2007_MarkingScheme

2007_MarkingScheme - n§%%* « Marking Scheme...

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Unformatted text preview: n§%%* « Marking Scheme fifififiififififi Hong Kong Examinations and Assessment Authority 2007$§%E¥M£E§% Hong Kong Advanced Level Examination 2007 aaamafie Eflfififig Mathematics and Statistics AS-Level 2K? fififififi§§fi§§k ’ EEWE$JF§IT7E§5 igi’éfigéfi ° %é fiifiJ‘Tifléfiiflfigf’:E19$51Efiifééi‘gflififfiiBfiE/JVDEE’I‘E ° This document was prepared for markers’ reference. It should not be regarded as a set of model answers. Candidates and teachers who were not involved in the marking process are advised to interpret its contents with care. * Jitfi‘Bfl/Elfififi‘ififiili 24 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. Markers 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 concept/technique had been used. Use of notation different from those in the marking scheme should not be penalized. In marking candidates’ 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 1 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 ((1). The symbol @should be used to denote 1 mark deducted for a . At most deducted 1 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. In the marking scheme, ‘r.t.’ stands for ‘accepting answers which can be rounded off to’ and ‘f.t.’ stands for ‘follow through’. Steps which can be skipped are shaded whereas alternative answers are enclosed with rectan les . 25 Solution Marks 1. (a) (i) (Hi) a r r(r—l) x 2 ~ =1+—x + (—) + 1 -- 1M for any two terms correct a 2! a =l+fl+r(r—21)x2 +--- a 2a So, we have f— = and Mr :1) = _—1 . a 3 2a 18 . 3 1 Solvmg, we have a = E— and r = E . 1A for both correct (ii) The binomial eXpansion is valid for 2% <1 . 1M can be absorbed . —3 3 3 Thus, the range ofvalues of x 15 7 < x < E . 1A accept |x | < —2- . x r 1 1 2 . (b) (l) 1~— =1——x——-x +-v- 1M forreplacmg x by —x a 3 18 .. . . . . . — 2x (11) The b1nom1a1expansron IS valid for ~3—— <1 . . —3 3 3 Thus, the range ofvalues of x is —2— < x < E . 1M accept lxl < E -------- --<6) 2. (a) Let u z 2:2 +50 . Then, we have d—u : 4t . 1A dt N : I—m—EOO’ 2 dt (21‘ +50) - . 200 = I —2—du 1M u —200 . i So, we have N = + C , where C 1s a constant. 1A u ': Therefore, we have N = —:i—2—~0—0—— + C . 2t +50 Using the condition that N = 4 -when t = 0 , we have 4 = —4 + C . 1M for finding C Hence, we have C = 8 . 1A Thus, we have N = 8 — —§i . 2t +50 26 Solution Marks Alternative Solution Let u=2v2 +50. Then, we have du = 4vdv . 1A 1 800 [Mg = L—f—L-z—dv (2v +50) 212+5o = J fldu 1M For substitution 1 50 1,2 212+50 {—200} 1A For —200 u 50 u .'.N~4=—_22—0-Q———_2—00 1M Forusing N(0)=4 2: +50 50 N zygfi . 1A 2t +50 200 (b) When N26,wehave 8— 2 =6. 1M 21 +50 So,wehave 1:5 . 1A The number ofbacteria will be 6 million 5 days after the start ofthe research. -------- --(7> 1~e4x 3. (a) y= 1+98X d 1 8x _44x_1_4x88x y #( +8 X e )8(‘ 2 e X e ) 1M for quotient rule or product rule dx (1+9 ”) When x=0 , wehave §y~=—2. 1A x Alternative Solution _1__e4x y— 1+28x lny = ln(1— e4x)~1n(1+ 68X) 1 dy _ —4e4x 898x —— _ 44 - 8_ 1M For log differentiation y dx 1—6 A 1+e A dy _ 1—8“ [—49%- 8e8x ] dx_l+e8x 4x 8x l—e 1+6 _ _4e4x _ 8(1_ e4x)e8x 1+e8x (1+68X)2 When x20, —=—— 2—2 1A 27 Solution Marks (iii) 1n( 22 + 1) + 32 = 2x 22 dx Therefore, we have ix- dz 2:0 Note that x=O when 2:0 . Also note that 9: = —2 . dx x20 d_y dz 2:0 (b) (i) Since (22+1)e32 =ea+flx , we have ln(22+l)+322a+flx . (ii) Since the graph ofthe linear function passes through the origin and the slope ofthe graph is 2 , we have a = 0 and fl = 2 . 1A for both correct 1A lM for chain rule :_3 1A _ 62+2ln(22+l) y: 1 e 2 l+el22+4|n(z +1) y_1—(22 +1)2e‘” _1+(22 +1)4e122 (1 + (22 +1)4e122)(—6(22 +1)2e62 —2(z2 +1)(2z)eéz) dy a (1 «(22 +1)2e6”)(12(22 + D4612” +4(z2 +1)3(2z)e'22) dz (1+(22+1)48122) 2 1M for quotient rule or product rule 1A 28 -------- --<7> 4. (a) (b) Solution Marks (i) Note that 5.1 < 5.3 and k 2 0 . Hence, we have 5.3 = k —1.2 . Thus, we have k = 6.5 . 5.3 =k-1.2 or 5.3 =5.1—k So, we have k = 6.5 or k = —0.2 (rejected as k 2 0 ). Thus, we have k = 6.5 . (ii) Stemgunits) Leafgtenthst 1 2 8 9 2 1 1 2 3 4 4 1M + 1A 3 6 7 9 4 7 5 1 6 5 (iii) The mean = 3.05 hours 1A (accept 3.0500 hours) The median = 2.4 hours 1A (accept 2.4000 hours) The revised mean is greater than the mean obtained in (a)(iii). 1A The revised median is the same as the median obtained in (a)(iii). 1A The change in the mean is positive. 1A There is no change in the median. 1A 29 Solution Marks 5. (a) P(A'nB) = P(Bl A’)P(A’) 20.3(1—(1) 1M ————————————————————— P(A’ n B) = P(A/ l B)P(B) = 0.6b """""""""""""""""" “II I Hence, we have 0.6b = 0.3(1~ a) . l Thus, we have a + 2b =1 . 1 either one (b) P(A m B') = 1“(l-Q'l/IN’M) = 0.7a " ———————————————————————— ——l P(AUB') =1— P(A’n B) = 1 — 0.6b 1M for complementary events Note that P(A o B’) = P(A) + P(B’) — P(A m B’) . So, we have 3a = 4b . Solving a + 2b = l and 3a : 4b , we have a = 0.4 and b = 0.3 . 1A for both correct P(A m B') = P(B'l A)P(A) = 0.7a ---------------------- " P(A U B’) =1— P(A’n B) 21— 0.3(1— a) 1M for complementary events = 0.7 + 0.3a Note that P(A u B’) = P(A) + P(B’) ~ P(A m B’) . Hence, we have 0.7+0.3a=a+1—b—0.7a 1M So, we have b : 0.3 . _________________________ _, By (a), we have 0+2(0~3) =1 - both correct Thus, we have a=0.4 . 1A ____________________ __. | | | | | I I I I I I E I Hence, we have 1—0.61): a+(1—b)—0.7a 1M : I I I I I I I I I I I I I l I (c) Since P(A m B) = P(A) — P(A mB’) , P(A) = 0.4 and P(A n B’) = 0.28 , we have P(A m B): 0.12 = (0.4)(0.3) = P(A)P(B) . 1M for relating P(A m B) and P(A)P(B) Thus, A and B are independent events. lA f.t. Since P(A) = a , we have P(A m B) = P(A) — P(A m B') : a —0.7a = 0.3a . With the help of P(B) : 0.3 , we have P(Afl B) = P(A)P(B) . ]M for relating P(A m B) and P(A)P(B) Thus, A and B are independent events. 1A f.t. Since P(A’l B): 0.6 , we have P(A | B) = l — P(A' | B) = 1 —0.6 = 0.4. With the help of P(A) = 0.4 , we have P(A | B): P(A) . 1M for relating P(A | B) and P(A) Thus, A and B are independent events. 1A f.t. --------- «(7) 30 6. (a) The required probability 1 4 3 ~(‘1‘6X3X§) _L 60 in 0016666667 z0.0167 Solution Marks 1A for numerator or denominator + 1A for all 1A a—l for r.t. 0.017 The required probability era 1 z E a 0.016666667 z0.0167 1A for IS! blacket + lA for all 1A a—l for r.t. 0.017 The required probability _ Cl C? Cl" C? 1 Z 6—0 a 0.016666667 z 0.0167 1A for 2nd blacket + 1A For all 1A a—l for r.t. 0.017 The required probability P310 1 z E z 0.016666667 z 0.0167 1A for numerator or denominator + 1A for all 1A a—l for r.t. 0.017 (b) The required probability 1 5 5 4 ::——+ — —- — 60 (10)(9)(8) _L 45 ‘z 0.155555556 z0.1556 leor (p+q+r+s)+ll\/lfor using (a) 1A a—i for r.t. 0.156 The required probability 1 C150,? = —— + 60 CllOC29 7 533 fz 0.155555556 z0.1556 leor (p + q + r + s) +1Mfor using(a) lA (1—1 for r.t. 0.156 3l Solution Marks The required probability 1 1315st 2 — + 60 P310 7 24—5 z 0.155555556 z 0.1556 1M for (p+q). + 1M for using (a) 1A (1—1 for r.t. 0.156 The required probability 4 3 6 2 4 5 : -— — — + — — -~ (10)(9)(8) (10)(9)(8) _l “ 45 z0.155555556 z0.1556 1Mfor(p+q+r+s) 2A (1—1 for r.t. 0.156 The required probability C§C16 Cf‘Cle : 10 8+ 10 8 C2 C1 C2 C] _l 45 #0155555556 z0.1556 1Mf0r(p+q+r+s) 2A a—l for r.t. 0.156 The required probability £24116 PEPJ‘PE _ 10 + 10 P3 P3 7 :E L 2 0155555556 z 0.1556 32 1Mfor(p+q+r+s) 2A a—l for r.t. 0.156 -------- --<6) Solution 4 8x~40 8‘3 7. (a) lim = lim x :8 x->iuo x+4 x—>:oo 1+3 X g l the equation ofthe horizontal asymptote to Cl is y = 8 . lim f(x)=+oo and lim f(x)=—oo X—>-—4_ x—>—4+ the equation ofthe vertical asymptote to Cl is x = — 4 . The x—intercept of Cl is 5. They-intercept of C1 is —10 . y 1A for all the asymptotes of C1 1A for all the intercepts of CI 1A for the shape of CI -------- --<3> i l l l 33 (b) (i) (ii) Solution 2 g(x):(x+4)8(x+5) g,(x) : 3(x +430: — 2) g'(x):0 when x¢—4 or x=2 ,, _3(x+l) g(x)— 4 g”(x) = 0 when x = ——1 x x<fi4 17—14 l g’(x) + lg"<x) l — g(x) 7| Since g'(2) = O and g”(2) > 0 , the coordinates ofthe relative minimum point are (2, _—27) . 2 Since g’(—4) = 0 and g"(—4) < 0 , the coordinates ofthe relative maximum point are (—4, 0) . <0 if x<—1 Since g"(x) =0 if xz—l , >0 if x>—1 —27 the coordinates ofthe point of inflexion are (—1,—74—) . C1 :y=f(x) , where f(x)= 8x—40 . x+4 2 _ C2 , where ' Note that f(x)=g(x) 8x—40~(x+4)2(x—5) x+4 _ 8 <:> 64(x—5)=(x+4)3(x—5) <:> (x—5)((x+4)3-64)=0 <:> x=0 or x=5 <:> So, the coordinates ofthe points ofintersection are (0, w 10) and (5, 0) . Also, they-intercepts of CI and C2 are —10. When f(x)=0 , we have x=5. When g()r)=0 , wehave x¢—4 or x25. So, the x-intercept of CI is 5. Also, the x—intercepts of C2 are —4 and 5. 34 Marks. 1M forjustification ————————— —— 1A either one 1 M for justification 1A Marks T_________----________---__ _____________________________ " 1A for the shape of C2 1A for all the extreme points and the point of inflexion 1A for all the points of intersection H ll 1 4; -------- --<8> (c) The required area 5 = [(f<x>—g<x>)dx 0 = J‘5(8x—4o_(x+4)2(x—5)]dx 0 x+4 8 5 3 2 if 8— 72 — L+§5——'3x—10 dx 0 x+4 8 8 5 4 3 2 =[8x—721n(x+4)—[i‘—+3L—3§——10x]] =£5_5_721n 2 32 4 =253—144ln 3 32 2 53395677443 z 33.9568 0 1M accept L(g(x) — f(x))dx 1M for division 1A for correct integration 1A (1—1 for r.t. 33.957 -------- --<4) 35 Solution Marks (a) (i) The total profit made by company A 6 , = J. f(t) dt 1A withhold 1A for omitting this step 0 ' ‘ z %( f(0) + f(6) + 2(f(1) + f(2) + f(3) + f(4) + f(5))) 11/1 for trapezoidal rule z 37.48705341 , z 37.4871 billion dollars lA a—l for r.t. 37.487 (ii) f(t)=1n(e’+2)+3 9112: 6' 1A dt 6[ + 2 d2f(t) dt2 _ (e1 +2)e’ —e’(e’) (e’ +2)2 I Z [28 2 1A (e +2) . d2f(t) . Smce d 2 >0 , f(t) lS concave upward for 03:36 . 1M t Thus, the estimate in (a)(i) is an over-estimate. 1A fit. ' -------- --(7) 1 1 :2 t4 1 1 2 1 4 b i =— 1+—+ +... =——+———t + 1+... 1A —1foromittin l) 0 40—:2 mi 40 1600 J 40 1600 64000 pp g (ii) Note that e’ =1+z‘+%t2 +éz‘3 +5251 +--- . Hence, we have 1M for any four terms correct 8 e’ 40—;2 :8 —1-+—1—t2+ I t4+~~ 1+t+i12+it3+—1—t4+--- 40 1600 64000 2 6 24 i:1t+ 21t2+ 231r3+ 263 t4+--- 1A pp-l foromitting‘ ’ 5 5 200 600 24000 (iii) The total profit made by company B 6 = j gde 0 6 z I i+lt+112+£t3+ 263 tgdt 1M 0 5 5 200 600 24000 6 = lt+ 1 t2+ 7 t3+ 23 14+ 263 t5 1A for correct integration 5 10 200 2400 120000 0 :41.8224 billion dollars 1A 61—] for r.t.4l.822 -------- --(6) (c) Since the estimate in (b)(iii) is an under—estimate, we have 1A 6 6 , J-f(t)dt<37.4871<41.8224< J-g(t)dt . 0 0 Thus, Mary’s claim is correct. 1A f.t. 36 Solution Marks 9. (a) A(t)=(—t2+5t+a)ekl+7 Since A(0)=3 , we have a+7=3 . Thus, we have a=—4 . 1A The required amount of water stored =(—12 +5—4)ek +7 : 7 million cubic metres 1A -------- --<2) (b) A(t) = (—t2 +5r—4)e’” + 7 dA(t) dt = (~2t + 5) 9’” + (—t2 + 52‘ — 4) (Item) 1M for product rule = (—ktz +(5k—2)t+5—4k )e’” Note that when t=2 , dAU) =0 . dt So,wehave 2k+l=0. 