# GCE Maths Examiner Feedback C3

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Examiner Feedback - C3

Core Mathematics 3

Examiner Feedback

Senior Examiners feedback on real student responses to some of the questions on the June 2011 and January 2012 examination papers. Supporting Edexcel GCE in Mathematics

Core Mathematics 3 (6665)

Contents

About this booklet

2

June 2011 - General examiner feedback

3

June 2011 Question 34

June 2011 Question 49

June 2011 Question 514

June 2011 Question 7

20

January 2012 - General examiner feedback

28

January 2012 Question 129

January 2012 Question 435

January 2012 Question 540

January 2012 Question 647

January 2012 Question 853

About this booklet

This booklet has been produced to support mathematics teachers delivering the GCE Mathematics specification, in particular the Core Mathematics 3 (C3) unit.

This booklet looks at some questions from the C3 June 2011 and January 2012 examination papers. It shows real student responses to these questions, and how the examining team follow the mark schemes to demonstrate how the students are awarded the marks.

How to use this booklet

Core Mathematics 3 (6665)

June 2011 - General Examiner Feedback

Candidates generally found this paper accessible with attempts being made by almost all students on all questions. The standard of calculus and algebra was pleasing with question 1 and question 7 being a useful source of marks. Questions that proved to be more demanding were 3(b), 4(b) and (d), 5(c), 6(b) and 8(c). The length of the paper was not an issue as there was little evidence of students failing to finish. Points which need to be addressed by candidates in future examinations are: Candidates would be well advised to state formulae before using them

Candidates should be more careful in their use and/or omission of brackets In show that questions, all necessary steps must be shown The rubric clearly states that answers without working may not gain full marks. This was applied in questions 5(c), 6(b)(ii),7(b) and 8(c).GCE Core Mathematics 3, June 2011

Question 3

For this question candidates were asked to transform a graph consisting of two straight line segments. Part (a) combined a translation and a stretch. Part (b) was more of a challenge and involved the modulus function.

In part (a), the majority of candidates achieved all three marks. A few candidates applied only one of the two transformations resulting in R at (0, 3) or (4, 6). Others applied a scale factor of instead of 2 resulting in R at (0, 1.5). Both branches of the V were required to cross the x-axis and a few candidates lost a mark because of this. Wrong notation for (0, 6) was a little too frequently seen, but the position on the graph often clarified the students intention.

Part (b) was more demanding for candidates who were generally conversant with the modulus function and sketched a W shape. The most common mistake was to have the W shape shifted to the right so that at least one of the vertices was on the positive x-axis. Some candidates gave incorrect coordinates even though their drawing suggested that their R was correct. Even when the diagram was correct, R was at times seen labelled as (3, 4), (3, 4) or (4, 3). Mark Scheme

Question NumberSchemeMarks

3. (a)V shapeVertex on y-axis and both

branches of graph cross x-axis

y co-ordinate of R is -6B1

B1

B1

(3)

(b)W shape

2 vertices on the negative x-axis.

W in both quad 1 and quad 2.

R = (4, 3)B1

B1dep

B1

(3)

6 Marks

Student Attempt A

Part (a)

B1 B1 B1 : This is a fully correct solution.

Part (b)

B1 B0 B0 : This is marked as a special case, scoring 1 mark when the candidate draws

QUOTE

instead of

QUOTE

Student Attempt B

Part (a)

B1: A V shape has been drawn.B1: The vertex is on the y-axis and their graph appears in both quadrants 1 and 2.B0: R is incorrectly given as (0, 3).Part (b)

B1: A W shape has been drawn.B1: The two vertices are on the negative x-axis and it appears in both quadrants 1 and 2.B0: R is incorrectly given as (3, 4).Student Attempt C

Part (a)

B1: A V shape has been drawn.B0: The vertex is not on the y-axis.B1: The y coordinate of R is correctly given as 6.Part (b)

B1: A W shape has been drawn.B0: The two vertices are on the positive x-axis, not the negative x-axis as required.B0: R is incorrectly given as (4, 3).GCE Core Mathematics 3, June 2011

Question 4

This was a fairly straightforward question that tested candidates knowledge of functions with respect to both

QUOTE

proved an additional challenge.

QUOTE

. Candidates usually struggle with the concept of domain and range. The negative in

QUOTE

and

In part (a) the vast majority of candidates knew how to find the inverse of a function but poor algebra was seen, revealing a lack of understanding in the basic laws of logarithms.

QUOTE

often led to .There were a variety of numerical values given in (b), but most just stated it as being any real number. Those who did manage to obtain 4, often wrote

, indicating that the idea of a domain was not fully understood.

QUOTE

or

or In part (c) the composite function was well attempted and simplified. It was fairly common however to see candidates getting to

QUOTE

and stopping.

The candidates in part (d) were more successful than in part (b), but there was little evidence of a realisation that the two answers were linked. A few graphs appeared, but many who correctly answered (c) could not establish the range of a negative quadratic. Mark Scheme

Question NumberSchemeMarks

4. (a)

oeM1

M1A1

(3)

(b)

B1(1)

(c)

M1

dM1A1

(3)

(d)

B1ft(1)

8 Marks

Student Attempt A

(a) M1 M1 A1: This is a perfectly acceptable response. (b)B0: The domain is incorrect.

(c) M1 M1 A1: The function fg(x) has been stated and fully simplified. The factorisation is correct but was not required for this question.

(d) B0: The range is not correct, nor does it follow through from the students answer to (b).Student Attempt B

(a) M1: The candidate makes an attempt to change the subject of the formula.

M0: The candidate fails to deal with the negative correctly and hence uses incorrect ln work.

A0: Incorrect answer.(b) B0: The domain is incorrect. This was a common answer.

(c) M1 M1 A1: The function fg(x) has been stated and fully simplified. There is no incorrect working.

(d)B1: The range is correct. The working is also correct but is not required for a B mark.

Student Attempt C

(a) M1 M1 A1: This is a fully correct solution.(b) B0: The domain is incorrect.

(c) M1 M0 A0 M1: The function fg(x) has been stated.

M0: The final answer is correct but the ln work is incorrect on line 2.

A0: This follows M0.

(d) B0: The range is incorrect nor does it follow through from their answer to (b).

GCE Core Mathematics 3, June 2011

Question 5

For this question, candidates needed to solve ln and exponential equations in a real life context. Part (b) involved proving the value of a constant from the information given. Part (c), involving a differential, was one of the more demanding aspects of the whole paper.

Part (a) was almost always answered correctly.In part (b) most candidates then correctly substituted and

QUOTE

. Only the better candidates, however, were able to show the given value of k without doubt. It was not uncommon to see students writing

resulting in the line and QUOTE

going straight on to In part (c) the derivative was not too difficult, but the numbers used made this question tricky. Many candidates, however, did end up with a coefficient of t in their differential. There were a pleasing number of completely correct solutions, by and large using the method shown on the mark scheme. A small number were able to proceed successfully with a change to powers of 3. A common difficulty was in the processing of the QUOTE

term caused by the lack of a bracket around the, meaning that

was sometimes processed in error. Mark Scheme

Question NumberSchemeMarks

5. (a)

B1

(1)

(b)

M1

M1

dM1

A1*

(4)

(c) ft on their p and k

4.15 or awrt 4.1M1A1ftM1A1

dM1

A1

(6)

11 Marks

Student Attempt A

(a) B1: Correct.(b) M1: 2.5 and 4 have been correctly substituted into .

M1: scores this mark.

M1: lns have been correctly taken to achieve .

A0: The given answer has been written down but it has not been shown to be true. If there

was an extra line with replaced by this would have been acceptable.

(c)M1 A1: The expression has been differentiated and given numerically (within a later equation).

M1 A1: They have set and achieved .

M1: The candidate has taken lns of both sides.

A0: The final answer is incorrect. The terms were not cancelled in the 4th line from the end.

Student Attempt B

(a) B1: Correct.(b) M1: 2.5 and 4 have been correctly substituted into .

M1: has been implied by the line .

M1: lns have been correctly taken to achieve .

A1: The given answer has been shown to be true. (is evidence of this).(c) M1: The expression has been differentiated to .

A1: This can be awarded as a result of their final answer. (Correct value of k used) .

M1 A0: They have set and achieved .

M1: The candidate has taken lns of both sides.

A0: The final answer is correct but cannot be awarded as does not exist.

Student Attempt C

(a) B1: Correct.(b) M1: 2.5 and 4 have been correctly substituted into .

M1: Can be awarded for the line .

M1: lns have been correctly taken to achieve .

A0: The given answer has not been shown to be true. going to needs to be fully justified.

(c) M1 A1: The expression has been differentiated correctly.

M0: The differential has been set to but the line constant is never reached.

A0 M0 A0: No more marks can be awarded.GCE Core Mathematics 3, June 2011

Question 7

Part (a) of this question involved the simplification of a rational expression. This was straightforward as long as the denominators were in the factorised forms. Using the given answer to part (a), candidates then had to differentiate using the quotient or chain rule to produce an equation of the normal to the curve.

The level of accuracy in part (a) was pleasing with a significant majority of candidates scoring full marks. Nearly all factorised x2 A large proportion of candidates answered part (b) correctly and efficiently. Nearly all knew the required set of steps to reach the answer, with algebraic and arithmetical errors rather the cause of dropped marks. Those who multiplied out the brackets, followed by an application of the quotient or the chain rule, were most successful. Errors arising from differentiating 5 to give 1, careless slips with the derivative of the denominator and missing brackets leading to Mark Scheme

Question NumberSchemeMarks

7. (a)

B1M1

M1A1

A1*

(5)

(b)f (x)

f (x)

f (1)

Use m1m2 = 1 to give gradient of normal

their

or any equivalent formM1M1A1

M1A1

M1

M1

A1

(8)

13 Marks

Student Attempt A

Part (a)

(a) B1 M1 M1 A1 A1: This is a completely correct response. Part (b)

(b) M1 M1 A1: These marks are awarded on line 4. The candidate subsequently multiplies out the bracket incorrectly but the award can be made on the previous line.

M1: The candidate substitutes into their (now incorrect) differential.

A0: Their answer of is incorrect.

M1: going to is a correct method for finding the gradient of the normal.

M1: There is a correct method for the equation of the normal. The point has been used in a linear equation with gradient .

A0: The answer is incorrect.

Student Attempt B

Part (a)

(a) B1 M1 M1 A1 A1: This is a completely correct response. Part (b)

(b) M1 M1 A1: The expression for is correct.

M0: The candidate never substitutes into to find a numerical gradient.

A0: Follows M0.

M0: No numerical value of the normal gradient has been attempted.

M0: No linear equation has been attempted using numerical values.

A0: Incorrect answer.Student Attempt C

Part (a)

(a) B1: x2 9 has been factorised as .

M0: The fractions have not been combined appropriately. The denominators need to be correct.

M0: The single fraction is not in the required form (as a result of the denominator).

A0 A0: Following the M0s.Part (b)

(b) M0: Neither the product or chain rule has been applied to f(x).

M1: It can be implied that the correct method has been used to differentiate

. Evidence used for this is .

A0: The value given for is incorrect.

M1: The candidate substitutes into their (incorrect) differential.

A0: Their value of the gradient, 15 is incorrect.

M1: The numerical value of the normal gradient is correct for their tangent gradient.

M1: A linear equation has been attempted using their normal gradient and .

A0: Incorrect answer.

Core Mathematics 3 (6665)

January 2012 - General Examiner Feedback

This paper proved to be very accessible to most of the candidates and there was very little evidence of candidates being short of time. The paper afforded a typical E grade candidate plenty of opportunity to gain marks across the majority of questions.

The standard of algebra seen was good, although a number of candidates made basic sign, bracket or manipulation errors.

There was little evidence in this paper of candidates working in degrees instead of radians or vice versa.

In summary, questions 1, 2, 3,6(a), 6(c), 6(d) and 7 were a good source of marks for the average candidate, mainly testing standard ideas and techniques; and questions 4, 5,6(b) and especially question 8 were discriminating questions at the higher grades.'

GCE Core Mathematics 3, January 2012

Question 1

This question tested the candidates ability to differentiate expressions using the product and quotient rules. An extra level of difficulty in both parts (a) and (b) involved the use of the chain rule in order to differentiate the terms ln (3x) and sin (4x).

The question was answered very well, with many candidates scoring full marks.In part (a) candidates are advised to write down the product rule before attempting to use it. Failure to do so, followed by incorrect expressions, does risk the loss of many marks in such questions. A common mistake in (a) was to differentiate ln (3x) to get instead of .

Some candidates also failed to simplify their answer and left it as 2x ln (3x) or even just as

2x ln (3x) .

Part (b) was equally encouraging with again, most students writing down the quotient rule. A common error was seen in differentiating sin (4x) to just cos (4x). Worryingly a large number of candidates simplified the denominator (x3)2 A minority of candidates chose to write the expression as x 3 sin (4x) and proceeded using the product rule.

Mark SchemeQuestion NumberSchemeMarks

1. (a) QUOTE

for any constant BM1

Applying vu( + uv(, M1, A1 A1

(4)

(b) Applying QUOTE

M1 A1+A1 A1

A1 (5)

(9 marks)

Student Attempt A

(a) M1 M1 A1 A1: The product rule can be implied by their u, u, v, v and the subsequent expression for. It is also simplified correctly as required by the question.

(b) M1: The quotient rule can be implied by their u, u, v, v and the subsequent expression for .

A1: The first term on the numerator is correct.

A1: The second term on the numerator is correct.

A1: The denominator is correctly simplified to x6.

A0: Their correct expression has not been simplified.

Student Attempt B

(a) M1: v = ln (3x) going to scores this mark.

M1: The product rule is stated and clearly applied correctly.

A1: The 2x ln 3x term is correct (and simplified).

A0: The other term is incorrect as is incorrect.(b) M1: The quotient rule is stated and correctly applied to the expression.

A1: The first term on the numerator is correct.

A1: The second term on the numerator is correct.

A0: The denominator has not been simplified to x6.

A0: Their expression (although correct) has not been simplified as required by the question.

Student Attempt C

(a) M1: This can be awarded for v = ln (3x) going to .

M0: The product rule has not been quoted. We therefore look at the expression. It could be implied from their answer that they think the product rule is vu + uv

A0 A0: Follows from M0.

(b) M1: The quotient rule is quoted and an attempt has been made to apply it (even though there is an error on the denominator). The written formula was crucial in awarding this mark.

A1: The first term on the numerator is correct.

A1: The second term on the numerator is correct.

A0: The denominator is incorrect.

A0: The final expression is incorrect.

GCE Core Mathematics 3, January 2012

Question 4

For this question candidates needed to find from a function expressed in the form

QUOTE

. In order to find an equation of the normal, they needed to use the numerical value of the normal gradient together with a correct point that lies on the line.

This question tested stronger candidates. It seems as though this area is misunderstood by many candidates with

QUOTE

being changed without reason to and

being swapped to. Many were able to differentiate tan

QUOTE

, or the loss of the

QUOTE

. Common errors seen were the loss of a factor of 2, introduction of a new coefficient of QUOTE

to QUOTE

term altogether. Candidates who achieved QUOTE

correctly, sometimes failed to invert it or just then used it as A very common error was

QUOTE

being rewritten as . Most candidates were aware that they had to substitute QUOTE

into their differentiated expression or the reciprocal of their differential but the evaluation was too frequently incorrect. The most common incorrect answers were 2, , 4,

Most candidates who lost all other marks did manage to get the mark for

.

As expected, the most common error was in using the wrong value for the gradient of the normal, and hence the equation of the normal. For most candidates this required two inversions of their derivative whereas a more simple approach would have been to write down their numerical value of

.

Mark Scheme4.

M1, A1

substitute QUOTE

into their M1, A1

When , awrt 3.46B1

their m(x their M1

A1

(7 marks)

Student Attempt A

M1 A1: The expression is differentiated correctly and set as. See line 2.

M1: is substituted into their .A0: The value of is incorrect (as a result of an error in inverting ).B1: is given.M1: The candidate uses their normal gradient with (, ) to produce an equation of the normal.

A0: The final equation is incorrect.

Student Attempt B

M1 A1: The expression is differentiated correctly and set as .M1 A1: is substituted into their and correctly stated as .B1: is given.M0: The candidate uses their tangent gradient with (,) to produce an equation. This is an incorrect method.

A0: The final equation is therefore incorrect.

Student Attempt C

M1 A1: The expression is differentiated correctly and set as .M0 A0: is never substituted into their or .B0: is never stated.M0: The candidate uses a normal gradient but it is never numerical. The normal equation must be in a linear form.

A0: The final equation is therefore incorrect.

GCE Core Mathematics 3, January 2012

Question 5

Questions on solving equations using QUOTE

are familiar to students from previous examinations and are usually straightforward. What made this one more demanding was the issue of having to deal with rather than

.

