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1990 A. Maths Paper1 Marking Scheme

# 1990 A. Maths Paper1 Marking Scheme - GA,3.2.r‘...

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Unformatted text preview: GA ,3 .2: .r‘ RL‘iﬂ‘R ESTRmMEDam EBB-CH: I. ‘1 . i ‘ z , «U' ‘. ﬁrmwwm ‘ uanwmlmWmmwmwmw_wm*LW.W~_“___MMWWMLWhA~" RESTRICTED W‘éISB'CH: , v.1 1990 HKCE Additional Mathematics 1 Solutions Marks Remarks 1' f'(x) = c/x:+k —gsin2x + sin2x—2./ x2 + k For product rule. dx dx = 2 x +rk cost + _§§l§£§ .lx’ + k f'(0) = 1 2 Jk-= 1 For substituting x = O in f'(x) 1 k‘: 2. (a) Kg = 6E - 5K Omit vector sign A A (pp - 1) =-i+2j 1A —a -+ c713=071+ﬁ 1M OB+BP _9 _* acceptable = 0A + tAB A A = -ti + (2t +5)j 1A . Solution 1A 5? : ﬁg = t . 1 — t 1M (pp - 1) A A A t(-i + 7j) + (1 — t) (Sj) A A —c1 + (2t + 5)j 1A Omit dot sign (fr—t) l/ .L‘. 1}; l { .-' / I]. {1: Q RESTRICTED mam: , Solutions Marks Remarks Multiplying 1 + itanG 1 + itanQ acceptable (a) 1 + itan9 = cosQ + isine cose + isine 1 — itanO c059 — isinG ' cosG + isinQ = (c0529 — sinZQ) + 231n9cosﬂi cos 9 + sin 9 “= c0329 + isinZG Alt. Solution ______._————- 1 + itanG _ cosQ + isinQ 1 - itanO _ c059 - isinO c059 + isine 3m = c0520 + isin29 Can be omitted _ 2n cos 3 + lain—5 Accept ? 8W cos—g + isin—E 4W 4W or cos-3 — isin—E etc. Alt. Solution 1+J§1=1+f§1 1+ﬂ1 1-B1 1-131 1+J§1 =lc-1+J‘51)=<:osz—1“-+:Lsim§-T 2 3 3 Vé . 75 (1 + fji) = (cosgl + isingz 1—f31 3 3 = co (3k + 1)2’"‘+ 181 (3k + 1)27} E 9 n 9 k = O, 1, 2 -2A 2A for k = O, 1, 2 0R 2W 2W = (:05-'—9 + 15111—6 cosgﬂ + isingz, 9 9 1A+1A Angles in degree cosl4ﬂ'+ isin14ﬁ'(or cos-4W + isin—AN) acceptable 9 9 9 9 .g .. . n «:3 .3 H dnHJ'iﬁ mile-What“: , 1A 29 f PW . "W-.- ,.,.-_~.,/.. .... 2... .... ”.4.- j -.....J.————————L———L————- RES l KIL I |:U WE‘DXFF Solutions , Marks Remarks (a) x+ = k+2 a f) - k } (*) P _ (b) (d + 1) ([3+ 2) = 4 ----- (1) u13+(1+ﬁ)+3(+2=4 k + k + 2 +<x + 2 = 4 " = -2k For eliminating/3. Subs. into the equation (-2k)2 — (k + 2) (-2k) + k - 0 _6k2 + 5k = 0 k = 0 or —5/6 Alt. Solution 1 Subs. >( = -2k into (*) { ~2k + ﬁ = k + 2 —2kﬁ=k _ k = 0 or {i=4 -2k-é=k+2 k - :2 6 Alt. Solution 2 Subs..x = —2k, ﬂ - 3k + 2 into (1) (-2k + 1) (3k +2 + 2) = 4 bkz + 5k = o k = O or -% Unit circle Axes or curves not labelled Correct centre (pp - 1) Horizontal straight line Separate diagrams (pp - 1) R'*’ Position correct Solve Circle and line touch at correct point. {(x — 3)2 + y2 = 1 1A The intersection is the complex no. 3 + i y = 1 Ans.: 3 + i 'A 33 z'nu-s-n' r .'. .- be" ' a, I" . ‘ '1'". V f ‘ / g 1/ ‘ I“ '3 S‘ ; 1’ I ' I 6. RESTRICTED WEBB‘EFF ‘ P4 Solutions Marks Remarks (x+2)2-8‘x+2[ +15>,o |x+2I2-8|x+2| +1sgo (|x+2|—3) (|x+2t—5);o Ix + 2| 2 5 o'r” Ix + 2|s3 Omit 'or' (pp -1) use 'and' (no mark) (x23 or xs—7) or —55 xél 1A+1A use ',' (pp - l) Alt. Solution Case (i) x 2 —2 (or x > —2) xa-L x<—2 (deduct no (x+2)2—8(x+2)+15;o mark) (x — 1) (x — 3)?-0 (2) Solve without stating range of x (no mark) Since Case (11) x (—2 (or x é-Z) (x + 2)2 + 8(x + 2) +1520 (x+ 5) (x+ 7)20 x2—5 or xS—7 Since x<—2, xS—7 or —2>x>/—5 Combining the 2 cases, x33 or xS—7 or —SSx\<l 2x + 4y + 4% + ioyg-é = 0 1M For implicit differentiation £1 8 -(x + 2x) dx 2x + 5y ' 1A —<x + 2y) , -_1 2x + 5y 2 1M y = 0 ' 1A Subs. into the equation, x = :1 1A The equations are x + 2y + 1 = 0 _ -1 I -l 1 y"'2'x+§ and y’T“? 1A+le+2y—1=0 7 i i 1TBEB£:;I”Eililﬁﬁi'ﬂll[IBTiFlﬂlﬁmlriﬂitﬂﬂib'l W....- 8. (a) (i) (11) (b) (i) (ii) (111) , . It I f 3 / 'f'i/ m.w,, .H ."w.~...m-_«{t;;;_._w..h.g*u. ‘ ' ‘ ,‘,__-__w RESTRICTED [79% '57:“: P5 Solutions Matks Remarks 35 . 2’ = [3?] IE" cosQ 1A Omit vector sign (pp - 1) = IZI c059 1A Omit dot sign » (pp - 1) )7). E’ = [fl '2' cosQ = I?! c039 II E N + D N g4 - m + ncosZQ --9 Y- (mi-(+1137) % “1 ll --y = mcosZG + n? . y = mcosZO + n From (1), m + ncosZG = mcosZG + n (m - n) (1 - cos29) = 0 , m ' n (l’coszeil) Using (a) (11) a blx )\+ l AK+ 1 Av:- IITKI “/32 +42 = 5 59-5 CB 25 3 Eli; ; mam Imummtn-h- m- 1A 1A IN 1A 1A 1M 1A IN IA IN IA Accept (m — n) (1-i.9’)=0 Accept omitting c0329#1 Rum-*5." (a) (i) (11) Alt. { 4‘ i- ~ »‘I n " I _. I.I - “near W. ..A__..__ Wm"~--'m~ - RESTRICTED Maﬁa 2 Solutions Marks f(x) = x2 + 4x + 1 = (x + 2)2 — 3 1A Vertex of Cl is (—2, -3) IA x2+4x"+1=-o x = -2 t J3' 1A P0 = (-2 + J3) - (-2- J33 1M =2JT 1A Solution (.m - ﬁ)2 = (u-Fﬁ )2 - 44;? 