Mark Scheme (Results) Summer Pearson Edexcel International GCSE in Further Pure Mathematics Paper 1 (4PM0/01)

Transcription

1 Mark Scheme (Results) Summer 04 Pearson Edexcel International GCSE in Further Pure Mathematics Paper (4PM0/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert service helpful. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: Summer 04 Publications Code UG0886 All the material in this publication is copyright Pearson Education Ltd 04

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 for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according 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 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. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. Types of mark o M marks: method marks o A marks: accuracy marks. Can only be awarded if the relevant method mark(s) has (have) been gained. o B marks: unconditional accuracy marks (independent of M marks) Abbreviations o cao correct answer only o ft follow through o isw ignore subsequent working o SC - special case o oe or equivalent (and appropriate) o dep dependent o indep independent o eeoo each error or omission No working If no working is shown then correct answers may score full marks If no working is shown then incorrect (even though nearly correct) answers score no marks.

4 With working If there is a wrong answer indicated always check the working and award any marks appropriate from the mark scheme. If it is clear from the working that the correct answer has been obtained from incorrect working, award 0 marks. Any case of suspected misread which does not significantly simplify the question loses two A (or B) marks on that question, but can gain all the M marks. Mark all work on follow through but enter A0 (or B0) for the first two A or B marks gained. If working is crossed out and still legible, then it should be given any appropriate marks, as long as it has not been replaced by alternative work. If there are multiple attempts shown, then all attempts should be marked and the highest score on a single attempt should be awarded. Follow through marks Follow through marks which involve a single stage calculation can be awarded without working since you can check the answer yourself, but if ambiguous do not award. Follow through marks which involve more than one stage of calculation can only be awarded on sight of the relevant working, even if it appears obvious that there is only one way you could get the answer given. Ignoring subsequent work It is appropriate to ignore subsequent work when the additional work does not change the answer in a way that is inappropriate for the question: eg. incorrect cancelling of a fraction that would otherwise be correct. It is not appropriate to ignore subsequent work when the additional work essentially shows that the candidate did not understand the demand of the question. Linear equations Full marks can be gained if the solution alone is given, or otherwise unambiguously indicated in working (without contradiction elsewhere). Where the correct solution only is shown substituted, but not identified as the solution, the accuracy mark is lost but any method marks can be awarded. Parts of questions Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another

5 General Principles for Further Pure Mathematics Marking (but note that specific mark schemes may sometimes override these general principles) Method mark for solving a term quadratic equation:. Factorisation: where where and. Formula: Attempt to use the correct formula (shown explicitly or implied by working) with values for a, b and c, leading to. Completing the square: Solving b x bx c + + = x ± ± q ± c where Method marks for differentiation and integration:. Differentiation. Integration: Power of at least one term decreased by. Power of at least one term increased by. Use of a formula: Generally, the method mark is gained by either quoting a correct formula and attempting to use it, even if there are mistakes in the substitution of values or, where the formula is not quoted, the method mark can be gained by implication from the substitution of correct values and then proceeding to a solution. Answers without working: The rubric states "Without sufficient working, correct answers may be awarded no marks". General policy is that if it could be done "in your head" detailed working would not be required. (Mark schemes may override this eg in a case of "prove or show...

6 Exact answers: When a question demands an exact answer, all the working must also be exact. Once a candidate loses exactness by resorting to decimals the exactness cannot be regained. Rounding answers (where accuracy is specified in the question) Penalise only once per question for failing to round as instructed - ie giving more digits in the answers. Answers with fewer digits are automatically incorrect, but the isw rule may allow the mark to be awarded before the final answer is given.

7 Question Answer Marks Number.(a) y = x+ B 7 R B () (b) y+ x= 7 B () (Shade region in or out but R must be indicated.) (c) +, + = 6 7 (, ) lies in R () [5] Notes (a) B for EITHER y = x+ drawn with a positive gradient, with a y intercept of (0, ) or just marked on the y axis, OR for y+ x = 7drawn with a negative gradient, with a y intercept of (0, 7) or just 7 marked on the y axis. Allow marks on y axis with no numbers provided they are complete for at least up to 7 graduations. B for BOTH lines drawn correctly (b) B for the correct region shaded in or out (no ft) (c) for substitution of x = and y = into y x+ and y+ x 7. (Verifying that 0for both x and y is NOT required) for conclusion; therefore (, ) lies within the region R (cso) (Accept eg., so YES) Special Case Allow y = x + and y + x = 7 in part (c) with correct substitution for A0

