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1 Mark Scheme (Results) Summer 06 Pearson Edexcel GCE in Core Mathematics (666/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at or Alternatively, you can get in touch with us using the details on our contact us page at Pearson: helping people progress, everywhere Pearson aspires to be the world s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 0 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 06 Publications Code 666_0_606_MS All the material in this publication is copyright Pearson Education Ltd 06

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 PEARSON EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 5. 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) 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

5 . All A marks are correct answer only (cao.), unless shown, for example, as 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. 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.. Ignore wrong working or incorrect statements following a correct answer.

6 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 the correct formula (with values for a, b and c).. Completing the square b Solving x bx c 0 : x q c 0, q 0, leading to x = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( x n x n ). Integration Power of at least one term increased by. ( x n x n )

7 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.

8 Question Number Scheme Marks. r, S 5 (a) Way (a) Way (a) a 5 Way S (b) or a or a a a a a a a a a 6 a * or.5a=5 a so a = 6* or 6 { S } ; 56 (c) D T T or Substituting r or 0.5 and n = into the formula for S n 5 a a 6 5 * Correct proof * (dp) a a a a Correct proof * Applying the formula for S n with r, n = and a as 6. Obtains 5 with no errors seen and concludes a = 6*. (their a) 6 S or * ; [] [] [] 56 cao [] Writes down either "6" or awrt 6. or 9 "6" or awrt.8, using 6 a or their a A correct expression for the difference i.e. T T using a 6 or their a. d or.60 cao []

9 . (a) Question Notes Allow invisible brackets around fractions throughout all parts of this question. There are three possible methods as described above. Note that this is a show that question with a printed answer. In Way this mark usually requires a = p/q where p and q may be unsimplified brackets from the formula (or could be 00/5 for example) as an intermediate step before the conclusion a = 6. Exceptions include a = 5/ * 56/5 i.e. multiplication by reciprocal rather than division or 5 = 5a/6 followed by the obvious a = 6 These also get In reverse methods such as Way we need a conclusion so a = 6 or some implication that their argument is reversible. Also a conclusion can be implied from a preamble, eg: If I assume a = 6 then find S= 5 as given this implies a = 6 as required This is a show that question and there should be no loss of accuracy. In all the methods if decimals are used there should not be rounding. If appears this is correct. If it is rounded it would not give the exact answer. 6( ) or.5 are each correct if they are rounded then treat this as incorrect e.g. Way :.5/0.5 = 5 so a = 6 is but /0.5 = 5 so a = 6 is A0 and /0.5 = 5 so a = 6 is A0 Yet another variant on Way : take a=6 then find the next terms as 8, 6, then add to get 5. Again need conclusion that a = 6 or some implication that their argument is reversible. Otherwise A0 (b) 6 (their a found in part ( a)) S or 56 cao (c) NB Using Sum of 0 terms minus Sum of 9 terms is NOT a misread Scores M0M0A0 8 9 Can be implied. Writes down either 6 or 6, using a 6 (or their a found in part (a)). Note Ignore candidate s labelling of terms. Note d Note Note and 6 = This is dependent on previous M mark and can be implied. Either or 6 6 or awrt 6. awrt.8, using 6 a (or their a from part (a)). st and nd can be implied by the value of their 8 "their a found in part (a)" difference "their a found in part (a)" Either or 6 6 is st, nd M0..60 or.60 cao (This answer with no working is ) But.6 with no working is M0M0A0 Note Special case 8 D T9 D (6) is st, nd Obtains awrt 6., then obtains awrt.8 but rounds to 6 5 when subtracting award A0

