# Mathematics Grade 11 P2 November Mock Exam 2 2026 Memo (Eng) > Grade 11 Mathematics Paper 2 memorandum from the November Mock Exam 2 2026 (English), 11 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/grade-11-mathematics-2026-november-mock-exam-2-memorandum-paper-2/8ad9bef0-f36c-42b3-8c8b-50123bc26f30 - Updated: 2026-10-29 - Subject: Mathematics - Grade: Grade 11 - Language: English - Document type: Memorandum - Year: 2026 - Exam period: November Mock Exam 2 - Paper: 2 - Publisher: Hlayiso - Curriculum: CAPS - Pages: 11 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 P2 November Mock Exam 2 2026 Memo (Eng)](https://www.hlayiso.com/library/grade-11-mathematics-2026-november-mock-exam-2-memorandum-paper-2/8ad9bef0-f36c-42b3-8c8b-50123bc26f30) ## About this document Mathematics · Grade 11 · November Mock Exam 2 · 2026 · English. Memorandum, 11 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 Mock hlayiso mock exams Practice Series: Grade 11 Exam Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 11 MATHEMATICS P2 NOVEMBER EXAMINATION 2026 (MOCK) MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 11 pages. Copyright reserved Please turn over ### Page 2 Mathematics/P2 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. 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 if the statement is correct). S/R Award a mark if the statement AND reason are both correct. QUESTION 1 500 1.1 x̄ = = 50 books ✓ 500 10 ✓ answer (2) 1.2 σ = 9,15 ✓ answer (1) 1.3 x̄ + σ = 59,15 ✓ 59,15 Values above 59,15: 60 and 70, i.e. 2 Saturdays ✓ answer (CA) (2) 2 × 100 = 20% 10 1.4 Ordered: 38; 41; 43; 45; 47; 49; 52; 55; 60; 70 47 + 49 ✓ median Median = = 48 2 ✓ Q1 and Q3 Q1 = 43; Q3 = 55 ✓ IQR (CA) (3) IQR = 55 − 43 = 12 (Accept quartiles by the position method: Q1 = 42,5; Q3 = 56,25; IQR = 13,75) 500 + x + (x + 12) 1.5 = 52 ✓ total 500 + 2x + 12 12 512 + 2x = 624 ✓ = 624 x = 56 ✓ answer (3) [11] Copyright reserved Please turn over ### Page 3 Mathematics/P2 3 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 2.1 Cumulative frequencies: 6 ; 20 ; 48 ; 78 ; 93 ; 100 ✓ first three ✓ last three (2) 2.2 100 ✓ grounding at (40 ; 0) ✓ plotting at upper limits 80 ✓ smooth shape (3) 60 40 20 v 40 50 60 70 80 90 100 Points: (40 ; 0), (50 ; 6), (60 ; 20), (70 ; 48), (80 ; 78), (90 ; 93), (100 ; 100) 2.3 70 ≤ v < 80 ✓ answer (1) 2.4 Reading at v = 75: cumulative frequency ≈ 63 ✓ reading at 75 Number faster than 75 km/h ≈ 100 − 63 = 37 (accept ✓ answer (CA from ogive) 35 – 39) (2) 2.5 Q1 at position 25: Q1 ≈ 62 km/h ✓ Q1 and Q3 Q3 at position 75: Q3 ≈ 79 km/h IQR ≈ 79 − 62 = 17 km/h (accept 15 – 19) ✓ answer (CA from ogive) (2) [10] QUESTION 3 3 − (−2) 5 1 3.1 mPQ = = = ✓ mPQ = 12 6 − (−4) 10 2 ✓ substitution of P or Q y − 3 = 12 (x − 6) ✓ simplification (3) y = 12 x − 3 + 3 y = 12 x Copyright reserved Please turn over ### Page 4 Mathematics/P2 4 Hlayiso/Mock 2026 NSC – Marking Guidelines 3.2 mRS = −2 (mRS × mPQ = −1) ✓ mRS = −2 y = −2x + 5 ✓ equation (2) 2 x = −2x + 5 ✓ equating 1 3.3 ✓x=2 2x = 5 ⇒ x = 2 5 ✓y=1 (3) y = 21 (2) = 1 S(2 ; 1) √ √ √ 3.4 PQ = (6 + 4)2 + (3 + 2)2 = 125 = 5 5 ✓ PQ √ √ √ RS = (2 − 0)2 + (1 − 5)2 = 20 = 2 5 ✓ RS √ √ ✓ area formula with Area △PQR = 12 × PQ × RS = 12 (5 5)(2 5) substitution = 25 square units ✓ answer (4) 5−3 1 3.5 mQR = =− 0−6 3 ✓ 161,57◦ Inclination of QR: tan β = − 13 ⇒ β = 161,57◦ ✓ 26,57◦ Inclination of PQ: tan α = 12 ⇒ α = 26,57◦ ✓ method ✓ answer PQ̂R = 180◦ − (161,57◦ − 26,57◦ ) = 45◦ OR √ ✓ SQ ✓ SQ = RS √ SQ = (6 − 2)2 + (3 − 1)2 = 20 = RS and ✓ method ✓ answer (4) RŜQ = 90◦ 180◦ − 90◦ ∴ PQ̂R = = 45◦ RS (or tan Q̂ = SQ = 1) 2 [16] Copyright reserved Please turn over ### Page 5 Mathematics/P2 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 ( ) −2 + 4 1 + 3 4.1 M ; = M(1 ; 2) ✓x=1 2 2 ✓y=2 (2) 4.2 M is also the midpoint of BD (diagonals of a parallelogram bisect each other) ✓ xD = −1 3 + xD −4 + yD ✓ yD = 8 (2) = 1 ⇒ xD = −1; = 2 ⇒ yD = 8 2 2 D(−1 ; 8) OR D = A + C − B = (−2 + 4 − 3 ; 1 + 3 + 4) 3−1 1 4.3 mAC = = ✓ mAC 4+2 3 ✓ mBD 8+4 mBD = = −3 ✓ product = −1 and −1 − 3 conclusion mAC × mBD = −1 ∴ AC ⊥ BD ABCD is a parallelogram with perpendicular diagonals ∴ ABCD is a√rhombus √ √ √ ✓ AB ✓ BC OR AB = 25 + 25 = 50; BC = 1 + 49 = 50 ✓ AB = BC and conclusion Parallelogram with a pair of adjacent sides equal ∴ (3) rhombus 8−3 4.4 mDC = = −1 ✓ mDC = −1 −1 − 4 ✓ y = −x + 7 y − 3 = −1(x − 4) ⇒ y = −x + 7 ✓ E(7 ; 0) (3) At E: 0 = −x + 7 ⇒ E(7 ; 0) 4.5 Inclination of CE: tan θ = −1 ⇒ θ = 135◦ ✓ 135◦ 0−1 1 ✓ mAE mAE = =− 7+2 9 ✓ 173,66◦ Inclination of AE: tan α = − 19 ⇒ α = 173,66◦ ✓ answer AÊC = 173,66◦ − 135◦ = 38,66◦ (ext ∠ of △) √ √ ✓ lengths ✓ cos rule OR cosine rule in △ACE: AC = 40, CE = 18, ✓ subst ✓ answer √ (4) AE = 82 82 + 18 − 40 cos AÊC = √ √ ⇒ AÊC = 38,66◦ 2 82 18 [14] Copyright reserved Please turn over ### Page 6 Mathematics/P2 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 5 5.1.1 tan A = − and sin A < 0: A in the fourth quadrant ✓ r = 13 (diagram) 12 y ✓ answer (2) A x 13 (12 ; −5) r2 = 122 + (−5)2 = 169 ⇒ r = 13 12 cos A = 13 5.1.2 sin(180◦ + A) . cos(−A) = − sin A . cos A ✓ − sin A ( )( ) 5 12 ✓ cos A =− − ✓ answer (3) 13 13 60 = 169 5.2.1 sin 302◦ = sin(360◦ − 58◦ ) = − sin 58◦ = − cos 32◦ = ✓ − sin 58◦ −k ✓ answer (2) √ 5.2.2 sin 32◦ = 1 − k2 ✓ tan 32◦ √ 1 − k2 ✓ answer (2) tan 212◦ = tan 32◦ = k (− sin x)(sin x) 5.3 = ✓ − sin x ✓ sin x (cos x)(tan x) ✓ cos x ✓ tan x − sin2 x − sin2 x ✓ answer (5) = = sin x sin x cos x . cos x = − sin x (1) sin 30◦ . tan 45◦ (1) 5.4 = = 2 ✓ 21 and 1 − cos 60◦ − 21 ✓ − 12 = −1 ✓ answer (3) sin x cos x 5.5 LHS = + ✓ tan x = cos sin x x cos x sin x ✓ common denominator sin2 x + cos2 x ✓ sin2 x + cos2 x = 1 (3) = sin x . cos x 1 = = RHS sin x . cos x Copyright reserved Please turn over ### Page 7 Mathematics/P2 7 Hlayiso/Mock 2026 NSC – Marking Guidelines [20] QUESTION 6 6.1 2(1 − sin2 x) − 5 sin x + 1 = 0 ✓ cos2 x = 1 − sin2 x 2 sin2 x + 5 sin x − 3 = 0 ✓ standard form (2 sin x − 1)(sin