Mathematics Grade 12 Paper 2 EC June 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2019 MATHEMATICS P2/WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 17 pages./ Hierdie nasienriglyn bestaan uit 17 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE 2019) QUESTION 1/VRAAG 1 1.1 Mean/Gemiddelde = 48  48 (1) 1.2 SD/SA = 22,08 22,08 Penalty of 1 mark for incorrect rounding Penaliseer 1 punt vir verkeerde afronding (2) 1.3 Girls performed better. Girls / Meisies The girls’ mean percentage is bigger than that of boys and the girls Reason / Rede standard deviation is smaller than that of boys Meisies het beter gevaar. Die meisies se gemiddelde persentasie is groter as dié van die seuns en die standaardafwyking is kleiner as dié van die seuns. (2) 1.4 51 – 48 = 3 3  each boy’s percentage must be increased by 3. elke seun se persentasie moet met 3 vermeerder word. (1) 1.5 Boys’ standard deviation will remain the same remain the same Die seuns se standaardawyking sal dieselfde bly. dieselfde bly (1) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2019) MATHEMATICS P2/WISKUNDE V2 3 QUESTION 2/VRAAG 2 2.1 Ages (in years) Frequency Cumulative Frequency Ouderdom (in jare) Frekwensie Kumulatiewe Frekwensie 18  x  28 4 4 frequency 28  x  38 10 14 frekwensie 38  x  48 14 28 cumulative 48  x  58 17 45 frequency 58  x  68 12 57 kumulatiewe 68  x  78 3 60 frekwensie (2) 2.2 Ogive - Ages of people registering to vote Ogief - Ouderdomme van mense wat registreer om te stem 70 Cumulative Frequency 60 50 grounding at / 40 anker by (0 ; 18) 30 Kumulatiewe Frekwensie 20 upper limits 10 boonste limiete 0 0 20 40 60 80 100 Ages (in years) shape / vorm Ouderdom (in jare) (3) 2.3 48  x  58 answer / antwoord (1) 2.4 60 – 49 = 11 senior citizens / senior burgers 49 answer / antwoord (2) 2.5 Q1 = 39 value of / waarde van Q1 Q2 = 50 value of / waarde van Q2 Q3 = 58 value of / waarde van Q3 (3) 2.6 minimum and maximum minimum en maksimum  box / boks (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE 2019) QUESTION 3/VRAAG 3 3.1.1 KL  ( x2  x1 ) 2  ( y2  y1 ) 2  (10  0 ) 2  ( 2  7 ) 2 substitution / vervanging answer / antwoord  125  5 5 (2) 3.1.2 y2  y1 mKM  x2  x1 47 substitution / vervanging  70 11 gradient of KM / gradiënt van KM  7 (2) 3.1.3 y2  y1 mLM  x2  x1 42  7  10 gradient of LM / gradiënt van LM 2  tan   2 tan   2 value of  / waarde van     63, 43 (3) 3.1.4 11 tan    11 7  tan    Re f   57,53 7 reference angle / verwysingshoek   122, 47 value of θ / waarde van θ  LMK  122, 47  63, 43    59, 04 value of LMK / waarde van LMK (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2019) MATHEMATICS P2/WISKUNDE V2 5 3.2 mKN  mLM y 7 2 x0 y  2x  7  y  2x  7 mMN  mKL y 4 1  x7 2 x 1 x 1 y    y   2 2 2 2 x 1 2x  7    2 2 x 13  y   2 2 4 x  14   x 1 value of x / waarde van x x 3 y 1 value of y / waarde van y OR/OF OR / OF -5 5  mKL  mKL = 10 10 -5 5 mNM =  mNM  10 10 Hence the coordinates of/ value of x / waarde van x Vervolgens die koördinate van N(-3;1) value of y / waarde van y OR/OF OR/OF  7 3 Midpointof KM/ Middelpunt vanKM  ;   Midpoint of