Mark Scheme (Results) Summer Pearson Edexcel International A-Level In Core Mathematics C12 (WMA01)

Transcription

1 Mark (Results) Summer 07 Pearson Edecel International A-Level In Core Mathematics C (WMA0)

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 07 Publications Code WMA0_0_706_MS All the material in this publication is copyright Pearson Education Ltd 07

3 PEARSON 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) 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.

4 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. 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.

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 ( 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. Integration Power of at least one term decreased by. ( n n ) 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 (a) ad a d () () ( marks) (a) For attempting adwith a = 6 and d =0. Alternatively lists 6, 6, 6, 6,... with 5 terms and picks out the 5th term 6 n For attempting a ( n ) d with n =0, a = 6 and d =0 n For attempting a l with n =0, a = 6 and l a 9d Alternatively lists (with or without the +signs) with 0 terms and adds 00 Do not allow misreads in either case. For eample, finding the 0 th term in part (a) or the sum to 5 terms in part is M0

7 (a) B () B () 8 8 (c) () ( marks) (a) B Accept eact alternatives like and 0. but not epressions such as 0. B Accept eact alternatives such as or (All forms must have a ) (c) Either the correct coefficient (accept or 0.5) or the correct power of (accept or ) Only accept or simplified equivalents such as 0.5 Do NOT accept for this mark.

8 y 6 y... B Equation of l is y c Substitutes, 5 into y c 5 c y ( marks) Alt Equation of l is y c B Substitutes, 5 into y c 65 c y 9 y ( marks) B States or implies the gradient of l is Alternatively accept l written in the form y... States or implies the gradient of l is the same as l Eg. y ' ' c. If the gradient of l is incorrect then you must see Evidence that the gradient used for l has been linked with the gradient of l For eample y 6 Gradient of l is so equation of l is y c Or a statement that the gradients are the same. y 6 The gradient of l is 6 so gradient of l is 6. You must see some evidence of the candidate using equal gradients. They cannot just write down a gradient for this mark. So for eample, given y 6 gradient of l is or equation of l is y c scores M0, as there is insufficient evidence of the candidate using equal gradients. In the alternative scheme the first two marks can be implied by stating that the new equation is of the form y c y 5 Substitutes, 5 into their y ' ' c c... Also score for ' ' oe It is for knowing how to find an equation of a line knowing the gradient with the point, 5 Hence follow through on an incorrect gradient, even a perpendicular one. y Accept forms like this y y ' '... y c 5 c c..

9 (a)(i) (ii) dy d y dy d, 8 Substitutes their into y 8 y 0, (c) Substitutes their into their Statement +reason. ie d y d d y minimum cso () ( marks) (5) (5)

10 (a)(i) For a correct power on any of the three terms including the 8 0. Two of the three terms correctly differentiated (can be unsimplified) You may accept 6 as.5 and as 8 (a)(ii) 0.5 Cao but remember to isw. Accept alternatives for the terms in such as dy Allow epressions given in the form of the question 6 Differentiating again. Scored for reducing any fractional power by one (seen once allowing follow through) d y 6 Cao. See part (i) notes for acceptable alternatives. Eg accept Sets (or implies that) their d y 0 d n Dependent upon the previous M. For forming an equation of the type A, following correct inde work. (Ignore any reference to ). Part (a) must be correct and both M's must have been scored. For substituting their solution (of d y 0 8 y 8 y... y 0 Part (a) must be correct and all three M's must have been scored. (c) For substituting their into their d y a numerical result. Alternatively, for substituting their into their and considering the sign. Eg When 6 0 d y CSO Requires a correct and a correct 6 A statement and a conclusion is required to score this mark. Allow the candidate to state that when If the candidate gives the numerical value to d y d y, and finding (or implying to find) d y minimum it must be correct. Accept 6 oe or awrt 8.5 Alternatives in part (c) Finding the value of y at, left of and right of. Alternatively finding the d y at, left of and right of A statement and a conclusion is required to score this mark. A sketch graph can be used instead of a statement. Numerical values must be correct.

11 5(a) Attempts f ( ) Remainder = Attempts f ( ) 6 Remainder = 0 ( ) is a factor * (c) Divides their f() by ( ) to get the quadratic factor (d) f ( ) 0 ( )( ) 0 f ( ) )( ( ) d () () (), B,B () (0 marks)

12 (a) Attempts to calculate f ( ) condoning slips. Algebraic division does not score this mark. (Remainder is) Accept sight of f() = for both marks Attempts to find f ( ) condoning slips. We must see some substitution or calculation 6 or 08 9 is OK. We cannot accept just sight of f() =0 Algebraic division does not score this mark. * It is a ''show that'' so requires both a statement and a conclusion. For eample f() = 6 0 ( ) is a factor. The conclusion could be QED or a tick following the working to f() = 0 It could also as a preamble before the working. (c) Divides f() by ( ) to get a quadratic. If part is done by division it can be awarded in (c) from work in. If inspection is used look for first and last terms. Eg f ( )... If division is used look for first two terms 6... d Dependent upon the previous M. For further factorisation of their Method and accuracy marks may be awarded if seen in part (d) f ( ) ( ), f ( ) ( ) f ( ) ( ) f ( ) oe You do not need to see the f ( ) If there is little or no working in (c) with no quadratic factor seen then.. award all marks if the answer is correct (See above for possible options) award A0 A0 by implication for award no marks for answers such as or equivalent (d) B B It is quite difficult to ascertain whether candidates are finding roots or solving the question so mark as follows Sight of either and/or seen in part (d). Both of, Do not accept this if candidates write down roots first and just later Accept set language Eg :

13 6(a) 0 6 cos BAC BAC Arc BD = r Perimeter = =. (m) d, (c) Area of sector BAD = r () () Area of triangle ABC ab sin C 0 sin Area of flowerbed BCD = 0 sin d =.8 /.8 (m ) () (9 marks) (a) Attempts use of the formula cos A or 0 6 cos BAC 0 The sides must be in the correct ''position'' within the formula. Condone different notation Eg. BAC awrt 0.5 The angle in degrees (awrt 9.9) is A0 Attempts arc formula: In radians uses Arc BD = r 0 ''0.5'' ''9.9'' In degrees uses Arc BD = r d Dependent upon the arc formula having been used. It is for calculating the perimeter as 8 + arc length. Perimeter = awrt.(m) (c) Attempts area of sector formula: Area of sector BAD = r 0 ''0.5'' ''9.9'' In degrees uses Area of sector BAD = r Attempts area of triangle formula: Area of triangle ABC = absin C 0 sin''0.5'' 0 6 You may see Herons formula used with S and A S( S 0)( S 6)( S ) Watch for other methods including the calculation of a perpendicular. d Dependent upon both correct formulae. It is scored for finding area of triangle - area of sector Allow awrt.8 or.8 (m )

14 7 (a) (8, 5) B y 7 B (c) 5 k 0 () () () (d) 0, 8 Shape B, 5 (0,-8) and (,-5) B y 0 Asymptote B () (7 marks) (a) B Accept (8, 5) or =8, y = 5 or a sketch of y f with a minimum point marked at (8, 5) B y 7. It must be an equation and not just '7' (c) Accept one ''side'' of the inequality condoning a misunderstanding of whether the boundary is included or not. Allow for k >5, k 5, k < 0, k 0 Condone a different variable for the cao 5 k 0. Allow k 5and k 0 k 5, k 0 5,0 k :5 k 0 (d) B Do not allow k 5or k 0 For a reflection of the original curve in the ais. Look for the shape shown in the scheme but be tolerant of slips at either end. 0, 8 B For the graph to have an intercept of and a (single) maimum point of, 5 Accept 8 being marked on the y- ais and the graph passing through this. Condone 8,0 as long as it is marked on the correct ais B For giving the equation of the asymptote as y 0 The graph must clearly be asymptotic but be tolerant of slips. See practice items for clarification.

