Oxford Cambridge and RSA Examinations GCEAdvanced GCE A2 7890 - 2Advanced Subsidiary GCE AS 3890 - 2 Mark Schemes for the Units January 2010 3890-2/7890-2/MS/10JMathematics 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, Diplomas, GCSEs, OCR Nationals, Functional Skills, 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 specifications 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 todays 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 2010 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG15 0DL Telephone:0870 770 6622 Facsimile:01223 552610E-mail:publications@ocr.org.uk CONTENTS Advanced GCE Mathematics (7890) Advanced GCE Pure Mathematics (7891) Advanced GCE Further Mathematics (7892) Advanced Subsidiary GCE Mathematics (3890) Advanced Subsidiary GCE Pure Mathematics (3891) Advanced Subsidiary GCE Further Mathematics (3892) MARK SCHEMES FOR THE UNITS Unit/ContentPage 4721 Core Mathematics 11 4722 Core Mathematics 25 4723 Core Mathematics 39 4724 Core Mathematics 413 4725 Further Pure Mathematics 117 4726 Further Pure Mathematics 221 4727 Further Pure Mathematics 324 4728 Mechanics 128 4729 Mechanics 231 4730 Mechanics 334 4732 Probability & Statistics 137 4733 Probability & Statistics 240 4734 Probability & Statistics 343 4736 Decision Mathematics 146 4737 Decision Mathematics 252 Grade Thresholds58 4721Mark SchemeJanuary 2010 14721 Core Mathematics 1 1 [ ]35 ) 6 (1 36 ) 6 (22 =+ xx B1 M1 A1 3 3 (x 6)2

q = 1 (their p)2

35 = q2(i) 01234-2 -1 0 1 2 3 4 B1 B1 2 For x < 0, straight line joining(2, 0) and (0, 4) For x > 0, line joining (0,4) to (2, 2) and horizontal line joining (2,2) and (4,2) (ii)Translation1 unit right parallel to x axis B1 B1 2 4 Allow: 1 unit right,1 along the x axis,1 in x direction,allow vector notation e.g.||.|

\|01, 1 unit horizontally 3 x xxy8 3dd2 = When x = 2, 4dd =xy Gradient of normal to curve = 41 ( ) 2411 = + x y 0 6 4 = y x M1 A1 M1 A1 B1 ft M1 A1 7 7 Attempt to differentiate (one of 3x2, 8x) Correct derivative Substitutes x = 2 into their xydd Must be numerical = 1 their m Correct equation of straight line through (2, 1), any non-zero numerical gradient Correct equation in required form 4721Mark SchemeJanuary 2010 24(i) m = 4B11 May be embedded (ii) 24 62= pp2 = 4 p = 2 or p = 2 M1 A1 A1 3 ()6p2 = 24or 36p4 = 576(iii) 2 45 252 4 21+= + == nnn M1 M1 A1 3 7 Addition of indices as powers of 5 Equate powers of 5 or 25 5 228 13 08 ( 8) 4 1 134 3or 2= + = = =k xk kk k 3 4 = k 2) 3 4 ( + = x or 2) 3 4 ( = x 3 8 19 = x or12 4 19 M1* M1 dep A1 A1 M1 M1 A1 7 7 Use a substitution to obtain a quadratic (may be implied by squaring or rooting later)orfactorise into 2 brackets each containing x Correct method to solve resulting quadratic 8 12 4 3 or212or 42= == k kk

Recognise the needto square to obtain x Correct method for squaringa + b(3 or 4 term expansion) 6(i) xxy2dd=When x = 1,2dd=xy B1* B1 dep 2 (ii) 3 . 216 52= +aa 0 ) 1 )( 3 . 1 (0 3 . 1 3 . 22= = + a aa a a =1.3 M1 A1 M1 A1 4 uses 2 12 1y yx x correct expression correct method to solve a quadratic or correct factorisation and cancelling to get a + 1 = 2.3 1.3 only 4721Mark SchemeJanuary 2010 3 Alternative method: Equation of straight line through (1,6) with m = 2.3 found then a2 + 5 = 2.3a + c seen M1 withc = 3.7A1then as main scheme (iii)A value between 2 and 2.3 B1 1 7 2 < value < 2.3 (strict inequality signs) 7(i)(a) Fig 3 (b) Fig 1 (c) Fig 4 B1 B1 B1 3 (ii) 2) 3 ( x 2) 3 ( = x yM1 A1 2 5 Quadratic expression with correctx2term and correct y-intercept and/or roots for their unmatched diagram (e.g. negative quadratic with y-intercept of 9 or root of 3 for Fig 2) Completely correct equation for Fig 2 8(i)Centre(3, 2) 17170 4 4 ) 2 ( 9 ) 3 (22 2=== + +rry x B1 M1 A1 3 Correct method to find r2 Correct radius(ii) 0 4 ) 4 3 ( 4 6 ) 4 3 (2 2= + + + + x x x x 0 4 18 102= + x x0 ) 2 )( 1 5 ( = + x x51= xorx = 2 523= y ory = 2 M1* A1 A1 M1 dep A1 A1 6 9 substitute for x/y or attempt to get an equation in 1 variable only correct unsimplified expression obtain correct 3 term quadratic correct method to solve their quadratic SRIf A0 A0, one correct pair of values, spotted or from correct factorisationwwwB1 9(i) f / (x) = 21221 x x M1 A1 A1 3 Attempt to differentiate 2 xor2121 kxwww Fully correct expression 4721Mark SchemeJanuary 2010 4 (ii) f // (x) = 233412+ x x f // (4) = 81.41423 +=161 M1 A1 ft A1 M1 A1 5 8 Attempt to differentiate theirf / (x) One correctly differentiated term Fully correct expression www in either part of the question Substitution ofx = 4 into their f // (x) oe single fraction www in either part of the question 10 0 25 4 ) 30 (2= k k 0 100 9002= kk = 3 ork = 3 M1 M1 B1 B1 4 4 Attemptsac b 42involving k States their discriminant = 0 11(i)P = 2 + x + 3x + 2 + 5x + 5x = 14x + 4 M1 A1 2 Adds lengths of all 4 edges with attempt to use Pythagoras to find the missing length May be left unsimplified (ii) Area of rectangle = 23 6 ) 2 ( 3 x x x x + = +Area of triangle 26 ) 4 )( 3 (21x x x = = Total area =x x 6 92+M1 A1 2 Correct method splitting or formula for area of trapezium Convincing working leading to given expression AG (iii)14x + 4 39 25 x x 6 92+< 99 33 2 32 + x x< 0 ) 3 )( 11 3 ( + x x < 0 x |.|

\|< 311< 3 25 x < 3 B1 ft B1 B1 M1 B1 M1 A1 7 11 ft on their expression for P from (i) unless restarted in (iii).(Allow > ) o.e. (e.g. 1435 ) soiby subsequent working Allow Correct method to find critical values x < 3 identified root from linear < x < upper root from quadratic Fully correct including inequality signs or exact equivalent in words cwo Total72 4722Mark SchemeJanuary 2010 54722 Core Mathematics 2 1(i)2(1 cos2x) = 5cos x 1M1Use sin2x = 1 cos2x 2cos2x + 5cos x 3 = 0A.G.A12Show given equation correctly

(ii)(2cos x 1)(cos x + 3) = 0M1Recognise equation as quadratic in cos x and attempt recognisable method to solve cos x = M1Attempt to find x from root(s) of quadratic x = 60oA1Obtain 60o or /3 or 1.05 rad x = 300oA14Obtain 300o only (or 360o their x) and no extra in range SR answer only is B1 B1 6 2(i)( ) x x x x 4 3 d 4 62 = }+ cM1*Attempt integration (inc. in power for atleast one term) A1Obtain 3x2 4x (or unsimplified equiv), with or without + cy = 3x2 4x + c 5 = 12 8 + cM1dep*Use (2, 5) to find c c = 1 Hence y = 3x2 4x + 1A14Obtain y = 3x2 4x + 1 (ii)3p2 4p + 1 = 5M1*Equate their y (from integration attempt)to 5 3p2 4p 4 = 0M1dep*Attempt to solve three term quadratic (p 2) (3p + 2) = 0 p = -2/3A13Obtain p = -2/3 (allow any variable) fromcorrect working; condone p = 2 still present,but A0 if extra incorrect solution 7 3(i)(2 x)7 = 128 448x + 672x2 560x3M1Attempt (at least) two relevant terms product of binomial coeff, 2 and x (or expansion attempt that considers all 7brackets) A1Obtain 128 448x A1Obtain 672x2 A14Obtain 560x3 (ii)560 (1/4)3 = -35/4M1Attempt to use coeff of x3 from (i), withclear intention to cube 1/4 A12Obtain -35/4 (w6),(allow 35/4 from +560x3 in (i)) 6 4722Mark SchemeJanuary 2010 6 4(i)M1Attempt y-coords for at least 4 of thecorrect 5 x-coords only ) 7 log 5 . 6 log 2 6 log 2 + + M1Use correct trapezium rule, any h, to findarea between x = 3 and x = 5 M1Correct h (soi) for their y-values 1.55A14Obtain 1.55 (ii)B1Divide by 2, or equiv, at any stage toobtain 0.78 or 0.77, 1.55following their answer to (i) 0.78B12Explicitly use log a = log a ona single term 65M1Attempt subtraction (correct order) at anypoint = (3 9 + 30) (9 1/3 +10)M1Attempt integration inc. in power for atleast one term = 24 182/3A1Obtain ( 1/3x3 + 10x) or 11x and 1/3x3 + x = 51/3M1Obtain remaining term of form kx-1 ORA1Obtain 9x-1or any unsimplified equiv [ ] [ ]313313119 11 x x x x + +M1Use limits x = 1, 3 correct order &subtraction = [(33 + 3) (11 + 9)] [(9 + 3) (1/3 + 1)]A17Obtain 51/3, or exact equiv = 16 102/3 = 51/3 7 6(i)f(3) = 0 54 + 9a 3b + 15 = 0M1Attempt f(3) and equate to 0, or equivmethod 3a b = 13A1Obtain3a b = 13, or unsimplified equiv f(2) = 35 16 + 4a + 2b + 15 = 35M1Attempt f(2) and equate to 35, or equivmethod 2a + b = 2A1Obtain2a + b = 2, or unsimplified equiv Hence a = 3, b = 4M1Attempt to solve simultaneous eqns A16Obtain a = 3, b = 4 (ii)f(x) = (x + 3)(2x2 3x + 5)M1Attempt complete division by (x + 3), orequiv A1Obtain 2x2 3x + cor2x2 + bx + 5, fromcorrect f(x) ie quotient is(2x2 3x + 5)A13Obtain 2x2 3x + 5(state or imply asquotient) 9+ + +}5 . 5 log 2 5 (log d ) 2 ( log21215310x xx x x x d ) 2 ( log d ) 2 ( log5310 21531021+ = +} } ( ) ( ) { } [ ]31331 1312 210 9 1 9 11 x x x dx x x + = + } 4722Mark SchemeJanuary 2010 77(i)132 = 102 + 142 2 10 14 cos M1Attempt to use correct cosine rule in ABC cos = 0.4536 = 1.10A.G.A12Obtain 1.10 radians(allow 1.1 radians) SR B1 only for verification of 1.10, unlesscomplete method (ii)arc EF = 4 1.10 = 4.4B1State or imply EF = 4.4cm (allow 4 1.10) perimeter = 4.4 + 10 + 13 + 6M1Attempt perimeter of region - sum of arcand three sides with attempt to subtract 4 from at least one relevant side = 33.4 cmA13Obtain 33.4 cm (iii)area AEF = x 42 1.1M1Attempt use of() r2, with r = 4 and = 1.10 = 8.8A1Obtain 8.8 area ABC = 10 14 sin 1.1M1Attempt use of ()absin, sides consistentwith angle used = 62.4A1Obtain 62.4 or better (allow 62.38 or62.39) hence total area = 53.6 cm2A15Obtain total area as 53.6 cm2 10 8(i)u5 = 8 + 4 3M1Attempt a + (n 1)d or equiv inc list ofterms = 20 A.G.A12Obtain 20 (ii)un = 3n + 5 ie p = 3, q = 5B1Obtain correct expression, possunsimplified, eg 8 + 3(n 1) B12Obtain correct 3n + 5, or p = 3, q = 5stated (iii)arithmetic progressionB11Any mention of arithmetic (iv)( ) ( ) ( ) ( ) 1256 3 1 16 3 1 2 162 22= + + N NN NM1Attempt SN, using any correct formula (inc (3n + 5) ) M1Attempt S2N , using any correct formula, 26N + 12N2 13N 3N2 = 2512 with 2N consistent (inc (3n + 5) ) 9N2 + 13N 2512 = 0M1*Attempt subtraction (correct order) andequate to 1256 (9N + 157)(N 16) = 0M1dep*Attempt to solve quadratic in N N = 16A15Obtain N = 16 only, from correct working

