Mathematics Grade 12 P1 North West June 2024 Memo (Afr)

PROVINCIAL ASSESSMENT PROVINSIALE ASSESSERING GRADE 12/GRAAD 12 MATHEMATICS P1/WISKUNDE V1 JUNE/JUNIE 2024 MARKING GUIDELINES/NASIENRIGLYNE MARKS: 150 TIME: 3 hours These marking guidelines consists of 14 pages. Hierdie nasienriglyne bestaan uit 14 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 2 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne NOTE:  If a candidate answered a question TWICE, only mark the FIRST attempt.  Consistent accuracy applies in ALL aspects of the marking guidelines. LET WEL:  Indien ’n kandidaat ’n vraag TWEE keer beantwoord, word slegs die EERSTE poging nagesien.  Volgehoue akkuraatheid is DEURGAANS op ALLE aspekte van die nasienriglyne van toepassing. QUESTION/VRAAG 1 1.1 2𝑥(4 − 𝑥 ) = 0 1.1.1 ∴ 𝑥 = 0 𝑜𝑟│ 𝑜𝑓 𝑥 = 4 𝑥 = 0 𝑥 = 4 (2) 1.1.2 𝑥(3𝑥 − 7) = 4  std form 3𝑥 2 − 7𝑥 − 4 = 0 −(−7) ± √(−7)2 −4(3)(−4) ∴𝑥=  subst 2(3) ∴ 𝑥 = 2,81 𝑜𝑟│ 𝑜𝑓 − 0,47  each answ/elke antw (4) 1.1.3 𝑥 + √𝑥 − 3 = 15 √𝑥 − 3 = 15 − 𝑥  isol √ ∴ 𝑥 − 3 = (15 − 𝑥 )2 𝑥 − 3 = 225 − 30𝑥 + 𝑥 2  square /kwadreer ∴ 𝑥 2 − 31𝑥 + 228 = 0  std form ∴ (𝑥 − 12)(𝑥 − 19) = 0  factors/faktore ∴ 𝑥 = 12 𝑜𝑟 │𝑜𝑓 𝑥 ≠ 19  𝑥 = 12; 𝑥 ≠ 19 (5) 1.1.4 𝑥 2 + 𝑥 − 12 > 0  factors/faktore ∴ (𝑥 + 4)(𝑥 − 3) > 0 + − +  critical values/kritieke w ∴ 𝑥 < −4 𝑜𝑟 │𝑜𝑓 𝑥 > 3  answer combo/antwoord (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 3 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne 1.2 3𝑥 − 𝑦 = 1 .............................................  𝑥 2 + 2𝑥𝑦 = 3𝑦 2 − 7 ................................ from / uit  : 𝑦 = 3𝑥 − 1 subst in :  y subject/onderwerp 𝑥 2 + 2𝑥 (3𝑥 − 1) = 3(3𝑥 − 1)2 − 7  subst ∴ 𝑥 2 + 6𝑥 2 − 2𝑥 − 27𝑥 2 + 18𝑥 − 3 + 7 = 0 ∴ −20𝑥 2 + 16𝑥 + 4 = 0 ∴ 5𝑥 2 − 4𝑥 − 1 = 0  std form ∴ (5𝑥 + 1)(𝑥 − 1) = 0  faktore/factors 1 ∴ 𝑥= − 𝑜𝑟 │ 𝑜𝑓 𝑥 = 1  both x-values/beide y-w 5 1 1 8 if / as 𝑥 = − 5 ∶ 𝑦 = 3 (− 5) − 1 = − 5  both y-values/beide x-w if / as 𝑥 = 1 ∶ 𝑦 = 3(1) − 1 = 2 (6) OR/OF 3𝑥 − 𝑦 = 1 . . . . . . . . . . . . . . . . ① 𝑥 2 + 2𝑥𝑦 = 3𝑦 2 − 7 . . . . . . . . . . . ② 𝑦+1 𝑥= . . . . . . . . . . . . . . . . . . . ③ 3 Subst. ③ into/in ②: 𝑦+1 2 𝑦+1  x subject/onderwerp ( ) + 2𝑦 ( ) = 3𝑦 2 − 7 3 3 𝑦 2 + 2𝑦 + 1 2𝑦 2 + 2𝑦 + = 3𝑦 2 − 7  subst 9 3 𝑦 2 + 2𝑦 + 1 + 6𝑦 2 + 6𝑦 = 27𝑦 2 − 63 20𝑦 2 − 8𝑦 − 64 = 0 5𝑦 2 − 2𝑦 − 16 = 0 (𝑦 − 2)(5𝑦 + 8) = 0 8  std form 𝑦=2 𝑜𝑟 │ 𝑜𝑓 𝑦= − 5  faktore/factors 1 𝑥=1 𝑜𝑟 │ 𝑜𝑓 𝑥 = −  both y-values/beide y-w 5  both x-values/beide x-w (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 4 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne 𝑎 𝑏 𝑐 1.3 √2 = 3; √3 = 5; √5 = 8 1 1 1 exp form ∴ 2𝑎 = 3; 3𝑏 = 5; 5𝑐 = 8 1 1 𝑐 1 𝑏 ∴ [(2 ) ] = 8 𝑎 subt 1 2𝑎𝑏𝑐 = 23 exp eq/eksp verg 1 =3 𝑎𝑏𝑐 1 answer/antwoord (4) ∴ 𝑎𝑏𝑐 = 3 [25] QUESTION/VRAAG 2 2.1 19; 8; −1; −8; . . . −11 − 9 − 7 − 5 2 2 2 2.1.1 𝑇5 = −13 −13 (1) 2.1.2 2𝑎 = 2; 𝑎 = 1 𝑎 = 1 3𝑎 + 𝑏 = −11 3(1) + 𝑏 = −11; 𝑏 = −14 𝑏 = −14 𝑎 + 𝑏 + 𝑐 = 19 1 − 14 + 𝑐 = 19; 𝑐 = 32  𝑐 = 32 ∴ 𝑇𝑛 = 𝑛2 − 14𝑛 + 32 answer/𝑎𝑛𝑡𝑤 (4) 2.1.3 𝑇26 = (26)2 − 14(26) + 32 subst = 344 𝑇25 = (25)2 − 14(25) + 32 subst = 307 ∴ 𝑇26 − 𝑇25 = 344 − 307 = 37 answer/antwoord OR/OF 𝑇𝑛 = 2𝑛 − 13  𝑇𝑛 𝑇25 = 2(25) − 13  𝑠𝑢𝑏𝑠𝑡 = 37  37 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 5 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne 2.1.4 𝑏 𝑛 = − 2𝑎 −14  method/metode = −( ) 2 =7 ∴ 𝑇7 = 72 − 14(7) + 32 subst into Tn = 53  53 (3) 2.2.1 3; 𝑏; 13; 18; . . 𝑏 − 3 = 13 − 𝑏  𝑇2 − 𝑇1 = 𝑇3 − 𝑇2 2𝑏 = 16 ∴ 𝑏 = 8 𝑏 = 8 (2) 2.2.2 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑 = 3 + (𝑛 − 1)(5) subst = 5𝑛 − 2 answer/antwoord (2) 2.2.3 𝑛 𝑆𝑛 = [2𝑎 + (𝑛 − 1)𝑑 ] 2 30 𝑆30 = [2(3) + 29(5)] subst 2 = 2 265 𝑆30 = 2 265 OR/OF 𝑇30 = 5(30) − 2 = 148 𝑛 𝑆𝑛 = (𝑎 + 𝑙 ) 𝑇30 = 148 2 30 = 2 (3 + 148) 𝑆30 = 2 265 (2) = 2 265 2.3 2𝑥 + 2𝑥+1 + 3.2𝑥 + 2𝑥+2 + . . . (15 𝑡𝑒𝑟𝑚𝑠) = 2𝑥 + 2.2𝑥 + 3.2𝑥 + 4.2𝑥 + . . . + 15.2𝑥 𝑎 = 2𝑥 ; 𝑑 = 2𝑥  𝑎 = 2𝑥 ; 𝑑 = 2𝑥 𝑛 𝑆𝑛 = [2𝑎 + (𝑛 − 1)𝑑 ] 2 15 𝑆15 = [2.2𝑥 + 14.2𝑥 ]  subst 2 = 15(8). 