Mathematics Grade 12 Paper 1 NSC November 2022 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE 12/GRAAD 12 MATHEMATICS P1/WISKUNDE V1 NOVEMBER 2022 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 21 pages. Hierdie nasienriglyne bestaan uit 21 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 2 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne NOTE:  If a candidate answers a question TWICE, only mark the FIRST attempt.  Consistent Accuracy applies in all aspects of the marking guidelines. LET WEL:  Indien 'n kandidaat 'n vraag TWEE keer beantwoord, sien slegs die EERSTE poging na.  Volgehoue akkuraatheid is DEURGAANS op ALLE aspekte van die nasienriglyne van toepassing. QUESTION1/VRAAG 1 1.1.1 (3x  6)( x  2)  0 x2 x  2 or x  2  x  2 (2) 1.1.2 2x 2  6x  1  0  b  b 2  4ac x 2a 6  6   4(2)(1)  correct substitution into 2 x correct formula 2(2)  2,82 x  2,82 or x  0,18  0,18 (3) 1.1.3 x  90  x 2 x2  x  90  0  standard form ( x  9)( x  10)  0 CV: x   9 or x  10  critical values –9 10 x   9 or x  10  x   9 or x  10 OR/OF (4) ( ;  9) or (10 ; ) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 3 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 1.1.4 x  7 x  12 x  12  7 x  isolating the root x  122  7 x  2  squaring both sides x 2  24 x  144  49 x x 2  25x  144  0  standard form ( x  16)( x  9)  0 x  16 or x  9  both answers (4) OR/OF OR/OF x  7 x  12  0  standard form  x  3 x  4  0 or let x k  factors x  3 or x 4  answers x  9 or x  16  both answers for x (4) 1.2 2x  y  2 y  2x  2 ………( 1 )  eq 1 xy  4 ………( 2 ) (1) in (2): x(2 x  2)  4  substitution 2x 2  2x  4  0  standard form x2  x  2  0 ( x  2)( x  1)  0 x  2 or x  1  x-values y2 y  4  y-values (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 4 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF OR/OF 2x  y  2 1 x y 1 ………( 1 )  eq 1 2 xy  4 ………( 2 ) (1) in (2): 1  y y  1  4 2   substitution 1 2 y  y40 2  standard form y2  2y  8  0 ( y  4)( y  2)  0 y   4 or y  2  y-values x  1 x2  x-values (5) OR/OF OR/OF 2 x  y  2 ………( 1 ) 4 y ………( 2 )  eq 2 x (2) in (1): 4  substitution 2x  2 x 2 x2  2 x  4  0  standard form x2  x  2  0 ( x  2)( x  1)  0 x  2 or x  1  x-values y2 y  4  y-values (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 5 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF OR/OF 2 x  y  2 ………( 1 ) 4 x ………( 2 )  eq 2 y (2)in (1): 4 2   y  2  substitution  y 8  y2  2 y  0 y2  2 y  8  0  standard form ( y  4)( y  2)  0 y   4 or y  2  y-values x  1 x2  x-values (5) 1.3 2.5 n  5n1  5n2  2.5 n  5n.51  5n.52  exp law  5 n 2  5  25  5 n 22 common factor 2  5n (11)  answer/explanation (3) OR/OF Any integer multiplied by an even number will be even 3 y 1 1.4  96 x 32 3 y 1 3 y 1 x   x   96  2   96 2 25 25 5x x 3 y 1.2 5  2 2 .3 2 5x x  3 y 1.2 5  2 2 .3 2 5x 5  2  x  2  x  2 x y 1  2 2 y 1  2  y  2  y  2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 6 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF OR/OF 3 y 1  96 x 32   2  3 y 1    2 2  3 y 1  2  5   (96) x  5   (96) x  2   2  32 y  2 10  25 x.3x 2 2 y  2 10 3 .2  25 x.3x  32 y 2.210  25 x.3x 10  5 x  x  2  x  2 2 y  2  2  y  2  y  2 (4) [25] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 7 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1.1 a  14 T6  14r 5  448  T6  14r 5  448 r 5  32 Answer only: full marks r  2 r=2 (2) 2.1.2 Tn  14(2)n1 14(26  1)  substitution into correct Sn  formula 2 1 S6  882  S6  882 114 674 – 882 = 113 792 113 792  896  2n  1 128  2n  128  2n n=7 7 (4) OR/OF OR/OF a( r n  1 ) Sn  r 1 14( 2 n  1 )  substitution into correct 114 674  2 1 formula 8 191  2  1 n 2 n  8 192  2 n  8 192 n  log 2 8 192 n = 13  n = 13  7 more terms must be added to the first 6 terms. 