8 Question Number. (a) tan θ =.5 Answer Marks θ = , θ = 8., 8., () (b) 6 cos θ + cosθ + 5 = 0 ( θ )( θ ) 6cos + 5 cos + = 0 5 cosθ = θ = 46.4, (4) 6 ( cosθ = no solution) [7] Notes (a) for any valid value of θ for 8.º for 8.º Use of Radians; allow for θ = Alternative tanθ tanθ using the correct double angle formula ( tan θ = tan θ tan θ = ) to form and solve a TQ tan θ + 4 tanθ = 0 tanθ = , or tan θ= for 8.º for 8.º (b) for attempting to form the TQ ( terms in any order) d for an attempt to solving their TQ as far as cos θ (see general principles for definition of an attempt) 5 for cosθ = 6 for θ = 46.4 (Note: 80 is outside of the range, do not penalise) (rounding subject to general principles) For any extra values within the given range deduct one mark for each up to a maximum of marks in (a), and mark in (b). Ignore extra values given outside of the range.

9 Question Number (a) e x y = x ( x ) ( x ) x x e e dy = dx Answer Marks x ( x ) ( x ) dy 4e = dx * cso (5) (b) (c) dy 8 = dx 9 B () x= 0 y = B 8 y+ = x 9 9y+ 8x+ = 0 oe () [9]

10 Notes (a) Method (quotient rule) d for an attempt at factorising LHS and rearranging to make y the subject. y must be the subject of the equation with e x only in the numerator for an attempt at Quotient rule. The denominator must be squared. There must be two terms in the numerator irrespective of order and signs with a clear attempt at some differentiation. OR, for an attempt at Product rule. There must be two terms irrespective of order and signs with a clear attempt at some differentiation, (see general guidance) for ONE term correct in the numerator, need not be simplified. for BOTH terms correct in the numerator, need not be simplified. for d y fully correct and simplified as shown. cso Note this is a show question; sufficient dx working must be seen to award marks. (a) Method (implicit differentiation) for the term y differentiated to give terms added irrespective of order and signs) and dy, an attempt at product rule on xy, (there must be d x e x differentiated to give e x e x. d for substituting a re-arranged expression for y, { y = } into the above in an attempt to x eliminate y for factorising the differentiated expression and making d x y dx dy e y =. d x (x ) for simplifying sufficiently to achieve (x ) in the denominator for a fully simplified d y as shown. cso Note this is a show question; sufficient working must dx be seen to award marks. (b) B for d y = 8 (no ft, cao) dx 9 (c) B for y = for using their x, their y, and their d y dx ( y y) = mx ( x), or use y = mx + c to achieve a value for c. 9y+ 8x+ = 0 oe. (integer coefficients)

11 Question Answer Marks Number 4(a) a+ d = 08 a+ d = 54 9d = 54 d = 6 () (b) a = 08 + = 0 () n = + (c) Sn ( 0 ( n ) ( 6) ) ( 0 ) ( 4 ) = n n+ = n n * cso () (d) n( n) 4 = 00 n 4n+ 400 = 0 ( n )( n ) 6 5 = 0 n = 6, 5 (4) []

12 Notes (a) for using Un = a+ ( n ) d, where U = 08, U =54 and n must be and respectively. for both equation in a and d correct for solving the simultaneous equations to give d = 6 Alternative U U for. Award even if the terms and n s are the wrong way around, but numerator must n n be a difference of U s and denominator must be difference of n s for or for d = 6. (b) B for a = 0. ( in epen) Other Alternatives Sight of correct values for d and a without working, from lists or any other working achieves full marks. (c) for attempting to use their a and d in S ( a ( n ) d) n n = + (the formula must be seen first if there are errors in substitution for the award of this mark) for a fully correct substitution (their values) and expression for S n, no simplification necessary for this mark for the fully correct simplified expression as shown cso NB. This is a show question sufficient working must be seen for the award of these marks. (d) for equating the given Sn to 00 to form an equation in n. for forming a correct TQ in n eg ( n 4n+ 400 = 0 or for an attempt at solving their TQ for n = 6, 5 cao n n = )

13 Question Number 5. Answer Marks dv = 7 dt B 4 V= πrh= πr B dv = 4π r dr dr dv dr = dt dt dv dr 7 = π r = = dt 4π [6] Notes Alternative (r is the variable) B for d V = 7 dt B for substituting h = 4 r into the formula for the vol of a cone in terms of the single variable r for an attempt at differentiating their V, (in terms of r only) dv for = 4π r dr for correct expression of chain rule (eg., d r d r d t = ) which can lead to d r dv dt dv dt for using r = (from h=4r) to achieve d r 0.67 dt, correct to sf Alternative (h is the variable) B for d V = 7 dt B for substituting h = 4 r into the formula for the vol of a cone in terms of the single variable h, h ( r = ) 4 for an attempt to differentiate their V dv π h for = dh 6 dr dr dh dv for a correct expression of chain rule eg., = which can lead to d r dt dh dv dt dt dr 6 for using h = to achieve = 7 = = 0.67 dt 4 π h π

