# Mathematics Grade 12 P1 Eastern Cape Mock 2021 Memo (Eng) > Grade 12 Mathematics Paper 1 memorandum from the Eastern Cape Mock Exam 2021 (English), 22 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/maths-p1-gr12-memo-sept-2021-english-sesotho/3e5f1e43-6f7f-4cd8-9af5-eea479d53fe6 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Language: English - Document type: Memorandum - Year: 2021 - Exam period: Eastern Cape Mock Exam - Paper: 1 - Curriculum: CAPS - Pages: 22 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 P1 Eastern Cape Mock 2021 Memo (Eng)](https://www.hlayiso.com/library/maths-p1-gr12-memo-sept-2021-english-sesotho/3e5f1e43-6f7f-4cd8-9af5-eea479d53fe6) ## About this document Mathematics · Grade 12 · Eastern Cape Mock Exam · 2021 · English. Memorandum, 22 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Matching paper and memo - [MATHS P1 QP GR12 SEPT 2021 English](https://www.hlayiso.com/library/maths-p1-qp-gr12-sept-2021-english/12f7db49-5bd7-414f-af58-65ed09d4c3e4) — QP · 2021 · Eng ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ SETIFIKEITI SE PHAHAMENG SA NAHA GRADE/KEREITE 12 SEPTEMBER 2021 MATHEMATICS P1/DIPALO P1 MARKING GUIDELINE/DITATAISO TSA HO TSHWAYA MARKS/MATSHWAO: 150 This marking guideline consists of 22 pages./ Tataiso ena ya ma tshwao ena le maqephe a 22. ### Page 2 Downloaded from hlayiso.com 2 MATHEMATICS P1/ DIPALO P1 (EC/LWETSE2021) NOTE/NEHA HLOKO: • If a candidate answers a question TWICE, mark the FIRST attempt ONLY. Haeba mohlahlubuwa o arabile ha bedi tshwaya leko ya PELE FEELA. • Consistent accuracy applies in ALL aspects of the marking guideline. Tumellano e nepahetseng e sebetsa ho di karolo KAOFELA ho tshwayo. • If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt. Haeba mohlahlubuwa o hlabile teko ya potso a seka bela a araba potso, tshwaya tefo e hlajiweng. • The mark for substitution is awarded for substitution into the correct formula. Le tshwao la sabustitshushini le fowa ho sabusitikhushini e nepahetseng ya meralo. QUESTION 1/POTSO 1 1.1.1 x 2 + 2 x − 15 = 0 ( x − 3)( x + 5) = 0  factors / fekethas x = 3 or / of x = −5  x = 3  x = −5 OR/KAPA OR/KAPA −b  b − 4ac 2 x= 2a  substitution / −2  22 − 4(1)(−15) = sabusithitshushini 2(1) −2  64 = 2 = 3 or / of − 5  x = 3  x = −5 (3) 1.1.2 3x + x − 1 = 0 2 −b  b 2 − 4ac x= 2a −1  12 − 4(3)(−1) =  substitution / 2(3) sabosititshushini −1  13 = Penalise 1 mark for incorrect 6 rounding = 0, 43 or / of − 0, 77 Penaliseer 1 punt vir verkeerde afronding  x = 0, 43  x = −0,77 (3) 1.1.3 x( x − 3)  −2 y x 2 − 3x + 2  0  standard form ( x − 1)( x − 2)  0 Formo etlwaelehileng 1 2 x  x  1 or / of x2  factorisation fekethoreseishini  x  1 or/of  x  2 (4) Copyright reserved/ Ditokelo di sireleditswe Please turn over/ phetla ### Page 3 Downloaded from hlayiso.com (EC/LWETSE2021) MATHEMATICS P1/DIPALO P1 3 1.1.4 √43 − 𝑥 − 𝑥 + 1 = 0 √43 − 𝑥 = 𝑥 − 1  isolating the surd 2 (√43 − 𝑥) = (𝑥 − 1)2 Ho itlhaola sede 43 − 𝑥 = 𝑥 2 − 2𝑥 + 1 𝑥 2 − 𝑥 − 42 = 0  squaring both sides (𝑥 − 7)(𝑥 + 6) = 0 Ho sekwera mahlakori ka ∴ 𝑥 = 7 or 𝑥 ≠ −6 bobedi  standard form / foromo etlwaelehilengg  factorisation / fekithorisashini  selection / kgetho e nepahetseng (5) Copyright reserved/Ditokelo di sireleditswe Please turn over/ Phetla ### Page 4 Downloaded from hlayiso.com 4 MATHEMATICS P1/ DIPALO P1 (EC/LWETSE2021) 1.2 2𝑦 − 𝑥 = 3(1) 𝑦 2 + 3𝑥 = 2𝑥𝑦(2) 𝑥 = 2𝑦 − 3(3) Substitute (3) into (2) 𝑦 2 + 3(2𝑦 − 3) = 2𝑦(2𝑦 − 3) 𝑦 2 + 6𝑦 − 9 − 4𝑦 2 + 6𝑦 = 0  x = 2y − 3 −3𝑦 2 + 12𝑦 − 9 = 0 𝑦 2 − 4𝑦 + 3 = 0 (𝑦 − 3)(𝑦 − 1) = 0  substitution /sabusithikhushini ∴ 𝑦 = 3 or 𝑦 = 1 𝑥 = 2(3) − 3 or 𝑥 = 2(1) − 3 = 3 = −1  standard form / foromo etlwaelehileng OR/KAPA  factorisation / fekhuthoriseshini 2𝑦 − 𝑥 = 3(1)  y-values / di y-velu 𝑦 2 + 3𝑥 = 2𝑥𝑦(2) 𝑥 3  x-values /di x-velu 𝑦= + (3) 2 2 Substitute(3) into (2) 𝑥 3 2 𝑥 3 OR/KAPA ( + ) + 3𝑥 = 2𝑥 ( + ) 2 2 2 2 𝑥 2 6𝑥 9 + + + 3𝑥 = 𝑥 2 + 3𝑥 x 3 4 4 4 3𝑥 2 6𝑥 9  y= + − + + =0 2 2 4 4 4 −3𝑥 2 + 6𝑥 + 9 = 0 𝑥 2 − 2𝑥 − 3 = 0 (𝑥 − 3)(𝑥 + 1) = 0 ∴ 𝑥 = 3 or 𝑥 = −1  substitution /sabosititshushini 3 3 1 3 𝑦 = ( + ) or 𝑦 = (− + ) 2 2 2 2 =3=1  standard form / foromo etlwaelehileng  factorisation / fekithorisashini  x-values / di x-velu  y-values / di y-velu (5) Copyright reserved/ Ditokelo di sireleditswe Please turn over/ phetla ### Page 5 Downloaded from hlayiso.com (EC/LWETSE2021) MATHEMATICS P1/DIPALO P1 5 1.3 5±√𝑝(6−𝑝)−9 𝑥= 2 For non-real roots: Δ<0 𝑝(6 − 𝑝) − 9 < 0  0 −𝑝2 + 6𝑝 − 9 < 0 y 𝑝2 − 6𝑝 + 9 > 0 (𝑝 − 3)2 > 0  standard form / foromo ∴ 𝑝 ∈ ℝ but 𝑝 ≠ 3 etlwaelehileng OR/KAPA 3 x  factorisation / fekithorisashini 5±√𝑝(6−𝑝)−9 𝑥= 2 For non-real roots:  answer / karabo Δ<0 𝑝(6 − 𝑝) − 9 < 0 OR/KAPA −𝑝2 + 6𝑝 − 9 < 0 y (3 − 𝑝)(𝑝 − 3) < 0 3 ∴ 𝑝 ∈ ℝ but 𝑝 ≠ 3 x  0  standard form / foromo etlwaelehilengng  factorisation / fekethorisaeshini  answer / karabo (4) [24] Copyright reserved/Ditokelo di sireleditswe Please turn over/ Phetla ### Page 6 Downloaded from hlayiso.com 6 MATHEMATICS P1/ DIPALO P1 (EC/LWETSE2021) QUESTION 2/ POTSO 2 2.1.1 −16 ; −16 ; −12 ; −4 0 4 8 4 4 8 8 (1) 2.1.2 2a = 4 a = 2  a=2 3a + b = 4 3(2) + b = 0 b = − 6  b = −6 a + b + c = −16 2 − 6 + c = −16  c = −12  c = −12  Tn = 2n 2 − 6n − 12 Tn = 2n − 6n − 12 2 (4) 2.1.3 T38 = 2(38) − 6(38) − 12 2  substitution / sabosetitshushini = 2 648  answer / karabo (2) Copyright reserved/ Ditokelo di sireleditswe Please turn over/ phetla ### Page 7 Downloaded from hlayiso.com (EC/LWETSE2021) MATHEMATICS P1/DIPALO P1 7 2.1.4 General term for first differences: Themo ya kakaretso ya phapang ya pele  Tn = 4n − 4 𝑇𝑛 = 4𝑛 − 4 400 = 4𝑛 − 4  Tn = 400 ∴ 𝑛 = 101 𝑇𝑛(linear) = (𝑇𝑛+1 − 𝑇𝑛 )(quadratic)  answer / karabo ∴ 𝑛 = 101 and + 1 = 102 (3) The terms are: 101 and 102 OR/KAPA OR/KAPA 2(𝑛 + 1)2 − 6(𝑛 + 1) − 12 − (2𝑛2 − 6𝑛 − 12) = 400 2𝑛2 + 4𝑛 + 2 − 6𝑛 − 6 − 12 − 2𝑛2 + 6𝑛 + 12 = 400 4𝑛 − 4 = 400  4n − 4 = 400  4𝑛 = 404 ∴ 𝑛 = 101 ∴ Between 𝑇101 and 𝑇102  answer / karabo (3) OR/KAPA OR/KAPA Trial and error 𝑇102 = 2(102)2 − 6(102) − 12 = 20184 𝑇101 = 2(101)2 − 6(101) − 12 = 19784 subst. for T101 and T102 Difference : 400 Ka T101 le T102 ∴ Between 𝑇101 and 𝑇102  400  answer / karabo (3) 2.2.1 Tn = a + (n − 1)d 89 = 2 + (n − 1)(3)  substitution / sabustikhushini 3n − 1 = 89 3n = 90 n = 30  answer / karabo (2) Copyright reserved/Ditokelo di sireleditswe Please turn over/ Phetla ### Page 8 Downloaded from hlayiso.com 8 MATHEMATICS P1/ DIPALO P1 (EC/LWETSE2021) 2.2.2 𝑘 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑡𝑜 30 𝑡𝑒𝑟𝑚𝑠. 𝑛 𝑆𝑛 = [𝑎 + 𝑙]  Sum formula / moralo wa samo 2 30  substitution / sabositikhushini = [2 + 89] 2 = 1365  answer / karabo OR / KAPA OR / KAPA n Sn =  2a + (n − 1)d  2  Sum formula / moralo wa samo 30 =  2(2) + (30 − 1)(3)  2  substitution / sabositikhushini = 1365  answer / karabo (3) [15] Copyright reserved/ Ditokelo di sireleditswe Please turn over/ phetla ### Page 9 Downloaded from hlayiso.com (EC/LWETSE2021) MATHEMATICS P1/DIPALO P1 9 QUESTION 3/ POTSO 3 3.1 T9 = ar 8 = 768 T13 = ar12 = 12 288 ar12 12 288 ar12 12 288 =  = ar 8 768 ar 8 768  r = 16 4 r = 2  r = 2 768 a= ( 2 ) 8 =3  value of a / velu ya a (3) 3.2.1 54 24 S2 = − 19 19 30 = 19  answer / karabo (1) 3.2.2 30 T1 + T2 = 19 30 a + ar = 30 19  a + ar = 30 19 a (1 + r ) = 19 30 a= 19(1 + r ) (1) 3.2.3 𝑎 54 𝑆∞ = = 1 − 𝑟 19 54(1 − 𝑟) ∴𝑎= 19 54(1 − r )  a= 30 19 𝑎= . . . . . . . . 𝑓𝑟𝑜𝑚 (3.2.2) 19(1 + 𝑟) 30 54(1 − 𝑟) ∴ = 19(1 + 𝑟) 19 30 (1 − 𝑟)(1 + 𝑟) = 54  equating / ho le kanntsha 2 5 1−𝑟 = 9 2 4 𝑟 = 9 2 ∴𝑟= 3 4  r2 = 9  answer / karabo (4) [9] Copyright reserved/Ditokelo di sireleditswe Please turn over/ Phetla ### Page 10 Downloaded from hlayiso.com 10 MATHEMATICS P1/ DIPALO P1 (EC/LWETSE2021) QUESTION 4/ POTSO 4 4.1 D(0 ; 1)  (0 ; 1) (1) 4.2 x = −2 ; y = −1  x = −2  y = −1 (2) 4.3 𝑥 ∈ ℝ but 𝑥 ≠ −2  𝑥 ∈ ℝ  x  −2 (2) 4.4 g ( x) = b x 8 = b −3  substitution / sabusitikhushini 1 8= 3 b 1 b3 = 8 1 b =  answer / karabo 2 (2) 4.5 −3 y= −1 x+2 −3 0= −1  substitution y = 0 / x+2 sabusitikhushini y = 0 −3 1= x+2 x + 2 = −3 x = −5  x = −5 A( −5; 0) −3 y= −1 0+2 5 =−  y=− 5 2 2 C ( 0; − 52 ) (3) Copyright reserved/ Ditokelo di sireleditswe Please turn over/ phetla ### Page 11 Downloaded from hlayiso.com (EC/LWETSE2021) MATHEMATICS P1/DIPALO P1 11 y 4.6 1 1 y x=   x=  2 2  y = log 1 x  y = log 1 x 2 2 OR/KAPA OR/ KAPA −x y=2  x = 2− y  x = 2− y y = − log 2 x  y = − log 2 x (2) 4.7.1 −5  x  −2 OR/KAPA x (−5 ; − 2)  answer / karabo (A) (2) 1 4.7.2 0 x OR/KAPA x  (0 ; 18 ]  answer / karabo (A) (2) 8 [16] Copyright reserved/Ditokelo di sireleditswe Please turn over/ Phetla ### Page 12 Downloaded from hlayiso.com 12 MATHEMATICS P1/ DIPALO P1 (EC/LWETSE2021) QUESTION 5/POTSO 5 5.1 −𝑥 2 − 2𝑥 + 8 = 0  f ( x) = 0 𝑥 2 + 2𝑥 − 8 = 0 (𝑥 + 4)(𝑥 − 2) = 0 ∴ 𝑥 = −4 or 𝑥 = 2 ∴ 𝑅(−4; 0) and 𝑆(2; 0)  factorisation /fekithorisashini ∴ RS = 6 units  values of x / velu tsa x  answer / karabo (4) Copyright reserved/ Ditokelo di sireleditswe Please turn over/ phetla ### Page 13 Downloaded from hlayiso.com (EC/LWETSE2021) MATHEMATICS P1/DIPALO P1 13 5.2 −4 + 2 x= 2  method / method = −1  x = −1 y = −(−1) − 2(−1) + 8 2 =9  y =9  T ( −1;9 ) OR/KAPA OR/KAPA f ( x) = − x − 2 x + 8 2 b x=−  − b 2a 2a  −2  = −   2(−1)  = −1  x = −1 y = −(−1) − 2(−1) + 8 2 =9  y =9  T(−1;9) OR/KAPA OR/KAPA  −2 x − 2 = 0 f '( x) = −2 x − 2 = 0 − 2x = 2 x = −1 y = −(−1) 2 − 2(−1) + 8 =9  x = −1  y = 9 (3)  T(−1;9) 5.3.1 f ( x) = − x 2 − 2 x + 8 f '( x) = −2 x − 2  f / ( x) −2 x − 2 = 2  f / ( x) = 2  