x + 3) = 0 ✓ factors 1 ✓ rejecting sin x = −3 sin x = or sin x = −3 (no solution) ✓ 30◦ and 150◦ 2 x = 30◦ + k.360◦ or x = 150◦ + k.360◦ , k ∈ Z ✓ +k.360◦ , k ∈ Z (6) sin x 3 6.2 = ⇒ tan x = 0,75 ✓ cos sin x 3 x = 4 cos x 4 ✓ tan x = 0,75 Reference angle = 36,87◦ ✓ 36,87◦ ✓ −143,13◦ (4) x = 36,87◦ or x = 36,87◦ − 180◦ = −143,13◦ [10] QUESTION 7 7.1 y f: (−90◦ ; 1) (60◦ ; 1) ✓ x-intercepts −30◦ ; 150◦ f and y-intercept 12 g 1 2 ✓ turning points (60◦ ; 1), 150◦ (−120◦ ; −1) ✓ end points (±180◦ ; − 21 ) x −30◦ (−180◦ ; − 21 ) (180◦ ; − 12 ) g: ✓ intercepts −180◦ ; 0◦ ; 180◦ (−120◦ ; −1) (90◦ ; −1) ✓ turning points (−90◦ ; 1), f : solid line; g: dashed line (90◦ ; −1) ✓ shape (6) 7.2 −180◦ ≤ x < −15◦ or 165◦ < x ≤ 180◦ ✓ −180◦ ≤ x < −15◦ OR x ∈ [−180◦ ; −15◦ ) ∪ (165◦ ; 180◦ ] ✓ 165◦ < x ≤ 180◦ (2) 7.3 h(x) = − cos(x − 60◦ ) + 1 ✓ − cos(x − 60◦ ) ✓ +1 (2) 7.4 2(−1) − 1 = −3 ✓ answer (1) Copyright reserved Please turn over ### Page 8 Mathematics/P2 8 Hlayiso/Mock 2026 NSC – Marking Guidelines [11] QUESTION 8 8.1 BÂD = β − α (ext ∠ of △ABD) ✓ answer (1) 8.2 In △ABD: AD̂B = 180◦ − β AD BD ✓ sine rule in △ABD = sin α sin BÂD ✓ substitution of β − α AD x ✓ making AD the subject (3) = sin α sin(β − α ) x sin α AD = sin(β − α ) AC 8.3 In △ADC: sin β = ✓ trig ratio in △ADC AD ✓ substitution of AD (2) x sin α . sin β AC = AD . sin β = sin(β − α ) 40 sin 32◦ . sin 51◦ 40 sin 32◦ . sin 51◦ 8.4 AC = = ✓ substitution sin(51◦ − 32◦ ) sin 19◦ ✓ sin 19◦ = 50,60 m ✓ answer (3) [9] QUESTION 9 9.1 V = π r2 h + 43 π r3 (two hemispheres = one sphere) 2,5 = π (0,5)2 h + 43 π (0,5)3 ✓ volume of sphere π ✓ volume of cylinder 2,5 = 0,25π h + ✓ = 2,5 6 2,5 − π6 ✓ answer (4) h= = 2,52 m 0,25π 9.2 SA = 2π (0,5)(2,52) + 4π (0,5)2 ✓ curved SA of cylinder = 2,52π + π ✓ SA of sphere = 11,06 m2 (accept 11,05 using unrounded h) ✓ answer (CA) (3) [7] Copyright reserved Please turn over ### Page 9 Mathematics/P2 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 10 10.1 A D 2 O B 1 C Construction: Join OB and OD. ✓ construction Ô1 = 2Â (∠ at centre = 2 × ∠ at circumference) ✓ S/R Ô2 = 2Ĉ (∠ at centre = 2 × ∠ at circumference) ✓S Ô1 + Ô2 = 360◦ (∠s around a point) ✓ S/R 2Â + 2Ĉ = 360◦ ✓ conclusion (5) ∴ Â + Ĉ = 180◦ 10.2.1 AĈB = 90◦ (∠ in semi-circle) ✓S ✓R (2) 10.2.2 AB̂C = 180◦ − 90◦ − 34◦ = 56◦ (sum of ∠s in △) ✓ AB̂C = 56◦ AD̂C = 180◦ − 56◦ = 124◦ (opp ∠s of cyclic quad) ✓ S/R (2) 10.2.3 DÂC = DĈA (∠s opp equal sides; AD = DC) ✓ S/R 180◦ − 124◦ ✓ answer (2) DÂC = = 28◦ (sum of ∠s in △) 2 10.2.4 OÂD = 28◦ + 34◦ = 62◦ OA = OD (radii) ∴ OD̂A = OÂD = 62◦ (∠s opp ✓ S/R (OA = OD, 62◦ ) equal sides) AÔD = 180◦ − 2(62◦ ) = 56◦ (sum of ∠s in △) ✓ AÔD = 56◦ = AB̂C AÔD = AB̂C = 56◦ ∴ OD ∥ BC (corresp ∠s equal) ✓R (3) OR AÔD = 2 AĈD = 2(28◦ ) = 56◦ (∠ at centre = 2 × ∠ at circumference) [14] Copyright reserved Please turn over ### Page 10 Mathematics/P2 10 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 11 11.1.1 AM = MB = 12 (16) = 8 units (line from centre ⊥ ✓S ✓R (2) chord) 11.1.2 OA = OC = 17 (radii) OM2 = 172 − 82 = 225 (Pythagoras) ⇒ OM = 15 ✓ OM = 15 CN = 21 (30) = 15 (line from centre ⊥ chord) ✓ CN = 15 ✓ ON = 8 ON2 = 172 − 152 = 64 (Pythagoras) ⇒ ON = 8 M, O, N collinear (OM and ON ⊥ parallel chords) ✓ MN = 23 (4) MN = 15 + 8 = 23 units 11.2.1 PA = PB (tangents from same point) ✓ S/R 180◦ − 48◦ ✓ answer (2) PÂB = PB̂A = = 66◦ (∠s opp equal 2 sides; sum of ∠s in △) 11.2.2 AĈB = PÂB = 66◦ (tan-chord theorem) ✓S ✓R (2) 11.2.3 AÔB = 2 AĈB = 132◦ (∠ at centre = 2 × ∠ at ✓S ✓R (2) circumference) OR AÔB = 360◦ − 90◦ − 90◦ − 48◦ = 132◦ (tan ⊥ radius; sum of ∠s in quad) 11.2.4 OA = OB (radii) ✓ S/R 180◦ − 132◦ ✓ answer (2) OÂB = = 24◦ (∠s opp equal sides) 2 OR OÂP = 90◦ (tan ⊥ radius); OÂB = 90◦ − 66◦ = 24◦ 11.3 OÂP = OB̂P = 90◦ (tan ⊥ radius) ✓S OÂP + OB̂P = 180◦ ∴ PAOB is a cyclic quad (converse: opp ∠s of cyclic ✓R (2) quad / opp ∠s suppl) OR P̂ + AÔB = 48◦ + 132◦ = 180◦ [16] Copyright reserved Please turn over ### Page 11 Mathematics/P2 11 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 12 12.1 PŜQ = TP̂Q = x (tan-chord theorem) ✓S ✓R PQ̂S = TP̂Q = x (alt ∠s; QS ∥ TV) ✓ S/R ∴ PQ̂S = PŜQ ∴ PQ = PS (sides opp equal ∠s) ✓R (4) 12.2 QR̂P = QŜP = x (∠s in the same segment) ✓ S/R PR̂S = PQ̂S = x (∠s in the same segment) ✓ S/R ∴ QR̂P = PR̂S, so RP bisects QR̂S ✓ conclusion (3) 12.3 PQ̂K = PQ̂S = x (proved in 12.1) ✓ S (both) QR̂K = QR̂P = x (proved in 12.2) ∴ PQ̂K = QR̂K ✓S ∴ PQ is a tangent to circle QKR (converse: tan-chord ✓R (3) theorem) 12.4 In △PQS: QP̂S = 180◦ − PQ̂S − PŜQ = 180◦ − 2x ✓S (sum of ∠s in △) ✓R (2) OR QR̂S = 2x (12.2) and QP̂S = 180◦ − QR̂S (opp ∠s of cyclic quad) [12] TOTAL: 150 Copyright reserved ## Related documents - [Gr 11 Math P2 (English) June 2023 Possible Answers](https://www.hlayiso.com/library/gr-11-math-p2-english-june-2023-possible-answers/7d433fdb-38a7-4d1e-b09f-472448486aa3) — Mathematics · Memo · 2023 · Eng - [MATHS P2 AB GR11 NOV 2023 Afri+English](https://www.hlayiso.com/library/maths-p2-ab-gr11-nov-2023-afri-english/eee869d2-4e88-4265-9665-6bcf4b2cfc27) — Mathematics · QP · 2023 · Eng - [Gr 11 Math P2 (English) June 2023 Question Paper](https://www.hlayiso.com/library/gr-11-math-p2-english-june-2023-question-paper/ab59de7d-dca0-4ed6-95a6-dcb8ce60d27d) — Mathematics · QP · 2023 · Eng - [Gr 11 Mathematics P2 (English) 2023 Question Paper](https://www.hlayiso.com/library/gr-11-mathematics-p2-english-2023-question-paper/b8f6a489-3cb4-476d-8c5a-3c2de035fcf8) — Mathematics · QP · 2023 · Eng - [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 - [MATHS P2 GR11 QP NOV2016 English](https://www.hlayiso.com/library/maths-p2-gr11-qp-nov2016-english/8e51b3ab-c405-4dab-b3be-458409636e04) — Mathematics · QP · 2016 · 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 - [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 - [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 ## 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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