Examiners’ Report

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Summer 2014

GCE Mathematics

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Core Mathematics C1 (6663)

General Introduction

In general there was a very wide range of mathematical ability displayed.

For a considerable number of candidates, working without a calculator really highlighted a lack of fractional dexterity and a lack of understanding as to how fractions work. Some language caused problems and students seem easily confused by mathematical words such as ‘discriminant’ and ‘differentiate’. The discipline of using mathematical notation or producing complete mathematical statements was not well demonstrated by this group of candidates.

Disappointingly, it was rare to see students attempting to check answers. When fractions were multiplied together it was very common for no cancellation to be attempted before multiplication. The resulting numerators and denominators after multiplication were very large (often wrong) and the subsequent cancellation of the final fraction usually failed.

For the more able however, the examination gave plenty of opportunities to use their mathematics effectively and with purpose, but also the questions had been written to allow progress through each question even if part of the question could not be done. There were some excellent attempts at the paper resulting in full marks in many questions.

Report on Individual Questions

Question 1

Most candidates (87.5%) achieved full marks on this question. A very few differentiated instead of integrating. Common errors were not simplifying the coefficient of x2, omitting the +C or leaving the integral sign in the final answer, so that 11% lost just one mark. Very few responses were left blank.

Question 2

In this question 56.5% of candidates gained full marks.

Part (a) was correctly answered by almost all candidates, however, many candidates failed to use their answer to solve part (b)

In part (b) poor bracketing and poor basic understanding of the laws of indices were the main reasons for an incorrect final answer. Many candidates are careless in the way they write fractions and answers are often ambiguous leaving examiners uncertain whether the answer is

x2 or ? The negative power created problems for many; for example, was often

rewritten as 4x–2 (this was the type of error also regularly seen in question 7). The responses were mixed with fairly equal numbers failing to apply the power to the 32 and failing to

achieve x-2. Common incorrect answers were , , x2 and 4x–2.

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Question 3

In this question 45% of candidates gained full marks with a further 25% dropping just one or two marks.

In part (a) most candidates scored full marks, though a small number changed the direction of the inequality. Other errors were mistakes in the rearranging (ending up with 2x > 10 or 4x > –4).

In part (b) most students correctly rearranged the inequality and got the first two marks for the critical values x = –3 and x = 12 by factorising and solving correctly. Only a few students erroneously gave x = –12 and x = 3. Some students used the quadratic formula to find these values, and mistakes substituting were made here. A large number made the ‘usual’ mistake of simply leaving x ≤ –3 and x ≤ 12 as their answer, whilst some of those who realised that they needed the inside region did enough to gain the method mark, but not the accuracy mark as a result of incorrect notation. Some students picked the inside region correctly but then wrote –3 < x < 12 with strict inequalities.

In part (c) the mark was gained only if students had worked correctly through the rest of the problem. As a result this was the least commonly gained mark and relied on accurate work and good understanding. Students generally realised the required method and often used a number line to correctly identify the region following their previous work, though many missed the subtlety of the ‘strictly greater than’ for the .

Question 4

In this graph question 65.5% of the candidates gained full marks or dropped just one mark.

In part (a), most candidates achieved –1 for x or wrote (–1, 0). Sometimes the –1 was found in the body of the question text.

In part (b) many graphs were well drawn; most candidates realised that it was a cubic graph and drew the correct shape. Negative cubic curves were quite common and there were also a few quadratic and hyperbolic graphs. Where the cubic was correct the commonest error was a curve crossing at (0, 0) and turning at (2, 0). Cubic curves crossing the x-axis 3 times were also quite common. A few candidates produced cubic graphs with a single point of inflexion at the origin.

Part (c) appeared to be the least well-answered part of the question. Several students wrote down the correct number of solutions but did not give a reason. Others wrongly gave a reason related to the number of times the graph crossed the x-axis instead of to the number of times the two curves crossed. Several candidates also tried to solve the resulting equation. There was some confusion between the words ‘intersection’ and ‘intercept’.

Question 5

This question was answered very well by most candidates with 77% gaining full marks.

Part (a) was usually correct with a good method shown. The main mistake was candidates simply substituting 1 into the formula to get the correct answer 2 fortuitously, which scored

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no marks. Generally when candidates made this mistake in (a), they then went on to find the third and fourth terms by substituting “3” and “4” into the formula in (b). The majority of candidates obtained full marks in part (b) and most applied the correct method. A small number of candidates used a wrong a1 from (a) and there were quite a few who wrote a4 = 160 where they had neglected to subtract the 3. A significant number of candidates wrongly used the arithmetic sum, using a = 2 and d = 5 with n = 4. Some also evaluated the first four terms but failed to add them up.

Question 6 58.5% of candidates gained full marks and a further 18.5% dropped one mark, usually the last mark.

Part (a) was done well by virtually all the candidates. The correct form with c = 4 was expressed almost universally, with c = 16 seen occasionally by those who were unsuccessful.

In part (b) there were some for whom placing surd in a context seemed to prevent them from starting. Most candidates applied method 1 and started very well with the appropriate expression for the width. Many mistakes arose with the multiplying out of the terms (1 + √5)(1 – √5) in the denominator to get 1 – 5 = 4 instead of –4. The numerator was expanded much more successfully. Those who had managed to obtain the correct numerator and denominator then typically simplified correctly but quite a few only divided one of the terms in the numerator by the 4 or –4. The alternative method of (p + q√5)(1 + √5) was successfully used by a minority and a few candidates achieved the correct answer either by inspection or using trial and improvement. Some errors in sign occurred at the final stage, the main error here being √5 – 5, which often went undetected by candidates, despite producing a negative width.

Question 7

This question provided some discrimination in part (b) and 53.7% gained full marks on the whole question.

A very high proportion of candidates scored full marks in part (a). Most expanded the brackets correctly and then differentiated with only a small minority using the chain rule, which is on an A2 module. Of those candidates who did not score full marks this was mainly because of an incorrect expansion of the brackets or an incorrect method for differentiation without expanding. Frequently the 2x was squared to 2x2 or there were often sign errors. Occasionally the constant of 1 was squared to 2 and there were a surprising number of misreads with the original expression often being written as (1 + 2x).

Part (b) was completed reasonably well by most candidates. Of those who did not score full marks, most did not manage to achieve the expression in the necessary correct form

of  . The most common error was to multiply the numerator by 2x–2 rather than by

x–2 and some added or subtracted 2x–2 creating a third term. Even those who divided both

terms by 2x2 made errors to obtain, for example, 2x2 for the first term or for the second.

Most who achieved – as a power correctly differentiated it to – .

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Common errors were due to the misconception that the numerator and denominator could be differentiated separately .There were quite a number of candidates who, having done the necessary work to change the expression into a suitable form for differentiation, then completely failed to do the differentiation. A few candidates (who had covered A2 material) attempted to treat the expression as a quotient and differentiate it directly. This approach had mixed success with the necessary formula often misquoted or misapplied.

Question 8

This question discriminated but provided access for all candidates. 29.2% gained full marks.

Part (a) was completed well by the majority of candidates. Most used the correct formula of a + (n – 1)d with a correct combination of a, n and d. Some used a list and a small group worked backwards, usually correctly, to find n = 8 and related that to 2007. Some who got a wrong answer of, for example, 210 just left it and didn’t try to correct their mistake.

Part (b) was again completed well by the majority of candidates. The majority quoted and used the correct formula with most errors coming either in their arithmetic or using n = 13 instead of n = 14. A surprising number could not multiply 7 and 430 to a correct value. Again in certain cases lists were seen, some of those with a correct total.

Part (c) caused numerous problems. Of those who had an idea of the overall method, common mistakes included taking d = +20 instead of d = –20, multiplying the wrong side of the equation by 3 or omitting the 3 altogether, and a large number of arithmetical errors. Some tried to use the sum to n terms formula, and some just put “= n” on one side of their formula. Even so, a majority of candidates achieved n = 10 and most of those who did went on to relate this to 2009 and achieve full marks in this part of the question. Many candidates generated comparative lists, usually containing correct values, and obtained the correct answer in this way, though they had not demonstrated sufficient mathematical thoroughness to be credited with full marks.

Question 9

40% gained full marks on this question and a further 26% lost one or two marks only.

Part (a) was well answered with the majority of candidates correctly identifying – as the gradient of line one and using m1 × m2 = –1 to find the gradient of line two. Some candidates did not divide the constant term by three when rearranging the equation into the form y = mx + c and this led to problems in part (b). A surprising number of candidates did not rearrange the original equation for line one and it was quite common to see the gradient of the original equation given as –2, although most knew to use the negative reciprocal to get the perpendicular gradient. Another common error was to substitute a point other than the origin (usually B) into their line equation, leading to an incorrect answer with a non-zero constant and making part (b) more difficult.

Part (b) proved to be quite challenging for some candidates and a number of candidates failed to attempt the question due to a lack of understanding. The majority, however, realised what was required. Most candidates realised that they had to find the coordinates of C and knew to use simultaneous equations to find the intersection of the two lines. However many arithmetic

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errors frequently ensued due to poor manipulation of fractions and an inability to simplify: many candidates made their subsequent calculations more difficult by failing to realise that, for example, 52 13 = 4, although many did get the correct coordinates of C, including some who made no further progress. Some candidates used the y-coordinate of B as 26 or 13 rather than 26/3.

Many missed the simplicity of the triangle in question and embarked on the more complicated methods given as alternatives in the mark scheme. Those who tried to work with OC and BC almost always made mistakes manipulating surds and many were unable to simplify their final answer to a fraction. Those using the easier method often used the y-coordinate (6) instead of the x-coordinate (4) in the base × height. Also, in finding B, many candidates found where line one intercepts the x axis and as a result tried to find the area of the incorrect triangle. Some forgot the in the area of a triangle formula. On the other hand, there were a number of concise and accurate solutions.

This question in particular tested the ability to deal with calculations involving fractions, and poor arithmetical skills led to lost accuracy marks which was disappointing to see.

Question 10

26.2% gained full marks and a further 28% lost one or two marks.

In part (a) the first three marks were obtained relatively easily by most candidates, but simplifying the fractional coefficients often proved to be a challenge, in particular ÷ 3

which often became , and dividing by in the second term also caused problems. Many candidates failed to go on and find c. Those that did often made arithmetical mistakes and so did not get the correct answer of 53. Those that did manage to achieve c = 53 failed to get the final mark as they did not simplify their coefficients, being seen frequently in the final answer or simplified incorrectly.

In part (b) some candidates substituted (4, 25) into their f(x) rather than f ′(x) which resulted in no marks for this part. Most, however, did substitute into the correct f ′(x). Many errors were made in the calculation and a significant minority die not achieve the correct gradient. Most were able to continue to find the gradient of the normal and then use (4, 25) to write down an equation of the normal with a few using y = mx + c to find c. A minority tried to find the equation of the tangent. Many candidates did not read this question properly with regard to integer coefficients being required, giving their final answer as y + 0.5x – 27 = 0. There were common mistakes made when re-arranging from y = – x + 27 to the form of the equation required.

Question 11 Full marks was scored by 25% of the candidates while 32% scored 4 marks or fewer out of the 10 available. This question provided good discrimination.In part (a) the discriminant was found correctly by most candidates. There were some arithmetic errors, and a few included a square root with the b2 – 4ac or used the whole quadratic formula. A few confused discriminant with derivative and differentiated.

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Part (b) proved a significantly tougher challenge and although a good percentage of candidates achieved full marks, there were frequent errors. Most candidates seemed to know that completing the square was required, however many struggled to deal with the coefficient of x2 not being unity. Bracketing errors were common and a large minority could not even get as far as 2(x + 2)2 + k.

Part (c) was again done with varying success. Most candidates opted to either use differentiation or set 2x2 + 8x + 3 = 4x + c. Those candidates who used differentiation were more likely to achieve a full solution that was fully correct though there were many who did not realize that they needed to set the gradient of the curve equal to the gradient of the line; commonly it was equated either to zero or the constant in the gradient function was erroneously believed to be the required value of c. Those who set 2x2 + 8x + 3 = 4x + c fared less well, with many candidates not knowing what to do after they had rearranged their equation to collect together the x terms.

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Core Mathematics 2 (6664)

General Introduction

This paper proved a good test of students’ knowledge and students’ understanding of Core Mathematics C2 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.

While standards of algebraic manipulation were generally good, some weakness in this area was seen in question 3(b) and question 6(c). Work on indices was sometimes problematical

and a significant minority of students in question 4 incorrectly wrote as and many

students were unable to deal with the work required in Q06(c), often writing as

Report on Individual Questions

Question 1

This was an accessible first question; the trapezium rule seemed to be well understood by the majority of students and most answered it very well. In part (a) some students lost the mark for giving the missing value as 1.6 or 1.600, although this did not stop them from being able to gain full marks in part (b).

A common error in part (b) was to use 0.2 as the width of each strip, using the number of ordinates rather than the number of strips. Others misinterpreted the strip width from its definition in the formula book. Most students clearly showed the correct structure of the y-values but there was omission of brackets. It was common to see expressions such as  × (1.414 + 2.236) + 2(1.601 + 1.803 + 2.016), which led to the wrong answer. If subsequent working indicated that the correct bracketing was intended this lapse was condoned, but this was usually not the case. Occasionally the method mark was lost by including an extra term in the second bracket.

Question 2

This question was well attempted by most students, with the majority of students able to score 5 or 6 marks. In part (a) most showed that f(2) = 0, but a large number did not write a conclusion and so lost the second mark. A number of students who ignored the rubric and used long division in this part of the question, lost both marks.

In part (b) most achieved the correct quadratic expression, either by long division or inspection. A small number equated coefficients with (x – 2)(Ax2 + Bx + C), usually doing so

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correctly. Most students then went on to factorise the quadratic successfully, usually by inspection and a small number used the quadratic formula, but this did not always lead to the correct factors. A few students neglected to show the full factorisation of f(x), after factorising the quadratic, and so lost the final mark.

Question 3

The majority of students scored well in part (a), with many scoring all four marks. The most successful approach was to use the formula for expanding (a + b)n. With this method the most common mistakes were forgetting to include the negative sign when using 3x (yielding +576x) and forgetting to square the coefficient of x in the third term (producing 720x2). Some students lost the final A mark due to leaving, for example, an unsimplified 576x. Pascal’s triangle was not used very often. A smaller number of students tried taking out a factor of 2 from the bracket, however it was quite common for these students to forget to raise this to the power of 6. Those making this error could score a maximum of 1 mark for part (a). Those

who took this approach, then had problems dealing with and made errors in

squaring .

Students generally found part (b) more challenging and many students were let down by

weak algebra. Many appreciated that they had to multiply their answer to part (a) by

and earned the method mark, although they then sometimes did not try to simplify their

answers. For those who did multiply by their expansion from part (a), the majority

were accurate in multiplying out the brackets but a lack of proficiency in algebra let some down in this part. Loss of accuracy marks here also resulted from an incorrect expansion from part (a). A significant number of students did not collect up their like terms after expanding. Interestingly, those who listed their expansion in part (a) were far more likely to leave their new expansion un-simplified. Good presentation aided students to collect like

terms effectively after the multiplication. A few students tried to expand and

multiply this by their answer to part (a). Another error that was seen occasionally was to

expand (2 – 3x) and then to raise the answer to the power 6.

Question 4

Many students found this question fairly challenging. The majority of students could

successfully integrate and unsimplified expressions such as were accepted although

this was sometimes simplified incorrectly as .

Integrating the second term caused greater difficulties however. It was fairly common for students to rewrite the second term incorrectly as 3x–2 and then to integrate to get –3x–1. Some

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then wrote this as but by then the error had already occurred and recovery was not

possible.

A few students tried to change the function prior to integrating, for example multiplying the

expression by x2 to give which presumably, they were far more comfortable

integrating. Another error that was sometimes seen was to integrate both the top and bottom,

so that was integrated to give and integrated to give . Some students

confused powers and denominators, changing to x3 – 6 and then integrating x–3.

Having attempted the integration, most students were able to handle the limits correctly, with only a few adding the two results after substituting 3 and 1 instead of subtracting them or subtracting them the wrong way round. Many students however then struggled to get the

correct final answer. Some left their answer as , despite the demand for the form

a + b3. Others made errors manipulating the surds. A surprising number of students did no integration at all and simply substituted in the limits, which gained no marks.

Question 5

Part (a) was well attempted with the majority of students getting the correct answer. Where errors were seen it was commonly the use of r2, r2, or r for the sector area, though occasional miscalculations from a correct formula did occur.

