Mark Scheme (Results) Summer Pearson Edexcel International A Level in Core Mathematics C34 (WMA02/01)

Transcription

1 Mark Scheme (Results) Summer 05 Pearson Edecel International A Level in Core Mathematics C4 (WMA0/0)

2 Edecel and BTEC Qualifications Edecel 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 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 05 Publications Code IA0408 All the material in this publication is copyright Pearson Education Ltd 05

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4 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.

5 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 5.. 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 and can be used if you are using the annotation facility on epen. 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) d or dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper or ag- answer given or d The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for eample, as ft to indicate that previous wrong working is to be followed through.

6 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. If you are using the annotation facility on epen, indicate this action by MR in the body of the script. 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.

7 General Principles for Core 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 Solving b c 0 : b ( ) 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 small 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.

8 Question Number. (a) (b) Scheme Marks dy d d 8 0 d y y y y d d B Either Way : Sets d y in each term in their differentiated epression d d 84y4y 0, y6 0* dd,* Or Way : Obtains d y 8 y (ft their differentiated epression) d y d 8 y, so y 6 = 0* y dd,* y into y y 6 B where A and B are constants Put y 6 or A or y or and obtains, or Ay Bor, both, and, and no etra solutions (9 marks) (6) ()

9 Notes for Question (a) Differentiating 4 y with respect to to obtain dy A By d Condone d y... at start. May not have lost the +5 d d dy B Sight of y y d d A fully correct derivative. Accept 8d ydydyyd 0(needs = 0) d depends on previous M mark and B mark so has at least two terms in d y and at least d two other terms. Way : Sets d y d in each term in their differentiated function. Or Way : May see algebra used to give d y (condone sign slips and slight copying d errors but not omission of terms) dd Dependent upon both previous M's. It is for proceeding to obtain an unsimplified (4 y) correct equation in and y equivalent to those in the scheme e.g. y * cso y6 0 no errors should have been seen - the solutions in the mark scheme would gain full marks (etra lines of working are not required) accept y = 6 (b) Substitutes y 6 or variable and reaches two term quadratic ( ) 0 5 y into the equation of curve C to form an equation in one 6 A B or Ay B for any values of A or B 5 4 y y y 5 0 y 5 6 but condone slips in the working 9 Either one correct pair of coordinates or both or both y values., and, ( two answers correct with no incorrect working implies ) Any etra solutions (obtained by substituting the variable found first into the quadratic, instead of the linear equation for eample) result in A Unsimplifed answers lose the final A mark e.g., and, 0 0 Allow (0.5, ) and (-0.5, -) need not be in brackets if pairing is clear but wrong pairing is A0. Both, y and, y is acceptable but, y is not sufficient

10 Question Number Scheme Marks (a) 4( 6) A( ) B( )( ) C( ) Way Let 40 5CC 8 d Let 5 6.5AA 4 A 4, C 8 Compare constants / terms in or substitute another value of into identity and conclude that B = 0 e.g. 4 4ABCB 0 * * (4) Way 4( 6) A( ) B( )( ) C( ) (a) Compare :so4 A B, :so04ab C, constants: so 4 4 A B C d So A4, C 8, and B = 0*, Way 4 6 4( ) 8( ) (b) ( )( ) 8 B ft Way (b) See ( )( )( ) ( )( )... ( )( ) or... ( )!! ( )( )( ) ( )( )... ( )( ) and... ( )!! 5 4( 4..) Or 4 6 ( ) ( )( ) See ( )( )( ) ( )( )... ( )( ) or... ( )!! ( )( )( ) ( )( )... ( )( ) and... ( )!! 4( 6)( 4..) d (5) B d (5) ( 9 marks)

11 Notes for question (a) Uses correct form 4( 6) A( ) B( )( ) C( ) allow sign errors. d Uses correct method, either substitution or equating coefficients of terms to find at least one constant (see Way and Way above) Both A4, C 8 (Correct answers with no working (cover up rule) imply ) * This needs a method as the answer is given and needs to conclude that B = 0. For method: May compare constants i.e. 4 = 4A + B + C with A = 4 and C = 8, or terms in i.e. 0 = 4A B C with A = 4 and C = 8, or term in :so4 A B with A = 4 or substitute another value for such as = 0 or = with A = 4 and C = 8. (b) Way Bft Writes their epression in the form There should be no B term. (This may be awarded for writing or using A ( ) C ft on values of A and C. A( ) C( ) ( ) C C ) This could appear in the separately in their solution and writing binomial epansions. Uses the binomial epansion correctly for one epansion, with power of outside the second bracket ignored. Allow missing brackets for this mark. Ignore constants outside the bracket for this method mark. One completely correct epansion in the bracket Both epansions in the brackets correct and unsimplified. Can be awarded for the two completely correct epansions in the brackets without mention of A and /or C or adding and with power of outside the second bracket ignored. d Multiplies out brackets and collects terms. Dependent upon previous M mark. Allow sign slips May be written as list or in reverse order. Implies previous M mark. Way 4 6 B Writes 4 6 ( ) ( )( ) 4. This could be given after the two binomial epansions. (It may be awarded for writing 4 6 ( ) and writing separately ( ) ) : Follow the scheme and the notes for Way d Must multiply out three brackets for this method May be written as list or in reverse order. Implies previous M mark. ( )( ) 4 N.B. ( ) ( ) ( ) ( ).. ( ) is an alternative 4 4 6! correct epansion and implies the B mark as well as contributing to the (one correct epansion) (two correct epansions)

12 Question Number Scheme (a) f'( ) e ( 5)e f'( ) 0 ()e 0 {Coordinates of A=, e } obtains y e ft (b) (c) Y e k 0 Marks (5) () 0,5 Shape including cusp B O 5,0 5,0 only B 0,5 only B () (0 marks)

13 Notes for Question (a) For applying the product rule correctly to get a form Ae ( 5)e with A a constant correct differentiation need not be simplified - isw Sets their f'( ) 0and proceeds to.. Correct answer in any equivalent form. (Ignore etra answers such as ln0 or even 0) ft For finding their correct eact y coordinate(ft their ).Allow even if positive. May not be given as coordinates. Allow y e for this mark (b) Uses their minimum y value as lower limit. Accept their e or their e (may use any letter for the M mark) e k 0 with both inequalities strictly less than this is cao (so e k 0 is A0). Need to use k this time, not y and need eact lower limit. (c) B Correct shape curve lies only in first two quadrants with maimum in first quadrant. Tends to ais as becomes large and negative. Crosses y ais and touches ais with discontinuous gradient cusp- (not a minimum point). Then gradient becomes steeper as becomes large and Positive. (Give bod if curve looks like straight line here but must not bend back on itself) B Correct coordinate may be on sketch or in the tet. Diagram takes priority over tet if there is a contradiction. Need both coordinates if in the tet but on the ais the.5 is sufficient (even allow (0,.5) on ais) Must be the only crossing point. B Correct y coordinate may be on sketch or in the tet. Need both coordinates if in the tet but on the y ais the 5 is sufficient (even allow (5,0) on y ais). Must be only crossing point.

14 Question Number Scheme Marks 0 4 (a) Uses ab. a b cos cos cos 0 c a b abcosc c a b a.b (b) Uses c c AB (c) 5 Uses any method (or no method) with their cos sin or 5 5 gives eact height Area of triangle OAB= 56 sin( AOB) 5 6 sin( AOB ) 5 5 (no evidence of calculator and clear working with surds) (See notes for other methods) * (7 marks) () () ()

15 Notes for Question 4 (a) Uses ab. a b cos to obtain cos.. 0 or writes arcos 65 0 Obtains cos or then isw (This answer implies ) 0 (isw if they go on to find the angle) 0 cos 5 6 earns A0 Special case: Uses printed answer to part (c) to find angle then deduces cos is M0A0 (b) Uses correct version of the cosine rule with their cos.. or.. to find AB or c (may include non eact angles or cosines) May be done by splitting into two right angles triangles correctly and using trigonometry and Pythagoras. Method must be completely correct. For correct eact answer only (c) Uses any method with their eact numerical cos to find eact value of sin Could use sin cos or Pythagoras' theorem on to find sin Use of angle = 48.9 or any non eact work is M0 5 Just writes down sin should be given the mark bod May find height of triangle by correct Pythagoras using 5 5cos h Uses a correct method - the formula 5 6sinC is most likely but may find height of triangle by trigonometry (or Pythagoras see first mark) and use b h, with values for b and h (usually 6 and 5sinC) For this mark non eact work may be seen. Completely correct eact work (without using calculator approimation) and states answer Sight of for sin or.76.. for height of triangle or the numerical value of the angle (48.) should be awarded A0 (c)alternative Method: A neat method is to use Area {( aa. )( bb. ) ( ab. ) } {(5)(6) (0) } This is M then for achieving the printed answer Alternative method for (c) using Heron's formula can get abc Attempts to use A s( sa)( sb)( s c) with s and a 5, b 6 and c their eact answer to part (b) Attempts to simplify their 5 6 by any means (even using a calculator) so this may be ineact Needs to show A= ( ) = 4 4 ( ) = (000) 5 5 (Seeing calculator approimation is A0 e.g. 4.) 4

16 Question Number Scheme Marks 5.(i) d y ( ) dy or y d ( ) ( ) d (see notes for further methods) or ( ) 4 or 4 ( ) 4, (4) (ii) t dt dt tlnt (+c) see notes for integration by parts. t t ln a t t ln 7 aln aaln a ln 7 a a 7 aln ln 7 a ln a or a = ln7 ln d (4) (8 marks)

17 Notes for Question 5 (i) Correct use of quotient, product, implicit differentiation OR chain rules may not be simplified accept any correct answer or correct formula quoted followed by slip. Correct answers are: ( )( ) (chain rule) ( ) ( ) ( ) (product rule) dy dy y y ( + ) = gives ( ) y so or d d ( ) (implicit differentiation) dy and proceeds to one of the listed correct equations in only. d 4 i.e. or ( ) 4 or 4 ( ) 4 Solves their quadratic by usual methods (see notes) to obtain two values for.. Need both answers - two correct answers for with no working implies the method here. Ignore y values if they are given as well. (ii) Writes as a sum of and t - and integrates this sum giving a sum with one correct term e.g. t + t - May use parts (see below for two variants on parts) If parts are used they must be used accurately (as given below). It is not the most efficient method here and usually results in errors and no marks. Both terms correct (ignore arbitrary constant) e.g. From parts may obtain tln t d Uses limits correct way round and sets I( a) I( a) ln 7 and also uses or states ln aln a ln leading to a.. This is dependent upon the previous M having been scored Correct answer. Accept any correct equivalent e.g. ln.5. ln7 ln is but if it is followed by ln5 this is A0 Parts in integration: Either: Or: for t t t t dt t t dt dt t t t t t d t ( t )ln t ln t d tfor t t t = lnt for then as before ( t)lnttlnt t for then as before

18 Question Number Scheme 6(a) 5e or equivalent decimal - Accept awrt 68 B (b) 50 0k 5e 0 k 0k 0k Way e Way e /e Way e e/ 0k ln -0k ln( / e) 0k ln(e / ) ln e ln ln k e No intermediate step 0 k needed 0 ln e k * * 0 Marks () (4) (c) Uses m = 0 and their numerical k so ' kt ' ' kt ' 0 5e e 0.8 o.e. ln0.8 t ' k ' d t awrt 40 (years) () (8 marks)

19 Notes for question 6 (a) B for 5e or for numerical answer, e.g allow awrt 68 (b) Uses t 0, m50, in m 5e kt 0 k to give 50 5e 0 k 0k 0k e (way ) or e /e o.e. (way ) or e e/ (way ) or 0 e k f( ) (variant on Way ) to give a correct equation e k B. Some solutions will move from one of these options to another by sound algebra this is acceptable. Taking logs correctly to give f(k) = ln B i.e. 0k ln (way ) or 0k ln( / e) o.e. (way ) or 0k ln(e / ) (way ) or 0k - = ln(/) (way 4*) (There are a number of correct alternatives but this line should follow directly from the previous one) This must be a correct equation. * cso- Needs both M marks, everything should have been correct and eact. Makes k the subject of the Formula. Needs an intermediate step for ways and but not for way. ln e ln ln(/e) e.g. k (Way or Way 4*) or k (Way) or straight to 0 0 ln e k (Way ) 0 e ln e ln Must conclude with the printed answer k or k e or k ln o.e.. Special Case Taking the mass as e in part (b) should be treated as misread. Can earn ln( e) A0A0 and obtains k 0 (c) Uses m 0 and their numerical k in m 5e kt ' kt ' e 0.8(NB k = ) ( ' k') t 0.97t NB e 0.8 is M0 (usually e 0.8 ) ln0.8 ln 5e 4 d Use of correct work to reach t or equivalent e.g. t ' k ' ' k ' Allow awrt 40 (may see 9.86 or 9.9). (Decimals are acceptable in part (c)) Do not allow -40 Special Case If the answer 40 appears with no working or after minimal working where no marks have been scored then award M0A0 special case. If the first M mark in (c) has been awarded and they give the answer 40 with no further working, then award

20 Question Number Scheme Marks 7(a) t 4 tan cot sec 0 4 ( t ) 0 t t B (b) So 4 4t ( t )( t ) 0and t 8t 0* * 4 t 8t 0 (t )( t ) 0 so t = tan ( t) or ,,, (4) (4) (8 marks)

21 Notes for Question 7 t tan sin (a) B Uses 4 tan 4 or 4tan 4. (B0 for tan ) t tan cos Uses either cot or tan t or sec tan or t (quoted correctly) Uses both cot or tan t and sec tan or t both quoted correctly cos [ This may also arise from the use of cot with sec sin cos and with cos sin to reach t ; so for all three of these identities and for reaching t result.] cot may be implied (by tancot = for eample) tan * As the answer is given, a fully correct intermediate line of working must be seen (Answer may have terms in a different order but must be equivalent correct equation in t) (b) Solves given quadratic in t. (may be one slip copying) Accept correct factorisation (shown), formula, completing square. Must then also square root. Need to get to t = This may be implied by just one answer for t or tan (from a graphical calculator) tan ( t) (need both plus and minus) or awrt (ignore further values of tan from their quadratic) N.B. tan is not enough. This is M0A0 For obtaining two answers for from their answers for t (must be in different quadrants if following through wrong t ) (check on your calculator) All 4 eact and correct and no etra values in the range If all four answers are correct but in degrees 0, 0, 50, 0 lose final A mark If the answers are given as decimals 0.54,.67,.6 and 5.76 lose final A mark.

22 Question Number Scheme Marks 8 (a) d ln tan y sec y dy tany cos y sin y cos y k sin ycos y 4 4 sinycosy sin 4 y * cso (b) Way Finds limits first Finds limits after removing lns (b) Way dy dy cossin4y cosd d sin4y B ln tan y sin c 4 Put 0, y ln tan sin 0 cc... ln or ln sin c Takes eponentials so tan y e tan y 8sin c e (so tan y 8sin Ae ) (bm on epen) Put 0, y, so A = or e c (bm on epen) 6 8sin tan y e (6) (0 marks) dy dy cossin4y cosd d sin4y B (4) (b) Way ln(cosec4 y cot 4 y ) sin ( c ) 4 Sub 0, y 6 ln(cosec cot ) sin 0 cc... ln or ln cos4y ln( ) ln(tan y) sin ln 4 sin4y 4 4 so tan y 8sin e Special case: Differentiates the answer. Marks available B0A0 B sin dy Bsin tan y Ae sec y ABcos e d dy Bcostanycos y d () dy Bcos sin ycos y d (4) B cos sin 4 y (4) B = 8 and A = 4/6 (6)

23 Notes for question 8 d (a) Uses chain rule to obtain ln tan sec d y y tany A y (A can even be ) Correct answer (unsimplified) sin y Uses identity for tan y and for sec y cos y cos y and reaches k sin y cos y or k sin y cos y NB Some use long alternative methods using sec tan sin y cos y sin y y y cos cos cos y y y k There needs to be a complete method leading to for the sin y cos y * cso Writes their epression in terms of sin 4y using the identity sin 4y sin ycos ythis is a given answer which must be stated and all aspects of the proof must be correct.( Need to multiply top and bottom of fraction by or use other convincing intermediate step). (b) B Separate terms. Accept without the integral as long as integration is implied by subsequent working. Use inverse of part (a) (integration the reverse of differentiation) as well as knowing cos dsinto produce Aln tan y Bsin ( c) (or quotes results) Correct answer, no need for ( c) (Correct answer implies the ) Subs 0, y into their integrated epression to find c = (must have c for this mark) 6 Uses correct ln work to find an unsimplified epression for tan y (must have c) (They may remove lns before finding a value for c) Correct answer and correct solution. Do not accept A= tan need A = Way A scheme is given but most will struggle to complete this method. The first 4 marks are readily accessible but showing the answer is difficult. If in doubt, send to review. B Separate terms as in first method Use standard integral on scheme and obtain A ln(cosec4 y cot 4 y) sin ( c) Correct answer, no need for ( c) Subs 0, y into their integrated epression to find c (must have c for this mark) 6 Uses definitions of cosec and cot together with double angle formulae to give tan y (must have c) Correct answer and conclusion with correct values for A and B Way (Special Case) This assumes the answer and differentiates. This is not a complete answer to the question as it merely shows that the given answer satisfies the differential equation. So this is treated as a misread and four of the si marks are available. B0 Not available as variables are not separated Implicit differentiation (There may be sign errors or wrong factors of ) A0 Not available Obtains epression in scheme (may be sign errors or wrong factors of ) B cos sin 4 y Uses definitions with double angle formulae to give (4) Correct answer and conclusion with correct values for A and B

24 Question Number Scheme 9(a) A and B are where y = 0 so t 9t 0 t( t 9) 0 t (0and ) When t, 5 A (,0) B Or Special case - uses answer - t t 5t (-5) When t, y 0 A (,0) B (b) dy y dt t d d t dt dy t 9 9 gradient = d t 4 d 9 Substitutes t into 9 Uses their 4 Marks and (5,0) to produce tangent equation 9 4 y 5 0 * * (5) () () (c) Substitutes, 9, into 9 4 y5 0 t t y t t 9( t ) t 4( t 9) t 5 0 4t 9t 54t 5 0 ( t 6t9)(4t5) 0 or ( t )( t )(4t 5) 0 d 5 t Coordinates of X are, 6 64 or 9 6 6, ddcso Accept awrt 6.56, 8.98 (5) ( marks)

25 Notes for Question 9 (a) Sets y = 0, so t 9t 0t. Uses t, to state the coordinate of B is 5 B States A (,0) (need not see working) must have both coordinates S.C. Sets t t 5t Uses t, to state the y coordinate of B is 0 B States A (,0) (need not see working) (b) Differentiates ()and t y() t ( allow one error) and calculates d y d by using dy dt - d dt dy t 9 d t Substitutes their t= into their d y to find the gradient of the tangent at B d Uses (5,0) and their non-zero numerical gradient to find an equation of the tangent. * Achieves given answer cso 94 y5 0 (c) Substitutes t, t y t 9t into9 4 y5 0 correctly to form a cubic equation in just t. (Need a correct equation) N.B. Any attempts to work in just or in just y are unlikely to achieve a cubic. If there seems to be any success send to review. Equations in a miture of, y and t score M0 d Attempts to solve a cubic polynomial = 0. Division by (t ) or by ( t ) Use of a graphical calculator is acceptable. Just seeing (4t + 5) is enough but any errors e.g. etra solutions or errors factorising) are penalised by the loss of the final A mark. May use other variables than t for this mark, when trying to factorise. 5 t (This answer with no working implies previous M mark if cubic has been seen) 4 dd Uses their value of t to find both the and y co-ordinates. It is dependent upon both the previous M's having been scored 05 5 cso Coordinates are, Accept awrt 6.56, Allow for two sets of coordinates if (5,0) is not rejected. But lose this mark if the correct point is rejected in favour of (5,0). Lose this mark for errors factorising cubic earlier, or for etra 05 5 wrong values of t found earlier. Allow for, y 6 64

26 Question Number 0. Scheme Marks da A (condone use of r throughout instead of ) B d da da d d d Uses dt d dt 0 dt dt dv V 6 8 B d dv dv d dv 9 Uses 8 d dt d dt dt 80 0 (6 marks) A V 6, A V 6 Way dv da A 9 B B Way dv dv da dv 4 dv 9 9 d t A t t t (6 marks) Uses d d d d 0 d A4 0 da Uses d V d V d d A d V 8 dt d da dt dt A First B d dv V 6 8 Second B d 0 Put = to give 9 0 d (6 marks) da Misunderstands area as Total Surface Area 4 8 B0 d da da d d d Uses 8 dt d dt 0 dt dt dv V 6 8 B d dv dv d dv 9 8 da0 dt d dt dt 0 40 da Misunderstands area as Curved Surface Area 4 B0 d d d Similar scheme to above with 4 0 dt dt dv V 6 8 B d dv dv d dv 9 8 ( or ) da0 dt d dt dt

27 Notes for Question 0 Must use calculus. Way B Correct statement d A d Uses correct chain rule d A d A d or equivalent e.g. d da da dt d dt dt dt d with d A and their d A dt 0 d to calculate d. NB If they correctly state the chain rule dt da da d then make an algebraic error they may be awarded this method mark dt d dt Obtain correct epression for d e.g. 0 then isw (award this mark in the two dt misread cases described where correct Curved Surface Area or Total Surface Area are used correctly) dv B Correct statement 8 d d Uses d V d V d with their d dt d dt dt and d V d to calculate d V, with = substituted. dt NB d V d should be in terms of one variable ( so of the form k and not kh ) 9 8 or 0.9π or or k =0.9 etc 0 0 Way Writes V in terms of A. Accept V.. A (first M on epen) This indicates way. dv A B 9 (both Bs on epen) da d This is dependent on the first method mark where V.. A. Uses d V d V d A with their d V dt da dt da and d A to find d V (second M on epen) dt 0 dt NB d V da should be in terms of one variable A ( so of the form ka ) 9 8 A or 0.9π or or k =0.9 etc (both As on epen) 0 0 Way Similar to Way but does calculation in one stage, not two see scheme above Condone use of r throughout, but if there is a miture of r, and h, then final accuracy mark may be with- held if full marks would have been gained. (Send to review if in doubt) NB: Misreads/ misunderstandings of cross section area- there are two eamples in the da scheme above- another possible is an open cylinder where area = 6 d d d dv then 6 0 dt dt and then V 6 8 which gives d dv dv d dv 9 8 These may be also combined with Way dt d dt dt 040 0

28 Question Number Scheme Marks (a) ( R.5. ) awrt.9- accept e.g..69 or 4 0 B. tan or 0.5π.5 t (b) H.9sin Hmin '.9' awrt.08 t "0.675" t t t (c) 4.9sin sin t t awrt. or.4 d 6 () (4) t dd t awrt 6.4 or Times are :0pm and 6:5pm or 6.4pm (4:0 and 8:5 or 8:4) allow hours 0minutes and 6 hours 5 or 4minutes or 40 minutes and 75 or 74 minutes Etra values in the range lose final A mark. (6) ( marks)

29 Notes for Question (a) B R awrt.9(dp) allow any equivalent e.g..69 or tan or tan.5. awrt (dp) also allow 0.5π (must be in radians) (b) States or attempts to calculate R with their value of R Hmin awrt.08. Or - 4 o.e.accept this for both marks as long as no incorrect 0 working is seen. t Attempts ' ' t... 6 ( Putting equal to is A0 (outside range) Putting equal to is M0A0 (wrong) ) (Allow method mark for using -90 or 70 degrees (not 90 degrees), if alpha was in degrees earlier) t awrt 0.9 (dp). Accept this for both marks as long as no incorrect working is seen. t 4 4 (c) d sin ' ', where (allow for degrees) 6 R R Dependent upon the previous M, using the correct order to find one value of t (allow consistent degrees) NB: sin (/.9) = Seeing is indication that sin instead of arcsin has been used. This indicates the wrong method and so M0. Accept either awrt. or.4 or awrt 6.4 or 6.5 do not need units ignore wrong units e.g. minutes and seconds for this mark. hours 0 minutes is correct here. Note that minutes 0 seconds could get this mark but would lose the final. dd Dependent upon the previous M, using the correct order to find a second value of t Accept awrt. or.4 and awrt 6.4 or 6.5- ignore wrong units. So 6 minutes 4 seconds could get this mark but would lose the net. Times are :0pm and 6:5pm (or 6.4pm) (4:0 and 8:5 (or 8.4)) (Need both times) allow hours 0minutes and 6 hours 5 or 4minutes or 40 minutes and 75 or 74 minutes. Etra values in the range lose final A mark Allow method marks for degrees, and accuracy marks if they converted to 4 4 sin 0 t '8.65', where and continued to correct answers. Using R R t 4 4 sin '8.65', where will lose the accuracy marks. 6 R R

30 Question Number Scheme Marks (i) 5 ( OP) or coordinates of P are 5,,6 6 B 5 OP ( 5 ) ( ) (6 ) Substitute their into their OP 6 d P has coordinates ( 4, 0.5,6.5) (5) (ii) Way 5 OA k or the coordinates of A are (5 k, k,4 k) 4 k 5 ( ) 4 or k 5 ( ) 4 o.e. Finds at least one value for k and substitutes into OA to give position d As k =, A has possible positions or (any notation) As k =, A has possible positions and (any notation) (5) 5 5 Way OA 4 5 or the coordinates of A are (,, ) ( ) or equivalent (see notes for variations) 5 5 Finds and substitutes into OA to give position d 4 As = so A has coordinates,, or,, As = so A has coordinates,, and,, 5 5 (5) Way (This is a common approach: see net page )

31 Question Number (ii) Scheme Marks Way Writes 5 4 y z This is equations, unknowns Writes y z This is the third equation in unknowns Eliminates two of the variables to obtain either, y, or z and uses it to find the other values d 4 A has coordinates,, or,, A has coordinates,, and,, 5 5 (5) Way 4 (minimal working): States 50 or Deduces position vectors or coordinates i j k or i j k i j k and i j k (5) (0 marks) Notes for Question (i) 5 B OP or coordinates of P are 5,,6. This may be implied by 6 its use in the scalar product. Attempts to use (may make slip copying) their OP. 0 obtaining an equation in and solving to give = They may use any multiple of i j + k (including (i j + k) or (i j + k) etc ) d Substitutes their into their OP. Dependent upon the previous M (ii) Way P has coordinates ( 4, 0.5,6.5). Accept it written in coordinate or in a vector form 4 either as column vector or as - 4i j +6.5k. Accept OP 5 Writing OA k or the coordinates of A as (5 k, k,4 k). 4 This mark may be implied by correct coordinates later. Correct equation in k (one variable) using D Pythagoras theorem for eample.

32 d (Dependent on first M mark) Solves their equation to give at least one value for k (their one variable, even allow use of μ here) and substitutes to find at least one possible position. a = OR. Allow coordinates ,, 5 5 or 4,, 5 5 a= 4 AND. Allow coordinates 5 5,, and,, If there are surds in the answer but it is otherwise correct then lose the final Way : This is described in the scheme where is the single variable used but could be used with 5 4 ( y, y, y) where y is 5, or with ( z, z, z) where z is in a similar way. 5 5 Way : The successful approach is described above in the scheme. Solving the equations may be lengthy, they may eliminate their variables one at a time and they may make errors. : This is for these correct equations in the scheme (Two equations three unknowns) : A correct further equation (third equation three unknowns) : Solves to obtain the three variables, as before (A common variation) Some write y z together with a scalar product giving 5y4z 50 This is two equations in three unknowns and usually stops there. This is marked M0A0M0A0A0 unless the candidate produces a third equation and makes progress towards the answer. Way 4: (Special Case) : Writes down 50 with little or no working (or or even 5) 5 d: Uses their correct fraction to find the coordinates (position vectors) as before

33 Question Number Scheme (a) awrt may be seen in the table B (b) e e Area = '0.799' 0 [The +0 is not required] B = awrt.055 (c) Way (c) Way (c) Way ln ln d ln d ln d ln ln d ln ln d d ln ln ( c) * ln d ln ln dln ln ln d ln ln ln d ln ln ln d Use u = ln substitution to get to (d) Volume = y d ln d ln ln ( c) * u ed u u u e u ue u du Marks u u u u e ue e du u u u u e ue e k ln d ln ( c) B () () (4) (4) (4) ln d 44ln (ln ) d Correct integration of at least two of their three terms (see notes) 4 4( ln ) (ln ) ln ( c) Volume 4 4( ln ) e (ln ) ln e dd e 5 e e(e 5) (6) (4 marks)

34 Notes for Question (a) B awrt should be in the table or given as answer not just appear in trapezium rule (b) B e e e +e For the strip width of or correct equivalent e.g. e, or e +e e Also accept awrt.4 o.e. This may be stated as h = the values above, or may be used correctly in the rule so e e, etc or.7 may be seen. 4 For correct application of the trapezium rule requires correct ft bracket... their answer to( a) 0 awrt.055 (c) Must use integration by parts or 0/4. Differentiating the answer is not acceptable. Way One correct application of integration by parts the correct way around. Aln Accept ln d ln d ln ln d A second application of integration by parts the correct way around Accept ln Aln Bd Accept the answer to the ln d part just written down as ln ln ln ( c) with or without 'c' d Way * Correct solution only One correct application of integration by parts the correct way around. Accept ln d ln ln dln ln ln d ln ln ln d d Way A second application of integration by parts the correct way around Accept ln ln ln Accept the answer to the ln d part just written down as ln and substituted into their epression ln ln ( c) with or without 'c' * Correct solution only d Uses substitution and performs one correct application of integration by parts the correct way around. u u u u e ue e du after second integration by parts Final integration and returns to ln ln ( c) with or without 'c' * Correct solution only (needs π and integral symbol) but not limits (d) B Volume = y d ln d and can condone missing d

35 to ln (ln ) where, and positive or negative constants. Needs attempt to multiply out to at least (ln ) where and are non zero positive or negative constants.and attempt to integrate. ( ln ) (ln ) ln with two of the three Multiplies out ln Look for are non zero terms integrated correctly (So if = 0 could score M0 here) Correct answer 4 4( ln ) (ln ) ln ( c) with or without c. Accept unsimplified and isw. dd Attempts to substitute both correct limits into the result of their integral. Both previous M's must have been scored. If there has been small slips simplifying the result of the integral before use of limits then allow Correct solution only e(e 5) Special case: For those who misunderstand/ misread and think that Volume = y d ln 4 d y (ln ) so - the first B may be eceptionally awarded. If anyone appears to make progress with a method for this integration (not rubbish) please send to review. This gains /6 marks

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