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2 Mark Scheme (Results) Summer 009 GCE GCE Mathematics (6668/0)

3 June Further Pure Mathematics FP (new) Mark Scheme Question Number Scheme Marks Q (a) = rr ( + ) r ( r+ ) r ( r+ ) B aef () (b) n n r = r = 4 = rr ( + ) r r+ = n n + n n + List the first two terms and the last two terms Includes the first two underlined terms and includes the final two = + ; underlined terms. n + n + + n + n + A = n + n + = = ( n + )( n + ) ( n + ) ( n + ) ( n + )( n + ) n + 9n + 6 n 4 n ( n + )( n + ) Attempt to combine to an at least term fraction to a single fraction and an attempt to take out the brackets from their numerator. = n + 5n ( n + )( n + ) = n(n + 5) ( n + )( n + ) Correct Result A cso AG (5) [6] 6668/0 GCE Mathematics June 009

4 Question Number Scheme Marks Q (a) z = 4 4 i, π < θ π y O α 4 arg z x 4 (4, 4 ) r = ( 4 ) + ( 4 ) = + = 64 = 8 4 π θ = tan ( 4 ) = 4 ( ( ) isin 4 ( 4) ) π π z = 8cos + ( ) π π 4 4 So, z = 8 cos + isin A valid attempt to find the modulus and argument of 4 4 i. Taking the cube root of the modulus and dividing the argument by. π π π π z = cos ( ( ) + isin( )) cos ( ( ) isin( )) + A Also, ( ( 7 π) ( 7 π z = 8cos + isin 4 4 )) or ( ( 9 π) ( 9 π z = 8cos + isin 4 4 )) Adding or subtracting π to the argument for z in order to find other roots. 7 7 ( isin ) = + π π z cos ( ( ) ( )) π π 4 4 Any one of the final two roots A and z = cos + isin Both of the final two roots. A π π Special Case : Award SC: AA0A0 for ALL three of ( + ) π π ( isin ) cos 7π 7π + and cos( ) isin( ) 4 4 ( ) +. cos isin, [6] Special Case : If r is incorrect (and not equal to 8) and candidate states the brackets ( )correctly then give the first accuracy mark ONLY where this is applicable. 6668/0 GCE Mathematics June 009

5 Question Number Scheme Marks Q dy sin x ycos x sin xsin x = d y cos sin sin y x = x x sinx sinx An attempt to divide every term in the differential equation by sin x. Can be implied. dy ycosx = sinx sin x Integrating factor cos x ± ( ) sin x cos x ± e or e d x sin x ln sin x = e = e ln sin x e or their P( x) ( ) ln cosec x e d A aef = sin x sin x or (sin x) or cosec x A aef dy ycosx sinx = x x x x sin d sin sin d y = sin x sinx sinx d ( y ) their I.F. = sin x their I.F d y = sinx cosx d y = cosx or sinx y cosx ( ) sin x = A y cosx sin x = y sin x K sin x = + A credible attempt to integrate the RHS with/without + K ddd y = sin x + Ksinx y sin x Ksin x = + A cao [8] 6668/0 GCE Mathematics June 009 4

6 Question Number Scheme Marks Q4 π A = + 0 ( a cosθ ) dθ π with Applies r ( dθ) 0 correct limits. Ignore dθ. B ( a + cos ) = a + 6acos + 9cos θ θ θ + cosθ = a + 6acosθ + 9 ± ± cosθ cos θ = Correct underlined expression. A π 9 9 = + 6 cosθ + + cosθ dθ A a a 0 Integrated expression with at least out of 4 terms of the form ± Aθ ± Bsinθ ± Cθ ± Dsin θ. * π 9 9 Ignore the θ 6 sinθ θ sinθ. Ignore limits. 4 0 a θ + 6asinθ + correct ft integration. A ft Ignore the. Ignore limits. = a + a + + = ( π a 0 9π 0) ( 0) = π a + 9 π π π a + A 9 Hence, 9π 07 π a + = π Integrated expression equal to 07 π a + = d* a = 49 As a > 0, a = 7 a = 7 Some candidates may achieve a = 7 from incorrect working. Such candidates will not get full marks A cso [8] 6668/0 GCE Mathematics June 009 5

7 Question Number Scheme Marks Q5 y = sec x = (sec x) (a) d d y (sec x) (sec xtan x) sec xtan x x = = Either (sec x) (sec xtan x ) or sec x tan x B aef Apply product rule: u = sec x v = tan x du dv = 4sec x tan x = sec x d y 4 = 4sec x tan x + sec x Two terms added with one of 4 either Asec xtan xor Bsec x in the correct form. Correct differentiation A 4 = 4sec x(sec x ) + sec x (b) d y Hence, 4 = 6sec x 4sec x dy y π = ( ) =, ( ) 4 d x π 4 = () = 4 Applies tan x = sec x leading to the correct result. Both dy y π = and 4 d x π 4 = 4 A AG B (4) d y π 4 ( ) ( ) 4 = 6 4 = 4 8= 6 π Attempts to substitute x = into 4 both terms in the expression for d y. d y = x x x x x x 4sec (sec tan ) 8sec (sec tan ) Two terms differentiated with 4 either 4sec x tan x or 8sec x tan x being correct 4 = 4sec x tan x 8sec xtan x π 4 ( ) ( ) d y 4 d y = 4 () 8 () = 96 6 = 80 = π 4 80 B π 6 π 80 π ( ) ( ) ( ) sec x + 4 x + x + x π π { ( ) ( ) ( ) } 40 π sec x + 4 x + 8 x + x Applies a Taylor expansion with at least out of 4 terms ft correctly. Correct Taylor series expansion. A (6) [0] 6668/0 GCE Mathematics June 009 6