In part (b) most responses correctly used the cosine rule but identification of the correct angle was more problematic. Students sometimes used BDC and others BCD. In such cases many students did realise the angle they were finding, and went on to find the area of BCD correctly in part (c), sometimes also recovering 0.943 when they proceeded to find the area of EAB. Truncating too early was not uncommon in such cases.

Where errors in the cosine rule were made it was usually due to mixing up the side lengths. A few students correctly reached the value of cos (DBC) but then failed to use inverse cosine to find the angle. A few instances of sine instead of cosine were seen.

Many students worked in degrees and converted to radians, mostly successfully. A handful of responses rounded to 0.94. Responses where the obtuse angle (2.198…) was found were

uncommon. A notable other fairly common incorrect attempt was in assuming ABE was

and using DBC = – – 1.4.

In part (c) the area of BCD was very well done. A few students incorrectly used 6.1 instead of 7.5. As noted in part (b), there were not infrequent attempts where one of the other angles had been found in part (b) but was used correctly in part (c) for this mark.

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The angle EBA was mostly found correctly but some responses used or or even 2

instead of . A common error was the assumption that triangle was a ‘3, 4, 5’ triangle based on its hypotenuse being 5 cm. Again, some students used degrees and converted to radians but not always correctly. A variety of methods were used to attempt to find the side lengths needed for the area of EAB with roughly equal proportions of each. Some found the third angle in the triangle and used the sine rule twice. Some attempted a ‘hybrid’ solution

with a mixture of degrees and radians, with expressions such as being

used, leading to a common error of EA = 4.004. When this was followed by attempts at Pythagoras then no marks could be gained, though a few students did pick up the method for including as part of their expression for the area of EBA.

Once the side lengths had been found, most, went on to find the area of EBA using a correct method. Errors causing the loss of this mark included use of an incorrect Pythagorean identity to find the third length, mixing degrees and radians (as noted above) and use of area = b h for the triangle.

Rounding too early to obtain 39.0 as their final answer caused several students to lose the final accuracy mark. The method marks, however, meant students who made a numerical mistake were not overly penalised.

Question 6

Part (a) was done very well with nearly all students obtaining the correct answer.

Part (b) was also attempted very well. A few students made errors evaluating their answers

and some used an incorrect sum formula such as and some mistakenly used

n = 20, which cost them both marks in this part. A few students used the formula arn – 1, thereby finding the 12th term instead of finding the sum. It was rare for students to try to find the sum by finding each of the 12 individual terms and then adding. Occasional truncation errors lost the accuracy mark.

Fully correct solutions to part (c) were uncommon. Although the majority managed to earn the first method mark, not many of the students could deal with the inequalities. It was

common for students to write as (17.5)N or to take logs prematurely. Of the students

able to get as far as , most were able to take logs to reach 43.2. However, not

many students managed to achieve the final accuracy mark, either because of errors with their inequality signs or because they gave their answers as N = 43.2 or N = 43. Many failed to

realise that is a negative number and therefore did not reverse the final inequality

when dividing through by this. This left them with N < 43.2 as their solution, even though they went on to state N = 44.

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Some students used trial and improvement in part (c). Although many of these reached the answer N = 44, insufficient working often meant that full marks were not awarded. Some students, for instance, failed to consider both S43 and S44 and others over-rounded their answers, thus being unable to show S – S44 was actually less than 0.5.

Question 7

In part (i) most students achieved the first mark, though a minority expanded as , which was the main reason that this mark was lost. Generally, most students

did proceed to find a correct solution for the equation, with 326.4 being the more common of the two. In many cases this arose from adding 360 to their value of – 60, with the second solution either not considered at all or attempts at 180 – (value of – 60). Adding 60 instead of subtracting was also occasionally seen, with no evidence of correct use of

first. Some students tried to obtain more solutions by adding 90 and thus lost the last mark despite having correctly found the two genuine solutions.

In part (ii) the major problem was the identity for tan x. Students sometimes incorrectly

quoted or replaced 2 tan x with , leading to the loss of all but potentially

the final mark, and it was a rarity for this to be gained in such cases. Students are advised to state identities of this type before attempting to apply them. Even where the correct identity was used, solutions were frequently lost by dividing through by sin x. Some who quoted

after factorising correctly, failed to find both solutions 0 and In many cases no

solutions to sin x = 0 were seen. Correctly obtaining proved more troublesome than

might be expected. There were some interesting attempts including squaring, obtaining an equation in or , the latter often leading to a correct solution. Some were not able to proceed further, or used incorrect identities such as cos x = 1 – sin x. Mostly these were unsuccessful with students getting lost in the algebra. In the few cases where they did get to

sin x = , the negative possibility was missing and no appreciation of the extra answers

generated were shown. Students who did get to were often not able to go on to get

both correct answers from this equation, with 0.84 and 2 – 0.84 the ones often given. Occasionally – 0.84 was given as the second solution. Some students chose to work in degrees and then convert to radians at the end. The final mark for the solutions to sin x = 0 was only scored by a small proportion of students, with most students who reached a solution having done so by dividing through by sin x. Even in cases where sin x = 0 was quoted after factorising their equation, many then ignored it, or only gave 0 and as the solutions. Overall, there were very few completely correct responses to this part. But those that were completely correct were often very well answered and presented.

Question 8

Many students scored full marks for the graph in part (a) but a wide variety of graphs were seen. Negative exponential graphs, the reciprocal function, parabolas and straight lines were fairly common. Some were very close to the x-axis, making it difficult to see if there were

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crossing or not and others appeared to have been erased and/or smudged which made them difficult to read. Mistakes were usually due to the curve passing through 3 on the y-axis, stopping at the y-axis or crossing the x-axis.

A number of students did not even attempt the graph in part (a) but did go on to make an attempt, sometimes successfully, in part (b).

Part (b) was well answered by the majority of students. The use of leading to a quadratic in y was the most common approach with letters other than y used occasionally.

was also frequently seen. Direct factorisation in ex was less common. Once the quadratic was identified there were very few cases of incorrect solutions but there were a number of students who stopped at the point of solving the quadratic and did not go on to find values for x. The most common errors in this part of the question were in attempting to use logs, for example or forming the quadratic . Many students believed that 9(3x) = 27x.

Question 9

Many students found this question challenging.

In Q10(a), many students did not appreciate that triangle OTQ was right-angled and so did not know where to start. Some of these unsuccessful students tried to work out the gradient of the tangent or tried to expand the general equation of the circle. Others assumed triangle OTQ was isosceles. Most students who did make use of the right angle at OTQ were able to reach OQ = 14 and then often went on to score full marks in part (a), although some could not square 6√5 accurately (30 was popular). Some tried to find (6√5)2 – 42. A small number of students used trigonometry to find a missing angle in triangle OTQ and then went on to find OQ = 14 and k = 5√3.

In part (b) many students scored the method mark for forming an equation of a circle using their value of k (or just the letter k), but some were reluctant to use k in their equation for the circle and left a gap instead (x – 11)2 + (y – )2 = 16. Some students did not know the correct form for the equation of the circle, for example quoting (x – a)2 – (y – b)2 = r2 or else (x – a)2(y – b)2 = r2, and some forgot to square one of the brackets or the radius. Another error was to mix up the 11 and k, giving the equation as (x – k)2 + (y – 11)2 = 16.

Question 10

This question involved several different areas of work, area, volume, algebraic manipulation and calculus, and although a significant number of students produced clear and well-structured solutions, this proved a taxing question for many students.

Part (a) was found to be challenging, with many students struggling to find the volume of the prism despite there being several possible methods. It was common to see 30x2y = 9600 derived with unconvincing or incorrect working, despite often being able to find the area of a trapezium correctly in part (b). Some students, realising that 30x2 must be the area of the trapezium, just used 6x × 5x. In part (b) there were many concise, correct and clearly set out solutions but it was very common to see several attempts and much crossing out, and extra terms slotted in at a late stage, presumably influenced by the required expression being given.

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Common errors included not finding areas of all 6 faces, finding too many areas, and combining dimensionally incorrect areas. Almost all students who attempted the question gained marks in part (c) with most differentiating at least one term correctly and many achieving the correct S and setting it equal to 0. Many, though, found it difficult to manipulate their equation correctly and of those who reached x3 = 64 many did not find x = 4. Common answers were x = 4 or x = 8. Many lost the last two marks by not realising that they had to use x to find the corresponding minimum value of S.

Part (d) was generally well attempted, even by those who did not complete Q10(c), with the majority remembering to compare Sʺ with 0. However, an incorrect Sʺ, which was quite common even for those with the correct S, did lose the final mark. A relatively small number of students set Sʺ = 0 and tried to solve for x, often concluding with x > 0x > 0 so minimum.

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Core Mathematics 3 (6665)

General Introduction

This paper was found to be accessible by most students sitting the paper. 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.

Overall the level of algebra was pleasing, although there were many examples of students not using brackets correctly.

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 rule (question 3) and quotient rules (questions 1 and 8(c)) in differentiation.

Report on Individual Questions

Question 1

Part (a) was generally done well. Most students used the Quotient Rule and were successful; the main errors that occurred were from bracketing mistakes. Students who did go wrong

tended to write .

Some of these were able to recover, helped by working towards the given answer but still losing an accuracy mark. A few used the Product Rule and even fewer used the chain rule. A minority of students found the inverse of f(x) rather than its derivative. In part (b) nearly all students were able to gain full marks here. Having equated the expression given in part (a) to –1, the majority of students multiplied out and solved a quadratic equation rather than adopting the shorter route of taking the square root of both sides. Having said this, almost everyone obtained the right quadratic equation and went on to find the correct coordinates. A few students did not find the y-coordinate; many found the coordinates of two points (5, 7) and (–1, 1), not noticing the domain x > 2. (Failure to reject the additional point was not penalised.)

Question 2

Part (a) was completed well with the majority of students able to reach the answer in its simplest form. The method of rearranging to ln (2x + 1) = 5 was the most popular and usually successful. Those who used the power law often lost the second mark because they left the answer in terms of √e10. Some students who used the power law correctly to give ln (2x + 1)2 = 10, then proceeded to take the square root of both sides to give ln (2x + 1) = √10. A minority of students attempted to expand (2x + 1)2 which almost always led to no further progress being made.

Part (b) was more challenging with the main issue being students failing to recognise that the LHS comprised of two functions multiplied together, and hence they did not apply the addition law. Expressions similar to ln 3x × 4x = 7 were commonly seen. Those who moved 4x to the RHS and then took in usually fared much better. Of those who did achieve the first

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two marks only a minority did not know how to collect terms and factorise to arrive at the correct answer.

Question 3

The mark in part (a) was scored by the vast majority of students, with only very occasional signs of students not knowing what to do. To verify that ‘P lies on C’, they had to substitute

into the equation. A very few substituted instead, or else there were occasional

incorrect statements given due to confusion in simplifying, such as

.

In part (b), most students demonstrated good knowledge of trigonometric differentiation and

applying the product and chain rules to find scoring the first three marks.

Of the students who failed to score these, the most common errors were: omitting the y in the y-term, giving 8 tan 2y + 16 sec2 2y; losing the 2 in one or other of the arguments in the expression, usually again the sec2 y

term, so having 8 tan 2y + 16 sec2 y; failing to use the product rule at all, and getting just on its own 16 sec2 2y (or

occasionally other single term expressions).

It was extremely rare to see any attempt to make the subject of the formula before differentiation. It was at this point of the question, after find the correct derivative, that errors

were most often made. Knowledge of finding a “numerical” value of or and then an

equation of a tangent at a particular point was good but the execution was poor.

The majority of students proceeded to try and find an expression for before attempting to

evaluate, and in so doing often made errors. The most common error seen was incorrectly

split fractions: = = + .

If not incorrectly split at this point, writing at a later point was also

common. Once a numerical value of had been found many students knew the form of the

tangent with a minority using a negative or negative reciprocal for the gradient. The rearrangement of the equation of the gradient resulted in a variety of errors due to the mixing of and fractions. Students lost the final mark through inaccuracy in algebraic simplification or not giving the equation in the required form ay = x + b.

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Question 4

This question proved to be accessible to most students and a large proportion found parts (a) and (b) accessible; part (c) was more testing and only more knowledgeable students scored the marks in this part.

Part (a) was generally well done with the majority of students drawing a W-shape in the correct quadrants. A few failed to have the left arm actually crossing the y-axis. In the majority of cases the points corresponding to P and Q were labelled correctly, although a few had (6, –1) instead of (6, 1) for the point corresponding to Q.

Part (b) proved to be more of a problem for some students, due to two transformations, although a high proportion still scored full marks. Most managed to draw a V-shape in the correct quadrants although a few did not have the right arm passing through the y-axis. Some students had trouble finding the correct y-intercept with (0, 28) being a common incorrect answer. The minimum point was more often correct although some did label it (6, 1) rather than (–6, 1).

Allowances were made for those students who mistakenly sketched either y = –2f(x) +3 or y = –2f(–x) + 3. Such students were allowed to score just one of the three marks. The number of students who did this was quite small.

Many students did not attempt part (c). Of those who did attempt it, many achieved b = 6, but only a few successfully reached a = 2. Often students got as far as ab = 12 but were unable to proceed any further.

Question 5

Part (a) was well tackled with the vast majority scoring full marks. Most students started by successfully factorising the quadratic x2 + x – 6 and proceeded to produce a single fraction with a quadratic numerator and denominator. Most were able to correctly perform the second factorisation and correctly cancel.

Students who started by using a cubic common denominator were much more likely to fail. Whilst the technique was correct they found it difficult to progress. Occasionally a step, such as failing to simplify the expression x2 – 2x + 6x + 3 before factorising, was omitted from the proof which resulted in the loss of the final mark. In a proof it is vital to show all steps.

Students found part (b) one of the most challenging parts of the paper and there were very few fully correct answers seen. A few students had a reliable method for finding the range. Methods usually involved substituting numbers from the domain with no real purpose. Many students were able to identify 4 as a limit but many wrote y > 4 instead of y < 4. Relatively few students were able to obtain the lower limit. The notation for the range was usually correct with either y or g(x) used.

Part (c) was well answered although not by the expected route. The majority of the students failed to spot that the intersection between a function and its inverse is always when y = x, therefore they went for the longer method using the inverse function. Virtually every student attempted to find g–1-(x), most successfully. They equated their answer to g(x) and obtained a quadratic which they solved, usually by correct use of the quadratic formula. A small, but

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significant number did not appreciate the meaning of exact and solved the final quadratic giving only decimal answers and some stopped having gained the correct quadratic, perhaps expecting it to factorise. Errors in manipulation and signs occasionally led to an incorrect 3TQ, but most students were able to gain full marks. A few students mistakenly equated g(x) to its reciprocal or to g(x). Although they did not have to for full marks, very few justified

their choice of , rejecting .

Question 6

This question was generally answered very well with many fully correct responses. Part (a) was successfully answered by most of the students, although some worked in degrees and were unable to understand why there was no change in sign. Partial credit was given to such a solution.

Part (b) was the one part where marks were most often lost, though the majority of students gained at least some marks. Where marks were lost, it was largely due to one of three reasons:

failure to differentiate at all. There were a small by significant number of attempts to rearrange the original equation into the one required, usually involving an attempt to replace cos2 x by 1 – sin2 x;

incorrect differentiation. Again a fairly common error to make, and either an incorrect sign or missing the “ ” in front of the sine term to blame. Students who differentiated at least obtained the 3x2 – 3 terms;

failure to explicitly set the derivative to 0 at any stage. There were a number of students who failed to make an explicit statement involving “ ” somewhere in the solution. This error lost the final two marks of this part for the students.

Part (c) was well answered with the majority of students gaining both marks here. The overriding common error was use of degrees, not radians, and strangely, this often even followed correct use of radians in part (a), an anomaly that is difficult to account for.

Question 7

Part (a) was found to be demanding. Those who started by using and then

combining to form a single fraction, were able to progress and generally went onto to achieve full marks. Students who used the double angle formula for tan before combining their fractions generally found it difficult to deal with the resulting algebra and failed to achieve more than the first mark or two.

In part (b) most students recognised the link with part (a) and successfully found two solutions. Common errors were 2 + 10 instead of 2 + 5 and using tan 60 instead of tan 30. There were a small number of students who attempted to use the tan addition formulae, but most gave up after only a couple of lines. The statement about ‘solutions based entirely on graphical or numerical methods not being accepted’ was clearly heeded, as solutions without working were rarely seen.

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Question 8

Part (a) was nearly always correct, and mostly just written down. For the ones who did make errors, they might be better advised to write down an intermediate line on two mark questions.

Part (b) highlighted some weaknesses in solving exponential equations. Most students knew that they had to reach a form e0.1t = A before taking logs, but there was a sizeable minority who insisted on taking logs early resulting in few, if any marks. Most students knew the meaning of the word integer and wrote their answer in an appropriate form but there were lots of cases where 0.1t = ln 5 ? t = 0.1 ln 5.