OR: alternative method is to use n/2 (a + l) = 1256 M1Attempt given difference as single summation with N terms M1Attempt a = uN+1 M1Attempt l = u2N M1Equate to 1256 and attempt to solve quadratic A1Obtain N = 16 only, from correct working 10 4722Mark SchemeJanuary 2010 8 9(i)M1Reasonable graph in both quadrants A1Correct graph in both quadrants B13State or imply (0, 6) (ii)9x = 150 M1Introduce logarithms throughout, or equivwith log 9 x log 9 = log 150M1Use log ab = b log a and attempt correctmethod to find x x = 2.28A13Obtain x = 2.28 (iii)6 5x = 9xM1Form eqn in x and take logs throughout (any base) log 3 (6 5x) = log 3 9x M1Use log ab = b log a correctly on log 5x orlog 9x or legitimate combination of thesetwo log 3 6 + x log 3 5 = x log 3 9M1Use log ab = log a + log b correctly on log(6 5x) or log 6 log 3 3 + log 3 2 + x log 3 5 = 2xM1Use log 3 9 = 2 or equiv (need base 3throughoutthat line) x (2 log 3 5) = 1 + log 3 2 5 log 22 log 133+= x A.G.A15Obtain 5 log 22 log 133+= xconvincingly(inc base 3 throughout) 11 4723Mark SchemeJanuary 2010 94723 Core Mathematics 3 1Obtain integral of form 1(2 7) k x M1any constant k Obtain correct 15(2 7) x A1or equiv Include + cB13at least once; following any integral 3_____________________________________________________________________________________ 2 (i)Usesin 2 2sin cos = B1 Attempt value ofsinfromsin cos 5cos k = M1any constant k; or equiv Obtain 512A13or exact equiv; ignore subsequentwork------------------------------------------------- (ii)Use 1cosecsin=or 2 2cosec 1 cot = + B1or equiv Attempt to produce equation involvingcos onlyM1using 2 2sin 1 cos = or equiv Obtain 23cos 8cos 3 0 + = A1or equiv Attempt solution of 3-term quadratic equationM1using formula or factorisation orequiv Obtain13 as only final value ofcos A15or exact equiv; ignore subsequentwork 8_____________________________________________________________________________________ 3(i)Obtain or clearly imply60ln x B1 Obtain ( 60ln 20 60ln10 and hence) 60ln 2 B12with no error seen ------------------------------------------------- (ii)Attempt calculation of form 0 1 2( 4 ) k y y y + + M1any constant k; using y-value attempts Identify k as 53A1 Obtain 53(6 4 4 3) + +and hence 1253 or 41.7A13or equiv ------------------------------------------------- (iii)Equate answers to parts (i) and (ii)M1providedln 2 involved Obtain 125360ln 2 =and hence 2536A12AG;necessary detail requiredincluding clear use of an exact valuefrom (ii) 7 4(i)Attempt correct process for compositionM1numerical or algebraic Obtain (7 and hence) 0A12 ----------------------------------------------- -- (ii)Attempt to find x-interceptM1 Obtain7 x A12or equiv; condone use of < ------------------------------------------------ -- (iii)Attempt correct process for finding inverseM1 Obtain 3(2 ) 1 y or 3(2 ) 1 x A1 Obtain correct 3(2 ) 1 x A13or equiv in terms of x ------------------------------------------------- (iv)Refer to reflection iny x = B11or clear equiv 8____________________________________________________________________________________ 4723Mark SchemeJanuary 2010 10 5 (i)Obtain derivative of form 2 7( 1) kx x + M1any constant k Obtain 2 716 ( 1) x x + A1or equiv Equate first derivative to 0 and confirm0 x =orsubstitute0 x =and verify first derivative zeroM1AG; allow for deriv of form 2 7( 1) kx x +Refer, in some way,to 21 0 x + = having no rootA14or equiv --------------------------------------------- ----- (ii)Attempt use of product rule*M1obtaining + form Obtain 2 716( 1) x ++ A1follow their 2 7( 1) kx x +Obtain + 2 2 6224 ( 1) x x + A1follow their 2 7( 1) kx x + ;or unsimplified equiv Substitute 0 in attempt at second derivativeM1dep *M Obtain 16A15from second derivative which iscorrect at some point 9_____________________________________________________________________________________ 6Integrate 3ex to obtain 3 13ex or 12ex to obtain 122ex B1or both Obtain indefinite integral of form 1231 2e exxm m+ M1any constants 1mand 2mObtain correct 123 13e 2( 2)exxk k A1or equiv Obtain 3ln 4e 64 =or 12ln 412e= B1or both Apply limits and equate to 185M1including substitution of lower limit Obtain 64 13 3( 2) 2( 2) 185 k k k k + = A1or equiv Obtain 172A17or equiv 7 _____________________________________________________________________________________ 7(a)Either:State or imply either d2dArr =ord250dAt= B1or both Attempt manipulation of derivatives to find ddrtM1using multiplication / division Obtain correct 2502 r A1or equiv Obtain1.6A14or equiv; allow greater accuracy

Or:Attempt to express r in terms of tM1using250 A t = Obtain 250tr= A1or equiv Differentiate 12ktto produce 12 12ktM1any constant k Substitute7.6 t =to obtain 1.6A1 (4) allow greater accuracy--------------------------------------------- -----4723Mark SchemeJanuary 2010 11(b)State d150 edktmkt= B1 Equate to( )3 and attempt value for tM1using valid process; condone signconfusion Obtain 1 1ln( )50 k k or1ln(50 ) kkor ln50 ln kk+A13or equiv but with correct treatment ofsigns 7_____________________________________________________________________________________

8(i)State scale factor is2 B1allow 1.4 State translation is in negative x-direction B1 or clear equiv by 32 unitsB13 ------------------------------------------------- (ii)Draw (more or less) correct sketch of2 3 y x = + B1starting at point on negative x-axis Draw (more or less) correct sketch of 3Nyx= B1showing both branches Indicate one point of intersectionB13with both sketches correct [SC: if neither sketch complete or correct but diagram correct for both in first quadrantB1] ---------------------------------------------- ---- (iii)(a)Substitute 1.9037 into 1 13 6(2 3) x N x= + M1or into equation 32 3Nxx+ = ; or equiv Obtain 18 or value rounding to 18A12with no error seen ---------------------------------------------- (b)State or imply 1 13 62.6282 (2 2.6022 3) N= + B1 Attempt solution for NM1using correct process Obtain 52A13concluding with integer value 11_____________________________________________________________________________________ 9(i)Identifytan55 astan(45 10 ) + B1or equiv Use correct angle sum formula fortan( ) A B + M1or equiv Obtain11pp+A13withtan 45 replaced by 1 ------------------------------------------------ - (ii)Either:Attempt use of identity fortan 2A *M1linking 10and5 Obtain 221tpt=A1 Attempt solution for t of quadratic equationM1dep *M Obtain 21 1 pp + +A14or equiv; and no second expression Or (1):Attempt expansion oftan(60 55 ) *M1 Obtain 111131 3pppp+++ A1follow their answer from (i) Attempt simplification to remove denominatorsM1dep *M Obtain 3(1 ) (1 )1 3(1 )p pp p + + +A1 (4)or equiv 4723Mark SchemeJanuary 2010 12Or (2):State or implytan15 2 3 = B1 Attempt expansion oftan(15 10 ) M1with exact attempt fortan15 Obtain 2 31 (2 3)pp + A2 (4) Or (3):State or imply 3 13 1tan15+ = B1or exact equiv Attempt expansion oftan(15 10 ) M1with exact attempt fortan15Obtain 3 1 33 1 3p pp p + + A2 (4)or equiv Or (4):Attempt expansion oftan(10 5 ) *M1 Obtain 1=+p ttpt A1 Attempt solution for t of quadratic equationM1dep *M Obtain 22 4 42pp + +A1 (4)or equiv; and no secondexpression------------------------------------------------- (iii)Attempt expansion of both sidesM1 Obtain3sin cos10 3cos sin10 + =7cos cos10 7sin sin10 + A1or equiv Attempt division throughout bycos cos10 M1or bycos(orcos10 ) only Obtain3 3 7 7 + = + t p pt A1or equiv Obtain3 77 3ppA15or equiv 12 _____________________________________________________________________________________ 4724Mark SchemeJanuary 2010 134724 Core Mathematics 4 1Long division methodCorrect leading term2xin quotientB1Evidence of correct div process M1Sufficient to convince (Quotient = ) 4 62 + x x A1 (Remainder= ) 9 11 + xA1 Identity method ( ) R x x Q x x x x + + + = + + + + 2 5 1 3 28 112 2 3 4M1 c bx x c bx ax Q + + + + =2 2or ; e dx R + =&ops 3 M1N.B. =1 a 1 of the 3 ops 9 11 4 6 1 = = = = = e , d , c , b , a(for all 5)A2S.R. B1 for 3 of these 4 2(i)Find at least 2 of( ) BA AB or, ( ) CB BC or, ( ) CA AC orM1irrespect of label; any notation Use correct method to find scal prod of any 2 vectorsM1oruse corr meth for modulus Use BC AB. = 0 or 0.=BC ABBC ABM1or use 2 2 2AC BC AB = +Obtain1 = p(dep 3 @ M1)A1 4 (ii)Use equal ratios of appropriate vectorsM1or scalar product method Obtain8 = p A1 2 6 3Use = x 2 cos b x a +2cos / x x2 2sin cos / x2sin 2 1 *M1Obtain x2sec + dep*M1 using reasonable Pythagattempt }+ = + x x x xtand sec2 A1( or may be 0 here/prev line) Obtain correct result x xtan2 A1no follow-through 1 361+ ISWA1exact answer required 5 4Attempt to connect du and dt or find tudd orutddM1not du = dt but no accuracyttu d1d =or t tu 1dd=or u tud e d2 = or2edd=uutA1 Indef int ( )} uud12A1no t or dt in evidence u1 = A1 Attempt to change limits if working with( ) u f M1or re-subst& use 1 and e 61 ISWA1ln e must be changed to 1, ln 1 to 0 6 4724Mark SchemeJanuary 2010 145(i)( )311 x += 1 +... x +31B1 291x B1 22182x acceptable (ii) (a)( )3116 8 x +=318 ( )312 1 x + B1 not( )31312116 x + ( )312 1 x += their (i) expansion with 2x replacing xM1not dep on prev B1 =294321 x x ++ A1 2188x acceptable Required expansion = 2 (expansion just found) B14accept equiv fractions N.B. If not based on part (i), awardM1 for( ) ( )2 323116 82 116 8318353231x..x . + + , allowing216xfor ( )216x , with3 @ A1 for 2...+29834x ... x , accepting equivalent fractions & ISW (ii) (b)21

21< < x or 21< x B1 1no equality 7 6 txtyxydddddd= M1quoted/implied t tx999dd = ISWB1 32233ddtttty =ISWB1Stating/implying 319332=ttt0 9 or93 2= = t t t A1WWW, totally correct at this stage t = 3 as final ans with clear log indication ofA2S.R. A1 if3 = t or3 = tinvalidity of3 ; ignore (non) mention of t = 0or (t = 3 & wrong/no indication) 6 7Treat ( ) y xx2dd as a productM1 ( )xyy yx dd3dd2 3= B1 xyy xyxyx xdd3 4dd2 32 2 2= + + A1Ignore=xydd if not used Subst (2, 1) and solve for xyddor vice-versaM1 xydd = 4 WWWA1 grad normal =xyddtheir 1A1stated or used Find eqn of line, through (2, 1), with either gradientM1using their xydd orxyddtheir 10 2 4 = + y x A1AEF with integral coefficients 84724Mark SchemeJanuary 2010 15 8(i) xxcose sin B1 1 (ii) } =x xx xcos cose d e sinB1 anywhere in part (ii) Parts with splitxx v , x ucose sind cos = =M1result f(x) +/ ( )} x x d g IndefInteg, 1st stage} x x xx xd e sine coscos cosA1accept } x xxd sin. ecos Second stage = x xxcos cose e cos + *A1 Final answer =1 dep*A26 7 9(i)P is |||.|

\|=|||.|

\| +|||.|

\|304211113B1 direction vector of is |||.|

\|211 and ofOPis their PB1 Use cos=b ab a. for |||.|

\|211 and their OPM1 3 35. =or better (0.615rad)A1 4 (ii)Use 02 113 211=|||.|

\|++|||.|

\|ttt. M1 ( ) ( ) ( ) 0 2 1 2 1 1 3 1 = + + + t t t A1 32t = A1 Subst. into |||.|

\|++t 2 1t 1t 3 to produce|||.|

\|313537 ISWA14 (iii)Use 2 2 2z y x + + where|||.|

\|zyx is part (ii) answerM1 Obtain975AEF, 2.89 or better (2.8867513.)A1 2 10 4724Mark SchemeJanuary 2010 1610 (i) x 331 .x 631B1+12 (ii)(a) Separate variables( )( )} }= t k xx xd d6 31M1or invert both sides Style: For the M1, dx & dt must appear on correct sides or there must be }sign on both sides Change( )( ) x x 6 31 into partial fractions from (i)B1 ( )} |.|

\| =x A xxA3 lnA1ord3 B1 or ( )}|.|

\| =xBB xxB- 6 ln1ord6 ( ) ( ) kt x x = + 6 ln313 ln31(+ c)A1f.t. from wrong multiples in (i) Subst (x = 0, t = 0) & (x = 1, t = 1) into eqn with cM1and solve for k Useln a + ln b = ln ab or ln a ln b = ln ba M1 Obtain k = 45ln 31with sufficient working & WWW A17AG (b)Substitute k = 45ln 31,2 = t& their value of c *M1 Reduce to an eqn of form =xx36dep*M1 whereis a const Obtainx =1727or1.6 or better (1.5882353) A2 4 S.R. A1 for 16 3=x 13 4725Mark SchemeJanuary 2010 174725 Further Pure Mathematics 1 1(i) ||.|

\| 0 32 4 aB1Two elements correct B12Remaining elements correct --------------------------------------------------------------------------------------------------------------------------------- (ii)4a 6 B1Correct determinant M1Equate det A to 0 and solve 23= a A13Obtain correct answer a. e. f. 5 2(i)1 3 32 3 + u u u B1Correct unsimplified expansion of 3) 1 ( uM1Substitute for x 0 8 9 6 22 3= + u u u A13Obtain correct equation ---------------------------------------------------------------------------------------------------------------------------------------- (ii)M1Use ad) (of new equation 4A1ft2Obtain correct answer from theirequation 5 3x iy B1Conjugate knownM1Equate real and imaginary parts 12 2 = + y x 9 2 = + y xA1Obtain both equations, OK with factorof i M1Solve pair of equations z= 2 + 5iA15Obtain correct answer as a complexnumber S.C. Solvingz + 2iz = 12 + 9i can getmax4/5, not first B1 5 4M1Express as sum of three series M1Use standard results ) 1 ( ) 1 2 )( 1 (61) 1 (412 2+ + + + n n n n n n n A1Obtain correct unsimplified answer M1Attempt to factorise A1Obtain at least factor of) 1 ( + n n) 7 3 )( 2 )( 1 (121 + + n n n n A16Obtain fully factorised correct answer 6 4725Mark SchemeJanuary 2010 185(i)B1Rotation 90o (about origin) B12Anticlockwise ---------------------------------------------------------------------------------------------------------------------------------------- (ii)EitherM1Show image of unit square after reflection in y = x ||.|

\|1 00 1A1Deduce reflection in x-axis B1ft Each column correct B1ft4ft for matrix of their transformation OrM1Post multiply by correct reflectionmatrix A1Obtain correct answer B1B1 State reflection, in x-axis S.C. If pre-multiplication, M0 but B1 B1 Available for correct description oftheir matrix 6 6(i)B1State or use 5 + i as a root M1Use = 6

x = 2A13Obtain correct answer ---------------------------------------------------------------------------------------------------------------------------------------- (ii)EitherM1Use = p8 = p A1ftObtain correct answer, from their root M1Use = q52 = q A1ft4Obtain correct answer, from their root OrM1Attempt to find quadratic factor M1Attempt to expand quadratic and linear A1A1 Obtain correct answers OrM1Substitute (5 i)into equationM1Equate real and imaginary parts A1Obtain correct answer for p A1ftObtain correct answer for q, ft their p 7 7(i)B11Obtain given answer correctly ------------------------------------------------------------------------------------------------------------------------------------- (ii)M1Express at least 1st two and last term using (i) A1All terms correct M1Show that correct terms cancel 2) 1 (11+nA14Obtain correct answer, in terms of n------------------------------------------------------------------------------------------------------------------------------------- (iii) 41B1Sum to infinityseen or implied B12Obtain correct answer S.C. - scores B17 4725Mark SchemeJanuary 2010 19|||.|

\|35 47 51aa 8(i)M1Attempt to equate real and imaginaryparts of (x + iy)2 & 5 12i52 2= y xand 6 = xy A1Obtain both results, a.e.f M1Obtain quadratic in x2 or y2 M1Solve to obtain3 ) ( = xor2 ) ( = y (3 2i) A15Obtain correct answers as complex nos --------------------------------------------------------------------------------------------------------------------------------------(ii)B1ftCircle with centre at their square root B1Circle passing through origin B1ft2nd circle centre correct relative to 1st

B14Circle passing through origin 9 9(i)M1Show correct expansion process for3 3 or multiply adjoint by A M1Correct evaluation of any 2 2 at anystage det A =6 6 = a A1Obtain correct answer A-1 = |||.|

\| + 6 3 32 1 2 14 1 1 31aa a M1Show correct process for adjoint entries A1Obtain at least 4 correct entries inadjoint B1Divide by their determinant A17Obtain completely correct answer ----------------------------------------------------------------------------------------------------------------------------------------- (ii)M1Attempt product of form A-1C or eliminate to get 2 equations and solve A1A1A1Obtain correct answer ft all 3 4S.C. if det now omitted, allow max A2 ft 11 10 (i) B1Correct M2 seen M2 = ||.|

\|1 04 1M3 =||.|

\|1 06 1 M1Convincing attempt at matrixmultiplication for M3 A13Obtain correct answer ----------------------------------------------------------------------------------------------------------------------------------------- (ii)Mn = ||.|

\|1 02 1 nB1ft1State correct form, consistent with (i) -------------------------------------------------------------------------------------------------------------------------------------- 4725Mark SchemeJanuary 2010 2010 (iii)M1 Correct attempt to multiply M & Mk

or v.v. A1Obtain element 2( k + 1 ) A1Clear statement of induction step, from correct working B14Clear statement of induction conclusion, following their working ----------------------------------------------------------------------------------------------------------------------------------------(iv)B1Shear DB1x-axis invariant DB13e.g. ( 1, 1 )( 21, 1 ) or equivalentusing scale factor or angles 11 4726Mark SchemeJanuary 2010 214726 Further Pure Mathematics 2 1(i)Get 0.876096, 0.876496, 0.876642B1For any one correct or from wrong answer; radians only B1All correct (ii)Subtract correctly (0.00023(0), 0.000084)Divide their errors as e4/e3 onlyGet 0.365(21)B1 M1 A1 On their answers May be implied Cao 2(i)Find f(x) = 1/(1+(1+x)2) Get f(0) = and f(0) = Attempt f(x) Correctly get f(0) = M1 A1 M1 A1 Quoted or derived; may be simplified or left as sec2y dy/dx = 1 On their f(0); allow f(0)=0.785 but not 45 Reasonable attempt at chain/quotient rule or implicit differentiation A.G. (ii)Attempt Maclaurin as af(0)+bf(0)+cf(0) Get + x x2

M1 A1 Using their f(0) and f(0) Cao; allow 0.785 3(i)Attempt gradient as f(x1)/(x2 x1) Equate to gradient of curve at x1 Clearly arrive at A.G. SC Attempt equation of tangentPut (x2 , 0) into their equationClearly arrive at A.G.M1 M1 A1 M1 M1 A1 Allow reasonable y-step/x-step Allow Beware confusing use of As y f(x1) = f(x1)(x x1) (ii)Diagram showing at least one more tangentDescription of tangent meeting x-axis, used as next starting value B1 B1 (iii)Reasonable attempt at N-R Get 1.60M1 A1 Clear attempt at differentiation Or answer which rounds 4(i) State r =1 and = 0. B1 B1 May be seen or implied Correct shape, decreasing r (not through O)

(ii)Use r2 d with r = e2seen or impliedIntegrate correctly as 1/8e-4 Use limits in correct orderUse r12 = e-4 etc.Clearly get k=1/8

M1 A1 M1 M1 A1 Allow e4 d In their answer May be implied 4726Mark SchemeJanuary 2010 22 5(i)Use correct definitions of cosh and sinh Attempt to square and subtract Clearly get A.G.Show division by cosh2 B1 M1 A1 B1 On their definitions Or clear use of first result (ii)Rewrite as quadratic in sech andattempt to solve Eliminate values outside 0 < sech 1 Get x = ln(2+3)Get x = ln(2+3) or ln(23) M1 B1 A1 A1 Or quadratic in cosh Or eliminate values outside cosh 1 (allow positive) 6(i)Attempt at correct form of P.F. Rewrite as 4= A(1 + x)(1 + x2) + B(1 x)(1 + x2) + (Cx + D)(1 x)(1 + x) Use values of x/equate coefficients Get A = 1, B = 1Get C = 0, D = 2 M1 M1 M1 A1 A1 Allow Cx/(x2+1) here; not C = 0 From their P.F. cwo SC Use of cover-up rule for A,B M1 If both correctA1 cwo (ii)Get Aln(1 + x) Bln(1 x) Get Dtan-1x Use limits in their integrated expressions Clearly get A.G. M1 B1 M1 A1 Or quote from List of Formulae 7(i)LHS = sum of areas of rectangles, area = 1x y-value from x = 1 to x = n RHS = Area under curve from x = 0 to n B1 B1 (ii)Diagram showing areas requiredUse sum of areas of rectanglesExplain/show area inequality with limits in integral clearly specifiedB1 B1 B1 (iii)Attempt integral as kx4/3

Limits gives 348(.1) and 352(.0) Get 350 M1 A1 A1 Allow one correct From two correct values only 4726Mark SchemeJanuary 2010 23 8(i) Get x = 1, y = 0 B1,B1 (ii)Rewrite as quadratic in x Use b2 4ac 0 for all real x Get correct inequality State use of k>0 to A.G.M1 M1 A1 A1 (x2y x(2y + k) + y = 0) Allow >, = here 4ky + k2 0 SC Use differentiation (parts (ii) and (iii)) Attempt prod/quotient ruleM1 Solve = 0 for x = 1A1 Use x =1 only (reject x=1), y = k A1 Fully justify minimumB1 Attempt to justify for all xM1 Clearly get A.G.A1 (iii)Replace y = k in quadratic in x Get x = 1 only x (1, k) x = 1 M1 A1 B1 B1 Through origin with minimum at (1, k) seen or given in the answer Correct shape (asymptotes and approaches) SC (Start again) Differentiate and solve dy/dx = 0 for at least one x-value, independent of kM1 Get x = 1 only A1 9(i)Rewrite tanh y as (ey e-y)/(ey + e-y) Attempt to write as quadratic in e2y

Clearly get A.G. B1 M1 A1 Or equivalent (ii)(a)Attempt to diff. and solve = 0 Get tanh x = b/a Use (1) < tanh x < 1 to show b < aM1 A1 B1 SC Use exponentials M1 Get e2x = (a + b)/(a b) A1 Use e2x > 0 to show b < aB1 SC Write x = tanh-1(b/a) M1 = ln((1 + b/a)/(1 b/a)) A1 Use ( ) > 0to show b < a B1 (b) Get tanh x =1/a from part (ii)(a) Replace as ln from their answer Get x = ln ((a + 1)/(a 1)) Use eln((a+1)/(a-1)) = ((a + 1)/(a 1)) Clearly get A.G. Test for minimum correctly

B1 M1 A1 M1 A1 B1 At least once SC Use of y = coshx(a tanhx) andcoshx = 1/sechx = 1/(1 tanh2x) 4727Mark SchemeJanuary 2010 244727 Further Pure Mathematics 3 1METHOD 1 line segment between 1land 2l=[4, 3, 9] B1For correct vector [1, 1, 2] [2, 3, 4] ( )[ 2, 0, 1] = = n M1* A1 For finding vector product of direction vectors distance = ( )2 2 2[4, 3, 9] [ 2, 0, 1]1752 0 1 =| |+ + |\ .. M1 (*dep) For using numerator of distance formula 0 , so skewA15For correct scalar product and correct conclusion METHOD 2lines would intersect where 1 3 22 1 34 2 5 4s ts ts t+ = + = +` + = +) 2 43 32 4 9s ts ts t = + = = B1 M1* A1 M1 (*dep) For correct parametric form for either line For 3 equations using 2 different parameters For attempting to solve to show (in)consistency contradiction, so skewA1For correct conclusion 5 2(i) ( )( )5 5 a b c d + +( ) 5 5 ac bd bc ad = + + + H M1 A12For using product of 2 distinct elements For correct expression (ii) (e = )1 OR 1 0 5 +B11For correct identity (iii) EITHER 1 55 5a ba b a b+ OR ( )( )5 5 1 a b c d + + = 5 10ac bdbc ad+ = + = inverse =2 2 2 255 5a ba b a b M1 A12For correct inverse as ( )15 a b+and multiplying top and bottom by 5 a b OR for using definition and equating parts For correct inverse.Allow as a single fraction (iv) 5 is primeOR 5 B11 For a correct property (or equivalent) 63 Integrating factor = 2d2e exx}=B1For correct IF ( )2de edx xyx= M1For( )3d. their IF e . their IFdxyx= 2e e ( )x xy c= + A1For correct integration both sides (0, 1) c = 2M1 A1 For substituting(0, 1) into their GS and solving for c For correct cf.t. from their GS 3 2e 2ex xy = + A16For correct solution 6 4(i)(z = )2, 2, 2i , 2i M1 A12 For at least 2 roots of the form{ } 1, i k AEF For correct values 4727Mark SchemeJanuary 2010 25 (ii) 1ww =2, 2, 2i , 2i M1For 1ww = any one solution from (i) 1zwz=+M1 For attempting to solve for w, using any solution or in general 23, 2 w =B1 A1 For any one of the 4 solutions For both real solutions 4 25 5i w = A15 For both complex solutions SR Allow B1 and one A1 from 2 k 75(i) 2 23 33, 0, 6 k(= AB , 3, 1, 0 k(= BC ,1 23 33, 1, 6 k(= CAB1 B1 For any one edge vector of ABC For any other edge vector of ABC 2 2 213 3 36, 18, 3 k(= n= 1221, 3, 2 k( M1 M1 For attempting to find vector product of any two edges For substituting A, B or C intor . n substitute A, B or C 1 22 33 2 3 x y z + + =A15 For correct equationAG SR For verification only allow M1, then A1 for 2 points and A1 for the third point (ii)Symmetry B1*For quoting symmetry or reflection in plane OABorOxzor0 y = B1 (*dep)2 For correct plane Allow in y coordinates or in y axis SRFor symmetry implied by reference to opposite signs in y coordinates of C and D, award B1 only (iii) 1 12 21 12 21, 3, 2 1, 3, 2cos1 3 1 3 ( ( =+ + + +. M1 A1 For using scalar product of normal vectors For correct scalar product 1 32 29 92 21 313 += = =M1 A14For product of both moduli in denominator For correct answer.Allow 13 116(i)( )216 0 m + = 4i m = M1For attempt to solve correct auxiliary equation(may be implied by correct CF) CF =cos 4 sin 4 A x B x + A12 For correct CF (AEtrig but not4i 4ie ex xA B+ only) (ii)dsin 4 4 cos 4dyp x px xx = +M1 A1 For differentiating PI twice, using product rule For correct ddyx 22ddyx= 8 cos 4 16 sin 4 p x px x A1For unsimplified 22ddyx.f.t. fromddyx 8 cos 4 8cos 4 p x x = M1For substituting into DE 1 p = A1For correct p ( ) cos 4 sin 4 sin 4 y A x B x x x = + + B16 For using GS = CF + PI, with 2 arbitrary constants in CF and none in PI 4727Mark SchemeJanuary 2010 26 (iii)(0, 2) A = 2B1For correct A.f.t. from their GS d4 sin 4 4 cos 4 sin 4 4 cos 4dyA x B x x x xx = + + + M1For differentiating their GS d0, 0 0dyx Bx= = = M1For substituting values for x and ddyx

to find B 2cos 4 sin 4 y x x x = + A14 For stating correct solution CAO includingy = 127(i) cos 6 0 = 126 k = M1 For multiples of 12seen or implied { }1121, 3, 5, 7, 9, 11 =A1 A13 A1 for any 3 correct A1 for the rest, and no extras in 0 < R>0 From correct R, F values 4728Mark SchemeJanuary 2010 29 4 ii 0.5g - T = 0.5a T - 0.4g = 0.4a a = 1.09 ms-2 T = 4.36 N M1 A1 B1 B1 [4] N2L for either particle no resolving, at least 1 unknown Formula round the pulley, M0A0. But award M1 for T-0.4g = 0.4 1.09 etc later Both equations correct 5 i 11 = 3 + 20a(a = 0.4) 8 = 3 + (11-3)t/20 t = 12.5 M1 M1 A1 [3] Uses v = u + at, no zero terms Their a>0.t/20 = (8-3)/(11-3) is M1M1 ii s(A,20) = 8 20 (=160) s(B,20) = (3 +11) 20/2 =3 20+0.4 202/2 (=140) 8T = (3+11) 20/2 + 11 (T-20) or (160 140) = 11t 8t T = 26 2/3B1 B1 M1 A1 A1 [5] Or s(A) = 8T or as stage of s(B)=(3+11)20/2 + 11(T-20) 3 part equation balancing distances Accept 26.6 or 26.7 iii

B1 B1 B1 [3] Linear rising graph (for A) starting at B's start Non-linear rising graph for B below A's initially. Accept 2 straight lines as non-linear.Single valued graphs graphs intersect and continue 6 i a = 2 0.006t - 0.18 a = 0.012t - 0.18 M1 A1 [2] Differentiates v (not v/t) Award for unsimplified form, accept +c, not +k ii 0.012t - 0.18 = 0 t = 15 0.006 152 - 0.18 15 + k = 0.65 k = 2AG M1* A1 D*M1 A1 A1 [5] Sets a = 0, and solves for t Substitutes t(v(min)) in v(t) iiis = 0.006t3/3 - 0.18t2/2+ 2t (+c) (s = 0.002t3 - 0.09t2 + 2t (+c)) t = 0, s = 0 hence c = 0 L = 0.002 28.43 - 0.09 28.42 + 2 28.4 L = 30.0 m M1A1 B1 M1 A1 [5] Integrates v (not multiplies by t). Award if +c omitted, accept kt Explicit, not implied (or uses limits 0, 28.4) Substitutes 28.4 or 14.2 in s(t), (and k=2) Accept a r t 30(.0), accept +c 4728Mark SchemeJanuary 2010 30 7 i (Fr =) 0.15 600gcos10 (Wt cmpt =) 600gsin10 600 0.11 = T - 0.15 600gcos10 - 600gsin10 (66 = T - 868.6 1021) T = 1960 N B1 B1 M1 A1 A1 [5] Implied by Fr = 0.15600gcos10 (=868.6..) N2L. T with at least 1 resolved forces and 600 0.11 1955.6.. ii a b a(up) = +/-(600gsin10+.15600gcos10)/600 a(up) = +/-3.15 ms-2AG UP v2 = 2 0.11 10 v = 1.48 when cable breaks t = 1.48/3.149 (t = 0.471 time for log to come to rest) s = 1.482/(2 3.149) s = 0.349 distance for log to come to rest DOWN a(down) = (600gsin10-0.15600gcos10)/600 10+0.349= 0.254t2/2 t = 9.025 T = (9.025 + 0.471) = 9.5 s M1 A1 [2] M1 A1 M1

M1 A1 B1 M1 A1 A1 [9] 2 resolved forces and 600a or unit mass Disregard sign, accept 3.149 Correct, need not be accurate Or 1.48 = 0 + 3.15t Correct, need not be accurate = 0.254 Needs a< 3.15, s>10.Or V2=20.254 (10+0.349) [ V= 2.29..], V=0.254t Correct, need not be accurate Accept 9.49 4729Mark SchemeJanuary 2010 314729 Mechanics 2 1759.840 B1Average Speed = 40120 (759.840)120M1(759.8)(Average speed) 245 W A1[3]32(i)v2 =29.83or29.81.8M1Kinematics or energy v1 =g 6 or8 . 58 or3057or 7.67 A1Speed of impact () v2 =g 6 . 3 or28 . 35 or2521or5.94 A1Speed of rebound ()I =) 67 . 7 94 . 5 ( 2 . 0 + M1 2.72A1ft[5]+ve, ft onv1 and v2 (ii)e = 5.94/7.67M1 0.775 or 515 A1ft[2]Allow 0.774, ft onv1 and v2

73(i) = 0.2(from vertex) or 0.8 or 0.1B1com of conical shell M10.5 = 0.2 + 0.30.65 A1 = 0.47 A1[4]AG (ii)s = 0.5B1slant height, may be implied M1 Tsin80 0.5 = 0.47 0.5 9.8 A1 T = 4.68 N A1[4]84 (i)D 400 = 700 0.5M13 terms D = 750 N A1[2] (ii)P = 750 12M1 9 000 Wor9 kW A1ft[2] (iii)P/35 = 400M1 14 000 W or 14 kW A1[2] (iv)D = 14000/12B1ftMay be implied M13 terms 3500/3 = 400 + 7009.8sin A1Their P/12 = 6.42 A1[4]104729Mark SchemeJanuary 2010 32 51612= 2x+3yM1 4=2x+3yA1aef .2(8)2 +.3(4)2 or .2x2 +..3y2 or .2(82 -x2) or.3(42 -y2) B1 .2(8)2 +.3(4)2 - .2x2 -...3y2 = 81M1 2x2 +3y2 = 14A1aef Attempt to eliminate x or y from a linear and a quadratic equation M1 15y2 - 24y - 12 = 0 or 10x2 - 16x - 26 = 0A1aef Attempt to solve a three term quadraticM1 x = -1 (or x = 2.6)A1 y = 2 (or y = -2/5)A1 x = -1 and y = 2 onlyA1 speeds 1, 2 away from each other A1[12]126(i)302 = V12 sin2 1 29.8250M1 V12 sin2 1 = 5800 AEFA1 m V12 = m 502 + m9.8250 V1 cos 1 = 40B1 V1 = 86.0A1AG 1 = 62.3 A1[5]AG (ii)0 = 5800 tp 4.9tp2 M130 = V1 sin 1 9.8t tp = 15.5A1 t = 4.71 -5800 = 30 9.8tq M1 tq = 10.8 A1[4] (iii)R = 4015.5M1 R = 621A1(620, 622) V2 cos 2 10.8 = 621B1V2 cos 2 = 57.4 0 = V2

sin 2 10.8 4.9 10.82M1 V2 sin 2 = 53.1 or 53.0 A1(52.9,53.1) Method to find a value of V2 or 2M1 2 = 42.8A142.6 to 42.9 V2 = 78.2 m s-1 or 78.1 m s-1 A1[8]or 78.117 7(i)cos = 3/5 or sin = 4/5 or tan= 4/3 or =53.1 B1 = angle to vertical Rcos = 0.29.8M1 R = 3.27 N or 49/15 A1[3] (ii)r = 4B1 M1 Rsin = 0.242 A1 = 1.81 rad s-1 A1[4] 4729Mark SchemeJanuary 2010 33 (iii) = 26.6 or 51sin = or 52cos = or tan = 0.5 B1 = angle to horizontal T = 0.98 or 0.1gB1 Ncos = Tsin + 0.29.8M1Vertically, 3 terms N3/5 = 0.438 + 1.96A1 N = 4.00A1may be implied Nsin + Tcos = 0.242 M1Horizontally, 3 terms 44/5 + 0.98cos 26.6 = 0.82 A1 = 2.26 rad s-1 A1[8]15 4730Mark SchemeJanuary 2010 344730 Mechanics 3 1 0.4(3cos60o 4) = -I cos (= -1) 0.4(3sin60o) = Isin (= 1.03920) M1 A1 A1 For using I = mv in one direction SR: Allow B1 (max 1/3) for3cos60o 4 = -I cosand 3sin60o = Isin[tan= -1.5 3 /(1.5 4); I2 = 0.42[(1.5 4)2 + (1.5 3 )2]] M1 For eliminating I or (allow following SR case) = 46.1 or I = 1.44A1Allow for (only) following SR case. M1For substituting for or for I (allow following SR case) I = 1.44 or = 46.1A1ft [7] ft incorrect or I; allow for (only) following SR case. Alternatively I2 = 1.22 + 1.62 21.21.6cos60oor V2 = 32 + 42 234cos60o

I = 1.44 ) 08 . 2 ( 1360 sin) 2 . 1 ( 3sinoror= or = = 120) 08 . 2 ( 1360 sin) 6 . 1 ( 4sinandoror M1 A1 M1 A1 M1 A1ft For use of cosine rule For correct use of factor 0.4 (= m) For use of sine rule must be angle opposite 1.6; ( = 73.9) ft value of I or V = 46.1A1 [7] 2 2a + 3b = 24 b a = 0.64 [2(b - 2.4) + 3b = 8] b = 2.56 v = 2.56 M1 A1 M1 A1 M1 A1 B1ft [7] For using the principle of conservation of momentum For using NEL For eliminating a ft v = b 3(i) 2W(a cos45o) = T(2a) W = 2 T M1 A1 A1 [3] For using mmt of 2W = mmt of T AG (ii)Components (H, V) of force on BC at B areH = -T/ 2 and V = T/ 2 -2W W(a cos) + H(2a sin) = V(2a cos) [W cos - T 2 sin = T 2 cos -4Wcos] T 2 sin = (5W - T 2 ) cos tan = 4 B1 M1 A1 M1 A1ft A1 [6] For taking moments about C for BC For substituting for H and V and reducing equation to the formX sin = Y cos 4730Mark SchemeJanuary 2010 35Alternatively for part (ii) anticlockwise mmt = W(a cos) +2W(2a cos + a cos45o) = T[2a cos( 45o) + 2a][5W cos + 2 W = T( 2 cos +2 sin) + 2] T 2 sin = (5W - T 2 ) cos tan = 4 M1 A1 A1 M1 A1ft A1 [6] For taking moments about C for the whole For reducing equation to the form X sin = Y cos 4(i)[-0.2(v + v2) = 0.2a] [v dv/dx = -(v + v2) [1/(1 + v)] dv/dx = -1 M1 M1 A1 [3] For using Newtons second law For using a = v dv/dx AG (ii) ln (1 + v) = -x (+ C) ln(1+ v) = -x + ln3 [(1 + dx/dt)/3 = e-x dx/dt = 3e-x -1 ex dx/dt = 3 ex] [-ex/(3 ex)] dx/dt = -1 M1 A1 A1 M1 A1 [5] For integrating For transposing for v and using v = dx/dt AG (iii)[ln(3 ex) = -t + ln2] ln(3 ex) = -t + ln2 Value of t is 1.96 (or ln{2 (3 e)} M1 A1 A1 [3] For integrating and using x(0) = 0 5(i) Loss of EE = 120(0.52 0.32)/(21.6)and gain in PE = 1.54 v = 0 at B and loss of EE = gain in PE (= 6) distance AB is 4m M1 A1 M1 A1 [4] For using EE = x2/2L and PE = Wh For comparing EE loss and PE gain AG (ii)[120e/1.6 = 1.5] e = 0.02 Loss of EE = 120(0.52 0.022)/(21.6) (or 120(0.32 0.022)/(21.6)) Gain in PE = 1.5(2.1- 1.6 0.02) (or 1.5(1.9 + 1.6 + 0.02) loss) [KE at max speed = 9.36 0.72 (or 3.36 + 5.28)] (1.5/9.8)v2 = 9.36 0.72 Maximum speed is 10.6 ms-1 M1 A1 B1ft B1ft M1 A1 A1 [7] For using T = mg and T = x/L ft incorrect e only ft incorrect e only For using KE at max speed= Loss of EE Gain (or + loss) in PE First alternative for (ii) x is distance AP [ (1.5/9.8)v2+ 1.5x + 120(0.5 x)2/3.2 =120 x0.52/3.2] KE and PE terms correct EE terms correct v2 = 470.4x 490x2 [470.4 980x = 0] x = 0.48 Maximum speed is 10.6 ms-1 M1 A1 A1 A1 M1 A1 A1 For using energy at P = energy at A For attempting to solve dv2/dx = 0 4730Mark SchemeJanuary 2010 36Second alternative for (ii) [120e/1.6 = 1.5] e = 0.02 [1.5 120(0.02 + x)/1.6 = 1.5 x /g] n =490 M1 A1 M1 M1 A1 For using T = mg and T = x/L For using Newtons second law For obtaining the equation in the formx = -n2x , using (AB L eequil) for amplitudeand using vmax = na.a = 0.48 Maximum speed is 10.6 ms-1 A1 A1 6(i)PE gain by P = 0.4g 0.8 sinPE loss by Q = 0.58g 0.8 (0.4 + 0.58)v2= g 0.8(0.580.4sin ) v2 =9.28- 6.4sinB1 B1 M1 A1ft A1 [5] For using KE gain = PE loss AEF (ii) 0.4g sin R = 0.4v2/0.8 [0.4g sin R = 4.64 3.2 sin ] R = 7.12 sin 4.64 M1 A1 M1 A1 [4] For applying Newtons second law to P and using a = v2/r For substituting for v2 AG (iii) R(1.53) = 0.01(48...), R(1.54) = -0.02(9...) or simply R(1.53) > 0 and R(1.54) < 0 R(1.53) R(1.54) < 0 1.53 < < 1.54 M1 A1 M1 A1 [4]For substituting 1.53 and 1.54 into R( ) For using the idea that if R(1.53) and R(1.54) are of opposite signs then R is zero (and thus P leaves the surface) for some value of between 1.53 and 1.54.AG 7(i) TAP = 19.6e/1.6 and TBP = 19.6(1.6-e)/1.6 0.5g sin30o + 12.25(1.6 e) = 12.25e Distance AP is 2.5m M1 A1 M1 A1ft A1 [5] For using T = e/L For resolving forces parallel to the plane(ii) Extensions of AP and BP are 0.9 + x and0.7 x respectively 0.5g sin 30o + 19.6(0.7 x)/1.6 19.6(0.9 + x)/1.6 = 0.5 x x = -49x Period is 0.898 s B1 B1ft B1 M1A1 [5] AG For stating k < 0 and using T = 2/k (iii) 2.82 = 49(0.52 x2) x2 = 0.09 x = 0.3 and -0.3 M1 A1ft A1 A1ft [4] For using v2 = 2(A2 x2) where 2 = -k ft incorrect value of k May be implied by a value of x ft incorrect value of k or incorrect value of x2 (stated) 4732Mark SchemeJanuary 2010 374732 Probability & Statistics 1Note: (3 sfs) means answer which rounds to ... to 3 sfs. If correct ans seen to > 3sfs, ISW for later rounding Penalise over-rounding only once in paper.1(i)attempts at threading indep prob of succeeding in threading const B1 B12 in context in context (ii) (a)0.74 0.3 = 0.0720 (3sf) M1 A12 Condone 0.072 (b)0.75

= 0.168 ( 3 sfs) M2 A13 or 1-(0.3+0.70.3+0.720.3+0.730.3 +0.740.3) M1 for one term omitted or extra or wrong or 1-0.7 or(0.3+...+0.70.3) or 0.3, 0.7 muddle or 0.74 or 0.76 alone. 0.6 not 0.7 M0 in (a) M1 in (b) 1/3,2/3 used M1in (a) M1 in (b) (iii)likely to improve with practice hence independence unlikely or prob will increase each time B1 B12 or thread strands gradually separate 1st B1 must be in context. hence independence unlikely or prob will decrease each time or similar Allow change Total[9] 2(i) (a)Use of correct midpts lf f(= 706 40) = 17.65 l2f(= 13050.5) 2" 65 . 17 "40"13050.5" (= 14.74) = 3.84 ( 3 sfs) B1 M1 A1 M1 M1 A16 11,14,18,25.5 lwithin class, > three lf seen [17.575,17.7] > three l2fseen 40,-mean,.Dep>0. (l-17.65)f, at least 3M1,40, M1,3.84 A1. 4 max B1M0A0M1M0A0 (b)mid pts used or data grouped or exact values unknownoe B11 not orig values were guesses (ii)20 5 = 4 M1 A12 condone 20 [4,5] or ans 5 (iii)20.5th value requd and 1st two classes contain 14 values16 20 M1 B12 condone 20th

oe or third classoe (iv)(a)increaseB11 (b)decreaseB11 Total[13] 3(i)Shm = 0.2412 Shh =0.10992 Smm = 27.212 r = ___Shm__ (ShhSmm) = 0.139 (3 sfs) B1 M1 A13 Allow x or 5 any one S correct ft their Ss (ii)Small, low or not close to 1 or close to 0oe pts not close to line oe B1 ft B1 1st B1 about value of r 2nd B1 about diag (iii)none or unchanged or 0.139oeB11 (iv)Larger oeB11 Total[7] 4732Mark SchemeJanuary 2010 384(i) (021) + 141 + 281 + 381 = 87or 0.875oe (021) + 141 + 2281 + 3281 (= 187) - (87)2 = 6471 or 1.11 (3 sfs)oe M1 A1 M1 M1 A15 > 2 non-zero terms seen If3 or 4 M0M0M1(poss) > 2 non-zero terms seen dep +ve result M1 all4 (x-0.875) terms seen. M1mult p,A1 1.11 (ii) Bin stated or implied 0.922 (3 sfs) M1 A12 Eg table or m n4341(n+m=10,n,m 1) or10C4or 5(or 4 or 6) terms correct (iii) n = 10 & p = 81 stated or implied 10C4481687 = 0.0230 (3 sfs) M1 M1 A13 condone 0.023 Total[10] 5(i) 123135146 3!oe = 18245or 0.247 (3 sfs)oe M1 M1 A13 6C1 5C1 3C1

14C3 With repl M0M1A0

(ii) 124135146 +123134145 +121132143 = 36431or 0.0852 (3 sf) M2 A13 6C3 + 5C3 + 3C3M1 for any one ( 14C3)M1 all 9 numerators correct. With repl M1(6/14)+(5/14)+(3/14) Total[6] 6(a)A: diag or explanation showing pts close to st line, always increasing B:Diag or expl based on r=1=>pts on st line =>r(s)=1

B1 B1 B1 3 . Diag or expl based on r(s) 1=>pts not on st line =>r 1 r=1=>pts on st line&r(s) 1=>pts not on st line B1B1 r=1=>r(s)=1 B2 (b)y = 2.4 4.5 + 3.7 = 14.5 4.5 = 0.4 14.5- c c = 1.3 a=x-by :-14.5 M1A1; then a=4.5-0.4x14.5=-1.3 M1A1 M1 A1 M1 A14 Attempt to sub expression for y x=0.96x+1.48-c oe sub x=4.5 and solve c=1.3 14.5 M1A1.(y-3.7)/2.4=0.4y-c and sub14.5 M1 c=1.3 A1 Total[7] 7(i) 25/37 B22B1 num, B1 denom25/37xp B1 (ii) 2315 seen or implied 5939 seen or implied = 1357585or 0.431 (3 sfs) oe M1 M2 A14 M1 num, M1 denom Allow M1 for 39/59xor + wrong p Total[6] 4732Mark SchemeJanuary 2010 39 8 (i) 5!/2

= 60 M1 A12 Allow 5P3 (ii)4! = 24 M1 A12 Allow 24! (iii) 2/5 3/4 or3/52/4 2 = 3/5 oe M1 M1 A13 allow M1 for 2/5 3/5 2 or 12/25

or (63!)(i) M2 or 3!(i),6(i),(6+6)(i),6k(i) or 66 or 36 or 1-correct answerM1 (k,integer5) Total[7] 9(i)p2 B11 (ii)(q2p)2oe =AGB11 (iii)r=q a/(1-r) used (S =)221 qp = 22) 1 ( 1 pp p/(2-p) AG B1 M1 A1 M1 A15 May be implied With a=pand r=qor q Attempt to simplify using p+q=1 correctly. Dep on r = q or q ) 1 )( 1 () 1 (2q qq+ or p/p(1+q) Correctly obtain given answer showing at least one intermediate step. P2Total[7] Total 72 marks 4733Mark SchemeJanuary 2010 404733 Probability & Statistics 2Penalise over-specified answers (> 6 SF) first time but only once per paper. Use A or C to annotate over-assertive or no context respectively 1 =x= 15.16 2 245 s =

= 1.363 B1 M1 M1 A14 15.16 or 15.2 as answer only Use 225xx[=1.0904]Multiply by 5/4, or equiv for single formula Final answer 1.36 or 1.363 only, not isw 2(i)Not all equally likely those in range 0 to 199 more likely to be chosen M1 A12 Not all equally likely stated or implied Justified by reference to numbers, no spurious reasons (ii)Ignore random numbers greater than 799, or 399 B11Any valid resolution of this problem, no spurious reasons 3B(60, 0.35) N(21, 13.65) ||.|

\| 65 . 1321 5 . 18 = (0.6767) = 1 0.7507 = 0.2493 M1 M1 A1 M1 A1 A16 B(60, 0.35) stated or implied N(21, )Variance or SD = 13.65 Standardise, their np and npq or npq, wrong or no ccBoth npq and cc correct Answer, a.r.t. 0.2494H0 : = 60; H1 : < 60 () 967 . 180 / 560 9 . 582 == z < 1.645 B2 M1 A1 B1 Both correct, B2 B1 for one error, but not x, t,x ortStandardise 58.9 & 80, allow or errors z, art 1.97 or p in range [0.024, 0.025] Explicit comparison with 1.645 or 0.05, or +1.645 or 0.95 if 1.967 or 0.976 used or: ()08 . 59805645 . 1 60 = = c 58.9 < 59.08 M1 B1 A1 60 z5/80, any z = 1, allow errors or , not just +; z = 1.645 and compare 58.9 59.1 or better,on wrong z Reject H0

Significant evidence that people underestimate time M1 A17 Correct first conclusion, needs essentially correct method including 80 or 80 Contextualised, uncertainty acknowledged SR: = 58.9: B0M1A0B1 max 2/7 SR: 2-tail: max 5/7 5(i)H0 : = 11.0H1 : > 11.0 ()P( 19) = 1 0.9823 = 0.0177 < 0.05 B2 M1 A1 B1 Allow . Both correct, B2 One error: B1, but not C, x etc Find P( 19) [or P(< 19) if later 0.95] art 0.0177[0.9823, ditto] Compare 0.05 [0.95 if consistent], needs M1 ()CR 18, P( 18) = 0.0322 19 > 18 M1 A1 B1 CR or CV 16/17/18/19 stated or clearly implied, but not 19): First B2 only. (ii)Cant deduce cause-and-effect, or there may be other factors B11Conclusion needed. No spurious reasons.If DNR in (i), couldnt deduce even if 4733Mark SchemeJanuary 2010 41 6(i)(a)Probabilities dont total1B11Equivalent statement (b)P(> 70) must be < P(> 50)B11Equivalent statement (c)P(> 50) = 0.3 < 50 P(< 70) = 0.3 > 70 B11Any relevant valid statement, e.g. P(< 50) = 0.7 but P(< 50) must be < P(< 70) (ii) = 60 by symmetry 110(0.7) 0.524(4)= = = 10/0.5243 = 19.084B1 M1 B1 A14 = 60 obtained at any point, allow from One standardisation, equate to 1, not 0.758 1 [0.524, 0.5245] seen in range [19.07, 19.1], e.g. 19.073 7(i) M1 A12 Horizontal lineEvidence of truncation [no need for labels] (ii) = 8 [ ]1153181115261t dt t =}[= 67] 82

= 3 B1 M1 B1 M1 A15 8 only, cwd Attempt }kt2 dt, limits 5 and 11 seenk = 1/6 stated or implied Subtract their (non-zero) mean2 Answer 3 only, not from MF1 (iii)N(8, 3/48) ||.|

\| 48 / 38 3 . 81= 1 (1.2) = 1 0.8848 = 0.1151 Normal distribution only approx. M1 A1 A1 M1 A1 B16 Normal stated or implied Mean 8 Variance their (non-zero) (ii)/48 Standardise, n, ignore sign or errors. cc: M0 Answer, art 0.115 Any equivalent comment, e.g. CLT used 8(i)P( 4) = 0.0473 Therefore CR is 4 P(Type I error) = 4.73% M1 B1 A13 P( r) from B(10, 0.7),r = 3/4/5, not N 4 stated, not just 4, nothing else Answer, art 0.0473 or 4.73%, must be stated(ii)B(10, 0.4) and find P(> 4) 1 P( 4) = 0.3669 M1 M1 A13 Must be this, not isw,on (i) Allow for 0.6177 or 0.1622 Answer, art 0.367 (iii)0.5 0.3669 = 0.18345 M1 A1 2 0.5 (ii) Ans correct to 3 SF, e.g. 0.184 from 0.367 115 4733Mark SchemeJanuary 2010 42 9(i)1 P( 7) = 1 0.9881= 0.0119 M1 A12 Allow for 0.0038 or 0.0335 Answer, a.r.t. 0.0119 (ii)Po(12) P( 14) P( 12)[0.7720 0.5760]= 0.196 M1 M1 A13 Po(12) stated or implied Formula, 2 consecutive correct terms, or tables, e.g. .0905 or .3104 or .1629Answer, art 0.196 (iii)Po(60) N(60, 60) 69.5 6060 | | |\ . = (1.226) = 0.8899 M1 A1 M1 A1 A15 N(60, ) Variance or SD 60 Standardise, & , allow or wrong or no cc and cc both correct Answer 0.89 or a.r.t. 0.890 (iv)(a)1 e3m(1 + 3m)M1 A12 M1 for one error, e.g. no 1 , or extra term, or 0th term missing; answer, aesf (b)m = 1.29, p = 0.89842 m = 1.3, p = 0.9008 Straddles 0.9, therefore solution between 1.29 and 1.3 M1 A1 A1 A14 Substitute 1.29 or 1.3 into appropriate fn Comp 0.90.10 1.29 0.8980.10158.00158 1.3 0.9010.09918.0008146 Explicit comparison with relevant value, & conclusion, needs both ps correctorMethod for iteration; 1.296 1.2965or better; conclusion stated M1A1 A1A1 Can be implied by at least 1.296 Need at least 4 dp for M1A2 4734Mark SchemeJanuary 2010 434734 Probability & Statistics 3 1(i) (ii) 0202 2d e d 15 5xax x+ =} } 2a/5+ 1/5=1 a = 2 -------------------------------------------------------- 022 02 2d e d5 5xx x x x+} } 202d5 5aax x= } -2x 2 202 1 1e d e e5 5 10x xx x x ((= + (( } = - 0.7 M1 A1 A1 3 ----------------------- M1 A1 M1 A1 A1 5 [8] Sum of probabilities =1 ------------------------------------------------------ x f(x) dx a By parts with 1 part correct Both parts correct CAO 2(i)

(ii)

(iii) 4 cartons: Total, Y ~ N(2016, 36) P(Y 2000)= (- 16/36) = 0.00383------------------------------------------------------- E(V ) = 0 Var(V) = 36 + 169 = 180 ------------------------------------------------------- 0.5 B1B1M1A1 4 -------------- B1 M1 A13 -------------- B11[8] Mean and variance ------------------------------------------------------ CWO ----------------------------------------------------- 3(i) (ii) (iii) Normal distribution Mean1 2 ; variance2.47/n1 +4.23/n2 -------------------------------------------------------------------------------------- H0: 1 = 2 , H1:1 2 (9.65 7.23)/(2.47/5+4.23/10) =2.527 > 2.326 Reject H0 There is sufficient evidence at the 2% significance level that the means differ ------------------------------------------------------- Any relevant comment. B1B1B1 3 -------------- B1 M1 B1 A1 M1 A1 6 ------------- B1 1 [10] ------------------------------------------------------ Or find critical region Numerator Compare with critical valueSR1:If no specific comparison but CV and conclusion correct B1. Same in Q5,6,7 SR2: From CI: 2.42z M1, correct z = 2.326 B1, (0.193, 4.647) A1 0 in not in CI ; reject H0 etc M1A1 Total 6 Conclusions not over-assertive in Q3, 5, 6 ----------------------------------------------------- e.g sample sizes too small for CLT to apply 4734Mark SchemeJanuary 2010 44 4(i)

G(y)=P(Y y)=P(1/(1+V) y) =P( V 1/y 1) = 1 F(1/y 1) =30 0,8 0 1/ 2,1 1/ 2.yy yy < > 224 0 1/ 2,g( )0 otherwise.y yy< = -------------------------------------------------------- 24y2/y2 dywith limits =12 M1 A1 A1 A1B1 M1A1 7 M1 A1 2 [9] Use ofF 8y3 obtained correctly Correct range. Condone omission of y 0 For G'(y) Correct answer with range ------------------------------------------------------------------ With attempt at integration 5(i) (ii) Use ps z(psqs/200) z=1.645 s = (0.1350.865/200) (0.0952, 0.1747) -------------------------------------------------------- H0: p1 p2 = 0, H1: p1 p2 > 0 1 127 / 200 8 / 10035 / 300 265 / 300 (200 100 ) + = 1.399 > 1.282 Reject H0. There is sufficient evidence at the 10% significance level that the proportion of faulty bars has reduced M1 B1 A1A1 4 -------------- B1 M1B1 A1A1 M1 A17 [11] Or /199 (0.095, 0.175) to 3DP ------------------------------------------------ Or equivalent Correct form. Pooled estimate of p = 35/300 Correct form of sd OR: P(z 1.399) = 0.0809 0.2 O-F: 0.60.40.20.10.30.20.40.3 20.3125 0.024107 D s = =(0.3125-0.2)/(0.024107/8) =2.049 >1.895 Reject Ho there is sufficient evidence at the 5% significance level that the reduction is more than 0.2 -------------------------------------------------------- 0.3125 t (0.024107/8) t = 2.365 (0.1827, 0.4423) B1 B1 M1B1 A1 M1A1M1A19 -------------- M1B1 A1 3 [12] B1 Use paired differences t-test Must have /8 OR: P(t 2.049)= 0.0398 5.991 Reject H0 , (there is sufficient evidence at the 5% that)vegetable preference and gender are notindependent ------------------------------------------------------------- (H0: Vegetables have equal preference H1: All alternatives) Combining rows: 483042E-Values: 404040 2 = (82+102 + 22)/40 = 4.2 4.2 < 4.605 Do not reject H0, there is insufficient evidenceat the 10% significance levelof a difference inthe proportion ofpreferred vegetables B1 M1 A1 M1 A1 A1 M1 A1

8 ------------- M1 A1 M1 A1 M1 A1 6

[14] For both hypotheses At least one correct All correct Correct form of any one All correct ART9.64 OR: P( 9.641)=0.00806 0.10 AEF in context 4736Mark SchemeJanuary 2010 464736 Decision Mathematics 1 TO BE ANSWERED ON INSERT 1(i) Path:A B C D E F Weight:9 M1 A1 B1 B1 B1 Evidence of updating at C, D, E or F All temporary labels correct, with no extras All permanent labels correct cao cao [5] (ii)Total weight of all arcs = 25 Only odd nodes are B and E. Least weight path joining B to E is B C E = 3. Weight: 28 Route: (example) ABDFECBCDE D C - A B1 M1 A1 B1 Total weight = 25(may be implied from weight) B to E = 3 28 (cao) A valid closed route that uses BC, CD and DE twice and all other arcs once [4] (iii)A B E F Graph is now Eulerian,so no need to repeatarcs B1 B1 cao Eulerian (or equivalent) [2] Total = 11

A3 B3D 115 51 C3E3F 2 | 3 3 1 | 0 . 4 | 5 65 5 | 6 76 6| 9 109 3 | 4 54 4736Mark SchemeJanuary 2010 47 2(i)A graph cannot have an odd number of odd vertices (nodes) B1Or equivalent (eg 35 = 15 7 arcs) Not from a diagram of a specific case [1] (ii)It has exactly two odd nodes eg C A B C D E A DB1 B1 2 odd nodes A valid semi-Eulerian trail [2] (iii)AE = 2AC = 3 AB = 5 CD = 7 Weight = 17 B1 B1 B1 Correct tree (vertices must be labelled) Order of choosing arcs in a valid application of Prim, starting at A (working shown on a network or matrix) 17 [3] (iv)Lower bound = 29 A E D F C B A = 34 F C A E D and F D C A - EVertex B is missed out B1 M1 A1 B1 29 or 12 + their tree weight from (iii) A E D F C 34, from correct working seenCorrectly explaining why method fails, need to have explicitly considered both cases [4] Total = 10 For reference (ii) (iii) (iv) ABCDE A-538 2 B5-6-- C36-7- D8 - 7-9 E2--9 - CF = 6DF = 6 B AC D

E A B C D E A B C F D E 2 9 6 6 7 3 8 5 6 ABC D E

4736Mark SchemeJanuary 2010 48 3(i)x = number of clients who use program X y = number of clients who use program Y B1 Number of clients on X and Y, respectively [1] (ii)Spin cycle: 30x + 10y < 180 3x + y < 18 Rower: 10x < 40 x < 4 Free weights: 20x + 30y < 300 2x + 3y < 30 B1 B1 B1 3x + y < 18, or equivalent, simplified x < 4, or equivalent, simplified 2x + 3y < 30, or equivalent, simplified Allow use of slack variables instead of inequalities [3] (iii)Both must take non-negative integer valuesB1Non-negative and integer Accept x + y < 12 as an alternative answer [1] (iv) 1 1 2 3 4 5510xy Checking vertices or using a profit line (4, 6) 72 (337,757) 7717or(24/7, 54/7) 7717 (0, 10) 60 (4, 0) 36 Checking other feasible integer points near (non-integer) optimum for continuous problem (3, 8) 75 Put 3 clients on program X, 8 on program Y and 1 on program Z B1 M1 A1 M1 M1 A1 Axes scaled and labelled appropriately (on graph paper) Boundaries of their three constraints shown correctly (non-negativity may be missed) Correct graph with correct shading or feasible region correct and clearly identified (non-negativity may be missed) (cao) Follow through their graph if possible x = 3.4, y = 7.7may be implied from (3, 8) Could be implied from identifying point (3, 8) in any form cao, in context and including program Z [3] [3]Total = 11 y 10 5 x 0 12344736Mark SchemeJanuary 2010 49 4(i)AAAAAAAAAA AAAAADDDDC CCBBBBBBBB Box 1AAAAA Box 2AAAAA Box 3AAAAA Box 4DDDDCCCB BBBBBBB Cannot fit all the items into box 4There is only room for one B in a box B1 M1 M1 A1 B1 15 As, 4 Ds, 3 Cs, 8Bs (but not just A D C B) Three boxes each containing A A A A A(or shown using weights) A box containing all the rest Completely correct, including order of packing into boxes Any identification of a (specific) volume conflict

[5] (ii)BBBBBBBBCC CDDDDAAAAA AAAAAAAAAA Box 1BDAA Box 2BDAA Box 3BDAA Box 4BDAA Box 5BAAAAAA Box 6BA Box 7B Box 8B Box 9CCC Box 5 is over the weight limit More than five As is too heavy for one box B1 M1 M1 A1 B1 8 Bs, 3 Cs, 4 Ds, 15 As (but not just B C D A) Four boxes each containing B D A A (in any order) Using exactly 9 boxes, the first eight of which each contain a B (with or without other items) and the ninth contains three Cs. Completely correct, including order of packing into boxes Any identification of a (specific) weight conflict [5] (iii)Items may be the wrong shape for the boxes eg too tall B1Reference to shape, height, etc. but not practical issues connected with the food [1] Total =11 For reference Item typeABCD Number to be packed15834 Length (cm)10402010 Width (cm)10305040 Height (cm)10201010 Volume (cm3)1 00024 00010 0004 000 Weight (g)1 000250300400 4736Mark SchemeJanuary 2010 50

5(i)Minimise 2a 3b + c + 18 minimise2(20-x) - 3(10-y) + (8-z) + 18 minimise -2x + 3y z maximise 2x 3y + z(given) a + b c > 14 (20-x) + (10-y) (8-z) > 14 x + y z < 8(given) -2a + 3c < 50 -2(20-x) + 3(8-z) < 50 2x 3z < 66(given) 10 + 4a > 5b 10 + 4(20-x) > 5(10-y) 4x 5y < 40(given) a < 20 20-x < 20 x > 0 b < 10 10-y < 10 y > 0 c < 8 8-z < 8 z > 0 B1 M1 A1 (Constant has no effect on slope of objective) Replacing a, b and c in objective to get-2x + 3y - z (Condone omission of conversion to maximisation here) Replacing a, b and c in the first three constraints to get given expressions Showing how a < 20, b < 10, c < 8 givex > 0, y > 0, z > 0 [3] (ii)PxyzstuRHS1-23-10000 011-11008 020-30106604-5000140 x and z columns have negative entries in objective row, but z column has no positive entries in constraint rows, so pivot on x col 81 = 8; 662 = 33; 404 = 10 Least ratio is 81, so pivot on 1 from x col New row 2 = row 2New row 1 = row 1 + 2(new row 2) New row 3 = row 3 2(new row 2) New row 4 = row 4 4(new row 2) PxyzstuRHS105-320016011-11008 00-2-1-2105000-94-4018 M1 A1 M1 A1 M1 A1 M1 A1 Constraint rows correct, with three slack variable columns Objective row correct Choosing to pivot on x column(may be implied from pivot choice) Calculations seen or referred to and correct pivot choice made (cao) Pivot row unchanged (may be implied) or follow through for their +ve pivot Calculations for other rows shown (cao) An augmented tableau with three basis columns, non-negative values in final column and value of objective having not decreased Correct tableau after one iteration (cao) [2] [2] [2] [2] x = 8, y = 0, z = 0 x = 8 a = 20 8 = 12 y = 0 b = 10 0= 10 z = 0 c = 8 0=8 B1 M1 A1 Non-negative values for x, y and z from their tableau Putting their values for x, y and z intoa = 20 x, b = 10 y and c = 8 z Correct values for a, b and c, from their non-negative x, y and z [3] (iii)x < 20,y < 10 and z < 8M1 A120, 10, 8 Correct inequalities for x, y and z [2] Total =16 4736Mark SchemeJanuary 2010 51 TO BE ANSWERED ON INSERT 6(i)10 n(n-1) B1 B1 10 1+2++(n-1) seen, or equivalent Check that sum stops at n-1 not n [2] (ii)(a)912 3 45 B1 M1 A1 Their 10 minus 1 1, 2 and 3 45 following from method mark earnedcao [3] (b)1+2+3++(N-1)= N(N-1), where N = n(n-1) = n(n-1)(n(n-1) 1) (given) M1 A1 1+2+3++(N-1) or N(N-1),where N = n(n-1) Convincingly achieving the given result [2] (iii) M1M2M3 M4 Vertices in tree Arcsin tree Vertices not in tree Sortedlist ABCDE DED | 2 | E A B CD | 2 | E D E AD | 2 | EB CA | 3 | E A | 3 | EA | 4 | C D E A CD | 2 | EBC | 5 | DA | 3 | EB | 6 | EA | 4 | CB | 7 | C DEACBD | 2 | EA | 8 | BA | 3 | EC | 9 | E A | 4 | C B | 6 | E M1 M1 M1 A1 (Order of entries in M1, M2 and M3 does not matter) ArcA | 3 | Eis added to M2, A is added to M1 and deleted from M3 ArcA | 4 | Cis added to M2, C is added to M1 and deleted from M3 ArcC | 5 | Dis not added to M2 and arc B | 6 | E is added to M2 cao (lists M1, M2 and M3 totally correct, ignore what is done in list M4). [4] (iv) 450030100| | |\ . = 18750 seconds M1 A1 Or equivalent cao, with units(312 min 30 sec or 5 hours 12 min 30 sec) [2] Total =13 4737Mark SchemeJanuary 2010 524737 Decision Mathematics 2 1(i) B1 Bipartite graph correct [1] (ii)

