Mathematics Grade 10 P1 Gauteng November 2020 Memo (Afr)

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION PROVINSIALE EKSAMEN NOVEMBER 2020 GRADE/GRAAD 10 MATHEMATICS PAPER 1 WISKUNDE VRAESTEL 1 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 50 7 pages/bladsye 1

Downloaded from hlayiso.com MATHEMATICS/WISKUNDE MEMORANDUM (PAPER/VRAESTEL 1) GRADE/GRAAD 10 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 has not redone the question, mark the crossed out version.  Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the second calculation error.  Assuming answers/values in order to solve a problem is NOT acceptable. OPMERKING:  As ʼn kandidaat ʼn vraag twee maal beantwoord, merk slegs die EERSTE poging.  As ʼn kandidaat ʼn poging van ʼn vraag doodgetrek het, en die vraag nie oorgedoen het nie, merk die deurgehaalde weergawe.  Konsekwente akkuraatheid is van toepassing op ALLE aspekte van die nasienmemorandum. Hou op nasien by die tweede berekeningsfout.  Veronderstelling van antwoorde/waardes om ʼn probleem op te los, is NIE aanvaarbaar NIE. 2

Downloaded from hlayiso.com MATHEMATICS/WISKUNDE MEMORANDUM (PAPER/VRAESTEL 1) GRADE/GRAAD 10 QUESTION/VRAAG 1 1.1 𝑥 = −5  answer/antwoord (1) 1.2 𝑥 = 15  answer/antwoord (1) 1.3 𝑥 = 11  answer/antwoord (1) [3] QUESTION/VRAAG 2 2.1 𝑥2 + 𝑥 − 6 𝑥 2 − 16 1 2 × 3 2 × 3𝑥 − 12𝑥 𝑥 − 2𝑥 𝑥+4  (𝑥 + 3)(𝑥 − 2) (𝑥 + 3)(𝑥 − 2) (𝑥 + 4)(𝑥 − 4) 1 (𝑥 + 4)(𝑥 − 4) = × ×  3𝑥(𝑥 − 4) 3𝑥(𝑥 − 4) 𝑥 2 (𝑥 − 2) 𝑥+4  𝑥 2 (𝑥 − 2) 𝑥+3 =  answer/antwoord 3𝑥 3 (5) 2.2 62−𝑎 . 12𝑎 2𝑎+1 (2.3)2−𝑎 . (3. 22 )𝑎 = 2𝑎+1 22−𝑎 . 32−𝑎 . 3𝑎 . 22𝑎 = 2𝑎+1 = 22−𝑎+2𝑎−𝑎−1 . 32−𝑎+𝑎 22−𝑎+2𝑎−𝑎−1 32−𝑎+𝑎 = 21 . 32 21 . 32 = 18  answer/antwoord (4) 3

Downloaded from hlayiso.com MATHEMATICS/WISKUNDE MEMORANDUM (PAPER/VRAESTEL 1) GRADE/GRAAD 10 2.3 (𝑎 + 𝑏)2 =𝑎2 + 2𝑎𝑏 + 𝑏 2 a2 + 2ab + b2 = 𝑎2 + 𝑏 2 + 2𝑎𝑏  rearrangement /herrangskikking = 8𝑎𝑏 + 2𝑎𝑏 = 10𝑎𝑏 answer/antwoord (3) [12] QUESTION/VRAAG 3 3.1 1 64 p 3  27  1  4p    1  4 1  3 =  4 p   16 p 2  p    3  3 9  4 1  16 p 2  p    3 9 (2) 3.2 24ky – 6ky – 24y + 96y . 3 3 = 6ky (4y2 – 1) – 24y (1 – 4y2)  6ky (4y2 – 1) – 24y (1 – 4y2) = 6ky (4y2 – 1) + 24y (4y2 – 1)  + 24y (4y2 – 1) = (6ky + 24y) (4y2 – 1)  (6ky + 24y) (4y2 – 1) = 6y (k + 4) (2y + 1) (2y – 1)  6y (k + 4) (2y + 1) (2y – 1) (4) [6] 4

