Mathematics Grade 12 Paper 2 Western Cape Mock 2018 Memo

Downloaded from hlayiso.com OVERBERG EDUCATION DISTRICT/ ONDERWYSDISTRIK GEMEENSKAPLIKE VRAESTEL/ COMMON PAPER GRAAD/GRADE 12 WISKUNDE V2 / MATHEMATICS P2 MEMORANDUM SEPTEMBER 2018 PUNTE/MARKS: 150

Downloaded from hlayiso.com NOTE:  If a candidate answers a question TWICE, mark only the first one.  Consistent accuracy applies in ALL aspects of the marking memorandum. LET WEL:  Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, merk slegs die eerste poging.  Volgehoue akkuraatheid is deurgaans op ALLE aspekte van die memorandum van toepassing. QUESTION/ VRAAG 1 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS 1.1 Punte / Marks Frekwensie Kumulatiewe / Frequency Frekwensie/ Cumulative Frequency 0 < 𝑥 ≤ 10 2 2  first 3 values / eerste 10 < 𝑥 ≤ 20 9 11 3 waardes 20 < 𝑥 ≤ 30 21 32 30 < 𝑥 ≤ 40 32 64  second 3 values / tweede 40 < 𝑥 ≤ 50 19 83 3 waardes 50 < 𝑥 ≤ 60 7 90 (2) 1.2 100 90  upper boundaries / Cumulative frequency/ 80 boonste grense 70 60 50  cumulative frequency / Kumulatiewe frekwensie 40 kumulaiewe frekwensie 30 20 10 0  anchor / anker 0 10 20 30 40 50 60 Punte/Marks (3) 1.3 90 – 74 = 16  90 – 74  16 Answer only: full marks Accept from 14 to 18 / Aanvaar: vanaf 14 tot 18 (2) [7] 2

Downloaded from hlayiso.com QUESTION/VRAAG 2 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS 2.1.1 𝑎 = 122,67  𝑎 = 122,67 𝑏 = −10,74  𝑏 = −10,74 𝑦 = 122,67 − 10,74𝑥 𝑦 = 122,67 − 0,74𝑥 (3) 2.1.2 𝑦 = 122,67 − 10,74(2,5)  substitution / substitusie = 95,82  answer / = 96 antwoord (2) 2.1.3 Dalvon. The gradient is steeper.  Dalvon Dalvon. Die gradient is steiler.  reason / rede (2) 2.2.1 1,50 kg since the standard deviation is 0  1,50 kg  reason / rede (2) 2.2.2 Andrew: 1,06 × 11 = 11,66 kg  11,66 kg Bongi: 0,77 × 16 = 12,32 kg  12,32 kg ∴ Bongi won the award  Bongi (3) [12] 3

Downloaded from hlayiso.com QUESTION/ VRAAG 3 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS 3.1.1 K(−1+5 ; −3−5)  substitution / substitusie 2 2  answer / antwoord (2) ∴ K (2; −4) 3.1.2 −1+𝑝 = 1 and −3+𝑞 −1+𝑝 =3  =1 2 2 2 −3+𝑞  =3 ∴𝑝=3 and 𝑞 = 9 2 (2) 3.1.3 9+5 𝑚AB = = −7 3−5 tan𝛼 = −7  tan𝛽 = −7 ∴ 𝛼 = 98,13°  𝛽 = 98,13° CA 3.1.2 (2) 3.1.4 𝛽 −3+5 𝑚BC = −1−5  substitution / substitusie −1 = 3 1 1 tan𝛽 = −  tan𝛽 = − 3 3 ∴ 𝛽 = 180° − 18,43° = 161,57°  𝛽 = 161,57° ̂ C = 63,44° ∴ AB ̂ C = 63,44°  AB CA 3.1.3 (4) 4

