GCE AS. Mathematics. Mark Schemes. January 2008

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1 GCE AS Mathematics January 008 Mark Schemes Issued: April 008

3 NORTHERN IRELAND GENERAL CERTIFICATE OF SECONDARY EDUCATION (GCSE) AND NORTHERN IRELAND GENERAL CERTIFICATE OF EDUCATION (GCE) Introduction MARK SCHEMES (008) 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

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

7 ADVANCED SUBSIDIARY (AS) General Certificate of Education January 008 Mathematics Assessment Unit C assessing Module C: AS Core Mathematics [AMC] WEDNESDAY 9 JANUARY, AFTERNOON MARK SCHEME AMCW

8 (a) f() = = 6 W (b) ( 3) = = (7 4 3)( 3) ( + 3)( 3) = 4 3 W W 8 (i) A y 3 x W (ii) y A x W (iii) y 4 A 3 x W 6 dy 3 (i) dx = 6x 0x + 6 MW4 (ii) x = => m = = 0 m N = 0 x = => y = = 4 W y 4 = (x ) W 0 x + 0y 4 = 0 AMCW

9 4 (i) 3, 5 4 (ii) m AC = = 3 W (iii) m BD = 3 y 5 = 3 x 3 Line BD 3y = x + 6 W (iv) 9y 3x = 8 4y 3x = 3 5y = 5 y = x = 3 W W D = ( 3, ) 9 5 (a) y = x x + (x ) = 0 x + 4x 4x + 0 = 0 5x 4x 9 = 0 (5x 9)(x + ) = 0 9 x = or x = 5 W 3 y = or y = 3 5 W (3 4 ) 3 x (b) = 3 W (3 3 ) x + 4x 3 3 6x + 3 = 3 3 4x (6x + 3) = 3 W 9 0x = 4 x = 5 W AMCW

10 6 (i) C = 5x(x + 4) + 0(4x + 8) W C = 5x + 60x + 80 (ii) C = 5x + 60x + 80 < 900 5x + 60x 80 < 0 x + x 364 < 0 W ± x = or (x 4)(x + 6) = 0 ± 600 x = x = 4 or x = 6 y 6 4 x 0 < x < 4 or 6 < x < 4 W 0 < x < (i) 6 + 4x x = (x 4x 6) = [(x ) 0] = 0 (x ) W (ii) Max value = 0 when x = MW (iii) x = 4x + 6 => x + 4x + 6 = 0 => (x ) + 0 = 0 => (x ) = 0 => (x ) = ± 0 => x = ± 0 W AMCW

11 (iv) y x + 0 MW (i) T = v C = v = 500v + v v v dc (ii) dv = v 3 W v 3 = = 500 v = v 3 v = 30 km/h d C = v 4 dv d when v = 30 C = +ve dv \ minimum W 9 Total 75 AMCW

13 ADVANCED SUBSIDIARY (AS) General Certificate of Education January 008 Mathematics Assessment Unit C assessing Module C: AS Core Mathematics [AMC] TUESDAY 5 JANUARY, MORNING MARK SCHEME AMCW

14 GCE MATHEMATICS 008 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). AMCW

15 ( + x) 0 = + 0(x) (x) (x) 3 3 W3 = + 0x + 80x + 960x 3 W 5 (i) Ratio = = 0.0 W (ii) = Sum to infinity = a l r = = = 5 99 = 7 33 WW 6 3 (i) Ú 4x + 3 x dx = x 3 x + c MW4 (ii) Ú 4x + 3 x dx W = 3 x x = 3 3 W = 7.5 W 9 4 (i) s = r =.5 W = 3 W (ii) Angle GOC = = radians 3 Using cosine rule GP = cos GP =.97 m W W (iii) Using sine rule sin sin = W = radians = 36.9 W 0 AMCW

16 5 (i) x tan x 0 tan 0 = tan = tan 6 = tan 4 = tan 3 =.7305 W3 Ú 3 tan x dx = 0 4 [ f(0) + f + f 6 + f 4 + f 3 ] = 0.70 W (ii) y x MW (iii) y x 9 AMCW

