# Mathematics Grade 12 Paper 1 Free State Mock 2018 Memo > Grade 12 Mathematics Paper 1 memorandum from the Free State Mock Exam 2018, 13 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/2018-grade-12-trial-memo-p1-fs/61ec36d6-afa8-4807-a756-7ea63166784b - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2018 - Exam period: Free State Mock Exam - Paper: 1 - Curriculum: CAPS - Pages: 13 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 Paper 1 Free State Mock 2018 Memo](https://www.hlayiso.com/library/2018-grade-12-trial-memo-p1-fs/61ec36d6-afa8-4807-a756-7ea63166784b) ## About this document Mathematics · Grade 12 · Free State Mock Exam · 2018. Memorandum, 13 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com VOORBEREIDENDE EKSAMEN GRAAD 12 WISKUNDE V1 SEPTEMBER 2018 PUNTE: 150 NASIENRIGLYNE Hierdie nasienriglyn bestaan uit 13 bladsye. Kopiereg voorbehou Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Wiskunde V1 2 VS/September 2018 Graad 12 Nasienriglyne 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. VRAAG 1 1.1 1.1.1 −3𝑥 2 + 2𝑥 + 2 = 0 −2 (2)2 − 4(− 3)(2) ✓ vervang met x= korrekte formule 2(− 3) ✓ 𝑥 = −0,55 x = −0,55 of x = 1,22 ✓ 𝑥 = 1,22 (3) 1.1.2 x + 2x − 3 = 5 2 x2 + 2x − 8 = 0 ✓ standaard vorm (x + 4)(x − 2) = 0 ✓ faktore x = −4 of x=2 ✓ beide x waardes (3) 1.1.3 x − 2 x − 15  0 2 (x − 5)(x + 3)  0 ✓ faktore CVs : −3 en 5 ✓ kritiese waardes Oplos sin g : −3  x  5 ✓ antwoord (notatsie) (3) 1.1.4 − 2x − 1 = 2 − x x − 2 = 2x − 1 x2 − 4x + 4 = 2x − 1 ✓ x 2 − 4x + 4 = 2x −1 x 2 − 6x + 5 = 0 ✓ standaard vorm (x − 5)(x − 1) = 0 ✓ faktore x = 5 of x = 1  x = 5 slegs ✓ x = 5 slegs (4) 1.1.5 7.3 x − 3 x +1 = 36 7.3 x − 3.3 x = 36 3 x (7 − 3) = 36 ✓ 3 x (7 − 3) = 36 3 x (4 ) = 36 3x = 9 ✓ 3x = 9 3 =3 x 2 x = 2 ✓ x=2 (3) Kopiereg voorbehou Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Wiskunde V1 3 VS/September 2018 Graad 12 Nasienriglyne 1.2 2 x −1 = y en x 2 − xy = 3x − 3 x 2 − x(2 x − 1) = 3x − 3 ✓ vervanging x 2 − 2 x 2 + x = 3x − 3 x 2 + 2x − 3 = 0 (x + 3)(x − 1) = 0 ✓ standaard vorm x = −3 of x = 1 ✓ faktore y = −7 of y = 1 ✓ beide x-waardes ✓ beide y-waardes (5) 1.3 5 x.5 x +1 5. 125 − 52 x 5 2 x .5 = 5. 25.5 − 2 x 5 ✓ 5. 5 . 5 = 5 5. 5 − 5 ✓ 5 = 25 − 5 ✓ antwoord = 20 (3) [24] Kopiereg voorbehou Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Wiskunde V1 4 VS/September 2018 Graad 12 Nasienriglyne VRAAG 2 2.1 3 ; − 2 ; − 7 ; − 12 ; ... 