Mark Scheme (Results) Summer Pearson Edexcel GCE Mathematics. Core Mathematics 1 (6663/01)

Transcription

1 Mark Scheme (Results) Summer 07 Pearson Edecel GCE Mathematics Ce Mathematics (6663/0)

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 f employers. F further infmation visit our qualifications websites at 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 wld s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, f all kinds of people, wherever they are in the wld. We ve been involved in education f over 0 years, and by wking across 70 countries, in 00 languages, we have built an international reputation f our commitment to high standards and raising achievement through innovation in education. Find out me about how we can help you and your students at: Summer 07 Publications Code * All the material in this publication is copyright Pearson Education Ltd 07

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 f what they have shown they can do rather than penalised f omissions. Eaminers should mark accding to the mark scheme not accding 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 wthy of credit accding 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. When eaminers are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out wk should be marked UNLESS the candidate has replaced it with an alternative response.

4 PEARSON EDEXCEL GCE MATHEMATICS General Instructions f Marking. The total number of marks f the paper is 7. The Edecel Mathematics mark schemes use the following types of marks: M marks: Method marks are awarded f 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 f crect ft cao crect answer only cso - crect solution only. There must be no errs in this part of the question to obtain this mark isw igne subsequent wking awrt answers which round to SC: special case oe equivalent (and appropriate) d dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper ag- answer given d The second mark is dependent on gaining the first mark 4. All A marks are crect answer only (cao.), unless shown, f eample, as A ft to indicate that previous wrong wking 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 . F misreading which does not alter the character of a question materially simplify it, deduct two from any A B marks gained, in that part of the question affected. 6. If a candidate makes me 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 none are crossed out, mark all the attempts and sce the highest single attempt. 7. Igne wrong wking increct statements following a crect answer.

6 General Principles f Ce Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark f solving 3 term quadratic:. Factisation ( b c) ( p)( q), where pq c, leading to = ( a b c) ( m p)( n q), where pq c and mn a, leading to =. Fmula Attempt to use the crect fmula (with values f a, b and c). 3. Completing the square b Solving b c 0 : q c 0, q 0, leading to = Method marks f differentiation and integration:. Differentiation Power of at least one term decreased by. ( n n ). Integration Power of at least one term increased by. ( n n )

7 Use of a fmula Where a method involves using a fmula that has been learnt, the advice given in recent eaminers repts is that the fmula should be quoted first. Nmal marking procedure is as follows: Method mark f quoting a crect fmula and attempting to use it, even if there are small errs in the substitution of values. Where the fmula is not quoted, the method mark can be gained by implication from crect wking with values, but may be lost if there is any mistake in the wking. Eact answers Eaminers repts have emphasised that where, f eample, an eact answer is asked f, wking with surds is clearly required, marks will nmally be lost if the candidate rests to using rounded decimals.

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9 Question Number Scheme. 3 d 4 Igne any spurious integral signs throughout Raises any of their powers by. 6 3 E.g. k k n n their n their n. Allow the powers to be un-simplified e.g k k. 3 Any one of the first two terms 6 4 crect simplified un-simplified. Two of: ,, 6 c Any two crect simplified terms. Accept 8 f but not 8 f. Accept 0. f 8 but 3 would clearly need to be identified as 0.3 recurring. All crect and simplified and including + c all on one line. Accept 8 f but not 8 f. Apply isw here. Marks A A A (4 marks) 9

10 Question Number Scheme. 4 n y Decreases any power by. Either n dy 4 d 3 3 dy d their n their n 4 0 f fractional n. Crect derivative, simplified unsimplified including indices. E.g. allow f and allow 3 f Attempts to substitute = 8 into their changed (even integrated) epression that is clearly not y. If they attempt algebraic manipulation of their dy/d befe substitution, this mark is still available. B: 8 seen implied anywhere, including from substituting = 8 into y. May be seen eplicitly implied from e.g A: cso 6 6 rational equivalents f and allow e.g. 3 6 Apply isw so award this mark as soon as a crect answer is seen. Marks A BA ( marks) 0

11 Question Number 3.(a) a (b) a 3 a 3 Scheme k k only B k(" k" ) " k " k 7 6 k F substituting their a into k( a ) a to find 3 a3 in terms of a just k k a 3 eact simplified equivalent such as a 3 k k ( k ) but not k Must be k seen in (a) but isw once a crect simplified answer is seen. Note that there are no marks in (b) f using an AP ( GP) sum fmula unless their terms do fm an AP ( GP). Writes + their a + their a3 = 0. 3 k k k a 0 " k" " " 0 r E.g. k 0. Must be a r k crect follow through equation in terms of k only. Solves their equation in k which has come from the sum of 3 terms = 0, 4k k 0 k... and reaches k = Condone po algebra but if a quadratic is obtained e.g. then the usual rules apply f 6k 7k 0 k... solving see General Principles. (Note that it does not need to be a 3- term quadratic in this case) 7 k eact equivalent e.g Do not allow k k 3 Igne any reference to k = 0. Allow.83 recurring as long as the recurring is clearly indicated e.g. a dot over the 3. A A Marks (3) (3) (6 marks)

12 Question Number 4. (a) (b) Scheme Marks ( ) d d... Uses ( ) d and proceeds as far as d... d Crect answer only can sce both 6 marks. A () n Attempts S a l n n S a ( n ) d with n =, n S a 40, l 06, d '6' WAY Or S 40 ( ) "6" n Attempts S a l n S "6" n S a ( n ) d with n =, n S 40 ( ) "6" a 40, l 06 '6', d '6' WAY If they are using n S a ( n ) d, the n must n be used consistently. S 076 WAY Crect sum (may be implied) A S 870 WAY Attempts to find ( ) 06 ( ) ( ) 06. Does not have to be ( ) consistent with their n used f the first Method mark. Attempts to find the total by adding the sum to terms with ( - ) Total "076" "840"... lots of 06 attempts to find the (WAY ) total by adding the sum to terms with ( - ) lots of 06. I.e. Total "870" "8446"... consistency is now required f this (WAY ) mark. Dependent on both previous dd method marks. 036 cao A () (7 marks)

13 Listing in (b): Week Bicycles Total : Attempts the sum of either terms of a series with first term 40 and their d up to 40 +d d. A: S = Then follow the scheme Special case in (b) Treats as single AP with n = S 40 ( ) "6" 36 Sces 000 n n S a ( n ) d with n =, a = 40, d = 6 A: 36 n : 3

14 Question Number.(a) f ( ) ( 4) 3 Scheme : f 4, 0 (where is a single number a numerical epression 0) A: Allow 4 3 and igne Marks A =(-4)^+3 (b) (0, 9) any spurious = 0 Allow a = -4, b = 3 to sce both marks (4, 3) B: U shape anywhere even with no aes. Do not allow a V shape i.e. with an obvious verte. B: P(0, 9). Allow (0, 9) just 9 marked in the crect place as long as the curve ( straight line) passes through touches here and allow (9, 0) as long as it is marked in the crect place. Crect codinates may be seen in the body of the script as long as the curve ( straight line) passes through touches here. If there is any ambiguity, the sketch has precedence. (There must be a sketch to sce this mark) B: Q(4, 3). Crect codinates that can be sced without a sketch but if a sketch is drawn then it must have a minimum in the first quadrant and no other turning points. May be seen in the body of the script. If there is any ambiguity, the sketch has precedence. Allow this mark if 4 is clearly marked on the -ais below the minimum and 3 is marked clearly on the y-ais and cresponds to the minimum, B B B () (3) 4

15 (c) PQ PQ 4 6 Crect use of Pythagas' Theem on points of the fm (0, p) and (q, r) where q 0 and p r with p, q and r numeric. Crect un-simplified numerical epression f PQ including the square root. This must come from a crect P and Q. Allow e.g PQ Allow A Cao and cso i.e. This must come PQ 4 7 from a crect P and Q. Note that it is possible to obtain the crect value f PQ from (-4,3) and (0, 9) and e.g. (0, 3) and (4, -3) but the A marks in (c) can only be awarded f the crect P and Q. A (3) (8 marks)

16 Question Number 6.(a) Replaces with states states y 7 y 8 0* Scheme Uses the addition power law of indices on +. E.g Cso. Complete proof that includes eplicit statements f the addition and power law of indices on with no errs. The equation needs to be as printed including the = 0. If they wk backwards, they do not need to write down the printed answer first but must end with the version in including = 0. The following are eamples of acceptable proofs. y y y 0. y A* Marks y 7 8 y 7 y8 0 y 7y y 7y 8 0 Sces A0 as Special Case: has not been shown eplicitly With without the multiplication signs and with no subsequent eplicit evidence of the power law sces A0 Eample of insufficient wking: y sces no marks as neither rule has been shown eplicitly. () 6

17 (b) y 7y 8 0 y y 8 0 y Solves the given quadratic either in terms of y in terms of See General Principles f solving a 3 term quadratic 7 Note that completing the square on e.g. y y 4 0requires y,8,8,3 7 y q 4 0 y... 4,8 Crect values A : Either finds one crect value of f their obtains a crect numerical epression in terms of logs e.g. f k > 0 log log k k k log A:, 3only. Must be values of. A (4) (6 marks) 7

18 Question Number 7.(a) (b) Scheme Attempts to substitutes 4 into f '(4) 30 f '( ) 30 their 4 algebraically manipulated f '( ) f '(4) 7 Gradient = 7 Attempts an equation of a tangent using their numeric f '(4) which has y( 8) " 7" ( 4) come from substituting = 4 into the given f '( ) their algebraically y " 7" c 8 " 7" 4 c manipulated f '( ) and 4, 8 with c... the 4 and -8 crectly placed. If using y = m + c, must reach as far as c = Cao. Allow y0 7and allow the y = to become detached but y 7 0 it must be present at some stage. E.g. y = Allow the marks in (b) to sce in (a) i.e. mark (a) and (b) together 6 : (these cases only) A: Any crect terms which can be simplified un-simplified. This f ( ) 30 6 c 0.. includes the powers so allow f and allow 3 f (With without + c) A: All 3 terms crect which can be simplified un-simplified. (With without + c) Igne any spurious integral signs Substitutes 4, f ( ) 8 into their f() (not f '( ) ) i.e. a changed 4,f ( ) 8 f '( ) containing +c and rearranges c c... to obtain a value numerical epression f c. f ( ) Cao and cso (Allow f and e.g. f ). Isw here so as soon as you see the crect answer, award this mark. Note that the f() = is not needed. A A Marks AA A (4) () (9 marks) 8

19 Question Number 8.(a) 4 Gradient of l oe Point P = (, 6) y "6" 4 y"6" 4 "6" c c y 49 0 Scheme States implies that the gradient of l 4. E.g. may be implied by a perpendicular gradient of. Do 4 not award this mark f just 4 rearranging to y... unless they then state e.g. d y 4 d States implies that P has codinates (, 6). y = 6 is sufficient. May be seen on the diagram. Crect straight line method using P(, 6 ) and gradient of. grad l Unless is being used as 4 4 the gradient here, the gradient of l clearly needs to have been identified and its negative reciprocal attempted to sce this mark. Accept any integer multiple of this equation including = 0 Marks B B A (4) 9

20 8(b) y y y and y Substitutes y = 0 into their l to find a value f substitutes y = 0 into l their rearrangement of l to find a value f. This may be implied by a crect value on the diagram. Substitutes y = 0 into their l to find a value f and substitutes y = 0 into l their rearrangement of l to find a value f. This may be implied by crect values on the diagram. (Note that at T, = 9.8 and at S, = -.) Fully crect method using their values to find the area of triangle SPT with vertices at points of the fm (, 6 ), (p, 0) and (q, 0) where p q Attempts to use integration should be sent to your team leader Method : "6" ST ('9.8' '.') '6'... Method : SP PT '.' '6' '9.8' '6' Note that if the method is crect but slips are made when simplifying any of the calculations, the method mark can still be awarded Method 3: Triangles ( '.') '6' ('9.8' ) '6'... Method 4: Shoelace method ( ) (must see a crect calculation i.e. the middle epression f this determinant method) Method : Trapezium + triangles ('.') '' "" "6" ("9.8" ') '6' cso oe e.g 369 0, 9 36, but not e.g Note that the final mark is cso so beware of any errs that have ftuitously resulted in a crect area. dd A (8 marks) (4) 0

21 Question Number 9.(a)(i) (a)(ii) Scheme B: Straight line with negative gradient anywhere even with no aes. (0, c) B: Straight line with an intercept at (0, c) just c marked on the positive y-ais provided the line passes through the positive y-ais. Allow (c, 0) as long as it is marked in the crect place. Allow (0, c) in the body of the script but in any ambiguity, the sketch has precedence. Igne any intercepts with the -ais. Either: F the shape of a y curve in any position. It must have two branches and be asymptotic hizontally and vertically with no obvious overlap with the asymptotes, but otherwise be generous. The curve may bend away from the asymptote a little at the end. Sufficient curve must be seen to suggest the asymptotic behaviour, both vertically and hizontally and the branches must approach the same asymptote Or the equation y = seen y = independently i.e. whether the sketch has an asymptote here not. Do not allow y =. B: Fully crect graph and with a hizontal asymptote on the positive y-ais. The asymptote does not have to be drawn but the equation y = must be seen. The shape needs to be reasonably accurate with the ends not bending away significantly from the asymptotes and the branches must approach the same asymptote. Igne = 0 given as an asymptote. Allow sketches to be on the same aes. B B B B Marks (4)

22 (b) c 0 c c b 4 ac ( c) 4 3 Sets 3 c, attempts to multiply by and collects terms (to one side). Allow e.g. > < f =. At least 3 of the terms must be multiplied by, e.g. allow one slip. The = 0 may be implied by subsequent wk and provided crect wk follows, full marks are still possible in (b). Attempts to use b 4ac with their a, b and c from their equation where a 3, b c and c. This could be as part of the quadratic fmula as b < 4ac as b > 4ac ( c) * as b 4ac etc. If it is part of the quadratic fmula only look f use of b 4ac. There must be no s. Completes proof with no errs increct statements and with the > appearing crectly befe the final answer, which could be from b - 4ac > 0. Note that the statement 3 c 0 starting A* If b with e.g. 3 cwould be an err. Note: A minimum f (b) could be, 3 c 3 c 0 () b 4 ac ( c) (A) 4ac is not seen then 43needs to be seen eplicitly. (3)

23 (c) ( c) c ( c) c 0c 3 0 c c " ", c " " : Attempts to find at least one critical value using the result in (b) by epanding and solving a 3TQ (See General Principles) (the = 0 may be implied) A: Crect critical values in any fm. Note that may be seen as 3. Chooses outside region. The 0 < can be igned f this mark. So look f c < their, c > their. This could be sced from c c. Evidence is to be taken from their answers not from a diagram. Crect ranges including the 0 < e.g. answer as shown each region written separately e.g. 0,,. The, 0 c, c critical values may be un-simplified but must be at least ,. Note that 0 c and c would sce A0. Allow the use of rather than c in (c) but the final answer must be in terms of c. A A (4) ( marks) 3

24 Question Number 0.(a)(i) (ii) k 3 7 c only Scheme (b) : Attempts to find the 'y' intercept. Accept as evidence 3with without the bracket. If they epand f() to polynomial fm here then they must then select their constant to sce this mark. May be implied by sight of 7 on the diagram. A: k 7. Must clearly be identified as k. Allow this mark even from an increct incomplete epansion as long as the constant k = 7 is obtained. Do not isw e.g. if 7 is seen followed by k = -7 sce A0. c oe (and no other values). Do not award just from the diagram must be stated as the value of c. 3 f ( ) ( ) ( 3) 4 0 ( 3) Attempts f() as a cubic polynomial by attempting to square the first bracket and multiply by the linear bracket epands ()( 3) and then multiplies by Must be seen used in (b) but may be done in part (a). Allow po squaring e.g. () 4 : Reduces powers by in all terms including any constant 0 A: Crect proof. Withhold this f '( ) 6 3* mark if there have been any errs including missing brackets earlier e.g. ( ) ( 3) 4 0 ( 3)... Marks A B A* (3) (3) 4

25 (c) Alt (b) Product rule. f '(3) '' ( 3)( 0) Substitutes = 3 into their f '( ) the given f '( ). Must be a changed function i.e. not into f(). Sets their f '( ) the given f '( ) = their f '(3) with a consistent f '. Dependent on the previous method mark equivalent 3 term quadratic e.g (A crect quadratic equation may be implied by later wk). This is cso so must come from crect wk i.e. they must be using the given f '( ). Solves 3 term quadratic by suitable method see General Principles. Dependent on both previous method marks. oe clearly identified. If = 3 3 is also given and not rejected, this mark is withheld. (allow -.6 recurring as long as it is clear i.e. a dot above the 6). This is cso and must come from crect wk i.e. they must be using the given f '( ). f( ) ( ) ( 3) f '( ) ( ) ( 3) 4( ) : Attempts product rule to give an epression of the fm p( ) q( 3)( ) : Multiplies out and collects terms A: f '( ) 6 3* d A cso dd A cso () ( marks) A*

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