Technical Mathematics Grade 11 P2 November Mock 2026 Memo

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 1.1 …equal. ✓ answer A (1) √ 1.2 AB = (4 − (−2))2 + (2 − (−1))2 ✓ SF √ A √ = 36 + 9 ✓ √ 45 CA √ √ = 45 = 3 5 units ✓3 5 CA (3) 2 − (−1) 3 1 1.3 mAB = = = ✓ SF A 4 − (−2) 6 2 ✓ answer CA (2) 8−2 6 1.4 mBC = = = −2 ✓ SF A 1 − 4 −3 1 ✓ mBC = −2 CA mAB × mBC = × (−2) = −1 ✓ product = −1 CA 2 ∴ AB ⊥ BC (3) ( ) −2 + 1 −1 + 8 1.5 M ; ✓ x-value A 2 2 ( ) ✓ y-value A 1 7 M − ; OR M(−0,5 ; 3,5) (2) 2 2 Copyright reserved Please turn over

Technical Mathematics/P2 3 Hlayiso/Mock 2026 NSC – Marking Guidelines 1.6 y − y1 = m(x − x1 ) ✓ substitution of m CA y − 2 = −2(x − 4) ✓ substitution of point CA y = −2x + 10 ✓ answer CA (3) 1 1.7 tan θ = mAB = ✓ tan θ = 12 CA 2 θ = 26,57◦ ✓ answer CA (2) 1.8 Diagonals of a rectangle bisect each other, so M is also ✓ method M the midpoint of BD: 4 + xD 1 2 + yD 7 ✓ x = −5 CA =− and = 2 2 2 2 ✓y=5 CA D(−5 ; 5) (3) OR B → A: (x − 6 ; y − 3), so C(1 ; 8) → D(−5 ; 5) √ √ 1.9 BC = (1 − 4)2 + (8 − 2)2 = 45 ✓ SF √ A 1 ✓ BC = 45 CA Area △ABC = × AB × BC (B̂ = 90◦ ) 2 ✓ area formula / substitution 1 √ √ CA = × 45 × 45 2 ✓ answer CA = 22,5 square units (4) 1 1.10 m = mAB = ✓ m = 21 CA 2 1 ✓ substitution of C CA y − 8 = (x − 1) ✓ answer CA 2 1 15 (3) y = x+ OR y = 0,5x + 7,5 2 2 [26] Copyright reserved Please turn over

Technical Mathematics/P2 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 2.1.1 cos(56,3◦ + 23,8◦ ) = cos 80,1◦ ✓ 80,1◦ A = 0,17 ✓ answer CA (2) tan 56,3◦ 2.1.2 = tan 56,3◦ × sin 23,8◦ ✓ cosec B = sin1 B A cosec 23,8◦ ✓ substitution A = 1,4994 . . . × 0,4035 . . . = 0,61 ✓ answer CA (3) 5 2.2.1 sin θ = , θ in the 2nd quadrant ✓ sketch in correct quadrant 13 y A 13 5 θ x −12 ✓ x = −12 CA x2 = 132 − 52 = 144, x = −12 ✓ answer CA 12 (3) cos θ = − 13 13 5 2.2.2 sec θ + tan θ = + ✓ sec θ = − 13 CA −12 −12 12 18 ✓ tan θ = − 12 5 CA =− ✓ answer CA 12 3 (3) =− 2 2.3 cos θ = −0,65 ✓ cos θ = −0,65 A Reference angle = 49,46◦ ✓ reference angle CA θ = 180◦ − 49,46◦ = 130,5◦ ✓ 130,5◦ CA or θ = 180◦ + 49,46◦ = 229,5◦ ✓ 229,5◦ CA (4) [15] Copyright reserved Please turn over

Technical Mathematics/P2 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 (− sin θ )(− tan θ ) 3.1 = ✓ − sin θ A (− cos θ )(tan θ ) ✓ − tan θ A sin θ · tan θ = ✓ − cos θ A − cos θ · tan θ ✓ tan θ A sin θ =− ✓ simplification CA cos θ ✓ answer CA = − tan θ (6) 1 3.2 LHS = − cos x cos x ✓ common denominator A 1 − cos2 x = ✓ sin2 x A cos x ✓ splitting CA sin2 x = ✓ tan x = cos sin x x CA cos x (4) sin x = sin x · cos x = sin x · tan x = RHS [10] QUESTION 4 4.1 y f: (90◦ ; 2) ✓ intercepts 0◦ , 180◦ , 360◦ A 2 f ✓ turning points A ✓ shape A g: x ✓ end points (0◦ ; 0), 90◦ 180◦ 270◦ 360◦ (360◦ ; 0) A g ◦ ✓ turning point (180 ; −2) A −2 ✓ shape A (180◦ ; −2) (270◦ ; −2) (6) 4.2 2 ✓ answer A (1) 4.3 360◦ ✓ answer A (1) Copyright reserved Please turn over

Technical Mathematics/P2 6 Hlayiso/Mock 2026 NSC – Marking Guidelines 4.4 −2 ≤ y ≤ 0 OR y ∈ [−2 ; 0] ✓ critical values A ✓ notation A (2) 4.5 180◦ < x < 360◦ OR x ∈ (180◦ ; 360◦ ) ✓ critical values CA ✓ notation CA (2) [12] QUESTION 5 5.1 BD2 = AB2 + AD2 − 2(AB)(AD) cos  ✓ cosine rule A BD2 = 502 + 702 − 2(50)(70) cos 65◦ ✓ SF A BD2 = 4 441,67 . . . BD = 66,65 m ✓ answer CA (3) 5.2 BD̂C = 180◦ − 72◦ − 48◦ = 60◦ ✓ answer A (1) BC BD 5.3 ◦ = ✓ sine rule CA sin 60 sin 48◦ 66,65 × sin 60◦ ✓ SF CA BC = ✓ answer CA sin 48◦ BC = 77,67 m (3) 1 5.4 Area △ABD = (50)(70) sin 65◦ = 1 586,04 m2 ✓ area △ABD A 2 1 ✓ SF area △BCD CA Area △BCD = (66,65)(77,67) sin 72◦ = 2 461,67 m2 ✓ area △BCD CA 2 Total area = 4 047,71 m2 ✓ total CA (4) (Accept 4 047,43 m2 if unrounded values were used.) [11] QUESTION 6 6.1 …a right angle OR 90◦ ✓ answer A (1) Copyright reserved Please turn over

Technical Mathematics/P2 7 Hlayiso/Mock 2026 NSC – Marking Guidelines 6.2.1 PQ̂R = 90◦ ✓S A (∠ in semicircle) ✓R A (2) 6.2.2 PR̂Q = 180◦ − 90◦ − 34◦ = 56◦ ✓S CA (sum of ∠s in △) ✓R A (2) 6.2.3 QÔR = 2 × QP̂R = 68◦ ✓S A (∠ at centre = 2 × ∠ at circumference) ✓R A (2) 6.2.4 PŜQ = PR̂Q = 56◦ ✓S CA (∠s in the same segment) ✓R A (2) 6.3 …supplementary OR add up to 180◦ ✓ answer A (1) 6.4 SR̂Q = 180◦ − 100◦ = 80◦ ✓S A (opp ∠s of cyclic quad) ✓R A (2) [12] Copyright reserved Please turn over

Technical Mathematics/P2 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 7.1 …the angle in the alternate segment. ✓ answer A (1) 7.2.1 BÂC = F ĈB = 48◦ ✓S A (tan-chord theorem) ✓R A (2) 7.2.2 AB̂C = AĈE = 70◦ ✓S A (tan-chord theorem) ✓R A (2) 7.2.3 AĈB = 180◦ − 70◦ − 48◦ = 62◦ ✓S CA (∠s on a straight line) OR (sum of ∠s in △) ✓R A (2) 7.2.4 BÔC = 2 × BÂC = 96◦ ✓S CA (∠ at centre = 2 × ∠ at circumference) ✓R A (2) 7.2.5 OĈF = 90◦ (radius ⊥ tangent) ✓ S/R A OĈB = 90◦ − 48◦ = 42◦ ✓ answer CA OR (2) OĈB = OB̂C (∠s opp equal radii) 180◦ − 96◦ OĈB = = 42◦ 2 7.3 …equal in length. ✓ answer A (1) 7.4.1 PA = PB (tangents from same point) ✓ S/R A 180◦ − 50◦ ✓ answer CA PÂB = PB̂A = = 65◦ 2 (2) 7.4.2 AĈB = PÂB = 65◦ ✓S CA (tan-chord theorem) ✓R A (2) Copyright reserved Please turn over

Technical Mathematics/P2 9 Hlayiso/Mock 2026 NSC – Marking Guidelines 7.4.3 AÔB = 2 × AĈB = 130◦ ✓S CA (∠ at centre = 2 × ∠ at circumference) ✓R A OR OÂP = OB̂P = 90◦ (radius ⊥ tangent); (2) AÔB = 360◦ − 90◦ − 90◦ − 50◦ = 130◦ (sum of ∠s in quad) 7.4.4 OÂP = 90◦ (radius ⊥ tangent) ✓ S/R A OÂB = 90◦ − 65◦ = 25◦ ✓ answer CA (2) [20] Copyright reserved Please turn over

Technical Mathematics/P2 10 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1 …bisects the chord. ✓ answer A (1) 1 8.2.1 AM = AB = 12 cm ✓S A 2 (line from centre ⊥ to chord) ✓R A (2) 8.2.2 OM = r − 8 ✓ answer A (1) 8.2.3 OA2 = AM 2 + OM 2 (Pythagoras) ✓ Pythagoras S/R A r2 = 122 + (r − 8)2 ✓ substitution CA r2 = 144 + r2 − 16r + 64 ✓ expansion CA 16r = 208 ✓ answer CA r = 13 cm (4) OR 4h2 − 4dh + x2 = 0: 4(8)2 − 4d(8) + 242 = 0 32d = 832, d = 26, r = 13 cm 8.2.4 OM = 13 − 8 = 5 cm ✓ answer CA (1) 8.2.5 d = 2 × 13 = 26 cm ✓ answer CA (1) [10] QUESTION 9 450 9.1.1 n= = 7,5 r/s ✓ n = 7,5 A 60 ω = 2π n = 2π (7,5) ✓ SF CA ✓ answer CA ω = 47,12 rad/s (3) 9.1.2 D = 300 mm = 0,3 m ✓ conversion to m A v = π Dn = π (0,3)(7,5) ✓ SF CA v = 7,07 m/s ✓ answer CA (3) OR v = ω r = 47,12 × 0,15 = 7,07 m/s Copyright reserved Please turn over

Technical Mathematics/P2 11 Hlayiso/Mock 2026 NSC – Marking Guidelines π π 9.2.1 110◦ × = 1,92 rad ✓ × 180 ◦ A 180◦ ✓ answer CA (2) 9.2.2 s = rθ = 1,8 × 1,92 ✓ SF CA s = 3,46 m ✓ answer CA (2) r2 θ (1,8)2 (1,92) 9.2.3 Area = = ✓ SF CA 2 2 ✓ answer CA = 3,11 m2 (2) 9.2.4 P = 2r + s = 2(1,8) + 3,46 ✓ 2r + s CA P = 7,06 m ✓ answer CA (2) 9.3 4h2 − 4dh + x2 = 0 ✓ formula A 4(0,4)2 − 4d(0,4) + (1,2)2 = 0 ✓ SF A 0,64 − 1,6d + 1,44 = 0 ✓ simplification CA 1,6d = 2,08 ✓ answer CA d = 1,3 m (4) 9.4.1 V = π r2 h = π (0,15)2 (0,8) ✓ SF A V = 0,06 m3 (0,05655 m3 ) ✓ answer CA (2) 9.4.2 Two hemispheres = one sphere ✓ SF A 4 V = π (0,15)3 3 ✓ answer CA V = 0,01 m3 (0,01414 m3 ) (2) 9.4.3 Total volume = 0,05655 + 0,01414 = 0,07069 m3 ✓ adding volumes CA Capacity = 0,07069 × 1 000 ✓ ×1 000 A = 70,69 litres ✓ answer CA (3) Copyright reserved Please turn over

Technical Mathematics/P2 12 Hlayiso/Mock 2026 NSC – Marking Guidelines 9.4.4 A = 2π rh + 4π r2 ✓ 2π rh + sphere A A = 2π (0,15)(0,8) + 4π (0,15)2 ✓ SF cylinder CA A = 0,754 + 0,283 ✓ SF sphere CA ✓ answer CA A = 1,04 m2 (4) [29] QUESTION 10 30 10.1 a= =5m ✓ answer A 6 (1) 8 + 10 10.2 m1 = = 9 ; m2 = 10,5 ; m3 = 11,75 ✓ mid-ordinates A 2 m4 = 12,25 ; m5 = 11,25 ; m6 = 9,75 ✓ mid-ordinates A AT = a(m1 + m2 + . . . + m6 ) ✓ SF CA AT = 5(9 + 10,5 + 11,75 + 12,25 + 11,25 + 9,75) AT = 5(64,5) = 322,5 m2 ✓ answer CA (4) [5] TOTAL: 150 Copyright reserved Please turn over