10 Question Number y x 8, 0 x Scheme Marks. (a) B cao [] 8 dx ;.5 "their" x (b) (c) Outside brackets or For structure of trapezium rule... for a candidate s y-ordinates o.e. 0.5 cao [] Area ( R) "0.5" (.5)() cao [] 6 Question Notes (a) B For only (b) B For using or or equivalent. Requires the correct... bracket structure. It needs the.5 stated but the 0 may be omitted. The inner bracket needs 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 nd bracket this may be regarded as a slip and the M mark can be allowed ( An extra repeated term forfeits the M mark however (unless it is 0)). M0 is awarded if values used in brackets are x values instead of y values 8 For 0.5 or fraction equivalent e.g. 0 or Note NB: Separate trapezia may be used : B for 0.5, for / h(a + b) used or times Then Special case: Common error: as before. Bracketing mistake 0.5 (.5 0) their 6 scores B A0 unless the final answer implies that the calculation has been done correctly (then full marks can be given). An answer of.5 usually indicates this error. Many candidates use and score B0 Then they proceed with.5 "their " and score This usually gives 6.6 for B0A0 (c) their answer to (b) area of triangle with base and height.5 or alternative correct method.5 e.g. their answer to (b).5 d 0 x x implied by a correct answer or by an answer where they have subtracted 5 from their answer to part (b). Must use answer to part (b). 5.5 or fraction equivalent e.g. 5 or B;

11 Question Number. P (, 8) and Q (0, ) Scheme (a) PQ 0 8 or 0 8 Applies distance formula. Can be implied. PQ or. (b) ( x ) ( y 8) or Way (b) Way (c) Way (c) Way x y x y Gradient of radius = or 0 Gradient of tangent = m 5 Marks [] ( x ) ( y 8) k, where k is a positive value. ( x ) ( y 8) oe [] x y x 6y c 0, where c is any value. x y x y oe [] This must be seen or implied in part (c). Using a perpendicular gradient method on their gradient. So Gradient of tangent = gradient of radius y ( x 0) 5 y (their changed gradient)( x 0) x 5y 95 0 x 5y 95 0o.e. dy ( x ) ( y 8) 0 dx d d (0 ) ( y y 8) 0 dx dx 5 Correct differentiation (or equivalent). Seen or implied Substituting both x 0 and y into a valid differentiation to find a value for d y dx y ( x 0) 5 y (their gradient)( x 0) x 5y 95 0 x 5y 95 0o.e. (c) Way 0x y ( x 0) 8( y ) 9 0 B B 0x y ( x 0) 8( y ) 9 0 B 0x y ( x 0) 8( y ) c 0 where c is any value M x y o.e. x 5y [] [] [] 8

12 (a) (b) Question Notes Allow for PQ 0 8 or for 5 PQ. Can be implied by answer. Need to see. You can ignore subsequent work so followed by 5.8 earns, but PQ 5 5.8, with no exact value for the answer given, earns A0. Allow this time. NB Some use equation of circle to find this distance Achieving gets Others find half of their. Do not isw here as it is an error confusing d with diameter. Give A0 Either of the correct approaches for equation of circle (as shown on scheme) Correct equation (two are shown and any correct equivalent is acceptable) (c) B st nd Common error A correct start to finding the gradient of the tangent (see each scheme) Complete method for finding the gradient of the tangent (see each scheme) Where implicit differentiation has been used the only slips allowed here should be sign slips. Correct attempt at line equation for tangent at correct point (0, ) with their tangent gradient. If the y = mx +c method is used to find the equation, this is earned at the point where the x- and y-values are substituted to find c e.g. = -/5 0 + c Accept any correct answer of the required format; so integer multiple of x 5y 95 0 or x 95 5y 0 or x 5y 95 0 (must include =0 ) e.g. 6x + 0y 90 = 0 earns Also allow 5y + x 95=0 etc dy ( x) ( y8) so (y-)=6(x-0) is marked B0 M0 A0 (Way ) dx

