NATIONAL SENIOR CERTIFICATE GRADE/GRAAD 12
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1 NATIONAL SENI CERTIFICATE GRADE/GRAAD MATHEMATICS P/WISKUNDE V EXEMPLAR 0/MODEL 0 MEMANDUM MARKS: 0 PUNTE: 0 This memandum consists of pages. Hierdie memandum bestaan uit bladse.
2 Mathematics P/Wiskunde V DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum NOTE: If a candidate answers a question/vraag TWICE, onl mark the FIRST attempt. Consistent accurac applies in all aspects of the marking memandum. LET WEL: Indien n kandidaat n vraag TWEE keer beantwod, merk slegs die EERSTE poging. Volgehoue akkuraatheid is DEURGAANS in ALLE aspekte van die memandum van toepassing. QUESTION/VRAAG.. 0 ( ) ± ( 6) ± ( 6) ( )( ) ( ) 0, 8,6 facts both answers subs into crect fmula 0,, 6 () () ( ) ( ) + 9 ± ( ) + 9 ± 0, 7,6 0,, 6 ().. 8 ; ( ) > 0 ( ) 8 ()
3 Mathematics P/Wiskunde V DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum ( ) 8 0 ; > 0) ( ) ( 0 0 > + ( ) 8 () facts 8 ().. ( ) 0 positive is alwas 0 < < < > 0 < (). ( ) ( )( ) and ( )( ) and 6 + subst 6 standard fm facts -values -values (6) standard fm facts -values -values Answer onl full marks
4 Mathematics P/Wiskunde V DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum. 0 ( 6)( + ) and.. No, there will be no intersection between the graphs. Min value of ( ) + is Nee, daar sal geen snding tussen die grafieke wees nie. Min waarde van ( ) + is ( ) + ( ) ( ) No, there will be no intersection between the graphs. + standard fm facts - values - values + + reason reason (6) (6) () () ()
5 Mathematics P/Wiskunde V DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum Nee, daar sal geen snding tussen die grafieke wees nie. ().. ( ( ) + + ) ( 6) ( )( ) < 0 No, there is no solution to the equation f() g() Nee, daar is geen oplossing vir die vergelking f() g() ( ) + + k ( ) k k > 0 f all real values of k > / vir alle reeele Answer onl full marks waardes van reason () () k 6 + k 0 ( 6) ( )( k) k k F real unequal roots k > 0 k > k > / Vir reeele ongelke wtels () []
6 Mathematics P/Wiskunde V 6 DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum QUESTION/VRAAG.... T n a + ( n ) d ( n ) n 6 6n 88 n 8 S n n [a + ( n ) d] 8 [(8) + 7(6)] 76 a 8 and d 6 T n 00 substitution in fmula () ().. Sum of all numbers from to 00 / Som van alle getalle van tot [() + 99()] 00(0) 0 Sum of numbers not divisible b 6 / Som van getalle wat nie deelbaar deur 6 is nie 0 ( ) , 8; ; r.. T n ar n 6 n n n+ ( ) S n > n 6 substitution ( ) r () (in an fmat) () S n > / n 6 ()
7 Mathematics P/Wiskunde V 7 DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum.. < S S n > > n > n ( r ) a r 6 < < ( n 6 a r 6 n n n n n ) Answer gets closer and closer to the me terms gets added together Antwod beweeg nader en nader aan hoe meer terme bmekaar getel wd S n > simplification n > / n 6 () substitution of a and r () epanding the series () [6]
8 Mathematics P/Wiskunde V 8 DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum QUESTION/VRAAG.. ; ; ; z.. T n n + 6 0; ; 8 0 nd difference a a a a ().. T n n + 6 d a a 0 a a st differences 0; ; 8 () a + b 0 a + b b 0 b a + b + c + + c a + b + c c T n n + n n + n T n ()
9 Mathematics P/Wiskunde V 9 DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum. Consider the sequence made up b the first facts of each term: Beskou die r wat deur die eerste fakte van elke term gevm wd: ; ; 9; ; 8 An arithmetic sequence / rekenkundige r: T n a + ( n ) d + ( n ) n To find the no. of terms: 8 n Aantal terme: n 8 n n T n no. of terms The second fact is me than the first fact / Tweede fakt is meer as die eerste fakt: n + T n n n T n Consider the sequence made up b the second facts of each term: Beskou die r wat deur die tweede fakte van elke term gevm wd: ; 6; 0; ; 8 Also an arithmetic sequence / rekenkundige r: T n a + ( n ) d + ( n ) Answer onl n full marks In sigma notation: (n )(n ) (n )(n ) (6n 0n + 6) n n n n T n answer in sigma notation () [0]
10 Mathematics P/Wiskunde V 0 DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum QUESTION/VRAAG f ( ) + f (0) ; ( ) ; 0 subst 0 ( 0 ; ) subs 0 ; 0 () () A(- / ; 0) 0 shape f - B(0;-) - both intercepts crect hizontal and vertical asmptote ().. ( + ) ( + ) () + k ( ) k + k ( ) + k ()
11 Mathematics P/Wiskunde V DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum.. f a. b a. b a. b a. b b b 0 + q ( ) [subs [subs ( 0 ; ) ( ; ).. A translation of units up and unit to the left. 'n Translasie van eenhede na bo en eenheid na links. ] ] subs q a b f ( ) units up unit to the left () () Dilation b a fact of and 7 units up. Verkleining deur fakt van en 7 eenhede na bo. dilation b fact 7 units up () []
