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

Transcription

1 . Mark Scheme (Results) Summer 05 Pearson Edexcel International GCSE Further Pure Mathematics (4PM0) Paper

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 for employers. For further information visit our qualifications websites at or 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 world s leading learning company. 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 05 Publications Code UG04559 All the material in this publication is copyright Pearson Education Ltd 05

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. 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. 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 correct answer only o ft follow through o isw ignore subsequent working o SC - special case o oe or equivalent (and appropriate) o dp dependent o indep independent o eeoo each error or omission No working If no working is shown then correct answers normally 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 in the body of the script 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 loses two A (or B) marks on that part, but can gain 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 ofthe 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 Power of at least one term decreased by.. Integration: 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 implicationfrom 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 Number. 4x 9 0 x or Scheme seen as upper limit B d 4 9 d 0 0 V y x x x 4 6x 7x 8dx 0 Marks x x x 0 d (units ) Notes [5] B for x = allow { x }. Award when seen anywhere in the question. d for a correct statement for the volume of revolution, which must include π and the function squared. Ignore limits for this mark. Ignore a missing dx. If π is seen at the end of the question, (you will see this) award this mark. for a fully correct expanded expression as shown for the volume of revolution with both correct limits. You may not see this expression. The mark can be awarded as implied by the correct integrated expression seen. for an attempt at integrating their expression for the volume, which must contain as a minimum, Ax 4 as their highest power of x, and π. Award for term in x, or even for their 8 8x. x n n x seen in one Note: this M mark is dependent on the first being awarded. 04 (units ) cao Do NOT accept an answer of 04 (units ) with no integration seen. If the volume is left as negative withhold this mark. If they change a negative to a positive (due to limits being wrong way around), then you can award this mark.

8 Question Number.(a) (b) Scheme Marks dy x x 8xe 8x e dx () dy x 8x e 8x e dx x x 8x e x x y x * cso ALT: Reverse argument: correct method; fully correct Notes () [5] for an attempt at product rule. There must be two terms added. There must be an attempt at differentiating BOTH terms (usual rules for differentiation) for ONE term correct, need not be simplified for BOTH terms correct, need not be simplified. Award when seen and isw any attempts at simplification. for multiplying their dy dx x x 8xe 8x e through by x on BOTH sides for a correct factorized expression. Note: this is a show question so look out for fudging of their work to achieve the given answer. ALT You will see attempts working from the given answer which is fine. For substituting x substituted into y x 4 e x to give; 4x e x 8x e 8x e x x x for multiplying their d y dx by x and comparing the result, to verify

9 Question Number Scheme Marks.(a) x x l x mx m n l 4 ml 8 m lm n n 7 () 4 x x 7 4 x ALT: () (b) (i) fx min Bft (ii) when x Bft () Notes [5] (a) Note: there is only one method mark in part (a). The method MUST be complete for award of this mark for setting the given expression or f(x) equal to l x m n and attempting to expand the for the values of l = 4, and m =. x m into terms, ie., x Amx m where A 0 Accept embedded values. If there is an error transferring the correct embedded value, isw. for the value of n = ALT for taking 4 as a common factor of the term in x and x, and attempting to complete the square (usual rules please refer to General Guidance) 4 x 7 for achieving correct. Penalise poor bracketing unless final answer is for the final answer { f(x)} = 4x. Accept answers embedded in the (b) expression. If there is an error transferring the correct embedded value, isw. Bft for minimum = (ft their value of n) Bft for x = (ft their value for m) There must be no transposition of the and the unless it is clear they write f( x) and x

10 Question Number Scheme Marks 4(a) a 8 B () (b) S a d 0 6 d d 4 () (c) (i) S n n n 0 50 (ii) 0n n 50 n 0n 5 0 Crit values n.07,.07 Greatest n (4) [7] (a) B a = 8 only (b) Notes for a a d (0 ) ft their a. The method must be complete. ALT (c) for d = 4 Note: d = 4 is M0A0 Method of differences for correct values of S n seen in a table and finding first and second differences. for at least three 4 s seen with no incorrect values. (i) for writing down the correct inequality as shown for forming a TQ with the correct inequality as shown (ii) for attempting to solve the TQ either by formula or completing the square. (Please refer to General Guidance for the definition of an attempt) Note: Attempts to factorise using integers are MO for n =. If they also offer n = then this is A0. Do not isw. Special Case: Answer only of n =, or using trial and error and giving n = is

11 Question Number Scheme Marks 5(a) B () (b) 6 7 B 7, 7 ft, (4) 7 (c) 4 4 (d) ALT: and Subtract and substitute Correct answer () () [9]

12 Notes NOTE: If they use the quadratic formula to answer any part of the question, award zero marks in that part. (a) B for simplification as shown cso. This is a show question so multiplication, ie., (b) must be seen. B for both sum and product (the sum need not be simplified to ) If they use the given answer in part (a), they must achieve an expansion that is as a minimum A where A 0 For an fresh attempt at an expansion and simplification of acceptable attempt; A ( ) where A 0 {Note: is the correct expansion}. Minimally Note: Their attempt must have been sufficiently simplified in order to substitute their sum in terms of and ft for substituting their values for the Sum and Product into their. (c) for the correct answer 7 oe for expanding ( ) B, where B or 0 AND for substituting their values for the Sum and Product OR for for answer as shown (d) for the correct algebra on the expansion of or to give either; or final answer as shown ALT uses the given equation and substitutes α and β, and subtracts to give 7 6 and and substitutes their values of leading to a value for for the correct answer as shown.

13 Question Number 6(a) 0 4 cos C 04 Scheme Marks C sin A sin A B ft (5) (b) APC 80 ( ) AP 4 sin sin AP () (c) Area 4sin 6.96 Area cm ALT: Use Heron's formula () [0]

14 (a) Notes uses a correct COS rule formula for any angle of the triangle. If there are errors in substitution, the correct formula must be seen first. for one of awrt C = 78.5, or B = 8.6, or A = 6.0 uses COS rule again, or SIN rule. If SIN rule is used ft their first angle for this mark. If there are errors in substitution, the correct formula must be seen first. for one of awrt C = 78.5, or B = 8.6, or A = 6.0 ft (b) uses 80 the two angles already found. Follow through their angles for this mark. If they use trigonometry, their angles must add to exactly 80. Both M marks must be scored for this A mark to be awarded. Working in radians is acceptable and correct, but angles must be to awrt sf. Angle A =.099 c Angle B = 0.67 c Angle C =.69 c for APC = 80 (their A + their B) ABP = 80 (their A + their C) for using either SIN or COS rule to find the length of AP. {Note PB 0 cm } (c) for the answer as shown correctly rounded. uses correct formula for Area of a triangle = sin ab C for answer as shown 7 (cm ). ALT using Heron s formula, 0 4 s 8 Area= Area = 7.7 = 7 (cm )

15 Question Number Scheme (a) x x x x 4!! Marks x x x oe for each coeff A (- ee) (b) x x x x 4!! () x x x oe for each coeff A (- ee) () (c) x B () (d) 4 x x x x x x x x 7 x 5 5 x x x x 5 x x x x x x oe for each coeff 6 7 ()

16 (e) x x x dx dx x x x x d 0.6 (4) [4] Notes (a) for using a correct binomial expansion at least up to the term in x. If there are errors in substitution, withhold this mark if the formula is not seen. Each term, must have at least, the correct power of x. The expansion must start with. (b) for + the term in x correct, and either term in x or x correct, need not be simplified. for the expansion correct as shown above. Accept equivalent fractions. for using a correct binomial expansion at least up to the term in x. If there are errors in substitution, withhold this mark if the formula is not seen. Each term, must have at least, the correct power of x. The expansion must start with. for + the term in x correct, and either term in x or x correct, need not be simplified. for the expansion fully correct as shown above. Accept equivalent fractions. (c) B for x or x

17 (d) for setting 4 4 x 4 x x x or their (a) their (b) {only need terms as far as x } If there is a or a 9 present, M0 for multiplying out their (a) (b). (Ignore the presence of a or a 9 for this mark) Check that they have multiplied out fully. There are six terms in total up to and including terms in x. for the answer as shown (oe) (e) for using their answer in part (d) to form an integral of a quadratic expression. Ignore limits and condone a missing dx. for a correct integration (no ft) d for correct substitution of 0.6. Allow missing cso If 0.6 is seen without working, no marks in part (e). ( Not hence obtain... )

18 Question Number 8(a)(i) Scheme Marks cos A cos A sin A sin A sin A, sin A*, () (ii) sin A sin Acos A B () (b) sin A sin Acos A cos Asin A sin A sin A sin Acos A sin sin sin sin A A A A sin A 4sin A (4) (c) sin A A 0, 0, 50 (any one) A 70, 0, 50 (4) (d) (i) sin d sin sin d 4 cos cos 4 (ii) cos cos cos 0 cos oe for 5,8, (5)

19 ALT for (d) sin d sin cos sin d (d) (i) cos cos (+ c) (ii) cos cos cos 0 cos = = 8 5 Notes (a) (ii) for the correct expression for cos A for using cos A + sin A =, and substituting into the expression for cos A for the correct identity as shown. Note this is show question! (ii) B for the correct identity for sin A dd [7] (b) for substituting sin A into their expression in part (a) to give sin (A +A) for using the given cos A and their sin A in their expression for sina for using cos A + sin A = for the final identity as shown. Note: this is a show question (c) 8sin A 6sin A sin A 4sin A sin A k, where for k for A equal to any one of 0º, 0º, 50º for any two correct angles for all three correct angles If there are extra angles outside of range, ignore. If there are extra angles within the range deduct one A mark for each up to a maximum of marks. (d) (i) for re-arranging the GIVEN expression for sin A to make sin A the subject for an attempt at integrating their re-arranged expression As a minimum for this mark, sin AdA cos A ( c) (+c not required) for the correct integrated expression as shown (+c not required) (ii) dd for substituting 4 and 0 into their integrated expression and attempting to evaluate (there must be a final answer given for this mark). for the final answer as shown ALT

20 (i) sin A sin A cos A sin A sin Acos A for finding for attempting to integrate. for (ii) For a minimum attempt you need to see; cos A cos A( c) sin AdA cos Asin AdA cos A cos A( c) cos Asin AdA cos A ( c) dd for substituting 4 and 0 into their integrated expression and attempting to evaluate (there must be a final answer given for this mark). for the final answer as shown ALT (Using substitution) (i) for writing the integral as sin sin sin cos du cos u and differentiating to achieve sin d and substituting du u for substituting and integrating sin u.d u du u d For definition of an attempt, see General Guidance for substituting u cos to give cos A cos A( c) (ii) dd for substituting 4 and 0 into their integrated expression and attempting to evaluate (there must be a final answer given for this mark). for the final answer as shown

21 Question Number 9 (a) (i) (b) (ii) (i) (ii) Y Scheme y B x B x or y or,0 0, B Marks () B () B shape (c) B asymptotes x B crossing points (d) x x x dy dx 7 0 dy 9 x dx 7 7 d 7 Grad l = 9 Eqnl: 7 y x 9 ft () (5)

22 (e) 7 x y x 9 x 8x 7 4x 77x ALT: 7 9 x 4x 7 4x 58x 48 0 x x (or use formula) solve linear eqn solve linear eqn x-coordinate of B is Accept x 7 correct answer (5) [7] Notes (a) (i) NOTE: If answers are transposed, award B0B If equations are on the graph, they must be written as equations. B answer as shown. y B (ii) (b) (i) B B (ii) answer as shown answer as shown answer as shown x x or,0 y or 0, (c) B B B for a rectangular hyperbola with one with branches in the correct quadrants. Please be generous on the shape of the curves. for the correct asymptotes f t their answers to part (a) There must be at least one branch of their graph for the award of this mark. for the correct intersections, ft their answers to part (b) (d) for attempting to differentiate x y. When using quotient rule, there must be an x attempt to differentiate and subtract the terms in the numerator; the denominator must be squared

23 ALT (using product rule) d for attempting to differentiate and add two terms. dy x x x ( ) x () dx x x for a fully correct differentiated expression for substituting expression x [ft their value from part (b) (i)] into their differentiated Check the differentiation as they can achieve 9 7 from incorrect calculus. for gradient of normal ft uses y y mx x (e) 7 m 9 to achieve an equation for the normal, where m is. f '(/ ) 7 for equating their straight line in the form y x with the equation of the 9 curve. Simplification is not required for this mark. for simplifying their equation to form a TQ for the correct TQ for an attempt to solve their TQ (please see General Guidance) for the coordinate of B = ALT for equating their straight line in the form y x with the equation of the 9 curve. 7 for attempting to form a linear equation in x 9 x for the correct linear equation for attempting to solve their equation (moves at least one term correctly to the other side of the equality) 4 for the coordinate of B = 7

24 Question Number Scheme Marks V 50 r h B 0(a) 50 r A r rh r r 00 A r * r () (b) da 00 4 r dr r da r dr r r () (c) d A 00 dr 4 r d A r min cso dr () (d) A min Amin 75 cso () []

25 (a) Notes B for equating the formula for the volume of a cylinder to the given volume of 50 for substituting their 50 = πr h into the formula for the surface area of a cylinder (b) S r rh for the answer as shown (this is a show question; beware of fudging answers) for an attempt to differentiate the GIVEN expression only for S for setting their d A dr = 0 Accept 00 4 r r... r for the answer as shown r =.996 (c) for attempting to find the second derivative of their d A dr (see General Guidance) for substituting their value of r into their da dr, but r must be positive. for d A > 0 hence minimum cso Their value for dr d A dr must be correct ( when r =.9966.) ALT 00 d A Also accept a conclusion by inspection. ie., 4 and are both positive hence > 0 r dr for substituting a value for r <.996 or r >.996 to test gradient d A dr When r <.996 d A < 0, when dr r >.996, d A dr > 0. for substituting a value for r <.996 and r >.996 to test d A dr. (d) around r =.996. for conclusion; that as r increases the gradient goes from negative to positive hence, minimum. for substituting their r into the GIVEN expression for A and evaluating, but their r must be a positive value. for A min = 75 cso

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