PhysicsAndMathsTutor.com. Mark Scheme (Results) January Pearson Edexcel International Advanced Level In Core Mathematics C34 (WMA02) Paper 01

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1 Mark Scheme (Results) January 07 Pearson Edecel International Advanced Level In Core Mathematics C (WMA0) Paper 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: January 07 Publications Code WMA0_0_70_MS All the material in this publication is copyright Pearson Education Ltd 07

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. When eaminers are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 EDEXCEL IAL 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. 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. 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. 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 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.

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. Formula Attempt to use correct formula (with values for a, b and c). mn a, leading to =. 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 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. Most candidates do show working, but there are occasional awkward cases and if the mark scheme does not cover this, please contact your team leader for advice.

7 Differentiate wrt dy dy d d B 6y y 0 Substitutes (, ) AND rearranges to get d y 7 d 0 7 ( y) ( ) so 7 +0y 7 = 0 0 (6) (6 marks) dy : Differentiates implicitly to include either d or y dy Accept d appearing as dy y or y d as y y dy : Differentiates y y and and 7 0 d B: Uses the product rule to differentiate y giving 6y y d term Do not penalise students who write d +6yd + d y +y d y = 0 : Substitutes, y into their epression (correctly each at least once) to find a numerical value for dy dy 6y (may be incorrect). Note that d d y : Use of ( y ) m( ) where m is their numerical value of d y d, and their m as far as c =.. Alternatively uses y m c with : Accept integer multiples of the answer i.e. 7k +0ky 7k = 0 for eample 0y 0 d y d d Note: If the gradient 7 0 just appears (from a graphical calculator) only M may be awarded

8 (a) f( ) 5 6 = 0 so awrt.65 awrt and.69 awrt (c) f (.75) , f (.65) Sign change (and as f ( ) is continuous) therefore a root lies in the interval.75,.65.7 ( dp) (a) Way : : Must state f() = 0 (or imply by writing 56 0) and reach 5 6 : completely correct with all lines including f() = 0 stated or implied (see above), and 56 oe with or without a = 5, b = 6. Isw after a correct answer If a candidate writes 5 6 () () () (7 marks) then they can score 0 for a correct but incomplete solution. Similarly if a candidate writes Way : : starts with answer, cubes and reaches a =.., b =. : Completely correct reaching equation and stating hence f() = 0 Ignore subscripts in this part, just mark as the first, second and third values given. : An attempt to substitute into their iterative formula. E.g. Sight of, or can be implied by awrt. : awrt. : awrt.65 and awrt.69 (c) : Choose suitable interval for, e.g..75,.65 and at least one attempt to evaluate f(). Evidence would be the values embedded within an epression or one value correct. A minority of candidates may choose a tighter range which should include (alpha to dp). This would be acceptable for both marks, provided the conditions for the A mark are met. Some candidates may use an adapted f ( ) 0, for eample g( ) 5 6 This is also acceptable even if it is called f, but you must see it defined. For your information g(.75) 0.00,g(.65) ( )0.00 If the candidate states an f (without defining it) it must be assumed to be f( ) 5 6 : needs (i) both evaluations correct to sf, (either rounded or truncated) (ii) sign change stated (>0, <0 acceptable as would a negative product) and (iii) some form of conclusion which may be.7 or so result shown or qed or tick or equivalent

9 Qu Scheme Marks (a) 9 A B ( )( ) A =, B = and attempt to find A or B, () ( )... ( )... 6 Attempts ''... ' ' (...) B B,... (6) (9 marks) (a) : For epression in markscheme or 9 + = A (+) + B ( ) and use of substitution or comparison of coefficients in an attempt to find A or B (Condone slips on the terms) : One correct value (this implies the ) : Both correct values (attached to the correct fraction). You do not eplicitly need to see the epression rewritten in PF form. B: Correct epansion ( )... with or without working. Must be simplified B: For taking out a factor of implied by the candidate multiplying their B by. Evidence would be seeing either or : For the form of the binomial epansion with n and a term of. before the bracket or could be... : Attempts to combine the two series epansions. Condone slips on signs but there must have been some attempt to combine terms (at least once) and to use both their coefficients : Two terms correct which may be unsimplified. : All four terms correct. (cao) Could be mied number fraction form. ISW after a correct answer To score it is sufficient to see just any two terms of the epansion. eg. Alternative use of binomial in line of scheme: ie. ( )() ( ) B: For seeing either or as the first term : It is sufficient to see just the first two terms (unsimplified) then marks as before Way : Otherwise method: Use of (9 )( ) ( ) : B B : as before Then : Attempt to multiply three brackets and obtain +.. : two terms correct : All four correct Way : Use of 9 or alternatives is less likely send to review

10 .(a) 0 < f() < 5 y = 5 (c) (d) ( 5) y 5y d y 5 f ( ) 0 5 fg( ) B o.e. () () () Uses and so deduce no real roots () (0 marks) (a) : One limit such as y 0 or y 0.8. Condone for this mark both limits but with (not y) or with the boundary included. For eample 0,0.8,0 0.8, 0 y 0.8 : Fully correct so accept 0 < f() < 5 and eact equivalents 0 < y < 0,0.8 5 : Set y = f() or f ( y) and multiply both sides by denominator. d:make (or a swapped y ) the subject of the formula. Condone arithmetic slips : o.e for eample y/f ( ) 5 5 or y - do not need domain for this mark. ISW after a correct answer. (c) Mark parts c and d together B: fg() = - allow any correct form then isw 5 (d) : Sets fg() = gf() with both sides correct (but may be unsimplified) and forms a quadratic in. Do not withhold this mark if fg or gf was originally correct but was ''simplified'' incorrectly and set equal to a correct gf : Correct TQ. It need not be all on one side of the equation. The =0 can be implied by later work : Attempts the discriminant or attempts the formula or attempts to complete the square. : Completely correct work (cso) and conclusion. If b ac has been found it must be correct ( 576 )

11 5.(a) 7 or equivalent e.g. 7 AND 8 9 or equivalent e.g. 9 8 B () (c)... B oe 7 9 h 0 0 '' '' '' '' =.9 (only) cos d = sin sin d = sin + cos (+c ) d () () sin cos = 5 (d) 5 (a) B: Both correct ( as above) Must be eact and not decimal B: See as part of trapezium rule or h form. : Correct structure of the bracket in the trapezium rule. 7 9 You may not see the zero's Eg '' '' '' '' () (9 marks) stated or used. This can be scored if h is in an unsimplified :.9 only. This may be a result of using the decimal equivalents. Sight of.9 will score all marks (c) : For a correct attempt at integration by parts to give an epression of the form sin sin d If you see such an epression you would only withhold the mark if there is evidence of an incorrect formula (seen or implied) Eg udv uv vdu d: For sin cos following line one : cso Allow all three marks for candidate who just writes down the correct answer with no working D I Watch for candidates who write down methods like this. cos sin 0 cos This is a commonly taught algorithm (They differentiate down the lh column and integrate on the rh column. The answer is found by D I D I where D is the second entry in the D column. This can score full marks for the answer sin cos but also pick up methods for slips. If they attempt D I D I it is M0 as they are implying an incorrect formula. Ask your TL if unclear. (d) : Using limits 5 and correctly in their answer to part (c) - substituting (seen correctly in all terms or implied in all terms) and subtracting either way around : or equivalent single term. CSO. It must have been derived from sin + cos and then write sin cos

12 6.(i) dy d 5 ln( ) 0 () d y (sin cos ) (cos sin ) d (sin cos ) d y (sin cos ) (cos sin ) (sin cos ) (cos sin ) d (sin cos ) (sin cos ) sin (ii) d y ( )sin ( )cos d sin B, B * * () (i) : Applies the Product rule to y 5 ln Epect d y A B ln( ) for this mark (A, B positive constant) d : cao- need not be simplified (ii) : Applies the Quotient rule, a form of which appears in the formula book, to Epect d y (sin cos ) ( cos sin ) d (sin cos ) for y sin cos (6 marks) Condone invisible brackets for the M and an attempted incorrect 'squared' term on the denominator Eg sin cos B: Denominator should be epanded to sin cos... and (sin cos ) B: Denominator should be epanded to... k sin cos k and ( k sin cos ) sin. For eample sight of (sin cos ) sin cos sin without the intermediate line on the Denominator is B0 B : cso answer is given. This mark is withheld if there is poor notation cos cos sin sin If the only error is the omission of (sin cos ) then this final * can be awarded.. Use of product rule or implicit differentiation needs to be applied correctly with possible sign errors differentiating functions for, then other marks as before. If quoted the product rule must be correct dy Product rule (sin cos ) (sin cos ) ( cos sin ) d dy Implicit differentiation sin cos y sin cos y cos sin d To score the B s under this method there must have been an attempt to write d y as a single fraction d

