PhysicsAndMathsTutor.com. Mark Scheme (Results) June GCE Core Mathematics 4 (6666/01)

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1 Mark Scheme (Results) June 03 GCE 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: June 03 Publications Code UA All the material in this publication is copyright Pearson Education Ltd 03

3 General Marking Guidance All candidates must receive the same treatment. Eaminers must mark the first candidate in eactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Eaminers should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Eaminers should always award full marks if deserved, i.e. if the answer matches the mark scheme. Eaminers should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and eemplification may be limited. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edecel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided. 3. 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 eample, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks. 5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer. 8. In some instances, the mark distributions (e.g. M, B and A) printed on the candidate s response may differ from the final mark scheme

5 General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving 3 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). 3. Completing the square b Solving + b+ c= 0 : ± ± q± c, q 0, leading to =... Method marks for differentiation and integration:. Differentiation n n Power of at least one term decreased by. ( ) PhysicsAndMathsTutor.com. Integration Power of at least one term increased by. ( n n + ) Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent eaminers reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are mistakes in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may be lost if there is any mistake in the working. Eact answers Eaminers reports have emphasised that where, for eample, an eact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals. Answers without working The rubric says that these may not gain full credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done in your head, detailed working would not be required.

6 Question Number. (a) e, st Application: e = e ( ) = e e e { } = e ( e e ) + c { e ( e e ) 0 } (b) (a) ( e (e e )) ( 0 e (0e e )) = Scheme du u = =, nd Application: dv = e v = e d du u = = dv = e v = e d { } e λe, λ > 0 M { } e e ± A e ± B e ± C e d Either { } ( ) or for ± K e { } ± K e e { } Marks A oe M ± A e ± Be ± C e M Correct answer, with/without Applies limits of and 0 to an epression of the form ± A e ± Be ± Ce, A 0, B 0 and C 0 and subtracts the correct way round. + c A = e e cso A oe Notes for Question M: Integration by parts is applied in the form e λe { } or equivalent. A: e e { } M: Either achieving a result in the form A e B e C e { } M: (where A 0, B 0 and 0) PhysicsAndMathsTutor.com M, where λ > 0. (must be in this form). ± ± ± (can be implied) ( ) C or for ± K e { } ± K e e { } ± A e ± Be ± C e (where A 0, B 0 and C 0) A: e ( e e ) or e e + e or ( + )e or equivalent with/without + c. (b) M: Complete method of applying limits of and 0 to their part (a) answer in the form ± A e ± Be ± Ce, (where A 0, B 0 and C 0 ) and subtracting the correct way round. Evidence of a proper consideration of the limit of 0 (as detailed above) is needed for M. So, just subtracting zero is M0. 0 A: e or e or + e. Do not allow e e unless simplified to give e. Note: that without seeing e or equivalent is A0. WARNING: Please note that this A mark is for correct solution only. So incorrect [...] leading to e is A0. 0 Note: If their part (a) is correct candidates can get MA in part (b) for e from no working. Note: from no working is M0A0 [5] [] 7

7 Question Number. (a) (b) + = ( + ) ( ) Scheme ( + ) ( ) B Marks 3 ( )( ) ( )( ) = ( ) + ( ) +...!! See notes M A A 3 = = See notes M Answer is given in = + + the question. A * [6] ie: so, (a) B: ( 6 ) ( ) + = = B = A cao ( + ) ( ) or Notes for Question ( ) ( ) + seen or implied. (Also allow ( ) M: Epands ( + ) to give any out of 3 terms simplified or un-simplified, Eg: + or ( )( ) ( )( ) + + or !! or epands ( ) to give any out of 3 terms simplified or un-simplified, Eg: + ( ) or PhysicsAndMathsTutor.com ( )( ) + +! 3 ( ) ( ) or 3 ( )( ) ( )! ( + )( ) ). 3 ( )( ) Also allow: ( ) for M.! 3 4 A: At least one binomial epansion correct (either un-simplified or simplified). (ignore and terms) 3 4 A: Two binomial epansions are correct (either un-simplified or simplified). (ignore and terms) Note: Candidates can give decimal equivalents when epanding out their binomial epansions. M: Multiplies out to give, eactly two terms in and eactly three terms in. A: Candidate achieves the result on the eam paper. Make sure that their working is sound. 3 Special Case: Award SC FINAL MA for a correct multiplied out with no errors to give either or or or leading to the correct answer of M [3] 9

