# Mathematics Grade 12 P2 Eastern Cape June 2021 Memo (Afr) > Grade 12 Mathematics Paper 2 memorandum from the Eastern Cape June 2021 (Afrikaans), 17 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/maths-p2-gr12-memo-june-2021-afr-engl/05f73b65-4b39-4fd4-91f8-a4661e8f632a - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2021 - Exam period: Eastern Cape June - Paper: 2 - Curriculum: CAPS - Pages: 17 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 P2 Eastern Cape June 2021 Memo (Afr)](https://www.hlayiso.com/library/maths-p2-gr12-memo-june-2021-afr-engl/05f73b65-4b39-4fd4-91f8-a4661e8f632a) ## About this document Mathematics · Grade 12 · Eastern Cape June · 2021. Memorandum, 17 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 NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2021 MATHEMATICS P2/WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN (EXEMPLAR/EKSEMPLAAR) MARKS/PUNTE: 150 This marking guideline consists of 17 pages. Hierdie nasienriglyn bestaan uit 17 bladsye. ### Page 2 2 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/2021) QUESTION 1/VRAAG 1 1.1 a = 62 ✓ value of a / waarde van a b = 51 ✓ value of b / waarde van b c = 42 ✓ value of c / waarde van c d = 35 ✓ value of d / waarde van d e = 31 ✓ value of e / waarde van e (5) 1.2 Skewed to the right / Skeef na regs ✓ answer / antwoord OR/OF Positively skewed / Positief skeef (1) 1.3 Yes / Ja ✓ Yes / Ja Q3 = 51 and the upper 25% is from Q3 above. ✓Reason / Rede Q3 = 51 en die boonste 25% is vanaf Q3 en op. (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 (EC/JUNE 2021) MATHEMATICS P2/WISKUNDE V2 3 QUESTION 2/VRAAG 2 2.1 Positive impact / Positiewe impak ✓ Positive impact Positiewe impak The number of learners obtaining lower marks decreased while ✓ Reason / Rede those obtaining higher marks increased in the Post Test. Die aantal leerders wat laer punte behaal het, het verminder terwyl die wat hoër punte behaal het in die na-toets vermeerder het. (2) 2.2 20  x  30 ✓ answer / antwoord (1) 2.3 Less / Minder ✓answer / antwoord (1) 2.4 Frequency / Frekwensie Frequency Cumulative Frequency ✓ Pre-Test / Voor Toets Marks Frekwensie Kumulatiewe Frekwensie ✓ Post Test / Na Toets Punte Pre-Test Post Test Pre-Test Post Test Voor Toets Na Toets Voor Toets Na Toets Cumulative Frequency 0 ≤ x < 10 16 5 16 5 Kumulatiewe Frekwensie 10 ≤ x < 20 28 18 44 23 ✓ Pre-Test / Voor Toets 20 ≤ x < 30 34 25 78 48 ✓ Post Test / Na Toets 30 ≤ x < 40 12 32 90 80 40 ≤ x < 50 0 10 90 90 (4) 2.5 Ogives for Pre and Post Tests Ogiewe vir Voor en Na Toetse 100 90 Cumulative Frequency ✓ grounding / anker 80 70 ✓ upper limits used 60 boonste limiete 50 Kumulatiewe Frekwensie 40 ✓ shape / vorm 30 20 10 0 0 10 20 30 40 50 60 Marks / Punte Pre-test Post-test Voor toets Na toets (3) 2.6 Pre: 90 – 78 = 12 learners obtained 60% and more ✓ 12 Post: 90 – 48 = 42 learners obtained 60% and more ✓ 42 Therefore, the teacher achieved the target. ✓ conclusion Voor - toets: 90 – 78 = 12 leerders het 60% en meer behaal gevolgtrekking Na - toets: 90 – 48 = 42 leerders het 60% en meer behaal Daarom het die onderwyser die doelwit behaal (3) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 4 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/2021) QUESTION 3/VRAAG 3 3.1 t=2 ✓ value of t / waarde van t (1) 3.2.1 PR = ( x2 − x1 ) + ( y2 − y1 ) 2 2 = (−5 − 2 ) 2 + ( 0 − 6 ) 2 ✓substitution / vervanging ✓answer / antwoord = 85 (2) 3.2.2 y2 − y1 0−6 mPR = = ✓substitution / vervanging x2 − x1 −5−2 6 ✓gradient of PR = 7 gradiënt van PR (2) 3.3  6 tan RPS = 7 6 ✓ tan  =  7 RPS = 40, 6   ✓ value of / waarde van PRQ 40, 6 + PRQ = 90 (ext.  of  / buite  van ) ✓ method / metode  ✓ reason / rede  PRQ = 49, 4  ✓value of / waarde van RPS (5) 3.4 6 mPR = 7 −4 − 0 −4 ✓ substitution / vervanging mPQ = = ✓ gradient of PR / gradiënt van PR 2 − (−5) 7 6  4 24 ✓ method / metode mPR  mPQ =   −  = −  − 1 7  7 49 ✓ conclusion / gevolgtrekking  PRQ is not right angled at R  PRQ is nie reghoekig by R nie OR / OF OR / OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 (EC/JUNE 2021) MATHEMATICS P2/WISKUNDE V2 5 PQ = (−5 − 2) 2 + (0 − (−4)) 2 ✓ substitution / vervanging = 65 RQ 2 =100 ✓ squares / vierkante PR 2 = 85 PQ 2 = 65 RQ 2  PR 2 + PQ 2 ✓ RQ2  RP 2 + PQ2  PQR is not right angled at P ✓ conclusion / gevolgtrekking  PRQ is nie reghoekig by R nie (4) 3.5 4 mnewline = mPQ = − ✓ gradient of new line 7 4 gradiënt van nuwe lyn y − 0 = − ( x − 0) 7 ✓ substitution / vervanging 4 y = − x 7 ✓equation / vergelyking (3) 3.6 SP = 7 units / eenhede ✓ length of SP / lengte van SP SR = 6 units / eenhede RQ = 10 units / eenhede ✓ length of RQ / lengte van RQ Area of SPR / Oppervlakte van SPR 1 ✓ Area of /Oppervlakte van SPR = 7 6 2 = 21 units 2 /eenhede 2 Area of RPQ / Oppervlakte van RPQ 1 ✓ Area of /Oppervlakte vanRPQ =  7  10 2 = 35 units 2 / eenhede2 Area of SPR Oppervlakte van SPR / Area of PRQ Oppervlakte van PRQ 21 = 35 3 = ✓answer/antwoord 5 (5) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 6 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/2021) QUESTION 4/VRAAG 4 4.1.1 x2 + y 2 + 6x − 6 y + 9 = 0 ✓ method / metode x2 + 6x + 9 + y 2 − 6 y + 9 = − 9 + 9 + 9 ✓ ( x + 3)2 + ( y − 3) 2 = 9  ( x + 3) 2 + ( y − 3) 2 = 9 ✓ coordinates of L L ( −3;3) and/en r = 3units/eenhede koördinate van L ✓ value of r / waarde van r (4) 4.1.2 − 5 + xM 5 + yM −3 = 3= ✓ method / metode 2 2 ✓ value of x / waarde van x xM = −1 yM = 1 ✓ value of y / waarde van y  M ( −1;1) (3) 4.1.3 5 −1 3 −1 ✓ mLM OR mKM mKL = = − 1 OR mLM = = −1 −5 + 1 −3 + 1 ✓ mtangent mtangent = 1 y − 1 =1( x − (−1) ) ✓ substitution / vervanging  y=x+2 ✓ equation / vergelyking (4) 4.1.4 ( x + 