# Mathematics Grade 11 P1 Gauteng November 2021 Memo (English) > Grade 11 Mathematics Paper 1 memorandum from the Gauteng November 2021 (English), 17 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr11-mathematics-p1-eng-nov-possible-answers/7e8d3b3d-853d-4943-a851-44ed3fc9fc98 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Document type: Memorandum - Year: 2021 - Exam period: Gauteng November - Paper: 1 - Curriculum: CAPS - Pages: 17 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 November 2021 Memo (English)](https://www.hlayiso.com/library/gr11-mathematics-p1-eng-nov-possible-answers/7e8d3b3d-853d-4943-a851-44ed3fc9fc98) ## About this document Mathematics · Grade 11 · Gauteng November · 2021. Memorandum, 17 pages. 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Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com PROVINCIAL EXAMINATION NOVEMBER 2021 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 1) 17 pages ### Page 2 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 INSTRUCTIONS AND INFORMATION A – ACCURACY C.A. – CONSISTENT ACCURACY NOTE: • If a candidate answered a question TWICE, mark only the first attempt. • If a candidate CROSSED OUT an answer and did not redo it, mark the crossed-out answer. • Consistent accuracy applies to ALL aspects of the marking guidelines. • Assuming values/answers in order to solve a question is UNACCEPTABLE. 2 ### Page 3 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 1 1.1 1.1.1 (𝑥 + 2)(2𝑥 − 1) = 0  simplification 1  both values for x 𝑥 = −2 or 𝑥 = 2 (2) 1.1.2 3− x = 0  x=3  answer (1) 1.2 1.2.1 2 x 2 − 3x − 4 = 0 −b  b 2 − 4ac x= 2a  correct substitution −(−3)  (−3) − 4(2)(−4) 2 x=  answers 2(2) x = 2,35 or x = −0,85 (3) 1.2.2 (2 x + 3)(3 − x)  4 −2 x 2 + 3 x + 9  4  simplify −2 x 2 + 3 x + 5  0  standard form 2 x − 3x − 5  0 2  factors  critical values (2 x − 5)( x + 1)  0  answer 5 −1  x  2 NOTE: Candidates can write the answer as separate inequalities but must state AND between the inequalities, not OR. (5) 1.2.3 x+2+x =4 x+2 = 4− x  squaring both sides x + 2 = 16 − 8 x + x 2 𝑎2 + 𝑏 2 = 𝑐 2  standard form  factors 0 = x − 9 x + 14 2  critical values 0 = ( x − 7)( x − 2)  exclusion of x = 7 x  7 or x = 2 (5) 3 ### Page 4 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 1.2.4 22 x +1 − 3(22 x −1 ) + 4 x = 12  3  22 x  2 − + 1 = 12  correct factorisation  2  3  simplification 2 2 x = 8 22 x   = 12  equating to like bases 2  answer 22 x = 8 2 2 x = 23 3 x= 2 (4) 1.3 1.3.1 f ( x) = x 2 + 8 x + 16 f (2) = 22 + 8(2) + 16 f (2) = 36  answer (1) 1.3.2 f ( x)  0 x 2 + 8 x + 16  0  factors ( x + 4)( x + 4)  0 ( x + 4) 2  0  critical value CV : − 4  answer x ; x  −4 NOTE: Both criteria must be presented for the answer mark. (3) 4 ### Page 5 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 1.4 𝑦 − 2𝑥 + 1 = 0 and 𝑥𝑦 = 2𝑦 + 𝑥 2 + 3𝑥 − 10 … (2) 𝑦 = 2𝑥 − 1 … (1)  y as subject in (1) sub (2) into (1)  substitution  standard form 2 𝑥(2𝑥 − 1) = 2(2𝑥 − 1) + 𝑥 + 3𝑥 − 10  factors 2𝑥 2 − 𝑥 = 4𝑥 − 2 + 𝑥 2 + 3𝑥 − 10  values for x 𝑥 2 − 8𝑥 + 12 = 0 (𝑥 − 6)(𝑥 − 2) = 0  values for y 𝑥 = 6 or 𝑥 = 2 if 𝑥 = 6 if 𝑥 = 2 𝑦 = 2(6) − 1 𝑦 = 2(2) − 1 𝑦 = 11 𝑦=3 NOTE: Any other valid method. (6) 1.5 ( xp ) 2 + 3 = 2 x( p − x) x 2 p 2 + 3 = 2 px − 2 x 2  standard form x p + 3 − 2 px + 2 x = 0 2 2 2 ( p 2 + 2) x 2 − 2 xp + 3 = 0  substitution = b 2 − 4ac = ( −2 p) 2 − 4( p 2 + 2)(3)  value for ∆ = 4 p 2 − 12 p 2 − 24 = −8 p 2 − 24  complete explanation p 2  0 for p  −8 p 2  0 for p  −8 p 2 − 24  0 for p   roots are non-real. (4) [34] 5 ### Page 6 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 2 1 2.1 2.1.1 −  16 x 2  2  −2   81x   interchange base AND 1 additive inverse of  81x −2  2 index = 2   16 x   answer 9 = 2 4x NOTE: Any correct alternative method. (2) 2.1.2 9 x + 32 x +1 18x.2− x  simplification of 9x and 32 x + 32 x +1 18x = 2x x −x 3 .2 .2  factorise 32 x (1 + 3)  answer = 2x 0 3 .2 =4 NOTE: Any correct alternative method. (3) 2.1.3 10 − 32 x = 3x − 2  squaring both sides 10 − 32 x = (3x − 2) 2  simplification 10 − 32 x = 32 x − 4.3x + 4  standard form 2.32 x − 4.3x − 6 = 0  factors 32 x − 2.3x − 3 = 0 (3x − 3)(3x + 1) = 0  value of x and rejection 3x = 3 or 3x = −1 x =1 no solution (3x  0) (5) 2.2 3 2a − 2 3a  surd and squaring = (3 2a − 2 3a ) 2  multiplication = 9. 4a 2 − 12 6a 2 + 4. 9a 2  simplification  simplification = 9.2a − 12 6a 2 + 4.3a 18a − 12a 6 + 12a = 30a − 12a 6 (4) [14] 6 ### Page 7 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 3 3.1 3.1.1 A=9  answer B = 11  answer (2) 3.1.2 Tn = 2n + 1  answer (2) 3.1.3 Tn = 2n + 1 Tn = 2(100) + 1  substitution Tn = 201  answer NOTE : Answer only, award FULL marks. (2) 3.1.4 Tn = 2n + 1 305 = 2n + 1  substitution 304 = 2n  answer n = 152 (2) 3.2 𝑇3 − 𝑇2 = 𝑇2 − 𝑇1 3𝑥 2 + 4𝑥 − 3 − (2𝑥 2 + 5𝑥 + 2)  substitution = 2𝑥 2 + 5𝑥 + 2 − (2𝑥 2 + 5𝑥 + 5 ) 3𝑥 + 4𝑥 − 3 − 2𝑥 2 − 5𝑥 − 2 2  simplification = 2𝑥 2 + 5𝑥 + 2 − 2𝑥 2 − 5𝑥 − 5 𝑥 2 − 𝑥 − 5 = −3  standard form 𝑥2 − 𝑥 − 2 = 0 (𝑥 − 2)(𝑥 + 1)  answers ∴ 𝑥 = 2 or 𝑥 = −1 (4) [12] 7 ### Page 8 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 4 4.1 -1 ; x ; 3 ; x+8 x-(-1) 3-x x+8-3  first differences 3-x-(x+1) x+5-(3-x)  second differences 3 − x − ( x + 1) = x + 5 − (3 − x)  equating 3 − x − x −1 = x + 5 − 3 + x 2 − 2x = 2x + 2  answer 4x = 0 x = 0 (4) 4.2 -1 ; 0 ; 3 ; 8  2nd difference 1 3 5  value for a 2 2  value for b 2a = 2 a =1  value for c 3a + b = 1  equation of Tn 3(1) + b = 1  complete the square of b = −2 Tn a + b + c = −1 1 − 2 + c = −1 c=0 Tn = n 2 − 2n Tn = n 2 − 2n + 1 − 1 Tn = (n − 1) 2 − 1 NOTE: If a candidate uses Tn = (n − 1) − 1 to 2 validate the quadratic pattern, award ZERO marks. (6) 8 ### Page 9 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 4.3 Tn = (n − 1) 2 − 1  substitution T65 = (65 − 1) 2 − 1  answer T65 = 4095 (2) 4.4 (n − 1) 2 − 1  9800  setting up inequality n 2 − 2n − 9800  0  factors (n + 98)(n − 100) CVs -98 or 100  choice of n = 100  answer n  100  n = 101 (4) [16] QUESTION 5 5.1 5.1.1 A = P(1 − i ) n  correct substitution A = 250000(1 − 0.1)5 into correct formula A = R147622,50  answer (2) 5.1.2 𝐴 = 𝑃(1 + 𝑖)𝑛  correct substitution 𝐴 = 250 000(1 + 0.08)5 into correct formula 𝐴 = 𝑅367 332,02  answer (2) 5.1.3 95 × 147 622.50 100 = 𝑅140 241,375  R140 241, 375 New Machine - Sale of Old Machine  answer 𝑅367 332,02 − 𝑅140 241,375 𝑅227 090,65 (2) 5.2 5.2.1  i  m 1 + ieff = 1 + nom   m   correct substitution 12  0.115  into correct formula 1 + ieff = 1 +   12,13%  12   conclusion ieff = 12,13% Bank B offers a better interest rate. NOTE: Award no marks if candidate concludes only. (3) 9 ### Page 10 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 5.2.2 𝐴 = 𝑃(1 + 𝑖)𝑛 0.115 24 0.115 36  value of i 𝑅280 000 = [𝑃 (1 + ) − 50 000] (1 + ) 12 12  n = 24 𝑃 = 𝑅197 759,51.  