GCE AS. Mathematics. Mark Schemes. January 2010

Transcription

1 GCE AS Mathematics January 00 Mark Schemes Issued: April 00

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3 NORTHERN IRELAND GENERAL CERTIFICATE OF SECONDARY EDUCATION (GCSE) AND NORTHERN IRELAND GENERAL CERTIFICATE OF EDUCATION (GCE) Introduction MARK SCHEMES (00) Foreword Mark Schemes are published to assist teachers and students in their preparation for examinations. Through the mark schemes teachers and students will be able to see what examiners are looking for in response to questions and exactly where the marks have been awarded. The publishing of the mark schemes may help to show that examiners are not concerned about finding out what a student does not know but rather with rewarding students for what they do know. The Purpose of Mark Schemes Examination papers are set and revised by teams of examiners and revisers appointed by the Council. The teams of examiners and revisers include experienced teachers who are familiar with the level and standards expected of 6- and 8-year-old students in schools and colleges. The job of the examiners is to set the questions and the mark schemes; and the job of the revisers is to review the questions and mark schemes commenting on a large range of issues about which they must be satisfied before the question papers and mark schemes are finalised. The questions and the mark schemes are developed in association with each other so that the issues of differentiation and positive achievement can be addressed right from the start. Mark schemes therefore are regarded as a part of an integral process which begins with the setting of questions and ends with the marking of the examination. The main purpose of the mark scheme is to provide a uniform basis for the marking process so that all the markers are following exactly the same instructions and making the same judgements in so far as this is possible. Before marking begins a standardising meeting is held where all the markers are briefed using the mark scheme and samples of the students work in the form of scripts. Consideration is also given at this stage to any comments on the operational papers received from teachers and their organisations. During this meeting, and up to and including the end of the marking, there is provision for amendments to be made to the mark scheme. What is published represents this final form of the mark scheme. It is important to recognise that in some cases there may well be other correct responses which are equally acceptable to those published: the mark scheme can only cover those responses which emerged in the examination. There may also be instances where certain judgements may have to be left to the experience of the examiner, for example, where there is no absolute correct response all teachers will be familiar with making such judgements. The Council hopes that the mark schemes will be viewed and used in a constructive way as a further support to the teaching and learning processes. iii

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5 CONTENTS Page Module C Module C 7 Module FP 3 Module 3 Module S 9 v

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7 ADVANCED SUBSIDIARY (AS) General Certificate of Education January 00 Mathematics Assessment Unit C assessing Module C: AS Core Mathematics [AMC] MONDAY JANUARY, MORNING Standardising Meeting Version MARK SCHEME

8 (i) y A = (3, 4) 4 3 x (ii) A = (, ) y x 4 60x + 0y + 0z = x + 30y + 5z = x + 40y + 5z = x + 60y + 0z = x + 0y + 0z = x + 40y + 00z = x + 90y + 5z = x + 40y + 5z = 600 0x + 50y + 00z = x + 00y = x + 00y = x + 000y = 3600 W x = 0 y = 0 z = 40 9 [Turn over

9 y 5 y 5 3 (i) m AB = = 3 y + 3 (ii) m BC = 4 y 5 4 y 5 y + 3 = or = 3 y (y 5)(y + 3) = y y 3 = 0 (y 3)(y + ) = 0 y = 3 or y = 7 4 3x + (a) [6x + 9x x 3 8x + ] 3x 3x + [6x x ] 3x 3x [6x x ] 3x + 3x [(3x + )(x )] 3x + (x )(3x ) (b) = 0 5 MW (c) 4x = (x ) 3x+ = 3x + = x = [Turn over

10 5 (i) x 8x + 7 = [(x 4) 6 + 7] = (x 4) 9 (i) Alternative solution x + px + p + q = x 8x + 7 p = 4 q = 9 (ii) Min value = 9 When x = 4 (iii) x 8x + 7 = 0 (x 7)(x ) x = 7 or x = W (iv) y x 0 6 (i) f() = = 0 (ii) ) 5x 3 x 5x 50x 4x x 50x 4 M 0 4x + 8 4x (x )(5x )(5x + ) (iii) (x )(5x )(5x + ) = 0 x = or x = 5 or x = 5 MW [Turn over

11 7 (a) (i) dy dx = 6x 8x MW3 (ii) x = m = 4 6 = 8 x = y = = 3 y + 3 = 8(x ) y = 8x 9 (b) P = 6t + 7t dp dt = 8t 7t W dp dt = 8t 7t. 0 8 t. 7 t t t , t, 0 8 3x = mx 5 3x mx + 3 = 0 b 4ac, 0 M m 36, 0 y 6 6 x 6, m, 6 MW 7 Total [Turn over

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13 ADVANCED SUBSIDIARY (AS) General Certificate of Education January 00 Mathematics Assessment Unit C assessing Module C: AS Core Mathematics [AMC] MONDAY 5 JANUARY, MORNING MARK SCHEME 7

14 (i) θ 3 3 θ sin = 3 θ = rad θ =.46 rad 4 or Cosine Rule 4 = cos θ θ =.46 rad (ii) Area Sector = r θ Μ = 9.46 = 6.57 unit (iii) Area triangle = ab sin c = 3 3 sin.46 = 4.47 unit or base ht = 4 3 = 4.47 area segment = =.0 unit (iv) (x a) + (y b) = r (x ) + (y ) = 9 W 8

15 4 (i) ( x + ) x + 4x + 4 = x x 4 = x B = C= 4 x (ii) ( x + ) x 4 d x = x x 3 x 4 = + 4x 3 x dx 8 = MW3 = (i) Let S n = a + [a + d] + [a + d] [a + (n )d] then S n = [a + (n )d] + [a + (n )d] a add S n = [a + (n )d] + [a + (n )d] [a + (n )d] = n[a + (n )d] (ii) a + a + d = a + d = S n = n [a + (n )d] a + 40d = 475 a + d = a + 80d = d = 948 d = a = 5 (iii) n Sn = a + ( n ) d S =0 f g 0 = 80 MW 4 9

16 ( x) 3 MW3 3 = 5360 x 4 5 h (i) Area h = 0 fst + last + othersg 0 Area = f ( )g = = 96 m (ii) Area 60 = ( x 60x ) x = 80 3 = 00 = 00m 30x 60 0 dx M MW (iii) The equation for y does not at all model the deep trough around x = 50 or x = 60 when y = 0 6 (a) q p A q p tan A = tan A = q q p p p p 0

17 (b) tan x sinx = 0 sin x sinx = 0 cos x sin x = 0 cos x sin x = 0 or cos x = MW x = 0, ±80, ±60 MW3 7 (a) log log 5 log 5 9 log log log 5 4 log log 5 3 log = = log5 9 log 5 5 log5 9 M (b) (i) P( ) t MW (ii) t 3 P( ) = P t 3 ( ) = t 3 log( ) = log 3 t log( ) = log log 5. t = log 05. = 8.3 years (9) 5 Total 75

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19 ADVANCED SUBSIDIARY (AS) General Certificate of Education January 00 Mathematics Assessment Unit F assessing Module FP: Further Pure Mathematics [AMF] WEDNESDAY 0 JANUARY, AFTERNOON MARK SCHEME 3

20 x + y x 4 = 0 x + y 6x 8y + 0 = 0 Subtract to give: 4x + 8y 44 = 0 x = y + Substitute into equation ( y + ) + y ( y + ) 4 = 0 4y 44y + + y + 4y 4 = 0 5y 40y + 75 = 0 y 8y + 5 = 0 (y 3)(y 5) = 0 y = 3, 5 Hence x = 6 + = 5, x = 0 + = W This gives the points (5, 3) and (, 5) 8 4

21 (a) A rotation of 90 clockwise about the origin O. MW3 (b) (i) det Q = 6 5 = (ii) When Q is used as a transformation matrix it will leave the area of a shape unchanged by the transformation. (iii) R = QP = = 5 3 (iv) = 5 3 x y x = y x y = 3 5 x y = 4 7 Hence A has coordinates (4, 7) Alternative Solution 5 3 x = y Hence = 5x + 3y Hence = x + y Hence Equation 3 Equation gives 4 = x Using gives y = 7 Hence A has coordinates (4, 7) 5

22 3 (i) det = a + 3 a 3 3 a = + ( + ) a a 3 3 = (3a + 6) (9 + ) + (a + )( 9 + a) = 6a + 9a 9 + a + a = a a 8 (ii) If a = 3 det = Hence there is one unique solution (iii) a = 4 det = = 0 Hence there is either no solution or an infinite number of solutions. x + y + 5 = 4 3x + 4y + = x 3y + 3 = 6 gives x 3y + 3 = which is inconsistent with equation. Hence there are no solutions. 0 6

23 4 (i) Zero rotation = Identity = I Rotation of 7 clockwise about centre = P Rotation of 44 clockwise about centre = Q Rotation of 6 clockwise about centre = R Rotation of 88 clockwise about centre = S (ii) I P Q R S I I P Q R S P P Q R S I Q Q R S I P R R S I P Q S S I P Q R MW5 (iii) MW3 (iv) G and H are isomorphic since they are both cyclic groups of order 5 7

