# Mathematics Grade 11 P1 Gauteng June 2024 Memo (Afrikaans) > Grade 11 Mathematics Paper 1 memorandum from the Gauteng June Exam 2024 (Afrikaans), 11 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr-11-mathematics-p1-afrikaans-june-2024-possible-answers/935b6195-9108-4f0f-89ab-74a80ebee3f4 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Language: Afrikaans - Document type: Memorandum - Year: 2024 - Exam period: Gauteng June Exam - Paper: 1 - Curriculum: CAPS - Pages: 11 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Mathematics](https://www.hlayiso.com/library/grade-11/mathematics) › [Mathematics Grade 11 P1 Gauteng June 2024 Memo (Afrikaans)](https://www.hlayiso.com/library/gr-11-mathematics-p1-afrikaans-june-2024-possible-answers/935b6195-9108-4f0f-89ab-74a80ebee3f4) ## About this document Mathematics · Grade 11 · Gauteng June Exam · 2024 · Afrikaans. Memorandum, 11 pages. Read online or download the PDF. 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Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com PROVINSIALE EKSAMEN JUNIE 2024 GRAAD 11 NASIENRIGLYNE WISKUNDE (VRAESTEL 1) 11 bladsye ### Page 2 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 1) GRAAD 11 INSTRUKSIES EN INLIGTING A – AKKURAATHEID CA. – KONTINUE AKKURAATHEID LET WEL: • Indien 'n kandidaat 'n vraag TWEE keer beantwoord, sien slegs die eerste poging na. • Indien 'n kandidaat 'n vraag DOODTREK en dit nie oordoen nie, sien die doodgetrekte antwoord na. • Deurlopende akkuraatheid is van toepassing op ALLE aspekte van die nasienriglyn. • Dit is ONAANVAARBAAR om waardes/antwoorde aan te neem om 'n probleem op te los. 2 ### Page 3 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 1) GRAAD 11 VRAAG 1 1.1 1.1.1 x9 ✓ antwoord (1) 1.1.2 x=9 ✓ antwoord (1) 1.1.3 (1) 2 + 4 9 −1 5 ✓ antwoord 8  Irrasionaal (1) 1.1.4 x=0 ✓ antwoord (1) 1.2 1.2.1 2 x 2 − 3x − k = 0 2 x 2 − 3x − 5 = 0 (2 x − 5)( x + 1) = 0 ✓ faktore 5 x = of x = −1 ✓ antwoorde 2 (2) 1.2.2 2 x 2 − 3x − 4 = 0 −(−3)  (−3) 2 − 4(2)(−4) ✓ substitusie x= 2(2)  x = 2,35 of x = −0,85 ✓✓ antwoorde LET WEL: Penaliseer 1 punt vir verkeerde afronding, SLEGS in die vraag. (3) 1.3 1.3.1 x−4= − x−2 16 − 8 x − x 2 = x − 2 ✓ kwadreer beide kante x 2 − 9 x + 18 = 0 ✓ standaardvorm ( x − 6)( x − 3) = 0 ✓ faktore x = 3 of x  6 ✓ kritieke waardes ✓ verwerping (5) 3 ### Page 4 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 1) GRAAD 11 1.3.2 x 2 + 4  3x + 2 x 2 − 3x + 2  0 ✓ standaardvorm ( x − 2)( x − 1)  0 ✓ faktore x  1 of x  2 OF ✓✓ antwoorde x  (−;1)  (2; ) (4) 1.4 y + 2 = 2x  y = 2x − 2 (1) ✓ y as onderwerp 2x2 = 2 − y 2 (2)  2 x 2 = 2 − (2 x − 2) 2 ✓ substitusie 2 x 2 = 2 − (4 x 2 − 8 x + 4) 2 x2 = 2 − 4 x2 + 8x − 4  6x2 − 8x + 2 = 0  3x 2 − 4 x + 1 = 0 ✓ standaardvorm (3 x − 1)( x − 1) = 0 1 ✓ faktore x = of x = 1 3 ✓ albei x- waardes 4 y =− of y = 0 3 ✓ albei y- waardes OF y + 2 = 2x OF y+2 x = (1) 2 ✓ x as onderwerp 2x2 = 2 − y 2 (2) ✓ substitusie y+2 2  2( ) = 2 − y2 2 y2 + 4 y + 4 2( ) = 2 − y2 4 y + 4y + 4 2 ( ) = 2 − y2 2  y + 4 y + 4 = 4 − 2 y2 2 ✓ standaardvorm 3y + 4 y = 0 2 ✓ faktore y (3 y + 4) = 0 4 ✓ albei y-waardes y =0 of y=− 3 1 x =1 of x= ✓ albei x-waardes (6) 3 1.5 1.5.1 Vir ongelyke wortels:   0 ✓ voorwaarde van  4 ### Page 5 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 1) GRAAD 11  2 x 2 − 4 x + (3 − p) = 0  = b 2 − 4ac   = (−4) 2 − 4(2)(3 − p) ✓ substitusie in   = 16 − 24 + 8 p (4)   = −8 + 8 p ✓ uitdrukking vir   −8 + 8 p  0 8p  8  p 1 ✓ antwoord 1.5.2 Uit 1.5.1 3− p  0 (a  0 en b  0) ✓ c0 − p  −3 ✓ antwoord 1  p  3 LET WEL: Enige ander geldige metode (2) 1.6 ( p 2 + 1) x 2 = −2 pqx − q 2 ✓ standaardvorm  ( p 2 + 1) x 2 + 2 pqx + q 2 = 0  = (2 pq) 2 − 4( p 2 + 1)q 2 ✓ substitusie in   = 4 p q − 4 p q − 4q 2 2 2 2 2 ✓ vereenvoudiging   = −4q 2 ✓ uitdrukking vir   q 2  0, q  ; q  0  −4q 2  0 ✓ verduideliking  Wortels is nie-reël (5) [35] 5 ### Page 6 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 1) GRAAD 11 VRAAG 2 2.1 2.1.1 (2 −1 + 3−1 ) 2 1 1 ✓ verander eksponente = ( + )2 2 3 3+ 2 =( ) 6 5 ✓ vereenvoudiging = ( )2 6 25 = ✓ antwoord 36 (3) 2.1.2 2 .4 x + 2 − 4 x − 3 2 x .2 x 2 .4 x .4 2 − 4 x .4 − 3 = 4x 1 4 x (32 − ) 64 ✓✓ faktore = x 4 1 = 32 − 64 2 048 − 1 = 64 2 047 = ✓ antwoord 64 63 LET WEL: Aanvaar 31 (3) 64 1 2.1.3 − 3 2 [ 12 + 3 (3 3)] 1 1 − = 3 2 [2 3 + (3 3) 3 ] ✓ vereenvoudig hakies 1 1 1 1 − = 3 [2.3 + 3 .3 ] 2 2 3 6 ✓ wortels met rasionale 1 1 1 − eksponente = 3 [2.3 + 3 ] 2 2 2 1 1 − = 3 .3 (2 + 1) 2 2 ✓ faktore = 30 (3) =3 ✓ antwoord (4) 6 ### Page 7 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 1) GRAAD 11 2.2.1 3 2 x = 16 4 3 x4 = 8 3 4 4   x 4 3 = 83 ✓ metode  x = (3 8 ) 4 ✓ x as onderwerp  x = 16 ✓ antwoord (3) NOTA: Enige ander geldige metode. 2.2.2 5 x +1 + 5 x = 150 5 x.51 + 5 x = 150  5 x (51 + 1) = 150 ✓ faktore  5 (6) = 150 x 5 x = 25 ✓ vereenvoudiging 5 = 5 x 2 ✓ 25 met grondtal 5 x = 2 ✓ antwoord (4) LET WEL: Enige ander geldige metode. 2.3 9 − 54 6 2 (9 − 9.2.3 ) 2 =  ✓  2 6 2 2 2 9 2 − 3 2 . 3. 2 = 6.2 9 2 − 3.2 3 ✓ vereenvoudiging = 12 3(3 2 − 2 3 ) ✓ faktorisering = 12 3 2 −2 3 = 4 (3) [20] 7 ### Page 8 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 1) GRAAD 11 VRAAG 3 3.1 T4 = −17 ✓ antwoord (1) 3.2 Tn = −7n + 11 ✓ antwoord (1) 3.3 Tn = −7n + 11  −227 = −7n + 11 ✓ vervang Tn − 238 = −7n ✓ antwoord  n = 34 (2) 3.4 2a = −7 7 ✓ waarde van a a=− 2 ✓ gevolgtrekking  maksimum (2) LET WEL: Enige ander geldige metode. [6] VRAAG 4 4.1 69 0 − 63 T4 − 69 − 63 − 57 6 6 volgende term (T4 ) is: − 120 ✓ antwoord (1) 4.2 Tn = an2 + bn + c T2 = a(2) 2 + b(2) + c = 4a + 2b + c T1 = a(1) 2 + b(1) + c = a + b + c T2 − T1 = 3a + b 2a = 6 ✓ 2de verskil a = 3 ✓ waarde van a 3a + b = −69 3(3) + b = −69 9 + b = −69  b = −78 ✓ waarde van b a + b + c = 69 3 + (−78) + c = 69 ✓ waarde van c  c = 144 Tn = 3n 2 − 78n + 144 (4) 8 ### Page 9 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 1) GRAAD 11 4.3 Kleinste term (minimum) is by die draaipunt van die kwadratiese vergelyking: − b − (−78) 78 ✓ vervang in sim.as n = = = = 13 2a 2(3) 6 ✓ waarde van n Tn = 3n − 78n + 144 2 T13 = 3n 2 − 78n + 144 T13 = 3(13) 2 − 78(13) + 144 ✓ substitusie T13 = −363 ✓ antwoord (4) [9] VRAAG 5 5.1 Die punte A( − 4 ; 0) en B(0 ; 8) lê op g. 8−0 ✓ vervang in mAB m AB = 0 − (−4) ✓ waarde van mAB m AB = 2 y = mx + c  8 = 2(0) + c  A(0;8) ✓ waarde van c c = 8  g ( x) = 2 x + 8 LET WEL: Enige ander geldige metode. (3) 5.2 f ( x) = a( x − x1 )( x − x2 ) f ( x) = a( x + 4)( x − 2) ✓ vervang punte A en C vervang (0;8) 8 = a (0 + 4)(0 − 2) 8 = a (−8) ✓ waarde van a a = −1 f ( x) = −( x + 4)( x − 2) f ( x) = −( x 2 + 2 x − 8) f ( x) = − x 2 − 2 x + 8 ✓ standaardvorm (A)  a = −1 ✓ waarde van b b = −2 ✓ waarde van c c =8 (5) 9 ### Page 10 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 1) GRAAD 11 5.3 f ( x) = − x 2 − 2 x + 8 − b − (−2) ✓ korrekte substitusie x = = 2a 2(−1)  x = −1  f (−1) = −(−1) 2 − 2(−1) + 8 ✓ korrekte substitusie  f (−1) = 9 LET WEL: Enige ander geldige metode. (2) 5.4 y9 ✓ antwoord (1) 5.5 mg = 2 1 ✓  mp = − mp 2 y − y1 = m( x − x1 ) 1 ✓ vervang punt C  y − 0 = − ( x − 2) 2 1  y = − x +1 2 ✓ antwoord (3) 5.6 f (−1) = 9 ✓ f (−1) = 9 en g (−1) = 6 g (−1) = 6  f (−1) − g (−1)  PQ = 9 − 6  PQ = 3 ✓ antwoord OF OF DP(−1;9) ✓ gebruik DP en g (−1) g (−1) = 6  PQ = 9 − 6 ✓ antwoord  PQ = 3 (2) 5.7 x  2 ; x  −4 ✓✓ antwoord (A) (2) [18] VRAAG 6 6.1 x = −3 ✓ antwoord y =1 ✓ antwoord (2) 10 ### Page 11 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (VRAESTEL 1) GRAAD 11 6.2 −4 f ( x) = +1 x+3 −4 0= +1 x+3 −4 −1 = x+3 − x − 3 = −4 − x = −1 x =1 ✓ x-afsnit −4 y= +1 0+3 −4 y= +1 3 1 ✓ y- afsnit y =− 3 (2) 6.3 y = −4 ✓ antwoord (1) 6.4 ✓ vorm van k ✓ k en h afsnitte ✓ asimptote k en h ✓ vorm van h (4) 6.5 p ( x) = 3h( x)  p( x) = 3(2 − x − 4)  p( x) = 3.2 − x − 12 asimptoot: y = −12 ✓ antwoord NOTA: Slegs antwoord, volpunte. (1) 6.6 q( x) = 2 x − 4 ✓ antwoord (1) 6.7 x=0 ✓ antwoord (1) [12] TOTAAL: 100 11 ## Related documents - [Gr 11 Mathematics P2 (Afrikaans) June 2024 Question Paper](https://www.hlayiso.com/library/gr-11-mathematics-p2-afrikaans-june-2024-question-paper/4934c5d3-e2a1-4d6b-acf6-d00e0623d46d) — Mathematics · QP · 2024 · Afr - [NW 11 MATHEMATICS P1 QP AFR NOV 2024](https://www.hlayiso.com/library/nw-11-mathematics-p1-qp-afr-nov-2024/f9574947-5433-4acd-b172-fba18e967389) — Mathematics · QP · 2024 · Afr - [Gr 11 Math P1 (Afrikaans) June 2023 Question Paper](https://www.hlayiso.com/library/gr-11-math-p1-afrikaans-june-2023-question-paper/3a6596d1-46af-4766-9815-0901dee27aaa) — Mathematics · QP · 2023 · Afr - [Gr11 Math P1 (Afrikaans) June 2019 Question Paper](https://www.hlayiso.com/library/gr11-math-p1-afrikaans-june-2019-question-paper/8a92fc2a-58f8-4c91-9d87-bc71603b08cd) — Mathematics · QP · 2019 · Afr - [MATHS P1 GR11 MEMO NOV 2020 English+Afrikaans](https://www.hlayiso.com/library/maths-p1-gr11-memo-nov-2020-english-afrikaans/89544f20-6bf0-4ca5-b1c6-bb6abeb0c3cd) — Mathematics · Memo · 2020 · Afr - [MATHS P1 GR11 MEMO NOV 2019 English+Afrikaans2](https://www.hlayiso.com/library/maths-p1-gr11-memo-nov-2019-english-afrikaans2/c5737a2c-1e54-41a6-8970-5bbb54f884f3) — Mathematics · Memo · 2019 · Afr ## More from Grade 11 Mathematics - [FS Maths Grade 11 June 2025 P1 only](https://www.hlayiso.com/library/fs-maths-grade-11-june-2025-p1-only/82015d8f-bcc0-478e-ad47-d3b295dabb01) — QP · 2025 - [FS Maths Grade 11 June 2025 P2 only](https://www.hlayiso.com/library/fs-maths-grade-11-june-2025-p2-only/5d0e7dd8-cb37-419c-8fe2-0aee27fbdc84) — QP · 2025 - [KZN Maths Grade 11 November 2025 P2 and Memo](https://www.hlayiso.com/library/kzn-maths-grade-11-november-2025-p2-and-memo/a2aac931-a969-41c8-bde2-85d23035d042) — QP + Memo · 2025 - [MATHS P2 ANSWER BOOK GR11 2024 ENG AFR D watermark](https://www.hlayiso.com/library/maths-p2-answer-book-gr11-2024-eng-afr-d-watermark/91fd4f6b-a252-4d30-b6b1-f8f546015336) — Addendum · 2024 · Afr - [NW 11 MATHEMATICS P2 AB ENGAFR NOV 2024](https://www.hlayiso.com/library/nw-11-mathematics-p2-ab-engafr-nov-2024/8ecc44b8-11a8-4044-8a89-cabd7fafe08c) — Addendum · 2024 - [ALTERNATIVE ANSWERS TO SOME QUESTIONS](https://www.hlayiso.com/library/alternative-answers-to-some-questions/589e5071-8eca-4868-96d4-89372a4d874b) — QP · 2023 ## Collection - [Grade 11 past papers](https://www.hlayiso.com/grade-11/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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