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 MOCK EXAMINATION 2026 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 12 pages. Copyright reserved Please turn over

Technical Mathematics/P2 2 Hlayiso/Mock 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 S/R Statement/Reason 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. • 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. QUESTION 1 √ 1.1 AB√= (3 − √(−3)) + (4 − 2) 2 2 ✓ SF A = 40 = 2 10 units ✓ answer CA (2) ( ) −3 + 3 2 + 4 1.2 M ; ✓x=0 A 2 2 ✓y=3 A M(0 ; 3) (2) 4−0 1.3 mBC = =2 ✓ SF A 3−1 ✓ answer CA (2) 1.4 tan θ = 2 ✓ tan θ = mBC CA θ = 63,43◦ ✓ answer CA (2) 1.5 m = 2 (parallel lines have equal gradients) ✓m=2 CA y − 2 = 2(x − (−3)) ✓ substitution of A CA y = 2x + 8 ✓ equation CA (3) Copyright reserved Please turn over

Technical Mathematics/P2 3 Hlayiso/Mock 2026 NSC – Marking Guidelines 2−0 1 1.6 mAC = =− ✓ mAC A −3 − 1 2 1 ✓ product = −1 CA mAC × mBC = − × 2 = −1 ✓ conclusion CA 2 ∴ AC ⊥ BC, i.e. AĈB = 90◦ (3) [14] QUESTION 2 2.1.1 r2 = (−2)2 + (4)2 = 20 ✓ substitution of T A x2 + y2 = 20 ✓ equation CA (2) √ √ 2.1.2 OS = 20 = 2 5 units ✓ answer CA (1) 4−0 2.1.3 mOT = = −2 ✓ mOT A −2 − 0 1 ✓ mtangent CA mtangent = (tangent ⊥ radius) ✓ substitution of T CA 2 1 ✓ equation CA y − 4 = (x + 2) 2 (4) 1 y = x+5 2 2.1.4 K(0 ; 5) ✓ answer CA (1) 4x2 y2 2.2.1 + =1 ✓ dividing by 16 A 16 16 x2 y2 x2 y2 ✓ standard form CA + = 1 OR + =1 (2) 22 42 4 16 Copyright reserved Please turn over

Technical Mathematics/P2 4 Hlayiso/Mock 2026 NSC – Marking Guidelines 2.2.2 y ✓ x-intercepts ±2 CA 4 ✓ y-intercepts ±4 and shape CA (2) x −2 2 −4 [12] QUESTION 3 180◦ 3.1.1 2,3 × = 131,78◦ ✓ answer A π (1) 1 3.1.2 sec 47◦ = ✓ cos147◦ A cos 47◦ = 1,47 ✓ answer CA (2) 3.1.3 sin(131,78◦ − 47◦ ) = sin 84,78◦ ✓ 84,78◦ CA = 1,00 (0,9959 . . .) ✓ answer CA (2) 3.2.1 y ✓ x = −5, y = 12, r = 13 in quadrant II A 13 12 ✓ answer CA (2) θ x −5 r2 = (−5)2 + 122 r = 13 5 cos θ = − 13 Copyright reserved Please turn over

Technical Mathematics/P2 5 Hlayiso/Mock 2026 NSC – Marking Guidelines 3.2.2 1 − sec(θ · sin) θ( ) = 1− − 13 12 ✓ sec θ = − 13 5 CA 5 13 ✓ sin θ = 13 12 CA 12 = 1+ 5 ✓ answer CA 17 = OR 3,4 (3) 5 3.3 sin x = −0,6 ✓ sin x = −0,6 A reference angle = 36,87◦ x = 180◦ + 36,87◦ = 216,87◦ ✓ 216,87◦ CA or x = 360◦ − 36,87◦ = 323,13◦ ✓ 323,13◦ CA (3) [13] QUESTION 4 4.1.1 cos x ✓ answer A (1) 4.1.2 − sin x ✓ answer A (1) (− sin x)(cos x) 4.1.3 + (− sin x)2 ✓ − sin x A − tan x sin x cos x ✓ cos x A = + sin2 x ✓ − tan x A sin x ✓ sin2 x A cos x = cos2 x + sin2 x ✓ tan x = cos sin x x CA =1 ✓ answer CA (6) 4.2.1 1 − cos2 θ = sin2 θ ✓ answer A (1) Copyright reserved Please turn over

Technical Mathematics/P2 6 Hlayiso/Mock 2026 NSC – Marking Guidelines 1 − cos θ 4.2.2 LHS = cos θ ✓ sec θ = cos1 θ A sin θ sin θ ✓ tan θ = cos θ A cos θ 1−cos2 θ 1 − cos2 θ cos θ ✓ cos θ CA = × cos θ sin θ ✓ sin2 θ CA sin2 θ (4) = sin θ = sin θ = RHS [13] QUESTION 5 5.1 y f: f ✓ intercepts at 90◦ , 270◦ and 2 (0 ; 2) A 1 g ✓ turning points (0◦ ; 2); (180◦ ; −2); (360◦ ; 2) A 90◦ 180◦ 270◦ 360◦ x ✓ shape A g: −1 ✓ x-intercepts at 0◦ ; 90◦ ; −2 180◦ ; 270◦ ; 360◦ A ✓ turning points at 45 ,◦ 135◦ , 225◦ , 315◦ (y = ±1) A ✓ shape A (6) 5.2.1 180◦ ✓ answer A (1) 5.2.2 2 ✓ answer A (1) 5.2.3 x = 90◦ or x = 270◦ ✓ 90◦ CA ✓ 270◦ CA (2) Copyright reserved Please turn over

Technical Mathematics/P2 7 Hlayiso/Mock 2026 NSC – Marking Guidelines 5.2.4 −1 ≤ y ≤ 3 OR y ∈ [−1 ; 3] ✓ answer CA (1) [11] QUESTION 6 6.1 D̂ = 180◦ − 35◦ − 115◦ = 30◦ ✓ answer A (1) BD 40 6.2 ◦ = ✓ sine rule A sin 35 sin 30◦ 40 sin 35◦ ✓ SF CA BD = sin 30◦ BD = 45,89 m ✓ answer CA (3) AB 6.3 tan 28◦ = ✓ trig ratio A 40 AB = 40 tan 28◦ = 21,27 m ✓ answer CA (2) CD 40 6.4 ◦ = ✓ rule A sin 115 sin 30◦ 40 sin 115◦ ✓ SF CA CD = = 72,50 m ✓ answer CA sin 30◦ OR (3) CD2 = 402 + 45,892 − 2(40)(45,89) cos 115◦ CD = 72,50 m 1 6.5 Area △BCD = (BC)(BD) sinCB̂D ✓ area rule A 2 1 ✓ SF CA = (40)(45,89) sin 115◦ ✓ answer CA 2 = 831,74 m2 Accept 831,7 – 831,8 (3) [12] Copyright reserved Please turn over

Technical Mathematics/P2 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 7.1 …twice the angle subtended by the same arc at the ✓ answer A circumference (of the circle). (1) 1 7.2.1 Â = BÔD = 65◦ ✓S A 2 (∠ at centre = 2 × ∠ at circumference) ✓R A (2) 7.2.2 BĈD = 180◦ − 65◦ = 115◦ ✓S CA (opp ∠s of cyclic quad) ✓R A (2) 7.3.1 PM = MQ = 8 cm ✓S A (line from centre ⊥ to chord) ✓R A OM 2 = OP2 − PM 2 (Pythagoras) OM 2 = 102 − 82 = 36 OM = 6 cm ✓ answer CA (3) 7.3.2 ON = OP = 10 cm (radii) ✓ ON = 10 (radii) A MN = 10 − 6 = 4 cm ✓ answer CA (2) [10] Copyright reserved Please turn over

Technical Mathematics/P2 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1.1 PR̂Q = T P̂Q = 62◦ ✓S A (tan-chord theorem) ✓R A (2) 8.1.2 PÔQ = 2 × 62◦ = 124◦ ✓S CA (∠ at centre = 2 × ∠ at circumference) ✓R A (2) 8.1.3 OP̂T = 90◦ (radius ⊥ tangent) ✓ S/R A OP̂Q = 90◦ − 62◦ = 28◦ ✓ answer CA OR (2) OP̂Q = OQ̂P (∠s opp equal radii) 180◦ − 124◦ OP̂Q = = 28◦ (sum of ∠s in △) 2 8.1.4 Ŝ = 180◦ − 75◦ = 105◦ ✓S A (opp ∠s of cyclic quad) ✓R A (2) 8.2.1 DĈE = BÂD = 80◦ ✓S A (ext ∠ of cyclic quad) ✓R A (2) 8.2.2 AB̂D = AD̂B (∠s opp equal sides) ✓ S/R A 180◦ − 80◦ ✓ answer CA AB̂D = = 50◦ (sum of ∠s in △) 2 (2) 8.2.3 DÂC = DB̂C = 35◦ ✓S A (∠s in the same segment) ✓R A (2) 8.2.4 BĈD = 180◦ − 80◦ = 100◦ (opp ∠s of cyclic quad) ✓ BĈD = 100◦ S/R CA BD̂C = 180◦ − 100◦ − 35◦ = 45◦ (sum of ∠s in △) ✓ answer CA (2) Copyright reserved Please turn over

Technical Mathematics/P2 10 Hlayiso/Mock 2026 NSC – Marking Guidelines 8.2.5 AĈB = AD̂B (∠s in the same segment) ✓ S/R A AĈD = AB̂D (∠s in the same segment) ✓ S/R A But AD̂B = AB̂D (∠s opp equal sides) ✓ S/R A ∴ AĈB = AĈD (= 50◦ ) ∴ AC bisects BĈD ✓ conclusion A (4) [20] QUESTION 9 AD AE 9.1.1 = ✓S A DB EC (line ∥ one side of △ / prop theorem; DE ∥ BC) ✓R A (2) 6 9 9.1.2 = ✓ substitution CA 4 EC EC = 6 units ✓ answer CA (2) 9.1.3 AC = 9 + 6 = 15 units ✓ answer CA (1) 9.2 In △ADE and △ABC: Â = Â (common) ✓ S/R A AD̂E = AB̂C (corresp ∠s; DE ∥ BC) ✓ S/R A AÊD = AĈB (corresp ∠s; DE ∥ BC) ✓ S/R A ∴ △ADE ||| △ABC (∠, ∠, ∠) (3) DE AD 9.3 = (||| △s) ✓ ratio CA BC AB DE 6 = ✓ answer CA 15 10 DE = 9 units (2) [10] QUESTION 10 6 000 10.1.1 n= = 100 rev/s ✓ answer A 60 (1) Copyright reserved Please turn over

Technical Mathematics/P2 11 Hlayiso/Mock 2026 NSC – Marking Guidelines 10.1.2 ω = 2π n = 2π (100) ✓ SF CA ω = 628,32 rad/s ✓ answer CA (2) 10.1.3 D = 180 mm = 0,18 m ✓ D = 0,18 m A v = π Dn = π (0,18)(100) ✓ SF CA v = 56,55 m/s ✓ answer CA (3) 10.1.4 40 = π D(100) ✓ SF CA D= 40 = 0,1273 m = 127,32 mm ✓ answer in mm CA 100π (2) π 3π 10.2.1 135◦ × ◦ = rad = 2,36 rad ✓ answer A 180 4 (1) 3π 10.2.2 s = rθ = 6 × ✓ SF CA 4 s = 14,14 m ✓ answer CA (2) r2 θ 62 × 34π 10.2.3 Area = = ✓ SF CA 2 2 = 42,41 m2 ✓ answer CA rs 6 × 14,14 (2) OR Area = = = 42,42 m2 2 2 10.2.4 Perimeter = s + 2r = 14,14 + 2(6) ✓ s + 2r CA = 26,14 m ✓ answer CA (2) 10.3 4h2 − 4dh + x2 = 0 4(0,4)2 − 4d(0,4) + (1,6)2 = 0 ✓ SF A 0,64 − 1,6d + 2,56 = 0 ✓ simplification CA 1,6d = 3,2 d=2m ✓ answer CA (3) [18] Copyright reserved

Technical Mathematics/P2 12 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 11 12 11.1.1 a= =2m ✓ answer A 6 (1) 3,0 + 3,4 11.1.2 m1 = = 3,2; m2 = 3,6; m3 = 3,7 ✓ any THREE mid-ordinates 2 m4 = 3,4; m5 = 3,0; m6 = 2,6 A AT = 2(3,2 + 3,6 + 3,7 + 3,4 + 3,0 + 2,6) ✓ all mid-ordinates A AT = 2(19,5) = 39 m2 ✓ SF CA ✓ answer CA (4) 11.1.3 Cost = 39 × R185 ✓ multiplying by 185 CA = R7 215 ✓ answer CA (2) 11.2.1 V = π (2)2 (6) = 24π ✓ SF A V = 75,40 m3 ✓ answer CA (2) 11.2.2 s2 = 22 + 1,52 = 6,25 ✓ Pythagoras A s = 2,5 m ✓ answer CA (2) 1 11.2.3 Vcone = π (2)2 (1,5) = 2π = 6,28 m3 ✓ SF cone A 3 Vtotal = 75,40 + 6,28 ✓ adding volumes CA = 81,68 m3 ✓ answer CA (3) 11.2.4 Area = 2π rh + π rs ✓ cylinder curved area CA = 2π (2)(6) + π (2)(2,5) ✓ cone curved area CA = 75,40 + 15,71 = 91,11 m2 ✓ answer CA (3) [17] TOTAL: 150 Copyright reserved