KM / Middelpunt van KM   7 ; 3   2 2 2 2  Midpoint of LN / Middelpunt van LN   ;  7 3  7 3 Midpointof LN/ Middelpunt vanLN  ;  2 2  2 2 x 10 7 y 2 3 value of x / waarde van x   and/en  2 2 2 2 value of y / waarde van y  N( 3;1) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE 2019) 3.3  1 mLM  mMN  2     product of gradients /  2 produk van gradiënte  1   LMN  90  conclusion / gevolgtrekking OR/OF LMN is a right angle / is ' n regtehoek (2) 3.4  KL  NM  5 5 KL  NM  5 5 KM  7 2  112  KM  170  170 ∧ ∧   LMN = 90°and/en LMK = 59,04°  LMN  90 and / en LMK  59, 04  KMN  30,96   KMN  90  56, 04  30,96 1 Area of KMN =  170  5 5  sin 30,96 2  Area of KMN  37,50 square units / vierkante eenhede (5) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2019) MATHEMATICS P2/WISKUNDE V2 7 QUESTION 4/VRAAG 4 4.1 M (2 ; 4) value of x / waarde van x value of y / waarde van y (2) 4.2 r  (5  2 )  ( 0  4 ) 2 2 2 ( x 2) 2  25  ( y  4 )2  ( x  2 )2  ( y  4 ) 2  25 25 (3) 4.3 8 0  mGH GH 1 5 8 4  mtan 6 3 3 substitution / vervanging 4 3 equation / vergelyking 8 1 4 3 35 ∴ 4 4 (4) 4.4 At/By J, y = 0 ( x  2 ) 2  ( 0  4 ) 2  25 y=0 ( x  2 )2  9 x23 substitution / vervanging x  5 or/of x   1 x=–1  J (  1;0 ) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE 2019) 4.5 HJG is a right angled triangle 8, 6 and 10 . So the rotation of J around M will complete a rectangle Hence, j 1 6; 0 8 J| 5; 8 value of x / waarde van x value of y / waarde van y HJG is ’n reghoekige driehoek (8, 6 en 10) Dus sal die rotasie van J om M die reghoek voltooi. Vervolgens is | 1 6; 0 8 | 5; 8 (2) 4.6 12 2 17 0 completing the square / 12 2 17 voltooiing van die vierkant 12 36 2 1 17 36 1 6 1 20 factorisation/faktorisering: x factorisation/faktorisering: y Distance between the centres: Afstand tussen die middelpunte: distance formula / 2 6 4 1 =5 afstand formule ∴ the centre lies on the original circle / die middelpunt lê op die omtrek van die oorspronklike sirkel conclusion / gevolgtrekking (5) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2019) MATHEMATICS P2/WISKUNDE V2 9 QUESTION 5/VRAAG 5 5.1.1 sin 42° 48 1 k  1 k 2 tan 42° √1  tan 42° 42 √ 1 k 2 (2) 5.1.2 sin 84  sin 2  42  2sin 42 cos 42 double angle/dubbelhoek expansion/identiteit/ontwikkeling  2. k . 1  k 2  2k 1  k 2 substitution/vervanging (3) 5.1.3 sin 3° 45° 42° 3° = 45° – 42° 4 5° 4 2° 4 5° 4 2° expansion/identiteit/ontwikkeling √2 √2 substitution/vervanging . 1 . substitution/vervanging 2 2 (4) 5.2 sin ( x  450 ). tan (180  x ).sin (90  x ) cos (  x )   cos x  cos x . tan x . cos x  tan x cos x  cos x sin x  cos x  cos x . cos x sin x   sin x cos x answer / antwoord (6) 5.3.1 cos (A + B )  cos A cos B  sin A sin B expansion/identiteit/ontwikkeling (1) 5.3.2 LHS/LK 3 cos 2α α compound angle identity saamgesteldehoek-identiteit cos 2 α cos α sin 2 α sin α 2 cos α 1 . cos α 2 sin α cos α . sin α cos double angle identity cos dubbelhoek-identiteit 2 cos α cos α 2 sin α cos α 2 cos α cos α 2 1 cos α cos α sin double angle identity sin dubbelhoek-identiteit 2 cos α cos α 2 cos α 2 cos α 4 cos α 3 cos α  (1  cos 2  ) RHS/RK (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE 2019) 5.4 2p cos   1 1 1 4 p2   1 4 p2 2p cos 2  cos   sin  2 2  cos 2  cos 2   sin 2    2 7 p  2 p  2 1 4 p2 substitution/vervanging 7 p  4 p2  1 4 p2 8 p 2  7 p 1  0 standard form 8 p  1 p 1  0 standaardvorm 1 p or / of p  1 values of p / 8 waardes van p OR/OF OR/OF cos 2  2 cos 2  1 7 p  2.  