13 Question Number. (a) (b) f( x) 6x x Scheme Marks Attempting f f 6 5 or f 5 cao [] Attempts f( ). f( ) 6( ) ( ) f( ) 0 with no sign or substitution errors 0, and so ( x ) is a factor. and for conclusion. [] f ( x) ( x ) (6x x ) ( x )(x )(x ) [] 8 Question Notes Note Long division scores no marks in part (a). The remainder theorem is required. Attempting f or f. 6 or 6 is sufficient 5 cao (c) (a) (b) Note Attempting f( ). (This is not given for f()) Must correctly show f( ) 0 and give a conclusion in part (b) only. No simplification of terms is required here. Stating hence factor or it is a factor or a tick or QED are possible conclusions. Also a conclusion can be implied from a preamble, eg: If f( ) 0, ( x ) is a factor. Long division scores no marks in part (b). The factor theorem is required. (c) st st nd Special cases Attempting to divide by ( x ) leading to a quotient which is quadratic with at least two terms beginning with first term of 6x + linear or constant term. Or f ( x) ( x ) ( 6x linear and/or constant term ) (This may be seen in part (b) where candidates did not use factor theorem and might be referred to here) (6x x ) seen as quotient or as factor. If there is an error in the division resulting in a remainder give A0, but allow recovery to gain next two marks if (6x x ) is used For a valid attempt to factorise their three term quadratic. ( x )(x )(x ) and needs all three factors on the same line. Ignore subsequent work (such as a solution to a quadratic equation). Calculator methods: Award for correct answer ( x )(x )(x ) with no working. Award A0A0 for either ( x )(x )(x ) or ( x )(x )(x ) or ( x )(x )(x ) with no working. (At least one bracket incorrect) Award for x,, followed by ( x )(x )(x ). Award M0A0M0A0 for a candidate who writes down x,, giving no factors. Award for 6( x )( x )( x ) or ( x )( x )(x ) or equivalent Award SC: A0A0 for x,, followed by ( x )( x )( x ).

14 Question Number Scheme 5. (a) 9x C ( 9 x) C ( 9 x), (b) f ( x) kx 9x A x Bx (a) First term of 6 in their final series B C... x or C... x Way At least one of (a) Way (a) (6) 88x 9x ( 9 x) ( 6x 8 x )( 6x 8 x ) Way x 6 x x x 96x x (6) 88x 9x ( 9 ) x 9 9 () 9 x x... (6) 88x 9x At least one of 88x or Both 88x and 9x 9x First term of 6 in their final series B Attempts to multiply a term quadratic by the same term quadratic to achieve either terms in x or at least terms in x. At least one of 88x or 9x Both 88x and 9x First term of 6 in final series B At least one of ()... x or... x At least one of 88x or Both 88x and Marks 9x 9x Parts (b), (c) and (d) may be marked together (b) A "6" Follow through their value from (a) Bft (c) (d) kx 9 x ( kx) 6 88 x 9 x... x terms: 88x 6kx x giving, 6k 56 k x terms: So, 9x 88kx B 9 88 ; May be seen in part (b) or (d) and can be implied by work in parts (c) or (d). k See notes 96 [] [] [] [] [] [] 9

15 (a) Ways and Way b B cao st nd Note Special Case 6 Question 5 Notes Correct binomial coefficient associated with correct power of x i.e C... x or C... x They may have and 6 or and () signs and brackets for the M marks. At least one of 88x or or even and as their coefficients. Allow missing 9x (allow +- 88x) Both 88x and 9x (May list terms separated by commas) Also full marks for correct answer with no working here. Again allow +- 88x If the candidate then divides their final correct answer through by 8 or any other common factor then isw and mark correct series when first seen. So (a) B.It is likely that this approach will be followed by (b) B0, (c) A0, (d) A0 if they continue with their new series e.g. 6x 8 x... (Do not ft the value as a mark was awarded for 6) Slight Variation on the solution given in the scheme ( 9 x) ( 9 x)( 9 x)( 6x 8 x ) ( 9 x)(8 08x86 x...) 6 6x 9x x 9x (6) 88x 9 x... First term of 6 B Multiplies out to give either terms in x or terms in x. At least one of 88x or 9x Both 88x and 9x (b) (c) Bft Note Parts (b), (c) and (d) may be marked together. Must identify A 6 or A their constant term found in part (a). Or may write just 6 if this is clearly their answer to part (b). If they expand their series and have 6 as first term of a series it is not sufficient for this mark. Candidate shows intention to multiply (+kx) by part of their series from (a) e.g. Just ( kx) 6 88 x... or ( kx) 6 88x 9 x... are fine for. This mark can also be implied by candidate multiplying out to find two terms (or coefficients) in x. i.e. f.t. their 88x 6kx N.B. 88kx x with no evidence of brackets is M0 allow copying slips, or use of factored series, as this is a method mark k o.e. so.5 is acceptable ( kx) 6 88x 9 x... to give exactly two terms (or coefficients) (d) Multiplies out their in x and attempts to find B using these two terms and a numerical value of k. 96 Note Award A0 for B 96x But allow for B 96x followed by B 96 and treat this as a correction Correct answers in parts (c) and (d) with no method shown may be awarded full credit.

