LEVEL 2 CERTIFICATE Further Mathematics

Transcription

1 LEVEL CERTIFICATE Further Mathematics Paper 860/ Calculat Mark scheme 860 June 07 Version:.0 Final

2 Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every associate understands and applies it in the same crect way. As preparation f standardisation each associate analyses a number of students scripts. Alternative answers not already covered by the mark scheme are discussed and legislated f. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a wking document, in many cases further developed and expanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this mark scheme are available from aqa.g.uk Copyright 07 AQA and its licenss. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges f AQA are permitted to copy material from this booklet f their own internal use, with the following imptant exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even f internal use within the centre.

3 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Glossary f Mark Schemes GCSE examinations are marked in such a way as to award positive achievement wherever possible. Thus, f GCSE Mathematics papers, marks are awarded under various categies. If a student uses a method which is not explicitly covered by the mark scheme the same principles of marking should be applied. Credit should be given to any valid methods. Examiners should seek advice from their seni examiner if in any doubt. M A B ft SC M dep B dep Method marks are awarded f a crect method which could lead to a crect answer. Accuracy marks are awarded when following on from a crect method. It is not necessary to always see the method. This can be implied. Marks awarded independent of method. Follow through marks. Marks awarded f crect wking following a mistake in an earlier step. Special case. Marks awarded f a common misinterpretation which has some mathematical wth. A method mark dependent on a previous method mark being awarded. A mark that can only be awarded if a previous independent mark has been awarded. Or equivalent. Accept answers that are equivalent. eg accept 0.5 as well as [a, b] Accept values between a and b inclusive. [a, b) Accept values a value < b. Accept answers which begin. eg.,.,.6 Use of brackets It is not necessary to see the bracketed wk to award the marks.

4 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Examiners should consistently apply the following principles Diagrams Diagrams that have wking on them should be treated like nmal responses. If a diagram has been written on but the crect response is within the answer space, the wk within the answer space should be marked. Wking on diagrams that contradicts wk within the answer space is not to be considered as choice but as wking, and is not, therefe, penalised. Responses which appear to come from increct methods Whenever there is doubt as to whether a student has used an increct method to obtain an answer, as a general principle, the benefit of doubt must be given to the student. In cases where there is no doubt that the answer has come from increct wking then the student should be penalised. Questions which ask students to show wking Instructions on marking will be given but usually marks are not awarded to students who show no wking. Questions which do not ask students to show wking As a general principle, a crect response is awarded full marks. Misread miscopy Students often copy values from a question increctly. If the examiner thinks that the student has made a genuine misread, then only the accuracy marks (A B marks), up to a maximum of marks are penalised. The method marks can still be awarded. Further wk Once the crect answer has been seen, further wking may be igned unless it gs on to contradict the crect answer. Choice When a choice of answers and/ methods is given, mark each attempt. If both methods are valid then M marks can be awarded but any increct answer method would result in marks being lost. Wk not replaced Erased crossed out wk that is still legible should be marked. Wk replaced Erased crossed out wk that has been replaced is not awarded marks. Premature approximation Rounding off too early can lead to inaccuracy in the final answer. This should be penalised by mark unless instructed otherwise. Continental notation Accept a comma used instead of a decimal point (f example, in measurements currency), provided that it is clear to the examiner that the student intended it to be a decimal point.

5 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Q Answer Mark Comments ( ) ( ) (a) 7 ( ) ( )8.5 ( ) 5 ( )0 Accept if both 0 and 0 are seen.5 Condone n.5 (b) n n n n not processed 0.5n not processed 5

6 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 6 B 6 in crect position in a by matrix (a) Condone missing brackets f B if numbers in a by array Brackets may be square curly etc Igne commas and fraction lines 6 followed by further wk 5k = k k = 6k k = 5 ( 6k) 8k = with no increct equation seen Any one crect equation (b) 5k k k 6k with no further crect wk M0 Igne subsequent attempt to convert 8 to a mixed fraction decimal Igne subsequent attempt to convert to an improper fraction 8 decimal Igne subsequent rounding truncation of.75 Answer only.7.8. M0 T & I is zero 6

