PMT. Mark Scheme (Results) Summer Pearson Edexcel GCE in Further Pure Mathematics FP2R (6668/01R)

Transcription

1 Mark Scheme (Results) Summer 04 Pearson Edecel GCE in Further Pure Mathematics FPR (6668/0R)

2 Edecel and BTEC Qualifications Edecel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Epert service helpful. Pearson: helping people progress, everywhere 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 04 Publications Code UA08876 All the material in this publication is copyright Pearson Education Ltd 04

3 General Marking Guidance All candidates must receive the same treatment. Eaminers must mark the first candidate in eactly 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. Eaminers 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. Eaminers should always award full marks if deserved, i.e. if the answer matches the mark scheme. Eaminers 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 eemplification may be limited. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edecel 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 eample, 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.

5 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 ( + b + c) = ( + p)( + q), where pq = c, leading to = ( a + b + c) = ( m + p)( n + q), where pq = c and mn = a, leading to =. Formula Attempt to use the correct formula (with values for a, b and c).. Completing the square b Solving + b + c = 0 : ± ± q ± c = 0, q 0, leading to = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( n n ). Integration n+ Power of at least one term increased by. ( n )

6 Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent eaminers 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. Eact answers Eaminers reports have emphasised that where, for eample, an eact 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.

7 Question Number. A B (a) = + 4r r+ r Scheme Marks ( ) ( ) = A r + B r+ A=, B= = 4r r r+ A () n n (b) () = r= 4r r= r r+ = = n n+ n+ Aft n + = n + (a) (b) n n = * A () r= 4r n+ Notes for Question complete method for finding PFs A both PFs correct Award A for both PFs seen correct w/o working. M0A0 otherwise showing fractions with their PFs. Min at start and at end. Must start at and end at n. Required sum may be used or sum Aft Identify non-cancelling fractions, follow through their PFs - sum or sum Acso correct final answer [5]

8 Question Number Scheme 5 = 0 (or <) or mult through by Marks 5 = 0 (or <) ( + )( ) = 0 or ( )( ) + = 0 CVs =, A = 0 B <, 0 < < or in set language (with curved brackets for A) A (5) Special case If used deduct final mark only. [5] Notes for Question obtaining two non-zero cvs by any valid method (not calculator) A non-zero cvs correct B = 0 deducing one appropriate range from their cvs A both ranges correct First marks award with inequalities or = A0 if strict inequality not used ALT: If multiplied through by : > 0: 5 + < 0 < ( )( ) (solve cvs =, = quad) 0, < < B,A < 0 5 > 0 cvs =, = < A

9 Question Number. (a) dy tan e cos + = 4 y Scheme Marks tan d lnsec e = e = sec or cos A dy 4 sec ytan sec e cos sec d d ( ysec ) sec = e Bft( 4 = e ( + ) 4 4 y c ysec ) e cos 4 4 y = + c oe A (6) (b) y, 0 = = = + c 4 c = 4 (a) (b) y ( e ) cos 4 4 = + oe A Notes for Question attempting the integrating factor, including integration of ()tan ln cos or ln sec seen A correct integrating factor sec or cos multiplying the equation by the integrating factor may be implied by the net line. Bft y their IF 4 attempting a complete integration of rhs Must include k e but 4 4e would imply differentiation. Constant not needed (Incorrect IF may lead to integration by parts, so integration must be complete) A correct solution in form y =... constant must be included using given initial conditions to obtain a value for c A fully correct final answer May be in the form ysec =... or 4 ysec =... () [8]

10 Question Number Scheme Marks 4. ( y = ) rsinθ = cos θsinθ dy = 4sin θ sinθ + cos θ cosθ dθ A sin θ sinθ cos θ cosθ = 0 4sin cosθ sin θ cosθ = 0 d ( ) ( 6sin θ ) cosθ = 0 ( cosθ = 0 no solutions in range) sinθ = dda 6 ALT for last marks above: cos θ cosθ = sin θ sinθ tanθ tanθ = / tan θ tanθ = 5 tan θ tanθ / 5 d(double angle formula) = = dda (sinθ = / 6 cosθ = 5/ 6 r sinθ = cos θsinθ sinθ = cos θ = sin θ = = 6 6 Eqn. l: 4 r sinθ = = 6 6 r 6 cosec θ 9 = oe ( 0 < θ < π) Must be seen in eact form Notes for Question 4 Using y = rsinθ = cosθsinθ differentiate rsin θ or rcosθ using product rule or cos θ = sin θ and chain rule A correct differentiation of r sinθ d equate their derivative to 0 and use cos θ = sin θ if not used prior to differentiation, or an appropriate double angle formula for their derivative. Depends on second M mark dd solve the resulting equation. Depends on second and third M mark A correct value for sinθ or tan θ or cosθ depending on the equation solved use their value for a trig function to obtain an eact value for cos θ and sinθ if needed now. May be implied by the net stage. use their values for sinθ and cos θ in r sinθ = cos θsinθ NB: These two M marks require 0 sinθ /, / cosθ, 0 tanθ 0 < θ < π not needed A correct equation in form r =.. ( ) A [9]

11 Question Number 5. (a) d y dy y = Scheme Marks dy d y seen d 4 d d d = + d d d d y y y y y y Alt: dy d y B ( and diff) AA B y y y y y y = 0 d d d d d d y d d d d d d y y y y d d d = 5 d d d A A (4) (b) At 0 d y 9 4 = = 4 = ( or.5) d y 9 7 = = 4 6 ( ) 5 or.5 A A 9 7 y = ! 6! (! or,! or 6) 9 7 y 8 96 = A (5) sf or better [9]

12 (a) (b) Notes for Question 5 B dy d y seen in the differentiation divide equation by y and differentiate wrt chain and product/quotient rules needed AA - for each error. Ignore any simplification following the d y differentiation and obtaining... = ALT: B as above differentiating before dividing d y AA rearrange to a correct epression for, - each error using values for and d y to obtain a value for d y d y A correct value for d y A correct value for Taylor's series formed using their values for the differentials, accept! or and! or 6. A correct series, must start y = (or end =y)

13 Question Number 6. (a) w= ( u i = ) + iy i y+ Scheme Marks ( y i) ( ) + iy w= ( u i = ) i y+ y i ( u i ) w= = = y y ( y ) ( y) ( ) y + y + i y y + + A + y y = y y y = A (5) Alternative : w z = wi u i + iy = i i ( u ) u i = A ui = i + u y = A (5) Alternative : w+ i = w z z + i = iz+ iz+ z z+ i z = iz+ iz+ A z i = z y = A (5)

14 Question Number (b) + i + i u+ iv= = i + i + i + Scheme Marks + i i u+ iv= i+ i u+ iv= + i A u u = v= ( ) 4 ( 4 + ) ( 4 + ) = = = 4 6 v u + v =, Centre O *,A (6) Alternative : + i iθ + iy w= u+ i v =, ( = re ) or i+ i+ y w + = 4, = + 4,A u + v = Centre O * A []

15 Question Number Scheme Marks Alternative : z i = z ( iw) w i w = iw iw iw w i i w w + = A w = u + v = Centre O A (6) Alternative : z w = iz + w u+ iv z = = iw u+ iv i ( ) Realise the denominator Correct result A Set imaginary part = and simplify epression u + v = Centre O A (6)

16 (a) (b) Notes for Question 6 substitute z = + iy multiply the numerator and the denominator by conjugate of the denominator A correct equation with real denominator on rhs use w = u - i and equate imaginary part in (their) equation to - A deducing y = Alternative re-arrange to z =... replace w with u - i or replace with u + iv and realise the denominator A correct rhs May still have v separate real and imaginary parts in (their) above equation. This may be implied by a correct answer. A deducing y = Alternative : A A Alternative : A A substitute z = + i/ or work with + iy multiply the numerator and the denominator by conjugate of the denominator A correct equation with real denominator on rhs use their u, v and find u + v simplify and cancel including use of Acso u + v = Centre O y = Alternative : as above find w substitute y = A w = deduce circle centre O Acso u + v =

17 Alternative : A A Alternative : A A

18 Question Number 7. (a) ( ) 5 Scheme cosθ + isinθ = cos5θ + isin 5θ B = cos θ + 5cos ( i sinθ) + cos θ( i sinθ)! θ( θ) + θ( θ) + ( θ)! 4! 4 5 cos isin cos isin isin Marks 5 4 = cos θ + 5i cos θ sinθ 0 cos θ sin θ 4 5 0i cos θ sin θ + 5cosθ sin θ + i sin θ A sin 5 5cos sin 0cos sin sin 4 5 θ = θ θ θ θ + θ ( ) ( ) 5 = 5 sin θ sinθ 0 sin θ sin θ + sin θ θ = θ θ + θ * A (5) 5 sin 5 6sin 0sin 5sin 5 (b) Let = sinθ = sin 5θ = 5θ = 0, 0, 570, 690, 90, 050, 90 (or in radians) A, A Or 0, 570, 90, 90, 650 θ = 4, 66, ( 4 ), ( 8 ), 86, 0, 58 (or in radians) Or 4, 4, 86, 58, 0 d(at least values) sinθ = 0.669, 0.94, 0.05, 0.5, A (5) π 4 5 (c) 4 ( 4sin θ 5sin θ) dθ = 4 ( sin 5θ 5sinθ) dθ 0 0 = cos5 θ 5cos θ π 4 5π π cos 5cos π A 5 4 = = 0 5 A (4) [4]

19 (a) (b) (c) Notes for Question 7 B applies de Moivre correctly uses binomial theorem to epand ( cosθ isinθ) 5 + May only show imaginary parts - ignore errors in real part A simplifies coefficients to obtain a simplified result with all imaginary terms correct equates imaginary parts and obtains an epression for sin 5θ in terms of powers of sinθ A cso correct result uses substitution = sinθ deduces that sin 5θ = ± AA gives a set of results for 5θ - A for useable results A for the remaining useable results (no repeats in the set of 5) at least values for θ A for the 5 different values of uses previous work to change the integrand A correct result after integrating - limits can be ignored substitute given limits and use numerical values for trig functions A final answer correct (oe provided in the given form)

20 Question Number 8. (a) = e z Scheme Marks z dz = e dy dy dy z dy = e dz A d y z dz dy z d y dz dy d y = e + e = + dz dz dz dz AA d y dy + y = ln dy d y dy + + y = z dz dz dz d y dy + y = z A (7) dz dz Alt: z = ln dy dy dz dy = = A dz dz d y dy d y dz dy d y = + = + AA dz dz dz dz d y dy + y = ln dy d y dy + + y = z dz dz dz d y dy + y = z A (7) dz dz

21 Question Number (b) Au eqn: m + m = 0 Scheme Marks ( m )( m ) + = 0 m =, A CF: z = e + e A z y A B PI: Try y = az + b dy d y = a = 0 dz dz ( ) a az + b = z a =, b= 4 Complete soln: z z y = Ae + Be z AA (6) 4 (c) y = A + B ln B ft () 4 (a) (b) (c) Notes for Question 8 differentiates = e z wrt y; chain rule must be used A correct differentiation d y differentiates again to obtain AA one mark for each correct term substitutes in the given equation Acso obtains the required equation ALT: Works with z = ln ; marks awarded as above forms and solves the auiliary equation A both values for m correct - may be implied by their CF A correct CF tries a suitable epression for the PF and obtains values for the constants in the PF AA shows the complete solution; one mark for each correct term in the PF Bft reverses the substitution to obtain the solution in the form y =... Follow through their complete solution from (b) [4]

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