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2000_MarkingScheme

2000_MarkingScheme - radiatorssss FOR TEACHERS’ USE ONLY...

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Unformatted text preview: radiatorssss FOR TEACHERS’ USE ONLY AS Mathematics and Statistics General Marking Instructions It is very important that all markers should adhere as closely as possible to the marking scheme. In many cases, however, candidates will have obtained a correct answer by an alternative method not specified in the marking scheme. In general, a correct answer merits all the marks allocated to that part. unless a particular method has been specified in the question. Makers should be patient in marking alternative solutions not specified in the marking scheme. In the marking scheme, marks are classified into the following three categories: ‘M‘ marks ' awarded for correct methods being used; ‘A' marks awarded for the accuracy of the answers; Marks without ‘M' or 'A' awarded for correctly completing a proof or arriving at an answer given in a question. In a question consisting of several parts each depending on the previous parts, ‘M‘ marks should be awarded to steps or methods correctly deduced from previous answers, even if these answers are erroneous. However, 'A’ marks for the corresponding answers should NOT be awarded (unless otherwise specified). For the convenience of markers, the marking scheme was written as detailed as possible. However, it is still likely that candidates would not present their solution in the same explicit manner, e.g. some steps would either be omitted or stated implicitly. In such cases, markers should exercise their discretion in marking candidates’ work. in general, marks for a certain step should be awarded if candidates' solution indicated that the relevant conceptftechnique had been used. Use of notation different from those in the marking scheme should not be penalized. In marking oandidates‘ work, the benefit of doubt should be given in the candidates‘ favour. Marks may be deducted for poor presentation (pp). The symbol .should be used to denote I mark deducted for pp . At most deducted I mark from Section A and 1 mark from Section B for pp . In any case, do not deduct any marks for pp in those steps where candidates could not score any marks. Marks may be deducted for numerical answers with inappropriate degree of accuracy (or). The symbol @should be used to denote 1 mark deducted for a. At most deducted I mark from Section A and 1 mark from Section B for a. In any case. do not deduct any marks for a in those steps where candidates could not score any marks. Marks entered in the Page Total Box should he the NET total scored on that page. ZW-AS-M at 5—13 Hfifigitfiili§lifl FOR TEACHERS’ USE ONLY HBEflBfié-éfifi FOR TEACHERS’ USE ONLY Solution Into) = i J" In x 1- ln y = i y ‘ y_xd_y 191+ y y__rfl lM difi'erentaition of in .r l id_y_ ad" (of d: = 1‘“ ) lM chain rule I y d’: J’- W y- ]M quotientlproduct rule 2 .d_” ._ I E y +1" cl: it} I Ch: EJi=fl 1 at least one step til! xy+ 34:J tM differentaition of a ' ! 1M chain rule 1M quotient rule ZOOO-AS-M & S—l9 stnsststtsss FOR TEACHERS’ USE ONLY mafia-€2,355 FOR TEACHERS’ USE ONLY ' HFRflEfi$E§ FOR TEACHERS, USE ONLY Sulutim Marks Solution l l l 1 3 1 —(—--—) —(——)r——) ' 1 1A ml: nd 2, (a) (MEI)! = l+-2—(2I)+2 22 (2.\-)2+2 231 2 (2x)5+v- IM anyStcrms , ,1, 5 8'71 i '_ 1 l 3. Amaofthe shaded region = Lawn-1 —e‘)dx mapt e“ —I—x-‘ =1+x~xl+§x1+n m 1A limits I . . (1+Bx“) 1 = 17%(Bx3h-v- IA (pp—I form1551nggx) a ,r ' 3 3 =1—4x"+-.- =|:x+:—:3—Ee§:| 1A for x+zx 1 —1 “ 1A fur—m;E __ J (b) “_2“4x2) 3 = M 36.2537 (or zs-se) m a—l forr.t.6.254 __ ..._5 “+2” 2 ( ) -1 '1 = (1+Bx’) 1 (l+2x)1 4, (a) The graph of Ex) is concave downward (or convex upward) when 3 1 2 l 3 1<x<3. (orlsrs: BIC.) IA :2 (1—4: +---)(1+x—Ex +31: +~-) MI I 7 l 3 3 (b) The points ofinflexlon are (1.46) and (3,0). IA s Inc—Ex +5.1- —4x +--- (c) l 1 7 3 =[1~x—;I —3x +... 1" Allmntlmlx, | I (l—2x+4x1)_3 = [pun—230]"? IA minimum point [A shape and coordinates of points [jag 2x0 —2::)]2 2| = I+[—%][—21(le2x)]+ ‘ [ex—3e) M +—-‘-——3-I-'———[—2X(l-21)]J +'-' 3 l 5 "'44) =1+:(l—2x)+§x*(1—2x)1+—2~x3(|—2x)’+-.- =1+x——]-x1—lx’+o-- m 2 (pp—1 for cxrra terms or missing ‘+---‘ in all cases) ----- (5) IMO-AS-M & 5—20 IOOO-AS-M & 3-21 Rmflfiffiiéfifl FOR TEACHERS’ USE ONLY 5-1 mflflffiggfifl FOR TEACHERS’ USE ONLY FOR TEACHERS’ USE ONLY Qfififlfiifi%fifi =_ 75kg 3.4;kg 89.5kg 535“” “”5“" ”Ska _ 75kg 55kg 91kg Q,= I‘m-H) thterrn rung) (55th) _m+2 wk; msg 9' 4 ““2"" {5.15m} (Hill) (15mm 15'5“ , rn—l +4 9,4 3 in". (313133;? 3353;?- IA median lM lst or 3rd quartile lA interquartile range OD 5kg v. Q, g?" and round ofi' to ill) 75kg 90 kg (sun) (to: 5 use) the nearest 05th term (pp—1 for wrong/missing unit) (1:) (Illustrating method n: my) 100 95 90 . 85 so” (3%) “titan 1A any correct box-and—whisker 75 , diagram with scale 1A alleorreot, same scale '70 65 60' 015 New weights weights (c) No. The dirt ms in b cannot show individual difference. or i. There is no substantial evidence for making the claim. ii. it is possible that the woman weighed 60 kg on completion of the programme is the won-ran weighed 99 kg when the ro ammo started. an counterexam le man ION-AS-M a 3-22 FOR TEACHERS’ USE ONLY Cliflflfifi$fiafi FOR TEACHERS’ Qiflfifiifigflfl Solution 5. (a) Probability of having no vanilla flavour ice-cream .3 1 2(3) P_a“ ' c; (””[6][5] 4 ‘ rr ) l = — or 0.2 5 i l 03) Probability of having exactly 1 cup ofvanilla flavour ice-cream 4 . ., 4 C; 6 5 4 P, 3 = — or 0 6 5 i J I The probability that there rs no The probability that there is at least people killed in a traffic accident 1 people killer! in a traffic accident = 2-0.1 (P) . (‘1’ re 0304837418 at 01195162582 The required probability The required probability = p5+5p‘(1-p) =(1—q)‘+5(1aql‘q a: 0.92547759l A; 0925477591 nu 0.9255 :5 0.9255 8. (a) The probability ofo customer winning a prize in 1 trial :iilWiil (h) Let .r and y be the probabilities