# Mathematics Grade 12 P2 Eastern Cape Prelim 2024 Memo > Grade 12 Mathematics Paper 2 memorandum from the Eastern Cape Prelim Exam 2024 (isiXhosa), 17 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/maths-p2-mg-gr12-sept-2024-eng-isixhosa/2bd15693-4f5a-4882-97b4-eebc9f2abacb - Updated: 2026-10-29 - Subject: Mathematics - Grade: Grade 12 - Language: Isixhosa - Document type: Memorandum - Year: 2024 - Exam period: Eastern Cape Prelim Exam - 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 Prelim 2024 Memo](https://www.hlayiso.com/library/maths-p2-mg-gr12-sept-2024-eng-isixhosa/2bd15693-4f5a-4882-97b4-eebc9f2abacb) ## About this document Mathematics · Grade 12 · Eastern Cape Prelim Exam · 2024 · isixhosa. Memorandum. 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 GRADE/IBANGA 12 SEPTEMBA 2024 MATHEMATICS P2/IMATHEMATIKA P2 MARKING GUIDELINE/ISIKHOKELO SOKUMAKISHA MARKS/AMANQ 150 AKU: This marking guideline consists of 17 pages. Esi sikhokelo sokumakisha sinamaphepha ayi17. Hosted by www.ecexams.co.za. ### Page 2 Downloaded from hlayiso.com 2 MATHEMATICS P2/IMATHEMATIKA P2 (EC/SEPTEMBA 2024) NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and not redone a question, mark the crossed out version. • Consistency accuracy applies in ALL aspects of the Marking Guideline. Stop marking at the second calculation error. • Assuming answers/values in order to solve a problem is NOT acceptable. GEOMETRY S A mark for a correct statement (A statement mark is independent of a reason) R A mark for the correct reason. (A reason mark may only be awarded only if the statement is correct S/R Award a mark if a statement AND a reason are both correct QAPHELA: • Xa umviwa ephendule umbuzo KABINI makisha kuphela ilinge LOKUQALA. • Xa umviwa elihlabile ilinge lombuzo akawuphinda umbuzo, makisha okuhlatyiweyo. • Iconsistency accuracy yenzeka kuzo ZONKE iinkalo zesikhokelo sokumakisha. Yeka ukumakisha kwimpazamo yesibini yekhalityhuleyishini. • Ukucingela iimpendulo/iiveliyu ukusombulula ingxaki AKWAMKELEKANGA. IJIYOMETRI S Inqaku ngesitheyithimenti esichanekileyo (Inqaku lesitheyithimenti lahlukile kwelesizathu) R Inqaku lesizathu esichanekileyo (Inqaku lesizathu linikwa kuphela xa isitheyithimenti sichanekile) S/R Nika inqaku ukuba isiteyithimenti NEsizathu ziyafana Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 3 Downloaded from hlayiso.com (EC/SEPTEMBA 2024) MATHEMATICS P2/IMATHEMATIKA P2 3 QUESTION 1/UMBUZO 1 1.1 82 64 55 50 41 71 78 88 98 96 63 66 80 84 88 1.1.1 88  answer / impendulo (1) 1.1.2 Range / Omvang = 98 − 41 = 57  answer / impendulo (1) 1.1.3 1104  1104 x= 15 = 73,60  answer / impendulo (2) 1.1.4  = 16,30  answer / impendulo (1) 1.1.5 x −  = 73, 60 − 16,30  73,60 − 16,30 = 57,30  57,30  There were 3 truck drivers.  