1M —1 Thus, we have k = 7 . lA -------- «(3) (c) (i) When A(t) 2 7 , we have — t2 + 51— 4 Z 0 1M accept setting quadratic equation 22 — 51‘ + 4 s 0 1£tS4 lA(acceptt=l—>t=4) Thus, the adequate period lasts for 3 months. i (ii) Note that A(t)=(—t2 +51—4)e2 +7 . 2 ;’ So, (Wt) = t——%+7 e2 dt 2 2 and (11:51) =0 when t=2 (rejected since [>4 ) or [=7 . 1A fort = 20r7 dA < 0 if 4 < t < 7 I do =0 if [=7 1Mfortesting+lA 1‘ >0 if 7<t512 So, A(t) attains its least value when t: 7 . The least amount of water stored = A0) a 6.456447098 z 6.4564 million cubic metres 1A a —l for r.t. 6.456 million cubic metres 37 Solution —I Naemm Ao)=c42+y—4)é?+7. 2 ‘_’ So, dAU) =[L—i+7]ez dt dAU) m and =0 when t=2 (rejectedsince t>4)or t=7. lAfort= 20r7 dZAU) dt2 dzAU) d 2 1M for testing + 1A 1‘ Therefore, we have (:7 Note that there is only one local minimum after the adequate period. So, A(t) attains its least value when t = 7 . The least amount of water stored = A(7) z 6.456447098 z 6.4564 million cubic metres 1A a —l for r.t. 6.456 million cubic metres dzA t (in) 2( ) either one (it 2 —I —/ I —t 9! — w : ——+———7— 82 +[l—2 e2 E 4 4 2 2 i l _ 2 i I = _l:_.. + 1.3.: __ 8 e 2 . _ _ _ _ _ _ _ _ _ n _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ __' 4 4 —t A t l 9 2 — . . . (iv) Since d ( ) = get—Sf —75)e 2 > 0 for 10 <t <12 , 1M for conSidering the Sign ------ —~, I A(t) increases, within that year, filter the adequate period has ended for 6 months. 1A f.t. either one 2 -' l . —1 13 41 ‘“ Since d A20) =—((t~—)2———je2 <0 for 10<t<12, ————————————————————————————— —~' dt 4 2 4 dAU) . . . decreases, Within that year, after the adequate period has ended for 6 months. 1A fit. * dt -------- "00) 38 10. Solution (3) The required probability _1 [2.4%‘2-4 + 2.4‘e‘2-4 2.42624] + 0! 1! 2! 0.430291254 z 0.4303 (b) Let $ X be the expense ofa customer. Then, X~N(375,1252) . The required probability = P(300 < X < 600) I P(300—375 <Z< 600—375) 125 125 = P(—0.6 < Z < 1.8) = 0.2257+0.4641 = 0.6898 (c) The required probability = (0.25)(0.6898) + (0.8)(0.5 — 0.4641) = 0.20117 z 0.2012 ((1) The required probability 3 —2.4 s .2445—(020117)3 2; 0.00170163 ,2 0.0017 (e) The required probability 4 —2.4 0.00170163+(0.20117)4 4! ~' " 0.430291254 ‘z 0.00443193‘1 z 0.0044 (1) Suppose that the revised least expense is $ x . Then, we have P(X2x)=0.33 . x—375 So, we have P(Z 2 ): 0.33 . 125 Therefore, we have xl—23575 = 0.44 . Hence, we have x = 430 . Thus, the revised least expense is $ 430 . 39 Marks 1M for complemeantary events + 1M for Poisson probability 1A a—l for r.t. 0.430 -------- --<3> 300—375 S Z S 600—375 125 125 1M(accept P( )) 1A a—1 for r.t. 0.690 -------- «(2) 1M for 0.25(b)+0.8p , 0 < p<0.5 1A a~l for r.t. 0.201 -------- --<2) 3 —2.4 1M for 3'45—(03 1A a—l for r.t. 0.002 -------- --(2> 1M for numerator using (c) and (d) + 1M for denominator using (a) 1A a—l for r.t. 0.004 -------- --<3) 1A 1A 1A -------- --<3> Solution 11. Let Xg be the net weight ofa can of brand D coffee beans. Then, X ~N(300,7.52) . ' (a) The required probability :P(X<283.5 or X>316.5) = p(Z <M or Z >w) IMWCCN HZ: 283.5—300 or 22 3105—300” 7.5 7.5 7.5 7.5 . = P(Z < —2.2 or Z > 2.2) = 2(0.0139) 20.0278 1A a—l for m. 0.028 ’ -------- --(2) (b) (i) The required probability = (1 ~ 0.0278)“(0.0278) M for (1_ p)“ p _____________ __,‘ z 0020387152 ' i 2 0.0204 1A a—l for r.t. 0.020 : I l (ii) The required probability i — C3°(1— 0 0278)29(0 0278) IM f C3°(1— 29 W for p z (a) _ 1 . . or 1 p) p "" " (either one) Q 0368195889 ‘ 20.3682 1A a—l for m. 0.368 5 I (iii) The required probability E z (1 — 0.0278)30 + 0.368195889 1M for ((1_ p)” W) + 1M for q = (moo-J z 0.797404575 z 0.7974 1A a—l for r.t. 0.797 -------- ——<8) (0) (i) The required probability z —;—(0.368195889) 1M for %((b)(ii)) z 0.184097944 20.1841 1A a~1forr.t. 0.184 (ii) The required probability N 0.184097944 ' 1M for numerator using (c)(i) 0.797404575 + 1M for denominator using (b)(iii) ‘z 0230871443 z 0.2309 1A a—l for r.t. 0.231 -------- --(5> 40 Solution Marks 12. (a) The required probability =léllél+léllél+l§l1él ‘15625 =(l327936 z 0.3279 (b) The required probability eououu--- ll 4 6' 04444444444 z 0.4444 (c) The required probability 4 5124 g: 0.262144 z 0.2621 1M for geometric probability lA (1—1 for r.t. 0.328 -------- -~(2) 1M must indicate infinite series and have at least 3 terms 1M for summing geometric sequence 1A (1—1 for r.t. 0.444 -------- --(3> 1M for numerator = (b) — (a) + 1M for denominator using (b) 1A a—l for r.t. 0.262 — The required probability 5124 : __15625 E 9 _ 4096 _15625 i=(1262144 z 0.2621 1M for complementary probability + 1M for denominator using (b) 1A a—l for r.t. 0.262 41 -------- --(3> Solution (d) (i) The required probability 1 3 1 — (5X7) + (3X1) 5 7 m 0.714285714 z 0.7143 Marks 1M for either case 1A a—l for r.t. 0.714 The required probability 1 4 =1_ _ _ (2X7) 5 7 z 0.714285714 z 0.7143 ll (ii) The required probability 5 = 1 —-— 7 285714286 3 7 0. 0.2857 z 1M for complementary probability 1A a—l for r.t. 0.714 1M for 1—(d)(i) 1A awl for r.t. 0.286 The required probability 1 4 = —— — l 1 (2)(7)()() (iii) The required probability 4 2 (3X3) z 4 2 4 2 (3)(;) + (1 — 6X1 — 7) _ 3‘. 33 4 02424242424 z 0.2424 42 1M for denominator = (2)(7) 1A a—l for r.t. 0.286 1M for p‘] p=<b) + 1M for { N o q=(d)(n) 1A a—l for r.t. 0.242 -------- --<7) pq+(1~p)(l -q) rl p=(d)(ii) q=(b) ...
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