Mark Scheme (Results) June 2013 - Edexcel
Mark Scheme (Results)June 2013GCE Core Mathematics 4 (6666/01)
Edexcel and BTEC QualificationsEdexcel and BTEC qualifications come from Pearson, the world’s leading learning company. Weprovide a wide range of qualifications including academic, vocational, occupational and specificprogrammes for employers. For further information, please visit our website atwww.edexcel.com.Our website subject pages hold useful resources, support material and live feeds from oursubject advisors giving you access to a portal of information. If you have any subject specificquestions about this specification that require the help of a subject specialist, you may find ourAsk The Expert email service helpful.www.edexcel.com/contactusPearson: helping people progress, everywhereOur aim is to help everyone progress in their lives through education. We believe in every kindof learning, for all kinds of people, wherever they are in the world. We’ve been involved ineducation for over 150 years, and by working across 70 countries, in 100 languages, we havebuilt an international reputation for our commitment to high standards and raising achievementthrough innovation in education. Find out more about how we can help you and your studentsat: www.pearson.com/ukJune 2013Publications Code UA035676All the material in this publication is copyright Pearson Education Ltd 2013
General Marking Guidance All candidates must receive the same treatment. Examiners must mark thefirst candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded forwhat they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to theirperception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should beused appropriately. All the marks on the mark scheme are designed to be awarded. Examinersshould always award full marks if deserved, i.e. if the answer matches themark scheme. Examiners should also be prepared to award zero marks if thecandidate’s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principlesby which marks will be awarded and exemplification may be limited. Crossed out work should be marked UNLESS the candidate has replaced it withan alternative response.
EDEXCEL GCE MATHEMATICSGeneral Instructions for Marking1. The total number of marks for the paper is 75.2. The Edexcel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for ‘knowing a method and attempting to apply it’,unless otherwise indicated.A marks: accuracy marks can only be awarded if the relevant method (M) marks havebeen earned.B marks are unconditional accuracy marks (independent of M marks)Marks should not be subdivided.3. AbbreviationsThese are some of the traditional marking abbreviations that will appear in the mark schemes: bod – benefit of doubtft – follow throughthe symbolwill be used for correct ftcao – correct answer onlycso - correct solution only. There must be no errors in this part of the question toobtain this markisw – ignore subsequent workingawrt – answers which round toSC: special caseoe – or equivalent (and appropriate)dep – dependentindep – independentdp decimal placessf significant figures¿ The answer is printed on the paperThe second mark is dependent on gaining the first mark4. All A marks are ‘correct answer only’ (cao.), unless shown, for example, as A1 ft toindicate that previous wrong working is to be followed through. After a misread however,the subsequent A marks affected are treated as A ft, but manifestly absurd answersshould never be awarded A marks.5. For misreading which does not alter the character of a question or materially simplify it,deduct two from any A or B marks gained, in that part of the question affected.6. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all theattempts and score the highest single attempt.7. Ignore wrong working or incorrect statements following a correct answer.8. In some instances, the mark distributions (e.g. M1, B1 and A1) printed on the candidate’sresponse may differ from the final mark scheme
General Principles for Core Mathematics Marking(But note that specific mark schemes may sometimes override these general principles).Method mark for solving 3 term quadratic:1. Factorisation( x 2 bx c) ( x p)( x q), where pq c , leading to x (ax 2 bx c) (mx p)(nx q), where pq c and mn a , leading to x 2. FormulaAttempt to use correct formula (with values for a, b and c).3. Completing the square2Solving x 2 bx c 0 :b x q c,2 q 0, leading to x .Method marks for differentiation and integration:1. DifferentiationPower of at least one term decreased by 1. ( x n x n 1 )2. IntegrationPower of at least one term increased by 1. ( x n x n 1 )Use of a formulaWhere a method involves using a formula that has been learnt, the advice given in recentexaminers’ reports is that the formula should be quoted first.Normal marking procedure is as follows:Method mark for quoting a correct formula and attempting to use it, even if there are mistakesin the substitution of values.Where the formula is not quoted, the method mark can be gained by implication from correctworking with values, but may be lost if there is any mistake in the working.Exact answersExaminers’ reports have emphasised that where, for example, an exact answer is asked for, orworking with surds is clearly required, marks will normally be lost if the candidate resorts tousing rounded decimals.Answers without workingThe rubric says that these may not gain full credit. Individual mark schemes will give details ofwhat happens in particular cases. General policy is that if it could be done “in your head”,detailed working would not be required.
