Mark Scheme (Results) Summer GCE Core Mathematics 4 (6666/01R)

Transcription

1 Mark Scheme (Results) Summer 0 GCE Core Mathematics 4 (6666/0R)

2 Edexcel and BTEC Qualifications Edexcel 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 Expert 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 0 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 0 Publications Code UA0679 All the material in this publication is copyright Pearson Education Ltd 0

3 General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. 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 7.. The Edexcel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes: bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for example, as 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.. 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 term quadratic:. Factorisation ( x + bx+ c) = ( x+ p)( x+ q), where pq = c, leading to x = ( ax + bx + c) = ( mx + p)( nx + q), where pq = c and mn = a, leading to x =. Formula Attempt to use correct formula (with values for a, b and c).. Completing the square b Solving 0 :, 0, leading to x =... x + bx+ c= x± ± q± c q Method marks for differentiation and integration:. Differentiation n n Power of at least one term decreased by. ( x x ). Integration Power of at least one term increased by. ( n n x x + ) Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners 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. Exact answers Examiners reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals. 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. Scheme x + A B C + + (x + )( x + ) (x + ) ( x + ) ( x + ) A =, C = At least one of A or C are correct. Breaks up their partial fraction correctly into three terms and both " A " = and " C " =. B Marks B cso x + A( x + ) + B(x + )( x + ) + C(x + ) x = =C C = x = + = A = A A= 4 4 Writes down a correct identity and attempts to find the value of either one A or B or C. M Either x : 0 = A + B, constant : = A + B + C x := A + B + C leading to B = x + So, + (x + )( x + ) (x + ) ( x + ) ( x + ) Correct value for B which is found using a correct identity and follows from their partial fraction decomposition. A cso [4] 4 Notes for Question BE CAREFUL! Candidates will assign their own A, B and C for this question. B: At least one of A or C are correct. B: Breaks up their partial fraction correctly into three terms and both " A " = and " C " =. M: Writes down a correct identity (although this can be implied) and attempts to find the value of either one of A or B or C. This can be achieved by either substituting values into their identity or comparing coefficients and solving the resulting equations simultaneously. A: Correct value for B which is found using a correct identity and follows from their partial fraction decomposition. Note: If a candidate does not give partial fraction decomposition then: the nd B mark can follow from a correct identity. the final A mark can be awarded for a correct B if a candidate goes writes out their partial fractions at the end. Note: The correct partial fraction from no working scores BBMA. Note: A number of candidates will start this problem by writing out the correct identity and then attempt to find A or B or C. Therefore the B marks can be awarded from this method.

7 Question Number. Scheme x + xy y + = 0 x x ln Marks B oe Differentiates implicitly to include either M* x ± λ x or ± ky. = ln+ y + x y = 0 d (ignore) y + y + x B d d d... y y y x y 0 A {(, ) } Substitutes x =, y = into their ) ln + + () () = 0 differentiated equation or expression. dm* ln = 0 + ln = + ln = dm* ( ln e ln ) ln ( e ) = lne to achieve ln ( e ) A cso [7] 7 Notes for Question B: Correct differentiation of x x x x. I.e. ln or x x = ( ) ( ) ln x ( x )ln ( x )ln x or = e lne or x x ( ) e ln x = = (ln)e ln M: Differentiates implicitly to include either ± λ x or ± ky. (Ignore = ). d B: xy y + y + x d x st d d A:... + y y y + x y 0 = Note: The st A0 follows from an award of the nd B0. Note: The " = 0" can be implied by rearrangement of their equation. x x ie: ln + y + x y leading to ln + y = y x will get A (implied). nd M: Note: This method mark is dependent upon the st M* mark being awarded. Substitutes x =, y = into their differentiated equation or expression. Allow one slip. rd M: Note: This method mark is dependent upon the st M* mark being awarded. Candidate has two differentiated terms in d y and rearranges to make d y the subject. Note: It is possible to gain the rd M mark before the nd M mark. x y + ln Eg: Candidate may write = before substituting in x = and y = y x nd A: cso. Uses Note: = to achieve ( ) =, ( ) lne ln e = lne needs to be seen in their proof. = ln µ e, λ = and µ = λ Notes for Question Continued

8 . Alternative Method: Multiplying both sides by x + xy y + = 0 Aliter Way x + xy y + = 0 x x ln B Differentiates implicitly to include either x = ln+ y + x M* ± λx or ± ky. 6y = 0 (ignore) xy + y + x B d d... + y y y + x 6y 0 = A d d {(, ) } ln + () + y y ()() 6() 0 = Substitutes x =, y = into their dm* differentiated equation or expression. ln = ln = 9+ ln + ln = = dm* ( ln e ln ) = + ( ln e ln ) ln ( e ) = + = Uses = lne to achieve ln ( e ) = A cso [7] 7 NOTE: Only apply this scheme if the candidate has multiplied both sides of their equation by. NOTE: For reference, d x y y + = ln 6y x NOTE: If the candidate applies this method then xy + y + x must be seen for the nd B mark.

9 Question Number. 0 4 Scheme d x, u = + (x + ) + (x + ) du Either = ± K(x + ) or = ± λ( u ) M du = (x + ) or u du du du Either = (x + ) or = ( u ) A du Correct substitution A d x = ( u ) du + (x + ) u (Ignore integral sign and d u ). = du u An attempt to divide each term by u. dm ± Au ± B ln u ddm = u lnu u lnu A ft { So [ u ln u] } = ( ln) ( ln) = + ln Notes for Question Applies limits of and in u or 4 and 0 in x in their integrated function and subtracts the correct way round. + ln M Marks A cao cso [8] 8 M: Also allow du =±λ or ( u )du =± λ ( u ) Note: The expressions must contain du and. They can be simplified or un-simplified. A: Also allow du = or ( u )du =± λ ( u ) Note: The expressions must contain du and. They can be simplified or un-simplified. A: ( u ) du. (Ignore integral sign and d u ). u dm: An attempt to divide each term by u. Note that this mark is dependent on the previous M mark being awarded. Note that this mark can be implied by later working. ddm: ± Au ± B ln u, A 0, B 0 Note that this mark is dependent on the two previous M marks being awarded. Aft: u lnu or ± Au ± B ln u being correctly followed through, A 0, B 0 M: Applies limits of and in u or 4 and 0 in x in their integrated function and subtracts the correct way round. A: cso and cao. + ln or + ln ( 0.6 ), = A + ln B, so A=, B = Note: ln is A0.

10 . ctd Notes for Question Continued Note: ( u ) du = u lnu with no working is nd M, rd M, rd A. u but Note: ( u ) d u = ( u )lnu with no working is nd M0, rd M0, rd A0. u

11 Question Number 4. (a) { } ( ) Scheme (8 9 x) = 8 9x Power of M Marks ( ) 9x 9x = 8 = 8 or B 8 8 ( )( ) ( )( )( ) = {} + ( kx) + ( kx) + ( kx) +...!! see notes M A 9x ()( ) 9x ()( )( ) 9x = {} ! 8! = x; x x See notes below! 9 4 = x; x x A; A [6] 700 = 0 7 = 0 (8 9 ), so x = 0. Writes down or uses x = 0. B (b) { x } 9 When x = 0., 4 (8 9 x) (0.) (0.) (0.) +... M 4 6 = =.9079 So, 700 = = 9.0 (4 dp) 9.0 cso A cao [] 9 Notes for Question 4 (a) M: Writes or uses. This mark can be implied by a constant term of 8 or. B: ( 8) or outside brackets or as candidate s constant term in their binomial expansion. M: Expands (... + kx) to give any terms out of 4 terms simplified or un-simplified, Eg: + ( ) kx ()( ) ()( )( ) ()( ) or ( kx) + ( kx) or ( kx)!!! ()( ) ()( )( ) or ( kx) + ( kx) where k are fine for M.!! ()( ) ()( )( ) A: A correct simplified or un-simplified + ( kx) + ( kx) + ( kx)!! expansion with consistent ( kx ). Note that ( kx ) must be consistent (on the RHS, not necessarily the LHS) in a candidate s expansion. Note that k. You would award BMA0 for ( x) because ( kx ) is not consistent. ( ) 9x ()( ) ()( )( ) 9x !! 8

12 4. (a) ctd Notes for Question 4 Continued 9x ()( ) 9x ()( )( ) 9x Incorrect bracketing = {} ! 8! 8 is MA0 unless recovered. A: For 4 x (simplified please) or also allow 0.7 x. Allow Special Case AA0 for either SC: ;... 8 x = or SC: 9 4 K x x x either a simplified fraction or a decimal. (where K can be or omitted), with each term in the [ ] 9 4 A: Accept only x x or 0.8x 0.778x 6 9x ()( ) 9x ()( )( ) 9x Candidates who write = where 8! 8! and not and achieve + x; x + x +... will get BMAA0A Note for final two marks: x; x x = + x x x +... scores final A0A x; x x = x x +... scores final A0A k = 8 Alternative method: Candidates can apply an alternative form of the binomial expansion. 8 ()( ) ()( )( ) { (8 9 x) } = ( 8 9 x) = (8) + (8) ( 9 x) + (8) ( 9 x) + (8) ( 9 x)!! B: ( 8) or M: Any two of four (un-simplified or simplified) terms correct. A: All four (un-simplified or simplified) terms correct. A: 4 x 9 4 A: x x 6 Note: The terms in C need to be evaluated, so 8 C0(8) C(8) ( 9 x) C(8) ( 9 x) C(8) ( 9 x) without further working is B0M0A0. (b) B: Writes down or uses x = 0. M: Substitutes their x, where A: 9.0 cao Be Careful! The binomial answer is x < into at least two terms of their binomial expansion. 9 and the calculated 700 is which is 9.00 to 4 decimal places.

13 Question Number Scheme Marks. (a) = 6.48 (dp) 6.48 or awrt 6.48 B [] (b) Area ; + ( their 6.48) +. B; M = = 49.7 ( dp) 49.7 or awrt 49.7 A [] (c) (d) (4 e t ) d e t t + t = t e t { } + t t t { t} = te 6e + 8 e t t 6e + t = 0 (8) (8) (0) (0) (8) e 6e (8) (0) e 6e = + + (0) 8 8 = 96e 6e e = Difference 8 ( ) t t { } ± At e ± B e d t, A 0, B 0 See notes. t t t M A B t e 6e A Substitutes limits of 8 and 0 into an integrated function of the form of either t t ± λ te ± µ e or t t ± λ te ± µ e + Bt and subtracts the correct way round e = 60 e 49.7 = =.46 ( dp).46 or awrt.46 B dm A [6] (a) Notes for Question B: 6.48 or awrt Look for this on the table or in the candidate s working. (b) B: Outside brackets or M: For structure of trapezium rule[... ]. Allow one miscopy of their values. A: 49.7 or anything that rounds to 49.7 Note: It can be possible to award : (a) B0 (b) BMA (awrt 49.7) Note: Working must be seen to demonstrate the use of the trapezium rule. Note: actual area is 0.88 Bracketing mistake: Unless the final answer implies that the calculation has been done correctly, their (nb: answer of 0.69). Award BM0A0 for ( ) []

14 Notes for Question Continued. (b) ctd Alternative method for part (b): Adding individual trapezia Area = B: and a divisor of on all terms inside brackets. M: First and last ordinates once and two of the middle ordinates twice inside brackets ignoring the. A: anything that rounds to 49.7 (c) M: For t t t { } 4e t ± Ate ± B e d t, A 0, B 0 A: For e t t te t e t (some candidates lose the 4 and this is fine for the first A mark). or 4e t t 4 e t t e t t t t t or t e e or t e e These results can be implied. They can be simplified or un-simplified. B: t or x (bod). Note: Award B0 for integrating to t (implied), which is a common error when taking out a factor of 4. Be careful some candidates will factorise out 4 and have t + + which would then be fine for B. 8 Note: Allow B for d t = 4 A: For correct integration of 0 4e t t to give This can be simplified or un-simplified. te 6e or 4 te 9e t t t t dm: Substitutes limits of 8 and 0 into an integrated function of the form of either t t or equivalent. t t ± λ te ± µ e or ± λ te ± µ e + Bt and subtracts the correct way round. Note: Evidence of a proper consideration of the limit of 0 (as detailed in the scheme) is needed for dm. So, just subtracting zero is M0. 8 A: An exact answer of 60 e. A decimal answer of without a correct answer is A0. Note: A decimal answer of without a correct exact answer is A0. Note: If a candidate gains MABA and then writes down 0.8 or awrt 0.8 with no method for substituting limits of 8 and 0, then award the final MA0. IMPORTANT: that is fine for candidates to work in terms of x rather than t in part (c). Note: The " t " is needed for B and the final A mark. (d) B:.46 or awrt.46 or -.46 or awrt Candidates may give correct decimal answers of or Note: You can award this mark whether or not the candidate has answered part (c) correctly.

15 Question Number 6. (a) Scheme a 6 l: r = b + λ c, OA = 7, 0 6 a 6 A is on l, so 7 = b + λ c 6 0 OB = 4 8 Marks (b) { AB } { λ } k :0 = 6 λ = 4 λ = 4 : + 6 = a + 6(4) = Substitutes their value of λ into a + 6λ = a = a = { i a λ } Finds the difference = 4 7 { BA } = 7 4 between OA and OB Ignore labelling. 4 4 AB = { BA } = 4 6 AB l AB d = 0 c = 4 + c = 0; c = 4 See notes. { } { } { λ } j : b + c = 7 b + ( 4)(4) = 7; b = See notes. (c) AB = or AB = ( 4) + ( ) + ( ) See notes. M B M A cao M M; A ft [] ddm; A cso cao [] So, AB = A cao 4 7 See notes for = + = 7 + ; = 0 alternative 6 6 methods. (d) OB { OA BA} Notes for Question 6 (a) B: λ = 4 seen or implied. M: Substitutes their value of λ into a + 6λ = A: a =. Note: Award BMA if the candidate states a = from no working. Alternative Method Using Simultaneous equations for part (a). B: For 60 6λ = 6 M: 60 6λ = 6 and a + 6λ = solved simultaneously to give a =... A: a =, cao. M;A cao [] []

16 6. (b) ctd Notes for Question 6 Continued M: Finds the difference between OA and OB. Ignore labelling. If no subtraction seen, you can award M for out of correct components of the difference. 6 6 M: Applies the formula AB c or BA c correctly to give a linear equation in c which is set equal 6 to zero. Note: The dot product can also be with ± k c. Aft: c = 4 or for finding a correct follow through c. ddm: Substitutes their value of λ and their value of c into b + cλ = 7 Note that this mark is dependent on the two previous method marks being awarded. A: b = M: An attempt to apply a three term Pythagoras in order to find AB, (c) so taking the square root is required here. A: cao Note: Don t recover work for part (b) in part (c). (d) M: For a full applied method of finding the coordinates of B. Note: You can give M for out of correct components of B. 7 A: For either 0 or 7i 0j 6k or ( 7, 0, 6) cao. 6 Helpful diagram! B 4 8 A BA = p B q r 4 BA = l

17 Notes for Question 6 Continued Acceptable Methods for the Method mark in part (d) Way 4 OB { = OA + BA} = (using their BA ) Way 4 OB { = OA AB} = 7 6 (using their AB ) Way 4 OB { = OB + BA} = (using their BA ) Way 4 4 OB { = OB AB} = 4 8 (using their AB ) Way Minus 4 Minus Minus 7 Minus 0, so OA + their BA 8 Minus 6 Minus 6 Way 6 OB { = OA OB} = Way 7 OB = i 4j+ 8 k, OA = i 7j+ 6k and OB = pi + qj+ rk, + p 4 + q 8 + r, 7, 6 =,, ( ) p = () = 7 q = 7() + 4 =0 r = 6() 8 = 6 M: Writing down any two equations correctly and an attempt to find at least two of p, q or r.

18 Question Number 7. (a) Scheme π = = x 7sec t, y tan t, 0 t d 8sec tsecttan t sec t = t =, sec t cost cos t = = = = 8sec t tan t 7sect tan t 7 tan t 7sin t π π sec ( 6 ) 4 At t =, = = π π = = 6 8sec tan (b) { t t} ( ) ( ) = + = = tan sec y x x 7 7 ( ) At least one of d x Both d x Applies their d y d or d y y correct. and d are correct. t divided by their d x 4 7 B B M; Marks A cao cso y x + = 9 + y = x y = x 9 * A * cso 9 9 a = 7 and b = 6 or 7 x 6 a = 7 and b = 6 B [] 7 7 (c) ( ) ( ) = π π V x 9 or x 9 = { π } 9 x x 7 = { π} () 9() (7) 9(7) { π} (( 87 ) ( 4.8 4) ) = 46π = or 847.π (a) B: At least one of d x B: Both d x and d y M: Applies their d y d 4 A: 7 or d y Notes for Question 7 For π ( x ) 9 or π ( x ) 9 Ignore limits and. Can be implied. Either ± Ax ± Bx or M B x oe M 9 x x oe A Substitutes limits of and 7 into an integrated function and subtracts the dm correct way round. 46π or 847.π correct. Note: that this mark can be implied from their working. are correct. Note: that this mark can be implied from their working. t divided by their d x, where both d y and d x or any equivalent correct rational answer not involving surds. Allow 0.0 with the recurring symbol. are trigonometric functions of t. A [4] []

19 (b) (c) (a) Way Notes for Question 7 Continued Note: Please check that their d x is differentiated correctly. Eg. Note that x = 7sec t = 7( cost) = 8( cos t) ( sin t) is correct. M: Either: Applying a correct trigonometric identity (usually + tan t = sec t ) to give a Cartesian equation in x and y only. Starting from the RHS and goes on to achieve 9tan t by using a correct trigonometric identity. Starts from the LHS and goes on to achieve 9sec t 9 by using a correct trigonometric identity. =. Note this result is printed on the Question Paper, so no incorrect working is allowed. B: Both a = 7 and b = 6. Note that 7 x 6 is also fine for B. A*: For a correct proof of y ( x ) 9 B: For a correct statement of π ( x ) 9 or π ( x ) 9. Ignore limits and. Can be implied. M: Either integrates to give ± Ax ± Bx, A 0, B 0 or integrates x correctly to give x oe A: 9 x x or. x 9x oe. dm: Substitutes limits of and 7 into an integrated function and subtracts the correct way round. Note: that this mark is dependent upon the previous method mark being awarded. A: A correct exact answer of 46 π or 847. π. Note: The π in the volume formula is only required for the B mark and the final A mark. Note: A decimal answer of without a correct exact answer is A0. Note: If a candidate gains the first BMA and then writes down 66 or awrt 66 with no method for substituting limits of and 7, then award the final MA0. Alternative response using the Cartesian equation in part (a) d y y = x 9 = x 9 x π At t =, 6 π x = = 6 7sec 4 = ( 4 ) 9 ( 4 ) ( x 9) =± Kx = ( x 9) x oe π Uses t = to find x and substitutes 6 their x into an expression for d y. So, = = 8 Note: Way is marked as M A dm A Note: For way the second M mark is dependent on the first M being gained. 8 M A dm A cao cso

20 Notes for Question 7 Continued 7. (b) Alternative responses for MA in part (b): STARTING FROM THE RHS For applying + tan t = sec t oe RHS = x 9 = 7sec t 9 = 9sec t 9 = 9tan t to achieve 9tan t Way { }( ) ( ) = tant = y{ = LHS} cso Correct proof from ( ) 9 M x to y. A* M: Starts from the RHS and goes on to achieve 9tan t by using a correct trigonometric identity. 7. (b) Alternative responses for MA in part (b): STARTING FROM THE LHS Way For applying + tan t = sec t oe { LHS = } y = tan t = ( 9tan t) = 9sec t 9 to achieve 9sec t 9 ( x ) x x = 9 9 = 9 9 = cso Correct proof from y to ( ) 9 M x. A* M: Starts from the LHS and goes on to achieve 9sec t 9 by using a correct trigonometric identity. 7. (c) Alternative response for part (c) using parametric integration Way V = π π 9 tan t( 8sec t sect tan t) tan t ( 8sec tsecttan t) B Ignore limits and. Can be implied. = { π} 79sec ttan tsecttan t { π} ( ) = 79sec t sec t secttan t 4 { π} ( ) = 79 sec t sec t secttan t 4 { π} ( ) = 79 sec t sec t secttan t = π { } 79 sec t sec { } t V = π = 79π 4 46π = or 847.π ± Asec t ± Bsec t M 79 sec t sec t Substitutes sect = and sect = into an integrated function and subtracts the correct way round. 46π or 847.π A dm A []

21 Question Number 8. k( M x) (a) Scheme =, where M is a constant is the rate of increase of the mass of waste products. Any one correct explanation. B Marks (b) (c) M is the total mass of unburned fuel and waste fuel (or the initial mass of unburned fuel) Both explanations are correct. = k or = M x km ( x) B ln ( M x) = kt { + c} or ln ( M x ) = t { + c } k See notes M A t 0, x 0 ln M 0 = k(0) + c See notes M { = = } ( ) ( ) c =ln M ln M x = kt ln M then either... kt ln M x ln M or... kt = ln M ln M x = ( ) ( ) kt ln M x M = kt = ln M M x e kt M x M = e kt = M M x Me kt = M x ( M x)e kt = M M x = Me kt leading to x = M Me kt kt or x = M( e ) oe B ddm A * cso k ln 4 x = M, t = ln4 M = M( e ) M kln 4 kln 4 = e e = kln4 = ln So k = A ln 9 x = M e dm x = M x = M A cso [] [6] [4]

22 Notes for Question 8 Continued 8. (a) B: At least one explanation correct. B: Both explanations are correct. is the rate of increase of the mass of waste products. or the rate of change of the mass of waste products. (b) (c) M is the total mass of unburned fuel and waste fuel or the initial mass of unburned fuel or the total mass of rocket fuel and waste fuel or the initial mass of rocket fuel or the initial mass of fuel or the total mass of waste and unburned products. 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. M: Both ± λln ( M x) or ± λln ( x M) and ± µ t where λ and µ are any constants. A: For ln ( M x) = kt or ln ( x M) = kt or ln ( M x ) t k = or ln ( x M ) = t k or ln ( km kx ) t k = or ln ( kx km ) = t k Note: + c is not needed for this mark. IMPORTANT: + c can be on either side of their equation for the st A mark. M: Substitutes t = 0 AND x = 0 in an integrated or changed equation containing c (or A or ln A, etc.) Note that this mark can be implied by the correct value of c. ddm: 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 both previous method marks being awarded. kt A: x = M Me kt kt M (e ) or x = M( e ) or x = or equivalent where x is the subject. kt e Note: Please check their working as incorrect working can lead to a correct answer. Note: = k( M x) = x = ln( km kx ) { + c} is B(Implied) MA. km kx k M: Substitutes x = M and t = ln 4 into one of their earlier equations connecting x and t. ln A: k =, which can be an un-simplified equivalent numerical value. i.e. k = is fine for A. ln 4 dm: Substitutes t = ln 4 and their evaluated k (which must be a numerical value) into one of their earlier equations connecting x and t. Note: that the nd Method mark is dependent on the st Method mark being awarded in part (c). A: x = M cso. Note: Please check their working as incorrect working can lead to a correct answer.

23 Notes for Question 8 Continued Aliter 8. (b) Way (b) = k B M x ln ( M x) = kt { + c} See notes M A ln ( M x) = kt + c M x = Ae kt { t = 0, x = 0 } M = A M 0 = Ae k (0) M M x = Me kt ddm So, x = M Me kt A [6] BMA: Mark as in the original scheme. M: Substitutes t = 0 AND x = 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. ddm: 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 both previous method marks being awarded. kt c ln M x kt c ln M x e e ln M x = e kt + A would be dddm0. Note: ( ) = + leading to ( ) = + or ( ) A: Same as the original scheme. Aliter x t 8. (b) = k Way 0 M x 0 B x t ( M x) [ kt] 0 ln = M A ( ) ( ln ) ( ) ln 0 ln M x M = kt Applies limits of M ln M x + M = kt and then follows the original scheme. (a) BMA: Mark as in the original scheme (ignoring the limits). ddm: Applies limits 0 and x on their integrated LHS and limits of 0 and t. MA: Same as the original scheme.

24 Aliter 8. (b) Way 4 Aliter 8. (b) Way Notes for Question 8 Continued = k = k M x x M B ln x M = kt + c { t 0, x 0 } = = ln 0 M = k(0) + c c = ln M ln x M = kt ln M then either... or... kt = ln x M ln M kt = ln M ln x M ln x kt = M kt = ln M As x < M M x M kt ln M x M = kt = ln M M x e kt M x M = e kt = M M x Me kt = M x ( M x)e kt = M M x = Me kt leading to x = M Me kt kt or x = M( e ) oe Modulus not required for st A. Modulus not required here! Understanding of modulus is required here! B: Mark as in the original scheme. MAM: Mark as in the original scheme ignoring the modulus. ddm: Mark as in the original scheme AND the candidate must demonstrate that they have converted ln x M to ln ( M x) in their working. Note: This mark is dependent on both the previous method marks being awarded. A: Mark as in the original scheme. Use of an integrating factor (I.F.) = k( M x) + kx = km I.F. = e kt B d ( e kt ) e kt x = km, kt kt e x = Me + c MA x = M + ce kt (0) { t 0, x 0 } 0 M ce k = = = + M c = M x = M Me kt ddma M A M ddm A * cso [6]

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