Mathematics Grade 12 Paper 2 Western Cape November 2017 Memo

Downloaded from hlayiso.com SEPTEMBER 2017 MATHEMATICS PAPER 2 / WISKUNDE VRAESTEL 2 MEMORANDUM 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. SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Marks QUESTION/ VRAAG 1 1.1 Percentage of Frequency Cumulative  frequency column/ income (x) Frequency Frekwensie kolom 20 < 𝑥 ≤ 30 10 10 30 < 𝑥 ≤ 40 15 25  first 3 points on 40 < 𝑥 ≤ 50 20 45 C. f column/ 1ste 3 punte op Kf 50 < 𝑥 ≤ 60 10 55 kolom 60 < 𝑥 ≤ 70 5 60  last 2 points on C. f column/ laaste 2 punte op Kf kolom (3) 1.2.1 60 − 45  45 = 15 𝑝𝑒𝑜𝑝𝑙𝑒 𝑠𝑝𝑒𝑛𝑡 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 50% 𝑜𝑛 𝑡ℎ𝑒𝑖𝑟 𝑖𝑛𝑐𝑜𝑚𝑒 𝑜𝑛 𝑒𝑑𝑢𝑐𝑎𝑡𝑖𝑜𝑛  Answer/ Antwoord (2) 1.2.2 𝑀𝑒𝑑𝑖𝑎𝑛 = ≈ 42 − 43 Answer/ Antwoord (2) 1.3 40 < 𝑥 ≤ 50  Answer/ Antwoord OR 𝑥 ∈ (40; 50] Accept/ Aanvaar: from 40 to 50 (1) / van 40 tot 50 [8] 1

Downloaded from hlayiso.com SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BE Marks SKRYWERS QUESTION/ VRAAG 2 2.1  Plotting 5 - 8 & points correct/ 2.3 Afsteek van 5 – 8 punte korrek  Plotting 9 -10 points correct/ Afsteek van 9 – 10 punte korrek 2.3 Regression line/ Regressielyn:  y -int:  29  Passing through  (30; 56) (2) (2) 2.2 𝑎 = 29,22 (29,21542 … ) 𝑎 𝑏 𝑏 = 0.89 equation / ∴ 𝑦 = 29,22 + 0,89𝑥 vergelyking (3) 2.3 On graph/ Op grafiek 2.4 r = 0,66 r Moderately positive relationship/ redelike (matige) positiewe  Moderately verwantskap positive/ Redelike positiewe (2) 2.5 559 559 Mean/ Gemiddelde, 𝑥̅ = 10 = 55,90  10  55,90 Standard deviation/ Standaardafwyking: σ = 11.36  σ = 11.36 (4) 2.6 44,54 67,26  Interval 55,90 Answer/Antw 6 scores lie within one σ/ 6 Punte val binne een σ (2) [15] 2

Downloaded from hlayiso.com SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Marks QUESTION/ VRAAG 3 3.1 4−0  Subst in gradient formula/ 𝑚 𝑇𝑄 = 0−3 Vervang in gradiënt formule −4 = 3  Answer/Antwoord (2) 3.2 RQ =√(10 − 3)2 + (7 − 0)2  Substitute/ Vervang = √49 + 49  Answer/Antwoord = √98 Units/ 𝑒𝑒𝑛ℎ𝑒𝑑𝑒 (2) 3.3 4−0 −4 𝑚 𝑇𝑄 = 0−3 = 3 or/of 𝑚 𝑇𝐹 = 𝑚 𝑇𝑄 𝑦= −4 𝑥+4  −8−4 = −4  eqn of/ vgl van TQ or gradients = 3 𝑘 3 −4  Subst/ vervang (k; -8) or gradient  −8 = (𝑘) + 4 −4𝑘 = 3(−12) 3 of TF −4 −12 = (𝑘)  Simplify/ Vereenv 𝒐𝒓 cross × 3 𝒌= 𝟗  Answer/Antwoord (4) 3.4 𝑺(𝟕: 𝟏𝟏)  𝑥= 7  𝑦 = 11 (2) 3.5 TŜR = TQ ̂R .opp  of ⃦ / ….teenoorst.  van ⃦ 𝑚 𝑚  TŜR = TQ ̂R −4 𝑚 𝑇𝑄 = 𝑚𝑅𝑄 = 1 3 −4  tan 𝛼 = 3 −4 tan 𝛼 = tan β = 1  tan β = 1 3 RA/ 𝑉𝐻 = 53,13° β = 45°  𝛼 = 126, 87° 𝛼 = 126,87°  β = 45° ∴ TŜR = 𝛼 – β  Answer/Antwoord (6) ∴ 𝐓𝐒̂𝐑 = 81, 87° 3.6 MQ = √(2)2 + (2)2 = √8  √8 But/Maar RQ = √98 … 𝑐𝑎𝑙𝑐. 𝑖𝑛 𝑞 3.2 √𝟖  √𝟖 √𝟗𝟖 ∴ 𝑹𝒂𝒕𝒊𝒐 = √𝟗𝟖 𝟐 =𝟕  Answer/ Antwoord (3) [19] 3

Downloaded from hlayiso.com SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Marks QUESTION /VRAAG 4 4.1 Radius to midpoint of chord/ radius na midpt van koord  Answer/ Antwoord (1) 4.2 3 𝑚𝑇𝑆 𝑚 𝑇𝑆 = −  −1 3  𝑚𝑃𝑁 = 𝑚 𝑇𝑆 × −1 … … TS  NP 𝑚𝑃𝑁  𝑚𝑃𝑁 = 1 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) OR/ OF 𝑦 = 𝑚𝑥 + 𝑐 Subst (-3; 8) ∴𝑦−8 =𝑥+3 8 = −3 + 𝑐 𝑐=5  Equation/ Vergelyking (5) ∴ 𝒚 = 𝒙 + 𝟏𝟏 4.3 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑃(0; 𝑦) 𝑦 = 0 + 11 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑃(0; 𝑦) 𝒚 = 𝟏𝟏  Answer/ Antwoord 𝑅𝑎𝑑𝑖𝑢𝑠: 11 − 5 = 6 𝑅𝑎𝑑𝑖𝑢𝑠  Answer/ Antwoord  𝒚 = −𝟏 (4) 4.4 Substitute M (𝑥: 0)/ 𝑆𝑡𝑒𝑙 𝑀(𝑥; 0)  Subst (x; 0) / vervang (x; 0) 0 = 𝑥 + 4  𝑥 = −11 𝑴 = (−𝟏𝟏; 𝟎)  Length of TO/ Lengte van TO  Length of MO/ Lengte van MO 𝑇𝑂 = 5 , 𝑀𝑂 = 11 𝑇𝑀2 = 52 + 112 …pyth  Answer/ Antwoord  𝑀𝑇 = √146 (4) 4.5 (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2 TM is the diameter of/ is die TM is the diameter/ is die middellyn ʘ TSM middellyn van ʘ TSM √146 146 𝑟 = 2  𝑟2 = 4 𝑟 = √146 2 Centre of / midpt ʘ TSM : −11 5 𝑥+𝑥1 𝑦+𝑦1  centre ( ; ) Midp. of TM = ( ; ) 2 2 2 2 −11+0 0+5 = ( ; ) 2 2 −11 5 𝟏𝟏 𝟓 𝟐 = ( ; 2)  (𝒙 + )𝟐 + (𝒚 − 𝟐) 2 𝟐 𝟏𝟏 𝟓 𝟐 146 73 146 73 ∴ (𝒙 + )𝟐 + (𝒚 − 𝟐) = 4 or 2 or 36,5  or 2 or 36,5 (5) 𝟐 4 [19] 4

Downloaded from hlayiso.com SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Marks QUESTION /VRAAG 5 5.1  figure 1 in correct quadrant; 3 1 𝑟 = 5 / figuur 1 in korrekte -4 A 𝐵 kwadrant; 𝑟 = 5  figure 2 in correct quadrant; -3 5 2√2 𝑥 = 2√2 / figuur 2 in korrekte kwadrant; 𝑥 = 2√2 2 −3 2√2  substitution / substitusie 5( )− 6( ) 5 3 8 = −3 − 6 (9) 25  Answer/ 𝐴𝑛𝑡𝑤𝑜𝑜𝑟𝑑 (5) =− 3 5.2.1 𝑐𝑜𝑠 (360 − 24°)  𝑐𝑜𝑠 24° = 𝑐𝑜𝑠 24° = p  Answer/ 𝐴𝑛𝑡𝑤𝑜𝑜𝑟𝑑 (2) 5.2.2 𝑐𝑜𝑠 2(24°)  𝑐𝑜𝑠2(24°) = 2𝑐𝑜𝑠 2 24° − 1 2𝑐𝑜𝑠 2 24° − 1 = 2p2 - 1  Answer/ 𝐴𝑛𝑡𝑤𝑜𝑜𝑟𝑑 (3) 5.3 1−𝑐𝑜𝑠2𝐴 𝐿𝐻𝑆 = tan 𝐴 𝑅𝐻𝑆 = 𝑠𝑖𝑛2𝐴 1−(1−2𝑠𝑖𝑛2 𝐴) = 2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴 1 − 2𝑠𝑖𝑛2 𝐴 2𝑠𝑖𝑛2 𝐴 = 2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴 2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐴 = 𝑐𝑜𝑠𝐴  simplify / vereenvoudig = 𝑡𝑎𝑛𝐴  Answer/ 𝐴𝑛𝑡𝑤𝑜𝑜𝑟𝑑 (4) 5.4 sin 20° . (cos 40°) + cos 20° . sin 40° 𝑐𝑜𝑠40° 𝑐𝑜𝑠20° = sin(20° + 40°)  𝑠𝑖𝑛40° = sin 60°  𝑠𝑖𝑛 60° √3 √3 =  (5) 2 2 5.5 2cos²θ + 5sinθ + 1 = 0 ∴ 2(1 − sin2 θ) + 5 sin θ + 1 = 0  (1 − 𝑠𝑖𝑛2 𝜃) ∴ 2 − 2sin2 θ + 5 sin θ + 1 = 0  standard form/ ∴ −2sin2 θ + 5sinθ + 3 = 0 Standaardvorm ∴ 2sin2 θ − 5sin2 θ − 3 = 0  factors faktore ∴ (2sinθ + 1)(sinθ − 3) = 0 1 sin 𝜃 = − 2 sin 𝜃 ≠ 3 1 ∴ sinθ = − ; sinθ ≠ 3 2  210° and/ en 330° θ = 180° + 30° + 360°n or /𝑜𝑓 θ = 360° − 30° + 360°n,  360°𝑛 , 𝑛 ϵ Z (7) ∴ θ = 210° + 360°n or /𝑜𝑓 θ = 330° + 360°n , n ϵ Z [26] 5

Downloaded from hlayiso.com SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Marks QUESTION /VRAAG 6 6.1 𝑎=2  Answer/ 𝐴𝑛𝑡𝑤𝑜𝑜𝑟𝑑 (1) 6.2  Max TPs /Maks DPe Min TP/Min DP x-intercepts / x-afsnitte (3) 6.3 180° 180° (1) 6.4 𝑥 = −90° or 𝑥 = 90°  𝑥 = −90° (2)  𝑥 = 90° 6.5 3 − 5.2 sin 𝑥 cos 𝑥  − 5.2 sin 𝑥 cos 𝑥 = 3 − 5 sin 2𝑥 sin 2𝑥 = 3 − 5 (-1) Answer/ Antwoord (3) =8 [10] 6

Downloaded from hlayiso.com SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Marks QUESTION /VRAAG 7 7.1 𝑟 𝑐𝑜𝑠 𝛼 = 𝐴𝐵  expression AB / 𝑟 uitdrukking AB 𝐴𝐵 = (1) cos 𝛼 7.2 𝑟 𝑐𝑜𝑠 𝛼 = 𝐴𝐶  expression AC / 𝑟 𝐴𝐶 = uitdrukking AC 𝑐𝑜𝑠 𝛼 𝐵𝐶 2 = 𝐴𝐵2 + 𝐴𝐶 2 − 2𝐴𝐵. 𝐴𝐶𝑐𝑜𝑠𝐵𝐴̂𝐶 cosine rule / cos reël 𝑟 2 𝑟 2 𝑟 𝑟 substitution / substitusie 𝐵𝐶 2 = ( ) +( ) − 2( )( ) 𝑐𝑜𝑠𝜃 𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠 𝛼 2𝑟 2 2𝑟 2 𝑐𝑜𝑠𝜃 𝐵𝐶 2 = 2 − 𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠2 𝛼 common factor / 2 2𝑟 2 gemeenskaplike faktor 𝐵𝐶 = (1 − 𝑐𝑜𝑠𝜃) 𝑐𝑜𝑠 2 𝛼 𝑟√2(1 − 𝑐𝑜𝑠𝜃) 𝐵𝐶 = 𝑐𝑜𝑠𝛼 OR 𝑟 𝑐𝑜𝑠 𝛼 = 𝐴𝐶  expression AC / 𝑟 𝐴𝐶 = uitdrukking AC 𝑐𝑜𝑠 𝛼 ∴ 𝐴𝐵 = 𝐴𝐶 cosine rule / cos reël 𝐵𝐶 2 = 2𝐴𝐵2 − 2𝐴𝐵2 𝑐𝑜𝑠 𝜃 substitution 𝐴𝐵 = 𝐴𝐶 / vervanging 𝐴𝐵 = 𝐴𝐶 2 2 𝐵𝐶 = 𝐴𝐵 (2 − 2𝑐𝑜𝑠 𝜃) 𝐵𝐶 = 𝐴𝐵√2 − 2𝑐𝑜𝑠𝜃 𝑟 common factor / 𝐵𝐶 = √2(1 − 𝑐𝑜𝑠𝜃) gemeenskaplike faktor (4) 𝑐𝑜𝑠𝛼 7