8 Question Number Scheme Marks Q6 z w =, z = i z + i (a) wz ( + i) = z wz+ iw= z iw= z wz i w i w= z( w) z = ( w) Complete method of rearranging to make z the subject. i w z = ( w) A aef z i w = = w Putting in terms of their z w = d iw = w w = w w = 9 w u + iv = 9 u + iv u + v = 9 ( u ) + v u + v = 9u 8u v 0= 8u 8u + 8v + 9 Applies w = u + iv, and uses Pythagoras correctly to get an equation in terms of u and v without any i's. Correct equation. dd A 0 = u u + v Simplifies down to u + v ± αu ± βv ± δ = 0. ddd ( ) u v + + = ( ) u v + = (b) 9 {Circle} centre ( ) v, 0, radius One of centre or radius correct. A 8 8 Both centre and radius correct. A Circle indicated on the Argand diagram in the correct position in follow through quadrants. Ignore plotted coordinates. Bft (8) O u Region outside a circle indicated only. B () [0] 6668/0 GCE Mathematics June 009 7

9 Question Number Scheme Marks Q7 y = x a, a > (a) y a Correct Shape. Ignore cusps. Correct coordinates. B B a O a x (b) x a a x a =, > () { x > a}, x a = a x x a = a x aef x + x a = 0 4()( a ) ± x = Applies the quadratic formula or completes the square in order to find the roots. ± + 8a x = Both correct simplified down solutions. A { x < a}, x + a = a x x + a = a x or x a = x a aef { x x = 0 x( x ) = 0} x = 0, x = 0 x = B A (6) (c) x a a x a >, > + 8a x < {or} + + 8a x > x is less than their least value x is greater than their maximum value B ft B ft {or} 0< x < For{ x < a}, Lowest < x < Highest 0< x < A (4) [] 6668/0 GCE Mathematics June 009 8

10 Question Number Scheme Marks Q8 d x t x = e, x = 0, = at t = 0. dt dt dt (a) AE, m + 5m + 6 = 0 ( m + )( m + ) = 0 m =,. So, mt mt Ae + Be, where m m. t t xcf = Ae + Be t t Ae + Be A d x x = ke = ke = ke dt dt t t t ke + 5( ke ) + 6ke = e ke = e k = t t t t t t { So, xpi = e t } Substitutes k e t into the differential equation given in the question. Finds k =. A So, x A B their x + their x * t t t = e + e + e CF PI dt t t t Ae Be e == Finds d x by differentiating dt their x and their x CF PI d* t = 0, x = 0 0 = A + B + t = 0, = = A B dt A + B = A B = A =, B = 0 Applies t = 0, x = 0 to x and t = 0, = to d x to dt dt form simultaneous equations. dd* So, x t = e + e t x t t = e + e A cao (8) 6668/0 GCE Mathematics June 009 9

11 Question Number Scheme Marks x t = e + e t (b) dt t t e e 0 = = Differentiates their x to give d x dt and puts d x equal to 0. dt = t e 0 t = ln A credible attempt to solve. d* t = ln or t = ln or awrt 0.55 A ln ln ln ln So, x = e + e = e + e x = + Substitutes their t back into x and an attempt to eliminate out the ln s. dd = + = = 9 uses exact values to give 9 A AG d x dt t = 9e + e t At t = d x ln, = 9e + e dt ln ln d x Finds dt d x and substitutes their t into dt d* 9 = 9() + = + = + As x = + = < 0 then x is maximum. d 9 dt 9 + < 0 and maximum conclusion. A (7) [5] 6668/0 GCE Mathematics June 009 0

12 Mark Scheme (Results) Summer 00 GCE Further Pure Mathematics FP (6668) Edexcel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

13 Edexcel is one of the leading examining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edexcel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on , our GCSE team on , or visit our website at If you have any subject specific questions about the content of this Mark Scheme that require the help of a subject specialist, you may find our Ask The Expert service helpful. Ask The Expert can be accessed online at the following link: Summer 00 Publications Code UA098 All the material in this publication is copyright Edexcel Ltd 00

14 June 00 Further Pure Mathematics FP 6668 Mark Scheme Question Number Scheme Marks (a) r r+ A () (b) n = = ( )( + ) n n+ Aft r r r n = = n+ (n+ ) * A () (c) Sum = f(000) f(99) = or A () 7 GCE Further Pure Mathematics FP (6668) Summer 00

15 Question Number Scheme Marks f() t = x cos x, f(0) = - B f ( t) = ( + sin x), dt f (0) = 0.5 A t t f ( t) = f (0) + t f (0) + f (0) + f (0) +...! = 0.5t 0.5 t t +... A 5 GCE Further Pure Mathematics FP (6668) Summer 00

16 Question Number Scheme Marks (a) ( x+ 4)( x+ ) ( x+ ) = 0, ( x+ )( x + 7x+ 0) = 0 so ( x+ )( x+ )( x+ 5) = 0 or alternative method including calculator Finds critical values and -5 A A Establishes x > - Aft Finds and uses critical value to give 5 < x < - A (6) (b) x > - Bft () 7 GCE Further Pure Mathematics FP (6668) Summer 00

17 Question Number Scheme Marks 4(a) Modulus = 6 B Argument = π arctan( ) = A () (b) z π π = 6 (cos( ) + i sin( )) = 6 (cos π + isin π ) =4096 or 6 A () (c) π π π π 4 4 w = 6 (cos( ) + isin( )) = (cos + isin ) 6 6 OR + i OR ior i ( = + i) Aft A(,0) (5) 0 GCE Further Pure Mathematics FP (6668) Summer 00

18 Question Number Scheme Marks 5(a).5 + sin θ = sin θ = 0.5 π 5π θ = or 6 6, A, π 5π and θ = or 8 8 A () (b) Area = = θ, - π, 9 5π 8 (.5 + sin θ ) d π 8 (.5+ sin θ + ( cos6 θ))dθ - 9 π 5π 8 π 8 = (.5θ cos θ + ( θ sin 6 θ)) 6 5π 8 π 8-9 π A 5π = A 4 6 (7) 0 GCE Further Pure Mathematics FP (6668) Summer 00

19 Question Number Scheme Marks 6(a) Imaginary Axis Re(z) = 0 6 Real axis Vertical Straight line Through on real axis B B () (b) These are points where line x = meets the circle centre (, 4) with radius 5. The complex numbers are + 9i and i. A A () (c) 0 0 z 6 z w 6 w = = 0 6w = 0 5 w = 5 A This is a circle with Cartesian equation ( u 5) v 5 + = A (5) 0 GCE Further Pure Mathematics FP (6668) Summer 00

20 Question Number Scheme Marks 7(a) d y d d = y z. and d y z dz dz = so d y. d z = z A Substituting to get dz z z x z = and thus d z ztanx = A. 4 tan (5) (b) tanxx d I.F. = e = lncosx e = cos x A d ( cos ) = z x cos x = zcos x cos x zcos x = = + + z tan x xsec x csec x (cos x + ) sin x = + x +c 4 A A (6) (c) y ( tan x xsec x csec x) = + + Bft () GCE Further Pure Mathematics FP (6668) Summer 00

21 Question Number Scheme Marks 8(a) Differentiate twice and obtaining dy d y = λsin 5x + 5λxcos5x and = 0λcos5x 5λxsin 5x A Substitute to give λ = A 0 (4) (b) Complementary function is y = Acos5x+ Bsin 5x or Pe 5ix 5ix + Qe A So general solution is y = Acos5x+ Bsin 5x+ sin 5 0 x x or in exponential form Aft () (c) y= 0 when x = 0 means A = 0 B dy 5B cos5x + x + xcos5x and at x = 0 d y 5 0 So B = A So y = sin 5x+ sin 5 0 x x A (5) (d) "Sinusoidal" through O amplitude becoming larger Crosses x axis at π π π 4π,,, B B () 4 GCE Further Pure Mathematics FP (6668) Summer 00

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23 Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fax Order Code UA098 Summer 00 For more information on Edexcel qualifications, please visit Edexcel Limited. Registered in England and Wales no Registered Office: One90 High Holborn, London, WCV 7BH

24 Mark Scheme (Results) June 0 GCE Further Pure FP (6668) Paper

25 Edexcel is one of the leading examining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edexcel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on or visit our website at If you have any subject specific questions about the content of this Mark Scheme that require the help of a subject specialist, you may find our Ask The Expert service helpful. Ask The Expert can be accessed online at the following link: June 0 Publications Code UA07968 All the material in this publication is copyright Edexcel Ltd 0

26 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.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark

27 June 0 Further Pure Mathematics FP 6668 Mark Scheme Question Number Scheme Marks. x= ( x 4)( x+ ) x 4x = 0 x=, x= 6 A both Other critical values are x=, x= 0 B, B < x<, 0< x< 6 A A st for ±( x 4x ) =0 not required. B marks can be awarded for values appearing in solution e.g. on sketch of graph or in final answer. nd for attempt at method using graph sketch or +/- If cvs correct but correct inequalities are not strict award AA0. (7) 7 GCE Further Pure Mathematics FP (6668) June 0

28 Question Number. (a) y Scheme x d y dy dy x dy = e y y e y y d y x d y dy dy = e y y y d (k = 4) Marks A A () d y 4 6 = + + = 0 d y 0 = e ( ) = 0 0 6x 0x y = + x+ + = + x+ x + 5x 6 0 (b) e ( ) (a) (b) st for evidence of Product Rule st A for completely correct expression or equivalent nd A for correct expression or k = 4 stated nd require four terms and denominators of and 6 (might be implied) A follow through from their values in the final answer. B B Aft (4) 7 GCE Further Pure Mathematics FP (6668) June 0

29 Question Number Scheme Marks. dy y ln x + 5 = x x 5 Integrating factor e x 5 x 5ln x e = e 5 = x A 4 x ln x x x ln x = x ln x x = 4 6 ( + C) A x ln x x ln x C x y = + C y = x 6x x A A st for attempt at correct Integrating Factor st A for simplified IF nd ln x for times their IF to give their x ln x x rd for attempt at correct Integration by Parts nd A for both terms correct rd A constant not required 4 th x 5 y = their answer + C (8) 8 GCE Further Pure Mathematics FP (6668) June 0

30 Question Number 4. (a) ( r ) ( r) ( r) ( r) Scheme + = (b) ( r ) ( r) ( r) ( r) ( r ) ( r ) r (c) Marks A= 8, B=, C = 6 A = + + = 4 + (*) Acso r = : = 4 + r = : 5 = 4 + : : ( ) ( ) r = n: n+ n = 4 n + A () () (a) (b) (c) Summing: (n+ ) = 4 r + ( ) ( ) = n B n r = n n+ n+ Acso 6 Proceeding to ( )( ) r= st require coefficients of,,, or equivalent st require,-,,- or equivalent st for attempt with at least, and n if summing expression incorrect. RHS of display not required at this stage. st A for, and n correct. nd require cancelling and use of 4r + Award B for correct kn for their approach nd A is for correct solution only (5) 9 GCE Further Pure Mathematics FP (6668) June 0 4