2 p  1 2  2 cos 2  1 7 p  8 p2  1 substitution/vervanging 8 p 2  7 p 1  0 8 p  1 p 1  0 standard form / standaardvorm 1 p  or / of p 1 8 values of p / waardes van p OR/OF cos 2  1  2sin  2 OR/OF    1 4 p2 2 7 p  1  2. 1 4 p2 7 p  1  2(1  4 p 2 )  1  2sin 2  7 p 1  2  8 p 2 substitution / vervanging 8 p 2  7 p 1  0  8 p  1 p 1  0 1 standard form p or / of p  1 standaardvorm 8 values of p / waardes van p (5) [25] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2019) MATHEMATICS P2/WISKUNDE V2 11 QUESTION 6/VRAAG 6 6.1 y [  1;1] OR/OF answer / antwoord 1 y 1 (1) 6.2 g: asymptotes at -90° and 90° asymptote vir -90° en -90° x-intercepts x-afsnitte shape / vorm (3) 6.3 180° 180° (1) 6.4 x = – 45° – 45° (1) 6.5 x [ 45 ; 90) critical values OR/OF kritiese waardes notation / notasie 45 x  90 (2) 6.6 h(x) =sin (x – 45°) + 1 h(x) =sin (x – 45°) + 1 (1) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE 2019) QUESTION 7/VRAAG 7 7.1 In Δ KLN : h tan w  h LN  LN  h tan w LN  tan w (1) 7.2 In NLM LM LN  =  correct sine rule sin N sin M korrekte sinusreël LM LN  substitution sin (180  ( y  z )) sin z vervanging LN.sin ( y  z )  LM  sin z isolating LM h isoleer LM But LN  tan w h sin ( y  z ) LM  tan w sin z answer / antwoord (4) 7.3 h sin ( y  z ) LM  and/en tan w sin z h  38 m, w  21o , y  52o and/en z  59 substitution 38. sin (52  59 ) vervanging  LM  tan 21 sin 59 answer / antwoord 107,82 m (2) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2019) MATHEMATICS P2/WISKUNDE V2 13 QUESTION 8/VRAAG 8 8.1 the centre of a circle / die middelpunt van die sirkel answer / antwoord (1) 8.2.1 AC = 10 cm (line from centre  chord) length of AC (lyn vanaf die middelpunt  op koord) lengte van AC Reason/rede (2) 8.2.2 ( x  5 )  10  x 2 2 2 radius = (x + 5) x  10 x  25  100  x 2 2 applying Pythagoras 10 x  75 theorem / toepassing  x  7,5 cm van Pythagoras se  radius = 7,5 cm + 5 cm stelling = 12,5 cm value of x / waarde van x length of radius / lengte van radius (4) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 14 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE 2019) QUESTION 9/VRAAG 9 9.1 interior opposite angle / teenoorstaande binnehoek answer / antwoord (1) 9.2.1   Statement / bewering R 2 = Q2  41 (s in the same seg)/(e in dieselfde seg.) Reason / rede   Statement / bewering P4 = Q1  41 (tan-chord theorem)/(raaklyn-koord stelling) Reason / rede   Statement / bewering T1 = P4  41 (s opp. = sides)/(e teenoor gelyke sye) Reason / rede OR/OF   T1 = R 2  41 (tan – chord theorem)/(raaklyn-koord stelling) (6)  9.2.2(a) T 2  34  78 ( tan  chord theorem )/(raaklyn - koord stelling )  Statement / bewering  T 2  44 Reason / rede (2)  9.2.2(b) 41  Q 2  44   34  180  ( opp.  s of a cyclic quad.)  