Mathematics Grade 11 P2 November Mock 2026 Memo (English)

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

Mathematics/P2 2 Hlayiso/Mock 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 1 370 1.1 x̄ = = 24,67 push-ups ✓ 370 15 ✓ answer (2) 1.2 Minimum = 12; Q1 = 18; Median = 23; Q3 = 28; ✓ min and max Maximum = 49 ✓ Q1 and Q3 ✓ median (3) 1.3 ✓ box with Q1 , median, Q3 ✓ whiskers to 12 and 49 (2) 10 20 30 40 50 1.4 Positively skewed (skewed to the right) ✓ answer (1) (mean 24,67 > median 23 / right whisker is longer) 1.5 σ = 9,17 ✓ answer (1) 1.6 (x̄ − σ ; x̄ + σ ) = (15,50 ; 33,84) ✓ interval Values from 17 to 33: 11 learners ✓ answer (2) [11] Copyright reserved Please turn over

Mathematics/P2 3 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 2.1 Cumulative frequencies: 6 ; 20 ; 42 ; 60 ; 67 ; 70 ✓ first three ✓ last three (2) 2.2 ✓ grounding at (0 ; 0) 70 ✓ plotting at upper limits 60 ✓ smooth shape (3) 50 40 30 20 10 t 0 30 60 90 120 150 180 Points: (0 ; 0), (30 ; 6), (60 ; 20), (90 ; 42), (120 ; 60), (150 ; 67), (180 ; 70) 2.3 Median at position 70 2 = 35 ✓ reading at 35 Median ≈ 80 minutes (accept 78 – 82) ✓ answer (2) 2.4 Midpoints: 15 ; 45 ; 75 ; 105 ; 135 ; 165 ✓ midpoints ∑ f x = 90 + 630 + 1 650 + 1 890 + 945 + 495 = 5 700 ✓ ∑ f x = 5 700 5 700 ✓ answer (3) x̄ = = 81,43 minutes 70 [10] Copyright reserved Please turn over

Mathematics/P2 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 √ √ √ 3.1 AC = (6 + 2)2 + (1 − 5)2 = 80 = 4 5 ✓ substitution ✓ answer (2) 7−5 1 3.2 mAB = = ✓ mAB 4+2 3 ✓ mBC 1−7 mBC = = −3 ✓ product = −1 (3) 6−4 mAB × mBC = 13 × (−3) = −1 ∴ AB ⊥ BC ( ) −2 + 6 5 + 1 3.3 M ; = M(2 ; 3) ✓x=2 2 2 ✓y=3 (2) 1−5 1 3.4 mAC = =− ✓ m = − 21 6+2 2 ✓ substitution of B y − 7 = − 2 (x − 4) 1 ✓ equation (3) y = − 12 x + 9 3.5 Inclination of AB: tan α = 13 ⇒ α = 18,43◦ ✓ 18,43◦ Inclination of AC: ✓ 153,43◦ tan β = − 12 ⇒ β = 180◦ − 26,57◦ = 153,43◦ ✓ method ✓ answer BÂC = 18,43◦ + (180◦ − 153,43◦ ) BÂC = 45◦ ✓ AB and BC ✓ equal OR √ ✓ method ✓ answer (4) AB = BC = 40 and B̂ = 90◦ , so ◦ ◦ BÂC = 180 2−90 = 45◦ 3.6 M is the midpoint of BD (diagonals of a rectangle bisect each other) ✓ xD = 0 4 + xD 7 + yD ✓ yD = −1 (2) = 2 ⇒ xD = 0; = 3 ⇒ yD = −1 2 2 D(0 ; −1) [16] Copyright reserved Please turn over

Mathematics/P2 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 6+2 4.1 mPQ = =2 ✓ mPQ = 2 1+3 ✓ mQR = mPQ k−6 mQR = =2 ✓ answer (3) 5−1 k − 6 = 8 ⇒ k = 14 4.2 Inclination of PQ: tan θ = 2 ⇒ θ = 63,43◦ ✓ 63,43◦ Inclination of y = −x + 4: tan β = −1 ⇒ β = 135◦ ✓ 135◦ Angle between the lines = 135◦ − 63,43◦ = 71,57◦ ✓ method ✓ answer (4) ( ) −3 + 1 −2 + 6 4.3 Midpoint of PQ: ; = (−1 ; 2) ✓ midpoint 2 2 ✓ m⊥ = − 21 1 ✓ substitution m⊥ = − 2 ✓ equation (4) y − 2 = − 12 (x + 1) y = − 12 x + 23 4.4 T lies on the perpendicular bisector of PQ: ✓ T on perpendicular −1 = − 12 t + 32 bisector ✓ substitution 1 5 2t = 2⇒ t =5 ✓ answer OR TP2 = TQ2 : (t + 3)2 + 1 = (t − 1)2 + 49 ✓ TP2 =TQ2 ✓ 8t = 40 ⇒ t = 5 simplification ✓ answer (3) [14] QUESTION 5 3 5.1.1 sin θ = − ; θ in the third quadrant ✓ x = −4 (diagram) 5 ✓ answer (2) y θ x 5 (−4 ; −3) x2 = 52 − (−3)2 = 16 ⇒ x = −4 −3 3 tan θ = = −4 4 Copyright reserved Please turn over

Mathematics/P2 6 Hlayiso/Mock 2026 NSC – Marking Guidelines 5.1.2 sin(−θ ) + cos(180◦ − θ ) = − sin θ − cos θ ✓ − sin θ ( ) ( ) 3 4 ✓ − cos θ =− − − − ✓ answer (3) 5 5 7 = 5 sin x . cos x . (− tan x) 5.2 = ✓ sin x ✓ cos x (− sin x) . cos x ✓ − tan x ✓ − sin x = tan x ✓ cos x ✓ answer (6) ( ) (−1) 12 5.3 = + sin2 17◦ + cos2 17◦ ✓ −1 and 12 − 12 ✓ − 12 = 1+1 ✓ sin 73◦ = cos 17◦ and =2 sin2 + cos2 = 1 ✓ answer (4) 1 1 − cos2 x 5.4.1 LHS = − cos x = ✓ common denominator cos x cos x ✓ sin2 x sin2 x ✓ cos sin x = x = tan x (3) cos x sin x = sin x . = sin x . tan x = RHS cos x 5.4.2 Undefined when cos x = 0: ✓ 90◦ x = 90◦ or x = 270◦ ✓ 270◦ (2) [20] Copyright reserved Please turn over

Mathematics/P2 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 6.1 (2 sin x + 1)(sin x − 1) = 0 ✓ factors 1 ✓ both equations sin x = − or sin x = 1 2 Reference angle = 30◦ ✓ 210◦ ✓ 330◦ x = 210◦ + k.360◦ or x = 330◦ + k.360◦ ✓ 90◦ or x = 90◦ + k.360◦ , k ∈ Z ✓ +k.360◦ , k ∈ Z (6) 6.2 Reference angle: x − 20◦ = 56,31◦ ✓ 56,31◦ x − 20◦ = 56,31◦ + k.180◦ ✓ +k.180◦ x = 76,31◦ + k.180◦ x = 76,31◦ or x = 256,31◦ ✓ 76,31◦ ✓ 256,31◦ (4) [10] QUESTION 7 7.1 y f: (180◦ ; 1) ✓ x-intercepts −180◦ ; 0◦ ; (−180◦ ; 1) (0◦ ; 1) (90◦ ; 1) 180◦ f ✓ turning points g ✓ shape g: x −180◦ −135◦ −45◦ 45◦ 135◦ 180◦ ✓ x-intercepts ±45◦ ; ±135◦ ✓ turning points ✓ end points (6) (−90◦ ; −1) (90◦ ; −1) f : solid line; g: dashed line 7.2 180◦ ✓ answer (1) 7.3 30◦ < x < 150◦ ✓ critical values ✓ notation (2) 7.4 −2 ≤ y ≤ 4 OR y ∈ [−2 ; 4] ✓ −2 ✓4 (2) [11] Copyright reserved Please turn over

