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- Chrystal Sparks
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12 NATIONAL SENIOR CERTIFICATE GRADE/GRAAD MATHEMATICS P/WISKUNDE V NOVEMBER 04 MEMORANDUM MARKS: 50 PUNTE: 50 Tis memorandum consists of pages. Hierdie memorandum bestaan uit bladsye. Copyrigt reserved/kopiereg voorbeou
13 Matematics P/Wiskunde V DBE/November 04 NOTE: If a candidate answers a question TWICE, only mark te FIRST attempt. Consistent accuracy applies in all aspects of te marking memorandum. LET WEL: Indien n kandidaat n vraag TWEE keer beantwoord, merk slegs die EERSTE poging. Volgeoue akkuraateid is DEURGAANS op ALLE aspekte van die memorandum van toepassing. QUESTION/VRAAG.... ( )(4 ) 0 or b ± b 4ac a ± ( ) 4()( 4) () ± 7 6,5 or/ of,85 4 () standard form/standaardvorm substitution into correct formula/ substitusie in korrekte formule answers/ antwoorde (4) ± ± 4,5 or/ of,85 for adding 9 on bot sides/tel 9 by aan beide kante ± 4 answers (4) Copyrigt reserved/kopiereg voorbeou
14 Matematics P/Wiskunde V DBE/November 04.. ( ) 0 common factor/gemeen. faktor simplification/ vereenvoudiging answer/antwoord (). ( ) common factor/gemeen. faktor simplification/ vereenvoudiging answer/antwoord () y...() answer/antwoord () 5y 4 6y...() () in () : ( y ) 5( y ) y 4 6y (4 y y 9) 0y 5y 4 6y y 6y 7 0y 5y 4 6y 0 y 5y 0 (y )( y ) 0 y or y or ( ) 0 or (0 ; ) (; ) substitution/substitusie simplification/ vereenvoudiging standard form/ standaardvorm factorisation/faktorisering y-values/y-waardes -values/-waardes (6) Copyrigt reserved/kopiereg voorbeou
15 Matematics P/Wiskunde V 4 DBE/November 04 y : ( ) 0 0 or y or y substitution/substitusie simplification/ vereenvoudiging standard form / standard vorm factors/faktore - values/- waardes y-values/y-waardes (6). ( )( ) < 6 < 6 4 < 0 ( )( 4) < 0 standard form/ standaardvorm factorisation/faktorisering 4 OR/ OF < < 4 or ( ; 4) 4 critical values in te contet of inequality / kritiese waardes in die konteks van die ongelykeid notation/notasie (4).4 k 4 0 k 4 0 k 4 answer/antwoord () [] Copyrigt reserved/kopiereg voorbeou
16 Matematics P/Wiskunde V 5 DBE/November 04 QUESTION/VRAAG. T 4 (). T5 a ( n ) d a and d 7 (5 )(7) subst. into correct formula /subt. in 75 korrekte formule 75 (). 5 general term/ (7n 5) algemene term n complete answer /volledige antwoord ().4 50 ( 7 p ) p 0 S S n n n 5 07 [ a l] [ 75] general term/ algemene term complete answer / volledige antwoord () substitution/substitusie 07 () n S n [ a ( n ) d] 5 [ () (5 )(7)] 07.5 Te new series/die nuwe reeks is ( n ) ( n ) 6 n n T is divisible by /is deelbaar deur 4 Ten T 7, T, T5,..., T5 are divisible by 4, tus eac 4 t term is divisible by 4. Daarna is T 7, T, T5,..., T5 deelbaar deur 4, d.w.s. elke 4 de term is deelbaar deur 4. 5 number of terms divisible by 4 will be aantal terme deelbaar deur 4 sal wees 6 4 Copyrigt reserved/kopiereg voorbeou substitution/substitusie 07 () generating new series divisible by 4/ vorming van nuwe reeks deelbaar deur 4 T n 75 6 (4) T is divisible by 4/ is deelbaar deur 4 identifying terms divisible by 4/ identifiseer terme deelbaar deur 4 reasoning/redenering 6 (4)