QuestionNumber1. (a)Scheme Marksdu u x2 2x dxndx 2 e x dx , 1st Application: , 2 Application: dv e x v e x dx x 2 e x 2 xe x {dx} )(b){ x e1 2( xe e ) 0xA1 oe or for K xe {dx} K ( xe e {dx} ) M1 Ax 2 e x Bxe x C e xM1Correct answer, with/without cA1x x 2 e x 2( xe x e x ) { c}xM1Either Ax 2 e x Bxe x C e x {dx} x 2 e x 2 xe x e x dx2 xdu 1 dx v ex x 2 e x λ xe x {dx} , λ 0 x 2 e x 2 xe x dx( u x dv e x dxxx[5]} (12 e1 2(1e1 e1 ) ) ( 02 e0 2(0e0 e0 ) )Applies limits of 1 and 0 to an expression of theform Ax 2 e x Bxe x Ce x , A 0 , B 0 and M1C 0 and subtracts the correct way round. e 2e 2 cso A1 oe[2]7Notes for Question 1(a) M1: Integration by parts is applied in the form x 2 e x λ xe x {dx} , where λ 0 . (must be in this form).A1: x 2 e x 2 xe x {dx} or equivalent. M1: Either achieving a result in the form Ax 2 e x Bxe x C e x {dx} (can be implied) ( (where A 0 , B 0 and C 0) or for K xe x {dx} K xe x e x {dx})M1: Ax e Bxe C e (where A 0 , B 0 and C 0)2 x(b)xxA1: x 2 e x 2( x e x e x ) or x 2 e x 2 x e x 2e x or ( x 2 2 x 2)e x or equivalent with/without c .M1: Complete method of applying limits of 1 and 0 to their part (a) answer in the form Ax 2 e x Bxe x Ce x ,(where A 0 , B 0 and C 0 ) and subtracting the correct way round.Evidence of a proper consideration of the limit of 0 (as detailed above) is needed for M1.So, just subtracting zero is M0.A1: e 2 or e1 2 or 2 e . Do not allow e 2e0 unless simplified to give e 2.Note: that 0.718. without seeing e 2 or equivalent is A0.WARNING: Please note that this A1 mark is for correct solution only.So incorrect [.]0 leading to e 2 is A0.1Note: If their part (a) is correct candidates can get M1A1 in part (b) for e 2 from no working.Note: 0.718. from no working is M0A0
QuestionNumber2. (a)Scheme 1 1 Marks11 22(1)(1)xx ( 1 )( 32 ) 1 ( 1 )( 12 ) 2 1 1 x 2x . 1 ( x) 2( x) 2 . 2!2! 2 2 x x 1113 1 x x 2 . 1 x x 2 . 2828 13111 1 x x 2 x x 2 x 2 .282481 2 1 x x21(1 x) 2 (1 x) 12B1See notes M1 A1 A1See notes M1Answer is given inA1 *the question.[6](b) 1 ( 261 ) 1 1 1 1 1 26 2 26 1 ( 26 ) ie:so,2M13 31405 5135270253 4056B170254056A1 cao[3]9Notes for Question 2(a)12B1: (1 x) (1 x)1 2or(1 x)(1 x) 121seen or implied. (Also allow ( (1 x)(1 x) 1 ) 2 ).1M1: Expands (1 x) 2 to give any 2 out of 3 terms simplified or un-simplified,Eg:1 ( 1 )( 12 ) 21 1 ( 1 )( 12 ) 2x or x 2xx or 1 . 222!2! 2 or expands (1 x)Eg: 12to give any 2 out of 3 terms simplified or un-simplified,( 1 )( 32 )( 1 )( 32 ) 1 1 ( x) 2 or 1 . 21 ( x) or ( x) 2( x)222!2!2 ( 12 )( 32 ) 2( x) for M1.2!A1: At least one binomial expansion correct (either un-simplified or simplified). (ignore x 3 and x 4 terms)A1: Two binomial expansions are correct (either un-simplified or simplified). (ignore x 3 and x 4 terms)Note: Candidates can give decimal equivalents when expanding out their binomial expansions.M1: Multiplies out to give 1, exactly two terms in x and exactly three terms in x 2 .A1: Candidate achieves the result on the exam paper. Make sure that their working is sound.1113 Special Case: Award SC FINAL M1A1 for a correct 1 x x 2 . 1 x x 2 . 2828 3 2 1 2 1 213 2 11 2multiplied out with no errors to give either 1 x x x x or 1 x x x x or848282811 2 11 215 2 11 211 x x x xor 1 x x x x leading to the correct answer of 1 x x 2 .242428282Also allow: 1 .