Downloaded from hlayiso.com MATHEMATICS/WISKUNDE MEMORANDUM (PAPER/VRAESTEL 1) GRADE/GRAAD 10 QUESTION/VRAAG 4 4.1 1 3𝑥+3 ( ) = 729 3 3−(3𝑥+3) = 36 3−(3𝑥+3) = 36 −3𝑥 − 3 = 6 −3𝑥 − 3 = 6 −3𝑥 = 6 + 3 ∴ 𝑥 = −3  𝑥 = −3 (3) 4.2 1−𝑥 −4 ≤ <3 3 −4 × 3 ≤ 1 − 𝑥 < 3 × 3 −12 ≤ 1 − 𝑥 < 9  −12 ≤ 1 − 𝑥 < 9 −12 − 1 ≤ −𝑥 < 9 − 1 −13 ≤ 𝑥 < 8  −13 ≤ 𝑥 < 8 ∴ −8 < 𝑥 ≤ 13 −8 < 𝑥 ≤ 13 𝑥𝜖(−8; 13] Interval notation/ Intervalnotasie (4) 4.3 number of children/aantal kinders: 𝑥 Adults/volwassenes : 150 − 𝑥 ∴ 6𝑥 + 10(150 − 𝑥) = 𝑅1100 6𝑥 + 10(150 − 𝑥) 6𝑥 + 1500 − 10𝑥 = 𝑅1100 = 𝑅1100 −4𝑥 = 1100 − 1500  simplifying/ ∴ 𝑥 = 100 vereenvoudiging ∴ 100 children/kinders answer/antwoord (4) [11] 5

Downloaded from hlayiso.com MATHEMATICS/WISKUNDE MEMORANDUM (PAPER/VRAESTEL 1) GRADE/GRAAD 10 QUESTION/VRAAG 5 5.1 𝑇𝑛 = 𝑑𝑛 + 𝑐 For/Vir 𝑛 = 1 −3 = 4(1) + 𝑐 ∴ 𝑐 = −7  𝑐 = −7 𝑇𝑛 = −3 + 4𝑛 − 4 ∴ 𝑇𝑛 = 4𝑛 − 7 4𝑛 − 7 OR/OF OR/OF 𝑇𝑛 = 4𝑛 − 7 answer only/slegs antwoord 4𝑛 −7 (2) 5.1.2 𝑇38 = 4(38) − 7 𝑇38 = 145  𝑇38 = 145 (1) [3] 6

Downloaded from hlayiso.com MATHEMATICS/WISKUNDE MEMORANDUM (PAPER/VRAESTEL 1) GRADE/GRAAD 10 QUESTION/VRAAG 6 6.1 𝑦 =𝑥+2 𝑦 = 1+2 𝐹𝐺 = 3 units/𝑒𝑒𝑛ℎ𝑒𝑑𝑒  𝐹𝐺 = 3 units/eenhede 𝑦 = 3𝑥 2 − 12 𝑦 = 3(1)2 − 12 𝐹𝐻 = 9 units/𝑒𝑒𝑛ℎ𝑒𝑑𝑒 𝐹𝐻 = 9 units/𝑒𝑒𝑛ℎ𝑒𝑑𝑒 𝐺𝐻 = 𝐹𝐺 + 𝐹𝐻 𝐺𝐻 = 12 units/𝑒𝑒𝑛ℎ𝑒𝑑𝑒  𝐺𝐻 = 12 units/𝑒𝑒𝑛ℎ𝑒𝑑𝑒 (3) 6.2.1 𝑥 + 2 = 3𝑥 2 − 12 2  𝑥 + 2 = 3𝑥 − 12 3𝑥 2 − 𝑥 − 14 = 0 standard form/ standaardvorm (3𝑥 − 7)(𝑥 + 2) = 0 factors/faktore 7 𝑥 = 3 and/𝑒𝑛 𝑥 = −2  both/beide 𝑦 =𝑥+2 7  substitute/vervang 3 7 𝑦 =3+2 13 = 3 7 13 7 13 𝐸=( ; ) (3 ; 3 ) 3 3 (5) 6.2.2 asymptote/asimptoot 𝑦 = 2 and/en  Asymptote/Asimptoot point/punt ( 1 ; 4) 𝑦=2 𝑦 = 𝑎𝑥 + 𝑞 4 = 𝑎1 + 2 4 = 𝑎1 + 2 4 − 2 = 𝑎1 ∴𝑎=2 𝑎 =2 ∴ 𝑦 = 2𝑥 + 2 (3) 6.2.3 𝑦 = −3𝑥 2 + 12  correct answer/korrekte antwoord (2) 6.3 𝑓(𝑥) = 3𝑥 2 − 12 𝑥 ≤ −2 or/of 𝑥 ≥ 2  𝑥 ≤ −2 𝑥 ≥2 (2) [15] TOTAL/TOTAAL 50 7