Mathematics Grade 12 P1 North West Prelim 2024 Memo (Eng)

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P1 SEPTEMBER 2024 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 18 pages and a cognitive grid of 2 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 2 NW/September 2024 NSC – Marking Guidelines NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent Accuracy applies in ALL aspects of the marking guidelines. QUESTION 1 1.1.1 (2 x − 6)( x + 5) = 0 2 x = 6 or x = − 5  x =−5 x=3  x=3 (2) 1.1.2 7 x − 11x + 3 = 0 2 − b  b 2 − 4ac x= 2a −(−11)  (−11) 2 − 4(7)(3)  substitution into = correct formula 2(7) 11  37 = 14  x = 1,22 x = 1, 22 or x = 0,35  x = 0,35 (3) 1.1.3 x  5x 2 x2 − 5x  0  standard form 0 5 x ( x − 5)  0  critical points  x0 or x5  x0  x5 (4) 1.1.4 3 x + 12 − x = 8 3 x + 12 = x + 8  isolate the root (3 x + 12 ) = ( x + 8) 2 2  square both sides 9( x + 12) = x 2 + 16 x + 64 9 x + 108 = x 2 + 16 x + 64 0 = x 2 + 7 x − 44  standard form 0 = ( x + 11)( x − 4) x = −11 or x = 4  both answers n.a.  select correct answer (5) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 3 NW/September 2024 NSC – Marking Guidelines 1.2 2y = 5 + x 2y − 5 = x  x = 2y − 5 y 2 + 3 xy = 2 x 2 + 50 y 2 + 3(2 y − 5) y = 2(2 y − 5) 2 + 50  substitution y 2 + 6 y 2 − 15 y = 2(4 y 2 − 20 y + 25) + 50  simplification = 8 y 2 − 40 y + 50 + 50 0 = y 2 − 25 y + 100  standard form 0 = ( y − 20)( y − 5) y = 20 or y = 5  both y-values x = 2 ( 20 ) − 5 or x = 2 ( 5 ) − 5 = 35 =5  both x-values (6) OR 5+ x 5+ x y=  y= 2 2 y + 3xy = 2 x 2 + 50 2 5 + x 5 + x 2  + 3x   = 2 x + 50  substitution 2   2   2   25 + 10 x + x 2   15 x + 3 x 2   simplification +  = 2 x + 50 2   4   2  25 + 10 x + x + 30 x + 6 x 2 = 8 x 2 + 200 2 0 = x 2 − 40 x + 175  standard form 0 = ( x − 35)( x − 5) x = 35 or x = 5  both x-values 5 + 35 5+5 y = or y = 2 2 = 20 =5  both y-values (6) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 4 NW/September 2024 NSC – Marking Guidelines 1.3 (2 2p −1 3 ) k 7 6p −3 2 = k  exponential laws  12  7    26 p .2− 3 = 1 (7 )k 2 81 81 =  1 23 23 (729) 2 81 34 = = k 7 = 27 23 (27) 23 . 33  3 = 8  answer (4) [24] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 5 NW/September 2024 NSC – Marking Guidelines QUESTION 2 2.1 T4 = 22  22 (1) 2.2 Tn = a + (n − 1)d = 4 + ( n − 1)(6)  substitution = 4 + 6n − 6 = 6n − 2 T50 = 6(50) − 2 = 298  answer (2) 2.3.1 k −1 Pk =  ( 6n − 2 ) n=0 0 P1 =  ( 6(0) − 2 ) = − 2 0  P1 =  ( 6(0) − 2 ) n=0 n=0 2 −1 P2 =  ( 6n − 2 ) = − 2 + 4 = 2  P2 = S 2 = − 2 + 4 n=0 3 −1 P3 =  ( 6n − 2 ) = − 2 + 4 + 10 = 12  P3 = S3 = − 2 + 4 + 10 n=0 (3) 2.3.2 2a = 6 a=3 a=3 3a + b = T2 − T1 3(3) + b = 4 b = −5 b=–5 a + b + c = T1 3 − 5 + c = −2 c=0 c=0 Pk = 3k 2 − 5k  answer (4) 2.3.3 P50 = 3(50) − 5(50) 2  substitution = 7 250  answer (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 6 NW/September 2024 NSC – Marking Guidelines 2.3.4 P50 + S m = Pq S m = Pq − P50 = 7 920 + m m  a = 298 Sm =  2(298) + (m − 1)(6)  substitution into correct 2 formula m 7 920 + m = 596 + 6m − 6  S m = 7 920 + m 2 15 840 + 2m = 590m + 6m 2 0 = 6m 2 + 588m − 15 840 0 = m 2 + 98m − 2 640  standard form 0 = (m − 22)( m + 120) m = 22 or m = −120  m = 22 n.a.  22 terms must be added to P50 . (5) OR Pq − P50 = 7920 + m Pq − 7 250 = 7 920 + m Pq = 15 170 + m q = 50 + m Pk = 3k 2 − 5k  k = 50 + m 15 170 + m = 3(50 + m) − 5(50 + m) 2  substitution into correct 15 170 + m = 3(2 500 + 100m + m 2 ) − 250 − 5m formula  Pk = 15 170 + m 0 = 3m 2 + 294m − 7 920 0 = m 2 + 98m − 2 640  standard form 0 = (m − 22)(m + 120) m = 22 or m = −120  m = 22 n.a.  22 terms must be added to P50 . (5) [17] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 7 NW/September 2024 NSC – Marking Guidelines QUESTION 3 3.1 17(0, 2) = 3, 4 m above the ground  answer (1) 2 3.2 4 4 radius : 15; 15   ; 15   ; ... 5 5 radius: Tn = ar n − 1 17 − 1 4 T17 = 15    substitution into Tn 5 = 0, 4222124651 V =  r 2h V17 =  (0, 4222124651) 2 (0, 2)  substitution: volume = 0,11 m3  answer (3) OR 2 4 4 radius : 15; 15   ; 15   ; ... 5 5 volume: V =  r 2 h 2   4  2  4  2  (15) (0, 2);  15     (0, 2);  15     (0, 2); ...  substitution: volume 2   5    5    Tn = ar n − 1 17 − 1  16  T17 =  (15) 2 (0, 2)    substitution into Tn  25  = 0,11 m 3  answer (3) 2   4  3.3 volume :  (15) 2 ( 0, 2 ) ;  15    (0, 2); ...   5   a =  (15) 2 ( 0, 2 ) = 45 a 2 4 r r =  5 a ( r n − 1) Sn = r −1  4 2 17    45     − 1   5     substitution into Sn S17 =   2 4   −1 5 = 392,50 m 2  answer (4) [8] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 8 NW/September 2024 NSC – Marking Guidelines QUESTION 4 −2 2 4.1 q=   substitution 5 25 =  answer (2) 4 x 2 4.2 g: y=  5 y −1 2 g ( x) : x =    swop x and y 5 y = log 2 x ; x  0  y = log 2 x (2) 5 5 4.3 x  (0 ; )  0 ;   notation (2) OR x  0; x  R   answer (2) 4.4 x  (0 ; 0,59]  0 ; 0,59  notation (2) OR  0 ; 0,59  notation (2) 0  x  0,59 −x + 2 5 5 4.5 k ( x) =   − 2 2 − ( x − 2) 5 5 =  − 2 2 ( x − 2) ( x − 2) 2 5 2 5 =  −    − 5 2 5 2 5 5 The graph g shift 2 units right and units down.  2 right  down (3) 2 2 [11] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 9 NW/September 2024 NSC – Marking Guidelines QUESTION 5 5.1 h( x ) = x − 1 0 = x −1  y=0 x =1  x=1 A(1 ; 0) (2) 5.2  x + x2 y1 + y2  D 1 ;   2 2  1 − 5 0 − 6  − 4 − 6 1−5 0−6 = D ;  = D ;     2 2   2 2  2 2 = D ( − 2 ; − 3) (2) 5.3 a g ( x) = −3 x+2 B (− 5 ; − 6) : OR A(1 ; 0) : a a  substitution of D into g −6 = −3 0= −3 −5 + 2 1+ 2  substitution of a a B(– 5 ; – 6) or A(1 ; 0) −3 = 3= −3 3 a=9  a=9 (3) 9  g ( x) = −3 x+2 5.4 f ( x) = ( x + p ) 2 + q = ( x − 1) 2 + q  ( x − 1) 2 + q D(− 2 ; − 3) : − 3 = (− 2 −1) 2 + q  substitution of D −3 = 9 + q q = −12  q = – 12 f ( x) = ( x − 1) 2 − 12 f ( x) = x 2 − 2 x + 1 − 12 = x 2 − 2 x − 11 (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 10 NW/September 2024 NSC – Marking Guidelines 5.5 0 = x 2 − 2 x − 11  y=0 −(−2)  (−2) 2 − 4(1)(−11) x= 2(1) x = 4, 46 or x = − 2, 46  x = 4, 46  x = − 2, 46 (3) OR 0 = ( x − 1) 2 − 12  y=0 12 = ( x − 1) 2  12 = x − 1 x = 4, 46 or x = − 2, 46  x = 4, 46  x = − 2, 46 (3) 5.6 ( − ; − 2,46) or (1 ; 4,46)  ( − ; − 2,46 )  (1 ; 4,46 ) (2) OR x  − 2,46 or 1  x  4,46  x  − 2, 46  1  x  4, 46 (2) 5.7 f // ( x) = 2  f // ( x) = 2 f // ( x) Maximum value of f ( x) + 14 at turning point of f ( x) + 14 2 = −12 + 14  – 12 =1  answer (3) 5.8 f ( x) = x 2 − 2 x − 11 f / ( x) = 2 x − 2 1 = 2x − 2  f / ( x) = 1 3 = 2x 3 3 x= = 1,5  x= = 1,5 2 2 f (1,5) = (1,5) 2 − 2(1,5) − 11 = −11,75  f (1,5) = −11,75 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 11 NW/September 2024 NSC – Marking Guidelines h( x ) = x − 1  x = −10,75 −11,75 = x − 1 x = −10,75  answer m1 = 10,75 + 1,5  f ( x) = − 1 = 12, 25 −1 = x 2 − 2 x − 11 0 = x 2 − 2 x − 10 x = 4,32 or x = − 2,32  answer (7) n.a.  4,32  m  12, 25 OR ( x + m) 2 − 2( x + m) − 11 = x − 1  standard form x 2 + 2mx + m 2 − 2 x − 2m −11 − x + 1 = 0 x 2 + (2m − 3) x + (m 2 − 2m − 10) = 0  substitution  = b 2 − 4ac = (2m − 3) 2 − 4(1)( m 2 − 2m − 10)  simplification = 4m 2 − 12m + 9 − 4m 2 + 8m + 40 = − 4m + 49  =0 To touch  = 0 0 = − 4m + 49 4m = 49 m= 49  answer 4  f ( x) = − 1 = 12, 25 −1 = x 2 − 2 x − 11 0 = x 2 − 2 x − 10 x = 4,32 or x = − 2,32  answer (7) n.a. [25] 4,32  m  12, 25 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 12 NW/September 2024 NSC – Marking Guidelines QUESTION 6 x (1 + i ) − 1 n 6.1 F =   i  0,075  20 12  0,075 3 000 1 + − 1  i=  12  12   =  n = 20x12 = 240 0,075  substitution 12 = R 1 661192,18  answer (4) m  i  6.2.1 1 + ieff = 1+ nom   m  12  0,075   substitution = 1+   12  = 1,077632599  simplification  ieff = 0,077632599  r = 7,7633%  answer (3) 6.2.2 Amount short on 31 Dec 2023 = R 3 500 000 − R1 661192,18 = R 1 838 807,83  R3 500 000 - F Amount short on 31 Dec 2021 A = P (1 + i ) n 212  0,075  1 838 807,82 = P 1 +   substitution  12  P = R 1 583 415,54  answer x (1 + i ) − 1 n F=   i x (1 + 0,077633) − 1 17  n = 17 1 583 415,54 =    substitution 0,077633 x = R 47 932, 46  answer (6) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 13 NW/September 2024 NSC – Marking Guidelines QUESTION 7 7.1 f ( x) = 5 x 2 + 2 x f ( x + h) = 5( x + h) 2 + 2( x + h)  substitution of (x + h) = 5( x + 2 xh + h ) + 2 x + 2h 2 2 = 5 x 2 + 10 xh + 5h 2 + 2 x + 2h  simplify f (x + h) f ( x + h) − f ( x) = (5 x + 10 xh + 5h + 2 x + 2h ) 2 2 − (5 x 2 + 2 x) = 10 xh + 5h 2 + 2h  simplification f ( x + h) − f ( x ) f / ( x) = lim h →0 h 10 xh + 5h 2 + 2h = lim h →0 h h(10 x + 5h + 2)  factors = lim h →0 h = lim (10 x + 5h + 2) h →0 = 10 x + 2  answer (5) OR f ( x) = 5 x 2 + 2 x f ( x + h) − f ( x ) f / ( x) = lim h →0 h [5( x + h) 2 + 2( x + h)] − (5 x 2 + 2 x)  substitution of (x + h) f / ( x) = lim h →0 h 5( x + 2 xh + h ) + 2 x + 2h − 5 x 2 − 2 x 2 2 = lim h →0 h 5 x + 10 xh + 5 h + 2 x + 2h − 5 x 2 − 2 x 2 2  simplify f (x + h) f / ( x) = lim h →0 h 10 xh + 5h + 2h 2  simplification = lim h →0 h h(10 x + 5h + 2)  factors = lim h →0 h = lim (10 x + 5h + 2) h →0 = 10 x + 2  answer (5) 7.2.1 f ( x) = 5 x − x + 2 x 4 3 f / ( x) = 20 x3 − 3x 2 + 2  20x3  − 3x 2  2 (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 14 NW/September 2024 NSC – Marking Guidelines 1 7.2.2 8x 2 + 4 f ( x) = 2 x3 1 2 8x 4 = 3 + 2x 2 x3 5 5 − − = 4 x 2 + 2 x−3  4x 2  2x − 3 7 7 − − f / ( x) = −10 x 2 − 6 x − 4  −10x 2  − 6x − 4 (4) 7.3 2 xt = 7 7 x=  x the subject 2t y = 4x2 − 3 2 7 y = 4  − 3  substitution  2t   49  = 4 2  − 3  4t  = 49t − 2 − 3  simplification dy = − 98t −3  answer (4) dt [16] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 15 NW/September 2024 NSC – Marking Guidelines QUESTION 8 8.1 f ( x) = x 3 + ax 2 + bx + c f / ( x) = 3x 2 + 2ax + b  f / ( x) = 3 x 2 + 2ax + b f // ( x) = 6 x + 2a  f // ( x) = 6 x + 2a  5  5 0 = 6  −  + 2a  f //  −  = 0  3  3 2a = 10 a=5 f / ( x) = 3x 2 + 2(5) x + b −16 = 3(− 2) 2 + 10(− 2) + b  f /( − 2 ) = −16 −16 = 12 − 20 + b b = −8 f ( x) = x 3 + 5 x 2 − 8 x + c 16 = (− 2)3 + 5(− 2) 2 − 8(− 2) + c  f ( − 2 ) = 16 16 = 28 + c c = −12 (5) 8.2 f ( x) = x 3 + 5 x 2 − 8 x − 12 f / ( x) = 3x 2 + 10 x − 8 0 = 3 x 2 + 10 x − 8  0 = 3 x 2 + 10 x − 8 0 = (3 x − 2)( x + 4) 2 x= or x = − 4  both x-answers 3 y = −14,81 y = 36  both y-answers 2  Q  ; −14,81 P( − 4 ; 36) (3) 3  8.3 f/  both x-intercepts  x-value of turning point  shape (3) [11] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 16 NW/September 2024 NSC – Marking Guidelines QUESTION 9 9.1 A(x) = L×B = ( x)( x 2 − 8 x + 17)  ( x)( x 2 − 8 x + 17) (1) = x − 8 x + 17 x 3 2 9.2 Maximum area : A / ( x) = 3x 2 − 16 x + 17  A / ( x) 0 = 3x 2 − 16 x + 17  A / ( x) = 0 x = 3,87 or x = 1, 46  x = 1,46 n.a. max area = (1, 46)3 − 8(1, 46) 2 + 17(1, 46)  substitution = 10,88unit 2  answer (5) 9.3 f ( x) = 2 x − 8 / p p m=− = 2x − 8  − = 2x − 8 x x p = − 2 x2 + 8x 1 area = bh 2 1  = ( x)(−2 x 2 + 8 x) substitution 2 = − x3 + 4 x 2 = − (1, 46)3 + 4(1, 46) 2  substitution = 5, 41  5, 41 unit 2  answer (4) OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 17 NW/September 2024 NSC – Marking Guidelines f ( x) = x 2 − 8 x + 17 f / ( x) = 2 x − 8  f / (1, 46) = − 5,08 = 2(1, 46) − 8 = − 5,08  f (1, 46) = 7, 4516 f (1, 46) = (1, 46) 2 − 8(1, 46) + 17 = 7, 4516 y − y1 = m( x − x1 )  y-intercept y − 7, 4516 = − 5,08( x − 1, 46) y = − 5,08 x + 14,8686  h = 14,8686 − 7, 4516 = 7, 42 1 Area = bh 2 1  answer = (1, 46)(7, 417) (4) 2 [10] = 5, 41 unit 2 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 18 NW/September 2024 NSC – Marking Guidelines QUESTION 10 10.1 P(Tom or Harry) = 1 − 0, 42 = 0,58 answer (1) 10.2 P(T and H) = P(T) × P(H)  P(T and H) = P(T)×P(H) P(T or H) = P(T) + P(H) − P(T and H) 0,58 = 0, 4 + P(H) − P(T) × P(H) 0,58 = 0, 4 + P(H) − (0, 4) × P(H)  substitution 0,18 = 0,6 P(H) 0,3 = P(H)  P(H) = 0,3  P(T and H) = P(T) × P(H) = (0, 4)(0,3) = 0,12  answer (4) 10.3 P(only T) = 0,4 – 0,12 = 0,28  answer (1) 10.4 P(H and D) = 0,16 P(H or D) = P(H) + P(D) – P(H and D) 0,3 = 0,3 + P(D) – 0,16  substitution P(D) = 0,16  answer (2) 10.5 No  No (1) Harry is always absent from the mathematics class on the same days that Dic is absent. [9] QUESTION 11 11.1 (10)( 9 ) = 90  (10)(9) = 90 (1) 11.2 2  2!  2!  6! = 5 760  2  2!  2!  2  2!  2!  6! = 5 760 (2) 11.3 5  8!  4  54 = 0, 22 10!  8! 5  8!  4  = 0, 22 10! (3) [6] TOTAL: 150 Copyright reserved

Downloaded from hlayiso.com Mathematics/P1 NW/September 2024 NSC – Marking Guidelines COGNITIVE LEVELS MATHEMATICS P1 COGNITIVE LEVELS TOPICS LEVEL 1 LEVEL 2 LEVEL 3 LEVEL 4 (20%) (35%) (30%) (15%) DIFFERENTIATION TOTAL MARKS ROUTINE COMPLEX QUESTION KNOWLEDGE PATTERNS FUNCTIONS PROBABILITY PROBLEM ALGEBRA FINANCE SOLVING PROCEDURES PROCEDURES 1.1.1 2 2 1.1.2 3 3 1.1.3 4 4 1.1.4 5 5 1.2 6 6 1.3 4 4 24 2.1 1 1 2.2 2 2 2.3.1 3 3 2.3.2 4 4 2.3.3 2 2 2.3.4 5 5 17 3.1.1 1 1 3.1.2 3 3 3.1.3 4 4 8 4.1 2 2 4.2 2 2 4.3 2 2 4.4 2 2 4.5 3 3 11 5.1 2 2 5.2 2 2 5.3 3 3 5.4 3 3 5.5 3 3 5.6 2 2 5.7 3 3 5.8 7 7 25 6.1 4 4 6.2.1 3 3 6.2.2 6 6 13 7.1 5 5 7.2.1 3 3 7.2.2 4 4 7.3 4 4 16 8.1 5 5 8.2 3 3 8.3 3 3 11 Copyright reserved

Downloaded from hlayiso.com Mathematics/P1 NW/September 2024 NSC – Marking Guidelines COGNITIVE LEVELS TOPICS LEVEL 1 LEVEL 2 LEVEL 3 LEVEL 4 (25%) (30 %) (30%) (15%) DIFFERENTIATION TOTAL MARKS ROUTINE COMPLEX QUESTION KNOWLEDGE PATTERNS FUNCTIONS PROBABILITY PROBLEM ALGEBRA FINANCE SOLVING PROCEDURES PROCEDURES 9.1 1 1 9.2 5 5 9.3 4 4 10 10.1 1 1 10.2 4 4 10.3 1 1 10.4 2 2 10.5 1 1 9 11.1 1 1 11.2 2 2 11.3 3 3 6 TOT 30 53 44 23 24 25 36 13 37 15 150 % 20% 35% 29% 15% Pol 20% 35% 30% 15% 25 25 35 15 35 15 150 Copyright reserved