2. (a) ctd(b)Notes for Question 2 Continued11 1113Note: If a candidate writes down either (1 x) 2 1 x x 2 . or (1 x) 2 1 x x 2 .2828with no working then you can award 1st M1, 1st A1.Note: If a candidate writes down both correct binomial expansions with no working, then you can award1st M1, 1st A1, 2nd A1.M1: Substitutes x 1into both sides of26 1 1 x 1 2 and 1 x xx 2271405(or better) andor equivalent fraction2513523 314053 35314055 1405 andor 0.6 3 andand 1orEg:or3 and 513525135213523 1352 are fine for B1.702514050182650A1:or any equivalent fraction, eg:oretc.40568112105456Special Case: Award SC: M1B1A0 for 3 1.732001972. or truncated 1.732001 or awrt 1.732002.7025 1.732001972. andNote that3 1.732050808.4056B1: For sight ofAliter2. (a)Way 2 1 1 x (1 x)(1 x) x (1 x)(1 x)1 (1 x 2 )2 2x (1)(1 x) 1 (1 x) 2 1 ( 1)( 2) 1 ( x 2 ) . 1 ( 1) ( x) ( x) 2 . 2! 2 1 1 x 2 . (1 x x 2 . )2 1 1 x x2 x221 2 1 x x21(1 x 2 ) 2 (1 x) 1B1See notes M1A1A1See notes M1Answer is given in theA1 *question.[6]Aliter2. (a)Way 212 2B1: (1 x ) (1 x) 1 seen or implied. 1 M1: Expands (1 x 2 ) 2 to give both terms simplified or un-simplified, 1 ( x 2 ) 2 1or expands (1 x) to give any 2 out of 3 terms simplified or un-simplified,( 1)( 2)( 1)( 2)( x) 2 or 1 . ( x)2Eg: 1 ( 1) ( x) or . ( 1) ( x) 2!2!A1: At least one binomial expansion correct (either un-simplified or simplified). (ignore x 3 and x 4 terms)A1: Two binomial expansions are correct (either un-simplified or simplified). (ignore x 3 and x 4 terms)M1: Multiplies out to give 1, exactly one term in x and exactly two terms in x 2 .A1: Candidate achieves the result on the exam paper. Make sure that their working is sound.1
Aliter2. (a)Way 3Aliter2. (a)Way 4Notes for Question 2 Continued1 1 x 1(1 x)(1 x)22 (1x)(1x)2B12 (1 x)(1 x)(1 x)(1 x) 1 x 1 Must follow (1 x) 1 x 2 . M1A1A1onfrom above.2 1 1 x x2dM1A12Note: The final M1 mark is dependent on the previous method mark for Way 3.Assuming the result on the Question Paper. (You need to be convinced that a candidate isapplying this method before you apply the Mark Scheme for Way 4). 1 1 x x (1 x)(1 x) 1 x 1 2x21 1 ( 1 )( 12 ) 2(1 x) 2 1 x 2x .2! 2 1( 1 )( 12 ) 1 (1 x) 2 1 ( x) 2( x) 22! 2 11 1 2 22(1)1(1) x x x x 2 B111 2 1 x x . ,28 11 . 1 x x 2 . 28 M1A1A111 1 11 RHS 1 x x 2 (1 x) 2 1 x x 2 1 x x 2 . 2228 11 21 2 1 2 1 x x x x x282211 2 1 x x2811So, LHS 1 x x 2 RHS28See notes M1A1 *[6]1 B1: (1 x) 1 x x 2 (1 x) seen or implied.2 M1: For Way 4, this M1 mark is dependent on the first B1 mark.12121Expands (1 x) 2 to give any 2 out of 3 terms simplified or un-simplified,Eg:1 ( 1 )( 12 ) 21 1 ( 1 )( 12 ) 2x or x 2x or 1 . 2x2!22! 2 1or expands (1 x) 2 to give any 2 out of 3 terms simplified or un-simplified,( 1 )( 12 )( 1 )( 12 ) 1 1 ( x)21 ( x) or ( x) 2( x) 2 or 1 . 