Mathematics Grade 12 Paper 1 NSC November 2023 Question Paper

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE 12/GRAAD 12 MATHEMATICS P1/WISKUNDE V1 NOVEMBER 2023 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 17 pages. Hierdie nasienriglyne bestaan uit 17 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 2 DBE/November 2023 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 memorandum. LET WEL:  Indien 'n kandidaat 'n vraag TWEE keer beantwoord, merk slegs die EERSTE poging.  Volgehoue akkuraatheid is DEURGAANS op ALLE aspekte van die memorandum van toepassing. QUESTION 1/VRAAG 1 1.1.1 x 2  x  12  0 x  3x  4  0 factors/formula x  3 or x   4 answer answer (3) 1.1.2 3x  2 x  6 2 3x 2  2 x  6  0  standard form  substitution into 2  22  43 6 correct formula x 23  answer x  1,79 or x  1,12  answer (4) 1.1.3 2x  1  x  1 2 x  1  x  1 2  squaring both sides 2x  1  x 2  2x  1 x 2  4x  0  standard form x x  4   0 x = 0 or x = 4  both x-values x  0 or x  4  valid answer (4) 1.1.4 x 2  2x  3 x 2  2x  3  0  standard form x  3x  1  0  critical values/factors CV’s: x = – 1 ; x = 3 –1 3 x  1 or x  3 answer (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 3 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 1.2 1 1  1 . . . (1) x y x  2  2y ... (2) x  2y  2  x  2y  2 1 1  1 2y  2 y substitution y  2y  2  2y2  2y 2y2  5y  2  0 standard form (2 y  1)( y  2)  0 y 1 or y2 y-values 2 x  1 or x2 x-values (5) OR/OF OR/OF 1 1  1 . . . (1) x y x  2  2y ... (2) x x y 1 y  1 2 2 1 1  1 substitution x x 1 2 1 2  1 x x2 x  2  2x  x 2  2x x2  x  2  0 standard form ( x  1)( x  2)  0 x  1 or x  2 x-values 1 y or y  2 y-values (5) 2 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 4 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 1.3 2 m1  2 m  3n2  3n 2 m (2  1)  3n (32  1)  factors 2 m (3)  3n (8)  2m (3)  3n (23 ) (same 2 m (3)  3n (2 3 ) bases) m  3 and n  1  m  3 and n  1 m  n  4 mn  4 (4) OR/OF OR/OF 2 m1  2 m  3n2  3n 2 m (2  1)  3n (32  1)  factors 2 m (3)  3n (8) 2 m (3)  3n (2 3 )  2m (3)  3n (23 ) (same 2 m3  3n1 bases) Only true if m – 3 = 0 and n – 1 = 0  m – 3 = 0 and n – 1 = 0 m  n  4 mn  4 (4) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 5 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1.1 7 + 12 + 17 + ….. Tn  a  n  1d d = 5 T91  7  91  15 substitution into correct formula T91  457 answer (3) OR/OF OR/OF d=5 d = 5 Tn  5n  2 substitution n = 91 T91  5(91)  2 T91  457 answer (3) 2.1.2 Sn  n 2a  n  1d  2 substitution into correct S 91  2  7  91  15 91 formula 2 answer (2) S 9  21 112 OR/OF OR/OF S n  a  l  n 2 substitution into correct S 91  7  457  91 formula 2 S 91  21 112 answer (2) 2.1.3 Tn  7  n  15 substitution into correct 5n  2  517 formula 5n  515 equate n  103 answer (3) 2.2.1 T1  3 ; T2  T1  9 and T3  T2  21 3 12 33 66 111 9 21 33 45  constant second diff = 12 12 12 12  first differences : 33 and T5  3  9  21  33  45  111 45 (2) OR/OF OR/OF 2a  12 a6  constant second diff = 12 3(6)  b  9 b  9 69 c  3 T5  6(5)2  9(5)  6 = 111 substitute 5 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 6 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 2.2.2 2a  12  2a  12 a6 3(6)  b  9 or 5  6  b  21  3(6)  b  9 / 5  6  b  21 b  9 69 c  3  69 c  3 (3) c6 Tn  6n 2  9n  6 2.2.3 Tn  12n  9  0  Tn  12n  9 / / 3 3 n n  4 4  Tn is increasing for n  N  increasing for n  N (3) OR/OF OR/OF 9 b 9 n b  n    2a 2(6) 2a 2(6) 3 n 3 n  4 4  min at n = 1 for n  N  Tn is increasing for n  N  increasing for n  N (3) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 7 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1.1 Tn  ar n1 Tn  32 n 1  Tn  32 n 1 (1) 3.1.2 k 3  2 .2  98 301 p 1 p k 3 p  .2  3  6  12  ... p 1 2 expansion n=k  3 2  1 k   98 301 n=k 2 1 substitution into correct 2k  32 768 formula 2 k  215 OR/OF k  log 2 32 768 k  15 k = 15 (4) 3.2 S 22  22 2a  21(3)  substitution into S n 2  S 22  22a  693 S 22  22a  693 a S  substitution into S 1 1 3 3a  2 3a  22a  693   734  S22  S  734 2 44a  1386  3a  1468 41a  82 a2  answer (5) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 8 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1 y=–4 y=–4 (1) 4.2 x – intercept: 0  2 x  4 y = 0 4  2x x2 x = 2 (2)  B2 ; 0 4.3 y  2 0  4  3 y = – 3  A(0 ;  3) y  mx  c 3 gradient m 2 k x   3 x3 equation (3) 2 3 k(1) k 1  1  3  4.4 3 2 2 f (1)  2  4  2 1 f(1) = – 2 Vertical distance     2  units 3 1 answer (3) 2 2 4.5 g x   f x   4 g x   2 x ; x  [2 ; 4)  g x   2 x (1) 4.6 1  Range of g : y   ; 16  4  1  1  1 x   ; 16  (2) Domain of g 1 : x   ; 16  or/of  x  16 4  4  4 4.7 g : y  2x g 1 : x  2 y swop x and y 1  g 1x   log 2 x , x   ; 16   equation 4  (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 9 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 (1 ; 8) x=1y=8 (2) 5.2 1 x = 0 y   (0  1) 2  8 2 1 7 2  answer (2)  15  C0 ;   2 5.3 d 8 substitution (1 ; 8) (1) 1 d  8 5.4 yR; y  0  y0 (1) 5.5  3  x  0 or x  5   3  x  0 x5 OR/OF (3) x [3 ; 0) [5 ; ) 5.6 8 8  2x  k    2x  k  x x  2 x  kx  8  0 2  standard form   (k ) 2  4 2 8 substitution into  k 2  64  0    0 or   0 CV : k  8 ; k  8  8  k  8 or/of k  (8 ; 8) inequality (5) OR/OF OR/OF g / ( x)  h / ( x) 8 8  2  2    2 x x2  8  2 x 2 x  2  x-values y  4  B(2 ; 4) and A(– 2 ; – 4)  y-values For tangents: h( x)  2 x  k or h( x)  2 x  k 4  2(2)  k  4  2(2)  k k 8 k  8 inequality  8  k  8 or/of k  (8 ; 8) (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 10 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 5.7 h( x)  2 x  8 8  2x  8  x  2 x  8x  8 2  2 x 2  8x  8  0 x 2  4x  4  0 ( x  2) 2  0 x  2 x2 15  f (2) f (2)  2 h(2)  4  h(2) 15 4  t 2 7 answer t   2 (4) OR/OF OR/OF 15 x2 f (2)  2  f (2) Tangent point of contact (2 ; 4) 1  4   (2  1) 2  8  t 2  h(2) 15 4  t 2 7 t   answer 2 (4) OR/OF OR/OF g ( x)  8 x 1 g / ( x)  8x 2 2  8x2 1 1  4 x2 x=2 x2 8 y 4 2 R(2 ; 4) 1 y   ( x  1)2  8  t 2  h(2) 1 4   (2  1)2  8  t  f(2) 2 7 t answer (4) 2 [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 11 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1.1 A  P1  i  n 6  r  n=6 19 319,48  18 5001    1200  substitution into correct  r  6 formula 1    1,04429...  1200  r  0,00725... 1200 r = 8,7% answer (3) 6.1.2 i  8,7  12 substitution into correct 1  1   formula 100  1200  r = 9,06% answer (2) 6.2.1 A  P1  in  0  10 0001  0,2n substitution into correct formula n5 answer (2) 6.2.2 F  x 1  i   1 n  i  8,7  60  i x 1    1  1200   n 20 000   substitution into correct 8,7 formula 1200 x  R267,26 answer (4) 6.3 P  x 1  1  i  n  i i   0,112  n  substitution into correct 20 0001  1    formula   12   1 600 000  0,112 12 n 56  0,112   1  1   75  12  n  0,112  19 1     12  75  19   n  log  0,112    correct use of logs 12    1  75    n  147,80   n  147,80 Tino will make 147 withdrawals of R20 000  n  147 (5) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 12 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1 f x  h   f x  f / x   lim h 0 h f / x   lim  4x  h    4 x 2 2   substitution into correct h 0 h formula  4 x  8 xh  4h 2  4 x 2 2  f (x+h) =  4 x 2  8xh  4h 2   f x  lim / h0 h  8 xh  4h 2 f / x   lim simplification h0 h h 8 x  4h  f / x   lim common factor h0 h f x   lim  8 x  4h  / h0 f / x  8x answer (5) OR/OF OR/OF f ( x  h)  4( x  h) 2  4 x 2  8xh  4h 2  f (x+h) =  4 x 2  8xh  4h 2 f ( x  h)  f ( x)  4 x 2  8xh  4h 2  (4 x 2 )  8xh  4h 2 simplification f x  h   f x  f / x   lim h 0 h  8 xh  4h 2 substitution into correct f x   lim / h0 h formula h 8 x  4h  f / x   lim h0 h common factor f x   lim  8 x  4h  / h0 f / x  8x answer (5) 7.2.1 f ( x)  2 x 3  3 x  6x 2 f / ( x)  6 x 2  3 – 3 (2) 7.2.2  Dx 7.3 x 2  2 x 5   23  2 D x 7.x  2 x 5   x3   14  1 derivative with rational exp  x 3  10 x 6   10 x 6 (3) 3 7.3  6x 2  8  0 8 x2  6 2 CV’s: x   2 or x  2 CV’s: x   3 3 3 2 2 Positive for :  x answer (3) 3 3 [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 13 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 f / ( x)  3x 2  12 x  9  f / ( x)  3x 2  12 x  9  3x 2  12 x  9  0  f / ( x)  0 x 2  4x  3  0 ( x  3)( x  1)  0  x  3 or x  1  both x-values f (3)  (3) 3  6(3) 2  9(3)  4  4 f (1)  (1) 3  6(1) 2  9(1)  4  0  both y-values (4)  turning points are: (3 ; 4) and (1 ; 0) 8.2 y (3 ; 4)  y-intercept 4 f  both x-intercepts  both turning points O 1 4 x  shape (4) 8.3 0k 4 or/of k  (0 ; 4) k between y-values of turning points (2) 8.4 f // ( x)  6 x  12  0  f // ( x)  6 x  12 x2  f / / ( x)  0 Max at (2 ; 2)  x-value  y-value f / (2)  3 gradient at x-value  y  2  3x  2 or 2  32  c g ( x)  3 x  4 g ( x)  3 x  4 equation of tangent (6) OR/OF OR/OF 3 1 3 1 Point of inflection: x =  2 2 x2  x-value Max at (2 ; 2)  y-value f / (2)  3 gradient at x-value  y  2  3x  2 or 2  32  c g ( x)  3 x  4 g ( x)  3 x  4 equation of tangent (6) 8.5 tan  3  gradient of g   71,57  answer (2) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 14 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 A D x 432 432 6 x x B C x+8 9.1 432  xb 432 432 b  b  x x  432   (x  8 ) A( x)  ( x  8)  6  x   432    6 3456  x  A( x)  432  6 x   480 x 3 456 (3) A( x)   6 x  480 x 9.2 A( x)  3 456 x 1  6 x  480  3456 x1  6 x  480 3 456 A / ( x)   6 3 456 x2  A / ( x)   6 x2 3 456  60 x2 3 456  6 x 2  x  576  24 cm answer (3) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 15 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 3  432   432  total area  2( x  8)(3)  2    4   A D  ( x) 4 x 4  x   x  432 x 4 4 B C 3 x+8 3 3 A D  432   432  x total area  2(4)   6   ( x)   6 x  x   x  432 432 6 4 x 4 x B C 3 3 x+8 3 3 A D  432   432  4 x 4 total area  4(4)(3)  2( x)(3)     x  2   (4)  x   x  432 x 4 4 B C 3 3 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 16 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1.1 P(A and B) = P(A) × P(B) 1 3 1 3     3 4 3 4 1 1   4 4 (2) 10.1.2 P(A or B) = P(A) + P(B) – P(A and B) 1 3 1 =    substitution 3 4 4 5 =  answer 6 (2) OR/OF OR/OF A B 1 1 1 12 4 2 P(A or B) = 1 1 1 5     substitution 12 4 2 6  answer (2) 10.2.1 0,72 B 0,05  branch 1 with probabilities S Not B 0,28  branch 2 with probabilities No S B 0,35 0,95  branch 3 with probabilities Not B 0,65 (3) o 10.2.2 P(NOT below 0 ) = P(S; NOT below 0o) + P(NS ; NOT below 0o)  value of = 0,05 × 0,28 + 0,95 × 0,65 P(S; NOT below 0o) = 0,6315  value of P(NS ; NOT below 0o)  answer (3) 10.3.1 n(S) = 10!  10! (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 17 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 10.3.2 4 Options; 2 ×8 ×7 × 6 ×5 ×4 ×1 ×3 ×2 ×1 = 80 640  (2  8!) 8 ×2 ×7 × 6 ×5 ×4 ×3 ×1 ×2 ×1 = 80 640 8 ×7 ×2 × 6 ×5 ×4 ×3 ×1 ×1 ×1 = 80 640 8 ×7 × 6 ×2 ×5 ×4 ×3 ×2 ×1 ×1 = 80 640 Total number of possibilities = 322 560 4(2 × 8!) or 322 560 322 560 4 322 560 P(5 learners in between) =   (4) 10! 45 n(S) OR/OF OR/OF 2 ×8 ×7 × 6 ×5 ×4 ×1 ×3 ×2 ×1  (2  8!) 4 possible starting positions  4(2  8!  1)  322 560 4(2 × 8!) or 322 560 8(8!) = 322 560 322 560 4 P(5 learners in between) =   322 560 10! 45 n(S) (4) [15] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou