Mathematics Grade 12 Paper 1 Eastern Cape Mock 2012 Memo

Downloaded from hlayiso.com Province of the EASTERN CAPE EDUCATION NATIONAL SENIOR CERTIFICATE GRADE/ GRADE 12 SEPTEMBER 2012 MATHEMATICS P1 (WISKUNDE V1) MEMORANDUM MARKS/ 150 PUNTE: This memorandum consists of 11 pages. Hierdie memorandum bestaan uit 11 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P1 (Memo) (SEPTEMBER 2012) QUESTION/VRAAG 1 1.1 1.1.1 a+ simplification vereenvoudiging a2 + 1 = 2a standard form a2 – 2a + 1 = 0 standaard vorm factorisation (a – 1)(a – 1) = 0 faktorisering value of a a=1 waarde van a (4) 1.1.2 1 – 2x + transpose and square both sides bring om en kwadreer beide kante 5x – 1 = 4x2 – 4x + 1 standard form standaard vorm 4x2 – 9x + 2 = 0  factorise (4x – 1)(x – 2) = 0 faktoriseer  both values of x x = or/of x = 2 beide waardes van x  correct value of x Correct solution/Korrekte oplossing: x = 2 korrekte waarde van x (5) 1.1.3 simplification – vereenvoudiging simplification vereenvoudiging  one mark for each value of x x 3 or/of x 6 een punt vir elke waarde van x (4) 1.2 y – 2x + 1 = 0 y = 2x – 1 y = 2x – 1 (1) Substitute/Vervang (1) in xy = 2y + x2 + 3x – 10 substitution x(2x – 1) = 2(2x – 1) + x2 + 3x – 10 vervanging 2x2 – x = 4x – 2 + x2 + 3x – 10 x2 – 8x + 12 = 0 standard form (x – 6)(x – 2) = 0 standaard vorm x = 6 or/of x = 2 and/en both values of x y = 2(6) – 1 or/of y = 2(2) - 1 beide waardes van x = 11 =3 both values of y beide waardes van y (5)

Downloaded from hlayiso.com (SEPTEMBER 2012) WISKUNDE V1 (Memo) 3 1.3 write in factors skryf in faktore simplification vereenvoudiging factorise faktorisering simplification vereenvoudiging answer antwoord (5) 1.4 equate values = vergelyk waardes answer antwoord (2) [25] QUESTION/VRAAG 2 2.1 1 ; 4 ; 11 ; 22 ; 37 ; 56 ; 79 1 mark for each term 1 punt vir elk term (2) 2.2 1 4 11 22 37 56 79 3 7 11 15 19 23 Identification of 2nd 4 4 4 4 4 difference Identifisering van 2de Tn = an2 + bn + c verskil and/en 2a = 4 a=2 value of a Tn = 2n2 + bn + c waarde van a n=1 T1 = 2(1)2 + b(1) + c 1=2+b+c (1) 1st equation -1 = b + c 1ste vergelyking n=2 T2 = 2(2)2 + b(2) + c 4 = 8 + 2b + c 2nd equation -4 = 2b + c (2) 2de vergelyking (2) – (1): -3 = b b = -3 value of c (1) -1 = -3 + c waarde van c c=2 formula Tn = 2n2 – 3n + 2 formule (6) [8]

Downloaded from hlayiso.com 4 MATHEMATICS P1 (Memo) (SEPTEMBER 2012) QUESTION/VRAAG 3 3.1 3.1.1 1st 3 terms 1ste3 terme T1 = 1 T2 = -2 T3 = -5 answer sequence is arithmetic/ry is meetkundig antwoord (2) 3.1.2 d = -3  value of d waarde van d (1) 3.1.3 Sn = substitution in correct formula -125 = vervanging in korrekte formule -125 = simplification -250 = n(-3n + 5) vereenvoudiging standard form -250 = -3n2 + 5n standaard vorm 3n2 – 5n – 250 = 0 factorization faktorisering (3n + 25)(n – 10) = 0 correct value of n or/of n = 10 korrekte waarde n = 10 n ℕ van n. (5) 3.2 1 + (0,2) + (0,2)2 + (0,2)3  expansion a=1,r= uitbreiding  ratio verhouding  substitution in correct formula vervanging in = korrekte formule = 1,25  answer antwoord (4) 3.3 3.3.1  ratio verhouding  substitution vervanging  answer antwoord (3)

Downloaded from hlayiso.com (SEPTEMBER 2012) WISKUNDE V1 (Memo) 5 3.3.2  ratio verhouding  substitution in correct formula vervanging in korrekte formule answer antwoord (3) = [18] QUESTION/VRAAG 4 4.1 4.1.1 Tn = arn-1 substitution in the correct formula Tn = 400 vervanging in korrekte formule T7 = 400(1,1)7-1 simplification = 400(1,1)6 vereenvoudiging = 708,62 answer 7 month/7de maand R708,62 th antwoord (3) 4.1.2 Sn = substitution into correct formula vervanging in korrekete formule simplification vereenvoudiging answer Saves in 7 months/Spaar in 7 maande R3 794,87 antwoord (3) 4.2 4.2.1 a = 1900  the value of a Alternative/Alternatief a = 1911,875 die waarde van a r=  the value of r r = 1,00525 die waarde van r = = R1 058 663,93  substitution in correct formula vervanging in korrekte formule = R1 058 663,93  answer antwoord (4)  answer 4.2.2 7,5% antwoord (1) 4.2.3 1+ =  substitution substitusie = 0,0776  Simplification vereenvoudiging Effective annual interest rate = 7,76%  Answer Effektiewe jaarlikse rentekoers = 7,76% antwoord (3) [14]

Downloaded from hlayiso.com 6 MATHEMATICS P1 (Memo) (SEPTEMBER 2012) QUESTION/VRAAG 5 5.1 y ℝ ;y 5  answer antwoord (1) 5.2 y = -3x + 2 (1)  equating f(x) and g(x) y= (2) vergelyk f(x) en g(x)  simplification (-3x + 2) = vereenvoudiging  standard form -3x2 – x + 2 = -3 + 5(x + 1) Standaard vorm  for both x values -3x2 – 6x = 0 Vir beide x waardes  for both y values -3x(x + 2) = 0 vir beide y waardes x = 0 or/of x = -2 for each correct coordinates 1 mark y = -3(0) + 2 or/of y = -3(-2) + 2 vir elke regte = 2 = 8 koördinaat 1 punt (0 ; 2) and/en (-2 ; 8) (7) 5.3 f(x) =  for description reflection about asymptote: vir beskrywing (1) refleksie in die asimptote: x = -1 or/of reflection about asymptote: refleksie in die asimptote: y=5 [9] QUESTION/VRAAG 6 6.1 (3;-2):  substituting (3;-2) (0;0) :  substitusie (0;0) 9a = 2 a=  value of a die waarde van a (3) 6.2 B(6 ; 0)  coordinate of B koördinaat van B (1) 6.3 g(x) is a one-too-many function; for every y-value there are  for answer/vir antw. two corresponding x-values.  substantiation/ g(x) is een-te-veel funksie; vir elke y-waarde is daar twee regverdiging ooreenstemmende x-waardes (2)

Downloaded from hlayiso.com (SEPTEMBER 2012) WISKUNDE V1 (Memo) 7 6.4 y  both x intercepts Beide x-afsnitte (0 ; 6)  turning point draaipunt . (-2 ; 3) (0 ; 0) x (3) 6.5 Range/Wydte : y 3 or/of y 3  range/wydte (1) 6.6 g: y =  correct equation -2 korrekte vergelyking (2) [14] QUESTION/VRAAG 7 7.1 For inverse/Vir inverse : x = : y = lo  for base vir basis  answer antwoord (2) 7.2 y g:  for curve vir kurwe  y-intercept y-afsnit y= 2-x . 1 g-1: O P .1 x  for curve vir kurwe y= lo  x-intercept x-afsnit (4) 7.3 see graph/sien grafiek  dotted line gebroke lyn (1) 7.4 y=x y=x (1) 7.5 see graph/sien grafiek (1) [9]

Downloaded from hlayiso.com 8 MATHEMATICS P1 (Memo) (SEPTEMBER 2012) QUESTION/VRAAG 8 8.1 Pv = 2 000 ; i = n = 18 Fv = 2860  correct value of i korrekte waarde van Pv(1 + i)n = Fv ⇛ i  substitution into correct formula Vervanging in korrekte formule  answer or/of 24% antwoord (3) 8.2 Fv =  value of i waarde van i  substitution into = correct formula vervanging in ( korrekte formule  answer (correct No, he will be R55,01 short conclusion) Nee, hy sal R55,01 kort antwoord (korrekte afleiding) (3) 8.3 8.3.1 The deposit/Die deposito = 15% of R125 000,00 = R18 750 The loan/Die lening = 125 000 – 18 750  value of loan = R106 250 waarde van or/of The loan/Die lening = 85% of R125 000 (1) = R106 250 8.3.2 Pv =  value of i waarde van i 106 250  substitution into correct formula 50,476x = 106 250 Vervanging in korrekte formule x=  answer = R2 104,96 antwoord (3) [10]

Downloaded from hlayiso.com (SEPTEMBER 2012) WISKUNDE V1 (Memo) 9 QUESTION/VRAAG 9 9.1  answer = 2 antwoord (1) 9.2  one mark for each correct term een punt vir elke korrekte term (2) 9.3 = 2a2x +  1 mark for each derivative 1 punt vir elke = 2a2x + afgeleide (2) 9.4 f(x) = x3  substitution into f´(x) = correct formula vervanging in korrekte formule =  simplification vereenvoudiging =  simplification vereenvoudiging  answer = 3x2 antwoord (4) [9] QUESTION/VRAAG 10 10.1 y = x(x2 – 27) = x3 – 27x  At turning points/By draaipunt:  both values of x Beide waardes 3x2 – 27 = 0 van x x2 = 9  both values of y x = 3 or -3 Beide waardes y = 33 – 27(3) or/of y = (-3)3 - 27(-3) van y = -54 = 54 Turning points/Draaipunte  1 mark for both (3 ; -54) or/of (-3 ; 54) coordinates 1 punt vir koördinate vir beide draaipunte. (4)

Downloaded from hlayiso.com 10 MATHEMATICS P1 (Memo) (SEPTEMBER 2012) 10.2 f(x) = 4 + 12x – 3x2 – 2x3 f´(x) = 12 – 6x – 6x2  1st derivative 1ste afgeleide f (x) = -6 – 12x  2nd derivative At point of inflection/By punt van infleksie: 2de afgeleide f (x) = 0 -6 – 12x = 0 - 12x = 6  value for x x= waarde van x y = 4 + 12(- =4–6-  value for y = -2 waarde van y  coordinate coordinate are/koördinate is koördinate (5) 10.3 3mx2 – 6x – 12  differentiation differensiasie 3mx2 – 6x – 12 = 0 3m(2)2 – 6(2) – 12 = 0  substitution 12m – 12 – 12 = 0 vervanging 12m = 24 m=2  value of m waarde van m y = 2x3 – 3x2 – 12x + n At x = 2 is y = -3  substitution of x -3 = 2(2)3 – 3(2)2 – 12(2) + n and y -3 = 16 – 12 – 24 + n vervanging van x -3 = -20 + n en y n = 17  value of n waarde van n (5) [14] QUESTION/VRAAG 11 11.1 Volume =  formula formule 10 000 =  substitution vervanging h (3) 11.2 Exterior area/Buite oppervlakte A= +2  formula formule =2  substitution of h vervanging van h = (2)

Downloaded from hlayiso.com (SEPTEMBER 2012) WISKUNDE V1 (Memo) 11 11.3 For minimum/Vir minimum:  differentiation of A differensiasie van A  simplification vereenvoudiging  value of r waarde van r and/en  value of h. = 23,35 waarde van h Allow/Laat toe: 23,33 or/of 23,34 (4) [9] QUESTION/VRAAG 12 12.1 2x + y 10  for 1st inequality vir 1ste ongelykheid x + 3y   2nd inequality x 0 2de ongelykheid y  x 0  y 0 (4) feasible region gangbare gebied 10 F : Feasible region line/lyn Gangbare gebied 2x + y line/lyn F x + 3y = 15 5 4 A 0 3 5 10 15 (6) 12.3 Minimum number of capsules and tablets: Minimum aantal kapsule en tablette tablets tablette 3 tablets/tablette Read at A./Gelees by A. capsules 4 capsules/kapsules kapsules (3) [13] TOTAL/TOTAAL: 150