Mark Scheme (Results) Summer Pearson Edexcel GCE In Further Pure Mathematics FP2 (6668/01)

Transcription

1 Mark Scheme (Results) Summer 017 Pearson Edexcel GCE In Further Pure Mathematics FP (6668/01)

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 150 years, and by working across 70 countries, in 100 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 017 Publications Code 6668_01_1706_MS All the material in this publication is copyright Pearson Education Ltd 017

3 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.

4 General Instructions for Marking EDEXCEL GCE MATHEMATICS 1. 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.

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: 1. 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: 1. Differentiation Power of at least one term decreased by 1. ( x n x n1 ). Integration Power of at least one term increased by 1. ( x n x n1 )

6 Question Number 1(a) (b) (c) Scheme Notes Marks r r r 1 r r 1 r r 1 r r 1 n r1 Correct proof (minimum as r 1 or r r 1 shown) ( Can be worked in either direction r r n n1 Terms of the series with r 1, r n and one of r, r n 1 should be shown. 1 1 Extracts correct terms that do not n 1 cancel n n 1 1 n n * n1 1 n 6r r r 1 n 1 n rn n n n 1 n 1 n n n 1 n 1 n n1 Correct completion with no errors Attempts to use f ( n) f ( n 1)or f ( n) may be missing Attempt at common denominator, Denom to be n n 1 or n 1 n 1 Numerator to be difference of quartics. may be missing B1 *cso d (1) () 4n 18n n n1 n 6r 1 1 r r 1 n n 1 rn 1 1 OR: n1 n1 n1 n n n1 4n 18n n n1 cao Alternative for part (c) Attempts the difference of terms (either difference accepted) may be missing Valid attempt at common denominator for their fractions may be missing cao cao () Total 7 d If (b) and/or (c) are worked with r instead of n do NOT award the final A mark for the parts affected. This applies even if r is changed to n at the end.

7 Alternative for (b) - by induction. NB: No marks available if result in (a) is not used. Assume true for n k k1 k r 1 kk 1 1 Uses together with the (k + 1)th r1 r r 1 k 1 k 1 k r1 term as fractions (see (a)) k k1 1.. k1 k 1 k 4k k1k k k k 1 Show true for n 1 Hence proved by induction Combines the fractions to obtain a single fraction. Must be correct but numerator need not be factorised. This must be seen somewhere Complete proof with no errors and a concluding statement.

8 Question Number. NB Scheme x 1 x xx Notes Question states "Use algebra..." so purely graphical solutions score max 1/9 (the B1). A sketch and some algebra to find CVs or intersection points can score according to the method used. Can use, or for the first 6 marks in all methods Marks x 1 0 x x x Collects expressions to one side. x x4 0 : Attempt common denominator x x : Correct single fraction x 0, Correct critical values B1 x x x x x Attempt to solve their quadratic as far as x =... x 4, 6 Correct critical values. May be seen on a sketch. : Attempt two inequalities using their 4 x, 0 x 6 4 critical values in ascending order. (dependent on at least one previous M with or < throughout mark) : All 4 CVs in the inequalities correct 4 x, 0 x 6 [ 4, ) (0,6] :Inequality signs correct Set notation may be used. or "or" but not "and" Alternative 1: Multiplies both sides by x x 4 4 x x x x x x x x x x x 4 0 x x x x x x x x x 6 x 4 0 x... Both sides x x May multiply by more terms but must be a positive multiplier containing x x d cao (9) Total 9 : Collects expressions to one side : Correct inequality x 0, Correct critical values B1 Attempt to solve their quartic as far as x x x x =... to obtain the other critical values Can cancel x and solve a cubic or x and x and solve a quadratic. x 4, 6 Correct critical values : Attempt two inequalities using their 4 x, 0 x 6 4 critical values in ascending order. (dependent on at least one previous M with or < throughout mark) d : All 4 CVs in the inequalities correct 4 x, 0 x 6 :Inequality signs correct Set notation may be used. or "or" but [ 4, ) (0,6] not "and" cao (9) Total 9

9 y Alternative : using a sketch graph (probably from calculator) Draw graphs of x 1 y and y 0 x x x x CVs x 0, (Vertical asymptotes of graphs.) B1 x 1 x x x Eliminate y xx 4 x x x x x : Obtains a quadratic equation : Correct equation Attempt to solve their quadratic as far as x =... CVs x 4,6 Correct critical values : Attempt two inequalities using their 4 x, 0 x 6 4 critical values in ascending order. d with or < throughout (dependent on at least one previous M mark) : All 4 CVs in the inequalities correct 4 x, 0 x 6 : All inequality signs correct cao (9) NB As above, but with no sketch graph shown: CVs x 0, must be stated somewhere. B1 Otherwise no marks available.