answer / impendulo Daar was 3 trokbestuurders (3) 1.2 let total mass of 8 people be x: laat dietotale massa van 8 mense x wees : number of people to be added be k: aantal mense wat by moet kom k wees : x x = 75  = 75 8 8 x = 600  mass of 8 people (600) 75k + 600 = 1000 Ubunzima babantu abayi8 (600) 1000 − 600 k =  equation / i-ikhweyizhini 75 ( 75k + 600 = 1000 ) k = 5,333 It will be approximately equal to 5 people Iqikeleleka ukulingana nabantu abayi5  answer / impendulo (4) [12] Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 4 Downloaded from hlayiso.com 4 MATHEMATICS P2/IMATHEMATIKA P2 (EC/SEPTEMBA 2024) QUESTION 2/UMBUZO 2 TEST A / UVAVANYO A 39 33 35 44 37 40 24 31 30 5 TEST B / UVAVANYO B 41 45 48 40 47 42 37 44 43 24 2.1 (44 ; 40)  44 in TEST A / kuVAVANYO A (1) 2.2 a = 25,48  a = 25,48 b = 0,49  b = 0,49 y = 25,48 + 0,49 x  y = 25,48 + 0,49 x (3) 2.3 y = 25,48 + 0,49(14) correct substitution / = 32 Isaphusitityhushini echanekileyo  answer / impendulo (2) 2.4 r = 0,79  r = 0,79 Strong positive correlation comment / ukuphawula (phawula) Correlation esitrongo ephozithivu (2) [8] Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 5 Downloaded from hlayiso.com (EC/SEPTEMBA 2024) MATHEMATICS P2/IMATHEMATIKA P2 5 QUESTION 3/UMBUZO 3 y B (3 ; 4)  E G O 45 0 F (5 ; 0) x A • H (1 ;–4) C 3.1 BF = (3 − 5)2 + (4 − 0)2  correct substitution Isaphusitityhushini = 20 echanekileyo (2)  answer / impendulo 3.2 4−0  correct substitution mBF = 3−5 Isaphusitityhushini = −2 echanekileyo  answer / impendulo (2) 3.3 tan  = −2  tan = mBF = −2  = 116,57o   = 116,57o  = 116,57o − 45o = 71,57o [ext  of a ]   = 71,57o (3) 3.4 tan 45 = mAB = 1 o  mAB = 1 1− 5 mHF =  correct substitution −4 − 0 Isaphusitityhushini =1 echanekileyo HF AB [ mAB = mHF = 1]  mHF = 1  Reason /Isizathu [ mAB = mHF ] (4) Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 6 Downloaded from hlayiso.com 6 MATHEMATICS P2/IMATHEMATIKA P2 (EC/SEPTEMBA 2024) 3.5 Kite / iKhayithi  answer / impendulo (1) 3.6 HC FC  correct ratio / = [line/lyn || to one side of a  / aan een sy van ] Iratio echanekileyo AH BF 2 FC correct substitution = Isaphusitityhushini 1 2 5 echanekileyo FC = 4 5  FC BC = 4 5 + 2 2 = 6 5 AC = 6 5 [adj.sides of a kite / aangr. sye van vlieër ]  AC = BC (4) 3.7 ˆ = 71,57 o  s opp = sides / e teenoor = sye  B̂ = A  Ĉ = 36,87o  Ĉ = 36,87 o Area of/van AOFC = Area of/van ΔABC − Area of/van ΔOBF 1  Area of/ka ∆ABC =  6 5  6 5  sin 36,87o − 12 2 = 41,99 answer / impendulo (3) [19] QUESTION 4/UMBUZO 4 y S O P x R Q T 4.1.1 3 x = 0 y = − ( 0) − 8 4 = −8  y- coordinate / y kho-odineyithi Q (0 ; − 8) (2) Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 7 Downloaded from hlayiso.com (EC/SEPTEMBA 2024) MATHEMATICS P2/IMATHEMATIKA P2 7 4.1.2 4 mQR = [tan ⊥ rad / raakl ⊥ rad ] 3  mQR 4 substituting mQR and y + 8 = ( x − 0) 3 Q (0 ; − 8) 4 y = x −8 ukusaphusitityhutha mQR 3 kunye no Q (0 ; − 8)  equation / i-ikhweyizhini (3) 4.1.3 4 x −8 = 0 y=0 3 x=6 x=6 P(6 ; 0) (2) 4.1.4 0+6 −8+ 0  correct