Mark Scheme (Results) Summer Pearson Edexcel GCE in Core Mathematics 3 (6665/01)

Transcription

1 Mark Scheme (Results) Summer 05 Pearson Edexcel GCE in Core Mathematics 3 (6665/0)

2 Edexcel and BTEC Qualifications Edexcel 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: Summer 05 Publications Code UA0499 All the material in this publication is copyright Pearson Education Ltd 05

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4 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.

5 PEARSON EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75. 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. 3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) 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 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.

6 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.

7 General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving 3 term quadratic:. Factorisation ( 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 the correct formula (with values for a, b and c). 3. Completing the square b Solving x bx c 0 : x q c 0, q 0, leading to x = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( x n x n ). Integration Power of at least one term increased by. ( x n x n )

8 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 small errors 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.

9 Question Number Scheme Marks.(a) tanθ p tanθ = = -tan θ -p Final answer MA () (b) cosθ= = = secθ + tan θ + p (c) cot( θ 45) ( ) Final answer + tan θtan 45 + p - = = = tan θ-45 tan θ-tan 45 p- Final answer MA MA () () (6 marks) (a) M Attempt to use the double angle formula for tangent followed by the substitution tan p. tan θ p For example accept tan θ= = tan θ p tan θ Condone unconventional notation such as tan θ= followed by an attempt to tan θ substitute tan p for the M mark. Recovery from this notation is allowed for the A. tan A tan B Alternatively use tan( AB) with an attempt at substituting tan A tanb tan A tan B p. The unsimplified answer p p is evidence p p p sin θ sin θcosθ p p It is possible to use tan θ= = = but it is cosθ cos θ- - p unlikely to succeed. p p A Correct simplified answer of tanθ = or -. p ( - p)( + p) Do not allow if they ''simplify'' to - p Allow the correct answer for both marks as long as no incorrect working is seen.

10 (b) M Attempt to use both cosθ = secθ and tan + θ = sec θ with tan p in an attempt to obtain an expression for cos in terms of p. Condone a slip in the sign of the second identity. Evidence would be cos θ = p Alternatively use a triangle method, attempt Pythagoras' theorem and use adj cos θ = hyp The attempt to use Pythagoras must attempt to use the squares of the lengths. + p P A cosθ= Accept versions such as + cosθ =, cosθ= p + p + p Withhold this mark if the candidate goes on to write cosθ= + p (c) M Use the correct identity cot( θ - 45) = tan( θ - 45) and an attempt to use the tan( A B ) formula with A=θ, B=45 and tan p. For example accept an unsimplified answer such as tan tan 45 p tan 45 tan tan 45 p tan 45 It is possible to use cos ( θ - 45) cot( θ - 45) = sin ( θ - 45) and an attempt to use the formulae for sin( A B) and cos( A B ) with A=θ, B=45.sin θ= p and cosθ = p p Sight of an expression p cos 45 sin 45 p p is evidence. p cos 45 sin 45 p p A Uses tan 45 orsin 45 cos 45 oe and simplifies answer. Accept + p or + - p p - Note that there is no isw in any parts of this question.

11 Question Number Scheme Marks Shape B.(ai) 5 and ln,0 0, 3 B y 5 B (3) Shape inc cusp Bft (aii) 5 and ln,0 0,3 y 5 Bft Bft (b) 5 x æ ö ln ç çè ø B ft (c) x æ3ö e - 5 =- ( x) = ln ç çè ø MA ( x) æ7ö = ln ç çè ø B (3) () (3) (0 marks)

12 (a)(i) B For an exponential (growth) shaped curve in any position. For this mark be tolerant on slips of the pen at either end. See Practice and Qualification for examples. ln 5,0 and 0, 3. Allow 5 ln and 3 being marked on the correct axes. 5 Condone 0,ln and 3,0being marked on the x and y axes respectively. B Intersections with the axes at 5 Do not allow ln,0appearing as awrt (0.9, 0) for this mark unless seen elsewhere. Allow if seen in body of script. If they are given in the body of the script and differently on the curve (save for the decimal equivalent) then the ones on the curve take precedence. B Equation of the asymptote given as y 5. Note that the curve must appear to have an asymptote at y 5, not necessarily drawn. It is not enough to have -5 marked on the axis or indeed x 5. An extra asymptote with an equation gets B0 (a)(ii) Bft Bft For either the correct shape or a reflection of their curve from (a)(i) in the x- axis. For this to be scored it must have appeared both above and below the x - axis. The shape must be correct including the cusp. The curve to the lhs of the cusp must appear to have the correct curvature Score for both intersections or follow through on both the intersections given in part (a)(i), including decimals, as long as the curve appeared both above and below the x - axis. See part (a) for acceptable forms y or follow through on an asymptote of y Cfrom part (a)(i). Note that the curve must appear to have an asymptote at y Cbut do not penalise if the first mark in (a)(ii) has been withheld for incorrect curvature on the lhs. Bft Score for an asymptote of 5 (b) æ ö Bft Score for x ln 5 ç, x awrt 0.9 or follow through on the x intersection in part (a) çè ø (c) x x e 5 or e 5 x..ln.. M Accept A B Allow squaring so x x e 5 4 e.. and.. x ln(..), ln(..) æ ö x ln 3 = ç or exact equivalents such as x = ln.5. You do not need to see the x. çè ø Remember to isw a subsequent decimal answer æ ö x ln 7 = ç çè or exact equivalents such as x = ln 3.5. You do not need to see the x. ø Remember to isw a subsequent decimal answer.5 If both answers are given in decimals and there is no working awrt.5, x award SC 00

13 Question Number Scheme Marks 3(a) 4cos sin R cos( ) R B arctan awrt 6.57 MA (b) 0 cos( θ- 6.6) = cos( θ- 6.57) = 0 M (3) ( θ ) = θ=... dm θ = awrt 5.8 A θ = '-77...' θ = -awrt 5.3 ddma (5) (c) k<- 0, k> 0 Bft either Bft both () (0 marks) You can marks parts (a) and (b) together as one. (a) B For R 0 5. Condone R 0 M For arctan or arctan leading to a solution of Condone any solutions coming from cos 4,sin Condone for this mark arctan.. If R has been used to find award for only 4 arcos arcsin ' R ' ' R ' A awrt 6.57

14 (b) =. their R æ ö θ their 6.57 = arccosç çètheirr ø cos θ their 6.57 = their R M Using part (a) and proceeding as far as cos( θ their 6.57) This may be implied by ( ) Allow this mark for ( ) dm Dependent upon the first M- it is for a correct method to find θ from their principal value Look for the correct order of operations, that is dealing with the ''6.57'' before the ''''. Condone subtracting 6.57 instead of adding. β their 6.57 cos( θ their 6.57 ) =... θ their 6.57 = β θ= A awrt θ = 5.8 ddmfor a correct method to find a secondary value of θin the range Either θ 6.57 = '- β' θ = OR θ 6.57 = 360-' β' θ = THEN MINUS 80 A awrt θ =-5.3 Withhold this mark if there are extra solutions in the range. Radian solution: Only lose the first time it occurs. FYI. In radians desired accuracy is awrt dp (a) 0.46 and (b) 0.90, 0.44 Mixing degrees and radians only scores the first M (c) Bft Follow through on their R. Accept decimals here including 0» awrt 4.5. Score for one of the ends k> 0, k<- 0 Condone versions such as g() θ > 0, y> 0 or both ends including the boundaries k 0, k - 0 B ft For both intervals in terms of k. Accept k> 0 or k<- 0. Accept 0 k > Accept k Î ( 0, )(-,- 0 ) Condone k> 0, k<- 0 k> 0and k<- 0 for both marks but - 0> k> 0 is B B0

15 Question Number 4(a) ( θ = ) 0 Scheme B Marks () - 40 (b) Sub t = 40, θ = = 0-00e λ (c) 40 λ e - = 0.5 MA ln λ = MA 40 ln 0. θ = 00 T = M - their ' λ ' T = awrt 93 A (4) () (7 marks) - 40 Alt (b) Sub t θ λ = 40, = 70 00e = 50 ln00-40λ = ln 50 MA ln00- ln 50 ln λ = = MA (4)

