Mark Scheme (Results) Summer GCE Further Pure Mathematics 2 (6668/01)

Transcription

1 Mark Scheme (Results) Summer 0 GCE Further Pure Mathematics (6668/0)

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 0 Publications Code UA0597 All the material in this publication is copyright Pearson Education Ltd 0

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. 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. 8. In some instances, the mark distributions (e.g., B and A) printed on the candidate s response may differ from the final mark scheme

5 General Principles for 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 correct formula (with values for a, b and c).. Completing the square b Solving + b+ c= 0 : ± ± q± c, q 0, leading to =... Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( n n ). Integration Power of at least one term increased by. ( n n+ ) 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 mistakes 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.

6 Question Number Scheme Marks. (a) (b) A B = + =, r+ r+ r+ r+ r+ r+ ( )( ) n+ n+ n + = = n+ + ( n ) n = n = n r+ r+ n+ n+ ( )( ) ( ),A () depa () [5] Notes for Question (a) for any valid attempt to obtain the PFs A for r+ r+ NB With no working shown award A if the correct PFs are written down, but M0A0 if either one is incorrect (b) for using their PFs to split each of the terms of the sum or of ( r+ )( r+ ) into PFs. At least terms at the start and at the end needed to show the diagonal cancellation resulting in two remaining terms. dep Acao for for simplifying to a single fraction and multiplying it by the appropriate constant n = n + NB: If r is used instead of n (including for the answer), only M marks are available.

7 Question Number (a) z = 5 5i= r( cosθ + isinθ ) Scheme Marks ( ) r = = 0 B () (b) 5 π π arg z = arctan = or ± nπ A () (c) w or 0. z = 0 = B () 5 (d) w π π 5π arg =, = z 6 5π or ± nπ,a () [6]

8 (a) B for z = 0 no working needed Notes for Question (b) for 5 argz = arctan ± 5, ( ) 5 tan arg z =±, 5 5 argz = arctan ± 5 5 tan ( arg z ) =± OR use their z with sin or cos used correctly 5 π π A for = or ± nπ (must be th quadrant) 6 6 or (c) B for w z = or or (d) w π for arg = arg z z using their arg z A for 5 π 5π or ± nπ Alternative for (d): w ( 6 ) + ( 6+ ) i Find = z 0 w tan arg = z 6+ 6 from their w z arg w 5π = z A cao Work for (c) and (d) may be seen together give B and A marks only if modulus and argument are clearly identified ie cos 5 π isin 5 π + alone scores B0A0 5

9 Question Number ( ) d y Scheme = 0 = sin0 = B d y dy + cos ( = 0) ( ) d y = 0 = cos0 = A 8 Marks d d d y = y ( ) y y y 0 0!! 0 0 (! or and! or 6) = ( y ) ( ) y 8 = + + A cao [5] Alt: y = + + a + b +... B 8 y = a+ 6 b+... Diff twice a+ b+ = + + a + b sin... A Correct differentiation and equation used a+ = 0 a = 6b+ = b= y 8 = + + Acao

10 B for d y = 0 wherever seen Notes for Question d y dy ± ± = oe for attempting the differentiation of the given equation. To obtain k cos ( 0) d y A for substituting = 0 to obtain = 0 y = f = f 0 + f 0 + f ( 0) + f ( 0) with their values! ()! d y d y for and. Factorial can be omitted in the term but must be shown eplicitly in the term or implied by further working eg using 6. for using the epansion ( ) ( ) ( ) Acao for y 8 allowed. Must include y =. = + + (Ignore any higher powers included) Eact decimals Alternative: B for y a b 8 = for differentiating this twice to get y = a+ 6 b+... (may not be completely correct) A for correct differentiation and using the given equation and the epansion of sin to get a+ 6 b+... = for equating coefficients to obtain a value for a or b A cao for y 8 = + + (Ignore any higher powers included)

11 Question Number (a) Assume true for n k Scheme k k = : z = r ( cos kθ + isin kθ ) Marks k k k n= k+ : z + = ( z z = ) r ( coskθ + isin kθ) r( cosθ + isinθ) ( θ θ θ θ ( θ θ θ θ) ) k + = r cos k cos sin k sin + i sin k cos + cos k sin ( ( ) θ ( ) θ ) k + = r cos k+ + isin k+ depacso if true for n = k, also true for n = k + k = z = r ( cosθ + isinθ ); True for n = true for all n Acso (5) Alternative: See notes for use of i r e θ form (b) π π w = cos + isin 5 5 5π 5π w = cos + isin 5 w = i = i or oe A () [7]

12 Notes for Question (a) NB: Allow each mark if n, n + used instead of k, k + k+ k k for using the result for n= k to write z ( = z z) = r ( coskθ + isin kθ) r( cosθ + isinθ) for multiplying out and collecting real and imaginary parts, using i = k + r cos kθ + θ + isin kθ + θ ( ) OR using sum of arguments and product of moduli to get ( ) ( ) dep for using the addition formulae to obtain single cos and sin terms k + OR factorise the argument r ( cosθ( k + ) + isinθ( k + ) ) Dependent on the second M mark. k + Acso for r ( cos( k + ) θ + isin ( k + ) θ ) Only give this mark if all previous steps are fully correct. Acso All 5 underlined statements must be seen Alternative: Using Euler's form i ( cosθ sinθ) e θ z = r + i = r May not be seen eplicitly ( ) k k iθ k iθ k ikθ iθ z + = z z = re re = r e re k + i( ) r e k + θ = dep on nd M mark ( ( ) θ ( ) θ ) ( ) k + = r cos k+ + isin k+ Acso k = z = r cosθ + isinθ True for n = true for all n etc A cso All 5 underlined statements must be seen (b) for attempting to apply de Moivre to w or attempting to epand w 5 and collecting real and imaginary parts, but no need to simplify these. Acao for i = i (oe eg 5 instead of )

13 Question Number 5 (a) Scheme Marks dy y + = d ln I F: e e ( ) = = dy y + = dep ( ) dep y = = + c y c = + Acso (5) (b) =, y = 5 c = y = + Aft () dy 8 (c) = dy 0, = = =± or ±,A y = + = Acao Alt: Complete square on y =... or use the original differential equation =±, y = A,A B shape (,) (,) B turning points shown somewhere (5) []

14 Notes for Question 5 (a) for dividing the given equation by May be implied by subsequent work. d ln for IF= e = e = ( ) must be seen together with an attempt at integrating this. ln must be seen in the integrated function. dep for multiplying the equation d y y + = by their IF dep on nd M mark dep for attempting the integration of the resulting equation - constant not needed. Dep on nd and rd M marks c Acso for y = + oe eg y = + c Alternative: for first three marks: Multiply given equation by to get straight to the third line. All M marks should be given. (b) for using =, y = 5 in their epression for y to obtain a value for c Aft for y = + follow through their result from (a) (c) for differentiating their result from (b), equating to 0 and solving for A for =± (no follow through) or ± No etra real values allowed but ignore any imaginary roots shown. Acao for using the particular solution to obtain y =. No etra values allowed. Alternatives for these marks: for making d y 0 = in the given differential equation to get y = and using this with their particular solution to obtain an equation in one variable OR complete the square on their particular solution to get y = + A for =± (no follow through) Acao for y = No etra values allowed B B for the correct shape - must have two minimum points and two branches, both asymptotic to the y-ais for a fully correct sketch with the coordinates of the minimum points shown somewhere on or beside the sketch. Decimals accepted here.

15 Question Number Scheme Marks 6 (a) + 6 5= = 0 + 5= 0 ( )( ) + 5 = 0 or by formula = 5, = A 6+ 5= = A = 0 = A (6) B line B quad curve (b) Bft (on - coords from (a)) () -5 - O (c) < 5, < < 0, > B,B,B () Special case: Deduct the last B mark earned if or used []

16 Notes for Question 6 (a)nb: Marks for (a) can only be awarded for work shown in (a): for + 6 5= 5 for obtaining a term quadratic and attempting to solve by factorising, formula or completing the square A for = 5, = for considering the part of the quadratic that needs to be reflected ie for oe A for a correct term quadratic, terms in any order + = 0 oe 6+ 5= 5 A for = 0 = NB: The question demands that algebra is used, so solutions which do not show how the roots have been obtained will score very few if any marks, depending on what is written on the page. Alternative: Squaring both sides: Square both sides and simplify to a quartic epression Take out the common factor A, a correct linear factor and a correct quadratic factor and linear factors A any two of the required values A all values correct (b) B B for a line drawn, with negative gradient, crossing the positive y-ais for the quadratic curve, with part reflected and the correct shape. It should cross the y-ais at the same point as the line and be pointed where it meets the -ais (ie not U-shaped like a turning point) Bft for showing the coordinates of the points where the line crosses the curve. They can be shown on the -ais as in the MS (accept O for 0) or written alongside the points as long as it is clear the numbers are the coordinates The line should cross the curve at all the crossing points found and no others for this mark to be given. (c)nb: No follow through for these marks B for any one of < 5, < < 0, > correct B B for a second one of these correct for the third one correct Special case: if or is used, deduct the last B mark earned.

17 Question Number 7 (a) dy Scheme Marks dv = v+ d y dv dv d v = + + A dv d v dv 8 v 8 v ( + ) = d v v + = d v + v= * A (6) See end for an alternative for (a) (b) λ + = 0 λ = oe A ( v ) Ccos Dsin = + ( ) P.I: Try v= k ( + l) ( i i or v Ae Be ) = + A dv d v = k = 0 ( ( )) 0+ k + l = dep k = = ( l 0) v= Ccos + Dsin + or v = Ae + Be + i i A (6) (c) y = Ccos + Dsin + i i or y= Ae + Be + Bft ()

18 Alternative for (a): y v = Question 7 continued dv dy = y d v d y dy dy = + y A d v d y dy = + y d v d y dy + v= 8 + 8y+ y= d v d v + = * A

19 Notes for Question 7 (a) for attempting to differentiate y = v to get d y - product rule must be used for differentiating their d y to obtain an epression for d y - product rule must be used A for d y dv dv d v = + + for substituting their equation in v and dy and d y and y = v in the original equation to obtain a differential for collecting the terms to have at most a term equation - terms only if a previous error causes d v to be included, otherwise terms d v Acao and cso for v + = * Alternative: (see end of mark scheme) y for writing v = and attempting to differentiate by quotient or product rule to get d v for differentiating their d v to obtain an epression for d v be used - product or quotient rule must A for d v d y dy dy = + y for multiplying their d v by for multiplying by and adding to the original equation. y to each side and equating to (as rhs is now identical Acao and cso d for d (b) for forming the auiliary equation and attempting to solve v v + = * A for λ = oe A for the complementary function in either form. Award for a correct CF even if λ = i only is shown.

20 Notes for Question 7 continued for trying one of v= k, k or v = k+ l and dv and d v v = m + k+ l as a PI and obtaining dep for substituting their differentials in the equation original equation is used. Dep on nd M mark of (b) d v + v =. Award M0 if the Acao for obtaining the correct result (either form) (c) Bft for reversing the substitution to get y = Ccos + Dsin + i i or y = Ae + Be + follow through their answer to (b)

21 Question Number 8 (a) ( y ) rsin asin sin Scheme = θ = θ θ Marks dy = a θ θ + θ θ dθ ( cos sin sin cos ) dy = a sinθ cosθ + cos θ dθ ( ) depa dy At P 0 sinθ 0 ( n/a ) or cos θ cos θ 0 dθ = = + = sinθ = 0 not needed cos θ = Acso cosθ = * (6) (b) r = asin θ = asinθ cosθ r a = = a A () (c) Area = φ φ r dθ = a sin θdθ 0 0 ( cos ) d 0 = a φ θ θ a sin = θ θ φ 0 A ( ( )) a φ sin φ cos φ cos φ = dep on nd M mark a arccos = dep (all Ms) a 9arccos + 6 * A (7) [5]

22 Notes for Question 8 (a) for obtaining the y coordinate y = rsinθ = asin θsinθ dep for attempting the differentiation to obtain d y Product rule and/or chain rule must dθ be used; sin to become ± cos ( cos to become ± sin ). The may be omitted. Dependent on the first M mark. dy A for correct differentiation eg a ( cosθ sinθ sinθcosθ) dθ = + oe for using sin θ = sinθ cosθ anywhere in their solution to (a) for setting d y 0 dθ = and getting a quadratic factor with no sin θ included. Alternative: Obtain a quadratic in sinθ or tanθ and complete to cosθ = later. Acso for cosθ = or cosφ = * Question 8 (a) Variations you may see: y = rsinθ = asinθsinθ y = asinθsinθ y = asin θcosθ y = a(cosθ cos θ) dy/dθ = a(cosθsinθ + sinθcosθ) = a(cosθsinθ + sinθcos θ) = asinθ(cosθ + cos θ) = asinθ(cos θ -) or = asinθ(cos θ sin θ) or = asinθ( sin θ) dy/dθ = a(sinθcos θ sin θ) = asinθ(cos θ sin θ) dy/dθ = a(-sinθ +sinθcos θ) = asinθ(cos θ -) At P: dy/dθ = 0 => sin θ = 0 or: cos θ sin θ = 0 cos θ = 0 sin θ = 0 tan θ = cos θ = / sin θ = / tan θ =± => cos θ = ± cos θ = ± 6 Sin θ = ± =± => cos θ = ± (b) for using sin θ = sinθ cosθ, cos θ + sin θ = and cosφ = in r = asin θ to obtain a numerical multiple of a for R. Need not be simplified. Acao for R= a Can be done on a calculator. Completely correct answer with no working scores /; incorrect answer with no working scores 0/

23 Notes for Question 8 continued (c) φ φ for using the area formula r dθ = a sin θdθ 0 0 for preparing sin θ dθ for integration by using cos = sin Limits not needed for attempting the integration: cos θ to become ± sin θ - the may be missing but inclusion of implies differentiation - and the constant to become kθ. Limits not needed. A for = a θ sin θ Limits not needed dep for changing their integrated function to an epression in sin θ and cosθ and substituting limits 0 and φ. Dependent on the second M mark of (c) dep for a numerical multiple of a for the area. Dependent on all previous M marks of (c) Acso for a 9arccos + 6 * This is a given answer, so check carefully that it can be obtained from the previous step in their working. Also: The final marks can only be awarded if the working is shown ie sin θ cannot be obtained by calculator.

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25 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 FN Telephone Fa Order Code UA0597 Summer 0 For more information on Edecel qualifications, please visit our website Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Esse CM0 JE