Examiner Feedback - C1 - Pearson qualifications



Core Mathematics 1

Examiner Feedback

Senior Examiner’s feedback on real student responses to questions on the June 2012 examination papers.

Supporting Edexcel GCE in Mathematics

Core Mathematics 1 (6663)

Contents

| | |

|About this booklet |2 |

|June 2012 - General examiner feedback |3 |

|June 2012 – Question 1 |4 |

|June 2012 – Question 2 |9 |

|June 2012 – Question 3 |14 |

|June 2012 – Question 4 |19 |

|June 2012 – Question 5 |24 |

|June 2012 – Question 6 |29 |

|June 2012 – Question 7 |37 |

|June 2012 – Question 8 |43 |

|June 2012 – Question 9 |52 |

|June 2012 – Question 10 |61 |

About this booklet

This booklet has been produced to support mathematics teachers delivering the GCE Mathematics specification, in particular the Core Mathematics 1 (C1) unit.

This booklet looks at questions from the C1 June 2012 examination paper. 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 1 (6663)

June 2012 - General Examiner Feedback

This paper proved a good test of candidates’ knowledge and candidates’ understanding of Core 1 material. There were plenty of easily accessible marks available for candidates who were competent in topics such as differentiation, integration, recurrence relations, arithmetic series and basic transformation of curves. Therefore, a typical E grade candidate had enough opportunity to gain marks across the majority of questions. At the other end of the scale, there was sufficient material, particularly in Q8 and Q9 to stretch and challenge the most able candidates.

While standards of algebraic manipulation were generally good, some weakness in this area was seen in Q5(c), Q8(a), Q9(b) and Q9(c). In Q6, it was disappointing to see a number of good candidates who made arithmetic errors in calculations such as 10 + 14(5) in part (a) and

[pic][ 2(10) + 59(5) ] or even 30 × 315 in part (b). Also a significant minority of candidates in Q7(b) incorrectly simplified [pic] to [pic].

In summary, Q1, Q2(a), Q4, Q5, Q6(a), Q6(b), Q9(a), Q9(b) and Q10 were a good source of marks for the average candidate, mainly testing standard ideas and techniques; and Q2(b), Q3, Q6(c), Q6(d), Q7, Q9(c) and Q9(d) were discriminating at the higher grades. Q9(e) was the most demanding question with only about 3% of the candidature able to correctly find the area of the quadrilateral ACBE.

GCE Core Mathematics 1, June 2012

Question 1

This question tests a candidate’s ability to apply the integration rule of [pic]

to three different types of term. It also tests a candidate’s understanding of indices, especially when integrating [pic] with respect to x.

About three-quarters of the candidature achieved all 4 marks in this question.

While most candidates were able to integrate both [pic] and 5 correctly, a significant minority struggled to integrate [pic] correctly, giving incorrect answers such as [pic] or [pic] or [pic]

or [pic].

Incorrect simplification of either [pic] to [pic] or [pic] to [pic]; or not simplifying [pic]to [pic] and the omission of the constant of integration were also other common errors.

It was pleasing to see very few candidates who differentiated all three terms.

Mark Scheme

|Question Number |Scheme |Mark |

|1. |[pic] |M1 A1 |

| |[pic] |A1; A1 |

| | | |

| | |4 Marks |

Student Attempt A

M1: For either [pic][pic] or [pic] or [pic].

A1: For either [pic] or [pic], which can be un-simplified or simplified.

A0: This mark is not earned as the [pic] term has become an [pic] term in the simplified version. This mark is for both the [pic] and [pic] terms correct and simplified on the same line.

A1: For [pic] appearing on one line.

Student Attempt B

M1: There are three ways this response could earn the M1 mark. This is for either [pic] becoming [pic]; or for [pic] becoming [pic] or for 5 integrating to [pic].

A1: For the correct un-simplified [pic] seen on first line.

A0: This mark is for both the [pic] and the [pic] terms correct and simplified on the same line, which is not achieved here.

A0: This mark is for [pic] which should appear on one line. The “ + c ” is missing.

Student Attempt C

M1: There are two ways this response could gain the M1 mark here. This is for either [pic] becoming [pic] or for 5 integrating to [pic].

A0: This response does not achieve either [pic] or [pic] which can be un-simplified or simplified. NOTE: Even though [pic] appears in this response - it is the incorrect result of combining the incorrect integration of [pic] with the incorrect integration of [pic].

A0: The first two terms are integrated incorrectly.

A1: For [pic] appearing on one line.

GCE Core Mathematics 1, June 2012

Question 2

This question tests a candidate’s application of the law of indices for rational exponents. In part (a), the equivalence of [pic] as [pic] is tested. Part (b) is more demanding and tests [pic], where [pic] is a bracketed term. In this case the candidate is required to reciprocate and square root all three elements in the brackets.

This question proved discriminating with about a third of the candidature gaining all 4 marks.

In part (a), the majority of candidates were able to evaluate [pic] as 8. Many of those who were unable to achieve 8, were able to score one mark by rewriting [pic] as either [pic] or [pic]. Those candidates who chose to cube 32 first to give 32768 were usually unable to find [pic]. Common errors in this part included rewriting [pic]as either [pic] or [pic] or evaluating [pic] as 6.

Part (b) proved more challenging than part (a), with the majority of candidates managing to obtain at least one of the two marks available by demonstrating the correct use of either the reciprocal or square root on [pic]. The most able candidates (who usually reciprocated first before square rooting) were able to proceed efficiently to the correct answer. The most common mistake was for candidates not to square root or not to reciprocate all three elements in the brackets. It was common for candidates to give any of the following incorrect answers:

[pic], [pic], [pic], [pic], [pic] or [pic].

Mark Scheme

|Question Number |Scheme |Mark |

|2. (a) |[pic] |M1 |

| |[pic] |A1 |

| | |(2) |

| | | |

|(b) |[pic] |M1 |

| |[pic] |A1 |

| | |(2) |

| | |4 Marks |

Student Attempt A

(a) M1: For [pic].

A1: Leading to a correct answer of 8.

(b) M1: The method mark is for use of the power of [pic] or the power of (1. Even though it is appreciated that the transition of [pic] fails to apply the power of [pic] to two of the three elements, the transition of [pic], shows that the power of (1 has been correctly applied to all three elements.

A0: Incorrect answer.

Student Attempt B

(a) M1 A1: For [pic] leading to a correct answer of 8.

(b) M0: [pic] does not show that the power of (1 or the power of [pic] has been fully applied.

A0: Incorrect answer.

Student Attempt C

(a) M1: For either [pic] or [pic].

A0: Correct answer of 8 is not achieved.

(b) M1: [pic] shows that the power of [pic] has been correctly applied on all three elements.

A0: Incorrect answer.

GCE Core Mathematics 1, June 2012

Question 3

This is an unstructured question testing the use and manipulation of surds. To solve this problem candidates have to apply two methods – one of rationalising the denominator and one of simplifying the surds [pic] and[pic]. These methods can be applied in any order, resulting in a number of different solutions.

This question proved discriminating, with just under a half of the candidature gaining all 5 marks. About a third of the candidature gained only two marks by either correctly rationalising the denominator or by correctly simplifying [pic] and [pic].

A significant number of candidates answered this question by multiplying [pic] by [pic]

to obtain [pic] and proceeded no further. A small number of candidates multiplied the denominator incorrectly to give either [pic] or [pic].

Those candidates who realised that [pic] and [pic]usually obtained the correct answer of [pic], although a few proceeded to the correct answer by writing [pic][pic] as [pic].

A number of candidates started this question by firstly simplifying [pic] and[pic]to give [pic], but many did not take the easy route of cancelling this to [pic] before rationalising. Although some candidates wrote [pic] as [pic], those candidates who rationalised this usually obtained the correct answer.

A small minority of candidates attempted to rationalise the denominator incorrectly by multiplying [pic] by either [pic] or [pic].

Mark Scheme

|Question Number |Scheme |Mark |

|3. |[pic] |Writing this is sufficient for M1. |M1 |

| |[pic] |For 12 ( 8. |A1 |

| | |This mark can be implied. | |

| |[pic] | |B1 B1 |

| |[pic] | |A1 cso |

| | | | |

| | | |5 Marks |

Student Attempt A

M1: For a correct method to rationalise the denominator.

A1: For the denominator correctly rationalised to give 12 ( 8 (this is incorrectly simplified to 2).

B1: For [pic].

B1: For [pic].

A0: Incorrect final answer.

Student Attempt B

M0: Multiplying by [pic] is an incorrect method for rationalising the denominator.

A0: Follows from M0. (Even though 12 ( 8 appears in the denominator).

B0: [pic] or [pic] is not seen or implied.

B0: [pic] or [pic] is not seen or implied.

A0: Follows from the earlier M0.

Student Attempt C

M0: No attempt to rationalise the denominator.

A0: Follows from M0.

B1: For [pic].

B1: For [pic].

A0: Follows from the earlier M0.

GCE Core Mathematics 1, June 2012

Question 4

This question tests a candidate’s ability to apply the differentiation rule of [pic] to four different types of term, as well as a candidate’s understanding of [pic]. It also tests a candidate’s understanding of indices, especially when differentiating [pic] and later [pic] with respect to x.

This question was well answered with about two-thirds of the candidature achieving all 6 marks.

In part (a), a small minority of candidates struggled to deal with the fractional power when differentiating [pic]. Some candidates incorrectly reduced the power by 1 to give a term in [pic]; whilst other candidates struggled to multiply (6 by [pic] or incorrectly multiplied (6 by [pic]. A few candidates did not simplify [pic] to give [pic]or integrated throughout, or added a constant to their differentiated expression.

In part (b), the most common error was for candidates to differentiate [pic] to give [pic]. A few candidates did not understand the notation for the second derivative, with some integrating their differentiated result in part (a) to achieve an answer similar to the expression given in the question.

Slips occurred with the omission of x from fractional power terms, with some candidates writing [pic] in part (a) or [pic]in part (b).

Mark Scheme

|Question Number |Scheme |Mark |

|4. (a) | [pic] | | |

| |[pic] | |M1 |

| |[pic] | |A1 A1 A1 |

| | | |(4) |

| | | | |

|(b) |[pic] | |M1 A1 |

| | | |(2) |

| | | |6 Marks |

Student Attempt A

(a) M1: For either [pic] or [pic] or [pic].

A1: For [pic].

A0: A simplified [pic]is not achieved.

A1: For the constant term of 2.

(b) M1 A1: For a correct un-simplified expression achieved on the first line of part (b).

Student Attempt B

(a) M1: There are two ways this response could gain the M1 mark here. This is for either [pic]becoming [pic] or for [pic] differentiating to 2.

A1: For [pic].

A0: [pic]is not achieved.

A1: For the constant term of 2.

(b) M0: The candidate’s [pic] is not differentiated correctly or the candidate does not achieve [pic].

A0: Follows from M0.

Student Attempt C

(a) M1: For either [pic] or [pic] or [pic].

A0: Incorrect [pic] term.

A0: Incorrect [pic] term.

A1: For the constant term of 2.

(b) M0: The candidate’s [pic] is not differentiated correctly or the candidate’s [pic] does not differentiate to [pic].

A0: Follows from M0.

GCE Core Mathematics 1, June 2012

Question 5

This question tests a candidate’s understanding and use of recurrence relations of the form [pic]. Because part (b) has its answer given, the candidate can see if they are applying the recurrence relation formula correctly. Part (c) is more demanding because it links together the topics of sigma notation, recurrence relations and inequalities.

This question was answered more successfully by candidates than similar ones in the past. The notation did not appear to be such a mystery with most candidates realising that this question tested the topic of recurrence relations and not arithmetic sequences. About two-thirds of the candidates gained at least 6 out of the 7 marks available.

Part (a) was generally very well answered and indeed most who got this part right went on to score most of the marks in the question. Those who were unsuccessful often tried to work back from the given [pic] to arrive at [pic].

In part (b), most candidates scored full marks. Occasionally problems were caused by the incorrect use of brackets.

In part (c), the majority of candidates were able to find [pic], sum the first four terms of the sequence and write their sum [pic]23 or = 23. Some candidates found [pic] incorrectly by omitting brackets resulting in [pic]. Instead of summing the first four terms, a number of candidates solved [pic] or put each term [pic]. A few candidates summed by adding either

[pic], [pic] and [pic], or [pic], [pic]and [pic] or even [pic], [pic], [pic] and [pic]. Some arithmetic errors were made in summing the four terms and a number of candidates miscopied [pic] as 27 – 7c. A number of candidates used the formula for the sum to n terms of an arithmetic series in order to sum their four terms. A significant number of candidates, who achieved[pic], did not know that dividing by a negative number reverses the sign of the inequality. Those who rearranged [pic] into [pic] were more successful in achieving the correct result.

Mark Scheme

|Question Number |Scheme |Mark |

|5. (a) |[pic] , c is a constant | |

| |[pic] |B1 |

| | |(1) |

|(b) |[pic] |M1 |

| | [pic] (*) |A1 cso |

| | |(2) |

|(c) |[pic] |M1 |

| |[pic] |M1 |

| |[pic] or [pic] |M1 |

| |[pic] or [pic] |A1 cso |

| | |(4) |

| | |7 Marks |

Student Attempt A

(a) B1: Correct answer of [pic].

(b) M0: The candidate does not substitute their [pic] correctly into [pic].

A0: Follows from M0. Note that the answer [pic] is given in the question.

(c) M1: For substituting [pic] correctly into [pic], (although [pic] is expanded incorrectly).

M1: For an attempt to sum their [pic], [pic], [pic] and [pic].

M1: For the sum of their four terms [pic].

A0: Incorrect answer.

Student Attempt B

(a) B1: Correct answer of 6 ( c .

(b) M1: For substituting [pic] correctly into [pic].

A1: Leading to [pic] with no errors seen.

(c) M0: There is no substitution of [pic] into [pic].