10 Scheme Notes Marks. z i 0 4 arg z or z r 4 4 8,, 4 8,, , or, or i i i z 4e, 4e, 4e i or 4e where... : Uses tan to find arg z 1 arctan, arctan, or seen. 6 Allow equivalent angles : Either of values shown Correct r seen anywhere (eg only in answers) B1 Divides by to obtain at least values of which differ by or 4. At least correct (and distinct) values from list shown : All correct and in either of the forms shown Ignore extra answers outside the range (6) Total 6

11 Question Number Scheme Notes Marks 4. (a) 1 y ln 1 x dy 1 d1 x : 1 x 1 y ln 1x ln1 ln 1 x Must use chain rule ie dy 1 k with k 1 needed. Minus sign 1x 1x 1 x may be missing. : Correct derivative OR d y 1 x 1 x 1 x dy 1 : 1 x 1 x 1 d 1 Must use chain rule. Minus sign may be missing. : Correct derivative d y d y 1 x 4 1 x 8 1 x 16 1 x Correct second derivative obtained from a correct first derivative. Correct third derivative obtained from correct first and second derivatives (4) (a) Alternative by use of exponentials and implicit differentiation 1 1 y y ln e 1 x 1 1x 1x d y y e 1 x Differentiates using implicit differentiation and chain rule. dy Correct derivative in either form. y e 1x or Equivalents accepted. 1 x If dy d x 1 x has been used from here, see main scheme for second and third derivatives

12 (b) ' '' ''' y y y y 0,, 4, x 16x y 0 x!! 8 y x x x Attempt values at x = 0 using their derivatives from (a) y need not be seen but other values 0 0 must be attempted. Uses their values in the correct Maclaurin series. Must see x term Can be implied by a final series which is correct for their values.!,! or and 6 Correct expression. 1 Must start y =... or ln... 1 x f(x) =... allowed only if f(x) is defined to be one of these. cao Alternative (b) 1 yln ln 1x Log power law applied correctly 1 x x x x Replaces x with -x in the expansion for ln(1 + x) (in formula book) 8 y x x x Correct expression cao () (c) 1 1 x 1 x Correct value for x, seen explicitly or B1 6 substituted in their expansion Substitute their value of x into their expansion. May need to check this is ln correct for their expansion and their x (Calculator value for ln is 0.405) = Must come from correct work cso 1 NB: ln ln or ln ln scores 0/ as 1 x must be < Answer with no working scores 0/ () Total 10

13 Question Number Scheme Notes Marks 5. d y dy 6sin x (a) m m 0 m 0, Solves AE 0 y A B x A B x CF or e or e e oe Correct CF (CF or y = not needed) PI or y acosx b sin x Correct form for PI (PI or y = not needed) dy d y a sin x b cos x, 9a cosx 9b sin x : Differentiates twice; change of trig functions needed, 1 or for coeffs for first derivative, 1, or 9 for second derivative (1/ etc indicates integration) : Correct derivatives 9a cosx 9b sin x 6a sin x 6b cosx 6sin x Substitutes and forms simultaneous 9a 6b 0, 9b 6a 6 a..., b... equations (by equating coeffs) and attempts to solve for a and b d Depends on the second M mark 4 a, b Correct a and b x 4 Forms the GS (ft their CF and PI) y A Be cosx sinx Must start y =... ft (8) (b) 4 Substitutes x = 0 and y = 0 into their 0 A B GS dy x Be 4sinx 6cosx 0 B 6 Differentiates and substitutes x = 0 and y = 0 (change of trig functions needed, 1 ALT for (a) or for coeffs ) 4 0 A B, 0 B 6 A.., B.. Solves simultaneously to obtain values for A and B Depends on the second M mark B1 d 1 A, B Correct values x 1 4 Follow through their GS and A and B y e cos x sin x ft (5) Must start y =... Total 1 d y dy dy 6 6sin x y cosx : c Integrates both sides wrt x : Correct expression x I e e Correct integrating factor B1 : Uses x x 6 6 ye e cos x c yi I cos x c d x : Correct expression 4 x x 1 x : Integration by parts twice = e cosx e sin x ce B : Correct expression 1 x 4 y c Be cosx sin x Must start y =...

