Mark Scheme (Results) October Pearson Edexcel International Advanced Level in Core Mathematics C34 (WMA02/01)

Transcription

1 Mark Scheme (Results) October 08 Pearson Edexcel International Advanced Level in Core Mathematics C (WMA0/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding bo. 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: October 08 Publications Code WMA0_0_80_MS All the material in this publication is copyright Pearson Education Ltd 08

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. When examiners 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 PEARSON EDEXCEL IAL MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 5. 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) 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 example, 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. 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 ( 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 =. Formula Attempt to use the correct formula (with values for a, b and c).. Completing the square a, leading to x = b Solving x + bx + c = 0 : x ± ± q ± c = 0, q 0, leading to x = Method marks for differentiation and integration:. Differentiation. Integration Power of at least one term decreased by. ( x n x n ) Power of at least one term increased by. ( x n x ) Use of a formula n+ 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. 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 Question Number (a) cosθ + sinθ = R cos( θ α) (b) Scheme Marks R = + = 7 α = arctan = awrt.6. 7 cos θ.6. cos θ.6 7 θ θ... d θ awrt. 0 or awrt 0.0 θ.6 '.75...'.. and '.75...' dd θ awrt.0 and 0.0 () (5) [8 marks] (a) For R = 7. Condone R = ± 7 For α = arctan ( ± ) or α = arctan ± leading to a solution of α It is implied by α = awrt 76 or awrt. rads Condone any solutions coming from cosα =, sin α = If R has been used to findα award for only α arccos = ± α = arcsin ± ' R ' ' R ' α = awrt.6

8 (b). Using part (a) and proceeding as far as cos θ their.6. their R Condone slips on the. and miscopying their.6. This may be implied by θ their.6 arccos their R.. Condone for this mark cos θ their.6 or cos θ their.6 their R their R. but cos θ their.6 is M0 and hence dm0 etc their R d Dependent upon the first. It is for a full method to find one value of θ within the range 0 to π from their principal value. Look for the correct order of operations, that is dealing with the ''.6'' before the ''''. Condone adding.6 instead of subtracting. β their.6 cos θ their.6... θ their.6 β θ= awrt θ. 0 or θ awrt 0.0 Only allow. if it is preceded by an answer that rounds to.0 dd For a correct method to find a second value of θ (for their α ) in the range 0 to π. Eg θ. 6 ' β' θ OR θ. 6 π ' β' θ THEN MINUS π awrt θ. 0 and θ awrt 0.0. Only allow. if it is preceded by an answer that rounds to.0 Withhold this mark if there are extra solutions in the range. Degree solution: Only lose the first time it occurs. FYI. In degrees only lose the first A mark awrt (a) α = and (b) θ = 7.5, θ =. Mixing degrees and radians only scores the first M in part (b) Answers without working. If 7 cos.6. θ is written down then all marks are available. ( marks for one correct answer) If there is no initial statement then score SC then 0,0, 0 for a maximum of, for each solution.

9 Question Number Scheme Marks. x xy x y 0 x x y 6y 0 7 Substitute (, ) = 8 Uses gradient of normal = x = d 8 y = ( x+ ) 8x 7 y+ 8 = 0, 7 [7 marks] Applies the product rule to xy xy x y Accept exact alternatives such as xy x y + and allow if recovered from poor bracketing. You may see xy x y Attempts the chain rule to y Ay d x You may see y Ay For correct differentiation on x + x+ y x + + 6y You may see x + x+ y x+ + 6yy d Substitutes (, ) into a differentiated form and attempts to find a numerical value of It is dependent upon the differentiated form having exactly two terms in d y, one from xy and one from y If the candidate attempts to rearrange and collect terms before substituting you may condone poor algebra. d Attempts to find a numerical value to the gradient of the normal. It is dependent upon the previous method mark and finding the negative reciprocal of the value of d Correct attempt at the form of the normal at (, ) Eg. x x= y = their '' ''( x+ ) m Condone one sign slip on the or the +. Condone, for this method, answers from poor differentiation (eg just having one d y term). This mark is for the method of finding the equation of a normal. If the form y = mx + c is used it is for proceeding as far as c =... + = Allow k( 8x 7y 8 0) 8x 7y = where k is an integer Note that the error xy x y can lead to d y = which in turn gives a normal of y = This can potentially score B0 d A0 for 5 out of 7

10 Question Number Scheme Marks (a) (b) secθ Accept secθ p p Attempts sin θ 90 sin θ cos90 cosθ sin 90 with cosθ = p used Alternatively uses sin θ 90 sin 90 θ sin 90 cosθ cos90 sin θ with cosθ = p used sin θ 90 sin 90 θ cosθ with cosθ = p used Similarly Or sin θ 90 p. sin θ 90 sin θ cos 90 cosθ sin 90 with cosθ = p used Allow p for both marks as long as no incorrect work is used to generate this answer. (c) States sinθ sin θ cosθ or sin θ sin θ cosθ sin θ cosθ Attempts sin θ + cos θ = in part (c) with cosθ = p to get sinθ in terms of p. Only accept sinθ = p if a version involving squares has been seen first. Allow sin θ + cos θ = sinθ = p as a slip. (We have seen the Pythagorean identity) You may see an attempt using a right angled triangle. The same scheme may be applied. sin θ = p p but NOT sin θ =± p p or equivalent such as sin p p θ =, p ( p)( p) + or p p Final answer (do not isw here). sin θ = p p = p( p) is A0..(a) secθ cosθ p (b) Alternatively attempts to use sin θ + cos θ = and oe sin θ 90 sin θ cos90 cosθ sin 90 p (c) sinθ sin θ cosθ Uses sinθ cos θ, sin θ ( ) = = p = p p, sin θ = cos θ = p cos θ cos θ = Usually p p via this method. () () () [6 marks]

11 Question Number (a) Sets x 8x xe 0 = (b) ( x 8 e e x x ) Scheme x Marks e = 8 x= ln8 x= ln8 = ln = + Sets 0 ( ) e x x 8 x 8 e ln 8 = x = ( + x) + x d, * (5) (c) 8 x = ln awrt 0.0 = + 0. x = awrt 0.7, x = awrt 0. () [0 marks] ()

12 (a) Attempts to solve e x = 8using a correct order of operations. (Eg. some may take first) Allow for x = ln 8 Condone a slip on the 8. It may be implied by answers awrt 0.69 x = ln Note that x = ln8 is A0 (b) Attempts to differentiate the xe x x x term to ± Axe ± Be using the product rule. If the rule is quoted it must be correct. Correct derivative ( x 8 e e x x = + ) with correct bracketing or x x 8 e x e = States or sets d y 0 = (which may be implied) and takes out a common factor of e x reaching a form (......) e x x... ± =... or e = (... ±...) d Dependent upon the both previous M's, it is scored for using correct ln work, moving from (......) e x... ± =... x = or e x = (......) x = ± 8 * Reaches x = ln + x, 8 x = ln x + or 8 x = ln oe with correct work and no errors x + or omissions (See scheme for necessary steps that need to be seen) x x x x 8 ( e x e ) e x e 8 = + + = is an example where there is missing step. (No d y 0 = ) 0 = x x x 8 xe +e = 8 e = is also an example where there is missing step. ( No x + attempt to show the factorised line ( ) e x + x = 8) 0 = x x x x 8 xe +e = 8 x+.e = 8 e = is also an example where there is a x + missing bracket (for factorisation).. If the first isn't scored for an attempt at the product rule a special case M0A0dM0 A0 may be awarded for setting their d y 0 = and proceeding to a form x e =... x For example, if 8 e x = it would be for proceeding to x 8 e = x (c) Calculates x from the given iterative formula. May be implied by ln awrt 0.0 Allow answer written as 0. awrt x = 0.7, x = 0.. NB. The subscripts are not important. or awrt 0.

13 Question Number Scheme 5. (a) x 5x A( x) B( x )( x) C( x ) (b)(i) (ii) + + = Sub x = C = x = A = any two constants correct Coefficients of x + ( x+ ) ( x) ( x) ( x+ )( x) = A B B= all three constants correct Marks () = ln ( x+ ) + ln ( x) + ( x) ( + c) oe ft x + 5x+ d x= ln ( x+ ) + ln ( x) + ( x) = ln + ln + 8 ln + ln+ ( ) 5 = ln = + ln (6) [0 marks]

14 (a) Writes x + 5x+ = A( x) + B( x+ )( x) + C( x+ ) This may be implied by the sight of two equivalent fractions or via work leading to the constants Substitutes x = or x = or equivalent and attempts to find the value of one constant. It can be scored after scoring B0. x + 5x+ = A x ( x) + B x+ x + C x+ or similar Eg condone the use of ( ) ( )( ) ( ) Alternatively attempts to equate coefficients of x, x and constant terms to produce and solve simultaneous equations to find the value of one constant. Any two constants correct All three constants correct (b)(i) For A B..ln( x+ ) and..ln( x) x+ x For C..( x) ( x) ft All three of their integrals correct, following through on incorrect constants (but not zero's) There must be some attempt to write in the simplest form. (Cannot leave ( x) for instance) (b)(ii) Substitutes both x = and x = 0 into their answer for (b)(i) which involves lns and subtracts (either way around). Uses correct ln work to combine their ln terms cao 5 = + ln Note that the decimal equivalent + ln 0.565is correct

15 (a) Question Number Scheme Marks 6. (a) (b) (c) Alt 6. (a) + x = ( + x) ( x) x ( )( ) ( )( ) = + ( x) + ( x) ( x) + ( x) +...!! = + x x x+ x = + x + x + x + x x = + x + x 8. * (6) ( 0 ) ( ) + = 0 Sub so, x= '' k'' = x = + x + x ( + x) = + x + x ( x) x 8 8 ( )( ) ( )( ) + ( x) + ( x) +... = + x + x + ( x) + ( x) +! 8! + x x +... = + x + x x x x x +... = + x x Hence true x () () [9 marks] * For writing the given expression in index form ( + x) ( ) It may be implied by working but it must be a form that can lead to the answer. Do not allow ( + x) ( x) for this mark unless the expanded ( x) Score for the form of the binomial expansion with index or = + (** x) + (** x) +...! Eg Correct unsimplified form for one expression = + (** x) + (** x) +...! or (6) is subsequently set with index -

16 The correct simplified form for both expressions seen. This mark may be implied if a correct final expression is found following correct working. There doesn t need to be an implication that these expressions are to be multiplied. For multiplying terms in the first expansion by terms in the second expansion. Expect to see an attempt to find the six terms required to produce the given solution. Allow terms in x and greater to be seen which don t need to be correct. Follow through on their expansions but condone ''changes'' in an attempt to reach the given solution. (See Practice Items) * Correct solution only (b) or statement = + x + x 8 k = seen in (b) (c) Sub x = into both sides of (a) '' k '' = or + 0 = Do not allow k = or exact equivalent. Condone which follows (b) = You may see a variety of solutions to part (a). Please consider carefully when marking. Example: Mark in this order + : x x + x x x = = + = + x x x x * x * x * : For one attempt at the binomial expansion x + = + + x x x condoning slips on the bracketing : x x Completely correct intermediate form. x + = + + x x x : For a second use of the binomial expansion. It is dependent upon a correct first use. * x * x * x * x * x + = + = ( ) + ( x ) Expect to see a correct use of the binomial expansion in both TERMS. ( x) ( x) ( x x ) ( x x ) * x x 9x x 9x x 8 8 = + x + x 8 : + = + ( x) ( x) = + ( + x+...) ( +...) :

17 Question Number 7 (a) (b) Scheme Marks sin x y ln( cos x) cos x sin xcos x sin xcos x =, = = cot x sin x sin x, ( ) cot x= tan x= π x= arctan x= 6 π y = ln cos = ln 6 or ln () () [8 marks] 7 (a)alt I sin xcos x y = ln( cos x) y = ln(sin x) = sin x cos x cot x = sin x =, () 7 (a)alt cos x y ln(sin ) ln ln sin 0 II = x y = + x = + sin x cos x cot x = sin x =, () 7 (a)alt y y y = ln( cos x) e = cos x e = sin x III Then as main scheme, () (a) For differentiating y = ln( cos x) to d y ± A sin x = cos x sin x = oe cos x Uses the double angle identities sin x= sin xcos x and cos x= sin x The double angle for cos x may be implied by sight of '' sin x '' on the denominator as we can condone the missing bracket. Simplifies to show that = cot x showing at least one correct intermediate line between d y sin x cos x sin xcos x = and cot x usually or cos x sin x sin x sin x ( )

18 In the alternative versions the double angle identity (or identities) are seen before the differentiation. For example. (i) This one shows incorrect differentiation and is scored M0 A0 A0 y = ln( cos x) y = ln(sin x) = sin x (ii) This one can be awarded the method mark for differentiation and is scored A0 A0 sin xcos x y = ln( cos x) y = ln(sin x) = = cot x sin x The two accuracy marks are linked and cannot really be awarded separately A0 would be difficult to score via this method. (b) Uses cot x = tan x and proceeds to find x π 5π x =. Ignore additional (incorrect) values such as x = 6 6 Do not accept 0 for this mark. Do not allow this for candidates who guess k = Substitutes their value of x in y = ln( cos x) π For y = ln or ln following x = or 0 6 Do not allow this for candidates who guess k = xy, given within the range Withhold this final mark if there is another value ( )

19 Question Number Scheme Marks 8(i) xsin x= xcos x+ cos x ( ) = xcos x+ sin x + c d (ii)(a) = secθ tanθdθ π = cos θ secθ tanθdθ x 0 π π = sin θ tanθ d θ = tan θ dθ cosθ 0 0 (b) x ( ) * d = tan θdθ= sec θ dθ= tanθ θ x π θ = π π π = [ tan θ θ] =tan θ = 0 = d x () () () [0 marks] (i) Attempts integration by parts the correct way xsin x= ± xcos x± cos x d Integrates again to xsin x= ± xcos x± sin x ( + c) xcos x sin x( c) cos xx. sin x( c) = + + with no need for + c = + + can be accepted Note that international centres sometimes teach a ''tabular method'' ''D-I method'' Diff Int + x sin x - cos x + 0 sin x If there is no evidence of an intermediate line, a tabular method, or integration by parts score d for a fully correct answer AND M0 dm0 A0 for an incorrect answer

20 (ii)(a) = secθ tanθdθ or equivalent. Eg accept d x secθ tanθ dθ = Substitutes x = secθ into and simplifies to sinθ, x cosecθ, tan θ secθ or tan θ sec θ This may be implied if the has been adapted x * Completes proof. This is a show that question and you should expect the secθ to be either replaced by to be cancelled as seen below tanθ secθ tanθd θ= tan θ dθ secθ (ii)(b) sin θ cosec θ or allow cosθ Expect to see the correct limits and correct notation within their solution. (Condone incorrect notation in jottings/working at the side of their solution) The limits can just appear without the need to see calculations. A notational error is tan θ being written tanθ without a bracket It is possible to do this question from rhs to lhs but marks in a similar way. tan θdθ= ± sec θ± dθ= ± tanθ ± θ ( ) tan θd θ = tanθ θ π d Uses the limit(s) (and 0) in a function of the form ± tanθ ± θ π cso

21 Question Scheme Marks Number 9 (a) P = 50 () (b) (c)(i) (ii) (a) ( P = )50 (b) t 900e t 5 = 5e = 5 t e t e = 7 t = ln 7 t t t t e 5e 900e e t dp 5e = = dt t t e e ( e ) 5e 900e e dp 5e = = =.7 dt t= 8 e e ( ) ( ) Substitutes P = 5, cross multiplies to reach a form t t t 5e = 5 or equivalent such as e = 7. Note e t D( D ) A e t = B oe = is correct 7 e = > 0 t =... using ln's. Allow equivalent working from e = E, E > 0 Equivalent work may be seen. ln 5 + ln 5... t = t = t = ln 7 or equivalent such as t = ln 9, ln 0 but not t 5 t = ln 5 () () [9 marks] (Scheme requires a ln k )

22 (c)(i) 900e t t Attempts to apply the quotient rule on P = with u = 900e, v= e to reach an t e expression of the required form (see below). Condone slips on the coefficients. t t t t e Ae 900e Be dp Score for =, AB, > 0 dt t e Award this mark if the candidate incorrectly multiples out before writing down their d P dt t t t e 5e 675e t t t t dp u = 900e, u = 5e, v= e, v = e = dt t e Alternatively applies the product rule on For the product rule look for e 5e 900e e dp = dt t e t t t t dp t t t t t = 5e e 675e e e dt t t t 900e e or chain rule on e t e t P Qe t e t e t ± oe which may be left unsimplified. which may be left unsimplified. 900 e t (c)(ii) Substitutes t = 8 into their d P dt and calculates a value for dp dt.7 only. Note that this is not awrt If the candidate subsequently writes.7 this is A0

23 Question Number Scheme Marks 0 (a) y < () (b)(i),0 (ii) 0, () (c) x ( x ) ( x ) 5x gg( x)= x = = = x x x ( x ) 5 x d () (d) Shape y = Intersects y - axis at 0, meets x- axis at,0 ft Asymptotes at x = and y = O x = () (e) x 8 = 8 x = 8x + x = x d () [ marks] (a) Accept y <, g( x ) <, g <, < y < (,) (b)(i),0 Allow candidate to state x=, y = 0 or state (b)(ii) x=, g( x ) = 0 0, Allow candidate to state x= 0, y = or state x= 0, g( x) = SC: For candidates who in(i) write just xor A= and in (ii) write just yor B= score B0 SC This SC should also be used for candidates who embed the values. So for example in (i) show x 0 0 = x = and in (ii) show y = y = as there is no explicit statement in (i) x = 0 and x 0 (ii) y = 0 (c)

24 x Attempts to substitute g into g. gg( x)= x is sufficient x x 5 5 An alternative is using g( x) = + gg( x) = + x 5 + x d Multiplies by all terms on the numerator and all terms on the denominator by ( ) fraction of the form ax + b Condone poor bracketing cx + d Al gg( x) = x x to form a (d) Correct shape with cusp (not a minimum) at A. The curve must appear to have the same asymptote as the original curve (at x = and y = ) It should have the correct curvature on the rhs and not appear to bend back on itself ft Intersects y - axis at 0, and meets x- axis at,0. Follow through on coordinates from (b). Allow this to be marked A, B as long as the coordinates of A and B were correct. Gives the equation of both asymptotes as x = and y = (e) For writing down a correct equation leading to a solution of g( x ) = 8 Allow x x x + = 8 oe so allow = 8, = 8, x x = 8 and = 6 x x x x x d Solves an allowable equation (see above) by cross multiplying, collecting terms to reach x =.. x x+ Do not allow a candidate to score this mark from = 8 = 8 This scores M0 x x+ 8 x = oee only If both values are found x 8, 8 =± x = then this mark is scored only when '' '' is deleted x or 8 is chosen as the answer... Solution from squaring in (c) x : = 6 x d: Ax + Bx + c = 0 and solves by usual methods FYI the correct quadratic is 55x 60x+ 560 = 0 ( x 8)( x ) = 0 8 : Selects x =

25 Question Scheme Marks Number 0 (a) l : r = 0 + µ 0 () (b)(i) A = (,, ) (c) (ii) B = (,5, 8) Uses a correct pair of gradients a = and b= k or vice versa ab. = a bcosθ + = 6 cosθ θ = awrt 68.5 d, (d) AB = OR AB = ' 6 ' Area = ' OA AB sin( c)' 6 sin 68.5 awrt 5( units ) = = () () () [0 marks] (a) (b)(i) Scored for the rhs of the equation with gradient k and containing the point 0 x Correct equation with lhs r = 0 + µ oe such as y = µ. 0 z Allow for gradient any multiple of Allow with any scalar parameter inc λ. k Condone another constant appearing so r = k + µ is acceptable k x It must be an equation so r = or y = must be on the lhs z (,, ) (b)(ii) A = Accept in vector notation (,5, 8) B = Accept in vector notation (c)

26 Attempts to use a correct pair of gradient vectors. Eg uses their vice versa, Condone one sign slip only d Correct method for finding acute angle from ab. = a bcosθ a = and their b = k or Expect to see an attempt proceeding to cos θ =... condoning one slip on aband. an attempt at squaring and adding for a & b If there is no method shown for a or b then expect at least one to be correct. It is dependent upon the previous M θ = awrt 68.5 Note that. is the radian answer and scores A0. Allow awrt 68.5 coming from 80.5 (d) + + or Uses a correct method of finding distance AB Accept ( x x) ( y y) ( z z) Uses a correct method of finding the area of OABD Accept ' OA AB sin( c)' or two triangles Note that ' OA AB sin(80 c)' 6 sin.5 awrt 5( units ) = = is also correct awrt 5 Accept the exact answer 9 5 and then isw... There are various other ways of finding are OABD B l A C D l O Eg by finding the perpendicular distance between AB and OD Uses a correct method of finding distance AB It must be a full method. The values are given for a check. (It is a method mark!) Sets up a point C on l with coordinates ( λ,+ λ, + λ) Uses the fact that OC and AB are perpendicular λ 7 + λ. = 0 λ = 6 + λ Finds point C =,, 6 6 and hence distance OC = 5 6 And then multiplies AB by OC AWRT 5 Note: It is possible to find the area by finding the perpendicular distance between AO and BD Please scan through the whole response and mark d where the first M is a full method to find the perpendicular distance. Similarly, it is possible to attempt AO OB sin AOB Use the same scoring as above. d

27 Question Number Qu (a) Scheme t = x=, y = 8 dt 9 t = = t dt 6 y 8= x x 7y+ 50 = 0 Equation of tangent is ( ) (b)(i) At A, x = 5; y = 0 t 9 t = t t + t = 0 (ii) ( ) ( )( ) At t = 0,, ( At =, = 7 5 = 58 ) t =, x = 7( ) 5 = 58 or t x ( ) At B, x = 58 y y dt dt t9 ttdt = ( 6 t ) d 5 6t t = ( + C) ( = t.8t 5 5 ( + C) ) (c) = = ( ) 5 Either ( ) 5 6t t =.8 = Or 5 ( ) ( ) ( ) 0 5 ( ) 5 6t t =.8.8 = Marks dd (5) () (5) [ marks]

28 (a) At t = x=, y = 8. Score if (,8 ) is used in the tangent equation Attempts dt t 9 t ) before dt differentiating or slips in the product rule (eg bracketing errors) but do not allow d y t dt 9 t = or exact equivalent. t 6 A valid attempt at a tangent to C at t = Allow y ''8'' = '' ''( x '' '') k x 7 y+ 50 = 0 where k is an integer Allow ( ) (b) A = Allow x = 5 and A = 5 ( 5,0) For attempting to solve t( t ) x= 7t 5 B = 58,0 Allow x = 58 and B = 58 ( ) 9 = 0to produce a non zero value for t and substitute the value into (c) Attempts area = y d t. Do not be concerned with limits dt = 6t t dt. Do not be concerned with limits but it must be multiplied out Area ( ) Alternatively (rare) accept ( ) the other 5 Integrates to a form At Bt ( c) t 9 t dtby parts with u being one of + +. Condone slips on the coefficients only t or ( 9 t ) dd A full method to find the area of R using their limits. Accept either [ ] 0... or [ ]... Dependent upon both M's 907. (units ) or equivalent 56/5 and v being

29 Attempts made from a Cartesian equation: (a) Allow slips, but the should be awarded for an attempt at the product and chain rules on functions of x+ 5 x+ 5 the type y = 9 or 7 7 ( x+ 5) ( 58 x) y = x+ 5 x or the chain rules on a function of this type ( ) ( ) Scored for ( x 5) ( x 5) (c) Alt I (c) ( x+ 5) ( 58 x) ( x+ 5) 9 = + + or = oe x+ 5 x+ 5 9 y 9 d, x ( x 5) ( x 5) = = , 5 6 x+ 5 x ( 5) ( 5) x+ x+ = dd Cartesian ( ) ( ) Alt II (c) Cartesian Via parts ( x+ 5) ( 58 x) = ( ) ( ) 58 x dv x+ 5 with u= =, ( x)( x+ ) ( x+ ) Integrates by parts to the form = ( 58 x)( x+ 5) ( x+ 5) = + = 907. dd A full attempt to get y in terms of x AND forms y A correct expression for the area and in a form that can be integrated. If by parts is used the correct selection must be made for u and v' 5 Raises powers correctly to a form P( x+ 5) + Q( x+ 5) 5 Or by parts C( 58 x)( x+ 5) + D( x+ 5) dd A full method to find the area of R using their limits. Dependent upon both M's

30 Question Number (a) (b) (c) dv dr Uses Scheme = π r dv dv dr 0 dr = = π r dt dr dt V( 0.05t+ ) dt dr 5 = dt π r 5 ( 0.05t+ ) dv 0 0dt = V dv = dt V 0. 05t 0. 05t ( + ) ( + ) Marks d, V 0(0.05t+ ) = + c 0. Sub V = when t = = 00 + c c = V = t+ (d) Sub 0 t = into ( ) Sub V = 0 into ( ) 00 V = 0 = 0 ( ) ( ) + V r r = π = awrt.6(m) d, (a) dv dv = π r Do not accept = π r dr dr (b) Uses d V d V d r = or equivalent correctly. (It is not enough to just state this) dt dr dt Follow through on their d V dr d Makes d r dt the subject and attempts to replace V with π r dr 5 dr A dr A = dt π r 5 ( 0.05t+ ) Allow = dt Bπ r 5 ( 0.05t+ ) or = dt 5 Bπ r 0 t+ where A and B are integers and A B cancels to 5 ( ) () () (6) () [ marks]

31 (c) 0dt Separates variables to achieve V dv = ( 0. 05t+ ) or equivalent (with or without the integral sign) Integrates lhs to a form av with or without a constant Integrates rhs to a form b(0.05t+ ) with or without a constant V 0(0.05t+ ) = + c or equivalent including a constant 0. Substitutes V = and t = 0 into an integrated expression of the form pv = q(0.05t + ) n + c, where n is an integer, to find the constant c V = 0 or equivalent V = 0 ( 0. 05t+ ) ( t + 0).. Note that there is a solution to part (c) via definite integration that marks similarly V t V t 0dt V 0( 0. 05t + ) V dv = = ( 0. 05t ) is as before V 0( 0. 05t + ) with the next scored when = + 00 oe 0. (d) Substitutes t = 0 into their equation for V which includes a numerical constant and finds a value for V or V (This cannot be scored for impossible values ie when V < 0 ) 6 6 d Substitutes their V into V = π r r =... Alt substitutes their V into V = π r r =... 9 cso r = awrt.6(m) in dr 5 Watch for candidates who solve (d) using the differential equation = dt π r 5 ( 0.05t+ ) and find r in terms of t. This is an acceptable method (although unlikely) but do consider this kind of solution carefully.

32 Pearson Education Limited. Registered company number 8788 with its registered office at 80 Strand, London, WCR 0RL, United Kingdom