Downloaded from hlayiso.com SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Marks 3.1.5 KL = √(1 − 2)2 + (3 + 4)2  correct substitution / korrekte substitusie KL= √50 or 5√2  𝑎𝑛𝑠𝑤𝑒𝑟 CA 3.1.1 (2) 3.1.6 AB = 2√50 or 10√2  AB = 2√50 or 10√2 midpoint theorem / middelpuntstelling  midpoint theorem / Middelpuntstelling CA 3.1.5 (2) 3.2 1 1 Area = × 5 × h = 10  2 × 5 × h = 10 2 ∴ℎ=4 ℎ=4 ∴ 𝑦P = - 4  𝑦𝑃 = - 4  PA = PB; equating distance √(𝑥 − 5)2 + (𝑦 − 0)2 = √(𝑥 − 7)2 + (𝑦 − 6)2 formules/ gelyk stel van 𝑥 2 − 10𝑥 + 25 + 𝑦 2 = 𝑥 2 − 14𝑥 + 49 + 𝑦 2 − 12 𝑦 + 36 afstand formules 4𝑥 + 12𝑦 = 60  expansion / uitbreiding 𝑥 + 3𝑦 = 15  simplify / vereenvoudig 𝑥 + 3(−4) = 15 𝑥 = 27  𝑥 = 27 P(27; −4) (7) [21] 5

Downloaded from hlayiso.com QUESTION /VRAAG 4 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS 4.1.1 C(0; −4) −4−0 −4−0 𝑚AC = 0+2 = −2  𝑚AC = 0+2 = −2 1 𝑚AC × 𝑚BC = −2 × 2 =  1  𝑚AC × 𝑚BC =  1 ∴ 𝐵𝐶 ⊥ 𝐴𝐶 (2) 4.1.2 AB is a diameter / (converse of angle in semi-circle)  AB is a diameter / AB is ‘n middellyn (omgekeerde van hoek in halfsirkel) AB is ‘n middellyn 1 0= 𝑥−4 2 𝑥=8 𝐵(8; 0) 𝑥=8 ∴ center of circle (3; 0) ; 𝑟 = 5 ∴ (𝑥 − 3)2 + 𝑦 2 = 25  (𝑥 − 3)2 + 𝑦 2  25 (4) 4.1.3 0+4 4  𝑚radius = 4 𝑚radius = = 3 3−0 3 3 3 ∴ 𝑚𝑡angent = − 𝑚tangent = − 4 4 3 Equation of tangent: 𝑦 = − 4 𝑥 − 4 3  𝑦 = −4𝑥 − 4 (3) 4.2 𝑥 2 + 8𝑥 + (4)2 + 𝑦 2 + 10y +(5)2 = −𝑘 + (4)2 +(5)2  completing square / ∴ (𝑥 + 4)2 + (𝑦 + 5)2 = −𝑘 + 41 vierkantsvoltooiing −𝑘 + 41 = 49  equation ito k / vergelyking i.t.v k 𝑘 = −8  −𝑘 + 41 = 49  𝑘 = −8 (5) 4.3.1  circle / sirkel  intersecting all 4 quadrants / sny al 4 kwadrante . Q (-1;2) (2) 4.3.2 (𝑥 + 1)2 + (𝑦 − 2)2 = 9  (𝑥 + 1)2 + (𝑦 − 2)2 9 (2) [18] 6

Downloaded from hlayiso.com QUESTION /VRAAG 5 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS 5.1.1 sin(900 − 𝜃)  length of OP = √41 = cos 𝜃  𝑐𝑜𝑠 𝜃 5 = 41  5 √ √41 (3) 5.1.2 4 In ∆OPR ; tan 𝜃 = 5 OR 5 5 tan(180° − 𝜃) = In ∆OQT ; tan 𝜃 = 𝑎 𝑎 5 −tan𝜃 = 4 5 𝑎 ∴ = 4 5 4 5 5 𝑎 − =  equate 5 = 𝑎 5 𝑎 4𝑎 = 25 ∴ 4𝑎 = −25 25 1 25 1 𝑎 = − = −6 𝑎=− = −6  answer / antwoord 4 4 4 4 (2) 5.2.1 cos( 𝜃−180°). cos (90°+𝜃) 2 tan (180− 𝜃)  − cos 𝜃 −cos 𝜃.−sin 𝜃 =  −sin 𝜃 −2 tan 𝜃 sin 𝜃 cos 𝜃  −2 tan 𝜃 = −2 𝑡an 𝜃 sin 𝜃 cos 𝜃 cos 𝜃 = × 𝑠𝑖𝑛𝜃 cos 𝜃 −2 sin 𝜃  × 𝑐𝑜𝑠𝜃 𝒐𝒓 ÷ sin 𝜃 cos2 𝜃 =  answer / antwoord −2 (5) 5.2.2 tan𝜃 = 0 or / of tan𝜃 undefined  𝜃 = 0° ∴ 𝜃 = 0° or / of 𝜃 = 180° or / of 𝜃 = 90°  𝜃 = 180°  𝜃 = 90° (3) 7

