Mark (Results) June 0 GCE Core Mathematics C (6664) Paper
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EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 7.. 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.. 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
June 0 Core Mathematics C 6664 Mark Question. f( x) = x 7x x + 4 Remainder = f () = 7 + 4= 6 Attempts f() or f( ). = 6 6 A [] Attempts f ( ). f( ) = ( ) 7( ) ( ) + 4 f( ) = 0 with no sign or substitution and so ( x + ) is a factor. A [] errors and for conclusion. (c) f( x) = {( x + ) }(x 9x + 4) A = ( x + )(x )( x 4) d A (Note: Ignore the epen notation of (should be (c)) for the final three marks in this part). [4] 8 for attempting either f() or f( ). Can be implied. Only one slip permitted. can also be given for an attempt (at least two subtracting processes) at long division to give a remainder which is independent of x. A can be given also for 6 seen at the bottom of long division working. Award A0 for a candidate who finds 6 but then states that the remainder is 6. Award A for 6 without any working. : attempting only f ( ). A: must correctly show f ( ) = 0 and give a conclusion in part only. Note: Stating hence factor or it is a factor or a tick or QED is fine for the conclusion. Note also that a conclusion can be implied from a preamble, eg: If f ( ) = 0, ( x + ) is a factor. Note: Long division scores no marks in part. The factor theorem is required. (c) st : Attempts long division or other method, to obtain ( x ± ax ± b), a 0, even with a remainder. Working need not be seen as this could be done by inspection. ( x ± ax ± b) must be seen in part (c) only. Award st M0 if the quadratic factor is clearly found from dividing f ( x) by ( x ). Eg. Some candidates use their (x x 0) in part (c) found from applying a long division method in part. st A: For seeing (x 9x + 4). nd d: Factorises a term quadratic. (see rule for factorising a quadratic). This is dependent on the previous method mark being awarded. This mark can also be awarded if the candidate applies the quadratic formula correctly. nd A: is cao and needs all three factors on one line. Ignore following work (such as a solution to a quadratic equation.) Note: Some candidates will go from {( x + ) }(x 9x + 4) to { x =, } x =,4, and not list all three factors. Award these responses AA0. Alternative: st : For finding either f (4) 0 f = 0. GCE Core Mathematics C (6664) June 0 = or ( ) st A: A second correct factor of usually ( x 4) or (x ) found. Note that any one of the other correct factors found would imply the st mark. nd d: For using two known factors to find the third factor, usually (x ± ). nd A for correct answer of ( x + )( x )( x 4). Alternative: (for the first two marks) st : Expands ( x + )( x + ax + b) {giving x + ( a + ) x + ( b + a) x + b} then compare coefficients to find values for a and b. st A: a = 9, b = 4 Not dealing with a factor of : ( x + )( x )( x 4) or ( x + )( x )(x 8) scores AA0. Answer only, with one sign error: eg. ( x + )(x + )( x 4) or ( x + )(x )( x + 4) scores AA0. (c) Award AA for Listing all three correct factors with no working.
. { } ( + bx) = () + C () ( bx) + C () ( bx) +... { } 4 = 4 + 40bx + 70 b x +... (coeff x) = coeff x (40 b) = 70b So, 80 b = b = 70 4 as a constant term seen. B 40bx B C... x C... x ( ) or ( ) 70b x or Establishes an equation from their coefficients. Condone on the wrong side of the equation. 70( bx) A [4] b = (Ignore b = 0, if seen.) A The terms can be listed rather than added. Ignore any extra terms. st B: A constant term of 4 seen. Just writing () is B0. nd B: Term must be simplified to 40bx for B. The x is required for this mark. Note 40 + bx is B0. : For either the x 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 b) may be wrong or missing. Allow binomial coefficients such as,,,, C, C. A: For either 70b x or 70( bx ). (If 70bx follows 70( bx ), isw and allow A.) Alternative: bx +, Note that a factor of can be taken out first: but the mark scheme still applies. Ignore subsequent working (isw): Isw if necessary after correct working: e.g. 4 + 40bx + 70 b x +... leading to 9 + bx + 0 b x +... scores BBA isw. Also note that full marks could also be available in part, here. Special Case: Candidate writing down the first three terms in descending powers of x usually get 4 4 4 ( bx) + C 4() ( bx) + C () ( bx) +... = b x + b x + 90 b x +... C... x 4 C... x So award SC: B0B0A0 for either ( 4 ) or ( ) for equating times their coefficient of x to the coefficient of x to get an equation in b, or equating their coefficient of x to times that of x, to get an equation in b. Allow this M mark even if the equation is trivial, providing their coefficients from part have been used, eg: (40 b) = 70b, but beware b = from this, which is A0. An equation in b alone is required: e.g. (40 bx ) = 70bx b= or similar will be Special Case SC: A0 (as equation in coefficients only is not seen here). e.g. (40 bx ) = 70bx (40 b) = 70b b= will get A (as coefficients rather than terms have now been considered). Note: Answer of from no working scores A0. bx Note: The mistake k +, k 4 would give a maximum of marks: B0B0A0, A Note: For 70bx in part, followed by (40 b) = 70b b =, in part, allow recovery A. [] 6 GCE Core Mathematics C (6664) June 0
. = 0 and log ( x ) = log0 x = or x = log0 log x { =.406768... } =.4 ( sf).4 A cao ( x ) = { } ( x ) = or x = + = 7 or or. or awrt. A [] oe log0 : for x = or x = log0. Also allow for x = log log.4.4 with no working (or any working) scores A (even if left as ). Other answers which round to.4 with no working score A0. Trial & Improvement Method: : For a method of trial and improvement by trialing f (value between.4 and.4) = Value below 0 and f (value between.4 and.) = Value over 0. A for.4 cao. Note: x = log0 by itself is M0; but x = log0 followed by x =.406768... is. : Is for correctly eliminating log out of the equation. Eg : log ( x ) = log ( ) x = only gets when the logs are correctly removed. Eg : log( x ) = log() log( x ) + log() = 0 log(( x )) = 0 0 ( x ) = only gets when the logs are correctly removed, but ( x ) = 0 would score M0. x x Note: log ( x ) = log = = would score M0 for incorrect use of logs. Alternative: changing base log 0( x ) = log 0( x ) = log0 log 0 ( x ) + log0 = 0 log0 0 log0 ( x ) = 0 ( x ) = 0. At this point is scored. A correct answer in without any working scores A. [] 4 GCE Core Mathematics C (6664) June 0
4. + + 4 = 0 x y x y {( x ) 4 ( y ) 0} + + = ( ±, ± ), see notes. Centre is (, ). (,). A cao [] ( x + ) + ( y ) = + + 4 So r = + + 4 r = 4 (c) When x = 0, y y = 0 ± ± y = = () ( ) 4()( ) 48 r = ± "" ± "4" A 4 or 6 (Award A0 for ± 4 ). [] Putting x = 0 in C or their C. y y = 0 or ( y ) =, etc A aef Attempt to use formula or a method of completing the square in order to find y =... A cao cso So, y = ± ± Note: Please mark parts and together. Answers only in and/or get full marks. Note in part the marks are now A and not BB as on epen. : for ( ±, ± ). Otherwise, for an attempt to complete the square eg. ( x ± ) ± α, α 0 or ( ), 0 y ± ± β β. A: Correct answer of (, ) stated from any working gets A. : to find the radius using, and 4, ie. r = ± "" ± "4". By applying this method candidates will usually achieve 6, 6, 8 or 4 and not 6, 6, 8 or 4. Note: ( x + ) + ( y ) = = 6 r = 6 = 4 should be awarded M0A0. Alternative: in part : For comparing with x y gx fy c [4] + + + + = 0 to write down centre ( g, f ) directly. Condone sign errors for this M mark. in part : For using r = g + f c. Condone sign errors for this method mark. ( x + ) + ( y ) = 6 r = 8 scores M0A0, but r = 6 = 8 scores A isw. (c) st : Putting x = 0 in either x + y + 4x y = 0 or their circle equation usually given in part or part. st A for a correct equation in y in any form which can be implied by later working. nd : See rules for using the formula. Or completing the square on a TQ to give y = a ± b, where b is a surd, b their and b > 0. This mark should not be given for an attempt to factorise. nd A: Need exact pair in simplified surd form of { y = } ±. This mark is also cso. Do not need to see (0, + ) and (0, ). Allow nd A for bod ( +, 0) and (, 0 ). Any incorrect working in (c) gets penalised the final accuracy mark. So, beware: incorrect ( x ) + ( y ) = 6 leading to y y = 0 and then y = ± scores AA0. Special Case for setting y = 0: Award SC: M0A0A0 for an attempt at applying the formula Award SC: M0A0A0 for completing the 4 ± ( 4) 4()( ) 4 ± 60 square to their equation in x which will usually x = = = ± () be x + 4x = 0to give a ± b, where b is a surd, b their and b > 0. Special Case: For a candidate not using ± but achieving one of the correct answers then award SC: A A0 for one of either y = + or y = or y = + or y =. 8 GCE Core Mathematics C (6664) June 0 4
. ( 6 ) π r θ = = 6 π or 8.8 or awrt 8.8 (cm) Using r θ (See notes) 6π or 8.8 or awrt 8.8 [] π r π sin = r sin or sin 0 = 6 6 r 6 6 r r = 6 r Replaces sin by numeric value d 6 r = r r = r = A cso [] (c) their area of sector π r Area = 6 π π() = π or awrt 6. (cm) π or awrt 6. A cao [] 7 : Needs θ in radians for this formula. Candidate could convert to degrees and use the degrees formula. A: Does not need units. Answer should be either 6π or 8.8 or awrt 8.8 Correct answer with no working is A. This A can only be awarded in part. π r : Also allow cos or cos 60 =. 6 r st π r : Needs correct trigonometry method. Candidates could state sin = 6 x and x + r = 6 or equivalent in their working to gain this method mark. d: Replaces sin by numerical value. 0.009... = r 6 r from working incorrectly in degrees is fine here for d. A: For r = from correct solution only. Alternative: st r r for OC = sin0 or OC = cos60. nd for OC = r and then A for r =. Note seeing OC = r is. Special Case: If a candidate states an answer of r = (must be in part ) as a guess or from an incorrect method then award SC: M0M0B. Such a candidate could then go on to score A in part (c). (c) : For their area of sector their area of circle, where r > 0 is ft from their answer to part. Allow the method mark if their area of sector < their area of circle. The candidate must show somewhere in their working that they are subtracting the correct way round, even if their answer is negative. Some candidates in part (c) will either use their value of r from part or even introduce a value of r in part (c). You can apply the scheme to award either M0A0 or A0 or A to these candidates. Note: Candidates can get by writing "their part answer π r ", where the radius of the circle is not substituted. A: cao accept exact answer or awrt 6. Correct answer only with no working in (c) gets A Beware: The answer in (c) is the same as the arc length of the pendant A GCE Core Mathematics C (6664) June 0
6. { ar = 9 and ar = 44 } 44 r = Attempt to eliminate a. (See notes.) 9 r = or 0.7 or 0.7 4 4 A a ( 0.7) = 9 (c) (d) 9 a = = 6 0.7 6 A [] S 6 a Applies 0.7 r correctly using both their a and their r <. So, { S = } 04 04 A cao [] n 6( (0.7) ) Applies S n with their a and r and uses 000 > 000 at any point in their working. (Allow with = or < 0.7 ). + ( r) n from S n formula. (Allow with = or > ). 000(0.) 6 (0.7) n < = Attempt to isolate 6 6 6 nlog(0.7) < log Uses the power law of logarithms correctly. 6 (Allow with = or > ). (See notes.) 6 log( 6 ) n > =.04704... n = 4 See notes and n = 4 log 0.7 ( ) : for eliminating a by eg. 9r = 44 or by either dividing ar [] A cso = 44 by ar = 9 or dividing ar = 9 by ar = 44, to achieve an equation in r or r Note that 44 r r = is M0. 