17 6 (i) A( 7, 5) B(, ) Centre of circle has coordinates 7 +, 5 = ( 3, ) Radius = ( 7 3) + (5 ) = 5 W (ii) Equation of circle (x + 3) + (y ) = 5 MW (iii) y = 3 4 x + 6 gradient of tangent = 3 4 gradient of CD = 4 3 y = 4 (x 3) 3 3y = 4x + 8 W 9 AMCW

18 7 (i) sin x = 0.5 for 0 x sin 0.5 = 6 radians x = 6, 5 6, 3 6, 7 6 x =, 5, 3, 7 radians MW (ii) sin x = 3 cos x sin x cos x = 3 tan x = 3 producing the angle of W x = 08, 88 W AMCW

19 8 (i) 3 x = 4 log 3 x = log 4 (x ) log 3 = log 4 W x = log 4 (or x log 3 log 3 = log 4) log 3 x = + log 4 log 3 x =.3 W (ii) log 6 (x 4) = log 4 (x 4) log 4 6 log 4 6 = (iii) log x log 4 (x 4) = 6 log 4 x log (x 4) = 4 log 4 x log 4 (x 4) = log 4 x log 4 (x 4) = log 4 x x 4 = x x 4 = 6 x 6x + 64 = 0 W W W (x 8) = 0 x = 8 W 5 Total 75 AMCW

21 ADVANCED SUBSIDIARY (AS) General Certificate of Education January 008 Mathematics Assessment Unit F assessing Module FP: Further Pure Mathematics [AMF] THURSDAY 7 JANUARY, AFTERNOON MARK SCHEME AMFPW

22 GCE Advanced/Advanced Subsidiary (AS) Mathematics Mark Schemes 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) (b) 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; 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). AMFPW

23 (i) If solution is not unique, then the determinant of coefficients is zero l fi = 0 W l Expand determinant fi l(l + 3) 8 = 0 fi l + 3l 8 = 0 fi (l + 6)(l 3) = 0 fi l = 6, 3 W W (ii) If l = 3 then the equations become 6x + 9y = k x + 3y = 5 There is no solution if k π 3(5) i.e. k π 5 W 8 (i) (ii) 7 3 = and 7 k π for k =, Hence period = 3 (iii) 8 = Hence 8 is self-inverse (iv) A subgroup is {, 4, 7} MW3 W W 8 3 (i) Determinant of A li = 0 l 0 0 fi 0 l 3 = l AMFPW fi ( l)[( l) 9] = 0 fi l = 0 or ( l) = 9 fi l = or l = 3 or l = 3 fi l = fi l = 5 Therefore the eigenvalues are l =,, 5 7 MW3

24 (ii) l = 0 0 fi x y z = x y z x = x fi y + 3z = y 3y + z = z 0 fi z z 0 fi x = 0 fi y = z y = z W 0 fi is the corresponding unit eigenvector W l = 0 0 x x fi 0 3 y = y 0 3 z z x = x x = x fi y + 3z = y fi y = 3z fi y = z = 0 3y + z = z 3y = z x fi 0 0 fi 0 W 0 l = x x fi 0 3 y = 5 y 0 3 z z AMFPW

25 x = 5x fi y + 3z = 5y x = 0 fi z = y 3y + z = 5z y = z 0 fi z z 0 fi fi 0 W (iii) (a) D is matrix of eigenvalues fi (b) U is matrix of unit eigenvectors fi MW 7 4 (i) det M = fi M = 0 3 (ii) Multiply through by M fi x y = fi x y = 0 Hence P is the point (0, ) (iii) A reflection in the y-axis MW AMFPW

26 (iv) R = NM fi R = fi R = 0 3 W (v) X Y = 3 0 x y fi X = x 3y Y = y fi X + 3Y = x The line y = x + then becomes Y = (X + 3Y) + fi Y = X 6Y + fi 7Y = X + Therefore the transformed line has the equation 7y + x = 3 5 (i) x + y + 8x + 4 = 0 centre is ( 4, 0) Gradient of radius = = Hence gradient of tangent is Equation of tangent at ( 3, ) is y + = (x + 3) giving y = x + W (ii) x + y + 8x + 4 = 0 and x + y 4y 4 = 0 subtract to give 8x + 4y + 8 = 0 fi x + y + 7 = 0 fi y = x 7 Substitute into the equation of first circle to give x + (x + 7) + 8x + 4 = 0 fi x + 4x + 8x x + 4 = 0 fi 5x + 36x + 63 = 0 fi (5x + )(x + 3) = 0 x = 3, 4. MW Therefore the points of intersection are ( 3, ) and ( 4.,.4) MW 3 AMFPW

