Version 1.0 Level 2 Certificate in Further Mathematics June 2013 Paper 1 8360/1 Final 1 Mark Scheme
Mark schemes are prepared by the Principal Examiner 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 examiners 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 examiner understands and applies it in the same crect way. As preparation f standardisation each examiner 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, examiners encounter unusual answers which have not been raised they are required to refer these to the Principal Examiner. 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 2013 AQA and its licenss. All rights reserved. Copyright AQA retains the copyright on all its publications. However, registered centres 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 centres to photocopy any material that is acknowledged to a third party even f internal use within the centre. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 3644723) and a registered charity (registered charity number 1073334). Registered address: AQA, Devas Street, Manchester 5 6EX. 2
Glossary f Mark Schemes These examinations are marked in such a way as to award positive achievement wherever possible. Thus, f these papers, marks are awarded under various categies. M A B M dep B dep ft SC 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. 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. Follow through marks. Marks awarded following a mistake in an earlier step. Special case. Marks awarded within the scheme f a common misinterpretation which has some mathematical wth. Or equivalent. Accept answers that are equivalent. 1 eg, accept 0.5 as well as 2 3
9 ( 1) 3 9 1 1a 10 SC1 f 8 1b 9 x 3 = 1 x 3 = 8 2 SC1 f 2 2a 4 + 2, 3 + 11 2 2 ( 1, 7) SC1 f one codinate crect (r 2 =) 3 2 + 4 2 (r 2 =) 25 (d 2 =) 6 2 + 8 2 (d 2 =) 100 ft their centre 2b (r = 5) ft SC1 f 10 2c (x + 1) 2 + (y 7) 2 = 25 B1ft ft their centre and radius 1 Accept 1, 1.5 2d 0.5 B1 2 2 2 3a 3b 4 = 2 BC 3 (BC =) 6 their 6 = 2 AB 3 (AB =) 9 ft eg follow through their 6 using a similar triangles/scale fact method (AP =) 5 ft 6 2 (= 36) 4 x = their 36 33 9 4
(x + 7 + x 3)(x + 7 x + 3) Allow one sign err (2x + 4) 10 5a 5b 10 2(x + 2) 20x + 40 Alternative method x 2 + 7x + 7x + 49 ( ) x 2 3x 3x + 9 Allow one err x 2 + 7x + 7x + 49 (x 2 3x 3x + 9) All terms crect x 2 + 7x + 7x + 49 x 2 + 3x + 3x 9 = 20x + 40 20(100 + 2) 204 10 11449 9409 seen 2040 6 81x 4 y 20 B2 B1 f two components crect 7 2y 3 10y 2 + 4y 3y 2 + 15y 6 2y 3 10y 2 + 4y 3y 2 + 15y 6 2y 3 13y 2 + 19y 6 ft ft from A0 Must have at least five terms with at least four crect Accept 4 x 3 10 x 8a 4x 3 10x (+ 0) B2 B1 f 4x 3 4 x 3 B1 f 10x 10 x 8b 4x 3 10x + something extra eg 4x 3 10x + 9 (when x = 2) (gradient =) 12 B1ft ft their answer to (a) (when x = 2) (y =) 5 B1 their 5 = their 12 2 + c y 5 = 12(x 2) y = 12x 19 ft ft their m and their 5 sces B1 5
x = 6 ± {6 2 4(1)(7)} 2(1) x = 6 ± 8 2 Allow one substitution sign err 8 = 2 2 ft F simplifying their surd (if possible to do so) 9 x = 3 ± 2 Alternative method (x + 3) 2... (x + 3) 2 9 + 7 (= 0) (x + 3) 2 2 (= 0) (x + 3) 2 = 2 x + 3 = (±) 2 x = 3 ± 2 dep dep a + 2x = n(a x) 10 a + 2x = na nx n x + 2x = na a x(n + 2) = na a x(n + 2) = a(n 1) f collecting the x terms on one side and the other terms on the opposite side Allow one sign err x = a(n 1) x = na a n + 2 n + 2 11a x + 5, x and x 3 B2 Any der B1 f any two of x, x + 5 x 3 B1 f x, x 5 and x + 3 f(x)= x(x + 5)(x 3) ft their three facts f(x)= x 3 + 2x 2 15x b = 2 and c = 15 ft ft their three facts, one of which must be x 11b Alternative method ( 5) 3 + b( 5) 2 + c( 5) = 0 and eg 25b 5c = 125 and 9b + 3c = 27 (3) 3 + b(3) 2 + c(3) = 0 Allow one err in total b = 2 and c = 15 f(x)= x 3 + 2x 2 15x 6
x 2 12 x 4y x 2 12 = 4x and x 4y = 8 These can still be in matrix fm 12 (x 6)(x + 2) (= 0) x = ( 4) ± {( 4) 2 4 1 ( 12)} 2(1) x = 6 and 2 ft ft their quadratic if possible x = 6 and y = 1 / 2 y = 1 and 2 1 / 2 5 2 2 ft ft from their x values x = 2 and y = 2 1 / 2 Alternative method x 2 12 x 4y x 2 12 = 4x and x 4y = 8 These can still be in matrix fm (4)(2y + 5)(2y + 1) (= 0) y = 12 ± 12 2 4 4 5 (8y + 20)(2y + 1) (= 0) 2(4) y = 48 ± 48 2 4 16 20 2(16) (2y + 5)(8y + 4) (= 0) y = 1 and 2 1 / 2 5 2 2 ft x = 6 and 2 ft ft their quadratic if possible y = 1 / 2 and x = 6 ft from their y values y = 2 1 / 2 and x = 2 7
(y =) 8 3 1 (y =) 8 ( 3 + 1) ( 3 1) ( 3 + 1) (y =) 8 3 + 8 3 1 (y =) 4 3 + 4 Alternative method 1 y 3 = 8 + y and 3y 2 = 64 + 16y + y 2 y 2 8y 32 = 0 2y 2 16y 64 = 0 2 3 + 2 from 8 3 + 8 3 + 1 and 3 + 1 from 8 3 + 8 9 1 both sce SC3 Re-arrange and square both sides, Allow one err 13 and (y =) 8 ± {8 2 4(1)( 32)} 2(1) (y =) 16 ± {16 2 4(2)( 64)} 2(2) Allow one substitution sign err (y =) 4 ± 4 3 (y =) 4 + 4 3 Solution with negative sign must be discounted Alternative method 2 (a + b 3)( 3 1) (=8) a 3 + 3b a b 3 a = b (y =) 4 + 4 3 8
14 Join BD Angle BDC = 2x Alternate segment theem Angle BDO = x Angle DBO = x Isosceles triangle BOD Angle BOD = 180 2x Angle sum of triangle BOD y = 360 90 90 (180 2x) y = 2x Must have at least two different reasons stated in the proof Alternative method 1 B1ft Angle sum of quadrilateral ABOD y = 2x clearly shown from simplification Angle OBC = 90 2x Tangent-radius property Angle OCB = 90 2x Isosceles OBC Angle OCD= x Isosceles OCD Angle BCD = 90 2x + x = 90 x hence Angle at centre = 2 angle at circumference Angle BOD = 180 2x y = 360 90 90 (180 2x) y = 2x Must have at least two different reasons stated in the proof B1ft Angle sum of quadrilateral ABOD y = 2x clearly shown from simplification 9
Alternative method 2 Angle OBC = 90 2x Tangent-radius property Angle OCB = 90 2x Isosceles OBC Angle OCD= x Isosceles OCD Angle BCD = 90 2x + x = 90 x hence Angle at centre = 2 angle at circumference Angle BOD = 180 2x Angle BOD = 360 90 90 y = 180 y hence y = 2x Must have at least two different reasons stated in the proof Alternative method 3 B1ft Angle sum of quadrilateral ABOD y = 2x clearly shown from comparison Angle OBC = 90 2x Tangent-radius property Angle OCB = 90 2x Isosceles OBC 14 Angle OCD= x Isosceles OCD Angle BCD = 90 2x + x = 90 x y = 360 90 (90 2x) (90 x) x 90 Angle sum of quadrilateral ABCD hence y = 2x y = 2x clearly shown from simplification Must have at least two different reasons stated in the proof B1ft Alternative method 4 Angle BOD = 180 y Angle sum of quadrilateral ABOD Angle OCD = x Isosceles OCD Angle OBC = 90 2x Tangent-radius property Angle BCO= 90 2x hence Angle BOD reflex = 360 (90 2x) (90 2x) x x = 180 + 2x 180 y + 180 + 2x = 360 hence y = 2x Must have at least two different reasons stated in the proof B1ft 10 Isosceles OBC Angle sum of quadrilateral BODC... this can also come from Angle BOC (4x) + Angle DOC (180 2x) Angles round a point y = 2x clearly shown from rearranging
2(x 2 6x)... 15 2(x 3) 2... dep 2((x 3) 2 9 ( 3.5)) 2(x 3) 2 18 ( 7) dep 2(x 3) 2 25 Alternative method x 2 + bx + bx + b 2 a = 2 12 = 2ab 12 = 4b and 7 = ab 2 + c 7 = 2b 2 + c 2(x 3) 2 25 7 9 1 = 9 64 2 x 3 9 = 64 ( 3 x) 2 9 = 64 B1 Can be done at any stage the reciprocals 2 1 x 3 64 = 9 1 = 64 ( 3 x) 2 9 16 3 (x 2 9 ) = 64 x = 3 9 2 64 _1 = 64 3 (x 2 ) 9 the reciprocals 1 = 64 3/2 x 9 3 x = _9 64 x 2 = _9 64 x = 3 3 8 1 = 8 3 x 3 3 (x =) ± _27 27_ _27_ 512 512 512 _1_ = 64 3 x 9 1 = 64 3 x 2 9 SC3 f 512 27 11