7 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Valid explanation eg Number of columns of B ds not equal number of rows of A by ( ) matrix means rows and column First matrix means B and second matrix means A (c) B columns A rows B rows A columns B columns A columns B rows A rows Number of rows in second matrix cannot be me than number of columns in first matrix B has column, A has rows A should only have row A has too many rows B only has one column B needs another column It is a x multiplied by a There s nothing to multiply the by It is a x multiplied by It is a by multiplied by a by B values can t multiply with all the A values They are not compatible Because the dimensions of A and B are different Can t wk it out this way round Can wk out AB but not BA B has to be a by matrix 7

8 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 (a) ( ) 55 5 ( ) 7 7 ( ) 95 ( ) 05 5 ( ) 9 ( ) 65 5 ( ) 9 ( ) 5 ( ) 7 ( ) eg 65 5 = 7 Any der Must be integers May be seen in a fact tree repeated division Any der Must be integers If using division the crect answer must be seen f Crect answer can be implied by wking lines eg ( ) 5 ( ) 9 with blank answer line Answer line crect Allow inclusion of f eg ( ) ( ) 55 (b) b(a ) b( a) a = 6 and b = square number > with wking f seen Implied by square numbers > used eg (6 ) eg 9(6 ) Must be in crect der Allow unprocessed squares eg a = 6 and b = 5 SC a = 6 and b = square number > without wking f seen b(a ) = 0 b(a ) with further wk Answer line takes precedence over wking lines Embedded answer eg 8(6 ) A0 8

9 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE x = 56 6x = = x equation in x () x 7 56 x = x = with no crect further wk M0 56 x = M0 Solving 56 = M0 x Answer with 7 not seen in wking Embedded solution A0 A0 9

10 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method a = a a a = a a + a = a a = a a = (a a) 6a a = a a + a = a a 6a = 8 = 6a a = a + a dep Allow eg a f 6a Terms collected 5 5a = Alternative method 8 6 a a = 0 6 a eg fractions inverted 8 6 a a = 0 8 a 0 6 a = 0 8 a 8a + a = 8 6a + a = 8 a + a = 8 6 5a = dep Allow eg a f 8a Terms collected Alternative method and continue on the next page 0

11 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method (8 6) + (a a) = (0 8) + ( a) Using PM = MQ 5a a + 6 (= 0) PQ = PM PQ = MQ (0 6) + ( a) = ((8 6) + (a a) ) 5a + 8a 6 (= 0) (0 6) + ( a) = ((0 8) + ( a) ) 5a 88a + 8 (= 0) 5 cont (5a )(a ) (= 0) ( ) 5 (5a )(a + ) (= 0) (5a )(7a ) (= 0) 88 ( 88) dep eg Crect attempt to solve their -term quadratic Allow recovery of brackets in fmula Allow eg f ( ) Implied by crect solutions to their -term quadratic seen Must reject solution 7 Terms must be collected but do not have to be processed f dep eg (Alt ) a + = 6a needs terms collecting to = 6a a f dep Rejection of solution is implied by only 5 on answer line

12 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 r sin 8 = 0 r cos (90 8) = 0 r sin 8 sin 8 r = = 0 sin 90 sin 90 0 Any letter 0 sin 8 0 cos (90 8) 0 sin 8 dep sin 90 M M 0 0 (0 cos 8) cos(8 ).(...) SC Angle VAC = 8 seen on diagram and answer 5.76( ) If trigonometry and Pythagas are used it must be a fully crect method that would lead to the crect value of r f M If cosine rule with angle (8 ) used it must be a fully crect method that would lead to the crect value of r f M Answer 5.76( ) 5.8 but angle VAC = 8 not seen Zero.( ) seen and angle 8 in crect place with further wk eg 0 sin 8 =. 0. = 5.8 Zero sin 8 0 (even if subsequently evaluates sin 760) opp Throughout, accept opp o f r eg sin 8 = 0 M r sin = 0 r sin = 0 (unless recovered) M0 Answer.( ) coming from scale drawing Answer coming from scale drawing.( ) seen with no further wk followed by answer M Zero M