substitution xR = ; yR = Ukusaphusitityhutha 2 2 okuchanekileyo xR = 3 ; y R = −4 (3)  xR = 3  yR = −4 r 2 = (0 − 3) + (− 8 + 4)  correct substitution 2 2 4.1.5 Ukusaphusitityhutha = 25 okuchanekileyo  r 2 = 25 (x − 3)2 + ( y + 4)2 = 25  equation / i-ikhweyizhini (3) 4.1.6 k = −4 + 5 or / of k = −4 − 5  method / imethodi k = 1 or / of k = −9  k = 1  k = −9 (3) 4.2 (x − sin  ) + ( y + 2 sin  ) = −2 + sin  + 4 sin  2 2 2 2  (x − sin  ) + ( y + 2 sin  ) 2 2 r = −2 + 5 sin  2 2  r 2 = −2 + 5 sin 2  For any value of θ maximum of sin 2  = 1  maximum of sin 2  = 1 Vir enige waarde van  is maksimum van sin  = 1 2 imaximum ka sin 2  = 1  r = −2 + 5(1)  r = − 2 + 5(1) = 3  answer / impendulo (5) [21] Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 8 Downloaded from hlayiso.com 8 MATHEMATICS P2/IMATHEMATIKA P2 (EC/SEPTEMBA 2024) QUESTION 5/UMBUZO 5 5.1 y  x O 3 2 S ( −4; q ) 5.1.1 ( ) q = − 3 2 − (− 4) 2 2 Pyth  correct substitution Ukusaphusitityhutha =− 2 okuchanekileyo  answer / impendulo (2) 5.1.2 sin ( + 45 ) = sin  .cos 45 + cos  sin 45 o o o  expansion / ukuexpanda − 2 2  −4  2  ratios of / iiratio zika = . + . 3 2 2 3 2  2 sin  & cos −1 − 2 2  special angles / iispecial engile =  answer / impendulo 3 2 (4) 5.1.3 ( cos 2 − 360o = cos 2)  cos 2  identity / iayidentithi = 2 cos 2  − 1  −4  2  ratio of / verhouding van cos  = 2  −1 3 2  7  answer / impendulo = 9 (4) 5.2 sin ( 90o −  ) .cos 480o + cos (180o −  ) .tan 45o  cos cos  .sin 390o − tan180o  − cos 60 o cos  . ( − cos 60o ) + ( − cos  ) ( tan 45o ) =  sin 30o cos  ( sin 30o ) − tan180o 1 − cos  − cos  (1)  special angles / iispecial engile = 2 1 cos    − 0 2 3 − cos  = 2 1 cos  2  answer / impendulo = −3 (5) Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 9 Downloaded from hlayiso.com (EC/SEPTEMBA 2024) MATHEMATICS P2/IMATHEMATIKA P2 9 5.3 cos x cos 2 x LHS / LK = − sin 2 x 2sin x  simplification / 2sin x cos x − cos 2 x sin 2 x = ukusimplifaya 2sin 2 x sin x sin 2 x − cos 2 x sin 2 x  sin 2 x = 2sin 2 x sin x  common factor / sin 2 x (1 − cos 2 x ) Ikhomoni fekhitha = 2sin 2 x sin x 1 − (1 − 2sin 2 x )  identity / iayidentithi = 1− 2 sin 2 x 2sin x 2sin 2 x 2 sin 2 x =  2sin x sin x = sin x (5) 5.4.1 cos 60o sin 60o − =2 sin x cos x cos 60o cos x − sin 60o sin x =2 sin x cos x cos ( x + 60o ) = 2sin x cos x  simplification / ukusimplifaya cos ( x + 60o ) = sin 2 x (  cos x + 60o ) cos ( x + 60 ) = cos ( 90 − x ) o o  sin 2 x (3) 5.4.2 cos ( x + 60 ) = cos ( 90 − 2 x ) o o  x + 60o = 90o − 2 x x + 60 =  ( 90 − 2 x ) + 360 .k o o o x + 60o = 90o − 2 x + 360o.k or / of x + 60o = −90o + 2 x + 360o.k 3x = 30o + 360o.k    3x = 30o + 360o.k or / of − x = −120o + 360o.k  or / of   − x = −120o + 360o.k  x = 10o + 120o.k or / of x = 120o − 360o.k , k     OR / OF  x = 10o + 120o.k    x + 60o = 90o − 2 x + 360o.k or / of x + 60o = 360o − 90o + 2 x + 360o.k  or / of   x = 120o − 360o.k  3x = 30 + 360 .k or / of − x = 240 + 360 .k o o o o   x = 10o + 120o.k or / of x = −240o − 360o.k , k    360o.k , k   (4) Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 10 Downloaded from hlayiso.com 10 MATHEMATICS P2/IMATHEMATIKA P2 (EC/SEPTEMBA 2024) 5.5  y = c2 − a2 Pyth Theorem / Stelling y y = c 2 − a 2 Pyth Theorem / Stelling OR/OKANYE correct diagram Idayagram echanekileyo c 67,5o c2 − a2 22,5o x a 2 o = sin 45o  sin 45 2 = 2sin 22,5o cos 22,5o  2sin 22,5o.cos 22,5o c2 − a2 a = 2. . c c a 2 + b2 − a 2 a  substitution / isaphusitityhushini = 2. . c c 2ab = 2 2 2  c 2 i.t.o./ i.t.v a & b c (5) [32] Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 11 Downloaded from hlayiso.com (EC/SEPTEMBA 2024) MATHEMATICS P2/IMATHEMATIKA P2 11 QUESTION 6/UMBUZO 6 6.1 Period / iPhiriyodi ngu 360o  answer / impendulo (1) 6.2 Min value / Min veliyu= –1 answer / impendulo (1) 6.3 −1  y  1 −1+1  y  1 +1 correct critical values/ 0 y2 Iikhrithikhali veliyu ezichanekileyo correct notation / inoteyishini echanekileyo (2) 6.4 120  x  180 o o  correct critical values Iikhrithikhali veliyu ezichanekileyo correct notation / inoteyishini echanekileyo (2) 6.5 g ( x ) = − sin ( x − 30 − 60 ) o o – (– sin x – 30° – 60°) = − sin ( x − 90o ) (  sin x − 90o )  − cos x = − cos x (3) 6.6  intercepts with the axes/ ii-intasephuthi nee-axes  turning points / iitheningi poyinti  shape / isheyiphu (3) [12] Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 12 Downloaded from hlayiso.com 12 MATHEMATICS P2/IMATHEMATIKA P2 (EC/SEPTEMBA 2024) QUESTION 7 / UMBUZO 7 B A θ 2x (16 − 4 x ) 120o C 7.1 1 Aof ABC =  2 x  (16 − 4 x )  sin120o correct substitution / 2 = (16 x − 4 x )  sin 60o 2 Ukusaphusitityhutha okuchanekileyo = 8 3x − 2 3x 2  sin 60o  answer / impendulo (3) 7.2 A = 0  derivative / iderivative = 0 8 3 − 4 3x = 0  8 3 − 4 3x answer / impendulo x=2 (3) [6] Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 13 Downloaded from hlayiso.com (EC/SEPTEMBA 2024) MATHEMATICS P2/IMATHEMATIKA P2 13 QUESTION 8/UMBUZO 8 O 1 2 B E D 8.1 Construction: Draw DO and OB  construction Proof: In ∆ODE and ∆OEB DE = EB [given]  first statement (radii) OD = OB  radii   other 2 statements OE = OE  common  ODE  OEB  SSS   reason for congruency Eˆ1 + Eˆ 2 = 180o  s on str line  R  Eˆ1 = Eˆ 2 = 90o [ODE  OEB ] Ikhonsitrakhishini: Zoba uDO noOB  ikhonsitrakhishini iPrufu: ku∆ODE naku∆OEB DE = EB [ gegee]  isiteyithimenti sokuqala OD = OB  radiusse (iirediyasi) OE = OE  gemeen  ezinye iziteyithimenti eziyi2 ODE  OEB  SSS   izizathu zecongruecy Eˆ1 + Eˆ 2 = 180o  e op reguitlyn   Eˆ1 = Eˆ 2 = 90o [ODE  OEB ] R (5) Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 14 Downloaded from hlayiso.com 14 MATHEMATICS P2/IMATHEMATIKA P2 (EC/SEPTEMBA 2024) 8.2 O x P Q R 2 S 8.2.1 ˆ = 90o OQP line from centre to the midpoint  S R [ilayini ukusuka embindini ukuya kwimidipoyinti] (2) 8.2.2 PQ = 4  PQ OP 2 = PQ 2 + OQ 2  Pyth  ( x + 2 ) = 42 + x 2 2  substitution into Pythagoras/ Ukusaphusitityhutha x 2 + 4 x + 4 = 16 + x 2 kwiPythagoras 4 x = 12 x=3  simplification / ukusimplifaya OP = OS = 5  radii / radiusse  x-value /x–veliyu  PO (5) [12] Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 15 Downloaded from hlayiso.com (EC/SEPTEMBA 2024) MATHEMATICS P2/IMATHEMATIKA P2 15 QUESTION 9/UMBUZO 9 D C A 70o 1 O B 2 9.1 Oˆ1 = 110o s on a str.line / e op reguitlyn  S/R Oˆ 2 = 350o  s around a point / e om ' n punt  S R Cˆ = 175o  at centre = 2   at circumf  Ô1 =110° Ô2 = 250° [ Middelpunts  = 2  Omtreks ] ⸫Ĉ = 125° S R (5) Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 16 Downloaded from hlayiso.com 16 MATHEMATICS P2/IMATHEMATIKA P2 (EC/SEPTEMBA 2024) 9.2 P Q R 1 2 2 y 1 x− y x− y y 1 2 y 3 T x S 9.2.1 Tˆ1 = y  tan chord theorem / raaklyn koord Stelling  S R Tˆ3 = y [alt s / verw. e, PR||TS] S R (4) 9.2.2 ˆ ˆ T2 = R1 [s subt by equal chords / e onderspan deur gelyke koorde]  S  R (2) 9.2.3 Tˆ2 = x − y  ext.  of a  / buite  van ' n   S Rˆ1 = Tˆ2 = x − y  proved / bewys  y + x − y + x = 180o  s in TRS / ein TRS  S/R 2 x = 180o x = 90o  x = 90o  TR is the diameter of the circle linesubt.90o  R TR is die middellyn van die sirkel [lyn onderspan 90o ] (4) [15] Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ### Page 17 Downloaded from hlayiso.com (EC/SEPTEMBA 2024) MATHEMATICS P2/IMATHEMATIKA P2 17 QUESTION 10 / UMBUZO 10 AC = 36 units/yunithi, AD = 48 units/yunithi and/aze BM = 24 units/yunithi B C 2 1 A 3 2 1 N 2 M 1 D 10.1 Aˆ = Aˆ common / gemeen S S R Bˆ 2 = Dˆ ext  of a cylic quad / buite  van koordev.  R 3rd angle/3rdengile Mˆ 3 = Cˆ ext  of a cylic quad / buite  van koordev. or / of 3rd / de  OR/OKANYE ΔABM|||ΔADC   R  (4) 10.2 BM AM S R = |||Δs DC AC but/maar AM = DC given / gegee BM DC  AM = DC = DC AC CD 2 = BM  AC (3) 10.3 CD = 24  36 = 864 2  length of CD /2 ubude buka CD 2 CN AM = line|| to oneside of a   CD AD [lyn aan een sy van ' n ] S R AM = CD  CN in terms of CD 2 / CD 2 CN ngokwethem zika CN = CD 2 AD  correct substitution 864 = Ukusaphusitityhutha 48 okuchanekileyo = 18  length of CN Ubude bukaCN (6) [13] TOTAL/EWONKE: 150 Copyright reserved/Amalungelo akhuselekile Please turn over/Nceda tyhila Hosted by www.ecexams.co.za. ## Related documents - [G12 Maths P2 Sept Free State 2024](https://www.hlayiso.com/library/g12-maths-p2-sept-free-state-2024/b81ee450-da0b-40b1-84a9-e4089f8e3d30) — Mathematics · QP · 2024 - [KZN GR 12 MATHS P2 SEPT 2024](https://www.hlayiso.com/library/kzn-gr-12-maths-p2-sept-2024/6db62b2f-3c0b-4c51-8d6b-e875f4b5d67a) — Mathematics · QP · 2024 - [Gr 12 EC Maths P2 Memo Sep 2020 Errata](https://www.hlayiso.com/library/gr-12-ec-maths-p2-memo-sep-2020-errata/e1e6c6e0-9ebe-45f8-ba36-fe3147f0640a) — Mathematics · Memo · 2020 - [Mathematics P2 Nov 2019 Memo Afr 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