Part (c) was seen to be challenging. A statement of the quotient rule would have helped many students. Quite often e0.1t was differentiated to 0.1te0.1t or t = 10 was substituted in first producing a constant, which was then differentiated. A lot of fully correct answers were seen however.

Part (d) was seen to be one of the most challenging questions on the paper. Incorrect answers such as ‘time cannot be negative’ through to ‘they cannot get bigger than 200’ were frequently seen. The most fruitful method involved substituting in and showing that

cannot be found by referring to the sign of either ln or exp.

Question 9

The large majority of students answered part (a) correctly, but parts (b) and (c) proved to be more challenging.

For part (a) most students scored all three marks. Almost everyone found the value of R correctly. When finding the angle, the most common errors involved either using degrees

instead of radians or writing .

In part (b)(i) a good proportion of the students achieved 104 for the maximum value, although some failed to square the value of R and gave the maximum as 4 + 5√20.

In part (b)(ii) many realised that an angle of was required, however it was very common

to see .

In part (c)(i) an answer of 104 was seen for the minimum value as well as the maximum, as students set the trigonometric expression to –1 instead of 0. A minimum value of –104 was also seen quite frequently. Few seemed to realise that setting the bracket equal to 0 solved the problem.

In part (c)(ii) the correct answer was rarely achieved. Some students scored a method mark for A significant minority of students attempted to use calculus in parts (b) and (c). Such methods could achieve all of the marks but students choosing this route were rarely totally successful.

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Core Mathematics 4 (6666)

General Introduction

This paper proved to be a good test of C4 material and discriminated well across students of all abilities. There was enough opportunity for grade E students to gain some marks in at least 6 of the 8 questions on this paper. There were some testing questions involving converting parametric equations to a Cartesian equation, differential equations, integration and vectors that allowed the paper to discriminate well across the higher ability levels.

There were a number of questions on this paper, some of which were unstructured, that tested students’ ability to think for themselves. Question 6(iii) required students to find a correct strategy for integrating with respect to y, as part of solving their differential equation. Access to solving Q8(e) and Q8(f) was gained by those students who drew a diagram and used it to help them in assimilating the information given.

The standard of algebra seen by examiners was generally good, although a number of students made basic sign or manipulation errors in questions 1(b), 2(a), 3(d), 4, 5(b), 6(iii) and 7. In summary, questions 1, 2, 3(a–c), 4, 6(i), 7(a) and 8(a–d) were a good source of marks for the average student, mainly testing standard ideas and techniques; and questions 3(d), 5, 6(ii), 6(iii) and 8(e) were discriminating at the higher grades. Questions 7(b) and 8(f) proved to be the most challenging questions on the paper.

Report on individual questions

Question 1

This question was well answered, with about 55% of students gaining full marks and about 70% of students gaining at least 6 of the 7 marks available.

Q01(a), many students were able to differentiate correctly, factorise out , and rearrange

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

A minority did not apply the product rule correctly when differentiating whilst a small number left the constant term of –20 on the left hand side of their differentiated equation and a few differentiated to give 0. Sometimes examiners saw slips in students’ differentiation

such as becoming , or similarly becoming

Q01(b), the majority of students were able to apply a full method for finding an equation for the tangent, although some were unsuccessful in finding the correct answer due to errors in their manipulation or because of their incorrect answer to part (a). Common mistakes

included making arithmetic errors in evaluating at finding the equation of the

normal instead of the equation of the tangent, and not leaving the equation of the tangent in the form , where a, b and c are integers.

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Question 2

This proved to be the most accessible question on the paper with about 75% of students gaining all 5 marks.

Students improved their chance of success in this question by writing out the expansion of

as before attempting to answer parts (a) and

(b). This strategy reduced the risk of making a sign error when equating their x coefficients in part (a).

In part (a), the majority of students wrote down which led to the correct answer of

, although some deduced that with no intermediate working. The most common

error was to find as a result of writing down (or equivalent).

In part (b), the majority of students substituted their k from part (a) into either or

a simplified in order to find the value of A. Students only achieved full marks in Q02(b)

if they found from a correct . Common errors in this part included using k

instead of using a binomial coefficient of or stating instead of .

Question 3

This question was well answered, with about 29% of students gaining full marks and about 60% of students gaining at least 8 of the 11 marks available. Part (d) discriminated well across the higher ability students.

In part (a), most students were able to find the y-value corresponding to . The most common error was to truncate 0.682116... to 0.68211 rather than rounding it to 0.68212. In

applying the trapezium rule in Q03(b), a small minority of students multiplied by

instead of by Whilst the table of values clearly shows an interval width of 1, the

application of a formula with instead of sometimes caused this error.

Other errors included the occasional bracketing mistake or rounding their answer incorrectly to give 2.5775.

Whilst many students in part (c) identified their estimate of R as an overestimate, many found some difficulty in articulating a reason for this. Those who used a diagram were most successful in showing clearly the extra area. Whilst some students gave no reason, others believed it was an underestimate. Some referred to a negative gradient but this alone was not a sufficient explanation.

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In part (d), the majority of students differentiated the substitution correctly. The most

common errors on substitution included ignoring to obtain , partially

substituting for x and thus writing an integral involving both x and u, or writing an integral of

the form . A number of students attempted to integrate

directly to give instead of cancelling the u’s to obtain Some

students made erroneous algebraic assumptions such as prior to their

integration. Although a number of students integrated incorrectly to give

most correctly obtained with a few giving .

The majority of students applied the changed limits of 2 and 1 correctly to an ‘integrated’ function in u and gave an exact final answer. A return to x limits would have been acceptable but was seldom seen and only occasionally x limits were used erroneously in a function in u.

Question 4

This question was generally well answered, with about 60% of students gaining full marks, although about 10% of students scored 0 marks. A number of cases were seen where was misread as 80.

The majority of students multiplied out and then found although on

occasion the second term in their was sometimes incorrect. A significant number of students, however, used the product rule with and , to find

Surprisingly the differentiation was sometimes incorrect and it tended to be as a result

of using the product rule. A significant number of students stopped after either finding their

in terms of h or after evaluating as when

At this stage the majority of students applied Chain Rule to correctly write down an equation

for . They divided by their and substituted to find a value for .

Common errors at this stage included applying applying

and leaving their final answer as

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Question 5

This question was very discriminating with 29% of students scoring full marks, 26% of students scoring 3 marks out of the 5 marks available and 28% obtaining 0 marks. It was surprising to observe that a significant minority of students could not apply or were unaware

of the identity

In Q05(a), the majority of students were able to prove in one of two ways.

About half of them expanded and added this to and achieved the correct

result. The other half expanded and deduced that which

also led to the correct result. The most common error was for students to expand

to give

Part (b) was found to be much more discriminating, with more than half of the students obtaining 0 marks. Some students used the result part (a) to write down (with a surprising number believing ) but could not progress no further, although a number tried in vain to find a strategy to eliminate the parameter t. Those students who applied to obtain usually went on to achieve the correct cartesian equation. A significant number, however, converted to which then led to an incorrect answer of A few students who deduced that and in without deriving the Cartesian equation were penalised the final mark in this part.

Question 6

This question discriminated well across students of all abilities, with about 17% of students gaining full marks and about 45% of students gaining at least 8 of the 12 marks available.

In part (i), most students recognised the need to use integration by parts and many fully

correct solutions were seen. A few students labelled u and the wrong way round and a

common error was to integrate to give either or even During the second stage

of the method it was not uncommon to see students integrating to give

In part (ii), successful students either applied the method of or

applied a substitution of Many achieved an answer in the form ,

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with some incorrectly giving as , , or Common erroneous integration led to

logarithmic answers and also to . There were a few students who differentiated

to give and proceeded to deduce the correct answer.

Part (iii) was found to be challenging by the majority of students. They were often able to separate the variables, but although almost all could integrate correctly, many struggled

with the integration of . The most common method was to use the identity

to give . Those who integrated this to give

usually went on to score full marks, but some proceeded to and often

made little further progress or believed that Some alternative methods

of integration were seen such as rewriting the product as a sum

which could be integrated using standard results. Those who attempted

to use integration by parts for often proceeded no further than just a first

application. A significant number of students substituted and into their

integrated equation containing , but in most cases previous integration errors prevented them achieving the correct answer.

Those students who failed to separate the variables from the outset either attempted to integrate the given expression by parts, despite being a function of both x and y, or substituted

the given values of and to obtain a value for

Question 7

This question provided discrimination across students of higher abilities, with about 9% of students gaining full marks and about 56% of students gaining at least 6 of the 15 marks available. While many struggled with this question, strong students often produced clear, concise and well-structured responses.

In part (a), the majority of students found , used it to find the linear equation of the normal

to the curve at , put 0y and solved for x. A significant number, however, were let down by their differentiation and the quality of their algebra. Although most students differentiated correctly, a significant minority struggled to differentiate

correctly. Some students used to give

before obtaining , whilst others applied the chain rule correctly to give

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Common errors in this part included finding as either or

simplifying a correct to give , or

finding the equation of the tangent instead of the equation of the normal. Some students

substituted an incorrect value of into their This was usually which was found

by prematurely setting . Few students found the Cartesian equation of C and used it successfully to find the x-coordinate of Q.

In part (b), a significant number of students were unable to find a volume of revolution by using the parametric equations. Those who adopted a Cartesian equation approach also made

little progress. Some stated the volume as but did not know how to continue or

rewrote the volume formula as , with being replaced by . Those who did not

apply gained little access to this question, and it was also disappointing to see

some students who attempted to apply an incorrect . Only a minority of students

applied to give a correct , with a significant number

using incorrect manipulation to give Those that reached a stage where they

were integrating a multiple realised the need for using . Whilst the double angle formula was generally quoted correctly, this did not always lead to a correct expression for integration as a result of sign or bracketing errors. After integrating most

students used the correct limits of and 0, and only a small minority achieved the correct

answer of Occasionally, however, incorrect limits such as 3 and 0 were used.

The majority of students used the hint and applied to find the volume of the

cone, although a small number attempted to find this volume by applying

Common errors for finding the volume of the cone included mixing up

the values for r and h or using rather than . A large number of students did not attempt find the volume of the cone, whilst others found the area of a triangle and subtracted this from their

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Question 8

This question discriminated well across students of all abilities, with about 7% of students gaining full marks and about 63% of students gaining at least 8 of the 15 marks available. Most students were able to score full marks in parts (a), (b) and (d), with parts (e) and (f) offering good discrimination for the more able students.

Part (a) was well answered with only a few students adding instead of applying . Only a few made arithmetic errors in applying

In part (b), most students were able to write down a correct expression for , but some students did not form a correct equation by writing

In part (c), most students were able to apply the scalar product formula using either

or , to obtain the correct answer of The most common

error was to apply the scalar product formula with either or , which

usually resulted in giving although the minus sign was dropped by a significant

number of students. A small minority of students applied the cosine rule correctly to achieve the correct answer. A number of students struggled with this part and usually applied the scalar product formula with non-relevant vectors such as .

Part (d) was generally very well answered, particularly if the student had already gained all the marks in parts (a) and (b). Few wrote down the line for with position vector and direction vector the wrong way round.

In part (e), a significant number of students over-complicated the problem by forming and solving an equation in (or ), to give with some solving their equation incorrectly. Those students who drew a clear diagram, quickly found the coordinates of C and D by applying .

Part (f) was challenging and not attempted by all. Those students who drew a diagram tended to be more successful in gaining marks. In order to make progress, it was necessary to find either the perpendicular height of the trapezium or the area of one of the triangles APB, APD or BCP. The majority of students were unable to do this, and did not score any marks for this part. Although many students knew the formula for the area of a trapezium, many erroneously assumed that AD was the perpendicular height.

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Further Pure Mathematics 1 (6667)

General Introduction

The general standard of work was high with a lot of well organised clear solutions. Diagrams would have helped students make better progress in question 6 and question 7 and more detail when substituting in question 2 for example would have been helpful.

The questions which students found most challenging were question 5(b), question 6(c), question 7(ii) and question 9.

Report on individual questions

Question 1

The majority of students made a successful start to this paper and multiplied the numerator and denominator by the conjugate of the denominator in Q01(a). There were common algebraic errors e.g. (2i)2 = 2i2. Some students wrote 3 rather than 5 for their simplified denominator, but still gained the method mark for showing use of i2 = –1 in the numerator. A number of students failed to write the final answer as real and imaginary parts.

In part (b) many solutions included i incorrectly in the calculation of the modulus, thereby losing marks. This often occurred with students who did not factor out the i in the final solution of part a.There were some mistakes in expanding and simplifying, leading to a 3 term quadratic. Common errors were not squaring the 5, or equating the modulus to 13 rather than 132.

Question 2

There were a number of arithmetic mistakes in this question with students unable to correctly evaluate and . Some students failed to make an adequate conclusion, or drew no conclusion, in respect of the ‘sign change’ and ‘hence the root’.

In part (b) there were many fully accurate attempts at differentiation although a significant number made an error with the middle term, having issues with both the coefficient and the power.

Those who made errors in part (b) typically lost marks in part (c) for insufficient working. On a number of occasions, students gave an answer with no indication that a correct Newton-Raphson formula was used and no evidence of substitution into their derivative. A number of students with accurate solutions failed to give the final answer to 3 decimal places as required in the question.

Question 3

Almost all students achieved full marks in parts (a) and (c) and labelling was clear on most Argand diagrams. Generally part (b) was also done well with many students obtaining full marks. Students who approached this question by attempting to write the cubic as a product of three factors and expand had the most success, with only a small number failing to do so correctly. Some students used the sum and product of the complex roots to find the quadratic

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factor. Of those who chose to substitute the root into the original equation students who substituted the real root, x = 2, often made no further progress. Those who did substitute two of the roots into the equation generally did so correctly, however they then failed to separate out into real and imaginary parts to solve for p and q. Some students opted to use long division for part (b), but this method was almost always unsuccessful.

Question 4

In part (i)(a) there were a lot of correct answers with almost all students gaining at least one mark, although there were some who made two arithmetic mistakes which lost both accuracy marks. A few students failed to get the correct dimensional order of the answer.

In part (b) explanations for why AB and BA are not equal were many and varied and often correct. There were lots of references to transformations and matrix multiplication not being commutative, but these attempts were not usually sufficient for this mark. The most common approach was a discussion about the dimensions of the resulting matrices.

Part (c) was generally very well done, with the majority of answers gaining full marks. There were some errors in the calculation of the determinant and also some errors in the positions and signs of the elements within the inverse matrix. A significant number of students took the time to either factorise the determinant or to multiply the determinant into the matrix, neither of which were necessary to get all the available marks.

Question 5

Most students made a good start to this question and correctly expanded the quadratic expression. The majority of students also used the standard results for and

correctly. Very few students failed to appreciate that , and those who did then tried to falsely manipulate the algebra to achieve the printed result. Those who correctly used

were usually successful in obtaining the printed results. Algebraic manipulation was

excellent in a lot of cases and students generally attempted to factor out early on in their

solution.

Part (b) was slightly less well answered although many students realised they needed to use 4n and 2n, but poor substitution and use of brackets led to incorrect answers, losing the last mark.

Question 6

Students used a variety of methods to find the gradient and then the equation of the normal to obtain the given result in part (a). Most students wrote y in terms of x and were able to differentiate and find the gradient in terms of t correctly. Some students used implicit differentiation or the chain rule with the parametric equations. All these methods were usually very well done. Some students did not show their method for finding the gradient and lost marks in part (a).

In part (b) the vast majority were able to find the coordinates where the tangent and normal

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met the x-axis by using the given equations although there were a number of students who incorrectly used the tangent for A and the normal for B. A few students found the intersection of both lines with the y-axis and unfortunately lost marks in part (b) and part (c) for this misinterpretation of the question.

In part (c) the majority of students made a good attempt to subtract the x-coordinates of A and B. For some attempts this resulted in a negative length and consequently a negative area. Full marks were available however for those who went on to give the positive value for the area. There were algebraic manipulation errors and sign errors seen in arriving at an answer which was unfortunate.

Question 7

Many students in part (i) could not find the matrices securely to represent the basic transformations in part (a) and part (b). The correct order of composite transformation in part (c) was poorly understood with many students attempting to multiply in the wrong order. Some students in part (b) and part (c) left the matrix in terms of the trigonometrical ratios.

In part (ii) the majority of students successfully calculated the value of the determinant. Of those who went on to calculate an area for the triangle many used 364 and their determinant and were successful in finding a value for k. Those who attempted to find the area of the transformed triangle in terms of k usually made no further progress.

