Mathematics Grade 12 Paper 2 NSC November 2026 Memo (English)

UNOFFICIAL Copyright reserved Please turn over Unofficial memo hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P2 NOVEMBER 2026 — UNOFFICIAL MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 13 pages.

UNOFFICIAL Mathematics/P2 2 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 IMPORTANT: These are unofficial marking guidelines prepared by Hlayiso for revision. They are not the DBE’s official memorandum; the official marking guidelines may differ in mark allocation and accepted alternatives. 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. • Answers only will NOT necessarily be awarded full marks; where an answer-only allocation applies it is stated in the row. • Rounding: penalise incorrect rounding (or no rounding where TWO decimal places are required) ONCE per question, not per item. 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. READINGS OF THE QUESTION PAPER • Q1: the ogive points are read as (20;0), (40;8), (60;20), (80;44) and (100;50). The ANSWER BOOK table is not visible; the classes are taken as 20 ≤x < 40, 40 ≤x < 60, 60 ≤x < 80, 80 ≤x < 100. • Q9.2: ˆO2 = 4x is read as the angle AOC on the side of D (the arc is drawn towards D); ˆO1 is the reflex angle AOC. QUESTION 1 1.1 50 vehicles ✓50 (1) 1.2 Median = 25th value: ≈64 km/h ✓answer Note: Accept 62–66 km/h read from the curve. (1) 1.3 50−20 = 30 motorists ✓30 (1) Copyright reserved Please turn over

UNOFFICIAL Mathematics/P2 3 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 1.4 Speed (km/h) Frequency Cum. freq. 20 ≤x < 40 8 8 40 ≤x < 60 12 20 60 ≤x < 80 24 44 80 ≤x < 100 6 50 ✓8 and 12 ✓24 and 6 (2) 1.5 ¯x = 30(8)+50(12)+70(24)+90(6) 50 = 3060 50 = 61,2 km/h ✓midpoints ✓∑fx = 3060 ✓61,2 km/h Note: CA from 1.4. (3) [8] QUESTION 2 2.1.1 20 40 60 80 100 hours % 0 2 4 6 8 10 (vertical axis starts at 20 for clarity) ✓1–4 points correct ✓5–8 points correct ✓all 10 points correct (3) 2.1.2 a = 79,94 b = −4,11 ˆy = 79,94−4,11x ✓a = 79,94 ✓b = −4,11 ✓equation Note: Exact: a = 79,936..., b = −4,109...; r = −0,79 (not asked). (3) 2.1.3 ˆy = 79,94−4,11(7) = 51,17% ✓substitution of x = 7 ✓51,17% Note: Accept 51,17–51,18%. CA from 2.1.2. (2) Copyright reserved Please turn over

UNOFFICIAL Mathematics/P2 4 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 2.1.4 σ = 16,20% (mean = 59,8%) ✓16,20 Note: Accept sample SD s = 17,07. (1) 2.2 T1 = 100, T2 = 98 = 100(0,98), so the hours form a geometric sequence with a = 100, r = 0,98 100(0,98)n−1 = 54,55 (0,98)n−1 = 0,5455 n−1 = log0,98 0,5455 = 29,998... ≈30 n = 31, i.e. in the 31st month OR: 98(0,98)n−2 = 54,55 ⇒n−2 = 29 ⇒n = 31 ✓r = 0,98 ✓100(0,98)n−1 = 54,55 ✓use of logs ✓n = 31 Note: Check: 100(0,98)30 = 54,548... ≈54,55. Answer only: 1/4. (4) [13] QUESTION 3 3.1 mCD = 1−(−9) 4−2 = 10 2 = 5 ✓substitution ✓5 (2) 3.2 F (4+2 2 ; 1+(−9) 2 ) = F(3; −4) ✓x = 3 ✓y = −4 (2) 3.3 mEF = −1 5 (EF ⊥CD) y−(−4) = −1 5(x−3) At E, y = 0: 4 = −1 5(x−3) ⇒x−3 = −20 x = −17 ∴E(−17; 0) ✓mEF = −1 5 ✓substitution of F ✓y = 0; x = −17 (3) 3.4 EF = √ (3+17)2 +(−4−0)2 = √ 416 = 4 √ 26 CD = √ (4−2)2 +(1+9)2 = √ 104 = 2 √ 26 Area △CDE = 1 2 CD.EF (EF ⊥CD) = 1 2 √ 104 √ 416 = 1 2(208) = 104 square units ✓EF ✓CD ✓1 2 CD.EF ✓104 Note: Other valid methods (e.g. area rule with CˆED) score full marks. (4) Copyright reserved Please turn over

UNOFFICIAL Mathematics/P2 5 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 3.5 mAB = mDC = 5 (opp sides of parm) Inclination of AB: tan−1 5 = 78,69◦ mBC = 1−4 4+1 = −3 5; inclination of BC = 180◦−30,96◦= 149,04◦ A ˆBC = 149,04◦−78,69◦= 70,35◦(ext ∠of △) OR: B ˆCD from inclinations of CB and CD (149,04◦ and 78,69◦): B ˆCD = 109,65◦; A ˆBC = 180◦−109,65◦= 70,35◦(co-int ∠s) ✓mAB = 5 and 78,69◦ ✓mBC = −3 5 ✓149,04◦ ✓70,35◦ Note: Accept 70,34◦–70,35◦; a cosine-rule method in △ABC is acceptable. (4) 3.6 Diagonals of a parm bisect each other: (x+4 2 ; y+1 2 ) = (−1+2 2 ; 4−9 2 ) ⇒A(−3; −6) OR: B→A is the same translation as C→D: (x;y) →(x−2;y−10) AB ∥CD and EF ⊥CD ⇒E ˆHA = 90◦ ∴EA is a diameter of the circle (chord subtends 90◦) Centre = midpoint of EA = (−17−3 2 ; 0−6 2 ) = (−10; −3) r2 = (−17+10)2 +(0+3)2 = 58 (x+10)2 +(y+3)2 = 58 ✓x = −3 ✓y = −6 ✓E ˆHA = 90◦⇒EA diameter ✓centre (−10;−3) ✓r2 = 58 ✓equation Note: Finding H ( −31 13;−38 13 ) and solving with three points is valid but long; full marks if correct. CA from A. (6) [21] QUESTION 4 4.1 r2 = AK2 = (−7+4)2 +(0+4)2 = 25 ∴(x+4)2 +(y+4)2 = 25 ✓substitution of K (1) 4.2 mAK = 0+4 −7+4 = −4 3 ✓−4 3 (1) 4.3 mtan = 3 4 (tan ⊥radius) y−0 = 3 4(x+7) y = 3 4x+ 21 4 ✓m = 3 4 ✓substitution of K ✓equation Note: CA from 4.2. (3) 4.4 x2 −2x+1+y2 −2y+1 = 3+1+1 (x−1)2 +(y−1)2 = 5 M(1; 1) ✓completing the square (x) ✓completing the square (y) ✓M(1;1) (3) Copyright reserved Please turn over

UNOFFICIAL Mathematics/P2 6 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 4.5 r = √ 5 units ✓ √ 5 (1) 4.6 B: y = 0: (x+4)2 +16 = 25 ⇒x+4 = ±3 x = −1 or x = −7 (K) ∴B(−1; 0) Reflection in y = −4: (x;y) →(x;−8−y) ∴ T(−1; −8) ✓y = 0 substituted ✓B(−1;0) ✓xT = −1 ✓yT = −8 (4) 4.7 mKA = −4 3 (4.2); mAT = −8+4 −1+4 = −4 3 mKA = mAT and A is a common point ⇒K, A, T collinear OR: midpoint of KT = (−7−1 2 ; 0−8 2 ) = (−4;−4) = A ✓mAT = −4 3 ✓equal gradients and common point OR: ✓midpoint ✓= A (2) 4.8 AM = √ (1+4)2 +(1+4)2 = √ 50 = 5 √ 2 ≈7,07 Touching externally: AM = R+rnew rnew = 5 √ 2−5 ≈2,07 % reduction = √ 5−(5 √ 2−5) √ 5 ×100% = 7,38% ✓substitution ✓AM = 5 √ 2 ✓rnew = 5 √ 2−5 ✓% formula ✓7,38% Note: Rounding rnew to 2,07 gives 7,43%: accept. CA from 4.4/4.5. (5) [20] Copyright reserved Please turn over

UNOFFICIAL Mathematics/P2 7 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 5 5.1.1 sin12◦= √ 1−cos2 12◦= √ 1−k2 (or from a sketch: x = k, r = 1, y = √ 1−k2) ✓identity/sketch ✓ √ 1−k2 (2) 5.1.2 tan(−12◦) = −tan12◦= − √ 1−k2 k ✓−tan12◦ ✓answer (2) 5.1.3 1−2sin2 39◦= cos78◦= sin12◦= √ 1−k2 ✓cos78◦ ✓ √ 1−k2 (2) 5.2 LHS = cosθ ·cos2 θ +sin2 θ(−sinθ) cosθ −sinθ = cos3 θ −sin3 θ cosθ −sinθ = (cosθ −sinθ)(cos2 θ +sinθ cosθ +sin2 θ) cosθ −sinθ = 1+sinθ cosθ = RHS ✓cosθ ✓cos2 θ ✓−sinθ (numerator) ✓−sinθ (denominator) ✓difference of cubes ✓sin2 θ +cos2 θ = 1 (6) 5.3 Undefined when sinθ −cosθ = 0 tanθ = 1 θ = 225◦ (θ ∈[180◦;360◦]) ✓sinθ −cosθ = 0 ✓tanθ = 1 ✓225◦ Note: θ = 45◦included: max 2/3. (3) [15] Copyright reserved Please turn over

UNOFFICIAL Mathematics/P2 8 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 6 6.1.1 cos25◦= cos(35◦−10◦) = cos35◦cos10◦+sin35◦sin10◦ ✓cos(35◦−10◦) ✓expansion (2) 6.1.2 cos35◦cos10◦+sin35◦sin10◦−2sin10◦sin35◦ = cos35◦cos10◦−sin35◦sin10◦ = cos(35◦+10◦) = cos45◦= √ 2 2 ✓substitution of 6.1.1 ✓cos45◦ ✓ √ 2 2 or 1 √ 2 (3) 6.2 LHS = 4sinθ cosθ(cos2 θ −sin2 θ) = 2sin2θ cos2θ = sin4θ = RHS ✓common factor ✓2sin2θ and cos2θ ✓sin4θ (3) 6.3.1 A = 53,13◦ ✓53,13◦ (1) 6.3.2 sinA = 4 5, cosA = 3 5 Divide by 5: 4 5 cosx−3 5 sinx = 1 5 sinAcosx−cosAsinx = 1 5 sin(A−x) = 1 5 (ref ∠= 11,54◦) A−x = 11,54◦+k·360◦or A−x = 168,46◦+k·360◦ x = 41,59◦+k ·360◦or x = −115,33◦+k ·360◦, k ∈Z OR: cos(90◦−A) = 4 5: 5cos(x+36,87◦) = 1 x+36,87◦= ±78,46◦+k ·360◦(same solutions) ✓cosA = 3 5 ✓dividing by 5 ✓sin(A−x) = 1 5 ✓both equations with k ·360◦ ✓41,59◦✓−115,33◦ Note: Accept 244,67◦+k ·360◦ for the second family. k ∈Z must be stated once. Check: 4cos41,59◦−3sin41,59◦= 1. (6) [15] Copyright reserved Please turn over