x = −2  x = −2  y = − (−2) 2 − 2(−2) + 8 =8  y =8 W (−2;8) (4) 5.3.2 𝑔(𝑥) = 𝑚𝑥 + 𝑐 1 𝑚 = − (⊥ lines) 2  gradient / keratiente 𝑐=8 1 ∴𝑦 =− 𝑥+8 2  equation / ekweishini (2) Copyright reserved/Ditokelo di sireleditswe Please turn over/ Phetla ### Page 14 Downloaded from hlayiso.com 14 MATHEMATICS P1/ DIPALO P1 (EC/LWETSE2021) 5.4 f ( x) = − x 2 − 2 x + 8 h( x) = − f ( x − 1)  − f ( x −1) = −  −( x − 1) 2 − 2( x − 1) + 8  substitution / sabosititshushini = −  −( x 2 − 2 x + 1) − 2 x + 2 + 8 = −  − x 2 + 2 x − 1 − 2 x + 2 + 8  simplifying / = x2 − 9  equation / ekweshini OR/KAPA OR/KAPA h( x) = ( x + 3)( x − 3) = x2 − 9  roots/ di ruthi 3 and/le–3  +( x + 3)(x – 3)  equation / ekweshini OR/KAPA OR/KAPA New turning point / theneng pointe entjha = (0 ; –9) y = x2 − 9  (0 ; –9)  equation / ekweshini (4) [17] QUESTION 6/ POTSO 6 6.1 A = P(1 − i ) n 5510 = 9 670(1 − i) 4  subst. into correct formula 5510 Sabosit ho moralo o i = 1 − 4 nepahetseng 9 670 = 0,131177  simplification Ho nolofatsa  r = 13,12%  answer / karabo (3) Copyright reserved/ Ditokelo di sireleditswe Please turn over/ phetla ### Page 15 Downloaded from hlayiso.com (EC/LWETSE2021) MATHEMATICS P1/DIPALO P1 15 6.2 End of December 𝑥[(1 + 𝑖)𝑛 − 1] 𝐹= 𝑖 0,087 144 600 [(1 + ) − 1] 12 ∴𝐹= 0,087 12 = 𝑅151 438,20  n = 144 End of January  subst. into correct formula 𝐴 = 𝑃(1 + 𝑖)𝑛 Sabust ho moralo o nepahetseng 0,087 = 151 438,20 (1 + ) 12 = 𝑅152 536,13 OR/KAPA  adding final month’s interest Aketsa tswala ya kgwedi ya ho qetela 𝑥[(1 + 𝑖)𝑛 − 1](1 + 𝑖) 𝐹=  answer / karabo 𝑖 0,087 144 0,087 600 [(1 + ) − 1] (1 + ) 12 12 OR/KAPA ∴𝐹= 0,087 12 = 𝑹152 536,13  n = 144  subst. into correct formula Sabost. ho moralo o nepahetseng  adding final month’s interest Aketsa tswala ya kgwedi ya ho qetela  answer / karabo (4) 6.3.1 𝑥[1 − (1 + 𝑖)−𝑛 ] 𝑃= 𝑖 0,093 −72 𝑥 [1 − (1 + ) ] 12 350 000 = 0,093 12 0, 093 350 000 × 0,093  i= and n = 72 ∴𝑥= 12 12 0,093 −72  substitution into correct formula [1 − (1 + ) ] 12 Sabusititshushini ho moralo o ∴ 𝑥 ≈ 𝑅6 361,18 nepahetseng  answer /karabo (3) Copyright reserved/Ditokelo di sireleditswe Please turn over/ Phetla ### Page 16 Downloaded from hlayiso.com 16 MATHEMATICS P1/ DIPALO P1 (EC/LWETSE2021) 6.3.2 Outstanding balance = 𝑥[1 − (1 + 𝑖)−𝑛 ] 𝑃= 𝑖 0,093 −32 6 361,18 [1 − (1 + ) ] 12 ∴𝑃= 0,093 12 = 𝑅179 667,32  𝑖 = 0,093 12 and = 32  subst. into correct