# Mathematics Grade 11 P1 Gauteng November 2022 Memo (English) > Grade 11 Mathematics Paper 1 memorandum from the Gauteng November 2022 (English), 19 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr-11-mathematics-p1-english-november-2022-possible-answers/21c9fe27-5f1a-43e6-a8c7-b90adb07756c - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Language: English - Document type: Memorandum - Year: 2022 - Exam period: Gauteng November - Paper: 1 - Curriculum: CAPS - Pages: 19 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 2022 Memo (English)](https://www.hlayiso.com/library/gr-11-mathematics-p1-english-november-2022-possible-answers/21c9fe27-5f1a-43e6-a8c7-b90adb07756c) ## About this document Mathematics · Grade 11 · Gauteng November · 2022 · English. Memorandum, 19 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Matching paper and memo - [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) — QP · 2022 · Eng ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com PROVINCIAL EXAMINATION NOVEMBER 2022 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 1) 19 pages ### Page 2 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 1 1.1 1.1.1 When x = 2 and when x = 3 3− x 3−3 3− 2 0 and 1 =0 =1  x = 2 or x = 3 ✓ ✓ answers (2) 1.1.2 3 − x = 2x – 3 3 – x  0 and 2x – 3  0 ✓ setting up correct inequalities – x  –3 and 2x  3 3  x  3 and x  ✓✓ answers 2 3  x3 2 NOTE: Answer can be written as separate inequalities. OR NOTE: If the candidates solve the equation, the 3 answers are: x = or x = 2, the answer 4 3 3 of x = must be rejected to obtain , but 4 3 2 if not rejected, award . 3 (3) 1.2 (3a – 8)(2b + 7) = 0  3a – 8 = 0 or 2b + 7 = 0 8 a= ✓ value of a 3 7 b=– ✓ value of b 2 (2) 2 ### Page 3 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 1.3 1.3.1 4x2 – 20x + 1 = 0 − b  b 2 − 4ac x= 2a − ( −20)  ( −20) 2 − 4(4)(1) ✓ substitution x= 2(4) 20  384 x= 8  x = 4,95 or x = 0,05 ✓✓ answers NOTE: Penalise one mark for incorrect rounding-off in this question ONLY. (3) 1.3.2 (x + 1)(x – 3) > 12 x2 – 2x – 3 > 12 x2 – 2x – 15 > 0 ✓ standard form (x – 5) (x + 3) > 0 ✓ factors  x > 5 or x < –3 ✓✓ answers (4) 1.3.3 x – 5+ x = 7 x–7= 5+ x (x – 7)2 = ( 5 + x ) 2 ✓ squaring both sides x2 – 14x + 49 = 5 + x ✓ simplification x2 – 15x + 44 = 0 ✓ standard form (x – 4)(x – 11) = 11 ✓ factors x = 11 or x  4 ✓ answers with rejection (5) 3 ### Page 4 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 1.4 1.4.1 x = 3 and y = a x–y=1 3–a=1 a=2 but x2 – 3xy + by2 = –5 ✓ value of a  32 – 3(3)(2) + b (2)2 = –5 ✓ substitute for x and y  9 –18 + 4b = –5  4b = 4 ✓ simplification b=1 ✓ value of b (4) 1.4.2 x – y = 1  x = y + 1……(1) ✓ x as subject x2 – 3xy + y2 = –5……(2)  (y + 1)2 – 3y(y + 1) + y2 = –5 ✓ substitute into (2)  y2 + 2y + 1 – 3y2 – 3y + y2 = –5  – y2 – y + 6 = 0 ✓ standard form  y2 + y – 6 = 0 ✓ factors (y + 3)(y – 2) = 0  y = – 3 or y = 2 ✓ y-values  x = – 2 or x = 3 ✓ x-values The other solution: (– 2;– 3) NOTE: Candidates do not have to write the answer in coordinate form. (6) 4 ### Page 5 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 1.5 1.5.1 (p + 1)x2 + 2px + (p + 2) = 0  = b2 – 4ac  = (2p)2 – 4(p + 1)(p + 2) ✓ substitution for   = 4p2 – 4p2 – 12p – 8  = – 12p – 8 ✓ expression for  For equal roots, =0 ✓ condition for  – 12p – 8 = 0 8 p=– 12 2 ✓ answer p=– 3 (4) 1.5.2  > 0 ✓ condition of  – 12p – 8 > 0 2 p<– 3 ✓ answer (2) [35] 5 ### Page 6 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 2 2.1 1 1 1 2.1.1 ( n−1 . n+1 ) n 3 3 1 = (3−n+1.3−n−1 ) n ✓ simplification 1 = ( 3− 2 n ) n = 3−2 ✓ simplification 1 = ✓ answer 9 NOTE: Any other valid method. (3) 5 2 2 2.1.2 – 3 27 2 − 2 + − −2 8 3 4 5 1 1 2 25 ✓ simplification = [(33 ) 2 ] 3 − + −2 −2 (2 3 ) 3 (2 2 ) 5 1 2 25 = 3 − −2 + 4 2 ✓ simplification 2 − 2 5 = 9 – 23 + 21 ✓ simplification =3 ✓ answer NOTE: Any other valid method. (4) 6 ### Page 7 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 2.2 32–x + 8 = 3x 32.3–x + 8 = 3x 9 + 8 = 3x ✓ simplification 3x 9 + 8.3x = 32x 32x + 8.3x – 9 = 0 ✓ standard form (3x – 9)(3x + 1) = 0 ✓ factors 3x = 9 or 3x = –1 3x = 32 or NA ✓ answer with rejection x=2 NOTE: Any other valid method. (4) [11] QUESTION 3 3.1 1 ; 2 ; 5 ; 10 +1 +3 +5 +2 +2 2 nd difference 2a = 2 3a + b = 1 a+b+c=1 ✓ value of a a=1 3(1) + b = 1 1–2+c=1 ✓ value of b b = –2 c=2 ✓ value of c  Tn = n2 – 2n +2 NOTE: No penalty if not written as an equation. (4) 3.2 T50 = 502 – 2(50) + 2 ✓ substitution T50 = 2402 ✓ answer (2) 3.3 ROW 11 ✓✓ answer (2) 7 ### Page 8 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 3.4 1 ; 3 ; 5 ; 7 +2 +2 +2 Tn = 2n – 1 ✓ correct equation 241 = 2n – 1 ✓ Tn = 241 n = 121 ✓ answer (3) [11] QUESTION 4 4.1 4 ; 9 ; x ; 37… 5 ; x − 9 ; 37 − x ✓ 1st differences x − 9 − 5 ; 37 − x − ( x − 9) ✓ 2nd differences x − 14 ; 46 − 2 x x – 14 = 46 – 2x ✓ equating  3x = 60  x = 20 ✓ answer (4) 4.2 5 ; 11 ; 17 ;… (first differences) Tn = 6n – 1 ✓ correct equation 599 = 6n – 1 ✓ equating 600 = 6n n = 100 ✓ value of n  Between 100th and 101th terms ✓ conclusion (4) 8 ### Page 9 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 4.3 4.3.1 4 ; 9 ; 20 ; 37 5 11 17 6 6 ✓ 2 nd difference 2a = 6 a=3  Tn = 3n2 + bn +c ✓ value of a Since the a-value is positive, Tn has a MINIMUM value. ✓ conclusion (3) 4.3.2 y = 3n2 – 4n +5 − (−4) x= 2(3) 4 x= ✓ value of x 6 2 x= 3  2  11 2  f  = ✓ value of f    3 3  3 11  range: y  ✓ answer 3 OR 11 y [ ; ) 3 NOTE: Correct brackets must be used to obtain the answer mark in option 2. This is a theoretical solution as it is understood that a number pattern is composed of terms which are indicated as natural numbers. (3) 9 ### Page 10 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 4.4 For n to be common to both patterns: 6n – 1 = 3n2 – 4n + 5 ✓ equating  0 = 3n2 – 10n + 6 − b  b 2 − 4ac n= 2a − ( −10)  ( −10) 2 − 4(3)(6) n= 2(3) 5 7 n= ✓ simplification 3  NO, n is not a Natural Number. ✓ conclusion (3) [17] QUESTION 5 5.1 1 + ieffective = (1 + inominal)n 0,135  inominal = ✓ inominal 12 0,135 12 ✓ correct substitution into  1 + ie = (1 + ) 12 correct formula  1 + ie = 1,14  ie = 0,14 ✓ simplification ie  ie = 14% ✓ answer (4) 10 ### Page 11 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 5.2 A = P(1 + in)  7 500 = 6 800(1 + 0,075n) ✓ substitution into correct formula 7 500  – 1 = 0,075n 6 800 0,075n = 0,10  n = 1,333 yrs ✓ value of n  n = 1,333(12)  n = 15,996  n  16 mnths ✓ answer NOTE: The answer mark is for 16 months. (3) 5.3 Compounded semiannually for 6 years.  n = 12 ✓ value of n and i 0,05 i= = 0,025 2 A = P(1 + i)n A = 5 000(1 + 0,025)12 ✓ substitution into correct formula  A = R6 724,44 ✓ answer (3) 5.4 5.4.1 A = P(1 + in) A = P + Pin  linear function ✓ answer (1) 11 ### Page 12 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 5.4.2 m = Pi P>0 i>0 m>0 A ✓m>0 ✓ shape (c > 0) P n 0 (2) 5.4.3 n = 0: A = P + Pi(0) A=P n = 1: A = P + Pi(1) A = P + Pi  an INCREASE in Pi ✓ answer NOTE: Answer only, full marks. (1) [14] 12 ### Page 13 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 6 6.1 x=2 ✓ answer y=1 ✓ answer (2) 6.2 y = –1 ✓ answer (1) 6.3 x g f ✓ shape of f x=1 ✓ intercepts of f x ✓ shape of g O ✓ intercept of g x = –1 x=2 (4) 6.4 x  R; x  2 or x  (–;2) or (2; ) ✓ answer NOTE: Must state both conditions in option 1. (1) 6.5 y > –1 or y  (–1;) ✓ answer (1) 6.6 Point of intersection of asymptotes: (2;1) y – y1 = m(x – x1) y – 1 = –1(x – 2) ✓ substitute m and pt (2 ; –1) y = –x +3 ✓ answer (2) 6.7 2 units to the right and 1 unit upwards ✓ units right ✓ units up (2) 13 ### Page 14 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 6.8 d = ( x2 − x1 ) 2 + ( y 2 − y1 ) 2 d = (2 − 0) 2 + (1 − 0) 2  (0;0)(2;1) ✓ correct substitution d= 5 d = 2,24 ✓ answer NOTE: Answer only, full marks. No penalty for rounding-off incorrectly. (2) 6.9 (3 ; 0) ✓ x-value ✓ y-value (2) 6.10 x (–;2) or x<2 ✓ answer (1) [18] QUESTION 7 7.1 y = a( x − x1 )( x − x2 ) 4 = a(0 − 1)(0 − 3) ✓ substitute roots and point (0 ; 4) 4 = 3a 4 a = ✓ value for a 3 4 y= ( x − 1)( x − 3) 3 4 2 y= ( x − 4 x + 3) 3 4 2 16 y= x − x+4 ✓ answer 3 3 (3) 14 ### Page 15 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 4 2 16 7.2 f ( x) = x − x+4 3 3 −b x= 2a 16 − (− ) x= 3 ✓ substitution into correct 4 formula 2( ) 3 x=2 ✓ x-value 4 2 16  f (2) = (2) − (2) + 4 3 3 4  f (2) = − ✓ y-value 3 4  D(2 ; − ) 3 NOTE: Answer does not have to be in coordinate form. (3) 0−4 7.3 mg = (0;4)( −8;0) ✓ substitute correctly into −8−0 gradient formula 1  mg = ✓ value of mg 2  m p = −2 ✓ value of mp 4 4  − = −2(2) + c  D(2;− ) ✓ substitute m and point D 3 3 4  c =4− 3 8  c= ✓ value of c 3 8  p( x ) = −2 x + 3 NOTE: Answer does not have to be an equation. (5) 15 ### Page 16 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 7.4 m = tan 1 1 = tan ✓ tan = 2 2   = 26,57° ✓ answer (2) 7.5 4 2 16 1 x – x+4= x+4 ✓ equating 3 3 2 4 2 35 x – x =0 3 6 4 35  x x −  = 0 ✓ factors 3 6 35 x = 0 of x = ✓ x-answers 8 1  35  y=   +4 2  8  99 y= ✓ y-answer from correct selection 16 of x-value  35 99  K ;   8 16  NOTE: Answer does not need to be in coordinate form. (4) 7.6 7.6.1 1 < x < 3 ✓ answer NOTE: Can be written as separate inequalities. (1) 7.6.2 x(–8 ; 1] or [3 ; ) ✓ answer ✓ answer NOTE: Penalise 1 mark if answer is not with correct brackets. (2) 16 ### Page 17 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 4 7.7 TP(2 ; – ) from Q. 7.2 3 4 4  f (x) = (x – 2)2 – ✓ equation of f 3 3 4 13 ✓ equation of h  h(x) = (x – 4)2 – 3 3 (2) 7.8 7.8.1 j(x) = ax –8  0 = –4a –8……pt(–4 ; 0) ✓ substitution  4a = –8  a = –2 ✓ answer (2) 7.8.2 The graph of g to j is a 90° anticlockwise ✓ answer rotation about the origin. NOTE: Accept, graphs are perpendicular and if g is shifted anti-clockwise by 90°, j is obtained. (1) [25] QUESTION 8 8.1 Total = 13 + 7 + 24 + 3 Total = 47 ✓ answer (1) 13 + 7 8.2 8.2.1 P(M) = 47 20 P(M) = …or…0,43 ✓ answer 47 (1) 7 8.2.2 P(M and S) = or 0,15 ✓ answer 47 (1) 3 8.2.3 P(not M or S) = or 0,06 ✓ answer 47 (1) 17 ### Page 18 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 13 + 7 + 24 8.2.4 P(M or S) = ✓ method 47 44 P(M or S) = or 0,94 ✓ answer 47 OR P(M or S) = P(M) + P(S) – P(M and S) 20 31 7 = + – ✓ method 47 47 47 44 P(M or S) = or 0,94 ✓ answer 47 (2) 8.2.5 13 + 24 P(M or only S) = ✓ method 47 37 P(M or only S) = or 0,79 ✓ answer 47 NOTE: Answer only, full marks. (2) 8.3 8.3.1 P(A or B) = P(A) + P(B) – P(A and B) = 0,95 + 0,98 – 0,94 ✓ correct substitution P(A or B) = 0,99 ✓ answer NOTE: Answer only, full marks. (2) 8.3.2 P(Not Detected) = 1 – 0,99 P(Not Detected) = 0,01 ✓ answer NOTE: Answer only, full marks. (1) [11] 18 ### Page 19 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 9 9.1 9.1.1 NO. The survey was done on ONE day of the ✓ NO month and only in the morning. ✓ reasonable motivation NOTE: Answer must be NO. (2) 9.1.2 The survey should be done: • At different TIMES of the day. ✓ answer • On different DAYS of the month, especially at the end of the month when most people do shopping. NOTE: Any other valid reason. (1) 9.2 26 25 24 26 25 24 P(black) =   ✓   52 51 50 52 51 50 2  P(black) = or 0,118 ✓ answer P(black) 17 26 25 24 P(red) =   52 51 50 2  P(red) = or 0,118 ✓ answer P(red) 17  P(3 black or 3 red) = 0,118 + 0,118 ✓ method (+) 59  P(3 black or 3 red) = 0,236 or ✓ answer P(3 black or 3 red) 250 (5) [8] TOTAL: 150 19 ## Related documents - [Gr 11 Mathematics P2 (English) November 2022 Question Paper](https://www.hlayiso.com/library/gr-11-mathematics-p2-english-november-2022-question-paper/130a0fe0-0914-40c3-aedd-67a7e297f1cd) — Mathematics · QP · 2022 · Eng - [Gr11 MATH P1 (AFR) June 2022 Possible Answers](https://www.hlayiso.com/library/gr11-math-p1-afr-june-2022-possible-answers/31eae199-9c0b-4a71-ab81-f769f84c7ccd) — Mathematics · Memo · 2022 · Afr - [LImp Maths Gr11 June 2022 P1 and Memo](https://www.hlayiso.com/library/limp-maths-gr11-june-2022-p1-and-memo/5b6991e9-7f6f-4156-90d2-cd1c745a1d53) — Mathematics · Memo · 2022 - [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) — Mathematics · QP · 2024 · Eng - [MATHS P1 QP GR11 NOV 2023 English](https://www.hlayiso.com/library/maths-p1-qp-gr11-nov-2023-english/11b74f81-704a-47bc-b151-d2eea1003c3c) — Mathematics · QP · 2023 · Eng - [1 MATHS P1 QP GR11 NOV 2023 English (28 09 2023)](https://www.hlayiso.com/library/1-maths-p1-qp-gr11-nov-2023-english-28-09-2023/54f88191-f662-4aed-b968-fcd43f78efee) — Mathematics · QP · 2023 · Eng ## 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 - [NW 11 MATHEMATICS P2 AB ENGAFR NOV 2024](https://www.hlayiso.com/library/nw-11-mathematics-p2-ab-engafr-nov-2024/8ecc44b8-11a8-4044-8a89-cabd7fafe08c) — Addendum · 2024 - [ALTERNATIVE ANSWERS TO SOME QUESTIONS](https://www.hlayiso.com/library/alternative-answers-to-some-questions/589e5071-8eca-4868-96d4-89372a4d874b) — QP · 2023 - [MATHS P2 ANSBK GR11 NOV 2023 EA D](https://www.hlayiso.com/library/maths-p2-ansbk-gr11-nov-2023-ea-d/b6683a8b-2232-4874-a625-d729a5ee335a) — QP · 2023 ## 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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