the continuity correction was either not used or was incorrect.

Question 6

This question proved quite challenging to many candidates. In part (a) candidates managed to get the correct equation but were unable to solve it. Of those who were able to solve it successfully the most common method used was to substitute x = d2 and thus produced a quadratic in x2.

In part (b) many candidates found f(d) but then went on to equate this to 0 rather than using . The majority of candidates either forgot to justify their answer was the mode or they

did not know that to justify their answer was the mode they needed to prove it was a maximum point.

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Question 7

Parts(a), (b), (c) and (e) were well answered with the majority of candidates gaining full marks. The most common error was to forget to give the range of values although a uniform distribution was identified.

Part (d) proved to be more challenging with the majority of candidates making little progress. Those that did make an attempt managed to find the correct value for E(X) but only a very small number found Var(X) correctly. Few candidates recognised that they then needed to use the formula E(X2) = Var(X) + [E(X)]2.

The most common error was to use E(R) = .

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Statistics S3 (R) (6691)

Introduction

On the whole this paper was very accessible with the required methods well known by the majority of the students. There were some issues with accuracy in the numerical parts of the questions, but the interpretation in context saw some good responses.

Report on Individual Questions

Question 1

This was a productive start to the paper for the majority of students with completely correct responses seen often in part (a).

The hypothesis test in part (b) usually had the correct critical value, but common errors were the hypotheses being badly formed in words. Weaker responses usually missed the final conclusion in context.

Part (c) was known to more students than might have been expected and if the correct terms were not known in part (d), a justification by describing skewness was not uncommon.

Question 2

Full marks were often awarded in part (a) as the methods were well understood and were calculated accurately. Some confusion was evident in the formulation of the hypotheses in part (b), but again full marks here was not unusual.

Question 3

Fully correct solutions were not uncommon in this question and the methods were clearly well understood. Occasionally, the calculation of the variance in part (a) caused problems with squares being missed or sign changes not applied. Some weaker students were easily confused when trying to find the probabilities in each part, with errors being made with the standardisation and the incorrect inequality being used.

Question 4

This question proved to be a challenge to those students who were reluctant to set out their working clearly. These solutions usually did not gain any credit beyond the first two marks for finding the expression for the sample mean. The use of an incorrect critical z value and l

or the use of 6 rather than on the denominator was also seen, but a final correct value

was found by many students.

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Question 5

This question proved to be very well answered and it was pleasing to see so many fully correct solutions to part (a) and part (b), including conclusions in sufficient context to be awarded the marks available.

Part (c) and part (d), however, often lacked detail with values being calculated accurately in part (d) with no mention of a list being required. Students often stated ‘at random’ rather than describing how this might be achieved.

Question 6

The responses to part (a) were poor and students did not realise that specific assumptions were required and simply listed all the requirements for a Binomial distribution. Despite this weak start to the question, students typically gained most of the marks on the next two parts.

Most students were awarded the marks for part (b) and combining groups accurately was not unusual.

Occasional errors were made with the degrees of freedom and some solutions offered lacked context in the conclusion to part (c).

Part (d) proved more challenging with confused or inaccurate reasons being offered.

The majority of students finished the question well with most marks being given in part (e) with only the occasional conclusion missing sufficient context for the final accuracy mark.

Question 7

This was the most challenging question on the paper, but typically students produced good responses. The confidence interval methods were well understood in both parts with only occasional accuracy errors being seen.

Occasionally, the wrong z value was chosen in part (b), but the value 11 was seen very often in the final answer.

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Statistics S4 (R) (6686)

Introduction

Candidates found this paper very accessible and scored well. They were able to make a reasonable attempt at the majority of questions with some excellent scripts submitted.

Report on individual questions

Question 1

The question proved to be a friendly starter for the majority, with many candidates gaining full marks. The most common errors were made in part (a) where they omitted to give the hypotheses and, although they knew something was normally distributed, few said it was the differences that were normally distributed.