27 6 (a) z 6z + 5 = 0 6 ± 36 4(5) fi z = fi z = 6 ± 64 fi z = 6 ± 8i Hence z = 3 + 4i and z = 3 4i (b) (i) The perpendicular bisector of ( 5, 3) and (, 5) i.e. passes through (, 4) MW MW (ii) The half line through (, 0) and with gradient Im MW M(, 4) x ( 5, 3) x (, 5) p 4 (, 0) Re (iii) Line joining ( 5, 3) and (, 5) has gradient of = 6 = 3 Gradient of perpendicular line is 3 Line passes through (, 4) giving an equation of y 4 = 3(x + ) fi y = 3x Therefore this locus lies entirely outside the first quadrant, whilst the nd locus lies entirely inside the first quadrant. W Hence the two lines cannot intersect and there is no common point for both loci. 6 Total 75 AMFPW8 35.0

29 ADVANCED SUBSIDIARY (AS) General Certificate of Education January 008 Mathematics Assessment Unit S assessing Module S: Statistics [AMS] TUESDAY JANUARY, AFTERNOON MARK SCHEME AMSW

30 From calculator Â fx = 4 Â fx = 77 n = 36 Mean = 4/36 = 3.9 W n = s =.7 W 4 (i) Fixed number of trials, only outcomes, constant probability of success, trials independent any two correct M (ii) X ~ B(n, p) X ~ B(8, 6 ) P(X = ) = 8 C ( 6 ) ( 5 6 )6 = = 0.60 W (iii) Probability (at most ) = P(X ) = P(X = 0) + P(X = ) + P(X = ) P(X = 0) = ( 5 6 )8 = P(X = ) = 8 C ( 6 ) ( 5 6 )7 = P(X = ) = P(X ) = W 9 3 (i) No brothers = 8 30 brother = 4 30 brothers = brothers 3 5 = 30 Number of brothers attending a match, x 0 3 P(X) = x MW4 (ii) E(X ) = = 3 30 W (iii) Var(X ) = E(X ) [E(X )] E(X ) = = 5 30 W Var(X ) = 5 30 (3 30 ) = W AMSW

31 4 (i) Let X be the discrete r.v number of parties served in an evening then X ~ Po(5), P(X = x) = e 5. 5 x x! P(X = 4) = e ! = 0.75 W (ii) P(4 X 6) = P(4) + P(5) + P(6) W 5 54 = e 4! ! ! W = W (iii) P(X < 6) = P(0) + P() + P() + P(3) = e ! + 5! + 5! ! W = 0.65 W P(X > 6) = ( ) = 0.38 W Expected Bonus = = 6.57 ( 6.6) W Expected total pay = =.6 4 AMSW

32 5 (i) Let X be the discrete r.v the diameters of ball bearings produced in a factory X ~ N(µ, ), Z ~ N(0, ) µ m.070 XX P(X >.070) = 3.5% = P(X <.070) = µ = µ = (0.965) =.8 W P(X < 0.960) =.5% = µ = µ = (0.05) = (0.985) =.70 W.8 =.070 µ.70 = µ Subtracting 3.98 = 0.0 = = µ = =.0 W (ii) P(X > ) = P(X ) = = ( 0.74) W = (0.74) = = 76.5% W AMSW

33 6 (i) Ú f(x) dx = Ú 5 a 5 6 [x 6 (x ) dx + Ú (a x) dx = W x x] + [ax 4 ] = 6 [ 5 (5)] 6 [() ()] + [a 4a 4 ] [5a 5 4 ] = 6 [4 ] + [a 5a ] = a 5a = 5 a 5a + 6 = 0 W (a 3)(a ) = 0 a = 3 (ii) E(X) = Ú xf(x) dx = Mean 5 Ú x(x ) dx Ú x(3 x) dx 5 5 Ú ( 6 x x) dx + 3 Ú È Í Î8 x 6 x 3 È Í Î 5 È Í Î8 (5)3 (5) 6 È 3x Í Î x3 6 È Í Î Î È Í (3x x ) dx È Í Î8 (3 ) () 6 È Í Î + È 3 Í Î (6) (6)3 6 È Í Î È 3 Í Î (5) (5)3 6 È Í Î = 3 3 W 3 AMSW