8 . (a) ctd (b) Notes for Question Continued 3 Note: If a candidate writes down either ( + ) = or ( ) = with no working then you can award st M, st A. Note: If a candidate writes down both correct binomial epansions with no working, then you can award st M, st A, nd A. M: Substitutes B: For sight of A: = into both sides of 6 + and (or better) and or equivalent fraction 5 35 Eg: 3 3 and or and or 3 3 and are fine for B. or any equivalent fraction, eg: or etc. or and 3 35 Special Case: Award SC: MBA0 for or truncated.7300 or awrt Aliter. (a) Way Note that 705 = and = + ( + )( ) ( ) = = = ( ) ( ) = ( + )( ) ( ) ( ) ( ) B ( )( ) = + ( ) ( )( ) + ( ) +...! See notes MAA... (... ) = = + + See notes M Answer is given in the = + + question. A * [6] Aliter. (a) B: ( ) ( ) seen or implied. Way M: Epands ( ) to give both terms simplified or un-simplified, + ( ) or epands ( ) to give any out of 3 terms simplified or un-simplified, ( )( ) ( )( ) + + or ( )!! 3 4 A: At least one binomial epansion correct (either un-simplified or simplified). (ignore and terms) 3 4 A: Two binomial epansions are correct (either un-simplified or simplified). (ignore and terms) M: Multiplies out to give, eactly one term in and eactly two terms in. A: Candidate achieves the result on the eam paper. Make sure that their working is sound. Eg: + ( )( ) or... ( )( ) ( )

9 Aliter. (a) Way 3 Aliter. (a) Way 4 Notes for Question Continued + ( + )( + ) = = = ( + )( ) ( )( + ) ( + )( ) B Must follow = ( + ) on from above. MAA = + + dma Note: The final M mark is dependent on the previous method mark for Way 3. Assuming the result on the Question Paper. (You need to be convinced that a candidate is applying this method before you apply the Mark Scheme for Way 4). + ( + ) = = + + ( + ) = + + ( ) ( ) ( )( ) ( + ) = = ,! 8 ( )( ) ( ) = + ( ) + ( ) +... = +...! 8 B MAA RHS = + + ( ) = = See notes M = + 8 So, LHS = + = RHS 8 A * B: ( + ) = + + ( ) seen or implied. M: For Way 4, this M mark is dependent on the first B mark. Epands ( + ) to give any out of 3 terms simplified or un-simplified, Eg: + or ( )( ) + +! or ( )( ) ! or epands ( ) to give any out of 3 terms simplified or un-simplified, Eg: + ( ) or ( )( ) ( ) ( ) ( )( ) + + or ( )!! 3 4 A: At least one binomial epansion correct (either un-simplified or simplified). (ignore and terms) 3 4 A: Two binomial epansions are correct (either un-simplified or simplified). (ignore and terms) M: For Way 4, this M mark is dependent on the first B mark. Multiplies out RHS to give, eactly two terms in and eactly three terms in. A: Candidate achieves the result on the eam paper. Candidate needs to have correctly processed both the LHS and RHS of ( + ) = + + ( ). [6]

10 Question Number Scheme Marks 3. (a).5470 B cao [] (b) π Area ; + ( their.5470) B; M π = = =.7787 (4 dp).7787 or awrt.7787 A [3] (c) π For π sec. V = π sec 0 Ignore limits and. Can be implied. B π ± λ tan M = { π} tan 0 tan or equivalent A = π π A cao cso [4] 8 Notes for Question 3 (a) B:.5470 correct answer only. Look for this on the table or in the candidate s working. (b) B: Outside brackets π π or or awrt M: For structure of trapezium rule[... ] A: anything that rounds to.7787 Note: It can be possible to award : (a) B0 (b) BMA (awrt.7787) Note: Working must be seen to demonstrate the use of the trapezium rule. Note: actual area is Note: Award BMA for π π ( +.444) + ( their.5470) = Bracketing mistake: Unless the final answer implies that the calculation has been done correctly, π Award BM0A0 for + + ( their.5470) (nb: answer of ). 6 π Award BM0A0 for ( +.444) + ( their.5470) (nb: answer of ). 6 Alternative method for part (b): Adding individual trapezia π Area = π B: and a divisor of on all terms inside brackets. 6 M: First and last ordinates once and two of the middle ordinates twice inside brackets ignoring the. A: anything that rounds to.7787

