Mathematics Grade 12 Paper 2 Challenge Paper 2026 Memo

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

Mathematics/P2 2 NSC – Marking Guidelines Hlayiso/Challenge 2026 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. In show that questions the given answer earns no mark: marks are for the working that leads to it. • Valid alternative methods that are not shown here must be marked on merit, using the same mark allocation. 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 1.1 395+x+y 11 = 46 (sum of the 9 known values is 395) 395+x+y = 506 ⇒x+y = 111 x+(x+25) = 111 ⇒2x = 86 x = 43 and y = 68 (Check: 37 ≤43 ≤45 and 56 ≤68 ≤71, so the order is kept.) ✓∑= 506 / equation ✓substitution y = x+25 ✓x = 43 and y = 68 (3) 1.2 Min = 25; Q1 = 33; Median = 45; Q3 = 56; Max = 71 20 30 40 50 60 70 ✓min and max (whiskers) ✓Q1 and Q3 (box) ✓median (3) 1.3 σ = 14,38 minutes (Population standard deviation. If s = 15,08 is used: max 1 mark.) ✓✓answer (2) 1.4 (¯x−σ ; ¯x+σ) = (31,62 ; 60,38) Values: 33 ; 37 ; 43 ; 45 ; 48 ; 52 ; 56 7 learners ✓interval ✓7 learners (2) Copyright reserved Please turn over

Mathematics/P2 3 NSC – Marking Guidelines Hlayiso/Challenge 2026 1.5 New mean = 1,1(46)−5 = 45,6 minutes New σ = 1,1(14,38) = 15,82 minutes (Subtracting 5 does not change the spread.) ✓45,6 ✓15,82 (CA from 1.3) (2) [12] QUESTION 2 2.1 a = −18,42 b = 2,96 ˆy = −18,42+2,96x ✓a ✓b ✓equation (3) 2.2 r = 0,99 Very strong positive (linear) relationship ✓r ✓description (2) 2.3 ˆy = −18,42+2,96(27) = 61,50 About 61 cold drinks (accept 61 or 62; calculator: 61,44) ✓substitution ✓answer (whole number)(2) 2.4 ∑y = 456; ¯y = 456 8 = 57 New mean = 57+3 = 60 456+y9 9 = 60 ⇒456+y9 = 540 y9 = 84 cold drinks ✓¯y = 57 ✓equation with 9 days ✓84 (3) [10] Copyright reserved Please turn over

Mathematics/P2 4 NSC – Marking Guidelines Hlayiso/Challenge 2026 QUESTION 3 3.1 mAB = 6−0 1+2 = 2 ⇒inclination of AB = 63,43◦ mAC = 6−0 1−3 = −3 ⇒inclination of AC = 180◦−71,57◦= 108,43◦ B ˆAC = 108,43◦−63,43◦ [ext ∠of △] B ˆAC = 45◦ ✓mAB = 2 and 63,43◦ ✓mAC = −3 ✓108,43◦ ✓45◦ (4) 3.2 mBD ×(−3) = −1 ⇒mBD = 1 3 [BD ⊥AC] y−0 = 1 3(x+2) y = 1 3x+ 2 3 ✓m = 1 3 ✓substitution of B ✓equation (3) 3.3 AC: y−0 = −3(x−3) ⇒y = −3x+9 1 3x+ 2 3 = −3x+9 ⇒x+2 = −9x+27 x = 5 2; y = −3 (5 2 ) +9 = 3 2 D ( 21 2 ; 11 2 ) ✓equation of AC ✓equating ✓D (3) 3.4 In △ABD: A ˆDB = 90◦and ˆA = 45◦ A ˆBD = 180◦−90◦−45◦= 45◦= ˆA ∴AD = BD [sides opp equal ∠s] ∴△ABD is isosceles OR: AD2 = (1−2,5)2 +(6−1,5)2 = 22,5; BD2 = (2,5+2)2 +(1,5)2 = 22,5 ∴AD = BD ✓A ˆBD = 45◦ ✓AD = BD (S/R) (OR ✓✓both lengths) (2) 3.5 BD = √22,5 DC = √ (3−2,5)2 +(0−1,5)2 = √2,5 Area △BDC = 1 2 BD·DC = 1 2 √22,5√2,5 = 1 2 √56,25 = 3,75 square units OR: Area △ABC = 1 2(5)(6) = 15; DC AC = √2,5 √ 40 = 1 4 Area △BDC = 1 4(15) = 3,75 (same height from B) ✓BD and DC ✓area formula ✓3,75 OR: ✓15 ✓1 4 ✓3,75 (3) Copyright reserved Please turn over

