# Mathematics Grade 11 P1 Gauteng June 2022 Memo (Afrikaans) > Grade 11 Mathematics Paper 1 memorandum from the Gauteng June 2022 (Afrikaans), 12 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr11-math-p1-afr-june-2022-possible-answers/31eae199-9c0b-4a71-ab81-f769f84c7ccd - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Document type: Memorandum - Year: 2022 - Exam period: Gauteng June - Paper: 1 - Curriculum: CAPS - Pages: 12 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 2022 Memo (Afrikaans)](https://www.hlayiso.com/library/gr11-math-p1-afr-june-2022-possible-answers/31eae199-9c0b-4a71-ab81-f769f84c7ccd) ## About this document Mathematics · Grade 11 · Gauteng June · 2022. Memorandum, 12 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 PROVINSIALE EKSAMEN JUNIE 2022 GRAAD 11 NASIENRIGLYNE WISKUNDE VRAESTEL 1 12 bladsye ### Page 2 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (Vraestel 1) GRAAD 11 VRAAG 1 1.1 x  −3 ; x  −1  antwoord ✓ uitsluiting (2) 1.2.1 ( x + 4)( x − 1) = 0 x = −4 or x = 1 ✓ antwoorde (1) 1.2.2 3x 2 − 2 x = 14 3x 2 − 2 x − 14 = 0  standaard vorm −(−2)  (−2) 2 − 4(3)(−14)  korrek x= 2(3) vervanging 1  43 x= 3 x = −1,85 or x = 2,52  antwoord  antwoord NOTE: Penaliseer 1 punt vir afronding slegs en hierdie (4) VRAAG. Enige geldige metode. 1.2.3 45 x 2 − 2 x = 18 − 2 x − 2x beperkingn  x 2 − 2 x  0 restrictio  metode let : k = x 2 − 2 x stel: 45  k = 18 − k  k = 18k − 45 2  k 2 − 18k + 45 = 0  (k − 15)( k − 3) = 0  waardes van k  k = 15 ofor  k = 3 – x2x− 2 2 x2 substitute vervang x in terug k.  int o  k . back  x 2 − 2 x = 15 or of  x − 2 x = 3 2  gelykstelling  x 2 − 2 x − 15 = 0 or of  x − 2 x − 3 = 0 2  ( x − 5)( x + 3) = 0 or of  ( x − 3)( x + 1) = 0  x = 5 or  x = −3 or of  x = 3 or  x = −1  waardes van x (4) NOTA: Enige ander geldige metode. 2 ### Page 3 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (Vraestel 1) GRAAD 11 1.2.4 5 − x +1 = −x 5 − x = −x −1 5 − x = x2 + 2 x + 1  metode 0 = x + 3x − 4 2  standaardvorm 0 = ( x + 4)( x − 1)  faktore  x = −4 or x  1  kritieke waardes NOTA: Enige ander geldige metode.  verwerping (5) 1.2.5 32 x +1 − 4.3x = −1 3.32 x − 4.3x + 1 = 0  standaardvorm (3.3 − 1)(3 − 1) = 0 x x  faktore 1 3x = or 3x = 1 of  gelykstelling 3 3x = 3−1 ofor 3x = 30 x = −1 of x = 0  antwoorde OF 32 x +1 − 4.3x = −1 let 3x = k  3k 2 − 4k = −1 3k 2 − 4k + 1 = 0  standaardvorm (3k − 1)(k − 1) = 0 1  faktore k = of or k = 1 3  gelykstelling 3x = 3−1 ofor 3x = 30  antwoorde x = −1 of x=0 (4) 3 ### Page 4 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (Vraestel 1) GRAAD 11 1.2.6(a) x 2 − 3x  0 x( x − 3)  0  faktore CVs 0 ; 3 0 x3  antwoord (2) 1.2.6(b) 3 x  6 ✓ kritieke waardes ✓ antwoord NOTA: Antwoord kan as aparte ongelykhede geskryf word. (2) 1.3 x + y = −10 xy = −600...(2)  opstel van x = −10 − y...(1) vergelykings sub (1) in (2) (−10 − y ) y = −600  substitusie −10 y − y 2 = − 600 y 2 + 10 y − 600 = 0  faktore ( y − 20)( y + 30) = 600  waardes van y y = 20 of of y = −30 as y = 20 as y = −30  waardes van x x = −30 x = 20 NOTA: Enige ander geldige metode. (5) [29] 4 ### Page 5 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (Vraestel 1) GRAAD 11 VRAAG 2 2.1.1 2 8 − 4 32 + 3 50 Punt is vir = 2 4.2 − 4 16.2 + 3 25.2 die eerste  vereenvoudig wortels = 2.2 2 − 4.4 2 + 3.5 2 lyn.  vereenvoudiging = 4 2 − 16 2 + 15 2  antwoord =3 2 (3) 1 2.1.2 − 3 2 [ 12 + 3 (3 3) ] 1 1 1 1 − 3 [( 2 .3) + (3.3 ) ] 2 2 2 2 3  vereenvoudig wortels 1 1 3 1 − 3 [21.3 + (3 ) ] 2 2 2 2 1 1 1 − 3 [2.3 + 3 ] 2 2 2 2.30 + 30 ✓ vereenvoudiging 2 +1 ✓ vereenvoudiging 3 (4)  antwoord 2.1.3 52006 − 52004 + 24 52004 + 1 52004(52 − 1) + 24 52004 + 1  faktoriseer 52004(24) + 24  vereenvoudig 52004 + 1 24(52004 + 1)  faktoriseer 52004 + 1  antwoord 24 (4) 2.2.1 x = 6 + 6 + 6 + .........  metode  x 2 = 6 + 6 + 6 + ...  x2 = 6 + x  antwoord (2) 2.2.2 x2 − x − 6 = 0 ✓ standaardvorm ( x − 3)( x + 2) = 0 ✓ faktore x = 3 x  −2 ✓ antwoord ✓ uitsluiting (4) 5 ### Page 6 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (Vraestel 1) GRAAD 11 2.3 d 2 = ( 8 − 1)2 + ( 8 + 1)2 Pythag  metode d 2 = 8 − 2 8 +1+ 8 + 2 8 +1 d 2 = 18  waarde van d 2 d =3 2  antwoord (3) [20] 6 ### Page 7 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (Vraestel 1) GRAAD 11 VRAAG 3 3.1.1 -7 0 9 20 7 9 11 2 2 ✓ tweede verskil e 2a = 2 a = 1  waarde van a 3(1) + b = 7 b = 4  waarde van b 1 + 4 + c = −7  c = −12  waarde van c (4) Tn = n + 4n − 12 2 3.1.2 Tn = n 2 + 4n − 12 128 = n 2 + 4n − 12 ✓ gelykstelling 0 = n 2 + 4n − 140 0 = (n − 10)(n + 14)  faktore  n = 10 n  −14  keuse van n = 10 (3) 3.1.3 Tn = 2n + 5 599 = 2n + 5 ✓ gelykstelling 594 = 2n n = 297  waarde van n Die verskil sal tussen terme 297 en 298  gevolgtrekking wees. (3) 3.2.1 93  antwoord (1) 3.2. 2 Tn = 10n − 3 ✓ Tn = 10n − 3 Tn = (10n − 2) − 1 ✓ metode Tn = 2(5n − 1) − 1 Thus : (10n − 2) = 2(5n − 1)  gelykstelling 10n − 2 EWE vir EVEN for  enige anyvan waarde n. of  n.  verduideliking value Tn is is  ODD ONEWE when : indien: Tn = (10n − 2) − 1 (4) [15] 7 ### Page 8 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (Vraestel 1) GRAAD 11 VRAAG 4 4.1 1 − p;2 p − 3; p + 5 T2 − T1 = T3 − T2 ✓ metode  (2 p − 3) − (1 − p) = ( p + 5) − (2 p − 3)  2 p − 3 −1+ p = p + 5 − 2 p + 3 3 p − 4 = − p + 8 ✓ vereenvoudiging  4 p = 12 ✓ antwoord p=3 (3) 4.2.1 T1 = 1 − p T1 = 1 − 3 ✓ antwoord T1 = −2 (1) NOTA: Slegs antwoord: volpunte 4.2.2 T2 = 2 p − 3 T2 = 2(3) − 3 T2 = 3 d = T2 − T1 d = 3 − (−2) ✓ antwoorde d =5 (1) NOTA: Slegs antwoord: volpunte 4.3 1- p ; 2p - 3 ; p + 5 = 1 - 3 ; 2(3) - 3 ; 3 + 5 = -2 ; 3 ; 8 ; 13 ; 18; … Al die terme eindig op ʼn 3 of ʼn 8, behalwe  verduideliking T1 terwyl volkome vierkante eindig op (2) 0; 1; 4; 5; 6; 9. [7] 8 ### Page 9 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (Vraestel 1) GRAAD 11 VRAAG 5 5.1 −3 x  2 ✓ antwoord NOTA: Kandidate kan die antwoord as aparte ongelykhede skryf. (1) 5.2 f ( x) = a( x − x1 )( x − x2 ) f ( x) = a( x + 3)( x − 2) − 8 = a(1 + 3)(1 − 2) (1;−8) ✓ vervang punt D en − 8 = a(4)(−1) x-afsnitte a = 2 ✓ waarde van a f ( x) = 2( x + 3)( x − 2) f ( x) = 2( x 2 + x − 6) ✓ vereenvoudiging van tweeterm f ( x) = 2 x 2 + 2 x − 12  a = 2 b = 2 c = −12 ✓ waardes van a, b, c (4) 5.3 b x=− 2a 2 x=− 2(2) 2 x=− 4 1 x=− ✓ waarde van x 2 1 1 1  f (− ) = 2( ) 2 + 2( ) − 12 2 2 2 ✓ vervang x in f(x) 1 1 f (− ) = − 1 − 12 2 2 1 1 f (− ) = −12 ✓ waarde van y 2 2 1 1  draaipunt(− ;−12 ) 2 2 NOTA: Antwoord hoef nie in (3) koördinaatvorm te wees word nie. 9 ### Page 10 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (Vraestel 1) GRAAD 11 1 5.4 x=− +7 2 1 x=6 2 ✓ antwoord NOTA: Slegs antwoord: volpunte. (1) 5.5 In g m = 6 1  In q m = − ✓ gradiënt van q 6  y = mx + c 1 0 = − ( −3) + c 6 1 0= +c 2 1 c=− ✓ waarde van c 2 1 1 ✓ antwoord y =− x− 6 2 (3) NOTE: Answer only: FULL marks [12] 10 ### Page 11 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (Vraestel 1) GRAAD 11 VRAAG 6 6.1 ✓ vorm van p met y-afsnit = 1 ✓ vorm van h ✓ x- en y-afsnitte van h ✓ draaipunt van h (4) 6.2 Sien skets in Vr. 6.1. ✓ vorm van f(x) met y-afsnit = 1 (1) 6.3 y = 2( x − 1 + 2) 2 − 8 y = 2( x + 1) 2 − 8 ✓ antwoord (1) 6.4 p( x) = 4 x 1 1 ( x+ )  p( x + ) = 4 2 2 1 1 ✓ uitbreiding p( x + ) = 4 x.4 2 2 1 1 p( x + ) = 4 .(2 ) x 2 2 ✓ vereenvoudiging 2 1 ✓ vereenvoudiging p( x + ) = 4 x.2 2 1  p ( x + ) = 2 p( x) 2 (3) [9] 11 ### Page 12 Downloaded from hlayiso.com WISKUNDE NASIENRIGLYNE (Vraestel 1) GRAAD 11 VRAAG 7 7.1 p=4  waarde van p  waarde van q q=2 a  f ( x) = +2 x−4 a  vervang punt (5 ; 3) 3= + 2 T (5;3) 5−4 3=a+2 a = 3− 2 a =1  antwoord (4) 7.2 1 ✓ antwoord h( x) = − +2 x−4 (1) 7.3 𝑦∈ℝ; y2  antwoord NOTA: Albei voorwaardes moet gestel word. (1) 7.4 Simmetrie-asse van f is: y =x-p+q en y=-x-p+q y =x-4+2 y=-x+4+2 ✓ metode y =x-2 y=-x+6 Kies: y = - x + 6 ✓ keuse c=6 (2) [8] TOTAAL: 100 12 ## Related documents - [Gr11 MATH P2 (ENG) June 2022 Possible Answers](https://www.hlayiso.com/library/gr11-math-p2-eng-june-2022-possible-answers/1a429955-227a-4c34-aec9-9e119d0c4bc5) — Mathematics · Memo · 2022 · Eng - [LImp Maths Gr11 June 2022 P1 and Memo](https://www.hlayiso.com/library/limp-maths-gr11-june-2022-p1-and-memo/5b6991e9-7f6f-4156-90d2-cd1c745a1d53) — Mathematics · Memo · 2022 - [Gr11 Math P1 June 2017 Possible Answers](https://www.hlayiso.com/library/gr11-math-p1-june-2017-possible-answers/19ca7dfb-d358-4c42-bbfc-c68d580cfe21) — Mathematics · Memo · 2017 - [Gr11 Math P1 (Afrikaans) June 2017 Possible Answers](https://www.hlayiso.com/library/gr11-math-p1-afrikaans-june-2017-possible-answers/b8504619-0cb6-476d-944d-f53d2ea22f72) — Mathematics · Memo · 2017 · Afr - [Gr11 Math P1 June 2016 Possible Answer](https://www.hlayiso.com/library/gr11-math-p1-june-2016-possible-answer/8abf9959-f0c5-4918-94fd-aa3ebabec0f5) — Mathematics · Memo · 2016 - [Gr11 MATH P2 (AFR) June 2022 Question Paper](https://www.hlayiso.com/library/gr11-math-p2-afr-june-2022-question-paper/1fcd90d5-8dba-4271-a9ed-137098dfc983) — Mathematics · QP · 2022 · 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 - [MATHS P1 QP GR11 NOV2024 ENG D watermark](https://www.hlayiso.com/library/maths-p1-qp-gr11-nov2024-eng-d-watermark/739068f1-b085-4977-8a45-18f9bb927644) — QP · 2024 · Eng - [Gr 11 Mathematics P1 (Afrikaans) June 2024 Question Paper](https://www.hlayiso.com/library/gr-11-mathematics-p1-afrikaans-june-2024-question-paper/1a0f4a9d-f18a-4aca-af77-5a81acfb8e0c) — QP · 2024 · Afr ## 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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