14 Question Number Answer = + + 5!! 6(a) ( 4x 5 ) 4x ( 4x ) ( 4x ) Marks = x + x x (4) (b) ( or 0< ) 4x, x < < < or x <, () f( x) kx = + x + x x (c) ( ) k 48 48k = + kx x x + x + x () (d) 48k 4 = k = () [] (a) (b) Notes for substituting the power and term in x into the correct formula. If there are errors in substitution and the correct formula is not seen do not award this mark. for a fully correct substitution for two correct algebraic terms in lowest terms for a fully correct expansion in lowest terms allow ( or 0 <) 4x < or 4x for the method mark for < x < or x < accept x or x (c) for setting ( + kx) their expansion in (a) for multiplying out ( + kx) their expansion in (a) for the correct expansion only as far as x 5 (ignore terms over x 5 ) (d) For setting their coefficients of x and x 5 equal k 48kx 4x = Note: = scores M for k = cao

15 Question Number 7(a) vp Answer Marks = 0 + t () (b) a = t () P = + + (c) x t t 4t ( c) Q t = 0, x = 0 c= 0 B () Q (d) t+ t = t t + t = 0 (oe) t t 4 ± 4 48 t = t = , T = 0.6. (allow t instead of T and awrt 0.6) (4) (a) for an attempt at differentiation of x p for the correct expression for v only ( v Notes p = 0 + t ) (b) for an attempt at differentiating their v p. for the correct expression for a p only ( a = t ) (c) for an attempt at integrating v Q Q B for substitution of t = 0 to give c = 0 for the correct expression only x = t t + 4t ( + c) (d) for setting x P = x Q for forming the TQ (no ft) for an attempt to solve their TQ (usual rules) as far as t = a value P t = ,(8.7...), so T = 0.6 or better. If both are given as values for T, withhold this [] mark. (Accept 4 7 or 4 8 ). Minimally acceptable accuracy of dp.

16 Question Number 8 (a) Answer Marks p 7 α + β = αβ = B (i) 49 αβ = Bft 9 p α + β = α + β αβ = (ii) ( ) ft (4) (b) αβ ( 8 ) 7 = + β β = β + 8β + 7= 0 ( β )( β ) = 0 β =, 7 7 α =, α = ( ) p = α + β = 4, 0 (5) Alt: Find α first (c) x α + β x 0 + = αβ αβ x x+ = (= 0 not needed) ft(for finding a numerical value of the sum) (must have = 0) () []

17 (a) B p for BOTH α + β = AND Notes 7 αβ = (i) Bft for α β = 9 (ii) for the correct algebra on ( α ) ft for Accept 49 p 4 + β to give α β ( α β ) αβ α + β = + oe (but need not be simplified) 9 p 4 + and even p = + oe β + 8 (b) for rearranging α β = 8 to give α = oe, sub into their αβ, and equate to their for forming a TQ in β AND attempting to solve it d for using their values of β ( β =, 7 ) to find both values of α ft their values 7 for both values of α =, cao for finding values of p (p = 0, 4) 7 Alternative finding α first (b) for rearranging α β = 8 to give β = α 8 oe, substitute into their αβ, and equate to for forming a TQ { 9α 4α + 7 = 0 } AND attempting to solve it { (α 7)(α ) = 0 } d 7 for using their values of α ( α =, ) to find both values of β ft their values for both values of β =, 7 cao for finding p, (p = 0, 4) 7

18 Alternative (finding β first) ± + p p 84 (b) for attempting to solve x + px 7 = 0 to give x = and substituting this into 6 p+ p + 84 p p + 84 p+ p p+ p + 84 α β = 8, to give; = 8 = for a fully correct expression for rearranging the above expression to give a TQ in p ( p 6 p 80 = 0 ) p 6 p 80 = 0 p+ p p+ p + 84 p+ 4 p + 84 = = p+ p + 84 = 4 p + 84 = 4 + p 4 p + 6 = p+ p = p 6 p 80 0 for attempting to solve their TQ (this is an A mark, but treat as M mark) for p = 0, and p = 4 Alternative Same as above but with roots reversed p D p+ D 8 p 6 = 6 D + p D = 48 fully correct expression (roots reversed, D = discriminant) for forming the TQ p 4 D 48 D 4 p p 6 p 80 0 = = + = for attempting to solve their TQ (this is an A mark, but treat as M mark) for p = 0, and p = 4 (c) for sum of roots α + β α β not required for the award of this mark) and product of roots α β substituted into an equation in x, (=0 for substituting their values from part (a) to find a numerical SUM ft their values from (a) for fully correct equation 58 9 x x+ = 0. cao Need not be simplified, but do not accept un-evaluated double fractions in final answer. (= 0 required for this mark).

19 Question Number 9 (a) Answer Marks q+ 6 p xd = ( = 8) p+ q q 5 + p yd = ( = 8) p+ q Solve to get p =, q= 4 (or equivalent numbers to give the same ratio) () ALT: p 6 Use differences in x coords (or y coords) to get or q = 8 4 So p =, q= 4 (or equivalent numbers to give the same ratio) (b) Grad. AB 5 = = 6 Grad perp. = ft = + = (4) Equation perp: y 8 ( x 8) ( y x 4) (c) y = 6 e= 9 B () (d) F is ( 7,0) BB () (or method shown and both answers incorrect BB0) (e) EF = + 4 = 5 Area kite = AB EF = = 5 () ALT: oe = ( ) = 5 () Alt: Find DF or DE Area = AB DF, = 5 accept awrt 5 (sf), []

20 Notes (a) for using the correct formula with substitution the correct way for x OR y coordinates For using the correct formula with correct substitution for BOTH x AND y coordinates for the ratio p : q = : 4 oe Accept p =, q = 4 Alternative for finding differences in similar triangles in x and/or y for finding the ratio p : q = : 4 so, p : q = : 4 or accept p =, q = 4 Alternative for finding the length of either AD or DB ( 5and 4 5 respectively) for both lengths AD and DB for the ratio AD:DB = 5: 4 5 (or :4) Special case: accept ratios of :7 or 4 : 7 for A0 provided a correct method is shown. (b) y y for finding the gradient of the line AB using m = to give m = x x ft for using m m = to give gradient of the perpendicular = for attempting to find the equation of the normal through D. The formula must be seen first if there are errors in substitution, using a changed gradient from the tangent and which must be a numerical value. Or, using a complete method using y = mx + c as far as achieving a value for c. y 8= x 8 y+ x= 4 oe for ( ) ( ) (c) B for e = 9 (d) B for correct coordinate of x (7) or y (0) B for both x and y correct. If both are incorrect and the correct method has been used (eg, they have followed through from an incorrect value for e in part (c), Award BB0 (e) for finding the length (ft their coordinates) EF or DF ( 5) or the length AB 7 5 d for using the formula for the area of a triangle base height or any other FULL method for the area of a kite for 5, accept awrt 5 Alternatives to find the area in the ms

21 Question Number 0 Answer (a) Vol cuboid = DE BE Vol pyramid = height DE BE Marks Same so height = 9 (cm) () (b) EC = AE = height + 4, AE = = 9.55 (cm), () 4 (c) EH = EC + CH = 4+ 9 EH = 50 = 7.07 (cm) () (d) height 9 tanθ = = BD 4 (or other appropriate trig ratio) ft θ = = 70.4 () (e) Angle between plane ABE and plane BCDE: 9 9 tanφ = = 4 ( φ = ) Angle between plane BCDE and plane BEIH tanψ = 4 Reqd angle 9 = tan + tan = = 4. 4 (5) [5]

22 Notes (a) for using Vol cuboid = DE BE AND Vol pyramid = height DE BE so DE BE = height DE BE and height = 9 (cm) Show question (b) for finding the length of EC or EC, 4or 4 respectively for attempting to find the length AE or AE for 9.55 correctly rounded to 9.55 (cm) (sf) (c) for using EH = EC + CH = 4+ 9 for EH = 50 = 7.07 (cm) rounded to sf. (unless already penalized in (b)) (d) for attempting to use any appropriate trigonometry to find the angle between AE and their ft 4. (ft their EC). for using their correct lengths throughout. for θ = = 70.4 rounded correctly to dp (e) for using any appropriate trigonometry to find the angle between the planes ABE and BCDE for finding the angle tan 9 or φ = for using any appropriate trigonometry to find the angle between the planes BCDE and BEIH for finding the angle tan 4 or ψ = for the required angle (φ + ψ ) correct to dp = 4.º

23 Alternative for attempting to find the length l of A to the midpoint of IH using Pythagoras theorem, (height of the triangle = 9 + cm, the base of the triangle = cm. l = + ), the length m of A to the midpoint of EB (height = 9 cm, base = cm, of EF( = + 4 ) for l = 7 or.655..and m = 85 or EF = 5 cm m = + 9 = 85 ) AND the length for attempting to use cosine rule with their l, m and EF. (If there are errors in substitution the correct formula must be seen first) for fully correct cosine rule with their values θ = cos = θ = cos = for the required angle (correct to the nearest 0.º) of 4.º cao Note A: there are two types of rounding in this question so; o Deduct mark for failing to round to significant figures o Deduct mark separately for failing to round to dp, both subject to a maximum of one each for repeated rounding errors. Note B: Please note General Guidance for rounding. For answers with rounding errors that arise as a result of accumulated rounding errors, General Guidance does not apply. The given answers must round to the required values. (ie., awrt etc.,)

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