Mathematics Grade 12 P1 Eastern Cape Mock 2023 Memo (Afr)

NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIORSERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2023 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 17 pages./ Hierdie nasienriglyn bestaan uit 17 bladsye.

2 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2023) NOTE/LET WEL: • If a candidate answers a question TWICE, mark the FIRST attempt ONLY. Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, merk SLEGS die EERSTE poging. • Consistent accuracy applies in ALL aspects of the marking guideline. Volgehoue akkuraatheid geld deurgaans in ALLE aspekte van die nasienriglyn. • If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt. Indien ʼn kandidaat ʼn poging vir ʼn 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 1.1.1 x 2 + x − 30 = 0 ( x − 5)( x + 6) = 0  factors / faktore x =5 or / of x = −6  x = 5  x = −6 OR/OF (3) OR/OF −b  b 2 − 4ac x= 2a −1  (1) 2 − 4(1)(−30) = 2(1)  substitution / vervanging −1  121 = 2 = 5 or / of − 6  x = 5  x = −6 (3) 1.1.2 x(2 x − 6) = −3 2x2 − 6x + 3 = 0  standard form / standaardvorm −b  b 2 − 4ac x= 2a −(−6)  (−6) 2 − 4(2)(3) =  substitution / vervanging 2(2) 6  12 3 3 = OR / OF 4 2 = 2,37 or / of 0, 63  x = 2,37 or/of  x = 0,63 (4) 1.1.3 𝑥 2 − 2𝑥 + 1 > 0 (𝑥 − 1)(𝑥 − 1) > 0  factors / faktore 𝑥 ∈ ℝ, 𝑥 ≠ 1 𝑐/𝑣: 𝑥 = 1 ∴ 𝑥 ∈ ℝ, 𝑥 ≠ 1 (Accuracy/Akkuraatheid) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P1/WISKUNDE V1 3 1.1.4 2x −1 = 4 − 5x ( ) 2 (2 x − 1) 2 = 4 − 5x  squaring both sides kwadreer beide kante (2 x − 1) 2 = 4 − 5 x 4 x2 − 4 x + 1 + 5x − 4 = 0 4x2 + x − 3 = 0  standard form / standaardvorm −1 (1)2 −4(4)(−3)  factors / formula (4 x − 3)( x + 1) = 0 or / of x = 2(4) faktore / formule  answers with selection  x = 34 or / of x  −1 antwoorde met seleksie/keuse (4) 1.2 y − 2 x = −1......................................(1) y 2 + 2 xy = 3x 2 .................................(2) y = 2 x − 1.........................................(3)  y = 2 x −1 Substitute / Vervang (3) into/in (2) : (2 x − 1) 2 + 2 x(2 x − 1) − 3 x 2 = 0  substitution / vervanging 4 x − 4 x + 1 + 4 x − 2 x − 3x = 0 2 2 2 5x2 − 6 x + 1 = 0  standard form / standaardvorm (5 x − 1)( x − 1) = 0  factors / faktore  x = or / of x = 1 1 5  x-values / x-waardes y =− 3 5 or / of y =1  y-values / y-waardes (6) OR/OF OR/OF y − 2 x = −1.......................................(1) y +1 y 2 + 2 xy = 3 x 2 .................................(2) x= 2 y +1 x= ...........................................(3) 2 Substitute / Vervang (3) into / in ( 2 ) 2  y +1   y +1  y + 2y 2  − 3  =0  substitution / vervanging  2   2   y2 + 2 y +1  y2 + y2 + y − 3 =0  4   standard form / standaardvorm 8y + 4 y − 3y − 6 y − 3 = 0 2 2  factors / faktore 5 y2 − 2 y − 3 = 0  y-values / y-waardes (5 y + 3)( y − 1) = 0  x-values / x-waardes y =− 3 5 or / of y =1 x = 1 5 or / of x =1 (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

4 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2023) 1.3 2 x 2 − px + 1 = 0 For real unequal roots: Vir ongelykereële wortels :  b2 − 4ac  0 b 2 − 4ac  0 (− p ) 2 − 4(2)(1)  0  substitution / vervanging p2 − 8  0  standard form / standaardvorm  p  − 8 or / of p− 8   answer / antwoord (5) [25] QUESTION 2/VRAAG 2 2.1.1 a + 9d = 21  a + 9d = 21 a + 16d = 49  a + 16d = 49 −7d = −28 d =4  value of d / waarde van d (3) 2.1.2 a + 9(4) = 21  a = −15 a = −15 T18 = T17 + 4 = 49 + 4 = 53  T18 = 53  T1 + T18 = −15 + 53 = 38  answer / antwoord (3) 2.2.1 T1 = 4(1) − 19 = −15 T2 = 4(2) − 19 = −11  all three terms / al drie terme T3 = 4(3) − 19 = −7 (1) 2.2.2 Sn = [2a + (n − 1)d ] n 2  substitution / vervanging Sm = m2 [2(−15) + 4(m − 1)] and/en = 1 189 1189 = m2 (−30 + 4m − 4)  standard form / standaardvorm 0 = 2m2 − 17m − 1189  method / metode (2m + 41)(m − 29) = 0 or / of −(−17)  (−17) 2 − 4(2)(−1189) m= 2(2)  m = 29 or / of m  − 412  answer / antwoord (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P1/WISKUNDE V1 5 2.3.1 −78 − 76 − 72 − 66  both answers / beide 2 4 6 antwoorde 2 2 – 58 ; – 48 (1) 2.3.2 Tn = an 2 + bn + c 2a = 2 3a + b = 2 a + b + c = −78 a =1 3(1) + b = 2 1 − 1 + c = −78  a =1 b = −1 c = −78  b = −1  c = −78 Tn = n − n − 78 2  answer / antwoord (4) 2.3.3 k  78  answer / antwoord (Accept/Aanvaar k  78) (2) [18] QUESTION 3/VRAAG 3 3.1 a = 81 ......................(1) a + ar 2 = 117 ...........(2)   a + ar 2 = 117 ( ) a 1 + r 2 = 117  substitution / vervanging 81(1 + r ) = 117 2 117 1+ r2 =  simplification / 81 vereenvoudiging 36 r2 = 81 2 r= 3  answer / antwoord (4) 3.2.1 9 x r= 3x 32 x = x 3  answer / antwoord = 3x (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

6 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2023) 3.2.2 a S = 1− r 1 3x  a = 3x & r = 3x = 2 1 − 3x 2.3x = 1 − 3x  substitution / vervanging 3.3x = 1 1 3x = 3 3 = 3−1 x  x = −1  answer / antwoord (3) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P1/WISKUNDE V1 7 QUESTION 4/VRAAG 4 4.1 2 f ( x) = +3  x=5 x −5  y =3 x=5 y =3 (2) 4.2 𝑦 ∈ ℝ but/maar y  3  y3 (1) 4.3 2 f ( x) = +3 x−5 x − intercept / x − afsnit: 2  substitution / vervanging +3= 0 x −5 2 = −3 x −5 − 3 x + 15 = 2 x = 133  x = 133 y − intercept / y − afsnit: 2 y= +3 0−5 = 135  y = 135  Intercepts/Afsnitte :( 133 ;0) and / en ( 0; 135 ) (3) 4.4  asymptotes / asimptote  y-intercept / y-afsnit  x-intercept / x-afsnit  shape and quadrants vorm en kwadrante (4) 4.5 f is reflected in the x-axis and shifted 2 units downwards. f is gereflekteer in die x-as en 2 eenhede afwaarts geskuif.  f(x) reflected / gereflekteer OR/OF  in the x-axis / in die x-as f is shifted 2 units upwards and then reflected in the  shift 2 units / skuif 2 eenhede x-axis. downwards/upwards f is 2 eenhede opwaarts geskuif en daarna afwaarts/opwaarts gereflekteer in die x-as. (3) – [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

8 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2023) QUESTION 5/VRAAG 5 5.1 f ( x) = log b x x = by 9 = b −2  substitution / vervanging 1 b2 = 9 1  answer / antwoord b= 3 (2) 5.2 y = log 1 x 3 x = log 1 y  swopping x and y 3 omruil van x en y x 1 y =   OR/OF y = 3− x  answer / antwoord  3 (2) 5.3   answer (Accuracy) 0  x 1 antwoord (Akkuraatheid) (2) 5.4 y=0   answer (Accuracy) antwoord (Akkuraatheid) (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P1/WISKUNDE V1 9 QUESTION 6/VRAAG 6 6.1 f ( x) = x 2 − 6 x + 11 = x 2 − 6 x + 9 − 9 + 11  completing the square vierkantsvoltooiing = ( x − 3) 2 + 2  ( x − 3)2 + 2  At TP : x = 3 and / en y = 2  values for x and y waardes van x en y OR/OF f ( x) = x 2 − 6 x + 11 b (−6) x=− =−  substitution / vervanging 2a 2(1) =3  value of x / waarde van x  y = 3 − 6(3) + 11 2  substitution / vervanging =2  value of y / waarde van y  At TP : x = 3 and / en y = 2 (4) 6.2 mg = tan 63, 44 o  mg = 2 =2 y − 2 = 2( x − 3)  substitution / vervanging y = 2x − 4  equation of g / vergelyking van g (3) 6.3 f ( x) = g ( x) x 2 − 6 x + 11 = 2 x − 4 CA only if g is linear  equating / gelyk stel VA slegs as g linieêr is x − 8 x + 15 = 0 2  standard form / ( x − 3)( x − 5) = 0 standaardvorm x = 3 or / of x = 5  y = 2(5) − 4  x values / x-waardes =6  S (5 ; 6)  S coordinates / S- koördinate (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

10 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2023) 6.4.1 1 x  5  answer / antwoord (2) 6.4.2 k  −2  answer / antwoord Accept / Aanvaar k  −2 for 1 mark / vir 1 punt (2) [15] QUESTION 7/VRAAG 7 7.1 A = P(1 + i ) n 4n  0, 091   0,091 4 166 433 = 97 000 1 +   4   substitution into correct formula 166 433  4 091  n vervanging in korrekte formule =  97 000  4 000  166 433  4n = log 4091 4000 97 000  correct use of logs = 24 korrekte gebruik van logs  n = 6 years / jaar  answer / antwoord (4) 7.2.1 A = P (1 − i ) n = 482000(1 − 0,147) 5  substitution into correct formula vervanging in korrekte formule = R 217666,80  answer / antwoord (2) 7.2.2 A = P(1 + i ) n = 482 000 (1 + 0, 081) 5  substitution into correct formula = R 711500,99 vervanging in korrekte formule  answer / antwoord (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P1/WISKUNDE V1 11 7.2.3 Required amount / Bedrag benodig: = R 711 501 − R 217 666,80 = R 493 834, 20  amount / bedrag x[(1 + i ) n − 1]  correct formula / F= korrekte formule i  0, 073 60  x  1 +  − 1 0,073  12    0, 073   n = 60 and / en i = 12 493834, 20 = 1 +  0, 073  12    0, 073  60  x  1 +  − 1 12   12    0, 073 0, 073 493 834, 20  12 x = 12 12 ) ( 0,073  0, 073 60   0.073    1+  1 +  − 1 1 +   12    12  = R6 803, 01  answer / antwoord (6) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

12 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2023) QUESTION 8/VRAAG 8 8.1 f ( x) = 1 − x 2 f ( x + h) − f ( x ) f / ( x) = lim h →0 h 1 − ( x + h) 2 − (1 − x 2 ) = lim  substitution / vervanging h →0 h 1 − x − 2 xh − h 2 − 1 + x 2 2  expansion / uitbreiding = lim h →0 h −2 xh − h 2 = lim  simplification / vereenvoudiging h →0 h h(−2 x − h) = lim  factorisation / faktorisering h →0 h = lim( −2 x − h) h →0  answer / antwoord = −2 x (5) OR/OF OR/OF f ( x) = 1 − x 2  substitution / vervanging f ( x + h) − f ( x) = 1 − ( x + h) 2 − (1 − x 2 ) = 1 − x 2 − 2 xh − h 2 − 1 + x 2  expansion / uitbreiding = −2 xh − h 2 f ( x + h) − f ( x )  simplification / vereenvoudiging f / ( x) = lim h →0 h −2 xh − h 2  factorisation / faktorisering = lim h →0 h h(−2 x − h)  answer / antwoord = lim h →0 h (5) = lim(−2 x − h) h →0 = − 2x 8.2.1  1  2  1   Dx ( x 2 + x −2 − 2 ) Dx  x −   = Dx  x 2 + 2 − 2   x    x   ( = Dx x + x − 2 2 −2 ) 2x and / en constant/konstante is 0  − 2x −3 (3) = 2 x − 2 x −3 8.2.2 x5 2 y= − 10 x −1 1 −1  2x 2 = x5 − 2 x 2 10 −3 dy 1 −3  12 x 4  x 2  = x4 + x 2 (3) dx 2 [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P1/WISKUNDE V1 13 QUESTION 9/VRAAG 9 9.1 f ( x) = −2 x 3 + ax 2 + bx + c. 9 = −2(2)3 + a(2) 2 + b(2) − 3 9 = −16 + 4a + 2b − 3 4a + 2b = 28.........................(1)  equation 1 / vergelyking 1 At TP / By DP : −6 x + 2ax + b = 0 2 − 6(2) 2 + 2a(2) + b = 0  f '( x) = −6 x 2 + 2ax + b 4a + b = 24......(2)  equation 2 / vergelyking 2 4a + 2b = 28 4a + b = 24  value of b / waarde van b (1) − (2) b= 4 4a + 4 = 24 4a = 20  value of a / waarde van a a=5 (5) 9.2 f ( x) = −2 x + 5 x + 4 x − 3 3 2 f '( x) = −6 x 2 + 10 x + 4  f '( x) = 0 At/By E : −6 x 2 + 10 x + 4 = 0 3x 2 − 5 x − 2 = 0 (3x + 1)( x − 2) = 0  values of x / waarde van x  x = − 13 or / of x=2 y = −2(− 13 )3 + 5(− 13 ) 2 + 4(− 13 ) − 3 100 =− 27  both coordinates /  E ( − 13 ; − 100 27 ) beide koördinate (3) 9.3.1 1 1 x− or / of x2  x− 3 3  x2 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

14 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2023) 9.3.2 f '' ( x ) = −12 x + 10  f '' ( x ) = −12 x + 10 −12 x + 10 = 0 5 x=  value of x / waarde van x 6 5 x   answer / antwoord 6 (3) OR/OF b OR/OF x=− 3a 5 =− 3( −2)  substitution / vervanging 5 =  value of x / waarde van x 6 5  x  answer / antwoord 6 (3) 9.4 f '( x) = −6 x + 10 x + 4 2  f '( x) = −6 x + 10 x + 4 2 m = f '(−1) = −6(−1) 2 + 10(−1) + 4 = −12 m  y = −12 x + c 0 = −12(−1) + c c = −12  substitution / vervanging y = −12 x − 12  answer / antwoord (4) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P1/WISKUNDE V1 15 QUESTION 10/VRAAG 10 10.1 3x 2 + 2 xy + 6 xy = 147 3x 2 + 8 xy = 147  3x 2 + 2 xy + 6 xy = 147 147 − 3x 2  simplifying / y = . vereenvoudiging 8x (2) 10.2 V = lbh = 3 x.x. y  3x.x. y  147 − 3 x 2   substitution / = 3x 2    8x  vervanging 3 441x 9 x = − 8 8 441 27 x 2 V '( x) = −  V '( x) = 0 8 8 441 27 x 2  − =0 8 8  simplification / 27 x 2 = 441 vereenvoudiging 441 x2 = 27 21  answer / antwoord x= ( = 4, 04 ) (5) 3 3 [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

16 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2023) QUESTION 11/VRAAG 11 WATCH SOCCER/ WATCH RUGBY/ TOTAL/ KYK SOKKER KYK RUGBY TOTAAL Female / Vroulik 72 a 120 Male / Manlik 54 36 90 Total / Totaal b 84 210 11.1.1 a = 48  a = 48 b = 126  b = 126 (2) 11.1.2 72 P( F and/enWS ) = 210  answer / antwoord (2) 11.1.3 (For independent events) / (Vir onafhanklike gebeurtenisse) P( M )  P( R) = P( M and / en R) 90 84 P( M )  P( R) =  210 210 90 84   6 210 210 = 35 0,17  answer / antwoord 36 P( M and / en R) = 210 36 6  = 210 35  The events are independent Die gebeurtenisse is onafhanklik  conclusion / gevolgtrekking (4) 11.2.1 26 25 24 10 9 8  method / metode 26 x 25 x 24 x 10 x 9 x 8  answer / antwoord = 11 232 000 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P1/WISKUNDE V1 17 11.2.2 5 x 25 x 24 x 9 x 8 x 3  5  25  24 = 648 000  98 3 P(Vowel / Factor of 9) / P(Vokaal / Faktor van 9) 648000 = 11232 000  11232 000 as denominator / 3 as noemer = 52  answer / antwoord (4) TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief