# Mathematics Grade 11 P1 November Mock 2026 Memo (English) > Grade 11 Mathematics Paper 1 memorandum from the November Mock Exam 2026 (English), 12 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/grade-11-mathematics-2026-november-mock-exam-memorandum-paper-1/470265ad-99e7-4e3e-af02-0c28a273b390 - Updated: 2026-10-29 - Subject: Mathematics - Grade: Grade 11 - Language: English - Document type: Memorandum - Year: 2026 - Exam period: November Mock Exam - Paper: 1 - Publisher: Hlayiso - Curriculum: CAPS - Pages: 12 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 November Mock 2026 Memo (English)](https://www.hlayiso.com/library/grade-11-mathematics-2026-november-mock-exam-memorandum-paper-1/470265ad-99e7-4e3e-af02-0c28a273b390) ## About this document Mathematics · Grade 11 · November Mock Exam · 2026 · English. Memorandum, 12 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 hlayiso mock exams Practice Series: Grade 11 Exam Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 11 MATHEMATICS P1 NOVEMBER EXAMINATION 2026 (MOCK) MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 12 pages. Copyright reserved Please turn over ### Page 2 Mathematics/P1 2 Hlayiso/Mock 2026 NSC – Marking Guidelines 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 the question, mark the crossed-out version. • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines. Stop marking at the second calculation error. • Assuming answers/values in order to solve a problem is NOT acceptable. • The mark for substitution is awarded for substitution into the correct formula. QUESTION 1 1.1.1 (x − 3)(2x + 1) = 0 ✓x=3 1 ✓ x = − 21 (2) x = 3 or x = − 2 1.1.2 2x2 − 7x − 3 =√ 0 ✓ standard form −(−7) ± (−7)2 − 4(2)(−3) ✓ correct substitution x= 2(2) √ ✓ x = 3,89 7 ± 73 x= ✓ x = −0,39 (4) 4 x = 3,89 or x = −0,39 Penalise 1 mark for incorrect rounding. 1.1.3 x2 + 3x − 10 ≤ 0 ✓ standard form & factors (x + 5)(x − 2) ≤ 0 CV: x = −5 or x = 2 ✓ critical values −5 2 −5 ≤ x ≤ 2 OR x ∈ [−5 ; 2] ✓ answer (3) √ 1.1.4 2x + 7 = x + 2 ✓ isolating the surd 2x + 7 = x2 + 4x + 4 ✓ squaring both sides x2 + 2x − 3 = 0 ✓ standard form/factors (x + 3)(x − 1) = 0 x = 1 or x ̸= −3 ✓ answer with selection (4) √ Check: x = −3: 1 − (−3) = 4 ̸= 2 Copyright reserved Please turn over ### Page 3 Mathematics/P1 3 Hlayiso/Mock 2026 NSC – Marking Guidelines 1.2 y = x + 2 …(1) ✓ y the subject 2x2 − xy + y2 = 8 …(2) 2x2 − x(x + 2) + (x + 2)2 = 8 ✓ substitution ✓ expansion 2x2 − x2 − 2x + x2 + 4x + 4 = 8 ✓ standard form 2x2 + 2x − 4 = 0 x2 + x − 2 = 0 (x + 2)(x − 1) = 0 ✓ both x-values x = −2 or x = 1 ✓ both y-values (6) y=0 y=3 1.3 ∆ = b2 − 4ac = (p + 2)2 − 4(2)(p) ✓ substitution into ∆ = p2 + 4p + 4 − 8p ✓ simplification = p2 − 4p + 4 ✓ (p − 2)2 = (p − 2)2 ✓ ≥ 0 and perfect square (4) (p − 2)2 ≥ 0 and is a perfect square for all rational p ∴ the roots are real and rational. 1.4 Let the breadth be x cm; length = (x + 3) cm x2 + (x + 3)2 = 152 (Pythagoras) ✓ Pythagoras equation 2x2 + 6x + 9 = 225 x2 + 3x − 108 = 0 ✓ standard form (x + 12)(x − 9) = 0 ✓ factors x = 9 (x ̸= −12, length cannot be negative) ✓ answer with selection (4) Breadth = 9 cm [27] QUESTION 2 2x+2 . 2x−1 . 3x−1 2.1.1 = ✓ 6x−1 = 2x−1 .3x−1 22x . 3x . 3−1 x+2+x−1−2x ✓ 12x = 22x .3x =2 . 3x−1−x+1 ✓ adding/subtracting = 21 . 30 exponents =2 ✓ answer (4) √ √ 5 2+7 2 √ √ 2.1.2 = √ ✓ 5√2 and 7 2 2 2 ✓2 2 √ 12 2 = √ ✓ answer (3) 2 2 =6 Copyright reserved Please turn over ### Page 4 Mathematics/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines ( 3 )2 2 3 2.2.1 x 2 = (27) 3 ✓ raising both sides to 32 ( )2 ✓ 33 x = 33 3 = 32 ✓ answer (3) x=9 2.2.2 2x .22 − 2x = 48 ✓ common factor 2x (4 − 1) = 48 2x = 16 = 24 ✓ 2x = 16 x=4 ✓ answer (3) √ 2.2.3 Let k = x: k2 − 5k + 4 = 0 ✓ substitution/quadratic (k − 1)(k − 4) = 0 form √ √ x = 1 or x=4 ✓ factors x = 1 or x = 16 OR ✓x=1 √ √ ( x − 1)( x − 4) = 0 … ✓ x = 16 (4) ( )x 2.3 9x+1 = 9x .9 = 32 .9 = 32x .9 ✓ 32x .9 = 5 × 9 = 45 ✓ answer (2) [19] Copyright reserved Please turn over ### Page 5 Mathematics/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 3.1.1 −9 ✓ answer (1) 3.1.2 d = −4 ✓ −4n Tn = −4n + 11 ✓ +11 (2) 3.1.3 −4n + 11 = −233 ✓ equating −4n = −244 n = 61 ∴ T61 ✓ answer (2) 3.2 (2x + 3) − x = 5x − (2x + 3) ✓ T2 − T1 = T3 − T2 x + 3 = 3x − 3 ✓ simplification −2x = −6 x=3 ✓ answer (3) 3.3 T20 = a + 19d: 8 + 19d = 103 ✓ equation d=5 ✓d=5 T50 = 8 + 49(5) = 253 ✓ answer (3) [11] Copyright reserved Please turn over ### Page 6 Mathematics/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1.1 43 ✓ answer (1) 4.1.2 First differences: 5 ; 9 ; 13 Second difference: 4 2a = 4 ⇒ a = 2 ✓a=2 3a + b = 5 ⇒ b = −1 ✓ b = −1 a + b + c = −1 ⇒ c = −2 ✓ c = −2 Tn = 2n2 − n − 2 ✓ Tn (4) 4.1.3 2n2 − n − 2 = 1 768 ✓ equating 2n2 − n − 1 770 = 0 ✓ standard form/factors (2n + 59)(n − 30) = 0 n = 30 (n ̸= − 59 ✓ answer (3) 2 ) ∴ T30 4.1.4 First differences: Tn+1 − Tn = 4n + 1 ✓ 4n + 1 4n + 1 = 161 ✓ n = 40 n = 40 ✓ answer (3) Between T40 and T41 4.2 Pattern: 5 ; y ; 25 First differences: y − 5 ; 25 − y ✓ first differences (25 − y) − (y − 5) = 6 ✓ equation 30 − 2y = 6 y = 12 ✓ answer (3) [14] QUESTION 5 5.1 f (x) = a(x − 1)2 + 4 ✓ substitution of TP Subst. B(3 ; 0): 0 = a(3 − 1)2 + 4 ✓ substitution of B 4a = −4 ⇒ a = −1 ✓ a = −1 (3) f (x) = −(x − 1)2 + 4 = −x2 + 2x − 1 + 4 f (x) = −x2 + 2x + 3 5.2 A(−1 ; 0) ✓ answer (1) 5.3 y ≤ 4 OR y ∈ (−∞ ; 4] ✓ answer (1) Copyright reserved Please turn over ### Page 7 Mathematics/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines 5.4 −x2 + 2x + 3 = x + 1 ✓ equating x2 − x − 2 = 0 ✓ standard form/factors (x − 2)(x + 1) = 0 x = 2 (x = −1 at A) y = 2+1 = 3 D(2 ; 3) ✓ coordinates of D (3) 5.5 −1 < x < 2 ✓ critical values ✓ notation (2) 5.6 −1 ≤ x ≤ 0 or x≥3 ✓ −1 ≤ x ≤ 0 ✓ or ✓x≥3 (3) 5.7 f (x) − g(x) = −x2 + 2x + 3 − (x + 1) ✓ f (x) − g(x) = −x2 + x + 2 ✓ −x2 + x + 2 ( )2 = − x − 12 + 94 ✓ answer (3) 9 Maximum distance = = 2,25 units 4 OR b 1 x = − = ; − 14 + 12 + 2 = 94 2a 2 5.8 3 −8 OR y ∈ (−8 ; ∞) ✓ answer (1) 7.1.2 y-intercept: f (0) = 21 − 8 = −6 (0 ; −6) ✓ (0 ; −6) x-intercept: 2x+1 = 8 = 23 ✓ 2x+1 = 23 x + 1 = 3 ⇒ x = 2 (2 ; 0) ✓ (2 ; 0) (3) 7.1.3 y ✓ asymptote ✓ intercepts ✓ shape (3) (2 ; 0) x (0 ; −6) y = −8 7.1.4 g(x) = −2x+1 + 8 ✓ answer (1) 7.1.5 f (−1) = 20 − 8 = −7; f (2) = 0 ✓ f (−1) = −7 f (2) − f (−1) 0 − (−7) 7 ✓ substitution Average gradient = = = ✓ answer (3) 2 − (−1) 3 3 7.2 (0 ; −1): a + q = −1 …(1) ✓ equation (1) ( )−1 ✓ equation (2) (−1 ; 1): a 12 + q = 1 ⇒ 2a + q = 1 …(2) ✓ a = 2 and q = −3 (3) (2) − (1): a = 2 q = −3 [14] QUESTION 8 8.1 A = P(1 − i)n ✓ formula A = 8 500(1 − 0,18)4 ✓ substitution A = R3 843,03 ✓ answer (3) Copyright reserved Please turn over ### Page 10 Mathematics/P1 10 Hlayiso/Mock 2026 NSC – Marking Guidelines ( ) 0,096 12 8.2 1 + ieff = 1 + ✓ substitution 12 ✓ 0,100339 . . . ieff = 1,100339 . . . − 1 ✓ answer (3) ieff = 10,03% p.a. 8.3 22 000 = 15 000(1 + i)5 ✓ substitution 22 (1 + i)5 = ✓ fifth root √ 15 22 ✓ answer (3) 1+i = 5 = 1,0796 . . . 15 i = 7,96% p.a. ( )12 ( )4 8.4 A3 = 40 000 1 + 0,075 4 − 10 000 1 + 0,075 4 ✓ 40 000(1 + 0,075 4 ) 12 = 49 988,66 − 10 771,36 = R39 217,30 ✓ −10 000(1 + 0,075 4 ) 4 ( )36 ( )24 ✓ ×(1 + 0,082 36 0,082 0,082 12 ) A6 = 39 217,30 1 + 12 + 5 000 1 + 12 ✓ 5 000(1 + 0,082 12 ) 24 = 50 113,04 + 5 887,79 ✓ adding = R56 000,82 ✓ answer (6) Accept R56 000,80 – R56 000,85 (rounding in steps). [15] Copyright reserved Please turn over ### Page 11 Mathematics/P1 11 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 9 9.1.1 No. P(A and B) = 0,135 ̸= 0 ✓ No ✓ reason (2) 9.1.2 P(A) × P(B) = 0,45 × 0,3 = 0,135 ✓ P(A) × P(B) = 0,135 = P(A and B) ✓ conclusion (2) ∴ A and B are independent. 9.1.3 P(A or B) = 0,45 + 0,3 − 0,135 ✓ substitution = 0,615 ✓ answer (2) 9.2.1 S ✓ x in centre and 5 outside T W ✓ 30 − x; 20 − x; 15 − x ✓ 5 + x; 20 + x; x (3) 30 − x 5+x 20 + x x 15 − x 20 − x x I 5 9.2.2 (5 + x) + (20 + x) + x + (30 − x) ✓ sum of all regions = 100 +(20 − x) + (15 − x) + x + 5 = 100 ✓ 95 + x = 100 (2) 95 + x = 100 x=5 9.2.3 Exactly two: 25 + 15 + 10 = 50 ✓ 50 50 1 ✓ answer (2) P(exactly two) = = = 0,5 100 2 9.3.1 4 ✓ first branches with 7 R (RR) probabilities 5 8 R ✓ second branches with 3 B (RB) probabilities 7 5 ✓ outcomes (3) 7 R (BR) 3 8 B 2 B (BB) 7 Copyright reserved Please turn over ### Page 12 Mathematics/P1 12 Hlayiso/Mock 2026 NSC – Marking Guidelines 9.3.2 P(same colour) = P(RR) + P(BB) 5 4 3 2 ✓ 85 × 47 = × + × 8 7 8 7 ✓ 38 × 27 20 6 26 13 ✓ answer (3) = + = = (or 0,46) 56 56 56 28 [19] TOTAL: 150 Copyright reserved ## Related documents - [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 - [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 - [MATHS P1 QP GR11 NOV 2019 English2](https://www.hlayiso.com/library/maths-p1-qp-gr11-nov-2019-english2/8790b67d-1823-4797-8ba6-78bccad65b0b) — Mathematics · QP · 2019 · Eng - [Gr11 Math P1 (English) June 2019 Question Paper](https://www.hlayiso.com/library/gr11-math-p1-english-june-2019-question-paper/9e010e98-da6e-4074-a4f9-ae653c922762) — Mathematics · QP · 2019 · 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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