16 Question Number 6. cos 0 5 (i) cos 5 NB Misread (ii) NB Misread 8, 5 5 ; < Scheme Rearranges to give cos or 5 Marks At least one of or 8 or or 96 or awrt.68 or awrt Both 5 5 Misreading 5 as 6 or (or anything else) treat as misread so A0 A0 is maximum mark cos x sinx 0, 0 x 60 ( sin x) sin x 0 Applies cos x sin x sin x sin x 0 Correct term, sin x sin x 0 x x Valid attempt at solving and sin x... sin x sin x 0 (sin x )(sin ) 0, sin... oe sin x, sin x sin x (See notes.) cso x awrt9.5, 5.5 At least one of awrt 9.5 or awrt 5.5 or awrt. or awrt 6.0 ft awrt 9.5 and awrt 5.5 [6] 9 Writing equation as cos x sinx 0 with a sign error should be marked by applying the scheme as it simplifies the solution (do not treat as misread) Max mark is /6 ( sin x) sin x 0 sin x sin x 0 A0 (sin x )(sin x ) 0, sin x... Valid attempt at solving and sin x... A0 sin x, sin x sin x (See notes.) x awrt65.5 ft Incorrect answers A0 []

17 (i) (ii) Note st st nd nd Note rd ft th Note Special Cases Question 6 Notes Rearranges to give cos 5 can be implied by seeing either or 60 as a result of taking cos (...). Answers may be in degrees or radians for this mark and may have just one correct answer Ignore mixed units in working if correct answers follow (recovery) Both answers correct and in radians as multiples of π 5 and 8 5 Ignore EXTRA solutions outside the range but lose this mark for extra solutions in this range. Using cos x sin x on the given equation. [Applying cos x sin x, scores M0.] Obtaining a correct three term equation eg. either sin x sin x 0 or sin x sin x 0 or sin x sin x or x sin sin x, etc. For a valid attempt at solving a TQ quadratic in sine. Methods include factorization, quadratic formula, completion of the square (unlikely here) and calculator. (See notes on page 6 for general principles on awarding this mark) Can use any variable here, s, y, x or sin x, and an attempt to find at least one of the solutions for sinx. This solution may be outside the range for sinx sin x BY A CORRECT SOLUTION ONLY UP TO THIS POINT. Ignore extra answer of sin x, but penalise if candidate states an incorrect result. e.g. sin x. sin x can be implied by later correct working if no errors are seen. At least one of awrt 9.5 or awrt 5.5 or awrt. or awrt 6.0. This is a limited follow through. Only follow through on the error sin x and allow for 65.5 special case (as this is equivalent work) This error is likely to earn A0A0 so /6 or A0A0A0 if the quadratic had a sign slip. awrt 9.5 and awrt 5.5 If there are any EXTRA solutions inside the range 0 x 60 and the candidate would otherwise score FULL MARKS then withhold the final mark. Ignore EXTRA solutions outside the range 0 x 60. Rounding error Allow A0 for those who give two correct answers but wrong accuracy e.g. awrt 9, 6 (Remove final for this error) Answers in radians: lose final mark so either or both of., 6.0 gets fta0 It is possible to earn A0 on the final marks if an error results fortuitously in sinx = -/ then correct work follows.