2.1.1 a = 3 en d = −5 ✓ korrekte T21 = 3 + 20(− 5) vervanging ✓ antwoord (2) T21 = −97 2.1.2 T n= a + (n − 1)d − 177 = 3 + (n − 1)(− 5) ✓ vervanging − 177 = 3 − 5n + 5 5n = 185  n = 37 ✓ n = 37 (2) 2.2 S n = n 2 − 2n 2.2.1 S13 = (13) − 2(13) ✓ vervanging 2 ✓ antwoord S13 = 143 (2) 2.2.2 S12 = (12) − 2(12) 2 S12 = 120 ✓ S12 = 120 T13 = S13 − S12 T13 = 143 − 120 = 23 ✓ antwoord (2) 2.3 ✓ Eerste x 6 9 y 24 differensie i.t.v. y ✓ Tweede 6 –x 3 differensie i.t.v. y ✓ gelykstelling van twee differensies ✓ y = 15 (33 − 2 y ) = ( y − 12) ✓ x + y = 21 3 y = 45  y = 15 x = 6 x + y = 21 (5) [13] Kopiereg voorbehou Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Wiskunde V1 5 VS/September 2018 Graad 12 Nasienriglyne VRAAG 3 3.1 S n = a + ar + ar 2 + ...... + ar n−2 + ar n−1 ✓ Sn rS n = ar + ar 2 + ar 3 .. + ar n−2 + ar n−1 + ar n ✓ rS n S n − rS n = a − ar n ✓ S n − rS n = a − ar n ( S n (1 − r ) = a 1 − r n ) ( ✓ S n (1 − r ) = a 1 − r n ) ( ) (4) a 1− r n  S n= 1− r 3.2 3.2.1 4 4 (1) 4; ; ✓ antwoord 5 25 3.2.2 a 1 S = ✓ r= 1− r 5 4 S = 1 ✓ vervanging met 1− korrekte formule 5 S = 5 ✓ antwoord (3) 3.2.3 S  − S n  0.0001 ✓ opstelling van die ongelykheid   1 n  4 1 −      5   ✓ vervang. S  ; a en r 5−   0.0001 1 1− 5 4 − 4 + 4(0,2 )  0,00008 n (0,2)n  0,00002 ✓ (0,2) = 0,00002 n log 0,00002  n log 0,2 ✓ korrekte gebruik n  6,722706232 van logaritmes Kleinste n = 7 (5) ✓ n=7 [13] Kopiereg voorbehou Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Wiskunde V1 6 VS/September 2018 Graad 12 Nasienriglyne VRAAG 4 4.1 0 = 4x + 8 ✓ y=0 − 4x = 8  x = −2 ✓ x = −2 (2) A(− 2 ; 0) 4.2 B(10 ; 0 ) ✓ x = 10 (2) ✓ y=0 4.3 h( x ) = a ( x − x1 )( x − x2 ) h( x ) = a ( x + 2 )( x − 10) ✓ vervang x1 = −2 en 32 = a(6 + 2 )(6 − 10) x2 = 10 32 = −32a ✓ vervang (6 ; 32) ✓ a = −1 a = −1  h( x ) = −1( x + 2 )( x − 10) h( x ) = − x 2 + 8 x + 20 ✓ vergelyking van h(x) (4) 4.4 F (0 ; 8) en C (0 ; 20 ) ✓ F (0 ; 8) en C (0 ; 20 ) FC = 12 eenhede ✓ antwoord (2) 4.5 h(4) = −(4) + 8(4) + 20 = 36 ✓ y = 36 2 Reeks: y  (− ; 36] OF y / y  R; y  36 ✓ antwoord (2) 4.6 x  (10;  ) OF x / x  R; x  10 ✓ kritiese waardes ✓ notasie (2) 4.7 4 x + k = − x + 8 x + 20 2 x 2 − 4 x + k − 20 = 0 ✓ standard vorm  = b 2 − 4ac ✓ vervang met   = (− 4) − 4(1)(k − 20) 2 0 = 16 − 4k + 80 ✓ =0 4k = 96 ✓ antwoord k = 24 OF ✓ −2𝑥 + 8 = 4 ℎ′ (𝑥) = 𝑓 ′ (𝑥) ✓𝑥 = 2 −2𝑥 + 8 = 4 ∴𝑥=2 ✓ℎ(2) = 32 ℎ(2) = −22 + 8(2) + 20 = 32 ∴ (2; 32) ✓𝑘 = 24 (4) 32 = 4(2) + 𝑘 24 = 𝑘 [18] Kopiereg voorbehou Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Wiskunde V1 7 VS/September 2018 Graad 12 Nasienriglyne VRAAG 5 5.1 2+ x f (x ) = x −1 x −1+ 3 x −1+ 3 f (x ) = ✓ x −1 x −1 x −1 x −1 3 f (x ) = + 3 ✓ + x −1 x −1 x −1 x −1 f (x ) = 3 +1 ✓ f (x ) = 3 x −1 +1 (3) x −1 5.2 x =1 x =1 y =1 y =1 (2) 5.3 2+ x ✓ y=0 0= x −1 0 = 2+ x x = −2 ✓ x = −2 (2) A(− 2 ; 0) 5.4 x=3 ✓ antwoord (1) [8] Kopiereg voorbehou Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Wiskunde V1 8 VS/September 2018 Graad 12 Nasienriglyne VRAAG 6 6.1 1 x f (x ) =   3 6.1.1 1 y 1 y x=  ✓ x=  3 3 f ( x ) = log 1 x OF −1 f −1 (x ) = − log3 x ✓✓antwoord (2) 3 Slegs Antwoord:Volpunte 6.1.2 1 𝑥 𝑓(𝑥) = ( ) : 3 ✓ vorm ✓ y-afsnit 𝑓 −1 (𝑥) = log 1 𝑥: (− 1;3) f 3 ✓ vorm ✓ x-afsnit 1 1 (3;–1) 𝑓 −1 (𝑥) (4) 6.2 p(− 3) = 10 en p / (x ) = −2 p ( x ) = −2 x + c ✓ m = −2 10 = −2(− 3) + c  c = 16 ✓ c = 16 p(x ) = y = −2 x + 16 ✓ omruil van x en y x = −2 y + 16 (x = −2 y + 16) p (x ) = − x + 8 −1 1 ✓ antwoord 2 (4) [10] Kopiereg voorbehou Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Wiskunde V1 9 VS/September 2018 Graad 12 Nasienriglyne VRAAG 7 7.1 n  i  0,096 1 + ieff = 1 + nom  ✓ i=  n  12 12  0,096  ✓ vervang met 1 + ieff = 1 +   12  korrekte formule  ieff = 0,100338694 ✓ antwoord (i of r) (3)  reff = 10,03% 7.2 𝐴 = 𝑃(1 − 𝑖)𝑛 0,88 0,88 𝑛 ✓ i= 60 000 = 150 000 (1 − ) 4 4 ✓ vervang met 0,4 = (0,978)𝑛 korrekte formule 𝑙𝑜𝑔0,4 ∴𝑛= 𝑙𝑜𝑔0,978 ✓ Korrekte gebruik 𝑛 = 41,1897 … 𝑘𝑤𝑎𝑟𝑡𝑒 = 10,3 𝑗𝑎𝑟𝑒 van logaritmes ✓ antwoord (3) 7.3 7.3.1 x[1 − (1 + i)− n ] P= 0.1 i ✓ i=   0,1  −120  12 x 1 − 1 +   ✓ n = 120   12   R 250 000 =  ✓ vervang met 0,1 korrekte formule 12 ✓ antwoord (4)  x = R3 303,77 7.3.2 Uitstaande balans (B):   0,1  −84  R3 303,77 1 − 1 +   ✓ opstel van formule   12   ✓ n = 84 B= 0,1 ✓ vervang met (4) 12 korrekte formule B = R199 008,09 ✓ antwoord OF Uitstaande Balans (B): OF 36  0,1  ✓ opstel van formule 3 303,77[1 +  − 1] 36  0,1   12  ✓ n = 36 B = 250 0001 +  −  12  0,1 ✓ vervang met (4) 12 korrekte formule B = R199 007,93 ✓ antwoord [14] Kopiereg voorbehou Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Wiskunde V1 10 VS/September 2018 Graad 12 Nasienriglyne VRAAG 8 8.1 f ( x ) = −3x 2 f ( x + h) − f ( x ) f ' ( x) = lim h→ 0 h − 3( x + h ) − (− 3 x 2 ) 2 f ' ( x) = lim h→ 0 h − 3 x − 6 xh − 3h 2 + 3 x 2 2 ✓ vervang met korrekte f ' ( x) = lim formule h→ 0 h − 6 xh − 3h 2 f ' ( x) = lim h→ 0 h ✓ − 6 xh − 3h 2 h(− 6 x − 3h ) h(− 6 x − 3h ) f ' ( x) = lim ✓ h→ 0 h h f ' ( x) = −6 x ✓ antwoord (4) 8.2 3 y = 7x4 − 5 x − 1 x 1 ✓ y = 7 x 4 − 5 x 2 − 3 x −1 y = 7 x 4 − 5 x 2 − 3 x −1 5 − 1 1 ✓ 28x 3 ✓− x 2 dy 5 − 2 = 28 x 3 − x 2 + 3 x −2 dx 2 ✓ + 3x −2 (4) 8.3 g (x ) = ax − 24 x + b 3 g ' ( x) = 3ax 2 − 24 ✓ g ' ( x) = 3ax 2 − 24 0 = 3a(− 2) − 24 2 ✓ vervanging x = −2 24 = 12a g ' ( x) = 0 a=2 ✓ waarde van a 17 = 2(− 2 ) − 24(− 2 ) + b 3 ✓ vervanging (− 2; 17 ) 17 = −16 + 48 + b in g ( x ) b = −15 ✓ b = −15 (5) [13] Kopiereg voorbehou Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Wiskunde V1 11 VS/September 2018 Graad 12 Nasienriglyne VRAAG 9 9. f (x ) = − x 3 + 10 x 2 − 17 x − 28 1 y-afsnit: (0;−28) ✓ (0;−28) ( x-afsnit: (x + 1) x 2 − 11x − 28 = 0) ✓✓ ( ) (x + 1) x 2 −11x − 28 = 0 (x + 1)(x − 7 )(x − 4) = 0 ✓ x = −1 of x = 7 of x = 4 (−1 ; 0); (7 ; 0); (4 ; 0) (x + 1)(x − 7 )(x − 4) = 0 (5) ✓ x-waardes 9. f ( x ) = − x 3 + 10 x 2 − 17 x − 28 2 0 = −3 x 2 + 20 x − 17 ✓ 0 = −3x 2 + 20 x − 17 0 = 3 x − 20 x + 17 2 ✓ faktore 0 = (3 x − 17 )( x − 1) 17 x = or x =1 ✓ beide x-waardes 3  17   17  ✓  ;14,8  Draaipunte:  ;14,8  en (1; − 36)  3   3  ✓ (1; − 36) (5) 9. 3 ✓ vorm ✓ draaipunt ✓ x- en y-afsnitte (3) 9. k − 3  14,8 or k − 3  −36 ✓ k  17,8 4 k  17,8 or k  −33 ✓ k  −33 (2) [15 ] Kopiereg voorbehou Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Wiskunde V1 12 VS/September 2018 Graad 12 Nasienriglyne VRAAG 10 10.1 DE DE tan 60 = ✓ tan 60 = x x  DE = x 3 ✓ DE = x 3 EF = y − 2 x ✓ EF = y − 2 x Oppvl van reghoek = basis  hoogte = ( y − 2 x )x 3 ✓ ( y − 2 x )x 3 (4) = 3xy − 2 3x 2 10.2 A = 3 xy − 2 3 x 2 dA dA = 3 y − 4 3x = 0 ✓ = 3 y − 4 3x dx dx dA  x= y ✓ =0 4 dx y ✓ x= 2  y  y MaksOppvl = 3   y − 2 3   4 4 4 y 3y2 2 3y2 ✓ Vervanging x = = − 4 4 16 3 2 ✓ antwoord (5) = y 8 [9] Kopiereg voorbehou Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Wiskunde V1 13 VS/September 2018 Graad 12 Nasienriglyne VRAAG 11 11.1 11.1.1 ∴ P(A) = 0,45 P(A of B) = P(A) + P(B) = 0,45 + 0,29 ✓ vervanging = 0,74 ✓ antwoord (2) 11.1.2 P(A) × P(B) = 0,45 × 0,29 ✓ 0,45  0,29 = 0,1305 ≈ 0,13 319 ✓ 0,1267 or P(A of B) = P(A) + P(B) – P(A en B) 2500 = 0,45 + 0,29 – 0,1276 = 0,0695 ≈ 0,61 ✓ vervanging ✓ antwoord (4) 11.2 4 s unny Sonskynplay s occer speel sokker 5 P 1 1 s unny no play S 3 Sonskyn geen spel 5 NP 2 P 5 Geen not son not sunny spelplay NS 2 3 PN 3 Geen not son geen sunny play spel occer 1 1 5 ✓  3 5 P(speel nie sokker nie) = 2 3 ✓✓  1 1 2 3 7 3 5  +  = 3 5 3 5 15 7 ✓ 15 (4) 11.3 Aantal verskillende kodes =26  9  8  7 = 13104 ✓ 26 ✓ 9  8 7 ✓ antwoord (3) [13] TOTAAL: 150 Kopiereg voorbehou Blaai om asseblief ## Related documents - [2018 GRADE 12 TRIAL EXAM P1 FS](https://www.hlayiso.com/library/2018-grade-12-trial-exam-p1-fs/7e95d633-2263-4d89-a9cf-6871a4e1be21) — Mathematics · QP · 2018 - [Mathematics P1 Feb Marcg 2018 Memo Afr Eng](https://www.hlayiso.com/library/mathematics-p1-feb-marcg-2018-memo-afr-eng/d3e8b946-b792-496e-82fb-3464b75569c9) — Mathematics · Memo · 2018 · Eng - [2018 GRADE 12 TRIAL EXAM P2 FS .pdf](https://www.hlayiso.com/library/2018-grade-12-trial-exam-p2-fs-pdf/f06a5592-1e1b-494c-b6ce-40234267eaea) — Mathematics · QP · 2018 - [MATHS P1 QP GR12 SEPT 2018 Afrikaans](https://www.hlayiso.com/library/maths-p1-qp-gr12-sept-2018-afrikaans-1/48ef1811-7641-496b-9091-7e47b50d9aae) — Mathematics · QP · 2018 · Afr - [2018 GRADE 12 MATH TRIAL EXAM PAPER 1 MP](https://www.hlayiso.com/library/2018-grade-12-math-trial-exam-paper-1-mp/fe550454-c635-4e05-ac77-fa2e68af9f1a) — Mathematics · QP · 2018 - [2018 GRADE 12 MATH TRIAL EXAM PAPER 1 NW](https://www.hlayiso.com/library/2018-grade-12-math-trial-exam-paper-1-nw/131ac19c-b932-4da8-9155-30097df7327a) — Mathematics · QP · 2018 ## More from Grade 12 Mathematics - [MATHS P1 ANSWER BOOK GR12 SEPT 2025 English Afrikaans watermark](https://www.hlayiso.com/library/maths-p1-answer-book-gr12-sept-2025-english-afrikaans-watermark/d78c3fd3-dcd5-4f3f-911b-e5635b4686cc) — Addendum · 2025 · Afr - [Maths Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-gauteng/1d6bf7fd-d288-4a01-99f7-2c87e247760c) — QP · 2025 - [Maths Grade 12 NSC P2 QP September 2025 Afr Limpopo](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-afr-limpopo/f2f8d481-5d87-4e89-bb21-6077bbcf403d) — QP · 2025 · Afr - [Maths Grade 12 NSC P1 QP September 2025 Afr Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p1-qp-september-2025-afr-northern-cape/f5b3a4ce-fa91-4475-96e0-0654a87e1405) — QP · 2025 · Afr - [Maths Grade 12 NSC P2 QP September 2025 Eng Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-eng-northern-cape/4ede8207-ca82-4419-bae1-fae734498cc7) — QP · 2025 · Eng - [MATHS P2 ANSWER BOOK GR12 JUNE 2024 Afr+English](https://www.hlayiso.com/library/maths-p2-answer-book-gr12-june-2024-afr-english/4a71e975-4ac4-4f55-8e23-0b8155a5b0c6) — Addendum · 2024 · Eng ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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