M0 M0: For applying the formula for the sum to n terms of an arithmetic sequence.

A0: Follows from M0.

Student Attempt C

(a) B1: Correct answer of 6 ( c .

(b) M0: The candidate does not substitute their [pic] correctly into [pic].

A0: Follows from M0. Note that the answer [pic] is given in the question.

(c) M0: [pic]is not correct and there is no correct method of substituting [pic] into [pic].

M1: For an attempt to sum their [pic], [pic], [pic] and [pic].

M0: Candidate has not put the sum of their four terms either ≥ 23 or = 23, but is trying to solve “their sum = 0”.

A0: Follows from M0.

GCE Core Mathematics 1, June 2012

Question 6

This question tests a candidate’s application of arithmetic series to real-life problems. Part (a) and part (b) test standard formulae. Part (c) is more demanding, requiring the candidate to unify the units and reduce the sum formula to an unfamiliar form. In part (d), the solution of [pic] although trivial, requires a certain mathematical maturity that some candidates may lack.

This question was both well answered and discriminating with about three-quarters of the candidature gaining at least 7 of the 10 marks available and one-quarter achieving full marks.

Part (a) was well answered with the majority using a + 14d and a small minority using 5n + 5 in order to find the 15th term. A few candidates listed each term and a number identified the 15th term correctly. A small number of candidates found the total amount saved over the 15 weeks.

In part (b), many candidates used [pic]to find the total amount paid over 60 weeks, although a few applied l = a + (n – 1)d and substituted the result into [pic]. It was not uncommon, however, for arithmetic errors such as mulitplying 315 by 30 or even adding 20 to 295. Other candidates who failed to get full credit mixed n = 15 and n = 60, used d = 5, or found [pic] instead of [pic].

In part (c), most candidates successfully created an expression in pence for [pic] and set it equal to either 63 or 6300. Of those that set the expression to 63, a large majority did not successfully convert from pounds to pence and as a result did not reach the answer given on the paper.

In part (d), a significant number of candidates were unable to deduce m = 35 by looking at the given result in part (c). Instead, a number of these candidates wasted unnecessary time in attempting to factorise and solve a quadratic equation to give m = (36, 35, with some not realising that they needed to reject the negative result.

Mark Scheme

|Question Number |Scheme |Mark |

|6. (a) |Boy’s Sequence: [pic] | |

| |[pic] or [pic] |M1 A1 |

| | |(2) |

| | | |

|(b) |[pic] |M1 A1 |

| | [pic] |A1 |

| | |(3) |

| | | |

|(c) |[pic] |M1 A1 |

| |[pic] |dM1 |

| |[pic] or [pic] | |

| |[pic] | |

| |[pic] (*) |A1 cso |

| | |(4) |

| | | |

|(d) |[pic] |B1 |

| | |(1) |

| | |10 Marks |

Student Attempt A

(a) M1 A1: Fully correct.

(b) M1 A1 A1: For a correct method leading to a correct answer of 9450 p. The later incorrect converted answer of £945 is ignored.

(c) M1 A1: For a correct expression for [pic] in £’s.

M1: For [pic] equated to 63 (pounds).

A1: For correct working leading to the given answer of m(m + 1) = 35 × 36.

(d) B1: For m = 35.

Student Attempt B

(a) M0: For finding the sum of 15 terms and not the [pic]term of an arithmetic series.

A0: Follows from M0.

(b) M1 A1 A1: For a correct solution.

(c) M1 A1 M1 A1: For a correct proof.

(d) B0: No attempt is made.

Student Attempt C

(a) M1: For applying of a + 14d, with at least one of a = 10 and d = 5 correct.

A1: For a correct answer of 80p. The later incorrect converted answer of £8 is ignored.

(b) M0: Incorrect formula [pic] is applied.

A0 A0: Follows from M0.

(c) M1 A1: For a correct expression for [pic] in pence.

M1: For [pic] equated to an allowable value of 63.

A0: Required expression not reached.

[The candidate has "63" (and not 6300) in their equation and then incorrectly achieves 1260 by multiplying one side by 10 and dividing the other side by 10; meaning that they could not achieve the final mark for correct solution only].

(d) B1: For m = 35.

GCE Core Mathematics 1, June 2012

Question 7

In this question, candidates are given the derived function [pic] and a point [pic]lying on [pic] Part (a) is not mathematically demanding but requires the candidate to have a clear understanding that it is the gradient function that they have been given. Part (b) is more familiar, requiring candidates to integrate [pic]to give an equation for [pic]containing c and then recognise that the given point provides the information required to calculate c.

This question discriminated well with just over half of the candidature gaining at least 6 of the 8 marks available. A significant number of candidates started by integrating f ′ (x) to give f (x).

