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 visit our qualifications websites at www.edexcel.com or www.btec.co.uk for our BTEC qualifications. Alternatively, you can get in touch with us using the details on our contact us page at www.edexcel.com/contactus. If you have any subject specific questions about this specification that require the help of a subject specialist, you can speak directly to the subject team at Pearson. Their contact details can be found on this link: www.edexcel.com/teachingservices. You can also use our online Ask the Expert service at www.edexcel.com/ask. You will need an Edexcel username and password to access this service. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We’ve been involved in education for over 150 years, and by working across 70 countries, in 100 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: www.pearson.com/ukJanuary 2013 Publications Code US034359 All the material in this publication is copyright © Pearson Education Ltd 2012 www.mymathscloud.com
General Marking Guidance •All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. •Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. •Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. •There is no ceiling on achievement. All marks on 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. •Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. •When examiners are in doubt regarding the application of the mark scheme to a candidate’s response, the team leader must be consulted. •Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. www.mymathscloud.com
General Principles for Core Mathematics Marking Method mark for solving 3 term quadratic: 1. Factorisationcpqqxpxcbxx=++=++where),)(()(2, leading to x = ... amncpqqnxpmxcbxax==++=++andwhere),)(()(2, leading to x = ... 2. FormulaAttempt to use correct formula (with values for a, b and c). 3. Completing the square Solving 02=++cbxx: 22(),0bxqcq±±± ≠, leading to x = ... Method marks for differentiation and integration: 1. DifferentiationPower of at least one term decreased by 1. (1−→nnxx) 2. IntegrationPower of at least one term increased by 1. (1+→nnxx) Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners’ 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 mistakes in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working 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, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals. Answers without workingThe rubric says that these may not gain full credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done “in your head”, detailed working would not be required. www.mymathscloud.com
January 2013 6663 Core Mathematics C1 Mark Scheme Question Number Scheme Marks 1.2(14)xx−B1 Accept 2(41)xx−+or2(41)xx−−or2(1 4 )xx−−+or even2144()xx−or equivalent quadratic (or initial cubic) into two brackets M1 (1 2)(1 2)or(2 1)(2 1)or (2 1)(2 1)xx x xx x xx x−+ −−+ −−−A1 [3]3 marks Notes B1: Takes out a factor of x or –xor even 4x. This line may be implied by correct final answer, but if this stage is shown it must be correct. So B0 for 2(14)xx+M1: Factorises the quadratic resulting from their first factorisation using usual rules (see note 1 in General Principles). e.g. x ( 1 – 4x) ( x – 1). Also allow attempts to factorise cubic such as2(2)(12)xxx−+ etc N.B. Should not be completing the square here. A1: Accept either (12 )(12 ) or(21)(21) or (21)( 21)xx x xx x xx x−+ −−+ −−−. (No fractions for this final answer)Specific situations Note:2(14)xx−followed by2(12 )xx− scores B1M1A0 as factors follow quadratic factorisation criteria And 2(14)xx−followed byx(1- 4x)(1+4x) B1M0A0. Answers with no working: Correct answer gets all three marks B1M1A1 : (21)(21)xxx−+gets B0M1A0 if no working as 2(41)xx−would earn B0Poor bracketing: e.g. 2(1 4 )xx−+ −gets B0 unless subsequent work implies bracket round the –x in which case candidate may recover the mark by the following correct work. www.mymathscloud.com
Question Number Scheme Marks 2.()232333(23)(82)2or 2xxxaxb++++==with a = 6 orb = 9 M1 692x+=3(23)2xor+=as final answer with no errors or () 69yx=+or 3(2x + 3) A1 [2]2 marks Notes M1: Uses382=, and multiplies powers 3(2x + 3). Does not add powers. ( Just 382=or1382=is M0 )A1: Either 692x+3(23)2xor+=or () 69yx=+or 3(2x + 3)Note: Examples: 632x+ scores M1A0 : ()2323332382 2xxx++++==gets M0A0Special case: : 6922x=without seeing as single power M1A0Alternative method using logs: 23(23) log882(23) log 8log 2log 2xyxxyy++=⇒ + = ⇒=M1 So ()6 9yx=+or 3(2x + 3)A1 [2] www.mymathscloud.com