Question 8

Many students approached this question well. In part (a) they were able to find the gradient in terms of k and hence go on to obtain the given equation of the straight line. A small minority tried to use calculus to find the gradient and paid heavily for this incorrect interpretation of the question. Some responses left the gradient in an unsimplified form which complicated the calculation of the equation of the line. Some of the algebraic manipulation seen in this question was very inefficient with a lot of students multiplying by the gradient in its fractional form rather than removing fractions early on.

In part (b) the majority found the perpendicular gradient, stated the focus and the directrix and used them to find the equation of the line. Some students were careless with signs and algebra when expanding brackets.

Question 9

Students typically showed an understanding of the principal of mathematical induction and were able to obtain the first two marks in this question. Almost all students showed the function was divisible by 6 for k = 1.

The majority of students then considered f(k + 1) – mf(k). Those who used m = 1 or m = 2 were the most successful. In a significant number of cases the students falsely, or incorrectly, manipulated the algebra to write their expression as a multiple of 6. Many got stuck after forming f(k + 1) – f(k), only gaining the first two marks. A significant minority of students lost accuracy marks by not explicitly expressing f(k + 1) explicitly as a multiple of 6. The majority of those who correctly manipulated the algebra, in general, followed through to produce a complete solution.

30

Further Pure Mathematics 2 (6668)

General 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. It was disappointing to see otherwise good students make basic errors when using mathematics learnt in earlier modules, for example trigonometric identities.

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 such as the series expansion in question 3 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.

Question 1

This was a straightforward testing of the method of differences to sum a simple finite series. The large majority of students scored full marks on this question. The method of differences seemed to be well practised and familiar to most students. The terms of the series were usually clearly listed although some students listed terms from r = 0 or beyond r = n. Most students could identify the remaining four fractions that do not cancel. Some silly mistakes in working towards a single fraction were made and some students attempted the leap to the final given correct answer too quickly without showing the required working. It is important for students to remember the need to show all steps when an answer is given.

Question 2

The large majority tackled this inequalities question successfully with most students scoring at least four of the possible six marks. Several included a graph to illustrate their thinking. Those who did so could see that there were more than two intersection points. Most students identified the four critical values correctly and a small minority failed to combine them in the correct pair of inequalities. It was noted however that some students tried to find a solution by simply considering 3x2 – 19x + 20 on its own and getting x = 5 and x = as critical values.

There were a few cases where a quartic was obtained by squaring both sides of the given inequality. A number of those students who attempted this method were successful in identifying their four linear factors.

The large majority of students understood that algebra had to be used in this question and avoided using the calculator simply to give them a quick answer.

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Question 3

This was a straightforward series expansion using Maclaurin’s method. The large majority of students recognised that they had to differentiate twice, the second time using the product rule. This and the subsequent substitution were on the whole carried out correctly.

A small number of students attempted to use the series expansion of ex = 1 + x + + … .

Those who tried were, on the whole, successful but not those who tried 2√2

.

Question 4

Again this was a classic textbook example, this time a test for the de Moivre’s theorem method. Being a ‘show that’ question, detailed working was required to gain full marks and this was seen from most students.

Most students answered the first part of this question successfully, navigating their way through real and imaginary parts and the expansion of brackets involving (1 – cos2 ). Some students used (cos2  – 1) by mistake compromising the last two marks of part (a) .It was surprising to see that having expanded (c + is)6 using the binomial expansion, most students then expanded (1 – cos2 )3 ‘long hand’, often getting into a bit of a muddle when doing so.

Some students began by expanding without giving any indication of what z might be

representing. Most of those who did replace zn + with 2 cos n were successful in

obtaining the desired result after some further work.

Most students reached the correct equation cos 6 = and solved this correctly. Those who ended up with the wrong equation (for example cos 6 = 1 or cos 6 = 0) could still gain two of the available five marks. Students were confident proceeding to obtain three correct solutions in the required range although a surprising number left their final answer with only one or two angles.

Question 5

This was a classic textbook example of a second order differential equation. A large number of students tackled this question successfully. Some preferred to give the solution in terms of Ae(–1 + 3i)x + Be(–1 – 3i)x rather than e–x (A cos 3x + B sin 3x). Some mistakes occurred in finding the Particular Integral by not using y = ke–x and finding k. A small number of students lost the final accuracy mark in both part (a) and part (b) by not writing their solutions with “y =”.

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Question 6

Few students managed to score full marks on this question in spite of the generous mark scheme. Several persisted in substituting x + iy for z rather than x + ix which involved complicated algebraic manipulations. Once x = y was spotted most students proceeded to achieve the right answer for part (a). However, some students attempted to multiply through by a ‘conjugate’ involving z, not realising that there was still an imaginary element to that conjugate, before making any substitution.

Part (b) proved more difficult if the correct answer to part (a) had not been reached. Indeed, very few attempted the method intended in the scheme. The question was clearly a different twist on what the students expected and they did not know how to adapt. Some students realised the importance of identifying the real and imaginary parts and scored the first method mark of part (b). Some proceeded by substituting into (u – 3)2 + v2; others by eliminating x to obtain an equation in terms of u and v which in the hands of some was simplified to that of a circle.

Some students ignored the instruction “hence” in part (b) and started afresh by finding z in terms of w, earning no marks for their efforts.

Question 7

Most students started this differential equation question by using . Since the correct answer had been given, most students navigated their way safely and correctly to it. There was a small number of students who started with the given answer and ended showing that the given differential equation was correct. Common errors included writing y = v3

instead of y = . Some differentiated to give rather than .

In part (b) most students found the integrating factor correctly and proceeded to find a general solution in the form Most mistakes occurred when students did not

realise that they had to find for that vital method mark.

Since the remaining marks depended on this mark, several students lost the last four marks for part (b). The absence of the integrating constant lost some students the last two marks. Another silly mistake that examiners were surprised to see made by Further Mathematics

students was using the reciprocal incorrectly as = becomes y3 = . The

relatively straightforward algebraic manipulation at the end of this question to get the solution in the required form seemed one challenge too many.

Question 8

The polar coordinates question that concluded the paper was a straightforward one. Very few blank pages were seen for this question suggesting that the length of the paper was right for most of the students.

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This question proved a valuable source of marks for most students. They demonstrated a

sound understanding that they needed to start with x and proceed to . Some students

complicated the differentiation of x = (1 + tan ) cos by not noticing that this was in fact . At this level it was surprising to see students treating by

squaring both sides first. A small number forgot to find their value of r having correctly identified the value of . The relatively few students who used and proceeded to differentiate ended up with no marks.

Part (b) involved a simple substitution of . A sizeable majority of students spotted this and proceeded to integrate and substitute limits successfully. Those who did not or opted for more exotic identities for and subsequent complicated integration ended up losing the vital last marks as these were dependent on that second method mark of part (b). No instances of slipping into decimals were seen when moving towards the final answer.

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Further Pure Mathematics FP3 (6669)

General 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.

Report on Individual Questions

Question 1

In part (a), the vast majority of the students correctly identified the direction vector of the line and thus confidently formed the vector equation of the line, usually in the form .

In part (b), the substitution of the line, once expressed in parametric form, into the given equation of the plane and a solution determined for , giving the required point, was well known by many of the students and most scored full marks in this part.

However, part (c), a significant minority did not use this point to determine the perpendicular distance of P from the plane as requested, with many resorting to using the result given in the formula book, which scored no marks.

Question 2

In part (a), nearly all of the students correctly attempted one of the many ways to test for orthogonality with the use of being the most popular method. It is important that in questions such as this, the students make a conclusion based upon the test being used.

In part (b), the vast majority of the students correctly formed the characteristic equation and thus found the three corresponding eigenvalues of the matrix .

Most of the students were then able, part (c), to use the equation to set up equations in, and and solve to find the required eigenvector. However, = 0 was a very common

incorrect solution for x and students clearly need to be aware that any eigenvectors found must be non-zero.

In part (d), the majority of students were able to convert the given line equation into parametric form and then pre-multiply by the matrix . However, a significant minority of the students assumed that the direction vector for the line had a zero i component and thus lost the two accuracy marks for this part of the question.

Question 3

In part (a), the vast majority of the students recognised the need to express the quadratic expression in completed square form and then use a standard integral given in the formula book and there were many correct solutions seen here.

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In part (b), most of the students substituted the exponential definition of sinh into the integrand and, on expansion, were able to complete the integration. Less success was achieved by those students who used integration by parts and with parts having to be used twice, very few students gained full marks using this particular method of integration.

Question 4

In part (a), the majority of the students began with the right-hand side and used the exponential definition of tanh and basic algebraic processing to successfully produce the left-hand side. The solution in which both sides of the given identity were processed to an equivalent form was occasionally seen and often produced full marks for this part. A significant minority used only hyperbolic identities rather than the requested exponential definitions to prove the identity and such a solution earned no marks.

In part (b), a large number of students were achieving full marks and such students were well rehearsed in this type of solution. A number of the students, in wishing to use hyperbolic identities, initially squared a rearrangement of the given equation to obtain a quadratic equation in terms of either sinh , cosh or tanh . However, the students using this method often went on to find values for x without using any exponential functions and thus gained no marks.

Question 5

The vast majority of the students identified the given function as being composite and applied

the chain rule in determining . Most of the students used the product rule to differentiate

and the algebraic processing of the terms involved within both this and the quotient

rule, when used, was generally of a good standard. Nearly all of the students correctly differentiated artanh but the final stages of the solution caused some problems and a number of errors were seen in processing the terms down to the printed result. There were a number of students who initially rearranged the given function to make tanh y the subject and then used implicit differentiation and achieved equal success with those students who used the chain rule.

Question 6

In part (a), the majority of the students formed an equation of the chord, scoring the first three marks, but then got lost in a mass of algebra, made worse by the use of the unfamiliar trigonometric factor formulae and demanding trigonometric identity work and the final mark in this part was thus very rarely gained.

In part (b), most students correctly formed the coordinates of the mid-point although there were some who subtracted the coordinates or who forgot to divide by two.

In part (c), the vast majority made very little or no progress at all and clearly did not appreciate that the equation of the locus of the chord’s mid-point needed to be found. Those

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that understood this and hence found or , managed to make good progress with this part.

However, it was relatively rare to see a completely correct solution to part (c).

Question 7

In part (a), nearly all of the students found an expression for , with most using implicit

differentiation to do so, from which they obtained the printed result. Of those using this method, the most common error was to differentiate to get .

In part (b), the vast majority of students were familiar with the formula for finding the area of a surface of revolution, as given in the formula book. However, there were a number of errors made with each of, the substitution for the function , the limits of the integration and the actual integration itself. A significant number of students incorrectly used limits of 0 and

and integrated to give .

In part (c) the relevance of the circle C was recognised by some students who could then write down the required arc length. The majority of the students, however, used both the answer part (a) and the formula for arc length given in the formula book to attempt the arc length, with varying degrees of success.

Question 8

In part (a), the majority of the students used, with much success, the cross product of two sides of the triangle to find its area although a significant minority incorrectly used two of the given vectors a, b and c and thus incorrectly used two sides of the triangle or .

In part (b), a great deal of success was achieved with the use of a 3 × 3 determinant often being used. part (c), a number of students were able to make a correct deduction about the vectors a, b and c but many showed an inability to recognise the relevance of the result part (b) and often wrote down contradictory statements about these vectors.

Question 9

Many of the students found part (a) the most challenging part of the paper. For those that made an attempt, nearly all of them used integration by parts although there were a significant number of students who, in using this method, used incorrect expressions for

and and hence made very little progress in their solution. For those students who initially

made the correct choice of labels for and , many then incorrectly went on to apply parts

for a second time but more often than not, the solution was abandoned at this stage. Indeed, very few of the cohort achieved full marks for this part of the question.

In part (b), the In + 1 term in the given reduction formula caused confusion for some students, who are used to seeing an In term on the left hand side and the incorrect substitution of n = 2

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was a very common error. There were some students who, having substituted in the correct value of n, could not deal with the integration required to find I1 and failed to recognise that it is given in the formula book.

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Mechanics M1 (6667)

General 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 very challenging, in particular questions 5(d) and 6. However, there were some parts of all questions which were accessible to the majority. The questions on equilibrium and use of constant acceleration formulae were generally well understood and full marks for these questions were commonly seen. The paper discriminated well at all levels including at the top end, and 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.

Question 1

Overall this was a well-answered question and got the students off to a nice start. Most attempted to resolve horizontally and vertically. A few students attempted to resolve along the string but were usually unsuccessful. The only common error here was sine and cosine interchange. A small number attempted to use the sine rule, but the angles were usually wrong, often just using the angles shown in the diagram, rather than using an appropriate vector triangle. Lami’s theorem was very rare, but generally done correctly.

Students who answered part (a) correctly often lost marks in part (b) due to premature rounding of their tension. The only other common error was to use Wg instead of W in part (b). A small number assumed that the angle at C was 90 and tried to resolve along the strings.

Question 2

There were many fully correct solutions to this question. The majority resolved correctly, although a small number did mix up sine and cosine, and some did omit one term from their equation of motion along the plane. Many wrote equations that initially equated weight and friction, but on realising that they needed an acceleration started again to give a correct solution. Omission of was sometimes seen or was included in the term. However, almost everybody included , which gained them at least one mark.

Almost everybody was able to gain at least the method mark in part (b). There seemed to be less of an issue with over accuracy than seen in previous sessions, so the message appears to be getting through. As in question 1, there was an issue with premature rounding of the

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answer to part (a), leading to an inaccurate answer of 2.54 in part (b) if given to 3 significant figures. Some also forgot to square root at the end.

Question 3

The question was notable for the rarity of an entirely correct solution. The vast majority of students produced decimal answers which were in the main rounded to 3 significant figures. It should be noted that, for questions where g is involved, students would be better served if they consistently wrote numerical answers in an exact form in terms of g or to 2 significant figures. Some students are still using 9.81 for g and at least one took g as 10.

In part (a), some assumed that the velocity on leaving the ground was the impact velocity reversed, failing to realise that this made a fall of 2.0 m and a rise of 1.5 m untenable. The responses to this part were fairly evenly split between correct and incorrect answers. In the second part, merely quoting a formula scored no marks but using with various sign combinations for non-zero and earned the method mark but many failed to earn the accuracy mark due to sign confusion. A few students did not appreciate that “magnitude” demanded a positive answer.

In contrast, part (c) was most often incorrect, with students failing to deal correctly with the opposing signs of the before and after velocities and so gaining only the method mark, or using which scored no marks.

There were very few correct graphs part (d) with the vast majority of students scoring one mark out of three. Many students did not have a negative velocity at any point and a number of students who had the correct shape for the graph lost the second mark because they included a continuous vertical line. Correctly-shaped graphs were seldom fully labelled. In the final part, students were able to use suvat equations to find the time to reach the ground and the time to rise to 1.5 m but not all proceeded to finding . Many of those who did lose the final accuracy mark (1.74 instead of 1.75) due to premature approximation.

Question 4 The first part of this question proved to be fairly straight forward compared to previous moments questions and many scored 6 marks for not very much work if they chose the best method. Having said that, many students made their life more difficult than necessary by not taking the easy resolving option and using two moments equations. Many took moments about B and C, when A was clearly the most obvious choice. The resulting simultaneous equations sometimes proved too difficult to solve. There were very few cases of students mixing up the tensions.

Part (b) was generally more demanding, although again full marks were scored by many. The most common mistakes were to use just rather than for the load, to use and and to assume that the tensions from part (a) still applied. The additional force inevitably meant that more students got lost in the algebra and failed to get to a correct final answer, but the question still gave plenty of scope to score well. The multitude of different possible solutions meant that this was the most difficult question to mark by far and this would have been helped by students making it clear what each equation was referring to and making the multiplication by a distance explicit in moments equations, even when the distance was 1.

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Question 5

In part (a) almost all students were able to use to find the correct vector for the acceleration and then its magnitude or to find the magnitude of the force and then apply

to find the magnitude of the acceleration. As this was a ‘show that’ question, students needed to earn their marks and show sufficient stages in their argument and all too frequently, stages were omitted resulting in loss of marks. In the second part several students scored no marks for using the force rather than acceleration vector and some contrived to use the scalar magnitude combined with an initial vector velocity which also lost all the marks.

Part (c) was probably the most successful for students but a number left their answer as a vector and so gained no marks. In the final part, many students calculated

to gain the first two marks. Few chose the simplest

explanation of motion being in the same direction by factorising and there were many responses referring to either the bearing or gradient of both velocities. A few students started with the parallel idea and found the time to be unique at .

Question 6

Part (a) required the setting up an equation to relate the magnitudes of two forces with the magnitude of their resultant and then part (b) involved finding the magnitude of their vector difference.

In the first part there were two possible approaches. The cosine (or sine) rule could be applied to the triangle of forces, or the forces could be resolved and the components squared and added. Both methods seemed almost equally popular; however, there were a significant number of students who were unable to identify any valid strategy. There was confusion evident between vectors and their magnitudes and a lot of working was seen which was either crossed out or difficult to follow. A fairly common error was to use the wrong triangle (with 120 angle rather than 60). Occasionally the cosine rule was quoted incorrectly (or Pythagoras was attempted), and errors in squaring and simplifying were not uncommon. Some students identified the components correctly but failed to make further progress by not squaring and adding. Both methods resulted in a quadratic equation in X. It should be remembered that if a calculator is used to solve it then full credit can be achieved if the equation and answer(s) are correct. However, any error will lose the method as well as the accuracy mark. The final answer (5.93) was required to 3 significant figures, as stated in the question.

Again, in part (b), either the cosine rule or use of components were valid approaches for determining the magnitude of the vector difference. Those who found X successfully in the first part sometimes gained no more credit by just subtracting the magnitudes of their vectors. On the other hand, some who made no valid progress in part (a) used their answer correctly to achieve 3 out of the possible 4 marks. A fairly common error was to use X – 20 cos 60 rather than X + 20 cos 60 as a component, or the wrong triangle with the cosine rule.Full marks for this question were rare with a significant number of students achieving little or no credit.

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Question 7

This question was well answered by the majority of students. In part (a) most identified correctly individual equations of motion for the two masses and then solved them simultaneously to find the acceleration. Since the answer was given, any potential sign errors tended to be rectified but occasionally the answer did not strictly follow from the working. Sometimes the values ‘3’ and ‘4’ were used rather than ‘ ’ and ‘ ’ as given in the question. This was penalised as accuracy errors here, but all subsequent marks for the rest of the question were available. The most common error in finding the tension was to omit ‘ ’ in the final answer despite it being included in the working.

In the second part virtually all students found the velocity correctly by using ; the only significant error seen was in using ‘ ’ rather than ‘ ’ showing a lack of understanding of the situation.

Part (c) required a similar approach to part (a) but with one different mass. Since the answer was not given this time, there were some arithmetic and sign errors, but generally it was well done. Those who used a value of the tension from part (a) achieved no credit, as did those who tried to somehow use constant acceleration formulae.

The majority of students used an appropriate constant acceleration formula in the final part to find the maximum height reached, using the values of velocity and acceleration from previous parts of the question. Occasionally ‘ ’ or ‘ ’ were used, again showing a lack of understanding of the mechanics. Most, but not all, added ‘0.7’ from the initial part of the motion to reach the final answer as required.

Full marks for this question were often achieved and much good working was seen.

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Mechanics M2 (6678)

Introduction

This paper offered all students an opportunity to demonstrate their knowledge and understanding. Much of the work seen was completed to a very good standard, but the standard of presentation varied considerably. There were many well laid-out answers in clear writing, but there were many that were difficult to read - students need to make a clear distinction between the figures 4 and 9, and their characters 3, 5 and 8 are often difficult to distinguish. On some occasions, even the students are confused by their own writing and they miscopy figures from one step to the next. Illegible work will not gain any credit. The best solutions were accompanied by clear, accurately labelled, diagrams.

Students need to be reminded to read the rubric and the questions very carefully. In all cases, where a value for g is substituted, the value should be 9.8 m . The use of 9.81 will be penalised as an accuracy error. The rubric on the paper gives students a very clear reminder about the accuracy expected after the use of 9.8, but many students 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. As part of the checking process, the students should make sure that they have actually found what the question asked for at each stage.

Question 1

The great majority of students found this question accessible and they were quick to analyse the system and create appropriate moments equations. The "obvious" route of taking moments about the x and y axes was followed by most. Occasionally moments were taken about axes through the centre of mass. Some students worked with a vector equation, but the majority favoured two separate equations. Most errors were due to slips in the algebra and arithmetic, of which the most common was .

Question 2

This question was answered well with a significant number of students gaining full marks. The majority of students integrated correctly, with only a small number forgetting the . Very few students made calculation errors, but there was some careless transcription where an index or sign changed before the final answer was given. part (b) most students equated the component of the velocity to zero and were able to solve the resulting equation. The most common error was to give the final answer as a vector when the question asks for the speed.

Question 3

Part (a) was usually answered well and there were many fully correct answers. Many successful students used a structured layout with the use of a table. Most considered the difference between two triangles, although some students divided their shape in to triangles and parallelograms. The latter approach resulted in lengthy working that was rarely successful.

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The most common errors involved incorrect mass ratios, and the addition rather than subtraction of the two triangles. Several students did not have the centres of mass of the triangles in the correct place, often because they were working from the base of the triangle rather than from the vertex. Some students incurred accuracy errors because the was in some but not all stages of their working.

In part (b) clearly labelled diagrams helped students to identify the required angle, and many did this correctly. Those with an incorrect answer in part (a) often earned the method marks here. Some students found the angle between AC and the vertical but did not go on to find the angle between AB and the vertical. A few assumed that the triangle ABC was equilateral, and lost the last two marks.

Students lost time by finding the coordinate of the centre of mass rather than using the symmetry. Use of the cosine rule tended to work well but this was not a common method. Students need to be reminded to read the questions carefully; several lost the final mark in this question because they did not give their answer "to the nearest degree".

Question 4

Many students gave fully correct answers to this question.

In part (a) the majority of students treated the truck and trailer as a ‘single system’ from the start but some considered the equations of motion for the truck and trailer separately and solved their two equations simultaneously to derive the equation of motion for the complete system. They went on to use correctly to find the speed. A common error was to give the final answer as 22.55 m either forgetting or not realising that having used the approximation a maximum of 3 significant figures would be accepted.

In part (b) most students realised that a new equation of motion for the truck was now required. Common errors were to subtract the wrong resistance from the driving force or to omit the weight component. Some students did not realise that the driving force on the truck would be unchanged at this instant and recalculated it, often incorrectly.

Question 5

In part (a) many students recognised that the speed of the particle after the first impact would

be , but when using this to find the impulse on the particle, the majority of students did

not take account of the change in direction and obtained an answer of rather than .

Some of the students with a correct statement of impulse gave a negative value for when the question had asked about the magnitude of the impulse, so a positive answer was expected.

In part (b) most students obtained the correct speed, , of the particle for the third stage of

the motion. Many students then averaged the velocities, without recognising the different durations of the journeys. Some were under the misconception that there were accelerations rather than instantaneous changes to new constant velocities. Some students appeared to be

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confused about the relationship between speed. distance and time. For those using the correct method, rearranging the fractions to make the subject proved difficult, either because they

could not simplify or because they made errors in adding their fractions.

Question 6

This question proved a challenge for many students, which was evident from the number who produced more than one solution. Without a method being suggested in the question, the majority of students opted for a suvat approach rather than the more elegant and brief energy approach.

Using suvat with no specific angle of projection given, many chose to either ignore direction, incorrectly applying in the form , or they used the given value for sin α from part (b) and in both cases scored no marks. There were, however, some correct suvat solutions where students derived the vertical component of velocity at B in terms of

, used Pythagoras’ theorem, and then applied to obtain correctly. These solutions tended to be well thought out and easy to follow.

There were those who separated the motion into two parts, from A to highest point then from the highest point to B. This rarely succeeded, with the many equations causing confusion.

Having struggled with part part (a), some students did not attempt part (b). However, many who had not been successful in part (a) realised that there was an opportunity to make progress here with a more familiar problem now that the value of was given.

Many worked correctly with the vertical motion to establish a quadratic in the time to reach and went on to use this time correctly in an expression for the horizontal distance. Students

should be reminded that it is sensible to show how they have solved the quadratic equation and not just write down the solutions from a calculator; with incorrect answers and no method shown no credit will be given.

In contrast to part (a), a few successfully found the time in an over-complicated way, by finding the time taken to get to the highest point and then the time taken to reach B from there.

A common error was to assume that the direction of motion of the ball at was the same as the initial angle of projection.

Question 7

In part (a) the stronger students identified the two equations independent of the horizontal force at A very quickly, and they had little difficulty in obtaining the required result. A few students chose alternative points to take moments about, although generally these attempts were not fruitful because they had introduced another variable and could not see how to eliminate it. Another cause of difficulties was the poor use of variables, for example using both as a force and a length, and then confusing the two.

In part (b) the first mark was achieved by the majority of students. A handful of students ignored the and assumed that the resultant force at should act along the rod. In

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general, students who answered part (a) were able to correctly resolve to find in terms of .

Some students who did not score many marks in part (a) nevertheless managed to answer part (b) successfully. Most students who had found the horizontal and vertical components found the resultant correctly.

Question 8

In part (a) the vast majority of students followed the instructions and attempted to use the work-energy principle to find the work done against friction. This was a very well answered question with only the weakest students making errors in the initial equation. The most common error was leaving the final answer as 84.58 J, which is inappropriate accuracy following the use of . Solutions that did not use the work-energy principle received no credit.

In part (b) there was no restriction on the method to be used but many students continued to use energy, rather than adopting the alternative suvat approach. The vast majority found R , used this to find the maximum friction, equated this to their value for F from part (a) and solved for . Accuracy marks were sometimes lost in part (b) when using rounded answers from part (a).

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Mechanics M3 (6679)

General Introduction

This was found to be an accessible paper for most students. A few stopped mid-calculation towards the end of question 7 but others had time to re-do questions as a check.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.

Question 1

Most students made a good attempt at an equation of motion towards C with a correct expression for the acceleration and followed this with a correct vertical resolution to give the required second equation. Many good clear diagrams were seen and these were often part of a full solution using trigonometry correctly to eliminate the angle from the pair of equations and find OC. Those who left their equation of motion in terms of the generic “ ” for the radius of the circular motion often incorrectly cancelled with 4r, the given radius of the hemisphere. Occasionally attempts were made to use an equation of motion towards O, the centre of the plane face of the hemisphere.

Question 2

The majority of the solutions to the first part of the question were sound with very little incorrect work seen. However, marks were lost by those who did not make it clear that they were working on the surface of the earth. Nearly all students employed the integration method using the equation of motion to solve part (b) and could cope with the integration. Very few students omitted a constant of integration. The mistake that prevented correct

solutions was usually the use of rather than when obtaining a value for the constant.

Question 3

In part (a) fully correct answers were not frequently seen. Despite being given the masses in the question, some students tried to find expressions for the volume of the wax and the surface area of the shell in order to determine the ratio of masses. Many students were able to use the formula book to find correct distances to the individual centres of mass but often did not give them from a common point in their tables or equations. It seemed that some students did not realise they could use the formula book and simply guessed the distances. The majority of students who had the correct masses and distances went on to gain the required answer.

In part (b) they needed to use the distance from the centre of the plane face and again, those that had a diagram for this part often also had the correct solution. As ever, some students used the tangent ratio the wrong way up.

Question 4

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Most students knew the correct method but a substantial minority used equilibrium equations with Hooke’s Law for both parts, meaning that the only mark available was for finding the reaction in part (b). The EPE formula was well known and rarely seen incorrect when it was used. A few wrote an equation in words featuring the word “Work” but didn’t include a distance. Their equation was therefore dimensionally incorrect and lost most of the marks.

Most errors arose from using a wrong mass (m instead of 2m) either in the GPE term (both parts) or in the reaction in part (b). Consistent use of throughout the question which would have been classed as a misread was rare. A relatively common, and quite surprising, error, was using R cos = 2mg in part (b), wrongly assuming vertical equilibrium. Students should show this equation for R as a separate statement. This makes it much easier to award follow through marks when they are available. Very few attempted an equation of motion for either parts (a) or (b) but those who did generally used the full correct method and used the same method for both parts.

Question 5

Part (a) was generally answered well. Most students knew that they should use the double angle formula for the integration and did so successfully. The substitution in of the limits was sometimes not sufficiently explicit for a ‘show that’ question, but there was nearly always some attempt at it. Most seemed to know what was required in part (b), with very few trying to work with a lamina. The responses were fairly evenly split between those who used the

double angle and separated to integrate on its own, and those that realised that they could

take as and use the integral from part (a). Most students knew that they needed to

use the double angle formula to prepare and for integration. There were some

errors with this usually where students lost the often because of the way they set out

their working. Typically they would use the double angle formula and multiply out the

bracket, writing down correctly, then go on to separately calculate

and forget about the when they went back to put all their separate bits

together. Most realised that they needed to do the final division, but inevitably some s were omitted, probably due more to carelessness than lack of understanding.

Question 6

In part (a) nearly everyone knew essentially what they needed to do, but the requirement to produce an inequality led many students to make fundamental errors which lost them many marks. Almost everybody knew that they needed to produce an energy “equation” and resolve to the centre at some point. This was mostly done at the top, but a significant number took both equations at the general position. Whilst more complicated, this was almost always done accurately, probably because this had been thoroughly learnt. Those students nearly always went on to use and had the added bonus just being able to slot in the necessary angles in part (b).

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The big problem with part (a) arose when trying to get an inequality. While the majority knew that they needed to make T 0, this was not always done correctly and a minority used v 0.The answer was given, meaning that directions were sometimes fudged at the end. A significant number set T = 0 (either explicitly, or by complete omission) and then vaguely attempted to justify the inequality (or not) at the end. The worst (and fairly common) mistake was to insert the inequality directly into the energy equation. This led to the required answer, but lost several marks.

Part (b) was generally answered better than part (a) and many students that lost most of the marks in part (a) managed to pick up marks here. The principle at work seemed to be understood by most, but sign errors often led to one incorrect tension. Some students mixed up kT and and failed to refer back to the question which told them that kT was the greatest tension and the least.

Question 7

Part (a) was almost always correct. In parts part (b) to (d), the difference between the best and the worst attempts was enormous. There were a lot of extremely competent, fully correct solutions but equally many failed to prove SHM fully. The common mistakes were using instead of not considering directions and apparently not realising that the negative is essential ignoring the weight, using Hooke's Law in T = ma with extension x and thinking they had proved it. Dimension errors caused by using the additional extension as xl finding the acceleration at a specific point (usually the lowest).

Students should be encouraged to write their equation of motion clearly as a first stage in part (b). It is quite common to see solutions which start with “ ...... “and then spend several lines of working fiddling around with the forces before eventually getting to “ ”. They should know that no marks will be available until they write an equation. On the other hand, those who did this usually had the correct signs and notation when they finally produced an equation. Several omitted the conclusion (they had been asked to prove SHM and need to say that they have done so), a very easy mark for just saying “therefore it is SHM”. Many failed to realise the importance of using the given letters throughout the question; was often dropped from both ω and the amplitude leading to a significant mark loss.

Part (c) was not infrequently done using an equation of motion at the lowest point (a repetition of the proof in part (b) for some). This method proved beneficial to those who had a dimensionally incorrect .

The times in part (d) were found perfectly by many good students but it was often impossible to follow the intended reasoning of less good attempts. Clear statements about which portion of the motion is being considered would not only make it easier to give credit for correct

reasoning but might also help them to avoid mistakes. The solution was

much less common than the one; in both cases it was surprising that the

fraction of the period was found far more often using a displacement than a fraction of T.

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Mechanics 4 (6680)

General Introduction

The majority of the students for this paper showed a good command of the whole of the specification - they offered responses to all of the questions, and much of the work was completed to a high standard. However, there was a minority of students for whom work at this level was either not sufficiently well prepared or too demanding.

The best work was clearly set out, and accompanied by clearly labelled diagrams. Students should be reminded of the need to make their work clear to the examiners; some handwriting is so small that it is difficult to read. It would be helpful if students took more care in writing figures; there needs to be a distinction between 4 and 9, and it is common to see students miscopying their own 3, 5 and 8.

When an expression or equation is given in a question, the students should be advised not to overrule the examiners by using their own alternative version. No marks will be earned for solving the wrong equation.

Students need to be reminded to read the rubric and the questions very 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 students a very clear reminder about the accuracy expected after the use of 9.8, but many students lose marks for giving too many significant figures in their final answers.

Report on Individual Questions

Question 1

Most students made good progress with this question. Both differentiation of the square of the distance and use of the scalar product were popular methods. Unnecessarily taking the square root before differentiating created problems for some students.

Some students tried to reverse the problem by using the given answer to find the time at which this occurred; this approach usually failed to show that a minimum distance had been obtained.

A minority of students wrote down the position vectors of the particles but made no further valid progress. Some attempted a scalar product of vectors which were not relevant in this context, and occasionally only the initial distance between the particles was calculated.

Question 2

In part (a) virtually all students produced an equation of motion and deduced the value of the acceleration correctly. However, since the answer was given it was important that the substitution of v = 20 was actually seen in order to earn full marks.

In part (b) most students produced a correct differential equation in terms of v and x, and separated the variables correctly. Several entirely correct solutions were seen, but many

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students could make little valid progress with the integration of . Although there

were a variety of possible approaches (partial fractions, use of the arctanh or the logarithm formula from the formula book, or substitution) most required division or re-arrangement first. Some attempted partial fractions with the expression unchanged and some just wrote down a logarithm term. Those who adopted a valid method sometimes lost accuracy marks through sign errors or losing/gaining a zero in the numerical terms.

It was obvious that some students were using numerical integration on calculators (this is very risky). An incorrect final answer (which was common), with no working shown, will lose all the marks.

Question 3

Most students correctly identified the components of the velocity (before and after collision) perpendicular and parallel to the wall, and wrote down two equations relating them, usually in terms of trigonometric ratios of the two angles. With no guidance on what to do next, several students made no further progress. Among those students who did square and add their equations to obtain a correct ratio, there was some confusion about where the angle of deflection is - only the more able students knew that it lies between the continuation of the initial direction of motion and the final direction of motion.

Question 4

In part (a) most students drew an appropriate vector triangle and used the sine rule to find a relevant angle. The standard of diagrams presented was sometimes very poor; the diagrams were too small and the angles involved far from clear, leading to errors in subsequent calculations. Having found a relevant angle, some students made errors in calculating the bearing. Students should take note when a question specifies a level of accuracy – in this case they were asked for an answer to the nearest degree.

In part (b) many correct solutions were seen; the majority of students continued with their vector triangle. Students completing the work on several stages need to be aware of the risks of errors in the final answer through premature approximation.

Some chose to work in components throughout this question, either parallel or perpendicular to relative velocity, or ‘N’ and ‘E’; sometimes they were successful but some were not able to handle the resulting equations properly.

Question 5

In part (a) although the majority of students produced appropriate momentum and restitution equations, the directions of their velocities were often not well defined nor clearly shown on a diagram. This led to frequent sign errors. Mistakes in solving the resulting simultaneous equations were quite common. The definition of impulse was generally known but occasionally speeds rather components of velocity (or a mixture of the two) were used.

In part (b) there was confusion between speeds and components of velocity in the calculation of kinetic energy. Although it was not necessary to find the perpendicular components in finding the loss in energy, it was necessary to use the whole speed in the calculation of initial

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energy. Some found the final energy (rather than loss) as a fraction of the initial energy, and

occasionally the energies for both spheres were combined. Some students used in

place of .

Question 6

In part (a) a small number of students secured all three marks. Although almost all knew the terms could be derived from the equation of motion, few handled the signs of these correctly. The best way to demonstrate the result was to start with a clear diagram showing the directions of , and and the corresponding directions of the tension and the resistance. Some students obtained an equation of motion with different signs from those given, and a few persisted with the wrong signs in part (b).

In part (b) many students solved the differential equation correctly to find x in terms of a and t. Most errors were due to the use of incorrect initial conditions (commonly ), and in a few cases students had an incorrect solution for the auxiliary equation.

In part (c) most students realised that the string was slack when . Although a number of students correctly found that tan t = – , some substituted a negative value of to find the speed. Some students overlooked the fact that they had been asked for the speed, not the velocity.

Question 7

In part (a) since the expression for the potential energy was given, it was important that this was derived clearly and correctly. Almost all knew and applied the formula for elastic potential energy although not all could relate ‘ ’ to the angle defined in the diagram. It would make the working clearer if an initial statement of the position of the zero P.E. level were given. It would also help if students did not omit constants between lines of working and carried them through to the final line. Sometimes this part of the question was omitted or abandoned fairly quickly.

In part (b) most students differentiated correctly and set the derivative equal to zero to investigate the equilibrium positions. There were three possible angles: 0.72 and 0; some identified one or two of these, but often not all three. Many spurious angles were found which bore no physical relation to the problem. Again, most found the second derivative to identify the nature of the equilibrium positions. However, it was important that no terms were dropped when investigating the sign and that the actual values were stated in support of the conclusions.

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Mechanics 5 (6681)

General 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. There were some parts of all questions that were accessible to the majority.

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 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.

Question 1

The most popular way to approach this question was using a scalar product. A number of students had difficulty in obtaining the direction of the force. For many, their direction vector was a variant involving 15 and 8. A sketch showing the line would have helped these students. Others also tried to get the 4 into the direction vector for the force. The remainder of the process was usually successfully completed. The alternative technique was to find the angle between the force and but this often led to the same problems as finding the force’s vector.

Question 2

The approaches to this question separated into two variations. Those who chose the integrating factor method were generally successful, although a few forgot to change their constant term when they multiplied through by the term across at the end. Many of those who attempted the complementary function and particular integral method did not recognise the form that their particular integral should take. Those who used e–t rather than te–t were heavily penalised. Those who only used pt for the i component also lost a number of marks. Some students recognised their error when trying to evaluate their constants and adjusted their particular integral accordingly.

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Question 3

This question was successfully answered by the majority of students, with only a few demonstrating that they knew little about the concept. There were a number of students who did not realise that they needed to demonstrate that the sum of the forces was zero. There were also a few errors in calculating the vector products and occasionally the magnitude of the couple was not found.

Question 4

part (a) was answered successfully. Students who corrected the signs in their work when arriving at the wrong answer should make sure that the final version is consistent. The second part was shown to be a standard method and was usually completed without any problems. part (c) was more discriminating. Many students took some time to arrive at an expression for

in terms of , but if they did, they mostly realised that they should integrate this with

respect to . Other students arrived at an expression for , knew that they needed to get

distance involved and so changed to the v version of acceleration. What followed often

showed a disregard for identifying variables, since students often integrated with respect to x, expressions involving or or both. Another source of confusion was identifying which of m, M, t and T were variables and which were constants.

Question 5

This question proved to be a good source of marks for the majority of students. Part (a) was successfully completed by most, although there were some who tried to use the parallel axes theorem starting from the end of the rod, rather than from its centre of mass.

The second part was also largely successful.

In part (c), those students who used arrived quickly at the answer. The students who differentiated their answer to part (b) were sometimes unaware of which variable they were differentiating with respect to.

Those students who correctly interpreted part (d) generally answered it successfully. Drawing the rod in such a position that was obtuse tended to give rise to sign errors. A minority of students decided that the weight had to be perpendicular to the rod, rather than the force at P, and so they thought the rod should be horizontal.

Question 6

The majority of students were well versed in the proof required in part (a) and were mostly successful. In the second part, some students found the moment of inertia about the axis perpendicular to the lamina and then used the perpendicular axes theorem. Others used the formula for the moment of inertia about a diameter as given in the formula book. A few used integration with the limits 2a and a. All three methods were acceptable. Some worked with

the axis perpendicular to the lamina, arrived at the answer and then doubted the

correctness of the printed answer.

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They did, however, use the printed answer in part (c). Another problem often seen in this part was difficulty in relating the masses of the large and small circles to that of the actual lamina. Errors seen in part (c) included omitting one of the terms in the angular momentum equation

or the KE loss or both. Some students used for both of the terms of the final KE. A

number of students managed to miss out the in the expression for the initial KE.

55

Statistics S1 (6683)

General Introduction

The paper proved accessible to most students and apart from questions 2, 7 and 8 the vast majority were able to make some progress. Part of Q03, Q07 and PART proved to be quite discriminating but there were plenty of very good solutions seen on these questions too. The examiners reported that a number of students seemed unaware of the instruction to work to an “appropriate degree of accuracy”. If students are using a calculator they would be well advised to first write down all the values on their display before rounding to 3 significant figures or whatever other accuracy is required. We usually will accept an answer which rounds to 3sf and so if an error is made in rounding, but a more accurate version is seen, then this can be given the credit.

Comments on individual questions

Question 1

Most students had a thorough understanding of box-plots and skewness and there were many fully correct answers.

In part (a) the correct values were usually given but and were common incorrect answers where students failed to read the “leaves” from right to left .

In part (b) most students calculated the upper limit of 91 for outliers correctly and there were few errors in reading the scale and plotting 25, correctly. The upper whisker did cause some problems. An upper whisker stopping at 91 (the upper limit for outliers) or stopping at 75 (the next non-outlier in the data set) and then an outlier clearly shown by a cross or * at 99 was all that was required. Common errors were to draw both whiskers or to plot the outlier at 91 or simply to extend the whisker to 99 and indicate the outlier there too.

In part (c) most comments about the skewness of Penville were correct but some did not realize that they needed to comment on Greenslax too. The reasons were usually based on a comparison of the differences in the quartiles and were often correct although a number still think that means negative skewness. Some students elected to calculate the means (and wasted some time) to compare with the medians, which was accepted if the means were correct, but others tried an argument based on the mode failing to realize that the distributions were bimodal and therefore this approach was invalid.

Question 2

Although a quarter of the students scored full marks here there were plenty of others who struggled some due to a lack of understanding about coding and others due to poor algebraic manipulation. Most appreciated that coding affected the mean, although they often substituted 60.8 for rather than . Those who started by writing down 60.8 = 1.4x – 20 often managed to reach the correct answer but a number could not solve this equation for correctly. Those who did not write this down knew that they had to add 20 and divide by 1.4 but could not get the order of operations the correct way around. Many thought that coding did not affect the standard deviation and others simply repeated the calculation for the mean

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with 6.60 rather than 60.8. Some students were not clear about the difference between standard deviation and variance and included factors such as 1.42 or √1.4. A few students gave their final answer as 4.7 rather than 4.71 and lost the final mark.

Question 3

The calculations were generally carried out very well here but the final 3 parts, requiring the students to engage with the context, were answered less well.

Despite mentioning the expectation of accuracy of 3 significant figures in previous reports, a number of students lost a mark here as the only evaluated answer they gave was 0.96 rather than 0.962.

In part (b) most gave a comment about strong correlation or commented that the value of was close to 1 but some misinterpreted the question and explained why linear regression was a useful tool rather than justifying its use in this situation.

In part (c) a number gave an answer of 0.74 rather than 0.740 as required but this value was allowed for the final equation in part (d) which many obtained correctly. It was rare to see a candidate give their equation in terms of and rather than and .

Part (e) was not answered very well. Many confused the gradient with correlation and simply said that as the number of visitors increased so did the amount of money spent. Those who did get the idea of rate often failed to give the correct numerical values or they had visitors and money the wrong way around, for example “for every £1000 spent the number of visitors increased by 740” . The better answers identified that represented the amount of money spent per visitor and gave simple answers such as “each visitor spends £740”.

The most common error in part (f) was for students to simply substitute 2 500 000 into their equation but those who realized that 2500 should be substituted usually arrived at the correct answer and often went on to affirm the reliability of their estimate by pointing out that 2500 was within the range of the given data in part (g).

Casually stating “it is within the range of the data” in part part (g) did not secure the marks unless the students made it clear that their “it” referred to the number of visitors and not the amount of money spent.

Question 4

This question was answered very well with over 40% of the students securing all 9 marks. The tree diagram was answered very well with only occasional errors on the branches for broken or not broken biscuits, for example P(BJ) = rather than .

In part (b) and part (c) were usually correct although there were a number of transcription errors such as 0.335 instead of 0.0335 in part (c).

Part (d) was a slightly more challenging conditional probability and it caused difficulties for some students. Some misinterpreted it and simply found whilst others found

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and the usual mistake of having a numerator of P(K) × P(B) rather than P(K B) was not uncommon.

Question 5

This was another accessible question with only part (b) troubling some and over half of the students scored at least 9 of the 10 marks here.

In part (a) most chose to write down the sum of the probabilities in terms of and then set this equal to 1 and deduce the value of . Some chose a “verification” route but often failed

to give the final statement “therefore k = ” and lost the final mark. Many students still do

not recognize the cumulative distribution function F(x) and there were many blank or incorrect responses here. Some confused F(5) with P(X = 5) and gave the answer 0 and others

gave . Finding E(X) and E(X 2) caused few problems although some students ignored the

instruction to give the exact values and rounded their decimal answers.

In part (e) most knew how to find Var (X), although a few forgot to square E(X), and many knew how to deal with the Vat (3 – 4X) formula with fewer than usual trying 3 – 4 Var (X).

Question 6

Many students are still unsure about calculating the heights of bars in histograms but the work on linear interpolation and even standard deviation seems to be improving.

Most were able to state the correct width of the bar but few used frequency densities correctly to find the height, some finding the frequency density of but then calculating

. Some identified that represented 10 customers but were then

unable to use this correctly to find the height.

Part (b) was answered well but some students had an incorrect class width because they did not realize that the lower class boundary was 70 not 69.5.

In part (c) the mean was usually correct, although a few weaker students divided 6460 by 6 (the number of groups), and there were fewer errors made in calculating the standard deviation: some forgot the square root and others forgot to divide by 85. A few

students ignored these given values and recalculated (often incorrectly!) and wasted valuable time.

In part (d) most were able to use the given formula with their values and most gave a sensible comment based on their evaluation.

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Question 7

Generally the standard of work on the normal distribution was better than we have seen sometimes. Most students can now standardize correctly although the notation, and especially the distinction between probabilities and z values, is still not handled well.

Part (a) was answered but a small minority are still unclear when and when not to subtract the value found in the tables from 1.

In part (b) many found P(H > 180) but they did not appreciate that part (b) wanted a conditional probability and so just left their answer as P(H > 180) instead of dividing this by their answer to part (a). Some of those who did identify the conditional probability could not interpret their numerator correctly and simply wrote

. part (c) was often not attempted but those who did could usually obtain a probability of 0.0528 for . Unfortunately a number of students then used a rounded probability of 0.05 and therefore a value of 1.6449 rather than the correct value of 1.62. Standardizing and forming a suitable equation for were usually accomplished correctly by those who reached this stage and the correct answer was seen from well over a quarter of the students.

Question 8

This question was not answered very well. A clear Venn diagram would have helped some to get started.

Those who tried to draw a Venn diagram could quickly find P(A) from 1 – 0.18 – 0.22 to answer parts (a) and (b) would either follow from P(A) + 0.22 or 1 – 0.18.

After this though the students needed a clear argument to answer part (c) that did not assume that A and were independent. Many weaker students simply wrote down 0.6 × 0.22 = 0.132 and made no further progress. Some set off in the right direction by

quoting the conditional probability formula but to make further

progress they needed a second equation with and which they could obtain by using the addition formula and their answers from part (a) and part (b) but many gave up at this point. The stronger students were able to successfully solve these two equations to find P(B) or . Of course a few astute students noted that since as well then and are independent, thus answering PART (d) first. They could then use the given information to obtain and then their Venn diagram to see that = 0.33.

Those who obtained from a more conventional route nearly always chose to answer part (d) by checking that and giving a correct conclusion.

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Statistics S2 (6684)

Introduction

On the whole this paper was well answered. Most questions on the paper seemed accessible to the students. Generally the work was quite well presented although students knew the statistics needed to answer the questions but were let down by poor algebraic skills.

The paper proved accessible to the majority of students and there was little evidence of there not being enough time to complete the paper. There were the usual arithmetic and algebraic errors, but students work was generally well presented. The main errors were to write down the probability corresponding to an adjacent position of the required answer or using the wrong ‘block’, for example looking up in B(30, 0.25) instead of B(25, 0.5).

Report on individual questions

Question 1

This was a good opening question. Students were confident in identifying Poisson distribution with correct parameter and were able to use it for finding the values of the probabilities required. The most common errors made by students were to work out

in part (a)(ii).

In part (b) was usually calculated correctly for the different time period. However, too many students did not understand that to work out the probability that “the next patient arrives” is the same as . Many calculated .

Question 2

This question differentiated very well between students, with part (d) proving particularly difficult for many students.

Most students produced a fully correct solution in part (a). However, a significant minority failed to show where the factor of 486 came from, and were penalised 2 marks. Nearly all students equated their definite integral to 1 at a sufficiently early stage to convince examiners that the result c = had genuinely been obtained. Very few cases of verification were seen, though these were usually done sufficiently well to secure full marks.

A few tried to treat this as a discrete distribution substituting every integer from 1 to 9 into the formula and adding them to equal 1.

Part (b) produced a mixed response. There were many perfect solutions, which mainly tended to use definite integration rather than evaluating the constant of integration. However, a significant minority of students used neither limits nor a constant of integration, and consequently gained no credit. Some students failed to completely define F(t), as given in the question, and so lost the second mark.