2𝑥 = 15.2𝑥+3 ∴ 𝑘 = 15; 𝑝 = 3  𝑘 = 15 𝑝 = 3 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 6 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne OR/OF 𝑛 𝑆𝑛 = (𝑎 + 𝑙 ) 2 15 𝑥 𝑆15 = (2 + 15.2𝑥 )  𝑎 = 2𝑥 ; 𝑙 = 15. 2𝑥 2 = 15.23 . 2𝑥  subst = 15.2𝑥+3 ∴ 𝑘 = 15; 𝑝 = 3  𝑘 = 15 𝑝 = 3 (4) [21] QUESTION/VRAAG 3 3.1.1 1 −4 + 2 − 1 + . . . + 32 1 𝑎 = −4; 𝑟 = −  𝑎 and 𝑟 /𝑎 en 𝑟 2 1 𝑛−1 ∴ 𝑇𝑛 = −4. (− )  𝑇𝑛 (2) 2 3.1.2 1 𝑛−1 𝑇𝑛 = −4. (− ) 2 1 1 𝑛−1 = −4. (− ) 32 2 1 1 𝑛−1 1 − = (− ) 𝑶𝑹 𝑛 − 1 = log 1 ( ) 128 2 2 128 1 7 1 𝑛−1 (− ) = (− ) 𝑛−1= 7 2 2 7=𝑛−1 𝑛=8 ∴ 𝑛=8 𝑛 = 8 8 1 𝑛−1 ∑ −4. (− ) 2  answer/antwoord (2) 𝑛=1 3.1.3 𝑎 𝑆∞ = 1−𝑟 −4 = subst 1 − (−0,5) 8 answer =− 3 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 7 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne 3.2.1 1 𝑛 𝑆𝑛 = 32 − 32 ( ) 2 1 5 𝑆5 = 32 − 32 ( ) 2  31 (1) = 31 3.2.2 1 𝑎 = 16; 𝑟 = 2 𝑎 (1 − 𝑟 𝑛 ) ∴ 𝑆𝑛 = 1−𝑟 1 𝑛 255 16 [1 − (2) ] subst = 1 8 1−2 255 1 𝑛 =1−( ) 256 2 1 𝑛 1 ( ) = 1 𝑛 1 2 256 ( ) = 2 256 2−𝑛 = 2−8 ∴ 𝑛=8  8 terms/terme (3) 3.2.3 𝑆5−𝑛 − 𝑆5+𝑛 1 5−𝑛 1 5+𝑛 = [32 − 32 ( ) ] − [32 − 32 ( ) ] 2 2  subst = 32 − 25 . 2−5+𝑛 − 32 + 25 . 2−5−𝑛 = −2𝑛 + 2−𝑛  −2𝑛 + 2−𝑛 1 = −𝑝 + 𝑶𝑹 𝑝  answer/antwoord 2 −𝑝 + 1 = 𝑝 (3) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 8 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne QUESTION/VRAAG 4 4.1 𝑥 = −3   each equation / elke vgl 𝑦 = −2 (2) 4.2 −6 ℎ (𝑥 ) = − 2 𝑥 + 3  (−6; 0)  (0; −4)  asymptotes/asymptote  shape/𝑣𝑜𝑟𝑚 (4) 4.3 𝑦=𝑥+𝑐  subst (-3; -2) ∴ −2 = −3 + 𝑐 ∴ 𝑐= 1 ∴𝑦=𝑥+1  equation/vergelyking (2) 4.4 𝑦 ≠ −2; 𝑦 ∈ ℝ  𝑦 ≠ −3  𝑦∈ℝ (2) 4.5 −6 ≤ 𝑥 < −3  interval  notation (2) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 9 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne QUESTION/VRAAG 5 𝑔 (𝑥 ) = 𝑎 𝑥 5.1 B(0; 1) (0; 1) (2) 5.2 𝑔 (𝑥 ) = 𝑎 𝑥 27 27 = 𝑎−3  subst (−3; 8 ) 8 27 1 = 𝑎3 8 8 𝑎3 = 27 2 2 𝑎= 3 ∴ 𝑎=3 (2) 5.3 2 −𝑥 2 −𝑥 ℎ(𝑥) = 𝑔(−𝑥) = ( )  (3) 3 OR 3 𝑥 3 𝑥 ∴ ℎ(𝑥) = (2)  (2) (2) 5.4 𝑦 = log 3 𝑥   answer/antwoord 2 (2) 3 5.5 0<𝑥≤2  interval  notation/notasie (2) 5.6 𝑥 > 3; 𝑥 ∈ ℝ  (2) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 10 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne QUESTION/VRAAG 6 𝑓 (𝑥 ) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 and/𝑒𝑛 𝑔(𝑥) = 𝑚𝑥 + 𝑞 6.1 𝑐 = −2;  𝑐 = −2 𝑚 = tan 135° = −1  𝑚 = −1 ∴ 𝑦 = −𝑥 − 2 (2) 6.2 𝑥 − intercept 𝑜𝑓 𝑔: 𝑦 = 0 𝑦=0 𝑥 = −2 ∴ 𝐵(−2; 0)  (−2; 0) (2) 6.3 𝑓: 𝑦 = 𝑎(𝑥 − 𝑥1 )(𝑥 − 𝑥2 ) 1 𝑦 = 𝑎(𝑥 − (−2))(𝑥 − 2)  subst roots/wortels 3 3 1 3 4 = 𝑎 (− + 2) (− − )  subst (− 2 ; 4) 2 2 2 1 4= . (−2). 𝑎 2 ∴ 𝑎 = −4  𝑎 = −4 1 ∴ 𝑓(𝑥) = −4(𝑥 + 2) (𝑥 − ) 1 2  −4(𝑥 + 2) (𝑥 − 2) = −4𝑥 2 − 6𝑥 + 4 (4) 6.4 1 𝑏 −(−6) 3 −2 + 2 3 𝑥=− = = − 𝑶𝑹 𝑥= = − 3 2𝑎 2(−4) 4 2 4  𝑥 = −4 3 3 2 3 25 𝑓 (− ) = −4 (− ) − 6 (− ) + 4 = 4 4 4 4  subst 3 25 ∴ 𝐴 (− ; ) coordinate/koördinaat (3) 4 4 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 11 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne 6.5 𝑓(𝑥 ) = 𝑔(𝑥)  𝑓(𝑥 ) = 𝑔(𝑥) −4𝑥 2 − 6𝑥 + 4 = −𝑥 − 2  std form 4𝑥 2 + 5𝑥 − 6 = 0 (𝑥 + 2)(4𝑥 − 3) = 0 3 𝑥 = −2 𝑜𝑟⁄𝑜𝑓 𝑥 = 4  x values/waardes 3 NA 𝑦 = − (4) − 2 11 =− 4 3 11  coordinate/koördinaat ∴ 𝐻( ; − ) 4 4 (4) 6.6 3 𝑥<− 4  answer/antwoord (2) 6.7 25 4<𝑘< 4  values/waardes  notation/notasie (2) 6.8 3 9  x-value/waarde 𝐷( ; − ) 2 4  y-value/waarde (2) [21] QUESTION/VRAAG 7 7.1 𝑓(𝑥 ) = 𝑥 2 + 4𝑥 + 1 𝑓 (𝑥 + ℎ ) = (𝑥 + ℎ )2 + 4(𝑥 + ℎ ) + 1 = 𝑥 2 + 2𝑥ℎ + ℎ2 + 4𝑥 + 4ℎ + 1 answer of 𝑓(𝑥 + ℎ) /antwoord van 𝑓(𝑥 + ℎ) 𝑓 (𝑥 + ℎ) − 𝑓(𝑥) 𝑓 ′ (𝑥 ) = lim ℎ→0 ℎ 𝑥 2+2𝑥ℎ+ℎ 2 +4𝑥+4ℎ+1−(𝑥 2+4𝑥+1) = lim  subst ℎ→0 ℎ 𝑥 2+ 2𝑥ℎ+ ℎ 2+ 4𝑥 + 4ℎ + 1− 𝑥 2− 4𝑥 − 1  simplify/ vereenv = lim ℎ→0 ℎ ℎ(2𝑥 + ℎ + 4)  common factor/gemene = lim ℎ→0 ℎ fakt = lim (2𝑥 + ℎ + 4) ℎ→0  2𝑥 + 4 (5) = 2𝑥 + 4 7.2.1 𝑓(𝑥 ) = −3𝑥 4 + 5𝑥 2  −12𝑥 3 𝑓 ′ (𝑥 ) = −12𝑥 3 + 10𝑥  10𝑥 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 12 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne 7.2.2 2 1 𝑦 = √𝑥 −  𝑥2 𝑥 1  2𝑥 −1 = 𝑥 2 − 2𝑥 −1 1 𝑑𝑦 1 − 1 1  2 𝑥− 2 = 𝑥 2 + 2𝑥 −2 𝑑𝑥 2 2𝑥 −2 (4) 3 7.3 ℎ(𝑥) = −𝑥 + 2𝑥 − 5 ℎ(−2) = −(−2)3 + 2(−2) − 5  subst = −1 ℎ′ (𝑥 ) = −3𝑥 2 + 2 ℎ′ (−2) = −3(−2)2 + 2  subst 𝑚 = −10  𝑚 = −10 ∴ 𝑦 − (−1) = −10(𝑥 − (−2)) ∴ 𝑦 = −10𝑥 − 21  equation/vergelyking (4) 7.4 𝑥>1 answer/antwoord (2) [17] QUESTION/VRAAG 8 𝑓 (𝑥) = 𝑥 3 − 8𝑥 2 + 5𝑥 + 14 8.1 (0; 14)  answer/antwoord (1) 8.2 𝑓(𝑥 ) = 𝑥 3 − 8𝑥 2 + 5𝑥 + 14  subst 𝑓(7) = 73 − 8(7)2 + 5(7) + 14 =0  𝑓 (7) = 0 (2) 8.3 𝑥 3 − 8𝑥 2 + 5𝑥 + 14  method/metode 𝑘𝑥 2 − 7𝑥 2 = −8𝑥 2  𝑥2 − 𝑥 − 2 = (𝑥 − 7)(𝑥 2 + 𝑘𝑥 − 2) 𝑘𝑥 2 = −𝑥 2 2 = (𝑥 − 7)(𝑥 − 𝑥 − 2)  factor/faktor (𝑥 − 2) 𝑘 = −1 = (𝑥 − 7)(𝑥 − 2)(𝑥 + 1)  factor/faktor (𝑥 + 1) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 13 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne 8.4 𝑓 ′ (𝑥 ) = 0 ∴ 3𝑥 2 − 16𝑥 + 5 = 0  𝑓 ′ (𝑥 ) = 0 ∴ (𝑥 − 5)(3𝑥 − 1) = 0  factors/faktore 1 ∴ 𝑥 = 5 𝑜𝑟│ 𝑜𝑓 𝑥 =  each x-value/elke x-w 3 ∴ 𝑓(5) = 53 − 8(5)2 + 5(5) + 14 (5; −36) = −36 1 1 3 1 2 1 ∴ 𝑓 ( ) = ( ) − 8 ( ) + 5 ( ) + 14 3 3 3 3 400 = 27 1 400 = 14,81 (3 ; 27 ) (6) 8.5 𝑓"(𝑥) = 0  𝑓"(𝑥) = 0 6𝑥 − 16 = 0 8 8 𝑥=3 𝑥= 3 8 8 3 8 2 8 𝑓 ( ) = ( ) − 8 ( ) + 5 ( ) + 14 3 3 3 3 286 =− 27 = −10,59 8 286 ∴( ; − )  coordinate/koördinaat (3) 3 27 [16] QUESTION/VRAAG 9  x-intercepts/afsnitte  turning points/draaipunte  shape/vorm  (2; -16) [4] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics/P1/Wiskunde V1 14 NW/June/Junie 2024 Grade/Graad 12 – Marking Guideline/Nasienriglyne QUESTION/VRAAG 10 10.1 ℎ(𝑡) = 18𝑡 − 4𝑡 2  subst ℎ(2) = 18(2) − 4(2)2 = 20 𝑚  20 m (2) 10.2 ℎ = 18𝑡 − 4𝑡 2 ℎ′ (𝑡) = 18 − 8𝑡  h’(t) ′( ℎ 1,5) = 18 − 8(1,5)  subst 1,5 = 6 m/s 6 (3) 10.3 ℎ ′ (𝑡 ) = 0 18 − 8𝑡 = 0  ℎ ′ (𝑡 ) = 0 ∴ 𝑡 = 2,25 𝑠  𝑡 = 2,25 𝑠 ℎ(2,25) = 18(2,25) − 4(2,25 )2  subst 2,25 = 20,25 𝑚  20,25 (4) [9] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou