Mark Scheme (Results) January Pearson Edexcel International GCSE In Further Pure Mathematics (4PM0) Paper 2

Transcription

1 Mark (Results) Januar 07 Pearson Edexcel International GCSE In Further Pure Mathematics (4PM0) Paper

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3 General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactl the same wa as the mark the last. Mark schemes should be applied positivel. Candidates must be rewarded for what the have shown the 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 ma lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriatel. All the marks on the mark scheme are designed to be awarded. Examiners should alwas 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 worth of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles b which marks will be awarded and exemplification ma 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 work should be marked UNLESS the candidate has replaced it with an alternative response.

4 Tpes of mark o M marks: method marks o A marks: accurac marks o B marks: unconditional accurac marks (independent of M marks) Abbreviations o cao correct answer onl 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 ma score full marks If no working is shown then incorrect (even though nearl correct) answers score no marks. With working Alwas check the working in the bod of the script (and on an diagrams), and award an 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. An case of suspected misread loses A (or B) marks on that part, but can gain the M marks. If working is crossed out and still legible, then it should be given an appropriate marks, as long as it has not been replaced b alternative work. Ignoring subsequent work It is appropriate to ignore subsequent work when the additional work does not change the answer in a wa that is inappropriate for the question: eg. Incorrect cancelling of a fraction that would otherwise be correct. Parts of questions Unless allowed b the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another. 4

5 General Principles for Further Pure Mathematics Marking (but note that specific mark schemes ma sometimes override these general principles) Method mark for solving a term quadratic equation:. Factorisation: x bx c x p x q, where pq c leading to x... ax bx c mx p nx q where pq c and mn a leading to x.... Formula: Attempt to use the correct formula (shown explicitl or implied b working) with values for a, b and c, leading to x.... Completing the square: b 0: 0, 0 leading to x... x bx c x q c q Method marks for differentiation and integration:. Differentiation n Power of at least one term decreased b. x. Integration: n Power of at least one term increased b. x Use of a formula: Generall, the method mark is gained b either n x n x 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 b implication from the substitution of correct values and then proceeding to a solution. Answers without working: The rubric states "Without sufficient working, correct answers ma be awarded no marks". General polic is that if it could be done "in our head" detailed working would not be required. (Mark schemes ma override this eg in a case of "prove or show...") Exact answers: When a question demands an exact answer, all the working must also be exact. Once a candidate loses exactness b resorting to decimals the exactness cannot be regained. Rounding answers (where accurac is specified in the question) Penalise onl once per question for failing to round as instructed - ie giving more digits in the answers. Answers with fewer digits are automaticall incorrect, but the isw rule ma allow the mark to be awarded before the final answer is given. 5

6 Jan 07 4PMO Further Pure Mathematics Paper Mark Question Number (a) x = = x +.5 R () 5 + x = 5 Region R shaded in or out () [4] (a) One mark per line. Coordinates of the points where the lines cross the axes needed in each case but equations need not be shown. "Dashes" on the axes do not count. Award, B0, B0B0 or B0B0B0 Correct region shaded, in or out, need not be labelled R. (NB Not ft, but lines w/o coordinates on the axes accepted if the lines look to be correctl placed.) 6

7 Question number. (a) 6 sin sin 5 or 6 6sin sin 5 6sin sin 0 * () sin sin 0 sin, , ,0 or other correct value 60., 85 (5) [7] (a) Eliminate cos b using the Pthagorean identit. Working must be shown. cso Correct given answer reached. Factorise the equation given in (a), before or after using a substitution eg 40 sin 40 or sin or A (if substitution used) Two correct values for An one correct value for 40 (Need not lead to in range 0 90 ) Must be exact NB or at least dp. Either correct value for 60. must be dp Second correct value Ignore additional answers outside the required range. Deduct one A mark ( last A marks onl deducted) for each additional answer within the range. dr 0.5 dt da A r r dr da da dr r 0.5, dt dr dt , (cm /s) cao (5) [5] Correct statement, seen explicitl or used in chain rule. Attempt the differentiation Correct derivative USE the chain rule (ie sub their derivatives, can have r) cao Must be sf 7

8 Question number 4(a)(i) tan x tan( x) tan x (ii) tan x tan x tan( x) tan( x x) tan xtan x tan x tan x tan x tan x tan( x) tan x tan x tan x tan x tan x tan x tan x tan x tan x tan x tan( x) tan x tan x tan x tan x or tan x tan x tan x tan x tan x tan x tan x tan x cso (6) cos For perpendicular = 8 8 tan 8 () ALT 8 sin 9 9 cos sin sin sin tan (c) ( ) ( 8) 0 tan( x) ( ) 4 ()[0] (a) (i) Correct expression onl but allow tan x tan x instead of tan x and tan x tan x instead of tan x Use the given formula to change to tan x and tanx Use their result from (i) to eliminate tanx. Either numerator or denom. must be correct. Write their numerator and denominator as single fractions or multipl numerator and denominator b " tan x" Correct unsimplified fraction (ii)cso Correct result obtained with no errors seen. Work shown must justif each M mark Showing a method for finding the perpendicular or find a value for sin Correct exact value for tan. Answer onl given: Correct / incorrect 0/ NB If no working shown (calculator?) award for correct exact answer, M0A0 otherwise (c) Substitute their value for tan (not nec exact) in the identit in (a) (ii) cao Correct answer obtained. Must be in the given form. (Calculator solution scores M0A0) 8

9 Question number 5(a) x x x d dx x x 6x x 9x x, * d,cso x x x x (5) d dx 6 Gradient of normal = 6 ft x, 6 6 x 9 0 oe, (6) [] (a) Attempt the differentiation using the product rule. Must have two terms added, NB on e-pen Either term correct Other term correct Power or square root form for both. NB: ma be missing as = d Write their two terms over a common denominator, depends on the first M mark. cso Simplif to the GIVEN answer with no errors seen. Correct value for d/dx at x = ft Correct gradient of normal, follow through their d/dx Correct value of at x = Substitute their gradient of normal and coordinates in mx c Use of value of d/dx scores M0 Correct values substituted Correct equation with integer coefficients, terms in an order (can have 4 terms) 9

10 Question number 6 (a) Mark parts (i) and (ii) together 987 a () d 987 a 0d 5 a (8 ) d 5 a 7d ( on e- PEN Solve simultaneous equations, b elimination or substitution 987 a0 d 47 a 0d 75 a47 d 5 a7d a 7, d 4 d (5) A 4 ft 7 4 B B () (c) n n Sn 7 ( n )4 or 7 4n n n 7 ( n ) n 000 n 5n d n n dd (5) [] (a) Attempt an equation using either piece of information NB: on e-pen. Two correct unsimplified equations d Solve the simultaneous equations to obtain a value for either a or d. Depends on the first M mark. One correct answer Both answers correct ft A = their value of d Use S Ar B (c) d dd n n with n and their values of A and a to obtain a value of B or with r n and n, their a and d and solve the simultaneous equations for either A or B or an other complete method. B = (no ft) Use either form of the sum of an arithmetic series to obtain an expression for Sn. Formulae used must be correct. Set up an inequalit or equation with their sum and 000 and obtain a TQ. Depends on the first M mark Correct TQ, terms in an order, can have > or = Can be a multiple of the one shown. Solve b formula or calculator. If formula used, the formula must be correct (can be b implication due to numbers substituted). Negative answer need not be shown. If b calculator award mark b implication if answer is 0.4 or better. Depends on both previous M marks. cao n (c) Solution b trial and error: As above Further marks depend on sight of an inequalit(d) Correct inequalit () Substitution of at least two values of n () n = obtained from correct working. 0

11 Question number 7(a) x x5 ( x) (x5) 7 x x x ( x) 9 (x6) (x5) x 6 x 5 x ,, 6 x (x4) x 4 x 4 8 ALTs for last marks (x6) (x5) x 6 x 5 x ( ),, 6 x (x4) x 4 x 4 4 ALT d,dd, (5) log log log log log log log or log Forming TQ: log 7 log 7log log log 7log 0 OR 7log log 0 (Let A = log ) A 7A 0 A 0 A, dd log ( ) log 8 (6) [] (a) Attempt to change power of 9 or 7 to a power of Correct unsimplified expression with powers of alone d Expand brackets in the powers and write with all powers as single terms, depends on first M mark dd Remove common factor in numerator, depends on both previous M marks Correct value of k obtained (need not be written explicitl as k = 6) Change base. Can change to base or base or both terms to an other (same) base d Obtain a TQ Depends on the first M mark. Term can be in an order but must be separate terms. dd Solve their TQ. Substitution shown not needed. Depends on both previous M marks. Correct values for log or A OR log One correct value for Second correct value for d

12 Question number 8(a) AB AO OB 5p + q +a= 6p+q (= (p + q)) (i) BC BO OC p+ p+q= p+ q (with conclusion; common direction,common point etc) (ii) AB BC AB : BC : oe (4) 5 MN AB or MN AB OAB : OAC : 4 (ratio of bases) OMN : OAB : 4 (similar s) quad ABNM : OAB : 4 ABNM OAB and OAB 4 OAC 4 quad ABNM 9 9:6 OAC dcso (7) [] B determinant method: 5 OM p q and ON p 5 / 5/ 5 Area ABMN = 0 0 / 8 oe Area OAC =, 0 0, Area ABMN : Area OAC = : 9 :6 8 d MN ALT (a) Attempt to find a vector joining an of the points A, B, C Attempt a second vector joining an of the points A, B, C AC AO OC 8p 4q (i) Both vectors to be correct and a reason for collinearit given eg AB AC (ii) Correct ratio seen ie : oe Stating that MN AB B mark so no proof needed either b vectors or use of midpoint theorem. This mark can be given b implication if the similar triangles are used. Attempting the ratio OAB : OAC using their ratio of bases Correct ratio Attempting ratio quad ABNM : OAB using their ratio Correct ratio quad ABNM : OAB d Multipling the two ratios to eliminate OAB Depends on both M marks above cso Obtain the GIVEN answer with no errors seen

13 ALT d cso Correct vectors Use the "determinant" formula with their coefficients. Can start at an point and proceed in order round the quadrilateral (either direction). First and last coefficients must be the same and / must be included. Correct area Similar for triangle Giving a ratio of the areas with their calculated areas. Depends on both M marks above. Correct GIVEN result. No errors seen There are other methods which ma be seen, usuall based on absinc formula and using angle OAC (and angle OMN). Send to review.

14 Question number ( ) x 5 ( ) 9 (a) x oe () 5 (i),, Gradient of perpendicular = x x (ii) (5) (c) (i) s = 0 (, 0) (ii) t = (, ) ft ft () (d) Length PQ = ALT Length SN = Length TN= d (4) [] PSQT is a quad with perpendicular diagonals: Area = 0 5 Length ST= Length PQ = (either)(both or SN or TN ) ALT Area = (units ) d B "determinant" method Eg Area = d 4

15 (a) Attempt an equation for PQ using an complete method Correct equation in an form, no simplification needed Attempt one of the coordinates of N either b using the formula for the coords of a point dividing a line in a given ratio or b diagram. If a diagram is used the method is complete if one of the coords is deduced (and correct). Both coords correct. NB If coords are written down without an working shown award if both correct or A0 if one is correct. Correct gradient of perp. Ma onl be seen in the equation of l. Use their gradient of the perpendicular and their coordinates of N to obtain an equation for l. If the gradient used is the same as their gradient of PQ award M0. Must be a complete method. Correct equation an form, need not be simplified (c) (i)ft s = 0 No working needed ft their equation of l (ii)ft t = No working needed ft their equation of l NB Award these marks if the coordinates of S and/or T are given rather than s = 0, t = - (d) Attempting one of the necessar lengths All three lengths correct d Attempting the areas of triangles PSQ and PTQ and adding their results Correct total area. ALT Attempt length of PQ or ST Correct lengths of PQ and one of SN, TN or ST d Using their lengths and the formula for the area the quad. Correct area obtained ALT Use the "determinant" formula with their coordinates for S and T. Can start at an point and proceed in order round the diagram (either direction). First and last coordinates must be the same and / must be included. All coordinates correct Attempt to evaluate Correct area obtained. Must be positive. 5

16 Question number 0(a) MD 6cos0 * cso () Height of triangle 6sin0 = cm ht 8 cm* cso () (c) 0 7 MG BG d () (d) AC MD 6 or CE 7 Angle ECA tan 7.6, Angle required = (e) 8 BE 7 o (Angle EAB ),cao () sin ABE sin0, angle ABE 4.7, (5) 8 7 [5] (a) Attempt the length of MD using sin or cos of 60 or 0 or sin/cosine rule in ABC to find AC and divide b cso Correct, GIVEN, result from a correct statement cso (c) d (d) cao (e) Attempt the height of the triangle using sin or cos of 60 or 0 or an other complete method and add to 8. (Addition must be seen) (cm) with no errors seen Attempt the length of MG using Pthagoras with a + sign Attempt the length of BG using Pthagoras with a + sign. Depends on the first M mark. Correct answer. Must be sf Correct (numerical) length of AC or CE decimals allowed Find angle ECA using sine, cosine or tangent. Cosine rule ma be used in triangle BCE to find angle BCE but method must be complete, so length BE must be found. Obtain the correct answer b adding 0 to 7.6 or from the cosine rule. Must be dp Attempt the length of BE using Pthagoras with a + sign. Correct length. NB: These marks can be awarded for work seen in (d) provided used here. An complete method for obtaining angle ABE (oe) Correct numbers in their choice of method Correct answer, must be dp unless alread penalised in (d) 6

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