12 Mathematics P/Wiskunde V DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum QUESTION/VRAAG. f ( ) b a ( ) 9 / 8 9 TP ( ; ) 8 + 6, + - ( + ) [( + ) ] 6 9 [( + ) ] 6 9 ( + ) + 8. m tangent tan f ( ) TP ( ; ) 8 b / a f ( ) 0 9 / 6, 8 () [( + ) ] 6 9 / 6, 8 () tan ( ) ( ) + 6 Point of contact: P( ; 6). Eq of g: m( ) 6 ( + ) + 6 substitute in equation. d > () () () [0]
13 Mathematics P/Wiskunde V DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum QUESTION/VRAAG 6 6. g ( ) 8a 6 a a a(8) subst (8 ; ) a 6. 0 answer 6. 0 answer ; 0 ; ( 8)( ) 8 when, LHS but RHS Hence 8 onl () () () interchange and answer () (squaring both sides) facts 8 selects < < 8 < 8 0 < () () []
14 Mathematics P/Wiskunde V DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum QUESTION/VRAAG Selling price / Verkooppr s 0, n [ ( + i) ] Pv i 0, , , Pv ,09 i n 0 R6 79,9 () () 0, F v 0 n [( + i) ] i 0,09 + 0, , Total interest paid / Totale rente betaal (6 79,9 0) R n [ ( + i) ] Balance i 0,09 679,9 + 0, ,7 0, ( + ) 0,09 i n 0 R6 79,9 (6 79,9 0) , 9 n R6 09,7 0 () () ()
15 Mathematics P/Wiskunde V DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum 0,09 A ,7 8 n [( + i) ] Fv i 0,09 679,9 + 0,09 796, , 7 n 8 Balance of loan 66,7 796, ,77 R6 09,77 () 0,09 Balance ,77 8 0,09 679,9 + 0,09 7. New value of bond: 0, ,7 + 0, , ,8 6 8,8 8 subs of and 679,9 n 8 R6 09,77 () 0, 09 R6 09,7( + ) R6 8,8/ R6 8,8 () , ,09 log (0,060) n log n 09,7 0 months 0,09 n 0, subs into crect fmula use of logs ()
16 Mathematics P/Wiskunde V 6 DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum n 0, ,8 0,09 n log (0,060) 0,09 + n 09,7 0 months 8 00 subs into crect fmula use of logs () [6]
17 Mathematics P/Wiskunde V 7 DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum QUESTION/VRAAG f ( + h) f ( ) f ( ) f ( + h) ( + h) lim h 0 6 f ( ) 6h + h 6h + h f ( ) lim h 0 h h(6 + h) lim h 0 h lim(6 + h) f h 0 6 ( + h) f ( ) h + 6h + h [ ( + h) ] ( ) lim h 0 h [ ( + h + h ) ] lim h 0 h [ + 6h + h ] lim h 0 h 6h + h lim h 0 h h( 6 + h) lim h 0 h lim 6 h 0 d 8 d ( + h) + + substitution of of + h simplification to 6h + h fmula taking out common fact fmula substitution of + h simplification 6 h + h to h taking out common fact 8 () () () [7]
18 Mathematics P/Wiskunde V 8 DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum QUESTION/VRAAG 9 9. ( ) is a fact of f / is n fakt van f. 9. f ( ) f ( ) 0 ( + )( )( ) ( )( ) ( )( + )( ) 0 -intercepts: ( ; 0); (;0); (;0) + 0 ( ) ( ; 0) (; 0) () f ( ) f ( ) f ( ) ( )( + ) At turning points TP's are (,8) ( ; 6) and ;,8 (; 0) () f ( ) 8 f ( ) 0 - value - value - values () (- ; 6) (0 ; 0) f and - intercepts shape turning points () (- ; 0) 0 ( ; 0) ( ; 0) (,67 ; -,8)
19 Mathematics P/Wiskunde V 9 DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum 9. f ( ) < 0 if < <, 67 (- ;,67) etreme values notation etreme values notation () () []
20 Mathematics P/Wiskunde V 0 DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum QUESTION/VRAAG 0 0. After t hours: BF 0t km and CD 0t km BF 0t BC 00 0t BC 00 0t FC ( 0t) + ( 00 0t) 900t 00t t + 600t Pthagas 8000t answer () 0. FC is a minimum when FC is a minimum. FC 00t 8000t dfc 000t dt 8000 t,6hrs (96 minutes) FC 00t 8000t ( ) 8000(.6) 60 The will be 60km apart FC 00t 8000t dfc 000t 8000 dt dfc 0 dt answer () subs into equation answer () [0]
21 Mathematics P/Wiskunde V DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum QUESTION/VRAAG. P(A B) P(A) + P(B) 0,7 P(A) + P(A) 0,7 P(A) P(A) 0,9 P(B) (0,9) 0,8.. P A,P P(A B) P(A) + P(B) P(A) 0,9 answer () A 9 Y P A,Y B,P first tier second tier probabilities outcomes () B.. P(AY) 9 Y B;Y ().. P(P) () []
22 Mathematics P/Wiskunde V DBE/0 NSC/NSS Grade Eemplar/Graad Model Memandum QUESTION/VRAAG. Number of different letter arrangements: Aantal verskillende letter rangskikkings wat gevm kan wd:! 0. S and T can be arranged in! different was. The remaining three letters can be arranged in! different was Total number of different letter arrangements having S and T as the first two letters!.! S en T kan op! verskillende maniere rangskik wd. Die letters wat obl kan op! verskillende maniere rangskik wd Totale aantal letterrangskikkings waarin S en T die eerste twee letters van die rangskikking sal wees!.!! 0!! () P(having S and T as first two letters)!.! answer () [] TOTAL/TOTAAL: 0
NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2014 MATHEMATICS P1/WISKUNDE V1 MEMORANDUM
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