Mathematics Grade 12 Paper 1 Free State Prelim 2024 Memo

Downloaded from hlayiso.com PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN GRADE/GRAAD 12 MATHEMATICS P1 WISKUNDE V1 SEPTEMBER 2024 MARKS/PUNTE: 150 MARKING GUIDELINES NASIENRIGLYNE These marking guidelines consist of 15pages. Hierdie nasienriglyne bestaan uit 15 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 2 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne NOTE: • If a candidate answered a question TWICE, mark only the FIRST attempt. • If a candidate has crossed out an attempt to answer a question and did not redo it, mark the crossed-out version. • Consistent accuracy applies in ALL aspects of the marking guideline. • Assuming answers/values to solve a problem is UNACCEPTABLE. NOTA: • As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na. • As 'n kandidaat 'n antwoord van 'n vraag doodtrek en nie oordoen nie, sien die doodgetrekte poging na. • Volgehoue akkuraatheid word in ALLE aspekte van die nasienriglyne toegepas. • Om antwoorde/waardes aan te neem om 'n probleem op te los, is ONAANVAARBAAR. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 3 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 1.1.1 𝑥 = 7 𝑜𝑟 𝑥 = −10 ✓ 𝑥=7 ✓ 𝑥 = −10 (2) 1.1.2 3𝑥(2𝑥 + 1) = 1 6𝑥 2 + 3𝑥 − 1 = 0 −𝑏 ± √𝑏 2 − 4𝑎𝑐 ✓ standard form 𝑥= 2𝑎 −3 ± √(3)2 − 4(6)(−1) 𝑥= ✓ substitution into 2(6) correct formula −3±√33 𝑥= 12 𝑥 = 0,23 or 𝑥 = −0,73 ✓ 𝑥 = 0,23 ✓ 𝑥 = −0,73 OR/OF (4) OR 2 6𝑥 + 3𝑥 − 1 = 0 1 1 1 1  standard form 𝑥2 + 𝑥 + = + 2 4 6 4 1 2 5 (𝑥 + ) = 1 5 2 12  (𝑥 + 2)2 = 12 −3 ± √33 𝑥= 12 ✓ 𝑥 = 0,23 𝑥 = 0,23 or 𝑥 = −0,73 ✓ 𝑥 = −0,73 (4) 2 1.1.3 6𝑥 + 7𝑥 + 2 ≥ 0 (3𝑥 + 2)(2𝑥 + 1) ≥ 0 ✓ factors/CV’s 2 1 C/V − & − 3 2 1 2 1 𝑥 ≤ − 3 𝑜𝑟 𝑥 ≥ − 2 And then max ✓✓ answers (combo 3 marks) (3) 1.1.4 2x + x = 2 ( ) = ( 2 )2 2 2x + x ✓ squaring both sides 2x + x = 4 ( 2x ) = ( 4 − x) 2 2 ✓ squaring both sides 2 x = 16 - 8 x + x 2 ✓ standard form x − 10 x + 16 = 0 2 ( x − 2 ) ( x − 8) = 0 ✓ factors x=2 or x =8 ✓ answers (with x8 selection) (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 4 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne 1.2 −2𝑦 + 𝑥 = 4 𝑎𝑛𝑑 𝑥 2 + 𝑥𝑦 − 2𝑦 2 = 0 𝑥 = 2𝑦 + 4 ✓ 𝑥 = 2𝑦 + 4 𝑥 2 + 𝑥𝑦 − 2𝑦 2 = 0 (2𝑦 + 4)2 + 𝑦(2𝑦 + 4) − 2𝑦 2 = 0 ✓ substitution 4𝑦 2 + 16𝑦 + 16 + 2𝑦 2 + 4𝑦 − 2𝑦 2 = 0 4𝑦 2 + 20𝑦 + 16 = 0 𝑦 2 + 5𝑦 + 4 = 0 ✓ standard form (𝑦 + 1)(𝑦 + 4) = 0 𝑦 = −1 𝑜𝑟 𝑦 = −4 ✓ both y values 𝑥 = 2(−1) + 4 𝑜𝑟 𝑥 = 2(−4) + 4 𝑥 = 2 𝑜𝑟 𝑥 = −4 ✓ both x values (5) 𝑚 2𝑚−1 ) 1.3 4 = 𝑝(2 +𝑝 2𝑚 𝑝. 2 22𝑚 = +𝑝 𝑝.22𝑚 2 ✓ 22𝑚 = +𝑝 2 𝑝. 22𝑚 22𝑚 − =𝑝 2 𝑝 22𝑚 (1 − ) = 𝑝 ✓ Factorisation 2 2𝑚 𝑝 ∴ 2 = 𝑝 ÷ (1 − ) 2 2𝑚 2𝑝 2 = ✓ simplification: 2−𝑝 2𝑝 2𝑚 = 𝑙𝑜𝑔2 ( ) ✓ log form 2−𝑝 1 2𝑝 𝑚 = 𝑙𝑜𝑔2 ( ) (4) 2 2−𝑝 [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 5 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1. ... ; ... ; 0 ; 7; 12 ; ✓First differences 11 9 7 5 –2 –2 –2 ✓ 𝑇1 = −20 𝑇1 = −20 (2) 2.2 2 a = −2  a = −1  𝑎 = −1 3a + b = 11 3(−1) + b = 11  b = 14  𝑏 = 14 a + b + c = −20 Answer only: full −1 + 14 + c = −20 marks  c = −33  𝑐 = −33  𝑇𝑛 = −𝑛2 + 14𝑛 − 33 Tn = − n + 14n − 33 2 (4) 2.3 Tn = −n + 14n − 33 2 −14 n=   method 2(−1) Answer only: full 𝑛=7 (3) n=7 Max value= 16 marks T7 =16 OR/OF T = −2n + 14 n / 0 = −2n + 14   method n = 7 𝑛=7 n=7 T7 =16 WILL HAVE THE HIGHEST VALUE (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 6 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 3.1.1 (2𝑥 + 4) − (10𝑥 + 6) = (4𝑥 − 8) − (2𝑥 + 4) equating −8𝑥 − 2 = 2𝑥 − 12 −10𝑥 = −10 𝑥=1 𝑥=1 (2) 3.1.2 16: 6: −4  sequence 𝑇10 = 16 + (10 − 1)(−10)  substitution = −74  answer (3) 3.1.3 99  substitution 𝑆99 = [2(16) + (99 − 1)(−10)] 2  answer = −46926 (2) 3.2.1 𝑟 =𝑘−5  Common ratio −1 < 𝑘 − 5 < 1  substitution 4<𝑘<6  answer (3) 3.2.2 1  a and r 𝑎 = −1 or 𝑟 = − 2 −1 𝑆∞ = 1 1 − (− )  substitution 2 2 =−  answer 3 (3) 3.3 𝑥; 3𝑥; 10𝑥 − 20  Sequence in terms of x 10𝑥 − 20 3𝑥 =  ratio 3𝑥 𝑥 10𝑥 − 20 = 9𝑥 𝑥 = 20  x = 20 20; 60; 200  numbers (4) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 7 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 4.1 g ( x) = k x  substitution 3 = k1  𝑘=3 k = 3 (2) 4.2 𝑥=0  𝑥=0 𝑦 = −1  𝑦 = −1 (2) 4.3 g ( x ) = 3x  interchange  x = 3y Answer only: full marks  g −1( x ) = log 3x  answer (2) 4.4 −2  𝑦=0 0= −1 𝑥 −2 1= 𝑥 𝑥 = −2  𝑥 = −2 (2) 4.5 𝑓(𝑥)  asymptote  x-intercept  shape (Q2 and Q4) 𝑔(𝑥)−1  x- intercept  increasing shape (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 8 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne 4.6 𝑦 = −(𝑥 + 𝑝) + 𝑞 𝑦 = −(𝑥 + 0) − 1  𝑦 = −𝑥 − 1 𝑦 = −𝑥 − 1 −2 −1 = −x −1  equating x −2 − x = − x 2 − x  x2 = 2 Simplification x =  2 both answers (4) OR/OF y = −x + c −1 = −(0) + c  c = −1  y = −x −1 −2  𝑦 = −𝑥 − 1 −1 = −x −1 x  equating −2 − x = − x 2 − x  x2 = 2  x 2 = 2 Simplification x =  2 both answers (4) 4.7 −2 < 𝑥 ≤ 1 ✓✓ answer (2) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 9 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1 −𝑥 2 − 𝑥 + 12 = 0  𝑓(𝑥) = 0 𝑥 2 + 𝑥 − 12 = 0 (𝑥 + 4)(𝑥 − 3) = 0  factors 𝑥 = −4 or 𝑥 = 3  both 𝑥 values (3) A(−4;0) B (3;0) 5.2 𝑏 −1 𝑥=− =− 2𝑎 2(−1) 1 =−  𝑥 − value 2 2 1 1 1 ∴ 𝑓 (− ) = − (− ) − (− ) + 12 2 2 2  𝑦 − value 49 = 4  Answer 49 ∴𝑦≤ 4 OR/OF −4 + 3 𝑥= 2 1  𝑥 − value =− 2 1 1 2 1 ∴ 𝑓 (− ) = − (− ) − (− ) + 12 2 2 2 49 =  𝑦 − value 4 49 ∴𝑦≤  Answer 4 (3) 49 5.3 𝑘(𝑥) = 3 4 −12  𝑠𝑢𝑏𝑠𝑡 = 30.25 = 1,32  Answer (2) 5.4 𝑥 < −4 or 0 < 𝑥 < 3  𝑥 < −4 0<𝑥<3 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 10 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne 5.5 −1 < 12 − 𝑘 < 0 Check 4 Afr 49  Answer 12 < 𝑘 < 4 (2) 5.6 5 49 5 2 ℎ(𝑥) = −(𝑥 − )2 + −(𝑥 − 2) 2 4 49  4 (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 11 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1 𝐴 = 150 000(1 − 0,13)6  correct substitution in correct formula = 𝑅65 043,93 𝑛 = 6 & 𝑖 = 0,13  Answer (3) 6.2.1 0.053 2  4 and 2 𝐴 = 300 000 (1 + )  correct substitution in 4 correct formula = 𝑅308 002,67  Answer (3) 6.2.2 0,053 −72  correct substitution in 𝑥 [1 − (1 + ) ] 4 correct formula 308 002,67 = 0,053 4 0,053 308 002,67 ×  Simplification 4 𝑥= −72 0,053 [1 − (1 + ) ] 4 𝑥 = 𝑅6 664,20  Answer (3) 6.3 12 0.079 1 + 𝑖𝑒𝑓𝑓 = (1 + ) 12 𝑖𝑒𝑓𝑓 = 0,01321000694 answer 𝑖𝑛𝑜𝑚 4 1 + 0,01321000694 = (1 + ) 4 answer 𝑖𝑛𝑜𝑚 = 0,01314506756 answer Quarterly deposit= 27562.50 × 0.11 = 3031,88 32 0,01314506756 32 3031,88 [(1 + 4 ) − 1]  Correct substitution into 𝐹= 0,01314506756 the future value annuity 4 𝐹 = 𝑅 102 128,44 Answer (6) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 12 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 7.1 𝑓(𝑥) = −1 + 4𝑥 2 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑓 ′ (𝑥) = lim ℎ→0 ℎ −1 + 4𝑥 2 + 8𝑥ℎ + 4ℎ2 − (−1 + 4𝑥 2 ) ✓𝑓(𝑥 + ℎ) = −1 + 4𝑥 2 + = lim 8𝑥ℎ + 4ℎ2 ℎ→0 ℎ 8𝑥ℎ + 4ℎ2 ✓ substitution = lim ✓ simplification ℎ→0 ℎ ℎ(8𝑥 + 4ℎ) = lim ✓ common factor ℎ→0 ℎ = lim (8𝑥 + 4ℎ) ℎ→0 ✓ answer = 8𝑥 OR/OF ✓𝑓(𝑥 + ℎ) = −1 + 4𝑥 2 + 𝑓(𝑥 + ℎ) = −1 + 4(𝑥 + ℎ)2 = −1 + 4𝑥 2 + 8𝑥ℎ + 4ℎ2 8𝑥ℎ + 4ℎ2 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑓 ′ (𝑥) = lim ℎ→0 ℎ −1 + 4𝑥 2 + 8𝑥ℎ + 4ℎ2 − (−1 + 4𝑥 2 ) ✓ substitution = lim ℎ→0 ℎ 8𝑥ℎ + 4ℎ2 = lim ✓ simplification ℎ→0 ℎ ℎ(8𝑥 + 4ℎ) = lim ✓common factor ℎ→0 ℎ = lim (8𝑥 + 4ℎ) ℎ→0 = 8𝑥 ✓ answer (5) 7.2.1 𝐷𝑥 [(2𝑥 3 + 5)2 ] = 𝐷𝑥 [4𝑥 6 + 20𝑥 3 + 25] ✓ expansion = 24𝑥 5 + 60𝑥 2 ✓ 24𝑥 5 ✓ 60𝑥 2 (3) 7 3 7.2.2 𝑦 = 3𝑥 4 − + 2. √𝑥 2 2 𝑥 2 ✓ −7𝑥 −1 + 2𝑥 3 𝑦 = 3𝑥 4 − 7𝑥 −1 + 2𝑥 3 ✓ 12𝑥 3 𝑑𝑦 3 −2 4 −1 = 12𝑥 + 7𝑥 + 𝑥 3 𝑑𝑥 3 ✓ 7𝑥 −2 4 1 ✓ 3 𝑥 −3 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 13 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne 7.3 𝑦 + 5𝑥 = 4 𝑦 = −5𝑥 + 4 𝑚𝑇 = 5 1 ✓ gradient of tangent 𝑦 = 2𝑥 2 + 3𝑥 + 1 𝑑𝑦 = 4𝑥 + 3 𝑑𝑥 1 1 4𝑥 + 3 = 5 ✓ 4𝑥 + 3 = 5 14 4𝑥 = − 5 −14 𝑥 = 20 ✓ x value 7 𝑥 = − 10 7 2 7 𝑦 = 2 (− 10) + 3 (− 10) + 1 3 𝑦 = − 25 ✓ y value 7 3 (4) (− 10 ; − 25) [16] QUESTION/VRAAG 8 8.1 𝑓(𝑥) = −3𝑥 3 + 15𝑥 2 − 21𝑥 + 9 x intercepts (𝑥 − 1)(−3𝑥 2 + 12𝑥 − 9) = 0 ✓ 𝑦=0 𝑥 − 1 = 0 𝑜𝑟 𝑥 2 − 4𝑥 + 3 = 0 𝑥 − 1 = 0 𝑜𝑟 (𝑥 − 1)(𝑥 − 3) = 0 ✓ factors 𝑥 = 1 𝑜𝑟 𝑥 = 3 ✓ Both x- values y intercept 𝑦=9 (0; 9) ✓ (0; 9) (4) 3 2 8.2 𝑓(𝑥) = −3𝑥 + 15𝑥 − 21𝑥 + 9 𝑓 ′ (𝑥) = −9𝑥 2 + 30𝑥 − 21 −9𝑥 2 + 30𝑥 − 21 = 0 ✓ −9𝑥 2 + 30𝑥 − 21 3𝑥 2 − 10𝑥 + 7 = 0 (3𝑥 − 7)(𝑥 − 1) = 0 ✓ f / ( x) = 0 7 ✓ both x- values 𝑥 = 𝑜𝑟 𝑥 = 1 3 7 7 3 7 2 7 𝑓 ( ) = −3 ( ) + 15 ( ) − 21 ( ) + 9 3 3 3 3 32 7 32 = 9 (3 ; 9 ) ✓ both y values 𝑓(1) = −3(1)3 + 15(1)2 − 21(1) + 9 (4) =0 (1; 0) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 14 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne 8.3 ✓y- intercept  both x -intercepts  both turning points  shape (4) 8.4.1 7 7 ✓✓ answer (combo (1; ) 𝑜𝑟 1 < 𝑥 < 3 3 mark) (2) 8.4.2 32 ✓✓ answer (combo 0<𝑘< 9 mark) (2) [16] QUESTION/VRAAG 9 9.1 let x be the number of R20 increases in price  𝑃𝑟𝑖𝑐𝑒 = 400 + 2𝑥 𝑃𝑟 𝑖 𝑐𝑒 = 400 + 20𝑥 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 = 200 − 4𝑥  𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 = 200 − 4𝑥 𝑃𝑟𝑜𝑓𝑖𝑡 = (400 + 20𝑥)(200 − 4𝑥)  Profit = 80 000 − 2 400𝑥 − 80𝑥 2 = 80 000 + 2 400𝑥 − 80𝑥 2 𝑑𝑟 =0 𝑑𝑥 𝑑𝑟 2400 − 160𝑥 = 0 ✓ 𝑑𝑥 = 2 400 − 160𝑥 𝑑𝑟 ✓𝑑𝑥 = 0 160𝑥 = 2 400 𝑥 = 15 𝑥 = 15 𝑝𝑟𝑖𝑐𝑒 = 400 + 20(15)  𝑅700 = 𝑅700 (7) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 15 FS/VS/September 2024 Grade/Graad 12 Prep. Exam./Voorb. Eksam. Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 10.1.1 𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) ✓ 0,8 = 𝑃(𝐴) + 0,4 0,8 = 𝑃(𝐴) + 0,4 𝑃(𝐴) = 0,4 ✓ answer (2) 10.1.2 𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴) × 𝑃(𝐵) 0,8 = 𝑃(𝐴) + 0,4 − 0,4𝑃(𝐴) ✓Substitution 0,6𝑃(𝐴) = 0,4 2 𝑃(𝐴) = 2 3 ✓𝑃(𝐴) = 3 (2) 10.2 ✓𝑥 𝑎𝑛𝑑 1 − 𝑥 ✓0,9 𝑎𝑛𝑑 0,1 ✓0,05 𝑎𝑛𝑑 0,95 𝑃(𝐷 𝑃) + 𝑃(𝑁𝐷 𝑃) = 0,067 0,9𝑥 + 0,05(1 − 𝑥) = 0,067 ✓Substitution 0,9𝑥 + 0,05 − 0,05𝑥 = 0,067 ✓Simplification 0,85 𝑥 = 0,017 𝑥 = 0,02 2% of the population has the disease. ✓Answer (6) 10.3.1 21 × 20 × 19 × 10 × 9 × 8 ✓ 21 × 20 × 19 × 10 × 9 × 8 = 5745600 (1) 10.3.2 NO REPEATS P(starting with a 6) ✓ 21 × 20 × 19 × 1 × 9 × 8 21 × 20 × 19 × 1 × 9 × 8 ✓ 21 × 20 × 19 × 10 × 9 × 8 = 1 21 × 20 × 19 × 10 × 9 × 8 ✓10 1 = 10 (3) REPEATS P(starting with a 6) 21 × 21 × 21 × 1 × 10 × 10 ✓ 21 × 21 × 21 × 1 × 10 × 10 = ✓ 21 × 21 × 21 × 10 × 10 × 10 21 × 21 × 21 × 10 × 10 × 10 1 1 = 10 ✓10 (3) [14] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief