ADVANCED GCE 757/01 MATHEMATICS (MEI) Further Applications of Advanced Mathematics (FP) FRIDAY JUNE 008 Additional materials (enclosed): None Additional materials (required): Answer Booklet (8 pages) Graph paper MEI Examination Formulae and Tables (MF) Afternoon Time: 1 hour 0 minutes 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 any three questions. You are permitted to use a graphical calculator in this paper. Final answers should be given to a degree of accuracy appropriate to the context. 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 7. 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 printed pages. OCR 008 [K/10/5] OCR is an exempt Charity [Turn over
Option 1: Vectors 1 A tetrahedron ABCD has vertices A (, 5, ),B(, 1, 7), C(7, 0, ) and D (5,, 8). (i) Find the vector product AB AC, and hence find the equation of the plane ABC. [] (ii) Find the shortest distance from D to the plane ABC. [] (iii) Find the shortest distance between the lines AB and CD. [] (iv) Find the volume of the tetrahedron ABCD. [] The plane P with equation x +5 = 0 contains the point B, and meets the lines AC and AD at E and F respectively. (v) Find λ and μ such that AE = λac and AF = μad. Deduce that E is between A and C, and that FisbetweenAandD. [5] (vi) Hence, or otherwise, show that P divides the tetrahedron ABCD into two parts having volumes in the ratio to 17. [] Option : Multi-variable calculus Youaregiveng(x, y, ) = x (x + y + ). (i) Find x, and y. [] A surface S has equation g(x, y, ) = 15. (ii) Find the equation of the normal line to S at the point P (7, 7.5, ). [] (iii) The point Q is on this normal line and is close to P. At Q, g(x, y, ) = 15 + h, whereh is small. Find the vector n such that PQ = hn approximately. [5] (iv) Show that there is no point on S at which the normal line is parallel to the -axis. [] (v) Find the two points on S at which the tangent plane is parallel to x + 5y = 0. [8] Option : Differential geometry The curve C has parametric equations x = 8t, y = 9t t,fort 0. (i) Show that ẋ + ẏ =(18t + 8t ). Find the length of the arc of C for which 0 t. [] (ii) FindtheareaofthesurfacegeneratedwhenthearcofC for which 0 t is rotated through π radians about the x-axis. [] (iii) Show that the curvature at a general point on C is t(t + 9). [5] (iv) Find the coordinates of the centre of curvature corresponding to the point on C where t = 1. [7] OCR 008 757/01 Jun08
Option : Groups A binary operation is defined on real numbers x and y by x y = xy + x + y. You may assume that the operation is commutative and associative. (i) Explain briefly the meanings of the terms commutative and associative. [] (ii) Show that x y = (x + 1 )(y + 1 ) 1. [1] The set S consists of all real numbers greater than 1. (iii) (A) Use the result in part (ii) to show that S is closed under the operation. (B) Show thats, with the operation, is a group. [9] (iv) Show that S contains no element of order. [] The group G ={0, 1,,, 5, } has binary operation defined by x y is the remainder when x y is divided by 7. (v) Show that =. [] The composition table for G is as follows. 0 1 5 0 0 1 5 1 1 0 5 0 5 1 1 5 0 5 5 0 1 5 1 0 (vi) Find the order of each element of G. [] (vii) List all the subgroups of G. [] [Question 5 is printed overleaf.] OCR 008 757/01 Jun08
Option 5: Markov chains This question requires the use of a calculator with the ability to handle matrices. 5 Every day, a security firm transports a large sum of money from one bank to another. There are three possible routes A, B and C. The route to be used is decided just before the journey begins, by a computer programmed as follows. On the first day, each of the three routes is equally likely to be used. If route A was used on the previous day, route A, B orc will be used, with probabilities 0.1, 0., 0.5 respectively. If route B was used on the previous day, route A, B orc will be used, with probabilities 0.7, 0., 0.1 respectively. If route C was used on the previous day, route A, B or C will be used, with probabilities 0.1, 0., 0. respectively. The situation is modelled as a Markov chain with three states. (i) Write down the transition matrix P. [] (ii) Find the probability that route B is used on the 7th day. [] (iii) Find the probability that the same route is used on the 7th and 8th days. [] (iv) Find the probability that the route used on the 10th day is the same as that used on the 7th day. [] (v) Given that P n Q as n,findthematrixq (give the elements to decimal places). Interpret the probabilities which occur in the matrix Q. [] The computer program is now to be changed, so that the long-run probabilities for routes A, B and C will become 0., 0. and 0. respectively. The transition probabilities after routes A and B remain the same as before. (vi) Find the new transition probabilities after route C. [] (vii) A long time after the change of program, a day is chosen at random. Find the probability that the route used on that day is the same as on the previous day. [] 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 008 757/01 Jun08
757 Mark Scheme June 008 757 (FP) Further Applications of Advanced Mathematics 1 (i) 10 AB AC = 8 5 = 5 1 110 (ii) ABC is x+ y 10z = 9+ 0 0 x+ y 10z+ 9= 0 Distance is 5+ 10 8+ 9 + + 10 0 8 = ( ) ( = ) 15 5 (iii) 0 1 AB CD = 8 = 0 [ = 0 ] 5 5 0 10 1 5. 1 Distance is AC.nˆ = 1 + + = (iv) Volume is 1 (AB AC). AD 8 1 = 1. 110 0 = ( ) (v) E is ( + 10 λ, 5 5 λ, + λ) ( + 10 λ) ( + λ) + 5 = 0 λ = 7 F is ( + 8 μ, 5 μ, + μ) ( + 8 μ) ( + μ) + 5 = 0 μ = Since 0< λ < 1, E is between A and C Since 0< μ < 1, F is between A and D B ft 5 Ignore subsequent working Give for one element correct SC1 for minus the correct vector For x + y 10z Accept x+ y 110z = 99 etc Using distance formula (or other complete method) Condone negative answer Accept a.r.t..58 Evaluating AB CD or method for finding end-points of common perp PQ or P (,11, ) & Q ( 71 55 8 18, 9, ) PQ = (,, ) or 9 9 9 Scalar triple product Accept a.r.t. 7. 1
757 Mark Scheme June 008 (vi) 1 VABEF = (AB AE). AF 1 = λμ (AB AC). AD = λμvabcd = V 1 ABCD Ratio of volumes is : 17 1 1 = :17 (i) g = z ( x + y + z) = x y x g = ( x + y + z) y g = x ( x + y + z) = 1y 18z z (ii) At P, = 1, =, = x y z 7 Normal line is r = 7.5 + λ 1 9 (iii) δg 1δx δy+ δz If PQ = λ 1, 9 δg 1( λ) ( λ) + (9 λ) ( = 9 λ ) (iv) (v) h = δg, so h 9λ h 1 PQ 1, so 1 9 n = 9 9 9 Require = = 0 x y x y = 0 and x+ y+ z = 0 x+ y = 0 and z = 0 g( x, y, z ) = 0 0 = 0 15 Hence there is no such point on S g Require = 0 z and = 5 y x x 8y 1z = 5( x y) ag ft ft 5 1 ( 1 ) ft if numerical Finding ratio of volumes of two parts SC1 for : 17 deduced from 1 without working Partial differentiation Any correct form, ISW Evaluating partial derivatives at P All correct Condone omission of r = Alternative: M for substituting x = 7+ λ, into g = 15 + h and neglecting λ ft for linear equation in λ and h for n correct Useful manipulation using both eqns Showing there is no such point on S Fully correct proof Implied by = λ, = 5λ x y This can be awarded for x y = 1 and x 8y 1z = 5
757 Mark Scheme June 008 y = z and x = 5z (5 zz ) (5 z) = 15 z =± 5 Points are (5, 7.5, 5) and ( 5, 7.5, 5) (i) x + y = ( t ) + (18t 8 t ) = 57t + t 88t + t = t + 88t + t = (18t+ 8 t ) Arc length is (18t+ 8 t ) dt 0 = 9t + t = 8 (ii) Curved surface area is π yds (iii) 0 0 5 7 8 0 = π (9t t )(18t+ 8 t )dt = π (t + 7t t )dt = π 81t + 1t t = 100 π ( 7 ) 0 xy xy ( t )(18 t ) (8 t)(18t 8 t ) κ = = ( x + y ) (18t+ 8 t ) 8 t (9 1t 18 + 8 t ) 8 t (9 + t ) = = 8 t (9+ t ) 8 t (9+ t ) = t(t + 9) ft ag ag 8 5 10 or z = y and x = y 1 or y = x and z = x 10 5 5 1 or x =, y =, z = 8 λ λ λ or x: y: z = 10: : Substituting into g( x, y, z ) = 15 Obtaining one value of x, y, z or λ Dependent on previous ft is minus the other point, provided all M marks have been earned Note (18 + 8 t ) dt = 18t+ t = 8 0 0 earns A0A0 Using d s = (18t+ 8 t )dt Correct integral expression including limits (may be implied by later work) Using formula for κ (or ρ ) For numerator and denominator Simplifying the numerator
757 Mark Scheme June 008 (iv) When t = 1, x = 8, y = 7, κ = 19 ρ = ( ) dy y 18t 8t 10 = = = dx x t 5 1 n ˆ = 1 1 8 5 19 1 c = + 7 1 1 19 5 Centre of curvature is (18, 19) (i) Commutative: x y = y x (for all x, y) Associative: ( x y) z = x ( y z) (for all x, y, z) (ii) 1 1 1 1 1 ( x+ )( y+ ) = xy+ x+ y+ = xy+ x+ y = x y (iii)(a) 1 1 If x, y S then x > and y > 1 1 1 1 x+ > 0 and y+ > 0, so ( x+ )( y+ ) > 0 1 1 1 1 ( x+ )( y+ ) >, so x y S (B) 0 is the identity since 0 x = 0+ x+ 0= x If x S and x y = 0 then xy + x + y = 0 x y = x + 1 1 1 y + = > 0 (x + 1) so y S 1 (since x > ) S is closed and associative; there is an identity; and every element of S has an inverse in S (iv) If x x = 0, x + x+ x = 0 x = 0 or 1 0 is the identity (and has order 1) 1 is not in S B ag 7 1 Finding gradient (or tangent vector) Finding direction of the normal Correct unit normal (either direction) Accept e.g. Order does not matter Give for a partial explanation, e.g. Position of brackets does not matter Intermediate step required 1 1 1 or x y ( + )( + ) = 0 1 1 or y + = ( x + 1) Dependent on
757 Mark Scheme June 008 (v) = 8 + + = 58 = 5 + = 7 8 + (vi) (vii) So = Element 0 1 5 Order 1 {0}, G {0, } {0,, 5} ag B Give B for correct for correct Condone omission of G If more than non-trivial subgroups are given, deduct 1 mark (from final ) for each non-trivial subgroup in excess of 5
757 Mark Scheme June 008 Pre-multiplication by transition matrix 5 (i) 0.1 0.7 0.1 P = 0. 0. 0. 0.5 0.1 0. (ii) 1 0.88 1 P = 0.815 1 0.89 P(B used on 7th day) = 0.815 (iii) (iv) 0.88 0.1+ 0.815 0. + 0.89 0. = 0.191 0.5 0.8 0.0 P = 0. 0.0 0.7 0.8 0.8 0. 0.88 0.5 + 0.815 0.0 + 0.89 0. = 0.7 (v) 0.89 0.89 0.89 Q = 0.81 0.81 0.81 0.895 0.895 0.895 0.89, 0.81, 0.895 are the long-run probabilities for the routes A, B, C (vi) 0.1 0.7 a 0. 0. 0. 0. b 0. = 0. 0.5 0.1 c 0. 0. (vii) 0.0 + 0.1 + 0.a = 0., so a = 0.55 0.1 + 0.0 + 0.b = 0., so b = 0 0. + 0.0 + 0.c = 0., so c = 0.5 After C, routes A, B, C will be used with probabilities 0.55, 0, 0.5 0. 0.1+ 0. 0. + 0. 0.5 = 0. B A Give for two columns correct 7 Using P (or P ) For matrix of initial probabilities For evaluating matrix product Accept 0.81 to 0.8 Using diagonal elements from P Correct method Accept a.r.t. 0.19 For evaluating P Using diagonal elements from P Correct method Accept a.r.t. 0. Deduct 1 if not given as a ( ) matrix Deduct 1 if not dp Accept equilibrium probabilities Obtaining a value for a, b or c Give for one correct Using long-run probs 0., 0., 0. Using diag elements from new matrix
757 Mark Scheme June 008 Post-multiplication by transition matrix 5 (i) 0.1 0. 0.5 P = 0.7 0. 0.1 0.1 0. 0. (ii) (iii) (iv) ( 1 1 1 ) P = ( ) 0.88 0.815 0.89 P(B used on 7th day) = 0.815 0.88 0.1+ 0.815 0. + 0.89 0. = 0.191 0.5 0. 0.8 P = 0.8 0.0 0.8 0.0 0.7 0. 0.88 0.5 + 0.815 0.0 + 0.89 0. = 0.7 (v) 0.89 0.81 0.895 Q = 0.89 0.81 0.895 0.89 0.81 0.895 (vi) (vii) 0.89, 0.81, 0.895 are the long-run probabilities for the routes A, B, C 0.1 0. 0.5 = a b c ( 0. 0. 0.) 0.7 0. 0.1 ( 0. 0. 0.) 0.0 + 0.1 + 0.a = 0., so a = 0.55 0.1 + 0.0 + 0.b = 0., so b = 0 0. + 0.0 + 0.c = 0., so c = 0.5 After C, routes A, B, C will be used with probabilities 0.55, 0, 0.5 0. 0.1+ 0. 0. + 0. 0.5 = 0. B A Give for two rows correct 7 Using P (or P ) For matrix of initial probabilities For evaluating matrix product Accept 0.81 to 0.8 Using diagonal elements from P Correct method Accept a.r.t. 0.19 For evaluating P Using diagonal elements from P Correct method Accept a.r.t. 0. Deduct 1 if not given as a ( ) matrix Deduct 1 if not dp Accept equilibrium probabilities Obtaining a value for a, b or c Give for one correct Using long-run probs 0., 0., 0. Using diag elements from new matrix 7