Mathematics Grade 12 P1 NSC November Mock 2026 Memo (Eng)

hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P1 MOCK EXAMINATION 2026 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 11 pages. Copyright reserved Please turn over

Mathematics/P1 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. QUESTION 1 1.1.1 x(2x − 5) = 0 ✓x=0 x=0 or x= 5 ✓ x = 25 (2) 2 1.1.2 3x2 − 5x − 4 =√ 0 −(−5) ± (−5)2 − 4(3)(−4) ✓ correct √ substitution x= √ 2(3) ✓ 73 5 ± 73 ✓ x = 2,26 x= ✓ x = −0,59 (4) 6 x = 2,26 or x = −0,59 Answer only: 2/4 1.1.3 x2 + 2x − 15 ≤ 0 ✓ factors (x + 5)(x − 3) ≤ 0 ✓ critical values CV: x = −5 or x = 3 −5 3 ✓ answer (3) −5 ≤ x ≤ 3 OR x ∈ [−5 ; 3] √ 1.1.4 x+7 = x+1 ✓ isolating the root x + 7 = x2 + 2x + 1 ✓ squaring both sides x2 + x − 6 = 0 ✓ standard form (x + 3)(x − 2) = 0 √ x = 2 or x ̸= −3 (check: 4 ̸= −2) ✓ answer with selection (4) 1.1.5 3x .3 ( + 3x ).3−1 = 90 ✓ exponential laws/factor 3x 3 + 13 = 90 3x = 27 = 33 ✓ 3x = 27 x=3 ✓ answer (3) Answer only: 1/3 Copyright reserved Please turn over

Mathematics/P1 3 Hlayiso/Mock 2026 NSC – Marking Guidelines 1.2 y = 2x − 3 …(1) x2 + xy − y2 = 5 …(2) x2 + x(2x − 3) − (2x − 3)2 = 5 ✓ substitution x2 + 2x2 − 3x − 4x2 + 12x − 9 = 5 ✓ expansion x2 − 9x + 14 = 0 ✓ standard form (x − 2)(x − 7) = 0 ✓ factors x = 2 or x = 7 ✓ both x-values y=1 y = 11 ✓ both y-values (6) 1.3 For non-real roots: ∆ < 0 (−k)2 − 4(1)(k + 3) < 0 ✓ ∆ < 0 with substitution k2 − 4k − 12 < 0 (k − 6)(k + 2) < 0 ✓ factors/critical values −2 < k < 6 ✓ answer (3) [25] QUESTION 2 2.1.1 48 ✓ answer (1) 2.1.2 First differences: 5 ; 9 ; 13 Second difference: 4 2a = 4 ⇒ a = 2 ✓a=2 3a + b = 5 ⇒ b = −1 ✓ b = −1 a+b+c = 4 ⇒ c = 3 ✓c=3 Tn = 2n2 − n + 3 ✓ Tn (4) 2.1.3 2n2 − n + 3 = 1 038 ✓ equating 2n2 − n − 1 035 = 0 ✓ standard form/factors (2n + 45)(n − 23) = 0 n = 23 (n ̸= − 452) ✓ answer (3) T23 2.1.4 Tn+1 − Tn = 4n + 1 ✓ general term of first 4n + 1 = 300 differences n = 74,75 ∈ /N ∴ no two consecutive terms differ by 300 ✓n∈ / N with conclusion (2) Copyright reserved Please turn over

Mathematics/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines 2.2 a=3; d=4 ✓ a and d n[ ] Sn = 2(3) + (n − 1)(4) = 1 830 ✓ substitution into Sn 2 n(2n + 1) = 1 830 2n2 + n − 1 830 = 0 ✓ standard form (2n + 61)(n − 30) = 0 n = 30 (n ̸= − 61 ✓ answer (4) 2) [14] QUESTION 3 24 1 3.1.1 r= = ✓ r = 21 48 2 ( )7 3 ✓ answer (2) T8 = 48 12 = = 0,375 8 48 3.1.2 S∞ = = 96 ✓ substitution 1 − 12 ✓ answer (2) ( ( )n ) 48 1 − 21 ( )n 3.1.3 Sn = = 96 − 96 12 ✓ Sn 1 − 21 ( )n S∞ − Sn = 96 12 < 0,01 ✓ inequality ( 1 )n 1 2 < (9 600 1 ) ✓ use of logs log 9 600 n> = 13,23 log 12 ✓ answer (4) n = 14 3.2 a=2; r=3 ✓ a and r 2 (3 p − 1) ✓ substitution into Sn = 6 560 3−1 3 − 1 = 6 560 p 3 p = 6 561 = 38 ✓ answer (3) p=8 3.3 r = 2x − 1 −1 < 2x − 1 < 1 ✓ −1 < r < 1 0 < x < 1 (also accept x ̸= 12 ) ✓ answer (2) [13] Copyright reserved Please turn over