This was another demanding question for weaker candidates although many fully correct solutions were seen. A lack of understanding was seen by some who replaced

altogether or by others who substituted

QUOTE

, by others who ignored the QUOTE

by However candidates who were able to use some form the identity

were able to obtain at least one correct value for

QUOTE

was almost universally recognised, and in general candidates who reached the stage of cosec 3 = 3

QUOTE

was almost always then obtained. The invalidity of cosec 3

or

QUOTE

. Correct factorisation leading to cosec 3 = 3 QUOTE

usually formed a valid quadratic equation in . About half the candidates getting to this stage went on to obtain 4correct solutions. The others often gave 2 or 3 correct solutions. Mark Scheme5.Uses the identity in

M1

A1

M1

A1

, = awrt 6.5M1, A1

, 53.5 M1,A1

M1

All 4 correct answers awrt 6.5,53.5,126.5 or 173.5A1

(10 marks)

Student Attempt A

M1 A1: A correct quadratic in cosec 3 is obtained. The = 0 can be implied by subsequent working.M1 A1: The value cosec3 = 3 is correctly obtained.M1 A1: The principal value = 6.5 is correctly obtained. M1 A1: A secondary value = 126.5 is correctly obtained.M1: A correct method is used to find a third value- 173.5 can be used as evidence.A0: All four values need to be correct. 53.8 is incorrect.Student Attempt B

This is an alternative but perfectly acceptable method.

M1: cot2 is replaced by , cosec is replaced by and cos2 is replaced by 1 sin2 to produce a quadratic equation in sin

A0 : An error on line 5 means that the equation in sin3 is incorrect.

M1: The use of the quadratic formula justifies this award.A0: This is for sin (3) = . It is not reached due to earlier error.M1: A correct method can be seen to find their principal value.A0: Their principal value is incorrect.M1: There is evidence that this can be awarded. 180 - their principal value is sufficient.

A0: Their second value is incorrect.M0: There is no method seen to find a third value.A0: Follows M0.Student Attempt C

M0: The candidate replaces cosec u with 1 = cot u. This is an incorrect statement and incorrect

method.A0: Follows M0.M0: An equation in cosec or sin is required. This mark is dependent upon the first being awarded. No more marks can be gained, hence A0 M0 A0 M0 A0 M0 A0.

GCE Core Mathematics 3, January 2012

Question 6

Questions on roots of equation and iterative methods are usually extremely well done by candidates. The twist in this one, making it a more challenging example, involved a turning point in part (b) and the root of the differential, rather than the usual f (x) , in part (d).

This type of question is accessible to most candidates with zero scores being rare. (a) This part was done well on the whole with most candidates picking up marks. Part (b) was a subtle variation on a usual theme, with both differentiation and rearrangement required to score full marks. A few unfortunate candidates tried without any success to rearrange f(x) = 0. (c) Marks were generally lost due to degrees rather than a lack of accuracy. Otherwise this part resulted in 2 or 3 marks for most candidates. (d) The majority of candidates gained the M mark for the correct interval. A large number lost the last two marks for substituting into f (x). A few candidates also tried repeated iteration which was not by using a suitable interval. Only a small number of students completed the last part correctly.

Mark SchemeQuestion NumberSchemeMarks

6. (a)f (0.8) = 0.082, f (0.9) = 0.089M1

Change of sign ( root (0.8,0.9)A1 (2)

(b)f (x)

M1 A1

Sets f (x) = 0 QUOTE

( M1 A1* (4)

(c)Sub x0 = 2 into

M1

x1 = awrt 1.921, x2 = awrt 1.91(0) and x3 = awrt 1.908A1, A1 (3)

(d)[1.90775, 1.90785]M1

f (1.90775) = 0.00016... AND f (1.90785) = 0.0000076...M1

Change of sign ( x = 1.9078 A1 (3)

(12 marks)

Student Attempt A

(a) M1 A1: Completely correct. Both values are correct, there is a reason and a conclusion.(b) M1 A1: The differentiation is correct.

M1 A1: is implied and the candidate correctly reaches the given answer.

(c) M1 A1 A1: All three solutions are correct.(d) M1 A1 A1: This is a perfectly acceptable proof using g(x) x rather than f (x). The method is equivalent.

Student Attempt B

(a) M1: Both f (0.8) and f (0.9) have been calculated.

A0: A valid reason has been offered (sign change) but the conclusion needs to mention either x or root.(b) M1 A1: The differentiation is correct.

M1 A1: is implied and the candidate correctly reaches the given answer.

(c) M1 A1 A1: All three solutions are correct.(d)M1: The correct interval has been chosen.

M0: The candidate calculates f (x) rather than f (x) as required. This was very common.

A0: Follows M0.Student Attempt C

(a) M1 A1: Completely correct. (b) M1 A1: The differentiation is correct.

M0: is neither stated nor implied by the setting of their .

A0: Follows M0. This is a given solution requiring a full reason.