34 7 poor quality 0.04 James Robert 0.0 poor quality 0.5 George 0.03 poor quality (i) P(poor quality) = W = = W (ii) P(Robert poor) = P(Robert poor)/p(poor) P(Robert poor) = = W P(Robert poor) = 0.006/0.035 = = 0.90 W (iii) Let G = good quality potato P(G) = 0.45 (0.96) (0.98) (0.97) 3 M = = W P(George G) = P(George G) P(G) = / = = 0.5 W Total 75 AMSW

35 ADVANCED SUBSIDIARY (AS) General Certificate of Education January 008 Mathematics Assessment Unit assessing Module : Mechanics [AM] THURSDAY 7 JANUARY, AFTERNOON MARK SCHEME AMW

36 GCE Advanced/Advanced Subsidiary (AS) Mathematics Mark Schemes 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 proves 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) (b) 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; 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). AMW

37 Resolve along 3 N 3 4 cos 55 = W Resolve along N 4 sin 55 =.766 W q R R = =.46 N W tan q = q = 6. 9 (i) N F 0.03g 30 MW (ii) Resolve perpendicular to slope N = 0.03g cos 30 = = 0.55 N W (iii) Resolve along slope F = 0.03g sin 30 W F = mn = 0.546m W 0.03g sin 30 = 0.546m m = W 9 3 (i) R N A B P mg Q 80g MW (ii) Taking moments about P mg + 80 g (.) = (7) m = 54 kg MW W AMW

38 (iii) If about to tilt R = 0 Resolving vertically N = 54g + 80g N = 33. = 30 N W 0 4 (i) R m T mg T m mg MW (ii) Using F = ma on m T = ma W m mg T = ma W Combining mg = 3ma a = 3 g m s (iii) v = u, u = u, a = 3 g, t = S v = u + at u = u + 3 gs S = 3u g seconds W 0 W 5 (i) F = ma t 0.5g = 0.5a W t = 0.5a a = t m s W (ii) Integrating a = t m s v = 0.8t 0.3t + c If t = 0, v = 0, hence c = 0 (iii) Integrating v = 0.8t 0.3t s = 0.4t 0.t 3 + d If t = 0 then s = 0, hence d = 0 W W AMW

39 (iv) If v = t 0.3t = 0.4 3t 8t + 4 = 0 (3t )(t ) = 0 t = 3, W If t, s = = = 0.8 m 6 (i) Using conservation of momentum M = ( )v W v = 6 m s W (ii) 78 N.55 kg a.55g N Using F = ma.55g 78 =.55a W a = 750 m s W (iii) Using v = u + as 0 = 36 + ( 750.)s W s =.4 cm W 7 (i) Velocity 5 m s T (ii) acceleration = gradient of speed line MW a = 5 0, deceleration =.5 m s W (iii) Distance travelled by bus = 5 T + 0 s 0 5 = 5T + 75 MW time AMW

40 (iv) Distance travelled by car s = ut + at s = 5 3 Distances equal 5T + 75 = 5 3 (T + 5) WW (T + 5) 90T = 5T + 50T + 5 5T 40T 35 = 0 T 8T 65 = 0 (T 3)(T + 5) = 0 T = 3 W W Distance = = 70 m 4 Total 75 AMW

GCE AS. Mathematics. Mark Schemes. Summer 2010

GCE AS. Mathematics. Mark Schemes. Summer 2010 GCE AS Mathematics Summer 010 Mark Schemes Issued: October 010 NORTHERN IRELAND GENERAL CERTIFICATE OF SECONDARY EDUCATION (GCSE) AND NORTHERN IRELAND GENERAL CERTIFICATE OF EDUCATION (GCE) Introduction