]Mark Scheme (Results) January Pearson Edexcel International GCSE In Mathematics A (4MA1) Higher Tier Paper 1HR

Transcription

1 ]Mark Scheme (Results) January 019 Pearson Edexcel International GCSE In Mathematics A (4M) Higher Tier Paper 1HR

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 f employers. F further infmation visit our qualifications websites at Alternatively, you can get in touch with us using the details on our contact us page at Pearson: helping people progress, everywhere Pearson aspires to be the wld s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, f all kinds of people, wherever they are in the wld. We ve been involved in education f over 150 years, and by wking across 70 countries, in 100 languages, we have built an international reputation f our commitment to high standards and raising achievement through innovation in education. Find out me about how we can help you and your students at: January 019 Publications Code 4M_1HR_1901_MS All the material in this publication is copyright Pearson Education Ltd 019

3 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 f what they have shown they can do rather than penalised f omissions. Examiners should mark accding to the mark scheme not accding 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 wthy of credit accding 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 wk should be marked UNLESS the candidate has replaced it with an alternative response.

4 Types of mark o M marks: method marks o A marks: accuracy marks o B marks: unconditional accuracy marks (independent of M marks) Abbreviations o cao crect answer only o ft follow through o isw igne subsequent wking o SC - special case o oe equivalent (and appropriate) o dep dependent o indep independent o eeoo each err omission No wking If no wking is shown then crect answers nmally sce full marks If no wking is shown then increct (even though nearly crect) answers sce no marks. With wking If there is a wrong answer indicated on the answer line always check the wking in the body of the script (and on any diagrams), and award any marks appropriate from the mark scheme. If it is clear from the wking that the crect answer has been obtained from increct wking, award 0 marks. If a candidate misreads a number from the question. Eg. Uses 5 instead of 55; method marks may be awarded provided the question has not been simplified. Examiners should send any instance of a suspected misread to review. If wking is crossed out and still legible, then it should be given any appropriate marks, as long as it has not been replaced by alternative wk. If there is a choice of methods shown, then no marks should be awarded, unless the answer on the answer line makes clear the method that has been used. If there is no answer on the answer line then check the wking f an obvious answer. Igning subsequent wk It is appropriate to igne subsequent wk when the additional wk does not change the answer in a way that is inappropriate f the question: eg. Increct cancelling of a fraction that would otherwise be crect.

5 It is not appropriate to igne subsequent wk when the additional wk essentially makes the answer increct eg algebra. Transcription errs occur when candidates present a crect answer in wking, and write it increctly on the answer line; mark the crect answer. Parts of questions Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another.

6 Apart from questions 6, 8, 13b and 4 (where the mark scheme states otherwise) the crect answer, unless clearly obtained from an increct method, should be taken to imply a crect method converts to improper fractions converts to fractions with the same common 1 1 denominat 53 5 = Shown Dep on M Alternative method = crect method to add proper fractions = Shown Dep on M

7 (= 45) 60( 15) 4 4 OR "45" 5 (= 7) 4 "15" (= 1) OR "7" "1" 60 OR 9 4 " " " " 0 0 F a complete method (= 96) 11 4 oe (= 11) "11" 4 (a = ) (=6) Method to find a e.g. "6" b Method to find b (30 6 )

8 (= 3.815) OR 60 8 (= 7.5) M f oe OR accept 9, x 10 x f equating the expressions f angle P and angle Q 3x x 5 10 x = 4 x 1 f isolating the terms in x y 180 ("1" 5) y 180 (3 "1" 10) y 180 ("1" 5) (3 "1" 10) f a complete method dep on M 7 eg f an appropriate scale fact, candidates may wk in either cm m eg f a complete method, candidates may wk in either cm m

9 8 eg 8x + 4y = x + y = 9 x = 3, 3 crect method to eliminate x y: x 4y = 9 4x 16y = 36 y = 1.5 coefficients of x y the same and crect 4(9 + 4y) + y = 9 operation to eliminate the selected variable (condone any one arithmetic err in multiplication) writing x y in terms of the other variable and crectly substituting eg y = 9 4x = 9 x = (dep) crect method to find second variable using their value from a crect method to find first variable f repeating above method to find second variable oe, dep first 9 (a) B1 (b) B3 f the crect answer B m n f an answer in the fm 3 5, where m and n are positive integers B1 f at least crect steps in repeated prime factisation ( including tree diagram) (c) B1 Accept ,

10 π 3 π (=113.) π (= ) π (= ) 1 π (= 5.1 ) eg (π 6 π 4 ) oe f a complete method

11 (= 66) (a) "66" 18 f a complete method 0 n 1 n f n ± k oe as the denominat (b) (n 1) = 4n 4n + 1 Proved 3 (n + 1) = 4n + 4n + 1 ft on n ± k (k non zero) oe 4(n n ) + 1 4n 4n 1 1 n n 4 4 4(n + n) + 1 4n 4n 1 1 n n 4 4 Conclusion

12 13 (a) 3x x 8 B (B1 f at least 1 crect non zero term) (b) 3x x 8 = 0 (3x + 4)(x ) (=0) 4, 3 3 Dep on at least B1, ft on M marks only dep on dy being a 3 term quadratic dx 100 x 3 x ( ) 4 3 ( 8) 3 (dep nd ) (c) At x =, y = (= 0) 4 at x, 3 y = ( 4 3 )3 ( 4 3 ) 8 ( 4 ) Shown 4 Substitutes at least one of their 3 3 answer from (b) into ( y ) x x 8x 1 must show that (,0) is a turning point on the curve and give concluding statement

13 14 (a) 97 1 B (b) Crect graph f at least 4 points plotted crectly at end of interval f all 6 points plotted consistently within each interval at the crect height accept curve line segments accept curve that is not joined to (0, 0) (c) 14 A line drawn at CF = 60 to meet at least one curve sight of ft candidate's CFD

14 15 (a) 8 0 B (B1 two terms crect in a product of 81x y 3 terms) (b) 4n ( n n 15) (4n 1n)(n + 5) (4n 0 n)( n 3) 3 4n 8n 60n F a crect partial expansion ( may be unsimplified e.g 4n ( n 5n 3n 15) ) (c) ( c 3d )(c 3d ) 1 B1 (d) (4 x)(3 x) x(4 x) ( x 4)( x 3) x(4 x) 3 x 3 f either numerat denominat x factised crectly f both numerat and denominat factised crectly oe

15 16 (a) (b) Any two of f any two crect f a complete method Alternative method and oe 6 41 both crect " f a complete method " SC B f an answer of 41 7 oe

16 17 (a) πr + πr r 6r (b) S.A. 6πr : 4πr = 3 : Shown 3 ft their answer from (a), must be in terms of r. Ratios could be seen as fractions throughout eg 3 Vc : Vs = πr 3 : 3 4 πr 3 = 3 : 4 = 3 : oe eg ratios could be 3 :1

17 Shown = (dep on M) Conclusion - need not state the value of n 19 Angle BCE = 73 Angle BDE = angles may be written on the diagram 1 1 Angle DEB = 73 and Angle DCB = (=107 ) Angle DCE = 34 Angle DEB = 73 and Angle DBE = (=34 ) eg Alternate segment theem Opposite angles of a cyclic quadrilateral sum to 180 Alternate angles are equal Angles in the Same segment are equal Angles in a triangle sum to 180 B f a full set of reasons relevant to their method (B1 f at least one relevant circle theem)

18 0 Let N be the midpoint of BC B1 f recognising that required angle is MAN (could be marked on a diagram) Let sides of cube have length a cm any a 0 (a could be a number a letter) AN 4a a (= 5a ) AM 4a a 4a (= 9a ) eg tan MAN a "5 a " sin MAN a "9 a " crect trig statement f angle MAN, any a 0 (a could be a number a letter) o x 5 y 5 y cos recognising need f the cosine rule ( y 1) 5 y 5y x x x 5 ( 1) 5 5 y y 1 5 y 5y x x x x f expansion of crect equation ( y 1) ( x 1) in a 5y y 5 1 y = 8 3x = 1 x = 7 f crect linear equation with crect isolation of terms

19 eg EX ED DC CX EX EF FA AX 4 a crect statement f EX DC b + a a CX b + a FA b + EX a + ( b + a) f a complete method which gives a crect but unsimplified expression f EX 3a b

20 3 (a) x k 3 f squaring and rearranging crectly to the y, x y x k x ( y 1) k x fm x ( y 1) k p k 1 k (dep) f f 1 (p) = k Alternative method p f ( k) p k k k (b) a k (gf( a) ) a (gf( x) ) x k x 3 ka a k (dep) f rearranging gf = k and isolating crectly the terms in a k k 1 oe eg k k 1

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