Mathematics Unit FP1

Comparison of key skills specifications 2000/2002 with 2004 standards

ask CODE "WHAT CODE?" X015461

ask DATE "WHAT DATE?" July 2004

ask ISSUE "WHAT ISSUE?" Issue 1

Examiners’ Report

Principal Examiner Feedback

Summer 2014 (R)

GCE Mathematics

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Summer 2014

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Core Mathematics C1 (R) (6663)

Introduction

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

Report on Individual Questions

Question 1

This question was generally very well answered. Most gave the answer in the form given in the main scheme. Some students only took out the x term then stopped and others lost two marks as their first line was

2

(925)

xx

-

with wrong signs. It was very unusual to see anyone misunderstanding the instruction to factorise and continuing their answer by ‘solving’ to x = …. This was a good opening question.

Question 2

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

In part (b) some stated incorrectly that

(

)

2

0.51.5

xx

-

=

or that

1

1

16

16

x

x

-

=

.

Others stated that

(

)

2

0.50.25

xx

-

=

or

0.25

x

. Another common error was not squaring the 4 at the start of the bracket and only dealing with the x term.

Question 3

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

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

Question 4

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

6

x

as a multiple of an index and then differentiating.

In part (b) the main difficulty was in the integration of

6

x

. There were a few but not many, who integrated their answer to part (a).

Question 5

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

210

x

=

or

10

2

x

=

. Other errors were, for example, replacing 20 by

45

and then by

25

, or replacing 10 by

52

.

Some rationalised the denominator

(

)

616

-

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

Question 6

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

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

In part (b) a wrong expression for area was less common - most divided up the inside region to obtain their expression. The resulting quadratic was usually solved correctly and most students chose the inside region, as required. A minority put

120

A

=

so did not have an inequality or put

120

A

>

and obtained the wrong inequalities. Credit was given to those giving the answer as

32.5

x

-<<

and also to those who realised that

x

was a length and so gave the answer as

02.5

x

<<

.

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

Question 7

This proved more challenging than the earlier questions.

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

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

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

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

Question 8

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

xx

. Those who gave it as

x

to the power

3

2

usually had no problem integrating and dividing by the fraction

5

2

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

Question 9

This question was a reasonable discriminator.

In part (a) the quadratic and linear graphs were generally well drawn. Marks were lost due to the omission of co-ordinates particularly the

3

k

-

.

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

d

d

y

x

was the most popular approach. Those who began correctly by this method putting the gradient expression for the curve equal to the gradient of the line, usually completed it to find x, then y, then k. Many who attempted instead to set the curve expression equal to the line expression obtained a quadratic but proceeded no further. Of those who continued with this method, use of the condition for equal roots, putting the discriminant equal to zero usually was more successful than completion of square methods.

Question 10

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

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

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

Question 11

This was a well answered question .

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

18

x

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

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

Core Mathematics C2 (R) (6664)

Introduction

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

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

Report on Individual Questions

Question 1

This question was generally well answered and responses showed that students could work confidently with binomial expansions. Although the majority of responses gained full marks the error of not squaring the denominator in the

x

term when expanding the bracket was seen occasionally, leading to an incorrect expansion of

23

112126756

xxx

+++

.

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

Question 2

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

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

4

5

6

æö

ç÷

èø

instead of the correct

3

5

6

æö

ç÷

èø

.

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

Question 3

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

(

)

(

)

1

2319271115

2

æö

´´++++

ç÷

èø

or

(

)

1

2319271115

2

æö

´´++++

ç÷

èø

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

Question 4

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

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

The method in part (c) was well known and most chose to evaluate

1

f

2

æö

ç÷

èø

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

Question 5

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

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

(

)

7

sin90sin2.1

DC

p

=

-

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

Question 6

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

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

32

13

4

84

xx

æö

-+

ç÷

èø

interpreted as

32

13

4

84

xx

-+

.

Question 7

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

1

sin2

3

q

=

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

In part (ii) the majority of students recognised the need to apply the appropriate trigonometric identity,

22

sin1cos

xx

=-

, although some incorrect identities were seen including

cos1sin

xx

=-

. Those who did obtain a quadratic in

cos

x

sometimes made errors when rearranging or made mistakes when solving the quadratic. Commonly students dealt with the constant but then ‘lost’ the negative to give

2

5cos– 2cos

xx

leading incorrectly to

cos 0.4

x

=

.

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

cos 0

x

=

or cancelled their quadratic, losing one factor altogether.

Question 8

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

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

The most common error was from those whose understanding of logs was weak, wrote

(

)

log15

x

+

as

loglog15

x

+

. Some credit was given for any evidence of understanding of either the power law or addition/subtraction laws and some students could gain at least one or two marks. Solving the quadratic involving √x was challenging for many and of those who chose to square x + 15, sometimes produced x2 + 225. More successful students substituted y = √x to help with the factorising and solving of the equation.

Question 9

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

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

3

2

x

term.

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

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

Question 10

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

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

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

RAQ

Ð

a common substitute for

ARQ

Ð

.

Mathematics C3 (R) (6665)

Introduction

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

Report on Individual Questions

Question 1

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

Question 2

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

d

d

y

x

=

0

.

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

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

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

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

Question 3

Part (i)(a) This question was generally tackled well, with most students knowing the trigonometrical identities for

for tan, cot and cosec

xxx

. A surprisingly frequent error in part (a) was the omission of cosec x from the original equation, but those who started with the correct form could usually follow an appropriate process, with some making minor arithmetical/algebraic errors.

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

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

1

2

l

=

, as they had difficulty with the

1

2

in the denominator.

Question 4

Part (i) Many students wrote down

d

d

x

y

correctly. Some did it implicitly, and a few used

2

1/cos2

y

and quotient rule. The most common errors seen were

2

d

2sec2tan2

d

x

yy

y

=

and

d

4sec2tan2

d

x

yy

y

=

.

Most then went on to find

d

d

y

x

by inverting their result, often not until after expressing

d

d

x

y

in terms of x.

A number used a triangle with ratios as expressions of ‘x’ to reach their final expression. Of course some just manipulated the given answer, but there were many completely correct solutions. It was important that students use the identity

tansec

yy

+=

22

122

and not

tansec

xx

+=

22

1

in proving their result.

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

x

1

2

was frequently seen. The substitution of

e

2

was often correct, but a considerable number of students made algebraic errors in manipulating the resulting expression.

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

Question 5

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

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

In part (c) it was particularly noticeable that many students reversed the sign of the wrong expression, so obtained an incorrect inequality. Many students did not realise that

EMBED Equation.DSMT4

3

4

x

>

. And

3

4

x

<

meant that

3

4

was excluded from the solution set.

Question 6

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

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

6

8

to obtain

2

.

Part (d) and part (e) were moderately successful, although many students failed to square both the 2 and the

x

when simplifying the expression. Many then lost marks in part (e) because of incorrect solutions to part (d). A few managed to restart from

22

eln22

xk

=

and solve correctly.

Question 7

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

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

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

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

2

52

t

p

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

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

Core Mathematics C4 (R) (6666)

Introduction

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

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

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

Report on individual questions

Question 1

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

Part (a), most students started by manipulating

1

(910)

x

-

to give

1

2

110

1

39

x

-

æö

-

ç÷

èø

, although a few incorrectly wrote

1

2

10

31

9

x

-

æö

-

ç÷

èø

. The majority were able to use a correct method for expanding a binomial expression of the form (1 + ax)n. A variety of incorrect values of a such as

10

,

9

9

10

-

or

9

10

and n such as

1

2

, –1 or –2 were seen at this stage. The majority of students expanded

1

2

10

1

9

x

-

æö

-

ç÷

èø

to give

2

525

1...

954

xx

+++

, but some forgot to multiply this by

1

3

to give the answer to part (a). Sign errors, bracketing errors and simplification errors were also seen this part.

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

Question 2

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

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

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

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

2

d

e

d

x

v

x

=

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

22

2ee

xx

x

-

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

Question 3

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

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

d

d

y

x

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

d4210

d224

yyx

xyx

--

=

+-

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

Part (b) proved a test for the higher ability students. Most recognised that the numerator of their answer to part (a) had to be set to zero and obtained x = 2y – 5 or

5

2

x

y

+

=

, but then a minority gave up at this point. Whilst most substituted their x = 2y – 5 (or equivalent) into

22

102410

xyxyxy

+++-=

a significant minority who had problems with the resulting algebra and found difficulty in reaching a correct

2

322350

yy

-+=

or equivalent. Those who progressed this far were usually able to solve the quadratic equation to give both correct values of y. A minority of successful students applied the alternative method (as detailed in the mark scheme), of finding both values of x, followed by using

5

2

x

y

+

=

to find both values of y.

Question 4

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

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

2

25

(21)

xx

+

in the correct form of

2

(21)

ABC

xxx

++

+

, although a significant number missed the

x

factor to give the incorrect form of

2

(21)

BC

xx

+

+

. Many students were successful in either substituting values and/or equating coefficients in order to find their constants.

Part (b), the majority of students were able to write down a correct expression for the volume formed, although a number of students omitted π and applied

2

d

yx

ò

. A number of weaker students did not make the link with part (a) and made no further progress.

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

A

x

and

2

B

x

correctly with a few integrating

2

B

x

to give

2

ln()

Bx

. The most common error was to integrate

(21)

C

x

+

to give

ln(21)

Cx

+

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

ln

abc

+

.

Question 5

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

Part (a), many students wrote down

2

d

4

d

V

r

r

p

=

and used the Chain Rule correctly to set up an equation for

d

d

r

t

. They applied 3 divided by their

d

d

V

r

and substituted r = 4 to find a value for

d

d

r

t

. Common errors in this part included applying

d

3their,

d

V

r

´

substituting r = 3 into their

d

d

r

t

, giving their final answer as

2

d3

d4

r

tr

p

=

, or incorrectly rounding their answer to give either 0.015 or 0.01.

Part (b), the majority of students applied

2

d

8their

d

r

r

t

p

´

and substituted r = 4 to give the correct answer for

d

d

S

t

. Some applied

2

8

r

p

divided by their

d

d

r

t

whilst others made no attempt at this part.

Question 6

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

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

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

Part (c), the majority of students applied the scalar product formula in order to find the angle ACB. The majority achieved the correct answer by applying the scalar product formula between

AC

uuur

and

BC

uuur

(or

CA

uuur

and

CB

uuur

). Some students applied the scalar product formula between the direction vectors of l1 and l2 which gave an obtuse angle. Only a handful, however, manipulated this angle to give the correct answer of 27.7°. Comparatively, few chose to use the cosine rule.

Part (d), the majority of students used

1

sin

2

abC

, with

aAC

=

uuur

and

bBC

=

uuur

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

Question 7

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

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

1

5000

N

-

to give

ln(5000)

N

-

and omitting a constant of integration “

c

+

”, whilst a number of students struggled to integrate

(1)

kt

t

-

. Some students did not show sufficient steps in order to progress from

ln(5000)ln

Nkttc

--=-+

to

5000e

kt

NAt

-

=-

. Other students struggled to remove logarithms correctly and gave an equation such as

ln

5000eee

kttc

N

-

-=++

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

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

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

Question 8

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

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

4sin

xtt

=-

in order to find k. Only a minority of students found and applied

2

t

p

=-

to give the correct answer of

4.

2

k

p

=-

The majority used values of t such as

2

p

and

3

2

p

leading to incorrect answers of

4

2

k

p

=-

and

3

4

2

k

p

=+

respectively.

Part (b), the majority of students were able to apply the process of parametric differentiation followed by substitution of their t into their

d

d

y

x

. Occasional sign errors were seen in the differentiation of both x and y and a number of students obtained

4cos

t

-

for

d

d

x

t

. Only a minority of students used

2

t

p

=-

to obtain a correct answer of

d

2.

d

y

x

=-

Most used

2

t

p

=

to give a final answer of

d

2

d

y

x

=

, when it was clear from the diagram that the gradient of tangent to the curve at A must be negative.

Part (c), many students achieved

4sin4cos1

tt

-=-

(or equivalent) after setting their

d1

d2

y

x

=-

. The majority, however, were not able to find a correct strategy for solving their trigonometric equation. Some students who squared

4sin4cos1

tt

-=-

(or equivalent) to give

22

16sin16cos1

tt

+=

.

The most popular method was to square both sides of

2sin1

14cos2

t

t

=-

-

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

sin22sincos

ttt

=

to achieve

15

sin2.

16

t

=

Few students correctly rewrote

4sin4cos1

tt

-=-

as

42sin1

4

t

p

æö

-=-

ç÷

èø

(or equivalent). The majority of students who used a correct strategy usually achieved the correct answer of

0.6077

t

=

.

Further Pure Mathematics FP1 (R) (6667)

Introduction

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

Report on individual questions

Question 1

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

1

2

z

=-

.

Question 2

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

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

Question 3

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

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

Question 4

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

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

Question 5

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

Question 6

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

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

Question 7

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

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

0

y

=

at the intersection of the two straight lines.

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

Question 8

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

Question 9

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

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

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

Further Pure Mathematics FP2 (R) (6668)

Introduction

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

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

If a student runs out of space in which to give his/her answer than he/she is advised to use a supplementary sheet – if a centre is reluctant to supply extra paper then it is crucial for the student to say whereabouts in the script the extra working is going to be done.

Report on Individual Questions

Question 1

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

Question 2

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

Question 3

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

Question 4

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

Question 5

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

2

d

d

y

x

æö

ç÷

èø

and some students lost marks because having differentiated perfectly, they then made no attempt to form an expression for

3

3

d

d

y

x

as required by the question.

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

Question 6

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

Question7

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

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

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

Question 8

Most students had some idea of how to tackle part (a) although many were let down by getting the differentiation wrong to achieve

2

2

d

d

y

x

usually by missing the

dz

d

x

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

e

z

if one was lacking. Some students differentiated

d

dz

y

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

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

Further Pure Mathematics FP3 (R) (6669)

Introduction

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

Report on Individual Questions

Question 1

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

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

Question 2

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

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

Question 3

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

2

d11

cosech

d2coth

y

x

xx

=´´-

followed by

d

cosech2

d

y

x

x

=-

. Examiners would expect some intermediate working to justify this given answer. Some students wrote the equation as

2

ecoth

y

x

=

and used implicit differentiation and were largely successful with this approach.