Mathematics/P2 5 NSC – Marking Guidelines Hlayiso/Challenge 2026 3.6 BˆPC = B ˆAC, so P lies on the circle through A, B and C [converse: ∠s in the same segment] Centre N on the perpendicular bisector of BC: x = 1 2 B ˆNC = 2×45◦= 90◦[∠at centre = 2×∠at circumference], so △BNC is a right-angled isosceles △and N is 1 2BC = 5 2 above BC: N (1 2 ; 5 2 ) r2 = NB2 = (5 2 )2 + (5 2 )2 = 25 2 ( x−1 2 )2 + ( y−5 2 )2 = 25 2 (y > 0) OR: substitute A, B, C into x2 +y2 +Dx+Ey+F = 0: 4−2D+F = 0; 9+3D+F = 0; 37+D+6E +F = 0 D = −1, F = −6, E = −5: x2 +y2 −x−5y−6 = 0 (same circle) ✓P on circle through A, B, C (reason) ✓centre (1 2 ; 5 2 ) ✓r2 = 25 2 ✓equation OR: ✓concept ✓3 equations ✓solving ✓equation (4) [19] QUESTION 4 4.1 x2 −4x+4+y2 +6y+9 = 12+4+9 (x−2)2 +(y+3)2 = 25 M(2 ; −3) r = 5 ✓completing squares ✓M ✓r = 5 (3) 4.2 mMP = 0+3 6−2 = 3 4 mTP = −4 3 [tan ⊥radius] y−0 = −4 3(x−6) y = −4 3x+8 ✓mMP ✓mtan ✓substitution of P ✓equation (4) 4.3 T(0 ; 8) TP = √ (6−0)2 +(0−8)2 = √ 100 = 10 units OR: MT2 = 4+121 = 125; TP2 = 125−25 = 100 [tan ⊥radius] ✓T(0 ; 8) ✓substitution ✓10 (3) Copyright reserved Please turn over

Mathematics/P2 6 NSC – Marking Guidelines Hlayiso/Challenge 2026 4.4 TQ = TP = 10 [tans from same pt] x2 +(y−8)2 = 100 ⇒x2 +y2 −16y−36 = 0 Circle: x2 +y2 −4x+6y−12 = 0 Subtract: 4x−22y−24 = 0 ⇒x = 11y+12 2 Substitute: (11y+12 2 )2 +y2 −2(11y+12)+6y−12 = 0 125y2 +200y = 0 ⇒25y(5y+8) = 0 y = 0 (point P) or y = −8 5 Q ( −14 5 ; −8 5 ) = (−2,8 ; −1,6) OR: Q is the reflection of P in MT. Foot of ⊥from P on MT: MT: y = −11 2 x+8; line through P ⊥MT: y = 2 11(x−6) intersection (1,6 ; −0,8) is the midpoint of PQ ⇒ Q(−2,8 ; −1,6) ✓TQ = 10 (S/R) ✓equation of circle centre T ✓x in terms of y ✓standard form ✓Q OR: ✓concept ✓MT ✓⊥line ✓midpoint ✓Q (5) 4.5 MˆPT = M ˆQT = 90◦ [tan ⊥radius] ∴MT is a diameter [converse: ∠in semi-circle] Centre = (2+0 2 ; −3+8 2 ) = ( 1 ; 5 2 ) ; r2 = MT2 4 = 125 4 (x−1)2 + ( y−5 2 )2 = 125 4 ✓MT diameter (reason) ✓centre ✓equation (3) 4.6 The parallel tangent touches the circle at P′, the other end of diameter PM: P′ = (2(2)−6 ; 2(−3)−0) = (−2 ; −6) −6 = −4 3(−2)+k ⇒k = −26 3 −26 3 < k < 8 OR: substitute y = −4 3x+k into the circle: 25x2 −(24k +108)x+9k2 +54k −108 = 0 ∆> 0 ⇒3k2 +2k −208 < 0 ⇒(3k +26)(k −8) < 0 ✓P′(−2 ; −6) ✓k = −26 3 ✓answer (CA from 4.2) OR: ✓standard form ✓factors ✓answer (3) [21] Copyright reserved Please turn over

Mathematics/P2 7 NSC – Marking Guidelines Hlayiso/Challenge 2026 QUESTION 5 5.1 sinAcosB+cosAsinB = 4 5 …(1) sinAcosB−cosAsinB = 1 5 …(2) (1)+(2): sinAcosB = 1 2; (1)−(2): cosAsinB = 3 10 tanA tanB = sinAcosB cosAsinB = 1 2 3 10 = 5 3 ✓both expansions ✓1 2 and 3 10 ✓sinAcosB cosAsinB ✓5 3 (4) 5.2 cos(540◦−2x) = cos(180◦−2x) = −cos2x sin(x−450◦) = sin(x−90◦) = −cosx; cos(−x) = cosx tan(x−180◦) = tanx; tan(540◦+x) = tanx Expression = −cos2x −cos2 x +tan2 x = cos2 x−sin2 x cos2 x + sin2 x cos2 x = cos2 x cos2 x = 1 ✓−cos2x ✓−cosx and cosx ✓tanx.tanx ✓cos2x cos2 x ✓cos2 x−sin2 x ✓1 (6) 5.3 (cos15◦−sin15◦)2 = cos2 15◦+sin2 15◦−2sin15◦cos15◦= 1−sin30◦= 1 2 cos15◦> sin15◦, so cos15◦−sin15◦= √ 1 2 = √ 2 2 OR: = √ 2(cos45◦cos15◦−sin45◦sin15◦) = √ 2cos60◦= √ 2 2 OR: cos(45◦−30◦)−sin(45◦−30◦) = √ 6+ √ 2 4 − √ 6− √ 2 4 = √ 2 2 ✓1−sin30◦ ✓1 2 ✓positive root √ 2 2 OR: ✓ √ 2cos45◦form ✓cos60◦✓answer (3) [13] QUESTION 6 6.1 LHS = 1−(1−2sin2 x)+2sinxcosx 1+(2cos2 x−1)+2sinxcosx = 2sin2 x+2sinxcosx 2cos2 x+2sinxcosx = 2sinx(sinx+cosx) 2cosx(cosx+sinx) = sinx cosx = tanx = RHS ✓1−2sin2 x and 2cos2 x−1 ✓2sinxcosx ✓factorising numerator ✓factorising denominator(4) Copyright reserved Please turn over

Mathematics/P2 8 NSC – Marking Guidelines Hlayiso/Challenge 2026 6.2 Undefined if cosx = 0 or sinx+cosx = 0, i.e. tanx = −1 cosx = 0: x = −90◦or 90◦ tanx = −1: x = −45◦or 135◦ ✓cosx = 0 and sinx+cosx = 0 ✓±90◦ ✓−45◦and 135◦ (3) 6.3 [ f(x)]2 −2 f(x)−3 = 0 ( f(x)−3)(f(x)+1) = 0 f(x) = 3 or f(x) = −1 f(x) = −1: tanx = −1 ⇒x = 135◦+k.180◦; but then sinx+cosx = 0 and f(x) is undefined (6.2). No solution. f(x) = 3: tanx = 3 x = 71,57◦+k.180◦, k ∈Z (If tanx = −1 is not rejected: max 4/6. Penalise 1 mark if k ∈Z is omitted.) ✓standard form ✓factors ✓tanx = −1 considered ✓rejected with reason ✓tanx = 3 ✓71,57◦+k.180◦, k ∈Z (6) [13] QUESTION 7 7.1 a = 1 b = 2 (period 180◦) p = −30◦ ✓a ✓b ✓p (3) 7.2 sin2x = cos(x−30◦) = sin(90◦−(x−30◦)) = sin(120◦−x) 2x = 120◦−x+k.360◦⇒x = 40◦+k.120◦ or 2x = 180◦−(120◦−x)+k.360◦⇒x = 60◦+k.360◦, k ∈Z x ∈{−80◦; 40◦; 60◦; 160◦} OR: cos(90◦−2x) = cos(x−30◦): 90◦−2x = ±(x−30◦)+k.360◦ ✓co-function ✓40◦+k.120◦ ✓60◦+k.360◦ ✓all four values (4) 7.3 40◦≤x ≤60◦or 160◦≤x ≤180◦ (accept strict inequalities at 40◦, 60◦, 160◦if ≥is not used: max 2/3) ✓40◦≤x ≤60◦ ✓160◦≤x ✓x ≤180◦ (3) Copyright reserved Please turn over

Mathematics/P2 9 NSC – Marking Guidelines Hlayiso/Challenge 2026 7.4 −sinx = cos(x+90◦) = cos((x+120◦)−30◦) = g(x+120◦) Translate g 120◦to the LEFT (OR 240◦to the right) ✓cos(x+90◦) ✓120◦left / 240◦right (2) [12] QUESTION 8 8.1 In △PQR: tanα = h QR ⇒QR = h tanα ; similarly QS = h tanα In △QRS: d2 = QR2 +QS2 −2QR.QScos2θ [cosine rule] d2 = 2QR2(1−cos2θ) = 2QR2(2sin2 θ) = 4QR2 sin2 θ d = 2QRsinθ (θ acute) h tanα = d 2sinθ ⇒h = d tanα 2sinθ OR: QR = QS, so QM ⊥RS and R ˆQM = θ; sinθ = d/2 QR , QR = d 2sinθ ; h = QRtanα ✓QR in terms of h ✓QR = QS ✓cosine rule ✓1−cos2θ = 2sin2 θ ✓QR = d 2sinθ OR: ✓QM ⊥RS ✓θ ✓sinθ ✓QR ✓h (5) 8.2 h = 80tan32◦ 2sin50◦= 32,63 m ✓substitution ✓answer (2) 8.3 QR = 80 2sin50◦= 52,22 m QM = QRcos50◦= 33,56 m (QM ⊥RS, isosceles △) tanϕ = h QM = 32,63 33,56 ⇒ϕ = 44,19◦ (OR tanϕ = tanα cosθ = tan32◦ cos50◦) ✓QM ✓tanϕ ✓44,19◦(accept 44,18◦–44,20◦) (3) [10] Copyright reserved Please turn over

Mathematics/P2 10 NSC – Marking Guidelines Hlayiso/Challenge 2026 QUESTION 9 9.1 Construction: Draw diameter AOD and join DB. D ˆAS = 90◦ [tan ⊥radius] ∴B ˆAS = 90◦−D ˆAB A ˆBD = 90◦ [∠in semi-circle] ∴A ˆDB = 90◦−D ˆAB [∠s of △] ∴B ˆAS = A ˆDB A ˆDB = A ˆCB [∠s in the same segment] ∴B ˆAS = A ˆCB OR: Construction: join OA and OB. Let B ˆAS = x. O ˆAS = 90◦[tan ⊥radius] ⇒O ˆAB = 90◦−x = O ˆBA [∠s opp equal radii] A ˆOB = 180◦−2(90◦−x) = 2x [∠s of △]; A ˆCB = 1 2A ˆOB = x [∠at centre = 2×∠at circumference] ✓construction ✓S/R ✓S/R ✓S ✓S/R ✓conclusion OR: ✓construction ✓S/R ✓S/R ✓2x (S) ✓S/R ✓conclusion (6) 9.2.1 A ˆBD = P ˆAD = 35◦ [tan-chord theorem] ✓S ✓R (2) 9.2.2 D ˆBC = A ˆBD = 35◦ [equal chords subtend equal ∠s] A ˆBC = 35◦+35◦= 70◦ ✓S/R ✓answer (2) 9.2.3 A ˆOC = 2A ˆBC = 140◦ [∠at centre = 2×∠at circumference] O ˆAC = O ˆCA [∠s opp equal radii] O ˆAC = 180◦−140◦ 2 = 20◦ OR: A ˆDC = 180◦−70◦= 110◦[opp ∠s of cyclic quad]; D ˆAC = 35◦[∠s opp equal sides]; P ˆAC = 70◦ and O ˆAP = 90◦[tan ⊥radius] ⇒O ˆAC = 20◦ ✓140◦(S/R) ✓R ✓20◦ (3) [13] QUESTION 10 10.1 Construction: Draw DG ∥BF with G on AC. In △ADG: AE = ED and EF ∥DG ∴AF = FG [midpt theorem / line ∥one side of △] In △CBF: DG ∥BF FG GC = BD DC = 3 2 [line ∥one side of △] Let FG = 3k, GC = 2k; AF = 3k AF : FC = 3k : (3k +2k) = 3 : 5 ✓construction ✓AF = FG (S/R) ✓S ✓R ✓3 : 5 (5) Copyright reserved Please turn over

Mathematics/P2 11 NSC – Marking Guidelines Hlayiso/Challenge 2026 10.2 In △ADG: EF = 1 2DG [midpt theorem] In △CBF: △CDG ||| △CBF [∠∠∠; DG ∥BF] DG BF = CD CB = 2 5 ⇒BF = 5 2DG BE = BF −EF = 5 2DG −1 2DG = 2DG BE : EF = 2 : 1 2 = 4 : 1 ✓EF = 1 2DG (S/R) ✓2 5 (S/R) ✓BE = 2DG ✓4 : 1 (4) 10.3 area △AEF area △ADC = 1 2AE.AFsin ˆA 1 2AD.ACsin ˆA = 1 2 × 3 8 = 3 16 area △ADC area △ABC = DC BC = 2 5 [same height] area △AEF area △ABC = 3 16 × 2 5 = 3 40 (CA from 10.1) ✓area rule with sin ˆA ✓3 16 ✓2 5 ✓3 40 (4) [13] QUESTION 11 11.1 C ˆAB = x [tan-chord theorem] O ˆCA = C ˆAO = x [∠s opp equal radii] (accept C ˆAD, C ˆAO for C ˆAB; accept A ˆCO) ✓S ✓R ✓S ✓R (4) 11.2 O ˆCD = 90◦ [tan ⊥radius] and ˆE = 90◦(given) ∴OC ∥AE [corresp ∠s =] E ˆAC = A ˆCO = x [alt ∠s; OC ∥AE] ∴E ˆAC = C ˆAB, i.e. AC bisects E ˆAB OR: A ˆCB = 90◦[∠in semi-circle]; E ˆCA = 180◦−90◦−x = 90◦−x [∠s on str line]; in △AEC: E ˆAC = 180◦−90◦−(90◦−x) = x [∠s of △] ✓OC ∥AE (S/R) ✓S ✓R OR: ✓90◦(S/R) ✓90◦−x ✓x (3) 11.3 In △DCB and △DAC: ˆD = ˆD [common] D ˆCB = D ˆAC = x [tan-chord theorem / 11.1] ∴△DCB ||| △DAC [∠∠∠] ✓S ✓S ✓R (3) Copyright reserved Please turn over

Mathematics/P2 12 NSC – Marking Guidelines Hlayiso/Challenge 2026 11.4 DC DA = DB DC [||| △s] ⇒DC2 = DB.DA 64 = 4DA ⇒DA = 16; AB = 12, OC = 6, DO = 10 In △DOC and △DAE: OC ∥AE, so △DOC ||| △DAE [∠∠∠] AE OC = DA DO ⇒AE = 6×16 10 = 9,6 units OR: sin ˆD = OC OD = 6 10 (in △OCD); AE = DAsin ˆD = 16×0,6 = 9,6 ✓DC2 = DB.DA (S/R) ✓DA = 16 and OC = 6 ✓ratio from ||| / sin ˆD ✓9,6 (4) [14] TOTAL: 150 Copyright reserved