D = T C = G B = S Andy =food Beth= science Chelsey =geography Dean = television Elly = history B1 M1 A1 B1 A new bipartite graph showing the pairings AF, BG, CT and EH but not DS This alternating path written down, not read off from labels on graph B = S, C = G and D = T written down A = F, E = Hwritten down [4] (iii)Andy = food Beth= television Chelsey =geography Dean = politics Elly =history Science did not arise B1 B1 A = F, C = G, D = P and E = H(cao) (B = T may be omitted) S(cao) [2] Total =7 AF B G C H D P

ES T A F B G C H D [ P ] E S T4737Mark SchemeJanuary 2010 53 2Add a dummy row PRST April30283225 May32343235 June40403938 Dummy40404040 Reduce rows 5370 0203 2210 0000 Columns are already reduced Incomplete matching, cross through zeros 5370 0203 2210 0000 Augment by 1 4260 0204 1100 0001 Complete matching PRST April4260 May0204 June1100 Dummy0001 April = Tall Trees2500 May= Palace3200 June= Sunnyside3900 Total cost = 9600 B1 M1 A1 B1 B1 B1 B1 Adding a dummy row of all equal values Substantially correct attempt to reduce matrix (condone 1 numerical slip) Correct reduced cost matrix from reducing rows first and statement of how table was formed, including reference to columns (cao) Cross through zeros using minimum number of lines Correct augmented matrix and statement of how table was formed (cao) A = T, M = P, J = S (cao) 9600 (cao) with units [3] [2] [2] Total =7 4737Mark SchemeJanuary 2010 54 3(i) M1 A1 Durations not necessary Correct structure, even without directions shown Activities must be labelled Completely correct, with exactly three dummies and all arcs directed [2] (ii) Minimum project completion time = 10 hours Critical activities A, B, D, E, H M1 M1 A1ft B1 M1 A1 Follow through their activity network if possible Substantially correct attempt at forward pass (at most 1 independent error) Substantially correct attempt at backward pass (at most 1 independent error) Both passes wholly correct 10 hours (with units) cao Either B, E, H or A, D, H (possibly with other critical activities, but C, F, G not listed). Not follow through. A, B, D, E, H (and no others) cao [3] [3] (iii)No. of workers 0 1 23 4 56 7 8 910 hours M1 A1 On graph paper A plausible resource histogram with no holes or overhangs Axes scaled and labelled and histogram completely correct, cao [2] (iv)1 hourB1Accept 1 (with units missing) cao[1] (v)No need to change start times for A, B, C, D and EActivities G and H cannot happen at the same time, so they must follow one another This causes a 2 hour delay F could be delayed until 1 hour before H starts H should be started as late as possible a maximum delay of 3 hours M1 A1 B1 B1 G and H cannot happen together (stated, not just implied from a diagram) 2 cao Diagram or explaining that for max delay on F need H to happen as late as possible 3 cao [2] [2] Total = 15 A(6)D(1) B(5) E(2) G(2) C(4)H(3) F(1) 6|6 A(6)D(1) 0|0 B(5) 5|5 E(2) 7|7 G(2) 10|10 C(4)H(3) 5|6 F(1)7|7 7 6 5 4 3 2 1 0 4737Mark SchemeJanuary 2010 55 4(i)(1; 0) 6 (2; 0) 9 7 10 8 (0; 0) (3; 0) 76 710 (1; 1) 8(2; 2) B1 M1 A1 Correct structure (vertex labels and graph correct) Assigning weights to their graph (no morethan 1 error or no more than 2 arcs missing/extra) Completely correct network [3] (ii)MaximinB1cao[1] (iii)Stage StateActionWorking Suboptimal maximin 001010 2101010 201010 0min(6,10) = 6 01min(7,10) = 7 12min(8,10) = 88 0min(6,10) = 6 11min(7,10) = 7 2min(8,10) = 88 000min(9,8) = 88 1min(7,8) = 7 Weight of heaviest truck = 8 tonnes Maximin route = (0; 0) (1; 0) (2; 2) (3; 0) B1 B1 B1 M1 A1 M1 A1 B1 B1 Four or five columns, including stage, state and action Stage and state columns completed correctly Action column completed correctly Min values correct for stage 1 Suboptimal maximin values correct for stages 2 and 1 (follow through their network if possible, no more than 2 arcs missing/extra) Min values correct for stage 0 Maximin value for stage 0 (follow through their network if possible, no more than 2 arcs missing/extra) 8, cao Correct route, or in reverse [3] [2] [2] [2] Total =13 SRSpecial ruling for working forwards (iii)Stage StateActionWorking Suboptimal maximin 10099 1077 00min(9, 6) = 66 1min(7, 6) = 6 210min(9, 7) = 77 1min(7, 7) = 7 20min(9, 8) = 88 1min(7, 8) = 7 0min(6,10) = 6 301min(7,10) = 7 2min(8,10) = 88 Weight of heaviest truck = 8 tonnes Maximin route = (0; 0) (1; 0) (2; 2) (3; 0) B1 B0 B0 M1 A0 M1 A0 B1 B1 Four or five columns, including stage, state and action No follow through from incorrect networks Min values correct for stage 2 and suboptmal maximin values correct for stages 1 and 2 (cao) No follow through from incorrect networks Correct min values for stage 3 and maximin value for stage 3 (cao) 8, cao Correct route, or in reverse [3] [2] [2] [2] Maximum = B1 M1 M1 B1 B1 = 5 marks out of 9 (2; 1)104737Mark SchemeJanuary 2010 56 5(i)Conan GHIrow min D -1 -42-4 RobbieE31 -4-4 F1 -11-1*col max312 * Play-safe for Robbie is fairy Play-safe for Conan is hag Robbie should choose the elf M1 M1 A1 A1 B1 Calculating row minima (cao) Calculating column maxima (or their negatives) (cao) Fairy or F (not just -1 or identifying row) Hag or H (not just + 1 or identifying column) Follow through their play-safe for Conan Elf or E [5] (ii) Dwarf: 31[(-1) + (-4) + (2)] = -1 Elf:31[(3) + (1) + (-4)] =0 Fairy: 31[(1) + (-1) + (1)] = 31 M1 A1 D = -1 or F = 31 or -3, 0, 1 All three correct [2] (iii)Goblin: 3p + (1-p) = 1 + 2p Hag: p (1-p) = 2p 1 Imp:-4p + (1-p) = 1 5p 2p 1 = 1 5p p = 72 M1 A1 M1 A1 Any one correct (in any form) All three correct (in any form) Appropriate equation seen for their expressions 72 or 0.286 (or better) from method seen [2] [2] (iv)4 is added throughout the table to make all the entries non-negative If Conan chooses the goblin, this gives an expected value (in the new table) of 3x + 7y + 5z B1 B1 Add 4 to remove negative values Expected value when Conan chooses the goblin [2] (v) z = 75 m < 5.571, m < 3.571, m < 3.571 m < 3.571 (374) (257) Hence, maximum value for M is 3.571 4 = -0.429 or -73 M1 M1 A1 Using z = 75 to find a value for m ( or implied) Subtacting 4 from their m value cao [3] Total =16 4737Mark SchemeJanuary 2010 57 6(i) = 12 litres per second = 15 litres per second B1 B1 12 15 [2] (ii)At least 3 litres per second must flow into A, so AC and AF cannot both have flows of 1 B1At least 3 flows along SA[1] (iii)At most 4 litres per second can flow into B,and at least 4 must flow out, so BC and BD must have flows of 2 Hence, only 2 litres per second flows into Dand at least 2 litres per second must flow out, soDE and DT must both be at their lower capacities B1 B1 At B: flow in < 4 (and flow out > 4) hence given flows in BC and BD Stating that flow into D is 2 and hence given flows in DE and DT [2] (iv)Flow across {S, A, B, C}, {D, E, F, G, T} > 11(so 10 litres per second is impossible) Minimum = 11 eg Maximum = 12 No more than 12 can cross cut and 12 is possible, eg augment flow shown above by 1 litre per second along SAFT M1 A1 M1 A1 M1 A1 Or any equivalent reasoning (eg flow through C) Wholly convincing argument 11 Showing that 11 is possible (check C) 12 Showing that 12 is possible but 13 is not [2] [2] [2] (v) B1 M1 A1 A correct reduced network (vertex E and all arcs incident on E deleted), including arc capacities Or putting Ein and Eout with a capacity of 0 between them Or giving CE, EG and DE upper and lower capacities of 0 On same diagram or a new diagram SA = 3, SC = 2, SB = 4, BC = 2 and BT = 2 (and nothing through E, if shown) A valid flow of 9 litres per second through the network [3] Total =14 A (1,4)F (3,4) (1,4) (3,6)(2, 3)(4,8)

S (2,4) C(2,5)G (0,5)T (2,5) (3,4) B (2,5)e.g A2F

31 32 7 S 2C 2G 0T 2 42 B A1 F 3 232 6 42 3 SCG T 2 3 340 E B2 D 2 58Grade Thresholds Advanced GCE Mathematics (3890-2, 7890-2) January 2010 Examination Series Unit Threshold Marks 7892 Maximum Mark ABCDEU Raw7256484134270 4721 UMS10080706050400 Raw7261534639320 4722 UMS10080706050400 Raw7251433629220 4723 UMS10080706050400 Raw7255473932250 4724 UMS10080706050400 Raw7262544638310 4725 UMS10080706050400 Raw7253463932250 4726 UMS10080706050400 Raw7255474033260 4727 UMS10080706050400 Raw7252443628210 4728 UMS10080706050400 Raw7256484134270 4729 UMS10080706050400 Raw7251443730240 4730 UMS10080706050400 Raw7254474033260 4732 UMS10080706050400 Raw7262534435260 4733 UMS10080706050400 Raw7258504235280 4734 UMS10080706050400 Raw7247403428220 4736 UMS10080706050400 Raw7251453933280 4737 UMS10080706050400 59Specification Aggregation Results Overall threshold marks in UMS (ie after conversion of raw marks to uniform marks) Maximum Mark ABCDEU 38903002402101801501200 38913002402101801501200 38923002402101801501200 78906004804203603002400 78916004804203603002400 78926004804203603002400 The cumulative percentage of candidates awarded each grade was as follows: ABCDEU Total Number of Candidates 389028.253.173.087.296.41001385 389239.261.779.292.597.5100126 789030.860.183.895.099.3100459 789221.160.584.210010010043 For a description of how UMS marks are calculated see: http://www.ocr.org.uk/learners/ums/index.html Statistics are correct at the time of publication. List of abbreviations Below is a list of commonly used mark scheme abbreviations. The list is not exhaustive. AEFAny equivalent form of answer or result is equally acceptable AGAnswer given (working leading to the result must be valid) CAOCorrect answer only ISWIgnore subsequent working MRMisread SRSpecial ruling SCSpecial case ARTAllow rounding or truncating CWOCorrect working only SOISeen or implied WWWWithout wrong working Ft or Follow through (allow the A or B mark for work correctly following on fromprevious incorrect result.) Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; 1 Hills Road, Cambridge, CB1 2EU Registered Company Number: 3484466 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: 01223 552552 Facsimile: 01223 552553 OCR 2010 OCR (Oxford Cambridge and RSA Examinations) 1 Hills Road Cambridge CB1 2EU OCR Customer Contact Centre 14 19 Qualifications (General) Telephone: 01223 553998 Facsimile: 01223 552627 Email: general.qualifications@ocr.org.uk www.ocr.org.uk For staff training purposes and as part of our quality assuranceprogramme your call may be recorded or monitored