18 Question Number Scheme 5 x x. (a) x x dx c 5 (b) 0 x x 0 x or 0 x x x... 5 x Area( S) x (9) (9) or Question Notes (a) Either x x or x x,, 0 5 Either x x or x x,, 0 Marks At least one term correctly integrated Both terms correctly integrated [] Sets y 0, in order to find the correct x or x = 9 Applies the limit 9 on an integrated function with no wrong lower limit. or. 0 st At least one term correctly integrated. Can be simplified or un-simplified but power must be simplified. Then isw. dd oe [] 6 (b) nd Both terms correctly integrated. Can be un-simplified (as in the scheme) but the n+ in each denominator and power should be a single number. (e.g. not +) Ignore subsequent work if there are errors simplifying. Ignore the omission of c. Ignore integral signs in their answer. st Sets y 0, and reaches the correct Just seeing x without the correct x or x = 9 (isw if x is followed by x ) x gains M0. May just see x = 9. Use of trapezium rule to find area is M0A0 as hence implies integration needed. dd This mark is dependent on the two previous method marks and needs both to have been awarded. Sees the limit 9 substituted in an integrated function. (Do not follow through their value of x) Do not need to see MINUS 0 but if another value is used as lower limit this is M0. This mark may be implied by 9 in the limit and a correct answer. Common Error or. 0 Common Error 0 x x x so x Then uses limit etc gains M0 A0 so /

19 Question Number Scheme 8(i) Two Ways of answering the question are given in part (i) Way b a log or log a b Applying the subtraction law of logarithms (i) Way (ii) Way See also common approach below in notes (ii) Way Marks b a or a Making a correct connection between b log base and to a power. 5 5 a 5 9b a b a b a or b oe [] In Way a correct connection between log base and to a power is used before applying the subtraction or addition law of logs Either log (b ) log ( a ) log or log (b ) log log ( a ) nd log (b ) log ( a ) log = log a or log(b ) log ( a ) st a 5 {b } b a 9 9 [] Five Ways of answering the question are given in part (ii) x x ( ) ( ) 0 Deals with power 5 correctly giving So, x x or y or awrt 0.9 log A valid method for solving x xlog log or x or x log log Or x k to achieve x... x awrt.9 x 5 x Begins with ( ) (for Way and Way ) (see notes below) Correct application of (x 5)log log xlog either the power law or addition law of logarithms Correct result after applying the power and addition laws of logarithms. xlog 5log log xlog log 5log x Multiplies out, collects x terms to achieve x... log x awrt.9 (ii) x 5 log x Way oe d d x 5 Evidence of log and either x 5 x x or ( ) log log ( ) x 5 log x oe. x x log 5 x log 5 Collects x terms to achieve x... x awrt.9 d [] [] []

20 (ii) Way Way 5 (similar to Way ) x 5 5 x x ( ) x log 5 log or log and either Evidence of log x 5 x 5 or log x 5 log oe. x d [] x log 5 Rearranges to achieve... x awrt.9 x 5 log x log ( ) is replaced by x 5 log x x 5 log x oe. x x log 5 Collects x terms to achieve x... x log 5 d x awrt.9 [] Question 8 Notes (i) st Applying either the addition or subtraction law of logarithms correctly to combine any two log terms into one log term. nd For making a correct connection between log base and to a power. (ii) 5 b a or 9 9 st st d nd Special Case in (i) Common approach to part (ii) Common Presentation of Work in ii Note a 5 a 5 a 5 b o.e. e.g. Accept b 9 but not a b nor b 9 9 First step towards solution an equation with one side or other correct or one term dealt with correctly (see five* possible methods above) Completely correct first step giving a correct equation as shown above Correct complete method (all log work correct) and working to reach x = in terms of logs reaching a correct expression or one where the only errors are slips solving linear equations Accept answers which round to -.9 If a second answer is also given this becomes A0 log (b ) b Writes and proceeds to and to correct answer- Give log ( a ) a M0 (special case) Let x y Treat this as Way They get y y 0 for and need to reach y for Then back to Way as before. Any letter may be used for the new variable which I have called y. If they use x and obtain x, this may be awarded A0M0A0 Those who get y y 0 or y y 0 will be awarded M0,A0,M0,A0 x 5 x Many begin with log log ( ) 0. It is possible to reach this in two stages correctly so do not penalise this and award the full marks if they continue correctly as in Way. If however the solution continues with (x 5)log xlog 0or with (x5)log xlog 0 (both incorrect) then they are awarded A0M0A0 just getting credit for the (x + 5) log term. N.B. The answer (+).9 results from algebraic errors solving linear equations leading to x and gets A0A0