ofgenerating games A and B respectively. Then 5ix+£y=£ 9 6 3 and x+y=l 5 5 2 —x+—l—I =— 9 6‘ ) 3 S S 2 —-——x=— 6 ill. 3 3 =- or 0.6 x 5 r r The probabilities of generating game A and game 3 are % (or 0.6) and % (oril.4) respectively. IDOU-AS-M 15; 5—23 QBEflEififafiféfi USE ONLY ‘lM denominator lM numerator IA 1M numerator lA m4n lM binomial (at least]! terms] ‘lM cases 0 and l M a— l for r.t. 0.925 -----(5) lA [A Hill IM 2 IA = 0.4 or y S ‘1 || 1A ----- (m FOR TEACHERS' USE ONLY qmflflméfiflfi FOR TEACHERS’ USE ONLY Ema-223629563 FOR TEACHERS’ USE ONLY Sulutiun - .026 .. 9. (I) (I) “fl'lfi’r‘i-W I (ii) Let u=-Jl+8x , then u1=l+3x. 2ndu=3dx r'cx)=4e’°1’*(:—o.2sx) {if WWIPW‘OfiBdV lhg(x)dx= 16+ 6: a: (or ElfidH-Ca . a u m «HT-Em :3 I: 0<:<4 ccptcunsidering _I7 16+ 620.: —l) l d" (or [16):]: 115012—01 u) M integrand ‘ f "‘ Hype-“Noggin ‘ | 8:: 4" + | 81.: 4 IA limits <0 1f x>4 1 2 f(x}$f(4) for x90. 1 followthrough - I] [—35:13 +4u-%Jdu (or 95+L[116u1~%]du "—2?n-n 1,13T 1331 ... 1.9l15220.85221l. 678421.886] 21.7301 21. IA currecituldp. =[fi" +2“ ‘5” l (“r 95+[fi“ ‘3"1 ) 1" WWW“ l9.1 2L9 . =llfil 4 H115 1A a—l for r.t. HG pp—l for wrong/missing unit =1%[IE+21.355|+2(19.1152+20.8523+2!.6634+2l.8361+21.7301)] 1M Tliccxprcled Increase m profit Is 116 hundredthousand dollars. t124 The “peeled increase in profit is 124 hundred thousand dollars. [A rr—l for r.t. 124 ('3) FTDI‘“ (3X1). “:35 “4) (= 3-3851 J [01' 1 3' U . pp—l for mnymissing unit i.e. for) is boundcd above by ['(4) . IM 6 From (b)(i). g(x) increases to infinity as 2: increases lo infinity. (*1) (i) g(x)=l6+ ‘ mp0 and g(x)>0 fur:>0, J] + 3:: th: area under tha graph of 50:) will be greater than that of {(x) as r— 61-! x increases indcfmitely. .( J 6 1+ 8x 241+” IA Plan G will cvrntually result innbiggcr profit. [A z = ' g [4- 3x _ 6(l+4x) ' —1 (1 +31)! > 0 for x > 0 . 1 gm is strictly increasing for x > 0 . Alternative-graph fiat! 002x12) . 6x EJx 11m 16+ = lim 16 + “w[ 1/1.“: ] x...- 1 14wa as 1—H” M in! 2000-A5uM Jr 5—24 IND-AS—M & 5-25 qgfiflflifiéjgfi FOR TEACHERS’ USE ONLY HBRflEifiéafla‘gl FOR TEACHERS’ USE ONLY Eaflfiaflflméfififi FOR TEACHERS’ USE ONLY Elwiiififiééfia FOR TEACHERS’ USE ONLY Solution Marks 10. (a) (i) MW!“ (h) (i) r(t)=2D—pe'“ lnr(!)=lna+bln1 m 1n[20—r(r)]=lnp—qr 1A “'7 m.“ , ¥m¥$ (in “ml- ' ' ' ‘ . - - mm] —--m [[111 — DE} LlI 1M scale and labelling IA points and line llM scale and labelling 1A ignoring the point at I: 5 When “:5, Inna 1.6] [M for either Lu H5) at 3 fi'om the graph. From the graph. [up a 3.05 r(5)~20-1 'A r(5)e[I9-l.2|-I] 1m..21.1 IA pe{20.1,23.31 In r(5) £[2.95, 105] In F! e [3.00, 3.15} = “EA . .0523 “0.525 1A 4210.513,o.5501 omarks for p or q if the ph is no! correfl. 2000AS-M & 5—26 IMO-AS-M a: 5—27 FOR TEACHERS’ USE ONLY Hmflflifiéfififi ‘ .‘ Hmflfiifififi FOR TEACHERS’ USE ONLY Marks (iii) The total number, in thousands. of bacteria after 15 days ofcuitivntion [1M definite integral ln55=a—£"3" 1:. [3) 11M adding 100 In 98 = a —e"“' Eiiminating a , w: have 2"” 4"” + in 911 ~1n55= o 9'"l uz‘z" +llng=0 is a [no—pr'flm +1011 r I! =l20£+£z"“] +IOO 1M forintegrntion ‘1 u =300+£e'”"'~£+l00 e q 51' 1 93 u —1lc 2 _ -2l‘ _ _= - 1 =260 + 100 IA [20_ pl?" ]d: E [255‘ 163] (e ) c + 1! In 55 O 1M quadrant: equauun E350 1A A1115 [355,363] 111,1111195 pp—l for wrong/missing unit 9721: = E 55 2 W: I 94 0.30635 or 0.159365 Lat Na) thousand be the total number ofbacieria | 1“0-305 0' 0-594 IA LL 0306 . 0-694 aRCr 1 days ufcultivalion. Then E k = 0.5915 [c a 0.1529 or NU) : Ifzo— pawn: 1M l M 4.3401 “5.11929 ! k = 0.59 k N 0.13 _ . . 1 _ .—. 201 + ~31: " + 1: IM for mtcgrallon : {a A, 4.84 0 {a z 539 (or 5-90) (2 lip-1 IA 1.1 | for more 1111111 I tip- Nw) = 100 1 1 _ 1 a. 0.59 100 =13”. | (b) Usmg a H 4 34 . InN(7) = 4.80. IN! r.t. 4.30 q . p , 11(1) 2: 121 . m. 121 c:100——- 1:60.04 1.11 as [5102,5145] | ”mg q 1 Using { . [nN(7) = 5.12. Lt. 5.12 7 5.14 Hence the {oral number, in thousands, of bacteria :1 a: 5'39 NU) :1 167. (or comparing 111170 a: 5.1358) r.l. 167 —1'1‘O afiar l5 days of cuitivalion is N(15)= mun—32"” +c =35o q 1M+1A N(15)E[355.363] {‘1 " ”"3 will make the model 111 for the known data. a u 5.89 N(r)= ‘hNflJ _ em NU) —! 8519 as 114m 361a3r—!w The total possible catch ofcoml fish in that area since January 1. 1992 is 361 thousand tonnes. IMO-AS-M & 3-29 IA follow through 1M 1A Lt. 361~365- pp—E for wrongirnissing unit ION-AS-M Jr. 5-18 FOR TEACHERS’ USE ONLY Hflfififiifififlfi Rmflfiifiafifi FOR TEACHERS’ U_SE ONLY Rfififififlififlfi Solution In N(r)-a—e"" 1: N11) = 3“" N‘(() = —e"*(—k)e""'* = 112"“ Ne) (11) wma km'(.-)e"*' nkNUkH'} 1000-A5-M 10. 5—10 = k1 Noel—u (EH1 4) >0 when “2% 1 =0 when .r=f <0 when (>1 N'U) ismaximum u'l mi. £55.56 The maximum rate of change ofthe total cash of coral fish in that area since January 1, 1992 Occurred in 1997. [11 14(6):: 4.9-1, 11(5) - 143.5 In N(5)1x 4.78 , N(S) - 119.7 The volume of fish caught in 1997 = [N(6) — N(5)] thousand tonnes as 24 thousand tonnes Rfiflflfiifiéfififi FOR TEACHERS’ USE ONLY 1A lM IA 1e [5.47. 5.56] [A In MG) 5 [4.97. 4.99] N(6) e [l43.6. 145.3] In 11(5) 6 [4.711. 4.30] 14(5) e [119.7, 122.01 lM 1A pp—l for wrong/missing unit FOR TEACHERS’ USE ONLY 12. (a) Let X~N(20.51) and Z~N(0. 1). (i) (ii) 01) (i! (ii) Rfiflflfiifififlfi Solution P(ri3ky but not hamduus l A) = PU: <X< 21') MIL” (2‘: 27—20 5 5 ) B P(-I,6 <3 2‘: 1.4) p: 0.4452 + 0.4!92 a: 0.8644 99151911111) = P(X> 12) = P(Z>—l.6) 1: 0.4452 + 0.5 = 11.9452 P(I1n72rduus | A) = P(X> 27) = P(Z> 1.4) :11 0.5 — 0.4 I92 1:: 0.0808 0.0808 9452 a: 0.0855 P (a risky bottle is hazardous | A) = P(risky) = 0.6 P(risky | A) + 0.4 ”risky I B) x: 0.6(0.9452) + 0.4(0.058) g. 0.59032 1:: 05903 (p) . , = P(risk '5‘me PCB and nsk)‘ I “3"” P(risky) a (0.053)(0.4) ”0.59031 9a 0.0393 P(l1azardous| B)P(I'3) P(risky) a (0.004)(0.4) 0.59032 at 0.0027] In 0.0027 MB and hazardous | risky) = (iii) PUicensc suspended) = 1—(1—p)’ —5p(l— p)‘ = 1—11 41.59032)5 — 5(o.591132)(1—u.59032)‘ 2 0.9053 2000-AS-M 1!: 5—31 Rmfififi%flfi FOR TEACHERS’ USE ONLY Marks IA [M IA 0—] for r.t. 0.36:3 IA IA IM IM IA 0—] far r.1.fl.590 IA numerator 1M Bayes! thaorern 1A numerator 1M Bayes' zhem-em IM binomiai IM complement of cases 0& I EM p from bfi) FOR TEACHERS’ USE ONLY Solution Solution Marks Possible teams: Bl G. . B. 071. B; G. and 3; GI. 14- (a) E . I' 'l . E' . | i' .1 . _ ' . 6- (b) The probability that B. G. can enter the second round ofthe contest frequency: '00: loo-Cf l ‘ i 'T = 0.9 x 0.3 x! 6 6 = 0.72 M missing values: 30.79 20.09 (or 20.09 20. IO EAHA 6.13 0.80 0.51 0.80 ) lA+l A rr—l for more than 2 do. (c) Probability required = i(03‘x0.8+t.l.§h¢0.0+(UNCLE+0.7‘M0J’5) IM 4 cases I = (:56 IA {is} The maximum error for Po“) is 2.79 while that for 3(6, 0) is |.l9. The binomial distribution is better. 1A awarded only if(a) is correct (d) Suppose B. G. and E; G: are formed to represent the school. (c) (i) The probability ofgetting at least 1 stamp in a box ofthc Chips (i) The probability that exactly one team can enter the second round I 5 EA robabili oflcase = 1___) !M = (0.9x0.5)(1*0.7x0.6)+(0.7x0.6)(l—0.9>¢0.8) p f” 5 lM summatton of Zeases m d fr ‘ bl - . ' rea out a e r l - 0.9 x 0.8 0.7 x 0.5 — l — 0.9 it 0.3 l - 0.7 X 0.5 = 0.665l020233 = 0.5352 IA = 0.665! (pl) 1A 2—! for r.1.0.665 m (ii) The probability that at least one team can enter the second round = 05352 * 0‘9 x 0'3 3" 0‘7 x 0'6 [M (ii) The probability ofgetling at least 1 stamp in buying r l- 1—0.9x0.Bl-0.7x0.6 notmorethan 3 boxes Ir0.9:0.B+0.7x0.6—0.9x0.8x0.7x0. a pl+(1—pi)p.+(|-p1)2pl {1M casesl.tmd3 R 0.8376 [A [M Geometric (p) ... (0.665 102)" + (1 4.065102) + u —0.665|02)2] =1 (19514389532 :1 0.9624 1A a—l for r.t. 0.962 (e) (i) If the two teams are farmed randomty, the probability that exactly one team can enter the second round ”V! the combination 13le .326. l I = -2—>:0.5352+E[(0.9x0.6)(1—0.Tx0.8)+(0.7x0.8)(l—0.9:<0.6)] IM multiplying by 1 (iii) The probability ofgetting exactly 5 stamps in 2 boxes with stamps 2 l 5 5 i . 5 1 1 2 5 ¢ 1 5 5 3 1M eombinations(l,4)and(2,3) 0 . . = l(0.5352+0.4952) = 1|:C'6[EIE] (340(3) [0] +C2 [E] (3] (3(3) [-5] ] 1M multiplying by 2 025152 1A 1M binomial distribution _ ' r 2 0.4019 x 0.0050 + 0.2099 x 0.0535 read from table (i i) If B. G1 and B; G. are formed to represent tilt: school. the l s 1 s 5 :- probability that at least one team can enter the second round . = Cfo[—] [E] + C? C311] [_J a 0.4952 + 0.9 H 0.3 x 0.7 x 0.6 IM 6 t5 6 6 Er |- it — 0.9 x 0.631 —0.'l 2: 0.81 _ zes'mxtsHSXm) -r 0.9K0,6+0.7x0.8—0.9:0.6x0.7x0. 7 6'2 = 0.7976 M as 002199430: ([11) From (dXii), the combination 3. GI 311d 3: G; WE” have 5 better The probability of getting stamps in both boxes in buying two boxes chance ofhaving at least one team that can enter the second = PI 1 _ 1M t. 1M round of the center: a 044236070! ( p; ) The required conditional probability = E ]M P3 k 0.02799430I 0.44236070] a: 0.0633 0.0632 if using figures in table ZOOO—AS-M at 5—32 2W_AS_M & 543 ...
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