Mathematics Grade 12 Paper 1 NSC November 2026 Memo (English)

UNOFFICIAL Copyright reserved Please turn over Unofficial memo hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P1 NOVEMBER 2026 — UNOFFICIAL MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 13 pages.

UNOFFICIAL Mathematics/P1 2 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 IMPORTANT: These are unofficial marking guidelines prepared by Hlayiso for revision. They are not the DBE’s official memorandum; the official marking guidelines may differ in mark allocation and accepted alternatives. 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. • Answers only will NOT necessarily be awarded full marks; where an answer-only allocation applies it is stated in the row. • Rounding: penalise incorrect rounding (or no rounding where TWO decimal places are required) ONCE per question, not per item. QUESTION 1 1.1.1 x2 +x−6 = 0 (x+3)(x−2) = 0 x = −3 or x = 2 ✓factors ✓x = −3 ✓x = 2 Answer only: full marks (3) 1.1.2 3x2 −2x−10 = 0 x = −(−2)± √ (−2)2 −4(3)(−10) 2(3) = 2± √ 124 6 x = 2,19 or x = −1,52 ✓standard form ✓substitution into correct formula ✓2,19 ✓−1,52 Note: Penalise 1 mark for incorrect rounding. Accept x = 1± √ 31 3 simplified first. Answer only: max 2/4. (4) 1.1.3 81·2x−4 = 3x 34 ·2x−4 = 3x 2x−4 = 3x−4 i.e. (2 3 )x−4 = 1 = (2 3 )0 x−4 = 0 ⇒x = 4 OR: 81·2x 16 = 3x ⇒ (3 2 )x = 81 16 = (3 2 )4 ⇒x = 4 OR (logs): xlog 3 2 = log 81 16 ⇒x = 4 ✓81 = 34 ✓2x−4 = 3x−4 / same base ✓x = 4 OR: ✓2x−4 = 2x 16 ✓ (3 2 )x = (3 2 )4 ✓x = 4 Answer only: 1/3 (3) Copyright reserved Please turn over

UNOFFICIAL Mathematics/P1 3 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 1.1.4 √ (x+6)(10−x) = 4 √ 3 (−6 ≤x ≤10) (x+6)(10−x) = 48 −x2 +4x+60 = 48 x2 −4x−12 = 0 (x−6)(x+2) = 0 x = 6 or x = −2 Check x = 6: √ 12· √ 4 = 4 √ 3 ✓ Check x = −2: √ 4· √ 12 = 4 √ 3 ✓ ∴x = 6 or x = −2 (both valid) ✓squaring both sides ✓standard form ✓both values ✓both accepted after checking / domain Note: If a valid root is rejected, max 3/4. Answer only: max 2/4. (4) 1.2 y = 2x+2 (2x+2)2 +3x2 −4 = 0 4x2 +8x+4+3x2 −4 = 0 7x2 +8x = 0 x(7x+8) = 0 x = 0 or x = −8 7 y = 2 or y = −2 7 OR: x = y−2 2 : y2 + 3(y−2)2 4 −4 = 0 7y2 −12y−4 = 0 ⇒(7y+2)(y−2) = 0 y = −2 7 or y = 2; x = −8 7 or x = 0 ✓y the subject ✓substitution ✓standard form ✓both x-values ✓both y-values (correctly paired) OR: ✓x subject ✓substitution ✓standard form ✓y-values ✓x-values Note: CA for y-values from incorrect x-values. (5) 1.3 Let the first number be x; the second number is x+2 x(x+2) < 15 x2 +2x−15 < 0 (x+5)(x−3) < 0 −5 3 + − + −5 < x < 3 ✓x and x+2 ✓inequality ✓standard form ✓critical values −5 and 3 ✓−5 < x < 3 Note: The numbers are not stated to be integers. A candidate who lists only the integers −4;−3;...;2 with correct working: max 4/5. Answer only: max 2/5. (5) [24] Copyright reserved Please turn over

UNOFFICIAL Mathematics/P1 4 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 2 2.1.1 d = −2 3 −(−1) = 1 3 ✓d = 1 3 (1) 2.1.2 Tn = −1+(n−1) (1 3 ) = 27 n−1 = 84 ⇒n = 85 ✓substitution into Tn with Tn = 27 ✓n = 85 (2) 2.1.3 Tn = −1+(n−1)1 3 = n 3 −4 3 Last 65 terms: T21 to T85 85 ∑ n=21 (n 3 −4 3 ) or 85 ∑ n=21 [ −1+(n−1)1 3 ] OR: T21 = 17 3 : 65 ∑ k=1 (17 3 +(k −1)1 3 ) = 65 ∑ k=1 k +16 3 OR (from 27 down): 65 ∑ k=1 ( 27−(k −1)1 3 ) ✓general term ✓lower limit 21 ✓upper limit 85 OR: ✓general term ✓1 ✓65 Note: CA from 2.1.2. Value (not required): 3185 3 . (3) 2.2.1 T1 = 44 = 256 T2 = 43 = 64 ✓256 ✓64 (2) 2.2.2 4−n+5 = 1 1024 4−n+5 = 4−5 −n+5 = −5 ⇒n = 10 i.e. the 10th term OR: 256 (1 4 )n−1 = 1 1024 ⇒ (1 4 )n−1 = (1 4 )9 ⇒n = 10 ✓equating ✓4−5 / same base (or logs) ✓n = 10 Answer only: 1/3 (3) 2.2.3 r = 64 256 = 1 4 (−1 < r < 1) S∞= 256 1−1 4 = 1024 3 = 341,33 ✓r = 1 4 ✓substitution into S∞ ✓1024 3 or 341,33 Note: CA from 2.2.1. (3) [14] Copyright reserved Please turn over

UNOFFICIAL Mathematics/P1 5 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 3 3.1 2a = −2 ⇒a = −1 Maximum at n = 8: −b 2a = 8 ⇒−b −2 = 8 ⇒b = 16 T8 = −(8)2 +16(8)+c = 44 c = −20 Tn = −n2 +16n−20 OR: Tn = a(n−8)2 +44 with a = −1 Tn = −(n2 −16n+64)+44 = −n2 +16n−20 ✓a = −1 ✓b = 16 ✓substitution of (8; 44) ✓c = −20 OR: ✓a = −1 ✓turning-point form ✓expansion ✓answer (4) 3.2 Let the amount in the 1st minute be a litres; d = 2 Tank half full after 36 min and empty after 54 min: S36 = 1 2S54 36 2 [2a+35(2)] = 1 2 · 54 2 [2a+53(2)] 36a+1260 = 27a+1431 9a = 171 ⇒a = 19 litres OR: S36 = S54 −S36 (second half drains in the last 18 min): 18(2a+70) = 9[2(a+72)+17(2)] ⇒36a+1260 = 18a+1602 ⇒a = 19 Check: S36 = 1944 ℓ, S54 = 3888 ℓ ✓d = 2 ✓S36 = 1 2S54 ✓S36 ✓S54 substituted ✓a = 19 litres Note: Assumes the tank was full at the start. Answer only: 0/5. (5) [9] Copyright reserved Please turn over

UNOFFICIAL Mathematics/P1 6 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 4 4.1 f(x) = 5−x = (1 5 )x and the base 0 < 1 5 < 1 ✓ (1 5 )x with 0 < b < 1 Note: Accept: as x increases, f(x) decreases; or f ′(x) < 0 for all x. (1) 4.2 f −1: x = 5−y −y = log5 x y = −log5 x (or y = log 1 5 x) ✓swop x and y ✓y = −log5 x (2) 4.3 x y O y = x f f −1 (0;1) (1;0) (−1;5) (5;−1) Asymptote of f: y = 0 (the x-axis) Asymptote of f −1: x = 0 (the y-axis) f: ✓decreasing shape through (0;1) ✓asymptote y = 0 shown/stated (graph does not cut the x-axis) f −1: ✓decreasing shape through (1;0) ✓asymptote x = 0 shown/stated (graph does not cut the y-axis) Note: Graphs must be symmetrical about y = x (not required to draw it). CA from 4.2. (4) 4.4 −log5 x ≥−1 log5 x ≤1 0 < x ≤5 or x ∈(0; 5] ✓0 < x ✓x ≤5 Note: x ≤5 only: 1/2. (2) [9] Copyright reserved Please turn over

UNOFFICIAL Mathematics/P1 7 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 5 5.1 T(−2; 0) (g(x) = 0: 2x+4 = 0) ✓T(−2;0) (1) 5.2 TP = 6 ⇒P(4; 0) y = a(x+2)(x−4) Substitute S(−1;−10): −10 = a(−1+2)(−1−4) = −5a a = 2 f(x) = 2(x2 −2x−8) = 2x2 −4x−16 ∴a = 2; b = −4; c = −16 OR: 4a−2b+c = 0; 16a+4b+c = 0; a−b+c = −10 solving simultaneously gives a = 2, b = −4, c = −16 ✓P(4;0) ✓y = a(x+2)(x−4) ✓substitution of S ✓a = 2 and expansion OR: ✓P ✓any two equations ✓third equation ✓solving (4) 5.3.1 f(x) > −18 Turning point of f: x = −−4 2(2) = 1; f(1) = 2−4−16 = −18 f(x) > −18 everywhere except at the turning point x ∈R, x̸ = 1 OR: 2x2 −4x+2 > 0 ⇒2(x−1)2 > 0 ⇒x̸ = 1 ✓f(x) > −18 ✓turning point (1;−18) ✓x ∈R, x̸ = 1 OR: ✓✓2(x−1)2 > 0 ✓ answer (3) 5.3.2 g(x)·g−1(x) = 0 ⇒g(x) = 0 or g−1(x) = 0 g(x) = 0: x = −2 g−1: x = 2y+4 ⇒g−1(x) = x 2 −2 g−1(x) = 0: x = 4 x = −2 or x = 4 (The x-intercept of g−1 is the image of the y-intercept (0;4) of g.) ✓x = −2 ✓g−1(x) = x 2 −2 or (4;0) ✓x = 4 (3) [11] Copyright reserved Please turn over

UNOFFICIAL Mathematics/P1 8 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 6 6.1 x = −2 ✓x = −2 (1) 6.2 y ∈R, y̸ = 3 or (−∞; 3)∪(3; ∞) ✓answer (1) 6.3 0 = 4 x+2 +3 3(x+2) = −4 ⇒x = −10 3 B ( −10 3 ; 0 ) ✓y = 0 ✓simplification ✓B ( −10 3 ;0 ) (3) 6.4 4 x+2 +3 = −4x+k 4+3(x+2) = (−4x+k)(x+2) 4x2 +(11−k)x+10−2k = 0 For a tangent: ∆= 0 (11−k)2 −4(4)(10−2k) = 0 k2 +10k −39 = 0 (k +13)(k −3) = 0 k = 3 or k = −13 OR (calculus): f ′(x) = −4(x+2)−2 = −4 (x+2)2 = 1 ⇒x = −1 or x = −3 f(−1) = 7; f(−3) = −1 7 = −4(−1)+k ⇒k = 3; −1 = −4(−3)+k ⇒k = −13 ✓equating ✓standard form ✓∆= 0 ✓substitution ✓k2 +10k −39 = 0 ✓both values of k OR: ✓f ′(x) ✓f ′(x) = −4 ✓both x-values ✓both y-values ✓substitution ✓both k (6) 6.5 Lines of symmetry pass through C(−2;3) with m = ±1 Negative gradient: y−3 = −(x+2) ⇒y = −x+1 Reflection in y = −x+1: (x;y) →(1−y; 1−x) D ( 1−0; 1+ 10 3 ) = D ( 1; 13 3 ) BD = √( 1+ 10 3 )2 + (13 3 −0 )2 = 13 √ 2 3 ≈6,13 units OR: perpendicular through B: y = x+ 10 3 ; meets y = −x+1 at M ( −7 6; 13 6 ) BD = 2BM = 2 √(13 6 )2 + (13 6 )2 = 13 √ 2 3 ≈6,13 ✓y = −x+1 ✓xD = 1 ✓yD = 13 3 ✓substitution into distance formula ✓6,13 units OR: ✓y = −x+1 ✓ perpendicular line ✓M ✓BD = 2BM ✓6,13 Note: CA from 6.3. (5) [16] Copyright reserved Please turn over

UNOFFICIAL Mathematics/P1 9 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 7 7.1 197700 = 465000(1−0,115)n (0,885)n = 197700 465000 n = log0,885 (197700 465000 ) = 7,00 After 7 years ✓substitution into A = P(1−i)n ✓use of logs ✓7 years (3) 7.2 A = 10000 ( 1+ 0,065 2 )6 ( 1+ 0,07 4 )8 After 3 years: 10000(1,0325)6 = R12115,47 After 5 years: 12115,47(1,0175)8 = R13919,25 ✓i = 0,065 2 and n = 6 ✓i = 0,07 4 and n = 8 ✓product of the two growth factors / using R12 115,47 ✓R13 919,25 (4) 7.3.1 n = 44×12 = 528; i = 0,075 12 F = 800 [( 1+ 0,075 12 )528 −1 ] 0,075 12 F = R3306959,91 ✓n = 528 ✓i = 0,075 12 ✓substitution into F ✓R3 306 959,91 (4) 7.3.2 i = 0,06 12 = 0,005; n = 240 3306959,91 = x [ 1−(1,005)−240] 0,005 x = R23692,09 ✓i = 0,005 and n = 240 ✓substitution into P with 7.3.1 as P ✓R23 692,09 Note: CA from 7.3.1. (3) [14] Copyright reserved Please turn over

UNOFFICIAL Mathematics/P1 10 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 8 8.1 f(x+h) = −(x+h)2+(x+h) = −x2−2xh−h2+x+h f(x+h)−f(x) = −2xh−h2 +h f ′(x) = lim h→0 −2xh−h2 +h h = lim h→0 h(−2x−h+1) h = lim h→0(−2x−h+1) = −2x+1 ✓f(x+h) ✓−2xh−h2 +h ✓formula with limit ✓common factor ✓−2x+1 Note: Incorrect notation (e.g. no “lim” or “=”): penalise 1 mark. Answer only: 0/5. (5) 8.2.1 g(x) = 3x 1 3 −3x2 g′(x) = x−2 3 −6x ✓3x 1 3 ✓x−2 3 ✓−6x (3) 8.2.2 y = x(4x2 −1) 2x−1 = x(2x−1)(2x+1) 2x−1 y = 2x2 +x dy dx = 4x+1 ✓factorisation ✓2x2 +x ✓4x+1 Note: Quotient rule (not in CAPS) leading to 4x+1: full marks. (3) [11] Copyright reserved Please turn over

UNOFFICIAL Mathematics/P1 11 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 QUESTION 9 9.1.1 f(1) = −1+6−9+4 = 0 ∴(x−1) is a factor ✓f(1) = 0 (1) 9.1.2 f(x) = (x−1)(−x2 +5x−4) = −(x−1)(x−1)(x−4) = −(x−1)2(x−4) (1; 0) and (4; 0) ✓quadratic factor ✓factorised fully ✓both x-intercepts (3) 9.1.3 f ′(x) = −3x2 +12x−9 = 0 x2 −4x+3 = 0 ⇒(x−1)(x−3) = 0 x = 1 or x = 3 f(1) = 0 (local minimum); f(3) = −27+54−27+4 = 4 (local maximum) 0 < k < 4 ✓f ′(x) ✓f ′(x) = 0 ✓x = 1 and x = 3 ✓y-values 0 and 4 ✓0 < k < 4 (5) 9.1.4 g′′(x) = f ′′(−x); f ′′(x) = −6x+12 < 0 for x > 2 and > 0 for x < 2 g is f reflected in the y-axis (and shifted up 2): the point of inflection moves from x = 2 to x = −2 and the concave-down part (x > 2) moves to x < −2 g′′(x) = 6x+12 > 0 ⇒g is concave up for x > −2 ✓critical value −2 ✓x > −2 Note: Vertical shift does not affect concavity. (2) 9.2 f ′(x) = 9x2 +2ax+b; f ′′(x) = 18x+2a f ′′(2) = 0: 36+2a = 0 ⇒a = −18 Tangent y = 3 is horizontal: f ′(2) = 0 36+4(−18)+b = 0 ⇒b = 36 f(2) = 3: 3(8)+(−18)(4)+36(2)+c = 3 24−72+72+c = 3 ⇒c = −21 ✓f ′(x) ✓f ′′(x) ✓a = −18 ✓b = 36 ✓f(2) = 3 ✓c = −21 (6) [17] QUESTION 10 10.1 Total surface area (open top): (2x)(x)+2(2x)(h)+2(x)(h) = 300 2x2 +6xh = 300 h = 300−2x2 6x = 150−x2 3x ( = 50 x −x 3 ) ✓surface area = 300 ✓h in terms of x (2) 10.2 V = (2x)(x)(h) = 2x2 · 150−x2 3x = 2x(150−x2) 3 = 100x−2 3x3 ✓V = 2x2h ✓substitution of h (2) Copyright reserved Please turn over

UNOFFICIAL Mathematics/P1 12 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 10.3 V ′(x) = 100−2x2 = 0 x2 = 50 ⇒x = 5 √ 2 ≈7,07 (x > 0) V ′′(x) = −4x < 0 ∴maximum Vmax = 100(5 √ 2)−2 3(5 √ 2)3 = 1000 √ 2 3 ≈ 471,40 cm3 ✓V ′(x) ✓V ′(x) = 0 ✓x = 5 √ 2 or 7,07 ✓471,40 cm3 Note: Using x = 7,07 gives 471,40; accept 471,4. CA from x. (4) [8] QUESTION 11 11.1.1 No. P(A and B) = 0,3̸ = 0 ✓No ✓reason P(A and B)̸ = 0 (2) 11.1.2 P(only B) = 1−0,16−0,3−0,28 = 0,26 P(B) = 0,26+0,3 = 0,56 OR: P(A or B) = 1−0,28 = 0,72; P(A) = 0,46 0,72 = 0,46+P(B)−0,3 ⇒P(B) = 0,56 ✓method ✓0,26 ✓0,56 OR: ✓0,72 and 0,46 ✓ substitution ✓0,56 (3) 11.2.1 (49−x)+4+2+8+x+5+(60−x)+14 = 100 142−x = 100 ⇒x = 42 ✓sum of all regions = 100 (1) 11.2.2 n(M) = (49−42)+4+8+42 = 61 ⇒P(M) = 0,61 n(L) = 42+8+5+18 = 73 ⇒P(L) = 0,73 P(M)×P(L) = 0,61×0,73 = 0,4453 P(M and L) = 42+8 100 = 0,5 P(M and L)̸ = P(M)×P(L) ∴taking Mathematics is NOT independent of taking Life Sciences ✓P(M) = 0,61 ✓P(L) = 0,73 ✓0,4453 ✓P(M and L) = 0,5 ✓conclusion (not independent) Note: Conclusion mark only with comparison of P(M and L) and P(M)×P(L). (5) 11.3.1 3+2+4 = 9 bottles: 9! = 362880 ✓362880 (1) Copyright reserved Please turn over

UNOFFICIAL Mathematics/P1 13 NSC – Unofficial Marking Guidelines Hlayiso/November 2026 11.3.2 Let n = number of bottles remaining. Energy drinks at both ends: 2×1 ways; the other (n−2) bottles: (n−2)! ways P = 2×(n−2)! n! = 2 n(n−1) = 1 21 n(n−1) = 42 ⇒n2 −n−42 = 0 (n−7)(n+6) = 0 ⇒n = 7 (n > 0) Bottles of juice that broke: 9−7 = 2 OR: P = 2 n × 1 n−1 = 1 21 ⇒n = 7 Check: 2×5! 7! = 240 5040 = 1 21 ✓2×(n−2)! ✓n! as total ✓equation = 1 21 ✓n = 7 ✓2 bottles Note: Trial and error that tests the number broken with full calculations (e.g. 2×5! 7! = 1 21) and concludes 2: full marks. Answer only: 1/5. (5) [17] TOTAL: 150 Copyright reserved Please turn over