13 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method (x-codinate of A =) 0 and (y-codinate of B =) 8 (x-codinate of P =) May be implied on diagram eg 0 written next to A and 8 written next to B their 0 their 0 must be their x-codinate of A May be seen on diagram their 0 0 (area of triangle OBP =) dep their 8 their their 8 must be their y-codinate of B 7 6 ft ft M Alternative method (x-codinate of A =) 0 and (y-codinate of B =) 8 May be implied on diagram eg 0 written next to A and 8 written next to B (area of triangle OAB =) their 0 their 8 0 (area of triangle OBP =) their 0 dep eg their 0 their 0 6 ft ft M Alternative methods and and continue on the next two pages

14 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method (x-codinate of A =) 0 and (y-codinate of B =) 8 May be implied on diagram eg 0 written next to A and 8 written next to B (area of triangle OAB =) their 0 their cont (y-codinate of P =) their 8.8 and (area of triangle OPA =) their 0 their.8 and (area of triangle OBP =) their 0 their dep their 8 must be their y-codinate of B y-codinate of P may be seen on diagram 6 ft ft M Alternative method and continue on the next page

15 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method (x-codinate of A =) 0 and (y-codinate of B =) 8 May be implied on diagram eg 0 written next to A and 8 written next to B (AB =) their 0 their ( ) and (BP =) 5.( ) their.8( ) and (angle OBP =) tan their 0 their 8 5.( ) their 0 must be their x-codinate of A their 8 must be their y-codinate of B 7 cont (area of triangle OBP =) their 8 their 5. sin their 5. dep 6 ft ft M their 8 must be their y-codinate of B A = 0 and B = 8 A (8, 0) and B (0, 0) is but can subsequently sce up to Mft (answer 6) A (0, 0) and B (8, 0) is but can sce up to Mft if uses x-codinate of A as 0 and y-codinate of B as 8 (answer 6) A (0, 8) and B (0, 0) is but can sce up to Mft if uses x-codinate of A as 8 and y-codinate of B as 0 (answer 6) Area triangle OBP may be seen as the sum of two right-angled triangles Area triangle OBP may be seen as area trapezium OBPX area triangle OPX X is on the x-axis with PX perpendicular to the x-axis Allow marks f valid wking seen even if not subsequently used 5.9( ) answer 6 Answer 5.9( ) marks MA0 5

16 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method (BC =) their = cos A = cos A = cos A 7 8 their [ 0.77, 0.7] 0.8 Allow as two 6s labelled on BC after perpendicular drawn from A Do not allow if their comes from use of Pythagas theem ie (BC =) is M ( ) 8 8 cos 7 8 their 7 8 dep [06, 06.] ft Only ft M May be implied by final answer Alternative method (BC =) (angle ABC =) cos and sin A = their 8 7 their sin their sin A = sin (their 0.96 ) their dep Allow as two 6s labelled on BC after perpendicular drawn from A eg wks out angle ACB (=.09( ).) and uses sine rule Do not allow if their comes from use of Pythagas theem ie (BC =) is M ( ) May be implied by final answer [06, 06.] ft Only ft M continues on the next page 6

17 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 8 cont cos cos ans ds not sce dep unless recovered F the dep must have crect rearrangement but allow arithmetic errs Answer outside range is A0 eg 06.( ) from cos ( 0.8) 7

18 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method 5 < x 5 5. < x 5 5 eg x 5 5 and x > 5 < x. < x 5 x,, 0, eg x and x > 5 6x x 7 x Collects terms x x.5 x < x <.5. < x.5 x.5 implies x, (,,...) with no other values given Must have gained 9 Alternative method Shows that satisfies either < 5x 5 6x + 7 x + eg < 0 5 5x = 0 and yes Shows that satisfies both < 5x 5 and 6x + 7 x + Shows that ds not satisfy eg > + 6x + 7 x + shows that ds not satisfy < 5x 5 Shows that ds not satisfy 6x + 7 x + and shows that ds not satisfy < 5x 5 with no other values given Must have gained Alternative methods and and continue on the next two pages 8

19 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method 5 < x 5 5. < x 5 5 eg x 5 5 and x > 5 9 < x. < x 5 x,, 0, Shows that satisfies 6x + 7 x + shows that ds not satisfy 6x + 7 x + Shows that satisfies 6x + 7 x + and shows that ds not satisfy 6x + 7 x + with no other values given eg x and x > 5 eg = 5 and + = Must have gained Alternative method and continue on the next page 9

20 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method 6x x 7 x Collects terms x x.5 x < x <.5 x, (,,...) Shows that satisfies < 5x 5 shows that ds not satisfy < 5x 5 eg < 0 5 5x = 0 and yes 9 Shows that satisfies < 5x 5 and shows that ds not satisfy < 5x 5 with no other values given Must have gained Allow eg max and min. f. < x, unless contradicted by a list of values Condone omission of non-critical values from lists eg,, Using = signs when solving inequalities can sce M marks only unless recovered Increct notation eg f < can sce M marks only If answers to trials evaluated they must be crect Choose the scheme that favours the student identified as the only integer with no valid wking Zero 0

21 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method x x sin 50 x Any letter(s) b c sin 50 = b c = x = sin50 x = x = (.0) x = (.0) dep eg x = sin 50 Must have either x = x = (.0) Any letter 0 5. Alternative method x x cos sin = x = cos sin x = (.0) (.0) dep Any letter Must have either x = (.0) Any letter 5. Do not allow 5 as a misread of 50 x can be b AB AC etc b and c can be a and b AB and AC etc

22 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Straight line between (, 7) and (0, ) Points (0, ) (, ) (, ) (, 0) (, 5) Crect smooth parabolic curve with maximum at (, ) Straight line between (, 5) and (5, 0) Tolerance of ± small square Allow line to be extended Tolerance of ± small square May be plotted seen in a table Points can be implied Tolerance of ± small square Allow (ruled) straight line between (, 0) and (, 5) Curve passing through all crect points within tolerance sces Tolerance of ± small square Allow line to be extended (a) Igne extra points plotted Tolerance of ± small square means it is on the edges of within the shaded area x Points only can sce a maximum of Ruled straight lines f curve apart from between (, 0) and (, 5) If all marks would be awarded but either (i) graph has a line a curve that extends beyond the individual domains (ii) the curve ds not meet a line at a cusp A0 marks

23 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 5 f(x) 7 7 f(x) 5 [ 5, 7] Bft Crect ft their graph in (a) f B ft their graph in (a) f ft 5 f(x) f(x) 7 on their own embedded within an interval f f(x) only 5 and 7 chosen eg 5 < f(x) < 7 Allow f(x) to be y f fx eg 5 y 7 eg f 7 B Allow as two inequalities f(x) 5 (and/) f(x) 7 B (b) ft their graph if incomplete eg no graph drawn f x < 0 but otherwise crect and answer 5 f(x) ft their graph if drawn f x values beyond [, 5] eg straight line from (, 8) to (6, ) and answer y 8 eg straight line from (, 8) to (6, ) and answer f(x) 8 Straight line from (, 9) to (6, 7) and answer 7 y 9 Straight line from (0, 9) to (5, ) and answer f(x) 9 Bft ( ft) can be awarded f a range beyond [ 7, 9] if it is clear from wking (eg a table of values) where the answer is from 5 to 7 inclusive is B whereas 5 to 7 is f a crect inequality embedded eg 5 < f(x) 7 eg 5 f(x) 0 eg y 7 F igne increct notation if only 5 and 7 chosen eg 5 x 7 eg 5 < x 7 eg 5 f(x) 7 eg 5, 7 Bft Bft ft Bft Bft { 5,,,,, 0,,,,, 5, 6, 7} Wking out a statistical range eg 5 to 7 =

24 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 (5 x ) (x 5) (5 + x)(5 x) (x + 5)(5 x) partial factisation eg (x + 5)(x 5) Brackets in either der Do not allow (x 75) (a) (5 + x)(5 x) ( x 5)(x 5) (x + 5)(x 5) (5 x)( x 5) ( x + 5) is equivalent to (5 x) etc Do not allow f increct notation in final answer eg (5 + x)(5 x) Do not allow f use of multiplication signs in final answer eg (5 + x) (5 x) Crect answer followed by increct further wk A0 A0 A0

