Technical Mathematics Grade 12 Paper 1 Gauteng Mock 2025 Memo

PREPARATORY EXAMINATIONS VOORBEREIDENDE EKSAMEN 2025 MARKING GUIDELINES/ NASIENRIGLYNE TECHNICAL MATHEMATICS/TEGNIESE WISKUNDE (PAPER/VRAESTEL 1) (11091) 12 pages/bladsye Marking Codes/Nasienkodes A Accuracy/Akkuraatheid CA Consistent accuracy/Volgehoue akkuraatheid M Method/Metode R Rounding/Afronding NPR No penalty for rounding/Geen penalisering vir afronding nie NPU No penalty for units omitted/Geen penalisering vir eenhede weggelaat nie S Simplification/Vereenvoudiging F Formula/Formule SF Substitution in correct formula/Vervanging in korrekte formule

11091/25 NOTES: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt to answer a question and did not redo it, mark the crossed- out version. • Consistent accuracy applies in all aspects of the marking guidelines, where applicable. LET WEL: • Indien ʼn kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na. • Indien ʼn kandidaat 'n antwoord deurgetrek het en nie weer beantwoord het nie, merk die deurgetrekte antwoord. • Volgehoue akkuraatheid is deurgaans van toepassing op alle aspekte van die nasienriglyne waar van toepassing. QUESTION/VRAAG 1 CL/DV 1.1.1 − 𝑥 (2 − 4𝑥) = 0 2 1 ✓𝑥= / A 1E 2 1 4 2 𝑥 = 4/ 2 or/of 𝑥=0 ✓x=0 A OR/OF 4𝑥 2 − 2𝑥 = 0 −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎 −(−2) ± √(−2)2 − (4)(4)(0) = 2(4) 2 ± √4 2 1 = ✓x = 4 / 2 A 8 1 ✓x = 0 A 𝑥=2 or/of 𝑥=0 (2) 1.1.2 𝑥 2 − 3 = 4 − 5𝑥 2M 𝑥 2 + 5𝑥 − 7 = 0 ✓ Std form/vorm A −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥= ✓ SF CA 2𝑎 −(5) ± √(5)2 − 4(1)(−7) 𝑥= 2(1) ✓ 𝑥 = 1,14 CA −5 ± √53 𝑥= ✓ 𝑥 = −6,14 CA 2 (4) 𝑥 = 1,14 or/of 𝑥 = −6,14 PR (Only once) AO: Full marks/Volpunte 2

11091/25 1.1.3 𝑥 2 + 5𝑥 ≤ 6 3M −(5) ± √(5)2 − 4(1)(−6) 𝑥= ✓ Factors/Faktore /SF A 2(1) −5 ± √49 𝑥= 2 OR/OF ✓ Both critical values/Beide (𝑥 + 6)( 𝑥 − 1) ≤ 0 kritieke waardes CA −6 ≤ 𝑥 ≤ 1 ✓ Correct notation/Korrekte OR/OF notasie CA (3) 𝑥 ∈ [−6; 1] Can represent it graphically/Kan grafies voorstel AO: Full marks/Volpunte 1.2 𝑥+𝑦 =3 3E ∴𝑥 =3−𝑦 ✓make 𝑥 subject/ 𝑥 2 + 𝑦 2 = 89 maak 𝑥 onderwerp A (3 − 𝑦)2 + 𝑦 2 = 89 9 − 6𝑦 + 𝑦 2 + 𝑦 2 = 89 ✓ Substitution/Substitusie 2𝑦 2 − 6𝑦 − 80 = 0 CA 𝑦 2 − 3𝑦 − 40 = 0 ✓ Std form/vorm CA −(−3) ± √(−3)2 − 4(1)(−40) 𝑦= ✓ Factors/Faktore/SF CA 2(1) 3 ± √169 𝑦= 2 ✓ Both 𝑦-values (𝑦 − 8)(𝑦 + 5) = 0 Beide 𝑦-waardes CA 𝑦 = 8 or/of 𝑦 = −5 ✓ Both 𝑥-values ∴ 𝑥 = 3 − (8) = −5 or/of 𝑥 = 3 − (−5) = 8 Beide 𝑥-waardes CA 𝑥 = −5 or 𝑥 = 8 (6) OR/OF OR/OF ∴𝑦 =3 −𝑥 2 2 ✓ make 𝑦 subject/ (3 𝑥 + − 𝑥) = 89 maak 𝑦 onderwerp A 𝑥 2 + 9 − 6𝑥 + 𝑥 2 = 89 ✓ Substitution/Substitusie A 2𝑥 2 − 6𝑥 − 80 = 0 𝑥 2 − 3𝑥 − 40 = 0 ✓ Std form/vorm CA −(−3) ± √(−3)2 − 4(1)(−40) ✓ Factors/Faktore/SF CA 𝑥= 2(1) 3 ± √169 𝑥= 2 ✓ Both 𝑥-values/ OR/OF Beide 𝑥-waardes CA (𝑥 + 5)(𝑥 − 8) = 0 ✓ Both 𝑦-values/ 𝑥 = −5 or 𝑥 = 8 Beide 𝑦-waardes CA ∴ 𝑦 = 3 − (−5) or 𝑦 = 3 − 8 (6) 𝑦=8 or 𝑦 = −5 3

11091/25 1.3.1 𝑇𝑐 𝑇 ✓ 𝑇𝑐 (Subject of formula/ 2M 𝜂 =1− ℎ 𝑇ℎ 𝑇𝑐 Onderwerp van formule) =1− 𝜂 M 𝑇ℎ ✓Answer/Antwoord A 𝑇𝑐 = 𝑇ℎ (1 − 𝜂) OR/OF 𝑇 𝑇𝑐 ✓− 𝑇𝑐 (Subject of formula/ 𝜂−1=− ℎ 𝑇ℎ Onderwerp van formule) 𝑇𝑐 M =1− 𝜂 𝑇ℎ ✓ Answer/Antwoord A 𝑇𝑐 = 𝑇ℎ (1 − 𝜂) OR/OF 𝑇𝑐 𝜂 =1− 𝑇ℎ 𝑇𝑐 𝜂−1=− 𝑇ℎ 𝑇𝑐 = −𝑇ℎ (𝜂 − 1) (2) 1.3.2 𝑇𝑐 = 𝑇ℎ (1 − 𝜂) 1E 𝑇𝑐 = 600 (1 − 0,4) ✓ Substitution/Substitusie ∴ 𝑇𝑐 = 360 𝐾 CA ✓ Answer/Antwoord CA (2) 1.4 (111012 + 13) − 1112 ✓ 11012 A 1E = (111012 + 11012 ) − 1112 ✓ 1010102 A = 1010102 − 1112 ✓Answer/Antwoord CA = 1000112 ✓ base/basis 2 A (4) 4 OR/OF AO: 4 (111012 + 13) − 1112 = (29 + 13) − 7 ✓ 29 A = 42 − 7 = 35 ✓ 35 CA = 1000112 ✓ Answer/Antwoord CA ✓ base 2 A (4) [23] 4

11091/25 QUESTION/VRAAG 2 CL/ DV 2.1 𝑥 2 − 𝑝𝑥 + 𝑝 + 3 = 0 4E ∆ = 𝑏 2 − 4𝑎𝑐 ∆ = (−𝑝)2 − 4(1)(𝑝 + 3) ✓ Substitution/Substitusie A ∴ ∆= 𝑝2 − 4𝑝 − 12 ✓ Answer/Antwoord CA (2) 2.2 ∆≥ 0 4E 𝑝2 − 4𝑝 − 12 ≥ 0 ✓ ∆≥0 A (𝑝 − 6)(𝑝 + 2) ≥ 0 ✓ Factors/Faktore/SF A 𝑝 ≥ 6 or/𝑜𝑓 𝑝 ≤ −2 ✓ Answer/Antwoord CA (3) OR/OF − b  b 2 − 4ac p= 2a − (−4)  (−4) 2 − 4(1)( −12) = 2(1) 4  16 + 48 = 2 p = 6 or/of p = –2 (p – 6) (p + 2) ≥ 0 p ≥ 6 or/of p ≤ –2 [5] QUESTION/VRAAG 3 CL/ DV 3.1.1 √25𝑥 2 − 16𝑥 2 2M ✓3 A = √9𝑥 2 ✓x A = 3𝑥 (2) AO: Full marks/Volpunte 3.1.2 (log 2 32)(log 2 4) ✓ 5 log 2 2 A 2E = (log 2 25 )(log 2 22 ) ✓ log 2 2 = 1 A = (5 log 2 2)(2 log 2 2) ✓ Answer/Antwoord CA = 5 × 2(log 2 2) OR/OF [5(1)][2(1)] (3) 1 = 10 AO:3 5

11091/25 3.2 23𝑥 × 3 3E 𝐴= 23𝑥 + 3. 23𝑥+1 23𝑥 × 3 ✓ 23𝑥 × 21 A 𝐴 = 3𝑥 3𝑥 2 + 3 × 23𝑥 × 21 ✓ 2 (common factor/gemeenskaplike 23𝑥 ×3 faktor) A 𝐴 = 23𝑥 (1+3×2) 3 ✓(1 + 3 × 2) CA = 1+6 3 𝐴=7 ✓ Answer/Antwoord CA (4) 3.3.1 𝑧 = 2 𝑐𝑖𝑠 315° 2E 𝑧 = 𝑟 (𝑐𝑜𝑠 𝜃 + 𝑖 𝑠𝑖𝑛 𝜃) ✓ Substitution/Substitusie A 𝑧 = 2 (𝑐𝑜𝑠 315° + 𝑖 𝑠𝑖𝑛 315°) ✓ 1,41/ 2 A 𝑧 = 1,41 − 1,41 𝑖 ✓ 1,41 i/ −√2 𝑖 A z= 2 − 2i AO: Full marks/Volpunte (3) 3.3.2 Fourth quadrant/Vierde kwadrant ✓ Answer/Antwoord A 1E (1) 3.3.3 𝑟=2 ✓ Answer/Antwoord A 1E (1) 3.4 (𝑥 − 𝑦𝑖) = 𝑖(4𝑖 − 3) 3E 𝑥 − 𝑦𝑖 = 4𝑖 2 − 3𝑖 ✓S A 𝑥 − 𝑦𝑖 = −4 − 3𝑖 ✓𝑖 2 = −1 A 𝑥 = −4 or/of 𝑦 = 3 ✓𝑥 = −4 CA ✓𝑦 = 3 CA (4) [18] 6

11091/25 QUESTION/VRAAG 4 CL/DV 4.1.1 𝑦 = −8 ✓ 𝑦 = −8 A 1E No marks for writing only -8 Geen punte vir die skryf van net -8 nie (1) 4.1.2 For y-intercept let/vir y-afsnit laat 𝑥 = 0 2E 𝑓(𝑥) = 2𝑥 − 8 ✓x=0 A 𝑦 = 20 − 8 𝑦 = 1−8 𝑦 = −7 ✓ 𝑦 = −7 CA For x-intercept let/vir y-afsnit laat 𝑦 = 0 2𝑥 − 8 = 0 ✓y=0 A 2𝑥 = 8 2𝑥 = 23 ✓x = 3 CA 𝑥=3 (4) 4.1.3 ✓ Shape/Vorm 𝑓 A 1M ✓ Asymptote/ Asimptoot CA ✓ 𝑥 intercept/ x-afsnit 𝑓 CA ✓ 𝑦 intercept/ y-afsnit 𝑓 CA (4) 7

11091/25 4.2.1 𝑔(𝑥) = −2𝑥 + 𝑑 (−2 ; 0) 2M 0 = −2(−2) + 𝑑 ✓ 𝐒𝐅 A 𝑑 = −4 ✓ 𝑑 = −4 CA (2) 4.2.2 𝐶(0; −4) ✓𝑥=0 A 1M (a) ✓ 𝑦 = −4 CA (2) (b) ℎ(𝑥) = − √16 − 𝑥 2 ✓ Answer/ Antwoord CA 3M (1) (c) −4 ≤𝑥 ≤4 ✓ Notation/Notasie A 2M OR/OF ✓ Critical values/ 𝑥 ∈ [−4; 4] Kritieke waardes CA CA from 4.2.1 CA vanaf 4.2.1 (2) 4.3.1 𝐾 (−6 ; 0) ✓ 𝐾 (−6 ; 0) A 1E 𝐿 (2 ; 0) ✓ 𝐿 (2 ; 0) A (2) 4.3.2 𝑓(𝑥) = 𝑥 2 + 4𝑥 − 12 ✓ Standard form/ 4E 𝑥= − 𝑏 OR/OF 𝑓 ′ (𝑥) = 2𝑥 + 4 OR/OF 𝒙= −𝟔+𝟐 Standaardvorm A 2𝑎 𝟐 4 ✓M A 𝑥 = − 2(1) 0 = 2𝑥 + 4 𝑥 = −2 𝑥 = −2 𝑥 = −2 𝑓(−2) = (−2)2 + 4(−2) − 12 = −16 ✓ 𝑥-value /waarde CA 𝑁 (−2 ; −16) ✓ 𝑦-value/waarde CA (4) 4.3.3 For y-intercept let 𝑥 = 0/Vir y-afsnit stel 𝑥 = 0 1E (a) 𝑓(0) = (0)2 + 4(0) − 12 ✓0 A 𝑓(0) = −12 ✓ −12 CA or/of (2) 𝑓(0) = (0 + 6) (0 – 2) 𝑓 (0) = −12 𝑀 (0 ; −12) (b) 𝑦 ≥ −16 or/of ✓ Answer/Antw CA 1D y ∈ (–16 ;  ) CA from 4.3.2 CA vanaf 4.3.2 (1) (c) 𝑥=0 ✓𝑥=0 A 2E 𝑦 = −16 ✓ 𝑦 = −16 CA (2) 8

11091/25 𝑎 4.3.4 𝑡(𝑥) = 𝑥 + 𝑞 2D 𝑎 ✓ 𝑞 = −16 CA 0 = −6 − 16 𝑎 ✓ Subst/Vervang (−6 ; 0) = − 16 CA 6 𝑎 = −96 ✓ a = –96 CA (3) 4.3.5 −6 < 𝑥 < 0 ✓ Notation/Notasie A 3M OR/OF ✓ Both critical values/ 𝑥 ∈ (−6; 0) Beide kritieke waardes CA (2) 4.3.6 𝑥 < −6 ✓ 𝑥 < −6 CA 4E 𝑥 >0 ✓𝑥 >0 A (2) [34] QUESTION/VRAAG 5 CL/DV 5.1.1 𝐴 = 𝑃(1 + 𝑖)𝑛 ✓ Formula/Formule A 2E 0.18 2×12 ✓ SF A 𝐴 = 150 000 (1 + ) 0.18 12 ✓𝑖 = 12 A 𝐴 = R214 425,42 ✓ Answer/Antwoord CA NPR (4) 5.1.2 𝑖 𝑚 ✓F A 2E 𝑖 = (1 + ) −1 𝑚 0,18 12 ✓ SF A 𝑖 = (1 + ) −1 12 𝑖 = 0,1956 ✓ 19,56% CA 𝑟 = 19,56% NPR (3) 5.2 𝐴 = 𝑃(1 − 𝑖𝑛) ✓F A 2M 𝐴 = 100 000(1 − 0,02 × 10) ✓ SF A 𝐴 = R80 000 ✓ Answer/Antwoord CA NPR (3) 9

11091/25 5.3 0,096 2×4 ✓ SF A 2D 𝐴 = 6 000 (1 + ) 0,096 4 ✓𝑖= 4 A 𝐴 = R7253,554918 0,106 4×2 ✓ SF CA 𝐴 = 7253,554918 (1 + ) 2 0,106 ✓ 2 A 𝐴 = R10 964,22 ✓ 𝑅10 964,22 CA NPR (5) [15] QUESTION/VRAAG 6 CL/DV 6.1 𝑓(𝑥) = −3𝑥 + 5 2D 𝑓(𝑥 + ℎ) = −3(𝑥 + ℎ) + 5 = −3𝑥 − 3ℎ + 5 ✓ Definition/ Definisie A 𝑓′ (𝑥) = lim 𝑓(𝑥+ℎ)−𝑓(𝑥) ✓ SF CA ℎ ℎ→0 ✓S CA ′ (𝑥) −3𝑥−3ℎ+5−(−3𝑥+5) 𝑓 = lim ℎ→0 ℎ = lim −3𝑥−3ℎ+5+3𝑥−5 ✓ Answer/ ℎ→0 ℎ Antwoord CA −3ℎ = lim ℎ 0 AO:4 (4) ℎ→0 = −3 1 6.2.1 𝑓(𝑥) = 4 √𝑥 + 2(3𝑥 −3 )0 ✓ 4𝑥 2 A 2D 1 𝑓(𝑥) = 4𝑥 + 2 2 ✓2 A 1 1 2 ✓Answer/Antwoord 𝑓 ′ (𝑥) = 2 𝑥 − 2 or/of 2 𝑥 or/of √ √𝑥 CA (3) 6.2.2 (𝑥+2)2 3M 𝐷𝑥 [ 𝑥 ] 2 𝑥 +4𝑥+4 = 𝐷𝑥 [ 𝑥 ] 𝑥2 4𝑥 4 = 𝐷𝑥 [ 𝑥 + 𝑥 + 𝑥] ✓S A = 𝐷𝑥 [𝑥 + 4 + 4𝑥 −1 ] ✓1 CA 4 ✓ −4𝑥 −2 CA = 1 − 4𝑥 −2 or/of = 1 – x2 (3) 10

11091/25 6.2.3 𝑑 (7𝑥 3 + 3𝜋 2 − 𝑥 − 4 ) 3E 𝑑𝑥 2 ✓ 21𝑥 A 2 −5 2 4 = 21 𝑥 + 4𝑥 or/of 21 𝑥 + 5 ✓ 4𝑥 − 5 A x (2) 6.3 𝑔(𝑥) = 𝑥 2 − 8𝑥 + 20 2D ′ 𝑔′(𝑥) = 2𝑥 − 8 ✓ 𝑔 (𝑥) = 2𝑥 − 8 A 4 = 2𝑥 − 8 ✓ 𝑔′(𝑥) = 4 A 2𝑥 = 12 𝑥 =6 ✓ x-value/𝑥-waarde 𝑔(6) = (6)2 − 8(6) + 20 = 8 CA Point on the curve/Punt op die kromme: (6 ; 8) ✓ 𝑦-value/y-waarde CA (4) [16] QUESTION/VRAAG 7 [17] CL 7.1 (𝑥 + 1) ✓ Answer/Antwoord 1M A (1) 7.2 2 2 𝑥= 𝑜𝑟/𝒐𝒇 𝑥 = −2 𝑜𝑟/𝑜𝑓 𝑥 = −1 ✓𝑥 = 3 A 1M 3 ✓𝑥 = −2 A ✓𝑥 = −1 CA (3) 7.3 (0 ; −4) ✓0 A 1M ✓–4 A (2) 7.4 𝑓(𝑥) = 3𝑥 3 + 7𝑥 2 − 4 ✓ Derivative/ 3E 𝑓 ′ (𝑥) = 9𝑥 2 + 14𝑥 Afgeleide A 9𝑥 2 + 14𝑥 = 0 ✓ Equating derivative to 𝑥(9𝑥 + 14) = 0 0/Stel afgeleide gelyk aan 0 A OR/OF ✓ Factors/Faktore/SF CA −(14) ± √(14)2 − 4(9)(0) ✓ 𝑥-values/x-waardes 𝑥= CA 2(9) 9𝑥 + 14 = 0 or 𝑥=0 ✓ 𝑦-values/y-waardes 14 CA 𝑥 = − 9 ≈ −1,56 (5) 14 14 14 400 𝑓 (− ) = 3(− )3 + 7(− )2 − 4 = ≈ 1,65 9 9 9 243 𝑓(0) = 3(0)3 + 7(0)2 − 4 = − 4 TP (0, −4) and/en TP (−1,56; 1,65) 11

11091/25 7.5 ✓ x-intercepts/ 1D x-afsnitte CA ✓ y-intercept/y-afsnit CA ✓ turning points/ draaipunte CA ✓ shape/vorm CA (4) 7.6 14 ✓ Notation/Notasie 3E − <𝑥 <0 9 CA or/of – 1,56 < x < 0 ✓ Critical values/ Kritieke waardes CA OR/OF (2) 14 𝑥 ∈ (− 9 ; 0) or/of x  (–1,56 ; 0) [17] QUESTION/VRAAG 8 CL/DV 8.1 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋𝑟 2 ℎ 2D 864 = 𝜋𝑟 2 ℎ ✓ Answer/ 864 𝑉 Antwoord A ℎ = 2 or/𝑜𝑓 2 𝜋𝑟 𝜋𝑟 (1) 8.2 2 2D 𝑆𝐴 = 2𝜋𝑟 + 2𝜋𝑟ℎ 2𝜋𝑟. 864 ✓SF CA 𝑆𝐴 = 2𝜋𝑟 2 + 𝜋𝑟 2 ✓S CA 1728 (2) = 2𝜋𝑟 2 + 𝑟 8.3 𝑆(𝑟) = 1728 𝑟 −1 + 2𝜋𝑟 2 ✓ Standard form/ 4E 𝑆 ′ (𝑟) = −1728𝑟 −2 + 4𝜋𝑟 Standaardvorm CA −1728𝑟 −2 + 4𝜋𝑟 = 0 at minimum/ten minste ✓ Derivative/ 1728 Afgeleide CA − = − 4𝜋𝑟 𝑟2 ✓ Equating derivative to 0/Stel afgeleide 1728 = 4𝜋𝑟 3 gelyk aan 0 A 1728 ✓S CA 𝑟3 = 4𝜋 3 432 𝑟 = √ 𝜋 or/of 5,16 ✓r value/waarde CA (5) [8] 12

11091/25 QUESTION/VRAAG 9 CL/DV Do not penalise if C is omitted in question 9.1.2 and 9.2.1./Moet nie penaliseer indien C weggelaat word by vraag 9.1.2 en 9.2.1 nie. 9.1.1 ∫ 𝑑𝑥 ✓𝑥 A 1E =𝑥+𝐶 ✓𝐶 A (2) 9.1.2 2 ∫ [𝑥 (𝑥 + 𝑥) + 𝑥] 𝑑𝑥 3 ✓S A 4E 𝑥3 3 ✓ 3 CA = ∫ [𝑥 2 + 2 + ] 𝑑𝑥 𝑥 ✓ 2𝑥 CA 𝑥3 ✓3𝑙𝑛𝑥 + 𝐶 A = + 2𝑥 + 3𝑙𝑛𝑥 + 𝐶 3 (4) 9.2.1 1 2E ∫(−𝑥 3 )𝑑𝑥 ✓− 4 𝑥 4 + 𝐶 A 1 (1) = − 4 𝑥4 + 𝐶 9.2.2 0 ✓ Area notation using 3E ∫ (−𝑥 3 )𝑑𝑥 −2 integrals/Area- 1 0 notasie met gebruik 𝐴1 = [− 4 𝑥 4 ] −2 van integrale 1 1 A = [− (4) × 0 ] − [− (4) × (−2)4 ] 4 ✓ SF CA = [0] − [−4] ✓𝐴1 value/waarde CA = 4 units/eenhede2 2 ✓ SF CA 1 𝐴2 = − [− 4 𝑥 4 ] ✓𝐴2 value/waarde CA 0 1 1 = − [− (4) × 2 ] − [− (4) × (0)4 ] 4 ✓M CA = −[(−4) − 0] ✓ Answer/ = 4 𝑢𝑛𝑖𝑡𝑠 2 /eenhede2 Antwoord CA Atotal/totaal = 4 + 4 = 8 𝑢𝑛𝑖𝑡𝑠 2 /eenhede2 (7) [14] TOTAL/TOTAAL: 150 13