Mark Scheme (Results) Summer Pearson Edexcel International A Level in Further Pure Mathematics 3 (WFM03/01)

Transcription

1 Mark Scheme (Results) Summer 5 Pearson Edexcel International A Level in Further Pure Mathematics (WFM/)

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 5 years, and by working across 7 countries, in 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 5 Publications Code IA4579 All the material in this publication is copyright Pearson Education Ltd 5

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4 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.

5 EDEXCEL IAL MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edexcel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for example, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks. 5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer.

6 General Principles for Further Pure Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving term quadratic:. 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 (with values for a, b and c).. Completing the square b Solving x bx c : x q c, q, leading to x = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( x n x n ). Integration Power of at least one term increased by. ( x n x n )

7 Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are small errors in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may be lost if there is any mistake in the working. Exact answers Examiners reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals. Answers without working The rubric says that these may not gain full credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done in your head, detailed working would not be required..

8 Question Number. Scheme Notes Marks sinh x 7sinh x 5 Replaces coshx by sinh x or replaces coshx with cosh x + sinh x and then cosh x with sinh x. There must be no incorrect identities used. sinh x 7sinh x 4 Correct quadratic A sinh x sinh x4 sinh x Attempt to solve TQ in sinhx (usual rules) sinh x,4 Both (allow un-simplified e.g. 7 9 ) 4 A arsinhx ln x x Use of the correct log form of arsinh This mark may also be gained by using the exponential form of sinhx and attempting to solve to give x in terms of ln A: One correct exact value of x. Allow equivalent exact answers which may be 5 un-simplified. x ln, ln A: Both values correct and exact and no incorrect values. Allow equivalent exact answers which may be un-simplified. Condone missing brackets. 5 Correct work giving x ln, ln would generally lose the final mark Alternative: x x x x e e e e : Substitutes the correct exponential 7 5 definitions for coshx and sinhx A: Correct expression 4x x x x : Multiplies by e e 7e e 7e x A: Correct quartic in e x Solves their quartic as far as e x =... and x x x x x e e e 8e e... then converts to give x in terms of ln. There must be a recognisable attempt to x... solve a quartic with at least 4 terms as e.g. the product of two TQ s in e x. A: One correct exact value of x. Allow equivalent exact answers which may be 5 un-simplified. x ln, ln A: Both values correct and exact and no incorrect values. Allow equivalent exact answers which may be unsimplified. Condone missing brackets A, A (7) Total 7 A A A, A

9 Question Number Scheme Notes Marks. x y (a) (b) a b Substitutes the given point into the 5 hyperbola with the and 5 correctly and b a a b 4 positioned and substitutes the given value of e into the correct eccentricity equation b a a or a... 6 a a Solves simultaneously to obtain a value 45 5 for a or a or b or b or bor b... b b a8, b Allow equivalents for e.g. 5or awrt 4.47.Do not allow ± in either case Both (follow through their a). Must be ae,, coordinates. Alternative to (a): a sec Substitutes the given value of e into the and b a 5 b tan 4 correct eccentricity equation and substitutes the given point into the b 5 b 5 5 correct parametric form and eliminates, cosec cosec a 4 a 5 a and b 5 cosec a... or b... Solves to obtain a value for a or b 5 A, A Bft (4) () Total 5 a8, b Allow equivalents for e.g. 5or awrt 4.47.Do not allow ± in either case A, A

10 Question Number Scheme Notes Marks (a) k 9 9 Correct statement 7 or Correct eigenvalue A () (b) : Uses y component to establish 749k 49 k 7 * an equation for k A* A*: Correct k () (c) : Correct characteristic equation method (allow sign errors only) or A: Correct equation in any form A (d) Way (d) Way 4 9 NB 4 A: or 5 5, 5 A: and 5 9 p q 9r 4 7 q p4q7r r pr q 9r 6 p4q7r p r 5 p =, q =, r = Multiplies by M to obtain a vector in terms of p, q and r Correct equations : Solves simultaneously to obtain at least one of p, q or r. Dependent on the previous method mark. A: Correct answers Correct equations followed by correct answers scores full marks in part (d) p 6 46 q M r p 46 q r 4 5 : An appreciation that pi+ qj+ rk=m 6ij5k A: Correct inverse : Multiplies their inverse by the given vector. Dependent on the previous method mark. A: Correct vector AA A da A da (4) (4) (4) Total

11 Question Scheme Notes Marks 4(a) n n cosh xdxcosh xcosh xdx Writes cosh n x as cosh x cosh n x B n n cosh cosh d sinh cosh n x x x x x n cosh x sinh x dx : Parts in the correct direction (if the method is unclear or formula not quoted A only allow sign errors) A: Correct expression n n sinh x cosh x n cosh x cosh x dx Writes sinh x as cosh x n sinh xcosh x n I n n In n ni sinhxcosh xn I n n n sinh x cosh x n cosh x dx n cosh x dx d Substitutes for In and In- dd n n x x n n In sinh cosh I n * n Correct answer with at least one intermediate line of working and no errors seen Condone omission of dx and the occasional invisible x throughout but the final answer must be as printed. 4 4 ln I5 sinh xcosh x I 5 or I sinh xcosh x I 5 One application of reduction formula (I 5 in terms of I or I in terms of I ) 4 ln 4 ln I5 sinh xcosh x I 5 and I sinh xcosh x I 5 Second application of reduction formula (I 5 in terms of I and I in terms of I ) ln I cosh xdx sinh x I 4 4 B ln 5 55 cosh xdx 5 Must be exact A NB I5 sinh xcosh x sinh xcosh x I could score BA (b) ln A* (6) (4) Total (a) n n n cosh xdx cosh xcosh xdx sinh xcosh xdx Way B Writes cosh n x = (+ sinh x) cosh n- x n n n sinh xsinh xcosh xdx sinh xcosh x cosh xdx n n A : Parts in the correct direction (if the method is unclear or formula not quoted only allow sign errors) A: Correct expression n n n n cosh xdx cosh xdx sinh xcosh x cosh xdx n n d Adds In- to their integration by parts sinh cosh n In In n x x n In Substitutes for In and In- dd n I n I sinh xcosh n x I n n n n x x n n In sinh cosh I n * Correct answer with at least one intermediate line of working and no errors seen A*