(c) M1 A1 A1: All three solutions are correct.(d) M0: The interval is incorrect. 1.90885 is given instead of 1.90785.

M0 A0: Follows the M0.

GCE Core Mathematics 3, January 2012

Question 8

The first part of this question tested the candidates ability to prove two trigonometrical identities. Part (c) was much more demanding, and required the candidate having to spot how to use the identity in (b) to solve a complex equation.

(a) Most candidates were able to gain the first 2 marks. Many did not then realise, that to progress, they needed to divide both the numerator and denominator by

. Most candidates then wrote down a method that was either incomplete or incorrect.

(b) Again most candidates were able to score 2 marks. Responses that gave as

QUOTE

often failed to show how their expression could be modified to the form required.

usually went on to score full marks but those that used (c) There were responses that gained full marks. However, it was also very common for candidates to score only the first mark for using the given identity to obtain

. At this stage many failed to realise that this could be used to write down a simple equation from which a value of could be found. Most candidates who didnt spot the conventional route then in effect restarted the question and tried to expand tan

. The majority then gained some credit.

Mark SchemeQuestion numberSchemeMarks

8. (a)

M1A1

QUOTE

M1

A1 * (4)

(b)

M1

M1

A1 * (3)

(c)

M1

M1

M1 A1

M1

A1 (6)

(13 marks)

Student Attempt A

(a) M1 A1: The candidate gains the first two marks on their 4th line of working.

M1: The sentence Divide top and bottom by cos A cos B justifies this award.

A0: This is a given solution and there needs to be evidence of how tan A and tan B appear. The candidate could have just written down the answer.

(b) M1 M1 A1: This is completely correct. Line 3 to line 4 is a reasonable jump.

(c) M1 M1 M1 A1 M1 A1: This is completely correct using the alternative method.

is equivalent to .Student Attempt B

(a) M1 A1: The candidate gains the first two marks on their 2nd line of working.

M1: Taking out a common factor of cos A cos B is equivalent to dividing by cos A cos B.

A0: This is a given solution and all parts need to be correct. The bracketing on the 2nd of last line is incorrect

(b) M1: The compound angle identity is used with A = and B = .

M1: tan is correctly substituted in this expression.

A0: Although a reason is seen, multiply by needed to be performed on both the numerator and denominator for this mark to be awarded.(c)M1 M1 M1 A1: One solution in the range, has been correctly found.

M0 A0: No further progress is made in finding a second value in the range.

Student Attempt C

(a) M1 A1: The candidate gains the first two marks on their 2nd line of working.

M0 A0: No further work is seen.(b) M1: The compound angle identity is used with A = and B = .

M1: tan is correctly substituted in this expression.

A0: This is an unreasonable jump. No evidence can be seen to justify the award of this mark. They could have just written down the final expression.

(c)M1: This is the alternative solution. tan(( ) is replaced by tan to produce a quadratic in tan .

M1: Terms are collected to produce a quadratic equation in tan .

M0: An error means that no quadratic formula is used, hence M0.

M0 A0 A0: No further marks can be awarded.

Pearson Education Ltd is one of the UKs largest awarding organisations,

offering academic and vocational qualifications and testing to schools, colleges, employers and other places of learning, both in the UK and internationally. Qualifications offered include GCSE, AS and A Level, NVQ and our BTEC suite of vocational qualifications, ranging from Entry Level to BTEC Higher National Diplomas. Pearson Education Ltd administers Edexcel GCSE examinations. Through initiatives such as onscreen marking and administration, Pearson is leading the way in using technology to modernise educational assessment, and to support teachers and learners.If you have any subject specific questions about the content of this booklet that require the help of a subject specialist, you may find our Ask The Expert email service helpful. Ask The Expert can be accessed online at the following link: http://www.edexcel.com/Aboutus/contact-us/This Booklet has been written to provide teachers with additional support for GCE Mathematics. It provides real student responses to past examination questions with tips and guidance from our senior examining team.

Other publications in this series:

Mechanics 1 (M1) Examiner Feedback

Acknowledgements

Edexcel would like to thank Alistair MacPherson (Principal Examiner) for contributing his time and expertise to the development of this publication.

April 2012

All the material in this publication is copyright

Pearson Education Ltd 2012

Introduction to the question on what is expected from the student

Examiner commentary on the overall performance on the question

Student response

Examiner commentary on the students response

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Blank question from the exam paper

Total number of marks the student was awarded.

Mark scheme for this question

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