GRADE /GRAAD NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT MATHEMATICS P/WISKUNDE V NOVEMBER 06 MEMORANDUM MARKS: 50 PUNTE: 50 Tis memoradum cosists of 0 pages. Hierdie memoradum bestaa uit 0 bladsye.
Matematics/P DBE/November 06 NOTE: If a cadidate aswers a questio TWICE, oly mark te FIRST attempt. Cosistet accuracy applies i all aspects of te markig memoradum. LET WEL: Idie ' kadidaat ' vraag TWEE keer beatwoord, sie slegs die EERSTE pogig a. Volgeoue akkuraateid is op ALLE aspekte va die memoradum va toepassig. QUESTION/VRAAG.. ( 7).... 0 0 or 7 6 + 0 6 ± ( 6) ()() () 6 ± 8 0,5 or 5,65 6 + 0 6 + 9 + 9 ( ) ( )( ) 7 ± 7 ± 7 0,5 or 5,65 + + + 0 0 or Bot aswers are valid 0 7 correct substitutio ito correct formula 0, 5 5, 65 ( ) 7 0, 5 5, 65 isolate + stadard form factors bot aswers () () () (5)
Matematics/P DBE/November 06 + Let k k k 0 k( k ) ( )( ) k k k k 0 0 k 0 or ; Bot aswers are valid isolate k stadard form factors bot aswers (5) + By ispectio : 0 or or isolate 0 (5).. + + ( ) 86 86 86 7 epasio commo factor 7 + +.. f ( ) ( ) 0 + + + 86 86 + 5 5 + ( + )( ) or commo factor ( ) + factors bot aswers ()
Matematics/P DBE/November 06.. + < 0 ( + )( ) < 0 OR/ OF + + -... < < ( ; ) + 0 f / ( ) 0 we Te turig poit occurs at ( y) 5( y)( y) f is icreasig + y ad 5y y 0y 6y y y or y or y ad 5y y 5( ) 5 6 or y or y < < + + substitutio of y 6y bot y values bot values substitutio of () () () bot values bot y values
Matematics/P 5 DBE/November 06 y y 5 + 5 5 + 8 y ad 8 6 or y or y 5y equatig 8 + 5 bot values bot y values [] QUESTION/VRAAG. T 7 7. T a + ( ) d 87 5 + ( )( ) 87 5 + 96 a 5 ad d 87 5 + ( )( ) () (). + 9 87 96 ; 7;...; 87 S S 990 [ a + T ] [ 87] + 9 + 9 87 a aswer () ()
Matematics/P 6 DBE/November 06 S S ; 7;...; 87 990 [ a + ( ) d] [ ( ) + ( )( ) ] a aswer (). All egative terms ca be writte dow ad added to get te aswer of 990./Alle egatiewe terme ka eergeskryf word e da bymekaar getel word om 990 te kry. Sum S (5 + ) [ 5 87] 6 990 5; 5; 5... d 0 T 0 + 5 Last term i te sequece divisible by 5 is:/laaste term i die ry deelbaar deur 5 is: 87 + () 75 a aswer () 6 aswer [ 5 87] d 0 T 0 + 5 () T 0 + 5 75 0 + 5 0 00 0 Tere will be 0 terms i te sequece tat is divisible by 5./Daar is 0 terme i die ry deelbaar deur 5. 75 0 + 5 0
Matematics/P 7 DBE/November 06 5 ; ; ;... ; 8 ; 87; ; 87 T + 9 87 + 9 96 09 Tere are 09 terms i te sequece./daar is 09 terme i die ry. T ; T6 ; T ; T6... are divisible by 5./is deelbaar deur 5. Te largest iteger value of k suc tat 5k 09 5k 05 0, 0 6 k k 5 ; ; 7 ; ; 75; 79 ; 8 ; 87 ( ) d ( )( ) ( ) T a + 80 75 5 06 + Number of terms divisible by 5 06 + 5 0 + 9 87 09 5k 09 k 0 d 75 + 9 06 0 []