9 44 9 4 9 44 Note also that any of r = or r = = or = or = are fine for the award of 9 44 r 44 r 9 44. Note: A candidate just writing r = with no reference to a can also get the method mark. 9 Note: ar = 9 and ar = 44 leading to r = scores A. This is because r is the ratio 4 between any two consecutive terms. These candidates, however, will usually be penalised in part. for inserting their r into either of the correct equations of either 9 = r or 44 ar = 44 or { a = }. No slips allowed here for. r 9 : can also be awarded for writing down 44 = a a A for a = 6 only. Note 6 from any working scores A. 4 Note: Some candidates incorrectly confuse notation to give r = or. in part (getting 4 A0). In part, they recover to write a = 9 for and then 6 for A. ar = or { a } 9 [4] 0 GCE Core Mathematics C (6664) June 0 6
(c) a : for applying correctly (no slips allowed!) using both their a and their r, where r <. r A: for 04, cao. In parts or or (c), the correct answer with no working scores full marks. (d) st : For applying S n with their a and either the letter r or their r and uses 000. nd : For isolating + ( r) n and not ( ar ) n, (eg. ( 9) n ) as the subject of an equation or inequality. + ( r) n must be derived from the S n formula. rd k : For applying the power law to λ = µ to give k log λ = log µ oe. where λ, µ > 0. or rd k : For solving λ = µ to give k = log λ µ, where λ, µ > 0. A: cso If a candidate uses inequalities, a fully correct method with inequalities is required here. So, an incorrect inequality statement at any stage in a candidate s working for this part loses this mark. Note: Some candidates do not realise that the direction of the inequality is reversed in the final line of their solution. Or A: cso Note a candidate can achieve full marks here if they do not use inequalities. So, if a candidate uses equations rather than inequalities in their working then they need to state in the final line of their working that n =.04 (truncated) or n = awrt.0 n = 4 for A. n = 4 from no working gets SC: M0M0A. n A method of Tn > 000 6(0.7) > 000 can score M0M0A0 for a correct application of the power law of logarithms. Trial & Improvement Method: For a = 6 and r = 0.7, apply the following scheme: 6( (0.7) ) Attempt to find either S or S 4. S = = 999.6766... 0.7 EITHER () S = awrt 999.7 or truncated 999 OR () S4 = awrt00.8 or truncated 00. 4 6( (0.7) ) S4 = = 00.744... Attempt to find both S and S 4. 0.7 BOTH () S = awrt 999.7 or truncated 999 AND () S4 = awrt 00.8 or A So, n = 4. truncated 00 AND n = 4. GCE Core Mathematics C (6664) June 0 7
Note: A similar scheme would apply for T&I for candidates using their a and their r. So,... st : For attempting to find one of the correct S n s either side (but next to) 000. nd : For one of these S n s correct for their a and their r. (You may need to get your calculators out!) rd : For attempting to find both of the correct S n s either side (but next to) 000. A: Cannot be gained for wrong a and/or r. Trial & Improvement Cumulative Approach: A similar scheme to T&I will be applied here: st : For getting as far as the cumulative sum of terms. nd : () S = awrt 999.7 or truncated 999. rd : For getting as far as the cumulative sum to 4 terms. Also at this stage S < 000 and S4 > 000. A: BOTH () S = awrt 999.7 or truncated 999 AND () S4 = awrt 00.8 or truncated 00 AND n = 4. 6 Trial & Improvement Method: for (0.7) n < = 0.047 6 rd : For evidence of examining both n = and n = 4. 4 (0.7) 0.077... (0.7) = 0.07879... Eg: { = } and { } A: n = 4 Any misreads, Sn > 0000 etc, please escalate up to your Team Leader. 7. sin( x + 4 ) = ; 0 x < 60 sin x + = 7cosx ; 0 x < π sin( x 4 ), so ( x 4 ) 4.80... + = + = ( α = 4.80... ) So, x + 4 = { 8.897..., 40.80... } and x = { 9.897..., 6.80... } ( cos x) 7 cos sin or awrt 4.8 c or awrt 0.7 x + 4 = either "80 their α " or "60 + their α " (α could be in radians). Either awrt 9. or awrt 6.8 Both awrt 9. and awrt 6.8 + = x Applies sin x = cos x + = Correct term, cos x 7cosx 4{ 0} x { } x = Valid attempt at solving and cos... cos x 7cos x 4 0 (cos x )(cos + 4) = 0, cos... A A + = A oe x = cos x =, { cos x = 4} cos x = (See notes.) A cso π β = π c π x = or.0479... Either or c awrt.0 B c x = π or.98... Either π or c awrt.4 or π their β (See notes.) B ft [6] 0 [4] GCE Core Mathematics C (6664) June 0 8
st : can also be implied for x = awrt. nd : for x + 4 = either "80 their α " or "60 + their α ". This can be implied by later working. The candidate s α could also be in radians. Note that this mark is not for x = either "80 their α " or "60 + their α ". Note: Imply the first two method marks or award A for either awrt 9. or awrt 6.8. Note: Candidates who apply the following incorrect working of sin( x + 4 ) = (sin x + sin 4) =, etc will usually score M0M0A0A0. If there are any EXTRA solutions inside the range 0 x < 60 and the candidate would otherwise score FULL MARKS then withhold the final aa mark (the final mark in this part of the question). Also ignore EXTRA solutions outside the range 0 x < 60. Working in Radians: Note the answers in radians are x = awrt.6, awrt 6. If a candidate works in radians then mark part as above awarding the A marks in the same way. If the candidate would then score FULL MARKS then withhold the final aa mark (the final mark in this part of the question.) No working: Award AA0 for one of awrt 9. or awrt 6.8 seen without any working. Award AA for both awrt 9. and awrt 6.8 seen without any working. Allow benefit of the doubt (FULL MARKS) for final answer of sin x and not x = awrt 9., awrt 6.8 { } { } GCE Core Mathematics C (6664) June 0 9
st : for a correct method to use sin x = cos x on the given equation. Give bod if the candidate omits the bracket when substituting for sin x, but cos x + = 7cosx, without supporting working, (eg. seeing sin x = cos x ) would score st M0. Note that applying sin x = cos x, scores M0. st A: for obtaining either cos x + 7cosx 4 or cos x 7cos x + 4. st A: can also awarded for a correct three term equation eg. cos x + 7cosx = 4 or cos x = 4 7cosx etc. nd : for a valid attempt at factorisation of a quadratic (either TQ or TQ) in cos, can use any variable here, c, y, x or cos x, and an attempt to find at least one of the solutions. See introduction to the Mark. Alternatively, using a correct formula for solving the quadratic. Either the formula must be stated correctly or the correct form must be implied by the substitution. nd A: for cos x =, BY A CORRECT SOLUTION ONLY UP TO THIS POINT. Ignore extra answer of cos x = 4, but penalise if candidate states an incorrect result e.g. cos x = 4. If they have used a substitution, a correct value of their c or their y or their x. Note: nd A for cos x = can be implied by later working. st π B: for either or c awrt.0 nd B: for either π or c awrt.4 or can be ft from π their β or 60 their β where β = cos ( k), such that 0 < k < or < k < 0, but k 0, k or k. If there are any EXTRA solutions inside the range 0 x < π and the candidate would otherwise score FULL MARKS then withhold the final bb mark (the final mark in this part of the question). Also ignore EXTRA solutions outside the range 0 x < π. Working in Degrees: Note the answers in degrees are x = 60, 00 If a candidate works in degrees then mark part as above awarding the B marks in the same way. If the candidate would then score FULL MARKS then withhold the final bb mark (the final mark in this part of the question.) Answers from no working: π π x = and x = scores M0A0M0A0BB, x = 60 and x = 00 scores M0A0M0A0BB0, π x = ONLY or x = 60 ONLY scores M0A0M0A0BB0, π x = ONLY or x = 0 ONLY scores M0A0M0A0B0B. No working: You cannot apply the ft in the Bft if the answers are given with NO working. π 9π Eg: x = and x = FROM NO WORKING scores M0A0M0A0B0B0. For candidates using trial & improvement, please forward these to your Team Leader. GCE Core Mathematics C (6664) June 0 0
8. { V } = x y = 8 xy= 8 { L = (x + x + x + x) + 4y L = x + 4y} 8 8 y = L = x + 4 x x { } GCE Core Mathematics C (6664) June 0 Making y the subject of their expression and substitute this into the correct L formula. B oe 6 So, L = x + AG x Correct solution only. AG. A cso [] 6 ±λ dl 4 4x Either x or = = x x dx x Correct differentiation (need not be simplified). A aef L = 0 and their x =± value dl 4 4 ; = = 0 x = ; = 7 x = or their x =± value dx x x = 7 or x = A cso Substitute candidate s value of x ( 0) into a formula for L. dd (c) { } (c) 6 { x =, } L = () + = 4 (cm) 4 A cao [6] Correct ft L and considering sign. d L 97 For x =, 0 Minimum 4 dx = x > 97 and > 0 and conclusion. A [] 4 x B: For any correct form of xy= 8. (may be unsimplified). Note that x = 8is B0. Otherwise, candidates can use any symbol or letter in place of y. : Making y the subject of their formula and substituting this into a correct expression for L. A: Correct solution only. Note that the answer is given. Note you can mark parts and (c) together. nd : Setting their d L = 0 and candidate s ft correct power of x = a value. The power of x must dx be consistent with their differentiation. If inequalities are used this mark cannot be gained until candidate states value of x or L from their x without inequalities. 4 L = 0 can be implied by =. x nd A: x = 7 x = ± scores A0. nd A: can be given for no value of x given but followed through by correct working leading to L = 4. rd : Note that this method mark is dependent upon the two previous method marks being awarded. : for attempting correct ft second derivative and considering its sign. 97 A: Correct second derivative of (need not be simplified) and a valid reason (e.g. > 0), and 4 x conclusion. Need to conclude minimum (allow x and not L is a minimum) or indicate by a tick that it is a minimum. The actual value of the second derivative, if found, can be ignored, although substituting their L and not x into L is A0. Note: marks can be scored from a wrong value of x, no value of x found or from not substituting in the value of their x into L. Gradient test or testing values either side of their x scores M0A0 in part (c). Throughout this question allow confused notation such as d y for d L. dx dx
9. Curve: y x x = + + 4, Line: y = x + 4 {Curve = Line} x + x + 4 = x + 4 Eliminating y correctly. B Attempt to solve a resulting x x 0 { = 0 } ( x )( x + 4) { = 0 } x =... quadratic to give x = their values. So, x =, 4 Both x = and x = 4. A So corresponding y-values are y = 9 and y = 0. See notes below. Bft [4] x x x + x + x = + + x + c { ( 4)d } 4 { x x + + 4 x =...... 4 ( ) ( ) n n : x x + for any one term. st A at least two out of three terms. nd A for correct answer. Substitutes and 4 (or their limits from part) into an integrated function and subtracts, either way round. AA 64 + + 0 + 6 96 = 0 8 = 6 Area of = (9)(9) = 40. Uses correct method for finding area of triangle. So area of R is 6 40. =. Area under curve Area of triangle.. A oe cao [7] d GCE Core Mathematics C (6664) June 0
st B: For correctly eliminating either x or y. Candidates will usually write x + x + 4 = x + 4. This mark can be implied by the resulting quadratic. : For solving their quadratic (which must be different to x + x + 4) to give x =... See introduction for Method mark for solving a TQ. It must result from some attempt to eliminate one of the variables. A: For both x = and x = 4. nd Bft: For correctly substituting their values of x in equation of line or parabola to give both correct ft y-values. (You may have to get your calculators out if they substitute their x into y = x + x + 4). Note: For x =, 4 y = 9 and y = 0 eg. ( 4, 9) and (, 0), award B isw. If the candidate gives additional answers to ( 4, 0) and (, 9), then withhold the final B mark. Special Case: Award SC: B0M0A0B for { A}( 4, 0). You may see this point marked on the diagram. Note: SC: B0M0A0B for solving 0 = x + x + 4 to give { A}( 4, 0) and/or (6, 0). Note: Do not give marks for working in part which would be creditable in part. st n n for an attempt to integrate meaning that x x + for at least one of the terms. Note that 4 4x is sufficient for. st A at least two out of three terms correctly integrated. nd A for correct integration only and no follow through. Ignore the use of a ' + c'. nd : Note that this method mark is dependent upon the award of the first mark in part. Substitutes and 4 (and not 4 if the candidate has stated x = 4 in part.) (or the limits the candidate has found from part) into an integrated function and subtracts, either way round. Allow one slip! rd : Area of triangle = x (their their )(their ) x x y or Area of triangle = x + 4{ d x}. x Where x = their 4, x = their and y = their yusually found in part. 4 th : Area under curve Area under triangle, where both Area under curve > 0 and Area under triangle > 0 and Area under curve > Area under triangle. rd A:. or 4 oe cao. GCE Core Mathematics C (6664) June 0
Aliter 9. Way Curve: Area of R y x x = + + 4, Line: y = x + 4 ( 4) ( 4) d 4 = x + x + x + x x x = + + 0x + { c} x x + + 0 x =...... 4 ( ) ( ) rd : Uses integral of ( x + 4) with correct ft limits. 4 th : Uses curve line function with correct ft limits. n n M: x x + for any one term. A at least two out of three terms Aft Correct answer (Ignore + c). A Substitutes and 4 (or their limits from part) into an integrated function and d subtracts, either way round. 64 + + 00 + 8 80 = 70 0 6 See above working to decide to award rd mark here: See above working to decide to award 4 th mark here: So area of R is =.. A oe cao [7] st n n for an attempt to integrate meaning that x x + for at least one of the terms. Note that 0 0x is sufficient for. st A at least two out of three terms correctly ft. Note this accuracy mark is ft in Way. nd A for correct integration only and no follow through. Ignore the use of a ' + c'. Allow nd x x x x x A also for + + 4x + 4x. Note that or 4x 4x only counts as one integrated term for the st A mark. Do not allow any extra terms for the nd A mark. nd : Note that this method mark is dependent upon the award of the first mark in part. Substitutes and 4 (and not 4 if the candidate has stated x = 4 in part.) (or the limits the candidate has found from part) into an integrated function and subtracts, either way round. Allow one slip! rd : Uses the integral of ( x + 4) with correct ft limits of their x and their x (usually found in part ) {where ( x, y ) = ( 4, 0) and ( x, y ) = (,9).} This mark is usually found in the first line of the candidate s working in part. 4 th : Uses curve line function with correct ft (usually found in part ) limits. Subtraction must be correct way round. This mark is usually found in the first line of the candidate s working in part. for this method mark. 4 Allow ( x + x + 4) x + 4 { dx} rd A:. oe cao. Note: SPECIAL CASE for this alternative method Area of R x x 64 ( x x 0)dx 0x 00 8 80 4 4 = = = + The working so far would score SPEICAL CASE AAM0A0. 4 The candidate may then go on to state that = 70 0 = 6 If the candidate then multiplies their answer by - then they would gain the 4 th and. would gain the final A mark. GCE Core Mathematics C (6664) June 0 4
Aliter Curve: y x x = + + 4, Line: y = x + 4 9. {Curve = Line} y = ( y 4) + ( y 4) + 4 Eliminating x correctly. B Way Attempt to solve a resulting y 9y{ = 0 } y( y 9) { = 0 } y =... quadratic to give y = their values. So, y = 0, 9 Both y = 0 and y = 9. A So corresponding y-values are x = 4 and x =. See notes below. Bft [4] nd Bft: For correctly substituting their values of y in equation of line or parabola to give both correct ft x-values. 9. Alternative Methods for obtaining the mark for use of limits: There are two alternative methods can candidates can apply for finding 6. Alternative : 0 ( x + x + 4)d x + ( x + x + 4)dx 4 0 0 x x x x = + + 4x + + + 4x 4 0 64 = ( 0) + 6 96 + + + 0 (0) = 0 8 = 6 Alternative : 6 6 ( x + x + 4)d x ( x + x + 4)dx 4 6 6 x x x x = + + 4x + + 4x 4 6 64 6 6 44 6 96 = + + + + 6 + 44 + + 0 ( 08) 8 = ( 08) 0 = 66 4 = 6 GCE Core Mathematics C (6664) June 0
Appendix List of Abbreviations d denotes a method mark which is dependent upon the award of the previous method mark. ft or denotes follow through cao denotes correct answer only aef denotes any equivalent form cso denotes correct solution only AG or * denotes answer given (in the question paper.) awrt denotes anything that rounds to aliter denotes alternative methods Extra Solutions 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. Question Aliter 4. (c) ( x ) ( y ) 6, + + = centre ( x, y ) = (,) and radius r = 4. d = 4 = Applying their r their x A aef Way Hence, y = ± Applies y = their y ± their d So, y = ± ± Special Case: Award Final SC: A A0 if candidate achieves any one of either y = + or y = or y = + or y =. A cao cso [4] Aliter 8. Way L x x = 8 4 6 x ( L x) = 6 L = x + x L x x = 8 4 Rearranges their equation to make y the subject. Correct solution only. AG. B oe A cso [] GCE Core Mathematics C (6664) June 0 6
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