GCE. Mathematics (MEI) Mark Schemes for the Units. June /7895-8/MS/R/09. Advanced GCE A Advanced Subsidiary GCE AS

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1 GCE Mathematics (MEI) Advanced GCE A Advanced Subsidiary GCE AS Mark Schemes for the Units June /7895-8/MS/R/09

2 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. OCR qualifications include AS/A Levels, GCSEs, OCR Nationals, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new syllabuses to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today s society. 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 009 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG5 0DL Telephone: Facsimile: publications@ocr.org.uk

3 75 Mark Scheme June 009 CONTENTS Advanced GCE Mathematics (MEI) (7895) Advanced GCE Further Mathematics (MEI) (7896) Advanced GCE Further Mathematics (Additional) (MEI) (7897) Advanced GCE Pure Mathematics (MEI) (7898) Advanced Subsidiary GCE Mathematics (MEI) (895) Advanced Subsidiary GCE Further Mathematics (MEI) (896) Advanced Subsidiary GCE Further Mathematics (Additional) (MEI) (897) Advanced Subsidiary GCE Pure Mathematics (MEI) (898) Unit/Content MARK SCHEME FOR THE UNITS Page 75 (C) Introduction to Advanced Mathematics 75 (C) Concepts for Advanced Mathematics 8 75 (C) Methods for Advanced Mathematics 75 (C) Applications of Advanced Mathematics (FP) Further Concepts for Advanced Mathematics 756 (FP) Further Methods for Advanced Mathematics Further Pure 758 Differential Equations 76 Mechanics 6 76 Mechanics 5 76 Mechanics MEI Mechanics Statistics Statistics Statistics Statistics Decision Mathematics Decision Mathematics Decision Mathematics Computation 90

4 75 Mark Scheme June Numerical Methods MEI Numerical Computation 97 Grade Thresholds 0

5 75 Mark Scheme June (C) Introduction to Advanced Mathematics Section A (0, ) and (/, 0) o.e. isw M for evidence of correct use of gradient with (, 6) eg sketch with stepping or y 6 (x ) seen or y x + o.e. or for y x + c [accept any letter or number] and for 6 + c; for (0, ) [c is not sufficient for ] and for (/, 0) o.e.; allow when x 0, y etc isw ( s ut) [ a ] o.e. as final answer t ( s ut) [condone [ a ] ] 0.5 t for each of complete correct steps, ft from previous error if equivalent difficulty [eg dividing by t does not count as step needs to be by t ] ( ) [ ] s a ut gets M only (similarly t other triple-deckers) 0 www for f() soi and for k or 7 k o.e. [a correct -term or -term equation] long division used: for reaching (9 k)x + in working and for + (9 k) o.e. equating coeffts method: M for (x )(x + x ) [+ ] o.e. (from inspection or division) x < 0 or x > 6 (both required) each; if B0 then for 0 and 6 identified; 5 (i) 0 www (ii) 80 www or ft 8 their (i) for 5 ( ) or 5 ( ) or for seen B for 80x ; for or (x) seen 6

6 75 Mark Scheme June any general attempt at n being odd and n being even 7 (i) n odd implies n odd and odd odd even n even implies n even and even even even (ii) (i) / www (ii) 0 9 (i) (x + ) www as final answer (ii) ft their ( a, b); if error in (i), accept (, ) if evidence of being independently obtained 0 (x 9)(x + ) M M0 for just trying numbers, even if some odd, some even or n(n ) used with n odd implies n even and odd even even etc [allow even odd even] or A for n(n )(n + ) product of consecutive integers; at least one even so product even; odd odd odd etc is not sufft for SC for complete general method for only one of odd or even eg n m leading to (m m) for 5 0 or for 5 /5 o.e. for /6 or for 8 or or 7 or 08 or for M for terms correct of soi, for terms correct for a, B for b or for 5 soi each coord.; allow x, y ; or for o.e. oe for sketch with and marked on axes but no coords given 5 9 or correct use of quad formula or comp sq reaching 9 and ; allow for attempt with correct eqn at factorising with factors giving two terms correct, or sign error, or attempt at formula or comp sq [no more than two errors in formula/substn]; for this first M or allow use of y etc or of x instead of x x 9 [or ] or ft for integers /fractions if first earned x ± cao must have x ; or for (x + )(x ); this may be implied by x ± A0 if extra roots if M0 then allow SC for use of factor theorem to obtain both and as roots or (x + ) and (x ) found as factors and SC for x + found as other factor using factor theorem [ie max SC] 0

7 75 Mark Scheme June 009 Section B i y x ii eqn AB is y -/ x + o.e. or ft x -/x + or ft x 9/0 or 0.9 o.e. cao y 7/0 oe ft their their x for grad AB or / o.e. 6 need not be simplified; no ft from midpt used in (i); may be seen in (i) but do not give mark unless used in (ii) eliminating x or y, ft their eqns if find y first, cao for y then ft for x ft dep on both Ms earned iii 9 ( + ) o.e and completion 0 to given answer or square root of this; for or soi or ft 0 0 their coords (inc midpt) or for distance cos θ and tan θ and for showing sinθ and 0 completion iv 0 for 6 + or 0 or square roots of these v 9 www or ft their a 0 for ½ 6 or 9 their

8 75 Mark Scheme June 009 ia expansion of one pair of brackets correct 6 term expansion eg [(x + )](x 6x + 8); need not be simplified eg x 6x + 8x + x 6x + 8; or M for correct 8 term expansion: x x + x x + 8x x x + 8, if one error allow equivalent marks working backwards to factorisation, by long division or factor theorem etc or for all three roots checked by factor theorem and for comparing coeffts of x ib cubic the correct way up x-axis:,, shown y-axis 8 shown G G G with two tps and extending beyond the axes at ends ignore a second graph which is a translation of the correct graph ic [y](x )(x 5)(x 7) isw or (x ) 5(x ) + (x ) + 8 isw or x x + 59x 70 if one slip or for [y ] f(x ) or for roots identified at, 5, 7 or for translation to the left allow for complete attempt: (x + )( x + )(x ) isw or (x + ) 5(x + ) + (x + ) + 8 isw (0, 70) or y 70 allow for (0, ) or y after f(x + ) used ii or long division of f(x) or their f(x) + by (x ) attempted as far as x x in working x x obtained ( ) ( ) ± [ x ] or (x ) 5 or correct long division of x 5x + x + by (x ) with no remainder or of x 5x + x + 8 with rem or inspection with two terms correct eg (x )(x.. ) dep on previous earned; for attempt at formula or comp square on their other factor ± 0 o.e. isw or ± 5 5 6

9 75 Mark Scheme June 009 i (5, ) 0 or 5 ii no, since 0 < 5 or showing roots of y y o.e. are not real 0 for ± 0 etc or ft from their centre and radius for attempt (no and mentioning 0 or 5) or sketch or solving by formula or comp sq ( 5) + (y ) 0 [condone one error] or SC for fully comparing distance from x axis with radius and saying yes iii y x 8 or simplified alternative for y (x 5) or ft from (i) or for y x + c and subst their (i) or for ans y x + k, k 0 or 8 iv (x 5) + (x) 0 o.e. subst x + for y [oe for x] 5x 0x + 5[ 0] or better equiv. obtaining x (with no other roots) or showing roots equal expanding brackets and rearranging to 0; condone one error; dep on first one intersection [so tangent] o.e.; must be explicit; or showing line joining (,) to centre is perp to y x + (, ) cao allow y alt method y ½ (x 5) o.e. x + ½ (x 5) o.e. x y cao showing (, ) is on circle alt method perp dist between y x 8 and y x + 0 cos θ where tan θ showing this is 0 so tgt line through centre perp to y x + dep; subst to find intn with y x + by subst in circle eqn or finding dist from centre 0 [a similar method earns first for eqn of diameter, nd for intn of diameter and circle each for x and y coords and last for showing (, ) on line award only if (, ) and (9, 0) found without (, ) being identified as the soln] x 5 0 sin θ x (, ) cao or other valid method for obtaining x allow y 5 7

10 75 Mark Scheme June (C) Concepts for Advanced Mathematics Section A using Pythagoras to show that hyp. of right angled isos. triangle with sides a and a is a completion using definition of cosine www a any letter or a number NB answer given 6 x + 5x value at value at M if one error ft attempt at integration only (i) 9 for (ii) divergent + difference between terms increasing o.e. (i). for. 8 (ii) 8 5 (i)sketch of cosx ; one cycle, sketch of cosx; two cycles, both axes scaled correctly D 80 for their (i) or π. 60 θ o.e. or for other rot 6π versions of 7.50 (ii) (-way) stretch parallel to y-axis sf D 5 6 y x x 5 use of y 0, s.o.i. ft x 5, c.a.o. x < or x > 5 ft 7 use of cos θ sin θ at least one correct interim step in obtaining sin θ sin θ 0. for two terms correct NB answer given 5 θ 0 and 80,.(7 ) r.o.t to nearest degree or better - for extras in range 5 8 attempt to integrate x 5 [y] x 5x + c subst of (, 6) in their integrated eqn c 0 or [y] x 5x + 0 A for two terms correct 5 8

11 75 Mark Scheme June (i) 7 (ii) 5.5 o.e. for at least one of 5 log 0 a or ½ log 0 a or log 0 a 5.5 o.e. Section B 0 i 0.6(0...), 0.8(5...), [],.(76...).(0...),.6(0...) points plotted correctly ft ruled line of best fit T P L Correct to d.p. Allow 0.6,. and.6 tol. mm ii b their intercept a their gradient - b -8 and a.5 iii to 5 m iv 9 logt 9 t [years] approx accept 5 to 59 v For small t the model predicts a negative height (or h 0 at approx.75) Hence model is unsuitable D ia or 6 ( ) or ( 0 60) 6 + ib correct use of AP formula with a 0 and d 0 n(5 + 5n) or 5n(n + ) or 5(n + n) or (5n + 5n) 0n + 0n c.a.o. Or better iia iib 555 for 5( ) or 5( 9 ) o.e. iic correct use of GP formula with a 5, r 5( n ) o.e. 65 D n 588 www 9 c.a.o. i 6. S need not be simplified ( ) ( ). 7 7 for. o.e. 9

12 75 Mark Scheme June 009 ( h 7) ( 7) ii ( + ) h numerator 6h + h 6 + h s.o.i. iii as h tends to 0, grad. tends to 6 o.e. f.t.from 6 +h iv y 6 (x ) o.e. y 6x 6 6 may be obtained from d y dx v At P, x 6/6 o.e. or ft At Q, x c.a.o. 0

13 75 Mark Scheme June (C) Methods for Advanced Mathematics Section A π π 6 sin x dx [ cosx] π cos + cos0 cao [] [ cosx] or [ cosu] substituting correct limits in ±k cos 0. or better. (i) 00 Ae 0 A A e 500k e 500k k ln 0.5 k ln (ii) 00e kt -kt ln 0.0 t -ln 0.0 k 9966 years [5] [] 50 A e 500k ft their A if used taking lns correctly or better ft their A and k taking lns correctly art 9970 π Can use degrees or radians reasonable shape (condone extra range) passes through (-, π), (0, π) and (, 0) - [] good sketches look for curve reasonably vertical at (, π) and (, 0), negative gradient at (0, π). Domain and range must be clearly marked and correct. g(x) x b 0 or (0, ) x 0 x, so a or (, 0) [] Allow unsupported answers. www x is A0 www

14 75 Mark Scheme June 009 5(i) e y + sin x e y d y cos x dx d y cos x y dx e [] Their e y e y o.e. cao dy dx (ii) y ln( + sin x) y ½ ln( + sin x) d y cosx dx + sinx cos x as before y e E [] chain rule (can be within correct quotient rule with dv/dx 0) /u or /( + sin x) soi www 6 x + + ff( x) x x + x x + + x x + x + x x * f (x) f(x) Symmetrical about y x. E [5] correct expression without subsidiary denominators e.g. x+ + x x x x+ x+ stated, or shown by inverting 7(i) (A) (x y)(x + xy + y ) x + x y + xy yx xy y x y * (B) (x + ½ y) + ¾ y x + xy + ¼ y + ¾ y x + xy + y E E [] expanding - allow tabulation www (x + ½ y) x + ½ xy + ½ xy + ¼ y o.e. cao www (ii) x y (x y)[(x + ½ y) + ¾ y ] (x + ½ y) + ¾ y > 0 [as squares 0] if x y > 0 then x y > 0 if x > y then x > y * E [] substituting results of (i)

15 75 Mark Scheme June 009 8(i) A: + ln x 0 ln x so A is (e, 0) x e B: x 0, y e 0 e so B is (0, e ) C: f() e e 0 g() + ln E E [5] SC if obtained using symmetry condone use of symmetry Penalise A e, B e, or co-ords wrong way round, but condone labelling errors. (ii) Either by invertion: e.g. y e x x y x e y ln x y + ln x y or by composing e.g. fg(x) f( + ln x) e + ln x e ln x x (iii) [ ] e x dx e x 0 0 (iv) e 0 e e d dx x ln x x. dx x x ln x x + c ln x dx ln x ( x) dx g( x)dx ( + ln x) d e e + ln e (v) Area x [ x x x x] [ x x ] ln ) e ln e ln(e ) e * f( x)dx g( x)d 0 x e f ( x )dx g( x) d 0 e ( e ) e e x E E [] cao [] cao ft D E [6] cao taking lns or exps e + ln x or + ln(e x ) [ e ] x o.e. or u x [ ] u e substituting correct limits for x or u o.e. not e 0, must be exact. parts: u ln x, du/dx /x, v x, dv/dx condone no c ft their x ln x x (provided algebraic ) substituting limits dep www Must have correct limits 0.6 or better.

16 75 Mark Scheme June 009 or or Area OCB area under curve triangle e ½ ½ e Area OAC triangle area under curve ½ e ½ e Total area (½ e ) e cao [] OCA or OCB ½ e 0.6 or better

17 75 Mark Scheme June 009 Section B 9(i) a [] or 0. or better (ii) d y (x ) x x. d x (x ) 6x x x (x ) x x (x ) x(x ) * (x ) E [] quotient rule www must show both steps; penalise missing brackets. (iii) d y 0 when x(x ) 0 dx x 0 or x, so at P, x (/) when x, y (/) 9 d y when x 0.6, dx d y when x 0.8, dx 0.6 Gradient increasing minimum x (iv) dx u x du dx x (u + ) 9 du u (u + ) u + u + du du 7 u 7 u ( u ) du * u Area x x d x When x, u, when x, u 7 7 ( u + + )du u [ ] u + u + u ln [ + + ln ) ( + ln) ] ( + 7 cao E [7] E if denom 0 also then M0. o.e e.g. 0.6, but must be exact. o.e e.g. 0., but must be exact -/6, or -0.9 or better 8/9 or 0.6 or better o.e. e.g. from negative to positive. Allow ft on their gradients, provided ve and +ve respectively. Accept table with indications of signs of gradient. ( u + ) 9 u (du) expanding o.e. condone missing du s [ u + u + lnu] substituting correct limits, dep integration 5

18 75 Mark Scheme June ln ( + ln ) 7 5 cao [7] o.e., but must evaluate ln 0 and collect terms. 6

19 75 Mark Scheme June (C) Applications of Advanced Mathematics Section A cosθ sinθ R cos( θ + α) Rcosθ cosα Rsinθsinα Rcosα, Rsinα R + 7, R 7. tan θ ¼ θ cos(θ + 0.5) cos(θ + 0.5) 7 θ , 5.57 θ 0.5, 5.8 x A B ( x + )(x + ) x + (x + ) x A(x + ) + B(x + ) x - - -A A x -½ -½ ½ B B - x ( x + )(x + ) x + (x + ) x dx dx ( x + )(x + ) x + (x + ) ln(x + ) ½ ln(x + ) + c d y x y dx dy x dx y ln y x + c when x, y, ln + c c - ln y x x y e [7] [7] [] correct pairs R 7. tan θ ¼ o.e. θ 0.5 θ arcos / 7 ft their R, α for method (penalise extra solutions in the range (-)) correct partial fractions substituting, equating coeffts or cover-up A B - ln(x + ) ft their A -½ ln(x + ) ft their B cao must have c separating variables condone absence of c c - o.e. o.e. When x 0, y V π 0 x dy π ( y)dy π 0 [ y ] y 0 π(6 8) 8π [5] must have integral, π, x² and dy s.o.i. must have π,their ( y), their numerical y limits y y 7

20 75 Mark Scheme June 009 d y 5 a( + t ).t dt dx at dt dy dy / dt at dx dx / dt at ( + t ) * t ( + t ) At (a, ½ a), t gradient 6 6 cosec θ + cot θ + cot θ cot θ * cot θ cot θ 0 (cot θ )(cot θ + ) 0 cot θ, tan θ ½, θ 6.57 cot θ -, tan θ -, θ 5 E [7] E [6] (+t²) - kt for method ft finding t clear use of + cot² θ cosec² θ factorising or formula roots, - cot /tan used θ 6.57 θ 5 (penalise extra solutions in the range (-)) 8

21 75 Mark Scheme June 009 Section B 7(i) AB 0 0 r 0 + λ 0 (ii) n cosθ 5 0 θ 7.57 (iii) 0. + cosφ 9 θ 5 * [] [5] E [] or equivalent alternative correct vectors (any multiples) scalar product used finding invcos of scalar product divided by two modulae 7 or better ft their n for method ±/ o.e. exact (iv) sin 7.57 k sin 5 k sin 7.57 / sin 5. 0 (v) r 0 + μ x -μ, z μ x + z - -θ + θ - θ, θ point of intersection is (-, -, ) distance travelled through glass distance between (0, 0, ) and (-, -, ) ( + + ) cm [] [5] ft on their 7.57 o.e. s.o.i. subst in x + z - www dep on μ 9

22 75 Mark Scheme June 009 8(i) (A) 60 5 CB/OB sin 5 CB sin 5 AB CB sin 5 * E [] AB AC or CB AOC 5 o.e. (B) cos 0 sin 5 cos 0 sin 5 sin 5 ( )/ sin 5 sin 5 * E [] simplifying (C) Perimeter AB circumference of circle > perimeter of polygon π > π > 6 E [] (ii) (A) tan 5 FE OF FE tan 5 DE FE tan 5 tan5 t (B) tan 0 tan 5 t tan 0 t t t t t + t 0 * E [] E [] ± + (C) t circumference < perimeter π < ( ) π < ( ) * E [] using positive root from exact working 0

23 75 Mark Scheme June 009 (iii) 6 < π < ( ).06 < π <.5 [].06,.5

24 75 Mark Scheme June 009 Comprehension. [ + + ( ) + ( ) ] 0. 5 *, E. (i) b is the benefit of shooting some soldiers from the other side while none of yours are shot. w is the benefit of having some of your own soldiers shot while not shooting any from the other side. Since it is more beneficial to shoot some of the soldiers on the other side than it is to have your own soldiers shot, b w. E > (ii) c is the benefit from mutual co-operation (i.e. no shooting). d is the benefit from mutual defection (soldiers on both sides are shot). With mutual co-operation people don t get shot, while they do with mutual defection. So c> d. E + ( ) ( n ) or equivalent (allow n, n+) n n 6000 so you have played 6000 rounds.,. No. The inequality on line, b + w < c, would not be satisfied since 6 + (-) >. b + w < c and subst No, > o.e. 5. (i) Round You Opponent Your score Opponent s score C D - D C - C D - D C - 5 C D - 6 D C - 7 C D - 8 D C - Cs and Ds in correct places, C-, D (ii) [ + ( ) ] 0. 5 D ft their, - 6. (i) All scores are increased by two points per round (ii) The same player wins. No difference/change. The rank order of the players remains the same. 7. (i) They would agree to co-operate by spending less on advertising or by sharing equally. (ii) Increased market share (or more money or more customers). D

25 755 Mark Scheme June (FP) Further Concepts for Advanced Mathematics Section A (i) M (ii) 9 x y 5 x 8, y z + z 7z 5 ( z )( z + z+ 5) ± 6 0 z + z+ 5 0 z z + j and z j (i) [] (ft) (ft) [] [5] [] Dividing by determinant Pre-multiplying by their inverse Show z is a root; may be implied Attempt to find quadratic factor Correct quadratic factor Use of quadratic formula or other valid method Both solutions Asymptote at x - Both branches correct (ii) x + x+ x + 7x+ 0 x or x 5 x or > x 5 w 6w + w w roots are, -, q 9 αβγ q 9 p αβ + αγ + βγ 6 p 0 A [5] A(ft) [6] Attempt to find where graphs cross or valid attempt at solution using inequalities Correct intersections (both), or - and -5 identified as critical values x > x 5 s.c. for x 5 or > x> 5 Use of sum of roots can be implied Correct roots seen Attempt to use relationships between roots s.c. for other valid method One mark each for p - and q 9

26 755 Mark Scheme June 009 5(i) (ii) 5r+ 5r+ 5r 5r+ 5r+ 5r 5 ( 5r+ )( 5r ) ( )( ) 0 0 r ( 5r )( 5r + ) 5 r ( 5r ) ( 5r + ) n 7 5n 5n 5n+ n 5 5n ( n ) [] Attempt to form common denominator Correct cancelling First two terms in full Last term in full Attempt to cancel terms 6 When n, ( 7 ) n n, so true for n [] Assume true for n k (7k ) k ( 7k ) (7( k + ) ) k( 7k ) + (7( k + ) ) [ k( 7k ) + (( k + ) 8) ] E Assume true for n k Add ( k +) th term to both sides 7 k k ( k + )( 7k + 6) ( k + ) ( 7( k + ) ) But this is the given result with k + replacing k. Therefore if it is true for k it is true for k +. Valid attempt to factorise c.a.o. with correct simplification Since it is true for n, it is true for n,, and so true for all positive integers. E E [7] Dependent on previous E and immediately previous Dependent on and both previous E marks Section A Total: 6

27 755 Mark Scheme June 009 Section B 7(i) 5 ( 0, 0 ),(, 0 ),, 0 [] (ii) x, x, y [] (iii) Large positive x, y (e.g. consider x 00 ) Large negative x, y (e.g. consider x 00 ) + [] Clear evidence of method required for full marks (iv) Curve branches of correct shape Asymptotes correct and labelled Intercepts correct and labelled [] 5

28 755 Mark Scheme June (i) z ( + j) Radius z ( + j ) or z j All correct [] (ii) ( ( )) arg z + j 0 Equation involving the argument of a complex variable Argument 0 All correct [] (iii) π a cos π b + sin + P + + j ( ) A [] Valid attempt to use trigonometry involving π, or coordinate geometry mark for each of a and b s.c. only for a.59, b. (iv) π > arg( z ( + j) ) > 0 and z + j < ( ) [] arg( z ( + j) ) > 0 arg( z ( + j) ) < π z ( + j) < Deduct one mark if only error is use of inclusive inequalities 6

29 755 Mark Scheme June 009 9(i) Matrix multiplication is associative 00 MN MN 0 0 QMN 0 [] (ft) [] Attempt to find MN or QM 0 or QM 0 (ii) M is a stretch, factor in the x direction, factor in the y direction. Stretch factor in the x direction Stretch factor in the y direction N is a reflection in the line y x. Q is an anticlockwise rotation through 90 about the origin. [] (iii) (ft) Applying their QMN to points. Minus each error to a minimum of 0. B [] Correct, labelled image points, minus each error to a minimum of 0. Give B for correct diagram with no workings. Section B Total: 6 Total: 7 7

30 756 Mark Scheme June (FP) Further Methods for Advanced Mathematics 5 x x x x (a)(i) ln( + x) x x x x x ln( x) x... 5 Series for ln( x) as far as x 5 s.o.i. + x ln ln( + x) ln( x) x Seeing series subtracted 5 x x x Valid for - < x < Inequalities must be strict + x (ii) x + x ( x) + x x Correct method of solution x x B for x stated ( ) + ( ) 5 ln Substituting their x into their series 5 in (a)(i), even if outside range of validity. Series must have at least two terms + + SR: if > correct terms seen in (i), 80 allow a better answer to d.p..096 ( d.p.) Must be decimal places (b)(i) y (ii) r + y r + r sin θ Using y r sin θ a r( + sin θ) ( + sin θ) + sin θ a (AG) r a y x + y (a y) G G G r(0) a, r(π/) a/ indicated Symmetry in θ π/ Correct basic shape: flat at θ π/, not vertical or horizontal at ends, no dimple Ignore beyond 0 θ π Using r x + y in r + y a Unsimplified x + y a ay + y ay a x A correct final answer, not spoiled a x y a 5 6 8

31 756 Mark Scheme June 009 (i) λ M λi 0 λ 0 0 λ det(m λi) Attempt at det(m λi) with all ( λ)[( λ)( λ)] + [( λ)] elements present. Allow sign errors ( λ)(λ ) + ( λ) Unsimplified. Allow signs reversed. Condone omission of 0 λ λ + λ det M -7 (ii) f (λ) λ λ + λ + 7 f (-) eigenvalue Showing - satisfies a correct characteristic equation f (λ) (λ + )(λ λ + 7) Obtaining quadratic factor λ λ + 7 (λ ) + so no real roots www (M λi)s 0, λ - (M λi)s (λ)s M0 below x y 0 0 z 0 x + y z 0 x + z 0 Obtaining equations relating x, y and z x -z y z x z + z 6z Obtaining equations relating two variables to a third. Dep. on first s 6 Or any non-zero multiple x y 0.6 Solution by any method, e.g. use of 0 z 0. multiple of s, but M0 if s itself quoted without further work x 0., y -0.6, z -0. A Give if any two correct 9 (iii) C-H: a matrix satisfies its own characteristic Idea of λ M equation M M + M + 7I 0 M M M 7I (AG) Must be derived www. Condone omitted I M M I 7M Multiplying by M M M + M I o.e. 5 8 (iv) M Correct attempt to find M or Using their (iii) SC for answer without working 9

32 756 Mark Scheme June OR Matrix of cofactors: 7 0 Adjugate matrix : det M 7 Finding at least four cofactors Transposing and dividing by determinant. Dep. on above 9 (a)(i) G Correct basic shape (positive gradient, through (0, 0)) (ii) y arcsin x sin y x sin y and attempt to diff. both sides d y d y cos y Or cos y dx dx dy dx cos y x 0 www. SC if quoted without working Positive square root because gradient positive Dep. on graph of an increasing function 0 π x dx arcsin x 0 arcsin function alone, or any sine substitution x, or dθ www without limits Evaluated in terms of π jθ jθ 5jθ (b) C + js e + e + e Forming C + js as a series of powers This is a geometric series Identifying geometric series and attempting sum to infinity or to n terms j with first term a e θ j, common ratio r e θ Correct a and r jθ a e Sum to infinity r e jθ jθ e e jθ jθ e e jθ j e ( e jθ jθ 9e e jθ jθ 9 e e + 9cos ( θ + jsinθ) cos ( θ jsinθ) 0 cos θ jsin θ cos θ + jsin θ ( ) ( θ jθ ) Sum to infinity ) * Multiplying numerator and jθ denominator by e o.e. Or writing in terms of trig functions and realising the denominator Multiplying out numerator and denominator. Dep. on * Valid attempt to express in terms of trig functions. If trig functions used from start, for using the compound angle formulae and

33 756 Mark Scheme June 009 6cosθ + j sinθ 0 6cos θ 6cosθ C 0 6cos θ cosθ 5 cosθ 6sinθ S 5 cosθ (AG) Pythagoras Dep. on * Equating real and imaginary parts. Dep. on * o.e. 9 u u e + e (i) cosh u cosh u u e + + e u u u ( ) u u e e + e + + e cosh u u e e u u u + e + e cosh u cosh u (AG) Completion www (ii) x arcsinh y sinh x y x x e e y Expressing y in exponential form (, must be correct) e x ye x 0 (e x y) y 0 (e x y) y + e x y ± y + e x y ± y + Reaching e x by quadratic formula or completing the square. Condone no ± Take + because e x > 0 Or argument of ln must be positive x ln(y + y + ) (AG) Completion www but independent of d x dx (iii) x sinh u cosh u and substituting for all elements du du x + dx sinh u + cosh u du Substituting for all elements correctly cosh u du cosh u + du Simplifying to an integrable form sinh u + u + c u u Any form, e.g. e e + u Condone omission of + c throughout sinh u cosh u + u + c x x x + + arcsinh + c Using double angle formula and attempt to express cosh u in terms of x x x + x + arcsinh + c (AG) Completion www

34 756 Mark Scheme June 009 (iv) t + t + 5 (t + ) + Completing the square t + t + 5 dt ( t + ) + dt x + dx Simplifying to an integrable form, 0 by substituting x t + s.o.i. or complete alternative method Correct limits consistent with their method seen anywhere x x + x + arcsin h 0 (v) 8 + arcsinh Using (iii) or otherwise reaching the result of integration, and using limits + ln( + ) (ln( + ) + ) (AG) Completion www. Condone 8 etc (i) If a, angle OCP 5 so P is ( cos 5, sin 5 ) P (, ) (AG) Completion www OR Circle (x ) + y, line y -x + (x ) + (-x + ) Complete algebraic method to find x x ± and hence P Q ( + (ii) cos OCP sin OCP, - ) a a + a + Attempt to find cos OCP and sin OCP in terms of a Both correct P is (a a cos OCP, a sin OCP) P a a a, (AG) Completion www a + a + OR Circle (x a) + y a, line y x + a x (x a) + x + a a+ ± a+ + a a a + a a Complete algebraic method to find x Unsimplified

35 756 Mark Scheme June 009 a x a ± and hence P a + Q a a a +, a + a + (iii) G G G Gft As a, P (0, ) Stated separately As a, y-coordinate of P Stated a a a + a as a 8 Locus of P ( st & rd quadrants) through (0, 0) Locus of P terminates at (0, ) Locus of P: fully correct shape Locus of Q ( nd & th quadrants: dotted) reflection of locus of P in y- axis Attempt to consider y as a Completion www (iv) POQ 90 Angle in semicircle o.e. Loci cross at 90 8

36 757 Mark Scheme June Further Pure (i) Putting x 0, -y + 0z 6, -y z 8 y -, z 0 8 Direction is given by Equation of L is r + λ 0 (ii) 7 6 AB d ft 5 A ft Finding coords of a point on the line or (, 0, -), (, -, - ) etc or finding a second point Dependent on Accept any form Condone omission of r Evaluating AB d Give ft if just one error Distance is 0 n ˆ ft Appropriate scalar product Fully correct expression (iii) 6 AB d For AB d Evaluating magnitude Distance is AB d d ft In this part, M marks are dependent on previous M marks 5 5 5

37 757 Mark Scheme June 009 (iv) k + 6 At D, + λ + μ 5 9 λ + μ 5 λ 9 μ λ, μ 5 + ( k + ) 6+ 0 k 50 D is (6 + μ, - + μ, 9 μ) i.e. (6, 8, ) Alternative solutions for Q (i) e.g. x y 6 x t, y t t (t ) z 8 x t, y t, z - t (ii) + 7μ λ PQ μ + λ 5 + μ -λ PQ d PQ AB 0 ( + 7μ λ ) + ( μ λ) (5 + μ + λ) 0 7( + 7μ λ ) ( μ λ) + (5 + μ + λ) ft 8 ft 5 ft ft Condone use of same parameter on both sides Two equations for λ and μ Obtaining λ and μ (numerically) Give for λ and μ in terms of k Equation for k Obtaining coordinates of D Eliminating one of x, y, z Two equations for λ and μ Expression for shortest distance 7 7 λ, μ ft PQ (iii) 6 + λ + AX d + λ λ (7 + λ ) + (λ ) ( λ) 0 λ AX 0 AX ft 5 5

38 757 Mark Scheme June 009 (iv) 7 k k 5 0 k k k k At D, + λ + μ λ 6+ μ λ + μ 5 λ 9 μ λ, μ 5 ft Appropriate scalar triple product equated to zero Equation for k Condone use of same parameter on both sides Two equations for λ and μ Obtaining λ and μ D is (6 + μ, + μ, 9 μ) i.e. (6, 8, ) 8 Obtaining coordinates of D (i) z ( x + y) + 9 x( x + y) 6x + x z 9 xx ( + y) y z z (ii) At stationary points, 0 and 0 x y 9 x( x+ y) 0 x 0 or y x If x 0 then y + 0 y ; one stationary point is (0,, 0) If y x then 6x + 0 x ± ; stationary points are (,, ) and (,, ) (iii) z z At P (,, 9),, 9 x y Normal line is r + λ 9 9 (iv) δ z z z x y x δ + y δ δx + 9δ y h k + 9h A 7 ft ft Partial differentiation Give if just one minor error If A0A0, give for x ± For normal vector (allow sign error) Condone omission of r 6

39 757 Mark Scheme June 009 (v) k h OR Tangent plane is x+ 9y z ( + k) + 9( + h) (9 + h) k h z z 7 and 0 x y 9 x( x+ y) 0 x 0 or y x If x 0 then y + 7 y, z 0 ; point is (0,, 0) d 0 If y x then 6x + 7 x ; there are no other points M ft 6 z (Allow for 7 ) x (i) dx dy [ a( cos θ)] ( asin θ) dθ dθ (ii) (iii) s acos θdθ asin θ + C a ( + cos θ ) a cos s 0 when θ 0 C 0 dy asinθ d x a(+ cos θ ) sin θcos θ a cos tan θ ψ θ, and so s asinψ ds ρ a cosψ dψ OR a cos θ θ θ ( ) a cos θ ρ a(+ cos θ)( acos θ) ( asin θ)( asin θ) a θ a θ a cos (+ cos θ) a cos θ 8 cos 8 cos a θ (AG) 6 ft (AG) ft (AG) Forming dx dy dθ + dθ Using half-angle formula Integrating to obtain k sin θ Correctly obtained (+C not needed) Dependent on all previous marks Using half-angle formulae Differentiating intrinsic equation Correct expression for ρ or κ 7

40 757 Mark Scheme June 009 (iv) When θ π, ψ π, x a( π + ), y a ρ a sinψ n ˆ cosψ a( π + ) c + a a 5 Centre of curvature is ( a( π ), a ) (v) Curved surface area is π yds π a a 0 π a 0 π 6 πa sin θ 0 6 π ( cos θ) cos θdθ 8π sin θcos θdθ π a 6 ft 5 Obtaining a normal vector Correct unit normal (possibly in terms of θ ) Accept (. a,.5 a) Correct integral expression in any form (including limits; may be implied by later working) Obtaining an integrable form Obtaining k sin θ or equivalent (i) 5 6 In G,, 6,, 5, [ or 5, 5 6, 5, 5 5, 5 6 ] (ii) (iii) (iv) 5 6 In H, 5 7, 5 7, 5, 5, [ or, 7, 7, 5, ] G has an element (or 5) of order 6 H has an element 5 (or ) of order 6 {, 6} {,, } G H OR G H ad() a() ad() a() ad() a(), so ad c da() d() da() d() B B All powers of an element of order 6 All powers correct in both groups Ignore {} and G Deduct mark (from B) for each proper subgroup in excess of two Give B for correct, B for correct, for correct Evaluating e.g. ad() (one case sufficient; intermediate value must be shown) For ad c correctly shown Evaluating e.g. da() (one case sufficient; no need for any working) 8

41 757 Mark Scheme June 009 da() d(), so da f (v) S is not abelian; G is abelian (vi) Element a b c d e f Order (vii) {e, c} {e, d} {e, f} {e, a, b} B or S has elements of order ; G has element of order or S is not cyclic etc Give B for 5 correct, B for correct, for correct Ignore { e } and S If more than proper subgroups are given, deduct mark for each proper subgroup in excess of Pre-multiplication by transition matrix 5 (i) P B (ii) P (iii) (iv) P , P Level (v) Expected number of levels including the next change of location is 5 0. Expected number of further levels in B is A ft Give for two columns correct Using P (or P ) Give for probabilities correct (Max if not at least dp) Tolerance ± For and Accept 0.7 to 0.7 Finding P(C) for some powers of P For identifying P For / ( 0.8) or 0.8 / ( 0.8) For 5 or For as final answer 9

42 757 Mark Scheme June 009 (vi) Q n Q A: B: 0.68 C: D: (vii) 0 0. a b c d a b c d 0. a 0.5, b 0.5, c 0.75, d A 5 A 5 Can be implied Evaluating powers of Q or Obtaining (at least) equations from Qp p Limiting matrix with equal columns or Solving to obtain one equilib prob or M for other complete method Give for two correct (Max if not at least dp) Tolerance ± Transition matrix and Forming at least one equation or a + b + c + d Give for two correct Post-multiplication by transition matrix 5 (i) P B Give for two rows correct (ii) ( ) P ( ) (iii) (iv) 0 ( ) P ( ( ) P ( Level (v) Expected number of levels including the next change of location is 5 0. Expected number of further levels in B is ) ) A ft Using P (or P ) Give for probabilities correct (Max if not at least dp) Tolerance ± For and Accept 0.7 to 0.7 Finding P(C) for some powers of P For identifying P For / ( 0.8) or 0.8 / ( 0.8) For 5 or For as final answer 0

43 757 Mark Scheme June 009 (vi) Q n Q (vii) A: B: 0.68 C: D: a b c d ( ) ( ) a b c d 0. a 0.5, b 0.5, c 0.75, d A 5 A 5 Can be implied Evaluating powers of Q or Obtaining (at least) equations from pq p Limiting matrix with equal rows or Solving to obtain one equilib prob or M for other complete method Give for two correct (Max if not at least dp) Tolerance ± Transition matrix and ( ) Forming at least one equation or a + b + c + d Give for two correct

44 758 Mark Scheme June Differential Equations (i) α Auxiliary equation α ±5 CF y Acos 5t + Bsin 5t F CF for their roots PI y at cos 5t + bt sin 5t y a cos 5t 5at sin 5t + bsin 5t + 5bt cos5t y 0a sin 5t 5at cos5t + 0b cos5t 5bt sin 5t Differentiate twice In DE 0 bcos5t 0a sin 5t 0cos5t Substitute and compare coefficients b, a 0 PI y t sin 5t GS y t sin 5t + Acos5t + Bsin 5t F 8 (ii) t 0, y A From correct GS y sin 5t + 0t cos5t 5Asin 5t + 5B cos5t Differentiate t 0, y 0 B 0 Use condition on y y t sin 5t + cos5t (iii) Curve through (0, ) Curve with zero gradient at (0, ) Oscillations Oscillations with increasing amplitude (iv) y sin 5t, y 0cos5t, y 50sin 5t y + y + 5y 50sin 5t + 0cos5t + 50sin 5t 0cos5t E α + α α ± i t CF e ( C cos t + Dsin t) t GS y sin 5t + e ( C cos t + Dsin t) F F 6 Substitute into DE Auxiliary equation CF for their complex roots Their PI + their CF with two arbitrary constants (v) Oscillations of amplitude or bounded oscillations; or both oscillate Compared to unbounded oscillations in first model o.e. or one bounded, one unbounded (i) dy sin x + y dx x x Rearrange dx I e x Attempting integrating factor ln x e

45 758 Mark Scheme June 009 (ii) x ( x y) xsin x d Correct and simplified dx Multiply and recognise derivative x y xsin x dx x cos x + cos x dx Integrate cos x + sin x + A All correct x cos x + sin x + A y F Must include constant x 9 ( x ) x + x x + A 6 Substitute given approximations y x F A + x Use finite limit to deduce A A 0 sin x x cos x y x Correct particular solution lim y Correct limit x 0 6 (iii) y 0 sin x x cos x 0 Equate to zero and attempt to get tanx tan x x E Convincingly shown (iv) dy + y x, multiply by I x dx x x 6 Rearrange and multiply by IF Same IF as in (i) or correct IF d ( x y) x x dx 6 Recognise derivative and RHS correct 5 x y x x + B 0 Integrate B y x + 0 x c.a.o Finite limit B 0 Use condition to find constant lim y E Show correct limit (or same limit x 0 7 as (ii)) (a)(i) α + 0 α Find root of auxiliary equation CF PI I a cos t + bsin t I asin t + b cos t Differentiate asin t + b cos t + a cos t + bsin t cos t Substitute a + b 0, b + a a b Compare coefficients and solve Ae t 8

46 758 Mark Scheme June 009 PI I (cost + sin t) 8 t e 8 GS I A + (cost + sin t) F 8 Their PI + their CF with one arbitrary constant (ii) t 0, l 0 0 A + A Use condition 8 8 t 8 I (cos t + sin t e ) c.a.o (iii) For large t, I (cos t + sin t) Consider behaviour for large t 8 8 Amplitude + 8 Curve with oscillations with constant amplitude Their amplitude clearly indicated (may be implied) (b)(i) (A) t 0, y 0 dy 0 (0) + e dt Substitute into DE Gradient dy 9 (B) At stationary point, 0, y 8 dt Substitute into DE 0 9 t + e e t Solve for t () 8 t ln dy (C) 0, e 0 dt Substitute into DE Giving 0 y + 0, so y 7 (ii) Curve through origin with positive gradient With maximum at (ln, 89 ) Follow their ln With y as x Follow their (C) t (i) x 7x + 6y 6e Differentiate t 7x + 6( x 0y + 5sin t) 6e Substitute for y t ( x x e 7 6 y ) y in terms of x, x, t x 7x 7x 0( x 7x e x + x + x e t + 0sin t t ) + 0sin t 6e t E 5 Substitute for y Complete argument (ii) t x ae 9cost + sin t t x ae + 9sin t + cost t x 9ae + 9cost sin t Differentiate twice In DE gives Substitute t 9ae + 9cost sin t + ( ae t + 9sin t + cost)

47 758 Mark Scheme June 009 t + + ( ae t 9cost + sin t) ae 0sin t So PI with a E Correct form shown a 7 AE α + α + 0 Auxiliary equation α, t + t t CF Ae Be F CF for their roots t t GS x Ae + Be + 7e 9cost + sin t F Their PI + their CF with two arbitrary constants 8 (iii) x ( x 7x e t ) y in terms of x, x, t 6 x Ae t Be t t e t t + 9sin t + cost t y Ae Be e + cost sin t F Differentiate GS for x Follow their GS c.a.o (iv) x sin t 9cost Follow their x y cost sin t Follow their y x y cost sin t sin t 9cost Equate 0cost 5sin t tan t Complete argument (v) Amplitude of x Attempt both amplitudes Amplitude of y One correct 5 Ratio is c.a.o (accept reciprocal) 6 5

48 76 Mark Scheme June Mechanics Q Mark Comment Sub (i) m Attempt to find whole area or If suvat used in parts, accept any t value 0 t 8 for max. c.a.o. (ii) 0.5 ( T 8) k 0 seen. Accept ±5 and ±0 only. If suvat used need whole area; if in parts, accept any t value 8 t T for min. Attempt to use k T 8. T c.a.o. [Award if T seen] (iii) m ft their 0. 6 Q Mark Comment Sub (i) so 6 N 0 arctan Using arctan or equiv. Accept arctan 0 or equiv..69 so.6 ( s.f.) Accept 57.. (ii) 0 0 W wj Accept and w wj (iii) T + T + W 0 Accept in any form and recovery from W wj. Award if not explicit and part (ii) and both k and w correct. k 0 Accept from wrong working. w Accept from wrong working but not. [Accept 0i or j but not both] 7 6

49 76 Mark Scheme June 009 Q Mark Comment Sub (i) The line is not straight Any valid comment (ii) 6t a Attempt to differentiate. Accept term 8 t correct but not. 8 (iii) a() 0 The sprinter has reached a steady speed We require t t 8 ( 8) F E Accept stopped accelerating but not just a 0. Do not ft a() 0. t t t 8 d Integrating. Neglect limits. m (.65 m) c.a.o. Any form. 8 One term correct. Neglect limits. Correct limits subst in integral. Subtraction seen. If arb constant used, evaluated to give s 0 when t and then sub t. [If trapezium rule used use of rule (must be clear method and at least two regions) correctly applied At least 6 regions used Answer correct to at least s.f.)] 7

