# Mathematics Grade 12 Paper 1 EC June 2019 Memo > Grade 12 Mathematics Paper 1 memorandum from the EC June 2019, 16 pages. Read online or get the free PDF download, question paper linked when available. - Page: https://www.hlayiso.com/library/maths-p1-memo-june-2019-ea/95a7c045-8037-4716-aa35-076f0ab5cfb9 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2019 - Exam period: EC June - Paper: 1 - Curriculum: CAPS - Pages: 16 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 EC June 2019 Memo](https://www.hlayiso.com/library/maths-p1-memo-june-2019-ea/95a7c045-8037-4716-aa35-076f0ab5cfb9) ## About this document Mathematics · Grade 12 · EC June · 2019. Memorandum, 16 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 NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2019 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 14 pages./ Hierdie nasienriglyn bestaan uit 14 bladsye. ### Page 2 Downloaded from hlayiso.com 2 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE/JUNIE 2019) NOTE/LET WEL:  If a candidate answered a question TWICE, mark the FIRST attempt ONLY. Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord het, merk SLEGS die EERSTE poging.  Consistent accuracy(CA) applies in ALL aspects of the marking guideline. Volgehoue akkuraatheid is op ALLE aspekte van die nasienriglyn van toepassing.  If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt. Indien ʼn kandidaat ʼn poging vir ʼn vraag deurgetrek het en nie die vraag weer beantwoord het nie, merk die poging wat deurgetrek is.  The mark for substitution is awarded for substitution into the correct formula. Die punt vir substitusie word toegeken vir substitusie in die korrekte formule. QUESTION 1/VRAAG 1 1.1.1 x2  7 x  0  both factors / beide faktore x( x  7)  0 x  0 or x  7  0  both -values / beide x-waardes x  7 (2) 1.1.2 Penalise 1 mark for 5 10 3 0 incorrect rounding off./ Penaliseer 1 punt vir b  b 2  4ac verkeerde afronding. x 2a (10)  (10) 2  4(3)(5) x  substitution / vervanging 2(3) 10  160 x 6  x  0, 44 or / of x  3, 77  -values / waardes (3) 1.1.3 (2 x  1)(4  x)  0 (2 x  1)( x  4)  0  critical values / kritieke waardes + – + ½ 4  answer (accuracy) / ½ 4 antwoord (akkuraatheid) 1 x4 2 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com (EC/JUNE/JUNIE 2019) MATHEMATICS P1/WISKUNDE V1 3 1.1.4 2 x3  x3  2 x  3    x  3 2 2  squaring both sides / kwadreer beide kante 4  x  3  x 2  6 x  9  multiplication / vermenigvuldiging 0  x 2  6 x  4 x  9  12 0  x 2  10 x  21  standard form / standaardvorm 0   x  3 x  7   factors / faktore  x  3 or x  7  answers / antwoorde (5) 1.2 y  3x  2.................(1) x 2  y  xy  x............(2)  substitution / vervanging (1) in (2) : x 2  (3x  2)  x(3 x  2)  x x 2  3 x  2  3x 2  2 x  x 4x2  6x  2  0  standard form / standaardvorm 2 x  3x  1  0 2  2 x  1 x  1  0  factors / faktore x  ½ or / of x  1  -values / waardes y  ½ or / of y  1  y-values / waardes (5) 1.3.1 3 45  2 80 x 125 3 9  5  2 16  5 x  factors under root / faktore onder wortel 25  5 9 5 8 5 x  simplification (this step must be seen) / 5 5 vereenvoudiging (hierdie stap moet gesien word) 1 x 5  answer / antwoord (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com 4 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE/JUNIE 2019) 1.3.2 5 a2  ..... 1 b3 a5 7 ....  2  b2 a5 From  2  : b 2  7 1  a5  2  equation (3) / vergelyking (3) b  .....  3   7  3  5 12  2  a    3 into 1 : a     5  substitution / vervanging 7    15 a 2 a2  3 5 7 2  simplification / vereenvoudiging 15 3 a a 2 2  5 7 2 19 3  simplification / vereenvoudiging a 2  5 7 2 2   3 19  isolating a / isolering van a a  5 72    a  19 25  343 (5) [26] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com (EC/JUNE/JUNIE 2019) MATHEMATICS P1/WISKUNDE V1 5 QUESTION 2/VRAAG 2 2.1.1 12 ; 9 ; 6 ; . . . . . a = 12 and/en d = –3  d = –3 Tn  12  (n  1)(3)  answer / antwoord  3n  15 (2) 2.1.2 Tn  3  40   15  substitution / vervanging  105  answer / antwoord (2) 2.1.3 40 S 40  12  (105)   substitution / vervanging 2  1860  answer / antwoord (2) 2.2.1 Quadratic Pattern / Kwadratiese Patroon :T1  10 1st difference pattern /1ste verskille patroon :  first differences / eerste verskille  7;  5;  3;...   answers / antwoorde (T1 and/en T2 ) T2  3 and / en T3  2 (3) 2.2.2 2a  2 3a  b  7 a  b  c  10  value of a / waarde van a  value of b / waarde van b a 1 3(1)  b  7 (1)  (10)  c  10  value of c / waarde van c b  10 c  19  Tn  n 2  10n  19 (can be awarded at formula / kan by formule toegeken word) (3) 2.2.3 n  10n  19  2019 2  equation / vergelyking n 2  10n  2000  0  n  50  n  40   0  factors / faktore n  50 or / of n  40  T50  2019  answer / antwoord (T50) (3) 2.3.1 S n  81  81 3 n T1  S1  81  81 3 1  answer / antwoord  54 (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com 6 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE/JUNIE 2019) 2.3.2 T2  S 2  S1  72  54  T2  18  18 T 18 1 1 r 2    r T1 54 3 3 Tn  a.r n 1 n 1  answer / antwoord 1  54   3 (3) 2.3.3 Yes/Ja.  YES / JA 1 r  1   1 ; r  0  reason / rede 3 (2) 2.3.4 a S  or / of S  81  81(3)  1 r   substitution / vervanging 54 1   81  81  1  13 3  answer / antwoord  81  81 (2) 2.4 3 12   2 x  3t     3(2)r 1   0 t 1 r 7 192  26  1  (2x + 3 + 2x + 6 + 2x + 9)  2x  3  2x  6  2x  9  0  sum of GS / som van MR 2 1 6 x  18  12 096  0  simplification / vereenvoudiging 6 x  12114  answer / antwoord x   2019 (4) [27] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com (EC/JUNE/JUNIE 2019) MATHEMATICS P1/WISKUNDE V1 7 QUESTION 3/VRAAG 3 3.1 (let / laat y  0) / (let / laat x  0)  substitution / vervanging (y = 0) 2 2 1  0 y 1 x 1 0 1  substitution / vervanging (x = 0) 2  1 y  2  1 x 1 2  x 1 y  1  coordinates of A / koördinate van A x  1 (1;0) (0;  1)  coordinates of B / koördinate van B (4) 3.2 x 1  answer / antwoord (1) 3.3 ∈ \1 OR/OF ∈ but y 1  answer / antwoord (1) 3.4  A  2  1;  2  1 and / en B 2  1; 2  1  coordinates of A and B AB   2  1  2  1   2  1  2  1 2 2 2 koördinate van A en B  substitution / vervanging  2 2   2 2  2 2  simplification / vereenvoudiging  16  answer / antwoord  AB  4 units / eenhede (4) 3.5 2 h( x )   1  – ve (reflection / refleksie)  x  5  (x – 5) (shift / skuif) (2) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com 8 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE/JUNIE 2019) QUESTION 4/VRAAG 4 4.1 y Parabola / Parabool  x-intercepts / x-afsnitte g  Turning point / Draaipunt (-½ ; ½)  Axis of symmetry / Simmetrie-as -1 O -½ 1 x  Shape / Vorm Straight line / Reguitlyn  x-intercept / x-afsnit f -3  y-intercept / y-afsnit (6) 4.2 From sketch / Vanaf skets 1  x  0 or / of x   1;0   answer / antwoord Algebraic solution / Algebraïese oplossing 2 x 2  2 x  0 2 x( x  1)  0 2 x( x  1)  0 -1 0 cv / kw : 1  x  0 (2) 4.3 ST  2 x 2  2 x  (3 x  3)  2 x 2  5 x  3  ST = –2x2 – 2x – (3x – 3) ST ( x)  4 x  5  0  ST’ = 0  4x  5 5 x= x 4 2  5  5 Max / Maks   2     5     3  4  4 49  1   / 6  answer / antwoord 8  8 (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com (EC/JUNE/JUNIE 2019) MATHEMATICS P1/WISKUNDE V1 9 QUESTION 5/VRAAG 5 5.1 f ( x)  2 x k  23  substitution / vervanging 1 k  answer / antwoord 8 (2) 5.2 f 1  x  : x  2 y  interchanging x and y omruil van x en y y  log 2 x  answer / antwoord g 1 ( x) : x  2 y  1  interchanging x and y x 1  2 y omruil van x en y y  12 x  12  answer / antwoord (4) 5.3 f ( x).g ( x)  0 f ( x) is always  ve  method / metode g ( x) is  ve for x  ½  ve for x  ½  answer / antwoord  x  ½ (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com 10 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE/JUNIE 2019) QUESTION 6/VRAAG 6 6.1 A  P 1  i  n 5  6,5   substitution / vervanging  150 000 1    100   answer / antwoord (2)  R 205513.00 6.2 A  P (1  i ) n 134 000  975000(1  i )7 134  substitution / vervanging (1  i )7  975  simplification / vereenvoudiging 1  i  7 134 975 i  0, 2468673864  answer / antwoord (3) rate / koers  24, 69% 6.3.1 n  i  1  ieff  1  nom   formula / formule  n  12  substitution / vervanging  6,5%  ieff  1   1  12  ieff  0,066971852  answer / antwoord (3) effectiverate / effektiewekoers  6,70% p.a. 6.3.2 60 20 60 20  6,5   7,5   6,5   7,5  x 1   1  2 000 000  1    1   1200   400   1200   400   equation / vergelyking (= 2 000 000) 2 000 000 x 60 20  6,5   7,5   making x subject of the formula / 1   1  1200   400  maak x die onderwerp van die formule x  R 997 500, 00  answer / antwoord (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com (EC/JUNE/JUNIE 2019) MATHEMATICS P1/WISKUNDE V1 11 QUESTION 7/VRAAG 7 Penalise 1 mark for incorrect notation in the question Penaliseer 1 punt vir verkeerde notasie in die vraag 7.1 5 5 5 5 5 2 5 5 5 10 5 5 5 5 10 5 5 5 ′ lim → 5 10 5 5 5 5 5  substitution / vervanging lim →  simplification / vereenvoudiging 10 5 5 lim → 10 5 5  common factor / gemene faktor lim → lim 10 5 5 →  answer / antwoord Answer ONLY: 0 marks (5) 10 5 SLEGS antwoord: 0 punte 7.2 x 1 y  2 x3 1 12 1 12 y  x  x 3  y  x  x 3 2 2 dy 1  12   x  3 x 4 dx 4 1  12  x  3x 4 4 (3) 7.3 f (1)  (1)3  1  f (1) and / en f (1) f (1)  (1)3  1 f ( x2 )  f ( x1 ) m x2  x1 1  (1)   substitution / vervanging 1  (1) 2  answer / antwoord  /1 2 (3) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com 12 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE/JUNIE 2019) QUESTION 8/VRAAG 8 8.1.1 f ( x)  (2 x  5)( x  2) 2  0 2 x  5  0 or / of x  2  0 5  x-intercepts / x-afsnitte  x or / of x  2 2  answer / antwoord AB  4,5 units / eenhede (2) 8.1.2 2 3 12 20 6 6 12 0  2 0 2 1 0  factors / faktore 2 0 or/ 1 0 2 or / 1  correct x-value for T / korrekte x-waarde vir T 1 ; (3) 8.1.3 m p  6(3) 2  6(3)  12  24  gradient / gradiënt y  y1  m( x  x1 )  substitution / vervanging y  11  24( x  3) y  24 x  61  answer / antwoord (3) 8.1.4 T(1 ; y)  max. value / maks. 2 1 3 1 12 1 20 waarde = 27  answer / antwoord ∴0 27 (accuracy/akkuraatheid) (3) 8.2.1 3 c( x)   x 2  6 x  0  equating to 0 / gelykstel 2 aan 0 x   32 x  6   0 x  0 or / of x  4  factors / faktore  0 x4  answer / antwoord (accuracy/akkuraatheid) (3) 8.2.2 04 c( x)  3 x  6  0 or / of x  method / metode 2  x-value / x-waarde x2 x2  c( x) is concave up for x  2 / c( x) is konkaaf opwaarts vir x  2  c( x) is concave down for x  2 / c( x) is konkaaf afwaarts vir x  2 conclusion/gevolgtrekking (4) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com (EC/JUNE/JUNIE 2019) MATHEMATICS P1/WISKUNDE V1 13 QUESTION 9/VRAAG 9 9.1 12  4 x h  answer / antwoord 3 (2) 9.2 Area  l  b  12  4 x   x   substitution / vervanging  3  4x2  4x   answer / antwoord 3 8 A( x )  4  x0 3  derivative / afgeleide 8  f ( x)  0  x  4 3 x  1, 5 m  answer / antwoord 12  4(1, 5) h  2m 3  answer / antwoord (6) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com 14 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE/JUNIE 2019) QUESTION 10/VRAAG 10 10.1.1 P(S) = 1 A B  0,3 & 0,25 0,3 0,25 0,35  0,35 & 0,25  0,1 0,1 (3) 10.1.2 P( A or/of B)  P( A)  P( B)  P( A and/en B)  method / metode  0,55  0, 6  0, 25  0,9  answer / antwoord or / of ( from sketch / vanaf skets) P( A or/of B)  0,3  0, 25  0,35  0,9 (2) 10.1.3 P  A and/en B   0,3  answer / antwoord (2) 10.1.4 No/Nee : P( A  B)  0, 25  0  answer / antwoord (1) 10.1.5 P ( A  B )  0,1  0  answer / antwoord No/Nee : or / of P ( A  B )  0,9  1 (1) 10.2.1 a  20  a = 20  b = (40 – x) (2) b  (40  x) 10.2.2 79  x  20  19  x  x  11  16  40  x  173  equation / vergelyking 2 x  173  185 2 x  12 x6  answer / antwoord (2) 10.2.3 P (at least 2 / ten minste 2)  adding correct values / 20  11  16  6  tel korrekte waardes op 173 53  answer / antwoord  / 0,31/ 30, 6% (2) 173 [15] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com    ### Page 16 Downloaded from hlayiso.com  ## Related documents - [Mathematics P1 Nov 2019 Memo Afr Eng](https://www.hlayiso.com/library/mathematics-p1-nov-2019-memo-afr-eng/eef87169-da21-443f-af20-6207010a2376) — Mathematics · Memo · 2019 · Eng - [Gr 12 Gauteng Sep 2019 P1](https://www.hlayiso.com/library/gr-12-gauteng-sep-2019-p1/a2deda82-cc1b-4103-84d4-6d1ce61c2914) — Mathematics · QP · 2019 - [2019 GRADE 12 MATH TRIAL EXAM PAPER 1 LIMP](https://www.hlayiso.com/library/2019-grade-12-math-trial-exam-paper-1-limp/831efe5b-dbdc-4b34-9f4d-0c7aadcfdef4) — Mathematics · QP · 2019 - [Mathematics P1 May June 2019 Eng](https://www.hlayiso.com/library/mathematics-p1-may-june-2019-eng/a645dde5-42e0-4a8d-85e6-9027775ee0b4) — Mathematics · QP · 2019 · Eng - [Maths P1 June 2016 Qp Eng Final2](https://www.hlayiso.com/library/maths-p1-june-2016-qp-eng-final2/bb0ceb33-a40b-46f9-a052-3c3aef3d7536) — Mathematics · QP · 2016 · Eng - [Mathematics P2 Nov 2019 Memo Afr Eng](https://www.hlayiso.com/library/mathematics-p2-nov-2019-memo-afr-eng/c35b9be5-727d-4eb3-acf4-8a0a42fcd3a1) — Mathematics · Memo · 2019 · Eng ## 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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