PMT. Mark Scheme (Results) Summer Pearson Edexcel GCE in Core Mathematics 4 (6666/01)

Transcription

1 Mark Scheme (Results) Summer 04 Pearson Edecel GCE in Core Mathematics 4 (6666/0)

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 04 Publications Code UA08467 All the material in this publication is copyright Pearson Education Ltd 04

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 General Instructions for Marking EDEXCEL GCE MATHEMATICS. 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 dm denotes a method mark which is dependent upon the award of the previous method mark. 4. 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.

5 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.

6 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 the correct formula (with values for a, b and c).. Completing the square b Solving + b + c = 0 : ± ± q ± c = 0, 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+ )

7 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 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. 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.

8 Question Number. (a) (b) At (, ), (a) (a) General Scheme + y y 0 = 0 Marks = + y + y = 0 M A B + y + ( y ) = 0 dm d + or = y y y A cso [5] m T = () + ( ) = ; = or ( ) () 6 P ( ) " " and either T: y = ( ) see notes M or ( ) = () + c c =..., T: y 9 0 K y 9 = 0 A cso = or ( ) [] 7 Alternative method for part (a) = + y + y = 0 M A B y + ( + y ) = 0 dm d + y = y or y y y Question s d Writing down y + y or = y from no working is full marks. y y Writing down d + = or y y y y y from no working is MA0B0MA0 A cso + y =, o.e. Few candidates will write + y + y = 0 leading to y This should get full marks.. (a) M Differentiates implicitly to include either or y ± ky. (Ignore = ). A and y 0 = 0 y = 0 B y y + If an etra term appears then award st A0. [5]

9 . (a) ctd dm + y + y + y = y will get st A (implied) as the " = 0" can be implied by rearrangement of their equation. dependent on the first method mark being awarded. An attempt to factorise out all the terms in d y as long as there are at least two terms in d y. ie ( y ) =... Placing an etra d y at the beginning and then including it in their factorisation is fine for dm. A y + y For or equivalent. Eg: y y cso: If the candidate s solution is not completely correct, then do not give this mark. isw: You can, however, ignore subsequent working following on from correct solution.. (b) M Some attempt to substitute both = and y = into their d y which contains both and y to find and mt either applies y ( their mt )( ) or finds c by solving ( m ) =, where mt is a numerical value. ( ) = their T () + c, where mt is a numerical value. Using a changed gradient (i.e. applying or is M0). their their A Accept any integer multiple of y 9 = 0 or 9 y = 0 or + y + 9 = 0, where their tangent equation is equal to 0. cso A correct solution is required from a correct d y. isw You can ignore subsequent working following a correct solution. Alternative method for part (a): Differentiating with respect to y. (a) M Differentiates implicitly to include either A B dm A (Ignore = ). and y or y = y = ± k or y y + dependent on the first method mark being awarded. An attempt to factorise out all the terms in d as long as there are at least two terms in d. y + y For or equivalent. Eg: y y cso: If the candidate s solution is not completely correct, then do not give this mark.

10 Question Number. Scheme ( 4)( 4 ) + = + + +! 4 ( k ) ( 4)( k) ( k)... (a) Either ( 4) k 6 = or ( k ) 4 leading to k = ( 4)( 5) (b) ( k ) + = + ( 4)( k) see notes M 6 k = or.5 or A 4 ( 4)( 5) Either or ( k ) or ( k ) M! ( 4)( 5) ( 4)( 5) k or ( k) M!! Either ( ) Marks [] ( 4)( 5) 45 A = A=! Question s In this question ignore part labelling and mark part (a) and part (b) together. ( 4)( 4 ) Writing down {( + k ) 4 } = + ( 4)( k) + ( k) +...! gets all the method marks in Q. i.e. (a) M and (b) MM 45 or.5 A [] 5 (a) M Award M for either writing down ( 4) k = 6 or 4k = 6 4 or epanding ( + k) to give + ( 4)( k) or writing down ( 4) k= 6 or ( 4 k) = 6 or 4k= 6 6 A k = or.5 or from no incorrect sign errors. 4 The M mark can be implied by a candidate writing down the correct value of k. Award M for writing down 4k = 6 and then A for k =.5 (or equivalent). 4 Award M0 for 4k = 6 (if there is no evidence that ( + k) epands to give + ( 4)( k) +...) + ( 4)( k) leading to ( 4) k = 6 leading to k = is MA0. ( 4)( 4 ) (b) M For either or ( 4)( 5) or 0 or ( k ) or!! ( 4)( 4 ) M Either ( k ) ( 4)( 5) or ( k ) ( 4)( 5) k!!! Candidates are allowed to use instead of! 45 A Uses k =.5 to give A = or.5 90 A = which has not been simplified is A Award A0 for A Allow A for A= 45 followed by A = 45 k =.5 leading to A = or.5 is A0. =. ( k ) or ( ) 45 ( 4)( 5) their k! or ( ) or 0k

11 Question Number. (a) { } Scheme 4 y y = + 5 Marks At =, y = 0.68 (5 dp) 0.68 B cao [] Outside brackets or their B aef (b) ( ) { } For structure of [...] M = (5.5489) = =.5774 (4 dp) anything that rounds to.5774 A [] (c) Overestimate and a reason such as {top of} trapezia lie above the curve a diagram which gives reference to the etra area concave or conve B d y > 0 (can be implied) bends inwards curves downwards [] du (d) { u = } = or u du = B 0 ±.u du Either ku ± k { du} or u + 5u αu ± βu ( ) { du } M u αu ± βu 5 ± λln(u + 5) or ± λln u +, λ 0 0 M 0 = du = ln(u + 5) with no other terms. u ln(u + 5) or 0ln u + A cso u Substitutes limits of and in u ln( u + 5) = 0ln ( () + 5 ) 0ln ( () + 5 ) (or 4 and in ) and subtracts M the correct way round. 9 0ln 9 0ln 7 or 0ln or 0ln 0ln 7 A oe cso 7 [6] Question s. (a) B 0.68 correct answer only. Look for this on the table or in the candidate s working. (b) B Outside brackets or or equivalent. M For structure of trapezium rule[... ] No errors are allowed [eg. an omission of a y-ordinate or an etra y-ordinate or a repeated y ordinate]. A anything that rounds to.5774 Working must be seen to demonstrate the use of the trapezium rule. (Actual area is.5448 )

12 . (b) contd ( ) their 0.68 = Award BMA for ( ) Bracketing mistake: Unless the final answer implies that the calculation has been done correctly award BM0A0 for ( their 0.68 ) (nb: answer of ). award BM0A0 for ( ) + ( their 0.68 ) (nb: answer of 4.685). Alternative method: Adding individual trapezia "0.68" "0.68" Area + + = B B: and a divisor of on all terms inside brackets. M M: First and last ordinates once and two of the middle ordinates twice inside brackets ignoring the. A A: anything that rounds to.5774 (c) B Overestimate and either trapezia lie above curve or a diagram that gives reference to the etra area eg. This diagram is sufficient. It must show the top of a trapezium lying above the curve. (d) d y or concave or conve or > 0 (can be implied) or bends inwards or curves downwards. Reason of gradient is negative by itself is B0. du B = or du = or du = or = udu or u du = o.e. ± ku ± k M Applying the substitution and achieving { du} or αu ± βu ( ) { du }, u αu ± βu k, α, β 0. Integral sign and du not required for this mark. 5 M Cancelling u and integrates to achieve ± λln(u + 5) or ± λln u +, λ 0 with no other terms. 0 A cso. Integrates to give 0 5 ln(u + 5) or 0ln u +, un-simplified or simplified. u BE CAREFUL! Candidates must be integrating or equivalent. u So d 0ln( 5) 5 u = u u + WOULD BE A0 and final A0. + M Applies limits of and in u or 4 and in in their (i.e. any) changed function and subtracts the correct way round. 0 9 A Eact answers of either 0ln 9 0ln 7 or 0ln or 0ln 0ln 7 7 or 0ln 7 or 9 ln 7 0 or equivalent. Correct solution only. You can ignore subsequent working which follows from a correct answer. A decimal answer of (without a correct eact answer) is A0.

13 Question Number 4. Scheme dv = 80π, V= 4 πhh ( + 4) = 4πh + 6 πh, dt dv ± αh ± β, α 0, β 0 M = 8πh + 6π dh 8πh + 6π A dv dh dv = ( ) d h dv dh 8πh + 6π = 80π Candidate's = 80π dh dt dt dt dh dt M oe dh dv dv dh dv = = 80π or 80π Candidate's dt dt dh dt 8πh + 6π d h d 80 When h = 6, h π = 80π = dependent on the previous M dm dt 8 π(6) + 6π 64π see notes dh -.5 (cms ) dt =.5 or or or A oe Alternative Method for the first MA u = 4π h v = h + 4 Product rule: du dv = 4π = dh dh dv = 4 π( h + 4) + 4πh dh ± αh ± β, α 0, β 0 M 4 π( h + 4) + 4πh A Marks [5] 5 Question 4 s M An epression of the form ± αh ± β, α 0, β 0. Can be simplified or un-simplified. A Correct simplified or un-simplified differentiation of V. eg. 8πh + 6π or 4 π( h + 4) + 4πh or 8 π ( h + ) or equivalent. Some candidates will use the product rule to differentiate V with respect to h. (See Alt Method ). dv does not have to be eplicitly stated, but it should be clear that they are differentiating their V. dh M dv dh dv Candidate's = 80π or 80π Candidate's dh dt d h Also allow nd dv dh dv M for Candidate's = 80 or 80 Candidate's d h d t d h Give nd dv dh dv M0 for Candidate's = 80 πt or 80k or 80 πt or 80k Candidate's dh dt d h dm which is dependent on the previous M mark. Substitutes h = 6 into an epression which is a result of a quotient of their d V and 80π (or 80) dh A.5 or or or (units are not required) π as a final answer is A0. 64π Substituting h = 6 into a correct d V gives 64π but the final M mark can only be awarded if this dh is used as a quotient with 80π (or 80)

14 Question Number 5. π = 4cos t +, 6 y = sin t Main Scheme (a) π π = 4 costcos sin tsin 6 6 Scheme π π + y = t t + t 6 6 So, { } 4 cos cos sin sin sin π π π cos t + costcos ± sin tsin Adds their epanded (which is in terms of t) to sint Marks M oe = 4 cost sin t + sin t = cost * Correct proof A * [] (a) Alternative Method π π π π π = 4 costcos sin tsin cos t + costcos ± sin tsin M oe = 4 cost sint = cost sint So, = cost y Forms an equation in, y and t. dm (b) + y = cost * Correct proof A * [] Main Scheme + y y + = ( + y) y + = 4 ( + y) + y = (b) Alternative Method ( + y) = cos t = ( sin t) = sin t So, ( + y) = y + + = ( y) y (b) Alternative Method ( + y) = cos t As then cos t + sin t = ( y) y Applies cos t + sin t = to achieve an equation containing only s and y s. ( y) y dm M + + = A a =, b = [] { } Applies cos t + sin t = to achieve an M equation containing only s and y s. ( + y) + y = A + + = M, A [] [] 5

15 Question 5 s 5. (a) M π π π cos t + costcos ± sin tsin or cos t + cost ± sin t If a candidate states cos( A + B) = cos Acos B ± sin Asin B, but there is an error in its application then give M. Awarding the dm mark which is dependent on the first method mark Main dm Adds their epanded (which is in terms of t) to sint Writing + y =... is not needed in the Main Scheme method. Alt dm Forms an equation in, y and t. π A* Evidence of cos 6 and sin π 6 evaluated and the proof is correct with no errors. π y t t, by itself is M0M0A0. 6 { + } = 4cos + + sin (b) M Applies cos t + sin t = to achieve an equation containing only s and y s. A leading ( + y) + y = SC Award Special Case BB0 for a candidate who writes down either ( + y) + y = from no working a =, b =, but does not provide a correct proof. Alternative method is fine for M A Writing ( + y) = cos t followed by cos t + a(4sin t) = b a =, b = is SC: BB0 Writing ( + y) = cos t followed by cos t + a(4sin t) = b states a =, b = and refers to either cos t + sin t = or cos t + sin t = and there is no incorrect working would get MA

16 Question Number 4 4 Scheme 6. (i) 4 4 e d e e 4 = { } e e 4 6 { c} 4 4 = + 8 8( ) (ii) = { + c} (iii) ( ) ()( ) { ( ) { c} } { } M ± α β α β > 4 4 e e d, 0, 0 8( ) ()( ) 4 4 e e { } A e e A 4 6 ± λ ( ) M or equivalent. A Marks = + {Ignore subsequent working}. [] e cosec y cosec y = π y = at = 0 6 Main Scheme = e cosec y cosec y or sin sin d e d y y y = B oe sin y cos y sin y d y = e d Applying or sin y sin ycos y M cosec y Integrates to give ± µ sin y M sin e y = { + c} sin ycos y sin y A e e B π sin π e 0 Use of y = and = 0 = + c or = c 6 M 6 8 in an integrated equation containing c c = giving sin y = e sin y = e A Alternative Method = e or sin sin d e d cosec y cosec y y y y = B oe ( cos cos ) d e y y y = d sin sin λcos λcos sin sin e y y = + c { } π π sin sin e = + c = giving c or = c sin y + sin y = e 6 y y ± y ± y M Integrates to give ± αsiny ± βsin y M e sin y sin y A e as part of solving their DE. B π Use of y = and = 0 in an 6 M integrated equation containing c sin y + sin y = e A 6 [] [7] [7]

17 Question 6 s (i) M Integration by parts is applied in the form αe βe { } ±, where α 0, β > 0. (must be in this form). 4 e e 4 or equivalent e e with/without + c. Can be un-simplified. 4 6 You can ignore subsequent working following on from a correct solution. dv 4 SPECIAL CASE: A candidate who uses u =, = e, writes down the correct by parts formula, but makes only one error when applying it can be awarded Special Case M. A { } A isw SC (ii) M A ± λ( ), λ 0. that λ can be. 8( ) or ( ) or with/without + c. Can be un-simplified. ()( ) ( ) You can ignore subsequent working which follows from a correct answer. (iii) B Separates variables as shown. and should be in the correct positions, though this mark can be implied by later working. Ignore the integral signs. Allow B for = e cosec y cosec y or sin y sin y = e M sin ycos y cosec y or sin y sin ycos y or sin sin cos cos seen anywhere in the candidate s working to (iii). M Integrates to give ± µ sin y, µ 0 or ± αsiny ± βsin y, α 0, β 0 A sin ycos y sin y (with no etra terms) or integrates to give sin y sin y B Evidence that e has been integrated to give e as part of solving their DE. M π Some evidence of using both y = and = 0 in an integrated or changed equation containing c. 6 that is mark can be implied by the correct value of c. A sin y = e or sin y + sin y = e or any equivalent correct answer. 6 You can ignore subsequent working which follows from a correct answer. Alternative Method (Using integration by parts twice) sin sin d = e d B oe cos ysin y sin y cos y = e + c { } cos ysin y sin y cos y = e Applies integration by parts twice to give ± αcos ysin y ± β sin y cos y M cos ysin y sin y cos y A (simplified or un-simplified) e e as part of solving their DE. B as in the main scheme M sin y + sin y = e A 6 [7]

18 Question Number 7. (a) (b) Scheme π = θ y = θ y = + θ θ < sec θ dθ =, = 8cosθ sinθ or = 4sin θ dθ dθ tan, 4cos or cos, 0. 8cosθsinθ 8 4 = cos sin sin cos = θ θ = θ θ sec θ π At P(, ), θ =, 4 8 π π = cos sin = 4 4 their d y dθ divided by their d dθ Correct d y Some evidence of π substitutingθ = into their d y 4 So, m ( N ) = applies " " Either N: y = ( ) " " or = () + c {At Q, y = 0, so, = ( ) } giving 5 = y = y dθ = { } (4cos θ ) sec θ { d θ} dθ m( N) = m( T) M Marks A oe M M see notes M 5 = or or awrt.67 A cso see notes M So, π y d = π (4cos θ) sec θ { dθ } see notes A y = 48cos θdθ 48cos θ { dθ } A π 4 Applies + cos θ = { 48} dθ = ( 4 + 4cos θ) dθ = + = + 4 { 48} θ sin θ { 4θ sin θ} π 4 π y = 48 θ + sin θ = = 6π π { } ( ) { } 4 {So V = π y = 6π + π } 0 5 6π Vcone = π () = 9 6π 9 Vol( S) = 6π + π Vol( S) = π + 6π 9 9 cos θ cos θ = M Dependent on the first method mark. For ± αθ ± β sin θ dm cos θ θ + sin θ 4 V cone Dependent on the third method mark. () π their ( ) ( a ) A dm = M 9 9 π π + 6 A 9 p =, q = 6 9 [6] [9] 5

19 Question 7 s 7. (a) st M Applies their d y dθ divided by their d dθ or applies d y dθ multiplied by their d θ SC Award Special Case st M if both d dθ and d y are both correct. dθ st A Correct d y 8cosθsinθ i.e. or 8 4 cos θ sin θ or sin cos θ θ or any equivalent form. sec θ nd π M Some evidence of substituting θ = or θ = 45 into their d y 4 For rd M and 4 th M, m( T ) must be found by using d y. rd M applies m( N) =. Numerical value for m( N) is required here. m( T) 4 th M Applies y ( their m )( ) =, where m( N) is a numerical value, N or finds c by solving = (their m ) + c, where m( N) is a numerical value, and m N = their m( T) or m N = or m N = their m( T ). their m( T) This mark can be implied by subsequent working. nd A 5 = or or awrt.67 from a correct solution only. (b) st M Applying as y with their d dθ dθ π outside integral. You can ignore the omission of an integral sign and/or dθ for the st M. Allow st M for (cos ) "their sec " d θ or 4(cos θ) "their sec θ " dθ st A Correct epression { π y d } = π (4cos θ) sec θ { dθ} N (Allow the omission of dθ ) IMPORTANT: The π can be recovered later, but as a correct statement only.. (Ignore dθ ). : 48 can be written as 4() for eample. nd A { } y d = 48cos θ { d θ } nd M Applies cos θ = cos θ to their integral. (Seen or implied.) rd dm* which is dependent on the st M mark. Integrating cos θ to give ± αθ ± β sin θ, α 0, β 0, un-simplified or simplified. rd A which is dependent on the rd M mark and the st M mark. Integrating cos θ to give θ + sin θ, un-simplified or simplified. 4 k k This can be implied by k cos θ giving θ + sin θ, un-simplified or simplified. 4 4 th dm which is dependent on the rd M mark and the st M mark. Some evidence of applying limits of 4 π and 0 (0 can be implied) to an integrated function in θ 5 th M Applies V π ( a ) cone = () their part ( ) answer. 5 Vcone = π 5 their Also allow the 5 th M for { }, which includes the correct limits. 4 th A π π π π A decimal answer of (without a correct eact answer) is A0. The π in the volume formula is only needed for the st A mark and the final accuracy mark.

20 7. Working with a Cartesian Equation 6 A cartesian equation for C is y = + 9 = ± ± ± or (a) st M λ( α β) = ± λ ( ± α ± β) st 7 A = 6( + 9) ( ) or = un-simplified or simplified. d ( + 9) nd dm Dependent on the st M mark if a candidate uses this method For substituting = into their d y 7() i.e. at P (, ), = = d ( + 9) From this point onwards the original scheme can be applied. ± λ (π not required for this mark) ± α ± β A For π 6 { } (π required for this mark) + 9 To integrate, a substitution of = tanθ is required which will lead to 48cos θdθ and so from this point onwards the original scheme can be applied. (b) st M For { } Another cartesian equation for C is β β (a) st M ± α = ± or ± α = ± y d y y 6 = 9 y 6 st A = y or 6 = y nd dm Dependent on the st M mark if a candidate uses this method For substituting = to find d y 6 d d i.e. at P(, ), () = y y... 4 = From this point onwards the original scheme can be applied.

21 Question Scheme Number 8. OA = i + 4j+ 7 k, OB = i + j+ 8 k & OP = 0i + j+ k AB = = ± ( i + j+ 8 k) ( i + 4j+ 7 k); = i j+ k (a) ( ) (b) { l : r } = 4 + λ or { } (c) (d) (e) 7 0 PB = OB OP = = or BP = AB PB 5 cos θ = = AB. PB () + ( ) + (). ( ) + () + (5) { } { cos θ} { r } Marks M; A r = + λ Bft 8 Applies dot product formula between their ( AB or BA) PB or BP. and their ( ) + 5 = = = Correct proof A cso l : = + µ 0 0 OC = + = or OD = = 4 { C(,, 4 ), D(,, ) } (f) h Way ( ) + () + (5) = sinθ 8 h = 7 sin( ) = 7 = 6 = awrt 4.9 ( ) Area 6 = 6 ( ) = 8 = 9 p + λd or p + µ d, p 0, d 0with either p= 0i + j+ k or d = their AB M M, or a M multiple of their AB. Correct vector equation. A ft Either OP + their AB or OP their AB M At least one set of coordinates are correct. A ft Both sets of coordinates are correct. A ft h = sinθ M their PB 8 7 sin( ) or 7. or 6or awrt 4.9 or equivalent ABCD = + ( h)( AB CD) A oe their their + their dm 9 A cao [] [] [] [] [] [4] 5

