Technical Mathematics Grade 12 P2 2026 Memo (Eng)

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

Technical Mathematics/P2 2 Hlayiso/Prediction 2026 NSC – Marking Guidelines MARKING CODES A Accuracy CA Consistent accuracy M Method R Rounding NPR No penalty for rounding NPU No penalty for units omitted S Simplification SF Substitution in correct formula ST Statement RE Reason S/R Statement with reason AO Answer only NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines where indicated. • No penalty for rounding (NPR) for ALL questions, unless stated otherwise. A tolerance range is given where rounding of an earlier answer changes the final answer. • In QUESTIONS 7, 8 and 9 (Euclidean geometry) a statement without the correct reason earns no mark for the reason. Accept any valid alternative proof. Copyright reserved Please turn over

Technical Mathematics/P2 3 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 1 1−5 −4 1.1 mAB = = ✓ SF A 2 − (−4) 6 2 ✓ answer CA mAB = − AO: full marks 3 (2) 2 1.2 tan θ = − ✓ tan θ = mAB CA 3 reference ∠ = 33,69◦ θ = 180◦ − 33,69◦ = 146,31◦ ✓ answer CA (2) √ 1.3 AB√= (2 − √(−4)) + (1 − 5) 2 2 ✓ SF A = 52 = 2 13 units ✓ answer in surd form CA (2) ( ) −4 + 2 5 + 1 1.4 M ; 2 2 ✓ x = −1 A M(−1 ; 3) ✓y=3 A (2) 1.5 mAB × m⊥ = −1 ✓ m1 × m2 = −1 A 3 ✓ m⊥ = 23 CA m⊥ = 2 ✓ substitution of M CA 3 y − 3 = (x − (−1)) ✓ equation CA 2 (4) 3 9 y = x+ OR y = 1,5x + 4,5 2 2 3 9 1.6 t = (1) + = 6 ✓t =6 CA 2 2 (1) [13] Copyright reserved Please turn over

Technical Mathematics/P2 4 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 2 2.1.1 r2 = (−3)2 + (6)2 = 45 ✓ substitution of T A x2 + y2 = 45 ✓ equation CA (2) √ √ 2.1.2 r= 45 = 3 5 units ✓ answer CA (1) 6−0 2.1.3 mOT = = −2 ✓ mOT A −3 − 0 1 ✓ mtangent CA mtangent = (tangent ⊥ radius) ✓ substitution of T CA 2 1 ✓ equation CA y − 6 = (x − (−3)) 2 (4) 1 15 y = x+ OR y = 0,5x + 7,5 2 2 2.1.4 K(0 ; 7,5) ✓ answer CA (1) 25x2 4y2 100 2.2.1 + = ✓ dividing by 100 A 100 100 100 x2 y2 x2 y2 + = 1 OR + =1 ✓ standard form CA 22 52 4 25 (2) 2.2.2 y ✓ x-intercepts ±2 CA 5 ✓ y-intercepts ±5 and elliptical shape CA (2) x −2 2 −5 [12] Copyright reserved Please turn over

Technical Mathematics/P2 5 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 3 180◦ 3.1.1 2,5 × = 143,24◦ ✓ answer A π (1) 1 3.1.2 cot(143,24◦ − 41,3◦ ) = ✓ tan 101,94 1 ◦ CA tan 101,94◦ = −0,21 ✓ answer CA (2) sin2 41,3◦ 0,4357 . . . 3.1.3 ◦ = ✓ substitution CA cos 143,24 −0,8011 . . . = −0,54 ✓ answer CA (2) 3.2.1 k2 + (−5)2 = 62 (Pythagoras) ✓ Pythagoras A k2 =√11 √ k = 11 (k > 0, R in quadrant IV) ✓k= 11 CA (2) √ 3.2.2 6 sin(θ + )11 tan θ ( ) 5 √ 5 ✓ sin θ = − 65 CA =6 − + 11 − √ 6 11 ✓ tan θ = − √511 CA = −5 − 5 = −10 ✓ answer CA (3) 3.3 5 cos x = −2 cos x = −0,4 ✓ cos x = −0,4 A reference ∠ = 66,42◦ ✓ reference angle CA x = 180◦ − 66,42◦ = 113,58◦ ✓ 113,58◦ CA or x = 180◦ + 66,42◦ = 246,42◦ ✓ 246,42◦ CA (4) [14] QUESTION 4 4.1.1 − cos x ✓ answer A (1) 4.1.2 − tan x ✓ answer A (1) Copyright reserved Please turn over

Technical Mathematics/P2 6 Hlayiso/Prediction 2026 NSC – Marking Guidelines (− cos x)(− tan x) 4.1.3 + (cos x)2 (tan x)2 ✓ − cos x A − sin x sin x ✓ − tan x A cos x · 2 ✓ − sin x A =− cos x + cos2 x · sin x sin x cos2 x ✓ cos2 x · tan2 x A = −1 + sin x2 ✓ −1 + sin2 x CA = − cos2 x ✓ answer CA (6) 4.2.1 1 + tan2 θ = sec2 θ ✓ answer A (1) sin θ (1 + tan2 θ ) 4.2.2 LHS = ✓ common factor A sec θ sin θ · sec2 θ ✓ sec2 θ A = ✓ sin θ · sec θ CA sec θ sin θ 1 ✓ cos θ = tan θ CA = sin θ · sec θ = sin θ · (4) cos θ = tan θ = RHS [13] QUESTION 5 5.1 y f: 3 x = 90◦ ✓ intercepts (0 ; 1), 45◦ and 135◦ g 2 A ✓ turning point (90◦ ; −1), f 1 end point (180◦ ; 1) A 45◦ 90◦ 135◦ 180◦ x ✓ shape A −1 g: −2 (90◦ ; −1) ✓ asymptote x = 90◦ A ✓ intercepts (0◦ ; 0) and −3 (180◦ ; 0) A ✓ shape (through (45◦ ; 1) and (135◦ ; −1)) A (6) 5.2.1 180◦ ✓ answer A (1) 5.2.2 x = 90◦ ✓ answer A (1) Copyright reserved Please turn over

Technical Mathematics/P2 7 Hlayiso/Prediction 2026 NSC – Marking Guidelines 5.3 45◦ < x < 90◦ or 135◦ < x < 180◦ ✓ 45◦ < x < 90◦ CA OR x ∈ (45◦ ; 90◦ ) ∪ (135◦ ; 180◦ ) ✓ 135◦ < x < 180◦ CA Penalise 1 mark if inclusive notation is used (2) 5.4 −3 ≤ y ≤ 1 OR y ∈ [−3 ; 1] ✓ −3 CA ✓ 1 and notation CA (2) [12] QUESTION 6 522 + 382 − 702 6.1 cos BĈD = ✓ SF in cosine rule A 2(52)(38) ✓ −0,19 CA cos BĈD = −0,1903 . . . ✓ answer CA BĈD = 100,97◦ (3) 1 6.2 Area △BCD = (BC)(CD) sin BĈD ✓ area rule A 2 1 ✓ SF CA = (52)(38) sin 100,97◦ ✓ answer CA 2 = 969,95 m2 Accept 969,9 – 970,0 (3) AB 6.3 tan 24◦ = ✓ trig ratio A 70 AB = 70 tan 24◦ = 31,17 m ✓ answer CA (2) AB 31,17 6.4 tan AĈB = = ✓ trig ratio A BC 52 AĈB = 30,94√◦ ✓ substitution CA OR AC = 522 + 31,172 = 60,63 m; ✓ answer CA 31,17 (3) sin AĈB = 60,63 [11] Copyright reserved Please turn over

Technical Mathematics/P2 8 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 7 7.1 …bisects the chord. ✓ answer A (1) 7.2.1 AM = 12 cm ✓S A (line from centre ⊥ to chord) ✓R A (2) 7.2.2 Let the radius be r: OA = OC = r (radii) OM = r − 8 ✓ OM = r − 8 A OA2 = AM 2 + OM 2 (Pythagoras) ✓ Pythagoras (R) A r2 = 122 + (r − 8)2 ✓ substitution CA r2 = 144 + r2 − 16r + 64 16r = 208 r = 13 cm ✓ answer CA (4) 7.2.3 ∠ in semicircle (CD is a diameter) ✓R A (1) 7.2.4 CD = 26 cm and AC2 = 122 + 82 = 208 ✓ AC2 = 208 or MD = 18CA AD2 = CD2 − AC2 = 676 − 208 = 468 AD = 21,63 cm ✓ answer CA OR MD = 5 + 13 = 18 cm; (2) AD2 = 122 + 182 = 468 √ AD = 21,63 cm (6 13) [10] Copyright reserved Please turn over

Technical Mathematics/P2 9 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 8 8.1 …the angle in the alternate segment. ✓ answer A (1) 8.2.1 AD̂B = T ÂB = 40◦ ✓S A (tan-chord theorem) ✓R A (2) 8.2.2 AĈB = AD̂B = 40◦ ✓S CA (∠s in the same segment) ✓R A OR (tan-chord theorem) (2) 8.2.3 BÂC = BD̂C = 58◦ (∠s in the same segment) ✓S A AB̂C = 180◦ − 58◦ − 40◦ (sum of ∠s in △) ✓R A AB̂C = 82◦ OR AD̂C = 40◦ + 58◦ = 98◦ (S) ✓ answer CA AB̂C = 180◦ − 98◦ = 82◦ (opp ∠s of cyclic quad) (R) (3) 8.2.4 V ÂD = AB̂D (tan-chord theorem) ✓ S/R A AB̂D = AD̂B (∠s opp equal sides; AB = AD) ✓ S/R A ∴ V ÂD = AD̂B ✓S A ∴ TV ∥ BD (alt ∠s equal) ✓ conclusion with R A OR (4) AB̂D = AD̂B = 40◦ (∠s opp equal sides) T ÂB = 40◦ = AB̂D (given / proven) ∴ TV ∥ BD (alt ∠s equal) 8.3.1 BÔC = 2 × 64◦ = 128◦ ✓S A (∠ at centre = 2 × ∠ at circumference) ✓R A (2) 8.3.2 OB̂T = 90◦ (tan ⊥ radius) ✓ S/R A OĈT = 90◦ (tan ⊥ radius) ✓ S/R A OB̂T + OĈT = 180◦ ∴ OBTC is a cyclic quadrilateral ✓ conclusion with R A (opp ∠s supplementary / converse opp ∠s of cyclic (3) quad) [17] Copyright reserved Please turn over

Technical Mathematics/P2 10 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 9 9.1 …in the same proportion. ✓ answer A (1) PS PT 9.2.1 = (line ∥ one side of △; ST ∥ QR) ✓S A SQ T R 6 8 ✓R A = 9 TR T R = 12 units ✓ answer CA (3) QU PT 8 9.2.2 = = (line ∥ one side of △; TU ∥ PQ) ✓ ratio CA UR T R 12 8 ✓ answer CA QU = × 30 = 12 units (2) 20 9.2.3 In △PST and △PQR: P̂ = P̂ (common) ✓S A PŜT = Q̂ (corresp ∠s; ST ∥ QR) ✓ S/R A PT̂ S = R̂ (corresp ∠s; ST ∥ QR) ✓ S/R or conclusion A ∴ △PST ||| △PQR (∠∠∠) (3) ST PS 9.2.4 = (||| △s) ✓ ratio A QR PQ 6 ✓ answer CA ST = × 30 = 12 units (2) 15 OR SQUT is a parallelogram (ST ∥ QU, TU ∥ SQ), so ST = QU = 12 units [11] Copyright reserved Please turn over

Technical Mathematics/P2 11 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 10 3 000 10.1.1 n= = 50 rev/s ✓ answer A 60 (1) 10.1.2 ω = 2π n = 2π (50) ✓ SF A = 100π ≈ 314,16 rad/s ✓ answer CA (2) 10.1.3 D = 250 mm = 0,25 m ✓ conversion to m A v = π Dn = π (0,25)(50) ✓ SF CA v = 39,27 m/s ✓ answer CA OR v = ω r = 314,16 × 0,125 = 39,27 m/s (3) 10.1.4 4h2 − 4dh + x2 = 0 ✓ formula A 4(60)2 − 4(250)(60) + x2 = 0 ✓ SF A 14 400 − 60 000 + x2 = 0 x2 = 45 600 ✓ x2 = 45 600 CA x = 213,54 mm ✓ answer CA (4) π 10.2.1 220◦ × = 3,84 rad ✓ answer A 180◦ (1) 10.2.2 s = rθ = 15 × 3,84 ✓ SF CA s = 57,60 cm ✓ answer CA Accept 57,6 (2) 10.2.3 140◦ = 2,4435 . . . rad ✓ 2,44 rad A s = 6 × 2,4435 = 14,66 cm ✓ answer CA Accept 14,64 – 14,66 (2) 10.2.4 Total length = 57,60 + 14,66 + 2(24,73) ✓ adding arcs and 2 × 24,73 = 121,72 cm CA ✓ answer CA (2) Copyright reserved Please turn over

Technical Mathematics/P2 12 Hlayiso/Prediction 2026 NSC – Marking Guidelines 10.2.5 Minor sector angle = 360◦ − 220◦ = 140◦ = 2,4435 ✓ 140◦ / 2,44 rad A rad ✓ SF CA r2 θ (15)2 (2,4435) ✓ answer CA Area = = 2 2 Accept 274,5 – 274,9 = 274,89 cm2 (3) [20] QUESTION 11 18 11.1.1 a= =3m ✓ answer A 6 (1) 4,2 + 5,1 11.1.2 m1 = = 4,65; m2 = 5,45; m3 = 5,9; ✓ mid-ordinates m1 –m3 A 2 m4 = 5,7; m5 = 5,0; m6 = 4,25 ✓ mid-ordinates m4 –m6 A AT = 3(4,65 + 5,45 + 5,9 + 5,7 + 5,0 + 4,25) ✓ SF CA = 3(30,95) = 92,85 m2 ✓ answer CA OR ( ) (4) 4,2 + 3,9 AT = 3 + 5,1 + 5,8 + 6,0 + 5,4 + 4,6 2 11.1.3 Turf needed = 92,85 × 1,1 = 102,14 m2 ✓ ×1,1 M Cost = 102,135 × R245 = R25 023,08 ✓ cost CA R25 023,08 > R24 000 ✓ comparison CA The budget is NOT sufficient. ✓ conclusion CA No marks for the conclusion without the 10% extra. (4) 1 4 11.2.1 V = π r2 h + × π r3 ✓ volume of cylinder A 2 3 2 ✓ volume of hemisphere A = π (15) (80) + π (15)3 2 3 = 18 000π + 2 250π = 20 250π ✓ answer CA ≈ 63 617,25 cm3 (3) Copyright reserved Please turn over

Technical Mathematics/P2 13 Hlayiso/Prediction 2026 NSC – Marking Guidelines 1 11.2.2 A = 2π rh + × 4π r2 ✓ curved surface of cylinder 2 = 2π (15)(80) + 2π (15)2 A = 2 400π + 450π = 2 850π ✓ surface of hemisphere A ≈ 8 953,54 cm2 ✓ answer CA (3) 11.2.3 30 × 63 617,25 = 1 908 517,5 cm3 ✓ conversion to m3 (or 2 m3 = 1,91 m3 to cm3 ) CA 1,91 m3 < 2 m3 ∴ there is enough concrete. ✓ conclusion CA (2) [17] TOTAL: 150 Copyright reserved Please turn over