Statement / bewering  Q 2  61 (teenoorst .  e van ' n koordevierhoek ) Reason / rede (2)  9.2.2(c) T 4  41  61 ( ext. s of a cyclic quad.)  Statement / bewering  T 4  102 (buite  van koordevierhoek ) Reason / rede OR/OF OR/OF  T 4  44  34  180 ( int . s of a  ) Statement / bewering  Reason / rede  T 4  102 (binne e van ' n  ) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2019) MATHEMATICS P2/WISKUNDE V2 15 ∧ 9.2.2(d) Statement /bewering W + 41°+ 41° = 180° ( int . ∠s of a Δ)/(binne e van ) ∧ ∴ 98°  Reason / rede (2) 9.2.3(a)   Q2  61 and/en T 2  44      Q2  T2  Q2  T 2  alt.  s are not equal  QR is not parallel to PT (alt. s are not equal) verw. e is nie gelyk nie QR is nie ewewydig aan PT nie (verw. e is nie gelyk nie) (2) 9.2.3(b)   R 2 + W  41  98    139  180  R2 +W  180  PRTW is not a cyclic quad. (Opp. s are not supp.)  PRTW is not a cyclic quad. PRTW is nie ' n koordevierhoek nie PRTW is nie ’n (teenoorst. e is nie supplementer ˆ nie) koordevierhoek nie (2) ∧ ∧ 9.2.3(c) R1 = T2 = 44°(∠s in same seg)/(∠e in dieselfde segment) ∧ ∧ R1 + R2 = 44° + 41°   = 95° ≠ 90°  R1 + R2  90 ∴ TQ is not a diameter (∠subt. by TQ is not a right angle) ∴TQ is nie ʼn middellyn nie (∠onder span deur TQ  TQ is not a diameter is nie ʼn reghoek nie) TQ is nie ’n middellyn nie (2) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 16 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE 2019) QUESTION 10/VRAAG 10 10.1 Construction: Draw heights h and l on KR and KS construction / konstruksie respectively. Join LS and MR Konstruksie: Teken hoogtes h en l op KR en KS onderskeidelik. Verbind LS en MR ratio of areas / verhouding van oppervlaktes Proof/Bewys: 1 Area of ΔKRS .KR.h KR 2 Area of ΔLRS 1 RL ratio of areas .RL.h 2 verhouding van oppervlaktes 1 Area of ΔKRS .KS. KS 2 Area of ΔMSR 1 SM .SM. 2 same base and same height But/Maar: area of/van ΔLRS area of/van ΔMSR dieselfde basis en hoogte Area of ΔKRS Area of ΔKRS Area of ΔLRS Area of ΔMSR  Area of ΔKRS = Area of ΔKRS Area of ΔLRS Area of ΔMSR KR KS ∴ RL SM (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNIE 2019) MATHEMATICS P2/WISKUNDE V2 17 10.2 10.2.1 AB = 5 units 5 (1) 10.2.2(a)  C is common / is gemeen Statement/bewering/   CBA = CFB (both/beide = 90°) Reason / rede   Statement/bewering/ CAB = CBF (sum of s of Δ)/(som van die e van ) Reason / rede  ΔCBA|||ΔCFB (, , ) Reason/Rede (3) 10.2.2(b) CB CA proportion / = (|||Δs) CF CB verhouding CB2  CF.CA reason / rede CB2  CB2  CF.CA  CF = CA (3) 10.2.3 CB2 substitution / CF = vervanging CA (12)2 CF= length of CF / lengte 13 van CF  11units/eenhede (2) 10.2.4 AF = 13 – 11 = 2 units/eenhede length of AF / lengte van AF (1) 10.2.5 CB CA = (prop. theorem/ verh . stelling ; D F || BA ) BD AF proportion / 12 13 verhouding = BD 2 reason / rede 24  BD  13 length of BD / CF CB lengte van BD = (prop. theorem/ verh . stelling ; DF || BA ) FE BD 11 12 proportion / = verhouding FE 24 13 22 length of FE / lengte  FE = units/ eenhede van FE 13 (5) [20] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

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