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

Transcription

1 Mark Scheme (Results) January 07 Pearson Edexcel International GCSE In 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 0 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: January 07 Publications Code 4PM0_0_70_MS All the material in this publication is copyright Pearson Education Ltd 07

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.

4 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 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. With working Always check the working in the body of the script (and on any diagrams), 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 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 any appropriate marks, as long as it has not been replaced by alternative work. 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. Parts of questions Unless allowed by the mark scheme, the marks allocated to one part ofthe question CANNOT be awarded in another. 4

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: 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 explicitly or implied by 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 by. x. Integration: n Power of at least one term increased by. x Use of a formula: Generally, the method mark is gained by 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 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..." 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.

6 Jan 07 4PMO Further Pure Mathematics Paper Mark Scheme Question Scheme number Mark both parts of this question together 8 r 6 r r r 4 r 8 r oe 7 ALT A rl 6 r 8 r Notes For the equation (or any equivalent) 8 r For the equation (or any equivalent) 6 M For dividing their two equations, eliminating and finding a value for r For r = 4 (cm) 9 For 7 oe ALT M Attempts to use the (correct) A rl formula to give 6 8 r Substitution of correct values of 6 and l 8 For r = 4 (cm) For the equation (or any equivalent) 8 their r 9 For 7 oe ALT using degrees For the equation r 8 r For the equation r 6 r M Divides their equations to eliminate to give 9r 6 For r 4 9 For 7 oe Marks M () M ()

7 Question number (a) (b) Scheme f (4) 4 p 4 4 4p p 4p p * f ( ) ( ) 9( ) ( ) 4 Marks M () M () (c) x 9x x x 4 x x ( x )( x ) M (d) x 9x x ( x 4)( x )( x ) ( x 4)( x )( x ) 0 x 4, x, x () M () (9) Notes (a) M For either f ( 4 ) or f (4), equating f (±4) = 0 and finding a value for p. For the award of this mark the method must be complete. p = (b) M For either f ( ) or f () and finding a value for f (±) using the given p. For the award of this mark the method must be complete. Division Divides by x and achieves at least x x k (complete method) f 4 or remainder of 4 using division (c) M Divides f (x) by x 4 or x (d) any method, achieves at least x ax b where a0, b 0, and attempts to factorise their TQ. (See general guidance for an acceptable attempt) x 9x x x x x Note: OR by inspection; x 4 and x are factors, hence third factor is x a For achieving x x ( x )( x ) or x x x x 4 For the correct factorisation of f (x) = ( x 4)( x )( x ) M For setting f (x) = 0 (can be implied by further work) and attempting to solve a factorised f (x) = 0. ie., ( x 4)( x '')('' x '') 0 x 4,' ',' ' For x 4, x, x Note: answers must be derived from correct algebra

8 Question number Scheme x 4x 6x x 0x 0 0 c.v s x, x x x M Marks M Inside region for their values x M () M M M Notes For multiplying out the given inequality and achieving a TQ. Min acceptable TQ is x bx c Allow; x bx c = 0, x bx c < 0, x bx c > 0 or use of or or even just x bx c For solving their TQ (see general guidance for the definition of an attempt) and finding two critical values For x, x For choosing the INSIDE region for their cvs. For a correctly defined region as shown x Accept x AND x Do not accept x OR x (This is MA0) Allow use of set language x x but not x x (This is MA0) NB: Cancelling through by x and stating x < is M0M0A0M0A0 The question states using algebra. There must be a minimal amount of working to award marks. For just x without evidence of algebra M0M0A0M0A0 Minimally acceptable attempt is as follows; x x OR x x x, or

9 Question number 4 (a) 4 8 ar ar, 8 Scheme 4 76 ar ar 40 Marks M (i) (ii) (b) 4 ar, ar 4 ar r ar 40 7 * a 6 S M M (6) M () (8)

10 Notes (a) M For setting up both equations for the sum and the difference. Accept any letter for the first term. M Adds or subtracts their equations to eliminate ar or ar 4 For both correct 4 4 ar and ar 40 (i) M Divides ar 4 by ar to achieve an equation for r For r Note: This is a given result and every step must be shown to achieve this mark (ii) For a oe ALT for part (a) (a) M Sets up both equations for the sum and the difference 8 76 ar ar 4 ar r ar ar 4 ar r 8 40 M Factorises and divides equations above to eliminate ar to give r 8 ar r ar r = 8 40 r Achieves a correct equation in r or r 4 r 840 rr 840 or 4 r 876 rr 876 (i) M Attempts to solve their equation in r as far as r = For r Note: This is a given result so every step must be seen. (ii) For a oe ALT for part (a) using t and t or any other letters e.g x,y (a) M Solves SE by elimination to give: t+ t= and t t= t OR t M 4 4 Award these marks when they identify t ar OR t ar and use t ar, t ar t ar AND t ar 40 Then follow ms for (i) and (ii). (b) M Uses correct formula for the sum to infinity of a geometric series a their a 6 S ' ' They must reach a value for S r for this mark For 6 S

11 Question number. (a) (b) Scheme BA BA BC cos0 44 AB AB 48 (4 ) ALT sin 0 AB 4 ( ) sin0 sin D sin sin D 8 8 sin D Marks M (M) () M ft () (c) ACD= 4.4 Area of ABC = 48 sin0 = Area of ADC = 8 sin( ) Area of ABCD= 40.8 = 40.6 cm (sf) ALT 8sin('4.76..') AD.74.. sin() M M (6) () Area of ABC = 48 sin0 = (M) Area ADC = sin( ) Area of ABCD = 40.8 = 40.6 cm (sf) (M) () (6) ()

12 Notes (a) M Uses a correct cosine rule to find length AB For AB 4 ALT (a) M For using a correct sine rule to find length AB For AB 4 ALT (a) M Divides triangle ABC into two congruent right angle triangles. 6 AB o sin 60 For AB 4 (b) M For using a correct sine rule to find ADC For the acute angle resulting from their sine rule = (accept minimum accuracy of 9.4 ) o For the correct obtuse angle ADC 0.6 The general principle of marking part (c) is; First M for triangle ABC, second M for triangle ADC (c) o ACD 4.76 (accept minimum accuracy of 4.4 ) M Area of ABC using correct formula for area of a triangle using 0 and their length AB or BC (but their AB = BC) Area ABC = (oe., accept minimum accuracy of 0.8) M Area of ADC using correct formula and their ADC and the given lengths cm and 8 cm. Area ADC = (accept minimum 9.8 ) Area of quadrilateral ABCD = 40.6 (cm ) ALT (c) For finding length AD =.74.. (accept minimum accuracy of.7) M Area of ABC using correct formula for area of a triangle using 0 and their length AB or BC (but their AB = BC) For substitution of correct values. [Area ABC = (oe., accept minimum accuracy of 0.8)] M Area of using correct formula and their AD and the given length cm and angle. For substitution of correct values. [Area ADC = (accept minimum 9.8 )] Area of quadrilateral ABCD = 40.6 (cm ) ALT (c) Divides triangle ABC into two congruent right angle triangles. (midpoint of AB is M) 6 BM o tan 60 accept M Area of ABC using correct formula for area of a triangle 6 ' ' ' ' Area ABC = (oe., accept minimum accuracy of 0.8) Areas of ADC and quadrilateral ABCD as above.

13 Useful Sketch A 0.7 cm 4 cm (6.98 cm) 0.6 D cm cm B 0 8 cm 4 cm (6.98 cm) C Area ABC = or cm Area of ADC = cm Total area = 40.6 cm Penalise rounding only once. If they their answer to (b) as awrt0.6 (e.g.0.64) deduct the A mark. If they then give their answer to (c) as 40.6 do not penalise.

14 Question number 6. (a) a =, b = Scheme Marks () (b) At intersection of the curve with the y-axis, x = 0 0 c c 7 y c 7 0 '' '' M () (c) At intersection of the curve with the x-axis, y = 0 '' x '7' '' x '7' 0 x s x '' Mft () (6) Notes (a) For a = or b = For a = and b = (b) M For using the given equation and setting x = 0 and y =. (oe). They must achieve a value for c for the award of this mark Follow through their values for a and b. If their b is incorrect or they even use the letter b allow b0 0. c = 7 (c) M Uses their values for a, b and c and sets y = 0. They must achieve a value for x for the award of this mark ft For 7 s