15 5 d 5 8 (a) c b 5 5 ( c) 6 b (6 60) ( b b 5 b) 6 () 5 0 * () b b b (c) b 0 B b b 5 0 ( b 5)( b ) 0 b 5, (8 marks) () (a) Raises the inde of any term in by one Two of the three algebraic terms correct (unsimplified). For eample accept cao including the +c Substitutes and b into their integrated epression, subtracts either way around and sets equal to 6 * Simplifies to the given solution. This is a given answer and therefore the intermediate line(s) must be correct. Minimum epectation for an intermediate line is 6 ( b b 5 b) 6 or equivalent with the bracket removed. (c) B b 0 Factorises/ attempts to solve the quadratic b 5,

16 (i) log log 5 0 log log only, y (ii) log p y log p y log p log p p y, y p y p y py p y p (5) (0 marks) (i) Use of the power law of logs For 'undoing' the logs by either setting log 0... log 0... or using the subtraction law and 0 log0 A correct simplified quadratic 5 0 A correct attempt to find a solution to a TQ of equivalent difficulty (ie no factors). Allow formula, completing the square and use of a calculator giving eact or decimal answers cso 5 9 or eact simplified equivalent without etra answers. (ii) Use of subtraction (or addition) law of logs For using log p p or equivalent in an attempt to get an equation not involving logs. log y log y y y p implies this and scores M0. p p y A correct equation in p and y not involving logs. Accept p y Score for an attempt to change the subject. This must include cross multiplication, collection of terms in y, followed by factorisation of the y term. p cso y p or equivalent such as p y p Special cases in (i): Case Allow the subtraction law either way around as the rhs of the equation will be Case log Special cases in (ii): 5 9 only will be awarded M0 A0 log p y y log p y log p y log log p p p p y y p y py p y will be awarded M0 A0 p (5)

17 0 (a) 0 OR 0 seen as the constant term B C C = a b a b 8 0a 56 a oe () () (c) 0b 60a 5 b () (8 marks) (a) B 0 OR 0 seen as the constant term For a correct attempt at the binomial epansion for ( a b) n with a =, b and n=0 8 Condone missing brackets. Accept any unsimplified term in as evidence Accept a power series epansion on 0 09 k 0k k condoning missing brackets. Again accept any unsimplified term in as evidence A completely correct unsimplified solution. Accept Accept C C ! Accept Can be listed with commas or appear on separate lines. Accept in reverse order. Sets their '0' a 56 a. Accept equivalents such as 0.5. Accept this for both marks (it can be done by substituting 0 into both sides of the epression) as long as it is not found from an incorrect method (c) Sets their '0' b their'60' a 5 b or 0.5

18 (a) Attempts U (tonnes) * (c) Attempts Attempts N U N S n N oe 6000 log N log.05 log N (0.) log.05 N n a( r ) with n 0 a 6000 / 0000 and r.05 ( r) (.05 ) 0000(.05 ) S 5 OR S (.05 ) (.05 ) Awrt 000 () (5) (a) () (0 marks) Attempts to use ar with a = 6000, r =.05. Accept r.5% Condone for this mark r =.5 or.005 Accept a list of terms with the same conditions * cso (tonnes). If candidate states U (tonnes) or 67.0 (or anything that rounds to 67) they don t need to round to the given answer. n Attempts to use ar n or ar with a = 6000, r =.05 or.5% condoning values of r being.5 or.005 n n For reaching the intermediate result or Allow to be rounded or truncated to. (to dp or better) Uses logs correctly to get n or n- This mark may be awarded from a sum formula This is scored for a correct (unrounded) answer. It may be left in log form. If the candidate has used n instead of n-, they will not score this unless they subsequently reach a final answer of. Allow for N or n. log Accept versions of n... (0.) or n...log (0.).05 log.05 or log. n... (0.5) log.05 (N ) = Do not accept N etc The two final A marks may be implied by finding n and adding to reach

19 So trial and improvement above and below Attempts U or 0 U Achieves either U 0 awrt 7960 or U awrt 8080 Attempts U and 0 U Achieves both U and U States (N ) = 0 awrt 7960 awrt 8080 Candidate uses a ''solve'' option on a calculator. N (N ) = 0. (to dp at least) states Candidates can attempt to write down 0 terms which would follow the same scheme as trial and improvement (c) n a( r ) Attempts S with n 0 a 6000 / 0000 and r.05/.5%/.5/.005 n ( r) Alternatively accept a list of ten terms starting with 0 000, 050, ,... added Or 6000, 6090, 68.5,.added A correct unsimplified answer to the cost for 0 years (.05 ) 0000(.05 ) Accept either S 5 OR S (.05 ) (.05 ) cso awrt ( ) 000

20 (a) dy y dy At = 8 6 Equation of normal is y 6 ( ) y 6 d* Sub in y (c) Normal meets ais at 6 B Area of triangle = Correct method for area = 8 + B* 6 6 d = 5.75 (5) () (6) ( marks) (a) d Two of the three terms correct (may be unsimplified). dy 8 6, need not be simplified. You may not see the d y Substitutes = into their d y The candidate must have scored both M's. It is for the correct method of finding the equation of a normal. Look for y 6 ( ) d y If the form y m c is used it is for proceding as far as c.. * cso y 6 Note that this is a given answer. y6is ok B* Either substitute Or substitute 0 in y or y y reach 8 8 by inspection or division and state

21 To mark consistently decide on the method first (c) Way One: Finding area under curve C Integrates 9 6 8with at least two terms correct (unsimplified) B or equivalent. May be unsimplified For sight of the normal meeting the ais at 6. May be embedded within a formula or on the diagram '6' 6 or correct integration with '6' limits of to their 6 8 Dependent upon the candidate having scored both M's. It is a fully correct method of finding the area of R. The limits of the integral(s) must be correct Correct method for area = ' 8 ' '6' 6 For correct method for the area of the triangle. Either d + cso 5.75 oe such as Alternatives in part (c). In some cases candidates may try to find the area between the line and the curve. If you see an attempt where this occurs score as follows. Way Two: Finding area between line and curve Integrates their either way around with at least two terms correct (unsimplified). Condone slips on the straight line (eg 8, 6 or on the subtraction) Completely correct integration of their epression either way around 8 8 You may follow through on their equation of the line or on their combined epression of line and curve. This is for correct integration following the award of the M mark. B For sight of the normal meeting the ais at 6. May be implied within a triangle formula of from the diagram For correct method for the area of the large triangle. The base length of the triangle is (their 6 -). The height of the triangle must be an attempt to find the y value on the normal at = Either by '6' y '6' '7.5' 56.5 NORM at or by correct integration with d '6' limits of to their 6 8 Dependent upon the candidate having scored both M's. It is a fully correct method of finding the area of R. The limits of the integral(s) must be correct Correct method for area = '6' cso 5.75 '7.5' ' 8 8 '

22 (a) sin 5cos sin tan 5cos sin cos 5cos cos sin 5cos cos cos 6cos cos * () 6cos k cos k cos k-)(cos k + )=0 cos k=, Either cos= 70.5, 89.7 Or cos= 0, 0 5.,.7, 60,0 (6) (0 marks) (a) Uses sin tan in the given equation. You may see it used in the form tan cos sin cos A correct equation (not involving fractions) in both sin and cos. Eg 5cos cos sin oe Replaces sin by cos to produce a quadratic equation/epression in just cos * Proceeds correctly to the given answer 6cos cos All notation should be consistent and correct including cos instead of cos and sin instead of sin

23 Attempts to factorise and solve epression of the form 6cos kcos k Accept use of formula or GC They may solve 6y y 0 which is fine Correct answers for quadratic Accept cos k=, with k = or or or even y, Correct order of operations to produce at least one value for in the range 0 60 arccos 80 arccos Look for cos =' ' or cos =' ' Two of awrt 5.,.7, 60,0 Any of these values would imply Correct order of operations to produce at least one other value for in the range 0 80from their principal value 60 arccos 80+arccos Look for cos =' ' or cos =' ' All four of awrt 5.,.7, 60,0with no additional solutions in the range. If the answers are given in radians 0.65,.56,.07,.09 just withhold the final Special Case : If candidates solve part giving values for (perhaps mistakenly thinking that the question was 6cos cos ) then you may award special case for sight of both solutions awrt 70.5,0 in the range 0 80

24 7 8 (a),, 6 () (a) (c) (7 ) (8 ) Or ('' ) ('6' ) Or (7 '') (8 '6') (Radius of circle ) = Equation of C is y r Attempts either value of r as ' ' ' 6' their r When r 5 When r 5 y y For an attempt at, May be implied by either correct coordinate, 6. No working is required, Correct answer scores both marks. Condone lack of brackets () () (8marks) Scored for using Pythagoras' theorem to find the distance between their centre and a point. Look for an attempt at ('' ) ('6' ) or similar. If the original coordinates are used then there must be some attempt to halve. = Correct answer scores both marks (c) For stating the equation of C is y r or algebraic 'r' Accept for any 'r' including an ( 0) ( y 0) r y k If a value of k is given then k must be positive Attempts either value of r Look for ' ' ' 6' their r Either of y or y 7 Allow for this mark variations like y Accept 0 0 r 6 Both of y and y 7. Equations must be simplified as seen here Any one correct equation will imply the first two M's.. Alt method to find equations using the intersections: : As above : Solves their y ' ' with their '' y '6' '' So this time the method is scored for either as before Intersections, and 6,9 '' '' or '6' '9'

25 5 (a) t H.5sin.75(m) 6 t H.5sin awrt 5.(m) 6 t (c) H.5sin 6 t sin 6 6 arcsin ' ' t sin t awrt 7. 6 B* () () And 6 arcsin ' ' t awrt 0.6 Times are 7:/7: am and 0:6/0:7 am (6) (9 marks) (a) B* This is a given answer. Score for sight of ( H ).5sin.75(m) 6 t Alternatively they can work backwards with H.75 t Hence (time) = a.m 6 6 For substituting t = into H and an attempt to calculate. Score for.5sin.. 6 awrt 5. (m) or 6 m following correct work..5sin (m) is M0A0 unless there is an eplanation that states that the period is hours so.5sin.5sin 6 6

26 (c) t t For substituting H = into H.5sin 6 WITH some attempt to make sin 6 the subject. t You may see the 6 being replaced by another variable which is fine for the first two marks t t sin 6 oe Condone awrt sin For a correct attempt to find one of the first two positive values of t using their arcsin t 6 arcsin So for sin 6 score for either t or 6 arcsin t 6 For sin 6 score for either t arcsin awrt. or awrt.6 t A relatively common answer following sin 6 is t.9.. This scores M0 as it not one of the first two positive values for a rhs of. One of t awrt 7. or awrt 0.6 One of these values would probably imply the previous M mark Can be implied one correct time 7:/7: am (07) or 0:6/0:7 am (06). Can be implied by one correct number of minutes / minutes and 66/67 minutes For a (correct) attempt to find both of the first two positive values of t using their Times are required for this mark only Accept both 7:/7: am and 0:6/0:7 am in hour times Allow 07/ 07 and 06/07 in hour times. Allow 07:/ 07: and 0:6/07 in hour times 07./ 07. and 0.6/0.7 in hour times. (The 0 must be present in 07./ 07. )... Some candidates may choose to do part (c) using degrees. This is fine and the scheme can be applied as long as 80 and the candidates are consistently using degrees sin 0t Attempts to find one value of t (but the units must be consistent) sin 0t 0t.8 or 8.9 t

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