Mathematics Grade 12 P1 Eastern Cape Mock 2021 Memo (Afr)

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2021 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 22 pages./ Hierdie nasienriglyn bestaan uit 22 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2021) 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  2 x  15  0 ( x  3)( x  5)  0  factors / faktore x  3 or / of x  5  x  3  x  5 (3) OR/OF OR/OF b  b 2  4ac x 2a 2  22  4(1)(15)  substitution / vervanging  2(1) 2  64  2  x  3  x  5  3 or / of  5 (3) 1.1.2 3x  x  1  0 2 b  b 2  4ac x 2a 1  12  4(3)(1)   substitution / vervanging 2(3) 1  13  Penalise 1 mark for incorrect 6 rounding  0, 43 or / of  0, 77 Penaliseer 1 punt vir verkeerde  x  0, 43  x  0,77 afronding (3) 1.1.3 x( x  3)  2 y x 2  3x  2  0  standard form ( x  1)( x  2)  0 standaardvorm 1 2 x  x  1 or / of x2  factorisation faktorisering  x  1 or/of  x  2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICS P1/WISKUNDE V1 3 1.1.4 43  x  x  1  0  isolating the surd 43  x  x  1 isolering van die wortelvorm  43  x    x  1 2 2  squaring both sides kwadreer beide kante 43  x  x 2  2 x  1  standard form / standaardvorm x 2  x  42  0 ( x  7)( x  6)  0  factorisation / faktorisering x  7 or / of x  6  selection / korrekte keuse (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2021) 1.2 2y  x  3 (1) y 2  3 x  2 xy (2) x  2y 3 (3)  x  2y  3 Substitute  3 into  2  / Vervang (3) in (2) y 2  3(2 y  3)  2 y (2 y  3)  substitution / vervanging y  6y 9  4y  6y  0 2 2 3 y 2  12 y  9  0  standard form / standaardvorm y2  4 y  3  0 ( y  3)( y  1)  0  factorisation / faktorisering  y  3 or / of y  1  y-values / y-waardes x  2(3)  3 or / of x  2(1)  3 3  1 OR/OF OR/OF 2y  x  3 (1) y 2  3 x  2 xy (2) x 3 y  (3) 2 2 Substitute  3 into  2  / Vervang (3) in (2) 2  x 3  x 3     3x  2 x    2 2 2 2 2 x 6x 9    3x  x 2  3x 4 4 4 2 3x 6 x 9    0 4 4 4 3 x 2  6 x  9  0 x2  2x  3  0 ( x  3)( x  1)  0  x  3 or / of x  1 3 3  1 3 y   or / of y     2 2  2 2 x 3 3 1  y  2 2  substitution / vervanging Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICS P1/WISKUNDE V1 5  standard form / standaardvorm  factorisation / faktorisering  x-values / x-waardes  y-values / y-waardes (5) 1.3 5  p (6  p)  9 x 2 For non-real roots: / Vir nie  reële wortels :  0 0 y p (6  p)  9  0  standard form / standaardvorm  p2  6 p  9  0 p2  6 p  9  0 3 x  factorisation / faktorisering ( p  3)  0 2 p but/maar p  3  answer / antwoord OR/OF OR/OF 5  p (6  p)  9 x 2 For non-real roots:/ Vir nie - reële wortels: 0 y  0 3 p (6  p)  9  0 x  standard form / standaardvorm  p2  6 p  9  0 (3  p)( p  3)  0  factorisation / faktorisering p but / maar p  3  answer / antwoord (4) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2021) QUESTION 2/VRAAG 2 2.1.1 16 ; 16 ; 12 ; 4 0 4 8 4 4 8 8 (1) 2.1.2 2a  4 a  2  a2 3a  b  4 3(2)  b  0 b   6  b  6 a  b  c  16 2  6  c  16  c  12  c  12 Tn  2n 2  6n  12  Tn  2n2  6n  12 (4) 2.1.3 T38  2(38)  6(38)  12 2  substitution / vervanging  2648  answer / antwoord (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICS P1/WISKUNDE V1 7 2.1.4 General term for first differences: Algemene term vir eerste verskille  Tn  4n  4 Tn  4n  4  Tn  400 400  4n  4 n  101 Tn  linear/ linieêr   (Tn 1  Tn )(quadratic/ kwadraties ) n  101 and / en n  1  102  answer / antwoord The terms are / Die terme is :101 and / en 102 OR/OF OR/OF 2(n  1)  6(n  1)  12  (2n  6n  12)  400 2 2 2n 2  4n  2  6n  6  12  2n 2  6n  12  400 4n  4  400  4n  4  400  4n  404  n  101  Between/Tussen T101 and / enT102  answer / antwoord OR/OF OR/OF Trial and error / Proefneming T102  2(102) 2  6(102)  12  20184 subst. for T101 and T102 T101  2(101) 2  6(101)  12  19 784 verv. vir T101 en T102 Difference/Verskil : 400  400 Between/TussenT101 and / enT102  answer / antwoord (3) 2.2.1 Tn  a  (n  1)d 89  2  (n  1)(3)  substitution / vervanging 3n  1  89 3n  90 n  30  answer / antwoord (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2021) 2.2.2 k is the sumto 30 terms / is die som tot 30 terme.  Sum formula / Som formule n Sn   a  l  2  substitution / vervanging 30   2  89  answer / antwoord 2  1365 OR / OF OR / OF n  Sum formula / Som formule Sn   2a  (n  1)d  2 30  substitution / vervanging   2(2)  (30  1)(3) 2  1365  answer / antwoord (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICS P1/WISKUNDE V1 9 QUESTION 3/VRAAG 3 3.1 T9  ar 8  768 T13  ar12  12 288 ar12 12 288 ar12 12 288    ar 8 768 ar 8 768  r  16 4 r  2  r  2 768 a  2  8 3  value of a / waarde van a (3) 3.2.1 54 24 S2   19 19 30  19  answer / antwoord (1) 3.2.2 30 T1  T2  19 30 a  ar  30 19  a  ar  19 30 a (1  r )  19 30 a 19(1  r ) (1) 3.2.3 a 54 S   1  r 19 54(1  r ) 54(1  r ) a   a 19 19 30 a ........ from / vanaf (3.2.2) 19(1  r ) 30 54(1  r )  equating / gelykstel   19(1  r ) 19 30 (1  r )(1  r )  54 1 r  9 2 5 r 2  94 4  r2  2 9 r  3  answer / antwoord (4) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2021) QUESTION 4/VRAAG 4 4.1 D(0 ; 1)  (0 ; 1) (1) 4.2 x  2 ; y  1  x  2  y  1 (2) 4.3 but/maar   x  2 (2) 4.4 g ( x)  b x 8  b 3  substitution / vervanging 1 8 b3 1 b3  8 1 b   answer / antwoord 2 (2) 4.5 3 y 1 x2 3 0 1  substitution y = 0 / x2 vervanging y = 0 3 1 x2 x  2  3 x  5  x  5 A(5;0) 3 y 1 02 5   y 5 2 2 C  0;  52  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICS P1/WISKUNDE V1 11 4.6 1 y 1 y x   x  2 2  y  log 1 x  y  log 1 x 2 2 (2) OR/OF OR/OF y  2 x  x  2 y  x  2 y y   log 2 x  y   log 2 x (2) 4.7 4.7.1 5  x  2 OR/OF x  (5 ;  2)  answer / antwoord (A) (2) 1 4.7.2 0 x OR/OF x  (0 ; 18 ]  answer / antwoord (A) (2) 8 [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2021) QUESTION 5/VRAAG 5 5.1  x2  2 x  8  0  f ( x)  0 x2  2x  8  0 ( x  4)( x  2)  0  factorisation / faktorisering  x  4 or / of x  2  values of x / waardes van x  R(4;0) and / en S(2 ; 0)  RS  6 units / eenhede  answer / antwoord (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICS P1/WISKUNDE V1 13 5.2 4  2 x 2  method / metode  1  x  1 y  (1)  2(1)  8 2 9  y 9  T  1;9  OR/OF OR/OF f ( x)   x  2 x  8 2 b b   x 2a 2a  2      2( 1)   x  1  1 y  (1) 2  2(1)  8 9  y 9  T(1;9) OR/OF OR/OF f '( x)  2 x  2  0  2 x  2  0  2x  2 x  1 y  (1)  2(1)  8 2 9  x  1  y  9  T(1;9) (3) 5.3.1 f ( x)   x  2 x  8 2 f '( x)  2 x  2  f / ( x) 2 x  2  2  f / ( x)  2  x  2  x  2  y   (2) 2  2(2)  8 8  y 8 W (2;8) (4) 5.3.2 g ( x)  mx  c 1 m ( lines / lyne)  gradient / gradiënt 2 c8 1  equation / vergelyking  y   x8 2 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 14 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2021) 5.4 f ( x)   x 2  2 x  8 h( x)   f ( x  1)   f ( x  1)    ( x  1) 2  2( x  1)  8  substitution / vervanging    ( x 2  2 x  1)  2 x  2  8  simplifying /     x  2 x  1  2 x  2  8 2 vereenvoudiging  x2  9  equation / vergelyking OR/OF OR/OF h( x)  ( x  3)( x  3)  roots/wortels 3 and/en –3  x 9 2  +( x + 3)(x – 3)  equation / vergelyking OR/OF OR/OF New turning point / Nuwe draaipunt = (0 ; –9)  (0 ; –9) y  x2  9  equation / vergelyking (4) [17] QUESTION 6/VRAAG 6 6.1 A  P(1  i ) n 5510  9 670(1  i) 4  subst. into correct formula 5510 vervanging in korrekte formule i  1 4 9 670  simplification  0,131177 vereenvoudiging  r  13,12%  answer / antwoord (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICS P1/WISKUNDE V1 15 6.2 End of December / Einde van Desember x (1  i ) n  1 F i  0, 087 144  600 1    1  12    n  144 F  0, 087  subst. into correct formula 12 vervanging in korrekte formule  R151438, 20 End of January / Einde van Januarie A  P(1  i ) n  0, 087   151438, 20 1    adding final month’s interest  12  voeg finale maand se rente by  R152536,13  answer / antwoord OR/OF OR/OF x 1  i   1 1  i  n F   i  0, 087 144   0, 087   n  144 600 1    1 1    subst. into correct formula  12    12  F  vervanging in korrekte formule 0, 087  adding final month’s interest 12 voeg finale maand se rente by  R152536,13  answer / antwoord (4) x 1  1  i   6.3.1 n P   i   0, 093  72  x 1  1    0, 093   12    i and n  72 350 000  12 0, 093  substitution into correct formula 12 vervanging in korrekte formule 0, 093 350 000  x  12   0, 093  72  1  1      12    answer / antwoord  x  R6361,18 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 16 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2021) 6.3.2 Outstanding balance = / Uitstaande balans = x 1  1  i   n P   i   0, 093 32  6361,18 1  1     i  0,093 12 and/en n  32   12    subst. into correct formula P  0, 093 vervanging in korrekte formule 12  R179 667,32  P = 179 667, 32 OR/OF OR/OF Outstanding balance / Uitstaande balans  A F  0, 093 40  6361,18 1    1  i  0,093 and/en n  40  12  40  0, 093   12  350 000 1     subst. Into correct formula  100  0, 093 vervanging in korrekte formule 12  R 476 628,84  R 296961, 79  R179 667, 05  P = 179 667, 32 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICS P1/WISKUNDE V1 17 6.3.3   0, 093   n  7 000 1  1      12   179 667,32   subst. into correct formula 0, 093 vervanging in korrekte formule 12 0, 093 179 667,32  n 12  1   1  0, 093    7 000  12  n  4 031   0,80108...      4 000  n 4 031  0,80108...  4 000 log 0,80108...  correct use of logs  n  4 031 korrekte gebruik van logs log 4 000  n  28, 73  = 28,73  n  28, 73 The number of months is 29.  n = 29 months / maande Die aantal maande is 29. OR/OF OR/OF   0, 093   n  7 000 1  1      12   179 667,32  0, 093  subst. into correct formula 12 vervanging in korrekte formule 0, 093 179 667,32  n 12  1   1  0, 093    7 000  12  n  4 031   0,80108...      4 000  n 4 031  0,80108...  4 000 log 0,80108...  correct use of logs  n  korrekte gebruik van logs 4 031 log 4 000  n  28, 73  n  28, 73  = 28,73 The number of months is 29.  n = 29 months / maande Die aantal maande is 29 (4) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 18 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2021) QUESTION 7/VRAAG 7 7.1 f ( x  h)  f ( x ) f '( x)  lim h 0 h  lim  5  2  x  h   5  2x2 2   substitution / vervanging h 0 h 5  2 x  4 xh  2h 2  5  2 x 2 2  lim  expansion / uitbreiding h 0 h 4 xh  2h 2  lim  simplification / vereenvoudiging h 0 h h(4 x  2h)  lim h 0 h  lim  4 x  2h   notation and lim (4 x  2h) h 0 h 0  4 x  answer / antwoord (5) 7.2.1 2x 2 y  7 x4  x 3 3  7 x4  2x 2  2x 2 dy 1   28 x 3  3x 2 1  28x3  3x 2 dx (3) 7.2.2  3x 2  7 x  6   Dx    x   Dx 3x  7  6 x  1  3x  7  6x 1  3  6 x 2  3 and differentiating constant 3 en afgeleide van konstante  6x 2 (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICS P1/WISKUNDE V1 19 QUESTION 8/VRAAG 8 8.1.1 f ( x)  2( x  x1 )( x  x2 )( x  x3 )  2  x  1  x  12   x  3  f ( x)  2  x  1  x  12   x  3   x  1 2 x  1 x  3 OR/OF    x  1 2 x 2  7 x  3   f ( x)   x  1 2 x  1 x  3  2 x  7 x  3x  2 x  7 x  3 3 2 2  exapnsion / uitbreiding  2 x3  5 x 2  4 x  3  simplifying / vereenvoudiging f ( x)  2 x 3  bx 2  cx  d  b  5, c   4, d  3 (4) 8.1.2 f '( x)  6 x 2  10 x  4 0  6 x 2  10 x  4  f '  x   6 x 2  10 x  4  0  3x 2  5 x  2  0  3x  1 x  2   0  factorisation / faktorisering  x   or / of x  2 1 3 N is at f (2) f (2)  2  2   5  2   4  2   3 3 2  chosing/kies : x  2  9 N  2 ;  9   y  9 (4) 8.1.3 (a)  x2 1 3  answer / antwoord (2) 8.1.3 (b) f ''( x)  12 x  10  f ''( x)  12 x 10 12 x  10  0  f ''( x)  0 12 x  10 x 5  answer / antwoord 6 OR/OF OR/OF 1 2 5 x 5 x 3  2 6 6 5 5  5 x  x  OR/OF x    ;  6 6  6 (3) 8.2 y  f (0) = 0  (m ; 0) (m;0)  shape / vorm x (3) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 20 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2021) QUESTION 9/VRAAG 9 9.1 A   12 15 x  8 x  2   15 xy    8 xy   17 xy   total surface area / totale buite-oppervlakte 5760  120 x 2  40 xy 5760  120 x 2  5760  120 x2  40 xy y  (2) 40 x 9.2 V  ( 12 b.h)  H V  12 15 x  8 x  y  substitution into V 1 5760  120 x 2 vervanging in V  15 x  8 x  2 40 x  substituing for y   60 x 144  3 x 2  vervanging van y  8640 x  180 x3 (2) 9.3 V '( x)  8640  540 x 2  V '( x)  8640  540 x2 V '( x)  0  V '( x)  0  8640  540 x 2  0 8640  540 x 2  simplification / x 2  16 vereenvoudiging x  4  answer / antwoord (4) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICS P1/WISKUNDE V1 21 QUESTION 10/VRAAG 10 10.1.1 P(B)  1  P(not/nie B)  1  0, 45  0,55  0,55 (1) 10.1.2 P(A and/en B)  P(A)×P(B)  P(A) P(B)  0, 2  0,55  0,11 P(A or/of B)  P(A)  P(B)  P(A and/en B)  substitution / vervanging  0, 2  0,55  0,11 16  0, 64 or / of  answer / antwoord 25 (3) 10.2 P(late/laat )  1 x  3 (1  x)  12 x  53 (1  x) 2 5 1 x  3 (1  x)  8  equating / gelyk stel 2 5 15 15 x  18(1  x)  16  substitution / vervanging 15 x  18  18 x  16  3 x  2 2  answer / antwoord x 3 (4) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 22 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2021) QUESTION 11/VRAAG 11 11.1 @@@ ###  20  22  21 20  22  21  10  10  10  10  10  10  9 240 000  answer / antwoord (3) 11.2 20  19  3  10  10  5  20  3  19  10  10  5  20  19  3  10  10  5 9 240000 1140 000  20  3  19  10  10  5   adding / optel 9 240000 19  9 240 000  or / of 0,12 or / of 12,34%  answer / antwoord 154 (5) [8] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief