Edexcel GCE In Mathematics (9MA0) Paper 02 Pure Mathematics 2

Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further MARK SCHEME 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. EDEXCEL GCE MATHEMATICS General Instructions for Marking 1. The total number of marks for the paper is 100. 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 have been earned. • B marks are unconditional accuracy marks (independent of M marks) • Marks should not be subdivided. MARK SCHEME 3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. • bod – benefit of doubt • ft – follow through • the symbol will be used for correct ft • cao – correct answer only • cso - correct solution only. There must be no errors in this part of the question to obtain this mark • isw – ignore subsequent working • awrt – answers which round to • SC: special case • oe – or equivalent (and appropriate) • dep – dependent • indep – independent • dp decimal places • sf significant figures •  The answer is printed on the paper • The second mark is dependent on gaining the first mark 4. 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. 5. Where a candidate has made multiple responses and indicates which response they wish to submit, examiners should mark this response. If there are several attempts at a question which have not been crossed out, examiners should mark the final answer which is the answer that is the most complete. 6. Ignore wrong working or incorrect statements following a correct answer. 7. Mark schemes will firstly show the solution judged to be the most common response expected from candidates. Where appropriate, alternatives answers are provided in the notes. If examiners are not sure if an answer is acceptable, they will check the mark scheme to see if an alternative answer is given for the method used. MARK SCHEME Pearson Education Limited. Registered company number 872828 with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom Question Scheme Marks AOs 1(a) 16+(21 1− )×d = 24 ⇒ =d ... M1 1.1b d = 0.4 A1 1.1b Answer only scores both marks. (2) (b) Sn = 1 2 n a n{2 + −( 1)d}⇒ S500= 1 2× 500 2 16{× + 499 "0.4× "} M1 1.1b = 57900 A1 1.1b Answer only scores both marks (2) (b) Alternative using Sn = 1 2 n a{ l} l =16 +(500 −1)×"0.4" = 215.6 ⇒ S500 = × 500{16 +"215.6"} M1 1.1b =57900 A1 1.1b (4 marks) MARK SCHEME Notes (a) M1: Correct strategy to find the common difference – must be a correct method using a = 16, and n = 21 and the 24. The method may be implied by their working. If the AP term formula is quoted it must be correct, so use of e.g. u a ndn =+ scores M0 A1: Correct value. Accept equivalents e.g. 8 , 4 , 2 etc. 20 10 5 (b) M1: Attempts to use a correct sum formula with a = 16, n = 500 and their numerical d from part (a) If a formula is quoted it must be correct (it is in the formula book) A1: Correct value Alternative: M1: Correct method for the 500th term and then uses Sn = 1 n a{ l}with their l 2 A1: Correct value Note that some candidates are showing implied use of u a ndn = + by showing the following: = (b) S500 = 1 × 500 2 16 × + 499× 218 (a) d = 55523.80952... 2 This scores (a) M0A0 (b) M1A0 Question Scheme Marks AOs 2(a) y B1 2.5 (1) (b) f 1.8( )= −7 2 1.8× 2 =0.52 ⇒gf 1.8( ) =g 0.52( ) = 3 0.52× 5×0.52−1 =... M1 1.1b gf 1.8( )= 0.975 oe e.g. A1 1.1b (2) (c) y= 3x ⇒5xy y− =3x ⇒x y(5 3− =) y 5x−1 M1 1.1b (g−1( )x =)5xx−3 A1 2.2a (2) (5 marks) Notes MARK SCHEME (a) B1: Correct range. Allow f (x) or f for y. Allow e.g. {y∈ : y }, −∞ y (−∞,7] (b) M1: Full method to find f (1.8) and substitutes the result into g to obtain a value. Also allow for an attempt to substitute x = 1.8 into an attempt at gf (x). E.g. gf ( )x = = =... A1: Correct value (c) M1: Correct attempt to cross multiply, followed by an attempt to factorise out x from an xy term and an x term. If they swap x and y at the start then it will be for an attempt to cross multiply followed by an attempt to factorise out y from an xy term and a y term. A1: Correct expression. Allow equivalent correct expressions e.g. −x , 1 + 3 3−5x 5 25x−15 Ignore any domain if given. Question Scheme Marks AOs 3 log 123(y+ −5) log 1 33( − y)= 2⇒ log3 121 3− y+ y 5 = 2 or e.g. 2 = log 93 B1 M1 on EPEN 1.1b log3 121− y 3 + y 5 = 2 ⇒ 121− y 3 + y 5 = 3 2 ⇒9 −27y =12y +5 ⇒y =... or e.g. log3(12y+ =5) log3(3 2 (1 3−y)) ⇒(12y 5+ =)3 2 (13y−) y⇒... M1 2.1 y = A1 1.1b (3) (3 marks) MARK SCHEME Notes B1(M1 on EPEN): Applies at least one addition or subtraction law of logs correctly. Can also be awarded for using 2 = log 93 . This may be implied by e.g. log ...3 = 2⇒ ...= 9 M1: A rigorous argument with no incorrect working to remove the log or logs correctly and obtain a correct equation in any form and solve for y. A1: Correct exact value. Allow equivalent fractions. Guidance on how to mark particular cases: log 123( y+ −5) log 1 33( − y)= 2⇒ log 1 333(( −y+y5))= 2 log 12 12y+5 ⇒ = 3 2 ⇒ −9 27y =12y 5+ ⇒y = 1 3− y B1M0A0 log 123( y+ −5) log 1 3 3( − y)= 2⇒ log 1 3 33(( −y+y5))= 2⇒ log3121 3−y+y5= log 12 12y+5 ⇒ = 3 2 ⇒ −9 27y =12y 5+ ⇒y = 1−3y B1M0A0 2 log3(12y+5)−log3(1−3y)= 2 ⇒ 121− y 3 + y 5 =3 2 ⇒ −9 27y =12y +5 ⇒y = B1M1A