16 (a) B Sight of ( θ = ) 0 (b) λ M Sub t =, θ = = - e λ and proceed to e = Awhere A is a constant. Allow sign slips and copying errors. 40 A e - λ = 0.5, 40 or e λ = or exact equivalent M For undoing the e's by taking ln's and proceeding to λ =.. ln 0.5 May be implied by the correct decimal answer awrt 0.07 or λ = -40 A ln cso λ = 40 Accept equivalents in the form ln a, ab, b Î such as λ = ln 4 80 (c) λt M Substitutes θ = 00 and their numerical value of λinto θ = 0-00e - and proceed ln 0. to T = their ' λ ' or ln 5 T = their ' λ ' Allow inequalities here. A awrt T = 93 Watch for candidates who lose the minus sign in (b) and use then reach T =-93and ignore the minus. This is M A0 ln λ = 40 in (c). Many

17 Question Number Scheme Marks 5.(a) p = π 4 or ( π) B dx dy MA π dx Sub y = into 4 y siny4 cosy dy dx =4 π dy (= 75.4) / = ( = 0.03) dy dx 4π M Equation of tangent y- = ( x-4π ) 4π M Using y - π = π ( x- 4π ) with x= 0 y= 3 cso M, A (b) x 4y siny 4y siny4 cos y 4π () (6) Alt (b) I 0.5 x 4y sin y x 4y sin y x 0.5x 4 cosy dy 0.5 d (7 marks) MA Alt (b) x II 6y 8ysin y sin y dy dy dy dy 3y 8siny 6ycosy 4sinycosy MA dx dx dx dx Or dx 3y dy 8sin y dy 6ycosy dy 4sin ycosy dy (a) B p = π 4 or exact equivalent ( π) Also allow x = 4 π (b)

18 M Uses the chain rule of differentiation to get a form A(4 y sin y) B C cos y, A, B, C 0 on the right hand side Alternatively attempts to expand and then differentiate using product rule and chain rule to a dx form x6 y 8ysinysin y PyQsinyRycos yssinycos y dy PQRS,,, 0 A second method is to take the square root first. To score the method look for a differentiated 0.5 expression of the form Px... 4 Qcos y A third method is to multiply out and use implicit differentiation. Look for the correct terms, condoning errors on just the constants. A dx 4 y siny4 cosy dy dy or dx 4ysiny4cosy with both sides correct. The lhs may be seen elsewhere if clearly linked to the rhs. In the alternative d x 3y 8sin y 6ycosy 4sin ycosy dy M Sub y into their d x dy x. Evidence could be minimal, eg dx y... dy dy It is not dependent upon the previous M but it must be a changed x 4y siny M Score for a correct method for finding the equation of the tangent at '4 ',. Allow for π ( ) ( y- = x their π ) their numerical d - 4 d Allow for æ πö y their numerical dx ç - ( ) = ( x-their 4π dy ) çè ø Even allow for π y ( ) ( x p ) their numerical d d It is possible to score this by stating the equation y = x+ c as long as '4 ', 4π is used in a subsequent line. M Score for writing their equation in the form y mx c and stating the value of 'c' Or setting x 0 in their y- = ( x-4π ) and solving for y. 4π Alternatively using the gradient of the line segment AP = gradient of tangent. y Look for y.. Such a method scores the previous M mark as well. 4 4 At this stage all of the constants must be numerical. It is not dependent and it is possible to score this using the ''incorrect'' gradient. A π cso y =. You do not have to see0, 3 3

19 Question Number Scheme Marks 6.(a) x+ x+ - 3= 7-x = 0- x M ( x+ ) ln = ln( 0 - x) x =... dm ln(0 x) x A* ln ln(0 xn ) (b) Sub x 0 3 into xn, x (awrt) MA ln x 3.080, x 3.08 (awrt) A 3 (c) A= ( 3.,3.9) cao M,A (3) (3) 6.(a)Alt 6.(a) () (8 marks) x+ x 0- x - 3= 7-x = M 0- x xln = ln x =... dm ln(0 x) x A* ln ln(0 x) x ( x + )ln = ln( 0 - x) M ln x+ backwards = 0- x dm x + Hence y = - 3 meets y= 7- x A* (3) (3)

20 (a) M Setting equations in x equal to each other and proceeding to make dm Take ln s or logs of both sides, use the power law and proceed to x =.. x+ the subject A* This is a given answer and all aspects must be correct including ln or log e rather than log 0 Bracketing on both ( x + ) and ln( 0- x) must be correct. ln( 0- x) Eg x+ ln = ln( 0-x) x = - is A0* ln x Special case: Students who start from the point + = 0-x can score M dma0* (b) ln(0 x n ) M Sub x 0 3 into x n to find x.. ln ln(0 3) Accept as evidence x, awrt x 3. ln ln(0 3) Allow x 0 3 into the miscopied iterative equation x to find x.. ln ln(0 3) Note that the answer to this, 4.087, on its own without sight of is M0 ln A awrt 3 dp x A awrt x 3.080, x Tolerate 3.08 for Note that the subscripts are not important, just mark in the order seen (c) Note that this appears as BB on e pen. It is marked MA M For sight of 3. Alternatively it can be scored for substituting their value of x or a rounded value of x from (b) into either x+ - 3 or 7 - x to find the y coordinate. 3.,3.9 A ( )

21 Question Number 7.(a) Applies vu ' uv' to x x 3 e x Scheme x g'( x) x x -e x 3x e Marks 3 x M A 3 g'( ) 5 e x x x x x A (b) Sets Sub, 3 x 3 x 5x x e 0 x 5x x 0 M x x 5x 0 x (0),, M,A 3 x 4 gx () xx e g, g() 4 dm,a 8e e Range - 4 g( x) A e 4 8e x into (c) Accept g( x ) is NOT a ONE to ONE function (3) (6) Accept g( x ) is a MANY to ONE function B - Accept g ( x ) would be ONE to MANY () Note that parts (a) and (b) can be scored together. Eg accept work in part (b) for part (a) (a) 3 x M Uses the product rule vu ' uv' with u x x and v e or vice versa. If the rule is quoted it must be correct. It may be implied by their u.. v.. u'.. v'.. followed by their vu ' uv'. If the rule is not quoted nor implied only accept expressions of the form 3 x e e x x x A BxCx condoning bracketing issues x 3 x Method : multiplies out and uses the product rule on each term of x e x e Condone issues in the signs of the last two terms for the method mark Uses the product rule for uvw = u ' vw+ uv ' w+ uvw' applied as in method 3 x Method 3:Uses the quotient rule with u x x and v e. If the rule is quoted it must be correct. It may be implied by their u.. v.. u'.. v'.. followed by their vu ' uv ' If the v x e rule is not quoted nor implied accept expressions of the form x e condoning missing brackets on the numerator and e x on the denominator. Ax Bx x x Ce 3 x x 3 x dy x Method 4: Apply implicit differentiation to ye x x e ye x 3x dx Condone errors on coefficients and signs (0 marks)

22 A A correct (unsimplified form) of the answer 3 g'( ) -e x x x x x x3x e by one use of the product rule x x 3 x x or g'( x) x -e xe x -e 3x e using the first alternative x x x or g '( x) x( x)e x e x ( x) e using the product rule on 3 terms x x x x x or g'( x) x e A Writes 3 g'( x) x 5x xe x e 3 e 3 x using the quotient rule.. You do not need to see f( x) stated and award even if a correct g'( x) is followed by an incorrect f( x ). If the f(x) is not simplified at this stage you need to see it simplified later for this to be awarded. (b) Note: The last mark in e-pen has been changed from a B to an A mark M For setting their f(x) =0. The = 0 may be implied by subsequent working. Allow even if the candidate has failed to reach a 3TC for f(x). Allow for f( x) 0or f( x) 0as they can use this to pick out the relevant sections of the curve M For solving their 3TC = 0 by ANY correct method. Allow for division of x or factorising out the x followed by factorisation of 3TQ. Check first and last terms of the 3TQ. Allow for solutions from either f( x) 0or f( x) 0 Allow solutions from the cubic equation just appearing from a Graphical Calculator A x,. Correct answers from a correct g'( x) would imply all 3 marks so far in (b) dm Dependent upon both previous M s being scored. For substituting their two (non zero) values of x into g(x) to find both y values. Minimal evidence is required x=.. y=.. is OK. A Accept decimal answers for this mark. 4 g =awrt AND g() =awrt e e A CSO Allow - 4 Range 4 e 8e, 4 é 4 ù - y, -, 4 e 4 4. Condone y- y 8e êë e 8eú 4 û e 8e Note that the question states hence and part (a) must have been used for all marks. Some students will just write down the answers for the range from a graphical calculator. Seeing just - 4 g( x) or g( x) special case e 8e They know what a range is! (c) - B If the candidate states NOT ONE TO ONE then accept unless the explicitly link it to g ( x). So accept It is not a one to one function. The function is not one to one g( x ) is not one to one If the candidate states IT IS MANY TO ONE then accept unless the candidate explicitly - links it to g ( x). So accept It is a many to one function. The function is many to one g( x ) is many to one If the candidate states IT IS ONE TO MANY then accept unless the candidate explicitly links it to g( x ) Accept an explanation like '' one value of x would map/ go to more than one value of y'' - - Incorrect statements scoring B0 would be g ( x) is not one to one, g ( x) is many to one and g( x) is one to many.