25 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method 9n + n + n + 9n + 6n + 9n n n + 9n 6n + Terms may be seen in a grid n with no increct wking Brackets can be recovered Alternative method (n + + n )(n + (n )) (n + + n )(n + n + ) n Brackets around n can be recovered (b) Alt n may come from increct wking eg n + 6n + (n 6n + ) = n eg 9n + n + (9n n + ) = n Alt Recovery of brackets eg 9n + 6n + 9n 6n + = n eg 9n + 6n + 9n 6n + = Alt Recovery of brackets eg (n + + n )(n + n ) = n eg (n + + n )(n + n ) = 0 Do not allow f use of multiplication signs in final answer eg n with no increct wking M0A0 M0A0 A0 M0A0 A0 5

26 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Single crect fraction with terms processed eg 600a 5 6a 00a 7a eg 50a 00a a 6 Only bracket allowed is (a + ) eg 50a a ( a ) 6a (sces M) Factises crectly using (a + ) Only needs to be seen once eg eg 8a a 6 8a ( a ) 5( a ) a 5a 5a 0 a 5a Award M f fully crect unprocessed expression with full cancelling seen eg 8 a ( a ) 5 ( a ) a 5 5a a 5 5a 50a 6 a a 50 a a A crect single fraction with (a + ) cancelled will be M eg 50a 5 50a eg a MA0 MA0 8a 5( a ) a 5a M0A0 a + 6 = (a + ) with no other valid wking M0A0 Brackets other than (a + ) may be seen 0a (5a 0) a 6 M0M0 Crect answer followed by increct further wk MA0 Allow one miscopy f up to MA0 6

27 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method 0.5 gradient of x + y = 7 Do not allow embedded May be implied (gradient =) their ft their Only ft a non-zero numerical value Implied by y = x + b a = () (y =) May be seen on diagram their 8 = their + b y their 8 = their (x ) dep dep on M b = 0 ft ft 8 their if M Alternative method 0.5 gradient of x + y = 7 Do not allow embedded May be implied (gradient =) their ft their Only ft a non-zero numerical value Implied by y = x + b a = () Crect method f elimination of y from x + y = 7 and y = their x + b dep eg x + (x + b) = 7 7x + b = 7 Substitutes x = into their equation dep eg + b = 7 b = 0 ft ft 8 their if M Alternative method and continue on the next page 7

28 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method 0.5 gradient of x + y = 7 Do not allow embedded May be implied (gradient =) their ft their Only ft a non-zero numerical value Implied by y = x + b a = () (y =) May be seen on diagram cont Crect method f elimination of x from x + y = 7 and y = their x + b and substitutes y = their 8 dep eg y = (7 y) + b 7y = 96 + b and 06 = 96 + b dep on M b = 0 ft ft 8 their if M y = x + 0 will gain full marks unless contradicted If an err is made in the constant term when rearranging x + y = 7 the can still be awarded f gradient = eg y = x + 9 and gradient = is (all other marks are possible) In alt and alt the mark f y = 8 will sometimes be the only mark awarded 8

29 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method wy = 8x y w = 8 x y w + = 8x y wy + y = 8x y(w + ) = 8x w 8x = y dep y term(s) collected eg wy y = 8x 8x M w 8x w 8x ( w ) 8x y = y = w 8x y = ( w ) 8x w x eg y = 0.5w 0.5 Must have y = 5 Alternative method 8 x y y = w w y 8x y + = w w 8x y( + ) = w w dep y term(s) collected 8x M w w y = 8x w w Must have y = y = 8x w in wking with 8x on answer line etc M w Allow multiplications signs and s throughout w = 8 x y y y with no further simplification M0 Crect answer followed by increct further wk MA0 9

30 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 b 6(a) 5 x+ 6(b) m 6(c) x ( )x Attempt at dy dx their (x x) = 0 dep Must have at least terms f their The = 0 can be implied by sight of a crect non-zero solution to their (x x) = 0 dy dx x = (and x = 0) ft ft M if their dy is a -term quadratic dx 7(a) (, 5) dy with crect expression f seen dx Condone y = x x etc Igne wking f second derivative testing f minimum point d y Stating = 0 is not sufficient f second M mark but may be implied by dx crect solution(s) seen 0

31 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method ( ) 6( ) + 7 = 0 with no increct evaluations seen = 0 Alternative method Must have = 0 7(b) (x + )(x 7x + 7) = 0 and (x + ) = 0 and x = ( ) 6( ) Allow ( ) f ( ) Allow recovery of brackets f ( ) eg = 0 eg = = 0

32 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method (x ) (x + ) seen (x + )(x 7x + c) c can be any non-zero value dep Implied by (x + )(x + bx + c) and b + = 6 b = 7 x 7x + 7 (= 0) 7 ( 7) 7 7 eg 7 Crect attempt to solve their -term quadratic Allow recovery of brackets Allow 7 f ( 7) Implied by crect solutions to their -term quadratic seen 7(c) 5.79 and. with x 7x + 7 (= 0) seen Must both be to dp Alternative method (x ) (x + ) seen x x x 7x 6x ( 0x) 7 dep x 7x + 7 (= 0) 7 ( 7) and. with x 7x + 7 (= 0) seen eg 7 Crect attempt to solve their -term quadratic Allow recovery of brackets Allow 7 f ( 7) Implied by crect solutions to their -term quadratic seen Must both be to dp is on the next page

33 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 7(c) cont Final mark can be awarded if both answers seen in wking with x 7x + 7 (= 0) seen but only one answer is written on answer line (x + ) followed by 5.79 and. without x 7x + 7 (= 0) seen (x ) instead of (x + ) can sce a maximum of M0M0A0A0 T & I on the cubic equation M0A0 M0A0 Zero x + (x) = (y) x + x = 6y x 6 = y 5x = 6y 5 x = 6 y x 5 = y 5 eg 5x = (y) Missing brackets may be recovered equation of the fm ax = by cx = dy x = eg x = 6y 5 ky ft if Pythagas used with only err being missing brackets 6 their y 5 their 6y their 5 dep their x (their x) dep on at least one M 8 y 6 y 6.y 5 5 5x = (y) with no further wk M0M0 x + x = y with answer 5 8 y M0A0 x + x = 6y with answer y M0A0 x + x = y with answer 8 y M0A0 (x) = (y) x x = 6y and 6y = y M0A0 y followed by further wk MA0 5

34 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 k 9(a) k = 0 k = etc k 9(b) k = 0 k = etc k + cos α = k eg ( + k)( k) k ( k)( k) 9(c) Answer k ± k A0 Crect answer followed by increct further wk A0 Answer k A0 Allow cos x cos θ etc cos c (cos α) f cos α Condone cos α f cos α cos(sin k) M0A0

35 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Angle in a semicircle (is a right angle) Angle at centre is 80, angle at circumference is half the angle at the centre (= 90 ) Angle at centre is 80, angle at centre is twice the angle at the circumference Angle subtended at circumference by a diameter 0(a) Do not allow half a circle to mean a semicircle Allow extra wds if not contradicty eg Angle at the circumference in a semicircle eg Angle inscribed in a semicircle Angle subtended by a diameter (no mention of at circumference) Angle in a hemisphere is 90 Angle at centre is 80 Angle at circumference is half the angle at the centre chds on diameter meet at 90 Triangle in a semicircle always has a right angle Angle in a semicircle is 80 Angle on a diameter is a right angle Because AB is a diameter 5

