NATIONAL SENIOR CERTIFICATE/ NASIONALE …

1 NATIONAL . SENIOR certificate /. NASIONALE . SENIOR SERTIFIKAAT. grade /GRAAD 12. JUNE/JUNIE 2016. mathematics P1/ wiskunde V1. memorandum . MARKS/PUNTE: 150. This memorandum consists of 12 Hierdie memorandum bestaan uit 12 bladsye. 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2016). NOTE/LET OP: If a candidate answered a question TWICE, mark the FIRST attempt ONLY. Indien kandidaat vraag TWEE keer beantwoord het, merk SLEGS die EERSTE poging. Consistent accuracy(CA) applies in ALL aspects of the memorandum . Volgehoue akkuraatheid geld deurgaans in ALLE aspekte van die memorandum . If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt. Indien kandidaat poging vir vraag deurgetrek het en nie die vraag weer beantwoord het nie, merk die poging wat deurgetrek is. The mark for substitution is awarded for substitution into the correct formula.

2 Die punt vir substitusie word toegeken vir substitusie in die korrekte formule. QUESTION 1/VRAAG 1. 2 2 7 = 0. (2 7) = 0 factorisation / faktorisering = 0 or/of 2 7 = 0 -values / waardes 7. =2. (3). 4. 4 + + 11 = 0. standard form / standaardvorm 4 2 + 11 + 4 = 0. 11 (11)2 4(4)(4). = Penalise 1 mark for substitution / vervanging 2(4). 11 57 incorrect rounding = 8 Penaliseer 1 punt vir = 0,43 or/of = 2,32 verkeerde afronding. -values / waardes 11 57. If stopped at : max. 2 marks 8. 11 57. As stop by : maks. 2 punte 8. (4). (2 1)( 3) > 0. + +. 1. 3 critical values with method 2 kritieke waardes met metode 1 3. 2. 1. < / > 3 answer / antwoord (3). 2. 3 . 3 +1 = 27 32 +1. 32 +1 = 33 33 . 2 + 1 = 3 equating / gelykstel =1 answer / antwoord (4). Kopiereg voorbehou Blaai om asseblief EC/JUNE/JUNIE 2016) mathematics P1/ wiskunde V1 3. 3 + = 2 . = 2 3 .. (1). 4 2 + 2 = 2 + 7.

3 4 2 + (2 3)2 = 2 (2 3) + 7 substitution / vervanging 4 2 + 4 2 12 + 9 = 4 2 6 + 7 removing brackets / verwyder 4 2 + 4 2 12 + 9 4 2 + 6 7 = 0 hakies 4 2 6 + 2 = 0 standard form / standaardvorm If formula is used, award 2 2 3 + 1 = 0. factor's mark for factors / faktore (2 1)( 1) = 0. 1 substitution. -values / waardes = 2 or/of = 1 As formule gebruik word, -values / waardes = 2 or/of = 1 ken faktore punt toe vir substitusie. (6). ( ) = ( 2)( 2 6 + 10). Consider the quadratic factor : ( 2 6 + 10). = 2 4 . = ( 6)2 4(1)(10) substitution into delta /. = 36 40 vervanging in delta = 4. => < 0, therefor NO Solutions answer / antwoord = 2 is the only solution. conclusion / gevolgtrekking (4). [23]. Copyright reserved Please turn over 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2016). QUESTION 2/VRAAG 2. 0 ; 1 ; 1 ; 6 ; 14 ; . 1 2 5 8 - 1st differences first differences / eerste verskille 3 3 3 - 2nd differences second differences / twee verskille (2).

4 Next term/volgende term = 25 answer/antwoord (1). 2 = 3 3 + = 1 + + = 0 3. =2. 3 3 3 11. =2 3 (2) + = 1 2. 2 + =0 = . 11. 11 2. = 2. =4 =4. 3 2 11. = +4. 2 2 answer / antwoord (4). 3 11 substitution / vervanging 30 = (30)2 (30) + 4. 2 2. = 1189 answer / antwoord (2). 3 2 = 4 3. 13 = 27 method / metode 2 = 40. = 20. values of a and b /. 13 = 7 waardes van a en b =6 (2). + ( 1) = formula / formule 6 + ( 1)(7) = 230 substitution / vervanging 7 = 231. = 33 answer / antwoord (3). 1 =. 1 . = 5. 5. 1 substitution / vervanging 1 < <1. 5. 5 < 1 < 5. 6 < < 4. answer / antwoord (3). 4 < < 6. (1 ). = + . method / metode 1 . 1 20. 16 (1 (2) ) substitution / vervanging 40 = + 20(3). 1. 1 2. = 31,99 + 60 simplification / vereenvoudiging 92 answer / antwoord (4). Kopiereg voorbehou Blaai om asseblief EC/JUNE/JUNIE 2016) mathematics P1/ wiskunde V1 5.. 1 1. 16 ( ) . 2. =1 =1. 1 1.

5 16 (2) (2).. =. 1 . 16 substitution / vervanging =. 1. 1 2. answer / antwoord (2). = 32. [25]. QUESTION 3/VRAAG 3. = 2 (horizontal asymptote / horisontale asimptoot) answer / antwoord (1). = 1 (vertical asymptote / vertikale asimptoot) answer / antwoord (1). 3. 2=0 y=0. 1. 3. =2. 1. 3 = 2 2. 5 = 2 . 5 x intercept / afsnit =. 2. 3. = 2. 0 1. y intercept / afsnit (3). = 5. asymptotes / asimptote x-and y-intercepts / x-en y-afsnitte shape / vorm (3). (4 ; 5) (2). [10]. Copyright reserved Please turn over 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2016). QUESTION 4/VRAAG 4.. =. 2 . ( 2). = x-coordinate / ko rdinaat 2( 1). = 1. = ( 1)2 2( 1) + 3 y-coordinate / ko rdinaat = 1 + 2 + 3. =4 ( 1 ; 4) coordinates of C / ko rdinate van C. (3). 2 2 + 3 = 0. 2 + 2 3 = 0 standard form / standaardvorm ( + 3)( 1) = 0 factors / faktore + 3 = 0 1 = 0. = 3 = 1 both values / beide waardes ( 3; 0) (1; 0) (3).

6 M=2 & c=6 value of m / waarde van m y = 2x + 6 value of c / waarde van c (2). 2 = (1)2 + (2)2 method / metode =5 substitution / vervanging = 5. answer / antwoord OR. C(-1 ; 4) and/en E(0 ; 6). = (0 + 1)2 + (6 4)2. = 5 (3). >1 answer / antwoord (2). [13]. Kopiereg voorbehou Blaai om asseblief EC/JUNE/JUNIE 2016) mathematics P1/ wiskunde V1 7. QUESTION 5/VRAAG 5. ( ) = . substitution / vervanging 27 3. 8. = . 3 3. ( ) = 3 answer / antwoord 2. 3 (2). = 2. 3 . =( ). 2. 3 swop and / ruil en om =( ). 2. = 3 answer / antwoord 2. (2). 3 = 1 OR other method (eg. sketch). 2. equating / gelykstel 3 1. =( ). 2. 2 answer / antwoord =. 3 (2). answer / antwoord (1). y asymptote / asimptoot negative y-intercept / negatiewe y-afsnit x shape / vorm q (3). [10]. Copyright reserved Please turn over 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2016). QUESTION 6/VRAAG 6.. 1 + = (1 + ) formula / formule.

7 8,5 4. = (1 + ) 1 substitution / vervanging 400. = 0,0877. / = 8,77% . answer / antwoord (3). = (1 + ) formula / formule 20. 8,5. = 12 000 (1 + ) substitusie / vervanging 400. = 18 273,54 answer / antwoord (3). = (1 ) . 12,4 substitution / vervanging 41 611,57 = 120 000 (1 ). 100. simplification / vereenvoudiging 0,3467630833 = 0,876 . correct use of logs / korrekte log(0,3467630833) gebruik van logs =. log(0,876). answer / antwoord 8 (4). 8 12 8 8 8 4. 20 000 = (1 + ) + (1 + ) + (1 + ) setting up equation / opstel 400 400 400. van vergelyking 8 12 8 8 8 4. 20 000 = [(1 + ) + (1 + ) + (1 + ) ]. 400 400 400. 20 000 x the subject of the formula /. = 12. 8 8 8 8 4 x die onderwerp van die formule [(1 + 400) + (1 + 400) + (1 + 400) ]. answer / antwoord = 5 678,05. (4). [14]. Kopiereg voorbehou Blaai om asseblief EC/JUNE/JUNIE 2016) mathematics P1/ wiskunde V1 9. QUESTION 7/VRAAG 7.

8 ( ) = 2 2 3 . ( + ) = 2( + )2 3( + ) substitute / vervang ( + ). = 2( 2 + 2 + 2 ) 3( + ). = 2 2 + 4 + 2 2 3 3 . ( + ) ( ) formula / formule ( ) = lim 0 Penalise 1 mark for incorrect use of formula. Must show ( ). 2 2 + 4 + 2 2 3 3 (2 2 3 ) Penaliseer 1 punt vir verkeerde = lim gebruik van formule. Moet ( ). 0 . toon. 2 2 + 4 + 2 2 3 3 2 2 + 3 . = lim 0 . 4 + 2 2 3 simplification / vereenvoudiging = lim 0 . (4 + 2 3). = lim common factor / gemene faktor 0 . = lim (4 + 2 3). 0. Answer ONLY: 0 marks answer / antwoord = 4 3 SLEGS antwoord: 0 punte (5). 3 Penalise 1 mark for incorrect notation. = 2 . 5 2 Penaliseer 1 punt vir verkeerde notasie. 1 3. = 2 2 1 1. 3. 5 2 2 5 1. 1 3. = 2 + 2 1. 3. 5 2 5 2. 1 3. = +. 5 2 (4). [9]. Copyright reserved Please turn over 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2016). QUESTION 8/VRAAG 8. ( ) = 3 4 2 11 + 20. ( ) = 3 2 8 11 = 0 ( ) = 0.

9 (3 11)( + 1) = 0. 3 11 = 0 + 1 = 0 factors / faktore 11. = (3,7) = 1 x-values / waardes 3. 400 y-values / waardes = ( 14,8) = 36. 27. ( 1 ; 36) &. 11. ( ; . 400. ) / (3,7 ; -14,8) coordinates / ko rdinate 3 27 (5). ( ) = 6 8 = 0 ( ) = 0. 6 = 8. 8 1. = 6 or/of 1 3 or/of 1, 33 answer / antwoord (2). OR. 11. 1+. 3. = 2. 4. = or/of 1,33. 3. = (2) = 3(2)2 8(2) 11. = 15 = 15. (2) = (2)3 4(2)2 11(2) + 30. =0 (2) = 0. 1 = ( 1 ). 0 = 15( 2) substitution / vervanging = 15 + 30 answer / antwoord (4). 36 units downwards answer / antwoord 14,8 units upwards answer / antwoord (2). = 9 answer / antwoord (1). = 1 = 5 answers / antwoorde (2). < 1 > 5 <1 >5 (2). [18]. Kopiereg voorbehou Blaai om asseblief EC/JUNE/JUNIE 2016) mathematics P1/ wiskunde V1 11. QUESTION 9/VRAAG 9. = 2 . 8 = 2 substitution / vervanging 8 answer / antwoord = (2). 2. = 2 2 + 4 2 2. 8 4 . = 2 2 + 4 ( ) No mark for the answer /.

10 2 substitution / vervanging 2 32 Geen punt vir die antwoord (3). = 2 + . ( ) = 2 2 + 32 1. ( ) = 4 32 2 = 0 ( ) = 0. 4 3 32 = 0 standard form / standaardvorm 4 3 = 32. 3 = 8 exponential law / eksponent wet = 23. =2 value of x / waarde van x dimensions / dimensies (5). [10]. QUESTION 10/VRAAG 10. ( and/en ) = ( ) ( ). = 0,4 0,5 rule / re l = 0,2 answer / antwoord (2). ( or/of ) = ( ) + ( ) ( and/en ) rule / re l = 0,4 + 0,5 0,2. = 0,7 answer / antwoord (2). ( and/en ) = 1 ( or/of ) rule / re l = 1 0,7. = 0,3 answer / antwoord (2). Copyright reserved Please turn over 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2016). first branch / eerste vertakking second branches / tweede vertakkings outcomes/uitkomste If probabilities not listed, maximum 1 mark /. Indien waarskynlikhede nie gelys is nie, maksimum 1 punt (3). 2. ( from )\ ( ) = answer / antwoord (1). 5. 3 5 3. 10.