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In part (c) a significant majority scored full marks by evaluating 1 – F(3). Other students worked from first principles and integrated the pdf between the limits of 3 and 9. The most common errors were to use the limits 4 and 9 instead of 3 and 9 or else to find either F(3) or 1 – F(2). Part (d) proved to be very challenging for the majority of students. The conditional probability eluded many students, and so they were unable to secure any marks. It was not uncommon to see ... given as the final answer, whilst others

multiplied When a quotient of probabilities was given, the two most common errors were giving an incorrect numerator and using F(3) as the denominator. However, a small minority of students were able to complete this part correctly.

Parts (d) and (e) produced a pleasing response from students. The most common errors were simply to square the answer to part (c) and the omission of the factor of 3.

Question 3

In part (a) many students had learnt when a Poisson distribution is suitable but too many did not use the context given in the question. It was pleasing to see that few students gave independent and random as separate reasons although some students lost the mark for “constant rate” by talking vaguely about rate or giving the conditions needed when approximating to a normal distribution.

Most students knew how to set about finding a critical region in part (b) and located the correct probabilities but gave the incorrect critical regions with being the most common error although were occasionally seen. Fewer students, than in previous years, gave the critical region using probability statements although a small minority gave critical values rather critical regions.

Part (c) was usually correct even if their answers to part (b) had failed to gain full credit.

In part (d) the majority of students stated whether 8 was or was not in their critical region and drew the correct contextual conclusion and managed to associate the “claim” with rather than H1. A few had not appreciated the connection with part (b) and redid the entire calculation. A minority of students did not appreciate that a statistical justification was required and said that 8 was double the expected number of emails and so was a significant result.

Part (e) was well done, even by students who had gained few marks up to this point. The first 3 marks were usually scored, although calculating were sometimes seen. A minority wrongly accepted H0 and did not include enough context their conclusion. A mistake sometimes seen was to compare the probability with 0.05 and incorrectly accept the null hypothesis.

Question 4

The majority of students were able to use the correct model in part (a)(i) and calculated correctly.

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In part (a) (ii) many students found the correct answer but not always by the quickest method which was using and finding . Others found

and used the formula for or even used the formula to calculate and

then subtract from 1. Some students incorrectly changed their model to .

Part (b) was frequently solved efficiently and completely correctly. However, a significant minority of students did not recognise that they should take to obtain instead choosing to use the tables to obtain an estimate for , usually obtaining 0.1. Other students incorrectly thought that 0.8 (or 0.2) was the value of and not .

Part (c) proved to be challenging and was very poorly done. Most students were able to see that the exact distribution was B(100, 0.975) for the number of times the cadet hits the target. Few realised that they should define B(100,0.025) as the model for the number of times the cadet misses the target and then follow this with an approximation to the Poisson obtaining

. Even when they used many were unable to change “the cadet hits at least 95” to “the cadet misses at most 5”.

Those that stayed with B(100, 0.975) approximated to the normal, saying things like “mean = 97.5, this is too big for the tables so approximate to the normal”

Question 5

Correct responses to part (a) were seen from the majority of students, although p "small" was sometimes offered. Students who stated that usually did so in addition to, rather than instead of, the only accepted response of n is large and p is close to 0.5.

Part (b) elicited a mixed response, ranging from a null response, through "too time consuming and too expensive" to the accepted answer that there would be no pea seeds left to sell. The words "it" and "they" sometimes were used without any reference as to context.

The correct hypotheses were correctly given by a large majority of students in part (c).

Sometimes they appeared in part (d), which was accepted, but occasionally different hypotheses appeared in part (c) and part (d). Occasionally either a letter other than was used or hypotheses for a 1–tailed test were given.

In part (d), a large majority of student correctly calculated the values of the two parameters of the normal distribution, although 121 sometimes appeared as both the mean and variance following a normal approximation to the Poisson distribution Po(121).

The application of a continuity correction caused problems to many students; some were unaware that one should be used, whilst others used 135.5 instead of 134.5. Students ought to be made aware that the omission of a continuity correction resulted in the loss of at least 4 marks. Nevertheless, a significant minority of students were able to obtain 0.0336. Few students chose to find the upper critical value. The standardisation was virtually always correctly done with the students' values.

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Although nearly all students were using a 2-tailed test, it was not uncommon to see students comparing 0.0336 with 0.05 (instead of 0.025), and hence arriving at the wrong conclusion. Nevertheless, a large majority of students knew that they were expected to give a statement and contextual conclusion. The latter was sometimes marred by a failure to give a complete conclusion; in particular, the words ‘company’s’ or ‘seeds’ were sometimes omitted.

Question 6

Students used a sound method, showed accurate and detailed working leading to a correct answer in part (a). One relatively common incorrect method involved integrating the pdf (as opposed to . Many of these students ended up with a final answer of 1. This is a powerful hint in such a question that something may be amiss.

The most common errors were to use only one of the three 'parts' of the pdf, add all three 'parts' before integrating, and integrating all three 'parts' successfully, but then dividing by three.

In part (b) was not technically difficult: but persistence and attention to detail were required, particularly regarding algebra and arithmetic. However, the real problem for many students was conceptual.

Almost all students could integrate all three parts of . However, a substantial number apparently did not address this central notion of cumulative probability: they failed to add

for the second part and for the third part. The students who used the method were able to use to gain the 2nd and 4th line of the c.d.f however they did not know what to use to calculate the C to gain the 3rd line; many simply used

for a second time.

It is pleasing to note that almost all students adopted the correct general strategy for part (c), which is to solve the equation . On this occasion, only a minority of students

chose the incorrect method of evaluating . However, a significant number of students

used the wrong part of their version of . Students would be advised to do a little bit of preliminary work, which consists of:

so use the third line of .

In part (d) very few students said there was little skewness or symmetrical.

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Statistics 3 (6691)

General Introduction

Overall this proved to be an accessible paper, giving the best students the opportunity to score the majority of the marks available. The required methods were well known by a majority of students; however there were some issues in the numerical parts with accuracy. Hypothesis testing questions were answered well with better use of context than in previous examinations. However many students are still confused about whether they have proved the null hypothesis to be true or have not shown the alternative hypothesis to be true for a particular test. Some students lost marks needlessly by failing to show intermediate working in the longer questions.

Report on individual questions

Question 1

This question was not answered particularly well, even by the best students. Problems and misunderstandings in describing how to take a sample were common and many students preferred the equal probability for each element approach in part (a). Marks were lost by students failing to be sufficiently precise in their descriptions in later parts.

Many students gave an example rather than the lack of a sampling frame in part (b) .Only weaker students suggested systematic or quota sampling methods.

In part (c)(i) many students failed to distinguish between the units and the list, whilst others were penalised for not specifying all the students. Generally the best answered part was part (c)(ii), however students need to pay attention to the detailed requirements.

Question 2

Most students understood the concept of a statistic and were able to score well in the first part. However the overall quality of work was poor in part (b), where too many students failed to engage with the question in any meaningful way. The mean was usually correctly stated by the better students, but the number of students correctly finding the variance was small.

Question 3

This was a routine question for the majority of students. Both the hypotheses and the result of the test were usually given in context and many fully correct solutions were seen here. Premature rounding proved costly in some solutions offered and typically these students did not show all their calculations

Question 4

This was a straightforward question which was a good source of marks for the great majority of students with all but the weakest students offering full solutions. Occasional errors were made in evaluating the variance, such as missing squares or sign changes and prevented some from gaining full marks, but less so than in question 2.

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Question 5

Many students failed to give Binomial conditions in context in part (a) but parts (b) and (c) were done well. Many students still wanted to include the probability in their hypotheses but most then went on to perform the test satisfactorily and arrived at the correct conclusion. There were very few students who failed to combine some classes, usually correctly. Also, most students managed to arrive at the correct degrees of freedom. Too many students lost marks at the end of this question by failing to show adequate working when their test statistic lacked accuracy.

Question 6

Very few students were awarded the mark for part (a) as they did not make an attempt to find the required expectation. However, typically they went on to gain all the marks in part (b), for the most part, reached a successful conclusion in part (c). Typical errors here were either choosing the wrong z value or calculating the width of the interval to be twice the correct answer.

Question 7

Marks were lost by some students in part (a) by failing to have consistency in the use of signs. The test was well done by all but the weaker students who usually failed to use the correct standard error. However the incorrect use of signs too often lost a mark in the actual test, but these students usually recovered to come to a correct conclusion. Most students gave the result well in context but some lost a mark by not using the key words, and/or not concentrating on the “less than” idea. Standardising correctly also proved to be an issue here, but many students scored well on this part.

Question 8

Part (a) was well done by the great majority but a mark was often lost carelessly by not giving the answer as required.

The test in part (b) was usually done well with most students speaking of positive correlation in their answer.

Ranking accurately was an issue in part (c), with many ranking height rather than weight, but the required technique was clearly evident.

The test in part (d) was also well done by most students with good answers in context using key words. There were more two tailed tests attempted here and a failure to relate their result to the question was more evident that earlier in the question.

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Statistics S4 (6684)

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. The presentation of the work was generally good and candidates were able to complete the paper within the time.

Report on individual questions

Question 1

The majority of candidates gained full marks for this question.

Question 2

The definitions of a type I and type II error were usually clearly written with many candidates giving the exact definition given in the mark scheme.

In part (b) the main error was to select the critical region, X rather than X < 2 since P(X) is closer to 5% than P(X ).

In part (c) most candidates were able to identify the correct probability required for a type II error following their CR in part (b) so gaining the method mark.

Question 3

The most able candidates gained full marks for this question. The most common error in

part (a)(i) is to make an arithmetical error in finding the value of or use rather than

.

Part (ii) was well answered. The main error is to use the (0.05) and (0.95) values instead of (0.025) and (0.975).

In part (b) many candidates know that they need to use but do not realise they need to

use the lowest value for from part (a)(ii) and the largest value for from part (a)(i).

Question 4

In part (a) a minority of candidates realise that it is the “differences” which need to be normally distributed and not the distributions themselves.

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In part (b) the most common error is an incorrect standard deviation and not writing hypotheses which matched the units they worked in. The majority of candidates are able to apply the method correctly and draw a conclusion in context.

Question 5

This question was accessible to all candidates although only the most able were able to do part (e).

The most common errors were to make a small error in the presentation of their solution in part (b) which meant it was not a fully correct solution. It is important that candidates get the brackets in the right place in their solutions.

In part (c) a minority of candidates gave the values to 3 decimal places rather than the 2 decimal places stated as the accuracy of the values in the table. A few candidates drew a straight line through the points in part (d)

In part (e) many candidates were able to identify the fact that the P(Type II error) 0.4 but this is where they stopped. A few candidates who got to a value of 0.23 got the inequality sign the wrong way round.

Question 6

Apart from part(a) where the majority of candidates were unable to explain what was meant by the Sampling distribution this question was well answered. The other main errors were the

use of for the Var(Xi) and in part (e)(ii) many candidates did not refer to the bias of

being the same as

Question 7

Part (a) was answered well with many candidates gaining full marks only a minority did not state what the assumption was that they needed to make.

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.

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Decision Mathematics D1 (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.

Report on individual questions

Question 1

This first question proved to be an excellent source of marks for many candidates with the mode being full marks obtained by 68.3% of candidates and only 23.3% scoring 3 marks or fewer. A number of candidates who used the matrix form of Prim’s algorithm lost marks by listing the arcs in the wrong order although the correct arcs had been selected in the table. Candidates would be advised to scan all labelled columns, circle the smallest value and then write down the corresponding arc immediately before going on to label the next column. Trying to write down the arcs selected in order after completing the algorithm is far more demanding. A common error was in the selection of the final two nodes or arcs with CD selected before BG, and only a few candidates lost marks by listing only the vertices in order instead of the required arcs. It was pleasing to note that only a small minority of candidates started from a different vertex than the required A, although some began at D, possibly due to arc DG being the shortest arc in the network. Finally, very few candidates appeared to reject arcs when applying Prim’s algorithm. If the candidate answered part (a) successfully then they typically answered parts (b) and (c) correctly. A number of candidates were able to recover from mistakes in part (a) to draw the correct minimum spanning tree and state a correct total.

Question 2

Many candidates made a good attempt at this question with the modal mark being 6 out of 7, 39.7% of candidates scored 6 or more marks and only 12.3% scored no marks. Candidates generally showed a good understanding of the process of constructing an activity network from a precedence table, using arcs drawn with arrows and labelled for activities. The final mark in part (a) was often lost due to missing arrows, but some candidates also added additional “activities”. Some scripts lacked a sink node at the end and a small number did not have a single source node. Some of the diagrams and labels were challenging to read, especially when they were very small and/or drawn with lines that crossed over. Some candidates were unsure about the placement of their dummies, putting them in ‘anywhere’ so that they had two dummies included. Some also had three dummies even though the question

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clearly stated that they were to use exactly two. A very small number of candidates put activity on node, and some failed to check that they had all activities present, with K being the activity that was missing most often.

A minority of candidates were able to give a competent description of why the first dummy was needed (to enable activities to be described uniquely in terms of their end events). Many candidates used the word “unique”, but gave an incorrect or incomplete explanation of what this meant. Common explanations were that, “D, F and G depend on A and B, and these start from the same place”. A significant number of candidates got the explanation for the second dummy correct by referring directly to the correct activities in the context of dependence. Some candidates used the word precedence to describe this dummy – and while some were confident with the terminology and were able to use this term accurately, more often than not candidates were confusing this word with dependence. A small number of candidates gave one vague general explanation for the two dummies, without reference to specific activities or they omitted to say that I depended solely on D.

Question 3

This question proved to be a good source of marks for many candidates with 49.8% of candidates scoring 6 marks or more and only 23.1% scoring 3 marks or fewer.

Part (a) was generally answered well by most candidates with the vast majority stating the correct three distinct pairings of the correct four odd nodes. 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. There were however many instances where the totals were incorrect. The majority of such mistakes occurred for the pairing of D with F. There were also some instances where no totals were given which lost candidates a significant number of marks. Candidates should be advised to be thorough when checking the shortest route between each odd pairing. Many candidates did not explicitly state the arcs that should be repeated instead stating that DE and FK should be repeated instead of the correct arcs DA, AE, FJ, JK. Furthermore, a number of candidates did not state the length of a shortest inspection route. Only a minority of candidates were able to answer part (b) correctly with the majority stating that J would appear 6 times (6 was the order of vertex J once the two additional arcs were added) rather than the correct answer of 3 times. In part (c) many candidates identified DE, DF and EF as the paths that needed to be considered, although they often missed stating the fact that DF was the shortest path that did not including vertex K. Many candidates, even with the correct selection of arcs, either did not state a route or gave an incorrect route. Some candidates misunderstood the reasoning altogether and focused on the fact that DK had the greatest weight of the previous pairings and therefore should be avoided and so EF should be repeated. Others still said that ‘FK is the least therefore start and end at F and K’. Those that did state DF usually went on to score the mark for stating the length of a shortest inspection route.

Question 4

This question proved to be an excellent source of marks for many candidates with the mode being full marks obtained by 51.6% of candidates and only 19% of candidates scoring 5 marks or fewer.

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The majority of candidates were able to successfully complete the bipartite graphs in parts (a) and (b) although there were a notable number of candidates who added arcs from F to G and from P to D to their bipartite graph in (a).

Part (c) was well attempted and most candidates were able to write down an alternating path from N or P to D. It is important that examiners can clearly identify the alternating paths so they should be listed (rather than drawn) separately (rather than left as part of a ‘decision tree’ of potential paths). A number of candidates are still not making the change status step clear. This can be done either by writing ‘change status’ or, more popularly, by relisting the path with the alternating connective symbols swapped over, this latter approach has the additional advantage of making the path very clear to examiners. A lack of change of status was penalised twice both in this part and in part (e). A significant number of candidates did not state the improved matching after stating their alternating path. If candidates are going to display their improved matching (or later their complete matching) on a diagram then it must be made clear that only a diagram with the exact number of required arcs going from one set to the other set will be accepted; examiners cannot accept diagrams with additional arcs even if they ‘appear’ to be crossed out.

The responses seen in part (d) were mixed. Candidates are clearly familiar with the type of explanation that is required and most are giving explanations which refer to all of the relevant nodes. However, some candidates provided an explanation which despite appearing to be true did not provide the required reasoning for why a complete matching was not possible. For example, the most common incorrect answer was ‘R is the only one who can play G and K’. Although this initially seems like a correct response, this is incomplete as it does not preclude others from either playing G or K. Others failed to home in on the relevant nodes stating for example, “D can only be done by J” or “because we can’t get to K”.

Part (e) was well done although less so than part (c) probably due to some candidates not updating their bipartite graph with the addition of the two further preferences given before the stem to part (e). Candidates that drew the new bipartite graph tended to do well in this part. Although those candidates who failed to state/draw an improved matching in part (c) usually did provide the complete matching in part (e).

