Mathematics Grade 12 P2 NSC November Mock 2026 Memo (Eng)

hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P2 MOCK EXAMINATION 2026 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 9 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 1.1 a = 32,07 ✓a b = 4,32 ✓b ŷ = 32,07 + 4,32x ✓ equation (3) 1.2 ŷ = 32,07 + 4,32(11) ✓ substitution ŷ = 79,62% (accept 79,59% – 79,62%) ✓ answer (2) 1.3 r = 0,97 ✓ answer (1) 1.4 Very strong positive relationship ✓ answer (1) 1.5 25 hours lies far outside the range of the data ✓ outside the data range (extrapolation). The predicted score would be ✓ score > 100% (2) 32,07 + 4,32(25) ≈ 140%, which is not possible. [9] Copyright reserved Please turn over

Mathematics/P2 3 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 2.1.1 30 ≤ t < 40 ✓ answer (1) 2.1.2 Midpoints: 15 ; 25 ; 35 ; 45 ; 55 ✓ midpoints ∑ f x = 4(15)+9(25)+14(35)+8(45)+5(55) = 1 410 ✓ ∑ f x = 1 410 x̄ = 1 410 = 35,25 minutes ✓ answer (3) 40 2.1.3 Cumulative frequencies: 4 ; 13 ; 27 ; 35 ; 40 ✓ cumulative frequencies CF ✓ plotting at upper limits 40 ✓ grounded at (10 ; 0) and smooth shape (3) 30 20 10 t 10 20 30 40 50 60 2.1.4 At t = 45: cumulative frequency ≈ 31 ✓ reading from ogive 40 − 31 = 9 learners (accept 8 – 10) ✓ answer (2) 2.2 New mean = 19 ✓ mean New standard deviation = 2,5 ✓ standard deviation (2) [11] Copyright reserved Please turn over

Mathematics/P2 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 √ √ √ 3.1 AB = (5 + 3)2 + (8 − 4)2 = 80 = 4 5 ✓ substitution ✓ answer (2) 8−4 1 3.2 mAB = = ✓ mAB 5+3 2 −4 − 4 mAC = = −2 ✓ mAC 1+3 mAB × mAC = −1 ∴ AB ⊥ AC ✓ product = −1 (3) ( ) 5+1 8−4 3.3 M ; = M(3 ; 2) ✓ x-value 2 2 ✓ y-value (2) 3.4 BÂC = 90◦ ∴ BC is a diameter (converse: ∠ in ✓ reason (1) semi-circle), so its midpoint M is the centre. 8+4 3.5 mBC = =3 ✓m=3 5−1 y − 4 = 3(x + 3) ✓ substitution of A y = 3x + 13 ✓ equation (3) 3.6 tan β = 3 ⇒ β = 71,57◦ (inclination of BC) ✓ 71,57◦ tan θ = 12 ⇒ θ = 26,57◦ (inclination of AB) ✓ 26,57◦ AB̂C = 71,57◦ − 26,57◦ = 45◦ ✓ subtraction ✓ answer OR √ √ AB = AC = 80 and BÂC = 90◦ ✓ AC = 80 ✓ isosceles 180◦ − 90◦ ✓ method ✓ answer (4) AB̂C = AĈB = = 45◦ 2 √ √ 3.7 AC = (1 + 3) 2 + (−4 − 4)2 = 80 ✓ AC √ √ Area = 12 × 80 × 80 = 40 square units ✓ answer (2) 3.8 D=B+C−A D(9 ; 0) ✓x=9 ✓y=0 (2) [19] Copyright reserved Please turn over

Mathematics/P2 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1 x2 − 6x + 9 + y2 + 4y + 4 = 12 + 9 + 4 ✓ completing the square (x) (x − 3)2 + (y + 2)2 = 25 ✓ completing the square (y) M(3 ; −2) ✓ centre r=5 ✓ radius (4) 2+2 4 4.2 mMP = = ✓ mMP 6−3 3 3 ✓ mtan mtan = − ✓ substitution of P 4 y − 2 = − 34 (x − 6) ✓ equation (4) y = − 34 x + 13 2 ( ) 4.3 T 0 ; 13 ✓ coordinates of T √ 2 ( )2 √ ✓ substitution PT = (6 − 0)2 + 2 − 13 = 56,25 2 ✓ answer (3) PT = 7,5 units 4.4 tan θ = − 34 ✓ tan θ = − 34 θ = 180◦ − 36,87◦ = 143,13◦ ✓ answer (2) 4.5 p>3 or p < −7 ✓ p>3 ✓ p < −7 (2) 4.6 Circles touch externally: distance between centres ✓ sum of radii = 10 = √5 + 5 = 10 ✓ distance formula (3 + 5)2 + (−2 − k)2 = 10 64 + (k + 2)2 = 100 ✓ (k + 2)2 = 36 (k + 2)2 = 36 ✓ both values (4) k = 4 or k = −8 [19] QUESTION 5 5.1.1 sin 203◦ = sin(180◦ + 23◦ ) = − sin 23◦ = −p ✓ reduction ✓ answer (2) √ √ 5.1.2 cos 23◦ = 1 − sin2√23◦ = 1 − p2 ✓ identity/diagram (or by diagram: x = 1 − p2 , r = 1) ✓ answer (2) Copyright reserved Please turn over

Mathematics/P2 6 Hlayiso/Mock 2026 NSC – Marking Guidelines 5.1.3 cos 46◦ = cos 2(23◦ ) = 1 − 2 sin2 23◦ = 1 − 2p2 ✓ double angle ✓ answer (2) (− sin x)(sin x)(− tan x) 5.2 ✓ − sin x ✓ sin x (cos x)(− sin x) − sin x tan x ✓ − tan x ✓ cos x = ✓ − sin x cos x sin x ✓ answer (6) = − tan x · = − tan2 x cos x [12] QUESTION 6 1 − (1 − 2 sin2 x) 6.1.1 LHS = ✓ 1 − 2 sin2 x 2 sin x cos x 2 sin2 x ✓ 2 sin x cos x = ✓ 2 sin2 x 2 sin x cos x sin x ✓ cos sin x x (4) = = tan x = RHS cos x 6.1.2 sin 2x = 0 or cos x = 0 ✓ 0◦ and 180◦ x = 0◦ ; 90◦ ; 180◦ ✓ 90◦ (2) 6.2 1 − 2 sin2 x + 3 sin x − 2 = 0 ✓ 1 − 2 sin2 x 2 sin2 x − 3 sin x + 1 = 0 ✓ standard form (2 sin x − 1)(sin x − 1) = 0 ✓ factors sin x = 12 or sin x = 1 x = 30◦ + k.360◦ or x = 150◦ + k.360◦ ✓ 30◦ + k.360◦ x = 90◦ + k.360◦ k∈Z ✓ 150◦ + k.360◦ ✓ 90◦ + k.360◦ , k ∈ Z (6) 6.3 cos 15◦ − cos 75◦ = cos(45◦ − 30◦ ) − cos(45◦ + 30◦ ) ✓ splitting into special = (cos 45 cos 30 + sin 45 sin 30) − (cos 45 cos 30 − angles sin 45 sin 30) ✓ both expansions ( √ ) ( ) √2 ✓ special values (3) = 2 sin 45◦ sin 30◦ = 2 22 12 = 2 [15] Copyright reserved Please turn over

Mathematics/P2 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 7.1 2 sin x cos x − cos x = 0 ✓ double angle cos x(2 sin x − 1) = 0 ✓ factors cos x = 0: x = −90◦ or x = 90◦ ✓ ±90◦ sin x = 12 : x = 30◦ or x = 150◦ ✓ 30◦ and 150◦ (4) 7.2 y f: (−135◦ ; 1) (45◦ ; 1) ✓ shape 1 ✓ turning points ✓ intercepts f x (0◦ ; ±90◦ ; ±180◦ ) −180◦ −90◦ 90◦ 180◦ g: ✓ shape and y-int (0; 1) (−45◦ ; −1) (135◦ ; −1) g ✓ x-intercepts and end points (±180◦ ; −1) (5) 7.3 30◦ ≤ x ≤ 90◦ or 150◦ ≤ x ≤ 180◦ ✓ 30◦ ≤ x ≤ 90◦ ✓ 150◦ ≤ x ✓ ≤ 180◦ (3) 7.4 −1 ≤ 2 cos x + 1 ≤ 3 ✓ −1 y ∈ [−1 ; 3] ✓ 3 (with notation) (2) [14] QUESTION 8 h 8.1 tan α = ✓ trig ratio BC h ✓ BC (2) BC = tan α BD BC 8.2 In △BCD: = ✓ sine rule sin x sin y BC sin x h sin x ✓ substitution of BC (2) BD = = sin y tan α . sin y 30 sin 65◦ 8.3 BD = ✓ substitution tan 40◦ . sin 50◦ BD = 42,30 m ✓ answer (2) [6] Copyright reserved Please turn over

Mathematics/P2 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 9 9.1 Construction: Join OB and OD. ✓ construction Ô1 = 2Ĉ [∠ at centre = 2 × ∠ at circumference] ✓ S/R Ô2 = 2Â [∠ at centre = 2 × ∠ at circumference] ✓S Ô1 + Ô2 = 360◦ [∠s around a point] ✓ S/R 2Â + 2Ĉ = 360◦ ∴ Â + Ĉ = 180◦ ✓ conclusion (5) 9.2.1 AĈB = 90◦ [∠ in semi-circle] ✓S ✓R (2) 9.2.2 TĈA = AB̂C = 28◦ [tan-chord theorem] ✓S ✓R (2) 9.2.3 TĈB = 90◦ + 28◦ = 118◦ ✓ 118◦ T̂ = 180◦ − 118◦ − 28◦ [sum of ∠s in △] ✓R T̂ = 34◦ ✓ answer (3) [12] QUESTION 10 10.1 Construction: Join DC and BE. Draw heights h (from ✓ construction E to AB) and k (from D to AC). area △ADE 1 AD.h AD ✓ AD DB ratio = 21 = area △DBE 2 DB.h DB area △ADE 1 AE.k AE ✓ AE EC ratio = 21 = area △EDC 2 EC.k EC ✓S Area △DBE = area △EDC ✓R [same base DE, same height; DE ∥ BC] AD AE ✓ conclusion (6) ∴ = DB EC AE AD 3 10.2.1 = = [line ∥ one side of △] ✓S ✓R (2) EC DB 2 10.2.2 In △ADG and △ABF: DÂG common; AD̂G = B̂ [corr ∠s; DE ∥ BC] ✓ ||| with reason △ADG ||| △ABF [∠∠∠] ✓ DG BF = 5 3 DG AD 3 = = ✓ answer (3) BF AB 5 DG = 35 (10) = 6 units Copyright reserved Please turn over

Mathematics/P2 9 Hlayiso/Mock 2026 NSC – Marking Guidelines 10.2.3 △ADE ||| △ABC [∠∠∠] ✓ ratio DE 3 = ⇒ DE = 18 ✓ DE = 18 BC 5 ✓ answer (3) GE = 18 − 6 = 12 units 1 area △ADE AD.AE. sin  10.2.4 = 21 ✓ area rule area △ABC 2 AB.AC. sin  3 3 9 = × = ✓ 53 × 35 5 5 25 ✓ answer (3) [17] QUESTION 11 11.1 In △PTA and △PBT: P̂ = P̂ [common] ✓S PT̂A = TB̂A [tan-chord theorem] ✓S ✓R ∴ △PTA ||| △PBT [∠∠∠] ✓R (4) PT PA 11.2 = [||| △s] ✓ S with R (1) PB PT ∴ PT = PA.PB 2 11.3 PT2 = 4 × 9 = 36 ✓ substitution PT = 6 units ✓ answer (2) 11.4 PT̂A = TB̂A [tan-chord theorem] ✓ S/R CT̂B = TB̂A [alt ∠s; TC ∥ PB] ✓S ✓R (3) ∴ PT̂A = CT̂B 11.5 TB̂A = CT̂B [alt ∠s; TC ∥ PB] ✓S ✓R TB̂A subtends chord TA and CT̂B subtends chord CB ✓S ∴ TA = CB [equal ∠s subtend equal chords] ✓R (4) ( )2 ( )2 area △PTA PT 6 4 ( )2 11.6 = = = ✓ 96 area △PBT PB 9 9 ✓ answer (2) [16] TOTAL: 150 Copyright reserved