NATIONAL SENIOR CERTIFICATE/ NASIONALE …

1 NATIONAL . SENIOR certificate /. NASIONALE . SENIOR SERTIFIKAAT. GRADE/GRAAD 12. JUNE/JUNIE 2017. mathematics P1/ wiskunde V1. memorandum . MARKS/PUNTE: 150. This memorandum consists of 14 Hierdie memorandum bestaan uit 14 bladsye. 2 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2017). 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.

2 The mark for substitution is awarded for substitution into the correct formula. Die punt vir substitusie word toegeken vir substitusie in die korrekte formule. QUESTION 1/VRAAG 1. 2 30 = 0. ( + 5)( 6) = 0 factors / faktore + 5 = 0 or/of 6 = 0 -values / waardes = 5 or/of = 6 (3). Penalise 1 mark for 3 2 + 1 = 0. incorrect rounding Penaliseer 1 punt vir (1) (1) 2 4(3)( 1). x verkeerde afronding. substitution / vervanging 2(3). 1 13. x . 6. x 0, 43 or / of x 0, 77 -values / waardes (3). x 2 2( x 4). x2 2x 8 0 factors / faktore ( x 2)( x 4) 0. critical values with method + +. kritieke waardes met metode 2 4 2 4. answer (accuracy) /. 2 4 / [ 2 ; 4] antwoord (akkuraatheid) (4). Copyright reserved / Kopiereg voorbehou Please turn over / Blaai om asseblief (EC/JUNE/JUNIE 2017) mathematics P1/ wiskunde V1 3.

3 3x 5 x 2. isolating/isoleer 5 x 3x 2 5 x squaring both sides / kwadreer albei kante 2. (3 x 2) 2 5 x 9 x 2 12 x 4 25 x 9 x 2 37 x 4 0 standard form / standaardvorm (9 x 1)( x 4) 0. answers / antwoorde 1. x or / of x 4. 9. Check :3 19 5 1. 9 2 3 4 5 4 2. 1. 3 53 2 12 10 2. testing and conclusion /. 2 4. 3 2 2 toets en gevolgtrekking x 4 is the only solution. OR/OF. 3x 5 x 2. 3x 5 x 2 2 0 standard form / standaardvorm 1. 3x 1 x 2 0. 1. 2. 1. 2. factors / faktore 3 x 2 1 or / of 1. x2 2. 1. 1. x2 . 1. or / of x 4 answers / antwoorde 3. 1. since x 2 must be >0. x 4 is the only valid answer conclusion / gevolgtrekking (5). y x 6 (1). ( x 3) 2 ( y 3) 2 (2). from / vanaf (1) : y x (3). substitution / vervanging (3) in (2) : ( x 3) 2 ( x 6 3) 2 18. removing brackets / verwyder hakies x 2 6 x 9 x 2 6 x 9 18.

4 2x2 0 standard form / standaardvorm x 0. -value / waarde y 0 6. 6 -value / waarde (5). Copyright reserved / Kopiereg voorbehou Please turn over / Blaai om asseblief 4 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2017). 1 7. 1 . 1 5. x . x adding denominator / optel van noemer 1 2. If candidate after step 3 concludes x 1 5. 2. x = 2, then max of (2/5). x As kandidaat na stap 3 aflei dat simplification / vereenvoudiging x 2. x = 2 is, dan maks van (2/5). x 1 5. 2. 2x2 2 5x standard form / standaardvorm 2 x2 5x 2 0. (2 x 1)( x 2) 0 factors or formula / faktore of formule x 1. 2 or / of x 2 answers / antwoorde (5). [25]. Copyright reserved / Kopiereg voorbehou Please turn over / Blaai om asseblief (EC/JUNE/JUNIE 2017) mathematics P1/ wiskunde V1 5. QUESTION 2/VRAAG 2.

5 1 ; 5 ; 12 ; 22 ; ; . 4 7 10 - 1st differences 3 3 - 2nd differences T5 = 35 and/en T6 = 51 answers / antwoorde (2). 2 = 3 3 + = 4 + + =1 3. =. 3 3 3 1 2. =2 3 (2) + = 4 2. 2+ =1 = . 11. 1 2. = 2 =0 =4. 3 1. = 2 2 2 . answer / antwoord (4). 3 2 1. = 3432. 2 2 equation / vergelyking 3 2 1. n n 3432 0. 2 2. standard form / standaardvorm 3n n 6864 0. 2. (3n 143)(n 48) 0 factors or formula / faktore of formule 143. n or / of n 48 answer n = 48 / antwoord n = 48. 3 (4). 2 1 = 3 2. + 2 = 3 2 method / metode 2 = 2 2. answer / antwoord = 2. OR/OF. 2 3 2 method / metode m . 2. 2 2. m . 2 answer / antwoord m 2. (2). T51 a 50d value of d / waarde van d 2 50 2 2 substitution into correct formula /. vervanging in korrekte formule 99 2. answer / antwoord (3). Copyright reserved / Kopiereg voorbehou Please turn over / Blaai om asseblief 6 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2017).

6 Terms between 50 and 500 divisible by 7. First term = 56 and Last term = 497 identification of first and last terms /. vasstel van eerste en laaste terme 56 (n 1)(7) 497 497 56 . 56 7 n 7 497. 1 substitution / vervanging 7 . OR/OF. 7n 448 64 terms / terme n 64 terms / terme answer / antwoord If /As: (500 50)/7 = 64,29 n = 64 max/maks. (1/3) (3). 1 2 2 2. a 2 & r OR / OF ; ;. 3 1 3 9. n 1. 1 2. Tn 2 Tn n 1 method / metode 3 3. n 1. 1 1 2. 2 =. 3 3 3 3 . n 1 substitution / vervanging n 1 . 6 = 6.(3) n 3 . n 1 1 . n = 6 answer / antwoord = 6 (3). 3 3 . 1 r 1 Yes Yes, because reason (2). 1 1 1. 3. 3 p S S4. 3 . p 2.. 2 1 13 4. substitution / vervanging 1 13 1 13. 80. 3p 3 simplification / vereenvoudiging 27. 81 80. 3p . 27 27. 1 exponential law / eksponensi le wet 3p.

7 27. 3 3 3. p answer / antwoord p 3 (5). 6. k expansion /. 1 1 (1 1) (1 1 1) (1 1 1 1) (1 1 1 1 1) (1 1 1 1 1 1). k 1 n 1. uitbreiding 1 2 3 4 5 6 answer / antwoord (2). 21. [30]. Copyright reserved / Kopiereg voorbehou Please turn over / Blaai om asseblief (EC/JUNE/JUNIE 2017) mathematics P1/ wiskunde V1 7. QUESTION 3/VRAAG 3. =5 answer / antwoord (1). ( x 5) 2 4 let / stel y = 0. square root / vierkantswortel x 5 2. x 3 or/of x 7 answer / antwoord (3). OR / OF. x 2 10 x 25 4 let / stel y = 0. x 10 x 21 0. 2. factors / faktore ( x 3)( x 7) 0. answers / antwoorde x 3 or/of x 7. y f x-intercepts / x-afsnitte 21. y-intercepts / y-afsnit turning point / draaipunt 3 7 x shape / vorm (5 ;- 4) (4). Range of f : [ 4; ] / 4 answer / antwoord (1). Reflection about the x-axis / answer / antwoord Refleksie in die x-as y ( x 5)2 4 equation / vergelyking (2).

8 Copyright reserved / Kopiereg voorbehou Please turn over / Blaai om asseblief 8 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2017). x 2 3 kx 1. equating / gelykstel x 2 kx 4 0. For g ( x) to be a tangent, roots are equal. standard form / standaardvorm b 2 4ac 0 0. ( k ) 4(1)(4) 0. 2. substitution / vervanging k 2 16 0. k 2 16 / k 4 k 4 0. answers / antwoord (5). k 4 / k 4 or/of k 4. [16]. QUESTION 4/VRAAG 4. p 1 and/en q 2 value of p / waarde van p value of q / waarde van q (2). a y 2. ( x 1). a substitution of point / vervanging van punt 4 2. 0 1. a 2 value of a / waarde van a 2 equation / vergelyking y 2. ( x 1) (3). Point of intersection of axes of symmetry of f is ( 1 ; 2) point of intersection / snypunt Point of intersection of axes of symmetry of g is: x 3 x 1 equating / gelykstel 2x 4.

9 X 2 x-value and y-value /. y 1 x-waarde en y-waarde Transformation is from ( 1; 2) (2; 1) method / metode 3 units to the right and 3 units down answer / antwoord (5). [10]. Copyright reserved / Kopiereg voorbehou Please turn over / Blaai om asseblief (EC/JUNE/JUNIE 2017) mathematics P1/ wiskunde V1 9. QUESTION 5/VRAAG 5. 5 1. a1 substitution / vervanging 6 2. 1 answer / antwoord a . 3 (2). 2. 1 1. p 9 12 substitution / vervanging 3 2 answer / antwoord (2). x 1 1. g:y . 3 2 answer / antwoord 3 2. x 1 (1). h( x) 3x answer / antwoord h log x h 1 : y log3 x or/of answer / antwoord h-1 (2). log 3. Points : (-2 ; 9 12 ) and 0 ; 3. 2 . coordinates of A / ko rdinate van A. y y m 2 1. x2 x1. 23. 19. substitution / vervanging 2. 2 0. 8.. 2. answer / antwoord (3). 4.

10 [10]. Copyright reserved / Kopiereg voorbehou Please turn over / Blaai om asseblief 10 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2017). QUESTION 6/VRAAG 6.. 1 + = (1 + ) formula / formule . 15 12. = (1 + ) 1 substitution / vervanging 1200. = 0,1607545177. / = 16,08% . answer / antwoord (3). = (1 + ). = 75 000(1 +12% 8) substitution / vervanging = 147 000. R 147 000 answer / antwoord . Monthly installment: 96 months R 1531, 25 answer / antwoord (3). A P (1 i ) n substitution / vervanging 147 000 75000(1 i )8. (1 i )8 1,96. simplification / vereenvoudiging 1 i 8 1,96. i 1, 087757306 1 making i subject of the formula /. maak i die onderwerp van die formule i 0, rate / koers 8, 78% answer / antwoord (4). 0, 07 . 6. 0, 05 . 42. 0, 05 . 24. A 60 000 1 1 5000 1 setting up equation /.