Technical Mathematics Grade 11 P2 2026 Memo (Eng)

Mock hlayiso mock exams Practice Series: Grade 11 Exam Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 11 TECHNICAL MATHEMATICS P2 NOVEMBER EXAMINATION 2026 (MOCK) 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 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. • In QUESTIONS 6, 7 and 8 (Euclidean geometry) a statement without the correct reason earns no mark for the reason. Accept any valid alternative method. QUESTION 1 3 − (−1) 4 1 1.1 mAB = = = ✓ SF A 5 − (−3) 8 2 ✓ answer A (2) √ 1.2 AB = (5 − (−3))2 + (3 − (−1))2 ✓ SF √ A √ √ = 64 + 16 = 80 ✓ √ 80 CA √ = 4 5 units ✓4 5 CA (3) ( ) −3 + 5 −1 + 3 1.3 M ; = M(1 ; 1) ✓x=1 A 2 2 ✓y=1 A (2) 1.4 m⊥ = −2 (mAB × m⊥ = −1) ✓ m = −2 CA y − 1 = −2(x − 1) ✓ substitution of M CA y = −2x + 3 ✓ answer CA (3) 1.5 Substitute x = −1: y = −2(−1) + 3 = 5, so C(−1 ; 5) ✓ substitution A lies on the line. ✓ y = 5 / LHS = RHS A The line is the perpendicular bisector of AB, so every point on it is equidistant from A and B: CA = CB. ✓ CA = CB with reason A ∴ △ABC is isosceles. (3) √ √ 2 2 OR CA √ = 2 +√6 = 40 and CB = 62 + 22 = 40, so CA = CB. Copyright reserved Please turn over

Technical Mathematics/P2 3 Hlayiso/Mock 2026 NSC – Marking Guidelines 5 − (−1) 6 1.6 mAC = = =3 ✓ mAC = 3 A −1 − (−3) 2 tan α = 3 ✓ answer CA α = 71,57◦ (2) 1 1.7 Inclination of AB: tan β = , β = 26,57◦ ✓ β = 26,57◦ CA 2 ✓ method α − β M CÂB = α − β = 71,57 − 26,57◦ ◦ ✓ answer CA = 45◦ √ (3) (Accept: √ cosine rule √ in △ABC with AC = 40, AB = 80, BC = 40.) 1.8 CM ⊥ AB, √ so CM is the perpendicular height. √ CM = (1 − (−1))2 + (1 − 5)2 = 20 ✓ SF √ A 1 1 √ √ ✓ CM = 20 CA Area = × AB ×CM = × 80 × 20 ✓ substitution CA 2 2 = 20 square units ✓ answer CA OR Area (4) 1 1√ √ = (AB)(AC) sinCÂB = 80 40 sin 45◦ = 20 2 2 √ 1.9 BE 2 = (x − 5)2 + (−3 − 3)2 = (2 13)2 ✓ SF A (x − 5)2 + 36 = 52 ✓ (x − 5)2 = 16 CA (x − 5)2 = 16 ✓ both values CA x − 5 = ±4 x = 9 or x = 1 ✓ selection x = 1 CA Since x < 5: x = 1 (4) [26] QUESTION 2 2.1.1 tan(2(48,6◦ ) − 117,3◦ ) = tan(−20,1◦ ) ✓ −20,1◦ A = −0,37 ✓ answer CA (2) sec 117,3◦ 1 2.1.2 ◦ = × tan2 48,6◦ ✓ reciprocals A 2 cot 48,6 cos 117,3◦ ✓ substitution A = (−2,1803 . . .) × (1,2865 . . .) = −2,81 ✓ answer CA (3) Copyright reserved Please turn over

Technical Mathematics/P2 4 Hlayiso/Mock 2026 NSC – Marking Guidelines 8 2.2.1 cos θ = and sin θ < 0: θ in the 4th quadrant ✓ sketch in 4th quadrant A 17 y 8 x 17 −15 ✓ y = −15 CA ✓ answer CA (3) y2 = 172 − 82 = 225, y = −15 15 tan θ = − 8 ( ) ( ) 15 15 2.2.2 8 tan θ − 17 sin θ = 8 − − 17 − ✓ tan θ substituted CA 8 17 ✓ sin θ = − 17 15 CA = −15 + 15 ✓ answer CA =0 (3) 5 2.3 4 cot θ = −5 ⇒ cot θ = − ✓ cot θ = − 54 A 4 4 ✓ tan θ = −0,8 CA tan θ = − = −0,8 5 Reference angle = 38,66◦ ✓ 141,3◦ CA θ = 180◦ − 38,66◦ = 141,3◦ ✓ 321,3◦ CA (4) or θ = 360◦ − 38,66◦ = 321,3◦ [15] Copyright reserved Please turn over

Technical Mathematics/P2 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 (− sin x)(tan x) 3.1 = ✓ − sin x A (− cos x)(1) ✓ tan x A sin x = · tan x ✓ − cos x A cos x ✓ tan 225◦ = 1 A = tan x · tan x ✓ cos sin x x = tan x CA = tan2 x ✓ answer CA (6) cos x sin x 3.2 LHS = − ✓ tan x = cos sin x A 1 − sin x cos x x ✓ common denominator A cos2 x − sin x(1 − sin x) = cos x(1 − sin x) ✓ sin2 x + cos2 x = 1 A cos x + sin2 x − sin x 2 = cos x(1 − sin x) ✓ simplification to cos1 x CA 1 − sin x (4) = cos x(1 − sin x) 1 = = sec x = RHS cos x [10] QUESTION 4 4.1 y f: f ✓ x-intercepts 90◦ , 270◦ A 2 ✓ end points and turning (120◦ ; 1) point A ✓ shape A g x g: 90◦ 180◦ 270◦ 360◦ ✓ x-intercepts 30◦ , 210◦ A (300◦ ; −1) ✓ turning points (120◦ ; 1), −2 (300◦ ; −1) A (180◦ ; −2) ✓ end points (0◦ ; − 12 ), (360◦ ; − 21 ) and shape A f : (0◦ ; 2), (90◦ ; 0), (180◦ ; −2), (270◦ ; 0), (360◦ ; 2) (6) g: (0◦ ; − 12 ), (30◦ ; 0), (120◦ ; 1), (210◦ ; 0), (300◦ ; −1), (360◦ ; − 12 ) 4.2 360◦ ✓ answer A (1) Copyright reserved Please turn over

Technical Mathematics/P2 6 Hlayiso/Mock 2026 NSC – Marking Guidelines 4.3 −2 ≤ y ≤ 2 OR y ∈ [−2 ; 2] ✓ critical values A ✓ notation A (2) 4.4 30◦ < x < 90◦ or 210◦ < x < 270◦ ✓ 30◦ < x < 90◦ CA OR x ∈ (30◦ ; 90◦ ) ∪ (210◦ ; 270◦ ) ✓ 210◦ < x < 270◦ CA ✓ notation (open intervals) CA (3) [12] QUESTION 5 5.1 PR2 = PQ2 + QR2 − 2(PQ)(QR) cos Q̂ ✓ cosine rule A PR2 = 4,22 + 5,62 − 2(4,2)(5,6) cos 105◦ ✓ SF A PR2 = 61,175 . . . ✓ answer CA PR = 7,82 m (3) sin QP̂R sin 105◦ 5.2 = ✓ sine rule CA 5,6 7,82 ✓ SF CA 5,6 sin 105◦ sin QP̂R = = 0,6916 . . . ✓ answer CA 7,82 (3) QP̂R = 43,76◦ 1 5.3 Area = (4,2)(5,6) sin 105◦ ✓ SF A 2 ✓ answer CA = 11,36 m2 (2) QS 5.4 In △PQS: sin QP̂S = ✓ trig ratio / area method CA PQ ✓ SF CA QS = 4,2 sin 43,76◦ ✓ answer CA QS = 2,90 m (3) 1 2(11,36) OR (PR)(QS) = 11,36 ⇒ QS = = 2,90 2 7,82 m [11] Copyright reserved Please turn over

Technical Mathematics/P2 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 6.1 …the interior opposite angle. ✓ answer A (1) 1 6.2.1 AD̂C = × 124◦ = 62◦ ✓S A 2 (∠ at centre = 2 × ∠ at circumference) ✓R A (2) 6.2.2 AB̂C = 180◦ − 62◦ = 118◦ ✓S CA (opp ∠s of cyclic quad) ✓R A (2) 6.2.3 OA = OC (radii) ✓ S/R radii / isosceles A OÂC = OĈA (∠s opp equal sides) ✓ R sum of ∠s in △ A 180◦ − 124◦ ✓ answer A OÂC = = 28◦ (sum of ∠s in △) (3) 2 6.2.4 BĈE = DÂB = 95◦ ✓S A (ext ∠ of cyclic quad) ✓R A (2) 180◦ − 62◦ 6.2.5 DÂC = DĈA = = 59◦ ✓ S/R DÂC = 59◦ CA 2 (∠s opp equal sides; sum of ∠s in △) ✓ answer CA (2) BÂC = 95◦ − 59◦ = 36◦ OR AD = DC ⇒ AĈD = DÂC (equal chords subtend equal ∠s), then same working. Accept any valid method. [12] QUESTION 7 7.1 …radius (diameter) … ✓ answer A (1) 7.2.1 AB̂C = CÂT = 32◦ ✓S A (tan-chord theorem) ✓R A (2) Copyright reserved Please turn over

Technical Mathematics/P2 8 Hlayiso/Mock 2026 NSC – Marking Guidelines 7.2.2 BÂC = 90◦ ✓S A (∠ in semicircle) ✓R A (2) 7.2.3 OÂT = 90◦ (tangent ⊥ radius) ✓ S/R OÂT = 90◦ A AÔT = 2 × AB̂C = 64◦ (∠ at centre = 2 × ∠ at ✓ S/R AÔT = 64◦ CA circumference) ✓ answer CA T̂ = 180◦ − 90◦ − 64◦ = 26◦ (3) OR In △TAB: T ÂB = 32◦ + 90◦ = 122◦ ; T̂ = 180◦ − 122◦ − 32◦ = 26◦ 7.2.4 AĈB = 180◦ − 90◦ − 32◦ = 58◦ (sum of ∠s in △) ✓S CA OR AD̂B = 90◦ − 32◦ = 58◦ (CA) ✓R A AD̂B = AĈB = 58◦ (∠s in same segment) (2) 7.3 In △TAC and △T BA: T̂ = T̂ (common) ✓S A T ÂC = T B̂A (= 32◦ ) (tan-chord theorem) ✓ S/R A T ĈA = T ÂB (sum of ∠s in △) ✓ S/R A ✓ conclusion with reason A ∴ △TAC ||| △T BA (∠∠∠) (4) 7.4.1 OM̂K = 90◦ and ON̂K = 90◦ (tangent ⊥ radius) ✓ S/R A OM̂K + ON̂K = 180◦ ✓ sum = 180◦ A ∴ KMON is a cyclic quad (opp ∠s supplementary / ✓R A converse opp ∠s of cyclic quad) (3) 7.4.2 M ÔN = 180◦ − 64◦ = 116◦ (opp ∠s of cyclic quad ✓ S/R M ÔN = 116◦ A KMON) Reflex M ÔN = 360◦ − 116◦ = 244◦ ✓R A 1 ✓ answer CA M P̂N = (244◦ ) = 122◦ (∠ at centre = 2 × ∠ at 2 (3) circumference) OR KM = KN (tangents from same point), K M̂N = 58◦ ; angle in the major segment = 58◦ (tan-chord); M P̂N = 180◦ − 58◦ = 122◦ (opp ∠s of cyclic quad) [20] Copyright reserved Please turn over

Technical Mathematics/P2 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1 …bisects the chord. ✓ answer A (1) 1 8.2.1 AF = AB = 24 cm ✓S A 2 (line from centre ⊥ to chord) ✓R A (2) 8.2.2 OE = 22 − x ✓ answer A (1) 8.2.3 OA2 = 242 + x2 (Pythagoras) ✓ OA2 A CE = 20 cm (line from centre ⊥ to chord) OC2 = 202 + (22 − x)2 ✓ OC2 CA OA = OC (radii): 576 + x2 = 400 + 484 − 44x + x2 ✓ equating (radii) A 44x = 308 ✓ simplification to x=7 44x = 308 CA (4) 8.2.4 r2 = 242 + 72 = 625 ✓ substitution CA r = 25 cm ✓ answer CA (2) [10] Copyright reserved Please turn over

Technical Mathematics/P2 10 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 9 72 × 1 000 9.1.1 72 km/h = m/s = 20 m/s ✓ conversion A 3 600 ✓ answer A (2) 9.1.2 v = π Dn ✓ formula A 20 = π (0,64)n ✓ SF CA n = 9,95 r/s ✓ answer CA (3) 9.1.3 ω = 2π n = 2π (9,947 . . .) ✓ SF CA ω = 62,50 rad/s ✓ answer CA v 20 (2) OR ω = = = 62,5 rad/s r 0,32 9.1.4 5 minutes = 300 s ✓ ×300 CA Revolutions = 9,947 . . . × 300 = 2 984,2 . . . ✓ answer CA 2 984 complete revolutions (accept 2 985 if n = 9,95 (2) was used; CA from candidate’s n in 9.1.2, NPR) r2 θ 9.2.1 Area = ✓ formula A 2 ✓ SF A 32 θ 6,75 = ✓ answer CA 2 (3) θ = 1,5 rad 180◦ ◦ 9.2.2 1,5 × = 85,94◦ ✓ × 180 π CA π ✓ answer CA (2) 9.2.3 s = rθ = 3 × 1,5 ✓ SF CA rs 3s ✓ answer CA s = 4,5 m OR Area = : 6,75 = , s = 4,5 m 2 2 (2) Copyright reserved Please turn over

Technical Mathematics/P2 11 Hlayiso/Mock 2026 NSC – Marking Guidelines 9.3 4h2 − 4dh + x2 = 0 ✓ SF A 4h2 − 4(2)h + (1,6)2 = 0 4h2 − 8h + 2,56 = 0 (h2 − 2h + 0,64 = 0) ✓ standard form CA ✓ factors / formula CA (h − 0,4)(h − 1,6) = 0 ✓ h = 0,4 or h = 1,6 CA h = 0,4 m (minor segment) or h = 1,6 m (major ✓ selecting h = 1,6 m (major segment) segment) CA Height of the tunnel h = 1,6 m (5) OR radius = 1 m ✓A; half-chord√ = 0,8 m ✓A; distance from centre to KL = 1 − 0,82 ✓A = 0,6 m 2 ✓CA; h = 1 + 0,6 = 1,6 m ✓CA 9.4.1 V = l × b × h = 1,2 × 0,8 × 0,75 ✓ SF A V = 0,72 m3 ✓ 0,72 m3 CA Capacity = 0,72 × 1 000 = 720 litres ✓ answer in litres CA (3) [ ] 9.4.2 TSA = 2 (1,2)(0,8) + (1,2)(0,75) + (0,8)(0,75) ✓ SF A = 2(0,96 + 0,9 + 0,6) = 4,92 m2 ✓ 4,92 m2 CA For 6 tanks: 6 × 4,92 = 29,52 m2 ✓ ×6 CA ✓ ×R185,50 CA Cost = 29,52 × R185,50 ✓ answer CA = R5 475,96 (5) [29] Copyright reserved Please turn over

Technical Mathematics/P2 12 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 10 36 10.1 a= =6m ✓ answer A 6 (1) 8+y y+9 10.2 m1 = 6,75; m2 = 7,75; m3 = ; m4 = ; ✓ numerical mid-ordinates A 2 2 m5 = 8,75; m6 = 7,75 ✓ m3 and m4 in terms of y A AT = a(m1 + m2 + . . . + m6 ) ( ) ✓ SF = 294 CA 8+y y+9 294 = 6 6,75 + 7,75 + + + 8,75 + 7,75 2 2 294 = 6(39,5 + y) ✓ answer CA 49 = 39,5 + y (4) y = 9,5 m ( ) 6+7 OR 294 = 6 + 7,5 + 8 + y + 9 + 8,5 = 2 6(39,5 + y), y = 9,5 m [5] TOTAL: 150 Copyright reserved Please turn over