23 Question Number 8(a) Scheme sin A seca+ tan A= + cosa cosa + sin A = cos A + sin A cos A = cos A - sin A cos A+ sin A+ sin Acos A = cos A-sin A (cos A+ sin A)(cos A+ sin A) = (cos A+ sin A)(cos A-sin A) cos A + sin A = cos A - sin A (b) sec θ+ tan θ = cosθ + sin θ = cosθ- sin θ Marks B M M M A* (5) cosθ+ sin θ= cosθ-sin θ tan θ =- 3 M A θ = awrt.80, 5.96 dma (9 marks) (a) B A correct identity for sec A = sin A OR tan A =. cosa cosa It need not be in the proof and it could be implied by the sight of sec A = cos A - sin A M For setting their expression as a single fraction. The denominator must be correct for their fractions and at least two terms on the numerator. + cosa tan A + sin A This is usually scored for or cos A cosa M For getting an expression in just sin A and cos A by using the double angle identities sina= sin Acos A and cos A= cos A- sin A, cos A - or - sin A. Alternatively for getting an expression in just sin A and cos A by using the double angle tan A sin A identities sin A = sin Acos A and tan A = with tan A =. - tan A cos A sin A For example = + cos A is BM0M so far cos A-sin A -sin A cos A M In the main scheme it is for replacing by cos A + sin A and factorising both numerator and denominator (4)

24 A* Cancelling to produce given answer with no errors. Allow a consistent use of another variable such as θ, but mixing up variables will lose the A*. (b) M For using part (a), cross multiplying, dividing by cosθ to reach tan θ = k Condone tan θ = k for this mark only A tan θ =- 3 dm Scored for tanθ = k leading to at least one value (with dp accuracy) for between 0 and π. You may have to use a calculator to check. Allow answers in degrees for this mark. A θ = awrt.80, 5.96 with no extra solutions within the range. Condone.8 for.80. You may condone different/ mixed variables in part (b)... There are some long winded methods. Eg. M, dm applied as in main scheme ( ) ( ) cosθ+ sin θ = cosθ-sin θ 4+ 4sin θ = -sin θ 3 sin θ =- is M ( forsin θ = k ) A 5 arcsin k θ =.80, 5.96for dm (for θ = ) A... æ 3 ö cosθ+ 3sin θ= 0 ( 0) cos( θ-.5) = 0 M for..cos( θ- α) = 0, α= arctan ç or çè 3 ø ) A θ =.80,5.96 dm A ( ) ( ) cosθ+ 3sin θ= 0 0 sin θ+ 0.3 = 0 M A θ =.80,5.96 dm A... cosθ=-3sinθ cos θ= 9sin θ sin θ= sin θ= ( ) M A 0 0 θ =.80,5.96 dm A 9 9 cos θ=-3sin θ cos θ= 9sin θ cos θ= cosθ= ( ) M A 0 0 θ =.80,5.96 dm A...

25 Question Number Alt I From RHS Alt II Both sides Scheme Marks cos A + sin A cos A + sin A cos A+ sin A = cos A-sin A cos A - sin A cos A+ sin A cos A+ sin A+ sin Acos A = cos A-sin A + sin A = cos A sin A = + cosa cosa (Pythagoras) M (Double Angle) M (Single Fraction) M = seca+ tan A B(Identity), A* cos A + sin A Assume true seca+ tan A= cos A - sin A sin A cos A+ sin A + = cosa cosa cos A- sin A + sin A cos A+ sin A = cos A cos A- sin A + sin A cos A cos A+ sin A = cos A -sin A cos A-sin A + sin Acos A (cos A -sin A) = cos A+ sin A cos A+ sin A B (identity) M (single fraction) M(double angles) + sin Acos A= cos A+ sin Acos A+ sin A= + sin Acos ATrue M(Pythagoras)A* Alt seca + tan A= + tan A cos A tan A = + cos A - tan A Very - tan A+ tan AcosA = difficult cos A( -tan A) - tan A+ tan A(cos A-sin A) = (cos A-sin A)( -tan A) sin A sin A - + (cos A-sin A) = cos A cos A æ sin A ö (cos A-sin A) - ç çè cos A ø cos A- sin A+ sin Acos A(cos A-sin A) cos A = (cos A-sin A)(cos A-sin A) (cos A- sin A))( + sin Acos A) = (cos A-sin A)(cos A-sin A) Final two marks as in main scheme (Identity) B (Single fraction) M (Double Angle and in just sin and cos ) M MA*

26 Question Number Scheme Marks 9.(a) x 3kx k ( x k)( x k) B ( x 5 k)( xk) ( x5 k) ( x k) ( x5 k) ( x k)( xk) ( x k) ( x k) M xk ( x k) A* (3) vu ' uv ' x k (b) Applies to y with u x kand v x k v x k ( x k) ( xk) f'( x) ( x k) 3k f'( x) ( x k) M, A A (3) Ck (c) If f'( x) ( x k) 3k f'( x) 0 ( x k) f(x) is an increasing function as f'( x) 0, for all values of x as negative negative = positive positive M A () (8 marks) (a) B For seeing x 3kx k ( x k)( x k) anywhere in the solution M For writing as a single term or two terms with the same denominator ( x 5 k) ( x k) ( x 5 k) Score for or ( x k ) ( x k ) ( x5 k)( xk) ( x k)( xk) ( x5 k)( xk) x k ( x k )( xk ) ( x k )( xk ) x 3kx k xk A* Proceeds without any errors (including bracketing) to ( x k)

27 (b) vu ' uv ' x k M Applies to y with u x kand v x k. v x k If the rule it is stated it must be correct. It can be implied by u x kand v x k with vu ' uv ' their u', v' and v xkxk If it is neither stated nor implied only accept expressions of the form f'( x) ( x k) The mark can be scored for applying the product rule to y ( x k)( x k) If the rule it is stated it must be correct. It can be implied by u x kand v x k with their u', v' and vu' uv' If it is neither stated nor implied only accept expressions of the form f'( x) x k xkx k x k 3k Alternatively writes y as y x k x k and differentiates to dy A d x ( x k) A Any correct form (unsimplified) form of f'( x ). ( x k) ( xk) f'( x) by quotient rule ( x k) f'( x) xk xkxk by product rule 3k and f'( x) by the third method ( x k) 3k 3k A cao f'( x). Allow f'( x) ( x k) x 4kx4k As this answer is not given candidates you may allow recovery from missing brackets (c) Note that this is B B on e pen. We are scoring it M A Ck M If in part (b) f'( x), look for f(x) is an increasing function as f'( x)/gradient 0 ( x k) Accept a version that states as k 0 Ck 0 hence increasing ( ) Ck If in part (b) f'( x), look for f(x) is an decreasing function as f'( x )/gradient<0 ( x k) Similarly accept a version that states as k 0 ( ) Ck 0 hence decreasing 3k A Must have f'( x) and give a reason that links the gradient with its sign. ( x k) There must have been reference to the sign of both numerator and denominator to justify the overall positive sign.

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