Mathematics Grade 12 Paper 2 Eastern Cape Mock 2014 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2014 MATHEMATICS P2/WISKUNDE V2 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 12 pages./ Hierdie memorandum bestaan uit 12 bladsye

Downloaded from hlayiso.com 2 MATHEMATICS P2/WISKUNDE V2 (SEPTEMBER 2014) QUESTION/VRAAG 1 1.1 16 answer/antwoord (1) 1.2  correct values Semi-IQR = korrekte waardes  substitution into = 5,75 correct formula / vervanging in korrekte formule  answer/antwoord (3) 1.3  Min and/en Max/Maks  Q1 & Q3  Q2 10 18, 5 25 30 40 (3) 1.4 The distribution is skewed to the left/Die verspreiding is skeef  answer/antwoord na links. (1) [8] Copyright reserved Please turn over

Downloaded from hlayiso.com (SEPTEMBER 2014) MATHEMATICS P2/WISKUNDE V2 3 QUESTION/VRAAG 2 2.1 80 70  plotting points/ 60 afsteek van punte Class/Klas B 50 40 30  regression line/ regressielyn (2.3) 20 10 0 0 10 20 30 40 50 Class/Klas A (2) 2.2 a = 29,22 (29,21542…..)  calculating value b = 0,89 of a and b  y = 29,22 + 0,89 bereken waardes van a en b  equation for the line/vergelyking vir die lyn (2) 2.3 On graph/Op grafiek (1) 2.4 OR/OF answer/antwoord ∑ ̅ √  Sy, Sx =√ ∑ ̅ √ =√  answer/antwoord ; (2) 2.5 Mean/Gemiddelde, ̅ 0 (Answer only: Full marks) answer/antwoord (Slegs antwoord: Volpunte) Standard deviation/Standaardafwyking, answer/antwoord (4) 2.6 55,9 – 11,36 and 55,9 + 11,36 interval/interval [44,54; 67,26] 6 scores/tellings answer/antwoord (2) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICS P2/WISKUNDE V2 (SEPTEMBER 2014) QUESTION/VRAAG 3 3.1 y=9  answer/antwoord (1) 3.2 AB = 11 units/eenhede  answer/antwoord OR/OF √ √ AB = 11 units/eenhede (1) 3.3 D(13; 9)  answer/antwoord (1) 3.4 Area of ∆ABC = ½ AB.CD (Oppervlakte van . . .)  length of CD / = ½(11  8) lengte van CD = 44 square units/vierkante eenhede Substitute into correct formula / Vervanging in korrekte formule answer/antwoord (3) 3.5 substitution into M=[ ] midpoint formula/ =[ ] vervanging in middelpunt formule = [6,5 ; 5] answer/antwoord (2) 3.6 =  mAC  correct gradient of perpendicular bisector/korrekte gradient van middelloodlyn  Substitution into correct formula/ Vervanging in korrekte formule answer/antwoord (4) 3.7 substitution into correct equation / y= vervanging in korrekte vergelyking y=  answer/antwoord No, it does not pass through B/Nee, dit gaan nie deur B nie. (justification/ regverdiging) (2) Copyright reserved Please turn over

Downloaded from hlayiso.com (SEPTEMBER 2014) MATHEMATICS P2/WISKUNDE V2 5 3.8 mBC = =  mBC = tan A ̂C =  answer/antwoord (3) 3.9  substitution/vervanging  answer/antwoord (2) [19] QUESTION/VRAAG 4 4.1 completing the square/voltooiing van vierkant =a+5 equating r2/gelyk stel aan r2  r2 = 25 (4) 4.2 M (1; -2)  answer in coordinate form/ antwoord in koördinaat vorm (1) 4.3 A(x; y)  substituting into correct  formula/vervanging in korrekte formule standard form/standaardvorm  y = -6 or/of y = 2  factors/faktore correct answer/korrekte antwoord (4) 4.4 mMA = mtan. = -  substituting into correct formula/vervanging in korrekte formule answer/antwoord  (4) Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICS P2/WISKUNDE V2 (SEPTEMBER 2014) 4.5 M(1; -2) , T(-1;-2) MT =√  substitution/vervanging =2 simplification/vereenvoudiging 2<√  answer/antwoord OR/OF T (-1; -2) LHS = LHS = 1 + 2 + 4 – 8 LHS = -1 LHS ≠ RHS -1 < 20  T lies inside the circle/T lê binne die sirkel (3) 4.6 (x – 1)2 + (y + 2)2 = 20  substitution into correct (x – 1 + 3)2 + (y + 2 – 1)2 = 20 formula/vervanging in korrekte (x + 2)2 + (y + 1)2 = 20 formule   simplification and answer (x + 2) and (y + 1) / vereenvoudiging en antwoord (x + 2) en (y + 1) (3) [19] QUESTION/VRAAG 5 5.1.1  equation/vergelyking =  answer/antwoord = m 1 √ 𝑚 62 m (2) 5.1.2  =  62 – 60 =  expansion/uitbreiding √ substitution/vervanging = √ (4) 5.2.1  - tan x  cos x  - sin x =  = 1  answer/antwoord (5) Copyright reserved Please turn over

Downloaded from hlayiso.com (SEPTEMBER 2014) MATHEMATICS P2/WISKUNDE V2 7 5.3  simplification/vereenvoudiging √   value of/waarde van θ  substitution of/vervanging van θ √  Cos 30 = √ √ √  Sin 60 =  tan 45 = 1 =1 (6) [17] QUESTION/VRAAG 6 LHS =  expansion in numerator/ uitbreiding van teller =  expansion in denominator/ uitbreiding van noemer =  = factorisation/faktorisering = [4] Copyright reserved Please turn over

Downloaded from hlayiso.com 8 MATHEMATICS P2/WISKUNDE V2 (SEPTEMBER 2014) QUESTION/VRAAG 7 7.1 B h A D C Let/Laat BD = h (height of triangle/hoogte van driehoek ABC) Let/Laat DC = x AD in terms of x  AD = (b – x) AD in terme van x Proof/Bewys: c2 = h2 + (b – x)2 = h2 + b2 – 2bx + x2  c2 (using/gebruik = b2 + (h2 + x2) – 2bx van Pythagoras) = b2 + a2 – 2bx a2 = h2+ x2  (h2 + x2) = a2 substitution of x/ x = a cos C vervanging van x  b + a – 2ab cos C 2 2 (4) 7.2.1 PSQ = 90  angle in semi- Q2 = 90 + α circle/hoek in semi- sirkel In ∆QSR: SR2 = SQ2 + QR2 – 2SQ.QR.cosSQR  use of cos rule/ = x2 + x2 – 2x.x.cos(90 + α) gebruik van cos-reël = 2x2 -2x2(-sin α)  correct = 2x2 + 2x2(sin α) substitution/ = x2(2 + 2sinα) korrekte vervanging  SR = x√  x2 (2 + 2sinα) (4) 7.2.2 √ √ Substitution/ √ √ Vervanging √ √  sinα = 0,5 α = 30 α = 30 In ∆PSQ  answer/antwoord = 10 units/eenhede (4) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com (SEPTEMBER 2014) MATHEMATICS P2/WISKUNDE V2 9 QUESTION/VRAAG 8 8.1 2  answer / (1) antwoord 8.2 f:  x- int/x-afsnit  shape/vorm g:  x-ints/x-afsnitte y-int/y-afsnit shape/vorm (5) 8.3 -120  critical values / kritiese waardes notation/notasie (2) 8.4 Translate f 30 to the right/Skuif f 30o na regs  correct translation/korrekte OR/OF translasie Translate f down by ½ / Skuif f ʼn ½ eenheid af  correct translation/korrekte translasie (2) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com 10 MATHEMATICS P2/WISKUNDE V2 (SEPTEMBER 2014) QUESTION/VRAAG 9 9.1 Const: Join O to C Konstr: Verbind O met C  Construction/ C konstruksie O 1 2 1 2  A and D (in terms of x and y) with reasons A en D in terme van x A en y met redes D Construction/Konstruksie: Join CO produced/Verbind CO  AOC (in terms of x) verleng. with a reason Proof/Bewys: AOC in terme van x met Let/Laat C1 = x and/en C2 = y rede A = x [ AO = OC= r] D = x + y [OD = OC= r]  equation for AOD/  AOC = 180 – 2x [Sum of angles of/Som v/d hoeke van ∆] gelykstel vir AOD  DOC = 180 – (2x + 2y) [Sum of angles of ∆ / “ ] But/Maar  AOD =  AOC –  DOC   AOD = 2y  AOD = 2y  AOD = 2  ACD (5) 9.2 9.2.1 MBC = MCB [MB = MC; radii] reason/rede  B3 = C2 = x [BO bisects/halveer MBC and/en B3 and/en C2 = x OC bisects/halveer MCB] and reason/en rede  O2 = 180o – 2x [angles of/hoeke van ∆]  answer/antwoord (4) 9.2.2  A = ½BMC [angle at centre = 2 angle at circ   A = ½BMC /middelpuntshoek = 2 omtrekshoek] (statement) and reason. = ½(180 – 4x) [angles of/hoeke van ∆] (stelling) en rede  A = 90 – 2x  substitution and answer/vervanging en Therefore/Daarom: antwoord  ABO = O2 – A [ext. angle of/buitehoek van   ABO = O2 – A ∆OAB] (statement) and reason = (180 – 2x) – (90 – 2x) (stelling) en rede = 90 substitution/ AO subtends a right angle/AO onderspan ʼn reghoek vervanging  AO is a diameter/AO is ʼn middellyn ABO = 90o  answer/antwoord (5) Copyright reserved Please turn over

Downloaded from hlayiso.com (SEPTEMBER 2014) MATHEMATICS P2/WISKUNDE V2 11 9.3 In ∆EHO, OE = x and OH = x – 2  OH = x – 2 EH2 = x2 – (x – 2)2 [Theorem of Pythagoras] Theorem of/Stelling EH2 = x2 – x2 + 4x – 4 van Pythagoras = 4(x – 1) EH  EH = √ EF [with reason/met  EF = √ line from centre of circle …/lyn van rede] middelpunt van sirkel ...] (4) [19] QUESTION/VRAAG 10 10.1  ATB = 90 (subtended by diameter/onderspan deur   ATB = 90 middellyn)   ATB =  EPB = 90   ATB =  EPB = 90 conclusion with reason/  TEPB is a cyclic quad/is ʼn koordevierhoek gevolgtrekking met rede (ext/buite  = int opp/teenoorst. binne  ) (3) 10.2 In ∆ATB and/en ∆APE statement with reason / A = A [common/gemeen] stelling met rede  ATB =  APE = 90   ATB =  APE = 90  ∆ATB /// ∆APE (  ,  ,  )  ∆ATB /// ∆APE with reason /met rede (3) 10.3  RTA =  TBA (tangent chord/raaklyn-koord)   RTA =  TBA (statement) with reason.  RTA =  ETP (vertically opposite/regoorst) (stelling) met rede   RTA =  ETP  TBA =  TEP (ext/buite  of cyclic quad/ (statement) with reason van koordevierhoek) (stelling) met rede statement with reason ∆TPE is Isosceles/gelykbenig. (  ETP =  TEP) (stelling) met rede ∆TPE is Isosceles/gelykbenig  TP = PE  reason/rede  conclusion/slotsom (6) 10.4  ATB =  EPB = 90   ATB =  EPB = 90  PTB =  A [tan chord/raaklyn-koord]   PTB =  A (statement) with reason  PTB =  PEB [subtended by BP/onderspan deur BP] (stelling) met rede   PTB =  PEB   A =  PEB (statement) with reason Hence/Vervolgens ∆ATB ///∆EPB [equiangular/ (stelling) met rede gelykhoekig]   A =  PEB  conclusion/slotsom (5) 10.5 (∆ATB /// ∆EPB) statement with reason stelling met rede AB.BP = BE.TB  AB.BP = BE.TB 2BP.BP = BE.TB [AO=OB=BP] 2BP.BP = BE.TB (statement) with reason 2BP2 = BE.TB (stelling) met rede answer/antwoord (4) [21] Copyright reserved Please turn over

Downloaded from hlayiso.com 12 MATHEMATICS P2/WISKUNDE V2 (SEPTEMBER 2014) QUESTION/VRAAG 11 11.1 Divides the two other sides proportionally/verdeel die twee  answer/antwoord (1) ander sye eweredig. 11.2 11.2.1  proportions / eweredighede [ED // LM] LM 2 = KL2 + KM 2 [Pythagoras]  Pythagoras = (3x)2 + (3y)2 = 9x2 + 9y2  answer/antwoord = 9 DL2 + 9 EM 2 ……….1 (4) 11.2.2 DM 2 + LE 2 = KM2 + KD2 + EK2 + KL2  statement/stelling = (3y)2 +(2x)2 + (2y)2 + (3x)2 = 9y2 + 4x2 + 4y2 + 9x2 = 13y2 + 13x2 substitution and = 13 EM2 + 13 DL2 simplification/ = 13(EM2 + DL2) vervanging en vereenvoudiging = 13 = use of 1/gebruik van 1  conclusion/slotsom (4) [9] TOTAL/TOTAAL: 150 Copyright reserved Please turn over