36 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 angle ABE = 90 x angle CBE = 90 + x angle DEB = 90 and angle DCB = 90 angle CDE = 90 x dep angle CED = 90 x angle DCE = x and all reasons given f their proof dep dep See guidance f acceptable wding f reasons 0(b) To award a particular mark, all previous marks must have been awarded First three B marks can be awarded with no increct reasons Do not mark any wking on the diagram statements are needed Increct angles sce eg angle ABE = 90 x angle DEC = 90 + x eg angle ABE = 90 x angle CDE = 90 x angle DCE = 90 + x Angle CDE and angle CDA are the same angle etc Angle EBA and angle ABE are the same angle etc Condone ABE f angle ABE etc Do not allow angle C f angle DCE etc CE must be proven to be a tangent if used in a response Reasons angle sum of triangle (is 80 ) angles in a triangle (add to 80 ) 80 in a triangle (adjacent) angles on a (straight) line (add to 80 ) 80 on a (straight) line exteri angle of triangle (= sum of opposite interi angles) (equal angles in an) isosceles (triangle) CD = CE (opposite angles in a) cyclic quadrilateral (add to 80 ) exteri angle of cyclic quadrilateral (= opposite interi angle) Degrees symbol may be omitted Abbreviations are allowed eg quad f quadrilateral 6

37 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 (0, 8) Answer line takes precedence over wking lines and diagram (a) Answer line blank with C labelled (0, 8) on diagram Answer line blank with 8 written next to C on diagram (8, 0) Answer 8 x x + x + 8 Allow one err but no omission Must have an x term Terms may be seen in a grid Implied by x + x + k k 0 ax + x + 8 a 0 x x + x + 8 x + x + 8 x x + x + 8 but an err in any collection of terms is A0 x + x + (x ) ( x) ft ft their quadratic in x with awarded (b) x with final answer (from substituting in x = 0) A0 Condone y = x f(x) = x in wking f dy If ( dx f (x) =) x on answer line also award final y = x f(x) = x on answer line A0 When marking (b), a maximum of A0 can be awarded from an expansion seen on the previous page if not contradicted by an expansion in (b) The final must be seen in (b) eg (b) no expansion seen with an answer of x + At top of previous page x + x + 8 eg (b) no expansion seen with an answer of x + 6 In (a) x + x + x + 8 = x + 6x + 8 Crect use of product rule and gradient function = x + A0 A0ft marks 7

38 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method (gradient of curve at C =) ft Crect ft their (b) when x = 0 May be implied their ft their Only ft a non-zero numerical value (gradient of nmal =) is y = (their )x + their 8 y their 8 = their (x 0) dep Must have used gradient of nmal not gradient of tangent Crect ft their 8 from (a) in the fm (0, k) 0 = (their )x + their 8 0 their 8 = their (x 0) dep Crect ft their 8 from (a) in the fm (0, k) x = 6 (c) x = 6 and BD = and AB = 6 with crect method seen Alternative method (gradient of curve at C =) ft Crect ft their (b) when x = 0 May be implied their ft their Only ft a non-zero numerical value (gradient of nmal =) is 0 their 8 = their x 0 x = their 8 their x = 6 x = 6 and BD = and AB = 6 with crect method seen Mdep Crect ft their 8 from (a) in the fm (0, k) 8

39 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method (x ) + (x + ) = 6 Eliminates y x x x + + x = 6 5x x (= 0) dep Expands both brackets crectly (5x + 6)(x ) (= 0) 6 eg ± 5 5 ( ) 5 5 Crect attempt to solve their -term quadratic Allow recovery of brackets in fmula Allow f ( ) Implied by crect solutions to their -term quadratic seen (x =). and (x =) (x =). and (y =). (x =) and (y =) 5 eg (x =) 5 6 and (x =) with 5x x (= 0) seen with 5x x (= 0) seen (.,.) and (, 5) with 5x x (= 0) seen eg ( 5 6, 5 7 ) and (, 5) with 5x x (= 0) seen 9

40 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method x x x + + y y y + = 6 Expands both brackets crectly x x x + + (x + ) (x + ) (x + ) + = 6 dep Eliminates y 5x x (= 0) (5x + 6)(x ) (= 0) 6 eg ± 5 5 ( ) 5 5 Crect attempt to solve their -term quadratic Allow recovery of brackets in fmula Allow f ( ) Implied by crect solutions to their -term quadratic seen (x =). and (x =) (x =). and (y =). (x =) and (y =) 5 eg (x =) 5 6 and (x =) with 5x x (= 0) seen with 5x x (= 0) seen (.,.) and (, 5) with 5x x (= 0) seen eg ( 5 6, 5 7 ) and (, 5) with 5x x (= 0) seen Alternative methods and and continue on the next two pages 0