Downloaded from hlayiso.com 7.3 50(√2(1 − 𝑐𝑜𝑠30°)  substituting values of BC, r 100 = and 𝜃 / substitusie van 𝑐𝑜𝑠𝛼 waardes vir BC, r and 𝜃 50(√2(1 − 𝑐𝑜𝑠30°) 𝑐𝑜𝑠𝛼 = 50(√2(1−𝑐𝑜𝑠30°) 100  𝑐𝑜𝑠𝛼 = 100 𝛼 = 75°  𝛼 = 75° (3) [8] GEOMETRY/ MEETKUNDE S: Statement/ Bewering R: Reason/ Rede S/R: Both Statement and Reason/ Beide Bewering en Rede Allow for alternative methods/ Kyk vir Alternatiewe metodes SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Marks QUESTION /VRAAG 8 8.1.1 90° S (1) 8.1.2 Supplementary/ supplementêr S (1) 8.2.1 ̂ E1 = 54° ... tan-chord thm/ raaklyn-koordstelling S R (2) 8.2.2 Ĉ1 = 36° ... angle in semi-circle/ hoek in halfsirkel S R OR angle subt. By diameter/ hoek onderspan deur midlyn (2) 8.2.3 Ĉ3 = 35° ... alt. s; AB//CD / verw.e; AB//CD S R (2) 8.2.4 AÊD = 55° ...opp s of cyclic quad/ teenoorst e S R (2) van kvh 8.2.5 ̂ E4 = 35° ... subt by AC/ ondersp deur AC S R Ê3 = 20° S (3) 8.3 ̂ C = 90° ... tangent- radius/ raaklyn- radius BA S R And/ en ̂C = 90° ... tangent- radius/ raaklyn- radius BE S ∴ BECA is a cyclic quad/ is ’n kvh ... opp.s suppl./ R teenoorst e suppl. ̂ E = AĈE... external  of cyclic quad/ ∴ DB  R can only get this mark if BECA is (5) Buite  van kvh proved a cyclic quad. [18] 8

Downloaded from hlayiso.com SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Marks QUESTION /VRAAG 9 9.1 D A 1 S * T E F B * C Constr./ Konstr: Draw ST with S on AB and T on AC such  Constr./ Konstr – can be that AS = DE and AT = DF/ Trek ST met S op AB en T op on diagram/ kan op diagram AC sodat AS = DE en AT = DF wees  S/R ; Congruency/ AST ≡ DEF …S, , S Kongruensie Ŝ1 = ̂ E … from/ vanaf congruency/ kongruensie  ̂S1 = E ̂ =̂ B … given / gegee ̂ =B  ST // BC …corresp. s =/ ooreenk. e =  S: ST // BC R AB AC  = AB AC AS AT  S: = AS AT AB AC  = (Constr/ Konstr: AS = DE and/ en AT = DF R (7) DE DF 9.2.1 Q1 = ̂ ̂ Q2 … given/ gegee  S: ̂ Q1 = ̂Q2 = N2 … alt. s; MN // QR / verw. e; MN // QR ̂ ̂ = N2 R MN = MQ … sides opp = s / sye teenoor = e R (3) QMN is isosceles/ gelykbenig 9.2.2 In PMN and/ en PQR: 1. P is common/ P is gemeen S ̂ ... corr./ ooreenk s; MN // QR ̂1 = R 2. N S R  R (, , ) or 3rd  given (4) PMN /// PQR … (, , ) or 3rd  given 9

Downloaded from hlayiso.com 9.2.3 From/ vanaf /// in 9.2.2: PM MN PN  Proportionality/ Eweredigheid = = PQ QR PR  Subst/verv PQ = 5 and/ en PM MN ∴ = QR = 7,5 5 7,5 PM 2 PM 5 2  MN = 3 ∴ = = MN 7,5 3 PM 2 PM 2  MQ = 3 ∴ = … MQ = MN from/ vanaf 9.2.1 MQ 3 PN 2 PN 2 ∴ = … Prop thm/ ewer.stelling; MN // QR or/ of  NR = 3 NR 3 line // to 1 side of  / lyn // aan 1 sy van  (5) [19] SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Marks QUESTION /VRAAG 10 A N B 1 2 2 1 T M P 10.1 ̂ T = 90° = BN BM ̂P S  BMPN is a cyclic quad/ is ’n kvh … ext  = int opp / buite = teenoorst binne R (2) 10.2 N1 = ̂ ̂ P1 …subtended by/ onderspan deur BM S R But/ Maar ̂ ̂ … tan-chord thm/ raakl-koordstelling P1 = A S R S R NM // AP … corresp/ ooreenk. s = TN TM ∴ = (6) AN MP [16] Total/Totaal: [150] 10