Thursday 14 June 01 Morning A GCE MATHEMATICS (MEI) 4757 Further Applications of Advanced Mathematics (FP3) QUESTION PAPER *471575061* Candidates answer on the Printed Answer Book. OCR supplied materials: Printed Answer Book 4757 MEI Examination Formulae and Tables (MF) Other materials required: Scientific or graphical calculator Duration: 1 hour 30 minutes INSTRUCTIONS TO CANDIDATES These instructions are the same on the Printed Answer Book and the Question Paper. The Question Paper will be found in the centre of the Printed Answer Book. Write your name, centre number and candidate number in the spaces provided on the Printed Answer Book. Please write clearly and in capital letters. Write your answer to each question in the space provided in the Printed Answer Book. Additional paper may be used if necessary but you must clearly show your candidate number, centre number and question number(s). Use black ink. HB pencil may be used for graphs and diagrams only. Read each question carefully. Make sure you know what you have to do before starting your answer. Answer any three questions. Do not write in the bar codes. You are permitted to use a scientific or graphical calculator in this paper. Final answers should be given to a degree of accuracy appropriate to the context. INFORMATION FOR CANDIDATES This information is the same on the Printed Answer Book and the Question Paper. The number of marks is given in brackets [ ] at the end of each question or part question on the Question Paper. 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. The total number of marks for this paper is 7. The Printed Answer Book consists of 0 pages. The Question Paper consists of 8 pages. Any blank pages are indicated. INSTRUCTION TO EXAMS OFFICER / INVIGILATOR Do not send this Question Paper for marking; it should be retained in the centre or recycled. Please contact OCR Copyright should you wish to re-use this document. OCR 01 [K/10/665] DC (NF) 47955/4 OCR is an exempt Charity Turn over
Option 1: Vectors 1 A mine contains several underground tunnels beneath a hillside. The hillside is a plane, all the tunnels are straight and the width of the tunnels may be neglected. A coordinate system is chosen with the z-axis pointing vertically upwards and the units are metres. Three points on the hillside have coordinates A(15, 60, 0), B( 75, 100, 40) and C(18, 138, 35.6). (i) Find the vector product AB AC and hence show that the equation of the hillside is x y + 5z = 650. [5] The tunnel T A begins at A and goes in the direction of the vector 15i + 14j k; the tunnel T C begins at C and goes in the direction of the vector 8i + 7j k. Both these tunnels extend a long way into the ground. (ii) Find the least possible length of a tunnel which connects B to a point in T A. [6] (iii) Find the least possible length of a tunnel which connects a point in T A to a point in T C. [6] (iv) A tunnel starts at B, passes through the point (18, 138, p) vertically below C, and intersects T A at the point Q. Find the value of p and the coordinates of Q. [7] Option : Multi-variable calculus You are given that g(x, y, z) = x + y z + xz + yz + 4z 3. (i) Find g x, g g and y z. The surface S has equation g(x, y, z) = 0, and P(, 1, 1) is a point on S. (ii) Find an equation for the normal line to the surface S at the point P. (iii) A point Q lies on this normal line and is close to P. At Q, g(x, y, z) = h, where h is small. Find the constant c such that PQ c h. [5] (iv) Show that there is no point on S at which the normal line is parallel to the z-axis. [5] (v) Given that x + y + z = k is a tangent plane to the surface S, find the two possible values of k. [8] OCR 01 4757 Jun1
