Mark Scheme (Final) - Pearson Qualifications

Mark Scheme (Final)Summer 2018Pearson Edexcel GCEIn Further Pure Mathematics FP1 (6667/01)

Edexcel and BTEC QualificationsEdexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding body.We provide a wide range of qualifications including academic, vocational, occupational andspecific programmes for employers. For further information visit our qualifications websitesat www.edexcel.com or www.btec.co.uk. Alternatively, you can get in touch with us usingthe details on our contact us page at www.edexcel.com/contactus.Pearson: helping people progress, everywherePearson aspires to be the world’s leading learning company. Our aim is to help everyoneprogress in their lives through education. We believe in every kind of learning, for all kindsof people, wherever they are in the world. We’ve been involved in education for over 150years, and by working across 70 countries, in 100 languages, we have built an internationalreputation for our commitment to high standards and raising achievement throughinnovation in education. Find out more about how we can help you and your students at:www.pearson.com/ukSummer 2018Publications Code 6667 01 1806 MSAll the material in this publication is copyright Pearson Education Ltd 2018

General Marking Guidance Allcandidatesmustreceivethesametreatment. Examiners must mark the first candidate in exactlythe same way as they mark the last. Mark schemes should be applied positively. Candidates mustbe rewarded for what they have shown they can do rather thanpenalised for omissions. Examiners should mark according to the mark schemenot according to their perception of where the grade boundariesmay lie. There is no ceiling on achievement. All marks on the markscheme should be used appropriately. All the marks on the mark scheme are designed to beawarded. Examiners should always award full marks if deserved,i.e. if the answer matches the mark scheme. Examiners shouldalso be prepared to award zero marks if the candidate’sresponse is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes willprovide the principles by which marks will be awarded andexemplification may be limited. When examiners are in doubt regarding the applicationof the mark scheme to a candidate’s response, the team leadermust be consulted. Crossed out work should be marked UNLESS thecandidate has replaced it with an alternative response.

QuestionNumber1.(a)(b)SchemeNotesMarksf ( z) 2 z 3 4 z 2 15z 13 ( z 1)(2 z 2 az b)At least one of eithera 2 or b 13 or seen as their B1coefficients.Both a 2 and b 13B1or seen as their coefficients.a 2, b 13[2] z 1 is a root1 is a root, seen anywhere. B113 2 2 2 z 2 z 13 0 z z 0 2 2 4 4(2)(13)2(2) z or 1 1 13 0 and z . z 2 4 2 or 2 z 1 Either2So, z 21 5 1 5 i, i2 2 2 2Correct method for solving a 3-termquadratic equation. Do not allow M1 M1here for an attempt at factorising. 1 13 0 and z .At least one of either1 51 5 i or i A12 22 2or any equivalent form.For conjugate of first complex root A1ft[4]Total 6

QuestionNumber2. (a)Schemef ( 3) NotesMarksAttempt both of37f ( 3) awrt 2.1 or trunc 2 or 2.0 or18and M1139f ( 2.5) awrt 1.2 or trunc 1.1 or 1202.05555555.f ( 2.5) 1.15833333.Sign change oe (and f ( x) is continuous) thereforea root {exists in the interval [ 3, 2.5]. }Both f ( 3) awrt 2.1 andf ( 2.5) awrt 1.2 , sign change and A1‘root’ or ‘ ’. Any errors award A0.[2](b)f ( x) 3x 3 24x Ax or Bx 223xM1or 2 x 5 2Calculus must be seen for this to beawarded.At least two terms differentiated correctly A1Correct derivative. A14 23x 2 " 2.055." " 7.148." Correct application of Newton-RaphsonM1using their values from calculus. 3 2.71243523. or 1047275or 2386386Exact value or awrt 2.712A1[5](c) 2.5 3 or"1.158.""2.055." 3 2.5 3 "2.055." " 2.055." "1.158." " 2.055." " 2.055." "1.158." 3 A correct linear interpolation statement 2.5 3 "2.055." M1with correct signs. "1.158."provided sign changed at the end.Do not award until is seen. (0.5) or "2.055." (0.5) or "3.213." 3 ( 2.5)(" 2.055.") 3("1.158.") " 2.055." "1.158." Achieves a correct linear interpolationstatement with correct signs for .dependent on the previous method mark.dM1 2.68020743. or 2.680 (3 dp)7873101or 211571157 2.680 : only penalise accuracy once in (b)A1 caoand (c), but must be to at least 3sf.

ALT (c)The gradient of the line between (-3, 2.055 ) and2.055. 1.158. 6.427.(-2.5, -1.158 ) is 3 2.5Equation of the line joining the pointsy 2.055. 6.427.( x 3)At y 0,0 2.055. 6.427.( x 3) x 2.680Correct attempt to find the equation of aM1line between the two points.Subs y 0 in their line and achieves x .dM1 2.680 : only penalise accuracy once in (b)A1 caoand (c), but must be to at least 3sf.[3]Total 10

QuestionNumber3. (i) (a)(b)Scheme1 1 3 A 2 3 1 2 Either 1 B A 1NotesMarks 1 3 or 1 2 Correct expression for A 111or 2 35M1A1[2](AB) 1 1 3 1 5 12 B 5 1 2 3 5 1 1 10 20 15 5 5 5 10 Writing down their A 1 multiplied by ABM1At least one correct row or at least two correct1 . A1columns of . (Ignore ).5 2 4 3 1 1 2 Correct simplified matrix for B A1[3]ALT (b) a bLet B d ec f 2a 3d 1a d 3 2c 3 f 12 2b 3e 5b e 5Writes down at least 2 correct sets ofsimultaneous equationsM1c f 1{ a 2, d 1, b 4, e 1, c 3, f 2 } 2 4 3 B 1 1 2 At least one correct row orA1at least two correct columns for the matrix BCorrect matrix for B A1[3](ii) (a)Rotation90 clockwise about the originRotation only. M1 90o or clockwise about the origin2 3 or 270o or (anti-clockwise) about the2 A1origin. -90o or (anticlockwise) about the2 origin. Origin can be written as (0, 0) or O.[2] 1(b) C C39 1 0 1 or C3 1 0 For stating C or C or ‘rotation of 270M1clockwise o.e. about the origin.Can be implied by correct matrix. 0 1 1 0 A1Correct answer with no working award M1A13o[2]Total 9

QuestionNumberScheme rn4. (a)2NotesMarks r 8 r 1At least one of the firstM1two terms is correct.Correct expression A1An attempt to factoriseout M1at least n.11 n(n 1)(2n 1) n (n 1) 8n62 1n (2n 1)(n 1) 3(n 1) 48 61n 2n2 3n 1 3n 3 48 61 n 2n2 50 62 n n2 25 61 n (n 5)(n 5)3 Achieves the correctA1answer.(b)n 5(c)k 2 2 1 k 1 22 (17 )(18 ) (3 )(2 ) (17)(22)(12) (2)( 3)(7) 43 4 3 6710 23409k 9k 1496 14 6710 k 29[4]5.Give B0 for 2 or more B1 caopossible values of n.[1]Applying at least one ofn 17 or n 2 to both1 2n (n 1)2 and their M141n (n 5)(n 5)3Applying n 17 andn 2 only to both1 2n (n 1)2 and their4M11n (n 5)(n 5) .3Require differences onlyfor both brackets.Sets their sum to 6710ddM1and solves to give k .2k or 0.2A1 cso9[4]Total 9

QuestionNumber5. (a)SchemeNotesdy k x 2dxdy 0or y xdxdy c2 x 2dxdyor (implicitly) y x 0dxdy1or (chain rule) ct 2 dxcy c 2 x 1 When x ct , mT or at P ct , ordy c 21 22dx (ct )tc dyyct1 2 , mT t dxxcttT: t 2 y ct x ctT: t 2 y x 2ct * 3c 8c t 2 2ct 5 5 3t 2 8 10t 3t2 10t 8 0 (t 4)(3t 2) 0 t .t 4, 2c 23c A 4c , , B c , 34 32 theirtheirM1dydtdxdtdy1 2dxt 1c1T: y 2 x ct tt(b)MarksA1c their mT x ct tM1where their mT has come from calculusAt least one line of working.Applies y Correct solution. A1 cso *[4] 8c 3c Substitutes , into tangent. M1 5 5 Correct 3TQ in terms of tA1Can include uncancelled c.Attempt to solve their 3TQ for t M1Uses one of their values of t to find A or B M1Correct coordinates.A1Condone A and B swapped or missing.[5]Total 9ALT 1(b)3c1 8c 2 x 5t 5 c 3c1 8c 2 ct t 5t 5 y 3t 2 10t 8ALT 2(b)then apply the original mark scheme. c c A ct1 , , B ct2 , t1 t2 t12 y x 2ct1c Substitutes ct , intot M13c1 8c 2 x their y 5t 5 Correct 3TQ in terms of t .A1Can include uncancelled c.Substitutes A and B intothe equation of the tangent, solves for M1x and yt22 y x 2ct2108t1 t2 , t1t2 333t 2 8 10tthen apply original schemeCorrect 3TQ in terms of t1 or t2 A1Can include uncancelled c.

QuestionNumber6. (a)SchemeMarks det M (8)(2) ( 1)( 4) det M 1212B1[1](b)Area T 216 18 12Area T h (1 k )18(k 1) 182(k 1) 184216their "det M "M1Uses (k 1) or (1 k ) in their solution. M1oror18(1 k ) 18 or2(1 k ) 18or4199{ 8h 18} h , k 1 222dependent on the two previous M marks112168(k 1) or 8(1 k ) 22their "det M "216ddM1or (k 1) or (1 k ) 4(their "det M ")216216,k 1 or h 4(their "det M ")4(their "det M ") k 5.5 or k 3.5At least one of either k 5.5 or k 3.5Both k 5.5 and k 3.5A1A1[5]ALT (b) 8 1 4 6 12 T 4 2 1 k 1 48 k95 31T or 18 seen 14 24 2k 46 48 k95311 31 2162 14 24 2k 46 141 4 6 12 4 18or21 k 1 11 744 62k 672 14k 2208 46k 2162 2280 190k 1330 1426At least 5 out of 6 elements are correct or 18M1seen.141their T 216 or2216 12 4k11 18Dependent on the two previousM marks. Full method ofevaluating a determinant.M1ddM114k 6 6 12k 12 4 1821196 96k 216 or 8 8k 1822So, 1 k 4.5 or k 1 4.5 k 3.5 or k 5.5At least one of either k 3.5 or k 5.5Both k 3.5 and k 5.5A1A1[5]Total 6

QuestionNumber7.(a)Scheme24a 2y 2 4ax , S (a,0), D a , , P(ak , 2ak )5 24a 24a 0 0 125ml 5 a a 2a5 24ay 5 x a or y 0 x a24a24aa a a a0 05512 x a 5 y 12 x 12a5l : 12 x 5 y 12a (*)l: y 0 ALT (a)y mx cAt S, 0 ma c24a ma cAt D,512a12 c ,m 551212ay x 12 x 5 y 12a *55NotesMarks24a Uses S a , 0 and D their " a ", to5 find an expression for the gradient of l orapplies the formulay y1x x1 y2 y1 x2 x1M1Can be un-simplified or simplified.Correct solution onlyleading to 12 x 5 y 12a A1 *No errors seen.[2]24a Uses S a , 0 and D their " a ", to5 M1find 2 simultaneous equations and solvesto achieve c ., m .Correct solution onlyleading to 12 x 5 y 12a A1*[2](b)2ak 2k mSP 2 2ak a k 1 ak 2 a 1 5 ml or mSP 12 ( 5 ) 12 2ak 5 2k 24k 5k 2 5So 2k 112 5k2 24k 5 0 (k 5)(5k 1) 0 k . As k 0, so k 5 (25a , 10a)Attempts to find the gradient of SP M1Some evidence of applying m1m2 1M1Correct 3TQ in terms of k in any form. A1Attempt to solve their 3TQ for k M1Uses their k to find P M1(25a , 10a) A1[6]

ALT 1(b)y 0 mSP ( x a)5SP: y 0 ( x a)121 5 ( 125 ) 12 12Can sub for x and achievey a5mSP 2 y 2 4ax 125 ( x a) 4ax25( x2 2ax a 2 ) 576ax25x2 626ax 25a2 0(25x a)( x 25a) 0 x .a5 a2a y a 2512 255 5x 25a y 25a a 10a 12 As k 0, (25a , 10a)M1M1Correct 3TQ in terms of a and x orA15 y 2 48ay 20a 2 0Attempt to solve their 3TQ for x M1x Uses their x to find y M1(25a , 10a)A1[6]ALT 2(b)0 mSP a c1 5 ( 125 ) 12 55y x a121255At P, 2ak ak 2 a1212then as part (b)mSP Subs S into y mSP x c to find c M1Some evidence of applying m1m2 1M1Correct 3TQ in terms of k A1Total 8

QuestionNumber8.SchemeNotesMarksf (n) 2n 2 32n 1 divisible by 7Shows f (1) 35f (1) 23 33 35 {which is divisible by 7}.{ f (n) is divisible by 7 when n 1 }{Assume that for n k ,f (k ) 2k 2 32k 1 is divisible by 7 for k B1.}f (k 1) f (k ) 2k 1 2 32( k 1) 1 (2k 2 32k 1 )Applies f (k 1) with at least 1 power correct M1f (k 1) f (k ) 2(2k 2 ) 9(32k 1 ) (2k 2 32k 1 )f (k 1) f (k ) 2k 2 8(32k 1 ) (2k 2 32k 1 ) 7(32k 1 )ork 2 8(22k 1 32 k 1 f (k ) 7(3or 8f (k ) 7(2) 7(2k 2(2k 2 32k 1 ) or f (k ); 7(32k 1 )or 8(2)k 22 k 1 3) or 8f (k ); 7(2k 2A1; A1))k 2)Dependent on at least one of the previousaccuracy marks being awarded. dM1Makes f (k 1) the subject f (k 1) 2f (k ) 7(32k 1 )or f (k 1) 9f (k ) 7(2k 2 ){ f (k 1) 2f (k ) 7(32k 1 ) is divisible by 7 asboth 2f (k ) and 7(32 k 1 ) are both divisible by 7}If the result is true for n k , then it is now truefor n k 1. As the result has shown to be truefor n 1, then the result is true for all n ( f(k 1) f(k ) 22k 1Correct conclusion seen at the end. CondoneA1 csotrue for n 1 stated earlier.).[6]ALTk 32 k 3 3f(k 1) f(k ) (2 )2f(k 1) f(k ) (2 )(2k 2k 2 (2k 2 3)Applies f (k 1) with at least 1 power correct M1 (9 )32 k 1 32k 1 ) 7.32k 1 or(2 )(2k 2 32k 1 ) or (2 )f (k );7.32k 1or A1;A1f(k 1) f(k ) (9 )(2k 2 32k 1 ) 7.2k 2(9 )(2k 2 32k 1 ) or (9 )f (k ); 7.2k 2NB: Choosing 0, 2, 9 willmake relevant terms disappear, but marksshould be awarded accordingly.Total 6