For the arc length part (b), the majority could make a start by using the correct formula. Most then opted to use the identity

22

1cosech2coth2

xx

+=

although a surprising number of students used

22

1+cosech2coth

xx

=

which would potentially lose them 3 marks. Those who reached an expression involving coth then often spotted the need for a natural logarithm although those with coth 2x sometimes missed the

1

2

.

Question 4

Students struggled to establish the reduction formula in part (a). Of those who made progress, the majority took u as (3 – x2)n and

d

d

v

x

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

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

Question 5

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

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

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

cos

sin(3cos)

3sin

yx

q

qq

q

-=--

and then just wrote down

3sincos3

yx

qq

+=

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

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

1

2

.

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

22

22

9144

11

4491

xy

xy

+=Þ+=

.

Question 6

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

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

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

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

Question 7

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

2

d

e2e1ed

d

xxx

y

Sx

x

p

---

=-\=+

ò

is not enough to score the 3 marks. It was expected that students should at least quote the general formula for surface area and then substitute their derivative.

Part (b) the majority of students knew how to use the substitution but a few failed to replace the

d

x

correctly and could make little progress. Those who did substitute correctly were confused by the minus when correctly obtaining

cosh

dd

e

x

u

xu

-

=-

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

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

sinh2

u

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

Question 8

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

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

Mechanics M1 (R) (6677)

Introduction

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

In calculations the numerical value of g which should be used is 9.8, as advised on the front of the question paper. Final answers should then be given to 2 (or 3) significant figures – more accurate answers will be penalised, including fractions.

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

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

Report on Individual Questions

Question 1

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

Question 2

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

Question 3

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

Question 4

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

Question 5

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

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

Question 6

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

Question 7

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

There were some sign errors in the second part but most students were able to make a good attempt at finding a value for

µ

. Part (c) most were able to find the component of the weight down the plane but many lost the rest of marks by using the original value of the friction. Also a good number of answer marks were lost due to the use of a rounded value of μ which gave a value of 4.66 N instead of 4.70 N for the limiting friction force.

Mechanics M2 (R) (6678)

Introduction

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

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

Report on Individual Questions

Question 1

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

Question 2

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

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

Question 3

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

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

Question 4

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

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

Question 5

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

Question 6

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

Part (b) Many candidates understood the method required to find the distance OB, but there were often sign slips in using

2

1

2

sutat

=+

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

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

Question 7

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

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

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

Mechanics M3 (R) (6679)

Introduction

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

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

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

Report on Individual Questions

Question 1

Being given an expression for

d

d

v

x

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

d

d

v

av

x

=

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

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

Question 2

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

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

Question 3

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

cossin

2

mgx

mgmg

a

mqq

=-

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

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

Question 4

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

cos

q

and then used their value to obtain V.

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

4

5

a

to obtain the distance AX.

Question 5

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

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

Question 6

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

The proof of the simple harmonic motion required in part (b) was, as usual, found difficult. Many students persist in using α for the acceleration instead of

x

&&

and so restrict themselves to a maximum of 3 of the available 5 marks, while others fail to give a concluding statement and so lose the last mark even if their work is fully correct. Those who achieved an equation from which ω could be deduced were usually able to obtain the amplitude as a multiple of l in part (c).

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

Mechanics M4 (R) (6680)

Introduction

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

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

Report on Individual Questions

Question 1

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

Question 2

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

Question 3

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

Question 4

Those students who started with a clear diagram showing the velocities of both spheres before and after the collision worked through part (a) with few problems other than algebraic slips. They all understood that if

T

was at rest before the collision then it would move along the line of centres.

Part (b) required the students to recognise that the largest value of

d

occurs when θ has the largest possible value. The majority of students made little progress with this part.

Question 5

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

In part (b) the method was well understood, but there were some slips in the algebra and the arithmetic, and at the very end some students gave no evidence in support of their conclusion. It is not sufficient to state

2

2

d

0

d

y

q

>

without showing an expression for the derivative which takes account of the constant factor 4mgl, and clearly satisfies

2

2

d

0

d

y

q

>

.

Question 6

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

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

In part (c) with only a small number of exceptions, the students understood that the greatest value of

x

occurs when

0

x

=

&

and gave a correct solution, often in exact form.

Mechanics M5 (R) (6681)

Introduction

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

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

In calculations the numerical value of g which should be used is 9.8, as advised on the front of the question paper. Final answers should then be given to 2 (or 3) significant figures – more accurate answers will be penalised, including fractions.

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

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

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

Report on Individual Questions

Question 1

The most popular way to approach this question was using