PMT. Mark Scheme (Results) Summer Pearson Edexcel GCE in Core Mathematics 1R (6663_01R)

Transcription

1 Mark Scheme (Results) Summer 0 Pearson Edecel GCE in Core Mathematics R (666_0R)

2 Edecel and BTEC Qualifications Edecel 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 0 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 UA08 All the material in this publication is copyright Pearson Education Ltd 0

3 General Marking Guidance All candidates must receive the same treatment. Eaminers must mark the first candidate in eactly 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. Eaminers 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. Eaminers should always award full marks if deserved, i.e. if the answer matches the mark scheme. Eaminers 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 eemplification may be limited. 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 7. The Edecel 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. All A marks are correct answer only (cao.), unless shown, for eample, 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, 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.

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 ( + b + c) = ( + p)( + q), where pq = c, leading to = ( a + b + c) = ( m + p)( n + q), where pq = c and mn = a, leading to =. Formula Attempt to use the correct formula (with values for a, b and c).. Completing the square b Solving + b + c = 0 : ± ± q ± c = 0, q 0, leading to = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( n n ). Integration n+ Power of at least one term increased by. ( n )

7 Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent eaminers 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. Eact answers Eaminers reports have emphasised that where, for eample, an eact 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. 9 ( 9 ) = B ( 9 ) =(+)(-) M 9 ( )( ) = + () B Take out a common factor, usually, to produce ( 9 ). Accept Must be correct. Other possible options are ( + )( ) or ( )(+ ) ( 0)( 9 ) ± or (9 ) M For factorising their quadratic term, usually ( 9 ) = (+)(-) Accept sign errors If ( ± ) has been taken out as a factor first, this is for an attempt to factorise ( ) cao ( + )( ) or any equivalent with three factors e.g. ( + )( + ) or (-)(--) etc including (+)(-) isw if they go on to show that = 0, ±

9 Question Number Scheme Marks.(a) (b) 8 = (8 ) = 9 or 8 = (8 ) = () M =79 6 ( ) = 6 or or equivalent M ( ) = 6 () () ( marks) (a) M Dealing with either the cube or the square root first. A correct answer will imply this mark. Also accept a law of indices approach 8 = 8 8 = 8 9 Cao 79. Accept ( ± )79 (b) M For correct use of power on both and the Cao = 6 term.

10 Question Number Scheme Marks.(a) ( a = ) k B () (b) a = (k ) M ar = k+ k + (k ) =.. k±... M r= k 8 = 66 k =... dm k = () ( marks) (a) B k cao (b) M An attempt to find a from iterative formula a = a. Condone bracketing errors for the M mark M Attempt to sum their a, a and a to get a linear epression in k (Sum of Arithmetic series is M0) dm Sets their linear epression to 66 and solves to find a value for k. It is dependent upon the previous M mark cao k =

11 Question Number Scheme Marks.(a) y = + 6 n n M dy d = 0 oe () (b) + d 6 + n n M 6 = + + c () (6 marks) (a) M n n For. ie. or or seen For or 6 (oe). (Ignore +c for this mark) For simplified epression 0 or 0 o.e. and no +c Apply ISW here and award marks when first seen. (b) M For n n. ie. ( ) 6 or or + seen Do not award for integrating their answer to part (a) 6 For either or 6 6 or simplified or unsimplified equivalents For fully correct and simplified answer with +c.

12 Question Number Scheme Marks Method = = 6 M, + 0 = 6 = 0 or a = and b = M = () Method = = M 0 0 = 0 = =, = oe M, () Method 6 M For multiplying both sides by allow a slip e.g = or =, where one term has an error or the correct (0 + 8) = NB = 6 is M0 A correct equation in with no fractional terms. Eg = 6 oe. M An attempt to solve their linear equation in to produce an answer of the form a or a 0 oe (accept 0 ) Method M For writing 8 as or 6 as A correct equation in with no fractional terms. Eg + 0 = or = oe. M An attempt to solve their linear equation in to produce an answer of the form a or a = 0 = =, = or = 0 = 00 = 0 = 0 or oe Accept 0

13 Question Number Scheme Marks 6(a). P= 0+ 6 o.e B 0+ 6 > 0 > M >.7 * Mark parts (b) and (c) together (b) A= (+ ) + (6+ ) = B () < M (c ) Try to solve their + = e.g. ( )( + ) = 0 so = M Choose inside region M < < or 0 < < ( as is a length ) 0.7 < < Bcao () () (9 marks) (a) B Correct epression for perimeter but may not be simplified so accept or ( 0 + ) or any equivalent M: Set P > 0 with their linear epression for P (this may not be correct but should be a sum of sides) and manipulate to get > * cao >.7. This is a given answer, there must not be any errors, but accept.7 < (b) Marks parts (b) and (c) together B Writes a correct statement in for the area. It need not be simplified. You may isw Amongst numerous possibilities are. (+ ) + (6+ ), 6 + 8, (6+ ) (+ ), (+)+(+) M Sets their quadratic epression < 0 and collects on one side of the inequality M For an attempt to solve a Term quadratic equation producing two solutions by factorising, formula or completion of the square with the usual rules (see notes) M For choosing the inside region. Can follow through from their critical values must be stated not just a table or a graph. Can also be implied by 0 < < upper value < <. Accept > - and <. or ( -,.) As is a width, accept 0 < < Also accept 0 or. instead of. would be MA0 Allow this final answer to part (b) to appear as answer to part (c) This would score final M in (b) then B0 in (c) (c ) Bcao.7 < <. Must be correct. [ This does not imply final M in (b)]

14 Question Number 7.(a) Method y y ( ) gradient = =, = 7 Scheme Method y y y y =, so = y y 6 8 Marks M, y = ( + ) or y+ = ( 7) or y = their ' ' + c M ± (y+ ) = 0 () Method : Substitute = -, y = and = 7, y = - into a + by + c=0 M -a +b + c = 0 and 7a - b + c = 0 Solve to obtain a =, b = and c = - or multiple of these numbers M () (b) Alternative for (b) Attempts gradient LM gradient MN = Or ( y+ ) = ( 7) equation p + p + M so = or = with = 6 substituted p+ = p =..., p = 8 So y =, y = 8 M, Attempt Pythagoras: ( p+ ) (6 + 8 ) = ( p ) + 7 M So p p p p = p =... M p = 8 Or use perpendicular line equations (c ) Either (y=) p + 6 or + p + through L and N and solve for y M (y = ) (a) M Uses the gradient formula with points L and M i.e. quote gradient = y y () () () (9 marks) and attempt to substitute correct numbers. Formula may be implied by the correct ( ) 7 or equivalent. 6 Any correct single fraction gradient i.e or equivalent 8 M Uses their gradient with either (-, ) or (7, -) to form a linear equation Eg y = their ' '( + ) or y + = their ' '( 7) or y = their ' ' + c then find a value for c by substituting (-,) or (7, -) in the correct way( not interchanging and y) Accept ± k(y+ ) = 0 with k an integer (This implies previous M) (b) M Attempts to use p + gradient LM gradient MN =. ie. = (allow sign errors) 6 7 Or Attempts Pythagoras correct way round (allow sign errors) M An attempt to solve their linear equation in p. cao p = 8 (c ) M For using their numerical value of p and adding 6. This may be done by any complete method (vectors, drawing, perpendicular straight line equations through L and N) or by no method. Assuming = 7 is M0 Accept for both marks as long as no incorrect working seen (Ignore left hand side allow k). If there is wrong working resulting fortuitously in give M0A0. Allow (8, ) as the answer.

15 Question Number 8. dy = 6 + d Scheme = B n n + M Marks 6 y = + ( + c), Use =, y =7 to give equation in c, 7 ( ) c = or equivalent eg. 0. ( y) = + + = + + c M (7 marks) B M =. This may be implied by n n in at least one case so see either + + oe in the subsequent work. or or both One term integrated correctly. It does not have to be simplified Eg. 6 or +. No need for +c Other term integrated correctly. See above. No need to simplify nor for +c. Need to see 6 + or a simplified correct version M Substitute =, y = 7 to produce an equation in c. Correctly calculates c = or equivalent e.g. 0. cso y = + +. Allow y = and accept fully simplified equivalents. e.g. y = ( ), y = + +

16 Question Number 9.(a) Scheme Marks (0, 8) U shaped parabola symmetric about y ais Graph passes through (0, 8) B B ( 0,k ) Shape and position for L M 9(b) Method : k,0 Equate curves k Both,0 Allow marks even if on the same diagram 8 k and ( 0, k ) + = + and proceed to collect terms on one side M k + (8 ) Method a Method b Uses 9 b = ac Attempt ( ) λ + 8 k dm 9 = (8 k) k =.. Deduce that k = 8 - λ dm k = o.e. Method : Attempts to set d y d M. = = Method a Method b Substitutes = "." into Substitutes = "." into y = + 8 y =...(.7) 8 + k dm Substitutes both their and y into y = + k to find k Finds k = dm k =. o.e. () (9 marks) () ()