41 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method y ( ) + (y ) = 6 Eliminates x y y y + y y y + = 6 + dep y Expands ( ) 5y 8y 5 (= 0) and (y ) crectly (5y + 7)(y 5) (= 0) 9 56 eg ± ( 8) Crect attempt to solve their -term quadratic Allow recovery of brackets in fmula Allow 8 f ( 8) Implied by crect solutions to their -term quadratic seen (y =). and (y =) 5 (x =). and (y =). (x =) and (y =) 5 eg (y =) 5 7 and (y =) 5 with 5y 8y 5 (= 0) seen with 5y 8y 5 (= 0) seen (.,.) and (, 5) with 5y 8y 5 (= 0) seen eg ( 5 6, 5 7 ) and (, 5) with 5y 8y 5 (= 0) seen Alternative method and continue on the next page

42 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method x x x + + y y y + = 6 Expands both brackets crectly y y y + y y y + = 6 + dep Eliminates x 5y 8y 5 (= 0) (5y + 7)(y 5) (= 0) 9 56 eg ± ( 8) Crect attempt to solve their -term quadratic Allow recovery of brackets in fmula Allow 8 f ( 8) Implied by crect solutions to their -term quadratic seen (y =). and (y =) 5 (x =). and (y =). (x =) and (y =) 5 eg (y =) 5 7 and (y =) 5 with 5y 8y 5 (= 0) seen with 5y 8y 5 (= 0) seen (.,.) and (, 5) with 5y 8y 5 (= 0) seen eg ( 5 6, 5 7 ) and (, 5) with 5y 8y 5 (= 0) seen Answers only (no valid wking) Both solutions from scale drawing Zero 5 marks (, 5) is often seen without seeing any crect method Zero Allow one miscopy f up to MA0A0

43 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method Replaces tan x with sin x cos x at least once in given expression eg sin cos x x sin x Crect steps leading to the single fraction cos x sin x cos x cos x x sin x sin cos x cos x sin x sin x dep sin sin x x cos x sin x = sin sin x x = Must see all steps leading to cos x = cos x sin x = sin x sin x = sin x Alternative method and continue on the next page

44 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method Replaces tan x with sin x cos x at least once in given expression eg sin x sin x cos x sin x sin x cos x Crect steps leading to the single fraction sin x (cos x ) sin x dep sin sin x x sin x (cos x ) = sin x sin x = sin x sin x = sin x Must see all steps leading to cont Allow cos θ etc cos c (cos x) f cos x etc Condone cos x f cos x etc Only substituting values f x Zero cos x etc with no wking Zero sin x Alt sin x cos x sin x sin x with no further wking M0A0 Any fully crect response that shows how the given expression is equal to is awarded marks eg sin cos x x sin x cos x = sin x sin x = sin x sin x sin x marks = sin x sin x sin x sin x = cot x cosec x = marks

45 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method (x 5x) (x.5) {(x.5).5 } (x.5) 75 dep eg {(x 5x) } (x 5x ) eg {(x.5).5 } (x.5) (x.5) 70 dep eg (x.5) x dep x 5 eg (x 5) 70 a = b = c = 5 d = 70 (5 x) 70 a = b = c = 5 d = 70 5

46 MARK SCHEME LEVEL TWO CERTIFICATE FURTHER MATHEMATICS 860/ JUNE 07 Alternative method (x 0x) (x 5) {(x 5) 5 } (x 5) 75 {(x 5) 5 } + 5 dep dep eg {(x 0x) } (x 0x ) eg {(x 5) 5 } eg {(x 5) } (x 5) dep eg (x 5) (x 5) 70 a = b = c = 5 d = 70 (5 x) 70 a = b = c = 5 d = 70 F M marks.5 may be seen as 5 F M marks (x.5) may be replaced by (.5 x) etc Expansion of given fm followed by trial and improvement eg (x 5) 70 ( a = b = c = 5 d = 70) eg Not fully crect 5 marks Zero 6