Mathematics/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1 9 = b−2 ✓ substitution 1 b2 = 9 ✓ b2 = 91 (2) 1 b= (b > 0) 3 ( 1 )y 4.2 f −1 : x = 3 ✓ swop x and y y = log 1 x ✓ answer (2) 3 4.3 y f: ✓ shape (decreasing, asymptote y = 0) ✓ (0 ; 1) and one other point (−1; 3) f −1 : (0; 1) ✓ shape (decreasing, (1; 0) f asymptote x = 0) x ✓ (1 ; 0) and one other point (3; −1) (4) f −1 4.4 log 1 x ≤ −2 ✓ critical value 9 3 x≥9 ✓ inequality (2) 4.5 g(x) = 3x − 3 y > −3 OR y ∈ (−3 ; ∞) ✓ answer (1) [11] QUESTION 5 5.1 p = −1 ✓ p = −1 q=2 ✓q=2 (2) a 5.2 3= +2 ✓ substitution of (4; 3) 4−1 a 1= ✓ simplification (2) 3 a=3 Copyright reserved Please turn over

Mathematics/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines 3 5.3 0= +2 ✓y=0 x−1 −2(x − 1) = 3 1 ( 1 ) ✓ answer (2) x=− −2 ; 0 2 5.4 y = −(x − 1) + 2 ✓ substitution of (1; 2) & y = −x + 3 m = −1 ✓ answer (2) 3 5.5 −x + k = +2 ✓ equating x−1 (x − 1)(−x + k − 2) = 3 x2 − (k − 1)x + (k + 1) = 0 ✓ standard form No intersection: ∆ < 0 (k − 1)2 − 4(k + 1) < 0 ✓ ∆ < 0 with substitution k2 − 6k − 3 <√0 6 ± 48 √ CV: k = = 3±2 3 ✓ answer (4) √ 2 √ 3 − 2 3 < k < 3 + 2 3 (−0,46 < k < 6,46) [12] Copyright reserved Please turn over

Mathematics/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 6.1 −x2 + 2x + 8 = 0 ✓y=0 x2 − 2x − 8 = 0 (x − 4)(x + 2) = 0 ✓ factors A(−2 ; 0) B(4 ; 0) ✓ both coordinates (3) b 2 6.2 x=− =− =1 ✓x=1 2a 2(−1) ✓y=9 (2) y = −(1)2 + 2(1) + 8 = 9 C(1 ; 9) OR f ′ (x) = −2x + 2 = 0 ⇒ x = 1 6.3 −x2 + 2x + 8 = 2x − 1 ✓ equating x2 = 9 ✓ x2 = 9 x = 3 or x = −3 ✓ both values (3) D: x = 3 E: x = −3 6.4 −3 ≤ x ≤ 3 ✓ critical values ✓ notation (2) 6.5 f (x) − g(x) = −x2 + 9 ✓ f (x) − g(x) Maximum distance = 9 units (at x = 0) ✓ answer (2) 6.6 f ′ (x) = −2x + 2 = 2 ✓ f ′ (x) = 2 x=0; y=8 ✓ point of contact (0; 8) y = 2x + 8 ✓ equation (3) [15] QUESTION 7 7.1 A = P(1 − i)n A = 320 000(1 − 0,14)5 ✓ substitution A = R150 536,65 ✓ answer (2) ( ) 0,09 12 7.2 1 + ieff = 1 + ✓ 0,09 12 12 ieff = 0,0938 . . . ✓ substitution into formula ieff = 9,38% p.a. ✓ answer (3) Copyright reserved Please turn over

Mathematics/P1 8 Hlayiso/Mock 2026 NSC – Marking Guidelines x [1 − (1 + i)−n ] 7.3.1 Pv = i[ ( )−240 ] ✓ i = 0,1125 12 0,1125 x 1 − 1 + 12 ✓ n = 240 1 200 000 = 0,1125 ✓ substitution into correct 12 formula (3) x = R12 591,07 [ ( )−180 ] 0,1125 12 591,07 1 − 1 + 12 7.3.2 Balance = 0,1125 ✓ n = 180 12 ✓ substitution into Pv = R1 092 647,98 ✓ answer OR ( )60 Balance = 1 200 000 1 + 0,1125 ✓ 1 200 000(1 + i)60 [( 12 )60 ] 0,1125 ✓ Fv with n = 60 12 591,07 1 + 12 −1 ✓ answer (3) − 0,1125 12 = R1 092 648,34 x [(1 + i)n − 1] 7.4 Fv = i[( )72 ] ✓ n = 72 and i = 0,072 12 0,072 x 1 + 12 −1 ✓ substitution into Fv 250 000 = 0,072 ✓ answer (3) 12 x = R2 786,30 [14] QUESTION 8 8.1 f (x + h) = 2(x + h)2 − 3(x + h) = ✓ f (x + h) 2x2 + 4xh + 2h2 − 3x − 3h ✓ 4xh + 2h2 − 3h f (x + h) − f (x) = 4xh + 2h2 − 3h ✓ formula with substitution 4xh + 2h2 − 3h ✓ lim(4x + 2h − 3) f ′ (x) = lim h→0 h ✓ answer (5) = lim (4x + 2h − 3) h→0 = 4x − 3 Penalise 1 mark for incorrect notation. y = 4x 2 − 3x−2 1 8.2.1 ( ) ✓ exponential form ✓ 2x− 2 1 dy 2 6 = 2x− 2 + 6x−3 1 =√ + 3 dx x x ✓ +6x−3 (3) Copyright reserved Please turn over

Mathematics/P1 9 Hlayiso/Mock 2026 NSC – Marking Guidelines (x − 2)(x2 + 2x + 4) 8.2.2 f (x) = ✓ factorisation x−2 ✓ simplification f (x) = x2 + 2x + 4 ✓ answer (3) f ′ (x) = 2x + 2 [11] QUESTION 9 9.1 f (3) = 27 − 27 − 27 + 27 = 0 ∴ (x − 3) is a factor ✓ f (3) = 0 f (x) = (x − 3)(x2 − 9) = (x − 3)2 (x + 3) ✓ factorisation x-intercepts: (3 ; 0) and (−3 ; 0) ✓ both intercepts (3) 9.2 f ′ (x) = 3x2 − 6x − 9 = 0 ✓ f ′ (x) = 0 x2 − 2x − 3 = 0 (x − 3)(x + 1) = 0 ✓ factors x = 3 or x = −1 ✓ x-values Turning points: (3 ; 0) and (−1 ; 32) ✓ y-values (4) 9.3 y ✓ x-intercepts and y-intercept (−1 ; 32) ✓ turning points 27 ✓ shape (3) f x −3 (3 ; 0) 9.4 f ′′ (x) = 6x − 6 > 0 ✓ f ′′ (x) > 0 x>1 ✓ answer (2) 9.5 0 < k < 32 ✓ critical values ✓ notation (2) [14] Copyright reserved Please turn over

Mathematics/P1 10 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 10 10.1 Length = 24 − 2x ; breadth = 15 − 2x ; height = x ✓ dimensions V = x(24 − 2x)(15 − 2x) ✓ V = x(24 − 2x)(15 − 2x) V = x(360 − 78x + 4x2 ) (2) V = 4x3 − 78x2 + 360x 10.2 V ′ (x) = 12x2 − 156x + 360 = 0 ✓ V ′ (x) = 0 x2 − 13x + 30 = 0 (x − 3)(x − 10) = 0 ✓ factors x = 3 (x ̸= 10, since 15 − 2x > 0) ✓ x = 3 with selection Vmax = 3(18)(9) = 486 cm3 ✓ maximum volume (4) [6] QUESTION 11 11.1 P(A and B) = P(A) + P(B) − P(A or B) = 0,4 + 0,35 − 0,61 = 0,14 ✓ P(A and B) = 0,14 P(A) × P(B) = 0,4 × 0,35 = 0,14 ✓ P(A) × P(B) = 0,14 P(A and B) = P(A) × P(B) ∴ A and B are independent ✓ conclusion (3) 11.2.1 0,1 ✓ first branches with Late probabilities Taxi 0,7 ✓ second branches (taxi) 0,9 Not late ✓ second branches (walk) with outcomes (3) 0,25 Late 0,3 Walk 0,75 Not late 11.2.2 P(late) = (0,7)(0,1) + (0,3)(0,25) ✓ sum of products = 0,07 + 0,075 = 0,145 ✓ answer (2) 11.3.1 8! = 40 320 ✓ answer (1) 11.3.2 Treat O and E as one unit: 7! ✓ 7! Vowels can swop: 2! ✓ ×2! 7! × 2! = 10 080 ✓ answer (3) Copyright reserved Please turn over

Mathematics/P1 11 Hlayiso/Mock 2026 NSC – Marking Guidelines 11.3.3 Consonants: P, R, B, L, M, S (6) ; vowels: O, E (2) n(E) = 6 × 6! × 2 = 8 640 ✓ 6×...×2 P= 8 640 = 3 ≈ 0,21 ✓ 6! 40 320 14 ✓ answer (3) [15] TOTAL: 150 Copyright reserved