n = 36  setting up equation OR  answer 𝑛 𝐴 = 𝑃(1 + 𝑖) 0.115 60 0.115 36 OR 𝑅280000 = 𝑃 (1 + ) − 50 000 (1 + ) 12 12  value of i 𝑃 = 𝑅197 759,51  n = 60  n = 36 OR  setting up equation 𝐴 = 𝑃(1 + 𝑖)𝑛  answer 𝑅280 000 = [𝑃(1 + 0,1213)2 − 50 000](1 + 0,1213)3 OR 𝑃 = 𝑅197 729,98  value of i  n=2 OR  n=3 𝐴 = 𝑃(1 + 𝑖)𝑛  setting up equation 𝑅280 000 = 𝑃(1 + 0,1213)5 − 50 000(1 + 0,1213)3  answer 𝑃 = 𝑅197 727.98 OR  value of i  n=5  n=3  setting up equation  answer (5) [14] 10 ### Page 11 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 6 6.1 3 f ( x) = −1 x+2 3  substitution 2= −1 k +2 3 3= k +2  answer 3k + 6 = 3 k = −1 (2) 6.2 x = −2  answer y = −1  answer (2) 6.3 3 f ( x) = −1 x+2 3 f (0) = −1 0+2 1 = 2  answer  1   0;   2 NOTE: Answer MUST be presented in coordinate form. (1) 6.4 3 f ( x) = −1 x+2 3 0= −1 x+2  simplification 3 1= x+2  answer x+2=3 x =1 (1;0) NOTE: Answer does NOT need to be in coordinate form. (2) 11 ### Page 12 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 6.5  shape  asymptotes  x- and y- intercepts (3) 6.6 m = −1 and (-2;-1) is on the axis of symmetry  y = mx + c  substitution for m −1 = −1(−2) + c AND the point  c = −3  answer  y = −x − 3 NOTE: Any other valid method. (2) 6.7 x > –2  answer (1) 6.8 3 f ( x) = −1 x+2 3 −y = −1  make y negative x+2  3   subtract 4 y = − − 1  x+2  −3  answer y= +1 x+2−4 −3 y= +1 x−2 (3) [16] 12 ### Page 13 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 7 7.1 q=5  answer (1) 7.2 f ( x) = − x 2 − 4 x + 5 x2 + 4x − 5 = 0  factors ( x + 5)( x − 1) = 0  critical values  answer x = −5 or x = 1  EP = 6 units (3) 7.3 −𝑏 𝑥= 2𝑎 −(−4) 𝑥=  x-value 2(−1) 𝑥 = −2 𝑓(−2) = 9  y-value ∴ 𝑁(−2; 9) OR OR 𝑓(𝑥) = −(𝑥 2 + 4𝑥) + 5  x-value 𝑓(𝑥) = −(𝑥 2 + 4𝑥 + 22 ) + 5 + 4  y-value 𝑓(𝑥) = −(𝑥 + 2)2 + 9 ∴ 𝑁(−2; 9) NOTE: Answer does NOT have to be in coordinate form. (2) 7.4 m of g = 1  m of h = −1  gradient of h = –1 y = mx + c 9 = −1(−2) + c  c=7 c=7  answer h( x ) = − x + 7 NOTE: Any other valid method. (3) 7.5 M(0;5) and N(-2;9) y −y mMN = 2 1 x2 − x1 9−5  correct substitution mMN = −2 − 0  answer mMN = −2 NOTE: NO marks for applying incorrect gradient formula. (2) 13 ### Page 14 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 7.6 At point N: f (−2) = 9  g(–2) = 3 At point T: g (−2) = 3  method ∴ distance of NT = 9 − 3  answer = 6 units OR OR NT = f ( x) − g ( x) = − x 2 − 4 x + 5 − ( x + 5) = − x2 − 5x  method x = −2  NT = −(−2) 2 − 5(−2)  substitution NT = 6 units  answer NOTE: Any other valid method. (3) 7.7 By symmetry: 𝑆(−4; 5)  answer OR OR At point S: y = 5 ∴ 5 = −𝑥 2 − 4𝑥 + 5  y=5 𝑥 2 + 4𝑥 = 0  equating 𝑥(𝑥 + 4) = 0 𝑥 = 0 or 𝑥 = −4  selection of –4 ∴ 𝑆(−4; 5) (3) 7.8 7.8.1 x=–5  answer OR  answer x=0 (2) 7.8.2 x 1  answer (1) [20] 14 ### Page 15 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 8 8.1 let x = 0 f ( x) = 3.2 x − 6 f (0) = 3.20 − 6 y = −3  answer  (0; −3) NOTE: Answer MUST be in coordinate form. (1) 8.2 let y = 0 0 = 3.2 x − 6 6 = 3.2 x  simplify both sides 2=2 x to base 2  answer x =1  (1;0) NOTE: Answer does NOT have to be in coordinate form. (2) 8.3  shape  asymptote  x- and y- intercepts (3) 8.4 𝑦 > −6 𝐎𝐑  answer 𝑦 ∈ [−6; ∞] (1) [7] 15 ### Page 16 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 9 NOTE: Candidates may present answers as a fraction, decimal fraction or percentage. 9.1 9.1.1 120 120 P(male)=  236 236 P(male)= 0,51  answer (2) 9.1.2 67 𝑃(Female and plays sport) = 236 67  𝑃(Female and plays sport) = 0,28 236  answer (2) 9.2 51 𝑃(Male and does not play sport) = 236  0,22 𝑃(Male and does not play sport) = 0,22  conclusion ≠0 NOT mutually exclusive OR OR  statement There are 51 male learners who do not play sport.  conclusion Therefore the two are NOT mutually exclusive. NOTE: Accept alternative answers expressed as a percentage or in fraction form. (2) 9.3 𝑃(male and does not play sport) = 0,22  P(male) x 𝑃(male)×P(does not play sport) P(does not play sport) 120 100 = 0,22 = × 236 236 = 0,22  equating ∴ 𝑃(male and does not play sport)  conclusion = 𝑃(male)×P(does not play sport) ∴ the events ARE independent (correct to two decimal places) (4) [10] 16 ### Page 17 Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 10 10.1 c = 84 − 75 c=9  answer (1) 10.2 f + 18 + x = 41  equating correctly f = 41 − x − 18  answer f = 23 − x (2) 10.3 f = 23 − x  x = 23 − f e = 17 − x  x = 17 − e  equating expressions of x  23 − f = 17 − e  equating to 75 but 40 + e + f + x = 75  simplification  e + f + x = 35  answer  17 − x + 23 − x + x = 35 40 − x = 35 x =5 (4) [7] TOTAL: 150 17 ## Related documents - [Gr11 Mathematics P2 (AFR) NOV Question Paper](https://www.hlayiso.com/library/gr11-mathematics-p2-afr-nov-question-paper/2cb86fbf-4570-48cc-a457-1097445cbdfb) — Mathematics · QP · 2021 · Afr - [Gr11 Mathematics P2 (ENG) NOV Question Paper](https://www.hlayiso.com/library/gr11-mathematics-p2-eng-nov-question-paper/9cf54202-491b-49e3-803e-677cd1be3b16) — Mathematics · QP · 2021 · Eng - [Gr 11 Mathematics P1 (English) November 2022 Question Paper](https://www.hlayiso.com/library/gr-11-mathematics-p1-english-november-2022-question-paper/35e1a85a-0c9e-4744-b794-2a0f7a262d46) — Mathematics · QP · 2022 · Eng - [Gr11 Math P1 November 2016 Question Paper](https://www.hlayiso.com/library/gr11-math-p1-november-2016-question-paper/159ea3da-41f3-4dcb-82d8-aeb9f1794ecd) — Mathematics · QP · 2016 - [Gr11 Math P1 (Afrikaans) November 2016 Question Paper](https://www.hlayiso.com/library/gr11-math-p1-afrikaans-november-2016-question-paper/e18ca4fc-229b-4d0b-8b3f-ec859a499bea) — Mathematics · QP · 2016 · Afr - [Maths Grade 11 November 2023 P1 and Memo](https://www.hlayiso.com/library/maths-grade-11-november-2023-p1-and-memo/59d9512d-30fd-4836-9eda-81b77787e852) — Mathematics · Memo · 2023 ## 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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