AQA LEVEL 2 CERTIFICATE FURTHER MATHEMATICS (8365/2) Paper 2

Transcription

1 AQA LEVEL CERTIFICATE FURTHER MATHEMATICS (8365/) Paper Mark scheme Specimen 00 Version 1.0

2 Principal Examiners have prepared these mark schemes f specimen papers. These mark schemes have not, therefe, been through the nmal process of standardising that would take place f live papers. Further copies of this Mark Scheme are available from aqa.g.uk Glossary f Mark Schemes AQA examinations are marked in such a way as to award positive achievement wherever possible. Thus, f these 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 within the scheme 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 1 [a, b] Accept values between a and b inclusive Allow answers which begin 3.14 eg 3.14, 3.14, Use of brackets It is not necessary to see the bracketed wk to award the marks. Copyright 018 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

3 is acknowledged to a third party even f internal use within the centre. 3

4 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 1 mark unless instructed otherwise. 4

5 x-codinate of Q = 6 3 may be implied seen on diagram their 3 dep 1 9 x + y = 100 x + y = 10 B B1 radius = 10 p = B1 3 r = 5 B1 4(a) x > 6 B1 4(b) x 4 x 4 B1 (, 0) B1 5(a) 5

6 6 B1 5(b) 4s + 5 = 1 7s 10 = t equation 6(a) s = 1.5 t = 0.5 ft ft 7 their s (b) (gradient =) = their c c = y 0 = their 0.5(x 4) 7 y = 0.5x y = 0.5(x 4) simplified equation 8(a) ab ad cd bc a c 6

7 Common denominat with at least one numerat crect eg 1 8x 6 x + 6x 1x 8x x 6x 8(b) 1 + 8x 6x x + 6 = (x 50) = 4x x 3x = 16 dep crect expansion and collection of terms 9 x = their 54 dep 3 ft ft their x with first and third gained 6x x x x 8 4 3x 5 1 x x x 4 15x 4 x 3 dep differentiates at least one term crectly 7

8 60x 3 + 6x dep differentiates their -term 9 dy dx crectly k = (14k + 30) crect equation with fractions eliminated k 8k 60 (= 0) dep equation (k + )(k 30) (= 0) 11 8 ± ( 8) crect attempt to solve their 3-term quadratic equation 14 ± and is A0 30x + 0x + 15x + 10x + 15x + y + y = 5 90x + y = 5 1(a) 5 90x y = must see wking f and y = 16 45x 8

9 1(b) 30x 15x + 0x (16 45x) 15x 10x + 0x (16 45x + 15x) 15x 10x + 0x (16 30x) 450x + 50x 900x = 50x 450x 150x + 50x 900x + 300x = 50x 450x 150x + 50x 600x = 50x 450x must see crect expansion of brackets x 1(c) their (50 900x) = 0 x =.8 dep 358 9

10 x x (a) 7 3(x 5) (b) 3x 9 = 3(x 3) sin x y = sin18 y sin x = sin sin x = [0.61, 0.6] sin 1 [0.61, 0.6] 38.(17 ) 38.() dep eliminates y a = 3 B = their 3 b 10

11 b = their their b = dep b =.5 ft ft B0M (numerat =) x(4x 5) 6x 4x 5 x 35 B1 (numerat =) x(x + 5)(x 5) (x + 5)(x 5) 6x x 35 B1 16 (ax + b)(cx + d) where ac = 6 and bd = ± 35 (3x + 7)(x 5) x + 5 3x

12 x 3x = 7 at least two terms crect x 3x 7 (= 0) 3-term quadratic equation 17 3 ± 3 65 ± 4 16 ( 3) 4 7 crect attempt to solve their 3-term quadratic equation and 1.7 is A0 18 B identifies there are 3 choices f first digit and 3 choices f second digit B3 B1 identifies there are 3 choices f first digit 18 identifies there is 1 choice f last digit 19(a) Identifies (x =) 3 1 may be implied 1

13 = = 0 must show four terms and equate to 0 Alternative method 1 19(b) (3x + 1)(x x ) x x 3 3x + 1 3x + 4x x 1 13

14 (3x + 1)(x x ) x x 3 3x + 1 3x + 4x x 1 (3x + 1)(x + 1)(x ) Alternative method f( 1) = 0 f() = 0 f( 1) = 0 and f() = 0 (3x + 1)(x + 1)(x ) (VM =) (DM =) = their 91 + their 73 their 91 their 73 cos VMD dep (cos VMD =) their 91+ their their 91 their 73 dep [66.8, 66.9] 67 dep on M may be implied dep on M3 4n + 6n + 6n + 9 allow one err 1 4n + 1n + 9 implied by 4n + 1n + k an + 1n

15 8n 3 + 1n + 4n + 36n + 18n + 9 8n n + 54n + 9 9n n + 54n + 9 9n n + 54n + 9 and 9(n n + 6n + 1) dep ft their 4n + 6n + 6n + 9 allow one err eg (9n n + 54n + 9) 9 = n n + 6n + 1 9n n + 54n + 9 and all cfficients are divisible by 9 15