Mark Scheme (Results) Summer Pearson Edexcel IAL in Further Pure Mathematics 2 (WFM02/01)

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Mark Scheme (Results) Summer 06 Pearson Edexcel IAL in Further Pure Mathematics (WFM0/0)

dexcel and BTEC Qualifications dexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding body. We provide a wide ange of qualifications including academic, vocational, occupational and specific programmes for employers. or further information visit our qualifications websites at www.edexcel.com or www.btec.co.uk. Alternatively, you can get in touch with us using the details on our contact us page at www.edexcel.com/contactus. 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. e ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, e have built an international reputation for our commitment to high standards and raising achievement hrough innovation in education. Find out more about how we can help you and your students at: ww.pearson.com/uk Summer 06 Publications Code WFM0_0_606_MS All the material in this publication is copyright Pearson Education Ltd 06

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.

PEARSON 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. 3. 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) d or dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper or ag- answer given or d The second mark is dependent on gaining the first mark

4. All A marks are correct answer only (cao.), unless shown, for example, as 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. 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.

General Principles for Further Pure Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving 3 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). 3. Completing the square Solving x bxc 0 : b x qc0, q 0, 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 )

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.

Question Number (a) (b) (c) Scheme Notes Marks r or 4 r r r r or equivalent or A B A, B 4r r r Correct partial fractions or correct values of A and B. Isw if possible so if correct values of A and B are found, award when seen even if followed by incorrect partial fractions. n... r 4r 3 n n Attempt at least first and last terms using their partial fractions. May be implied by e.g. n or or n Correct expression n 4n n n * Correct completion with no errors * Allow a different variable to be used in (a) and (b) but final answer in (b) must be as printed i.e. in terms of n. 5 5f(5) f(8) f(5) f(8) where f n r9 5 4r 5 8 5 5 5 7 5 cao Correct answer with no working in (c) scores both marks. n n B () (3) () Total 6

Question Number Scheme Notes x 9 x (ignore use of < instead of = when finding cv s) x 9xx x00 x... Attempts to solve or x 9 x OR x 9 x x 9xx x80 x... to obtain two non-zero values of x One correct pair of values. Allow the 44 irrational roots to be at least as given x OR x, 4 here or or awrt.3, -4.3 or truncated.3, -4.3 Attempts to solve x 9xx x80 x... x 9 x AND x 9 x to obtain four non-zero values of x Both pairs of values correct. Allow 44 the irrational roots to be at least as x AND x, 4 given here or or awrt.3, -4.3 or truncated.3, -4.3 For allow For allow x 4 or x 44 Allow alternative notation e.g., for allow One correct inequality. 44 but must be exact here., 4,, 4 x, x and x4, x, x and x x 4 and x 44 Allow alternative notation e.g., for allow Both inequalities correct. 44 but must be exact here., 4,, 4 x, x and x4, x, x and x B B Marks (6) Total 6

Q Alternative by squaring (ignore use of < instead of = when finding cv s) 4 x 9 x x 8x 84x 4x 4 x x x x 4 80 0... 44 x or x, 4 44 x and x, 4 x 4 or x x 4 Squares and attempts to solve a quartic equation to obtain at least two values of x that are non-zero. One pair of values correct as defined above : Obtains four non-zero values of x. : Both pairs of values correct as defined above See notes above and See notes above x In an otherwise fully correct solution, if any extra incorrect regions are given, deduct the final B mark. B B

3 Scheme Notes Marks dy dx x ky x x k dy ky x x dx x x k dx k x x Ie k y x x x dx k Divides by ( + x) including the ky term d: Attempt integrating factor. k dx x I e is sufficient for this mark but must include the k. Condone omission of dx. : x k Reaches k y their I x x their I dx 3 5 x x dx x x c 3 5 or by parts Correct integration 3 5 4 x x dx x x x c 3 5 3 5 x x c y 3 5 k x or e.g. 3 4 5 x x x c y 3 5 k x or e.g. 3 k 4 5 k k y x x x x c x 3 5 or e.g. 3 5 0x x4x c y k 5 x or e.g. 3 k 5 k k y x x x x c x 3 5 Correct answer with the constant correctly placed. Allow any equivalent correct answer. d (6) Total 6

Question Number 4(a) (b) f f f f f Scheme Notes Marks 3 x sin x : Attempt first 4 derivatives. Should be ' sincossincos sin. I.e. ignore 3 3 x cos x signs and coefficients. '' 9 3 x4sin x ' 3 3 '' 9 3 :f cos x and f 4sin x ''' 7 3 x 8 cos x ''' 7 3 '''' 8 3 :f 8 cos x and f 6 sin x '''' 8 3 x 6 sin x Allow un-simplified e.g. f '' 3 3 3. sin x ' '' 9 ''' '''' 8 y, y 0, y, y 0, y 3 3 3 4 3 3 6 Attempts at least derivative at x 3 d: Correct use of Taylor series. I.e. f 3 x x x 4 9 7 f x x x 8 3 8 3 f 0.485 3 f f 3 3 f 3 3...! Evidence of at least one term of the correct n f n 3 structure i.e. x 3 and not a n! Maclaurin series. Dependent on the previous method mark. : Correct expansion. Allow equivalent single fractions for 9 7 and/or 8 8 and allow decimal equivalents i.e..5 and 0.09375. Ignore any extra terms. : Attempts f 3 or states x 3 : 0.485 cao A d (6) () Total 8

Question Number 5 z Scheme Notes Marks 3w z w 3w 3 uiv w uiv 3 3 u v u v 0 3 u v u 0 8 8 u v 5 5 3 8 64 8 v 5 49 5 7 u,0, 8 64 8 8 : Attempt to make z the subject as far as z = : Correct equation Uses z and introduces w = u + iv : Correct use of Pythagoras. Condone missing brackets provided the intention is clear and allow e.g. (3v) = 3v but there should be no i s. Attempt to complete the square on the equation of a circle. I.e. an equation where the coefficients of u and v are the same and the other terms are in u, v or are constant. (Allow slips in completing the square). Dependent on all previous M marks. 5 : Centre,0 8 : Radius 7 8 dddd Alternative for the first 3 marks 3w : Attempt to make z the subject z w : Correct equation 5u3u v 3 uiv 3u3iv uiv 7v xiy uiv uiv uiv u v u v 5 3 7 u u v v x y u v u v Introduces w = u + iv, multiplies numerator and denominator by the complex conjugate of the denominator and uses x y correctly to obtain an equation in u and v. i (7) Total 7

Question Number Scheme Notes Marks d y dy 3 y 3x x dx dx 6(a) Correct roots (may be implied by their m 3m 0 m, CF) B x x y Ae : CF of the correct form Be : Correct CF y ax bx c Correct form for PI B dy d y ax b, a a 3 ax bax bx c3x x dx dx : Differentiates twice and substitutes into the lhs of the given differential equation and puts equal to 3x x or substitutes into the lhs of the given differential equation and compares coefficients with 3x x. For the substitution, at least one of y, y or y must be correctly placed. 3 a 7 7 : Solves to obtain one of b or c 6ab b c 4 : Correct b and c x x 3 7 7 Correct ft (their CF + their PI) but y Ae Be x x 4 must be y = Bft (9) (b) 7 Substitutes x = 0 and y = 0 into their 0 A B 4 GS dy x x 7 7 Ae Be 3x 0 AB dx Attempts to differentiate and substitutes x = 0 and y = 0 7 7 Solves simultaneously to obtain 0 A B, 0 AB A.., B.. 4 values for A and B 3 A, 4 B 5 Correct values 3 x x 3 7 7 Correct ft (their CF + their PI) y e 5e x x 4 4 but must be y = Bft (5) Total 4

Question Number Scheme Notes Marks 7. 3 9 C: r cos, C : r 3 3 cos (a) 3 9 cos 3 3 cos... Puts C = C and attempt to solve for or or r Eliminates cos θ and solves for r cos r 3 3 3 r r... 3 3 3 Correct θ or correct r. or r 6 4 Allow θ = awrt 0.54, r = awrt.3 3 3 3 3 Correct r and θ (isw e.g., ) r and 6 4 4 6 Allow θ = awrt 0.54, r = awrt.3 (3)

7(b) 9 3 3 3 cos d or cos d Attempts to use correct formula on either curve. The ½ may be implied by later work. 9 8 8cos 3 3 cos 7 7 3 cos cos 7 7 3 cos 4 4 Expands to obtain an expression of the form abcos ccos and attempts to use cos cos 9 97 8 3 3 cos d 7 3sin sin 8 6 : Attempts to integrate to obtain at least two terms from, sin, sin 97 8 : Correct integration with or without the ½ (NB 7 ) 8 8 6 97 8 97 8 7 3 sin sin. 7 3.sin sin. 0 8 6 0 8 6 6 6 6 : Uses the limits 0 and their 6 If the substitution for = 0 evaluates to 0 then the substitution for = 0 does not need to be seen but if it does not evaluate to 0, the substitution for = 0 needs to be seen. cos d cos d sin 6 3 9 9 9 3 6 6 6 3 4 cos : Uses cos, integrates to obtain at least k sin and uses the limits of their and to find the other area 6 NB can be done as a segment : Allow 3 3 sin 4 3 4 3 3 3 their sin their 4 6 4 6 97 35 3 9 3 05 45 3 96 64 6 3 4 3 8 : Adds their two areas both of which are of the form a b 3 : Correct answer (allow equivalent fractions for 05 45 and/or 3 8 ) (8) Total

Special Case Uses C C (b) 9 3 3 3 cos cos d C C. The ½ may be implied by later work. Attempts to use correct formula on cos 3 3 6cos 7 36 3 cos 36cos 7 36 3 cos 36 Expands to obtain an expression of the form abcos ccos and attempts to use cos cos 3 3 6 cos d 45 36 3 sin 9sin Attempts to integrate to obtain at least two terms from, sin, sin No more marks available

Question Number 8(a) WAY Scheme 5 5 3 z z 5z 0z Notes : Attempt to expand z 5 z 0 5 3 5 z z z z : Correct expansion with correct powers of z. z cos isin z cos z May be implied B 5 3 z 5 z 0 z cos5 0cos3 0cos 5 3 z z z Uses at least one of z 5 cos5 or z 3 cos3 5 3 z z 5 5 5 z 3cos B z 5 5 5 cos cos5 cos3 cos Correct expression 6 6 8 WAY (Using e e e 5e 0e 0e 5e e i i 5i 3i i i 3i 5i i e ) : Attempt to expand i i e e 5 : Correct expansion i i cos e e May be implied B 5i 5i 3i 3i i i e e 5 e e 0 e e cos5 0cos3 0cos 5i 5i 3i 3i Uses one of e e cos5 or e e cos3 5 i i 5 e e 3cos B 5 5 5 cos cos5 cos3 cos Correct expression 6 6 8 Marks (6) WAY 3 (Using De Moivre on cos 5θ and identity for cos 3θ) 5 5 4 3 3 3 4 4 5 5 cos isin c 5ic s0ci s 0ci s 5ci s i s : Attempts to expand. NB may only consider real parts here. : Correct real terms (may include i s) (Ignore imaginary parts for this mark) 5 3 4 cos5 cos 0cos sin 5cossin Correct real terms with no i s B cos 5 0cos 3 cos 5cos cos 5 3 6cos cos5 0cos 5cos 3 cos3 4cos 3cos Uses sin cos to eliminate sin θ Correct identity for cos 3θ B 5 6cos cos5 5cos3 0cos 5 5 5 cos cos5 cos3 cos Correct expression 6 6 8 (6)

(b) 5 5 5 5 cos5 cos3 cosd sin 5 sin 3 sin 6 6 8 80 48 8 : Attempt to integrate Evidence of cos n sin n where n 5 or 3 n ft: Correct integration (ft their p, q, r) 5 5 3 5 5 5 5 5 5 sin 5 sin 3 sin sin sin sin sin sin sin 80 48 8 80 3 48 8 3 80 6 48 8 6 6 Substitutes the given limits into a changed function and subtracts the right way round. There should be evidence of the substitution of and into their changed function 3 6 for at least of their terms and subtraction. If there is no evidence of substitution and the answer is incorrect, score M0 here. Allow exact equivalents e.g. 49 3 03 03 60 480 4.9 3 6 30 If they use the letters p, q and r or their values of p, q and r, even from no working, the M marks are available in (b) but not the A marks. ft (4) Total 0

Question Number 9(a) (b) Scheme Notes Marks Centre Cx, y is at 3.5,.5 c c 3 3 r.5.5 Max 5 z OC r 3.5.5 r z 5 arg z 4 A circle or an arc of a circle anywhere. Allow dotted or dashed. A circle or an arc of a circle (allow dotted or dashed) passing through or touching at and 5 on the positive real axis. (Imaginary axis may be missing) Fully correct diagram with and 5 marked correctly with no part of the circle below the real axis. It must be a major arc and not a semi-circle. The imaginary axis must be present and the arc must not cross or touch the imaginary axis. (3) May be implied and may appear on the diagram. Can score anywhere e.g. from finding the equation of the B circle in part (a) or as part of the calculation for OC. r or equivalent work e.g. yc.5.5 r, r or 3 cos 45 sin 45 seen 58 3 Oe e.g 58 3 4.5 4.5, Special Case correct work with arc below the real axis:, 3.5,.5 B0 Centre Cx y is at c c 3 3 r.5.5 Max z OC r 3.5.5 r r or equivalent work e.g. yc.5.5 r, r or 3 cos 45 sin 45 seen 58 3 Oe e.g 58 3 4.5 4.5, Total 7 (4)

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