17 Matematics P/Wiskunde V 6 DBE/November 04 Position of terms divisible by 4: ; 7 ; ; ; 47; 5 4n 5 T n 4n 5 n 6 generating sequence involving position of terms/vorming van reeks i.t.v. posisie van terme T n 5 6 (4) [] Copyrigt reserved/kopiereg voorbeou
18 Matematics P/Wiskunde V 7 DBE/November 04 QUESTION/VRAAG.. ; 7 ; ; p ; p p ( 4) p 5 p p 5 p () ; 7 ; ; p ; p first differences/ eerste verskille.. p p a a p () a a b 6 () b 6 b 9 b 9 a b c 9 c c 7 T n n 9n 7 c 7 answer/antwoord (4) T n T ( n ) d ( n )( 6) ( n )( n ) ( n )( n )( ) n 6n 4 6n 6 n 9n 7 d formula/formule substitution of first and second differences/substitusie van eerste en tweede verskille simplification/vereenvoudiging answer/antwoord (4) Copyrigt reserved/kopiereg voorbeou
19 Matematics P/Wiskunde V 8 DBE/November 04 7; ; 7 ; ; p ; p T0 7 c a a a b 6 b 9 T n 9n 7 n a () Tn n bn c T b c...() T 7 4 b c 7...() () () : b 6 b 9 sub in () : c 7 T n 9n 7 n c-value/c-waarde a-value/a-waarde b-value/b-waarde answer/antwoord a-value/a-waarde b-value/b-waarde c-value/c-waarde answer/antwoord (4) (4) Copyrigt reserved/kopiereg voorbeou
20 Matematics P/Wiskunde V 9 DBE/November 04.. Te sequence of first differences is/die reeks van eerste verskille is: 6 ; 4 ; ; 0 ;... 6(n )() 96 n 5 two terms are/twee terme is: T 5 9(5) 7 4 T (5) 7 9 Te sequence of first differences is/die reeks van eerste verskille is: 6 ; 4 ; ; 0 ;... Te formula for te sequence of first differences/die formule vir die reeks van eerste verskille is T n n 8 st difference/ ste verskil: n 8 96 n 04 n 5 two terms are/twee terme is: T 5 9(5) 7 4 T (5) 7 9 6(n )() (4) n (4) T T n 9n n n [ 7] ( n 9n 7) ( n ) 9( n ) n 9n 7 n n 06 n 5 T 5 T (5) 7 4 9(5) 7 9 T 96 n T n (4) n [( n ) 9( n ) 7] [ n 9n 7] n n 9n 9 7 n Tn T 96 T n Tn n 7 96 n 04 n 5 5 T5 5 9(5) T 5 9(5) (4) Copyrigt reserved/kopiereg voorbeou
21 Matematics P/Wiskunde V 0 DBE/November T 6 a 6 and r 4 4 or S 0 4, S 0 4, or or T 4 T 5 5 T 4 4 5,,... is an aritmetic sequence wit T98 (98 ) a and d subst. into correct formula/ subt in korrekte formule answer/antwoord substitution into correct formula /substitusie in korrekte formule () answer/antwoord () substitution into correct formula /substitusie in korrekte formule answer/antwoord () improper fractions/ onegte breuke 00 or answer/antwoord (4) 99 giving te first tree terms / gee die eerste drie terme answer /antwoord (4) [9] Copyrigt reserved/kopiereg voorbeou
22 Matematics P/Wiskunde V DBE/November 04 QUESTION/VRAAG p q 0 Reflect (0 ; ) across y to get T( ; 0) Reflekteer (0 ; ) om y om T( ; 0) te kry p value /waarde q value /waarde 0 () ( ) reflect across/reflekteer om y 4. Sifting g five units to te left sifts ( ; 0) five units to te left. 6 answer/antwoord () 4.4 equating bot graps/stel grafieke gelyk y OS OS since at S, > 0,7... y 6 6,45 units/ eenede and y OS 6 () answer/antwoord (5) Copyrigt reserved/kopiereg voorbeou
23 Matematics P/Wiskunde V DBE/November 04 Translate g one unit down and one unit to te rigt/transleer g een eeneid af en een eeneid na regs Te new equation/die nuwe vergelyking : p( ) S / ; / Terefore te image of S is ( ) Daarom is die beeld van S nou S ( ; ) / Now translate p back to g/transleer p terug na g: S ; ( ) ( ) ( ) OS OS 6,45 units/eenede 4.5 k < will give roots wit opposite signs/ k < sal wortels met teenoorgestelde tekens gee p( ) coord. of/koörd. van coord. of/koörd. van S / S answer/antwoord (5) k < () [] Copyrigt reserved/kopiereg voorbeou
24 Matematics P/Wiskunde V DBE/November 04 QUESTION y loga log subt. ; a a a or a a a log y swop and y/ruil en y : y g( ) log answer/antwoord answer/antwoord () () () OR/ OF g( ) log answer/antwoord () OR/ OF g( ) log answer/antwoord () 5.4 > 0 y y answer/antwoord () answer/antwoord () answer/antwoord () 5.5 ( 0 ; ) log 7 7 answer/antwoord () eponential form/ eksponensiële vorm simplification/vereenvoudiging answer/antwoord () [9] Copyrigt reserved/kopiereg voorbeou
25 Matematics P/Wiskunde V 4 DBE/November 04 QUESTION/VRAAG y 0, ( - coordinate of S is positive), () 6. ( 0 ; 6) QT f ( ) g( ) (4 QT 4 6 Deravitive of QT 8 or or 0,5 4 6) or Ma/ Maks QT 4 6 6,75units/eenede () correct formula/ korrekte formule substitution/substitusie derivative/afgeleide derivative equal to 0/ afgeleide gelyk aan 0 8 -value/-waarde substitution/substitusie () answer/antwoord (6) [] Copyrigt reserved/kopiereg voorbeou
26 Matematics P/Wiskunde V 5 DBE/November 04 QUESTION/VRAAG 7 7. n A P( i) ( i) i ,94... Rate of interest/rentekoers is,94 % p.a./p.j. substitution/substitusie writing in terms of i erskryf in terme van i answer/antwoord () ( i) 5 5 i i 0,94 Rate of interest/rentekoers is,94 % p.a./p.j. n [ i ] P i 7.. ( ) 0, , 0, , R5505,4 40 substitution/substitusie writing i.t.o i answer () 0, i n 40 substitution into correct formula answer/antwoord (4) Copyrigt reserved/kopiereg voorbeou
27 Matematics P/Wiskunde V 6 DBE/November 04 n [ ] 7.. ( i) P i 0, , n 0,0 (,0) 6000 n 5 (,0) 6 log n 6 log,0 n 80,07 Melissa settles te loan in 8 monts 7.. Samuel He is paying off is loan over a longer period tus more interest will be paid./hy betaal sy lening oor 'n langer tydperk af, dus sal y meer rente betaal. n 6000 substitute into correct formula/substitusie in korrekte formule use of logs/gebruik van logs answer/antwoord (4) Samuel reason/rede () Samuel He will pay/hy betaal R5505,4 40 R R8 0,0 Se will pay between/sy sal tussen R and/en R ,00 betaal. Samuel reason/rede () [] Copyrigt reserved/kopiereg voorbeou
28 Matematics P/Wiskunde V 7 DBE/November 04 Copyrigt reserved/kopiereg voorbeou QUESTION/VRAAG 8 8. ( ) lim ) ( lim lim ) ( ) ( lim ) ( ) ( ) ( ) )( ( ) ( ) ( f f f f f f f f f ) ( ) ( lim ) ( 0 0 ) ( lim 0 ) )( ( lim 0 ) )( ( lim 0 lim ) ( lim 0 ( ) 0 lim OR simplifying/vereenvouding formula/formule subst. into formula/subst. in formule factorization/faktorisering answer/antwoord (5) formula/formule subst. into formula/subst. in formule simplifying/vereenvoudiging factorization/faktorisering answer/antwoord (5)
29 Matematics P/Wiskunde V 8 DBE/November 04 f ( ) f ( ) f ( ) lim 0 ( ) lim 0 ( )( lim 0 ( ) lim 0 lim 0 ( ) ) formula/formule subst. into formula/subst. in formule factorization/faktorisering simplifying/vereenvoudiging answer/antwoord (5) 8. f ( ) 4 4 () 8. y 6 simplification/vereenvoudiging dy 5 derivative/afgeleide d 5 6 ( ) factors/faktore 5 y () 8.4 f() 4 / first derivative/eerste afgeleide f () second derivative/tweede // f () 4 afgeleide // // f is concave up wen/is konkaaf op as f () > 0 f ( ) > 0 4 > 0 > 4 > > (4) [4] Copyrigt reserved/kopiereg voorbeou
30 Matematics P/Wiskunde V 9 DBE/November 04 QUESTION/VRAAG 9 9. f / ( ) 8 0 ( )( ) 0 or y or y 8 or 8, y Turning points are/draaipunte is ; and (;0) 7 9. derivative/afgeleide derivative/ afgeleide 0 factors/faktore -values/waardes eac y- values/elke y-waarde (6) y ;8, 5 8 ( ; 0) O ( ; 0) -intercepts/afsnitte y-intercept/afsnit turning points/ draaipunte sape/vorm 9. < or 0 < < OR ( ; ) (0;) < (4) bot critical points/ beide kritieke-punte notation/notasie () Copyrigt reserved/kopiereg voorbeou
31 Matematics P/Wiskunde V 0 DBE/November 04 QUESTION/VRAAG 0 0. l 40 l b 00 b 50 V lb V ( 40 )(50 ) 0. V (50 )(40 ) V V ± ( 80) 4(6)(000) (6) 7,86 or 8,80 for a bo as large as possible, 8,80cm vir die grootste moontlike boks 8,80 cm answer () b 00 b 50 volume formula () simplifying/vereenvoudig derivative / afgeleide -values in any form / -waardes in enige vorm answer/antwoord (5) [9] QUESTION/VRAAG.. 8 P(male/manlik) or 0,46 or 46,% 80 answer/antwoord.. P(not game park/nie wildreservaat) P(game park/wildreservaat) or 0,66 or 65,56% answer/antwoord () () P(not game park/nie wildreservaat) or 0,66 or 65,56% answer/antwoord () Copyrigt reserved/kopiereg voorbeou
32 Matematics P/Wiskunde V DBE/November 04. Events are independent if /Gebeure is onafanklike indien P(male) P(ome) P(male and ome) P(manlik) P(uis) P(manlik en uis) 8 P(male/manlik) 80 0 and/en P(ome/uis) 80 or 0, or,% P(m) P() and teir values/en ulle waardes answer of product P(male/manlik) P(ome/uis) ,05 or 5,% P(male and ome/manlik en uis) 80 P(m and/en ) value/waarde 0,07 or 7,% Terefore P(male) P(ome) P(male and ome) Dus P(manlik) P(uis) P(manlik en uis) Tus te events are not independent./dus is die gebeure nie onafanklik nie conclusion/afleiding (4) Home/Huis Not Home/ Nie uis M 70 8 F P(female/vroulik) P(not ome/nie uis) , or 47,90% P(female and not ome/vroulik en nie-uis) ,5 or 50% Terefore P(female) P(not ome) P(female and not ome) Tus te events are not independent. Dus P(vroulik) P(nie-uis) P(vroulik en nie-uis) Dus is die gebeure nie onafanklik nie. Copyrigt reserved/kopiereg voorbeou P(f) P(not ) and teir values/en ulle waardes answer of product P(f and/en not ) value/waarde conclusion/afleiding (4) [7]
33 Matematics P/Wiskunde V DBE/November 04 QUESTION/VRAAG () 6 6! 6! P (6 5)!! ( )(5 4 ) 70 formula/formule answer/antwoord () 4 44 () product/produk () () Te five 'units' can be parked in 5 4 ways./die vyf 'eenede' kan op 5 4 maniere geparkeer word. Te tree silver cars can be parked in ways./die drie silwer motors kan op maniere parkeer word. So tere are ( )(5 4 ) 70 ways to park te cars./dus is daar ( )(5 4 ) 70 maniere om die motors te parkeer. Suppose for te moment te silver cars are at one end./veronderstel die drie silwer motors is op die punt. Te cars can be arranged in 6 ways./die motors kan op 6 maniere gerangskik word. For eac of tem te remaining four cars can be arranged in 4 4 ways./die 4 oorblywende motors kan op 4 4 maniere rangskik word. So ways if all cars at one end./dus is daar maniere as die motors op die punt is () Togeter, te silver cars can only occupy 5 different positions amongst te 7 positions../saam kan die silwer motors slegs 5 verskillende posisies ê tussen die 7 moontlike posisies. Total ways/totale getal maniere () [9] TOTAL/TOTAAL: 50 Copyrigt reserved/kopiereg voorbeou
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55 NATIONAL SENIOR CERTIFICATE GRADE/GRAAD MATHEMATICS P/WISKUNDE V NOVEMBER 04 MEMORANDUM MARKS/PUNTE: 50 Tis memorandum consists of pages. Hierdie memorandum bestaan uit bladsye. Copyrigt reserved/kopiereg voorbeou
56 Matematics P/Wiskunde V DBE/November 04 NOTE: If a candidate answers a question TWICE, only mark te FIRST attempt. If a candidate as crossed out an attempt of a question and not redone te question, mark te crossed out version. Consistent accuracy applies in ALL aspects of te marking memorandum. Assuming answers/values in order to solve a problem is NOT acceptable. NOTA: As 'n kandidaat 'n vraag TWEEKEER beantwoord, merk slegs die EERSTE poging. As 'n kandidaat 'n poging om die vraag te beantwoord, doodgetrek et en nie dit oorgedoen et nie, merk die doodgetrekte poging. Volgeoue akkuraateid word in ALLE aspekte van die nasienmemorandum toegepas. Aanvaarding van antwoorde/waardes om 'n probleem op te los, is ONaanvaarbaar. QUESTION/VRAAG σ 8,4 answer/antw. (68 8,4 ; 68 8,4) (49,58 ; 86,4) 6 candidates ad a mark witin one standard deviation of te mean/6 kandidate et 'n punt binne een standaardafwyking vanaf die gemiddelde..4 a,88...,8 b 0, ,66 y ˆ 0,66, 8 yˆ,8 0, 66.5 y ˆ 0,66, 8 y 0,66(60),8 6,4...% 6% 6,69% 6% interval answer/antw () () () value of a/ waarde van a value of b/ waarde van b equation/vgl () subs of 60 into equation answer/antw () answer/antw ().6 (8 ; 6) answer/antw () [] Copyrigt reserved/kopiereg voorbeou
57 Matematics P/Wiskunde V DBE/November 04 QUESTION/VRAAG. 50 < < 60 between 50 and 60/tussen 50 en 60.. Class Klas.. Frequency Frekwensie 0 < 0 0 < < < < < < < Cumulative frequency Kumulatiewe frekwensie answer/antw 8 55 () () grounding at (0 ; 0)/ anker by (0 ; 0) plotting at upper limits/ plot by boonste limiete smoot sape of curve/gladde kurwe Cumulative Frequency/ Kumulatiewe frekwensie Speed in km per our/ Spoed in km per uur (accept/aanvaar 4 45) motorists/motoriste (accept/aanvaar 0 motorists/motoriste) 44 () () [8] Copyrigt reserved/kopiereg voorbeou
58 Matematics P/Wiskunde V 4 DBE/November 04 QUESTION/VRAAG S y K L N M(5 ; 4) O A B P. r MN 5 answer/antw. ( 5) ( y 4) 5 equation/vgl. A( ; 0) ( 5) (0 4) 5 ( 5) (0 4) ( 5) ( 5) 9 ( 8)( ) 0 ( 5) ± 8 or/of 8 or/of A( ; 0) A( ; 0).4. m MB () () substitute into eq/ vervang in vgl y 0 standard form/ standaardvorm or perfect square form/kwadr vorm answer/antw () subst M and B into form/vervang M and B in form 4 m MB () Copyrigt reserved/kopiereg voorbeou
59 Matematics P/Wiskunde V 5 DBE/November m MB mpb (tangent radius/ rkl radius) m PB 4 y 4 c y y 4 ( ) m MB mpb 4 m PB 0 4 (8) c y 0 4 ( 8) y 4 6 y y K y M r 4 5 y 9.6 At/By L: L(0 ; 9).7 L(0 ; 9) ML ( (0 5) ) ( y (9 4) or / of 5 y 0 ) ML ( (5) (5) 50 ) (5) (9 ) or / of ( y.8 MK KL M K ˆ L 90 (radius tangent/radius rkl) ML is a diameter as it subtends a rigt angle/ml is middellyn ML 50 5 r or 7,9 Centre of circle midpoint of ML/Midpt van sirkel midpt v ML ,5 y 6, 5 Centre/midpt: (,5 ; 6,5) Equation of te circle KLM /Vgl van sirkel KLM: 50 5 (,5) ( y 6,5) 6, y 0 ) equation/vgl () 9 equation/vgl () equating simultaneously simplification () correct subst into distance formula/ korrekte subst in afstandformule answer in surd form/antw in wortelvorm () S value of/waarde van r, 5 y 6, 5 answer in correct form/ antw in korrekte vorm (5) Copyrigt reserved/kopiereg voorbeou
60 Matematics P/Wiskunde V 6 DBE/November 04 MK KL M K ˆ L 90 (radius tangent/radius rkl) ML is a diameter as it subtends a rigt angle/ml is middellyn Centre of circle midpoint of ML/Midpt van sirkel midpt v ML ,5 y 6, 5 Centre/midpt: (,5 ; 6,5) Equation of te circle KLM /Vgl van sirkel KLM: (,5) ( y 6,5) r subst (5 ; 4): ( 5,5) (4 6,5) r 6,5 r 50 5 (,5) ( y 6,5) 6, 5 4 By symmetry about LM/deur simmetrie om LM: MK KL M K ˆ L 90 (radius tangent/radius rkl) ML is a diameter as it subtends a rigt angle/ml is middellyn ML is a diameter /ML is 'n middellyn ML 50 5 r or /of 7,9 Centre of circle midpoint of ML/Midpt van sirkel midpt v ML ,5 y 6, 5 Centre/midpt: (,5 ; 6,5) Equation of te circle KLM /Vgl van sirkel KLM: 50 5 (,5) ( y 6,5) 6, 5 4 S, 5 y 6, 5 value of/waarde van r answer in correct form/antw in korrekte vorm (5) S value of/waarde van r, 5 y 6, 5 answer in correct form/antw in korrekte vorm (5) [] Copyrigt reserved/kopiereg voorbeou
61 Matematics P/Wiskunde V 7 DBE/November 04 QUESTION/VRAAG 4 y F B( ; 5) D 45 M A E O 4. y 0: E ; 0 E ; 0 4. tan DÊO mde DÊO 7, ,57 DÂE 7, ,57 4. m AB tan 6,57 y c y y ( ) 5 () c y 5 ( ) y 4 y 9 y-value/waarde -value/waarde tan DÊO 7,565 () 6,57 () m AB tan 6, 57 m AB subst of m and ( ; 5)into formula/ subst m en ( ; 5) in formule equation/vgl (4) Copyrigt reserved/kopiereg voorbeou
62 Matematics P/Wiskunde V 8 DBE/November Solve y 9 0 and y 8 simultaneously: (8) y ( ) 8 y y 5 D( 5 ; 5 4 ) y 5 subst/vervang -value/waarde subst/vervang y-value/waarde (4) y 9 y (y 9) 8 y 6 y y 5 4 ( ) D( ; ) 5 5 subst/vervang y value/waarde subst/vervang -value/waarde y ( ) 8 y ( ) y y D( ; ) (4) equating/gelyk stel value/waarde subst/vervang y-value/waarde (4) Copyrigt reserved/kopiereg voorbeou
63 Matematics P/Wiskunde V 9 DBE/November 04 y 9 () 6 y 6...() () (): y 9 y ( 5 ) 8 adding/optelling -value/waarde D( 4 y 5 4 ; ) 5 5 y 8.() 6y 7.() () (): 5y 9 4 y D( ; 5 4 ) 5 4 y 5 4 ( ) subst/vervang y-value/waarde (4) subtracting/aftrekking y-value/waarde subst/vervang -value/waarde (4) Copyrigt reserved/kopiereg voorbeou
64 Matematics P/Wiskunde V 0 DBE/November 04 F 8 D( ; ) 5 45 B( ; 5) M(0 ; 4 ) G 4 A( 9 ; 0) 6,57 E( ; 0) C O 4.5 area DMOE area AMO area ADE A (0) 9 A( 9 ; 0) area AMO area ADE. AO. OM. AE. yd (9)(4 ). (AO EO). yd 4 0,5 9 5,0 area ADE AD.AE.sin D ÂE 9 5 ( ).6.sin 6,57 5,0 correct metod/ korrekte metode A 9 (9)(4 ) AE y D AD 5 AE 6 area DMOE 8, square units/vk een answer/antw (6) Copyrigt reserved/kopiereg voorbeou
65 Matematics P/Wiskunde V DBE/November 04 area DMOE area rectangle DCOG area DMG area DEC ( ) ( ) ( )( ) , square units/vk een correct metod/ korrekte metode ,7 9 5 answer (6) area DMOE area EDO area ODM ( EO yd ) ( OM D ) or/of 8 or/of 8, square units/vk een area DMOE area EOF area DMF ( EO OF) ( OF OM)( D ) or 8 or 8, square units/vk een correct metod/ korrekte metode 9 4 y D or EO D OM or 4 answer/antw correct metod/ korrekte metode y F 8 EO 8 D 5 7 FM answer/antw (6) (6) Copyrigt reserved/kopiereg voorbeou
66 Matematics P/Wiskunde V DBE/November sq units/vk een area EOM ( EO OM) ED and DM or 4, or, area EDM ( ED DM sin EDˆ M) sin 5 0 or, area DMOE area EOM area EDM 6, or/of 8 or/of 8, square units/een area EOM 9 0 ED DM 0 area EDM correct metod/ korrekte metode answer/antw (6) [9] F 8 D( ; ) 5 45 B( ; 5) M(0 ; 4 ) G 4 A( 9 ; 0) 6,57 E( ; 0) C O Copyrigt reserved/kopiereg voorbeou
67 Matematics P/Wiskunde V DBE/November 04 QUESTION/VRAAG 5 C 4 A y 8 P 4 D 5. CP sin CÂP AP 4 sin 8 60 sin 90 sin sin C PˆA DPˆA 0 ( AP bisects DPˆ C) AD AP DP.AP.DP.cos APˆD 8 4 (8)(4) cos (8)(4)( ) 4,57... AD 4,96 correct sine ratio/ korrekte sin-ver () correct sine ratio/ korrekte sin-ver D PˆA 0 () correct subst into cosine rule/ korrekte subst in cos-reël 4, ,96 (4) Copyrigt reserved/kopiereg voorbeou
68 Matematics P/Wiskunde V 4 DBE/November sin DÂP sin APˆD DP AD sin y sin 0 4 4,96 4sin 0 sin y 4,96 0,40... AD 4 y,78 AP 8 DP (4,96) 8 (4,96) 4 cos y (8)(4,96) cos y 0, y,8.ap.dp.cosdâp (8)(4,96).cos y correct subst into sine rule/ korrekte subst in sin-reël sin y subject,78 () correct subst into cosine rule/ korrekte subst in cos-reël cos y subject,8 () [9] Copyrigt reserved/kopiereg voorbeou
69 Matematics P/Wiskunde V 5 DBE/November 04 QUESTION/VRAAG 6 6. cos (80 ) tan( 80 )sin(70 ) cos ( cos ) [ ( tan )]( sin )(cos ) cos cos sin cos sin cos ( sin )(cos ) 6. sin( α β ) cos[ 90 ( α β )] cos[( 90 α ) β ] cos( 90 α)cos β sin(90 α) sin β sinα cos β cosα sin β sin( α β ) cos[ 90 ( α β )] cos[( 90 β ) ( α )] cos( 90 β )cos( α ) sin(90 β )sin( α ) ( sin β )cosα cos β ( sinα) sinα cos β cosα sin β 6. y sin 76 cos 76 ( cos 76 sin 76 ) cos (76 ) cos 5 ( cos 8 ) cos (90 6 ) cos 8 ( sin 6 ) cos (90 6 ) sin 6 sin 6 y sin 76 cos 76 sin 76 sin 76 cos 76 cos 76 sin 76 cos 4 cos 76 sin 4 sin (76 4 ) sin 6 ( cos ) or cos tan or ( tan ) sin sin tan cos cos sin (5) rewrite as/erskryf cos[( 90 α ) β ] epansion/ uitbreiding simpl/vereenv () rewrite as/erskryf cos[( 90 β ) ( α )] epansion/ uitbreiding simpl/vereenv () ( cos 76 sin 76 ) recognition of cos double angle cos 5 cos 8 (4) cos 4 sin 4 recognition of sine compound angle sin(76 4 ) y sin 76 cos 76 cos 4 sin 4 cos (4 ) cos 8 sin 6 Copyrigt reserved/kopiereg voorbeou (4) cos 4 sin 4 recognition of cos double angle cos 8 (4) []
70 Matematics P/Wiskunde V 6 DBE/November 04 QUESTION/VRAAG y or y [0 ; ] critical values/ kritieke waardes notation/notasie () 7. sin cos sin sin sin sin 0 sin (sin ) 0 7. sin (sin ) 0 sin 0 or sin sin st form/st vorm sin 0 or sin () k.60 or 80 k.60 k.80, k Z 0 k.60 or 0 k.60, k Z f 0 ; 80 k.80 0 ; 0 k.60, k Z (4) y-intercept/afsnit -intercepts/afsnitte min/ma points/ min/maks punte g f () g() at/by: 0 ; 0 ; 80 ; 0 f ( 0 ) g( 0 ) at/by: 60 ; 0 ; 50 ; Series will converge if/reeks sal konvergeer as: < r < < cos < < cos < () 0 ; 0 ; 80 ; 0 60 ; 0 ; 50 ; 80 < r < r cos < cos < () 0 < < 60 or (0 ; 60 ) 0 < < 60 (5) [9] Copyrigt reserved/kopiereg voorbeou
71 Matematics P/Wiskunde V 7 DBE/November 04 QUESTION/VRAAG A O C y B ( at centre at circumference/ by midpt by omtrek) 8.. Bˆ (sum of s in / som v e in ) Ĉ y Bˆ 4 ( s opp sides/ e teenoor sye) S R S S () () 8. A B O C F 0 D 8.. Fˆ 90 (line from centre to midpt cord/ lyn vanaf midpt na midpt kd) 8.. A Bˆ C 50 (opposite s of cyclic quad/ tos e v koordev ) S R S R () () Copyrigt reserved/kopiereg voorbeou
72 Matematics P/Wiskunde V 8 DBE/November A E C 7 B 8.. (a) tangent radius/diameter / raaklyn radius/middellyn R 8.. (b) tangents from common pt OR tangents from same pt / raaklyne v gemeensk pt OF raaklyne vanaf dies pt 8.. AB² BC² AC² ( 7) (Teorem of/stelling vanpytagoras) ( 5)( ) 0 5 ( ) R () () AB² BC² AC² ( 7) standard form answer (4) [4] Copyrigt reserved/kopiereg voorbeou
73 Matematics P/Wiskunde V 9 DBE/November 04 QUESTION/VRAAG 9 9. A D k E B C 9.. Same base (DE) and same eigt (between parallel lines) Dieselfde basis (DE) en dieselfde oogte (tussen ewewydige lyne) 9.. AD DB AE k EC k But/Maar area DEB area DEC (Same base and same eigt/dieselfde basis en dieselfde oogte) area ADE area ADE area DEB area DEC AD DB AE EC same base/dies basis between lines/ tussen lyne () S S S R S (5) Copyrigt reserved/kopiereg voorbeou
74 Matematics P/Wiskunde V 0 DBE/November A F D E G M B C 9.. EM FD (Line parallel one side of AM AD OR prop t; EF BD) (Lyn ewewydig aan sy v EM OF eweredigst; EF BD) AM CM AM (diags of parm bisect/oekl parm alv) CM AM 7 (from 9../vanaf 9..) ME ME 9.. of FDC of BDC (AD BC) FD. area FDC area BDC BC. FD (opp sides of parm ) AD (tos sye v parm ) 7 S R answer/antw S R answer/antw AD BC subst into area form/ subst in opp formule S answer/antw () () (4) area FDC area ADC FD AD 7 (same eigts) (dieselfde oogtes) S R But Area ADC Area BDC (diags of parm bisect area) (oekl v parm alv opp) area FDC area BDC 7 S answer/antw (4) [6] Copyrigt reserved/kopiereg voorbeou
75 Matematics P/Wiskunde V DBE/November 04 QUESTION/VRAAG 0 P S X W 4 R y T 4 Q Y 0.. Tangent cord teorem/raaklyn-koordstelling R 0.. Tangent cord teorem/raaklyn-koordstelling R 0.. Corresponding angles equal/ooreenkomstige e gelyk R 0..4 s subtended by cord PQ OR s in same segment R e onderspan deur dieselfde koord OF e in dieselfde segment 0..5 alternate s/verwisselende e ; WT SP R 0. RW RT (Line parallel one side of OR RS RP prop t; WT SP) WR.RP RT (Lyn ewewydig aan sy v OF RS eweredigst: WT SP) S R () () () () () () 0. RTW RPS ( ; ; ) RW RT RS RP ( RTW RPS) RW.RP RT RS y Tˆ Rˆ (tan cord teorem/rkl-koordst) y Rˆ Qˆ ( s in same segment/ e in dieselfde segment) S S S R S R () (4) Copyrigt reserved/kopiereg voorbeou
76 Matematics P/Wiskunde V DBE/November 04 P S y X W y 4 R y T 4 y y Q Y 0.4 Qˆ PŜR (et of cyc quad/buite v kdv) P ŜR Ŵ (corresp s/ooreenk e ; WT SP) Qˆ Ŵ Qˆ ( s in same segment/ e in dies segment) Qˆ 80 ( y) ( s on straigt line/ e op reguitlyn) Ŵ 80 ( y) ( s of WRT/ e v WRT ) Qˆ Ŵ 0.5 In RTS and RQP: Rˆ Rˆ y (proven above/ierbo bewys) Ŝ Pˆ ( s in same segment/ e in dies segment) R Tˆ S RQˆ P ( rd angle of ) RTS RQP ( ; ; ) S R S R S S S S/R S ( ; ; ) () () () Copyrigt reserved/kopiereg voorbeou
77 Matematics P/Wiskunde V DBE/November RT RS ( RTS RQP) RQ RP RS RS RT RS RP RP RQ RP RS RP RT RP RW RS RW RQ RS RQ RS RQ RT RS ( RTS RQP) RQ RP WR.RP But RT (proven in 0./bewys in 0.) RS WR.RP RT RQ WR.RP WR RQ RQ.RS RQ.RS RS RP RS RP (proven in 0./bewys in 0.) S S RS on bot RP sides RT RS RP RQ WR.RP RT RS multiplication/ vermenigvuldig () () RT RQ ( RTS RQP) RS RP RT.RP RQ RS RT.RS and WR (proven in 0./bewys in 0.) RP RT.RS WR RP RQ RT.RP RS RT.RS RS RP RT.RP RS RP S RT.RS WR RP simplification/ vereenvoudiging () [0] TOTAL/TOTAAL: 50 Copyrigt reserved/kopiereg voorbeou
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