Mathematics Grade 12 Paper 2 EC June 2016 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2016 MATHEMATICS P2 / WISKUNDE V2 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 8 pages. Hierdie memorandum bestaan uit 8 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2016) QUESTION/VRAAG 1 1.1 Time (hours)/Tyd (ure) Cumulative Frequency Kumulatiewe Frekwensie 0 ≤ t < 20 30 20 ≤ t < 40 69 40 ≤ t < 60 129 60 ≤ t < 80 157 80 ≤ t < 100 167 100 ≤ t < 120 172 Time spent playing sport Tyd spandeer vir sport speel 200 180 Cumulative frequency 160 140 120 100 Kumulatiewe Frekwensie 80 60 40 20 0 0 20 40 60 80 100 120 140 Time in hours/Tyd in ure (4) 1.2 40 ≤ t < 60  answer/antwoord (1) 1.3 172 answer/antwoord (1) 1.4 (72;148) 148 ∴ 172-148 = 24 learners/leerders 24 (2) 1.5 Frequency/Frekwensie: 30; 39; 60;28;10;5 frequency/ 30 × 10 + 39 × 30 + 60 × 50 + 28 × 70 + 10 × 90 + 5 × 110 frekwensie 172 midpoints/ 7880 middelpunte = 7880 172  172 = 45,81 answer/antwoord (4) [12] QUESTION/VRAAG 2 2.1 6772 6772 𝑥̅ = 𝑥̅ = 20 20 answer/antwoord 𝑥̅ = 338,6 𝑚𝑙 (2) 2.2 2,71 ml  (2) answer/antwoord 2.3 [338,6 – 2,71; 338,6 + 2,71]  [335,89; 341.31] interval/interval (2) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2016) MATHEMATICS P2/WISKUNDE V2 3 QUESTION/VRAAG 3 3.1 1 𝑚𝐴𝐶 = − S 3 𝑚𝐵𝐷 = 3 [Diagonals of a rhombus]/ R [Hoeklyne van ruit/rombus]  subst, m = 3 & 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) (3;9) in eqn. / 𝑦 − 9 = 3(𝑥 − 3) verv. m = 3 & 3𝑥 − 𝑦 = 0 (3;9) in verg. (3) 3.1.2 𝑥 + 3𝑦 = 10 … … … . (1) 3𝑥 − 𝑦 = 0 … … … … (2)  equating two 3𝑥 + 9𝑦 = 30 … … … (3) (1) × 3 eqns / gelyk stel 10 𝑦 = 30 (3) − (2) van twee vergs. 𝑦 =3 simplification / 𝑥 + 3(3) = 10 vereenvoudiging 𝑥=1 y = 3 K(1; 3) x = 1 (4) 3.1.3 𝑥+3 𝑦+9 method using [ ; ] = [1; 3] midpoint / metode 2 2 𝑥+3 𝑦+9 gebruik middelpunt =1 =3  simplification / 2 2 𝑥+3=2 & 𝑦+9=6 vereenvoudiging 𝑥 = −1 & 𝑦 = −3  coordinates of B / B(-1; -3) koördinate van B (3) 3.1.4 AD =√(3 − 𝑥)2 + (9 − 𝑦)2  subst into eqn dist AD / verv. in verg. ∴ √50 = √9 − 6𝑥 + 𝑥 2 + 81 − 18𝑦 + 𝑦 2 afstd AD ∴ 50 = 𝑥 2 − 6𝑥 + 𝑦 2 − 18𝑦 + 90  subst/verv. But/Maar: 𝑥 = 10 − 3𝑦 AD = √50 ∴ (10 − 3𝑦)2 − 6(10 − 3𝑦) + 𝑦 2 − 18 𝑦 + 90 = 50 subst/verv. ∴ 100 − 60𝑦 + 9𝑦 2 − 60 + 18𝑦 + 𝑦 2 − 18𝑦 + 90 x =10 - 3y = 50  simplification ∴ 10𝑦 2 − 60𝑦 + 80 = 0 /vereenvoudiging ∴ 𝑦 2 − 6𝑦 + 8 = 0 standard form ∴ (𝑦 − 4)(𝑦 − 2) = 0  𝑦 = 4 or/of 𝑦 = 2 values for y 𝑥 = 10 − 3(4) or/of 𝑥 = 10 − 3(2) values for x 𝑥 = −2 or/of 𝑥 = 4 coordinates A(-2;4) C(4; 2) (8) 3.2.1 8−2 6  subs into eqn / 𝑚PQ = = verv. in verg 5 − (−3) 8 3 =4  answer/antwoord (2) 3.2.2 3  tan 𝜃 tan 𝜃 = 4 𝜃 = 36,9  answer /antwoord (2) 3.2.3 3 3 𝑦 = 𝑥+𝑐 subst/verv. m = 4 4 3 0 = (8) + 𝑐  subst/verv. (8:0) 4 𝑐 = −6 3  answer/antwoord 𝑦 = 𝑥−6 (3) 4 [25] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2016) QUESTION/VRAAG 4 4.1 A(0; y) midpt equation / vergelyking 0+8 ∴𝑝= =4 coordinates of D / 2 ∴ D(4; 4) koördinate van D (2) 4.2 𝑦+7 Ay = 2 = 4 y-coordinate of A y-koördinaat van A A(0; 1) A(0; 1) ∴ (𝑥 − 0)2 + (𝑦 − 1)2 = 𝑟 2 subst (0;1)and (4;4) ∴ (4 − 0)2 + (4 − 1)2 = 𝑟 2 into equation. ∴ 16 + 9 = 𝑟 2 verv. (0;1) en (4;4) in ∴ 𝑥 2 + (𝑦 − 1)2 = 25 vergelyking. ∴ 𝑥 2 + 𝑦 2 − 2𝑦 − 24 = 0 r2 (5) 4.3 𝑚𝐴𝐵 × 𝑚𝐹𝐷𝐸 = −1 [ tan radius]/[raaklyn radius] S/R 7−4 3 3 𝑚𝐴𝐵 = = mAB 4 8−4 4 4 4 𝑚𝐹𝐷𝐸 = − 3 ∴ 𝑚𝐹𝐷𝐸 = − subst m and (4;4) into 3 4 eqn / verv. m en (4;4) in (𝑦 − 4) = − (𝑥 − 4) 3 vergelyking. 4 28 ∴𝑦= − 𝑥+ 3 3 answer/antwoord (5) 4.4 𝑥2 + 𝑦2 = 𝑟2 subst (8;7) into eqn (8)2 + (7)2 = 𝑟 2 verv. (8;7) in vergelyking ∴ 𝑥 2 + 𝑦 2 = 113 answer/antwoord (2)  [14] QUESTION 5 5.1.1 sin 𝑥. cos 𝑥. tan 𝑥. cos 𝑥 sin x = cos x sin 𝑥. cos 𝑥. (−𝑡𝑎𝑛𝑥) tan x = − cos 𝑥 cos x sin x cos x -tan x -cos x (8) 5.2 sin 𝑥 1 + cos 𝑥 2 + = 1 + cos 𝑥 sin 𝑥 sin 𝑥 sin2 𝑥+(1+cos 𝑥)(1+cos 𝑥)  denominator / LHS/LK: sin 𝑥 (1+cos 𝑥) noemer  numerator / sin2 𝑥+1+2 cos 𝑥+cos2 𝑥 teller sin 𝑥(1+cos 𝑥)  simplification / 2 + 2 cos 𝑥 vereenvoudiging sin 𝑥 ( 1 + cos 𝑥) 2(1+1 cos 𝑥) sin 𝑥 (1+cos 𝑥)  identity / 2 identiteit = sin 𝑥 = RHS/RK  factorisation/ faktorisering (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2016) MATHEMATICS P2/WISKUNDE V2 5 5.3 cos 2𝑥 = cos(𝑥 + 𝑥)  expansion / = cos 𝑥. cos 𝑥 − sin 𝑥. sin 𝑥 uitbreiding = cos2 𝑥 − sin2 𝑥 identity / = cos2 𝑥 − (1 − cos2 𝑥) identiteit =2 cos2 𝑥 − 1 (2) 5.4 cos 2𝑥 + 3 sin 𝑥 = 2 1 − 2 sin2 𝑥 + 3 sin 𝑥 = 2  identity/identiteit 2 sin2 𝑥 − 3 sin 𝑥 + 1 = 0  standard form / (2 sin 𝑥 − 1)(sin 𝑥 − 1) = 0 standaardvorm 2 sin 𝑥 = 1 or/of sin 𝑥 − 1 = 0 factors/faktore sin 𝑥 = 2 1 or/of sin 𝑥 = 1  sin x = ½ sin x = 1 (both 𝑥 = 30° + 𝑛. 360 𝑥 = 90° + 𝑛. 360 eqns/albei vergls) 𝑥 = 150 + 𝑛. 360 ; 𝑛 ∈ 𝒁  x = 30°  x = 150°  x = 90° (7) 5.5 sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵 = sin(𝐴 + 𝐵)  identity/identitieit = sin 90° subst/vervanging =1  answer/antwoord (3) [25] QUESTION/VRAAG 6 6.1 shape g vorm g intercepts afsnitte min & max value/waarde shape f vorm f asymptotes asimptote intercept afsnit (6) 6.2 −180° < 𝑥 ≤ −90° or/of 0° ≤ 𝑥 ≤ 90° critical values / kritieke waardes  notation / notasie (2) [8] QUESTION/VRAAG 7 7.1 ̂N MN2 = PN2 + PM 2 − 2PN. PMcos MP correct subst into cos 2 2 rule / korrekte verv. in = 12 + 10 − 2(12)(10) cos 126,9° = 144 + 100 − 240(−0,6) cos-reel = 388,1  simplification / MN = 19,7 𝑚 vereenvoudiging MN = AD = 19,7 𝑚  answer / antwoord (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2016) 7.2 1 MN. PT = 1 PN. PM sin MP ̂N [Both equal Area of Equating Area form / 2 2 Gelykstel van Opp. ∆PMN] / [Albei is gelyk aan oppervlakte van ∆PMN] formules 1 1  correct subst / (10)PT = (12)(10) sin 126,9° korrekte verv. 2 2  simplification / PT = 12. sin 126,9° vereenvoudiging PT = 9,596 m  answer / antwoord = 9,6 m (4) [7] QUESTION 8 8.1 Supplementary / Supplementêr answer/ antwoord (1) 8.2.1 EF̂O = 90° [tan radius]/[raaklyn radius]  S R ̂ EGO = 90° [tan radius]/[raaklyn radius] S R EF̂O + EG ̂ O = 180° ∴ FOGE is cyclic quad [opp angles supplementary] is ʼn koordevierhoek [ teenoorst. hoeke supplementêr.] R (5) 8.2.2 ̂ G1 = ̂H = 𝑥 [ tan chord]/[raaklyn koord] S R ̂ K1 = H ̂ = 𝑥 [ corresp angles/ooreenkmst hoeke; EK || FH] S R ∴Ĝ1 = K ̂1 R EG is a tangent / is ʼn raaklyn [angle between line and chord]/ [hoek tussen lyn en koord] (5) 8.2.3 ̂ O1 = 2H ̂ [ angle at centre]/[middelpuntshoek] S R = 2x ̂ ∴ FEG = 180° − 2𝑥 [ opp angles of cyclic quad]/ R [teenoorst. hoeke van koordevierhoek] (3) [14] QUESTION/VRAAG 9 9.1 ̂ B2 = Â2 = x [angles opp equal sides]/[hoeke teenoor gelyke sye]  S/R ̂ ̂ A1 = B2 = x [ tan chord]/[raaklyn koord]  S/R ̂S1 = A ̂1 = x [corresp/ooreenkomstig; AD||SC]  S/R 𝐵̂3 = 𝐴̂2 = x [ alt int./verwissellend; AD|| SC]  S/R AD ̂C = B̂ 3 = x [ ext angle of cyclic quad]/  S/R [buitehoek van koordevierhoek] (5) 9.2 ̂ A1 = AD ̂C [from/vanaf 8.1] S AS || DC [alt angles equal] / [verw. hoeke gelyk]  S/R DA || CS [given/gegee] S ASCD is a/ʼn parallelogram [opp sides ||]/[teenoorst. sye ||] R (4) 9.3 ∆SAB ||| ∆ADB [ A,A,A]/[H,H,H]  ∆SAB (1) 9.4 SA SB S = [from/vanaf 8.3.1] AD AB ∴ AD. SB = SA. AB S But/Maar AD = SC [ ASCD is a/ʼn parallelogram] S/R and/en AB = SA[sides opp equal angles]/[sye teenoor gelyke hoeke] S/R = DC [ASCD is a/ʼn parallelogram] R ∴ SC. SB = DC2 (5) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2016) MATHEMATICS P2/WISKUNDE V2 7 QUESTION/VRAAG 10 10.1.1 In proportion / eweredig  Answer (1) 10.1.2 A k h D E B C 𝐴𝐷 𝐴𝐸 RTP: 𝐷𝐵 = 𝐸𝐶 PROOF/BEWYS: 1  ratio of area of / area ∆ADE .AD.ℎ = 2 verhouding van 1 area ∆ BDE .DB.ℎ opp. van 2 AD ∆ADE: ∆BDE = AD BD  BD 1 ratio of area of area ∆ADE 2 . AE. 𝑘 verhouding van = area ∆CED 1 . EC. 𝑘 opp. van 2 AE ∆ADE : ∆CED = AE EC  EC But/𝑀𝑎𝑎𝑟 Area ∆BDE = Area ∆ CED (same base and equating two areas same height)/(dieselfde basis en dieselfde hoogte) / gelykstel van twee area ∆ADE area ∆ADE ∴ = oppervlaktes area ∆BDE area ∆ CED AD AE ∴ = (5) DB EC 10.2.1 QT QW QT QW = prop thm/eweredigh stelling; TW || VR  = TP WR TP WR 15 𝑥+4 R ∴ =  substitution / 𝑥+2 𝑥 𝑥 2 + 6𝑥 + 8 = 15𝑥 vervanging 𝑥 2 − 9𝑥 + 8 = 0 ∴ (𝑥 − 8)(𝑥 − 1) = 0  std form/vorm ∴ 𝑥 = 8 or/of 𝑥 = 1 factors/faktore  both values for x/ beide waardes van x (6) 10.2.2 PV = PT  S/R prop thm/eweredigh stelling ; TV || QR VR TQ PV 10  substitution / = 18 15 vervanging PV = 12 𝑢𝑛𝑖𝑡𝑠/eenhede  answer/antwoord (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2016) QUESTION/VRAAG 11 11.1 𝐷2 = 90° [ line from centre to midpoint of chord]/[lyn  S/R vanaf die middelpt van sirkel na middelpt van koord] In ∆ ABC & ∆ DOC  S/R ̂ = D2 [ both equal to 90°]/[beide gelyk aan 90°] i) A S ii)Ĉ = Ĉ [common/gemeen] ∴ ∆ABC |||∆DOC [AAA/𝐻𝐻𝐻] (3) 11.2 𝑂𝐶 = 𝐴𝐶 𝐷𝐶 ∆ABC |||∆DOC S/R 𝐵𝐶 𝐷𝐶. 𝐵𝐶 𝑂𝐶 = 𝐴𝐶 (1) 11.3 2 AC = BC − AB 2 2 [Pythagoras]  S/R = 302 − 182  subst in eqn / = 576 verv. in verg. AC = 24 cm 𝐷𝐶. 𝐵𝐶  AC = 24 𝑂𝐶 = 𝐴𝐶 15 × 30 = 24 = 18,8 𝑐𝑚  subst / vervang.  answer / antwoord (5) [9] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief