Mathematics Grade 12 P2 Eastern Cape June 2022 Memo (Afr)

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2022 MATHEMATICS P2/WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 15 pages. Hierdie nasienriglyn bestaan uit 15 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2022) QUESTION/VRAAG 1 1.1 B ✓ answer/antwoord (1) 1.2 B ✓ answer/antwoord (1) 1.3 75% ✓ answer/antwoord (1) 1.4 Nothing. It remains the same. No change in standard deviation. ✓reason/rede Niks. Dit bly dieselfde. Geen verandering in standaardafwyking. (1) 1.5 75 − 30 Semi − IQR: IKV = ✓ answer/antwoord 2 Semi − IQR: IKV = 22,5 (1) [5] QUESTION/VRAAG 2 2.1 Time taken No. of pupils Cumulative frequency Tyd geneem Aantal leerlinge Kummulatiewe frekwensie 60 ≤ t ≤ 90 3 3 ✓ for values 90 ≤ t ≤ 120 6 9 vir waardes 120 ≤ t ≤ 150 7 16 150 ≤ t ≤ 180 8 24 180 ≤ t ≤ 210 6 30 (1) 2.2 ✓ anchor point/ ankerpunt (60;0) C ✓ (120; 9) ✓ (150; 16) A ✓ (210; 30) B (4) 2.3 2.3.1 See diagram above / Sien diagram hierbo ✓A (1) 2.3.2 See diagram above / Sien diagram hierbo ✓B (1) 2.3.3 See diagram above / Sien diagram hierbo ✓C (1) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) MATHEMATICS P2/WISKUNDE V2 3 QUESTION/VRAAG 3 3.1 Median score / Mediaan telling = 2𝑥 ✓ answer/antwoord (1) 3.2 ∑𝑥 Mean/Gemiddelde = 𝑛 4(𝑥 + 3) + 3(2𝑥) + 2(𝑥 − 1) + 2(6) ✓ substitution/vervanging = ✓ simplification/vereenvoudiging 11 ✓ answer/antwoord 12𝑥 + 22 = 11 (3) 3.3 Use of a calculator where the four values are as follows: Gebruik van ’n sakrekenaar waar die vier waardes soos volg is: 8 ; 10 ; 4 and/en 6 ✓ four values/vier waardes 𝑠𝑑(  ) = √5 ✓ answer/antwoord (2) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2022) QUESTION/VRAAG 4 𝑦 𝐷 𝐴(−4; 5) 𝐸 𝐵(4; 1) 𝑥 𝐶(−1; −4) 4.1 5−1 ✓ subst. into gradient formula/ 𝑚𝐴𝐵 = verv. in gradiënt formule −4 − 4 4 = −8 1 ✓ 𝑚𝐴𝐵 = − 2 1 𝑚𝐴𝐵 = − 2 ∴ 𝑚𝐶𝐷 = 2 (2) 4.2 ∴ 𝑚𝐶𝐷 = 2 ✓ 𝑚𝐶𝐷 = 2 ̅̅̅̅ 𝐶𝐷 ∶ 𝑦 = 2𝑥 + 𝑐 ✓sub of point/verv. van punt −4 = 2(−1) + 𝑐 (−1; −4) −2 = 𝑐 ✓ −2 = 𝑐 ̅̅̅̅ 𝐶𝐷 ∶ 𝑦 = 2𝑥 − 2 ✓ equation of CD / vergelyking van CD (4) 4.3 1 ̅̅̅̅ 𝐴𝐵 ∶ 𝑦 = − 𝑥+𝑐 2 1 ✓ substitution of point (4; 1) 1 = − (4) + 𝑐 2 vervanging van punt (4;1) 3=𝑐 ✓ equation of ̅̅̅̅ 𝐴𝐵 1 vergelyking van ̅̅̅̅ 𝐴𝐵 𝑦 = − 𝑥+3 2 ✓ equating of AB and CD 1 gelykstel van AB en CD − 𝑥 + 3 = 2𝑥 − 2 2 −𝑥 + 6 = 4𝑥 − 4 ✓ x –value/waarde 5𝑥 = 10 ✓ substitution of/vervanging van 𝑥=2 𝑥 − value/waarde 𝑦 = 2(2) − 2 ✓ 𝑦 − value/waarde 𝑦=2 𝐸(2; 2) (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) MATHEMATICS P2/WISKUNDE V2 5 4.4 𝐷(𝑥; 𝑦) 𝑥−1 𝑦−4 =2 =2 2 2 𝑥−1=4 𝑦−4=4 ✓𝑥=5 𝑥=5 𝑦=8 ✓𝑦=8 (2) 𝐷(5; 8) 4.5 5+4 𝑚𝐴𝐶 = ✓ substitution into gradient −4 + 1 𝑚𝐴𝐶 = −3 formula/vervanging in Line parallel to AC has same gradient. gradiënt formule Lyn ewewydig aan AC het dieselfde gradient. ✓ 𝑚𝐴𝐶 = −3 𝑦 = −3𝑥 + 𝑐 8 = −3(5) + 𝑐 ✓ 𝑐 = 23 𝑐 = 23 𝑦 = −3𝑥 + 23 ✓ equation of line/vergelyking van lyn (4) 4.6 𝑥 intercept of CD ∶ 𝑥 𝑎𝑓𝑠𝑛𝑖𝑡 𝑣𝑎𝑛 CD 2𝑥 − 2 = 0 𝑥=1 ✓𝑥=1 𝑚𝐵𝐶 = 1 ✓ 𝑚𝐵𝐶 = 1 Equation of Altitude/Vergelyking van hoogtelyn 𝑦 = −𝑥 + 𝑐 ✓ gradient of altitude −1 5 = −(−4) + 𝑐 gradiënt van hoogtelyn −1 𝑐=1 ✓ substitution of point ∴ 𝑦 = −𝑥 + 1 vervanging van punt 𝑥 intercept of Altitude / x-afsnit van hoogtelyn ✓equation of altitude 𝑥=1 vergelyking van hoogtelyn ✓𝑥=1 𝑥 intercept of CD = 𝑥 intercept of altitude x-afsnit van CD = x-afsnit van hoogtelyn (6) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2022) QUESTION/VRAAG 5 5.1 𝑥 2 + 𝑦 2 = 4𝑦 − 2𝑥 + 44 𝑥 2 + 2𝑥 + 1 + 𝑦 2 − 4𝑦 + 4 = 44 + 1 + 4 ✓✓ completing the square (𝑥 + 1)2 + (𝑦 − 2)2 = 49 voltooiing van die vierkant ✓ factorizing / faktorisering ∴ 𝑀(−1; 2) ✓ 𝑀(−1; 2) (4) 5.2 𝑦−2 𝑚𝑀𝐴 = 𝑥+1 ✓ gradient of MA/gradiënt van MA 𝑦−2 ✓ equating it to 1/stel dit gelyk aan 1 =1 𝑥+1 ✓ making 𝑦 or 𝑥 the subject. 𝑦−2=𝑥+1 maak y of x die onderwerp 𝑦 =𝑥+3 ✓ equating the two linear functions. 𝑥 + 3 = −𝑥 + 5 gelykstel van twee lineêre funksies 2𝑥 = 2 ✓ coordinates of 𝐴(1; 4) 𝑥=1 koördinate van 𝐴(1; 4) ∴ 𝑦=4 𝐴(1; 4) (5) 5.3 (𝑥 + 1)2 + (𝑦 − 2)2 = 𝑟 2 (1 + 1)2 + (4 − 2)2 = 𝑟 2 ✓ substitution of A/vervanging van A 8=𝑟 2 ✓ 8 = 𝑟2 (𝑥 + 1)2 + (𝑦 − 2)2 = 8 ✓ equation of the circle/vergelyking van . die sirkel (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) MATHEMATICS P2/WISKUNDE V2 7 5.4 𝐾(5; 0) ✓ 𝐾(5; 0) (1) 5.5 𝐴𝐾 = √32 ✓ 𝐴𝐾 = √32 1 Area of/van ∆ 𝐴𝑀𝐾 = 2 𝐴𝐾 × 𝐴𝑀 ✓ 𝐴𝑀 = √8 1 Area of/van ∆ 𝐴𝑀𝐾 = 2 √32 × √8 Area of/van ∆ 𝐴𝑀𝐾 = 8 units 2 /𝑒𝑒𝑛ℎ𝑒𝑑𝑒 2 ✓ 8 units 2 /𝑒𝑒𝑛ℎ𝑒𝑑𝑒 2 (3) [16] QUESTION/VRAAG 6 6.1 𝑦 𝑝 √𝑝2 − 1 26° 1 𝑥 6.1.1 √𝑝2 − 1 sin 26° = ✓ sketch in Quadrant 1 𝑝 skets in kwadrant 1 ✓ √𝑝2 − 1 (3) ✓ answer/antwoord 6.1.2 cos 52° = cos 2(26° ) ✓cos 2(26° ) = 2 𝑐𝑜𝑠 2 26° − 1 ✓2 𝑐𝑜𝑠 2 26° − 1 1 2 = 2( ) − 1 𝑝 2 = 2−1 ✓ answer/antwoord 𝑝 (3) 2 ° 6.1.3 𝑡𝑎𝑛 64 × (𝑝 + 1) 2 1 =( ) × (𝑝 + 1) 2 1 √𝑝2 − 1 ✓ ( ) √𝑝2 −1 1 = 2 × (𝑝 + 1) 𝑝 −1 ✓ 1 1 𝑝2 −1 = × (𝑝 + 1) ✓ (𝑝 − 1)(𝑝 + 1) (𝑝 − 1)(𝑝 + 1) 1 = ✓ answer/antwoord 𝑝−1 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2022) 6.2 sin(−β) + sin(360° − β) sin(180o − β) + sin 180° − sin 𝛽 + (− sin 𝛽) ✓ − sin 𝛽 = ✓ − sin 𝛽 sin 𝛽 + 0 −2 sin 𝛽 ✓ sin 𝛽 = ✓ simplification sin 𝛽 vereenvoudiging = −2 ✓ answer/antwoord (5) 6.3 𝜃 2𝑝 tan ( ) = sin(2𝜃) 2 82° ✓ substitution/vervanging 2𝑝 tan ( ) = sin(2 × 82° ) 2 sin 162° ✓ simplification/vereenvoudiging 𝑝= ✓ answer/antwoord (3) 2 tan 41° 𝑝 = 0,16 6.4 4 𝑠𝑖𝑛 𝜃 . 𝑐𝑜𝑠 3 𝜃 − 4 𝑐𝑜𝑠 𝜃. 𝑠𝑖𝑛3 𝜃 = 𝑠𝑖𝑛 4𝜃 ✓ common factor/gemene faktor 𝐿𝐻𝑆/𝐿𝐾 = 4 𝑠𝑖𝑛 𝜃 . 𝑐𝑜𝑠 3 𝜃 ✓ 2 × 2 sin 𝜃 cos 𝜃 − 4 𝑐𝑜𝑠 𝜃. 𝑠𝑖𝑛3 𝜃 ✓ (cos 2𝜃) = 4 sin 𝜃 . cos 𝜃 (𝑐𝑜𝑠 2 𝜃 − 𝑠𝑖𝑛2 𝜃) ✓sin 2 𝜃. = 2 × 2 sin 𝜃 cos 𝜃 (cos 2𝜃) ✓ 2. sin 2 𝜃. cos 2𝜃 = 2. sin 2 𝜃. cos 2𝜃 ✓ answer/antwoord = sin 4𝜃 = 𝑅𝐻𝑆/𝑅𝐾 (6) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) MATHEMATICS P2/WISKUNDE V2 9 QUESTION/VRAAG 7 7.1 cos 3𝑥 = sin(𝑥 − 30° ) cos 3𝑥 = cos[90° − (𝑥 − 30° )] ✓ Co- ratio cos 3𝑥 = cos[120 − 𝑥] Ko-verhoud. 3𝑥 = 120 − 𝑥 (ref angle) ✓ ref angle verwys.  3𝑥 = 120° − 𝑥 + 𝑘. 360° OR/OF 3𝑥 = 360° − [120° − ✓ quadrant 1 𝑥]𝑘. 360° kwadrant 1 4𝑥 = 120° + 𝑘. 360° 2𝑥 = 240° + 𝑘. 360° ✓ quadrant 4 ° 𝑥 = 30 + 𝑘. 90 ° 𝑥 = 120° + 𝑘. 180° kwadrant 4 ✓ 30° 𝑥 = 30° ; −60° ; 120° ✓ −60° ✓ 120° (7) 7.2 𝑔 𝑓 ✓ Shape of 𝑓. ✓ 𝑥 intercepts ✓ start and end points Vorm van f x afsnitte begin en eindpunte ✓ Shape of 𝑔. ✓ 𝑥 intercepts ✓ start and end points Vorm van g x afsnitte begin en eindpunte (6) 7.3 −30° < 𝑥 < 30° ° ✓✓ −30 < 𝑥 < 30 ° 𝑶𝑹/𝑶𝑭 30° < 𝑥 < 90° ✓ 30° < 𝑥 < 90° 𝑶𝑹/𝑶𝑭 150° < 𝑥 < 180° ✓ 150° < 𝑥 < 180° (4) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2022) QUESTION/VRAAG 8 P Q R 𝑑 S 1 2 𝛽 𝛼 𝛼 12 3 T 8.1 𝑄𝑇̂𝑅 = 𝛽 − 𝛼 ✓ answer/antwoord (1) 8.2 𝑆̂ = 90° − 𝛽 ✓ answer/antwoord (1) 8.3 𝑆̂ = 90° − 𝛼 ✓ answer/antwoord (1) 8.4 In ∆ 𝑅𝑆𝑇 𝑑 𝑅𝑇 ✓ use of sine rule = gebruik van sinusreël sin 𝛼 sin(90° − 𝛽) 𝑑 cos 𝛽 ✓ use of co-function 𝑅𝑇 = gebruik van ko-funksie sin 𝛼 ✓ answer/antwoord (3) 8.5 𝑃𝑅 𝑅𝑇 = sin 𝛽 sin(90° − 𝛼) ✓ use of sine rule 𝑅𝑇 sin 𝛽 gebruik van sinusreël 𝑃𝑅 = ✓ sub of RT cos 𝛼 𝑑 cos 𝛽 sin 𝛽 vervanging van RT 𝑃𝑅 = ✓ answer/antwoord (3) sin 𝛼 cos 𝛼 [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) MATHEMATICS P2/WISKUNDE V2 11 QUESTION/VRAAG 9 9.1 centre / middelpuntshoek ✓ answer/antwoord (1) A 𝑥 O • 𝑥 3 B C ̂ C = 2𝑥 9.2 BO [angle at centre = 2 × angle at circum] ✓ statement / stelling (S) [Middelpuntshoek = 2  Omtrekshoek] ✓ reason/rede (R) x ✓statement / stelling (S) BĈO = 3 [angles opp = sides ; OB = OC] ✓reason/rede (R) [hoeke teenoor = sye ; OB = OC] x x ✓ S/R ∴ 3 + 3 + 2𝑥 = 180° [sum of angles of ∆] [som van hoeke van  ] 8𝑥 = 540° 𝑥 = 67,5° ✓ answer/antwoord (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2022) 9.3 A 2 1 M O 1 .K B 2 2 3 1 C 9.3.1 ̂ A1 = 30° BÔ C = 60° [angle at centre = 2 × angle at circum] ✓ S/R [middelpuntshoek = 2 x omtrekshoek] ✓ S/R ̂ ̂ ̂ B2 = C1 + C2 [angles opp = sides ; OB = OC] ✓ S/R [hoeke teenoor = sye ; OB = OC] ✓S ̂ 2 = 180 − 60° [sum of angles of ∆] ∴ 2B ° [som van hoeke van  ] ̂ B2 = 60° ̂1 = 60° [BO bisects AB ̂ C] / [BO halveer AB ̂ C] ✓ S/R B (5) 9.3.2 ̂ A1 = 30 ° ̂ ̂2 + M B1 + B ̂ = 180° [opp angles of cyclic quad] [teenoorst. hoeke van koordevierhoek] ✓ S/R ̂ = 180° − 120° M ✓S ̂ = 60° M ✓ answer/ ̂ =2×̂ ∴ M A1 antwoord (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) MATHEMATICS P2/WISKUNDE V2 13 QUESTION/VRAAG 10 R 1 2 3 X 1 2 P 1 2 2 Y 1 3 Q S 10.1 R1 = ̂ ̂ R2 [YR bisects PR ̂ Q] / [YR halveer PR ̂ Q] ✓ S/R R2 + ̂ ̂ R3 = ̂ Y2 [angles opp = sides; RS = YS] ✓S ✓R [hoeke teenoor = sye ; RS = YS] ✓S ✓R R1 + ̂ ̂ P=̂ Y2 [ext angle of ∆]/[buitehoek van  ] ✓S ̂ ∴ R3 = P̂ ∴ SR = tangent/raaklyn [converse tan − chord theorem] [omgekeerde raaklyn-koord stelling] (6) 10.2 Q2 = ̂ ̂ R3 [alt angles = ; SR ∥ QX] ✓ S/R [verw. hoeke = ; SR || QX] ̂ P=̂ R3 [proven] / [bewys] ✓S Q2 = ̂ ̂ P ∴ QR = tangent [converse tan − chord theorem] QR = raaklyn [omgekeerde raaklyn-koord stelling] ✓S (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 14 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2022) QUESTION/VRAAG 11 A F ℎ 𝑘 G D ≫ E B ≫ C 11.1 Proof / Bewys: Construct Perpendicular heights DG ( 𝑘) and EF (ℎ) in ∆ 𝐴𝐷𝐸. Join ✓ constr. BE and DC. konstruk. Teken loodregte hoogtes DG (k) en EF (h) in ∆ 𝐴𝐷𝐸 Verbind BE en DC ✓ S/R 1 Area of ∆ADE 2 AD. h AD = = Area of ∆BDE 1 BD. h BD 2 1 ✓S/R Area of ∆ADE 2 AE. k AE = = ✓S ✓ R Area of ∆CED 1 CE. k CE 2 But/Maar Area ∆BDE = Area ∆CED (same base DE, same height) ✓S (dieselfde basis DE, dieselfde hoogte) DE ∥ BC Area of ∆ADE Area of ∆ADE = Area of ∆BDE Area of ∆CED AD AE ∴ = BD CE (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) MATHEMATICS P2/WISKUNDE V2 15 11.2 K 1 2 Y1 2 O 1 ≫ 1 P 2 4 3 2 1 2 M L ≫ X 11.2.1 In ∆KOP and/𝑒𝑛 ∆KLM ̂=K K ̂ [common] / [gemeen] ✓ S/R ̂ ̂ O1 = L [corresponding angles = ; OP ∥ LM] [ooreenkomstige hoeke = ; OP ∥ LM] ✓S/R ̂ ̂ P1 = M [corresponding angles =, OP ∥ LM] [ooreenkomstige hoeke = ; OP ∥ LM] ✓R ∴ ∆ KOP ||| ∆KLM [ A; A; A] (3) 11.2.2 KO OP = [similarity ∶ 𝑔𝑒𝑙𝑦𝑘𝑣𝑜𝑟𝑚𝑖𝑔] KL LM ✓ S/R KO KY ✓ S/R = [ line parallel to one side of ∆] KL KX [ lyn ewewydig aan een sy van  ] KY OP ∴ = KX LM (2) 11.2.3 Area of ∆ KOP = Area of Quad/𝑉𝑖𝑒𝑟ℎ𝑜𝑒𝑘 OLMP ∴ Area of ∆ KLM = 2 × Area of ∆ KOP ✓S 1 1 ✓S × LM × KX = 2 × × OP × KY 2 2 1 OP. KY OP KY = but/𝑚𝑎𝑎𝑟 = 2 LM. KX LM KX ✓S OP 2 1 = ✓S LM 2 2 OP 1 = ✓S LM √2 OP KO = [∆ KOP |||∆KLM] ✓ S/R LM KL KO 1 ∴ = (6) KL √2 [17] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief