Mathematics Grade 12 P1 NSC November Prediction 2026 Memo

hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P1 PREDICTION EXAMINATION 2026 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 10 pages. Copyright reserved Please turn over

Mathematics/P1 2 Hlayiso/Prediction 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. QUESTION 1 2 1.1.1 x= or x = −4 ✓ x = 32 3 ✓ x = −4 (2) 1.1.2 3x2 − 6x − 5 =√ 0 √ ✓ standard form −(−6) ± (−6)2 − 4(3)(−5) 6 ± 96 ✓ substitution x= = 2(3) 6 ✓ both answers (3) x = 2,63 or x = −0,63 1.1.3 x2 − 2x − 3 − 5 > 0 x2 − 2x − 8 > 0 ✓ standard form (x − 4)(x + 2) > 0 ✓ critical values x < −2 or x > 4 ✓ x < −2 ✓ x > 4 (4) √ 1.1.4 x − 4 = 2x + 7 ✓ isolate surd x2 − 8x + 16 = 2x + 7 ✓ squaring both sides x2 − 10x + 9 = 0 (x − 9)(x − 1) = 0 ✓ factors x = 9 or x = 1 √ Check x = 1: 1 − 9 = −2 ̸= 4 ∴ x = 9 only ✓ rejection of x = 1 (4) 1.1.5 (2x )2 − 6 · 2x − 16 = 0 ✓ 2x+1 = 2 · 2x and Let k = 2x : k2 − 6k − 16 = 0 4x = (2x )2 (k − 8)(k + 2) = 0 ✓ standard form in k 2x = 8 ⇒ x = 3 or 2x = −2 (no solution) ✓ factors ✓ x = 3 and rejection (4) Copyright reserved Please turn over

Mathematics/P1 3 Hlayiso/Prediction 2026 NSC – Marking Guidelines 1.2 Let length = x and breadth = y 2x + 2y = 34 ⇒ x = 17 − y ✓ x + y = 17 x2 + y2 = 132 = 169 ✓ x2 + y2 = 169 (17 − y)2 + y2 = 169 ✓ substitution 2y2 − 34y + 120 = 0 ⇒ y2 − 17y + 60 = 0 ✓ standard form (y − 12)(y − 5) = 0 The garden is 12 m by 5 m ✓ 12 m and 5 m (5) 22023 (23 + 2) 8 + 2 10 1.3 = = = ✓ common factor 22023 in 22023 (22 − 1) 4 − 1 3 numerator ✓ common factor in denominator ✓ 10 3 (3) [25] QUESTION 2 2.1.1 Tn = 2 + (n − 1)(7) = 7n − 5 ✓ Tn = 345 7n − 5 = 345 ⇒ n = 50 ✓ n = 50 (2) 50 2.1.2 S50 = [2 + 345] = 8 675 ✓ S50 2 20 ✓ S20 S20 = [2(2) + 19(7)] = 1 370 ✓ answer 2 50 OR: ✓ 142 and 30 terms ∑ (7n − 5) = S50 − S20 = 7 305 ✓ substitution ✓ answer (3) n=21 30 OR: T21 = 142; 30 terms: [142 + 345] = 7 305 2 2.1.3 Terms that are even numbers: 2 + 16 + 30 + . . . ✓ a = 2 and d = 14 (a = 2, d = 14) ✓ last term 338 Last even term ≤ 345: 338 ✓ n = 25 2 + (n − 1)(14) = 338 ⇒ n = 25 ✓ answer (4) 25 S25 = [2 + 338] = 4 250 2 2.2.1 First differences: 5 ; 9 ; 13; second difference = 4 2a = 4 ⇒ a = 2 ✓a=2 3a + b = 5 ⇒ b = −1 ✓ b = −1 a+b+c = 3 ⇒ c = 2 ✓c=2 (3) Tn = 2n2 − n + 2 Copyright reserved Please turn over

Mathematics/P1 4 Hlayiso/Prediction 2026 NSC – Marking Guidelines 2.2.2 First difference between Tn and Tn+1 : ✓ 4n + 1 5 + (n − 1)(4) = 4n + 1 ✓ n = 99 4n + 1 = 397 ⇒ n = 99 ✓ T99 and T100 (3) Between T99 and T100 [15] QUESTION 3 3.1 a = 80; r = 0,9 ✓ r = 0,9 T5 = 80(0,9)4 = 52,49 cm ✓ answer (2) 3.2 80(0,9)n−1 < 10 ✓ inequality (0,9)n−1 < 0,125 n−1 > log 0,125 = 19,74 ✓ use of logs log 0,9 ✓ 21st swing (3) n > 20,74 ∴ the 21st swing [ ] 80 1 − (0,9)15 3.3 S15 = = 635,29 cm ✓ substitution 1 − 0,9 ✓ answer (2) 80 3.4 S∞ = = 800 cm = 8 m ✓ S∞ formula 1 − 0,9 Sn < S∞ for every n, so the total distance gets closer ✓ 800 cm and closer to 8 m but never exceeds it. ✓ conclusion with reason (3) Thabo is NOT correct. [10] Copyright reserved Please turn over

Mathematics/P1 5 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 4 1 4.1 = b−1 ⇒ b = 3 ✓ substitution (1) 3 4.2 x = 3y ✓ swop x and y y = log3 x ✓ answer (2) 4.3 y ✓ f : shape and (0; 1) f ✓ g: shape and (1; 0) ✓ one other point on each graph, e.g. (1; 3) and (3; 1)(3) (1; 3) g (0; 1) (3; 1) x (1; 0) 4.4 h(x) = 3x−1 − 3 ✓ equation of h 3x−1 = 3 ⇒ x − 1 = 1 ⇒ x = 2 ✓x=2 (2) 4.5 g(x) ≤ 0 for 0 < x ≤ 1 and g(x) ≥ 0 for x ≥ 1 ✓ critical values 1 and 2 h(x) < 0 for x < 2 and h(x) ≥ 0 for x ≥ 2 ✓0<x≤1 0 < x ≤ 1 or x ≥ 2 ✓x≥2 (3) [11] QUESTION 5 5.1 p=1 ✓p q=3 ✓q (2) a 5.2 1= +3 ✓ substitution of (1 ; 1) 1+1 a ✓ simplification (2) = −2 ⇒ a = −4 2 5.3 y = x + c through (−1 ; 3): 3 = −1 + c ⇒ c = 4 ✓ substitution of (−1; 3) y = x+4 ✓ answer (2) Copyright reserved Please turn over

Mathematics/P1 6 Hlayiso/Prediction 2026 NSC – Marking Guidelines −4 1 5.4 A: +3 = 0 ⇒ x = ✓ x = 31 x+1 3 1 ✓ x < −1 x < −1 or 0 ≤ x ≤ ✓ 0 ≤ x ≤ 13 (3) 3 5.5 Reflection in the line x = −1 (the vertical asymptote of ✓ reflection f) ✓ in the line x = −1 (2) −4 4 Check: f (−2 − x) = +3 = + 3 = g(x) −x − 1 x+1 (Also accept: rotation through 90◦ about (−1 ; 3).) [11] QUESTION 6 b 6 6.1 x=− =− =3 ✓x=3 2a 2(−1) ✓ y = 16 (2) f (3) = −9 + 18 + 7 = 16 D(3 ; 16) 6.2 y ≤ 16 or y ∈ (−∞ ; 16] ✓ answer (1) 6.3 −x2 + 6x + 7 = 0 ⇒ (x − 7)(x + 1) = 0 A(−1 ; 0) ✓ A(−1; 0) and m = 1 7−0 ✓ equation (2) m= = 1 g(x) = x + 1 6 − (−1) 6.4 f (x) ≥ g(x) for −1 ≤ x ≤ 6; f increasing for x < 3 ✓ −1 ≤ x −1 ≤ x < 3 (accept −1 ≤ x ≤ 3) ✓x<3 (2) 6.5 MN = f (x) − g(x) = −x2 + 5x + 6 ✓ MN expression x=− =− b 5 = 5 (OR: ✓ x = − 2a b / derivative = 0 2a 2(−1) 2 ✓x= 2 5 dx (MN) = −2x + 5 = 0) d ✓ 12,25 (4) 25 25 49 MNmax = − + + 6 = = 12,25 units 4 2 4 6.6 Tangent is g moved up by the maximum length MN: ✓ gradient 1 / + 49 4 y = x+1+ 49 = x+ 53 ✓ answer (2) 4 4 OR: f ′ (x) = −2x + 6 = 1 ⇒ x = 52 , f ( 25 ) = 63 4, 53 y = x+ 4 [13] Copyright reserved Please turn over

Mathematics/P1 7 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 7 √ = 680 000(1 − i) ✓ substitution into 7.1 301 719,61 5 301 719,61 A = P(1 − i)n 1−i = 5 = 0,85 ✓ 0,85 680 000 i = 0,15 i.e. 15% p.a. ✓ 15% (3) 7.2 New taxi: 680 000(1,062)5 = R918 610,69 ✓ R918 610,69 Amount needed: ✓ R616 891,08 918 610,69 − 301 [ 719,61 = R616] 891,08 ✓ substitution into Fv x (1,008)60 − 1 ✓ answer (4) 616 891,08 = 0,008 x = R8 050,90 0,114 7.3.1 i= = 0,0095; n = 240 ✓ i and n 12 [ ] ✓ substitution into Pv x 1 − (1,0095)−240 950 000 = ✓ answer (3) 0,0095 x = R10 065,70 7.3.2 Balance after [ 60 instalments ](180 remaining): ✓ n = 180 10 065,70 1 − (1,0095) −180 ✓ R866 358,46 = R866 358,46 0,0095 ✓ substitution with R12 500 12 500 [1 − (1,0095)−n ] ✓ n = 113,61 (logs) 866 358,46 = 0,0095 (1,0095)−n = 0,34157 . . . ⇒ n = 113,61 ✓ 66 months (5) 114 payments instead of 180 Paid off 180 − 114 = 66 months sooner OR (balance): [ ] 10 065,70 (1,0095)60 − 1 950 000(1,0095)60 − = 0,0095 R866 358,83 (Accept balances from R866 358,46 to R866 358,83.) [15] QUESTION 8 8.1 f (x + h) = (x + h)2 − 4(x + h) = x2 + 2xh + h2 − 4x − 4h ✓ f (x + h) f (x + h) − f (x) = 2xh + h2 − 4h ✓ 2xh + h2 − 4h 2xh + h2 − 4h ✓ formula with limit f ′ (x) = lim = lim (2x + h − 4) h→0 h h→0 ✓ common factor = 2x − 4 ✓ answer (5) Copyright reserved Please turn over

Mathematics/P1 8 Hlayiso/Prediction 2026 NSC – Marking Guidelines (x − 2)(x2 + 2x + 4) 8.2.1 y= = x2 + 2x + 4 ✓ factorising difference of x−2 dy cubes dx = 2x + 2 ✓ x2 + 2x + 4 ✓ answer (3) ( ) √ 2 2 = x − 4x− 2 + 4x−2 1 8.2.2 x− ✓ expansion in index form x ✓ 1 + 2x− 2 3 Dx = 1 + 2x− 2 − 8x−3 3 ✓ −8x−3 (3) [11] QUESTION 9 9.1 f ′ (x) = k(x − 1)(x − 3) ✓ form of f ′ 9 = k(−1)(−3) ⇒ k = 3 ✓k=3 f ′ (x) = 3x2 − 12x + 9 = 3ax2 + 2bx + c ✓ equating coefficients (3) ∴ a = 1 ; b = −6 ; c = 9 OR: c = f ′ (0) = 9; f ′ (1) = 0: 3a + 2b = −9; f ′ (3) = 0: 27a + 6b = −9 ⇒ a = 1, b = −6 9.2 f (x) = x3 − 6x2 + 9x − 4 ✓ d = −4 f (1) = 1 − 6 + 9 − 4 = 0 (1 ; 0) ✓ (1; 0) f (3) = 27 − 54 + 27 − 4 = −4 (3 ; −4) ✓ (3; −4) (3) 9.3 y ✓ x-intercepts 1 and 4 ✓ y-intercept −4 ✓ turning points ✓ shape (4) (1; 0) 4 x −4 (3; −4) (1; 0) is a turning point on the x-axis ⇒ f (x) = (x − 1)2 (x − 4) 9.4 f ′′ (x) = 6x − 12 = 0 ⇒ x = 2 ✓x=2 f (2) = 8 − 24 + 18 − 4 = −2 (2 ; −2) ✓ y = −2 (2) (OR: x = 1+3 2 = 2, midway between turning points) Copyright reserved Please turn over

Mathematics/P1 9 Hlayiso/Prediction 2026 NSC – Marking Guidelines 9.5 m = f ′ (2) = 12 − 24 + 9 = −3 ✓ m = −3 y + 2 = −3(x − 2) ⇒ y = −3x + 4 ✓ equation tan θ = −3 ⇒ θ = 180◦ − 71,57◦ = 108,43◦ ✓ tan θ = −3 ✓ 108,43◦ (4) 9.6 −4 < k < 0 ✓ −4 and 0 ✓ strict inequality (2) [18] QUESTION 10 12 − h r 10.1 By similar triangles: = ✓ ratio from similar triangles 12 6 12 − h = 2r ⇒ h = 12 − 2r ✓ h = 12 − 2r V = π r2 h = π r2 (12 − 2r) = 12π r2 − 2π r3 ✓ substitution into V = π r2 h (3) 10.2 V ′ (r) = 24π r − 6π r2 = 0 ✓ V ′ (r) = 0 6π r(4 − r) = 0 ⇒ r = 4 (r ̸= 0) ✓r=4 V (4) = 12π (16) − 2π (64) = 64π ≈ 201,06 cm3 ✓ 64π or 201,06 cm3 (3) [6] Copyright reserved Please turn over

Mathematics/P1 10 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 11 60 40 1 11.1.1 P(T and I) = P(T ) × P(I) = × = ✓ product rule 120 120 6 1 ✓ substitution (2) n(T and I) = × 120 = 20 6 11.1.2 S ✓ 8 and 10 T W ✓ equation in y 24 16 32 ✓ y = 16 12 ✓ completed Venn diagram 8 10 (24 ; 32 ; 8 outside) (4) 10 8 I T and I only: 20 − 12 = 8; only I: 40 − 12 − 8 − 10 = 10 Let T and W only = y: only T = 40 − y, only W = 48 − y 8 + (40 − y) + 10 + (48 − y) + 8 + y + 10 + 12 = 120 136 − y = 120 ⇒ y = 16 11.1.3 Exactly one: 24 + 32 + 10 = 66 ✓ 66 P= 66 = 11 = 0,55 ✓ answer (2) 120 20 11.2.1 7 × 6 × 5 × 4 × 3 = 2 520 ✓ answer (1) 11.2.2 First digit 5 or 7: 2 × (5 × 4 × 3) × 3 = 360 ✓ case first digit 5 or 7 First digit 6: 1 × (5 × 4 × 3) × 2 = 120 ✓ case first digit 6 Total: 360 + 120 = 480 ✓ 480 (3) 11.2.3 Positions for the pair: 4 adjacent pairs of positions; 2 ✓ 4×2 orders (12 or 21) ✓ 480 Remaining 3 positions: 5 × 4 × 3 = 60 ✓ answer (3) 4 × 2 × 60 = 480 480 4 P= = ≈ 0,19 2 520 21 [15] TOTAL: 150 Copyright reserved Please turn over