22!222! A1: At least one binomial expansion correct (either un-simplified or simplified). (ignore x 3 and x 4 terms)A1: Two binomial expansions are correct (either un-simplified or simplified). (ignore x 3 and x 4 terms)M1: For Way 4, this M1 mark is dependent on the first B1 mark.Multiplies out RHS to give 1, exactly two terms in x and exactly three terms in x 2 .A1: Candidate achieves the result on the exam paper. Candidate needs to have correctly processed both111 the LHS and RHS of (1 x) 2 1 x x 2 (1 x) 2 .2 Eg:
QuestionNumber3. (a)(b)SchemeMarks1.154701B1 cao[1]1 πArea ; 1 2 (1.035276 their 1.154701) 1.414214 2 6 π12 6.794168 1.778709023. 1.7787 (4 dp)B1; M11.7787 or awrt 1.7787 A1[3](c)V π π02 x For π sec . 2 Ignore limits and dx .Can be implied. x λ tan 2 x 2 tan or equivalent 2 2π2 x sec dx 2 π x 2 {π } 2 tan 2 0 2π(a)(b)2B1M1A1A1 cao cso[4]8Notes for Question 3B1: 1.154701 correct answer only. Look for this on the table or in the candidate’s working.1 ππB1: Outside brackets oror awrt 0.2622 612M1: For structure of trapezium rule [ . ]A1: anything that rounds to 1.7787Note: It can be possible to award : (a) B0 (b) B1M1A1 (awrt 1.7787)Note: Working must be seen to demonstrate the use of the trapezium rule. Note: actual area is 1.762747174 ππ(1.035276 their 1.154701) 1.778709023.126Bracketing mistake: Unless the final answer implies that the calculation has been done correctly,1 πAward B1M0A0 for 1 2 (1.035276 their 1.154701) 1.414214 (nb: answer of 7.05596.).2 61 πAward B1M0A0 for (1 1.414214) 2 (1.035276 their 1.154701) (nb: answer of 5.01199.).2 6Alternative method for part (b): Adding individual trapeziaπ 1 1.035276 1.035276 1.154701 1.154701 1.414214 Area 1.778709023.6 222 Note: Award B1M1A1 forπ(1 1.414214) and a divisor of 2 on all terms inside brackets.6M1: First and last ordinates once and two of the middle ordinates twice inside brackets ignoring the 2.A1: anything that rounds to 1.7787B1:
Notes for Question 3 Continued3. (c) x B1: For a correct statement of π sec 2 2 x or π sec 2 or π 2 (1cos ( 2x ) )2{dx} .Ignore limits and dx . Can be implied. x2 Note: Unless a correct expression stated π sec would be B0. 4 x M1: λ tan from any working. 2 1 x x A1: 2 tan or 1 tan from any working.(2) 2 2 A1: 2π from a correct solution only.Note: The π in the volume formula is only required for the B1 mark and the final A1 mark.Note: Decimal answer of 6.283. without correct exact answer is A0. Note: The B1 mark can be implied by later working – as long as it is clear that the candidate has applied π y 2in their working. Note: Writing the correct formula of V π y 2 {dx} , but incorrectly applying it is B0.
QuestionNumberSchemex 2sin t ,4.(a)y 1 cos 2tMarks{ 2sin t} ,2 π2-t-πdxdydy 2cos t , 2sin 2t or 4sin t cos tdtdtdtdy 2sin 2t 4cos t sin t 2sin t dx 2cos t 2cos t 2π 2sin π dy 6 ; 1At t , 6 dx π 2cos 6 So,(b)2dxdycorrect. B1ordtdtdxdyB1are correct.BothanddtdtdydxApplies theirdivided by theirdtdtM1;πdyand substitutes t into their .dx6At least one ofCorrect value fory 1 cos 2t 1 (1 2sin 2 t )dyof 1 A1 cao csodx[4]M1 2sin 2 t22 x2 x x So, y 2 or y or y 2 2 1 2 2 2 Either k 2 or 2 - x - 2(c)Range: 0 - f ( x) - 2 or 0 - y - 2 or 0 - f - 2y x2or equivalent. A1 cso isw2B1[3]See notes B1 B1[2]9Notes for Question 4(a)dxdyorcorrect. Note: that this mark can be implied from their working.dtdtdxdyB1: Bothandare correct. Note: that this mark can be implied from their working.dtdtdydxπdyM1: Applies theirand attempts to substitute t into their expression for.divided by theirdtdtdx6This mark may be implied by their final answer.dy sin 2t1 followed by an answer of would be M1 (implied).Ie.dx 2cos t2A1: For an answer of 1 by correct solution only.dy 1 from incorrect methods.Note: Don’t just look at the answer! A number of candidates are findingdxdxdydy dy dxNote: Applyingdivided by their .is M0, even if they statedtdtdx dt dtdxdydy 2sin 2t Special Case: Award SC: B0B0M1A1 for 2cos t , 2sin 2t leading todtdtdx 2cos tπdy 1which after substitution of t , yields6dxNote: It is possible for you to mark part(a), part (b) and part (c) together. Ignore labelling!B1:At least one of
4. (b)Notes for Question 4 ContinuedM1: Uses the correct double angle formula cos 2t 1 2sin 2 t or cos 2t 2cos 2 t 1 orcos 2t cos 2 t sin 2 t in an attempt to get y in terms of sin 2 t or get y in terms of cos 2 tor get y in terms of sin 2 t and cos 2 t . Writing down y 2sin 2 t is fine for M1.x2or un-simplified equivalents in the form y f(x). For example:222 2x24 x2 x2 x x y or y 2 or y 2 2 1 or y 1 444 2 2 and you can ignore subsequent working if a candidate states a correct version of the Cartesian equation.IMPORTANT: Please check working as this result can be fluked from an incorrect method.Award A0 if there is a c added to their answer.B1: Either k 2 or a candidate writes down 2 - x - 2 . Note: 2 - k - 2 unless k stated as 2 is B0.Note: The values of 0 and/or 2 need to be evaluated in this partB1: Achieves an inclusive upper or lower limit, using acceptable notation. Eg: f ( x ) . 0 or f ( x) - 2B1: 0 - f ( x) - 2 or 0 - y - 2 or 0 - f - 2Special Case: SC: B1B0 for either 0 f ( x) 2 or 0 f 2 or 0 y 2 or (0, 2)Special Case: SC: B1B0 for 0 - x - 2 .IMPORTANT: Note that: Therefore candidates can use either y or f in place of f ( x)Examples:0 - x - 2 is SC: B1B00 x 2 is B0B0x . 0 is B0B0x - 2 is B0B0f ( x) 0 is B0B0f ( x) 2 is B0B0x 0 is B0B0x 2 is B0B00 . f ( x) . 2 is B0B00 f ( x) - 2 is B1B00 - f ( x) 2 is B1B0.f ( x) . 0 is B1B0f ( x) - 2 is B1B0f ( x) . 0 and f ( x) - 2 is B1B1. Must state AND {or} 2 - f ( x) - 2 is B0B0f ( x) . 0 or f ( x) - 2 is B1B0.A1: Achieves y (c)Aliter4. (a)Way 2f ( x) - 2 is B1B01 - f ( x) - 2 is B1B00 - f ( x) - 4 is B1B00 - Range - 2 is B1B00 Range 2 is B0B0.Range - 2 is B1B0f ( x) . 2 is B0B01 f ( x) 2 is B0B00 f ( x) 4 is B0B0Range is in between 0 and 2 is B1B0Range . 0 is B1B0Range . 0 and Range - 2 is B1B0.[0, 2] is B1B1(0, 2) is SC B1B0dxdy 2cos t , 2sin 2t ,dtdtπ dxdy π 2π 2cos 3 , 2sin At t , 36 dtdt 6 6 dyHence 1dxSo B1, B1.So implied M1, A1.
Notes for Question 4 ContinuedAliter4. (a)Way 3y Correct differentiation of their Cartesian equation. B1ft1 2dy xx 2dxFindsdy x , using the correct Cartesian equation only. B1dxFinds the
June 2013 GCE Core Mathematics 4 (6666/01) Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world’s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at www.edexcel.com. Our website subject pages hold .
CSEC English A Specimen Papers and Mark Schemes: Paper 01 92 Mark Scheme 107 Paper 02 108 Mark Scheme 131 Paper 032 146 Mark Scheme 154 CSEC English B Specimen Papers and Mark Schemes: Paper 01 159 Mark Scheme 178 Paper 02 180 Mark Scheme 197 Paper 032 232 Mark Scheme 240 CSEC English A Subject Reports: January 2004 June 2004
Matthew 27 Matthew 28 Mark 1 Mark 2 Mark 3 Mark 4 Mark 5 Mark 6 Mark 7 Mark 8 Mark 9 Mark 10 Mark 11 Mark 12 Mark 13 Mark 14 Mark 15 Mark 16 Catch-up Day CORAMDEOBIBLE.CHURCH/TOGETHER PAGE 1 OF 1 MAY 16 . Proverbs 2—3 Psalms 13—15 Psalms 16—17 Psalm 18 Psalms 19—21 Psalms
H567/03 Mark Scheme June 2018 6 USING THE MARK SCHEME Please study this Mark Scheme carefully. The Mark Scheme is an integral part of the process that begins with the setting of the question paper and ends with the awarding of grades. Question papers and Mark Schemes are d
Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. All marks on the mark scheme should be used appropriately. All marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme.
Volume 29, Issue 21 Virginia Register of Regulations June 17, 2013 2526 PUBLICATION SCHEDULE AND DEADLINES June 2013 through June 2014 Volume: Issue Material Submitted By Noon* Will Be Published On 29:21 May 29, 2013 June 17, 2013 29:22 June 12, 2013 July 1, 2013 29:23 June 26, 2013 July 15, 2013 29:24 July 10, 2013 July 29, 2013
the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme.
Mark scheme for Paper 1 Tiers 3-5, 4-6, 5-7 and 6-8 National curriculum assessments ALL TIERS Ma KEY STAGE 3 2009. 2009 KS3 Mathematics test mark scheme: Paper 1 Introduction 2 Introduction This booklet contains the mark scheme for paper 1 at all tiers. The paper 2 mark scheme is printed in a separate booklet. Questions have been given names so that each one has a unique identifi er .
2013 Subject Report -June 307 2014 Subject Report -June 324 2015 Subject Report -January 341 2015 Subject Report -May/June 351 2016 Subject Report -January 366 CSEC Social Studies Extra Information: Cut Scores 381 Mark Scheme Paper 02 January 2013 383 Mark Scheme Paper 032 January 2013 404 Specific Objectives 411