7 (4) 1 1 2.1.3 r OR 448r 5  14 r  2 2 1 r  2 a S  1 r 448 S   substitution 1 1 2 S   896  answer (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 8 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne k 1 1 1   3 p  6   20 6 2.2 p 0 1 1 1 3 1 T1  T2     T1  6 3 6 6 6 3 1 1 d   d 6 6 3 121 n   1   1   2   n  1   substitution 6 2  6  3  121 1 1 1  n  n   3 3 3 3 121 1 2  n 3 3 121  n 2 n = 11  value of n k  10  value of k (5) OR/OF OR/OF k 1 1 1   3 p  6   20 6 p 0 1 1 a  a 6 6 1 1 l k 3 6 l n  k 1  n  k 1 S n  a  l  n 2 121 k  1  1 1 1    k  6 2 6 3 6 121 k  1  1 1   k  6 2 3 3 121 k  1  k  1  6 2  3  121 (k  1) 2  6 6 121 (k  1) 2   k  1   121 6 6 k  1  11 k  10  value of k (5) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 9 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1 3a  b  7  3a  b  7 3b  7 3+b=7 (2) b4 OR/OF OR/OF T2  T1  7  T2  T1  7 4  2b  9  (1  b  9)  7  substitution (2) b=4 3.2 Tn  n 2  4n  9 T60  ( 60 ) 2  4( 60 )  9  substitution Answer only: full marks  3849  answer (2) 3.3 14 ; 21 ; 30 ; 41; First difference: 7 ; 9 ; 11 ; … first difference Common 2nd difference: 2 2 Tp  2 p  5 Answer only: full marks 2p  5 (3) OR/OF OR/OF First difference: 7 ; 9 ; 11 ; … first difference Tn  a  n  1d T p  7   p  12 2 Tp  2 p  5 2p  5 (3) 3.4 157  2 p  5  157  2 p  5 p  76  p  76  Between T76 and T77  T76 and T77 (3) OR/OF OR/OF Tn 1  Tn  157  Tn1  Tn  157 (n  1) 2  4(n  1)  9  (n 2  4n  9)  157 n 2  2n  1  4n  4  9  n 2  4n  9  157 2n  152 n  76  n  76  Between T76 and T77  T76 and T77 (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 10 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1.1 p = – 1 and q = 2 p=–1 q=2 (2) 4.1.2 1 20 =0 x 1  2x  2  1 1 x 2 1   ; 0  answer (2) 2  1 4.1.3 x  3  –3 2 5 5  Answer only: full marks  x (2) 2 2 4.1.4 y  xt subst (1 ; 2) 2  1 t t 1 t 1 (2) 1 4.1.5 2 x 1 Answer only: full marks 1 1  20 20 x 1 x 1 1 x 2 1  x 1 x  or x  1 (3) 2 OR/OF  1 x     ;  or 1 ;    2 4.2.1 y = –5  answer (1) 4.2.2  b  (4) x  2 x2 2a 2(1) f (2)  22  4(2)  5   9 y9 (2)  D(2 ;  9 ) OR/OF OR/OF f / ( x)  2 x  4 2x  4  0 x2 x2 f (2)  22  4(2)  5   9 y9 (2)  D(2 ;  9 ) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 11 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 4.2.3 q  5  q  5  9  a ( 2)  5 2  substitution of (2 ; – 9)  4  4a a  1  a  1 (3)  g ( x)  2 x  5 4.2.4 y  ( ;  5) OR y  5 ; y  R answer (1) 4.2.5 k 9 – 9  k  9 (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 12 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 g( x )  2 x  6 y6 y=6 (1) 5.2 y  2x  6 x  2y  6  swop x and y Answer only: Full marks 1  equation y  x3 (2) 2 5.3 1 x  3  2x  6 equating 2 x  6  4 x  12 3 x  18 x = – 6 x  6 y= –6 (3) A ( 6 ;  6) OR/OF OR/OF 2x  6  x equating x6 x = – 6 y6 y = – 6 (3) 5.4 AB  (6) 2  (12) 2 substitution  180  6 5  13,42 answer (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 13 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 