MATHEMATICS HL - May 1999 P1$

MATHEMATICS HL - May 1999 P1$ - 15 pages INTERNATIQNAL I...

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Unformatted text preview: 15 pages INTERNATIQNAL I BACCALAUREATE BACCALAUREAT INTERNATIONAL - BACPHLLERATO INTERNACIONAL MARKS CHEME May 1999 MATHEMATICS Higher Level Paper 1 M9915 10/1-1(1)M —6- M99/510/H(1)M By the remainder theorem, f(—1)=6—ll-—22—a+6 (M1) = -20 (M1) <=> a = -l (Az) Answer: a = —] (C4) Using a tree diagram, 3 4 4 3 = __ __ _ _ M1 M1 p(BGorGB) [9x8]+[9x8) ( )( ) 1 1 =_ _ AI 1 =_ A1 3 ( J 4 3 OR p(BG or GB) = 2 x a x E (M1)(MI) 1 =— 2 3 (A) Answer: % (C4) — 7 — M991510/H(1)M Let a be the first term and d be the common difference, then a+d=7 and SA=%(2a+3d)=12 (M1) a+d=7 => 4a+6d=12} (MU => a=15,d=—8 (A2) Answer: a=15 (C2) d=-8 (CZ) Substituting gives, 2(2}L+4)+3(—Av2)-(3}L+2) =2 (M1) <=> 41+8—3A—6—3A—2=2 (M1) 4: #21 :2 ,1 :4 (A1) Intersection is (2, -1, — 1) (A1) Answer: Intersection is (2, — 1, — 1) (C4) (l—i)z =1-3i <2 z =% (M1) —1 (z) z = 1'3} xi? (M!) 1—1 1+1 st: 2 =2-i (A2) 0R Letz=x+iy (I—i)(x+iy) =1—3i (M1) x+y-i(x—y) =l—3i x+y =1 m» =>x=2,y=—1 (A2) Answer: x = 2 (C2) y=_1 (62) Note: Award (C4) for z = 2 ~— i — s — M99/510/H(1)M The system of equations will not have a unique solution if the determinant of the matrix representing the equations is equal to zero. 4 —1 2 Therefore, 2 3 0 = 0 (MI) I -—2 a v: 4x3a+2a+2x(—4—3)=0 (M1) 4:: 14a =14 (MI) a :1 (A1) Answer: a: 1 (C4) (242)042) Notes: The graph of y2 is yl shifted k units to the right. Award (A2) for the correct graph. Award (A1) for indicating each point of intersection with the x-axis lie. (m + k, 0) and (n + k, 0) Answer: See graph (C4) Note: Award (C4) if the graph of y; is drawn correctly and correctly labelled with m + k and n + k. 8. Using a tree diagram, Let p(S) be the probability that the pupil speaks Spanish. Let p(A) be the probability that the pupil is Argentine. Then, from diagram, PSM)=% 4 5 0R p(S|A)— MA) _2 n 21 21 OR E6) 12 p(SlA) =T5‘ 4 5 Answer: p(S 1A) = % Spanish (15) _p(Sm1) oo 2 6 3(15) Argentine (3) Argentine (12) M99/SlO/H(1)M (M2) (A1) (A1) (M1) (MIX/11) (A1) (M 2) (A1) (A1) (C4) 9. 10. filo- By implicit differentiation, d 2 2 dy -—-2 —3 =2 =>4x-—6—= dx(x y ) ydx c, d_y=i£ (:1): 6y Whenx=5,50-3y2=2 @ y2=l6 <=> y=i4 Therefore 93 = ii dx 6 Note: This can be done explicitly Answers: 9.2: dx :1: 6 (a) A perpendicular vector can be found from the vector product ij JPXJQ= 1 —3 2 =F—3j—5E (b) Answers: Area AOPQ = 3‘ 5P“ 0b (a) (b) ~21—1 1 sine, where e is the angle between 5P and 5Q -§ —> =§i0Px 09‘ J3 2 i" — 3} — 51; (or any multiple) «IE 2 M99/510/H(1)M (M1) (A1) (M1) (A1) (CZXCZ) (MIXAI) (M1) (A1) (C2) (C2) — 11 — M99/510/H(1)M 11. Given y =arccos(l—2x2) then d—y =—_I—Tx—4x (M1) d1 (l-(I—2x2)2) d 4x —y = 2 4 112 (M1) dx (1—(1—4x +4x )) dy 4x __ =___ A2 dx (4x2 _4x4)U2 ( ) 0R cosy =1—2x2 (M1) dy —sm — ——4x y dx 3 — 4x —L (MD dx -siny 1—(1—2x2)2 d_y : 4x (A2) dx 4x2—4Jc4 Answer: dy 4x dy - 4x (C4) _=__._—or__—__ dx (4x2—4x4)l"'2 dx 4x2_4x4 — :2 — M99/510/H(1)M 12. Let f(x)=euc2 +bx+c where a: 1, b =(2-k) and c=k2. Then for a> 0, f(x)> o for all real values ofx if and only if b2 —4ac < 0 (M1) <=> (2 —-k)2 -4k2 < 0 (A1) 4:: 4—4ch2 —4/Ic2 <0 <2 31c2 + 4k — 4 > o <=> (3k—2)(k+2)>0 (M1) <=> k>%,k<—2 (A1) —2 E k 3 2 Answer: k < —2, or k > 3 (C2)(C2) 13. Let the volume of the solid of revolution be V. V: nj:((ax+2)2 —(x2 +2)2)dx (M1) = nj:(a2x2 + 4a): +4 — x" —4x2 -4)dx (M1) 1 2 3 2 1 5 4 3 a = n -—a +2 —— —— [3 x ax 5x 3 x 0 (M1) = n(-12Ea5 +§asj units3 (A1) 231': 2 = +5 15 (a ) Note: The last line is not required 2031: 2 . I Answer: V: (a +5) or equwalent (C4) 15 u l3 — M99/510/H(1)M 14. Letuzlx+lmx=2(u—l)=$£=2 2 du l 112 Then Jx[§x+l) dx =_[2(u—1)xu”1x2du (M1) =4'I-(uy2 —u”2 )du (A1) =4[%um -——§-u3’2]+C (M1) SH 312 = l:%(%x+1] —§-[—;—x+l] ]+C (A1) SIZ 312 312 Answer: 4[3(lx+l] ———2-[%x+1] J+C or £5[lx+1) [Ex—ZJ'l-C (C4) 15. The locus defined by |z — 4 — 3i | = I z — 2 + i | is the perpendicular bisector of the line joining the points 4 + 3i and 2 — i in the complex plane. (M2) Correct diagram (see below). (A2) OR Letz=x+yi then,[z—4—3i|=|z—2+i| :5 (x—4)2+(y—3)2 =(x—2)2+(y+l)2 (MI) <=> -8x—-6y+25=—4x+2y+5 <=> x+2y=5 (M1) Therefore the equation of the locus is x + 2y = 5. Correct diagram (see below). (A2) Award (AI) each for any two of the following (maximum of [2 marksj): Perpendicular line, midpoint (3,1), y—intercept, x-intercept, gradient. Answer: Correct locus (see diagram) (04) 16. 17. 18. -14.. Given (1+x)5(1+ax)6 = 1+bx+10x2 +...+a‘5x” <=> (1+5x+10x2 +...)(l+6a.wc+15a3x2 +.._)= ] +bx+le2 +...+afix” ¢=> 1+(6a+5)x+(15a2+30a+10)x2+...=1+bx+10x2+...+a“x” <=> 6a+5=b 0) l5a2+30a+10=10 ® (2) => 15a2+30a=0 <=> 15a(a+2)=0 => a=—2 Substitute into 0) b = —7 Note: a at 0 since a E 2' Answers: a: —2 b =—7 Let X be the number of counters the player receives in return. EtXJ=2pmxx=9 a £~:—><41+£%><5J+(%x15)+(ex0=9 <2} --—n:3 10 :) n=30 Answer: n = 30 Let cm) = 0017 then c1>(—z) :1. 0.017 = 0.983 2 =—2.12 x—p_1—l.02 0' 0' Butz= where x =1 kg 1—1.02 0' Therefore = -—2.12 <=> 0' = 0.009 43 kg = 9.4 g (to the nearest 0.1 g) Answer: 0' = 9.4 g (or equivalent) M99/510/H(1)M (M1) (M1) (M1) (A1) (C2) (C2) (M1) (MINAI) (A1) (C4) (M1) (A1) (M1) (A1) (C4) 19. 20. -15.. (3) Given 2:40, dz dv -— =—k d ¢=> v I t <=> lnv =—kt+C ¢=> V =Ae'k' (A = 3C) <=> v=v0e v b Put =—9— () v 2 then v30 : voe'k'r ¢=> %=e"“ 1 <=> ln— =—kt 2 e k Note: Accept equivalent forms, e. g. t = —: Answers: (a) v=VOe-’“ ln2 b t=———- ( ) k Given v = (3S + 2) (2s — 1) . dv dv ds dv then acceleration a = — = — x — = _ x v dt d: dt ds therefore a = W X (33 + 2) (25- 1) (25 — 1) c) a 3 —7(35 + 2) (23 — 1)3 therefore when 5 = 2, a = i6 27 56 Answer: acceleration = -E M995 1 01m 1 )M (M1) (A1) (M1) (A1) (C2) (C2) (M1) (M1) (M1) (A1) (C4) ...
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This note was uploaded on 05/06/2010 for the course MATH 1102 taught by Professor Chang during the Winter '10 term at Savannah State.

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MATHEMATICS HL - May 1999 P1$ - 15 pages INTERNATIQNAL I...

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