14 Question Number Scheme Notes Marks 6. r6 asin A 1 6 asin d 6 asin 6 1asin a sin 1 cos 6 asin 6 1asin a 1 a a 6 1 cos sin a 4 a 6 1 r Limits not needed. Use of dθ Can be gained if 1 appears later B1 : Squares ( 6 k sin, where k a or a as min) and attempts to change : sin to an expression in cos : Correct expression d: Attempt to integrate 1 cos sin Limits not needed : Correct integration limits not needed Correct area obtained from correct integration and correct limits. No need to simplify but trig functions must be evaluated. Set their area = 97 and attempt to d a 97 6 a... a 5 solve for a (depends on both M marks above) If 1 omitted from the initial formula and area set = 97, give the B1 by implication as well as this mark. cao and cso a 5 or a 5 scores A0 Alternatives: Splitting the area and so using integrals with different limits. Marks the same as the main scheme. 1 Limits 0 to (area above initial line) and limits to (area below initial line) and add the two results. Limits 0 to and to Twice the sum of the results needed. dd cso Total 8

15 Question Number Scheme Notes 7. dy cos x ysin x cos xsin x 1 (a) dy 1 Divides by cos x y tan x cos xsin x LHS both terms divided cos x RHS min 1 term divided : Attempt integrating factor tan d e x x needed tan x lnsec x I e e sec x : Correct integrating factor, 1 sec x or cos x Multiply through by their IF and integrate ysec x sin xcos x sec x LHS (integration may be done later) yi their RHS I : Attempt integration of at least one term on RHS (provided both sides have been multiplied by their IF.) 1 OR sec x K tan x y sec x cos x tan x c 1 : cos x or equivalent integration of sin xcos x( sin x or cos x) : tan x constant not needed. 1 y Include the constant and deal with it cos x tan x c cos x correctly. Must start y =... y cos x tan x ccos x Or equivalent eg y sin x tan x ccos x y 1 cos x cos x sin x c cos x Follow through from the line above (b) NB x 5... c... 4 x y... Substitutes y 8 or y Marks d ft Substitutes for x and y and solves for c (If substitution not shown award for at least one term evaluated correctly.) x to find a value for y 6 Must be in given form. Equivalent fractions allowed.... (b) There may be no working shown due to use of calculator. In such cases: Final answer correct (and in required form with no decimals instead of seen), score /. Final answer incorrect (or decimals instead of applies whether (a) is correct or not. seen), score 0/. This cao (8) () Total 11

16 Question Number 8. (a) Scheme Notes Marks z i w 1 iz w i : Attempt to make z the subject z oe 1 iw : Correct equation w i z 1 1 w i 1 wi 1 iw Uses z 1and introduce u + iv (or x i y) for w u iv i u iv i 1 u v u v 1 Correct use of Pythagoras on either side. v 1 oe v 1or y 1 Alternative 1 for (a) 1 i : Maps one point on the circle eg w1 i using the given transformation 1 i :Correct mapping eg i wi i Maps a second point on the circle v 1 oe : Forms Cartesian equation using their points : v 1or y 1 (5) Alternative for (a) w i : Attempt to make z the subject z oe 1 iw : Correct equation w i z 1 1 w i 1 wi Uses z 1 and changes to form 1iw w... w... or draws a diagram wi w i w i Perpendicular bisector of points 0, and 0, 1 v 1 oe v 1or y 1 Uses a correct geometrical approach

17 Alternative for (a) Let z x i y, z 1 x y 1 y z i x iy i xi w 1 iz 1+i x iy 1 y ix x i y 1 y ix w 1 y ix 1 y ix 1 i i 1 i 1 y ix1 y ix1 y i x x y x y y x y w Substitute z x iy and multiply numerator and denominator by complex conjugate of their denominator 1 i 1 1 y x x y x y x y y w w x xy xy x i x y y y 1 y y x w 4x i1 y y Applies : Multiply out and collect real and imaginary parts in numerator. Denominator must be real. : all correct x 4xi y 4x w i y y y 1 y 1 or v 1 y 1 (b) z i w 5 5 z i 5 1 iz 1 iz x i y+i 5 x iy i x y x y 7 x y y x y a 0, b, c 6 6 Uses w 5 and introduce x + iy : Correct use of Pythagoras Allow 5 or 5 : Correct equation Attempt circle form or attempt r from the line above. : correct : All correct, (6) Total 11 Or, for the last marks: 7 5 z 0 i 6 6 If 0 not shown score A0 No need to list a, b, c separately if answer in this form.

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