5.5 BC  62  62  72  6 2  BC AB = AC = 180 symmetry of g and g 1  AB = AC /midpoint (3 ; 3) 2  72   h  ( 180 )   2     h (A)  2   162  9 2 1 area of ABC  BC  h  substitution 2 1   72  162  54 units 2  answer (A) (5) 2 OR/OF OR/OF tan BD̂C  2  BD̂C  63,43  BD̂C  63,43 1 tan DĈA  2  DĈA  26,57  DÂC  36,86  DÂC  36,86 (ext angle triangle) AC = 180 Area of ABC  1 2  180   180 sin 36,86  substitution into the correct formula  53,99 units 2  answer (A) (5) OR/OF OR/OF Area of ∆ABC = Area of ∆BDC + Area of ∆ADC  Areas (∆BDC+∆ADC) 1 1 1  DC.BO  DC.height  DC.BO 2 2 2 1 1   9  6    9  6  1  DC.height 2 2 2  54 units 2  substitution  answer (A) (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 14 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1 A  P(1  i) n 8  m  8 13 459  12 0001    400   subst into correct  m  8 formula 1    1,121...  400  m 8 1  1,121.. 1  m 8  1,121.. 400 400 m  0,0144... 400 m  5,78%  5,78 % (4) 6.2 F  x (1  i)  1 n  i 0,075  0,075 12   1 0001    1 12  12    12 F 0,075 12  R12 421,22  answer He won't be able to buy the computer because  conclusion R13 000 – R12 421,22 = R578,78 (4) OR/OF He won't be able to buy the computer because R12 421,22 < R13 000 6.3.1 Loan amount = 85% × R250 000 = R212 500  answer (1) OR/OF OR/OF Loan amount = R250 000 – (15% × R250 000) = R212 500  answer (1) 6.3.2  0,13  5  0,13  5 A  212 5001    A  212 5001    12   12  A  224 262,53 answer P    x 1  1  i)  n i   0,13  67   x 1  1  )    12   substitution into 224 262,53  0,13 correct formula 12  – 67  x  R4 724,96  answer (5) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 15 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1 f x   x 2  x f x  h   f x  f / x   lim h 0 h ( x  h)  ( x  h )  ( x 2  x ) 2 f x   lim / substitution into h 0 h the formula x  2 xh  h  x  h  x 2  x 2 2  x 2  2 xh  h2  x  h f / x   lim h 0 h 2 xh  h  h 2 = lim h 0  2 xh  h 2  h h h(2 x  h  1)  lim common factor h 0 h  f x   2 x  1 / answer (5) OR/OF OR/OF f x   x 2  x f x  h  ( x  h) 2  ( x  h)  x 2  2 xh  h 2  x  h  x 2  2 xh  h2  x  h f x  h  f x   x 2  2 xh  h 2  x  h  x 2  x  2 xh  h 2  h  2 xh  h 2  h f x  h   f x  f / x   lim h 0 h 2 xh  h 2  h substitution into = lim the formula h 0 h h(2 x  h  1)  lim common factor h 0 h  f x   2 x  1 / answer (5) 7.2 f ( x)  2 x  3x  8 x 5 4  10x 4 f / x   10 x 4  12 x 3  8   12x 3 8 (3) 7.3 g ( x)  ax 3  3x 2  bx  c  g ( x)  3ax  6 x  b / 2 g / ( x)  3ax 2  6 x  b  g // ( x )  6ax  6 g // (1)  6a(1)  6  0 g // (1)  6a(1)  6  0 a 1 a  1 For concave up g // ( x)  0 6x  6  0 x>–1 (4) x>–1 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 16 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF Min gradient at (– 1 ; – 7) implies: at x = – 1 - point of inflection and g will be positive cubic hence a > 0 Since g is concave up x  1 OR/OF . (– 1 ; – 7 ) . Since g is concave up (– 1 ; y) x  1 Since g is concave up x  1 1 Answer only: 4 OR/OF Min gradient at (– 1 ; – 7) implies: at x = – 1 - point of inflection and g will be positive cubic hence a > 0 Since g is concave up x  1 OR/OF . (– 1 ; – 7 ) . Since g is concave up (– 1 ; y) x  1 Since g is concave up x  1 1 Answer only: 4 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 17 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF OR/OF Min gradient at (– 1 ; – 7) implies: at x = – 1 - point