Mark Scheme (Results) Summer 04 Pearson Edexcel GCE in Further Pure Mathematics FPR (6668/0R)
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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.
EDEXCEL GCE 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 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.
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 0 : x q c 0, 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. 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.
Question Number. A B (a) 4r r r Scheme Marks A r B r A, B 4r r r () n n (b) () r 4r rr r... 5 5 7 n n n ft n n (a) (b) n n * () r 4r n Notes for Question complete method for finding PFs both PFs correct Award 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 x sum ft Identify non-cancelling fractions, follow through their PFs - sum or x sum cso correct final answer [5]
Question Number Scheme x 5 0 x (or <) or mult through by x Marks x 5x 0 (or <) x x x 0 or x x x x 0 CVs x, x = 0 B x, 0 x or in set language (with curved brackets for ) (5) Special case If used deduct final mark only. [5] Notes for Question obtaining two non-zero cvs by any valid method (not calculator) non-zero cvs correct B x = 0 deducing one appropriate range from their cvs both ranges correct First marks award with inequalities or = A0 if strict inequality not used ALT: If multiplied through by x: x > 0: x 5x x x 0 (solve cvs x, x quad) 0, x B, x < 0 x 5x 0 cvs x, x x
Question Number. (a) dy tan e cos 4x y x x Scheme Marks tan d lnsec e x x x e sec x or cos x dy 4x sec x y tan xsec x e cos xsec x d d ysec x e Bft( 4x 4 4x ysec x e c ysec x) e cos 4 4x y c x oe (6) (b) y, x 0 c 4 c 4 (a) (b) y e cos 4 4x x oe Notes for Question attempting the integrating factor, including integration of ()tan x lncos or lnsec seen correct integrating factor sec x or cos x multiplying the equation by the integrating factor may be implied by the next line. Bft y their IF 4 attempting a complete integration of rhs Must include k e x but 4 4e x would imply differentiation. Constant not needed (Incorrect IF may lead to integration by parts, so integration must be complete) correct solution in form y =... constant must be included using given initial conditions to obtain a value for c fully correct final answer May be in the form ysec x... or 4ysec x... () [8]
Question Number Scheme Marks 4. ( y) rsin cos sin dy 4sin sin cos cos d sin sin cos cos 0 4sin cos sin cos 0 d 6sin cos 0 cos 0 no solutions in range sin dd 6 ALT for last marks above: cos cos sin sin tan tan / tan tan 5 tan tan / 5 d(double angle formula) dd ( sin / 6 cos 5/ 6 r sin cos sin sin cos sin 6 6 Eqn. l: 4 r sin 6 6 r 6 cosec 9 oe 0 Must be seen in exact form Notes for Question 4 Using yrsin cos sin differentiate rsin or rcos using product rule or cos sin and chain rule correct differentiation of rsin d equate their derivative to 0 and use cos sin 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 correct value for sin or tan or cos depending on the equation solved use their value for a trig function to obtain an exact value for cos and sin if needed now. May be implied by the next stage. use their values for sin and cos in r sin cos sin NB: These two M marks require 0 sin /, / cos, 0 tan correct equation in form r =.. 0 not needed [9]
Question Number 5. (a) d y dy y Scheme Marks dyd y seen d 4 d d d d d d d y y y y x y x x y x Alt: dyd y B ( and diff) B y y y y y y 4 0 x x x x x x d d d d d d y d d d d d d y y y x y x x d d d 5 d d d (4) (b) At 0 d y 9 4 x 4 or.5 d y 9 7 4 6 5 or.5 9 x 7 x y x... 4! 6! (! or,! or 6) 9 7 y x x x 8 96... (5) 0.5.5 0.854 sf or better [9]
(a) (b) Notes for Question 5 B dyd y seen in the differentiation divide equation by y and differentiate wrt x chain and product/quotient rules needed - for each error. Ignore any simplification following the d y differentiation and obtaining... ALT: B as above differentiating before dividing d y rearrange to a correct expression for, - each error using values for x and d y to obtain a value for d y d y correct value for d y correct value for Taylor's series formed using their values for the differentials, accept! or and! or 6. correct series, must start y = (or end =y)
Question Number 6. (a) w ui x iy ixy Scheme Marks x iy yix w u i ix y y ix w u i y y y x x xy xy i y y x x y x y y x y y x y (5) Alternative : w z wi u i xiy u i i u i ui i u y (5) Alternative : wi w z z i iz iz z z i z iz iz zi z y (5)
Question Number (b) x i x i u iv ix i ix Scheme Marks x i xi u iv ix xi uiv xi x 4 x 4 u x x v 4 x x 4 4 u 4 4x 4x x x x x x 4 6 v u v, Centre O *, (6) Alternative : x i i xiy w u i v, re or ix ix y w x 4, x 4, u v Centre O * []
Question Number Scheme Marks Alternative : zi z w i iw w iw iw iw w i i w w w u v Centre O (6) Alternative : z w iz w u iv z iw u iv i Realise the denominator Correct result Set imaginary part and simplify expression u v Centre O (6)
(a) (b) Notes for Question 6 substitute z = x + iy multiply the numerator and the denominator by conjugate of the denominator correct equation with real denominator on rhs use w = u - i and equate imaginary part in (their) equation to - deducing y = Alternative re-arrange to z =... replace w with u - i or replace with u + iv and realise the denominator correct rhs May still have v separate real and imaginary parts in (their) above equation. This may be implied by a correct answer. deducing y = Alternative : Alternative : substitute z = x + i/ or work with x + iy multiply the numerator and the denominator by conjugate of the denominator correct equation with real denominator on rhs use their u, v and find u v simplify and cancel including use of cso u v Centre O y Alternative : as above find w substitute w y deduce circle centre O cso u v
Alternative : Alternative :
Question Number 7. (a) 5 Scheme cos isin cos5 isin 5 B 5 4 5 4 cos 5cos i sin cos i sin! 5 4 5 4! 4! 4 5 cos isin cos i sin i sin Marks 5 4 cos 5i cos sin 0cos sin 4 5 0i cos sin 5cos sin isin sin 5 5cos sin 0cos sin sin 4 5 5 5 sin sin 0 sin sin sin 5 sin 5 6sin 0sin 5sin * (5) 5 (b) Let x sin 6x 0x 5x sin 5 5 0, 0, 570, 690, 90, 050, 90 (or in radians), Or 0, 570, 90, 90, 650 4, 66, 4, 8, 86, 0, 58 (or in radians) Or 4, 4, 86, 58, 0 d(at least values) sin 0.669, 0.94, 0.05, 0.5, 0.978 (5) 4 5 (c) 4 4sin 5sin d 4 sin 5 5sin d 0 0 cos5 5cos 4 5 4 5 cos 5cos 5 4 5 4 4 5 0 5 4 4 4 5 5 6 0 5 (4) [4]
(a) (b) (c) Notes for Question 7 B applies de Moivre correctly uses binomial theorem to expand cos isin 5 May only show imaginary parts - ignore errors in real part simplifies coefficients to obtain a simplified result with all imaginary terms correct equates imaginary parts and obtains an expression for sin5 in terms of powers of sin cso correct result uses substitution x sin deduces that sin 5 gives a set of results for 5 - for useable results for the remaining useable results (no repeats in the set of 5) at least values for for the 5 different values of x uses previous work to change the integrand correct result after integrating - limits can be ignored substitute given limits and use numerical values for trig functions final answer correct (oe provided in the given form)
Question Number 8. (a) x e z Scheme Marks z dz e dy dy dy z dy e dz d y z dz dy z d y dz dy d y e e dz dz x dz dz d y dy x x y ln x dy d y dy x x y z x dz x dz x dz d y dy y z (7) dz dz Alt: z ln x dy dy dz dy dz x dz d y dy d y dz dy d y x dz x dz x dz x dz d y dy x x y ln x dy d y dy x x y z x dz x dz x dz d y dy y z (7) dz dz
Question Number (b) Aux eqn: m m 0 Scheme Marks m m 0 m, CF: z y Ae B z e 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 (6) 4 (c) y Ax Bx ln x B ft () 4 (a) (b) (c) Notes for Question 8 differentiates x e z wrt y; chain rule must be used correct differentiation d y differentiates again to obtain one mark for each correct term substitutes in the given equation cso obtains the required equation ALT: Works with z ln x; marks awarded as above forms and solves the auxiliary equation both values for m correct - may be implied by their CF correct CF tries a suitable expression for the PF and obtains values for the constants in the PF 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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