Technical Mathematics Grade 12 P1 2026 Memo (Eng)

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

Technical Mathematics/P1 2 Hlayiso/Mock 2026 NSC – Marking Guidelines MARKING CODES A Accuracy CA Consistent accuracy M Method R Rounding NPR No penalty for rounding NPU No penalty for units omitted S Simplification SF Substitution in correct formula NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines where indicated. • No penalty for rounding (NPR) for ALL questions, unless stated otherwise. • AO = answer only. Where AO is indicated, the stated marks are awarded for a correct answer without working. QUESTION 1 1.1.1 x(3x + 8) = 0 ✓x=0 A x=0 or x=− 8 (x ≈ −2,67) ✓ x = − 38 A 3 (2) 1.1.2 2x2 − 6x − 4 =√ 0 OR x2 − 3x − 2 = 0 ✓ standard form A −(−3) ± (−3)2 − 4(1)(−2) ✓ SF CA x= √ 2(1) 3 ± 17 ✓ x ≈ 3,56 CA x= ✓ x ≈ −0,56 CA 2 x ≈ 3,56 or x ≈ −0,56 (4) 1.1.3 x2 − 2x − 8 > 0 ✓ factors A (x − 4)(x + 2) > 0 ✓ both critical values CA Critical values: x = 4 and x = −2 x −2 4 ✓ correct notation CA x < −2 or x>4 AO: full marks (3) 1.2.1 y = 5 − 2x ✓ y = 5 − 2x A (1) Copyright reserved Please turn over

Technical Mathematics/P1 3 Hlayiso/Mock 2026 NSC – Marking Guidelines 1.2.2 x(5 − 2x) − 2 = 0 ✓ substitution CA 5x − 2x2 − 2 = 0 2x2 − 5x + 2 = 0 ✓ standard form CA (2x − 1)(x − 2) = 0 ✓ factors CA x= 1 or x = 2 ✓ both x-values CA 2 ✓ both y-values CA y=4 y=1 (5) 1.3.1 V 2 =√PR ✓ V 2 =√PR A V = PR ✓ V = PR CA (2) √ √ 1.3.2 V = (60)(240) = 14 400 ✓ substitution CA V = 120 V ✓ answer CA NPU (2) 1.4 45 = 1011012 ✓ answer A (1) 1.5 1011012 + 11102 ✓ 59 OR binary addition M = 45 + 14 = 59 ✓ answer CA = 1110112 (2) [22] Copyright reserved Please turn over

Technical Mathematics/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 2.1.1 ∆ = b2 − 4ac = (−3)2 − 4(2)(5) ✓ SF A ∆ = −31 ✓ answer CA (2) 2.1.2 Roots are non-real (∆ < 0) ✓ non-real CA (1) 2.2 For equal roots: ∆ = 0 ✓∆=0 A (−p)2 − 4(2)(8) = 0 ✓ SF A p2 = 64 p = ±8 ✓ p = ±8 CA (3) 2.3.1 p+1 = 0 p = −1 ✓ p = −1 A (1) 2.3.2 2p − 4 < 0 ✓ p<2 A p < 2 (and p ̸= −1) (1) [8] Copyright reserved Please turn over

Technical Mathematics/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 √ ✓ 2x2 3 3.1.1 8x6 = 2x2 A (1) 2x+2 · 22x 3.1.2 ✓ prime bases A 23x · 21 = 2x+2+2x−3x−1 ✓ adding/subtracting = 21 = 2 exponents CA ✓ answer CA (3) √ 3.1.3 log 8 + log 125 − log3 9 + log2 2 1 = log(8 × 125) − log3 32 + log2 2 2 ✓ log 1 000 A = log 1 000 − 2 + 21 ✓2 A = 3 − 2 + 21 ✓ 21 A = 32 OR 1,5 ✓ answer CA (4) [ ] 3.2 log2 (x + 2)(x − 2) = 5 ✓ product law A x 2 − 4 = 25 ✓ exponential form CA x2 = 36 ✓ x2 = 36 CA x = 6 (x ̸= −6, since x − 2 > 0) ✓ answer with rejection CA (4) 3.3 3x · 3 − 3x = 54 3x (3 − 1) = 54 ✓ common factor A 3x = 27 = 33 ✓ 3x = 33 CA x=3 ✓ answer CA AO: full marks (3) 3.4.1 z = (1 + 2i)(i − 2) = i − 2 + 2i2 − 4i ✓ expansion A = −2 − 3i − 2 = −4 − 3i ✓ answer CA (2) 3.4.2 z̄ = −4 + 3i ✓ answer CA (1) Copyright reserved Please turn over

Technical Mathematics/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines 3.4.3 Im ✓ point (−4 ; −3) in quadrant 3 CA −4 Re ✓ line from origin/labels CA (2) −3 (−4 ; −3) OR z √ 3.4.4 r= (−4)2 + (−3)2 = 5 ✓r=5 CA −3 ✓ tan θ CA tan θ = −4 reference angle = 36,87◦ ✓ θ = 216,87◦ CA θ = 180◦ + 36,87◦ = 216,87◦ ✓ polar form CA z = 5 cis 216,87◦ OR 5 cis(−143,13◦ ) (4) [24] QUESTION 4 4.1.1 y = −3 OR (0 ; −3) ✓ answer A (1) 4.1.2 x2 − 2x − 3 = 0 ✓ factors A (x − 3)(x + 1) = 0 x = 3 or x = −1 ✓ both x-values CA (2) b −2 4.1.3 x=− =− =1 ✓x=1 A 2a 2(1) ✓ substitution CA y = (1)2 − 2(1) − 3 = −4 OR ✓ y = −4 CA 4ac − b2 4(1)(−3) − 4 y= = (3) 4a 4 Turning point: (1 ; −4) Copyright reserved Please turn over

Technical Mathematics/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines 4.1.4 y g f: ✓ x-intercepts CA ✓ y-intercept CA f ✓ turning point & shape CA g: 1 ✓ x-intercept −1 A −1 x ✓ y-intercept 1 & straight 3 line A (5) −3 (1 ; −4) 4.1.5 y ≥ −4 OR y ∈ [−4 ; ∞) ✓ answer CA (1) 4.1.6 −1 < x < 3 OR x ∈ (−1 ; 3) ✓ critical values CA ✓ notation CA (2) 4.2.1 x=0 and y = −2 ✓x=0 A ✓ y = −2 A (2) a 4.2.2 h(x) = − 2 x a ✓ substitution of P A 2= −2 −1 ✓ answer CA a = −4 (2) 4.2.3 k(x) = bx − 2 7 = b2 − 2 ✓ substitution of R A b2 = 9 b = 3 (b > 0) ✓ answer CA (2) −4 4.2.4 0= −2 ✓ h(x) = 0 CA x 2x = −4 x = −2 B(−2 ; 0) ✓ answer CA (2) Copyright reserved Please turn over

Technical Mathematics/P1 8 Hlayiso/Mock 2026 NSC – Marking Guidelines 4.2.5 y > −2 OR y ∈ (−2 ; ∞) ✓ answer A (1) 4.2.6 x ∈ R , x ̸= 0 ✓ answer A (1) [24] QUESTION 5 5.1 A = P(1 − i)n ✓ formula A A = 480 000(1 − 0,12)5 ✓ SF A A = R253 311,32 ✓ answer CA (3) ( ) 0,096 12 5.2 ieff = 1 + −1 ✓ 0,096 12 A 12 = 0,100338 . . . ✓ SF A ieff = 10,03% ✓ answer CA (3) 5.3 A = P(1 + i)n 20 000 = 12 000(1 + i)6 ✓ SF A (1 + i)6 = 5 ✓ simplification CA √3 ✓ 6th root CA 1 + i = 6 53 = 1,08887 . . . i = 0,0889 ✓ answer CA r = 8,89% per annum (4) ( ) 0,08 8 5.4 A2 = 25 000 1 + = R29 291,48 ✓ i = 0,08 4 ;n=8 A 4 ( ) 0,09 ✓ i = 12 ; n = 12 A 0,09 12 A3 = 29 291,48 1 + = R32 039,23 ✓ +5 000 CA 12 ✓ n = 24 ( ) CA 0,09 24 ✓ answer CA A5 = (32 039,23 + 5 000) 1 + 12 Accept answers in the range A5 = R44 314,23 R44 314 – R44 315 OR (5) A = 25 000(1,02)8 (1,0075)36 + 5 000(1,0075)24 = R44 314,23 [15] Copyright reserved Please turn over

Technical Mathematics/P1 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 6.1 f (x + h) = 2(x + h)2 − 3 = 2x2 + 4xh + 2h2 − 3 ✓ f (x + h) expanded A f (x + h) − f (x) = 4xh + 2h2 ✓ 4xh + 2h2 CA 4xh + 2h2 ✓ formula with limit f ′ (x) = lim h→0 h notation A = lim (4x + 2h) ✓ 4x + 2h CA h→0 f ′ (x) = 4x ✓ answer CA (5) dy 6.2.1 = 20x3 − 3 ✓ 20x3 A dx ✓ −3 A (2) f (x) = 6x−2 − x 3 1 6.2.2 ✓ exponential form A 1 2 f ′ (x) = −12x−3 − x− 3 ✓ −12x−3 CA ✓ − 31 x− 3 2 3 CA (3) x2 − 9 (x − 3)(x + 3) 6.2.3 = = x−3 ✓ factorising A x+3 x+3 Dx [x − 3] = 1 ✓ x−3 CA ✓ answer CA (3) 6.3.1 g′ (x) = 3x2 − 4 ✓ answer A (1) 6.3.2 g′ (1) = 3(1)2 − 4 = −1 ✓ answer CA (1) 6.3.3 g(1) = (1)3 − 4(1) = −3 ✓ y-value A y − (−3) = −1(x − 1) ✓ substitution CA y = −x − 2 ✓ equation CA (3) Copyright reserved Please turn over

Technical Mathematics/P1 10 Hlayiso/Mock 2026 NSC – Marking Guidelines 6.3.4 3x2 − 4 = 8 ✓ g′ (x) = 8 CA x2 = 4 x = 2 or x = −2 ✓ both values CA (2) [20] QUESTION 7 7.1 C(0 ; 27) ✓ answer A (1) 7.2 f (−3) = (−3)3 − 3(−3)2 − 9(−3) + 27 ✓ substitution of −3 A = −27 − 27 + 27 + 27 = 0 ✓ = 0 with conclusion A ∴ (x + 3) is a factor of f (x) (2) 7.3 f (x) = (x + 3)(x2 − 6x + 9) ✓ quadratic factor CA = (x + 3)(x − 3)2 ✓ (x − 3)2 CA x = −3 or x = 3 A(−3 ; 0) and B(3 ; 0) ✓ both coordinates CA (3) 7.4 f ′ (x) = 3x2 − 6x − 9 = 0 ✓ f ′ (x) A x2 − 2x − 3 = 0 ✓ f ′ (x) = 0 CA (x − 3)(x + 1) = 0 ✓ factors CA x = −1 or x = 3 ✓ x-values CA f (−1) = −1 − 3 + 9 + 27 = 32 D(−1 ; 32) and B(3 ; 0) ✓ y = 32 CA (5) 7.5 −1 < x < 3 OR x ∈ (−1 ; 3) ✓ critical values CA ✓ notation CA (2) 7.6 f ′ (0) = 3(0)2 − 6(0) − 9 = −9 ✓ answer CA (1) [14] Copyright reserved Please turn over

Technical Mathematics/P1 11 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1 V (0) = 200 kilolitres ✓ answer A (1) 8.2 V ′ (t) = −3t 2 + 12t + 15 ✓ −3t 2 + 12t A ✓ 15 A (2) 8.3 V ′ (2) = −3(2)2 + 12(2) + 15 ✓ substitution CA = 27 kilolitres per hour ✓ answer CA (2) 8.4 −3t 2 + 12t + 15 = 0 ✓ V ′ (t) = 0 CA t 2 − 4t − 5 = 0 (t − 5)(t + 1) = 0 t = 5 (t ̸= −1) ✓t =5 CA V (5) = −(5)3 + 6(5)2 + 15(5) + 200 V (5) = 300 kilolitres ✓ answer CA (3) [8] Copyright reserved

Technical Mathematics/P1 12 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 9 ∫ 3 9.1.1 dx = 3 ln x +C ✓ 3 ln x A x ✓ +C A (2) ∫ 9.1.2 (4x2 − 4x + 1) dx ✓ expansion A 3 4x3 ✓ 4x3 CA = − 2x2 + x +C ✓ −2x2 + x +C CA 3 (3) ∫ ( ) 1 9.1.3 x +2 2 x dx ✓ exponential form A 3 2 3 2x ✓ 32 x 2 CA = x2 + +C x 3 ln 2 ✓ ln2 2 +C A (3) 9.2.1 −x2 + 2x + 3 = 0 ✓ factors A x2 − 2x − 3 = 0 (x − 3)(x + 1) = 0 x = −1 and x = 3 ✓ both values CA (2) ∫ 3( ) 9.2.2 Area = −x2 + 2x + 3 dx ✓ integral with limits CA [ 3 −1 ]3 ✓ integration CA x ✓ substitution of limits CA = − + x + 3x 2 ( 3 ) −1( ) ✓ simplification CA 27 = − +9+9 − 1 +1−3 ✓ answer CA (3 ) 3 (5) 5 = 9− − 3 32 = ≈ 10,67 square units 3 [15] TOTAL: 150 Copyright reserved