1M = (—4)! - 4 - 12 1A mew-(3| -2J§ 1A (b) (i) (ii) Alt. (x' - P10! (111) (c) (i) (ii) Vertex of C2 is (-2, —3 a m) [M ng) = (x + 2)2 — (3 + m) 1A x2 + 4x + l - m I 0 is an; .3 ‘3 ,3 ECTDﬁl‘t-‘mﬂﬂa rhﬂl'ﬂ' Jim“ i 57‘. . \ Remarks Answerin decimal — no mark For subtraction Accept PQ = 12 Accept PQ = d -}3 x2 + 4x + 1 - m x = -2 t./m + 3 1A P'Q' = 2.Jm + 3 1A Accept P'Q' = 44m +I2 Solution ﬁ')2=4m+12 - 1A = |;x' - p 'l- 2./m + 3 1A 21—sz - 2(2Jﬁ) 1A m = 9 1A Vertex of C3 is (-2 + n, -3) 1M h(x) = (x + 2 - n)2 — 3 1A h(0) = 0 O = (2 - n)2 — 3 1M n - 2 t./§ 1A+1A 3.73, 0.268 10. (a) (b) (C) _8m.m_ <1) ix: dx —2(x2 +'k ~ 2) (x2 + 2)2 { ' f h~ §= , ./ If— -! , 4 m- -. - .-»—.—. . RESTRICTEE’W‘E 3U? Marks Solutions 2(x2 + 2) - 2x(2x + l) (x2 + 2)2 = —2(x2 + x - 2) (x2 + 252 < 0 x2+x-2>0 x 71 (11) -2(x2 + x - 2) Curve C or x (—2 - 0 (x2 + 2)2 1 or -2 1, y - l (1, 1) is a maximum point. -2, y = 2% (-2, :%) is a minimum point. : Shape Curve C2 : Intercepts End points Turning points Shape Intercepts End points Turning points (-2, 1%) , (l, 0) I 1A 1A 1A ihd 1A . 1A 1A 8 1A 1A 1A 1A ___3_ 1A 1A 1A 1A ___£_ ., _. ,.;..».n.l:c;r_n,~|n1u:mmnﬁ-m ? Remarks < 0. no mark. Curve not labelled but position correct - deduct 1 mark only. Pure plotting without part (a) — no mark. W. V . ~ i‘ f I ‘ .i' ‘ ﬂ/ 3 4}: . RESTRICTED may“: . Candidate Number Centre Number Seat Number 10. If you attempt Question Total Marks on this page 10, fill in the details in the first three boxes above and tie this sheet into your answer book. SO-CE-A MATHS |——7 y (5%)0N 26.6%) __ 6 _ Go on to the next page 11. i" (a) ' I RESTRICTED WE Solutions By Sine Law, x I 3 2‘ 2F 2? sin(:.- 3 — 9) sin—3 w. x = 2‘f3ein G; — 9) (b) S = %3xsin9 = 3.]? sintg - 9)sin9 dS —. d9 3 ijcosesin 3 - 3.]331n(% - 29) d5 A II -— . 0 when 9 - —— (3 d9 dzs 6 — - -6 J3cog(‘—'3’- 29) d92 dzs 357 9:“ :—6J'3_ Z . A. .2 max. Alt. Solution for checking maximum d3 T __ < 6 d9 > O for 0 < 9 6 is. < 0 for E < 9 < 1? d9 6 3 9 - EIis a maximum S max |‘ Lil tr 6 - 3 J3 sin(§_' %)sin .11}— 4. .E 6 ._- ...—.... 4 .. IL- 9) - sinGcos(W - 9)] ”(33(45— ‘ P_Q Marks Remarks IN IA x = 3cosO - chsinG 2 1A 1A S = 2sinOcosO - 2 , z 3"lisin 9 2 --3f§+3/T ' 4 2 cos(a - 29) 3 l 1M+1A Accept omitting O S 953% 1A 1M Awarded only when the 2nd derivative is correct. 1A for correct ranges of 9 [M for slope change from +ve to -ve. Only awarded if 1A if max. is 8 checked. g»: .p 54. :R ESTRJGT qudazmuh f E / g f ‘4 ‘ . - I >T. ‘ g" ’ .v r 1 - W- - 7w __.__._ _-;."::... __:::,:::;._;:_;::.—_’m__._.::::;.f:_:.;;:~_;::;:;.;. ,. W117,“ J‘ M". / RESTRICTED WEBB'U‘F P-IO Solutions Remarks (c) (1) x = 2 fjsin(3:' — dx dt - -2 chosﬂ— - O)-— . dx /3' Omit -ve sign Since a? = _ 3 (no mark) 92 1 dt 6cosGL - 9) (11) o: 96737 l Scos(Z — 9)§'1 2 3 d9 _ 1 greatest value of a? - 3 d9 1 least value of a? = E w: 1 wSWIFQ'“INIVFFJIEI'I m-tﬂfi‘lﬂ: GA29 i} rum m” V. “ “km". “NH “mmm....__.._._ V ‘ ‘ 1" ‘ ~. .7}; { . if T/ . . C I x! {1 I" i : h {I E i 4 xi 3 3 RESTRICTED WEEK“: p.” Solutions Marks Remarks 12. (a) Let z = x-+ yi (1) 22 = (x + yi) (x - yi) - x2 + y2 ,’,real 1 (11) z + E = (x + yi) + (x — yi) - 2x - 2Re(z) 1 H 2 (b) (i) (1) BY (a) (11) _ l - ': Re(pr) = 3(pr + pr) 1A 1 - = §(pr + pr) 1A - 0 1 (2) P 1 ' f Rd?) - 30;. + (13)) 1A Re(§) = Reerf) 1A a l 2 I2 _ 59422). 2(r + f) ‘ rf 1A = l E} + ﬁr 1A = O l 2 r? , V = O 1 Alt. Solution Let p = a + bi, r = c + d1 (1) pi + Br = (a + b1) (c - d1) + (a — b1) (c + d1) = ac + bd s 0 1A Re(pf) = ac + bd 1A - 0 1 (2) B a a + b1 c - d1 1M r c + d1 ' c - d1 = (ac + bd) + (be ~ ad)i c2 + d2 c + bd 1A a “(5) = 27117 (b) t 3‘: C ‘RVEsll'RICTED marsh Solutions (ii) Method 1 " B a Re(r) O arg(1::) = tilt 1 22 arg p — arg r = :2 or t 2 \$.0A_1_0C r‘,0ABC is a rectangle Method 2 lACl2 = (p - r) (p - r) = pﬁ'— p? — pr + r? l = pp + r? ( l I | IOBI2 = qﬁ = (p + r) (p + r) = pp + rr AC 3 QB .‘»0ABC is a rectangle Alt. Solution Method 1 Slope of CC = E (p = a + bi, r = c + di) c Slope of 0A - g Product of slope - ﬂ . E c a = _ac/ac (from (1)) = —1 OCJ_0A I, OABC is a rectangle Method 2 0A2 = 32 + bz, co2 = c2 + d2 AC2 = (a — c)2 + (b - d)2 = a2 + c2 + b2 + d2 — 2(ac + bd) = (a2 + b2) + (c2 + d2) = 0A2 + 002 OA_LOC (Converse of Pythagoras Theorem) .1 OABC is_a rectangle. Marks 1A 1A 1A 1A IA 1A 1A 1A Remarks Accept omitting t sign Accept omitting t if 2 Accept the negligence of considering a = 0 or c = 0 (b) (iii) 9 , .1 u‘ 5 '1; ' GA29 -" "y‘la_.wu m.“ an”, ”m. - _.. “-7.— R'Eé'TEiEED mm Solutions .Marks p=2ri E — r = 2ri — r p + r 2r1 + r =‘1+21 1A 1 + 21 _ 3 4 _ 5 + g1 _ 1A ILLLJE = arg(p + r) 9 1A tan 9 = gég 1M = 4 3 1A 14 ~ ‘V'i'RESTRICTED mmth P13 Remarks 3 + 41 5 Accept arg (p - r) - arg (p + r) = 9 (can be omitted) ...
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1990 A. Maths Paper1 Marking Scheme - GA,3.2.r‘...

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