3 Option 3: Differential geometry 3 A curve has parametric equations x = a(1 cos 3 θ), y = a sin 3 θ, for 0 where a is a positive constant. θ π 3, The arc length from the origin to a general point on the curve is denoted by s, and ψ is the acute angle defined by tan ψ = dy dx. (i) Express s and ψ in terms of θ, and hence show that the intrinsic equation of the curve is s = 3 a sin ψ. [9] (ii) For the point on the curve given by θ = π, find the radius of curvature and the coordinates of the centre 6 of curvature. [9] (iii) Find the area of the curved surface generated when the curve is rotated through π radians about the y-axis. [6] OCR 01 4757 Jun1 Turn over
4 Option 4: Groups 4 (i) Show that the set P = {1, 5, 7, 11}, under the binary operation of multiplication modulo 1, is a group. You may assume associativity. A group Q has identity element e. The result of applying the binary operation of Q to elements x and y is written xy, and the inverse of x is written x 1. (ii) Verify that the inverse of xy is y 1 x 1. [] Three elements a, b and c of Q all have order, and ab = c. (iii) By considering the inverse of c, or otherwise, show that ba = c. [] (iv) Show that bc = a and ac = b. Find cb and ca. (v) Complete the composition table for R = {e, a, b, c}. Hence show that R is a subgroup of Q and that R is isomorphic to P. The group T of symmetries of a square contains four reflections A, B, C, D, the identity transformation E and three rotations F, G, H. The binary operation is composition of transformations. The composition table for T is given below. A B C D E F G H A E G H F A D B C B G E F H B C A D C F H E G C A D B D H F G E D B C A E A B C D E F G H F C D B A F G H E G B A D C G H E F H D C A B H E F G (vi) Find the order of each element of T. (vii) List all the proper subgroups of T. [5] OCR 01 4757 Jun1
5 Option 5: Markov chains This question requires the use of a calculator with the ability to handle matrices. 5 In this question, give probabilities correct to 4 decimal places. A random walk is modelled as a Markov chain with five states A, B, C, D, E representing the possible positions, from left to right, of an object. At each step the object moves as follows. If the object is at A, it moves one place to the right (to B). If the object is at E, it moves one place to the left (to D). Otherwise, the probability that the object moves one place to the left is 0.4, and the probability that it moves one place to the right is 0.6. Steps occur at intervals of one minute, and the time taken to move may be neglected. The object starts at A, so after the first step (one minute later) the object is at B. (i) Which of the five states are reflecting barriers? [1] (ii) Write down the transition matrix P. [] (iii) State the possible positions of the object after 10 steps, and give the probabilities that the object is in each of these positions. (iv) Find the probability that after 15 steps the object is in the same position as it was after 13 steps. (v) Find the number of steps after which the probability that the object is at D exceeds 0.69 for the first time. (vi) Find the limits of P n and P n +1 as the positive integer n tends to infinity. (vii) For the interval of 100 minutes between the 00th step and the 300th step, find the expected length of time for which the object is at each of the five positions. (viii) At a certain instant, the object arrives at D. Find the expected number of successive occasions that the object moves to E (and then back to D). Hence find the expected time after this instant when the object first moves to C. OCR 01 4757 Jun1
14 4 (iv) 4 (v) R e a b c e a b c OCR 01
4757 Mark Scheme June 01 1 (i) 90 3 1464 Evaluation of vector product 160 198 1464 [ 73 ] 0 15.6 18300 5 A Give for one component correct x y5z (15) ( 60) 5(0) For x y5z d Equation of ABC is x y5z 650 E1 Evidence of substitution required [5] 1 (ii) 90 15 600 Appropriate vector product AB d A 160 14 10 0 3660 A Give if one error AB d 600 10 3660 A for any non-zero multiple correctly obtained d A 15 14 AB d A Distance is d A Distance is 180 m [6] OR 15 15 75 15 60 14 100 14 0 0 40 Obtaining a value of Distance is (10) ( 13) ( 4) Distance is 180 m 5
4757 Mark Scheme June 01 1 (iii) 15 8 14 Vector product of direction vectors d 14 7 14 [ 7 ] 7 1 A Give if one error 3 198 Appropriate scalar product ˆ 15.6 1 374.4 AC. d Fully correct method for finding 1 3 distance Distance is 14.8 m [6] 1 (iv) 15 15 75 93 60 14 100 38 0 40 p 40 Must use different parameters 15 15 75 93 60 14 100 38 Both equations correct 5, 5 Obtaining value of or 0 50 40 5( p 40) Or other method for finding p p 6 Q is (15 375, 60 350, 0 50) Q is (390, 90, 30) [7] (i) g x z x g 4y z y g z x y 4 z 6
4757 Mark Scheme June 01 (ii) g g g At P,,, 4 x y z 1 Normal line is r 1 1 1 (iii) If Q is (, 1, 1 ) g g g h g x y z x y z ( )( ) ( )( ) ( 4)( ) [ 1 ] (iv) FT PQ ( ) ( ) ( ) For direction of normal line 6 FT Allow 6 6 6 c A0 for c 1 1 [5] g g Require 0 x y x z, y 1 z 1 FT Requires some substitution in RHS or h on LHS z z z z z 4z3 0 Obtaining equation in one variable 5z 8z6 0 Discriminant is 64 10 56 0 Hence there are no such points E1 Correctly shown [5] Or 5x 8x60 or 10y 8y3 0 or 14 0 Dependent on quadratic with negative discriminant Condone omission of r = Allow for x etc g ( ) z 7
4757 Mark Scheme June 01 (v) g g g Require ( ) Allow if 1 x y z x z 4yz zxy 4 FT x y, y z (4z 4) (z) z... 4z3 0 Obtaining equation in one variable 5z 8z3 0 Or 5x 8x0 or 5y 4y 0 Or 4 0 Points (0, 0, 1) and ( 1.6, 0.8, 0.6) Obtaining at least one point k 001 or k 1.60.80. 6 Obtaining a value of k k 1, 1.8 [8] 3 (i) dx dy 3acos sin, 3asin cos d d dx dy (3a sin cos ) (cos sin ) d d (3asin cos ) Implies previous if values of x, y, z not seen s 3asin cosd FT requires workable integral form 3 asin ( c) s 0when0 c 0 Or dy 3asin cos tan dx 3a cos sin tan Hence and s 3 asin E1 Correctly shown [9]... used Required for final E1 0 8
4757 Mark Scheme June 01 3 (ii) ds 3asin cos d OR 1 3 When,, 3a 6 6 Radius of curvature is 3 3 4 a Condone use of even if 6 not established in (i) 9a 3 3a When, x, y 6 8 8 Obtaining second derivatives 3 3a 15a x, y 8 8 May be implied by later work 3 ( x y ) xy xy 3 81a 7a 64 64 9a 15a 3 3a 3 3a 8 8 8 8 Applying formula for or Radius of curvature is 3 3 4 a 1 sin Obtaining gradient or normal vector Normal vector is n ˆ cos 3 Correct normal vector 3 3 1 a1 8 3 3 c a Must use unit normal here 1 4 3 a 8 3 3 5a Centre of curvature is a 1, 4 4 [9] Not necessarily unit vector 9
4757 Mark Scheme June 01 3 (iii) Surface area is xds 3 3 a(1 cos )(3asincos )d FT Correct limits required 0 3 4 6 a (sincos sincos )d 0 1 1 5 6a sin cos 5 3 0 87 a 80 [6] Integrating For 1sin and 1cos 5 5 4 (i) P 1 5 7 11 1 1 5 7 11 5 5 1 11 7 7 7 11 1 5 11 11 7 5 1 Table shows closure Identity is 1 All elements are self-inverse Condone no mention of inverse of 1 4 (ii) 1 1 ( xy)( y x ) 1 1 Or ( y x )( xy) 1 1 1 1 1 1 x( yy ) x xex xx e E1 So y x is the inverse of xy [] 4 (iii) 1 1 1 1 1 1 a a, b b, c c, c ( ab) b a For any one of these Hence c ba E1 [] At least one trigonometric term Or equivalent 10