11 3. (c) Notes for Question 3 Continued B: For a correct statement of π sec or sec π or Ignore limits and. Can be implied. Note: Unless a correct epression stated π sec would be B0. 4 M: ± λ tan from any working. A: tan or tan from any working. ( ) A: π from a correct solution only. π. ( cos( )) { } Note: The π in the volume formula is only required for the B mark and the final A mark. Note: Decimal answer of without correct eact answer is A0. Note: The B mark can be implied by later working as long as it is clear that the candidate has applied in their working. Note: Writing the correct formula of V π y { } =, but incorrectly applying it is B0. π y

12 Question Number Scheme π π = sin t, y = cost = sin t, t 4. { } (a) (b) cost =, dy sint = or d y 4sintcost = So, d y sin t 4cos t sin t = = = sint cost cost π sin π dy 6 At t =, = ; = 6 π cos 6 y t t At least one of d Both d y or d y y correct. and d are correct. Applies their d divided by their d π and substitutes t = into their d y 6. Correct value for d y of = cos = ( sin ) M = sin t PhysicsAndMathsTutor.com B B M; Marks A cao cso So, y = or y = or y = y = or equivalent. A cso isw Either k = or B [3] (c) Range: 0 f ( ) or 0 y or 0 f See notes B B [] 9 Notes for Question 4 (a) B: At least one of d or d y correct. Note: that this mark can be implied from their working. B: Both d and d y are correct. Note: that this mark can be implied from their working. M: Applies their d y divided by their d π and attempts to substitute t = into their epression for d y 6. This mark may be implied by their final answer. dy sint Ie. = followed by an answer of would be M (implied). cost A: For an answer of by correct solution only. Note: Don t just look at the answer! A number of candidates are finding dy = from incorrect methods. Note: Applying d divided by their d y is M0, even if they state d y d y d =. Special Case: Award SC: B0B0MA for d cost =, dy dy sint = sint leading to = cost π which after substitution of t =, yields d y 6 = Note: It is possible for you to mark part(a), part (b) and part (c) together. Ignore labelling! [4]

13 Notes for Question 4 Continued 4. (b) M: Uses the correct double angle formula cos t = sin t or (c) cos t = cos t sin t in an attempt to get y in terms of or get y in terms of sin t and PhysicsAndMathsTutor.com cos t. Writing down y cos t cos t = or sin t or get y in terms of = sin t is fine for M. cos t A: Achieves y = or un-simplified equivalents in the form y = f(). For eample: 4 or or y = y = y= or y= and you can ignore subsequent working if a candidate states a correct version of the Cartesian equation. IMPORTANT: Please check working as this result can be fluked from an incorrect method. Award A0 if there is a + c added to their answer. B: Either k = or a candidate writes down. Note: k unless k stated as is B0. Note: The values of 0 and/or need to be evaluated in this part B: Achieves an inclusive upper or lower limit, using acceptable notation. Eg: f ( ) 0 or f ( ) B: 0 f( ) or 0 y or 0 f Special Case: SC: BB0 for either 0 < f ( ) < or 0 < f < or 0 < y < or (0, ) Special Case: SC: BB0 for 0. IMPORTANT: Note that: Therefore candidates can use either y or f in place of f ( ) Eamples: 0 is SC: BB0 0 < < is B0B0 0 is B0B0 is B0B0 f( ) > 0 is B0B0 f ( ) < is B0B0 > 0 is B0B0 < is B0B0 0 f( ) is B0B0 0 < f ( ) is BB0 0 f( ) < is BB0. f( ) 0 is BB0 f( ) is BB0 f ( ) 0 and f ( ) is BB. Must state AND {or} f( ) is B0B0 f ( ) 0 or f ( ) is BB0. f( ) is BB0 f( ) is B0B0 f( ) is BB0 < f ( ) < is B0B0 0 f( ) 4 is BB0 0 < f ( ) < 4 is B0B0 0 Range is BB0 Range is in between 0 and is BB0 0 < Range < is B0B0. Range 0 is BB0 Range is BB0 Range 0 and Range is BB0. [ 0, ] is BB (0, ) is SC BB0 Aliter 4. (a) Way cost =, dy = sint, So B, B. π At t =, d π dy π = cos = 3, = sin = Hence d y = So implied M, A.

14 Aliter 4. (a) Way 3 dy = = y Notes for Question 4 Continued Finds d y Correct differentiation of their Cartesian equation. Bft =, using the correct Cartesian equation only. B π π At t =, d Finds the value of when t = y π = sin and substitutes this into their d y M = Correct value for d y A Aliter y = cost = (cos t ) 4. (b) M Way y y y = cos t cos t = sin t = y = (Must be in the form y = f ( ) ). Aliter 4. (b) Way 3 Aliter 4. (b) Way 4 y = = sint t = sin So, y cos sin = y = t t = y t = y So, =± sin cos ( y) cos cos cos ( ) A Rearranges to make t the subject and substitutes the result into y. y cos sin = Rearranges to make t the subject and substitutes the result into y. M A oe So, y cos sin y cos sin A oe Aliter dy sint y c 4. (b) dy y c M Way 5 π π Eg: when eg: t = 0(nb: t ), Full method of finding y = = 0, y = = 0 c = 0 y = π π using a value of t: t A M Note: dy = sint = y =, with no attempt to find c is MA0.

15 Question Number 5. (a) { } = u = u or du Scheme du = or d u = B udu M ( ) u (u ) Marks = = du A * cso u(u ) [3] (b) A B A(u ) Bu u(u ) u + (u ) + See notes M A u = 0 = A A= u = = B B = 4 So 4 du = + du u(u ) u (u ) lnu + ln(u ) So, [ ] 3 = ln u + ln(u ) = ( ln 3+ ln((3) ) ) ( ln+ ln(() ) ) = ln3+ ln5 (0) 5 = ln 3 M N Integrates +, M 0, N 0 to u (u ) M obtain any one of ± λ ln u or ± µ ln(u ) At least one term correctly followed through A ft lnu + ln(u ). A cao Applies limits of 3 and in u or 9 and in in their integrated function and subtracts the correct way round. 5 ln 3 M A cso cao Notes for Question 5 (a) B: u du du = or d u = or du = M: A full substitution producing an integral in u only (including the du ) (Integral sign not necessary). (b) The candidate needs to deal with the, the ( ) and the and converts from an integral term in to an integral in u. (Remember the integral sign is not necessary for M). A*: leading to the result printed on the question paper (including the du ). (Integral sign is needed). M: Writing A B + or writing P Q + and a complete method for u(u ) u (u ) u(u ) u (u ) finding the value of at least one of their A or their B (or their P or their Q). A: Both their A = and their B = 4. (Or their P = and their Q = with the multiplying factor of in front of the integral sign). M N M: Integrates +, M 0, N 0 (i.e. a two term partial fraction) to obtain any one of u (u ) ± λ ln u or µ ln(u ) PhysicsAndMathsTutor.com ± or ± µ ln ( u ) Aft: At least one term correctly followed through from their A or from their B (or their P and their Q). A: lnu + ln(u ) Notes for Question 5 Continued 5. (b) ctd M: Applies limits of 3 and in u or 9 and in in their (i.e. any) changed function and subtracts the [7] 0

16 correct way round. Note: If a candidate just writes ( ln3+ ln((3) ) ) oe, this is ok for M. 5 A: ln correct answer only. (Note: 5, 3 3 a = b = ). Important note: Award M0A0MAA0 for a candidate who writes du du lnu ln(u ) u(u ) = u + (u ) = + AS EVIDENCE OF WRITING AS PARTIAL FRACTIONS IS GIVEN. u(u ) Important note: Award M0A0M0A0A0 for a candidate who writes down either du = lnu + ln(u ) or du = lnu + ln(u ) u(u ) u(u ) WITHOUT ANY EVIDENCE OF WRITING as partial fractions. u(u ) Important note: Award MAMAA for a candidate who writes down du = lnu + ln(u ) u(u ) WITHOUT ANY EVIDENCE OF WRITING as partial fractions. u(u ) Note: In part (b) if they lose the and find du we can allow a maimum of u(u ) MA0 MAftA0 MA0.

17 Question Number dθ = λ 0 θ, θ 00 dθ = λ 0 θ 6. ( ) (a) or Scheme dθ = B λ( 0 θ) Marks M A; ln ( 0 θ) ; = λt + c or ln ( 0 θ ) ; = t + c See notes λ M A { t = 0, θ = 0 } ln ( 0 0 ) = λ(0) + c See notes M c =ln00 ln( 0 θ) = λt ln00 then either... or... λt ln 0 θ ln00 λt = ln00 ln 0 θ = ( ) ( ) 0 θ λt = ln 00 e λt 0 θ = λt = ln 0 θ λt 00 e = 0 θ ( θ ) λt 00e = 0 θ λt 0 e = 00 0 θ = 00e leading to θ = 0 00e (b) { λ = 0.0, θ = 00 } PhysicsAndMathsTutor.com e λt dddm A * 0.0t = M 0.0t 0 00 Uses correct order of operations by 00e = t = ln 0.0t 00 moving from 00 = 0 00e 0 00 to give t =... and t = Aln B, t = ln where B > 0 dm t = ln = 00ln t = = 6 (s) (nearest second) awrt 6 A [8] [3]

18 (a) Notes for Question 6 B: Separates variables as shown. dθ and should be in the correct positions, though this mark can be implied by later working. Ignore the integral signs. Either or M: dθ ± Aln ( 0 θ ) dθ ± Aln( 0 θ ), A is a constant. 0 θ λ(0 θ) A: dθ ln( 0 θ ) ( ) ( ) 0 θ d θ ln 0 θ (0 ) or ln 0 λ λθ λ θ λ λ, M: λ λt t A: λ λt + c or t + c The + c can appear on either side of the equation. IMPORTANT: + c can be on either side of their equation for the nd A mark. M: Substitutes t = 0 AND θ = 0 in an integrated or changed equation containing c (or A or ln A ). Note that this mark can be implied by the correct value of c. { Note that ln00 = }. dddm: Uses their value of c which must be a ln term, and uses fully correct method to eliminate their logarithms. Note: This mark is dependent on all three previous method marks being awarded. A*: This is a given answer. All previous marks must have been scored and there must not be any errors in the candidate s working. Do not accept huge leaps in working at the end. So a minimum of either: λt 0 θ (): e λt λt = 00e = 0 θ θ = 0 00e 00 λt 00 λt λt λt or (): e = ( 0 θ ) e = 00 0 θ = 00e θ = 0 00e 0 θ is required for A. Note: dθ ln( 0λ λθ) is ok for the first MA in part (a). (0 λ λθ) λ (b) M: Substitutes λ = 0.0and θ = 00 into the printed equation or one of their earlier equations connecting θ and t. This mark can be implied by subsequent working. 0.0t dm: Candidate uses correct order of operations by moving from 00 = 0 00e to t =... Note: that the nd Method mark is dependent on the st Method mark being awarded in part (b). A: awrt 6 or awrt minutes 4 seconds. (Ignore incorrect units). Aliter 6. (a) dθ = λ Way 0 θ B ln ( 0 θ) = λt + c See notes M A; M A ln ( 0 θ) = λt + c ln ( 0 θ) = λt + c 0 θ = Ae λt θ = 0 Ae λt 0 { t = 0, θ = 0 } 0 = 0 Ae M A = 0 0 = 00 So, θ = 0 00e dddm A * [8]

19 (a) Aliter 6. (a) Way 3 Notes for Question 6 Continued BMAMA: Mark as in the original scheme. M: Substitutes t = 0 AND θ = 0 in an integrated equation containing their constant of integration which could be c or A. Note that this mark can be implied by the correct value of c or A. dddm: Uses a fully correct method to eliminate their logarithms and writes down an equation containing their evaluated constant of integration. Note: This mark is dependent on all three previous method marks being awarded. λt Note: ln ( 0 θ) = λt + c leading to 0 θ = e c λt + e or 0 θ = e + A, would be dddm0. A*: Same as the original scheme. Note: The jump from ln ( 0 θ) = λt + c to 0 θ = Ae λt with no incorrect working is condoned in part (a). dθ = λ dθ = λ 0 θ θ 0 B ln θ 0 = λt + c Modulus required M A for st A. M A { t = 0, θ = 0 } ln 0 0 = λ(0) + c Modulus not required here! M c = ln00 ln θ 0 = λt ln00 then either... or... λt = ln θ 0 ln00 λt = ln00 ln θ 0 θ 0 λt = ln 00 0 θ λt = ln 00 e λt 0 θ = λt = ln θ 0 As θ λt = ln 0 θ λt 00 e = 0 θ ( θ ) Understanding of modulus is required here! dddm λt 0 e = 00 λt 00e = 0 θ λt 0 θ = 00e A * leading to θ = 0 00e λt [8] B: Mark as in the original scheme. M: Mark as in the original scheme ignoring the modulus. A: dθ lnθ 0. (The modulus is required here). 0 θ MA: Mark as in the original scheme. M: Substitutes t = 0 AND θ = 0 in an integrated equation containing their constant of integration which could be c or A. Mark as in the original scheme ignoring the modulus. dddm: Mark as in the original scheme AND the candidate must demonstrate that they have converted ln 0 ln 0 θ in their working. Note: This mark is dependent on all three previous method θ to ( ) marks being awarded. A: Mark as in the original scheme.

20 Aliter 6. (a) Way 4 Use of an integrating factor Notes for Question 6 Continued dθ dθ = λ ( 0 θ) + λθ = 0λ IF = e B d ( e λt ) 0 e λt θ = λ, MA λt λt e θ = 0λe + k MA θ = 0 + Ke λt M { t = 0, θ = 0 } 00 = K θ = 0 00e MA

21 Question Number Scheme Marks y + y + 7 = 0 (a) dy dy dy = + 4y y = 0 M A B dy + 4 y + (4 + y) = 0 dm dy 4y y = = 4+ y + y A cso oe [5] (b) 4 + y = 0 M y = = y A + 4 ( ) + ( ) + 7 = 0 y + 4 y y + y + 7 = 0 M* = y + = = 9 y = 36 dm* = 3 y = 6 A (a) When = 3, y =( 3) When y = 6, = (6) ddm* y = 6 = 3 A cso Notes for Question 7 dy dy dy M: Differentiates implicitly to include either 4 or ± ky. (Ignore = ). dy and... + y + 7 = 0 + y = 0 d. Note: If an etra term appears then award A0. Note: The " = 0" can be implied by rearrangement of their equation. A: ( ) ( ) dy dy dy dy i.e.: + 4y y leading to 4 + y = 4y will get A (implied). dy dy B: 4y + 4 or 4 y + or equivalent d dm: An attempt to factorise out d y as long as there are at least two terms in d y. dy dy ie (4 + y) =... or... + ( + y) =... Note: This mark is dependent on the previous method mark being awarded. 4y + + 4y ( + y) y A: For or equivalent. Eg: or or 4 + y 4 y 4 + y + y cso: If the candidate s solution is not completely correct, then do not give this mark. [7]

22 (b) Notes for Question 7 Continued M: Sets the denominator of their d y equal to zero (or the numerator of their d equal to zero) oe. dy A: Rearranges to give either y = or = y. (correct solution only). The first two marks can be implied from later working, i.e. for a correct substitution of either y = into y or for = y into 4y. M*: Substitutes y = ± λ or or = ± µ y or y = ± λ ± a or = ± µ y ± b ( λ 0, µ 0) into + 4y + y + 7 = 0 to form an equation in one variable. dm*: leading to at least either = A, A > 0 or y = B, B > 0 Note: This mark is dependent on the previous method mark (M*) being awarded. A: For = 3 (ignore = 3) or if y was found first, y = 6 (ignore y = 6) (correct solution only). ddm* Substitutes their value of into y = ± λ to give y = value or substitutes their value of into + 4y + y + 7 = 0 to give y = value. Alternatively, substitutes their value of y into = ± µ y to give = value or substitutes their value of y into + 4y + y + 7 = 0 to give = value Note: This mark is dependent on the two previous method marks (M* and dm*) being awarded. A: ( 3, 6) cso. Note: If a candidate offers two sets of coordinates without either rejecting the incorrect set or accepting the correct set then award A0. DO NOT APPLY ISW ON THIS OCCASION. Note: = 3 followed later in working by y = 6 is fine for A. Note: y = 6 followed later in working by = 3 is fine for A. Note: = 3,3 followed later in working by y = 6 is A0, unless candidate indicates that they are rejecting = 3 Note: Candidates who set the numerator of d y equal to 0 (or the denominator of their d equal to zero) can dy only achieve a maimum of 3 marks in this part. They can only achieve the nd, 3 rd and 4 th Method marks to give a maimum marking profile of M0A0MMA0MA0. They will usually find ( 6, 3) { or even (6, 3) }. Note: Candidates who set the numerator or the denominator of d y equal to ± k (usually k = ) can only achieve a maimum of 3 marks in this part. They can only achieve the nd, 3 rd and 4 th Method marks to give a marking profile of M0A0MMA0MA0. Special Case: It is possible for a candidate who does not achieve full marks in part (a), (but has a correct denominator for d y ) to gain all 7 marks in part (b). Eg: An incorrect part (a) answer of d y 4 y = can lead to a correct ( 3, 6) in part (b) and 7 marks. 4+ y

23 Question Number 8. (a) { } Scheme 3 p l : r = 8 + λ, A( 3,, 6), OP = 0 p 3 p p 3 Finds the difference PA = 0 { AP } = 0 between OA and OP. 6 p p 6 Ignore labelling. M 3 + p 3 p = = 6 p p 6 Correct difference. A 3+ p = 6 + p p = 0 6 p See notes. M p = A cso (b) AP = 4 + ( ) + 4 or AP = ( 4) + + ( 4) See notes. M So, PAor AP = 36 or 6 cao A cao It follows that, AB "6" { PA } = = or PB "6 " { PA } = = See notes. B ft {Note that AB = "6" = (the modulus of the direction vector of l) } 3 OB = ± or 6 Uses a correct method in order to find both possible sets of M 3 3 coordinates of B. OB = 8 3 and OB = = and 6 Both coordinates are correct. A cao 4 8 Marks Notes for Question 8 8. (a) M: Finds the difference between OA and OP. Ignore labelling. If no subtraction seen, you can award M for out of 3 correct components of the difference. 3 + p 3 p A: Accept any of or (3 + p) i j+ (6 p) k or or ( 3 p) i + j+ ( p 6) k 6 p p 6 [4] [5] 9

24 8. (a) (b) Notes for Question 8 Continued M: Applies the formula PA or AP correctly to give a linear equation in p which is set equal to zero. Note: The dot product can also be with ± k. Eg: Some candidates may find = 0, for instance, and use this in their dot product which is fine for M. 6 5 A: Finds p = from a correct solution only. Note: The direction of subtraction is not important in part (a). M: Uses their value of p and Pythagoras to obtain a numerical epression for either AP or PA or AP or PA. Eg: PA or AP = 4 + ( ) + 4 or ( 4) + + ( 4) or or PA or AP = 4 + ( ) + 4 or ( 4) + + ( 4) or A: AP or PA = 36 or 6 cao or AP = 36 cao Bft: States or it is clear from their working that AB "6" { their evaluated PA } { PA } PB = "6" = (their evaluated ). = = or Note: So a correct follow length is required here for either AB or PB using their evaluated PA. Note: This mark may be found on a diagram. Note: If a candidate states that AP = AB and then goes on to find AP = 6 then the B mark can be implied. IMPORTANT: This mark may be implied as part of epressions such as: { AB } { AB } { PB } { PB } (0 ) (0 ) ( 5 ) 6 or (0 ) (0 ) ( 5 ) 36 = + λ + + λ + λ = = + λ + + λ + λ = or = (4 + λ) + (8 + λ) + ( λ) = 6 or = (4 + λ) + (8 + λ) + ( λ) = 7 M: Uses a full method in order to find both possible sets of coordinates of B: Eg : OB = ± Eg : OB = 8 3 and OB = Note: If a candidate achieves at least one of the correct (7,, 4) or (, 6, 8) then award SC M here. 3 3 Note: OB = 3 and OB = 7 is M A: For both (7,, 4) and (, 6, 8). Accept vector notation or i, j, k notation. Note: All the marks are accessible in part (b) if p = is found from incorrect working in part (a). Note: Imply MAB and award M for candidates who write: OB earlier working. 3 = ±, with little or no 6

25 8. Helpful Diagram! Notes for Question 8 Continued AB = λ + λ λ AB = 0+ λ 5 λ λ = 5 λ = λ B 8 + λ λ l λ = 7 3 A λ B 8 + λ λ PA = λ PB = 8+ λ λ PB = 9λ + 90λ p P 0 = 0 p 8. (b) Way : Setting AB { } = "6" or AB = "36" Note: It is possible for you to apply the main scheme for Way. AB = "6" AB = "36" (0 + λ) + (0 + λ) + ( 5 λ) = "36" Bft could be implied here. 8. (b) Way 3: Setting PB { } 9λ + 90λ + 5 = 36 9λ + 90λ + 89 = = 0 ( + 3)( + 7) = 0 λ = 3, 7 Then apply final M A as in the original scheme. λ λ λ λ... M A = "6 " or PB = "7" Note: It is possible for you to apply the main scheme for Way 3. PB PB λ λ λ = "6" = "7" (4 + ) + (8 + ) + ( ) = "7" Bft could be implied here. 9λ + 90λ + 6= 7 9λ + 90λ + 89 = = 0 ( + 3)( + 7) = 0 λ = 3, 7 Then apply final M A as in the original scheme. λ λ λ λ... M A

26 8. (b) 8. (b) Notes for Question 8 Continued (You need to be convinced that a candidate is applying this method before you apply the Mark Scheme for Way 4). Way 4: Using the dot product formula between PA and PB, PA PB ie: cos 45 =. PA. PB λ PA PB = 8+ λ = λ 6 4λ 4 4λ = 36 4 λ 36 { cos 45 = } = 6 9λ + 90λ = 9λ + 90λ + 6 9λ + 90λ + 6= 7 9λ + 90λ + 89 = 0 λ λ λ λ = 0 ( + 3)( + 7) = 0 λ =3, 7 For finding PA as before. M 36 or 6 A cao PB = 9λ + 90λ + 6 B oe Then apply final M A as in the original scheme.... M A (You need to be convinced that a candidate is applying this method before you apply the Mark Scheme for Way 5). Way 5: Using the dot product formula between AB and PB, AB PB ie: cos 45 = AB. PB cos 45 = = 0 + λ 4 + λ 0+ λ 8+ λ 5 λ λ 9λ + 90λ + 5 9λ + 90λ + 6 Attempts the dot product formula between AB and PB. Correct statement with AB and PB simplified as shown. Either AB = 9λ + 90λ + 5 or PB = 9λ + 90λ + 6 M A B { cos 45 } { cos 45 } λ + 8λ + 4λ λ + 6λ + 4λ λ + λ + λ = = 9λ + 90λ + 5 9λ + 90λ + 6 9λ 90λ = = 9λ + 90λ + 5 9λ + 90λ + 6 (9λ + 90λ + 5) = λ + λ + λ + λ + (9 90 5)(9 90 6) (9λ + 90λ + 5) = (9 90 6) λ + λ + 9λ + 90λ + 6 = (9λ + 90λ + 5) 9λ + 90λ + 89 = 0 λ λ λ λ = 0 ( + 3)( + 7) = 0 λ =3, 7 Then apply final M A as in the original scheme.... M A

27 8. (b) Way 6: 4 PA = = 4 So, PA = d Notes for Question 8 Continued and direction vector of l is d = or PA = d A correct statement relating these distances (and not vectors) M A B Note: PA = d Apply final M A as in the original scheme.... M A with no other creditable working is M0A0B0... Note: PA = d, followed by OB correct coordinates. 3 = ± is MABM and the final A mark is for both sets of 6

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