50 76 Mark Scheme June 009 Q Mark Comment Sub (i) cosα t (ii) cosα 5.8 ft their x. so 60cosα.8 and co sα 0.8 E Shown. Must see some working e.g cosθ.8 60 or 60 cosθ.8. If seen then this needs a statement that hence cosθ 0.8. (iii) sinα 0.96 Need not be explicit e.g. accept sin(7.7 ) seen. either 0 ( 0.96) 9.8 s Allow use of u, g ± (0, 9.8, 9.8). Correct substitution. s 8.88 so 8. m ( s. f.) c.a.o. or Time to max height is given by T 0 so T.9. Could use ½ total time of flight to the horizontal. y 0.96 t.9 t² Allow use of u, g ± (0, 9.8, 9.8) May use ( u+ v) s t. putting t T, y 8.88 so 8. m ( s. f.) c.a.o. 7 Q 5 Mark Comment Sub (i) v i+ ( t) j Differentiating r. Allow error. Could use const accn. v() i 5j F Do not award if 6 is given as vel (accept if v given and v given as well called speed or magnitude). (ii) a j Diff v. ft their v. Award if j seen & isw. Using NL F.5 ( j) Award for.5 (± a or a) seen. so j N c.a.o. Do not award if final answer is not correct. [Award for j WW] (iii) x + t and y t t Must have both but may be implied. Substitute t x so y ( x ) ( x ) c.a.o. isw. Must see the form y. [ ( x )(5 x) ] 8 8

51 76 Mark Scheme June 009 Q 6 Mark Comment Sub (i) Up the plane T gsin5 0 Resolving parallel to the plane. If any other direction used, all forces must be present. Accept s c. Allow use of m. No extra forces. T so 6.6 N ( s. f.) (ii) Down the plane, ( + m)g sin No extra forces. Must attempt resolution in at least term. Accept s c. Accept Mg sin5. Accept use of mass. Accept Mgsin5 m 8.07 so 8.07 ( s. f.) (iii) Diagram Any of weight, friction normal reaction and P present in approx correct directions with arrows. All forces present with suitable directions, labels and arrows. Accept W, mg, g and 9.. (iv) Resolving up the plane Or resolving parallel to the plane. All forces must be present. Accept s c. Allow use of m. At least one resolution attempted and accept wrong angles. Allow sign errors. P cos5 term correct. Allow sign error. P cos5 0 g sin5 0 Both resolutions correct. Weight used. Allow sign errors. ft use of P sin 5. All correct but ft use of P sin 5. P so 7.9 N ( s. f.) 5 (v) Resolving perpendicular to the plane May use other directions. All forces present. No extras. Allow s c. Weight not mass used. Both resolutions attempted. Allow sign errors. R + P sin5 g cos5 0 Both resolutions correct. Allow sign errors. Allow use of Pcos5 if Psin5 used in (iv). F All correct. Only ft their P and their use of Pcos5. R 5.79 so 5.7 N c.a.o. 6 If there is a consistent s cerror in the weight term throughout the question, penalise only two marks for this error. In the absence of other errors this gives (i) 5.5 (ii).69 (iv) (v).688 For use of mass instead of weight lose maximum of. 9

52 76 Mark Scheme June 009 Q 7 Mark Comment Sub With the. N resistance acting to the left (i) NL F. 8 Use of NL (allow F mga). Allow. omitted; no extra forces. All correct F 7. so 7. N c.a.o. (ii) The string is inextensible E Allow light inextensible but not other irrelevant reasons given as well (e.g. smooth pulley). (iii) One diagram with all forces present; no extras; correct arrows and labels accept use of words. Both diagrams correct with a common label. (iv) Method () For either box or sphere, F ma. Allow omitted force and sign errors but not extra forces. Need correct mass. Allow use of mass not weight. Box NL 05 T. 8a Correct and in any form. Sphere NL T a Correct and in any form. [box and sphere equns with consistent signs] Adding 5 a Eliminating variable from equns in variables. a.5 so.5 ms E Substitute a.5 giving T Attempt to substitute in either box or sphere equn. T 7.8 so 7.8 N Method () a For box and sphere, F ma. Must be correct mass. Allow use of mass not weight. a.5 E Method made clear. For either box or sphere, F ma. Allow omitted force and sign errors but not extra forces. Need correct mass. Allow use of mass not weight. either: box NL 05 T. 8a or: sphere NL T a Correct and in any form. Substitute a.5 in either equn Attempt to substitute in either box or sphere equn. T 7.8 so 7.8 N [If AG used in either equn award for that equn as above and for finding T. For full marks, both values must be shown to satisfy the second equation.] 7 50

53 76 Mark Scheme June 009 (v)(a) g downwards Accept ±g, ±9.8, ±0, ±9.8 (B) (C) Taking + ve, s.8, u and a 9.8 Some attempt to use s ut + 0.5at with a ±9.8 etc so.8 T.9T s ±.8 and u ±. Award for a g even if answer to (A) wrong. and so.9t T.8 0 E Clearly shown. No need to show +ve required. Time to reach ms is given by 0+.5t so t. remaining time is root of quad Quadratic solved and + ve root added to time to break. time is s Allow [Award for answer seen WW] Total.85 so.9 s ( s. f.) c.a.o. With the. N resistance acting to the right (i) F +. 8 so F.8 The same scheme as above (iii) (iv) (v) The same scheme with +. N instead of. N acting on the box Method () Box NL 05 T +. 8a Sphere as before Method () a These give a. and T 8. 0 The. N force may be in either direction, otherwise the same scheme Allow.5 substituted in box equation to give T 96. If the sign convention gives as positive the direction of the sphere descending, a.. Allow substituting a.5 in the equations to give T.8 (sphere) or 6. (box). In (C) allow use of a. to give time to break as s. and total time as.76 s 5

54 76 Mark Scheme June Mechanics Q Mark Comment Sub (a)(i) before u u P m kg Q km kg (ii) (iii) (iv) after v u u mu kmu mv + km PCLM applied Either side correct (or equiv) k v u E Must at least show terms grouped Need v < 0 E Accept k > without reason so k > [SC: v 0 used and inequality stated without reason] u v u u Use of NEL u so v E u k u so k.5 c.a.o. (b)(i) V Use of PCLM Use of mass 8 in coalescence Use of I Ft V E (ii) i cpt Allow wrong sign j cpt unchanged May be implied c.a.o. [Award / if barrier taken as.5 New velocity ms ] Q Mark Comment Sub (a)(i)(a) Yes. Only WD is against E Accept only WD is against gravity or conservative forces. no work done against friction. 5

55 76 Mark Scheme June 009 (B) (ii) Block has no displacement in that direction E gx 5gx Use of WE with KE. Allow m 5. Use of 50 At least GPE term GPE terms correct signs x so 0.8 m ( s.f.) c.a.o. (iii) V WE equation with WD term. Allow GPE terms missing 5 (b) Both KE terms. Accept use of 5. 0g 5g 80 Either GPE term 80 with correct sign V.6095 so.6 ms c.a.o. Force down the slope is Both terms. Allow mass not weight g sin0 Weight term correct Using P Fv P ( g sin0).5 F ft their weight term P so 8770 W ( s.f.) c.a.o

56 76 Mark Scheme June 009 Q Mark Comment Sub (i) c.w. moments about A Moments equation. 5R B 85 so R B 5 giving Accept no direction given 5 N Either a.c. moments about B or resolve R A so N F Accept no direction given (ii) c.w. moments about A Moments with attempt to resolve at least one force. Allow s c. 85 cosα 7. 5sinα 0 Weight term Horiz force term 85 5 so tanα E Must see some arrangement of terms or equiv (iii) A S N R F α 85 N B All forces present and labelled a.c. moments about B Moments with attempt to resolve forces and all relevant forces present 85 cosα +.5 5S sinα 0.5 All other terms correct. Allow sign errors. S 7. All correct Resolving horizontally and vertically Either attempted S F sinα 0 so F 7. E R 85 cosα 0 R 0 need not be evaluated here [Allow for the two expressions if correct other than s c ] Using F μr 7. μ so 0.07( s.f.) c.a.o

57 76 Mark Scheme June 009 Q Mark Comment Sub (i) Taking y-axis vert downwards from O Allow areas used as masses k πσ 8 + πσ 8 k Method for c.m. used 6π k k used (ii) (iii) ( πσ πσ ) 8 + 8k y Masses correct so 6 + k y 6 + k k gives OG as 5. and mass as 0πσ 0πσ 5. πσ 8 E Must see some evidence of simplification Need no reference to axis of symmetry Allow for either. Allow σ + Method for c.m. combining with (i) or starting again One term correct Second term correct ( 0πσ + 6πσ ) y y 6 E Some simplification shown 6 5 O G G above edge of base 6 5 seen here or below 8 seen here or below θ 8 tanθ 5 Accept and 8. θ so 5.7 ( s. f.) c.a.o. 5 8 or attempts based on 6 (iv) Slips when μ tanθ Or <.5 so does not slip There must be a reason

58 76 Mark Scheme June Mechanics (i) mv (. ) m 9.8(.6.6cos θ ) v cosθ v cosθ (ii) v cosθ R cosθ R 0.5( cos θ) 6.7cosθ R..7cosθ R 9.cosθ. (iii) Leaves surface when R 0. 9 cos θ ( ) ( θ 6.9 ) 9. 9 v E Equation involving KE and PE v Radial equation involving r Substituting for v Dependent on previous Special case: R. 9.cosθ earns A0SC ( ft if R a + bcosθ and 0 < b a < ) Dependent on previous Speed is. ms (iv) Tsinα + Rcosα Tcosα Rsinα Resolving vertically ( terms) Horiz eqn involving v r or rω OR. T mg sin α m cosα.. mg cosα R m sin α.. 5 sin α, cos α ( α 67.8 ).6 Solving to obtain a value of T or R Tension is 6.0 N Normal reaction is.09 N 8 Dependent on necessary s ( Accept 6,. ) Treat ω. as a misread, leading to T 6. 7, R 0.76 for 7 / 8 56

59 76 Mark Scheme June 009 (i) 5000x 00 Compression is 0.89 m (ii) Change in PE is (7.5 +.)sinθ J Change in EE is J Since Loss of PE Gain of EE, car will be at rest (iii) WD against resistance is 7560( + x) Change in EE is 5000x Change in KE is 00 0 Change in PE is ( + x) OR Speed 7.75 ms when it hits buffer, then WD against resistance is 7560x Change in EE is 5000x Change in KE is Change in PE is x 7 7 x ( + x) ( + x) ( + x) 500x ( + x).0( + x) x 7 0.( + x) x +.8x E F Equation involving EE and KE Accept 5 Or sinθ and 00.5 Or can also be given for a correct equation in x (compression): 500x 560x 6 0 Conclusion required, or solving equation to obtain x. ( x ) ( 500x ) ( ) ( x ) ( 500x ) ( 000 ) ( 560x ) Equation involving WD, EE, KE, PE Simplification to three term quadratic x. 0 57

60 76 Mark Scheme June 009 (a)(i) [ Velocity ] [ Force ] L T M L T [ Density ] ML Deduct mark for ms etc (ii) α β γ MLT (ML ) (LT ) (L ) α β α + β + γ γ (b)(i) π. ω π ω (.6). θ.6 ( ) Angular speed is 0.09 rads 5 F 5 ( ft if equation involves α, β andγ ) Using ω ( A θ ) For RHS ( b.o.d. for v 0.09 ms ) OR θ 0.08ωcos ω t Or θ ( ) 0.08ωsin ω t cos F ( ) sin (ii) θ 0.08sinωt When θ 0.05, 0.08sinωt 0.05 ω t t 0.6 cao or θ 0.08cosωt Using θ ( ± )0.05 to obtain an equation for t above can be earned in (i) or t 0.6 from θ 0.08cosωt or t.57 from θ 0.08cosωt Time taken is s cao 5 Strategy for finding the required time ( 0.6 or. 0.6 or ) Dep on first For θ 0.05sinωt, max B0A0M0 ( for sinωt ) 58

61 76 Mark Scheme June 009 (a) Area is ln 0 x x e dx e ln x xydx xe dx 0 x x e e ln x ln 0 ln 0 Integration by parts x x For x e e ln x ln ww full marks (B) Give B for 0.65 ln x d (e ) 0 y x dx y ln x e 0 9 For integral of For e x x (e ) If area wrong, SC for ln x and y area area (b)(i) a Volume is 6 πy dx π dx x π may be omitted throughout a π π x a 6 πxy dx π dx x x a a π π x π xy dx π y dx a 9 8 π a ( a a) a 8 π a (ii) Since a >, a > 8 so a a < a 8 ( a a) Hence x < a 8 i.e. CM is less than units from O OR As ( a ) a, x 8a Since x increases as a increases, x is less than E 6 E E Condone instead of Fully acceptable explanation Dependent on > throughout a Accept x, etc a ( for x stated, but requires correct justification ) 59

62 76 Mark Scheme June MEI Mechanics d dt (i) ( mv) mg (ii) dm dv v + m mg dt dt mg dv v + m mg v + dt ( ) Seen or implied Expand Use dm v dt mg v ( + ) dv v v + ( ) ( ) g g dt v + v + v + dv g v + dt Separate variables (oe) dv g v + dt E dv + g dt v Integrate v ln v + gt + c LHS t 0, v 0 ln c Use condition v ln v + gt ln t ( v ln v + + ln ) g v 0 t.68 (iii) As t gets large, v gets large dv So g (i.e. constant) Complete argument dt mg V mg a sin θ + a asin θ a GPE Reasonable attempt at EPE EPE correct dv mg mga cosθ + ( a sin θ a) a cosθ dθ a Differentiate mga cosθ + mgacosθ sin θ (i) ( ) ( sin ) ( ) m ga cosθ θ E Complete argument (ii) dv 0 dθ Set derivative to zero cos θ 0 or sin θ Solve π or π Both dθ θ 6 d V mga cosθ θ ( π)( mga) < ( cosθ) mgasin θ( sin ) Second derivative (or alternative method) V 0 unstable Consider sign One correct conclusion validly

63 76 Mark Scheme June 009 shown ( ) V π ( ) mga > 0 6 stable Complete argument (i) Mass of ring πrδrρ May be implied I C a 0 r a [ πρ r ] πa ρ 0 πρr dr Set up integral All correct Integrate M πa ρ Use relationship between ρ and M I C Ma E Complete argument (ii) ( ) (iii) (iv) I I M Use parallel axis theorem A C + a 0 Ma + Ma Ma E Convincingly shown I A θ Mg a sin θ 0 LHS RHS g θ sin θ 5. a Expression for θ θ small sin θ θ Use small angle approximation g θ θ, i.e. SHM E Complete argument and conclude 5. a SHM Period π 5. a. 5 g a F Follow their SHM equation e.g (v) C A P Show PAC in straight line (in any direction) m g 9 a M g a 0 0 Moments or ( m) x mx (oe) m 9 M Method may be implied 9 ( ) 0.5Ma m a 0 I + 0.6Ma E Convincingly shown I ω 0.6Ma ) KE ( ω Attempt to find KE 0.Ma ω C n π 0.Ma ω Work-energy equation Correct equation 0.Ma ω C nπ 5 5 6

64 76 Mark Scheme June 009 (i) At terminal velocity, F 0 k 60 90g Equilibrium of forces (ii) g E Convincingly shown k 0 dv NL 90v 90g gv 0 dx 90v dv dx Separate and integrate 90g gv ln 90g gv x + c 0 LHS g 90 gx e gv A gx 0 v 90g Ae 800 Rearrange, dealing properly with g constant x 0, v 0 A 90g Use condition 9x v 600 e 800 E Complete argument (iii) WD against R kv dx gx 90g e 800 dx gx g x + e 800 Integrate g g ( + e ) g x 800 v 600( e ) g Loss in energy g ( e ) g 90 GPE KE g 6000( g + e ) WD against R E Convincingly shown (including signs) (iv) g v 60 e (v) dv 90 90g 90v dt NL dv 0 dv dt g v t or t g v 0 Separate and integrate ln g v t + c t 0, v c.9598 Use condition (or limits) v 0 t ln Calculate t 5.5 s

65 766 Mark Scheme June Statistics Q (i) (ii) Median Mode cao cao S labelled linear scales on both axes H heights 0 0 Number of People (iii) Positive TOTAL 5 Q (i) 5 5 different teams 50 for 5 5 cao (ii) for either combination for product of both cao TOTAL 5 Q (i) 6 Mean 0.5 for mean S xx s.85 (ii) New mean New s for attempt at S xx cao ANSWER GIVEN ft (iii) On average Marlene sells more cars than Dwayne. Marlene has less variation in monthly sales than Dwayne. E E ft TOTAL 8 6

66 766 Mark Scheme June 009 Q (i) (ii) Q5 (i) E(X) 5 because the distribution is symmetrical. Allow correct calculation of Σrp E(X ) Var(X) Distance freq width f dens E ANSWER GIVEN for Σr p (at least terms correct) dep for 5² cao TOTAL for fds cao Accept any suitable unit for fd such as eg freq per 50 miles. L linear scales on both axes and label W width of bars 5 H height of bars (ii) Median 600th distance for 600 th Q6 (i) (ii) Estima te / (A) P(at most one) (B) P(exactly two) for attempt to interpolate cao TOTAL 8 aef for (0++)/00 aef P(all at least one) 0.5 for dep for product of next correct fractions cao TOTAL 6 6

67 766 Mark Scheme June 009 Q7 (i) a 0.8, b 0.85, c 0.9. (ii) P(Not delayed) for any one for the other two for product cao P(Delayed) for P(delayed) ft (iii) P(just one problem) one product correct three products sum of products cao (iv) P(Just one problem d elay) (v) P(Just one problem) P(Delay) P(Delayed No technical problems) Either for numerator 0.88 for denominator ft for for second term cao Or Or Or (using conditional probability formula) P(Delayed and no technical problems) P(No technical problems) for product for product cao for all products for sum of all products cao for numerator for denominator cao (vi) Expected number for product ft TOTAL 8 65

68 766 Mark Scheme June 009 Q8 (i) X ~ B(5, 0.) 5 (A) P ( V ) ( ) p q cao Or from tables Or: M for cao (B) P ( X ) P(X ) -P(X ) cao (C) E(X) np (ii) (A) Let p probability of a randomly selected child eating at least 5 a day H 0 : p 0. H : p > 0. (B) H has this form as the proportion who eat at least 5 a day is expected to increase. (iii) Let X ~ B(5, 0.) P(X 5) P(X ) > 0% P( X 6) P(X 5) < 0% So critical region is {6,7,8,9,0,,,,,5} 7 lies in the critical region, so we reject null hypothesis and we conclude that there is evidence to suggest that the proportion who eat at least five a day has increased. for product cao for definition of p in context for H 0 for H E for 0.6 for 0.06 for at least one comparison with 0% cao for critical region dep on and at least one dep for comparison 6 dep for decision and conclusion in context TOTAL 8 66

69 767 Mark Scheme June Statistics Question (i) (ii) EITHER: Sxy xy x y n S xx ( x) x n 50 S yy y ( y) n r S xy 87.8 S xxs yy OR: xy cov(x, y) xy n rmsd(x) S xx 8.8 n rmsd(y) r cov( rmsd( x S yy n x, y) )rmsd( y ) H0: ρ 0 H: ρ 0 (two-tailed test) where ρ is the population correlation coefficient For n 50, 5% critical value for method for S xy for method for at least one o f S xx or S yy for at least one of S xy, S xx or Syy correct for structure of r (AWRT 0.8) for method for cov( x,y) for method for at least one msd for at least on of cov( x,y), rmsd(x) or rmsd(y) correct for structure of r (AWRT 0.8) for H 0, H in symbols for defining ρ FT for critical value 5 (iii) (iv) Since 0.77 > we can reject H 0 : There is sufficient evidence at the 5% level to suggest that there is correlation between oil price and share cost Population The scatter diagram has a roughly elliptical shape, hence the assumption is justified. Because the alternative hypothesis should be decided without referring to the sample data and there is no suggestion that the correlation should be positive rather than negative. for sensible comparison leading to a conclusion for result FT for conclusion in context 6 elliptical shape E conclusion E E TOTAL 6 67

70 767 Mark Scheme June 009 Question (i) Meteors are seen randomly and independently There is a uniform (mean) rate of occurrence of meteor sightings (ii) (A) Either P(X ) Or P(X ) e 0.5! (B) Using tables: P( X ) P(X ) (iii) λ 0. P(X 0) e ! (iv) Mean no. per hour Normal approx. to the Poisson, X ~ N(78, 78) for appropriate use of tables or calculation for appropriate probability calculation for mean for calculation CAO for Normal approx. for correct parameters (SOI) (v) P(X 00) P Z > 78 P(Z >. ) Φ(.) Either P(At least one) e λ e 0.0 λ λ ln 0.0, so λ.605. t.605, so t.5 Answer t Or 0 λ e ! t, λ., P(At least one) e t, λ. 6, P(At least one) t, λ.9, P(At least one) t, λ 5., P(At least one) Answer t λ. e.6.9 e e 0.99 for continuity corr. for correct Normal probability calculation using correct tail CAO, (but FT wrong or omitted CC) formation of equation/inequality using P(X ) P(X 0) with Poisson distribution. for correct equation/inequality for logs for.5 for t (correctly justified) at least one trial with any value of t correct probability. trial with either t or t correct probability of t and t for t TOTAL

71 767 Mark Scheme June 009 Question (i) X ~ N(70,90 ) P(X <700) P Z < 90 P( Z < 0.) Φ( 0. ) Φ(0.) (ii) P( of below 700) (iii) Normal approx with μ np σ npq P (X 0) P Z P(Z 0.970) Φ(0.970) (iv) H 0 : μ 70; H is of this form since the consumer organisation suspects that the mean is below 70 μ denotes the mean intensity of 5 Watt low energy bulbs made by this manufacturer. (v) Test statistic Lower 5% level tailed critical value of z >.65 so not significant. There is not sufficient evidence to reject H 0 There is insufficient evidence to conclude that the mean intensity of bulbs made by this manufacturer is less than 70 for standardising use of tables (correct tail) CAO NB ANSWER GIVEN for coefficient for FT (min sf) for correct continuity corr. for correct Normal probability calculation using correct tail CAO, (but FT wrong or omitted CC) E for definition of μ must include 0 FT for.65 No FT from here if wrong. Must be.65 unless it is clear that absolute values are being used. for sensible comparison leading to a conclusion. FT only candidate s test statistic for conclusion in words in context TOTAL

72 767 Mark Scheme June 009 Question (i) H 0 : no association between type of car and sex; H : some association between type of car and sex; EXPECTED Male Female Hatchback Saloon People carrier WD Sports car A for expected values (to dp) (allow for at least one row or column correct) CONTRIBUTION Male Female Hatchback.98.8 Saloon People carrier WD Sports car.96. X.6 Refer to Ξ Critical value at 5% level > 9.88 Result is significant There is evidence to suggest that there is some association between sex and type of car. for valid attempt at (O E) /E for all correct NB These marks cannot be implied by a correct final value of X for summation for X CAO for deg of f CAO for cv sensible comparison leading to a conclusion NB if H 0 H reversed, or correlatio n mentioned, do not award first or final (ii) In hatchbacks, male drivers are more frequent than expected. In saloons, male drivers are slightly more frequent than expected. In people carriers, female drivers are much more frequent than expected. In WDs the numbers are roughly as expected In sports cars, female drivers are more frequent than expected. E E E E E TOTAL

73 768 Mark Scheme June Statistics Q W ~ N(, 0. 55) G ~ N(, 0.9 ) When a candidate s answers suggest that (s)he appears to have neglected to use the difference columns of the Normal distribution tables penalise the first occurrence only. (i) (ii) (iii) 5 P( G < 5) <. For standardising. Award once, P Z 0.9 here or elsewhere c.a.o. W + G ~ N( + 58, Mean. σ ) Variance. Accept sd (.057 ). P(this > 60) P Z > c.a.o. H W W7 + G G6 ~ N(96, Mean. σ ) Variance. Accept sd (.65). P(960 < this < 965) Two-sided requirement P < Z < ( ) c.a.o. Now want P(B(, 0.675) ) Evidence of attempt to use binomial. ft c s p value Correct terms attempted. ft c s p value. Accept P( ) c.a.o. 7 (iv) D H H ~ N(0, Mean. (May be implied.) ) Variance. Accept sd (.756). Ft c s from (iii). Want h s.t. P( h < D < h) 0.95 Formulation of requirement as - sided. i.e. P(D < h) h For.96. c.a.o

74 768 Mark Scheme June 009 Q (i) H 0 : μ Both hypotheses. Hypotheses in H : μ < words only must include population. where μ is the mean weight of the cakes. For adequate verbal definition. Al low absence of population if correct notation μ is used, but do NOT allow X... or similar unless X is clearly and explicitly stated to be a population mean. x s n 0.07(55) s n but do NOT allow this here or in construction of test statistic, but FT from there Allow c s x and/or s T n. est statistic is 0.07 Allow alternative: + (c s.895) ( ) for subsequent comparison with x. (Or x ( c s.895) (.00677) for comparison with.).68(). c.a.o. but ft from here in any case if wrong. Use of x scores A0, but ft. Refer to t 7. No ft from here if wrong. P(t <.68()) Single-tailed 5% point is.895. Must be minus.895 unless absolute values are being compared. No ft from here if wrong. Not significant. ft only c s test statistic. Insufficient evidence to suggest that the cakes are ft only c s test statistic. 9 underweight on average. (ii) CI is given by ± ± (0.896(),.08(5)) c.a.o. Must be expressed as an interval. ZERO/ if not same distribution as test. Same wrong distribution scores maximum B0A0. Recovery to t 7 is OK. (iii) x ± n Structure correct, incl. use of Normal..96. All correct. 7

75 768 Mark Scheme June 009 (iv) n < 0.05 Set up appropriate in equation. Condone an equation..96 n > Attempt to rearrange and solve. So take n 8 c.a.o. (expressed as an integer). S.C. Allow max (c.a.o.) when the factor is missing. (n > 6.879) 9 7

76 768 Mark Scheme June 009 Q (i) For a systematic sample she needs a list of all staff with no cycles in the list. All staff equally likely to be chosen if she chooses a random start between and 0 then chooses every 0 th. Not simple random sampling since not all samples are possible. E E E E E 5 (ii) (iii) Nothing is known about the background population... E Any reference to unknown distribution or non-parametric situation.... of differences between the scores. E Any reference to pairing/differences. H 0 : m 0 H : m 0 where m is the population median difference for the scores. Both hypotheses. Hypotheses in words only must include population. For adequate verbal definition. Diff Rank For differences. ZERO in this section if differences not used. For ranks. ft from here if ranks wrong. W (or W ) Refer to tables of Wilcoxon paired (/single No ft from here if wrong. sample) statistic for n. Lower (or upper if 5 used) ½ % tail is (or 65 if 5 used). i.e. a -tail test. No ft from here if wrong. Result is not significant. ft only c s test statistic. No evidence to suggest a preference for one of the uniforms. ft only c s test statistic

77 768 Mark Scheme June 009 Q (i) x ( x ) for 0 < x < λ, λ > 0 λ f f(x ) > 0 for all x in the domain. E λ λ x x λ dx 0 λ λ λ 0 Correct integral with limits. Shown equal to. (ii) λ M λ x x / λ μ dx 0 λ λ P( X μ x x < μ) dx 0 λ λ μ λ / 9 λ λ 9 which is independent of λ. 0 μ 0 Correct integral with limits. c.a.o. Correct integral with limits. Answer plus comment. ft c s μ provided the answer does not involve λ. (iii) Given σ E( X λ λ 9 ) λ λ 8 Use of Var(X) E(X ) E(X). c.a.o. (iv) Probability Expected f X Calculation of X. 6.97() c.a.o. Probs 50 for expected frequencies. All correct. Refer to χ. Allow correct df ( cells ) from wrongly grouped table and ft. Otherwise, no ft if wrong. P(X > 6.97) Upper 5% point is No ft from here if wrong. Not significant. ft only c s test statistic. Suggests model fits the data for these jars. ft only c s test statistic. But with a 0% significance level (cv 6.5) a different conclusion would be reached. E Any valid comment which recognises that the test statistic is close to the critical values

78 769 Mark Scheme June Statistics Q Follow-through all intermediate results in this question, unless obvious nonsense. (i) P(X ) θ θ ( θ) ( θ) [o.e.] n [ L [ θ] 0 θ( θ)] n [( -θ) ] n 0 + n θ ( θ n n0 n ) n n 0 n (ii) ln L ( n ) ln θ + ln ( θ) M 0 + n ( n n 0 n ) d ln L dθ n0 n n n0 n θ θ 0 ( - θˆ ) ( n 0 + n ) θˆ ( n n 0 n ) ˆ n0 + n θ n n0 (iii) x E(X) x θ( θ) x 0 θ {0 + ( θ) + ( θ + ( θ) + } θ θ ) So could sensibly use (method of moments) ~ ~ θ θ given by ~ X θ θ ~ + X A Product form Fully correct BEWARE PRINTED ANSWER Divisible, for algebra; e.g. by GP of GPs BEWARE PRINTED ANSWER BEWARE PRINTED ANSWER 5 6 (iv) To use this, we need to know the exact numbers of faults for components with two or more. x 0 00 ~ θ Also, from expression given in question, ~ ( 0 877) Var( θ ) E + 6 CI is given by (0 87, 0 97) ± 96 x For For.96 For

79 769 Mark Scheme June 009 Q (i) (ii) (iii) (iv) z tz tz Mgf of Z E ( e ) e dz π Complete the square z tz ( z t) + t t t ( z t e ) t For taking out factor e e dt e For use of pdf of N(t,) π For pdf Pdf of N(t,) t F or final answer e 8 Y has mgf M Y (t) t ( bt Mgf of + b is E[e ay + b) ay ] For factor e ( at) Y bt ( at) Y bt e E[e ] e M ( at) For factor E[e ] Y For final answer X μ Z, so X σz + μ σ M X ( t) e X μt.e ( σt ) e σ t μt + W e k X k kx E( W ) E[(e ) ] E(e ) M X (k) For factor μt e ( σt) For factor e For final answer X k For E[(e ) ] For E(e kx ) For M X (k) E( W ) M X () e σ μ+ E( W ) M X Var( W ) e () e μ+ σ e μ+ σ μ+ σ μ+ σ σ [ e (e )] 8 77

80 769 Mark Scheme June 009 Q (i) x 6 s s y 9 s s H 0 : μ A μ B H 0 : μ A μ B here μ, μ are the population means. W A B Pooled s [ 9 506] Test statistic is Refer to t if all correct. [No mark fo r use of sn, which are 8 57 and respectively.] D o not accept X Y or similar. No FT if w rong (ii) Double-tailed 0% point is 75 Not significant No evidence that population mean concentrations differ. There may be consistent differences between days (days of week, types of rubbish, ambient conditions, ) which should be allowed for. E E N o FT if wrong 0 A ssumption: Normality of population of differences. Differences are [ d, s 86 ( s 95)] Use of s ( 6) is not acceptable, even in a n denominator of s n / n ] Can be awarded here if NOT awarded in part (i) 0 Test statistic is 6 86/ 9 Refer to t 8 Double-tailed 5% point is 06 Significant Seems population means differ No FT if wrong No FT if wrong 0 78

81 769 Mark Scheme June 009 (iii) Wilcoxon rank sum test Wilcoxon signed rank test H 0 : media n A median B H : mediana median B [Or more formal statements] Q (i) Description must be in context. If no context given, mark according to scheme and then give half-marks, rounded down. Clear description of rows. And columns As extraneous factors to be taken account of in the design, wit h treatments to be compared. Need same numbers of each Clear contrast with situations for completely randomised design and randomised trends. E E E E E E E E E 9 (ii) e ~ ind N (0, σ ) Allow uncorrelated ij For 0 For σ αi is population mean effect by which ith treatment differs from overall mean (iii) Source of Variation Between Treatments SS df MS MS ratio Residual Total Refer to F, 5 No FT if wrong Upper % point is 89 No FT if wrong Significant, seems treatments are not all the same 0 79

82 77 Mark Scheme June Decision Mathematics Question (i) and 6, and 5, and (ii) Vertex 6 Vertex A 0 to edges ( each edge error) Vertex 5 Vertex Vertex Vertex (iii) identification sketch Question. (i) A's c takes, leaving. You have to take. A's c takes one and you lose. (ii) A's c takes leaving. Then as above. (iii) A's c takes leaving. You can then take, leading to a win. 80

83 77 Mark Scheme June 009 Question. (i) (0, ) 8 (5, 0) 55 (, 7) 6 (8, 0) 5 B lines shading 6 at (, 7) (ii) Intersection of x+5y60 and x+y8 is at (0,8) optimisation 8

84 77 Mark Scheme June 009 Question. (i) e.g. 0 exit 5 9 other vertex (ii) e.g. A E A A B A B A B E B A E A A B A B A E A 5 A B E B 6 A B A B A B E B 7 A B A B E B 8 A E A 9 A B E B 0 A E A process with exits 0.5, 0.5,.9 (Theoretical answers: /, /, ) (Gambler's ruin) (iii) e.g. 0 exit 5 next vertex in cycle 6 8 other vertex 9 ignore and re-draw probabilities duration ignore D conditionality equal prob efficient (iv) e.g. A B A B A E A A C A E A A E A A B C B C E C 5 A E A 6 A C A B EB 7 A E A 8 A B C E C 9 A E A 0 A E A A 0.7, 0., 0. (Theoretical probs are 0.5, 0.5, 0.5) (Markov chain) 8

85 77 Mark Scheme June 009 Question 5. (i) e.g. A 7 B 5 and/or 0 E (ii) A D 5 B 0 5 C connectivity lengths A connected tree ( each error) D C E Order: AE; AD; DB; DC or AD; AE; DB; DC or AD; D B; AE; DC Length: 65 km OR B A D C E Order: AD; DB; DE; DC Length: 66 km (iii) A B X D C A connected tree ( each error) E (iv) Length: 5 km Advice: Close BC, AE and BD facility (e.g. anglers) distances (e.g. B to C) B 8

86 77 Mark Scheme June 009 Question 6. (i)&(ii) D 60 A B C 50 time 0 minutes critical A; B; E; F; H E G 0 F 0 0 H sca (activity on arc) single start & end dummy rest forward pass backward pass cao (iii) e.g. H G F E D C B A cascade A Joan/Keith Least time 0 mins Minimum project completion times assumes no resource constraints. 8

87 77 Mark Scheme June Decision Mathematics Question. (a) e.g. "It is easy to overestimate the effect that your contribution will make." (b) e.g. A remove double negatives same meaning B C machine A combinatorial "ands" negations "ors" one for each alternative (c) e.g. ((a b) (~a c)) (~b c) ~ ((~a ~c) (~b ~c)) lines a, b, c negations "and"s "or"s 85

88 77 Mark Scheme June 009 Question. (i) revised 60marks score D D D S +0 6 S D S S + 66 (ii) D worth revise D S worth D worth 60 6 revise S S worth revise D 65 do not ask M S 66. M predicts D w orth 60 revise S ask M M predicts revise D S wo rth revise S Worth of Michael's help 0.8 marks D worth S worth chance node D worth S worth 60 6 D worth S worth D worth S worth 60 6 decision node chances decisions revise ask/don't 86

89 77 Mark Scheme June 009 Question. (i) a is the number of acres of land put to crop A, etc a + b 0 is equivalent to a + b c + d Given that a + b + c + d 0, the maximisation will ensure that a + b + c + d 0 (and it' s ea sier to solve using simplex). (ii) P a b c d s s RHS acres to A and 0 acres to C, giving profit of 800 ( iii) Max 50a + 0b + 0c + 0d st a + b 0 a + b + c + d 0 a + b + c + d 0 A P a b c d s s sur art R Minimise A (to zero) then drop A row and art column and continue normally OR P a b c d s s sur art R M 0 0M M M M M Proceed as per simplex, regarding M as a large fixed number. OR new obj surplus artificial constraints surplus artificial 87

90 77 Mark Scheme June 009 Question. (a) (i),(ii) and (iii) distance to 5 etc rest route Shortest distance from to is 0 ( st row and rd column of distance matrix) Not part of the question Not pa rt of the question Not part of the question 0 changed dists ' s in r of route rest of route 88

91 77 Mark Scheme June 009 Shortest route is followed by route matrix (,) follow ed by route ma trix (,) follo wed by rou te mat rix (, ) (iv) (v) (5) (5) (5) 5 () (0) Total length 96 5 ( ) Finds a (hopefully short) r oute visiting every ver tex a nd returning to the start, or, upper bound to th e TSP 89

92 77 Mark Scheme June Decision Mathematics Computation Question. (i ) n+ n+ n B B + (0 B ) (ii) Oscillation:,,,,,,,, (iii) B n+ B n+ + B n 0,,,, 0.5,,, 0.009, Oscillatory convergence (iv),,.5,.5,.65,,, 0.000, Faster and uniform convergence (v) Auxiliary eqn: x x + 0 x B n ( ) n A + Bn( ) n A + B giving B 6 n n B n ( )( ) ( )( ) + 6n or + n ( 6n) ( ) n + or ( + n)( ) n "the same" 90

93 77 Mark Scheme June 009 Question. (i) e.g reversal 0 rest

94 77 Mark Scheme June 009 (ii) {S, A, C, D, E} / {B, T} (iii) e.g. Max SA + SE st SA + BA + CA AB AC 0 AB + CB BA BC BT 0 AC + DC + BC CA CB CD 0 SE + DE ED 0 CD + ED DC DE DT 0 SA < SE < 0 AB < 0 BA < 0 AC < 0 CA < 0 BC < 0 CB < 0 CD < 0 DC < 0 ED < 8 DE < 8 BT < 0 DT < 0 end variables objective balancing capacities forwards backwards (i v) OBJECTIVE FUNCTION VALUE ) VARIABLE VALUE REDUCED COST SA SE BA CA AB AC CB BC BT DC CD DE ED DT Solution as per part (i) running 9

95 77 Mark Scheme June 009 Question. (i) e.g etc. if(rand()<0.55,, ) + A (ii) repeat ing until a player is ruined repeating 0 times estimating the probability (theoretical value is 0.68) "if" or equivalent accumulation (iii) e.g etc. if(rand()<0.5,, ) + A change of parameter count to ruin repetitions estimatin g the run length The theoretical value is 50, so there should be some long runs seen. (iv) e.g etc if(rand()< 0.55,, ) + A termination condition repetitions + probability estimate (theoretical answer ) (v) As above How can one tell when a simulation is not emptying the pot? 9

96 77 Mark Scheme June 009 Question. (i) max sm0+7sm+8s m+9sm+sm+sm0 +6sm+0sm+sm+sm+sm0 +7sm+8sm+sm+5sm st sm0+ sm+sm+sm+sm sm0+ sm+sm+ sm+sm sm0+ sm+sm+sm+sm sm+ sm+sm+sm+sm+sm+sm +sm +sm+sm+sm+sm end int 5 B (ii) LP OPTIMUM FOUND AT STEP OBJECTIVE VALUE NEW INTEGER SOLUTION OF AT BRANCH 0 PIVOT RE-INSTALLING BEST SOLUTION... OBJECT IVE FUNCTION VALUE ) VARIABLE VALUE REDUCED COST SM S SM SM SM SM S SM SM SM SM S SM SM SM ROW SLACK OR SURPLUS DUAL PRICES ) ) ) ) Invest million at site, million at site and million at site. Revenue million. B (iii) (a) million at site and million at site. Revenue million. (b) million at site, million at site and million at site (or m, m and m). Revenue million. 9

97 776 Mark Scheme June Numerical Methods (i) f(x).6(x 0.)(x )/(-0.)(-) +.x(x ) /0.(0. ) +.8x(x 0.)/( 0.) [,,] (x.x + 0.) 0(x x) + (x 0.x) -x +.x +.6 [] [] (ii) Newton's formula requires equally spaced data [E] [TOTAL 7] x x + /x -.5 (change of sign so root) [] f(x) x + /x so f '(x) x /x hence NR formula [] r 0 x r [] [TOTAL 7] (i) term mpe X X+Y X Y 0X + 0Y [ ] (ii) term X Y XY X/Y mpre [ ] [TOTAL 8] (i) to 6 dp: sin A sin B LHS RHS [] (ii) (iii) It is an approximate equality. LHS involves subtraction of nearly equal numbers. LHS involves trig functions, RHS just. Subtraction of nearly equal quantities is a bigger problem as the difference decreases. RHS involves no such problem. [EE] [EE] [TOTAL 6] r x r [G] [] cobweb diagram showing spiralling out from root [] [TOTAL 8] 6(i) x f(x) T.0876 M [] 95

98 776 Mark Scheme June T.98 T [] values [,,,] M.056 [subtotal 6] (i i) S.7959 (a.g.) [] S.806 [] [subtotal ] (i ii) S ( M + T ) /. 800 [] [subtotal ] (iv) M diffs ratio approx 0. 5 [] S dif fs 5.77E-05.99E-06 ratio (approx 0.065) [] Reasoning to: integral is secure as.80( 0) [] [subtotal 7] [ TOTAL 8] 7(i) x f(x) st diff n d diff [] f( x) (-0.7)(x ) / 0. +.(x )(x.) /( (0.) ) [ 0.6.5x x - x + 8 5x - 6.5x +. [ ] [subtotal 8] (ii) f '(x) 0x 6.5 f '(.) [ ] Central difference: (0. 0.6)/(. ) -0./ [] Suggests central difference is acc urate for quadratics. [E] [subtotal 5] (i ii) f '() [] Fo rward difference: ( )/(. ) -0.7/ [] Shows that forward difference is not exact for quadratics. [E] Quadratic estimate (-6.5) is likely to be more ac curate. (Allow comments [E] saying that we cannot be sure.) [subtotal 5] [TOTAL 8] 96

99 777 Mark Scheme June MEI Numeric al Computation (i) - < g'(α) < [] E.g. Multipl y both sides of x g (x) by λ and add ( λ)x to both sid es. [] Derivative of rhs set to zero at root: λg'(α) + λ 0 [] algebra to o btain given result [] In practice use an initial estima te x 0 in place of α [] (i ii) x x sinx [subtotal 7] [G ] Roots approximately 0.5,. [] Eg: r x r x r x r x r x r x r No convergence in each case [] L et g( x) sin x 0. 5 Then g' (x) cosx So λ / ( cosα) [] Smaller root: λ Larger root: λ (approx -0.5) (approx 0.) [] r x r r x r NB: must be using relaxatio

100 777 Mark Scheme June 009 n [] [] [subtotal 7] [TOTAL ] (i) f(x) h a + b [] f(x) x, x give 0 0 [] f(x) x h / aα [] f( x) x h 5 /5 aα [] Convincing algebra to verify given re sults [] [subtotal 8] (ii) L R m h function values setup: weights [MA] integral [] L R m h function value s w eights integral function value s repeat: [M weights ] integral [] Either repea t with h halv ed to verify that is correct to 6 dp [] Or observe that the meth od is converging so rapidly that will be correct to 6d p or [E] [subtotal ] (iii) Use routine known to deliv er 6dp and vary k: k.657 L R m h function values weights integral modify function values [] weights integral k find k integral [] h ence k.66 [subtotal ] [TOTAL ] 98

101 777 Mark Scheme June 009 (i) Use central difference formulae for nd and st derivatives to obtain first given result [] Hence obtain y h y [] Use central differenc e to obtain y y h H ence given re sult for y [] [] [subtotal 8] (i i) h x y

102 777 Mark Scheme June setup [M] numbers graph [A] [A] [subtotal 9] (ii) Obtain formula y a h + 0.5h [] Modify rout ine [] Trial on a to obtain a -. or -.5 [G] h x y 0. a