4757 Mark Scheme June 01 4 (iv) bc b( ba) 1 Or ba c a b c Any correct first step bc ea a E1 ac a( ab) eb b E1 cb a, ca b 4 (v) R e a b c e e a b c a a e c b b b c e a c c b a e R is closed Hence R is a subgroup E1 No need to mention identity or inverses Same pattern as P; hence R and P are isomorphic E1 Dependent on (only) 4 (vi) Eleme A B C D E F G H nt B3 Give for 3 correct; B for 6 correct Order 1 4 4 4 (vii) Ignore { E } and T in the marking { E, A}, { E, B}, { E, C}, { E, D}, { E, G } B Give for 3 correct { E, F, G, H } { E, A, B, G } { E, C, D, G } [5] Deduct 1 mark (from this B) for each subgroup of order given in excess of five Deduct 1 mark (from this ) for each subgroup of order 3 or more given in excess of three 11
4757 Mark Scheme June 01 5 Pre-multiplication by transition matrix Allow tolerance of 0.0001 in probabilities throughout this question (i) A and E are reflecting barriers [1] 5 (ii) 0 0.4 0 0 0 1 0 0.4 0 0 P 0 0.6 0 0.4 0 0 0 0.6 0 1 0 0 0 0.6 0 B Give for three columns correct [] 5 (iii) 0.1378.... 10 9 0.... For P or P 10 P 0.4689.... 10 Using first or second column of P or 0.... 9 0.3933.... P Possible positions are A, C, E P( A) 0.1378, P( C) 0.4689, P( E) 0.3933 5 (iv) 0........ 0.316.... 0.64... 13 0...,..... 13 P P Using first or second column of P or 0.6838...... 0.84. P, and diagonal elements from P 0........ 0.3160.64 0.6838 0.84 Probability is 0.7768 Allow for 0.1301 0.4 0.4651 0.48 0.4048 0.6 ( 0.518 ) 1
4757 Mark Scheme June 01 5 (v)................ Considering powers of P 15 17 P...., P.... 0.688... 0.6904... Appropriate element exceeding 0.69........ After 17 steps 5 (vi) n Limit of P is 0.131 0 0.131 0 0.131 0 0.3077 0 0.3077 0 0.4615 0 0.4615 0 0.4615 Give for any non-zero element B correct to 3 dp 0 0.693 0 0.693 0 0.4154 0 0.4154 0 0.4154 n1 Limit of P is 0 0.131 0 0.131 0 0.3077 0 0.3077 0 0.3077 0 0.4615 0 0.4615 0 Give for any non-zero element n n1 B SC If P and P interchanged, correct to 3 dp award 0.693 0 0.693 0 0.693 0 0.4154 0 0.4154 0 5 (vii) Expected time at A is 50 0.131 Condone 100 0.131 for A: 6. B: 15.4 C: 3.1 D: 34.6 E: 0.8 (min) A Give for one correct 5 (viii) 1 Expected number is 1 0.4 For 1 0.4 1.5 Expected time is 11.5 For 1.5 4 minutes 13
4757 Mark Scheme June 01 5 Post-multiplication by transition matrix Allow tolerance of 0.0001 in probabilities throughout this question (i) A and E are reflecting barriers [1] 5 (ii) 0 1 0 0 0 0.4 0 0.6 0 0 P 0 0.4 0 0.6 0 0 0 0.4 0 0.6 0 0 0 1 0 B Give for three rows correct [] 5 (iii) 0.1378 0 0.4689 0 0.3933..... 10 P..... 10 For P or 9 P.......... Possible positions are A, C, E P( A) 0.1378, P( C) 0.4689, P( E) 0.3933 10 Using first or second row of P or 9 P 14
4757 Mark Scheme June 01 5 (iv) 0 0.316 0 0.6838 0..... 13 P............... 13 Using first or second row of P or..... P, and diagonal elements from P. 0.64... P........ 0.84...... 0.3160.64 0.6838 0.84 Allow for 0.1301 0.4 0.4651 0.48 0.4048 0.6 ( 0.518 ) Probability is 0.7768 5 (v)... 0.688.... 0.6904........... 15 17 P....., P..... Considering powers of P.......... Appropriate element exceeding 0.69.......... After 17 steps 15
4757 Mark Scheme June 01 5 (vi) n Limit of P is 0.131 0 0.4615 0 0.4154 0 0.3077 0 0.693 0 0.131 0 0.4615 0 0.4154 Give for any non-zero element B correct to 3 dp 0 0.3077 0 0.693 0 0.131 0 0.4615 0 0.4154 Limit of P n1 is 0 0.3077 0 0.693 0 0.131 0 0.4615 0 0.4154 0 0.3077 0 0.693 0 Give for any non-zero element n n1 B SC If P and P interchanged, correct to 3 dp award 0.131 0 0.4615 0 0.4154 0 0.3077 0 0.693 0 5 (vii) Expected time at A is 50 0.131 Condone 100 0.131 for A: 6. B: 15.4 C: 3.1 D: 34.6 E: 0.8 (min) A Give for one correct 5 (viii) 1 Expected number is 1 0.4 For 1 0.4 1.5 Expected time is 11.5 For 1.5 4 minutes 16