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

Transcription

1 Mark Scheme (Results) Summer 06 Pearson Edexcel GCE in Further Pure Mathematics (6668/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding body. We rovide a wide range of qualifications including academic, vocational, occuational and secific rogrammes for emloyers. For further information visit our qualifications websites at or Alternatively, you can get in touch with us using the details on our contact us age at Pearson: heling eole rogress, everywhere Pearson asires to be the world s leading learning comany. Our aim is to hel everyone rogress in their lives through ecation. We believe in every kind of learning, for all kinds of eole, wherever they are in the world. We ve been involved in ecation for over 0 years, and by working across 70 countries, in 00 languages, we have built an international reutation for our commitment to high standards and raising achievement through innovation in ecation. Find out more about how we can hel you and your students at: Summer 06 Publications Code 6668_0_606_MS All the material in this ublication is coyright Pearson Ecation Ltd 06

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 alied ositively. Candidates must be rewarded for what they have shown they can do rather than enalised for omissions. Examiners should mark according to the mark scheme not according to their ercetion of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used aroriately. 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 reared to award zero marks if the candidate s resonse is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will rovide the rinciles by which marks will be awarded and exemlification may be limited. When examiners are in doubt regarding the alication of the mark scheme to a candidate s resonse, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has relaced it with an alternative resonse.

4 PEARSON EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the aer is 7. The Edexcel Mathematics mark schemes use the following tyes of marks: M marks: Method marks are awarded for knowing a method and attemting to aly 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 (indeendent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will aear 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 art of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: secial case oe or equivalent (and aroriate) d or de deendent inde indeendent d decimal laces sf significant figures The answer is rinted on the aer or ag- answer given or d The second mark is deendent on gaining the first mark

5 4. All A marks are correct answer only (cao.), unless shown, for examle, as ft to indicate that revious wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft.. For misreading which does not alter the character of a question or materially simlify it, dect two from any A or B marks gained, in that art of the question affected. 6. If a candidate makes more than one attemt at any question: If all but one attemt is crossed out, mark the attemt which is NOT crossed out. If either all attemts are crossed out or none are crossed out, mark all the attemts and score the highest single attemt. 7. Ignore wrong working or incorrect statements following a correct answer.

6 General Princiles for Further Pure Mathematics Marking (But note that secific mark schemes may sometimes override these general rinciles). Method mark for solving term quadratic:. Factorisation ( x bx c) ( x )( x q), where q c, leading to x = ( ax bx c) ( mx )( nx q), where q c and mn a, leading to x =. Formula Attemt to use the correct formula (with values for a, b and c).. Comleting the square Solving x bx c 0: b 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 )

7 Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners reorts is that the formula should be quoted first. Normal marking rocere is as follows: Method mark for quoting a correct formula and attemting 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 imlication from correct working with values, but may be lost if there is any mistake in the working. Exact answers Examiners reorts have emhasised that where, for examle, 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.

8 Question Number NB Scheme Notes Marks Question states "Using algebra..." so urely grahical solutions (using calculator?) score 0/6. A sketch and some algebra to find CVs or intersection oints can score according to the method used. (No B marks here so CVs -, - without working score 0) x x x x( x )( x ) ( x ) ( x ) Multily through by ( x) ( x ) ( x )( x )( x( x ) ( x )) 0 Collect terms and attemt to factorise ( x )( x )( x ) 0 Sketch need not be seen NB Alts NB Any values correct(a0); all CVs:,, correct() with no extras Any one correct interval (A0); all x x correct() with no extras Set notation may be used or or but not and,, Penalise final A only if used in final intervals. All marks available if above working is done with = instead of < in lines and x 0 x x x( x ) ( x ) 0 ( x)( x) ( x )( x ) x 0 or ( x )( x ) ( x )( x ) Attemt common denominator Simlify numerator x, x (errors such as qualify) All marks available if above working is done with = instead of < in lines and,, (6) Total 6 Alts x( x )( x ) ( x ) ( x ) Multily through by ( x) ( x ) 4 x x 6x 4 0 Obtain a 4 term quartic equation and attemt to solve to obtain at least non-zero values for x x,, Solution by calculator requires all 4 values correct ( or A0A0)

9 Question Number Scheme Notes Marks y Draw sketch of grahs of x y and y x x ALT: y x y x x x showing the area where they intersect. Grahs do not need to be labelled vertical asymtotes and intersection oints needed. Only award if followed by some algebra to find the x coordinates of the oints of intersection. Must obtain a quadratic eg x x x 0 x x x x x x 0 Eliminate y x Solve to x =... need not be seen until the intervals (CVs),,, are formed x x Any one correct interval (A0), all correct(), Set notation may be used NB Finding CVs for xx x w/o a sketch scores M0 and ossibly If all 4 CVs are given (ie, included) score and ossibly

10 Question Number.(a) ALTs (b) Scheme Notes Marks r( r )( r ) ( r )( r ) r ( r ) ( r)( r) r 7r ** ( r)( r) ) Start with RHS and use artial fractions ) Start with RHS, divide and then use artial fractions on the remainder. For either: comlete method as described; `.. n r ( r ) n( n ) n n or ( r ) n n 4 r. Attemt common denominator with at least two correct exressions in the numerator. Denominator must be seen now or later. No errors seen and at least one intermediate ste shown. cso No errors seen and at least one intermediate ste shown. Use formula for sum of the natural numbers from to n and n or either formula for the sum of an AP. If general formula not quoted the sub must be correct. (See general rules on "Use of a formula" age 7) Method of differences with at least lines shown, ( at start and at end or at start and at end). Last Line may be missing n cso () Ignore extra terms at start e to n n including the "(r - )" term in each line (or some lines). n n Extract the remaining terms. n Second M mark only needed. Attemt the correct common n n( n )( n ) 6 n( n ) n denominator using all their terms. r ( n ) deendent uon revious M (but n not the first). The numerators must n( n )( n ) 6 n( n ) n be changed. Or r ( n ) Denominator to be resent now or later. n( n n 9) a, b 9 ( n ) Need not be shown exlicitly. d cso () Total 7

11 Question Number (a) z 4 Scheme Notes Marks or z 4 z 8( i) Give B for either 6 or seen anywhere B arg z arctan 6 Accet 0.4 B 6 (b) ALT: r 4 6 r 4,,, ,,, Roots are Range not secified, you may see 7 4,,, Clear attemt at both r and with at least different values for their arg z, ie rincial arg n r 4 their 6, 4 all 4 correct distinct values of cao. 7,,, scores All in correct form cao i i i i i i i 7i e,e,e,e e,e,e,e scores i B: All 4 radius vectors to be the same length (arox) and erendicular to each other. Circle not needed. Radius vector lines need not be drawn. If lines drawn and marked as erendicular, accet for B B: All in correct osition relative to axes. Points marked must be close to the relevant axes. At least one oint to be labelled or indication of scale given. Obtain one value - usually e 4 - and lace on the circle. Position the other by sacing evenly around the circle., () BB () Total 7

12 Question Number 4.(i) NB (i) (a) Scheme Notes Marks If candidates aear to be considering any/all of, q, r to be non-ositive, send the attemt to review. By use of integrating factor: dx q r x dt e q dt xe xe qt qt e qt r e r e q qt qt c dt t 0, x 0, c xe qt r e q qt q t r q r q q dt qt Obtain IF e e, multily through by it and qt integrate LHS. Accet re dt for RHS Integrate RHS e ke. Constant of integration may be missing. Deendent on the first M mark. Substitute x = 0 and t = 0 to obtain c. Deendent on both M marks above. r r x e oe Change to x =... q q qt qt d dd ALT: (i) (a) By searating the variables: x d dt r qx Attemt to searate variables ln( r qx) t c q Use t 0, x 0 q t Integrate to give ln Constant of integration may be missing. Substitute x = 0 and t = 0 to obtain their constant. r r x e Oe q q d dd (4) (b) q t e t, 0, ( x ) r Cao B q ()

13 4(ii) ye e sin d ye e cos e cos d Or e sin e cos d e cos e sin e sin d Or e sin e cos e sind ye ALT: NB ye e cos e sin 4e sin d Or e sin e cos e sin d 4 4 ye e cos e sin 4ye c Multily through by IF of the e form and integrate LHS (RHS to have integral sign or be integrated later). IF = e and all correct so far. Use integration by arts once (signs may be wrong ) Use arts a second time (Sim conditions to revious use) Must rogress the roblem - not just undo the first alication RHS correct Relaces integral on RHS with integral on LHS (can be ye or e sin d ) and Or d ye e sin e cos ye c uses 0, y 0 to obtain a 4 4 value for the constant. Deends on the second M mark ye e sin d e (sin cos ) c 0, y 0 C cso y (sin cos ) e oe (7) By aux equation method: m 0 m Attemt to solve aux eqn CF y Ce oe PI y sin cos PI of form shown oe dy cos sin d cos sin sin cos sin Diff and subst into equation, 0, Both, Use 0, y 0 to obtain a d 0, y 0 C value for the constant cso(7) y (sin cos ) e Must start y =. Total If the equation is differentiated to give a second order equation and an attemted solution seen send to review.

14 Question Number Scheme Notes Marks. sin a sin b sin c sin (a) Seen anywhere "z" can be n i isin z or isin n z oe cos isin or e or z n z z i See below for use of e B : Attemt to exand owers of z z z z z z z z : Correct exression oe. A single ower of z in each 0z z term. No need to air. Must be numerical values; ncr s eg C score A0 sin sin 0sin 0sin At least one term on RHS correct no need to simlify. All terms correct oe = 6 sinq - 6 sinq + 8 sinq Decimals must be exact cso equivalents. a, b, c need not () be shown exlicitly. Must be in this form. Use of i isin i i e e oe B e i sin i e i e i e i e i 0e i e i sin i sin isin 0isin sin sin 0sin 0sin ALTs: Way = 6 sinq - 6 sinq + sinq cso 8 sin De Moivre on sin Im cos isin Im cos isin 4 cos sin 0cos sin sin sin sin 0 sin sin sin sin 0sin 6sin Also: sin cos sin sin sin 4sin B: sin Im cos isin Eliminate cos from the exression using cos sin on at least one of the cos terms. : Correct term exression B Thus: 6sin sin 0sin sin sin sin sin sin : Use their exression for sin to eliminate sin

15 Way Way sin sin 0sin :cso Correct result with no sin sin sin sin errors seen. De Moivre on sin and use of comound angle formulae sin B: Im cos isin Im cos isin sin Im cos isin cso () 4 cos sin 0cos sin sin 0 cos sin sin sin sin 4 : Use sin sin cos 0 sin sin sin sin cos cos sin sin cos sin 4 sin sin sin sin cos sin sin sin sin sin sin sin sin sin sin 4 6 sin sin sin cso cso Working from right to left: sin Imcos isin Im cos isin B sin Im cos isin Im cos isin b b c 4 a sin sin sin 0a sin sin asin sin sin sin sin : Find the imaginary arts in terms of sin and sub for sin, sin in RHS : Correct (unsimlified) exression a 0a a : Comare coefficients to 0a 0a b b 0 obtain at least one of the a b c 0 equations shown a, b, c cso cso B

16 (b) OR:(b) 0 sin d 0 cos cos 0cos 0 NB: Penultimate A mark has been moved u to here sin n cos n : n for n or ft: terms correctly integrated ft: Third term integrated correctly. :Substitute both limits in a changed function to give numerical values. Incorrect integration such as ncos n could get M0A0A0A0 sin 0 ** cso, no errors seen. sin a sin b sin c sin 0 a b sin d cos cos c cos 0 Or their a,b.c letters used or random numbers chosen sin n cos n : n for n or ft: Correct integration of their exression oe :Substitute both limits, no trig functions A0 A0 (s imossible here) ft ft cso () Total 0

17 Question Number 6(a) f ( x) tan x f ( x) sec x or cos 4 s ec x 4s ec x tan x f, f, 4 4 f 4, f Scheme Notes Marks B x f ( x) sec x(sec x tan x) Use of Chain Rule (may use roct sec xtan x rule) f ( x) s ec x(s ec x) : Attemt the third derivative tan x(s ec x(sec x tan x) ) : Correct third derivative, any equivalent form. tan x x x x 4 Use in f x and in their 4 derivatives. : Attemt a Taylor series u to x using their derivatives. 4 tan x not needed and coeffs need not be simlified but must be numerical. cso: A correct Taylor series. Equivalent fractions and factorials allowed. Must start tan x... or y =... or f(x) = rovided y or f x have been defined to betan x. cso Some alternative derivatives: (7) Using sin and cos sin x f x cos x cos x(cos x) cos x( sin x) f x cos x cos x cos x0 cos xsin x sin x f x 4 cos x cos x cos x(cos x) sin x cos xsin x cos x 6sin x f x 6 4 cos x cos x alternative third derivatives (relacing sec x with +tan x in second derivative then chain rule) f x sec xtan x f x tan xtan x tan x tan x f x sec x 6tan xsec x

18 (b) tan tan Sub x in their art (a) change to in at least 4 6 one term of their exansion or just use 6 to be seen exlicitly ie second 6 term does not qualify Must start tan and justify use of tan ** ie either seen 4 6 searately or a term of the exansion changed from k to k 4 6 n n cso () Total 9

19 Question Number Scheme Marks 7 (a) u dx u u dx x e e or e or x or dx dx x dy dy u dy e dx dx d y u dy u d y u dy d y e e e dx dx dx (a) d cso d y dy x x y x dx dx dy d y dy e e e e y e u u u u u d y dy u y e * obtaining d y using chain rule here or seen later (may not be shown dx exlicitly but aear in the substitution) correct exression for d y any equivalent form (again, may not be seen dx until substitution) d y obtaining using roct rule (enalise lack of chain rule by the A dx mark) d y a correct exression for any equivalent form dx substituting in the equation to eliminate x Only u and y now Deends on both revious M marks. Substitution must have come from their work obtaining the given result from comletely correct work. d cso (6)

20 ALTERNATIVE u dx u x e e x dy dy dx dy x dx dx d y d d d d d d x y x y x x y x y dx dx dx dx d y d y dy x dx d y dy dy x y x x d y dy u y e * dcso (6) dc so obtaining d y using chain rule here or seen later correct exression for d y here or seen later d y obtaining using roct rule (enalise lack of chain rule by the A mark) d y Correct exression for any equivalent form As main scheme ALTERNATIVE : u ln x dx x dy dy dy dx dx x d y dy d y dy d y dx x x dx x x dy d y dy x x y x x x x d y dy u y e * Deends on both revious M marks dcso

21 There are also other solutions which will aear, either starting from equation II and obtaining equation I, or mixing letters x, y and u until the final stage. d cso obtaining a first derivative with chain rule correct first derivative obtaining a second derivative with roct rule (Chain rule errors are enalised through A marks) correct second derivative with or variables resent Either substitute in equation I or substitute in equation II according to method chosen AND obtain an equation with only y and u (following sub in eqn I) or with only x and y (following sub in eqn II) Obtaining the required result from comletely correct work Question Number (b) m m m 0, CF Ae u Be y e u dy e d y 4 e u u Scheme Notes Marks u : Forms AE and attemts to solve to m =...or values seen in CF : Both values correct. May only be seen in the CF CF correct oe can use any (single) variable u u PI of form y e (or y ue if m is a solution of the aux equation) and differentiate PI twice wrt u. Allow with x instead of u (c) 4e 6e e e u u u u y Ae Be e u u u y Ax Bx x ln x ln x Or y Ae Be e ln x d substitute in the equation to obtain value for Deendent on the second A comlete solution, follow through their CF and PI. Must have y = a function of u Allow recovery of incorrect variables. Reverse the substitution to obtain a correct exression for y in terms of x No ft here or x x Must start y = d Bft B (7) () Total 4

22 Question Number 8(a) (b) Scheme Notes Marks 7cos cos cos 4... Solve to... : Angles correct, Decimal for to be sf minimum P(, ) and Q(, ) or, where arccos or cos d 49 cos d 49 sin OR Area of sector sin r 7 sin (cos ) d 9 cos cos d 9 cos cos d 9 9 8sin ( sin ) 9 Area 9 8sin ( sin ) ( sin ) or : r to be, or. 4 4 Need not be in coordinate brackets : Use of for C d r leading to cos k d OR Area of a segment : Correct integration of 49cos Ignore any limits shown. OR correct exression for the area of the sector : Set u the integral for C and reach k cos cos d d: Correct integration of the trig functions cos sin, cos sin Fully correct integration with limits, 0 or d: Deends on all M marks above Combine areas correctly to find the required area. Use of correct limits required. cso Correct answer, exact or awrt. cao () d d cso (7) Total 0 NB The area can be found by area of circle area of crescent : Send to review.

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