formula OR/KAPA sabosit ho moralo o nepahetseng Outstanding balance  P = 179 667, 32 =𝐴−𝐹 (3) 0,093 40 0,093 40 6 361,18 [(1 + ) − 1] 12 = 350 000 (1 + ) − OR/KAPA 100 0,093 12 = 𝑅476 628,84 − 𝑅296 961,79 = 𝑅179 667,05  𝑖 = 0,093 12 and = 40  subst. Into correct formula Sabosit. Ho moralo o nepahetseng  P = 179 667, 32 (3) Copyright reserved/ Ditokelo di sireleditswe Please turn over/ phetla ### Page 17 Downloaded from hlayiso.com (EC/LWETSE2021) MATHEMATICS P1/DIPALO P1 17 6.3.3 0,093 −𝑛 7 000 [1 − (1 + ) ] 12 ∴ 179 667,32 = 0,093 12  subst. into correct formula 0,093 −𝑛 179 667,32 × 12 − 1 = − (1 + 0,093) Sabositi. Ho moralo o 7000 12 nepahetseng 4 031 −𝑛 −0,80108. . . = − ( ) 4 000 −𝑛 4 031 ∴ 0,80108. . . = 4 000 𝑙𝑜𝑔 0 , 80 108. . . ∴ −𝑛 = 4 031 𝑙𝑜𝑔 4 000 −𝑛 ≈ −28,73 ∴ 𝑛 ≈ 28,73 The number of months is 29. Le nani la dikgwedi ke 29.  correct use of logs Tsela e nepahetseng ya ho OR/KAPA sebedisa dilogo 0,093 −𝑛 7 000[1−(1+ 12 ) ] 179 667,32 = 0,093 12  = 28,73 0,093 −𝑛 179 667,32 × 12 − 1 = − (1 + 0,093) 7 000 12 4 031 −𝑛  n = 29 months /dikgwedi −0,80108. . . = − ( ) 4 000 −𝑛 4 031 OR/KAPA ∴ 0,80108. . . = 4 000 𝑙𝑜𝑔 0 , 80 108. . . ∴ −𝑛 = 4 031 𝑙𝑜𝑔 4 000 −𝑛 ≈ −28,73 ∴ 𝑛 ≈ 28,73  subst. into correct formula ∴The number of months is 29. Sabosit. Ho moralo o Le nani la dikgwedi 29 nepahetseng  correct use of logs Tsela e nepahetseng ya ho sebedisa di logo  = 28,73  n = 29 months / kgwedi (4) [17] Copyright reserved/Ditokelo di sireleditswe Please turn over/ Phetla ### Page 18 Downloaded from hlayiso.com 18 MATHEMATICS P1/ DIPALO P1 (EC/LWETSE2021) QUESTION 7/ POTSO 7 7.1 f ( x + h) − f ( x ) f '( x) = lim h →0 h = lim ( 5 − 2 ( x + h ) − 5 − 2x2 2 )  substitution / sabosititshushini h →0 h 5 − 2 x − 4 xh − 2h 2 − 5 + 2 x 2 2 = lim  expansion / ho oketsa h →0 h −4 xh − 2h 2 = lim  simplification / ho nolofatsha h →0 h h(−4 x − 2h) = lim h →0 h = lim ( −4 x − 2h )  notation and lim (−4 x − 2h) h →0 h →0 = −4 x  answer / karabo (5) 2 2x 7.2.1 y = 7 x4 + x 3 3 = 7 x4 + 2 x 2  2x 2 dy 1  = 28 x 3 + 3 x 2  28x3  3x 2 1 dx (3) 7.2.2  3x 2 − 7 x − 6  = Dx    x  = Dx 3x − 7 − 6 x  −1  3x − 7  −6x −1 = 3 + 6 x −2  3 and differentiating constant 3 le difarenshiething konstent  +6x −2 (4) [12] Copyright reserved/ Ditokelo di sireleditswe Please turn over/ phetla ### Page 19 Downloaded from hlayiso.com (EC/LWETSE2021) MATHEMATICS P1/DIPALO P1 19 QUESTION 8/ POTSO 8 8.1.1 f ( x) = 2( x − x1 )( x − x2 )( x − x3 ) = 2 ( x + 1) ( x − 12 ) ( x − 3)  f ( x) = 2 ( x + 1) ( x − 12 ) ( x − 3) = ( x + 1)( 2 x − 1)( x − 3) OR/KAPA ( = ( x + 1) 2 x − 7 x + 3 2 )  f ( x) = ( x + 1)( 2 x −1)( x − 3) = 2 x − 7 x + 3x + 2 x − 7 x + 3 3 2 2  expansion / ho oketsa = 2 x3 − 5 x 2 − 4 x + 3  simplifying / ho etsa bobebe f ( x) = 2 x 3 + bx 2 + cx + d (4)  b = −5, c = − 4, d = 3 8.1.2 f '( x) = 6 x 2 − 10 x − 4 0 = 6 x 2 − 10 x − 4  f ' ( x ) = 6 x2 −10 x − 4 = 0  3x 2 − 5 x − 2 = 0 ( 3x + 1)( x − 2 ) = 0  factorisation / fekithorisashini  x = − or / of x = 2 1 3 ∴N is at f (2) f (2) = 2 ( 2 ) − 5 ( 2 ) − 4 ( 2 ) + 3 3 2  chosing/ho kgetha : x = 2 = −9 N ( 2 ; − 9)  y = −9 (4) 8.1.3 (a) − x2 1 3  answer / karabo (2) 8.1.3 (b) f ''( x) = 12 x − 10  f ''( x) = 12x −10 12 x − 10  0  f ''( x)  0 12 x  10 5  answer / karabo x 6 OR/KAPA OR/KAPA − +2 5 1 x= = 3 5 2 6 x= 6 5 x 5  5 6  x  OR/KAPA x   − ;  6  6 (3) 8.2 y  f (0) = 0  (m ; 0) (m;0)  shape / seemo x (3) [16] Copyright reserved/Ditokelo di sireleditswe Please turn over/ Phetla ### Page 20 Downloaded from hlayiso.com 20 MATHEMATICS P1/ DIPALO P1 (EC/LWETSE2021) QUESTION 9/VRAAG 9 9.1 A = ( 12 15 x  8 x  2 ) + (15 xy ) + ( 8 xy ) + (17 xy )  total surface area / 5760 = 120 x 2 + 40 xy Totale safeisi eria 5760 − 120 x 2 y =  5760 = 120 x2 + 40 xy 40 x (2) 9.2 V = ( 12 b.h)  H V = 12 15 x  8 x  y 1 5760 − 120 x 2  substitution into V = 15 x  8 x  Sabosititshushioni ho V 2 40 x = 60 x (144 − 3 x 2 )  substituing for y = 8640 x − 180 x 3 Sabositikhushini ya y (2) 9.2 V '( x) = 8640 − 540 x 2 V '( x) = 0  V '( x) = 8640 − 540 x 2  8640 − 540 x 2 = 0  V '( x) = 0 8640 = 540 x 2 x 2 = 16  simplification / x = 4 Ho nolofatsa  answer / karabo (4) [8] Copyright reserved/ Ditokelo di sireleditswe Please turn over/ phetla ### Page 21 Downloaded from hlayiso.com (EC/LWETSE2021) MATHEMATICS P1/DIPALO P1 21 QUESTION 10/POTSO 10 10.1.1 𝑃(𝐵) = 1 − 𝑃(not 𝐵) = 1 − 0,45 = 0,55  0,55 (1) 10.1.2 𝑃(𝐴 and 𝐵) = 𝑃(𝐴) × P(𝐵) = 0,2 × 0,55  P(A)×P(B) = 0,11 𝑃(𝐴 or 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 and 𝐵) = 0,2 + 0,55 − 0,11 16 = 0,64 or 25  substitution / sabosititshushini  answer / karabo (3) 10.2 P(late/laat ) = 1 x + 3 (1 − x)  12 x + 53 (1 − x) 2 5 1 x + 3 (1 − x) = 8  equating / ho le kantsha 2 5 15 15 x + 18(1 − x) = 16  substitution / sabosititshushini 15 x + 18 − 18 x = 16 − 3 x = −2 2 x= 3  answer / karabo (4) [8] Copyright reserved/Ditokelo di sireleditswe Please turn over/ Phetla ### Page 22 Downloaded from hlayiso.com 22 MATHEMATICS P1/ DIPALO P1 (EC/LWETSE2021) QUESTION 11/POTSO 11 11.1 @@@ ###  20  22  21 20  22  21  10  10  10  10  10  10 = 9 240 000  answer / karabo (3) 11.2 20  19  3  10  10  5 + 20  3  19  10  10  5  20  19  3  10  10  5 9 240000 1140 000  20  3  19  10  10  5 =  adding / ho aketsa 9 240000 19  9 240 000 = or / of 0,12 or / of 12,34%  answer / karabo 154 (5) [8] TOTAL/KAOFELA: 150 Copyright reserved/ Ditokelo di sireleditswe Please turn over/ phetla ## Related documents - [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 - [Gr12 Mathematics P1 (English) 2021 Preparatory Question Paper](https://www.hlayiso.com/library/gr12-mathematics-p1-english-2021-preparatory-question-paper/37860cd4-80b8-4144-af87-6a1797b9bda3) — Mathematics · QP · 2021 · Eng - [MATHS P1 QP Gr12 September 2014 English](https://www.hlayiso.com/library/maths-p1-qp-gr12-september-2014-english/bc7b1c66-4908-4bdc-8143-7ee244f10867) — Mathematics · QP · 2014 · Eng - [Maths P1 Memo](https://www.hlayiso.com/library/maths-p1-memo/7d83c9c7-adb9-4818-93b0-08773c9e1945) — Mathematics · Memo · 2016 - [MATHS P1 GR12 QP JUNE 2023 English](https://www.hlayiso.com/library/maths-p1-gr12-qp-june-2023-english/245dbccd-9e3b-4add-924a-72f6c39ee078) — Mathematics · QP · 2023 · Eng - [Mathematics P1 (English) QP](https://www.hlayiso.com/library/mathematics-p1-english-qp/11462bfe-d0db-4e5e-8cc4-7aa8f32251ac) — Mathematics · QP · 2023 · Eng ## More from Grade 12 Mathematics - [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 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 - [MATHS P2 GR12 QP JUNE 2024 English](https://www.hlayiso.com/library/maths-p2-gr12-qp-june-2024-english/f19338aa-ae27-4fb7-9f1b-e60b2b91eb8c) — QP · 2024 · Eng - [FS Gr 12 Sept 2024 Afr en Eng P1](https://www.hlayiso.com/library/fs-gr-12-sept-2024-afr-en-eng-p1/4d7be3dc-5f53-4da3-a833-8fa9c4d73be1) — QP · 2024 · Eng - [G12 Maths P2 Sept Free State 2024](https://www.hlayiso.com/library/g12-maths-p2-sept-free-state-2024/b81ee450-da0b-40b1-84a9-e4089f8e3d30) — QP · 2024 ## 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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