Mark (Results) Summer 014 Pearson Edexcel GCE in Core Mathematics R (6664_01R)
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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.
PEARSON EDEXCEL GCE MATHEMATICS General Instructions for Marking 1. 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) 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 aliter alternative method aef any equivalent form 4. All A marks are correct answer only (cao.), unless shown, for example, as A1 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. 7. 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. 8. Ignore wrong working or incorrect statements following a correct answer.
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: 1. 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: 1. Differentiation Power of at least one term decreased by 1. ( x n x n 1 ). Integration n+1 Power of at least one term increased by 1. ( x n x )
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.
1. 1 + 1x 3 8(7) 3x 8(7)(6) 3x... + + +...! 3! 3 8 3x 8 3x 3... + C + C +... 8 3x 1 + 1 Both terms correct as printed (allow 1x but not 1 8 ) 8(7) 8(7)(6) 3... x or... x! 3! or 8 C... x or 8 C... x 3 ( ) ( 3 ) : For either the x 3 term or the x term. Requires correct binomial coefficient in any form with the correct power of x, but the other part of the coefficient (perhaps including powers of and/or 3 or signs) may be wrong or missing. Special Case: Allow this only for an attempt at a descending expansion provided the equivalent conditions are met for any term other than the first 7 6 3x 8(7) 3x... + 8 ( 1) + ( 1 ) +...! e.g. 7 6 8 3x 8 3x... + C1 + C +... B1 A1: Either 63x or A1: Both 63x and Terms may be listed but must be positive 3... + 63x + 189 x +... Note it is common not to square the in the denominator of 1 1 16 756 3 + x+ x + x. This could score B1A0A0. 3 3 189x 3 189x 3x and this gives 8 4 3x 8 3 3x Note... + C 1 + C3 1 +... + + would score M0 unless a correct method was implied by later work A1A1 [4] Total 4
. (a) (b) (c) { r } S = 6a a 6a 1 r = a Either 6a 1 r = 6a or a 1 r = 6 or 1 r = 1 5 1 = 6(1 ) r = * cso A1* 6 a a a Allow verification e.g. 6a 6a 6a 6a 6a 5 1 1 r = 1 = = = 3 { 4 } { } 6 6 T = ar = 6.5 a 5 6.5 3 5 6 (Correct statement using = 6.5 6 the 4 th 4 5 term. Do not accept 6.5 6 = ) a = 108 108 A1 their S = a 6(their a) or 648 5 = 1 6 30 5 1 6 { } 5 30 ( 6 ) { } 108 1 ( ) S = = 645.701573... { } S S.7984... 3 Correct method to find S 30 5 ( their a) 1 6 : S30 = 5 1 6 (Condone invisible brackets around 5/6) A1ft: Correct follow through expression (follow through their a). Do not condone invisible brackets around 5/6 unless their evaluation or final answer implies they were intended. [] [] A1ft = awrt.73 A1 30 [4] (c) Alternative: 30 5 30 108 ar 6 Difference = = =.7984... 1 r 5 1 6 30 ar : For an attempt to apply 1 r. A1ft: ( ) 30 their a r with their ft a. 1 r A1: awrt.73 Total 8
3. (a) { 7 and 15 1 (b) Area ( R) ; 3 + ( 7 + 11 + 15) + 19} Both 7 and 15. Allow awrt.65 and 3.87 B1 Outside brackets 1 or 1 (may be implied) B1; For structure of{...} Note decimal values are 1 1 ; { 3 + 19 + ( 7 + 11 + 15)} = ; { 6.0909.. + 19.6707... } requires the correct structure for the y values. It needs to contain first y value plus last y value and the second bracket to be multiplied by and to be the summation of the remaining y values in the table with no additional values. If the only mistake is a copying error or is to omit one value from (..) bracket this may be regarded as a slip and the M mark can be allowed (nb: an extra repeated term, however, forfeits the M mark). M0 if any values used are x values instead of y values. Bracketing mistakes: e.g. 1 ( 3 + 19 ) + ( 7 + 11 + 15 ) ( ) 1 3 + 19 + 7 + 11 + 15 Both score B1 Alternative: Separate trapezia may be used, and this can be marked equivalently. 1 1 1 1 ( 3 + 7) + ( 7 + 11) + ( 11 + 15) + ( 15 + 19) B1 for 1, for correct structure = 1 5.76166865... = 5.76166... = 5.76 (dp) 5.76 A1 cao [3] (c) underestimate Accept under, less than etc. B1 [1] Total 5 [1]
4. (a) (a) Way (a) Way 3 (b) (c) (c) Way f () = 3 + 4a + 18 18 = 0 4a = 3 a = 8 ( )( ) f( x) = x px + qx + r ( ) ( ) = + + 3 px q p x r q x r 3 f ( x) = 4x + ax + 9x 18 r = 9 q= 0 also p= 4 a= p= 8 Attempts f () or f ( ) cso A1 Compares coefficients leading to -p = a a = 8 cso A1 ( 4x 3 + ax + 9x 18) ( x ) Attempt to divide ±f(x) by (x Q= 4x + ( a 8) x+ a 7 ) to give a quotient at least of R= 4a 3 the form ± 4x + g( ax ) and a remainder that is a function of a 4a 3 = 0 a = 8 cso A1 Attempts long division or other method, to obtain ( 4 x ± ax ± b), b 0, even f ( x) = ( x )( 4x + 9) with a remainder. Working need not be seen as this could be done by inspection. d: A valid attempt to factorise their quadratic see General Principles. This is = ( x )(3 x)(3 + x) dependent on the previous or equivalent e.g. method mark being awarded, but there must have been no = ( x )(x 3)(x+ 3) da1 remainder. or A1: cao must have all 3 = ( x )(x 3)( x 3) factors on the same line. Ignore subsequent work (such as a solution to a quadratic equation.) [3] 1 1 Attempts f ( ) or f ( ) 1 1 1 1 f = 4 + 8 + 9 18 = 1 Allow A1ft for the correct A1ft 8 4 numerical value of their a 14 4 [] ± ( 4x 3 + 8x + 9x 18) ( x 1) Q= x + x+ R = 1 3 6 : Attempt long division to give a remainder that is independent of x A1: Allow A1ft for the correct numerical value of their a 14. 4 A1ft [] Total 7
5(a) (b) Length DEA = 7(.1) = 14.7 Angle CBD = π.1 Both 7 cos( π.1) and 7sin( π.1) or Both 7 cos(.1) Both 7sin(.1) π and ( ) or 7 7cos( π.1) π and ( ) 7 7sin( π.1) Or equivalents to these : 7.1 only A1: 14.7 May be seen on the diagram (allow awrt 1.0 and allow 180 10). Could score for sight of Angle CBD = awrt 60 degrees. A correct attempt to find BC and BD. You can ignore how the candidate assigns BC and CD. 7cos( π.1) can be implied by awrt 3.5 and 7sin( π.1) can be implied by awrt 6. Note if the sin rule is used, do not allow mixing of degrees and radians unless their answer implies a correct interpretation. Dependent on the previous method mark. Note that.1 radians is 10 degrees (to 3sf) which if used gives angle CBD as 60 degrees. If used this gives a correct perimeter of 31.3 and could score full marks. P = 7 cos( π.1) + 7sin( π.1) + 7 + 14.7 their BC + their CD + 7 + their DEA Dependent on both previous method marks = 31.764... Awrt 31.3 A1 A1 d dd [] [4] Total 6
6. 1 3 x x 8 4 3 4 4 3 3 x + x dx = + + c { } 4 3 1 : n n x x + on either term 4 3 x x A1: +. Any correct 3 4 simplified or un-simplified form. (+ c not required) A1 x x 16 8 56 ( 64) + = + + 3 4 3 4 3 4 4 or d 4 3 0 4 3 4 3 4 3 x x ( 4) ( 4) x x ( ) () + = ( 0) + added to + = + ( 0 ) 3 4 3 4 3 4 3 4 4 0 Substitutes limits of and 4 into an integrated function and subtracts either way round. Or substitutes limits of 0 and 4 and and 0 into an integrated function and subtracts either way round and adds the two results. 1 1 = or 10.5 A1 At x = 4, y = 8 + 1 = 4 or at x =, y = 1 + 3 = 4 { } Area of Rectangle = 6 4 = 4 or Area of Rectangles = 4 4 = 16 and 4 = 8 Evidence of ( 4 ) their y 4 or ( 4 ) their y or Evidence of 4 their y 4 and their y ddd: Area rectangle integrated answer. Dependent on all previous method marks 1 7 So, area(r) = 4 = and requires: Rectangle > integration > 0 ddda1 A1: 7 or 13.5 [7] Total 7
[ ] 4 1 3 ± "their4" + 8 4 4 x 3 x 3 4 3 x x dx = 4x { + c} Alternative: 16 8 56 ( 64) = 8 16 3 4 3 4 Line curve. Condone missing brackets and allow either way round. 1 : n n x x + on either curve term 4 3 x x A1ft: "." Any correct 3 4 simplified or un-simplified form of their curve terms, follow through sign errors. (+ c not required) nd Substitutes limits of and 4 into an integrated curve and subtracts either way round. 3 rd for ±("8" " 16" ) Substitutes limits into the line part and subtracts either way round. nd A1 for correct ± (underlined expression). Now needs to be correct but allow ± the correct expression. 4 th 1 st,1 st A1ft nd, 3 rd nd A1 7 = A1: 7 or 13.5 3rd A1 If the final answer is -13.5 you can withhold the final A1 If -13.5 then becomes +13.5 allow the A1
7.(i) (ii) 1 sin θ = 3 sin θ (4sin θ 1) = 1; 0 θ < 180 { θ = { 19.471..., 160.588... }} sin θ = k where 1< k < 1 Must be θ and not θ. A1: Either awrt 9.7 or awrt 80.3 θ = { 9.7356..., 80.644... } A1: Both awrt 9.7 and awrt 80.3 Do not penalise poor accuracy more than once e.g. 9.8 and 80. from correct work could score A1A0 If both answers are correct in radians award A1A0 otherwise A0A0 Correct answers are 0. and 1.4 Extra solutions in range in an otherwise fully correct solution deduct the last A1 5(1 cos x) cos x 5 0 =, 0 x < π. 5sin x cos x 5 0 A1 A1 = Applies sin x = 1 cos x Cancelling out cos x or a valid attempt 5cos x + cos x = 0 at solving the quadratic in cosx and cos x(5cos x + ) = 0 d giving cosx = Dependent on the cos x =... previous method mark. awrt 1.98 or awrt 4.3(0) Degrees: 113.58, 46.4 A1 or their α and their π α, where Both 1.98 and 4.3(0) π A1ft α. If working in degrees allow 360 their α π 3π awrt 1.57 or and 4.71 or or 90 o and 70 o These answers only but ignore other answers outside the range B1 [3] [5] π 3π NB: x = awrt 1.98, 4.3(0), 1.57 or, 4.71 or Answers in degrees: 113.58, 46.4, 90, 70 Could score A0A1ftB1 (4/5) 8
8. (i) (ii) y 5 y = 8 y log 5 = log8 y log 5 = log8 or y = log5 8 log8 = = 1.90... awrt 1.9 A1 log5 Allow correct answer only 1 ( x + ) = log 15 4 log x 1 Applies the power law of logarithms log( x + 15) 4 = log x seen at any point in their working x + 15 Applies the subtraction or addition law log 1 = 4 of logarithms at any point in their x working x + 15 4 1 = x 1 x 16x + 15 = 0 or e.g. x + 5 = 6x ( )( ) x 1 x 15 = 0 x =... { x = 1, 15} x = 1, 5 ( x ) Alternative: log x + 15 8 = log ( ) Obtains a correct expression with logs removed and no errors Correct three term quadratic in any form A valid attempt to factorise or solve their three term quadratic to obtain x =... or x =... Dependent on all previous method marks. Both x = 1 and x = 5 (If both are seen, ignore any other values of x 0 from an otherwise correct solution) log + 15 8 = log x Applies the power law of logarithms ( x + 15) log = 8 x ( x + 15) 8 = x x + 30x + 5 = 56x x 6x + 5 = 0 ( )( ) x 1 x 5 = 0 x =... x = 1, 5 x Applies the subtraction law of logarithms Obtains a correct expression with logs removed Correct three term quadratic in any form A valid attempt to factorise or solve their 3TQ to obtain x =... Dependent on all previous method marks. Both x = 1 and x = 5 (If both are seen, ignore any other values of x 0 from an otherwise correct solution) A1 [] ddd A1 [6] Total 8 A1 ddd A1
9. (a) (b) (c) and (d) can be marked together { } π x 1 A = xy + + x sin 60 : An attempt to find 3 areas of the form: xy, pπ x and qx A1 A1: Correct expression for A (terms must be added) ( π ) 50 = πx 3 x 50 πx 3 x 50 x xy + + y = y = 8 4 x 8 4 x 8 + 3 * Correct proof with no errors seen π x P x y { = } + + π x 50 x P = + x + ( π + 3) x 8 Correct expression for P in terms of x and y Substitutes the given expression for y into an expression for P where P is at least of the form αx+ βy π x 100 π 3 100 3 x π P = + x + x P = + x + x x x 4 x 4 ( π 8 3) A1 * 100 x P = + + Correct proof with no errors seen A1 * x 4 π + (Note dp π + 8 3 = 100x + dx 4 π + 8 3 100x + = 0 x=... 4 8 3 4 = 1.919 ) 100 ±λ : Either µ x µ or x x A1: Correct differentiation (need not be 100x + awrt1 9 simplified). Allow ( ) Their P = 0 and attempt to solve as far as x =. (ignore poor manipulation) 400 400 or awrt 7. and no other x = = 7.180574... π + 8 3 A1 π + 8 3 values x = P = 7.708... (m) awrt 7.7 A1 { 7.18..., } d P 00 dx = > 0 Minimum x 3 n n 1 : Finds P ( x x allow for constant 0 ) and considers sign 00 A1ft: (need not be simplified) and 3 x > 0 and conclusion. Only follow through on a correct P and a single positive value of x found earlier. B1 A1 A1ft [3] [3] [5] [] Total 13
10(a) (b) 9 15 8 10 A +, = A 3, 1 ( ) : A correct attempt to find the midpoint between P and Q. Can be implied by one of x or y-coordinates correctly evaluated. 3, 1 A1: ( ) ( 9 3) + ( 8 + 1) or ( 9 3) + ( 8 + 1) or ( 15 3) + ( 10 + 1) or ( 15 3) + ( 10+ 1) Uses Pythagoras correctly in order to find the radius. Must clearly be identified as the radius and may be implied by their circle equation. Or ( 15 + 9) + ( 10 8) or ( 15 + 9) + ( 10 8) Uses Pythagoras correctly in order to find the diameter. Must clearly be identified as the diameter and may be implied by their circle equation. This mark can be implied by just 30 clearly seen as the diameter or 15 clearly seen as the radius (may be seen or implied in their circle equation) Allow this mark if there is a correct statement involving the radius or the diameter but must be seen in (b) + + = ( ( ) ) ( x 3) ( y 1) 5 or 15 ( x 3) ( y 1) 5 ± α + ± β = where A( α, β) ( x ) ( y ) k and k is their radius. + + = Allow ( x 3) + ( y + 1) = 15 A1 Accept correct answer only Alternative using x + ax + y + by + c = 0 Uses A( α, β) e.g. ( ) ( ) ± ± and x + ax + y + by + c = 0 x + 3x+ y + 1y+ c= 0 Uses P or Q and x + ax + y + by + c = 0 e.g. ( ) ( )( ) ( ) ( )( ) 9 + 3 9 + 8 + 1 8 + c= 0 c= 15 x x y y A1 6 + + 15 = 0 A1 [] [3] (c) Distance = 15 10 = (their r) 10 or a correct method for the distance e.g. 1 10 their r cos sin ( their r ) { = 15 } = 5 5 5 5 A1 []
(d) 0 sin ( ARQ ) = or 30 1 10 ARQ = 90 cos 15 sin ( ARQ ) = 0 or 10 ( their r) their r or 1 10 ARQ = 90 cos their r or 1 Part ( c) ARQ = cos their r or 1 Part ( c) ARQ = 90 sin their r 0 = 15 + 15 15 15cos ARQ or or ( ) ( ) ( ) 15 = 15 + 10 5 15 10 5 cos ARQ A fully correct method to find ARQ, where their r > 10. Must be a correct statement involving angle ARQ ARQ = 41.8103... awrt 41.8 A1 [] Total 9
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