31 Question Number 5. (a) x ( y ) + = 4 Scheme A Marks () (b) : Sketch of circle A: Evidence of correct centre and radius A (c) ( x+ i y) + i x+ i( y+ ) w = = + i( x + i y) ( y) + ix ( ) ( ) ( ) ( ) x + i y+ y ix = y + ix y ix On x-axis, so imaginary part = 0: ( y )( y) x + = 0 A ( y + )( y) x = 0 x + ( y ) = 4, so Q is on C Acso () (5) 9 Alt. (c) Let w= u+ iv: z + i u = + iz (since v = 0) u i z = ui d u i i u ( u i) z i = = ui ui A u + z i = =, so Q is on C u + Acso (a) Use of z = x + iy and find modulus (b) Award A0 if circle doesn t intersect x - axis twice (c) st M for subbing z = x + iy and collecting real and imaginary parts nd M for multiply numerator and denominator by their complex conjugate rd M for equating imaginary parts of numerator to 0 Award A for equation matching part (a), statement not required. GCE Further Pure Mathematics FP (6668) June 0 5

32 Question Number 6. Scheme Marks 5 π + cosθ = θ = B ( ) cos ( 4 4 cos cos ) + θ d = d + θ + θ θ sinθ θ = 4θ 4sinθ A Substituting limits 9π π = Area of triangle = ( )( ) cos sin 5 r θ r θ = = 4 A Area of R = π π = st for use of r dθ and correct attempt to expand nd for use of double angle formula - sin θ required in square brackets rd for substituting their limits 4 th for use of base x height 5 th area of sector area of triangle Please note there are no follow through marks on accuracy. A (9) 9 GCE Further Pure Mathematics FP (6668) June 0 6

33 Question Number 7. (a) Scheme Marks sin 5θ Im(cosθ i sin θ) cos θ (i sin θ) + 0 cos θ(i sin θ) + i sin θ 4 5 = i(5cos θ sinθ 0cos θsin θ + sin θ) A 5 Im(cosθ + i sin θ) 5 = 5sin θ( sin θ) 0sin θ( sin θ) + sin θ 5 = + B ( ) 5 sin 5 6sin 0sin 5sin θ = θ θ + θ (*) Acso (5) (b) 6sin 5 θ 0sin θ 5sinθ 5( sinθ 4sin θ) + = 5 6sin θ 0sinθ = 0 sin θ = 8 θ =.095 A Inclusion of solutions from sinθ = Other solutions: θ =.046, 4.7, 5.88 A sinθ = 0 θ = 0, θ = π (.4) B 4 5 (a) Award B if solution considers Imaginary parts and equates to sin 5θ st for correct attempt at expansion and collection of imaginary parts nd for substitution powers of cos θ (b) st M for substituting correct expressions nd M for attempting to form equation Imply rd M if 4.7 or 5.88 seen. Award for their negative root. Ignore π but nd A0 if other extra solutions given. (6) GCE Further Pure Mathematics FP (6668) June 0 7

34 Question Number Scheme Marks 8. (a) m + 6m+ 9= 0 m= C.F. t x = ( A+ Bt) e A P.I. x = Pcos t+ Q sin t B x = Psint+ Qcost x = 9Pcost 9Qsint ( P t Q t) ( P t Q t) ( P t Q t) 9 cos 9 sin + 6 sin + cos + 9 cos + sin = cos 9P+ 8Q+ 9P= and 9Q 8P+ 9Q= 0 P = 0 and Q = 8 A x = ( A+ Bt) e + sint Aft 8 t (b) t = 0: x = A = B t t x& = ( A + Bt)e + Be + cost 8 4 t = 0: x& = A + B + = 0 B = A 6 4t t x = + e + sin t A 8 (c) 59π t ( 0.9) 6 B x 8 Bft (a) (b) st Form auxiliary equation and correct attempt to solve. Can be implied from correct exponential. nd for attempt to differentiate PI twice rd for substituting their expression into differential equation 4 th for substitution of both boundary values st for correct attempt to differentiate their answer to part (a) nd for substituting boundary value (8) (5) () 5 GCE Further Pure Mathematics FP (6668) June 0 8

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37 Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fax Order Code UA07968 June 0 For more information on Edexcel qualifications, please visit Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Essex CM0 JE

38 Mark Scheme (Results) Summer 0 GCE Further Pure FP (6668) Paper

39 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert service helpful. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 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 0 Publications Code UA040 All the material in this publication is copyright Pearson Education Ltd 0

40 Summer Further Pure FP Mark Scheme 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.

41 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.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for example, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks.

42 General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving term quadratic:. Factorisation ( x + bx + c) = ( x + p)( x + q), where pq = c, leading to x =... ( ax + bx + c) = ( mx + p)( nx + q), where pq = c and mn = a, leading to x =. Formula Attempt to use correct formula (with values for a, b and c), leading to x =. Completing the square Solving + bx + c = 0 x b : ( ) x ± ± q ± c, 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+ ) Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are mistakes in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may be lost if there is any mistake in the working. Exact answers Examiners reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals. Answers without working The rubric says that these may not gain full credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done in your head, detailed working would not be required.

43 Summer Further Pure Mathematics FP Mark Scheme Question Number Scheme Marks. x 4 = x and x 4 = x, or graphical method, or squaring both sides, leading to x =... ( x = 4, x = ) x =, x = 4 anywhere Using only critical values to find an inequality x < x > 4 Notes y seen both strict, ignore and B B d A (5) x st accept or Require modulus of parabola and straight line with positive gradient through origin for graphical method. st B for x=, nd B for x=4 nd dependent upon first A0 for error in solution of quadratic leading to correct answer.

44 Question Number Scheme Marks. y = r sin θ = sinθ + sinθ cosθ B dy = cosθ + cos θ dθ 4cos θ + cosθ = 0 Aoe ± + cosθ = A OP = r = + = A 4 4 Notes B for or st for use of Product Rule or Chain Rule (require or condone ½) st A equation required nd Valid attempt at solving term quadratic (usual rules) to give (7) 7 nd A for exact or dp or better ( and ) rd for +x their rd A for any form A0 if negative solution not discounted.

45 Question Number Scheme Marks. (a) r = ( ) + ( ) = 4 B tanθ = (Also allow M mark for tan θ = ) M mark can be implied by π θ = ± or θ = ± π π θ = A (b) Finding the 4 th root of their r: r = 4 4 ( = ) For one root, dividing their θ by 4: π θ = 4 = π 6 For another root, add or subtract a multiple of π to their θ and divide by 4 in correct order. (cos θ + i sin θ ), where π π π θ =,,, 6 6 A A (a) Notes Accept A Accept awrt.. A0 if in degrees. (b) nd for awrt 0.5 st A for two correct values nd A for all correct values values in correct form and no more () (5) 8

46 Question Number Scheme Marks 4. m + 5m + 6 = 0 m =, C.F. ( x ) Ae Be t t = + A P.I. x = Pcos t + Qsin t xɺ = Psint + Qcost B xɺɺ = Pcost Q sint ( P cost Qsin t) + 5( Psin t + Q cost) + 6( P cost + Qsin t) = cost sin P + 5 Q + 6P = and Q 5P + 6Q =, and solve for P and Q P = 0 and Q = 0 A A x = Ae t + Be t + 0 cost + 0 sin t B ft (9) 9 Notes st form quadratic and attempt to solve (usual rules) st B Accept negative signs for coefficients. Coefficients must be different. nd for differentiating their trig PI twice rd for substituting x, xɺ and xɺɺexpressions 4 th Form equations in two unknowns and attempt to solve st A for one correct, nd A for two correct nd B for x=their CF + their PI as functions of t Condone use of the wrong variable (e.g. x instead of t) for all marks except final B.

47 Question Number Scheme Marks 5. (a) d y dy dy x + = + y (Using differentiation of product or quotient and also differentiation of implicit function) d y dy x + ( y) = **ag** A cso (b) d y d y x d d x x B d y dy... ( y) 0 = A At x = : d y = 4 B d y d y = 7 = B, B f ()( x ) f ()( x ) ( y = )f() + f ()( x ) y = + 4( x ) + ( x ) + ( x ) (or equiv.) A ft (a) (b) Notes Finding second derivative and substituting into given answer acceptable st for differentiating second term to obtain an expression involving d y and dy BBB for 4,7, seen respectively nd require f () or,f () etc and x-and at least first terms A for 4 terms following through their constants Condone f(x)= instead of y= () (8) 0

48 Question Number Scheme Marks 6. (a) (b) = =, r( r + ) r r + r r + 4 r = : r = : 4 r = : 5 r = n : n n + r = n : A n n + Summing: n = + = r( r + ) n + + n r B,Boe A () (c) ( n + )( n + ) ( n + ) ( n + ) n(n + 5) = = ( n + )( n + ) 4( n + )( n + ) Acao n n(6n + 5) = r= r( r + ) 4(n + )(n + ) Boe n(6n + 5) n(n + 5) Sn Sn = 4(n + )(n + ) 4( n + )( n + ) n(6n + 5)( n + ) n(n + 5)(n + ) = 4( n + )( n + )(n + ) n(6n + 7n + 0 6n n 5) n(4n + 5) = = 4( n + )( n + )(n + ) 4( n + )( n + )(n + ) (*ag*) (a) st and nd B Any form is acceptable (b) st must include at least 4 out of 5 of (r=),, and n-, n st A require all terms that do not cancel to be accurate nd Summed expression involving all terms that do not cancel nd A Correct expression rd for attempt to find single fraction (c) st for expression for Sn Sn A cso (6) ()

49 Question Number Scheme Marks 7. (a) dy dv = v + x seen B (b) + (**ag**) dv x v v x = x + v x v dv = v x dv v x = v A cso () ln( v ) = ln x ( + C) A ln( v ) = ln x + ln A Ax v y = x Ax y = x Bx or equivalent dy (c) Using y = at x = : = + 8 d y At x =, = 4 (a) (b) (c) Notes for substituting y and d y obtaining an expression in v and x only st for separating variables nd for attempting to integrate both sides st Aboth sides required or equivalent expressions. (Modulus not required.) rd Removing logs, dealing correctly with constant 4 th y dep on st M. Substitute v = and rearranging to y = f ( x) x for finding a numerical value for d y A for correct numerical answer oe. d Acso A (6) ()

50 Question Number Scheme Marks 8. (a) x + iy 6i = x + iy [ x ) ] x + x + y y + 6 = 4x 4x y x + y 4x + y = 0 ( x 4) + ( y + ) = 0 + ( y 6) = 4 ( y A Centre (4, ), Radius 0 = 5 = awrt 4.47 A A 4 y (6) (b) circle quad. 4 x Centre in correct quad for their Passes through O centre in 4 th Acao gradient 0) 6 8 Half line with positive Correct position, clearly through (6, B B (4) (c) Equation of line y = x 6 B Attempting simultaneous solution of ( x 4) + ( y + ) = 0 and y = x 6 x = 4 ± 0 A ( 4 0) + i( 0) Acao Notes (a) st M Substituting z = x + iy oe nd M implementing modulus of both sides and squaring. Require Re plus Im on both sides & no terms in i. Condone instead of 4 here. rd for gathering terms and attempting to find centre and / or radius (4) 4

51 Question Number Alt 8(c) Scheme nd A for centre, rd A for radius For geometric approach in this part. Centre (4,-) on line, can be implied. B Use of Pythagoras or trigonometry to find lengths of isosceles triangle x = 4 0 A ( 4 0) + i( 0) Acao Marks

52

53 Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fax Order Code UA040 Summer 0 For more information on Edexcel qualifications, please visit our website Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Essex CM0 JE

54 Mark Scheme (Results) Summer 0 GCE Further Pure Mathematics (6668/0R)

55 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert service helpful. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 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 0 Publications Code UA05974 All the material in this publication is copyright Pearson Education Ltd 0

56 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.

57 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.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes: bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for example, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks. 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. 8. In some instances, the mark distributions (e.g., B and A) printed on the candidate s response may differ from the final mark scheme

58 General Principles for Pure Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving term quadratic:. Factorisation ( x + bx+ c) = ( x+ p)( x+ q), where pq = c, leading to x = ( ax bx c) ( mx p)( nx q), where pq c and mn a, leading to x + + = + + = = =. Formula Attempt to use correct formula (with values for a, b and c).. Completing the square b Solving x + bx+ c= 0 : x± ± q± c, 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+ ) Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are mistakes in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may be lost if there is any mistake in the working. Exact answers Examiners reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals. Answers without working The rubric says that these may not gain full credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done in your head, detailed working would not be required.

59 Question Number. z = x Scheme Marks x + i w = A ix i w = + i ix u+ iv= i+ x u = v= x w is on the line v + = 0 A 4 Marks NOTES for replacing at least one z with x to obtain (ie show an appreciation that y = 0 ) A x + i w = ix for writing w as u+ iv and equating real or imaginary parts to obtain either u or v in terms of x or just a number A for giving the equation of the line v + = 0 oe must be in terms of v

60 Question Number Q - ALTERNATIVE : Scheme Marks w = x + iy + i i ( x + iy) Replacing z with x+iy x + iy + i y ix w = y + ix y ix w = ( x + i( y + ) )( y ix) y x i w = y + x ( x + y + y) + x x ix w = = i Using y = 0. This is where the first may be x x awarded. A if correct even if expression is unsimplified but denominator must be real A v =, A as in main scheme A Q - ALTERNATIVE : i z = Writing the transformation as a function of w iw z = i z = i ( u + iv) ( v ) i + iu ( v ) iu ( v ) iu z = u + i( v ) u = ( v ) + u ( v ) + u + i ( v ) ( ) v + u ( v ) ( v ) + u = 0 or simply ( v + ) = 0 Using y = 0. This is where the first may be awarded. A if correct even if expression is unsimplified but denominator must be real A v =, A as in main mark scheme above A

61 Question Number Scheme. NB Allow the first 5 marks with = instead of inequality Marks 6x > x x + ( )( ) ( ) ( ) 6x x x+ x x+ > 0 ( x)( x )( x x x) > 0 ( x)( x )( x )( x ) + + > 0 dep Critical values, B and, A, A Use critical values to obtain both of < x < < x< Acso 7 Marks NOTES for multiplying through by ( x + ) ( x) OR: for collecting one side of the inequality and attempting to form a single fraction (see alternative in mark scheme) dep for collecting on one side of the inequality and factorising the result of the above (usual rules for factorising the quadratic) OR: for factorising the numerator - must be a three term quadratic - usual rules for factorising a quadratic (see alternative in mark scheme) Dependent on the first M mark B for the critical values, - A A NB: for either or for the second of these the critical values need not be shown explicitly - they may be shown on a sketch or just appear in the ranges or in the working for the ranges. using their 4 critical values to obtain appropriate ranges e.g. use a sketch graph of a quartic, (which must be the correct shape and cross the x-axis at the cvs) or a table or number line Acso for both of < x<, < x<

62 Notes for Question Continued Set notation acceptable i.e. anywhere then A0.,, All brackets must be round; if square brackets appear If both ranges correct, no working is needed for the last marks, but any working shown must be correct. Purely graphical methods are unacceptable as the question specifies Use algebra. Q ALTERNATIVE : 6x > 0 x x + ( + ) ( x) ( x)( x+ ) 6x x > 0 ( x )( x+ ) ( x)( x+ ) > 0 dep Critical values, B and, AA Use critical values to obtain both of < x < < x< Acso 7 Marks

63 Question Number (a) ( )( ) A B = + r+ r+ r+ r+ Scheme Marks ( ) Br ( ) = Ar+ + + = r+ r+ r+ r+ ( )( ) A N.B. for M mark you may see no working. Some will just use the cover up method to write the answer directly. This is acceptable. () (b) = r+ r+ r+ r+ ( )( ) = n n+ n n+ n+ n+ Aft = + n+ n+ ( n+ )( n+ ) ( n+ ) ( n+ ) 6( n+ )( n+ ) = = 5n + 5n+ 0 n 0 ( n+ )( n+ ) 6 n( 5n+ ) ( n+ )( n+ ) = 6 * A (4) (c) = 0 ( ) = = = = = 0.55 A () 8 Marks

64 Notes for Question Question a for attempting the PFs - any valid method A for correct PFs ( )( ) = r+ r+ r+ r+ N.B. for M mark you may see no working. Some will just use the cover up method to write the answer directly. This is acceptable. Award A if correct, M0A0 otherwise. Question b If all work in r instead of n, penalise last A mark only. for using their PFs to list at least terms at the start and terms at the end so the cancelling can be seen. Must start at r = and end at r = n Aft for correct terms follow through their PFs for picking out the (4) remaining terms and attempting to form a single fraction (unsimplified numerator with at least terms correct) Acso for n( 5n+ ) ( n+ )( n+ ) 6 end in the mark scheme.) * (Check all steps in the working are correct - in particular rd line from NB: If final answer reached correctly from + n+ n+ onwards) give 4/4 (even without individual terms listed) (i.e. working shown from this point

65 Notes for Question Continued Question c for attempting scores M0 using the result from (b) (with numbers substituted) Use of Acso for sum = 0.55

66 Question Number 4(a) d y dy y + 5y 0 + = Scheme Marks d d d d d d y d d d d d d y y y y y y = 0 x x x x x x d d d d y = y y y y 5 A,,0 (4) Q4a ALTERNATIVE : y ( ) 5y d d d y dy = = 5 x y y x d d y dy dy d y = y y A,,0 (b) When x = 0 dy = and y = d y ( ) = 0 4 = 7 A d y = = A 7 6 y x x x!! = () 7 8 y = + x x + x A (5) 9 Marks

67 Question Number Alternative: y x ax bx Scheme Marks = ( + x + ax + bx )( a + 6bx) + ( + ax + bx...) + 5 ( + x+ ax + bx ) = 0 Coeffs Coeffs x: 0 x : 7 4a+ 4+ 0= 0 a = A 8 4a+ b+ 8a+ 0= 0 b= A 7 8 y x x x = + + A NOTES Accept the dash notation in this question Question 4a for using the product rule to differentiate d y y. d x for differentiating 5y and using the product rule or chain rule to differentiate dy A,,0 for dy dy d y 5 d y = Give AA if fully correct, AA0 if one error and A0A0 if y more than one error. If there are two sign errors and no other error then give AA0. Do NOT deduct if the two Alternative to Q4a d y terms are shown separately. Can be re-arranged first and then differentiated. for differentiating, product and chain rule both needed (or quotient rule as an alternative to product rule) A,,0 for d y dy dy d y = y y A0A0 if more than one error Give AA if fully correct, AA0 if one error and

68 Notes for Question 4 Continued Question 4b for substituting d y d y A d y for 7 A d y for obtaining the correct value, 6, for for using the series x x y = f ( x) = f ( 0) + x f ( 0) + f ( 0) + f ( 0 ) +... (! or,! or 6) (The!! general series may be shown explicitly or implied by their substitution) A 7 8 for y = + x x + x oe Must have y =.. and be in ascending powers of Alternative to Q4b for setting y = + x + ax + bx for( x ax bx )( a bx) ( ax bx ) ( x ax bx ) = 0 A for equating constant terms to get 7 a = A for equating coeffs of x to get A 8 b = for 7 8 y = + x x + x

69 Question Number 5(a) I.F. Scheme Marks tanx lnsecx = e = e = sec x A ysec x sec xsin x = sin xcosx ysec x = = tan x cos x ysec x= ln sec x ( + c) depa lnsecx + c y = sec x π (b) y =, x = π lnsec + c = π sec = ln( ) 4 + c Aft (6) c = 8 ln A π lnsec 8 ln π + 6 x= y = 6 π sec 6 ln + 8 ln y = 4 y = 8 ln 6 ln 6 ln 4 + = + = 4 A (4) 0 Marks

70 Question Number Scheme Alternative: c may not appear explicitly: π sec π π 6 y sec 6 sec = ln sec π y 8= ln 4 A Marks NOTES Question 5a y = 8 ln 6 ln 6 ln 4 + = + = 4 A tanx for the e or not sec x e tan x and attempting the integration - e ( ) lnsec x should be seen if final result is A for IF = sec x for multiplying the equation by their IF and attempting to integrate the lhs dep for attempting the integration of the rhs sin x= sin xcos x and on the second M mark sec x = needed. Dependent cos x Acao for all integration correct ie ysec x= ln sec x ( + c) constant not needed Aft for re-writing their answer in the form y =... Accept any equivalent form but the constant ln ( Asec x) must be present. eg y =, y = cos x ln ( sec x) + c sec x

71 Notes for Question 5 Continued Question 5b for using the given values y =, π x = in their general solution to obtain a value for the constant of integration A for eg c = 8 ln or e 8 A = (Check the constant is correct for their correct answer for (a)). 4 π π Answers to significant figures acceptable here and can include cos for using their constant and required form. or sec π x = in their general solution and attempting the simplification to the 6 Acao for y = 6 ln 4 or Alternative to 5b for finding the difference between y sec 6 π and sec π (or equivalent with their general solution) A for π sec π π 6 y sec 6 sec = ln sec π for re-arranging to y =... and attempting the simplification to the required form Acao for y = 6 ln 4 or

72 Question Number 6(a) z + z i θ i = e + e n n n n θ Scheme Marks = cosnθ + isin nθ+ cos nθ isin nθ = cosnθ * A () (b) ( ) 5 5 z z cos θ + = B ( ) 5 z z z z z z z z + = A ( z z ) ( z z ) ( z z ) cos θ = = cos5θ + 0cosθ+ 0cosθ 5 ( ) cos θ = cos 5θ + 5cos θ + 0 cosθ * A 6 (5) (c) cos5θ + 5cosθ+ 0cosθ = cosθ 5 6cos θ = cosθ A 4 ( ) cosθ 8cos θ + = 0 4 8cos θ + = 0 no solution B cosθ = 0 π π θ =, A (4) Marks

73 Notes for Question 6 Question 6a n n for using de Moivre's theorem to show that either z = cos nθ + isin nθ or z = cos nθ isin nθ n n A for completing to the given result z + z = cosnθ * Question 6b B for using the result in (a) to obtain ( ) 5 5 z z cos θ + = Need not be shown explicitly. for attempting to expand ( ) 5 z z + by binomial, Pascal's triangle or multiplying out the brackets. If C r is used do not award marks until changed to numbers z+ z = z + 5z + 0z+ 0z + 5z + z n A for a correct expansion ( ) 5 for replacing ( 5 5 ) ( z z, z z ), ( z z ) revised expression to their result for ( ) with cos5 θ, cos θ, cosθ and equating their 5 z+ z = cos θ Acso for cos θ ( cos 5θ 5cos θ 0 cosθ) Question 6c = + + * 6 5 A for attempting re-arrange the equation with one side matching the bracket in the result in (b) Question states "hence", so no other method is allowed. 5 for using the result in (b) to obtain 6cos θ = cosθ oe B 4 for stating that there is no solution for 8cos θ + = 0 oe eg or "ignore" but cosθ = 4 without comment gets B cos θ cos θ + > 0 A π π for θ = and and no more in the range. Must be in radians, can be in decimals (.57..., sf or better)

74 Question Number 7(a) y= λt e t Scheme Marks dy dt λ e λ e t t = t + t A d y dt t t t t = λ + λ + λ + λ A e 6 te 6 te 9 t e λe t + 6λte t + 6λte t + 9λt e t λte t 8λt e t + 9λt e t = 6e t dep λ = Acso (5) NB. Candidates who give λ = without all this working get 5/5 provided no erroneous working is seen. (b) m 6m+ 9 = 0 ( m ) = 0 y A Bt = + e t A C.F. ( ) ( ) y= A+ Bt e + t e Aft G.S. ( ) t t () (c) t = 0 y = 5 A = 5 B dy t t t t Be ( A Bt) e 6te 9t e dt = dy 4 4 B 5 dt = = + dep B = A y t t Solution: ( ) t t = 5 e + e Aft (5) Marks

75 Notes for Question 7 Question 7a for differentiating y = λt e t wrt t. Product rule must be used. A for correct differentiation ie dy dt = λte + λt e t t A for a correct second differential d y dt = λe + 6λte + 6λte + 9λt e t t t t dep for substituting their differentials in the equation and obtaining a numerical value for λ Dependent on the first M mark. Acso for λ = (no incorrect working seen) NB. Candidates who give λ = without all this working get 5/5 provided no erroneous working is seen. Candidates who attempt the differentiation should be marked on that. If they then go straight to λ = without showing the substitution, give A if differentiation correct and A0 otherwise, as the solution is incorrect. If λ then the M mark is only available if the substitution is shown. Question 7b for solving the term quadratic auxiliary equation to obtain a value or values for m (usual rules for solving a quadratic equation) y = A+ Bt e t A for the CF ( ) ( ) Aft for using their CF and their numerical value of λ in the particular integral to obtain the general solution ( ) t e e t y = A+ Bt + t Must have y =... and rhs must be a function of t. Question 7c B for deducing that A = 5 for differentiating their GS to obtain d y =... The product rule must be used. dt dep for using d y 4 dt = and their value for A in their d y dt previous M mark (of (c)) to obtain an equation for B Dependent on the Acao and cso for B = Aft for using their numerical values A and B in their GS from (b) to obtain the particular solution. Must have y =... and rhs must be a function of t.

76 Question Number π 9 A = 4 cosθ dθ 0 8 (a) ( ) 4 Scheme Marks A(limits for A mark only) sin θ = 8 π 4 π 9 sin 0 = 9 0 A (4) (b) ( ) r = cosθ ( ) r sinθ = cos θ sinθ d ( r sinθ ) = ( cos θ) sin θ sinθ ( cos θ) + cosθ dθ depa At max/min: sinθ sinθ + cos cosθ = 0 ( cos θ ) ( θ) sin θ sinθ = cos θ cosθ ( ) sin cos = sin cos θ θ θ θ ( θ) cosθ 4sin = 0 ( cosθ = 0) sin θ = 4 π sinθ =± θ =± A 6 π π r sin = cos = 6 4 B length PS =, (length PQ = 6)

77 Question Number Shaded area = Scheme 6 9, = 9 9 oe Marks,A (9) Marks NOTES Question 8a for ignore any shown. with or 9 A= r dθ = αcosθ dθ α = 4 or and limits not needed for this mark - A = π 9 π π π 4 cosθ dθ Correct limits 0, 0 with multiple 4 or, needed. 4 or may be omitted here, provided it is used later. A for ( ) 4 with multiple for the integration cos θ to become needed ± sin θ Give M0 for ± sinθ. Limits and 4 or not Acso for using the limits and 4 or as appropriate to obtain 9 ALTERNATIVES ON FOLLOWING PAGES