13 7 (a)(i) Substitute (0, 5) to give b = 5 so b = ±5 B Substitute,0to give a + b=0 so a = 5 (ii) Gives equation as y 5 5 or y 5 5 B y () Q P V shape correct way up P at, Q at (0, 8) B B B () O (7 marks) (a) (i) B: For both b = ±5 not just b= 5 : Substitute,0to give a + b=0. This mark is implied by a value of b : a = 5 corresponding to b 5 and a 5 corresponding to b 5 If they write down an equation rather than giving values of a and b then Just y 5 5 or y 55 scores B0,, A0 Both y 5 5 and y 55 scores B,, Linear equations y 5 5 and/or y 5 5(without the modulus) only score B0 A0 (ii) Note that this is an mark on e-pen 5 5 B: y = or y = 5 5 or allow equations such as for this mark only f ( ) 5 5 If candidates don t state (i), (ii) and write down just y 5 5 they would score (i) B0 A0 (ii) B There must be a sketch to score any of these marks. B: V shape the correct way up any position but not on the - ais. Accept V's that don't have symmetry, Score if the coordinates are stated within the tet OR marked on the aes. If they appear in both B: P at then the graph takes precedence. B: For crossing the y- ais at (0, 8). Accept 8 marked on the correct ais. Condone (8,0) marked on the correct ais

14 8 (a) tan tan tan( ) tan tan tan tan tan tan tan tan d tan tan tan tan tan ( tan ) OR tan tan tan tan tan tan oe So tan tan tan * tan *cso Put tan tan tan tan so tan tan tan ( tan ) tan 8tan So tan = or 0.. d () awrt 6.6, 6.6, 0 (5) (9 marks) (a) tan tan : Epands tan( ) condoning sign errors tan tan tan d: Uses the correct double angle formula both times within their epression for tan(+) tan : Multiplies both numerator and denominator by tan to obtain a correct intermediate line Eg tan tan ( tan ) or similar. tan tan Alternatively they write both numerator and denominator as single correct fractions. They cannot just write down the final given answer for this mark *: Correct printed answer achieved with no errors and all of the lines in the markscheme (c.s.o.) Withhold the final for candidates who use poor notation or mied variables. tan Eamples of poor notation would include tan tan tan tan tan tan : Attempts to use the given identity and multiplies by tan. Condone slips : Obtain tan 8tan or equivalent. Accept tan 8 for this mark d: Obtains one value of from tan =... using a correct method for their equation. The order of operations to find must be correct but can be scored from tan 0 0 : Either one of 6.6 or 6.6 or in radians 0.6 : CAO awrt 6.6, awrt 6.6, 0 (do not need degrees symbol) with no etras within the range Note: Answers only scores 0 marks. Answers from a correct cubic/quadratic scores d (implied) then as scheme

15 9(a) u d du ( ) u du d B u u du d ln u u dd 7 u u 7 8 ln ln 7 ( ln ) ln9 9 V d 0 = 8 ln ln * * (a) B: States or uses d u d or equivalent such as d 0.5 du or d du : Epression of the form u k d u and allow missing du and/or missing integral sign u d : Splits into the form... u... u and again allow missing du and/or missing integral sign. (7) () (9 marks) Alternatively they could use integration by parts at this stage u u du a b du u u A small u number of candidates will also use partial fractions that gives the same answer as the main scheme. u dd: For...ln u... u or 'obviously...ln u... u or by parts a bln u u : ln u u This answer or equivalent such as ln u u or by parts u ln u u : Applies limits of 7 and to the result of integrating their function in u, subtracts the correct way around and combines the ln terms correctly. Alternatively usingu applies the limits = and 0 to the result of their adapted function subtracts the correct way round and combines the ln terms correctly. *: given answer achieved correctly without errors. The only omission that would be allowed could be the d line which could be implied. You need to see an intermediate step with correct ln work before the final answer is reached. 9 : Attempts to use part (a) to find the eact volume. Accept d Condone only the omission of or 8 or a bracket for this M mark so accept 8 ln or ln ln 8 as evidence : Any correct eact equivalent in terms of ln and Accept for eample A correct answer implies both marks. Remember to isw after a correct answer. 8 ln 8 or 9

16 It is possible to do 9(a) by parts or via partial fractions without using the given substitution. This does not satisfy the demands of the question but should score some marks. A fully correct solution via either method scores 5 out of 7 9(a) By Parts By Partial Fractions B0 d d d d One term of ln ln Both terms of ln d Attempts limits ln ln 0 0 dd Correct un simplified answer ln 7 ln ln 7 ln Collects log terms ln ln 9 9 ln A0* 9 (7)

17 0.(a) When t = 0 N = 5 B () Puts t = 0 so N = 56.6 (accept 56 or 57 ) () (c) (d) dn dt e t ln t 9 0.t 8 e awrt t t = awrt 6. 0.t 0.t ( 0.) 00 ( ) 7e ( 7e ) =.8 so insects per week d () cso () (0 marks) (a) B: 5 cao 00 : Substitutes t = 0 into the correct formula. Sight of N 7e : Accept 56 or 57 or awrt These values would imply the M. (c) 0.t : Substitutes 8 and proceeds to obtain e C : For 0.t 7 0.t t e oe e oe Accept decimals Eg 0.0 is fine Condone slips on the power e awrt 0.09 or 0.t e awrt d: Dependent upon previous M, scored for taking ln's (of a positive value) and proceeding to t = : awrt 6. Accept 6 (weeks), 6.5 (weeks), 6 weeks days or 7 weeks following correct log work and acceptable accuracy. Accept 9 t 5ln oe for this mark 5 0.t It is possible to answer this by taking ln s at the point 9e 5 ln(9) 0.t ln(5) d As scheme (d) dn : Differentiates to give a form equivalent to ke t t ( 7e ) (may use quotient rule) dt dn 0.t 0.t : Correct derivative which may be unsimplified 00e ( 7e ) dt : Obtains awrt following a correct derivative. This is cso

18 . (a) R 7 B tan 5 awrt 0.0 () 7 sin R arcsin their"0.0" their "7" awrt 0.85 or awrt.95 awrt 0.85 and awrt.95 () (c)(i) Find 7000 y ( '' R'') = 5 (c)(ii) =.90 () ( marks) (a) B: R 7 no working needed. Condone R 7 5 : tan or tan 5 with an attempt to find alpha. Accept decimal attempts from tan awrt 0. 5 or tan awrt.9 If R is used allow sin OR cos R R with an attempt to find alpha : awrt 0.0. Answers in degrees are A0 : (Uses part (a) to solve equation) sin 7 theirr : operations undone in the correct order to give =... Accept sin k arcsin k : one correct answer to within required accuracy. Allow 0.7 or Condone for this mark only both 0.0 and : both values (and no etra values in the range) correct to within required accuracy. Allow 0.7,0.98 (c) (i) : For an attempt at : 5 (c)(ii) 7000 ( '' R'') : Uses their n to find This may be implied by.57 their 0. stated or calculated (dp) : Awrt.90 but condone.9 for this answer Answers in degrees, withhold the first time seen, usually part (a). FYI (a) ,68.9 (c)(ii) 08.9

19 . (a) Way : Uses kt or t = c and =.5 when t = so k = or c = Way : Uses kt +c with = 0, t = 0 and with =.5 when t = so k = t t () t = B () (c) d d t () so separate variables to give ( )d dt t( c) or ( ) t( c) so t = (When t = 0, = 0 so c = ¼ or c =0) so t () (d) Uses =.5 when t = to give = 5 8 B () (e) t = 6. hours later so 0.pm or. Mark (a) and together (a) : Uses correct = kt or t = c and =.5 when t = to find their constant (may not be k or c) This may be the result of a differential equation d k d t : t oe such as t or even 0.75 B: t = t Just this with no working is Mark (c),(d) and (e) together (c) : Correct separation but condone missing integral signs : Correct form for both integrals- may not find c or even include a c : Obtains a correct answer for t in terms of and by using either 0, t 0 t t oe. Alternatively uses.5, t 8 5 t oe Condone correct responses where c seems to have been either cancelled out or ignored (d) B: = 5 8 or decimal i.e..875 or () (9 marks) (e) : Substitutes into their epression for t. Implied by : 0.pm or : only t '' ''

20 (a) Puts = 0 and obtains θ 6 B Substitutes their θ to obtain y 0 or 0 0, dy dy d 5sec tan d d d sec 5sin cos cos B 5 sin or 5 oe (c) Puts d y 0 d and obtains and calculates and y or deduces correct answer Obtains (, 5) (d) y tan and sec 5 Uses tan sec to give y " " " " 5 () () () y so 5 y * * 5 ( marks) Alt (d) y 5 tan, 5, 5 y * * () () Alt (d) Assume y k and sub both tan and y 5sec 5sec k tan 5sec k sec k = 5 AND conclusion ''hence true'' * (a) B: For or 0 or awrt -0.5 but may be awarded for cos or sec if is not eplicitly 6 found : Substitutes their (or their cos or sec ) found from an attempt at 0 to give y 0 : cao. Accept y Correct answer with no incorrect working scores all marks. Note that also gives y 6 0 but scores B0 A0 ()

21 dy : Attempts to differentiate both and y wrt and establishes dy d d d d : Correct derivatives and correct fraction 5 B: For either (seen eplicitly stated or implied) or use of sec cos An alternative to seeing sec is cos cos sec : Fully correct solution showing all relevant steps with correct notation, no mied variables and no errors. tan cannot just be written as sin without an intermediate line of working sec cos sin cannot just be written as sin without an intermediate line of working sec However it is acceptable to write down tancos as sin due to this being a version of sin tan cos (c) : Sets their d y 0 and proceeds to find (, y) from their d : for (, 5) or =, y = 5 (d) y : Attempt to obtain tan and sec in terms of and y respectively. Allow tan sec 5 : Uses tan sec with their epressions for tan and sec in terms of and y respectively *: Obtains printed answer with no errors and with k = 5 only You do not need to see k eplicitly stated as 5, it is fine to be embedded within the formula

22 (a) Attempts BA a b i j8k or BC c b i j either way around Finds OD a b c i j8k i j 6k i 5j k () BA a b i j8k and BC c b i j cos ( ) ( 8) ( ) 7 So angle is. radians or 70.5 degrees (c) Area = 7 sin 5.or oe d () () (d) Area = "5." = 67.9 or 8 oe ( marks) () (a) : For attempting one of b a or a b or c b or b c. It must be correct for at least one of the components. Condone coordinate notation for the first two M marks : For attempting d ab +c It must be correct for at least one of the components. : cao. Correct answer no working scores all marks. It must be the vector (either form) and not a coordinate Note this can be attempted by finding the mid point E of AC and then using d =b+ BE but it must be a full 0,,,, 6,, method Attempts mp and uses... Attempts AC : Uses correct pair of vectors, so kbaand kbc. Each must be correct for at least one of the components d: A clear attempt to use the dot product formula to find cos k, k. It is dependent upon having chosen the correct pair of vectors. Allow for arithmetical slips in both their dot product calculation and the moduli, but the process must be correct. 7 7 It could also be found using the cosine rule. 7 ( is for attempt at all three lengths, so BA, BC, AC and d correct angle attempted using the correct formula) : For / or / or equivalent - may be implied by 70.5 or 09.5 or. radians or.9 radians : cso for awrt 70.5 degrees or. radians. (Note that invcos( /)=09.5 followed by 70.5 is A0... unless accompanied by a convincing argument that the angle 09.5 is the eterior angle, and therefore the interior angle is It is not awarded for simply finding the acute angle. A diagram with correct angles would be ok ) (c) : Uses correct area formula for parallelogram. You may see the area of the triangle ABC doubled which is fine. : Obtains awrt 5.. Allow this from an angle of 09.5 (d) : Realises connection with part (c) and uses.5 times answer to the area of ABCD (It can be implied by 67.9) : awrt 67.9

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