15 Question number 7. (a) Scheme x y Marks () (b) (c) Correct points plotted ln(x ) x ln(x ) x Line yx drawnx.6 or.7 () MM () (d) (x) e x x ln(x ) x ln(x ) Line yxdrawn on graphx 0.9 MM M (4) () Notes (a) For any two of three correct values, correctly rounded For all three correct values, correctly rounded NB: Accept for B0 three values which all round to the correctly rounded values. (b) ft Their points plotted correctly to within half of one square ft Their points joined up in a smooth curve from x = onwards. Allow a straight line between x = 0 and. Note: these follow through marks are from their table only. (c) M For forming the linear equation ln(x) x or for identifying that the line with equation yx is required. This can be implied from a correct line drawn. M For drawing their yx Coordinates of the correct line yx are (0,), (,), (,4), (,) etc For x =.6 or.7 (Note: must be dp) (d) M For taking natural logarithms of both sides of the given equation to give xln x M For forming the linear equation ln(x) x or for identifying that the line with equation yx is required. This can be implied from a correct line drawn. M For drawing their yx. Coordinates of the correct line yx are (0,), (,4) For x = 0.9 Do not penalise rounding in (d) if penalised in (c). The value in (d) must round to 0.9.

16 Question number 8. (a) (i) (ii) Scheme x x ( )( 4) x ( )( 4)( ) x ( )...!! x x x 4 x M (4) Marks (b) x x x.. 8 so, A, 8 B () (c) 4x x x x x 9x 4 x... x M () (d) 0. 4x dx = 0 x 0. 0 x 9x x x x dx MdM () () Notes (a) (i) M For an attempt at a binomial expansion. There must be as a minimum; the expansion must start with ; there must be a minimum of 4 terms (accept a list); the power of x must be correct; the factorial denominator must be correct. x must be seen at least once. Two terms in x simplified and correct x x x Fully correct as shown ie., 4 (ii) For x or x (b) x For A OR B or embedded as x OR x For A AND B or embedded as x AND (c) M For expanding 4xtheir Atheir expansion from (a) at least as far as x x 9x Fully correct as shown ignore further terms (d) M For attempting to integrate their answer to part (c) (minimum of two terms) For an attempt to integrate, see general guidance dm For substituting 0. (0 not required) into their integrated expression. For a value of only Note: If there is no evidence of integration in (d) M0M0A0

17 Question number 9. (a) (i) (b) (c) (ii) Scheme () Marks MM () M oe (integer multiples) x x x x M () (0) Notes (a) 4 (i) For the sum 6 (ii) For the product oe (b) M For the correct algebra to find M e.g., Their final expansion must be given in a form such that they can substitute their sum and product directly. For substituting their values for the sum and product into their Note 9 For Note: This is a show question. Every step must be correct for 7 the award of this mark. 0

18 (c) ALT (c) M M M M First Final For the correct algebra on the sum their and. For the correct sum of 8 allow 7 and substitution of 7 4 For the correct product of For using their sum and their product correctly to form an equation. x sum x product 0 (condone missing = 0) For the correct equation as shown. Accept any integer multiples. e.g 08x x 4 0 etc Attempts to form the equation as follows. Must be ve sum, + ve product x x x x 0 x x x x x x x x 7 8 x x Correct algebra only x x x x x x 0 4x 76x 7 0 oe with integer multiples 7

19 Question number 0. (a) dv a t t dt 8 Scheme Marks M () (b) t 8t 0 t t 0 t, M () (c) 4 t 4t t s t 4t t t c 4 When t 0, s c M 4 t 4t t s t 4 So s = 8 m dm () ALT 4 t 4t t s t 4t t dx t 8 m For correct substitution and evaluation {M dm} {()} (9)

20 Notes (a) M For an attempt to differentiate the given v. See general guidance for the definition of an attempt For the correct a = t 8t (b) M Sets their a = 0 and attempts to solve their TQ. They must achieve values only for t for the award of this mark. For t, Please check the whole method in part (c) before you begin to award marks. (c) M Attempts to integrate the given v. See general guidance for the definition of an attempt. Award this mark if the constant of integration is not seen. For the correct integrated expression for s, which must include +c. For c = (Or any other letter given for the constant of integration) dm For substituting the value of t = into an integrated expression For s = 8 ALT (c) M Attempts to integrate the given v. See general guidance for the definition of an attempt. The limits of integration not required for this mark For the correct integrated expression For + dm For substituting their limits of integration. For s = 8 Note: if their limits were the wrong way around they will achieve s 8. Even if they give the final answer as s = 8 this is A0. ALT Only apply this scheme when see they have added the additional displacement of m at t = 0 M Attempts to integrate the given v. See general guidance for the definition of an attempt. The limits of integration not required for this mark For the correct integrated expression +c not required dm For substituting the value of t = into an integrated expression 6 For achieving s For adding to their s to achieve s oe

21 Question number. (a) (b) Scheme Mark parts (i) and (ii) together f '( x) p qx 0 p q() 0 p 6q 0 9 () () 9 9 p q p q p q Solves simultaneous equations by substitution or elimination 6p q 0 p q q q p 6 (i) q (ii) f ''( x) negative constant so point is a maximum x x x x x x x x ( )( ) 0, M Marks M M (7) MM () (c) Volume x 6x dx x 0 dx M 4 Volume = x x 6x x 0x 00 dx x 4 x x 0x 00x (or integrate without simplification) M M V () ()

22 Notes (a) M Attempts to differentiate the given equation for curve C, equates to 0, and substitutes in x = to form an equation in p and q. M Substitutes (,9) into the given equation to form an equation in p and q. For both correct equations; p6q 0 and p q or any equivalent to either equation. M Attempts to solve the simultaneous equations by any method. For p = 6. This is a show so check that the method is correct. For q Finds the second derivate, substitutes the value of q and finds f ''( x) with a conclusion hence maximum. E.g. Minimally acceptable hence maximum OR Completes the square to show that the maximum value of y is 9 when x = y x x x x with a conclusion that the maximum value of y = 9 occurs when x = (b) M Sets the equation of l = equation of c with their values of p and q and forms a TQ. M Attempts to solve their TQ by any method, but must achieve two values of x. For x, Marks in part (c) are dependent on their method being dimensionally correct and complete (c) Method (Combined integration) M For a statement using the correct formula for the volume of rotation V y dx, using the equation for C with their value of q, minus the equation M ddm for line l rearranged to make y the subject. Ignore missing dx and ignore limits for this mark. must be present and the equations must be squared. For integrating their statement for V. Their limits of integration found in (b) must be shown, the correct way around for the award of this mark. 4 The highest power of x must be a term in x. Ignore missing for this mark. For the correct integrated expression for V, complete with limits. It need not be simplified for this mark and ignore missing for this mark. For substituting in both of their values from (b) and subtracting them. For the correct volume in terms of only of V or 66.6 oe. isw erroneous attempts to simplify after 66.6 oe seen

23 Method (Integration of curve and volume of truncated cone) M For a statement using the correct formula for the volume of rotation V y dx, using the equation for C with their value of q. must be present M and the equations must be squared. Evidence of an attempt to find the volume of a truncated cone must be seen for this mark. For integrating their statement for V. Their limits of integration found in (b) must be shown, the correct way around for the award of this mark, and substituted into their integrated expression. 4 The highest power of x must be a term in x. Ignore missing for this mark. V 9.6 ( ) For the correct volume for C ddm For a correct method to find the volume of a truncated cone using their values of x from (b) to find y and substitute into the volume of a truncated cone. When x, y and x, y 8 V '8' '8' '' '' (=9 ) For the correct volume in terms of only of V or 66.6 oe. isw erroneous attempts to simplify after 66.6 oe seen Method (Integration of curve and line separately) M For a statement using the correct formula for the volume of rotation V y dx, using the equation for C with their value of q. Ignore missing dx and ignore limits for this mark. must be present and the equation must be squared. AND For a statement using the correct formula for the volume of rotation V y dx, using the equation for l with their values of p and q. Ignore missing dx and ignore limits for this mark. must be present and the equation must be squared. M For integrating their statements for V. Their limits of integration found in (b) must be shown, the correct way around for the award of this mark. 4 The highest power of x in C must be a term in x, and x in l. Ignore missing for this mark. For the correct integrated expressions for C and l, complete with limits. They need not be simplified for this mark and ignore missing for this mark. ddm For substituting in their values from (b) and subtracting them. AND subtracting the volume of the truncated cone fromm the volume of the curve. For the correct volume in terms of only of V or 66.6 oe. isw erroneous attempts to simplify after 66.6 oe seen NOTE: Volume of revolution of C = 9.6 Volume of truncated come = 9

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