In most cases they realised that this was valid work for part (b) and so relabelled their initial work as (b).

In part (a), the majority of candidates evaluated f ′(4) and used a correct line formula in order to find the equation of the tangent. A few made arithmetic errors when evaluating [pic], some incorrectly manipulated y + 1 = 2(x – 4) into y = 2x ( 7 and some found the equation of the normal. Some candidates rewrote [pic] as [pic] and used this throughout the question and others deduced a tangent gradient of [pic] by looking at the coefficient of x in the f ′ (x) expression.

A significant minority of candidates started part (a) by differentiating f ′ (x) and substituting

x = 4 into the resulting expression. Other candidates incorrectly used (4, (1) by setting their expression for f ′ (4) equal to (1, which resulted in a meaningless equation.

In part (b), most candidates integrated f ′(x) to find f (x) but unfortunately a significant minority failed to find the value of c using (4, (1). Of those who used (4, (1) to find c, a number made arithmetic errors. Other candidates found f (4) in terms of c and equated their result to 0 instead of

(1. A small number of candidates failed to integrate 3 correctly and some candidates either incorrectly simplified [pic] to [pic] or [pic] to [pic]. A very small number of candidates found the value of c correctly but failed to include this evaluated c in an expression for f (x).

Mark Scheme

|Question Number |Scheme |Mark |

|7. (a) |[pic] lies on C where [pic] | |

| |[pic] |M1 A1 |

| |T: [pic] |dM1 |

| |T: [pic] |A1 |

| | |(4) |

| | | |

| |[pic] or equivalent | |

|(b) | | |

| | |M1 A1 |

| |[pic] |dM1 |

| | [pic] | |

| |So, [pic] |A1 cso |

| |[pic] |(4) |

| | |8 Marks |

| | | |

Student Attempt A

(a) M1 A1: For substituting [pic] into [pic] and finding the correct value of 2.

M1: For substituting [pic][pic] and [pic] into [pic] and finding a value for c.

A0: Although the correct value of c is seen, the equation of the tangent is not given.

(b) M1: For a clear attempt to integrate [pic] with at least one correct application of [pic]to [pic]

A1: For correct integration, in an un-simplified form, seen on the second line of working.

M1: For using [pic] and [pic] in an integrated equation to form a linear equation in c.

A0: Due to the error in simplifying the integrated [pic]term, the candidate’s final answer for

[pic]is incorrect.

Student Attempt B

(a) M1: For substituting [pic] into [pic] Note that the candidate believes [pic] is the same as [pic] and substitutes [pic] into their version of [pic].

A0: Incorrect gradient of 13.

M1: For substituting [pic][pic] and their [pic] into [pic] and finding a value for c.

A0: Incorrect answer.

(b) M1: For a clear attempt to integrate [pic] with at least one correct application (in this case two) of [pic]to [pic]

A0: Incorrect integration.

M0: No attempt to find the constant of integration.

A0: Follows from M0.

Student Attempt C

(a) M1 A1: For substituting [pic] into [pic] and finding the correct value of 2.

M0: For finding the equation of the normal instead of the equation of the tangent.

A0: Follows from M0.

(b) M1: For a clear attempt to integrate [pic] with at least one correct application of [pic]to [pic]

A0: Incorrect integration.

M0: For not substituting [pic] into their integrated equation. They are applying [pic] and not [pic]

A0: Follows from M0.

GCE Core Mathematics 1, June 2012

Question 8

This is the first time in the current specification that a negative quadratic has been being tested.

Part (a) tests the method of completing the square and is made more demanding because it is applied to a negative quadratic. Part (b) tests finding the discrimininant and is made more difficult by the fact that the quadratic is given in the form [pic]. In part (c), candidates can use the information gleaned from earlier parts in order to sketch [pic].

This question was poorly answered with about 10% of the candidature gaining all 8 marks.

In part (a), the [pic] term and the order of the terms in [pic] created problems for the majority of candidates. The most popular (and most successful) method was to complete the square. A large number of candidates struggled to deal with the [pic] term with bracketing errors leading to incorrect answers such as [pic] or [pic]. A successful strategy for some candidates involved negating the quadratic to get [pic] which was usually manipulated correctly to [pic]. Whilst a good number negated this correctly to [pic] some candidates wrote down incorrect results such as [pic], [pic] or [pic].

Whilst a number of candidates stopped after multiplying out [pic], those who attempted to use the method of equating coefficients were less successful.

In part (b), most candidates wrote down [pic] for the discriminant and the majority achieved the correct answer of (4, although some incorrectly evaluated 42 – 4((1)((5) as 36. The most common error was for candidates to substitute the incorrect values of a = 4, b = (5 and c = (1 into [pic]. Those candidates who applied the quadratic formula gained no credit unless they could identify the discriminant part of the formula.

In part (c), a majority of candidates were able to draw the correct shape of the graph and a number correctly identified the y-intercept as (5. Only a small minority were able to correctly position the maximum turning point in the fourth quadrant and some labelled it as (2, (1). A number of correct sketches followed from either candidates using differentiation or from candidates plotting points from a table of values. A number of candidates after correctly identifying the discriminant as (4 (including some who stated “that this meant no roots”) could not relate this information to their sketch with some drawing graphs crossing the x-axis at either one or two points. Common errors included drawing graphs of cubics or positive quadratics, drawing negative quadratics with a maximum at (5, 0) or with a maximum at (0, (5).

Mark Scheme

|Question Number |Scheme |Mark |

|8. (a) |[pic] p, q are integers. | |

| |[pic] |M1 |

| |[pic] |A1 A1 |

| | |(3) |

| | | |

|(b) |[pic] |M1 |

| |[pic] |A1 |

| | |(2) |

| | | |

|(c) | | |

| | | |

| | | |

| | | |

| | | |

| | | |

| | | |

| | | |

| | | |

| |Correct [pic] shape |M1 |

| |Maximum within the 4th quadrant |A1 |

| |Curve cuts through -5 or [pic]marked on the y-axis |B1 |

| | |(3) |

| | |8 Marks |

Student Attempt A

a) The candidate uses an alternative method and attempts to complete the square on

[pic].

M1: For (x ( 2)2 – ((2)2 – 5 .

A1 A1: Correct answer of (1 – (x – 2)2 achieved after multiplying (x – 2)2 + 1 by (1.

(b) M1 A1: For achieving the correct value of the discriminant.

(c) M1: Correct shape.

A1: The maximum is within the 4th quadrant.

B1: The curve cuts through (5 which is marked on the y-axis.

Student Attempt B

(a) M0: Incorrect method of completing the square.

A0 A0: Follows from M0.

(b) M0: Their discriminant = [pic] does not follow from a correct method. Although the discriminant 16 ( 20 is seen within the quadratic formula, is not identified by the candidate.

A0: Follows from M0.

(c) M0: Incorrect shape.

A0: Follows from M0.

B0: Curve does not cut (5 on the y-axis.

Student Attempt C

a) The candidate uses an alternative method and attempts to complete the square on

[pic].

M1: (x – 2)2 + 1 implies a correct method of completing the square on [pic]. (Ignore “ = 0”).

A0: Incorrect value of p = 2 or result is not written in the form –k – (x – 2)2 . If the candidate had instead stated p = (2 they would have gained the A1 mark by implied

working.

A0: Incorrect value of q.

(b) M1 A1: For achieving the correct value of the discriminant.

(c) M1: Correct shape.

A0: It is not clear whether the maximum is within the 4th quadrant.

B1: The curve cuts through (0, (5) which is marked on the y-axis.

GCE Core Mathematics 1, June 2012

Question 9

This structured coordinate geometry question tests finding a point on a line; finding a line which is perpendicular to a given line; finding the intersection of two lines; and using surds to find the distance between two points. Part (e) requires synthesising information, drawing a diagram and using higher level problem solving skills in order to find the required area.

This question proved discriminating across all abilities with about a quarter of the candidature gaining at least 12 out of the 15 marks available. A significant number of candidates gave up on this question before they reached part (e).

Part (a) was well answered by the majority of candidates. After the substitution y = 4, most were able to obtain [pic], although some simplified this to 8.5.

Again, part (b) was well answered with many candidates rearranging [pic] into the form

y = mx + c in order to find the gradient of [pic]. Occasionally, the use of two points on [pic] was seen as an alternative approach to finding the gradient of [pic], whilst some preferred to differentiate their [pic] after rearranging. Most candidates were able to use the perpendicular gradient rule to write down the gradient of [pic] and use this gradient to find an equation of [pic]. Methods of approach were roughly equally divided between those using [pic] or y = mx + c . The majority of candidates were able to simplify their equation into a correct form of [pic], although some rearranged [pic] incorrectly to give [pic]. Common errors in this part included candidates incorrectly finding the gradient of [pic] by finding the gradient between A and C or stating the gradient as 2 from looking at the coefficient of x in [pic].

In part (c), a large number of those with a correct equation of [pic] found the correct coordinates of D, with a few, fortunately, using their correct un-simplified version of [pic] rather than their incorrect rearrangement. The majority of candidates without a correct part (b) were able to demonstrate that they could solve the equations for [pic] and [pic] simultaneously and received some credit for this. There were a number of candidates who equated their equations for [pic] and [pic] to give [pic]. Some manipulated this into [pic]and then gave up; whilst others continued to set x = 0 to find a value for y and similarly set y = 0 to find a value for x.

In part (d), it was pleasing to see many candidates able to make a good attempt at finding the distance between the points C and D. Some drew diagrams and others quoted a correct formula. Relatively few candidates got mixed up when determining the differences in the x-values and the differences in the y-values, although a few used incorrect formulae such as [pic] or [pic]. Some candidates lost the final mark in this part by being unable to correctly manipulate fractions and surds, whilst others did not provide sufficient working to arrive at the answer given on the paper.

Part (e) was the most challenging question on the paper with the majority of candidates not attempting it and many of those that did only able to offer incomplete solutions. A significant number of candidates did not draw a clear diagram, which is essential in understanding the nature of this problem. Those that were successful usually summed up the area of two relevant triangles (usually triangle ABC and triangle ABE ) or found half the product of AB and CE, although a significant number of candidates used the incorrect method of finding the product of AB and CE .

A few candidates used other more elaborate methods to find the correct area of 45. Some candidates attempted to find lengths of various lines without any apparent purpose and gave no indication of finding an area. A small number thought quadrilateral ACBE was a trapezium.

Mark Scheme

|Question Number |Scheme |Mark |

|9. (a) |[pic] | |

| |[pic] |B1 |

| | |(1) |

| | | |

|(b) |[pic] |M1 A1 |

| |[pic] |B1ft |

| |[pic] |M1 |

| |[pic] or [pic] |A1 |

| | |(5) |

| | | |

|(c) |[pic] [pic] or [pic] |M1 |

| |[pic] |A1 A1 cso |

| | |(3) |

| | | |

|(d) |[pic] |“M1” |

| |[pic] |A1 ft |

| | [pic] (*) |A1 cso |

| | |(3) |

| | | |

|(e) |[pic] | |

| |[pic] Finding the area of any triangle |M1 |

| |[pic] | |

| |[pic] |B1 |

| |[pic] |A1 |

| | |(3) |

| | |15 Marks |

Student Attempt A

(a) B1: Correct value of p.

(b) M1 A1 B1 M1 A1: For finding the correct equation of [pic] leaving it in the required form.

(c) M1 A1 A1: For achieving the correct coordinates for D.

(d) M1 A1 A1: For correctly proving [pic].

(e) M1 B1 A1: For finding the correct area of 45 by adding the area of triangle ABC to the area of triangle ABE.

Student Attempt B

(a) B1: Correct value of p.

(b) M1 A1: 4y + 3 = 2x rearranged and correct gradient of [pic] found.

B0: Incorrect method for finding a perpendicular gradient.

M1: For [pic] where [pic] is any value.

A0: Incorrect answer.

(c) M1: For using [pic] and their [pic] to form a linear equation in one variable.

A0 A0: Incorrect values for x and y.

(d) M1 A1 ft: Correct follow through expression of [pic] for CD using their [pic] .

A0: As D is incorrect it is not possible to find the correct length of CD.

(e) M0 B0 A0: No attempt made.

Student Attempt C

(a) B1: Correct value of p.

(b) M1 A1 B1 M1 A1: For finding the correct equation of [pic] leaving it in the required form.

(c) M0: No attempt made to solve the equations of [pic] and [pic] simultaneously.

A0 A0: Follows from M0.

(d) M1 A1ft: Correct follow through expression for CD using their [pic].

A0: As D is incorrect it is not possible to find the correct length of CD.

(e) M1: For a correct method of applying [pic](AB)(CE).

B0: No method seen for correctly removing the square roots.

A0: Incorrect answer.

GCE Core Mathematics 1, June 2012

Question 10

In this question candidates are given the graph of y = f (x) = x2 (9 – 2x) .

Part (a) is straightforward and requires candidates to write down the coordinates where y = f (x) crosses the x-axis.

Part (b) is familiar to candidates and tests the effect of simple transformations on y = f (x) .

Part (c) is discriminating and requires candidates to find the value of k where (3, 27) on y = f (x) is mapped onto (3, 10) on y = f (x) + k .

This question was both well answered and discriminating with about two-thirds of the candidature gaining at least 6 of the 8 marks available and about one-third achieving full marks.

In part (a), most candidates solved f (x) = 0 to find the correct x-coordinate of [pic] for A. Some candidates, however, applied x = 0 on 9 – 2x and arrived at an incorrect value of 9. Other common incorrect values for x were 6 or 27.

In part (b), most candidates were able to give the correct shape for each of the transformed curves.

In part (i), most translated the graph of y = f (x) in the correct direction. Very few candidates translated y = f (x) to the right, and even fewer translated y = f (x) in a vertical direction. Some labelled the y-intercept correctly as (0, 27) but erroneously drew their maximum point slightly to the right of the y-axis in the first quadrant. Most realised that the transformed curve would cut the positive x-axis at “their x in part (a) – 3”. Other candidates, who gave no answer to part (a), labelled this x-intercept as A ( 3 or some left it unlabelled. Occasionally the point ((3, 0) was incorrectly labelled as (3, 0) although it appeared on the negative x-axis.

In part (ii), most graphs had their minimum at the origin and their maximum within the first quadrant. Many realised that the transformed curve would cut the x-axis at [pic]. Other candidates, who gave no answer to part (a), labelled this intercept as [pic], whilst some left it unlabelled. Some misunderstood the given function notation and stretched y = f (x) in the x-direction with scale factor 3 resulting in a maximum of (9, 27) and an x-intercept at (13.5, 0). Very occasionally a stretch of the y-direction; or a two way stretch; or even a reflection of y = f (x) was seen. In a few cases there was an attempt by some candidates to make the graph pass through both (0, 0) and (0, 27).

In part (c), a significant number of candidates wrote down k = (17 whilst some left this part unanswered. A number of candidates wrote down the equation y = f (x) + k = x2 (9 – 2x) + k and substituted in the point (3, 10) to find the correct value of k. Common incorrect answers included

k = 17 (from 27 ( 10) or k = 7 (following 3 + k = 10).

Mark Scheme

|Question Number |Scheme |Mark |

|10. (a) |{Coordinates of A are} (4.5, 0). |B1 |

| | |(1) |

| | | |

|(b)(i) | | |

| | | |

| | | |

| |Horizontal translation |M1 |

| |-3 and their ft 1.5 on postitive x-axis |A1 ft |

| |Maximum at 27 marked on the y-axis |B1 |

| | |(3) |

| | | |

| | | |

| | | |

|(b)(ii) | | |

| | | |

| | | |

| | | |

| |Correct shape, minimum at [pic] and | |

| |a maximum within the first quadrant. |M1 |

| |1.5 on x-axis |A1 ft |

| |Maximum at [pic] |B1 |

| | |(3) |

| | | |

|(c) |{k = } (17 |B1 |

| | |(1) |

| | |8 Marks |

| | | |

Student Attempt A

(a) B1: For (4.5, 0).

(b)(i) M1: For translating the original curve horizontally.

A1: For ((3, 0) and (1.5, 0) marked where the transformed curve meets the x-axis.

B1: The maximum at (0, 27) is marked on the y-axis.

(b)(ii) M0: Correct shape but the minimum is not at (0, 0).

A0: Follows from M0.

B0: Maximum at (0, 27) is incorrect.

(c) B1: For k = (17.

Student Attempt B

(a) B0: (6, 0) is incorrect.

(b)(i) M1: For translating the original curve horizontally.

A1 ft: For ((3, 0) and a correctly followed through (3, 0) (from their (6, 0)), marked where the transformed curve meets the x-axis.

B0: The coordinates (0, 27) are correct but the maximum is clearly not on the y-axis.

(ii) M1: Correct shape with minimum at (0, 0) and maximum within the first quadrant.

A0: Incorrectly followed through (6, 0) (should be (2, 0)), marked on the x-axis.

B0: Maximum at (3, 9) is incorrect.

(c) B1: For k = (17.

Student Attempt C

(a) B0: (7, 0) is incorrect.

(b)(i) M1: For translating the original curve horizontally.

A1 ft: For ((3, 0) and a correctly followed through (4, 0) (from their (7, 0)), marked where the transformed curve meets the x-axis.

B1: The maximum at (0, 27) is marked on the y-axis.

(b)(ii) M0: Correct shape but the minimum is not at (0, 0).

A0: Follows from M0, even though [pic] is followed through correctly from part (a).

B1: Maximum at (1, 27) is correct.

(c) B0: k = 7 is incorrect.

Pearson Education Ltd is one of the UK’s largest awarding organisations,

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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:

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:

▪ Core Mathematics 2 (C2) Examiner Feedback

▪ Core Mathematics 3 (C3) Examiner Feedback

▪ Core Mathematics 4 (C4) Examiner Feedback

▪ Mechanics 1 (M1) Examiner Feedback

▪ Statistics 1 (S1) Examiner Feedback

Acknowledgements

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

November 2012

All the material in this publication is copyright.

© Pearson Education Ltd 2012

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Introduction to the question on what is expected from the student

Paper / series

Student response

Question number

Blank question from the exam paper

Examiner commentary on the overall performance on the question

Total number of marks the student was awarded.

Examiner commentary on the student’s response

Mark scheme for this question

3/4

2/4

2/4

3/4

2/4

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O

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1.5

27

O

- 3

y

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1.5

[pic]27

O

y

x

5/8

4/8

4/8

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