ADDITIONAL MATHEMATICS PAPER 1

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1 000-CE A MATH PAPER HONG KONG EXAMINATIONS AUTHORITY HONG KONG CERTIFICATE OF EDUCATION EXAMINATION 000 ADDITIONAL MATHEMATICS PAPER 8.0 am 0.0 am ( hours This paper mus be answered in English. Answer ALL quesions in Secion A and an THREE quesions in Secion B.. All working mus be clearl shown.. Unless oherwise specified, numerical answers mus be eac. 4. In his paper, vecors ma be represened b bold-pe leers such as u, bu candidaes are epeced o use appropriae smbols such as u in heir working. 5. The diagrams in he paper are no necessaril drawn o scale. 香港考試局保留版權 Hong Kong Eaminaions Auhori All Righs Reserved CE-A MATH

2 FORMULAS FOR REFERENCE sin ( A ± B sin A cos B ± cos Asin B cos ( A ± B cos Acos B sin Asin B an A ± an B an ( A ± B an A an B A + B A B sin A + sin B sin cos A + B A B sin A sin B cos sin A + B A B cos A + cos B cos cos A + B A B cos A cos B sin sin sin Acos B sin ( A + B + sin ( A B cos A cos B cos ( A + B + cos ( A B sin A sin B cos ( A B cos ( A + B 000-CE-A MATH 保留版權 All Righs Reserved 000

3 Secion A (4 marks Answer ALL quesions in his secion.. Solve >. ( marks d. Find (a sin, d d (b sin ( +. d (4 marks. (a Show ha. + ( + ( + + d (b Find ( d from firs principles. (5 marks 4. P (, is a poin on he curve ( + ( + 5. Find (a he value of d d a P, (b he equaion of he angen o he curve a P. (5 marks 5. (a Solve. (b B considering he cases and >, or oherwise, solve. (5 marks 000-CE-A MATH Go on o he ne page 保留版權 All Righs Reserved 000

4 + i 6. Epress he comple number in polar form. + i Hence find he argumen θ of principal values < θ. + i + i α and β are he roos of he quadraic equaion + ( p + p 0, where p is real. (a Epress α + β and α β in erms of p., where θ is limied o he (6 marks (b If α and β are real such ha α + β, find he value(s of p. (7 marks 8. B D In Figure, O C A OA i, OB j. C is a poin on OA produced such ha AC k, where k > 0. D is a poin on BC such ha BD : DC :. + k (a Show ha OD i + j. Figure (b If OD is a uni vecor, find (i k, (ii BOD, giving our answer correc o he neares degree. (7 marks 000-CE-A MATH 4 保留版權 All Righs Reserved 000

5 Secion B (48 marks Answer an THREE quesions in his secion. Each quesion carries 6 marks. 9. F E C B D A O Figure In Figure, OAC is a riangle. B and D are poins on AC such ha AD DB BC. F is a poin on OD produced such ha OD DF. E is a poin on OB produced such ha OE k(ob, where k >. Le OA a and OB b. (a (i Epress OD in erms of a and b. (ii Show ha OC a + b. (iii Epress EF in erms of k, a and b. (5 marks (b I is given ha OA, OB and AOB 60. (i Find a. b and b. b. (ii Suppose ha OEF 90. ( Find he value of k. ( A suden saes ha poins C, E and F are collinear. Eplain wheher he suden is correc. ( marks 000-CE-A MATH 5 4 Go on o he ne page 保留版權 All Righs Reserved 000

6 0. Le 7 4 f (. + (a (i Find he - and -inerceps of he curve f (. (ii Find he range of values of for which f ( is decreasing. (iii Show ha he maimum and minimum values of f ( are 4 and respecivel. (9 marks (b In Figure, skech he curve f ( for 5. ( marks (c Le 7 4 sinθ p, where θ is real. sin θ + From he graph in (b, a suden concludes ha he greaes and leas values of p are 4 and respecivel. Eplain wheher he suden is correc. If no, wha should be he greaes and leas values of p? (4 marks 000-CE-A MATH 6 5 保留版權 All Righs Reserved 000

