Mark Scheme (Results) Summer GCE Core Mathematics 1 (6663/01R)

Transcription

1 Mark Scheme (Results) Summer 03 GCE Core Mathematics (6663/0R)

2 Edecel and BTEC Qualifications Edecel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Epert service helpful. Pearson: helping people progress, everywhere 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 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 03 Publications Code UA03566 All the material in this publication is copyright Pearson Education Ltd 03

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 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. 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. 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) 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 4. All A marks are correct answer only (cao.), unless shown, for eample, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks. 5. 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. 8. In some instances, the mark distributions (e.g., B and A) printed on the candidate s response may differ from the final mark scheme.

5 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:. 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 correct formula (with values for a, b and c). 3. Completing the square b Solving 0 :, 0, leading to =... + b+ c= ± ± q± c q Method marks for differentiation and integration:. Differentiation n n Power of at least one term decreased by. ( ). Integration Power of at least one term increased by. ( n n + ) 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 mistakes in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may be lost if there is any mistake in the working. 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. Answers without working The rubric says that these may not gain full credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done in your head, detailed working would not be required.

6 . 3 dy y = + 4+ = d substitute = 3 gradient = 3 ( ) n n : including 0 A: Correct differentiation (Do not allow 4 0 unless 0 = is implied by later work) : Substitutes = 3 into their d y (not y) d Substitutes = 3 into a changed function. They may even have integrated. A: cao A A [4]

7 = = : Attempts to multiply numerator and denominator by 3. This may be implied by a correct answer. A: = 3 3 B = A Way = = Note that = Correct answer only scores full marks Terms combined correctly with a common denominator (Need not be simplified) : Attempts to multiply numerator and denominator by 3. This may be implied by a correct answer. A: = A = 3 3 = = 6 3 is quite common and 3 3 scores A0BA0 (i.e. 5 3 is never seen) A B A [4] [4]

8 3. 4 = d or 4 : n n + 4 If they write 3 here. for either term. as 4 allow 3 A: 3 or 4 (one correct term 3 which may be un-simplified) 3 A: 3 and 4 (both terms 3 correct which may be un-simplified) Note that A0A is not possible 3 3 = + + c + + c Fully correct simplified answer with + c all appearing on the same line.,a,a A [4]

9 4.(a) 3 4+ y 3= 0 y = + Attempt to write in the form y = gradient = Accept any un-simplified form and allow even with an incorrect value of c A (a) Attempt to differentiate Way dy Alternative: 4+ = 0 dy d Allow p± q = 0, p, q 0 d gradient = Accept any un-simplified form A Answer only scores A (b) Using m N = m T Attempt to use m N = gradient from ( a) [] y 5 = ' '( ) or Correct straight line method using a changed gradient and the point Uses y = m + c in an attempt to find c (, 5) y= + 4 Cao (Isw) A (3) [5]

10 5.(a) y = 8 y = 3 Cao (Can be implied i.e. by 3 ) B y (Alternative: Takes logs base : log = log8 ylog = 3log y = 3) () (b) 3 8= Replaces 8 by 3 (May be implied) 4 + = ( ) + or ( + ) Replaces 4 by correctly. (b) Way = 3+ = 3 = : Adds their powers of on the lhs and puts this equal to 3 leading to a solution for. A: = or = 0.3 or awrt = Obtains 4 + in terms of 4 correctly A 4 = 8 Combines their and 4 correctly : Solves 8 = k leading to a 4 8 = 8 8 = = 3 solution for. A: = or = 0.3 or awrt A (4) [5]

11 6.(a) = k Accept un-simplified e.g. - k B (b) k k k 3 = ( ) ( ) Attempt to substitute their into 3 = ( ) kwith their in terms of k. = 3k + k * Answer given A* (c) 3k + k = Setting (d) ( k 3 k = 0 ) k(k 3) = 0 k =.. 3 k = n = Or = + ' ' n= ( k ) 3k k + = Solving their quadratic to obtain a value for k. Dependent on the previous. Cao and cso (ignore any reference to k = 0) Writing out at least 3 terms with the third term equal to the first term. Allow in terms of k as well as numerical values. Evidence that the sequence is oscillating between and k. This may be implied by a correct sum. 50 or or 50 ( ) = 5 An attempt to combine the terms correctly. Can be in terms of k here e.g 00 50k Allow an equivalent fraction, e.g. 50/ or 00/4 Note that the use of ( ) na+ l is acceptable here but ( ( ) ) n a+ n d is not. (3) d A A () () (3) Allow correct answer only [9]

12 7.(a) (b) (c ) Uses a+ ( n ) d with a=500, d=00 U 0 = ( 0 ) 00 and n = 9,0 or =( )300 A If the term formula is not quoted and the numerical epression is incorrect score M0. A correct answer with no working scores full marks. Mark parts (b) and (c) together : Attempt to use n S = { a+ ( n ) d} n { ( n ) 00} = 6700 with, A S = 6700, a= 500 and d = 00 n A: Correct equation If the sum formula is not quoted and the equation is incorrect score M0. : An attempt to remove brackets and collect terms. Dependent on the n + 4n 67 = 0 previous A: A correct three term equation in any form E.g. allow n + 4n = 67, n = 67 4n, 67 4n n = 0, 00n + 800n = etc. Replaces 67 with 4 8 with the equation as printed (including = 0) with n + 4n 4 8= 0* no errors. (= 0 may not appear on the last line but must be seen at some point) ( n 4)( n+ 8) = 0 n =.. or nn ( + 4) = 4 8 n=.. 4 Allow correct answer only in (c) Solves the given quadratic in an attempt to find n. They may use the quadratic formula. States that n = 4, or the number of years is 4 da A A () (5) () [9]

13 Ignore any references to the units in this question 8.(a) length is + 4 May be implied B + ( ± 4) > 9.and proceeds to > > 9. >..... (Accept invisible brackets) Attempts widths + lengths > 9. leading to >... E.g > 9. >.9 scores B0A0 >.8 * Achieves >.8 with no errors A(*) Mark parts (b) and (c) together (b)(i) + ( 4) < Cao B b(ii) Multiply out lhs, produce 3TQ = 0 and + 4 < 0 attempt to solve leading to =... according ( + 7)( 3) < 0 =... to general guidelines : Attempts the inside for their critical values (may be from a TQ here) Either -7 < < 3 or 0 < < 3 A: Accept either -7 < < 3 or 0 < < 3 or ( > -7 and < 3) or ( > 0 and < 3) but not e.g. ( > -7, < 3) or ( > -7 or < 3) (There is no specific need for them to realise > 0) A Note that many candidates stop here (c ).8 < < 3 ( 7)( + 3) < 0, = 7, = 3 3< < 7or0< < 7.8 < < 7 Scores B0A0Bft ( 4) < 4 < 0 Follow through their answers to (a) and (b) Provided their 3 >.8 Eamples < < ( )(+ ) < 0, =± < < or0 < <.8 < < Bft (3) (4) () [8] Scores B0M0A0B0

14 9.(a) (b) f( ) = ( + )( ) = + + = + 3 ( )( 4 4) 3 4 : Either stating or writing down that ( ± ) or ( ± ) is a factor may be implied by their f() A: Both ( + ) and ( - ) are factors - may be implied by their f() B: y or f()= ( + )( - ) : Multiplying out a quadratic to get 3 terms and then multiplying by the linear term to form a cubic. 3 A: 3 + 4or a = -3, b = 0, c = 4 AB A (5) Same shape and position (ignore any coordinates) with the maimum on the y-ais B (a) Way = 0, y = 4 c = 4 =, y = 0 + a b+ c = 0 =, y = a+ b+ c = 0 a b = 3 4a+ b = a =.. or b =... y intercept = 4 or their c coordinates at - and 4 or marked as coordinates. Allow (0, -) and (0, 4) if they are marked in the correct position. The curve must cross or at least stop at = - Uses (0, 4) to obtain c = 4 (can be just stated) Uses both (-, 0) and (, 0) in 3 y = + a + b + c to form simultaneous equations. Allow the equations to contain c here. Solves simultaneously with a value for c to obtain a value for a or a value for b Bft B B (3) [8] Either a = -3 or b = 0 A Both a = -3 and b = 0 A

15 9.(a) Way 3 dy 3 a b d n n : at least once including c 0 dy = 0 = 0 b = 0 d Correct value for b A = 0, y = 4 c = 4 Uses (0, 4) to obtain c = 4 (can be just stated) B 3() + a() + b = 0 or 3 ( ) + a( ) + b( ) + 4= 0 Obtains an equation in a a = 3 Correct value for a A Special case: A common incorrect approach is to assume the cubic is of the form e.g. f( ) = ( ± )( ± ) + 4 This scores B only for c = 4 (5) [8]

16 0.(a) (b) Alternative to part (b) f'( ) = = + = f( ) = + 9 ( + c) 3 3 ( 9) ( 9) c = 0 c = f( ) = : Correct attempt to split into separate terms or fractions. May be implied by one correct term. Divides by A: or multiplies by 9. + or equivalent : Independent method mark for n n+ on separate terms A: Allow un-simplified answers. No requirement for + c here Substitutes = 9 and y = 0 into their integrated epression leading to a value for c. If no c at this stage M0A0 follows unless their method implies that they are correctly finding a constant of integration. There is no requirement to simplify their f() so accept any correct un-simplified form. A A + 9 Sets f'( ) = = 0 and multiplies + 9 f'( ) = = 0 + 9= 0 by. The terms in must be in the + 9 numerator. E.g. allow = 0 They must be setting either the original f'( ) = 0or an equivalent correct epression = 0 Correct attempt to solve a relevant ( 9)( ) = 0 =... 3TQ in leading to solution for. d Dependent on the previous. = 8, = Note that the = solution could be just written down and is Bbut must come from a correct equation. A, B + 9 = + + = ( ) ( 8)( ) = 0 =.. = 8, = + 9 Sets = 0, squares and multiplies by. They must be setting either the original f'( ) = 0or an equivalent correct epression = 0 Correct attempt to solve a relevant 3TQ leading to solution for. Dependent on the previous. Note that the = solution could be just written down and is Bbut must come from a correct equation. A d A, B (6) (4) [0]

17 . (a) Alternative to part (a) (b) y = = 4( ) 35 Substitute y = ± ± into + 4y = 35to obtain an equation in only Alternative: = ( + ) or = ( + ) = 0 Multiply out and collects terms producing 3 term quadratic in any form. Solves their quadratic, usual rules, as (5+ 9)( ) = 0 =.. far as =... Dependent on the first i.e. a correct method for eliminating d y (or see below) 9 =, = 5 Both correct A for both 9, 3 : Substitutes back into either given y = y = 5 equation to find a value for y Coordinates are ( ) and (, 3) Correct matching pairs. Coordinates need not be given eplicitly but it must A be clear which goes with which y (6) = y ( y ) + 4y ( y ) = Substitutes + 4y = 35 Multiply out, collect terms producing 3 term quadratic in any form. Solves their quadratic, usual rules, as far as y =... Dependent on the first i.e. a correct method for d eliminating 9 y =, y = 3 5 Both correct A for both 9, : Substitutes back into either given = = 5 equation to find a value for 5y 6y 7 = 0 (5y+ 9)( y 3) = 0 y =.. Coordinates are ( ) and (,3) d = 9 9 ( ) + (3 ) 5 5 d = ( ) + (3 ) or Correct matching pairs as above. : Use of d = ( ) + ( y y ) or d = ( ) + ( y y ) where neither ( ) nor ( y y ) are zero. A Aft Aft: Correct ft epression for d or d (may be un-simplified) 4 d = 5 Allow 4.8 Acao (3) [9]

18 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code UA03566 Summer 03 For more information on Edecel qualifications, please visit our website Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Esse CM0 JE