Question 5

This question also proved to be an excellent source of marks for many candidates with the mode being full marks obtained by 30.8% of candidates and only 20% of candidates scoring 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 nodes S, L and/or Y). The route was usually given correctly and most candidates realised that whatever their final value was at Y this was therefore the value that they should give for their route.

Part (b) was done well although many candidates failed to read the question carefully and simply calculated the length of the new shortest route without stating the increase in the distance travelled. Some failed to realise that the final value at vertex M added to the length

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of the arcs ML and LY would give them the required length of the new shortest route, instead many added up the length of each arc from scratch.

Question 6

This question discriminated well leading to a good spread of marks. The modal mark was 11, 19.2% scored full marks and 25% scored 7 marks or fewer.

The inclusion of as an unknown value threw a significant number of candidates but the larger percentage took it in their stride. A few chose a value for at the beginning (usually 23) and then forgot it was a variable, using the same value throughout. One or two left out completely or felt it needed to be isolated in a bin of its own.

Part (a) was answered well with many correct responses seen. A common error was to put the 6 in the wrong bin. Only a few candidates just worked with as a chosen value within the range at this stage.

In part (b) most candidates’ selected middle-right pivots and many were able to carry out the sort correctly. Errors cropped up in the ordering of the sublists after the first (and subsequent) iterations. The most common occurrence of this tended to be that and 25 were interchanged after the first pass. Other errors included failing to select the 9 as a pivot for the fifth pass probably because the sublist of length two after the fourth pass appeared to already be in order. Very occasionally, candidates selected only one pivot for each iteration or failed to sort the list into (values greater than the pivot), (the pivot), (values less than the pivot) after the first iteration. There were only a few cases where candidates selected the first or last items as the pivot. Pivots were usually chosen consistently although the spacing and notation on some solutions made these difficult for examiners to follow. Some candidates over complicated the process by insisting on using a different ‘symbol’ to indicate the pivots for each pass. Those candidates who sorted into ascending order usually remembered to reverse their list at the end to gain full credit although a number of candidates left their list in ascending order and then went on to apply first-fit increasing in part (c).

Part (c) challenged many candidates with most only stating one correct allocation. Often candidates did not even seem to consider a second possible allocation despite the question clearly saying there were two. Some candidates did not know how to handle the value of in this part and used a value instead. A common error was omitting the values of 8 or 9 from bin 2 and putting in the value of 6 instead.

Many candidates could only score one mark in part (d) due to the lack of two correct allocations stated in part (c). Of those that did have two correct allocations and both answers for , most also supported their answers with relevant calculations, and so were able to gain full credit.

Question 7

This question also discriminated well leading to a good spread of marks. The modal mark was 12, only 4.8% scored full marks and 29.5% scored 6 marks or fewer.

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Part (a) was poorly answered on the whole. Many definitions of total float muddled the terms ‘events’ and ‘activities’. A large majority of candidates gave incomplete answers, for example, not clearly stating earliest start time or latest finish time (either saying start and finish time or early and late times). Some candidates tried to state the mathematical formula but did not fully explain all the symbols they used. The majority of candidates understood the idea that it was the time an activity could be delayed for but in many cases, candidates thought it was the total of the floats for all activities in a network.

In part (b) the dummy between events 6 and 7 caused the most difficulty, with a number of candidates going via K and getting a late event time of 25 rather than the correct value of 21. Most could gain the first mark by having only one rogue value and the early event times were largely correct. In general more mistakes were seen in the bottom half of the boxes. Candidates should be advised to take time checking their values as a significant number of subsequent marks can be lost if errors are made in this part.

The majority of students successfully identified the correct critical activities in part (c).

In part (d) the majority of candidates knew the method for the float calculation and showed it clearly. Others just gave the calculation of 21 – 14 which could have been the latest finish time – the earliest start time for the event between G and L rather than the correct calculation 21 – 11 – 3. Finding the lower bound in part (e) 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 (30) to the number of activities (13)) in their calculation.

In part (f) quite a few candidates drew a Gantt chart instead of a scheduling diagram, and so scored no marks. There were also quite a few instances where this part was left blank. Those that did schedule tended to make errors on activity H, which needed to take place after activity D. There were also minor errors in duration lengths seen meaning few scored full marks in this part. It would be advisable for candidates to check their working carefully to ensure that preceding activities are completed and that activities do not start before their earliest start time or continue beyond their latest finish time. Also it was common for at least one activity to be missing from the scheduling diagram.

Question 8

There was possibly some evidence of candidates having insufficient time to complete the paper as there were a number of blank (or unfinished) solutions to this question. However, this was also a demanding and discriminating question which is an alternative and feasible reason for blank/incomplete responses. Only 3.9% of candidates scored full marks, the modal mark was 3 (gained by 23.9% of candidates) and only 10.1% scored 5 or more marks. Part (a) was generally well attempted and most candidates were able to obtain at least one mark in this part. Some candidates incorrectly used strict inequalities and there were also errors in the directions of the inequalities.

Quite a few blanks responses were seen in part (b). Those that did attempt this part usually began by stating or calculating the exact coordinates of vertex D. However, many did not find the exact coordinates for vertex A, and instead either rounded their answers or read them directly from the graph. The majority of candidates then (incorrectly) went on to compare the

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expressions representing the value of the objective function, in terms of , at these two vertices. As a result, many scored only 1 or 2 marks in part (b). Even candidates who successfully found all 4 coordinates often compared vertices incorrectly and unnecessarily. Some compared the values from each pair of vertices in turn, achieving the method mark almost by ‘trial and error’. Other methods, based on an objective line approach, were rarely seen. When they were, candidates were more often successful, although some struggled to use ‘steeper’ and ‘shallower’ in the context of negative gradients. Others could ‘see’ the solutions but omitted to show how the gradients of the lines and the objective function were related to . Furthermore, the negative values and the negative reciprocal of in the inequalities caused some difficulties and some used incorrect algebra to obtain an answer that looked reasonable and it was not uncommon to see the correct final answer following incorrect working. The question stated that they should make their method clear but not all candidates were able to do this.

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Decision Mathematics D2 (6690)

Introduction

The 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.

Candidates should be advised to read questions carefully and answer as required. For example, marks were lost in question 1, by candidates minimising rather than the required maximisation.

Poor quality of handwriting causes a minority of candidates to lose many marks, particularly in misreading their own written numbers and capital letters. There were many instances of candidates losing marks through poor basic arithmetic, evident in questions 1, 2, 3, 5 and 7.

Most candidates were well prepared for the exam and there were very few blank pages. It was evident though that in the final question some candidates ran out of time, a few making no attempt, and many stopping mid-solution.

Question 1

This first question proved to be a good source of marks for many candidates with the mode being full marks obtained by 50.8% of candidates and only 24.1% scored 4 marks or fewer. A significant proportion of candidates were very well prepared for this question and provided perfect or near perfect solutions. It is clearly a well understood topic for many candidates. Unfortunately however, a smaller yet still significant proportion of candidates were confused by the combination of a maximisation problem with empty cells. There was evidence of confusion in approach and some candidates entered large values in the empty cells before minimising, replaced the empty cells with zeros after converting to a minimisation problem, or dealt with them midway through their solutions (usually after row/column reduction). Some candidates did not fully undertake row and column reduction, in some cases prematurely augmenting their tables. There were many candidates who began their solution as a minimisation problem and then realised their mistake and restarted. The number of arithmetic errors was small but significant, sometimes caused by candidates misreading their own handwritten numbers. Despite the issues apparent for some candidates, this was in general a well attempted question and many candidates were able to apply the algorithm correctly.

Question 2

This question also proved to be a good source of marks for many candidates with the mode again being full marks obtained by 51.9% of candidates and only 23.5% scoring 6 marks or fewer.

Most candidates found both nearest neighbour routes in part (a) but some failed to return to A, or made arithmetic errors in calculating the lengths of these two routes and some incorrectly doubled the length of their routes to obtain an upper bound.

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In part (b), most found a residual spanning tree, but unfortunately lost marks in this part as they did not use the correct two least lengths incident to F. A number of candidates incorrectly used both arcs of 88. A common mistake for the RST was to include arc CD. Some candidates calculated a Nearest Neighbour route for the vertices A to E rather than finding a minimum spanning tree.

In part (c) most candidates, who had obtained upper and lower bounds earlier, wrote down an interval containing these two values, although a significant number lost marks through poor notation, including writing  471. Those candidates with incorrect bounds became creative with their interval, particularly when their lower bound was greater than their upper bound.

Question 3

This question proved to be a good discriminator and gave rise to a good spread of marks. The mode was again full marks (obtained by 30.8% of candidates), 57.2% of candidates scored 10 marks or more although 11.1% of candidates scored no marks.

In part (a) the majority of candidates correctly identified the correct pivot and went on to divide the pivot row by 2. However a number incorrectly used the 3 in the top row as the pivot and some used the –1 in the third row. A small number of candidates decided to pivot on the x column or the z column instead of y, despite the question stating that the most negative number in the profit row should be used. A small number failed to change the basic variable in the pivot row. Having divided through, most candidates stated the correct row operations and applied them successfully to the table, although some numerical errors crept in. Most then went on to correctly identify the second pivot and to divide through again. Those that had a correct or virtually correct first iteration generally went on to state and apply the correct second set of row operations, ending with a correct optimal solution. However those that had made more significant errors in their first iteration often did not have the correct second set of row operations.

From time to time it was difficult to read candidates’ work due to crossings out/corrections and candidates should be reminded to make sure their work is clearly set out. Many candidates made use of only two tables, however, a significant number of candidates used several tables, often writing and rewriting elements within the table a number of times which must have taken candidates a considerable amount of time.

Part (b) was generally less successfully attempted. Of those candidates who provided an answer to this part many were able to correctly write down at least some of the values from the value column rather than from the profit row of the table. However a significant number lost marks because they did not write down all of the variables (often giving only the basic variables) or they did not write down P explicitly.

Surprisingly there was a significant number who did not attempt this part of the question or who wrote down only P + 43x + 27s + 4t = 47750.

Question 4

This question also gave rise to a good spread of marks and proved a good discriminator. The mode was again full marks gained by 34.4% of the candidates, 21.8% scored 5 or fewer marks.

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In part (a) candidates found the correct row minimums and column maximums and then deduced that the row maximin (-3) was not equal to the column minimax (1), though a few had 2 for the latter.

In part (b) the majority of candidates correctly defined and then used a dominance argument to eliminate one of the options for player B but a significant number deleted column 4 instead of column 2. Those that didn’t use a dominance argument usually went on to correctly form expressions for A’s expected winnings if B played each of its four options. In this second part most went on to set up their three probability expressions correctly (though some had errors when simplifying these expressions) and they then went on to draw a graph with 3 lines; a few candidates just attempted to solve 3 pairs of simultaneous equations, scoring no marks. It was noted that many graphs

were poorly drawn without rulers, went beyond the axes at , had uneven or missing scales on the vertical axes, or were so cramped that it was difficult for candidates and examiners to identify the

correct optimum point.

Most candidates then attempted to solve the pair of equations for which they considered to be their optimal point. Those that solved the correct pair usually went on to list the correct options for player A. Many candidates also stated the value of the game to A at the end of the solution despite this not being required.

Question 5

This question proved to be a good discriminator and gave rise to a good spread of marks. The mode was 8 scored by 21.3% of candidates, 20.2% scored full marks and only 15.6% scored 5 marks or fewer.

The vast majority of candidates stated the correct initial flow and completed the flow diagram in parts (a) and (b). Most then went on to find one or more flow augmenting routes, although a significant number failed to obtain the maximum flow of 70. A small number tried to increase the flow by more than 8, generally not realising that only 2 units could flow along BA. A number of candidates incorrectly tried to find flow augmenting routes starting SA… or made statements about decreasing the flow in particular arcs.

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 70. 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 nodes A or B.

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 70, 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.Question 6

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Candidates answered this question well with 18.7% scoring full marks and 58.1% scoring 5 marks or more.

The majority of candidates failed to make a clear definition of their and then use it consistently throughout the question. Common errors included omitting the word “number” or using P, Q, R in the definition and 1, 2, 3 elsewhere. Some candidates defined as being equal to 1 or 0 as in an allocation formulation. Most candidates correctly stated the objective function and “minimise”, although a small number stated “maximise” and there were some slips, either with the coefficients or suffices. There were a variety of errors made with the constraints, with some not having unit coefficients and commonly the non-negativity constraint for was absent. Other errors included slips with suffices or values and a mixture of equations and inequalities. A small number of candidates incorrectly wrote all their constraints as ≥ and some equated them to one like in an allocation formulation. Some candidates changed their notation between the objective function and constraints, for example using PA in the objective function and in the constraints.

Question 7

This question produced a variety of responses, from perfect ‘textbooks’ solutions (18.9% scored full marks) to minimal or blank attempts (19.7% scored no marks). A number of candidates showed a clear grasp of how to use the given information to work backwards through their table, from one stage to the next using the correct relevant values at each stage to find a correct final solution. Many candidates crossed working out and then attempted to squeeze in alternative answers, making it very difficult for examiners to actually mark their work. There were a number of errors made when choosing the correct elements to include in calculations or in the arithmetic. Nearly all correctly started backwards from July and most were able to obtain values for July, either opting to add storage costs in the current month or the next. A number of candidates forgot to carry forward previous optimal values but most were able to continue correctly. Very few extra rows were seen but a significant number of candidates lost marks for deleting or omitting state 2 (equivalent to having two aircrafts in stock at the beginning of the month) in the stages for May and/or April. Candidates who omitted these states appeared to be considering the demand required in an earlier month, and they therefore concluded that it would not be possible to have two aircrafts in storage at these stages. In essence these candidates were working forwards and not backwards; this is a common error when applying the principles of dynamic programming. Candidates should be advised that in decision mathematics they must rigorously apply the algorithm, rather than introduce their own logic or common sense. Some candidates gave up part way through but again this could well have been a timing issue. Most candidates did show the necessary working as requested. Those candidates who were eligible usually scored the final two B marks. A small minority of candidates started from March, attempting to apply the algorithm going forwards. As in previous series this approach was penalised with the loss of all dependent accuracy marks.

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Grade Boundaries: May/June 2014 GCE Mathematics ExaminationsThe 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 64 57 50 44 386664 Core Mathematics C2 60 53 46 40 346665 Core Mathematics C3 60 54 48 42 366666 Core Mathematics C4 59 52 46 40 346667 Further Pure FP1 62 54 46 38 316668 Further Pure FP2 65 59 53 47 416669 Further Pure FP3 55 48 41 35 296677 Mechanics M1 58 52 46 40 356678 Mechanics M2 64 56 48 40 336679 Mechanics M3 65 57 49 41 336680 Mechanics M4 57 50 43 36 296681 Mechanics M5 64 56 49 42 356683 Statistics S1 60 53 46 39 326684 Statistics S2 61 53 46 39 326691 Statistics S3 64 57 50 44 386686 Statistics S4 65 56 47 38 306689 Decision Maths D1 59 53 47 42 376690 Decision Maths D2 61 52 43 34 26

Distinction MeritAdvanced Extension 68 48

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Provisional Pass Rate Statistics: May/June 2014 GCE Mathematics ExaminationsThe 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 40.2 58.9 72.8 81.5 87.86664 Core Mathematics C2 34.5 48.9 62.0 71.5 79.56665 Core Mathematics C3 37.9 57.2 72.4 82.5 89.66666 Core Mathematics C4 26.6 40.2 54.1 66.4 76.56667 Further Pure Mathematics FP1 58.6 74.7 84.3 90.6 94.36668 Further Pure Mathematics FP2 38.3 57.4 71.4 80.6 86.66669 Further Pure Mathematics FP3 38.3 58.8 73.5 82.6 89.36677 Mechanics M1 35.2 52.6 66.2 76.0 81.86678 Mechanics M2 40.7 57.9 71.4 82.1 88.96679 Mechanics M3 41.6 61.4 73.7 81.8 87.76680 Mechanics M4 38.2 53.4 69.2 80.7 88.16681 Mechanics M5 46.8 61.0 71.4 81.4 87.46683 Statistics S1 28.0 44.1 59.1 71.8 81.96684 Statistics S2 33.3 55.6 70.3 80.7 87.86691 Statistics S3 43.7 71.2 85.1 91.1 95.06686 Statistics S4 66.2 46.4 63.0 73.6 81.66689 Decision Maths D1 27.4 46.4 63.0 73.6 81.66690 Decision Maths D2 58.7 82.1 92.1 96.2 97.9

20.5% of candidates gained 90 UMS or more on Core Mathematics C314.4% of candidates gained 90 UMS or more on Core Mathematics C4

Distinction MeritAdvanced Extension 11.0 42.8

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