Question 2

Part (a) was generally well answered but part (b) proved to be quite challenging for candidates. The main errors were to work out the probability of a type II error rather than work out the power and not realising the probability of accepting a sample changes for the second sample.

Question 3

Part (a) was answered well with many candidates gaining full marks. In part (b) although a pooled estimate of variance was worked out correctly by many candidates they then failed to use the square root of it in their calculations of t.

In part (c) candidates knew that to carry out the test in part (b) the variances needed to be equal but few commented on the fact that this has been established in part (a).

Question 4

The first couple of parts of this question were generally well answered.

Part (c) proved to be more demanding with only a minority of candidates managing to find the probability of a type II error successfully. Of those who were unsuccessful it was because they did not realise they needed to find P( .

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Question 5

The majority of candidates gained full marks for part (a)(i) and part (a)(ii).

In part (b) many candidates did not realise that they needed to use the highest value from part (a)(i) and the lowest value from part (a)(ii).

Question 6

The first 4 parts of this question were well answered but the majority of candidates found parts (e) and (g) quite demanding. In part (e) candidates knew what was required but few were able to clearly establish that Var(T) < Var(S). In part (g) few candidates realised they needed to use their answer to part (f) and substitute it into the formula for Var(T) found in part (d).

51

Decision Mathematics D1 (R) (6689)

Introduction

The paper proved accessible to the majority of candidates and there was little evidence of there not being enough time to complete the paper. The questions differentiated well, with most giving rise to a good spread of marks. All questions contained marks available to the E grade candidate and there also seemed to be sufficient material to challenge the A grade candidate also.

Candidates are advised to make their method clear; ‘spotting’ the correct answer, with no working, rarely gains any credit. Candidates are further reminded that they should not use methods of presentation that depend on colour or highlighters, but are advised to complete diagrams in (dark) pencil.

Candidates should ensure that they use technical language correctly. This was a particular problem in questions 4 and 6.

Report on individual questions

Question 1

This question provided an excellent start to the paper for the vast majority of candidates with 46.4% scoring full marks and only 23.8% scoring 8 marks or fewer. Candidates were clearly very well prepared for this topic and were familiar with all the components required for a complete solution.

Parts (a) and (c) were well attempted with the majority of candidates gaining full marks. From time to time the value of 19 was misplaced in part (a) and occasionally the values of 12 and 10 were interchanged in part (c).

Many correct solutions were seen in part (b), but a minority produced an ascending list and failed to reverse it, leading to marks being lost later in the question. A number of candidates did not choose their pivots consistently, switching between middle-left and middle-right pivots during the course of the quick sort algorithm. A very small number of candidates lost an item or changed one, and very few cases were seen where only one pivot was chosen per iteration. Some candidates did not indicate that their sort was complete. This could have been achieved either by having at the end a ‘list sorted’ statement, or every item in the original list being used as a pivot or the final list being rewritten at the end. A common error was the 10 and 12 being interchanged in the 1st pass; candidates should be reminded that items should remain in the order from the previous pass as they move into sub-lists.

The most common explanation provided in part (d) was to consider a lower bound calculation which many candidates did correctly. A very small minority argued on the basis that there were five items which exceeded half the size of a bin although some of these arguments were not quite precise enough to gain both marks. Some candidates failed to relate their argument or calculation back to part (c) and lost a mark as a result.

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Question 2

This question proved to be a good source of marks for many candidates with the mode being 6 marks, 23.3% gained full marks and only 20.9% scored 3 marks or fewer.

Part (a) represented the usual challenge with this type of question. Part (a) was worth two marks and as a result many candidates were able to obtained at least one mark. The vast majority provided an explanation about Hugo and tasks 1 and 3. It was common to see “Only Hugo can do 1 and 3” although it was interesting to note that it was fairly common for candidates to cover every eventuality and write “only Hugo can do 1 and 3” followed by “1 and 3 can only be done by Hugo”. Far less common but equally valid were the longer, more elaborate arguments involving more than one employee, for example, an argument based on employees A, C, J and P and tasks 2, 4 and 5. A minority of candidates did not realise what was required in this part and argued along the lines of “because A can only do 2” or similar.

In part (b) the overwhelming majority correctly selected Janelle for training. Although most struggled to provide a fully acceptable reason or indeed in some cases any reason at all. Often candidates argued that Janelle should be chosen because 5 can already be done by 2 employees.

Part (c) was answered extremely well but there was nonetheless the usual loss of marks for some candidates due to lack of change of status being stated or shown and/or failing to state the complete matching – in some cases candidates may have drawn the complete matching on a diagram which were not clear due to multiple lines being drawn from individual vertices. Change of status errors and lack of complete matching errors seem to be occurring less each session but are unfortunately still evident.

Question 3

This question also proved to be a good source of marks for many candidates with the mode being 8 marks, 15.8% gained full marks and only 23.9% scored 5 marks or fewer.

Part (a) was usually very well done with most candidates applying Dijkstra’s algorithm correctly. The boxes at each node in part (a) were usually completed correctly. When errors were made it was either an order of labelling error (some candidates repeated the same labelling at two different nodes) or working values were either missing, not in the correct order or simply incorrect (usually these errors occurred at D, G and/or T). The route was usually given correctly and most candidates realised that whatever their final value was at T this was therefore the value that they should give for their route.

Part (b) was also well attempted with many stating the correct path from S to T via E and the correct corresponding time. However, many candidates neglected to correctly write down the effect on the journey. Most stated simply that “the time taken increases” without quantifying the change. As the quickest route from S to T from part (a) (found using a shortest path algorithm) did not include E it shouldn’t have been a huge surprise that the time taken had increased in the second part.

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Question 4

This question discriminated well leading to a good spread of marks. The modal mark was 7, only 3.4% scored full marks and 70.5% scored 7 or more marks.

Part (a) on the application of the Route Inspection algorithm was generally done extremely well by nearly all candidates. Unfortunately though there were a few candidates who only gave two pairings of the four odd nodes or who gave several pairings but not three distinct pairings. However, most candidates stated the correct three distinct pairings of the correct four odd nodes. It was relatively common though to see errors in some of the totals as candidates did not always find the shortest route between their pairings. Most went on to state the correct length of a shortest inspection route.

In part (b) most candidates were able to correctly calculate the time taken for the inspection although some candidates simply wrote down 120/15 = 8.

Part (c) was challenging for most candidates and it was rare to see both marks awarded. The majority of candidates obtained no marks here due to arguments along the lines of “because that is the method that gives the shortest route” or fairly long-winded arguments to do with the order of nodes and the number of times a node can be entered/left without ever reaching the crux of the correct explanation. Of the two marks available a small minority scored a mark for conveying at least the idea of finishing at an odd vertex. However, it must be noted that some candidates wrote down a perfect explanation seemingly with ease.

Part (d) also proved to be fairly discriminating for candidates. Some candidates felt they needed to avoid B and D altogether and so repeat FI, others felt they should repeat BI but not because it was the least but rather because it would avoid repeating DF which was the largest pairing. There were some very thorough solutions, however, who considered the implications of starting at B and D in turn although some failed to explicitly identify that BI was repeated because it was the least pairing.

Those who correctly decided to repeat BI were usually able to write down the correct new length but there were often errors in calculating the new minimum total time; many candidates obtained an answer of 3272 – including a 2 second pause at the end of the test.

Question 5

This question also discriminated well leading to a good spread of marks. The modal mark was 9 scored by 28.0%, 4.7% of candidates scored full marks and 34.1% scored 5 marks or fewer.

Most candidates were able to draw the required lines correctly in part (a) although some were unable to draw lines sufficiently accurately (some drew lines without a ruler) or sufficiently long enough. As mentioned in previous examiners’ reports the following general principle should always be adopted by candidates:

lines should always be drawn which cover the entire graph paper supplied in the answer book and therefore,

lines with negative gradient should always be drawn from axis to axis.

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The rationale behind this is that until all the lines are drawn (and shaded accordingly) it is unclear which lines (or parts of lines) will define the boundary of the feasible region. If candidates only draw the line segments that they believe define the boundary of the feasible region then examiners are unaware of the order in which the lines were drawn and therefore it is unclear to examiners why some parts of the lines have been omitted. In general the lines 7x + 8y = 840, x = 25 and y = 25 were correctly drawn and were errors occurred they tended to be with the other two lines. Furthermore, a significant number of candidates were unable to select the correct feasible region.

In part (b) most candidates were able to draw a correct objective line, occasionally a line of reciprocal gradient was seen although this was fairly rare. It is worth noting that some candidates do not make their objective line clear and it when it is drawn deep into the feasible region it can sometimes be difficult to identify. Candidates should be reminded to ensure that their objective line is labelled or distinct from their constraint lines. The majority of candidates successfully labelled the optimal vertex.

In part (c) some candidates did not demonstrate any working to find the exact coordinates of vertex V. Candidates are possibly relying on calculators to do this for them and these candidates need to be reminded of the advice to candidates on the front cover of the question paper that ‘answers without working may not gain full credit’.

Candidates found part (d) to be quite discriminating. The majority of candidates did not test the integer points around their optimum vertex with the correct inequalities and often those who did attempt this testing did not demonstrate that sufficient testing had been undertaken. This is clearly an area of weakness for many candidates.

Question 6

This question proved to be a good source of marks for nearly all candidates. The mode was full marks, gained by 20.9% of the candidates, only 23.8% scored 3 marks or fewer.

Errors where they occurred were often due to lack of arrows or labels on arcs and sometimes extra activities or extra dummies appeared, for example, from the end of E to the sink node. Candidates sometimes found it difficult to have one finish and sometimes added extra activities after A and J in order to finish at one node. Only a small proportion of candidates attempted activity on node diagrams. The two required dummies were most often dealt with correctly.

Part (b) was met with varied success. There were the usual reasons for loss of marks including failing to include all relevant activities in the dependence argument stating, for example, ‘that D and G both depend on A but D depends on something else’. Candidates were also often too vague with regard to ‘uniqueness’ for the ‘uniqueness’ dummy (candidates are reminded that all activities are unique and that mention must be made to the fact that activities cannot share the same start and end events) and a significant number tried to explain in terms of dependence on D and F.

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Question 7

This question gave rise to a good spread of marks and proved a good discriminator. The mode was full marks gained by 28.6% of candidates, and 73.1% scored 8 or more marks.

The early and late event times were successfully completed by the vast majority of candidates. Errors where they arose included at the late event time at the end of C, the late event time at the end of A and/or at the end of G.

The float calculation is clearly well understood by the majority of candidates and very many got at least one mark in part (b). The lower bound in part (c) had more variable success; some did not do a calculation and tried to argue for a lower bound based on scheduling despite the question asking for a calculation. Others made either basic arithmetical errors or conceptual errors (the most common being calculating the ratio of the earliest possible finish time to the number of activities) in their calculation.

Part (d) in which the candidates were asked to schedule the activities was often well attempted although full marks were rare. Usually candidates were able to plot the critical activities correctly. Common errors included: not plotting all 11 activities, drawing a cascade chart, too many workers being used, the length of activities E, I and/or J being too long, errors in precedence of activities, errors in the start times of certain activities, for example, E and/or I.

Question 8

While 13.2% scored no marks on this question, only 12.8% scored all 6 marks and the modal mark was 3 (scored by 18.5% of candidates). Whilst the objective function was found correctly on many occasions, the absence of the word ‘minimise’ meant that the first mark could not be awarded. The first constraint (based on a total of at least 1000 litres of yoghurt) was usually correct. The other two constraints were either dealt with very well by candidates or not attempted at all. Simplified inequalities were not always seen and, on occasion, coefficients were left as fractions rather than integers.

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Decision Mathematics D2 (R) (6691)

Introduction

The vast majority of candidates demonstrated sound knowledge of all topics, and were able to produce well-presented solutions, making good use of the tables and diagrams, printed in the answer book.

Report on individual questions

Question 1

The first question provided an excellent start to nearly all candidates with 58.0% scoring full marks and 81.2% scoring at least 9 marks. Nearly all applied the north-west corner method correctly in part (a) and completed Table 1 correctly with the missing improvement indices in part (b). The majority went on to give the correct improved solution in part (c), although some had an extra theta in either cell AP or BQ, not understanding the requirement for balance both across rows, and down columns. A significant number of candidates had an additional zero in AQ, the exiting cell, and some did not state the entering or exiting cell. The vast majority correctly stated the cost of the improved solution in part (d).

Question 2

This question also proved to be a good source of marks to candidates with over half of candidates scoring full marks and 74.4% scoring at least 8 marks. The majority of candidates correctly stated in part (a) the difference between the classical and practical travelling salesperson problem although some incorrectly referred to the number of times an arc needed to be visited or argued along the lines of, “the practical TSP is more practical and gives a better answer”. In part (b) the application of the Nearest Neighbour algorithm to find an upper bound and the lower bound calculation in part (c) were nearly always done correctly. Nearly all candidates could then go in part (d) to write down the smallest interval which contained the optimal length of the route.

Question 3

Full marks were obtained by 47.8% of candidates and 75.3% of candidates scored 8 marks or more on this question on game theory. Nearly all candidates successfully reduced columns and they then went on to set up the three correct probability equations (although, on occasion, arithmetical errors occurred when candidates simplified these expressions). It was unfortunate that some graphs were poorly drawn, some without rulers, with uneven scales or so cramped that it was difficult for candidates to identify the correct optimal point. Most candidates then attempted to solve the correct pair of equations for what they considered to be their optimal point. While many had the correct probabilities for playing columns 2 and 3 a significant number of candidates did not state the player B should never play column 1. Most went on to give the correct value of the game to player B in part (b).

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Question 4

This question gave rise to a good spread of marks and proved a good discriminator. The mode was full marks scored by 36.2% of candidates and 75.2% scored 7 or more marks.

Part (a) was extremely well answered with nearly all candidates applying the simplex algorithm correctly by

choosing the correct pivot in both iterations,

dividing the pivot row correctly and changing the basic variables,

using the new (or old) pivot rows to update the other rows and,

stating the correct row operations.

Parts (b) and (c) caused more difficulty to candidates. It was common in part (b) for candidates to write either or instead of the correct and in part (c) many candidates did not give a fully correct reason why the new tableau was not optimal – it was insufficient to solely state that there were negatives in the profit row as the question explicitly asked the candidates to consider the profit equation and not the tableau.

Question 5

This question proved to be an excellent discriminator and gave rise to a good spread of marks. Only 29.0% scored full marks and only 68.1% of candidates scored 9 marks or more. In part (a) many candidates did not add a supersource or supersink to both diagram 1 and diagram 2 which meant that a lot of subsequent marks were lost in later parts. Most candidates who did add the supersource and supersink went on to score full marks in parts (a) and (b). In part (c) many increased the flow by 21 and listed their flow augmenting routes correctly. Most candidates went on to attempt the final flow diagram in part (d), although a significant number of candidates did not gain full marks as they did not have a flow of 102. A number of errors were often present such as two numbers on some or all of the arcs and a significant number either left one arc blank or had an inconsistent flow pattern, most notably at B or D.

In part (e), many gained the method mark, for a cut, but some candidates, who had been successful up to this point, attempted a cut not equal to 102, or they failed to quote the ‘maximum flow – minimum cut’ theorem. It is also advisable for candidates to draw the cut on the diagram showing their maximal flow pattern rather than stating the arcs that the cut passes through. Those that quoted the theorem without a cut lost both marks. Candidates should be reminded to refer to the original diagram containing the flow capacities, when considering possible cuts, rather than their optimal solution.

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Question 6

This question also proved to be an excellent discriminator and gave rise to a good spread of marks. The mode was 3 (out of 7) scored by 24.6% of candidates, 18.8% scored full marks and 10.1% scored no marks. The majority of candidates correctly defined the values taken by

and the set of values that and could take. However, the majority of candidates did not define the objective function correctly as many decided to simply ignore xC4 and xD1 rather than making the coefficients of these two variables sufficiently large. A significant number of candidates made errors with their constraints, by writing them as inequalities, ignoring xC4

and xD1, using coefficients other than 1 (or having right hand sides not equal to 1) or with inconsistent notation.

Question 7

While 23.2% scored 2 or fewer marks, 56.5% scored all 13 marks and 73.8% scored 5or more marks. Of those that scored no marks a majority left this part blank or had little understanding of how to use dynamic programming to solve this particular type of problem. Of those candidates who genuinely attempt this dynamic programming question the only errors seen were usually down to misreading values from the original table or from the candidates own table. Occasional arithmetical errors occurred and sometime the candidate missed out rows from stage 2 (Dolls house) but on the whole candidates answered this question well.

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Grade Boundaries:

May/June 2014 GCE Mathematics Examinations (R)The international (R) versions were strictly for candidates outside the UK.The tables below give the lowest raw marks for the award of the stated uniform marks (UMS).

Module 80 70 60 50 406663 Core Mathematics C1 (R) 64 57 50 43 376664 Core Mathematics C2 (R) 58 51 44 38 326665 Core Mathematics C3 (R) 60 53 47 41 356666 Core Mathematics C4 (R) 60 53 46 40 346667 Further Pure FP1 (R) 62 54 46 38 306668 Further Pure FP2 (R) 63 56 50 44 386669 Further Pure FP3 (R) 63 56 49 42 366677 Mechanics M1 (R) 59 53 47 41 366678 Mechanics M2 (R) 63 55 47 40 336679 Mechanics M3 (R) 58 51 44 37 316680 Mechanics M4 (R) 66 59 52 45 386681 Mechanics M5 (R) 60 52 44 37 306683 Statistics S1 (R) 61 53 45 37 296684 Statistics S2 (R) 58 51 44 37 306691 Statistics S3 (R) 67 59 52 45 386686 Statistics S4 (R) 60 52 44 36 286689 Decision Maths D1 (R) 60 54 48 43 386690 Decision Maths D2 (R) 62 53 44 36 28

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Provisional Pass Rate Statistics: May/June 2014 GCE Mathematics Examinations (R)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).

Module 80 70 60 50 406663 Core Mathematics C1 (R) 54.3 70.6 81.7 89.4 93.16664 Core Mathematics C2 (R) 54.7 71.9 81.7 87.9 92.16665 Core Mathematics C3 (R) 39.9 55.4 66.9 75.4 81.76666 Core Mathematics C4 (R) 43.4 60.5 70.9 77.6 83.36667 Further Pure FP1 (R) 53.9 71.4 82.3 91.1 94.76668 Further Pure FP2 (R) 25.1 42.9 56.9 70.4 78.26669 Further Pure FP3 (R) 24.6 37.1 49.7 67.1 75.46677 Mechanics M1 (R) 47.1 62.9 75.0 83.5 87.36678 Mechanics M2 (R) 37.1 56.6 71.6 79.9 86.56679 Mechanics M3 (R) 30.0 38.8 48.0 61.3 70.66680 Mechanics M4 (R) 31.6 42.1 78.9 84.2 94.76681 Mechanics M5 (R) 66.7 66.7 66.7 66.7 66.76683 Statistics S1 (R) 40.0 60.2 76.5 86.2 92.16684 Statistics S2 (R) 34.0 52.8 63.0 72.3 78.56691 Statistics S3 (R) 35.5 68.1 79.4 90.8 95.06686 Statistics S4 (R) 42.9 100.0 100.0 100.0 100.06689 Decision Maths D1 (R) 32.4 50.7 67.0 77.0 85.66690 Decision Maths D2 (R) 54.6 80.2 91.1 95.5 97.4

21.8% of candidates gained 90 UMS or more on Core Mathematics C3 (R)23.7% of candidates gained 90 UMS or more on Core Mathematics C4 (R)

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