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.

2 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. 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).

3 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. 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.

4 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. 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).

5 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. 1 ; 5 ; 12 ; 22 ; ; . 4 7 10 - 1st differences 3 3 - 2nd differences T5 = 35 and/en T6 = 51 answers / antwoorde (2).

6 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).

7 Copyright reserved / Kopiereg voorbehou Please turn over / Blaai om asseblief 6 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2017). 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.

8 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 . 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).

9 ( 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). Copyright reserved / Kopiereg voorbehou Please turn over / Blaai om asseblief 8 mathematics P1/ wiskunde V1 (EC/JUNE/JUNIE 2017).

10 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.