24 5 (i) M λi = 0 4 λ 0 8 λ = λ (4 λ)[(8 λ) (4 λ) ] +[ (4 λ)] = 0 (4 λ)[3 λ + λ 4] = 0 (4 λ)[λ λ + 7] = 0 (4 λ)(λ 9)(λ 3) = 0 λ = 4, 3, 9 (ii) If λ = 3 then 4 0 x 8 y = x y 4x y 3x x y x + 8y + = 3y y + 4 = 3 y Check with nd equation 4y + 8y y = 3y 3y = 3y Hence an eigenvector is 8

25 (iii) = Since this equals 4 0 we find that 0 is an eigenvector Also 8 5 = 45 = Hence 5 is an eigenvector (iv) P =

26 6 (a) 3 i = p 3+i 3 i = 3 i = 3 i 3 (b) (i) Straight half line through the point (0, 3) with gradient (ii) Perpendicular bisector of the line joining (4, ) and ( 4, 3) MW3 MW3 y (ii) (i) ( 4, 3) (0, 3) (0, ) (4, ) x 0

27 (iii) First line has equation y = x + 3 Second line: Gradient of line joining (4, ) and ( 4, 3) is given by 3 + = 4 4 Hence gradient of perpendicular is Line passes through midpoint which is (0, ) Hence equation is y = x + Alternative solution for second line w 4 + i = w + 4 3i When w = x + yi then (x 4) + (y + ) = (x + 4) + (y 3) x 8x y + y + = x + 8x y 6y + 9 8y = 6x + 8 Hence equation is y = x + To find point of intersection x + 3 = x + Hence x = which gives y = 5 Hence point of intersection is (, 5) 6 Total 75

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29 ADVANCED SUBSIDIARY (AS) General Certificate of Education January 00 Mathematics Assessment Unit assessing Module : Mechanics [AM] WEDNESDAY 0 JANUARY, AFTERNOON MARK SCHEME 3

30 F = ma a = 0 T Fr = 0 Fr = μr = 0. 3 g T = 0. 3 g = 5.88 N 6 (i) A B T T 0 X 0.5g MW (ii) 0.5g = T sin 30 + T sin 40 Μ T =.9 N T cos 30 = X + T cos 40 X = 0.9 N 9 4

31 3 + A B v 0.6 momentum before = momentum after = 0.4 v W = 0.4 v v = 0.7 speed 0.7 m s direction of motion reversed 6 4 (i) R T θ 6000g MW (ii) F = ma T g sin θ = 6000 MW3 T = 000 T = N 7 5

32 5 (i) v = 3t 4 t t = v = 3 4 = m s (ii) s = 3t 4t dt = t 3 t + c at t = 0 s = 3 c = 3 s = t 3 t + 3 (iii) v = 3t 4 t v = 0 4 stops at t = 0 or at t = s = 3 3 s = 7 4 t = [] t = t = t = 0 3 total distance is 0 7 m 6 (i) R T m g A T B MW m g (ii) A T = m B m g T = m T = 8m m = 8m m 8 = = 4 7 m 6

33 7 (i) t = 0 t = t = 0 m 5 m A B C A B u = u S = ut + at t = s = 0 0 = u + a a = a A C u = u S = ut + at t = s = = u + a a = a 35 = u + a 40 = u + a 5 m s = a deceleration 5 m s 0 = u 5 u =.5 m s (ii) u =.5 v = 0 v = u + at a = 5 0 =.5 5t t =? t = 4.5 s further time seconds 7

34 N 8 (i) C X T MW A Y θ B 0 (ii) T sin θ = 0 T =.5 N W (iii) X + 40 sin θ =0 X = 0.0 N Y = T cos θ = 5.77 N mag n resultant = =.5 N (iv) C 50 A θ B 0 W A 0 + W = 50 sin θ W W = 0 N 7 Total 75 8

35 ADVANCED SUBSIDIARY (AS) General Certificate of Education January 00 Mathematics Assessment Unit S assessing Module S: Statistics [AMS] WEDNESDAY 7 JANUARY, AFTERNOON MARK SCHEME 9

36 GCE ADVANCED/ADVANCED SUBSIDIARY (AS) MATHEMATICS Introduction The mark scheme normally provides the most popular solution to each question. Other solutions given by candidates are evaluated and credit given as appropriate; these alternative methods are not usually illustrated in the published mark scheme. The marks awarded for each question are shown in the right hand column and they are prefixed by the letters M, W and MW as appropriate. The key to the mark scheme is given below: M W indicates marks for correct method. indicates marks for working. MW indicates marks for combined method and working. The solution to a question gains marks for correct method and marks for an accurate working based on this method. Where the method is not correct no marks can be given. A later part of a question may require a candidate to use an answer obtained from an earlier part of the same question. A candidate who gets the wrong answer to the earlier part and goes on to the later part is naturally unaware that the wrong data is being used and is actually undertaking the solution of a parallel problem from the point at which the error occurred. If such a candidate continues to apply correct method, then the candidate s individual working must be followed through from the error. If no further errors are made, then the candidate is penalised only for the initial error. Solutions containing two or more working or transcription errors are treated in the same way. This process is usually referred to as follow-through marking and allows a candidate to gain credit for that part of a solution which follows a working or transcription error. Positive marking: It is our intention to reward candidates for any demonstration of relevant knowledge, skills or understanding. For this reason we adopt a policy of following through their answers, that is, having penalised a candidate for an error, we mark the succeeding parts of the question using the candidate s value or answers and award marks accordingly. Some common examples of this occur in the following cases: (a) a numerical error in one entry in a table of values might lead to several answers being incorrect, but these might not be essentially separate errors; (b) readings taken from candidates inaccurate graphs may not agree with the answers expected but might be consistent with the graphs drawn. When the candidate misreads a question in such a way as to make the question easier only a proportion of the marks will be available (based on the professional judgement of the examining team). 30

37 (i) Advantage: More manageable amount of data Disadvantage: Possibility unrepresentative (ii) Advantage: More concise than using raw data or frequency table Disadvantage: Loss of actual values may lead to inaccuracies (iii) Midvalues: From calculator n = 56 fx = 455 fx = 500 x = 8.5 m = 8.3 (3sf) σ n = 5.77 m σ n = 7.39 m (7.3m ) 9 (i) k k = k = 0.4 (ii) E(X) = ( ) ( ) = 0.3 E(X ) = ( ) ( ) =.6 Var(X) = E(X ) [E(X)] = =.07 (iii) E(Y) = 4 (E(X)) = = 0. Var (X) = ( 4) Var (X) = 6.07 = 33. = 33. (3sf) 3

38 3 (i) Let X be r.v. No. of vehicles during a one-minute period then X ~ Po (8) e P(X = 6) = 8 6 6! = 0. (3sf) (ii) Let Y be r.v. No. of vehicles during a 5 second period then Y ~ Po () P(Y ) = P(Y < ) = P(Y = 0 or ) [ ] e = 0 e + W 0!! = 3e = = (3sf) (iii) Passing singly / Random occurance / Independent events 8 4 (i) Let X be r.v. No. of sixes scored then X ~ B (8, 0.5) 8 3 P(X = 3) = ( ) (0.5)3 (0.75) 5 = 0.08 (3sf) (ii) P(X 3) = P(X > 3) = P(X = 0,, ) [ ( ) ( ) = 8 (0.5) 0 (0.75) (0.5) (0.75) ( (0.5) (0.75) 6 MW3 ) ] = [ ] = 0.3 (3sf) P(X = 5 X 3) P(X = 5) (iii) P(X = 5 X 3) = = P(X 3) P(X 3) ( ) 8 P (X = 5) = (0.5) 5 (0.75) 3 = P(X = 5 X 3) = = (3sf) 3

39 5 Let X be r.v. the mass, in grams, of bags of potatoes X ~ N (μ, 40 ) (i) P(X > 678.4) = 0.05 P(X < 678.4) = ( ) μ P Z < = μ 40 = Ф (0.975) =.96 μ = = 600 ( ) (ii) P(540 < X < 60) = P < Z < = P(.5 < Z < 0.5) W = Ф (0.5) Ф (.5) = Ф (0.5) ( Ф (.5)) = Ф (0.5) + Ф (.5) = W = = 0.53 (3sf) 33

40 6 (i) E(X) = 6 0 x (6x x 3 ) dx = (6x3 x 4 ) dx M [ ] 6 0 3x = 4 x = = (ii) P(X < ) = 0 (6x x 3 ) dx 08 [ ] = x 3 x 4 = (6 4) =, (iii) P( < X < 4) = 4 (6x x 3 ) dx 08 x [ 4 ] 4 = x [ ] = (4) = = (0.48 3sf) ( ) 3 (iv) P(X > 4) = + = E(charge) = [(charge) P(charge)] 3 = =

41 7 M Milk chocolate P Plain chocolate (i) P(MP or PM) = = + = (ii) P(Transfer MM) = = P(Transfer PP) = = P(Transfer MM then choose P) = = P(Transfer MP or PM then choose P) = = P(Transfer PP then choose P) = = P(Choose P) = + + = = Total 75 35

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