FSMQ. Additional Mathematics. Mark Scheme for the Unit. June 2008 ADVANCED FSMQ Oxford Cambridge and RSA Examinations

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1 FSMQ Additional Mathematics ADVANCED FSMQ 699 Mark Scheme for the Unit June 008 Oxford Cambridge and RSA Examinations

2 OCR (Oxford, Cambridge and RSA Examinations) is a unitary awarding body, established by the University of Cambridge Local Examinations Syndicate and the RSA Examinations Board in January 998. OCR provides a full range of GCSE, A level, GNVQ, Key Skills and other qualifications for schools and colleges in the United Kingdom, including those previously provided by MEG and OCEAC. It is also responsible for developing new syllabuses to meet national requirements and the needs of students and teachers. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 008 Any enquiries about publications should be addressed to: OCR Publications PO Box 00 Annersley NOTTINGHAM NG 0DL Telephone: Facsimile: publications@ocr.org.uk

3 CONTENTS Additional Mathematics FSMQ (699) MARK SCHEME FOR THE UNIT Unit/Content Page 699 Additional Mathematics Grade Thresholds 0

4 699 Mark Scheme June Additional Mathematics Section A (i) v= u+ at with v= 0, u = 0, t = 0 Must be used 0a = 0 a = or decel = are a = wrong Deceleration is ms E.g. v = u + as with v = 0, u = 0, a = 6s = 900 s = 0 Distance is 0 m Alternatives: u+ v s = t with v = 0, u = 0, t =0 s = 0 = 0 Allow alternatives (i) Or: s = 00 0 = 0 Or: s = vt at with v = 0, t = 0, a = s = 0 ( 0) = 0 x y + = 6 8 x+ y = Any correct equation will do. Usual answer y = x+ 8 SC. Omission of y = : give A0 s = ut + at with u = 0, t = 0, a = Midpoint is (, ) Gradient is equation is y = ( x ) y = x+ 7 SC. Omission of y = : give A0 soi isw soi E Gradient Any valid method In form ax + by = c N.B. Drawing of graph is 0. -ve reciprocal of their gradient Use their gradient plus their midpoint In form ax + by = c N.B. Drawing of graph is 0.

5 699 Mark Scheme June 008 x + x+ = Attempt at substitution. ( ) x + x + 0x+ 69 = 0 6x + 0x+ 6 = 0 x + x+ 6= 0 ( x+ )( x+ ) = 0 x =, y =, Points of intersection (, ),(, ) SC: For each pair obtained from accurate graph or table of values, or trial, (i) if more 0 0 than one Allow, or sig figs in both parts term Apply tmsf or tfsf otherwise. (i) dy y = x x+ = x dy = 0 when x =±, giving (, ) and (,) d y d y = 6 x; when x=, > 0 giving minimum at x = d y when x=, < 0 giving maximum at x= Any alternative method OK. soi soi soi 6 Expansion of (x + ) Solve term quadratic Either both x or one pair Either both y or other pair p and power Ans coeff powers mult (p correct) ans Correct derivative Setting their derivative = 0 Both x or one pair Both y or other pair (y values could be seen in ) Identify one turning point Both correct E General shape including axes and turning points At their x values. (but don t worry about intercepts on the axes.) This does require a scale on the x axis. Curve to be consistent in (i)

6 699 Mark Scheme June (i) dv Diffn a = = 0.7t 0.07t dt Each term 0 s = ( 0.6t 0.0t ) dt= 0.t 0.006t 0 = 0 60 = 60m 0+ N.B. Watch s = 0 = 60 7 (i) AC tan 0 AC = 0 tan m VC = = Alt forms for AC acceptable. 0sin 0 0 i.e. AC = = sin 0 tan 0 Angle C = = 70 AB AC = sin C sin B sin70 AB = 8.9 = 9.0 m sin60 8 (i) ( sin x) = sin x sin x+ sin x = 0 ( x )( x+ ) sin sin = 0 sin x = 0 0 x = 0,0 SC. sin x= x= 0,0 A0 A0 F 9 roots are,, allow ±, ±, ± Equation is (x )(x )(x ) = 0 Giving x 6x +x 6 = F F soi Int the given fn Both terms Deal with def.int Tan function Correct To find AB Must be s.f. Use of pythag.to change cos All working - answer given Solve quad in sin x or s etc ½ seen 0 seen 80 ans (only one extra angle) Factor form. Condone no = 0 Expand to give cubic i.e. a = 6, b = (Can be seen in cubic. Alternative method. f() = 0 a + b = f() = 0 a + b = 8 Solve to give a and b, isw 6

7 699 Mark Scheme June 008 Section B 0 (i) 0 0, v v+ Gavin's time minus Simon's time is mins = hr 0 0 = v v+ 0( v+ ) 0 v = v( v+ ) ( ) = vv+ v + v = 800 ( ) (iii) ± v = 0.7 or 0. Gavin:.77 hrs, Simon. hrs Gavin takes hrs 6 mins (66 mins) Simon takes hrs mins ( mins) SC For v = 0 68, give full marks but - tfsf soi F ¼ hr Subtract Clear fractions 700 Solve in decimals (ignore anything else) Convert (only one needs to be seen) Or give for both in decimals This is for one less than the other (i) = 6λ λ = 8 dy x E Correct derivative from =. x = 8 their λ or leaving it in dy When x =, = Sub x = Tangent at T is y = ( x ) y = x When y = 0, x= So B is (, 0) (iii) x x Area under curve = = Area of triangle = x = = Shaded area = 0 N.B. Area under (curve line) from 0 to only D 6 (numeric gradient to give tangent) Int. Function Sub limits for int and subtract triangle 7

8 699 Mark Scheme June 008 (i) Worker hours for tables = x Worker hours for chairs = 6y x + 6y 0 = 960 x+ y 60 Must see x and 6y 0x + 0y 800 ( x + y 80) Does not have to be simplified y x (iii) E Each line For y x E Must be a region including the y axis as boundary (iv) (v) N.B. Intercepts on axis must be seen N.B. Ignore < instead of We wish to maximise the profit. Profit per table = 0, profit per chair = i.e. P = 0x + y Greatest profit will occur where the lines y = x and x + y = 80 intersect. This is at (0, 90). Allow even if shading for y x is wrong. Something that connects 0 with x 0 ± 90 ± But answers must be integers. SC: Trying all corners without the corect answers SC: Drawing an O.F. line without the right answer 8

9 699 Mark Scheme June 008 (i) Angles on straight line means α = 80 β And cos(80 β ) = cos β ( a ) ( a ) x + c cosα =. x + x + a c = = ax ax (iii) x + a b cos β = ax x + a c N.B. also ax (iv) x + a b x + a c = ax ax x + a b = x + a c x a c x + a b = x a + c 8x + a = b + c ( ( ) ( ) x + a = b + c ) Must make reference to the figure of the question Correct cosine formula. Condone missing brackets. Use of (i), and (iii) Clear fractions Simplify (v) a = 6, b = 9, c = 7 gives x + 6 = (9 + 7 ) gives x = 6 i.e. x = 6 S.C. Use of cosine formula in large triangle to get an angle (C = 6., B =.) Then use of cosine formula in small triangle to get x = 6, only if the answer is 6. Can be substituted in any order SC: Scale drawing gets 0. 9

10 Grade Thresholds FSMQ Advanced Mathematics 699 June 008 Assessment Series Unit Threshold Marks Unit Maximum A B C D E U Mark The cumulative percentage of candidates awarded each grade was as follows: A B C D E U Total Number of Candidates Statistics are correct at the time of publication 0

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