FREE-STANDING MATHEMATICS QUALIFICATION Advanced Level 699/01 ADDITIONAL MATHEMATICS FRIDAY 6 JUNE 008 Additional materials: Answer Booklet (16 pages) Graph paper You are not allowed a formulae booklet in this paper. Afternoon Time: hours INSTRUCTIONS TO CANDIDATES Write your name in capital letters, your Centre Number and Candidate Number in the spaces provided on the Answer Booklet. Read each question carefully and make sure you know what you have to do before starting your answer. Answer all the questions. You are permitted to use a scientific or graphical calculator in this paper. Final answers should be given correct to three significant figures where appropriate. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 100. You are advised that an answer may receive no marks unless you show sufficient detail of the working to indicate that a correct method is being used. This document consists of 7 printed pages and 1 blank page. OCR 008 [100/58/0] OCR is an exempt Charity [Turn over
Section A 1 A driver of a car, initially moving at 0 m s 1, applies the brakes so that the car comes to rest with constant deceleration in 10 seconds. (i) Find the value of the deceleration. [] (ii) Find the distance travelled in this time. [] Thepoints A and B have coordinates (0, 8) and (6, 0) respectively. (i) Find the equation of the line AB. [] (ii) Find the equation of the line perpendicular to AB through its midpoint. [] Find the points of intersection of the line y = 5x + 1 with the circle x + y = 1. [5] Glass marbles are produced in two colours, red and green, in the proportion 7 : respectively. From a large stock of the marbles, 5 are taken at random. Find the probability that (i) all 5 are red, [] (ii) exactly are red. [] 5 (i) Use calculus to find the stationary points on the curve y = x x + 1, identifying which is a maximum and which is a minimum. [6] (ii) Sketch the curve. [1] 6 A speedboat accelerates from rest so that t seconds after starting its velocity, in m s 1,isgivenbythe formula v = 0.6t 0.0t. (i) Find the acceleration at time t. [] (ii) Find the distance travelled in the first 10 seconds. [] OCR 008 699/01 Jun08
7 A pyramid stands on a horizontal triangular base, ABC, as shown in Fig. 7. The angles CAB and ABC are 50 and 60 respectively. The vertex, V, is directly above C with VC = 10 m. The angle which the edge VA makes with the vertical is 0. 0 V 10 C A 50 60 B Fig. 7 (i) Calculate AC. [] (ii) Hence calculate AB. [] 8 It is required to solve the equation cos x = 5sinx 1. (i) Show that this equation may be written as sin x + 5sinx = 0. [] (ii) Hence solve the equation cos x = 5sinx 1 for values of x in the range 0 x 60. [] 9 The cubic equation x + ax + bx 6 = 0 has positive, distinct, integer roots. Find the values of a and b. [5] OCR 008 699/01 Jun08 [Turn over
Section B 10 Simon and Gavin drive a distance of 10 km along a motorway, both at constant speed. Simon drives at 5 km per hour faster than Gavin. Let Gavin s speed be v km per hour. (i) Write down expressions in terms of v for the times, in hours, taken by Gavin and Simon. [] Simon completes the journey in 15 minutes less than Gavin. (ii) Explain why 10 v 10 v + 5 = 1 and show that this equation reduces to the equation v + 5v 800 = 0. [5] (iii) Solve this equation to find v andhencefindthetimestakenbysimonandgavin. Giveyour answers correct to the nearest minute. [5] 11 The side of a fairground slide is in the shaded shape as shown in Fig. 11. Units are metres. y T O B Fig. 11 x The curve has equation y = λx. T has coordinates (, ). The line BT is a tangent to the curve at T. It meets the x-axisatthepointb. (i) Find the value of λ. [1] (ii) Find the equation of the tangent BT and hence find the coordinates of the point B. [6] (iii) Find the area of the shaded portion of the graph. [5] OCR 008 699/01 Jun08
1 A furniture manufacturer produces tables and chairs. 5 In each week the following constraints apply. There are workers, each working for 0 hours (i.e. there are 960 worker-hours available). Thereisamaximumof 1800availableforthepurchaseofmaterials. Each table requires 0 worth of materials and 1 worker-hours. Each chair requires 10 worth of materials and 6 worker-hours. It is necessary to make at least times as many chairs as tables. Let x be the number of tables produced each week and y be the number of chairs produced each week. (i) Show that the worker-hour constraint reduces to the inequality x + y 160. [] (ii) Find the inequality relating to the cost of materials constraint and the inequality relating to the numbers of tables and chairs. [] (iii) Plot these three inequalities on a graph, using 1 cm to represent 10 tables on the x-axis and 1 cm to represent 10 chairs on the y-axis. Indicate the region for which these inequalities hold. You should shade the region which is not required. [] When finished, each table is sold for a profit of 0 and each chair is sold for a profit of 5. (iv) The manufacturer wishes to maximise the profit. Explain why the objective function is given by P = 0x + 5y. [1] (v) Find the number of tables and chairs that should be made in order to maximise the profit. [] [Question 1 is printed overleaf.] OCR 008 699/01 Jun08
1 In the triangle shown in Fig. 1, M is the midpoint of BC. 6 A c x b B a M a C Fig. 1 (i) Explain why cos α = cos β. [] (ii) Using the cosine rule in the triangle BMA, show that cos α = x + a c. [] ax (iii) Find a similar expression for cos β. [1] (iv) Using the results in parts (i), (ii) and (iii), showthatx + a = (c + b ). [5] (v) A triangular lawn has sides 6 m, 9 m and 7 m. Find the distance from the midpoint of the longest side to the opposite corner. [] OCR 008 699/01 Jun08
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8 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (OCR) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. OCR is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. OCR 008 699/01 Jun08
FSMQ Additional Mathematics ADVANCED FSMQ 699 Mark Scheme for the Unit June 008 Oxford Cambridge and RSA Examinations
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 1998. 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 5050 Annersley NOTTINGHAM NG15 0DL Telephone: 0870 870 66 Facsimile: 01 55610 E-mail: publications@ocr.org.uk
CONTENTS Additional Mathematics FSMQ (699) MARK SCHEME FOR THE UNIT Unit/Content Page 699 Additional Mathematics Grade Thresholds 10
699 Mark Scheme June 008 699 Additional Mathematics Section A Q. Answer Marks Notes 1 (i) v= u+ at with v= 0, u = 0, t = 10 Must be used 10a = 0 a = or decel = are a = wrong Deceleration is ms (ii) E.g. v = u + as with v = 0, u = 0, a = 6s = 900 s = 150 Distance is 150 m Alternatives: u+ v s = t with v = 0, u = 0, t =10 s = 15 10 = 150 Allow alternatives (i) (ii) Or: 1 s = 00 150 = 150 Or: 1 s = vt at with v = 0, t = 10, a = s = 0 ( 150) = 150 x y + = 1 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 = 10, a = Midpoint is (, ) Gradient is equation is y = ( x ) y = x+ 7 SC. Omission of y = : give A0 soi isw soi E1 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.
699 Mark Scheme June 008 Q. Answer Marks Notes x + 5x+ 1 = 1 Attempt at substitution. ( ) x + 5x + 10x+ 169 1 = 0 6x + 10x+ 156 = 0 x + 5x+ 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) 5 7 0.168 10 (ii) 5 7 0.087 0 if more 10 10 than one Allow, or 5 sig figs in both parts term Apply tmsf or tfsf otherwise. 5 (i) dy y = x x+ 1 = x dx dy = 0 when x =± 1, giving (1, 1) and ( 1,) dx d y d y = 6 x; when x= 1, > 0 dx dx giving minimum at x = 1 d y when x= 1, < 0 giving maximum at x= 1 dx Any alternative method OK. soi 5 soi soi 6 Expansion of (5x + 1) 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 (ii) ) Identify one turning point Both correct (ii) E1 1 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) 5
699 Mark Scheme June 008 Q. Answer Marks Notes 6 (i) dv Diffn a = = 0.7t 0.07t dt Each term (ii) 10 s = ( 0.6t 0.0t ) dt= 0.1t 0.006t 0 = 10 60 = 60m 0+ 1 N.B. Watch s = 10 = 60 7 (i) AC tan 0 AC = 10 tan 0 8.9 m VC = = Alt forms for AC acceptable. 10sin 0 10 i.e. AC = = sin 50 tan 50 (ii) Angle C = 180 50 60 = 70 AB AC = sin C sin B sin70 AB = 8.9 = 9.10 m sin60 8 (i) (1 sin x) = 5sin x 1 sin x+ 5sin x = 0 (ii) ( x )( x+ ) sin 1 sin = 0 1 sin x = 0 0 x = 0,150 1 SC. sin x= x= 10,0 A0 A0 F1 9 roots are 1,, 1 allow ±1, ±, ±1 Equation is (x 1)(x )(x 1) = 0 Giving x 16x +1x 6 = 0 10 0 F1 F1 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 180 ans (only one extra angle) Factor form. Condone no = 0 Expand to give cubic i.e. a = 16, b = 1 (Can be seen in cubic. Alternative method. f(1) = 0 a + b = 5 f() = 0 a + b = 18 Solve to give a and b, isw 5 6
699 Mark Scheme June 008 Section B Q. Answer Marks Notes 10 (i) 10 10, v v+ 5 (ii) 1 Gavin's time minus Simon's time is 15 mins = hr 10 10 1 = v v+ 5 10( v+ 5) 10 v = v( v+ 5) ( ) = vv+ v + v = 800 ( 5) 5 800 0 (iii) 5 ± 5 + 800 v = 50.7 or 50.5 Gavin:.77 hrs, Simon.5 hrs Gavin takes hrs 6 mins (166 mins) Simon takes hrs 1 mins (151 mins) SC For v = 50 168, 15 give full marks but -1 tfsf soi 5 F1 5 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 15 less than the other Q. Answer Marks Notes 11 (i) 1 = 16λ λ = 8 1 (ii) dy 1 x E1 Correct derivative from =. x = dx 8 their λ or leaving it in dy When x =, = 1 Sub x = dx Tangent at T is y = 1( x ) y = x When y = 0, x= So B is (, 0) (iii) x x Area under curve = dx = 8 0 0 Area of triangle = x = = Shaded area = 0 N.B. Area under (curve line) from 0 to only D 6 5 (numeric gradient to give tangent) Int. Function Sub limits for int and subtract triangle 7
699 Mark Scheme June 008 Q. Answer Marks Notes 1 (i) Worker hours for tables = 1x Worker hours for chairs = 6y 1x + 6y 0 = 960 x+ y 160 Must see 1x and 6y (ii) 0x + 10y 1800 ( x + y 180) Does not have to be simplified y x (iii) E1 Each line For y x E1 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 = 5 i.e. P = 0x + 5y Greatest profit will occur where the lines y = x and x + y = 180 intersect. This is at (0, 90). Allow even if shading for y x is wrong. 1 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
699 Mark Scheme June 008 1 (i) Angles on straight line means α = 180 β And cos(180 β ) = cos β ( a ) ( a ) (ii) x + c cosα =. x 1 + 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 ) 1 5 Must make reference to the figure of the question Correct cosine formula. Condone missing brackets. Use of (i), (ii) and (iii) Clear fractions Simplify (v) a = 6, b = 9, c = 7 gives x + 6 = (9 + 7 ) gives x = 56 i.e. x = 16 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 = 16, only if the answer is 16. Can be substituted in any order SC: Scale drawing gets 0. 9
FSMQ Advanced Mathematics 699 June 008 Assessment Series Unit Threshold Marks Unit Maximum A B C D E U Mark 699 100 68 58 8 8 9 0 The cumulative percentage of candidates awarded each grade was as follows: A B C D E U Total Number of Candidates 699 6. 6.7 6.5 56.0 6.7 100 761 Statistics are correct at the time of publication 10
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