17 (a) B Shape for C. Approimately Symmetrical about the y ais B Coordinates of (0, 8) There must be a graph. Accept graph crossing positive y ais with only 8 marked. Accept (8,0) if given on y ais. M Shape for L. A straight line with positive gradient and positive intercept Coordinates of (0, k) and (-k/, 0) or k marked on y ais, and k/ marked on ais or even Accept (k, 0) on y ais and (0, -k/) on ais (b) Either Methods M Equate curves 8 k + = + and proceed to collect terms on one side and (8 k) terms together on the same side or on the other side Achieves an epression that leads to the point of intersection e.g + k (8 ) Method a dm (depends on previous M mark) Uses the fact that b = ac or ' b ac' = 0is true dm (depends on previous M mark) Solves their b = ac, leading to k=.. cso k = Accept equivalents like. etc. Method b dm (depends on previous M mark) Uses completion of the square as shown in scheme dm (depends on previous M mark) Uses k=8 - λ cso k = Accept equivalents like. etc. Methods M Equate d y = Not given just for derivative d Solves to get =. Method a dm Substitutes their. into equation for C to give y coordinate dm Substitutes both their and y into y = + k to find k cso k = Accept equivalents like. etc. Method b dm Substitutes their. into 8 + k dm Finds k cso k = Accept equivalents like. etc.

18 Question Number Scheme Marks 0(a). Attempts to use a+ ( n )" d" with a=a and d =d+ and n = M A+ ( d + ) = A+ d + * * (b) Calculates time for Yi on Day = ( A ) + (d ) M Sets times equal A+ d + = ( A ) + (d ) d =... M d = cso n (c ) Uses { A ( n )( D) } + with n =, and with D=d or d + M A+ '( d + )' = 78 A=... dm A = 0 Attempts to solve { } () () () (8 marks) (a) M Attempts to use a+ ( n ) d with a=a and d =d+ AND n = * cao This is a given answer and there is an epectation that the intermediate answer is seen and that all work is correct with correct brackets. The epressions A+ ( d + ) and A+ d + should be seen N.B. If brackets are missing and formula is not stated e.g. A+ d + A+ d + or A+ () d + A+ d + then this is M0A0 If formula is quoted and a = A and d = d + is quoted or implied, then M A0 may be given So a+ ( n ) d followed by A+ () d + = A+ d + achieves MA0 (b) M States a time for Yi on Day = ( A ) + (d ) M Sets their time for Yi, equal to A+ d + and uses this equation to proceed to d = cso d = Needs both M marks and must be simplified to (not 9/) [NB Setting each of the times separately equal to 0 leads to d = this will gain M0A0] n + with n = and D = d + or allow D = d (usually or ) n NB May use { A+ ( A+ D )} with n = and and D = d + or allow D = d (usually or ) A+ '' = "78" A=... (Must use their d + this time) (c ) M Uses the sum formula { A ( n )( D) } dm Attempts to solve { } Allow miscopy of 78 cao A = 0

19 Question Number Scheme.(a) 8 Substitutes = into y = 0 and gets B dy 8 = + d M dy Substitute = = then finds negative reciprocal (-) d dm Method Method States or uses y = ( ) or Or: Check that (, ) lies on the y = - + c with their (, ) line y = + 7 dm to deduce that y = + 7 * Deduce equation of normal as it has the same gradient and passes through a common point * (b) 8 Put 0 = + 7 and simplify to give + 8 = 0 M 7 y 8 Or put y = 0 to give y y 6= 0 7 y ( 9)( ) = 0 so = or (y ) (y + ) = 0 so y = dm 9 =, y =, Marks ( marks) (6) () PTO for notes on this question.

20 (a) B Substitutes = into epression for y and gets cao (must be in part (a) and must use curve equation not line equation) This must be seen to be substituted. M For an attempt to differentiate the negative power with. dy 8 Correct epression for = +, accept equivalents d dm Dependent on first M Substitutes = into their derivative to obtain a numerical gradient and find negative reciprocal or states that = (Method ) dm Dependent on first M Finds equation of line using changed gradient (not their ½ but -/, or -) e.g. y "" = ""( ) or y = - + c and use of (, ) to find c= * CSO. This is a given answer y = + 7 obtained with no errors seen and equation should be stated (Method )- checking given answer dm Uses given equation of line and checks that (, ) lies on the line * CSO. This is a given answer y = + 7 so statement that normal and line have the same gradient and pass through the same point must be stated (b) M Equate the two given epressions, collect terms and simplify to a TQ. There may be sign errors when collecting terms But putting for eample 0 8 = + 7 is M0 here Correct TQ = 0 (need = 0 for A mark) + 8 = 0 dm Attempt to solve an appropriate quadratic by factorisation, use of formula, or completion of the square (see general instructions). 9 = oe or y = - (allow second answers for this mark so ignore = or y = ) 9 9 Correct solution only so both =, y = or, If =, y = is included as an answer and point B is not identified then last mark is A0 Answer only with no working send to review. The question stated use algebra

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