Mathematics/P2 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1 BC2 = 122 + 92 − 2(12)(9) cos 58◦ ✓ cosine rule BC2 = 110,54 . . . ✓ substitution BC = 10,51 m ✓ answer (3) sin AB̂C sin 58◦ 8.2 = ✓ sine rule 9 10,51 ✓ substitution sin AB̂C = 0,7260 . . . ✓ answer (3) AB̂C = 46,55◦ 8.3 Area ABCD = 12 (12)(9) sin 58◦ + 12 (9)(7) sin 40◦ ✓ area △ABC = 45,79 + 20,25 ✓ area △ACD = 66,04 m2 ✓ answer (3) [9] QUESTION 9 9.1 s2 = 32 + 42 = 25 ✓ Pythagoras s = 5m ✓ answer (2) 9.2 V = π (3)2 (10) + 13 π (3)2 (4) ✓ volume of cylinder = 90π + 12π = 102π ✓ volume of cone = 320,44 m3 ✓ answer (3) 9.3 Curved surface area = π (3)(5) = 15π ✓ substitution = 47,12 m2 ✓ answer (2) [7] Copyright reserved Please turn over

Mathematics/P2 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 10 10.1 A 1 2 O 1 2 B C D ✓ construction Construction: Draw AO and produce it to D. In △AOB: OA = OB (radii) ✓ S/R ∴ Â1 = B̂ (∠s opp equal sides) ✓ S/R Ô1 = Â1 + B̂ (ext ∠ of △) ∴ Ô1 = 2Â1 ✓ similarly Similarly, in △AOC: Ô2 = 2Â2 ✓ adding (5) BÔC = Ô1 + Ô2 = 2(Â1 + Â2 ) = 2 BÂC 10.2.1 Ô1 = 2D̂ = 128◦ (∠ at centre = 2 × ∠ at ✓S ✓R (2) circumference) 10.2.2 AB̂C = 180◦ − 64◦ = 116◦ (opp ∠s of cyclic quad) ✓S ✓R (2) 10.2.3 OA = OC (radii) ✓ S/R OÂC = OĈA (∠s opp equal sides) 180◦ − 128◦ ✓R OÂC = = 26◦ (sum of ∠s in △) ✓ answer (3) 2 10.2.4 BÂC = BĈA (∠s opp equal sides; AB = BC) ✓ S/R 180◦ − 116◦ ✓ answer (2) BÂC = = 32◦ (sum of ∠s in △) 2 [14] Copyright reserved Please turn over

Mathematics/P2 10 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 11 11.1.1 AM = MB = 12 (line from centre ⊥ chord) ✓S ✓R OM2 = OA2 − AM2 (Pythagoras) ✓ Pythagoras with = 132 − 122 = 25 substitution OM = 5 units ✓ answer (4) 11.1.2 ON = 13 (radii) ✓ ON = 13 (radii) MN = 13 − 5 = 8 ✓ MN = 8 AN2 = AM2 + MN2 = 122 + 82 = 208 (Pythagoras) ✓ substitution √ √ AN = 208 = 4 13 = 14,42 units ✓ answer (4) 11.2.1 AB̂C = TÂC = 70◦ (tan-chord theorem) ✓S ✓R (2) 11.2.2 BÂC = AB̂C = 70◦ (∠s opp equal sides; AC = BC) ✓S ✓R (2) 11.2.3 PÂB = 180◦ − 70◦ − 70◦ = 40◦ (∠s on a str line) ✓S ✓R (2) 11.2.4 AĈB = PÂB = 40◦ (tan-chord theorem) ✓ AĈB = 40◦ with reason [or: sum of ∠s in △ABC] AD̂B = 180◦ − 40◦ = 140◦ (opp ∠s of cyclic quad) ✓ answer with reason (2) [16] Copyright reserved Please turn over

Mathematics/P2 11 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 12 12.1 AĈB = x (tan-chord theorem) ✓S ✓R AD̂E = x (alt ∠s; DE ∥ TS) ✓S ✓R (4) 12.2 AD̂E = BĈE = x (both equal to x) ✓S AD̂E is an exterior angle of quad DBCE ✓ S (ext ∠ = int opp ∠) ∴ DBCE is a cyclic quad (converse: ext ∠ of cyclic ✓R (3) quad) 12.3 AÊD = AB̂C (ext ∠ of cyclic quad DBCE) ✓S ✓R (2) 12.4 AD̂E = AĈB (proved in 12.1) ✓ S (both substitutions) AÊD = AB̂C (proved in 12.3) AD̂E = AÊD (given) ✓ S B̂ = Ĉ ∴ AB̂C = AĈB ✓R (3) ∴ AB = AC (sides opp equal ∠s) [12] TOTAL: 150 Copyright reserved