7 Candidae Number Cenre Number Sea Number Toal Marks on his page If ou aemp Quesion 0, fill in he firs hree boes above and ie his shee ino our answer book. 0. (b (coninued O 5 Figure 000-CE-A MATH 7 6 Go on o he ne page 保留版權 All Righs Reserved 000

8 This is a blank page. 000-CE-A MATH 8 7 保留版權 All Righs Reserved 000

9 . (a Le w cosθ + i sinθ, where 0 < θ <. I is given ha he 5 comple number w + is purel imaginar. w Show ha cos θ + 5 cosθ 0. Hence, or oherwise, find w. (8 marks (b A and B are wo poins in an Argand diagram represening wo disinc non-zero comple numbers z and z respecivel. Suppose ha z wz, where w is he comple number found in (a. z z (i Find and arg (. z z (ii Le O be he poin represening he comple number 0. Wha pe of riangle is OAB? Eplain our answer. (8 marks 000-CE-A MATH 9 8 Go on o he ne page 保留版權 All Righs Reserved 000

10 . Consider he funcion f ( 4m (5m 6m +, where (a Show ha he equaion f ( 0 has disinc real roos. m >. ( marks (b Le α, β be he roos of he equaion f ( 0, where α < β. (i Epress α and β in erms of m. (ii Furhermore, i is known ha 4 < β < 5. ( Show ha 6 < m <. 5 ( Figure 4 shows hree skeches of he graph of f ( drawn b hree sudens. Their eacher poins ou ha he hree skeches are all incorrec. Eplain wh each of he skeches is incorrec. f ( O 4 5 Skech A Figure CE-A MATH 0 9 保留版權 All Righs Reserved 000

11 . (b (ii ( (coninued f ( O 4 5 Skech B f ( O 4 5 Skech C Figure 4 (coninued ( marks 000-CE-A MATH 0 Go on o he ne page 保留版權 All Righs Reserved 000

12 . N R θ B Q m A 00 m m 00 m P Figure 5 Two boas A and B are iniiall locaed a poins P and Q in a lake respecivel, where Q is a a disance 00 m due norh of P. R is a poin on he lakeside which is a a disance 00 m due wes of Q. (See Figure 5. Saring from ime (in seconds 0, boas A and B sail norhwards. A ime, le he disances ravelled b A and B be m and m respecivel, where Le ARB θ. (a Epress an ARQ in erms of. Hence show ha 00 (00 + anθ. (4 marks (b Suppose boa A sails wih a consan speed of m s and B adjuss is speed coninuousl so as o keep he value of ARB unchanged. (i Using (a, show ha (ii Find he speed of boa B a 40. (iii Suppose he maimum speed of boa B is m s. Eplain wheher i is possible o keep he value of ARB unchanged before boa A reaches Q. ( marks END OF PAPER 000-CE-A MATH 保留版權 All Righs Reserved 000

13 000 Addiional Mahemaics Paper Secion A. 0 < <. (a sin cos (b 6 sin( + cos( +. (b 4. (a 5 (b (a or (b 6. cos + i sin, (a p, p (b 8. (b (i 5 (ii 48 保留版權 All Righs Reserved 000

14 Secion B Q.9 (a (i a + b OD OA + OC (ii OB + b a + OC OC a + b (iii EF OF OE OD kob a + b ( kb a + ( k b (b (i a. b a b cos AOB (cos60 b. b b 4 (ii ( OE. EF 0 kb.[ a + ( k b ] 0 ka. b + k( k b. b 0 k + 4k( k 0 7k 4k 0 k 0 (rejeced 7 k 4. or 7 k 4 保留版權 All Righs Reserved 000

15 7 ( Pu k : 4 7 EF a + ( b a b 4 4 CE OE OC Since 7 b ( a + b 4 a + b 4 CE µ EF, C, E, F are no collinear. The suden is incorrec. 保留版權 All Righs Reserved 000

16 Q.0 (a (i Pu 7 7 0, he -inercep is. Pu 7 7 0, he -inercep is. 4 4 (ii f ( is decreasing when f ( 0. 4( f ( + (7 4 ( ( ( + ( + ( (iii f ( is increasing when f ( 0, i.e. 4 or. f ( 0 when 4 or. As f ( changes from posiive o negaive as increases hrough, so f ( aains a maimum a. A, 4 he maimum value of f ( is 4. As f ( changes from negaive o posiive as increases hrough 4, so f ( aains a minimum a 4. A 4, he minimum value of f ( is. 保留版權 All Righs Reserved 000

17 (b (, 4 7 (, 5 f( O (4, (5, 7 7 4sinθ (c Pu sinθ,f (sinθ p. sin θ + The range of possible value of sin θ is sinθ. From he graph in (b, he greaes value of f( in he range is 4. he greaes value of p is 4 and he suden is correc. From he graph in (b, f( aains is leas value a one of he end-poins. f (, f (. he leas value of p is and he suden is incorrec. 保留版權 All Righs Reserved 000

18 Q. (a w cosθ + i sinθ w cos θ + i sin θ w w cosθ + i sin θ cos( θ + i sin( θ cosθ i sinθ 5 + w cos θ + i sin θ + 5(cosθ i sinθ cos θ + 5 cosθ + i(sin θ 5 sinθ Since w + 5 is purel imaginar, w cos θ + 5 cosθ 0 ( cos θ + 5cosθ 0 cos θ + 5 cosθ 0 cos θ or cosθ (rejeced θ ( 0 < θ < Imaginar par sin 5 sin 0 w cos + i sin z (b (i w z z arg( arg( z w 保留版權 All Righs Reserved 000

19 z z (ii z z z z i.e. OA OB. AOB arg( z arg( z z arg( z Since OA OB, OAB is isosceles. OAB OBA ( OAB is equilaeral. 保留版權 All Righs Reserved 000

20 Q. (a f ( 4m (5m 6m + Discriminan ( 4m + 4(5m 6m + 6 m 4 m + 4 4(9 m 6 m + 4(m > 0 ( m > he equaion f ( 0 has disinc real roos. (b (i 4 m ± m ± (m Since α < β, α m (m m + β m + (m 5m (ii ( Since 4 < β < 5, 4 < 5m < 5 5 < 5m < 6 6 < m < 5 ( Skech A : Since he coefficien of in f( is posiive, he graph of f( should open upwards. However, he graph in skech A opens downwards, so skech A is incorrec. Skech B : 6 Since α m and < m <, 5 6 > m > 5 0 > α > 5 In skech B, α is less han, so skech B is incorrec. 保留版權 All Righs Reserved 000

21 Skech C : 4m (5m 4m (5m 6m + 6m + 4m (5m 6m 0 (* Discriminan ( 4m + 4(5m 6m 6m 4m m(m 6 Since < m <, > 0. 5 As > 0, equaion (* has real roos, i.e. f( and alwas have inersecing poins. However, he line and he graph in skech C do no inersec, so skech C is incorrec. 保留版權 All Righs Reserved 000

22 Q. (a 00 an ARQ 00 anθ an ( ARQ+ QRB an ARQ + an QRB (an ARQ (an QRB ( ( ( (b (i A 0, anθ PQ RQ (ii Since ARB remains unchanged, 00( d (00 (00 00( d d (00 d 0000 (00 d d (00 A 40, d d ( he speed of boa B a 40 is 5 m s. 9 保留版權 All Righs Reserved 000

23 (iii From (ii, d d (00 d d ( ( d When > 00 (, >. d So i is impossible o keep ARB unchanged before boa A reaches Q. 保留版權 All Righs Reserved 000

24 000-CE A MATH PAPER HONG KONG EXAMINATIONS AUTHORITY HONG KONG CERTIFICATE OF EDUCATION EXAMINATION 000 ADDITIONAL MATHEMATICS PAPER.5 am.5 pm ( hours This paper mus be answered in English. Answer ALL quesions in Secion A and an THREE quesions in Secion B.. All working mus be clearl shown.. Unless oherwise specified, numerical answers mus be eac. 4. The diagrams in he paper are no necessaril drawn o scale. 香港考試局保留版權 Hong Kong Eaminaions Auhori All Righs Reserved CE-A MATH

25 FORMULAS FOR REFERENCE sin ( A ± B sin A cos B ± cos Asin B cos ( A ± B cos Acos B sin Asin B an A ± an B an ( A ± B an A an B A + B A B sin A + sin B sin cos A + B A B sin A sin B cos sin A + B A B cos A + cos B cos cos A + B A B cos A cos B sin sin sin Acos B sin ( A + B + sin ( A B cos A cos B cos ( A + B + cos ( A B sin A sin B cos ( A B cos ( A + B 000-CE-A MATH 保留版權 All Righs Reserved 000

26 Secion A (4 marks Answer ALL quesions in his secion.. Find + d. (4 marks. Epand 7 ( + ( in ascending powers of up o he erm. (5 marks. Q( 4 cosθ, sinθ P( 4, 0 O R E Figure Figure shows he ellipse E : +. P ( 4, 0 and 6 9 Q ( 4 cosθ, sinθ are poins on E, where 0 < θ <. R is a poin such ha he mid-poin of QR is he origin O. (a Wrie down he coordinaes of R in erms of θ. (b If he area of PQR is 6 square unis, find he coordinaes of Q. (6 marks 000-CE-A MATH Go on o he ne page 保留版權 All Righs Reserved 000

27 4. Prove, b mahemaical inducion, ha n n n ( n ( n ( for all posiive inegers n. (6 marks 5. A(, P O B(, 0 Figure In Figure, he coordinaes of poins A and B are (, and (, 0 respecivel. Poin P divides AB inernall in he raio : r. (a Find he coordinaes of P in erms of r. r (b Show ha he slope of OP is. + r (c If AOP 45, find he value of r. (6 marks 000-CE-A MATH 4 保留版權 All Righs Reserved 000

28 6. C A O + Figure d The slope a an poin (, of a curve C is given b +. The d line + is a angen o he curve a poin A. (See Figure. Find (a he coordinaes of A, (b he equaion of C. (7 marks 7. (a B epressing cos sin in he form r cos ( + θ, or oherwise, find he general soluion of he equaion cos sin. (b Find he number of poins of inersecion of he curves cos and + sin for 0 < < 9. (8 marks 000-CE-A MATH 5 4 Go on o he ne page 保留版權 All Righs Reserved 000

29 Secion B (48 marks Answer an THREE quesions in his secion. Each quesion carries 6 marks. 8. (a Find cos cos d. ( marks sin 5 sin (b Show ha 4 cos cos. sin Hence, or oherwise, find sin 5 d. sin (4 marks (c Using a suiable subsiuion, show ha 4 sin 5 cos 5 d d sin cos 6 4. (4 marks (d C O C (, 4 Figure 4 In Figure 4, he curves cos 5 sin 5 C : and C : cos sin inersec a he poin (,. 4 Find he area of he shaded region bounded b C, C and he line. (5 marks 000-CE-A MATH 6 5 保留版權 All Righs Reserved 000

30 9. Given a famil of circles F : + + (4k (k + (8k + 8 0, where k is real. C is he circle + 0. (a Show ha (i C is a circle in F, (ii C ouches he -ais. (4 marks (b Besides C, here is anoher circle C in F which also ouches he -ais. (i Find he equaion of C. (ii Show ha C and C ouch eernall. (7 marks (c L C C C O Figure 5 Figure 5 shows he circles C and C in (b. L is a common angen o C and C. C is a circle ouching C, L and he -ais bu i is no in F. (See Figure 5. Find he equaion of C. (Hin : The cenres of he hree circles are collinear. (5 marks 000-CE-A MATH 7 6 Go on o he ne page 保留版權 All Righs Reserved 000

31 0. P C (, B O M L A (, L Figure 6(a Figure 6(a shows a parabola P : 4. A, and C, are ( ( wo disinc poins on P, where < 0 <. L and L are angens o P a A and C respecivel and he inersec a poin B. Le M be he midpoin of AC. (a Show ha (i he equaion of L is + 0, (ii he coordinaes of B are, +, ( (iii BM is parallel o he -ais. (7 marks 000-CE-A MATH 8 7 保留版權 All Righs Reserved 000

32 (b P C B O D L A L Figure 6(b Suppose L and L are perpendicular o each oher and D is a poin such ha ABCD is a recangle. (See Figure 6(b. (i Find he value of. ( (ii Show ha he coordinaes of D are + +, +. (iii Find he equaion of he locus of D as A and C move along he parabola P. (9 marks 000-CE-A MATH 9 8 Go on o he ne page 保留版權 All Righs Reserved 000

33 . (a O h + r Figure 7(a In Figure 7 (a, he shaded region is bounded b he circle + r, he -ais, he -ais and he line h, where h > 0. If he shaded region is revolved abou he -ais, show ha he volume of he solid generaed is ( r h h cubic unis. (4 marks (b O C A B + 89 Figure 7(b In Figure 7 (b, A and C are poins on he -ais and -ais respecivel, AB is an arc of he circle + 89 and BC is a segmen of he line. A mould is formed b revolving AB and BC abou he -ais. Using (a, or oherwise, show ha he capaci of he mould is 88 cubic unis. ( marks 000-CE-A MATH 0 9 保留版權 All Righs Reserved 000

34 (c 4 Po θ G Molen gold Mould Figure 7(c Figure 7(d A hemispherical po of inner radius 4 unis is compleel filled wih molen gold. (See Figure 7 (c. The molen gold is hen poured ino he mould menioned in (b b seadil iling he po. Suppose he po is iled hrough an angle θ and G is he cenre of he rim of he po. (See Figure 7 (d. (i Find, in erms of θ, ( he disance beween G and he surface of he molen gold remaining in he po, ( he volume of gold poured ino he mould. (ii When he mould is compleel filled wih molen gold, show ha 8 sin θ 4sinθ + 0. Hence find he value of θ. (0 marks 000-CE-A MATH 0 Go on o he ne page 保留版權 All Righs Reserved 000

35 . (a c B b a θ ( C A O Figure 8(a In Figure 8 (a, a riangle ABC is inscribed in a circle wih cenre O and radius r. AB c, BC a and CA b. Le BCA θ. (i Epress cos θ in erms of a, b and c. (ii Show ha c r. sinθ (iii Using (i and (ii, or oherwise, show ha r abc. 4a b ( a + b c (7 marks (b In his par, numerical answers should be given correc o wo significan figures. Building Pedesrian walkwa Figure 8(b 000-CE-A MATH 保留版權 All Righs Reserved 000

36 . (b (coninued 5 m B C 5m P m 8m 0 Q A Figure 8(c Figure 8 (b shows a pedesrian walkwa joining he horizonal ground and he firs floor of a building. To esimae is lengh, he walkwa is modelled b a circular arc A B C as shown in Figure 8 (c, where A denoes he enrance o he walkwa on he ground and C he ei leading o he firs floor of he building. P and Q are he fee of perpendiculars from B and C o he ground respecivel. I is given ha A P 5 m, PQ m, B P 5 m, C Q 8 m and A PQ 0. (i Find he radius of he circular arc A B C. (ii Esimae he lengh of he walkwa. (9 marks END OF PAPER 000-CE-A MATH 保留版權 All Righs Reserved 000

37 000 Addiional Mahemaics Paper Secion A. ( + + c, where c is a consan (a ( 4 cosθ, sinθ (b (, + r r 5. (a (, + r + r (c 5 6. (a (, (b (a n, where n is an ineger (b 4 保留版權 All Righs Reserved 000

38 Secion B Q.8 (a cos cos d (cos 4 + cos d sin 4 + sin + c, where c is a consan 8 4 (b sin 5 sin sin cos sin sin ( sin cos cos sin 4 cos cos sin 5 d ( 4 cos cos d sin + sin 4 sin + 4 ( + + c, where c is a consan sin 4 + sin + c (c Pu : θ 4 6 sin 5( θ sin 5 d 4 ( dθ sin sin( θ cos 5θ d θ cosθ 4 cos 5 d cos 4 (d Area of shaded region cos 5 sin 5 ( d cos sin cos 5 sin 5 d d cos sin 4 sin 5 sin 5 d d (using (c sin sin [ + sin 4 + sin ] [ + sin 4 + sin ] 4 保留版權 All Righs Reserved 000

39 Q.9 (a (i Pu k ino F, he equaion becomes + + ( ( + ( i.e C is a circle in F. (ii Co-ordinaes of cenre of C (0,. radius of C Since he -coordinae of cenre is equal o he radius, C ouches he -ais. (b (i Pu 0 in F : + (4k + 4 (8k Since he circle ouches he -ais, (4k (8k k + 64k ( k + ( k + 0 k (rejeced or k, he equaion of C is + + [4( + 4] + [( + ] [( 8 + 8] (ii Co-ordinaes of cenre of C (0,, radius. Co-ordinaes of cenre of C (4, 4, radius 4. Disance beween cenres ( (4 5 sum of radii of C and C C and C ouch eernall. 保留版權 All Righs Reserved 000

40 (c Le radius of C be r and coordinaes of is cenre be (a, r. (0, (4,4 (a,r O Considering he similar riangles, r r r + 5r 0 r 6 a r a 0 he equaion of C is ( 0 + ( 保留版權 All Righs Reserved 000

41 Q.0 (a (i 4 4 d d d d A poin A, d d Equaion of L is 0 + (ii Equaion of L is ( 0 ( 0 ( ( : 0 ( ( + + ( + he coordinaes of B are, ( +. (iiithe coordinaes of M are (,, + + i.e. (., + + As he -coordinaes of B and M are equal, BM is parallel o he -ais. (b (i (Slope of L (Slope of L ( ( 保留版權 All Righs Reserved 000

42 (ii Since ABCD is a recangle, mid-poin of BD coincides wih mid-poin of AC, i.e. poin M ( Since BD is parallel o he -ais, he -coordinae of D -coordinae of B +. ( he coordinaes of D are + +, +. (iii Le (, be he coordinaes of D ( + ( + + he equaion of he locus is 0. 保留版權 All Righs Reserved 000

43 0 h Q. (a Volume d 0 ( r h d 0 [ r ] h ( r h h cubic unis (b Pu h, r 89 : Using (a, 89 capaci of he mould [ ( ( ] 88 cubic unis (c (i ( Disance 4 sinθ. ( Pu r 4, h 4 sinθ. Using (a, amoun of gold poured ino he po [4 (4 sinθ (4 sinθ ] 64 (64sinθ sin θ (ii When he mould is compleel filled, (64sinθ sin θ 64sin θ 9 sinθ sin θ 4sinθ + 0 (* Pu sin θ : 8sin θ 4sinθ + 0. sin θ is a roo of (* ( sinθ (4 sin θ + sinθ 0 ± 80 sinθ or sinθ (rejeced 8 sin θ θ 6 保留版權 All Righs Reserved 000

44 Q. (a A c O B b a θ ( θ C D (i a cosθ + b c ab (ii Consider ABD : AD r BDA BCA θ ABD 90 sin θ r c r c sinθ (iii sin θ + cos θ c ( r a + ( + b c ab c ( a + b c 4 r 4 a b r r a b c 4 a b ( a + b c abc 4 a b ( a + b c (b (i Consider A B C : A B ( A P + ( PB B C ( PQ + ( QC PB + ( 保留版權 All Righs Reserved 000

45 ( A Q ( A P + ( PQ ( A P ( PQ cos A PQ 5 + (5 ( cos0 40 A C ( A Q + ( QC Using (a (iii, pu a 450, b 465, c 50 : r 50 4(450 ( ( m (correc o sig. figures he radius of arc A B C is 9 m. (ii B N φ A O C Le O be he cenre of he circle passing hrough A, B and C, φ be he angle subended b arc A B C a O. Consider O A N (N is he mid-poin of A C A C sin φ r 465 φ sin φ.07 Lengh of walkwa lengh of A B C rφ 8.86 ( m (correc o sig. figures he lengh of he walkwa is 60 m. 保留版權 All Righs Reserved 000