UNOFFICIAL Mathematics/P2 9 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 7 7.1 y ∈[0; 2] or 0 ≤y ≤2 ✓answer (1) 7.2 180◦ ✓180◦ (1) 7.3 x y −180◦−150◦−120◦ −90◦ −60◦ −30◦ -2 -1 1 2 f g (−180◦;0) (−90◦;−1) (0◦;−2) ✓shape (increasing to −2 as x →0◦) ✓endpoints (−180◦;0) and (0◦;−2) ✓(−90◦;−1) (3) 7.4.1 AB = f(x)−g(x) = cos2x+1+cosx+1 = 2cos2 x−1+cosx+2 = 2cos2 x+cosx+1 ✓f(x)−g(x) ✓cos2x = 2cos2 x−1 (2) 7.4.2 AB = 2 ( cosx+ 1 4 )2 +1−1 8 = 2 ( cosx+ 1 4 )2 + 7 8 Minimum AB = 7 8 = 0,875 units, when cosx = −1 4 (x = −104,48◦, which lies in [−180◦;0◦]) OR: d d(cosx): 4cosx+1 = 0 ⇒cosx = −1 4; AB = 2 ( 1 16 ) −1 4 +1 = 7 8 ✓completing the square / cosx = −1 4 ✓7 8 (2) [9] QUESTION 8 8.1 x y C(0;0) B(a;0) A(bcosC;bsinC) b c A(bcosC; bsinC) and B(a; 0) c2 = AB2 = (bcosC −a)2 +(bsinC −0)2 = b2 cos2C −2abcosC +a2 +b2 sin2C = a2 +b2(sin2C +cos2C)−2abcosC = a2 +b2 −2abcosC ✓coordinates of A and B ✓distance formula ✓expansion ✓sin2C +cos2C = 1 (4) Copyright reserved Please turn over

UNOFFICIAL Mathematics/P2 10 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 8.2.1 (4p)2 = (2p)2 +42 −2(2p)(4)cosθ 16p2 = 4p2 +16−16pcosθ cosθ = 16−12p2 16p = 4−3p2 4p ✓cosine rule substitution ✓making cosθ the subject (2) 8.2.2 AˆEB = 180◦−θ (∠s on a str line) In △ABE: ( √ 5p)2 = (2p)2 +22 −2(2p)(2)cos(180◦−θ) 5p2 = 4p2 +4+8pcosθ p2 = 4+8p (4−3p2 4p ) = 4+8−6p2 7p2 = 12 ⇒p = √ 12 7 = 1,31 Height of the tree = AD = p = 1,31 m ✓180◦−θ ✓cosine rule in △ABE ✓cos(180◦−θ) = −cosθ ✓substitution of 8.2.1 ✓1,31 m Note: θ = 102,60◦. Otherwise: DB = 2p, DE = √ 3p, DC = √ 15p (Pythagoras, AD ⊥ plane), then the cosine rule at E in △DBE and △DEC with supplementary angles gives 28p2 = 48, the same p. (5) [11] QUESTION 9 9.1.1 ˆP1 = ˆT2 = 40◦ [∠s in same seg] ✓S ✓R (2) 9.1.2 P ˆRS = 180◦−(35◦+40◦) = 105◦ [opp ∠s of cyclic quad] ✓S ✓R (2) 9.2 ˆO1 = 360◦−4x [∠s round a pt] ˆD = 1 2 ˆO1 = 180◦−2x [∠at centre = 2∠at circumference] ˆD = C ˆBE = 2x+10◦ [ext ∠of cyclic quad] 180◦−2x = 2x+10◦⇒x = 42,5◦ ˆD = 2(42,5◦)+10◦= 95◦ OR: A ˆBC = 1 2 ˆO2 = 2x [∠at centre = 2∠at circumference] 2x+2x+10◦= 180◦ [∠s on a str line] x = 42,5◦; ˆD = 180◦−A ˆBC = 180◦−85◦= 95◦ [opp ∠s of cyclic quad] ✓S/R ✓S ✓R ✓S/R ✓x = 42,5◦ ✓95◦ OR: ✓S ✓R ✓S/R ✓x ✓S ✓R (95◦) Note: Check: ˆO2 = 170◦, A ˆBC = 85◦, C ˆBE = 95◦. (6) [10] Copyright reserved Please turn over

UNOFFICIAL Mathematics/P2 11 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 10 10.1 HK KC = EL LC = 1 3 [prop theorem; KL ∥HE] HK = 1 3(12) = 4 cm ✓S ✓R ✓4 cm (3) 10.2 AH HC = BE EC = 2 4 [prop theorem; HE ∥AB] HC = 16 ⇒AH = 8, AC = 24 cm △BDE ||| △BAC [∠∠∠; DE ∥AC] BD BA = BE BC = 2 6 = 1 3 (DE = 8 cm) Area △BDE = 1 2 BD.BE.sin ˆB = 1 9 (1 2 BA.BC.sin ˆB ) = 1 9 area △ABC Area DECA = 8 9 area △ABC Area △BDE Area DECA = 1 8 ✓BD BA = 1 3 (S/R; via ||| or prop th) ✓area rule / ratio (1 3 )2 ✓area DECA = 8 9△ABC ✓1 8 Note: A candidate who uses DE = 8 and AC = 24 with heights in the ratio 1 : 3 scores full marks. (4) [7] QUESTION 11 11.1 R S T G H P K L Constr: Mark G on RS and H on RT so that RG = PK and RH = PL. Join GH. In △RGH and △PKL: RG = PK; RH = PL [constr]; ˆR = ˆP [given] ∴△RGH ≡△PKL [SAS] ∴R ˆGH = ˆK = ˆS ∴GH ∥ST [corresp ∠s =] ∴RG RS = RH RT [line ∥one side of △/ prop theorem; GH ∥ST] ∴PK RS = PL RT, i.e. RS PK = RT PL ✓construction ✓SAS (S/R) ✓R ˆGH = ˆS ✓S/R (GH ∥ST) ✓R (prop theorem) ✓conclusion with RG = PK, RH = PL Note: Construction mark is also given if G, H are placed on PK, PL (in the smaller triangle) correspondingly. (6) Copyright reserved Please turn over

UNOFFICIAL Mathematics/P2 12 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 11.2.1 Line from centre ⊥to chord (OL ⊥PT) ✓R (1) 11.2.2 In △PML and △OMP: ˆM = ˆM [common] PˆLM = 90◦ [given PT ⊥KM] OˆPM = 90◦ [tan ⊥radius] ∴△PML ||| △OMP [∠∠∠] ✓S ✓S ✓S/R ✓R (4) 11.2.3 PL OP = ML MP [||| △s] PL = 1 2PT [11.2.1]; OP = 1 2PS [radii; POS diameter] 1 2PT 1 2PS = ML MP ⇒PT PS = ML MP ∴PT ML = PS MP ✓S ✓R ✓PL = 1 2PT ✓OP = 1 2PS ✓substitution and conclusion (5) 11.2.4 From △PML ||| △OMP: MP OM = ML MP ⇒MP2 = OM.ML In △OLP and △PLM: OˆLP = PˆLM = 90◦; L ˆOP = LˆPM [= 90◦−ˆM, sum of ∠s in △] ∴△OLP ||| △PLM [∠∠∠] ⇒PL ML = OL PL ⇒PL2 = OL.ML MP2 PL2 = OM.ML OL.ML = OM OL MP2 1 4PT2 = OM OL ⇒MP PT = 1 2 √ OM OL ✓MP2 = OM.ML ✓||| (S/R) ✓PL2 = OL.ML ✓MP2 PL2 = OM OL ✓PL = 1 2PT and conclusion Note: Alternative: OP2 = OL.OM (△OLP ||| △OPM) combined with 11.2.3; mark similarly. (5) [21] TOTAL: 150 Copyright reserved Please turn over

UNOFFICIAL Mathematics/P2 13 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 HOW DID OUR PREDICTION DO? The real November 2026 P2 compared with the Hlayiso 2026 Maths P2 Prediction paper Q Topic Real paper asked Our closest prediction item(s) Verdict Covered 1 Ogive Read n, median, “more than 60”; frequencies from ogive; grouped mean Q2: cumulative freq., ogive, “more than R220”, estimated mean  8/8 2 Regression; sequence Scatter plot, regression line, predict, SD (9); geometric decrease of battery hours (4) Q1.1, 1.3, 1.5 (regression, prediction, SD). No sequence (a P1 skill) in our P2 / 9/13 3 Analytical geometry (lines) Gradient, midpoint, perpendicular line to E, area △CDE, A ˆBC via inclinations, 4th vertex of parm, circle through 3 points with a 90◦angle Q3.2–3.6: gradient/line, inclination, A ˆBC, area, circle through A, B, C with a right angle (diameter)  19/21 4 Circles Circle eq., tangent at K, centre/radius from general form, reflection in y = −4, collinearity, % reduction to touch externally Q4.1–4.4: centre/radius, tangent, touching externally (distance between centres). Reflections were listed as “unlikely” / 16/20 5 Trig: reduction cos12◦= k; reduction identity using a difference of cubes; when undefined Q5.2 (sin34◦= m), Q5.3 (reduction), Q6.1.2 (undefined)  13/15 6 Trig: compound angles cos(35◦−10◦) evaluation; sin4θ identity; 4cosx−3sinx = 1 general solution Q5.1.3, Q6.1 (compound/double angles), Q6.2 (general solution of a different form) ≈ 9/15 7 Trig graphs Range, period, sketch −cosx−1; minimum vertical distance AB Q7.1–7.2 (period, sketch). Vertical distance was listed as “unlikely” ≈ 5/9 8 Cosine rule / 3D Proof of the cosine rule (4); cosine rule twice with 180◦−θ to find a height Q8 (3D tower with the sine rule). Rule proofs were listed as “unlikely” /≈ 2/11 9 Circle geometry Same seg, opp ∠s cyclic quad; centre ∠+ ext ∠cyclic quad with x Q9.2 (centre ∠, opp ∠s, ext ∠of cyclic quad)  10/10 10 Proportionality Prop theorem with ∥lines; area ratio △BDE : DECA Q10.2 (two ∥lines, prop theorem, area ratio)  7/7 11 Similarity Equiangular ⇒sides in proportion proof; similar △s with tangent/diameter; ratio proofs Q10.1 (same theorem proof, same wording); Q11.3–11.4 (similar △s and product/ratio)  16/21 Marks of the real paper covered by our prediction 114/150 A learner who mastered our prediction could have answered about 114/150 (76%) of this paper with confidence. Much of that comes from topics that appear every year, so the extra value of the prediction lies in the specific calls (the proof, the right-angle circle, the area ratio). What we got right • The theorem proof: equiangular triangles, sides in proportion (6 marks), our “most overdue” pick, set almost word for word. • The overall structure and weighting: ogive + regression, about 40 marks of analytical geometry, 50 of trigonometry and 38 of Euclidean geometry. • Circle through three points using a 90◦angle (diameter), angle via inclinations, area of a triangle. • Circle-geometry and proportionality riders: ext ∠ of cyclic quad, centre angle, area ratio with ∥lines, similar triangles with a ratio proof. What we missed • Three items we explicitly listed as “unlikely” all came up: the proof of the cosine rule, reflection of a point, and the minimum vertical distance between two graphs (about 12 marks). • 4cosx−3sinx = 1 (reducing acosx+bsinx to a single ratio), a rare and hard type. • A P1 geometric sequence inside the statistics question (4 marks). • The hardest geometry: 11.2.4 (two similarities, PL2 = OL.ML), and the 3D cosine-rule system in 8.2.2. Rating: 7/10. The core skills and the theorem proof were right, but the “unlikely” list cost about 12 marks and the hardest 15–20 marks went beyond our paper. Copyright reserved