22 8. (f) Helpful Diagram! Area APB = A 4 7 B 8 θ l 8 h = 6 =. = l DA = PB = 5 PA = CB = 4 D 0 P C 4 8. (f) Way Way PA = CB = and AB =, so BC AB 4 h = CB = ( ) + () + (4) = 4 = 6 = ( ) Candidates do not need to prove this result for part (f) Attempts PA or CB PA = CB = 4 M A oe Area ABCD = 4 + or h( their AB their CD) + dm oe = 9 9 A cso Finds the area of either triangle APB or APD or BCP and triples the result. Area APB = sinθ Attempts (their AB)(their PB)sinθ M = ( ) sin ( ) ( ) sin( ) or A or awrt 4.4 or equivalent ABCD = Area of APB dm 8. (f) ( ) Area ( ) = 9 9 A cso [4] [4]

23 Question 8 s 8. (a) M Finding the difference (either way) between OB and OA. If no subtraction seen, you can award M for out of correct components of the difference. A i j+ k or or (,, ) or benefit of the doubt (b) Bft { r } = 4 + λ or { r } = + λ, with AB or BA correctly followed through from (a). 7 8 r = is not needed. (c) M An attempt to find either the vector PB or BP. If no subtraction seen, you can award M for out of correct components of the difference. AB or BA PB or BP. M Applies dot product formula between their ( ) A Obtains { cos } and their ( ) θ = by correct solution only. AB PB If candidate starts by applying correctly (without reference to cos θ =...) AB. PB Award the final A mark if candidate achieves { cos } they can gain both nd M and A mark. θ = by either taking the dot product between (i) and or (ii) and. Ignore if any of these vectors are labelled incorrectly. 5 5 Award final A0, cso for those candidates who take the dot product between (iii) and or (iv) and 5 5 They will usually find { cosθ } = or may fudge{ cos θ } =. If these candidates give a convincing detailed eplanation which must include reference to the direction of their vectors then this can be given A cso (c) Alternative Method : The Cosine Rule 0 PB = OB OP = = or BP = PB = 7, AB = and PA = 4 ( ) = ( ) + ( ) ( )( ) cosθ Mark in the same way as the main scheme. Applies the cosine rule the correct way round M M oe cosθ = = Correct proof A cso 8 []

24 8. (c) Alternative Method : Right-Angled Trigonometry 0 PB = OB OP = = or BP = Either ( 4 ) + ( ) = ( 7 ) or So, Mark in the same way as the main scheme. Confirms PAB is right-angled M AB PA = = + 4 = 0 4 AB cosθ = cosθ = = Correct proof A cso PB 7 (d) M Writing down a line in the form p + λd or p + µ dwith either 0 a = or d = their AB or a multiple of their AB found in part (a). 0 0 Aft Writing + µ or + µ d, where d = their AB or a multiple of their AB r = is not needed. Using the same scalar parameter as in part (b) is fine for A. M [] d = their AB, found in part (a). (e) M Either OP + their AB or OP their AB. This can be implied at least two out of three correct components for either their C or their D. Aft At least one set of coordinates are correct. Ignore labelling of C, D Aft Both sets of coordinates are correct. Ignore labelling of C, D You can follow through either or both accuracy marks in this part using their AB from part (a). (f) M h Way : = sinθ their PB Way : Attempts PA or CB Way : Attempts (their PB)(their AB)sinθ Finding AD by itself is M0. A Either h = 7 sin( ) or PA = CB = 4 or equivalent. (See Way and Way ) or the area of either triangle APB or APD or BDP = ( ) sin ( ) o.e. (See Way ). dm which is dependent on the st M mark. A full method to find the area of trapezium ABCD. (See Way, Way and Way ). A 9 from a correct solution only. A decimal answer of (without a correct eact answer) is A0.

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