12 Question Number 5(a) (b) (b) Way Scheme Notes Marks x mx c 5 9 9x 5 m x cmxc 5 5m 9 x 5cmx5c 5 Uses E and L to obtain an equation in one variable Correct quadratic with terms collected b 4ac 5cm 4 5m 9 5c 5 Use of b 4ac 5cm 4 65cm 565m 5c 5 5c 565m 5 c 5m 9* Achieves printed answer with no errors A* Use of the unproved general case a m + b = c scores no marks (4) See end of scheme for alternatives c 4m 4m 5m 9 Uses the point (, 4) and solves simultaneously to obtain an equation in one variable 6m 4m7 : Solves their quadratic to 7 4m4m7 m, obtain values for m 4 4 A: Correct values A 7 7 m c, m c Finds at least one value for c 7 7 y x, y x Correct equations A (5) Uses the point (, 4) and solves 4c 4c m c 5 9 simultaneously to obtain an equation in one variable 6c c48 : Solves their quadratic to 7 4c74c c, obtain values for c 4 4 A: Correct values A 7 7 Finds at least one least one c m, c m value for m 7 7 y x, y x Correct equations A Generally if candidates assume (, 4) lies on the ellipse they score no marks A (5) Total 9

13 Question Number 6 Scheme Notes Marks x cos cos, y sin sin (a) dx dy sin sin, cos cos Correct derivatives B, B d d dx dy d d 4 sin 8sin sin 4 sin 4 cos 8 cos cos 4 cos Squares and adds their derivatives 88 cos cos sin sin 88cos 8 cos * : Use of at least one correct trig identity A*: Correct proof with no errors (b) dx dy S y d d d Substitutes y sin sin and S sin sin 8cosd 8 cos into a correct formula sin cos 8 cos d 8 sin cos d 8 cos Processes to reach an integrand of the form k sin cos Integrates to obtain an expression of the form 5 cos. Dependent on the previous method mark. (May be done by substitution) Use of limits and π and subtracts Dependent on all previous method marks. 8 5 Allow equivalents e.g. 5.6π A A* d ddd (5) (5) Total

14 Question Number 7(a) Scheme Notes Marks x y z,,, 4 i j +4k, i + j k correct vectors lying in i j k 4 4 : Attempt normal vector using vectors lying in. If the method is unclear, at least components should be correct. A: Correct normal (any multiple) B A Attempt scalar product with a point lying in the plane. Dependent on the previous method mark. d xy 4z * Correct equation A* (b) 4 or 4 4 d 4 or 4 d 4 4 Attempt scalar product between and their or (a point in the plane ) and 4 their 4 Use of correct distance method. Dependent on the previous method mark. (Modulus not needed here) d Note: 4 d could score the first method marks 4 Set their distance =. Dependent on both previous method marks dd 5 Correct equations (must see ± for this mark but may still score one of the final A marks if ± is missing). May be implied and allow un-simplified. A, 4 cso A, A (6) Total (5)

15 Question Scheme Notes Marks Number 8(a) dx cosh u Correct expression B du 4 9 x 6 sinh u d x. coshudu 9 6x sinh u9 A : A complete substitution attempt. A: Correct expression k sinh udu A cosh u sinh u 9 8 Use of sinh u cosh u May be implied by their integration cosh udu A (b) 4 x u, x u arsinh cosh u du sinh uu 4 arsinh 64x 9 dx sinhu u 6x ln 9 9 Correct limits Attempt integration of the form sinhu + u Substitute u limits and subtract the right way round. Dependent on the previous M (Condone omission of ) B d 9, ln cao A, A (5) Total Alternative changes back to x 9 arsinh 4 I sinh u cosh u u 94 x 6 x 4 arsinh x I ln 9 9 Use of 4 x Substitute x limits and subtract the right way round. Dependent on the previous M (Condone omission of ) B d 9, ln A, A (6)

16 Alternatives to 5(a) Way Tangent at 5cos,sin is : Full attempt at a general tangent cos A y x A: Correct tangent 5sin sin 9 9cos c 5m 5 sin 5sin Substitutes their c and m into c 5m c 5m 9* Achieves printed answer with no errors A* Way x y x y dy dx d y 9 9 x m x dx 5y 5y y x mx c x mc and y c 5m 9 5m 9 5m : Differentiates implicitly, uses d y m and y mx cto obtain x and y in terms of dx m and c 5mc 9c A: x and y 9 5m 9 5m x y 5m c 9c 5 9 Substitutes their x and y in terms of m m m and c into E 5mc 9c 9 5m c 5m 9* Achieves printed answer with no errors A A*

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