of inflection  point of inflection and g will be positive cubic hence a > 0  a > 0 Since g is concave up x  1  x  1 (4) OR/OF OR/OF . (– 1 ; – 7 ) pos graph . (– 1 ; y) point of inflection Since g is concave up x  1 Since g is concave up x  1  x  1 (4) 1 Answer only: 4 [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 18 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 f / x   mx 2  nx  k  1  1  f / x   m x  x  1  substitution of   ; 0   3  3  and (1 ; 0)  1 1  m 0  0  1  substitution of (0 ; 1)  3 1 1  m 3 m  3  m  3  1 f / x   3 x  x  1  3  1 f / x   3 x 2  x   2  1  f / x   3 x 2  x   2  3 3  3 3 f x   3x  2 x  1 / 2 n  2 n  2 k  1 k 1 (6) OR/OF OR/OF k 1 k 1 1 1 0  m  n  1 and m  n 1  0 9 3 m  n  1 (1)  m  n  1 m  3n  9 (2)  m  3n  9 (1) – (2) 4n  8  4n  8 n  2 n  2 m  2  1  m  3  m  3 (6) 8.2.1 f x    x 3  x 2  x  2  1  49 f     1 ,81  x-coordinates of the TP  3  27  1 49  T.P  ;   1 49   3 27   T.P  ;   3 27  f 1  3 T.P1 ; 3  T.P1 ; 3 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 19 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 8.2.2 f x    x 3  x 2  x  2  x3  x 2  x  2  0    x  2  x 2  x  1  0 x  2 or no solution x2 one x-intercept (1 ; 3) 2  1   two turning points   ;1,81  3  2 y-intercept  shape: neg cubic (5) 1 8.3.1  1 a 3 2 1   answer (1) 3 OR/OF OR/OF f / x   3x 2  2 x  1 f / / x    6x  2 f / /  a    6a  2  0  6a  2 1 a  answer (1) 3 4 4 8.3.2 b  units  3 3 4  b (2) 3 [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 20 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION9/VRAAG 9 9.1 Any point on f : x ; x 2     x ; x2  distance  ( x  10) 2  ( x 2  2) 2  substitution  x 2  20 x  100  x 4  4 x 2  4  simplification  x 4  3x 2  20 x  104  answer For min distance d 4 dx x  3x 2  20 x  104  0 4 x3  6 x  20  0  4 x3  6 x  20 ( x  2)(4 x 2  8x  10)  0  derivative = 0   82  4(4)(10)  96  no roots x = 2 x = 2 d  24  3(2) 2  20(2)  104  2 17  8,25 answer (A) (8) ( x ; x2 ) A B(10 ; 2) 9.2 y2  y1 x 2  2 mAB    mAB x2  x1 x  10 x  10 x  10  mtan g    mtan g   x2  2 x2  2 f / ( x)  2 x  f / ( x)  2 x x  10  2x    equating x2  2 2 x3  4 x  x  10 2 x3  3x  10  0  standard form x2 x = 2 y  (2) 2  4  AB  (2  10) 2  (4  2) 2 substitute into distance  2 17  8, 25 answer (A) (8) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 21 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 A C 0,15 0,05 0,05 x 0,1 0,2 0,183 y 33 B 10.1.1(a) y = 1 – 0,893 = 0,107 (0,11)  y = 1 – 0,893 (1) 10.1.1(b) x = 0,893 – 0,733  x = 0,893 – 0,733 = 0,16 (1) 10.1.2 P(at least 2 events) = 0,1 + 0,15 + 0,16 + 0,2  values = 0,61  answer Answer only: Full Marks (2) 10.1.3 P(B) = 0,643  P(B) = 0,643 P(C) = 0,56  P(C) = 0,56 P(B and C) = 0,36  P(B and C) = 0,36 P(B) × P(C) = 0,643 × 0,56 = 0,36  P(B) × P(C) = 0,36  P(B and C) = P(B) × P(C)  independent because  B and C are independent P(B and C) = P(B) × P(C) (5) 10.2.1 7  6  5  4  840  4 7  7  6  5  4  840 (3) 10.2.2 start with 5, 7, 9 or start with 6 or start with 8  3  5  1 3  45 3  5 1 3  1 5 1 2  1 5 1 2  1 5  1 2  10  45  10  10  1 5  1 2  10  65 65 65 13  answer P   0,08 840 168 (5) OR/OF OR/OF ends in 4 or ends in 6 or ends in 8  5  5  1 1  25 5  5 11  4  5 11  4  5 11  4  5  1 1  20  25  20  20  4  5  1 1  20  65 65 65 13  answer P   0,08 840 168 (5) [17] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou