Mark Scheme (Results) Summer GCE Core Mathematics C4 6666/01 Original Paper

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1 Mark Scheme (Results) Summer 0 GCE Core Mathematics C 6666/0 Original Paper

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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 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 or AG: The answer is printed on the paper dm denotes a method mark which is dependent upon the award of the previous method mark. ddm denotes a method mark which is dependent upon the award of the previous method marks. dm* denotes a method mark which is dependent upon the award of the M* mark.. 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 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. Misreads A misread must be consistent for the whole question to be interpreted as such. These are not common. In clear cases, please deduct the first A (or B) marks which would have been lost by following the scheme. (Note that marks is the maximum misread penalty, but that misreads which alter the nature or difficulty of the question cannot be treated so generously and it will usually be necessary here to follow the scheme as written). Sometimes following the scheme as written is more generous to the candidate than applying the misread rule, so in this case use the scheme as written. If in doubt, send the response to Review. 5

6 Question Number Scheme ** represents a constant (which must be consistent for first accuracy mark). (a) ( x) x ( ) 8x 8x = (9 + 8 ) = 9 + = (9) or outside brackets B Marks ( )( ) = + + +! ( )(** x); (** x)... with ** 8x ( )( ) 8x = + ; ! 9 8 = x; x... 8 = x; x... Expands ( + ** x) to give a simplified or an un-simplified + ( )(** x) ; A correct simplified or an un-... expansion with candidate s followed **x simplified [ ] through ( ) M; A Award SC M if you see ()( ) (** ) (** ) x + x or! ( )( ) (** x)! + x;... 9 A oe 8 or SC K + 9 x 8x x A [5] = (9 + 8 x) x = 8 + = = = 5 5 Notes on Question B: Writes down or uses M: Substitutes their x, where expansion. A: Either 7 79 or. 5 5 x = oe. 9 x < into at least one of the x or 8 x = Substitutes their x into their binomial expansion B oe M 7 or A [] 8 x term of their binomial 6

7 Question Number. (a) (c)(i) (ii) (c)(ii) Scheme Marks x 0 y 0 e e e e e or awrt 0.7 B [] Outside brackets or 0.5; B Area ( R) For structure of trapezium ; 0 + e + e + e + e rule{...} M Correct expression inside brackets A = = =.8 (dp).8 A cao [] du x u = x = xe dv x x = e v = e d x Use of integration by e x xe = e x parts formula in the M* correct direction. Correct expression. A aef 0 x x = xe e + c xe = xe e x x x 0 () () (0) (0) = ()e e (0)e e ( 8e e ) ( 0 ) = x x ( c) ± λ xe ± µ e + M Correct answer with/without + c Substitutes limits of and 0 and subtracts the correct way round. = e a=, b= or e A dm* Notes on Question M: SC: Allow either an extra term or one missing term in e + e + e. dm: Complete method of applying limits of and 0 and subtracting the correct way round. Evidence of a proper consideration of the limit of 0 is needed for M. So, just subtracting zero is M0. A [6] 7

8 Question Number. (a) (a) Scheme x = t + 5, y = + t =, = t So, t = = t = t At ( 9, 5 ), t = Candidate s d y divided by candidate s d x Correct d y M A Marks When () Substitutes their found t into their d y t =, = = () = M So, d y A cso [] x 5 t = y = + x 5 y = + 8 x 5 ( x 5) + 8 y = x 5 x 7 y = x 5 x 5 An attempt to eliminate t. Achieves a correct equation in x and y only. a =, b = 7, c = and d = 5 or x 7 x 5 Notes on Question Note: Part (a) and part can be marked together. Alternative Method for part (a) 8 M for ± λ( x 5) where λ 0 y = + = + 8( x 5) = 8( x 5) x 5 A for 8( x 5) At ( 9, 5 ), = 8(9 5) M for substituting x = 9 into their d y So, d y = A for = by correct solution only M Aoe A oe [] 7 Award MA for either 8 x = 5 y + or x 5 = y or equivalent. 8

9 Question Number Scheme Marks. 9 5 l : r = 8 + µ 5 (a) A(, 0, ) correct coordinates B [] 5 OA = 0, d = andθ is angle 5 Applies dot product 0 OA formula between OA and d cos θ = = OA. d () + (0) + ( ). (5) + ( ) + ( ) M (c) cos θ = = Correct ft expression or () + (0) + ( ). (5) + ( ) + ( ) ( )( 5 ) equation. 8 cos θ = or or OB = OA = 0 = 0 5 l : r = 0 + λ (d) OB = () + (0) + ( ) 8 or or isw In the form of their OB + λd with any one of either d or their ft OB correct. Correct equation and r = A ft A cao = 8 = B ft [] (e) So, OX OX = sin θ sin their OB = θ M M Aft oe cos θ = sinθ = Converts cos θ into an expression for sinθ M oe 5 5 [] [] 9 OX = = = OX = awrt.55 A [] 0 9

10 Notes on Question Note: Obtaining cos θ = is MAA0. 5 (e) Note: nd M mark can be awarded instead for candidate using sin ( awrt 7 ) (e) Alternative Method for part (e) λ d =, OX = 0 + λ = λ λ + 5λ 5 OX d = 0 λ = 5 + 5λ+ 6λ λ = 0 λ leading to 50λ + = 0 λ = 5 Position vector OX = 0 5 = OX = + + = M: Applies OX d = 0 and solves the resulting equation to find a value for λ. dm: Substitutes their value of λ 5 into 0 + λ. Note: This mark is dependent upon the previous M mark if a candidate uses this alternative method. A: For OX = awrt.55 (e) Alternative Method for part (e) BX BX = cos θ BX = = M: cos 5 5 their OB = θ M: Subtracts using Pythagoras to So, OX = ( ) (. ) find OX. OX = A: For OX = awrt.55 0

11 Question Number 5. (a) Scheme sin( π y) y x y = 5 π π y xy + x = cos( ) 0 Differentiates implicitly to include either ± kcos( π y) or. (Ignore = ) ( ) M ( sin( πy) ) πcos( π y), d x A y 5 0 ( ) and ( ) d Marks ± xy ± x y M π cos( π y) x = xy Grouping terms and factorising out d y. dm xy xy = cos( y) x πcos( πy) x A oe ( ) ( π π ) ( ) [5] At (, ), Substituting x = & y = into an ()() = ; = ( cos( ()) () ) equation involving d y π π π 5 ; M; y = mt ( x ) with T: y = ( x ) M π 5 their TANGENT gradient ; Cuts x-axis y = 0 = ( ) π 5 x Setting y = 0 in their tangent M equation. π+ 5 π+ π+ 5 So, x = + = + A oe cso [] 9 Notes on Question 5 Note: nd M can be implied for = ( ) π 5 x or = if no equation of tangent is x π+ 5 given. Note: Award nd M0 where m in y = m( x ) is either a changed tangent gradient or a normal gradient.

12 Question Number Scheme Marks 6. (i)(a) 7 = A + B ( x + )(x ) ( x + ) (x ) 7 x A(x ) + B( x+ ) Forms the correct identity. B When x =, A =. Substitutes either x = or x = When x =, B =. into their identity and correctly finds M one of either A or B. 7x Hence, = + ( x + )(x ) ( x + ) (x ) Correct partial fraction. A [] 7x = + ( x + )(x ) ( x + ) (x ) Either ± aln( x + ) or ± bln(x ) M = ln( x + ) + ln(x ) + c At least one ln term correct A ft Correct integration with + c A [] (ii) d x, u = x x + x du u du u u du du = x B oe Attempt to substitute u = xand du =.u du u u + u du = x to give an expression to be integrated which is in M terms of u only. = u u du du u + u + A (ii) = ln ( u + ) + c ln = x + + c Notes on Question 6 Note: st M can be implied by.u if the du is missing. u + u ( u ) ± λ ln + Correct answer in x with or without + c. M A [5]

13 Question Number Scheme 7. (a) x = tan θ, y = + cosθ,, 0 θ < π { } V = π ( + cos θ).sec θ dθ attempt at V = π y M Correct expression ignoring limits and π. B Marks { } V = ( π ) (+ (cos θ )) sec θ dθ { } V = ( π ) (cos θ ) sec θ dθ { } V = ( π ) (6cos θ 8cos θ + )sec θ dθ { } V = π (6 cos θ 8 + sec θ) dθ change limits: when x = = tanθ θ = and when π x = = tanθ θ = + cosθ π + θ θ ( ) 6 8 sec d = ( π ) 8+ 8cosθ 8+ sec θ dθ = ( π ) 8cosθ + sec θ dθ 8sinθ = ( π ) + tanθ π So, V ( ) ( ) π 8sinθ = π 8cosθ + sec θ d θ = ( π) + tanθ = π + + ( ) ( ) Using either cos θ = cos θ or cos θ = sin θ or cos θ = cos θ sin θ π Manipulates to give the final answer where k = π Evidence of changing both limits. Using the identity cos θ = cos θ to substitute for cos θ. π π M A * B M Either ± sinθ or tanθ M 8sinθ + tanθ A Substitutes limits of π and π and subtracts the correct way round. = ( 5) π ( ) (a) Note: The use of { } 5 π ddm Notes on Question 7 y dθ (i.e. an expression for area and not volume) is the st M0, st B0. dθ Note: For the st B, the correct expression of ( + cos θ ).sec θ must be stated on one line. Note: Award nd + cosθ M0 for applying cos θ = to give an expression in terms of cos θ. Note: The π in the volume formula is only required for the st M mark and the A mark. A [5] [5] 0

14 Question Number 8. (a) Scheme Marks dv = π h V = π (0) h { = 600π h} V = π (0) h B dv 600 dh dv 600 dh Bft d d d = h V dv dh dh dv = π h = ( ± π h ) their 600π dh M So, d h = 0.0 h Correct proof. A * cso [] dh 0.0 h = Attempt to separate variables. Integral signs not necessary. B h dh = 0.0 h = 0.0t + ( ) ( c) Separates variables and integrates to give ± αh ( ) =± βt + c Correct integration with/without + c M A t = 0, h = = 0.0(0) + c c = 0 h = = 0.0t + 0 Uses boundary conditions for t and h to find c. Then uses h with found c to form an equation in order to find t. M So, 0.0t = 0 50 t = = t = 9 (minutes) (nearest minute) awrt 9 A cso Notes on Question 8 (a) Note: Use of V = π r h is st B0 until r = 0 is substituted. Note: Award final A0 for dividing by 60 after achieving t = Note: The final A mark is for correct solution only. If the candidate makes a sign error then award final A0. [5] 9

15 (a) Notes on Question 8 continued Alternative Method for part (a) d d ( π0 h) = π h BB: ( π0 h) = π h dh π h = M: Simplifies to give an expression for d h π 0. So, d h = 0.0 h * A: Correct proof. Alternative Method for part 50 T dh 00 h = h dh = 0.0 h ( ) T [ t] = 0.0 T 0 B: Attempt to separate variables. Integral signs and limits not necessary. M: ± αh ( ) =± βt + c A: Correct integration with/without limits = 0.0T M: Attempts to use limits in order to find T. So, 0.0T = 0 50 T = = T = 9 (minutes) (nearest minute) A: A correct solution (with a correct application of limits) leading to awrt 9. 5

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