21 Question Number 9. (a) x FEA x Area( ) ; Scheme x or 0 x simplified or unsimplified 60 x Parts (b) and (c) may be marked together Attempt to sum areas (at least one correct) A x sin 60 x xy Correct expression for at least two terms of A x x 500 x x 000 xy y x x Correct proof. * y x (b) Marks [] (c) x P x x y x y x y Correct expression in x and y for their θ measured in rads Bft [] (d) (e) 500 x... y x Substitutes expression from (b) into y term. x 000 x 000 x P x x P x x x x 000 x P 6 Correct proof. * x Parts (d) and (e) should be marked together dp 6 dx 000 x ; 0 000() x ( ) 6 000() x x Correct differentiation ; (need not be simplified). Their P 0 or awrt (may be implied) 000 (6.6...) P 6 P (m) awrt 0 (6.6...) d P 000 dx 0 Minimum x Finds P and considers sign. 000 (need not be simplified) and 0 and conclusion. x ft Only follow through on a correct P and x in range 0 < x < 5. [] [5] [] 5

22 (a) (b) (c) (d) (e) st nd * Bft * st st nd nd rd ft Question 9 Notes 0 Attempts to use Area( FEA) x (using radian angle) or x (using angle in 60 degrees) x cao (Must be simplified and be their answer in part (a)) Answer only implies. N.B. Area( FEA) x 0 is awarded M0A0 An attempt to sum areas consisting of rectangle, triangle and sector (allow slips even in dimensions) but one area should be correct Correct expression for two of the three areas listed above. Accept any correct equivalents e.g. two correct from x sin or x, x, xy This is a given answer which should be stated and should be achieved without error so all three areas must have been correct and their sum put equal to 000 and an intermediate step of rearrangement should be present. Correct expression for P from arc length, length AB and three sides of rectangle in terms of both x and y with y (or y + y), x (or x + x) (or x + x + x), and x clearly listed. Allow addition after substitution of y. NB but allow use of their consistent in radians (usually ) from parts (a) and (b) for this mark. 0x or 60x do not get this mark. 500 x Substitutes y or their unsimplified attempt at y from earlier (allow x slips e.g. sign slips) into y term. This is a given answer which should be stated and should be achieved without error 000 Need to see at least x x Correct differentiation of both terms (need not be simplified) Not follow through. Allow any correct equivalent. dp dp e.g. 000x Also allow 000x awrt.6 dx dx Check carefully as there are many correct equivalents and some have two terms in xπ to differentiate obtaining for example 8 instead of Setting their d P 0 dx. Do not need to find x, but if inequalities are used this mark cannot be gained until candidate states or uses a value of x without inequalities. May not be explicit but may be implied by correct working and value or expression for x. May result in x 0 so A0 There is no requirement to write down a value for x, so this mark may be implied by a correct value for P. It may be given for a correct expression or value for x of 6.6, 6. or Allow answers wrt 0 but not Finds P and considers sign. Follow through correct differentiation of their P (not just reduction of power) 000 Need and 0 (or positive value) and conclusion. Only follow through on a correct P x and a value for x in the range 0 < x < 5 (need not see x substituted but an x should have been found) If P is substituted then this is awarded A0

23 Special case (d) Some candidates multiply P by to simplify If they write dp 000x 6 ; 0 then solve they will get the correct x and P They dx should be awarded A0 in part (d). If they then do part (e) writing d P Minimum They should be awarded A0 (so lose marks in all) dx x d( P) If they wrote 000x 6 ; 0 etc they could get full marks. dx

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