Mathematics Grade 12 P2 NSC November Prediction 2026 Memo

hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P2 PREDICTION EXAMINATION 2026 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 11 pages. Copyright reserved Please turn over

Mathematics/P2 2 Hlayiso/Prediction 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 1.1 a = 86,70 ✓a b = −7,57 ✓b ŷ = 86,70 − 7,57x ✓ equation (3) 1.2 r = −0,97 ✓ answer (1) 1.3 ŷ = 86,70 − 7,57(3,5) ✓ substitution ŷ = 60,21% (accept 60,20% – 60,22%) ✓ answer (2) 1.4 12 hours lies far outside the range of the given data ✓ reason (1) (extrapolation). OR: The line predicts 86,70 − 7,57(12) = −4,14%, a negative mark, which is impossible. 1.5 σ = 13,45 ✓ answer (1) (population SD σ ; if s = 14,17 is used: 0 marks in 1.5, then CA in 1.6) 640 1.6 x̄ = = 64 ✓ mean 64 10 (x̄ − σ ; x̄ + σ ) = (50,55 ; 77,45) ✓ interval Marks in the interval: 76 ; 71 ; 68 ; 62 ; 58 ; 55 6 learners ✓ answer (3) (CA from 1.5: interval (49,83 ; 78,17) still gives 6) [11] Copyright reserved Please turn over

Mathematics/P2 3 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 2 2.1 Cumulative frequencies: 6 ; 20 ; 45 ; 63 ; 74 ; 80 ✓ first three ✓ last three (2) 2.2 ✓ grounded at (0 ; 0) 80 ✓ plotting cumulative 60 frequency against upper limits 40 ✓ smooth shape 20 (Penalise 1 mark if plotted at midpoints/lower limits) 0 100 200 300 (3) Points: (0 ; 0), (50 ; 6), (100 ; 20), (150 ; 45), (200 ; 63), (250 ; 74), (300 ; 80) 2.3 At x = 220: cumulative frequency ≈ 67 ✓ reading 67 Number who spent more than R220 ≈ 80 − 67 = 13 ✓ answer (2) customers (accept 12 – 14 if read from candidate’s ogive) 2.4 Midpoints: 25 ; 75 ; 125 ; 175 ; 225 ; 275 ∑ f x = 6(25) + 14(75) + 25(125) + 18(175) + ✓ ∑ f x = 11 600 11(225) + 6(275) = 11 600 ✓ answer (2) 11 600 x̄ = = R145 80 [9] Copyright reserved Please turn over

Mathematics/P2 4 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 3 −3 − 1 3.1 mAB = = −2 ✓ mAB −1 − (−3) 5−1 1 mAC = = ✓ mAC 5 − (−3) 2 1 ✓ product = −1 (3) mAB × mAC = −2 × = −1 2 ∴ AB ⊥ AC, i.e. BÂC = 90◦ 5 − (−3) 4 3.2 mBC = = ✓ gradient 5 − (−1) 3 4 ✓ substitution y − 5 = (x − 5) ✓ equation (3) 3 4 5 y = x− 3 3 4 3.3 tan θ = ✓ tan θ = mBC 3 θ = 53,13◦ ✓ answer (2) 3.4 Inclination of AB: ✓ 116,57◦ tan β = −2 ⇒ β = 180◦ − 63,43◦ = 116,57◦ ✓ method AB̂C = β − θ [ext ∠ of △] ✓ answer AB̂C = 116,57◦ − 53,13◦ = 63,43◦ OR √ OR: tan AB̂C = AC 80 = √ = 2 (right-angled △ at A) ✓ AB and AC ✓ tan = 2 AB 20 ✓ answer (3) ⇒ AB̂C = 63,43 ◦ √ √ √ 3.5 AB = √(−1 + 3)2 + (−3 − 1)2√= 20 √=2 5 ✓ AB AC = (5 + 3)2 + (5 − 1)2 = 80 =√ 4 5√ ✓ AC Area △ABC = 2 × AB × AC = 2 (2 5)(4 5) = 20 1 1 ✓ answer (3) square units 3.6 BÂC = 90◦ ⇒ BC is a diameter [converse: ∠ in semi-circle] ✓ centre (2 ; 1) ( = midpoint of ) Centre BC ✓ r = 5 or r2 = 25 −1 + 5 −3 + 5 ✓ equation (3) = ; = (2 ; 1) 2 √ 2 r = 12 BC = 12 62 + 82 = 5 (x − 2)2 + (y − 1)2 = 25 OR: any correct method (simultaneous equations in x2 + y2 + Dx + Ey + F = 0 / perpendicular bisectors) giving centre (2 ; 1) and r2 = 25: same mark allocation Copyright reserved Please turn over

Mathematics/P2 5 Hlayiso/Prediction 2026 NSC – Marking Guidelines 3.7 AE2 = AB2 ✓ equation (k + 3)2 + (3 − 1)2 = 20 (k + 3)2 = 16 ✓ k = 1 or k = −7 k + 3 = ±4 k = 1 or k = −7 E in the first quadrant ⇒ k = 1 ✓ selection k = 1 (3) [20] QUESTION 4 4.1 x2 + 2x + 1 + y2 − 4y + 4 = 40 + 1 + 4 ✓ completing the squares (x + 1)2 + (y − 2)2 = 45 M(−1√ ; 2) √ ✓ M(−1 √ ; 2) r = 45 = 3 5 (≈ 6,71) ✓ r = 45 (3) 8−2 4.2 mMP = =2 ✓ mMP 2 − (−1) 1 ✓ mtan mtan = − [tan ⊥ radius] ✓ substitution of P 2 y − 8 = − 2 (x − 2) 1 ✓ equation (4) y = − 12 x + 9 √ 4.3 MR = (8 + 1)2 + (2 − 2)2 = 9 ✓ MR = 9 MK̂R = 90◦ [tan ⊥ radius] RK2 = MR2 − MK2 = 81 − 45 = 36 ✓ Pythagoras RK = 6 units ✓ answer (3) √ 4.4 Circles touch √ externally: √ MN √ ✓ MN = 4 5 = r1 + r2 = 3 5 + 5 = 4 5 ✓ distance formula (k + 1)2 + (6 − 2)2 = 80 ✓ (k + 1)2 = 64 (k + 1)2 = 64 k + 1 = ±8 ✓ both values (4) k = 7 or k = −9 4.5 The parallel tangent touches the circle at the other end of the diameter through P. Let it be P′ (x ; y): 2+x = −1 and 8+y =2 ✓ P′ (−4 ; −4) 2 2 ✓ m = − 21 and substitution P′ (−4 ; −4) ✓ equation (3) y + 4 = − 21 (x + 4) y = − 12 x − 6 Copyright reserved Please turn over

Mathematics/P2 6 Hlayiso/Prediction 2026 NSC – Marking Guidelines 4.6 The line cuts the circle at two points when it lies between the two parallel tangents: ✓ −6 and 9 −6 < p < 9 ✓ correct inequality (2) [19] QUESTION 5 5.1.1 a2 + (−8)2 = 172 ✓ Pythagoras a2 = 225 a = −15 (P in the third quadrant) ✓ answer (2) 5.1.2 cos(2θ = ) cos2 θ(− sin)2 θ ✓ double angle expansion −15 2 −8 2 ✓ substitution = − 17 17 ✓ answer (3) 225 − 64 161 = = 289 289 ( 64 ) 161 OR: 1 − 2 sin2 θ = 1 − 2 289 = 289 5.1.3 ◦ ) = sin θ . cos 30◦ + cos θ . sin 30◦ sin(θ + 30( ✓ expansion ( ) √ ) ( )( ) −8 3 −15 1 ✓ substitution = + ✓ answer (3) 17 2 17 2 √ −8 3 − 15 = 34 5.2.1 cos 236◦ = cos(270◦ − 34◦ ) = − sin 34◦ = −m ✓ reduction OR: cos(180◦ + 56◦ ) = − cos 56◦ = − sin 34◦ = −m ✓ answer (2) 5.2.2 cos 68◦ = cos 2(34◦ ) = 1 − 2 sin2 34◦ = 1 − 2m2 ✓ double angle ✓ answer (2) 2(− sin x)(cos x) 5.3 = ✓ − sin x and cos x (− sin x)(− sin x) + (− cos x)(cos x) −2 sin x cos x (numerator) = 2 ✓ (− sin x)(− sin x) sin x − cos2 x ✓ (− cos x)(cos x) − sin 2x = ✓ − sin 2x − cos 2x = tan 2x ✓ − cos 2x ✓ answer (6) [18] Copyright reserved Please turn over

Mathematics/P2 7 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 6 2 sin x cos x − sin x 6.1.1 LHS = ✓ 2 sin x cos x 1 + (2 cos2 x − 1) − cos x ✓ 2 cos2 x − 1 sin x(2 cos x − 1) = ✓ factorising numerator 2 cos2 x − cos x ✓ factorising denom. (4) sin x(2 cos x − 1) = cos x(2 cos x − 1) sin x = = tan x = RHS cos x 6.1.2 Undefined if cos x = 0 or 2 cos x − 1 = 0, i.e. cos x = 12 ✓ 90◦ and 270◦ x = 90◦ ; 270◦ ; 60◦ ; 300◦ ✓ 60◦ and 300◦ (2) 6.2 sin(3x − x) = cos x ✓ compound angle sin 2x = cos x ✓ 2 sin x cos x 2 sin x cos x − cos x = 0 ✓ common factor cos x(2 sin x − 1) = 0 cos x = 0 or sin x = 12 ✓ 90◦ + k.180◦ x = 90◦ + k.180◦ ✓ x = 30◦ + k.360◦ x = 30◦ + k.360◦ or x = 150◦ + k.360◦ , k ∈ Z ✓ x = 150◦ + k.360◦ , k ∈ Z OR: sin 2x = sin(90◦ − x): 2x = 90◦ − x + k.360◦ ⇒ x = 30◦ + k.120◦ OR: or 2x = 180◦ − (90◦ − x) + k.360◦ = ✓ sin(90◦ − x) 90◦ + x + k.360◦ ⇒ x = 90◦ + k.360◦ , k ∈ Z ✓ 2x = 90◦ − x + k.360◦ (Penalise 1 mark if k ∈ Z is omitted. Dividing by cos x ✓ x = 30◦ + k.120◦ loses solutions: max 3 marks.) ✓ 2x = 90◦ + x + k.360◦ ✓ x = 90◦ + k.360◦ ✓k∈Z (6) [12] Copyright reserved Please turn over

Mathematics/P2 8 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 7 7.1 180◦ ✓ answer (1) y f 1 0,5 x −180◦ −150◦ −120◦ −90◦ −60◦ −30◦ 0 30◦ 60◦ 90◦ 120◦ 150◦ 180◦ −0,5 −1 g 7.2 ✓ x-intercepts g: x-intercepts −135◦ ; −45◦ ; 45◦ ; 135◦ ; turning ✓ turning points and end points (−180◦ ; 1), (−90◦ ; −1), (0◦ ; 1), (90◦ ; −1), points (180◦ ; 1) ✓ shape (3) 7.3 sin(x + 30◦ ) = cos 2x = sin(90◦ − 2x) ✓ co-function x + 30◦ = 90◦ − 2x + k.360◦ ⇒ x = 20◦ + k.120◦ ✓ 20◦ + k.120◦ or x + 30◦ = 180◦ − (90◦ − 2x) + k.360◦ ⇒ x = ✓ −60◦ + k.360◦ −60◦ − k.360◦ ✓ all four values (4) x ∈ {−100◦ ; −60◦ ; 20◦ ; 140◦ } OR: cos 2x = cos(60◦ − x): 2x = ±(60◦ − x) + k.360◦ 7.4 20◦ < x < 140◦ (accept x ∈ (20◦ ; 140◦ )) ✓ critical values ✓ correct inequality (2) 7.5 h(x) = sin(x − 120◦ + 30◦ ) = sin(x − 90◦ ) ✓ sin(x − 90◦ ) h(x) = − cos x ✓ − cos x (2) [12] QUESTION 8 8.1 In △BCD: CB̂D = 180◦ − x − (90◦ + x) = 90◦ − 2x ✓ 90◦ − 2x [∠s of △] ✓ sine rule BC = CD [sine rule] ✓ cos x sin(90◦ + x) sin(90◦ − 2x) ✓ cos 2x d sin(90◦ + x) d cos x ✓ AB = BCtan y (5) BC = = sin(90◦ − 2x) cos 2x AB In △ABC: tan y = BC d. cos x. tan y AB = BC tan y = cos 2x Copyright reserved Please turn over

Mathematics/P2 9 Hlayiso/Prediction 2026 NSC – Marking Guidelines 25 cos 20◦ tan 35◦ 8.2 AB = = 21,47 m ✓ substitution cos 40◦ ✓ answer (2) 25 cos 20◦ 8.3 BC = = 30,67 m ✓ substitution into area rule cos 40◦ Area △BCD = 21 (25)(30,67) sin 20◦ = 131,11 m2 ✓ answer (2) 25 sin 20◦ OR: BD = = 11,16 m; Area cos 40◦ = 12 (25)(11,16) sin 110◦ = 131,11 m2 (accept 131,08 – 131,12) [9] QUESTION 9 9.1.1 BÂC = BĈE = 32◦ [tan-chord theorem] ✓S ✓R (2) 9.1.2 OĈE = 90◦ [tan ⊥ radius] ✓S ✓R OĈB = 90◦ − 32◦ = 58◦ ✓ answer OR: BÔC = 2BÂC = 64◦ [∠ at centre = 2 × ∠ at circumference] OR: OĈB = OB̂C [∠s opp equal radii] ✓ 64◦ S/R 180◦ − 64◦ ✓R OĈB = = 58◦ 2 ✓ answer (3) 9.2.1 BÂD = 12 BÔD = 68◦ [∠ at centre = 2 × ∠ at ✓S ✓R (2) circumference] 9.2.2 BĈD = 180◦ − 68◦ = 112◦ [opp ∠s of cyclic quad] ✓S ✓R (2) 9.2.3 AB̂D = AD̂B [∠s opp equal sides] ✓R 180◦ − 68◦ ✓ answer (2) AB̂D = = 56◦ 2 9.2.4 BĈE = BÂD = 68◦ [ext ∠ of cyclic quad] ✓S ✓R (2) OR: BĈE = 180◦ − 112◦ = 68◦ [∠s on a str line] [13] Copyright reserved Please turn over

Mathematics/P2 10 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 10 10.1 Construction: Mark G on DE and H on DF such that ✓ construction DG = AB and DH = AC. Join GH. In △ABC and △DGH: AB = DG, AC = DH [construction];  = D̂ ✓ ≡ (SAS) [given] ✓ DĜH = Ê ∴ △ABC ≡ △DGH [SAS] ✓ S/R ∴ DĜH = B̂ = Ê ✓ S/R ∴ GH ∥ EF [corresp ∠s =] ✓ conclusion (6) DG DH ∴ = [line ∥ one side of △] DE DF AB AC ∴ = DE DF AE AD 2 10.2.1 = = [line ∥ one side of △ / prop theorem; ✓S EC DB 3 DE ∥ BC] ✓R (2) 10.2.2 BDEF is a parallelogram [both pairs of opp sides ∥] ✓ BF = 8 (S/R) BF = DE = 8 [opp sides of parm] CF CE 3 = = [line ∥ one side of △; EF ∥ AB] ✓ CF = 12 (S/R) FB EA 2 ✓ answer CF = 32 (8) = 12 BC = 8 + 12 = 20 units OR: OR ✓ ||| (S/R) △ADE ||| △ABC [∠∠∠; DE ∥ BC] ✓ 52 DE AD 2 = = [||| △s] ✓ answer (3) BC AB 5 BC = 52 (8) = 20 units 1 area △ADE AD.AE. sin  2 2 4 10.2.3 = 21 = × = ✓ method (area rule) area △ABC 2 AB.AC. sin  5 5 25 ✓ answer (2) (AE : AC = 2 : 5 from 10.2.1) [13] Copyright reserved Please turn over

Mathematics/P2 11 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 11 11.1 PR̂Q = x [tan-chord theorem] ✓S✓R PŜT = x [alt ∠s; ST ∥ KL] ✓S✓R (4) (accept QR̂P, QR̂T, TR̂Q for PR̂Q; accept TŜP for PŜT) 11.2 PŜT = PR̂Q (= x) ✓S ∴ QSTR is a cyclic quadrilateral [ext ∠ of quad = ✓R (2) int opp ∠] 11.3 In △PST and △PRQ: P̂ = P̂ [common] ✓S PŜT = PR̂Q [proven] ✓S PT̂S = PQ̂R [sum of ∠s in △ / ext ∠ of cyclic quad] ∴ △PST ||| △PRQ [∠∠∠] ✓R (3) PS PT 11.4 = [||| △s] ✓ S/R PR PQ ∴ PS.PQ = PT.PR ✓ conclusion (2) 11.5 PQ = 4 + 5 = 9 ✓ PQ = 9 4 × 9 = 3 × PR ✓ substitution PR = 12 ✓ answer (3) TR = 12 − 3 = 9 units [14] TOTAL: 150 Copyright reserved