Matematics/P 8 DBE/November 06 QUESTION/VRAAG.. ; ; ; + 8 ;... +8 + + 5 + + + + 0 0.. First differeces/eerste verskille: ; ; 5 ; S [ ( ) + ( )( ) ] + ; ad + 5 calculatig secod differeces + + 0 S 50 < S > 50 > 50 > 5,8 > 5, 8 Te sum of te 6 first differeces will be greater ta 50. Terefore te 7 t term of te quadratic umber patter is te first satisfyig tis coditio./die som va 6 eerste verskille 7 sal groter as 50 wees. Gevolglik sal die 7 de term va die kwadratiese getalpatroo die eerste wees wat aa die voorwaarde voldoe. + 0,85 + 0,85 +... 7 T ar 0 ; r 0,85 or 9 0 T0 ( )( 0,85) substitutio ito correct,86 cm formula aswer ().. ( ) ( ).. a( r ) S S 5 r 0,85 7,77 5 ( ( 0,85) ) 5 7,77 Area of te page 0 60 Percetage of paper covered i grey ik: 7,77 00% 60 0,8% 60 0,8 [5]
Matematics/P 9 DBE/November 06 QUESTION/VRAAG. y 0 y 0. R(0 ; ) aswer. y a 9 a a. DP b y b 8 b b DP ( ) 6 uits.5 ( + ) + k 0 ( + ) k 0 < k < 8 < k < 0 8 substitutio a b 8 or use of logs b DP 6 uits k < or 8 < k < 0 8 () () () k > 8 () []
Matematics/P 0 DBE/November 06 QUESTION/VRAAG 5 5. f ( ) f ( ) 0 + 0 y ( ) + ( ) B( ; ) + or ( ) + 0 or ( ) y ( ) + ( ) () + + ( )( ) 0 0 0 or + ( ) + ( ) y B( ; ) 5. Rage/Waardeversamelig : y Rage/Waardeversamelig : y ( ; ] 5. or > ( ; ] ( ; ) 5. ( p) ( y + t) Vertical asymptote of () /vertikale asimptoot at Traslatio uits to te left / Traslasie eeede liks is te equatio of te vertical asymptote of ( + ) is die vergelykig va die vertikale asimptoot + y ( ) + ( ) y ( ; ] critical values or > critical values or > () () () () () ()
Matematics/P DBE/November 06 + + + + ( ) is te equatio of te vertical asymptote / is die vergelykig va die vertikale asimptoot 5.5 ( p)( y + t) ( y + t) ( p) y t p B( ;) Poit of itersectio of te asymptotes Syput va die asimptote p t t 5.6 -itercepts of f / -afsitte va f : + 0 ( )( ) 0 or / g ( ) < 0 for R ; Hece f ( ) < 0 or ; ; ( ] [ ) t p p t bot critical values or () []
Matematics/P DBE/November 06 QUESTION/VRAAG 6 6. y g g: sape: icreasig curve ( ; 0): oly o log grap 0 f f: ( ; 0) (0 ; ) 6. y log g : log y 6. ( ) log + y itercage ad y y + () Reflect te grap of g about te lie y to obtai g ad determie te poit of itersectio of f ad g./ Reflekteer die grafiek va g om die ly y e bepaal die syput va f ad g poit of itersectio of f ad g () 6. aswer () [0]
Matematics/P DBE/November 06 QUESTION/VRAAG 7 7. A P( + i ) 05, 50000 R56 89, 06 substitutig i ad values i correct formula [ ] 7. ( + i) P i 6 05, 56 89,06 05, 6 05, 0, 6 R 7 59, 79 per mot aswer 0, 5 i 6 substitutio ito correct formula aswer () 7. 50000 R 7 59,79 0,5 0,5 0,5 0,5 9 000 56 89,06 0,5 0,5 0,6097 0,5 log log 0,6097 5,87568 mots/ maade 6 paymets are required 6 paaiemete moet betaal word Tabiso will pay is loa off 0 mots sooer./tabiso los sy leig 0 maade vroeër af. 6 0, 5 i 6 substitutio ito correct formula aswer 9 000 substitute ito correct formula use of logs 5, 0 mots (5)
Matematics/P DBE/November 06 0,5 5689,06 0,5 0,65 0,5 9000 5796,8675 log,55695 0,5 0,5 9000 0,5 0,5 9000 9000 5,87568 6 paymets are required 6 paaiemete moet betaal word Tabiso will pay is loa off 0 mots sooer./tabiso los sy leig 0 maade vroeër af. 7. Te balace of is loa after te 5 t paymet was made: Die balas va sy leig adat die 5 ste paaiemet betaal is: 5 0,5 9000 5 0,5 Balace 5689,06 0,5 R 75,5 9 000 substitute ito correct formula use of logs 5, 0 mots 0,5 5689,06 0,5 9000 0,5 5 5 (5) Fial istalmet [ ( + i) ] P i Fial istalmet 0,5 9 000 R 78, 0,5 75,5 R 78, 0,5 0,87568 0,5 0, 5 75, 5 aswer 0,87568 0,5 9 000 0,5 0,5 aswer 0,87568
Matematics/P 5 DBE/November 06 0,5 Balace 5689,06 R 5 7,86 6 0,5 9000 0,5 6 0,5 5689,06 6 0,5 9000 0,5 6 Fial paymet 9 000 57,86 R 78, 9 000 57, 86 aswer QUESTION/VRAAG 8 [5] 8. f ( + ) ( + ) ( + ) ( + + ) + 6 + f ( + ) f ( ) 6 + + 6 + 6 + 6 f ( + ) f ( ) lim 6 + lim (6 + ) lim lim 6 ( + ) 6 f ( + ) f ( ) f ( ) lim ( + ) lim + 6 + lim 6 + lim (6 + ) lim lim 6 ( + ) f ( ) f ( + ) f ( ) f ( ) lim lim( 6 + 0 ) 6 (5) f ( + ) f ( ) lim / ( + ) 6 + lim( 6 + 0 ) f ( ) 6 (5)
Matematics/P 6 DBE/November 06 8. + lim g( ) a aswer aswer () 8. y y dy d 5 5 + 5 8. f ( ) + a + b + 8 f / ( ) + a + b At, m 8 ta / f ( ) ( ) + a ( ) + b 8 8 + a + b 8 a + b...() ( ) ( ) 8 ( ) y g f + 0 + + + 8 + a b a b 7...() a b 5 5 + a + b / f ( ) 8 or ( ) + a ( ) + b 8 + a + b + 8 a b (5) [6]
Matematics/P 7 DBE/November 06 QUESTION/VRAAG 9 9. f ( ) ( )( + ) 9. 9. // f ( ) 6 + 8 OR 6 + < 0 < < 0 + 8 0 8 or [ ; ] ; 9. f ( 0) d 8 f ( ) a f f ( ) ( ) 8 a a a + b + 8 or + c 8 + b + c b 8 b ( ) + 8 f f ( ) + 8 By itegratio/deur itegrasie f + + d ( ) f (0) d 8 a b c c equatig derivative to zero factors values 6 + 8 < () () < () () [ ; ] ; () d 8 f a + b + ( ) c a b c f ( ) + + d d 8 a b c (5) (5) []
Matematics/P 8 DBE/November 06 QUESTION/VRAAG 0 0. M( t) t + t + 7t M( ) ( ) + ( ) + 7( ) 6 6 molecules/molekules M t t + t + 7t / M t t + 6t + 7 0. ( ) ( ) / M ( ) ( ) + 6( ) 7 + 7 7 molecules per our/molekules per uur M t t + t + 7t / M (t) t + 6t + 7 // M ( t) 0 6t + 6 0 t Maimum rate of cage of te umber of molecules of te drug i te bloodstream is after our./maksimum tempo va veraderig va die getal molekules i die bloedstroom is a uur 0. ( ) M( ) ( ) + ( ) + 7( ) 6 / M ( t ) t + 6t + 7 M / ( ) 7 M // ( t ) M // ( t ) 0 aswer () () () [8]
Matematics/P 9 DBE/November 06 QUESTION/VRAAG. Watces TV durig eams Do ot watc TV durig eams Total Male 80 a 80+a Female 8 60 Total b 60 a + a 0 b 80 + 8 8. No 0 P ( M ad ot watcig TV) 0 60 8 60.. P ( watcig TV) 0,8 80% 5 60.. P ( female ad ot watcig TV) 0,075 7,5% 0 a 0 b 8 No reaso 8 60 60 () () () () [8]
Matematics/P 0 DBE/November 06 QUESTION/VRAAG. We wat to create codes tat are eve umbers greater ta 5000. Te digit 6 ca be used i oe of two places i tese codes ad terefore tis presets two scearios. Os wil kodes kry wat ewe getalle groter as 5000 is. Die syfer 6 ka i twee posisies i die kode gebruik word e twee opsies is mootlik: CASE : Te first digit is a 6./Die eerste syfer is 6. 6 5 5 0 Number of codes startig wit 6./Getal kodes wat met 6 begi. 5 0 CASE : Te first digit is a 5 or 7./Die eerste syfer is 5 of 7. 5 7 5 6 Number of codes ot startig wit 6./Getal kodes wat ie met 6 begi 5 0 Terefore total umber of possible codes./die totale getal mootlike kodes 0 + 0 60. ( 5 ) + ( 5 ) + ( 5 ) 60 60 + 60 ( 5 ) ( 5 ) 80 0 60 + 0 5 0 60 ( 5 ) ( 5 ) ( 5 ) 60 ( 5 ) ( 5 ) 60 [5] [5] [5] TOTAL/TOTAAL: 50