3) + ( y − 3) = 9 2 2 ( 0 + 3) 2 + ( y − 3) 2 = 9 ( y − 3) 2 = 0 y =3 ✓ value of x / waarde van x ✓ value of y / waarde van y N ( 0;3) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 (EC/JUNE 2021) MATHEMATICS P2/WISKUNDE V2 7 4.2.1 L ( – 3 ; 3) ✓ value of x / waarde van x L/ (2 ; –4) ✓ value of y / waarde van y (2) 4.2.2 − 4 −1 5 ✓ mML/ mML/ = =− 2 +1 3 5 y − (1) = − ( x + 1) ✓substitution / vervanging 3 5 2  y=− x− ✓equation / vergelyking 3 3 Not passing through the origin ✓ conclusion / gevolgtrekking Gaan nie deur die oorsprong nie (4) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 8 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/2021) QUESTION/VRAAG 5 5.1.1 5 3 sin  = − tan  = − 13 and/en 4 ✓ value of x / waarde van x x = −12 r =5 ✓ value of r / waarde van r sin( +  ) = sin  cos  + cos  sin  ✓ expansion / uitbreiding  −5  −4   −12  3  =    +    ✓ substitution / vervanging  13  5   13  5  16 =− 65 ✓ answer / antwoord (5) 5.1.2 cos 2 =1 − 2sin  2 2 ✓ expansion / uitbreiding 3 ✓ substitution / vervanging = 1− 2   5 7 = ✓ answer / antwoord 25 (3) 5.1.3 tan (− −180) = − tan (180 +  ) = − tan  ✓ reduction / reduksie  −5  5 = − =−  −12  12 ✓ substitution / vervanging (2) 5.2.1 1 − cos  = 0 or sin  = 0 cos  =1 or sin  = 0 ✓ method / metode  =180. k (k  ) ✓ answer / antwoord OR / OF  = 360.k or  =180+ 360. k (k  ) (2) 5.2.2 sin  cos  LHS/LK = − 1 − cos  sin  sin 2  − cos (1 − cos  ) ✓ common denominator = sin  (1 − cos  ) gemene noemer sin 2  − cos + cos 2 = ✓ simplification / vereenvoudiging sin  (1 − cos  ) 1 − cos  ✓ identity / identiteit = sin  (1 − cos  ) ✓ simplification / vereenvoudiging 1 = sin  = RHS/RK (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 (EC/JUNE 2021) MATHEMATICS P2/WISKUNDE V2 9 5.3 sin ( x − y ) LHS/LK = cos x.cos y sin x cos y − cos x sin y ✓ identity / identiteit = cos x.cos y sin x cos y cos x sin y = − ✓method / metode cos x.cos y cos x.cos y = tan x − tan y ✓ simplification / vereenvoudiging = 3k − 2k = k ✓substitution / vervanging (4) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 10 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/2021) QUESTION 6/VRAAG 6 6.1.1 1 ✓1 (1) 6.1.2 120° ✓ 120° (1) 6.2 f ( x) = g ( x) cos ( x − 60) = sin 3 x ✓ cos(x − 60) = cos(90− 3x) cos ( x − 60) = cos (90 − 3 x) ✓ without cos sonder cos  ( x − 60) = 90 − 3 x + k .360 (k  ) ✓ simplification vereenvoudiging x − 60 = 90 − 3x + k .360 or / of − x + 60 = 90 − 3x + k .360 4 x =150 + k .360 or / of 2 x = 30 + k .360 ✓ x = 30 + k.90 or / of x =15 + k.180 x = 37,5 + k .90 or / of x = 15 + k .180 ✓ two values twee waardes x {37,5, − 52,5;15 ; 127,5} ✓ two remaining values twee oorblywende waardes (6) 6.3 f: ✓ endpoints / eindpunte ✓ both intercepts beide afsnitte ✓ shape / vorm g: ✓ both intercepts beide afsnitte ✓ shape / vorm (5) 6.4 x =− 30 or/of x =150 ✓both values of x beide waardes van x (1) 6.5 f ( x) = cos( x − 60+15) ✓h(x) = cos (x – 45°) h(x) = cos (x – 45°) (1) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 (EC/JUNE 2021) MATHEMATICS P2/WISKUNDE V2 11 QUESTION 7/VRAAG 7 52 = 42 + 52 − 2(4)(5) cos A ✓ substitution into cosine rule / vervanging in die cosinusreël 2 cos A = ✓ value of cos A / waarde van cos A 5 42 = 52 + 52 − 2(5)(5) cos B ✓ substitution into cosine rule / vervanging in die cosinusreël 17 cos B = ✓ value of cos B / waarde van cos B 25 2 17 cos A − cos B = − 5 25 7 = −  − 0, 28 ✓ value of cos A – cos B / waarde van cos A – cos B 25 (5) [5] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 12 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/2021) QUESTION 8/VRAAG 8 OM ⊥ CD (line from centre which bisects the chord) ✓ S/R (lyn vanaf die middelpunt wat koord halveer ) AO = OB = OD =11cm ✓S In  OMD: ✓ use of Pythagoras theorem gebruik van Pythagoras - MD 2 + 7 2 =112 (Pythagoras theorem/stelling ) stelling MD = 6 2 ✓ length of MD/lengte van MD CD =12 2 ✓ length of CD/lengte van CD (5) [5] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 (EC/JUNE 2021) MATHEMATICS P2/WISKUNDE V2 13 QUESTION 9/VRAAG 9 9.1   R 2 = Q2 ( s in the same segment/e in dieselfde segment ) ✓ S/R   U 4 = U1 ( vert .opp. s / regoorst. e) ✓ S/R   S2 = T 2 ( s in the same segment / e in dieselfde segment ) ✓ S/R  RUS ||| QUT () (3) 9.2   ✓ S/R S2 = T 2 = x (s in the same seg / e in dieselfde segment)   ✓S U3 =S2 = x (tan-chord theorem / raaklyn-koord stelling) ✓R   U 6 = U3 = x (vertically opp. s / regoorst. e) ✓S/R (4)  9.3.1 R1 + x + 90 − x = 180 ( sum of s of  / som van die e van  ) ✓S   ✓ value of R 1 /  R1 = 90  waarde van R 1 QT is a diameter (QT subtends a right angle) ✓ QT is a diameter/ is ’n middellyn QT is ' n middellyn (QT onderspan ' n reghoek ) ✓R (4) 9.3.2  P = 90 ( in thesemicircle /  in ' n semisirkel ) ✓S OR / OF ✓R  P + 90 =180 (opp. s of a cyclic quad / teenoorst. e van ' n koordevierhoek )   P = 90 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 14 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/2021) 9.4.1   Q2 = T 2 = x (s opp. equal sides / e teenoor gelyke sye) ✓ S/R    Q 2 = S2 RS|| QT (Alt. s are equal / Verw. e is gelyk ) ✓R (2) 9.4.2   U 2 = Q2 = x ✓S VW is a tangent to circle passing through QUT ✓R (Converse of tan-chord theorem) VW is 'n raaklyn aan die sirkel wat deur QUT gaan (Omgekeerde van die raaklyn − koord stelling ) (2) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 (EC/JUNE 2021) MATHEMATICS P2/WISKUNDE V2 15 QUESTION 10/VRAAG 10 10.1 Construction: Draw diameter GOE. Join EH ✓ construction / konstruksie Konstruksie: Trek middellyn GOE. Verbind EH Proof/Bewys: ✓ S/R   G1 + G 2 = 90 (tangent ⊥ diameter) / ✓ S/R (raaklyn ⊥ radius)  EHG = 90 ( in thesemi circle) / ✓S ( in semi - sirkel )   G 2 + E = 90 (sum of s of Δ) / (som van die e van ' n ) ✓ S/R      G1 + G 2 = G 2 + E   G1 = E   But/Maar: E = F (s in thesamesegment)/ (e in dieselfde segment )    DGH = F (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 16 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/2021) 10.2 10.2.1 Kite / Vlieër ✓ answer / antwoord (1) 10.2.2  KLO = 90 (tan ⊥ rad.) / (raaklyn ⊥ radius) ✓S ✓R  KPO = 90 (tan ⊥ rad.) / (raaklyn ⊥ radius) ✓S  MLP = 90 (s in thesemi circle) / (ein ' n semi - sirkel ) ✓S ✓R (5) 10.2.3   KLO + KPO = 90 + 90 = 180 ✓S KLOP is a cyclic quad. (Opp. s are supp.) ✓R KLOP is ' n koordevierhoek (Teenoorst. e is suppl.) (2) 10.2.4   K + LOP =180 (Opp. s of cyclic quad.) ✓S✓R  But M = 67 ( s opp.equalsides)  ✓ S/R  LOP = 67 + 67 (Ext.  of ) ✓ value of LOP =134 ˆ waarde van LOP    K + 134 =180 ✓value of K / waarde    K = 46 van K OR/OF  OR/OF M = 67 ( s opp. = sides)  ✓S✓R LOM = 46 ( s of ) ✓value of / waarde van    K = 46 (ext.  of cyclic quad.) LOM ✓value of / waarde van  K ✓reason / rede (5) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 (EC/JUNE 2021) MATHEMATICS P2/WISKUNDE V2 17 QUESTION 11/VRAAG 11 AP 3 = (Prop.theorem; PQ||BC ) ✓ S/R PB 5 (Verhouding stelling ; PQ||BC) ✓S AP 3 = PR + BR 5 ✓ simplification / vereenvoudiging 5AP = 3PR + 3BR BR 1 = (Prop.theorem; RS||AC ) RA 3 ✓ S/R (Verhouding stelling ; RS||AC) AP 1 ✓S = AP + PR 3 ✓ simplification / vereenvoudiging 3BR = AP + PR 5AP = 3PR + AP + PR ✓ substitution / vervanging 4AP = 4PR AP = PR (7) [7] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ## Related documents - [Mathematics P2 Nov 2020 ANSBK Afr Eng (2) final](https://www.hlayiso.com/library/mathematics-p2-nov-2020-ansbk-afr-eng-2-final/199754f9-9f9b-4bd9-8890-beebee25863f) — Mathematics · QP · 2021 · Eng - [Mathematics P2 SEPT2021 ANSBK SESOTHO ENG DM](https://www.hlayiso.com/library/mathematics-p2-sept2021-ansbk-sesotho-eng-dm/ae538fae-90db-4237-9a93-c7eb83053862) — Mathematics · QP · 2021 · Sesotho - [Mathematics P2 Sept2021 QP SESOTHO ONLY D](https://www.hlayiso.com/library/mathematics-p2-sept2021-qp-sesotho-only-d/4d50cd68-a960-4bf3-b891-1fcbcd8ce495) — Mathematics · QP · 2021 · Sesotho - [MATHS P2 QP GR12 JUNE 2022 Afrikaans](https://www.hlayiso.com/library/maths-p2-qp-gr12-june-2022-afrikaans/49feb536-3d0c-4e63-b857-c88def50a8c1) — Mathematics · QP · 2022 · Afr - [MATHS P1 GR12 MEMO SEPT 2021 English IsiXhosa](https://www.hlayiso.com/library/maths-p1-gr12-memo-sept-2021-english-isixhosa/6e14ca2d-44f7-417b-b314-e40962f91676) — Mathematics · Memo · 2021 · Eng - [Maths-April-QP Memo-EC-2021-Gr12](https://www.hlayiso.com/library/maths-april-qp-memo-ec-2021-gr12/e1469b3f-b9fe-40c2-9c23-509439908f72) — Mathematics · Memo · 2021 ## 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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