Downloaded from hlayiso.com SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS 5.3 sin2𝜃 LHS = sin𝜃  2cos 𝜃 sin𝜃 OR 2sin𝜃. cos𝜃  2cos 𝜃 = 4cos2 𝜃−[(2cos²𝜃−1)+1] sin𝜃 RHS = 𝑐𝑜𝑠𝜃 = 2cos𝜃 4cos2 𝜃 − 2cos2 𝜃 + 1 − 1 = cos2𝜃+1 cos𝜃 RHS = 4cos 𝜃 − 𝑐𝑜𝑠𝜃 4cos2 𝜃 − 2cos²𝜃 2 =  2cos2 𝜃 − 1 2cos 𝜃 − 1 + 1 cos𝜃 = 4 cos 𝜃 − cos𝜃  2cos 𝜃 2cos2 𝜃 = 4cos𝜃 − 2cos𝜃 =2cos𝜃 = = 2cos𝜃 cos𝜃 ∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆 (4) 5.4.1 5 sin(A − B) = 3 sin(A + B) 5 (sinA cosB – cosA sinB) = 3(sinA cosB + cosA sinB)  expand sin(A − B) 5 sinA cosB − 5cosA sin𝐵 = 3sinA cosB + 3cosA sinB  expand sin(A + B) 2 sinA cosB = 8 cosA sinB  simplify / vereenvoudig ∴ sinA cosB = 4 cosA sinB (3) 5.4.2 sin A cos B = 4 cos A sin B sin A = 4 cos A tan B  sin A = 4 cos A tan B tan A = 4 tan B  tan A = 4 tan B 1 1 but tan B = 2 ∴ tan A = 4 ( 2 ) = 2  2 (3) [23] 8

Downloaded from hlayiso.com QUESTION /VRAAG 6 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS 6.1 𝑎 = 2  2 (1) 6.2  shape / vorm Mark f only / Merk slegs vir f  turning points / draaipunte  𝑥 & 𝑦 intercepts / x & y afsnitte (30°; 2) (0°; 1,732) 𝑓 (−150°; −2) (3) 0) 6.3 2 cos(𝑥 − 30 = 2sin 𝑥 cos(𝑥 − 300 ) = sin 𝑥 sin [90 − (𝑥 − 300 )] = sin 𝑥  co-function 0 (− 𝑥 + 120 ) = 𝑥  𝑥 = (− 𝑥 + 1200 ) 1200 – 𝑥 = 𝑥 𝑜𝑓/𝑜𝑟 1200 – 𝑥 = 1800 – 𝑥 2𝑥 = 1200 + 𝑘3600  2𝑥 = 1200 + 𝑘3600 or/of 𝑥 = 600 + 𝑘1800 𝑥 = 600 + 𝑘1800  𝑥 = −1200 , 600 ∴ 𝑥 = −1200 ; 600 OR OR cos(𝑥 − 300 ) = sin 𝑥 cos (𝑥 − 300 ) = cos(90° − 𝑥)  co-function 𝑥 − 30° = 90° − 𝑥  𝑥 − 30° = 90° − 𝑥 2𝑥 = 1200 + 𝑘3600  2𝑥 = 1200 + 𝑘3600 𝑥 = 600 + 𝑘1800 or/of 𝑥 = 600 + 𝑘1800 ∴ 𝑥 = −1200 , 600  𝑥 = −1200 , 600 (5) 6.4 k=3  3 (1) 6.5 𝑓 shifted 600 to the right  600  to the right (2) [12] 9

Downloaded from hlayiso.com QUESTION /VRAAG 7 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS 45 45 7.1 AC: sin 430 =  sin 430 = AC 𝐴𝐶 AC 𝑠𝑖𝑛 430 = 45 45 AC = sin 430 AC = 65,98 m  Answer 45 45 AD: sin 500 =  sin 500 = AD 𝐴𝐷 AD sin 500 = 45 45 AD = sin 500 AD = 58,74 m  Answer (4) 7.2 CD2 = AC2 + AD2 – 2(AC)(AD)cos 𝐴 using cos rule / gebruik van cos-reël CD2 = 65,982 + 58,742 – 2(65,98)(58,74)cos 690  substitution / substitusie CD = √5025,919618 CD = 70,89 𝑢𝑛𝑖𝑡𝑠 (3)  Answer [7] 10

Downloaded from hlayiso.com GEOMETRY/ MEETKUNDE S: Statement/ Bewering R: Reason/ Rede S/R: Both Statement and Reason/ Beide Bewering en Rede Allow for alternative methods/ Maak voorsiening vir alternatiewe metodes QUESTION /VRAAG 8 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS  8.1.1 CAB  x [∠s opp = sides /∠𝑒 𝑡𝑒𝑒𝑛𝑜𝑜𝑟 = 𝑠𝑦𝑒. ] S/R (1)  S 8.1.2 A O B  2 x [∠ at centre = 2∠ at circumf / midpts ∠ = 2 omtreks ∠] R (2)  180  2 x 8.2 B1  [sums AOB & ∠s opp = sides / R 2 𝑠𝑜𝑚 𝑒 𝐴𝑂𝐵 & ∠𝑒 𝑡𝑒𝑒𝑛𝑜𝑜𝑟 = 𝑠𝑦𝑒. ]  B1  90  x  90° − 𝑥  A M B  90 [sums AMB / 𝑠𝑜𝑚 𝑒 AMB ] R (3) 8.3 AM = 6 𝑢𝑛𝑖𝑡𝑠 [line from centre ⊥ chord/ 𝑙𝑦𝑛 𝑣𝑎𝑛𝑎𝑓 𝑚𝑖𝑑𝑝𝑡 ⊥ 𝑘𝑜𝑜𝑟𝑑] SR (2) [8] 11

Downloaded from hlayiso.com QUESTION /VRAAG 9 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS 9.1.1 tan-chord theorem/ raaklyn-koordstelling R (1) 9.1.2 angles opposite equal radii/sides / hoeke teenoor gelyke radii/ sye R (1) 9.2 AĈB = 90° [ in semi-circle/  in halwe sirkel]  S/R  AĈB = TO ̂A S CTOB is a cyclic quad/ is ’n koordevierhoek [ext  = opp int/ R buite  = teenoorst. binne ] OR/ OF AĈB = 90° [ in semi-circle/  in halwe sirkel]  S/R  AĈB = TO ̂B S CTOB is a cyclic quad/ is ’n koordevierhoek [opp s supplementary/ R teenoorst. e supplementêr] OR/ OF  S/R AĈB = 90° [ in semi-circle/  in halwe sirkel] OT̂A = 65° S ̂ C = 65° [s in ABC] and/ en OB CTOB is a cyclic quad/ is ’n koordevierhoek [opp s supplementary/ R teenoorst. e supplementêr] OR/ OF AĈB = 90° [ in semi-circle/  in halwe sirkel]  S/R OT̂C = 115° [ext  of /  buite van ] ̂ C = 65° [s in ABC] and/ en OB S CTOB is a cyclic quad/ is ’n koordevierhoek [ext  = opp int/ R buite  = teenoorst. binne ] ̂ 𝐊 as ext  / Kan ook 𝐂𝐁 Can also use 𝐂𝐁 ̂ 𝐊 as buite  gebruik (3) 9.3  SR COB = 50° [ at centre = 2 ×  at circumf./ middelpts  = 2 × omtr. ] OR [ exterior ∠ of ∆ / 𝑏𝑢𝑖𝑡𝑒 ∠ van ∆] (2)  9.4 K = 40° [s in ACK] S   S = 50° [s in SOK] S    S = COB S OK is a tangent / is ’n raaklyn [conv. tan-chord thm/ omg. raakl- koordst]  R (4) [11] 12

Downloaded from hlayiso.com QUESTION /VRAAG 10 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS 10.1 P k\ h S T Q R  Constr/ Konstr Constr/ Konstr: Draw SR and TQ/ Trek SR en TQ Proof/ Bewys: Area ∆PST  = Area ∆QST 1 Area ∆PST 12 PS×h PS 2 PS×h =1 = [equal altitudes ; 𝑔𝑒𝑙𝑦𝑘𝑒 ℎ𝑜𝑜𝑔𝑡𝑒𝑠] 1 SQ×h Area ∆QST 2 SQ×h SQ 2 Area ∆PST Area ∆PST 12 PT×k PT Area ∆STR =1 = [equal altitudes ; 𝑔𝑒𝑙𝑦𝑘𝑒 ℎ𝑜𝑜𝑔𝑡𝑒𝑠] PS Area ∆STR 2 TR×k TR = SQ Area QST = Area STR [ same base, height;/ dieselfde basis, hoogte; SR ST||QR] Area ∆PST Area ∆PST S ∴ = Area ∆QST Area ∆STR PS PT ∴ = (6) SQ TR 13

Downloaded from hlayiso.com SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS 10.2 10.2.1 Ĉ2 = x [alternate/ verw. s; EC || AD] S ̂ E1 = x [corresp./ ooreenk. s; EC||AD] S  any 1 correct R (3) 10.2.2 BD BA = AE [line || to 1 side of / lyn || aan 1 sy van ] OR/ OF SR DC [Proportional Theorem/ Eweredigheidstelling; EC||AD] BD BA  EA = AC ∴ DC = AC [EA = AC; sides opp. equal s/ sye teenoor gelyke e] R (4) [13] 14

Downloaded from hlayiso.com QUESTION /VRAAG 11 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ Marks BESKRYWERS 11.1 ̂ 1 = 𝑥 [tan-chord theorem/ raakl-koord stelling] D SR ̂ 3 = 𝑥 [alt. s; AB || ED / 𝑣𝑒𝑟𝑤𝑖𝑠. 𝑒 ; AB||ED] D SR ̂ 2 = 𝑥 [equal chords, equal angles / 𝑔𝑒𝑙𝑦𝑘𝑒 𝑘𝑜𝑜𝑟𝑑𝑒, 𝑔𝑒𝑙𝑦𝑘𝑒 ℎ𝑜𝑒𝑘𝑒.] D (6) SR 11.2 In 𝛥DEA 𝑎𝑛𝑑 𝛥DBC 1) ̂ D1 = ̂D3 = 𝑥 [ proven / re𝑒𝑑𝑠 𝑏𝑒𝑤𝑦𝑠] S ̂ 2) B2 = E [ext.  of cycl.quad / 𝑏𝑢𝑖𝑡𝑒ℎ𝑜𝑒𝑘 𝑣 𝑘𝑣ℎ. ] SR S&R ̂1 = A 3) C ̂1 [s of ∆ / e van ∆] ∴ 𝛥DEA ||| 𝛥DBC [,,] OR In 𝛥DEA 𝑎𝑛𝑑 𝛥DBC 1) ̂ D1 = ̂D3 = 𝑥 [ proven / re𝑒𝑑𝑠 𝑏𝑒𝑤𝑦𝑠] S ̂ 2) B2 = E [ext.  of cycl.quad / 𝑏𝑢𝑖𝑡𝑒ℎ𝑜𝑒𝑘 𝑣 𝑘𝑣ℎ. ] SR ∴ 𝛥DEA ||| 𝛥DBC [,,] S&R (4) 11.3 DE EA SR = [ 𝛥DEA ||| 𝛥DBC ] DB BC ∴ DE. BC = EA. DB AB = BD [equal sides opp equal s ; / gelyke sye teenoor gelyke e] SR AB = AE [given / gegee] ∴ BD = AE S ∴ DE. BC = EA. EA (5) 11.4 ̂ 1 + 3𝑥 A = 180° [opp s of cycl quad. / 𝑡𝑒𝑒𝑛𝑜𝑜𝑟𝑠𝑡. 𝑒 𝑣𝑎𝑛 𝑘𝑣ℎ] SR 75° + 3𝑥 = 180° 3𝑥 = 105°  Answer/ 𝑥 = 35° (3) Antw [18] Total/Totaal: 150 15