Mathematics Grade 12 Paper 2 Eastern Cape Mock 2020 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2020 MATHEMATICS P2/WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 13 pages. Hierdie nasienriglyn bestaan uit 13 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2020) QUESTION 1/VRAAG 1 1.1 a  4,1536  a  4,1536 b  0,958  b  0,958 y  4,1536  0,958x  y  4,1536  0,958x Answer Only: Full Marks (3) 1.2 r  0,98  r  0,98 (1) 1.3 Very strong positive correlation/ answer / antwoord Baie sterk positiewe korrelasie (1) 1.4 y  4,1536  0,958(51) substitution / vervanging y  45% answer / antwoord Answer Only: Full Marks (2) 1.5 x  60,8 Standard deviation / Standaardafwyking = 17, 51  Standard deviation/  60,8 17,51 ; 60,8  17,51 Standaardafwyking = 17, 51  43, 29 ; 78,31   43, 29 ; 78,31 6 learners / leerders 6 learners / leerders (3) [10] QUESTION 2/VRAAG 2 2.1 Frequency Cumulative Frequency Speed/Spoed Frekwensie Kumulatiewe Frekwensie ( km/h) (f) 0 < x  10 10 10 freq column / frek. kolom 10 < x  20 20 30 20 < x  30 45 75 30 < x  40 72 147 cum freq column 40 < x  50 23 170 kum frek kolom (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2020)) MATHEMATICS P2/WISKUNDE V2 3 2.2 OGIVE/OGIEF 200 Cumulative Frequency starting pt 150 / beginpunt 100 end point Kumulatiewe Frekwensie eindpunt 50 0 shape 0 10 20 30 40 50 60 vorm Speed/Spoed in km/h (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2020) 2.3 Q1 = 23 (accept / aanvaar 22 – 24)  Q1 Median / Mediaan = 31 (accept / aanvaar 30 – 32)  Median Mediaan (2) 2.4  for / vir Q3 = 37 (accept / aanvaar 36 – 38)  correct shape / korrekte vorm (2) 2.5 170 – 110 = 60 cyclists / fietsryers  answer / (accept / aanvaar 59 – 61) antwoord (1) [10] QUESTION 3 / VRAAG 3 3.1  2  (4) 1 substitution / vervanging mQR   answer / antwoord 06 3 (2) 3.2 mPQ  3  mPQ  3 1 1 mPQ  mQR  3   1  mPQ  mQR  3   1 3 3 ˆ  90  PQR (2) 3.3 Sub/Verv: y   x  2 into/in 3x  y  2  0  3x  ( x  2)  2  0  substitution / vervanging 3x  x  2  2  0  x – coordinate / x – 4x  4 koördinaat x 1  y – coordinate / y – y 1 koördinaat  P(1;1) (3) 3.4 QR = √(0 − 6)2 + (−2 − (−4))2 substitution in correct f vervanging in korrekte f QR  2 10 OR/OF 40 OR/OF 6, 32 answer / antwoord Answer Only: Full Marks (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2020)) MATHEMATICS P2/WISKUNDE V2 5 3.5 PR is the diameter (angle subtended by diameter =  for the statement PR is the 90 ) / PR is die middellyn (hoek onderspan deur diameter / vir stelling PR is middellyn = 90 ) die middellyn 7 3  Midpoint of PR Midpoint of / Middelpunt van PR  ;  Middelpunt van PR 2 2 1  6  1  4 2 2 PR  PR  50 50  for the radius / r 2 vir die radius 2  2 7  3 2  50   x     x       equation / vergelyking  2  2  2  OR/OF 2   x  3,5  x  1,5   50  2 2  2  (5) 3.6 ˆ  1 tan PNX ˆ  1  tan PNX ˆ  135  PNX  PNXˆ  135 ˆ 3 tan PMX ˆ 3  tan PMX ˆ  71,57  PMX ˆ  71,57  PMX   135  71,57  63,43  answer / antwoord (5) 3.7 1 formula / formule A   PQ×QR 2 1 A PQR   10  40 √10 2 APQR  10 square units / vierkante eenhede answer / antwoord OR/OF 1 A PQR   PQ×PR  sin 63, 43 2 1 A PQR   10  50  sin 63, 43 2 APQR  10 square units / vierkante eenhede (3) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2020) QUESTION 4 / VRAAG 4 4.1 x 2  6x  y 2  4 y  9  0 completing square x 2  6 x  9  y 2  4 y  4  9  9  4 voltooiing van vierkant x  32   y  22  4 standard form / C (3; 2) and / en r  2 standaardvorm 3 2 (4) 4.2 mtan  2 1 m BV  1  m BV  2 2 1 y  2  ( x  3) substitution / vervanging 2 1 1 answer / antwoord y  x 2 2 (3) 4.3 y4 answer / antwoord (1) 4.4 TA = 4 units / eenhede length of TA / lengte van TA TB = TA (tangents from the same point) S R (raaklyne vanaf dieselfde punt) answer / antwoord TB = 4 units / eenhede (4) 4.5 T(1; 4) y  2 x  k 4  2(1)  k substitution / vervanging k 2 answer / antwoord (2) 4.6 ˆ  2 tanSTA ˆ  2  tanSTA ˆ  116,57 STA ˆ  116,57  STA ˆ  116,57 (ext.angle of a c.q.)  ACB answer / antwoord (buitehoek van koordevierhoek) reason / rede OR/OF OR/OF Draw/Trek: CE || OX ; then/dan ˆ  1  tan VCE ˆ 1 2 tan VCE ˆ  26,57 2  VCE ˆ  26,57  VCE ˆ  180  (90  26,57)  116,57 answer / antwoord  ACB reason / rede (  s on straight line / e op reguitlyn) (4) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2020)) MATHEMATICS P2/WISKUNDE V2 7 QUESTION 5/VRAAG 5 5.1.1 cos 158   cos 22   cos 22  p  p (2) 5.1.2 sin 112  sin(90  22)  cos 22  cos 22  p p (2) 5.1.3 sin 38  sin60  22  sin60  22  sin 60 cos 22  cos 60 sin 22  expansion / uitbreiding 3 1 3 1  p 1 p2  p  1 p2 2 2 2 2 (4) 5.2 sin P  sin 2P sin P  sin 2P  0  standard form / sin P  2sin Pcos P  0 standaardvorm sin P(1  2cos P)  0  expansion / uitbreiding 1  factorisation / faktorisering sin P  0 or/of cos P  2  all correct values of P P  0 ;60 ;180 ;300 ;360 alle korrekte waardes van P OR/OF  P  2P  360k P  2P  360k or/of P  180  2P  360k ,  P  180  2P  360k 𝑃∈ℤ  P  60  120k P  360k or/of 3P  180  360k  all correct values of P P  60  120k alle korrekte waardes van P P  0 ;60 ;180 ;300 ;360 (4) 5.3 A  B  C  180 A  B  180  C  A  B  180  C cos(A  B)  cos(180  C)  cos(A  B)  cos(180  C) cos(A  B)   cos C (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2020) 5.4 cos 2 x  cos x  sin 2 x 1 1   2sin x cos x  sin x tan x sin x cos x  cos x  sin 2 x 2 LHS / 𝐿𝐾 = 2sin x cos x  sin x   cos x  cos x  1  cos 2 x 2   1 cos 2 x sin x(2 cos x  1)  sin x(2 cos x  1) 2 cos 2 x  cos x  1   2 cos 2 x  cos x  1 sin x(2 cos x  1)  2 cos x  1cos x  1 (2 𝑐𝑜𝑠 𝑥 + 1) (𝑐𝑜𝑠 𝑥 − 1) sin x(2 cos x  1) cos x 1 1 1 cos x 1     = RHS / RK   sin x sin x tan x sin x sin x sin x (5) 5.5 4  7 cos   cos 2  0 4  7 cos   2 cos 2  1  0  2 cos 2  1 2 cos 2   7 cos   3  0  standard form / (2 𝑐𝑜𝑠 𝜃 + 1) (𝑐𝑜𝑠 𝜃 + 3) = 0 standaardvorm 1  factors / faktore cos    or/of cos   3 (N/A) 2 1   120  360.k or/of   240  360.k , 𝑥 ∈ ℤ  cos    or/of cos   3 2    120  360.k    240  360.k (6) [25] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2020)) MATHEMATICS P2/WISKUNDE V2 9 QUESTION 6 / VRAAG 6 6.1 1 answer / b antwoord 2 (1) 6.2 A (30;1)  30 1 (2) 6.3 g (90)  cos(90  30)  cos 60 1  90  1 2   1 2 Q  90;   2 (2) 6.4 x  160  x  160 (1) 6.5 1  y  3 yR OR/OF answer / y   1;3 antwoord (2) [8] QUESTION 7 / VRAAG 7 7.1 ˆ LNM  180  2 p (angles opp. = sides) answer / antwoord (hoeke teenoor gelyke sye) reason / rede (2) 7.2 LM d   for applying the sine rule sin(180  2 p) sin p LM d gebruik van sinusreël  sin 2 p sin p  sin 2P d sin 2 p LM  sin p (2) 7.3 h tan q  h LM  tan q  h  LM tan q LM d sin 2 p h . tan q d sin 2 p sin p h  . tan q 2d sin p cos p tan q sin p h sin p 2d sin p cos p tan q h  2d cos p tan q h  sin p (3) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2020) QUESTION 8 / VRAAG 8 8.1 bisects the chord / halveer die koord  answer/antwoord (1) 8.2 EB  8  y  for/vir EB In AEB : 10  x  (8  y) ………….(1) 2 2 2 Pythagoras in AEB Eqn of the circle / Verg. van die sirkel: equation of the circle x 2  y 2  64 vergelyking van sirkel x 2  64  y 2 ………………..(2) Subst./Verv. (2) into/in (1)  substitution 100  64  y 2  64  16 y  y 2 vervanging 100  128  16 y 16 y  28 7 answer / antwoord y 4 7  OE  4 (5) 8.3 Double the size of the angle subtended by the same answer / antwoord arc. Dubbel die grootte van die hoek wat deur dieselfde boog onderspan word. (1) 8.4.1 ˆ (  at centre = 2  at the circumf) Ô 2 = 2B statement / stelling 2 reason / rede ( Middelpunts  = 2 x Omtrekshoek) (2) 8.4.2 Ĉ3 = D̂1 + D̂ 2 (  s opp = sides)/ (e teenoor = sye) statement / stelling reason / rede (2) 8.4.3 B̂1 + B̂2 = 180 – ( D̂1 + D̂ 2 ) (opp. s of a cyclic quad) statement / stelling reason / rede (teenoorst. e van ’n koordevierhoek) (2) 8.4.4 D̂1 = Ĉ1 (  s in the same segment) statement / stelling reason / rede (e in dieselfde segment) (2) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2020)) MATHEMATICS P2/WISKUNDE V2 11 QUESTION 9 / VRAAG 9 9.1 ˆ  80 (  s opp = sides)/(e teenoor = sye) PCQ statement / stelling ˆ  100 (  s on a straight line) reason / rede PCB statement / stelling (e op ’n reguitlyn) reason / rede  BC is not a diameter (angle between the tangent conclusion / and BC is not equal to 90 ) gevolgtrekking BC is nie ’n middellyn nie. (hoek tussen die raaklyn en BC is nie gelyk aan 90 nie) (5) 9.2 Pˆ1 = B ˆ (alt  s, PQ || AB) / (verw.e, PQ || AB) statement / stelling reason / rede B̂ = Cˆ (  s opp = sides; radii) 3 statement / stelling (e teenoor = sye: radiusse) reason / rede ˆC = Cˆ ( vert. opp. angles) / (regoorst. hoeke) 3 1 statement / stelling  P̂ = Cˆ 1 1 statement and reason  PQ = QC (sides opp = angles) stelling en rede (sye teenoor = hoeke) (6) 9.3 ˆ ˆ (ext.  of a cq) / (buite van koordev.) A=E statement / stelling 2 reason / rede D  180  Eˆ (co–interior  s; BE || CD) ˆ 2 statement and reason (ko-binne e : BE || CD) stelling en rede ˆ  180 D̂+A  ACDF is a cq (opp  s supplementary) / statement and reason is ’n koordev (teenoorst. e is suppl.) stelling en rede reason / rede OR/OF ˆ ˆ (corres.  s; BE || CD) D=E 1 statement / stelling (ooreenk. e : BE || CD) ˆE  180  Eˆ (  s on a straight line) reason / rede 2 1 ( e op ’n reguitlyn) statement and reason Aˆ  180  Eˆ (opp  s of a cq) 1 stelling en rede (teenoorst. e is suppl.) statement and reason D̂+A ˆ  180 stelling en rede  ACDF is a cyclic quad./ is ’n koordevierhoek (opp  s of a quad. supplementary) reason / rede (teenoorst. e van koordev. is supplementêr) (5) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2020) QUESTION 10 / VRAAG 10 10.1 M area / oppervlakte k h S T O N MS MT RTP/TE BEWYS: = SO TN Construction: Join SN, and OT, and construct perpendicular heights / Konstruksie: Verbind SN en OT, trek loodregte hoogtes Proof / Bewys: 1  MS  h area of the two triangles area ΔMST 2 MS   area van twee driehoeke area ΔOST 1  SO  h SO MS 2  1 SO  MT  k area ΔMST 2 MT   area of the two triangles area ΔTNS 1  TN  k TN area van twee driehoeke 2 MT  But / Maar MST is common / gemeen TN And / En  statement and reason area of OST = area of TNS (same base, stelling en rede same height) area van OST = area van TNS (dies. basis en dies. hoogte) MS MT  = SO TN (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2020)) MATHEMATICS P2/WISKUNDE V2 13 10.2.1 In APS and/en BRS Pˆ4 = Rˆ 1 (tan – chord theorem) statement and reason (raaklyn-koord Stelling) stelling en rede ˆ A = ˆB2  90 (given) / (gegee) statement / stelling APS ||| BRS (AAA) / () 3rd  / 3de  OR/OF reason for similarity rede vir gelykvormigheid (3) 10.2.2 AP PS AS = = (similar triangles)  for the statement BR RS BS (gelykvormige driehoeke) vir die stelling  AP.RS = BR.PS (1) 10.2.3 P̂  90 (  s in a semi – circle)  P̂2  90 2 ( e in ’n semi-sirkel) (  in a semi – circle) / Let/Laat: P̂4  x ( e in ’n semi-sirkel)  Ŝ1  90  x Ŝ1  90  x (  s of APS) / (e van APS)  Q̂  90  x  Q̂  90  x (ext  of a cq) / (buite van kv)  R̂ 2  x  R̂ 2  x (  s of QPR) / (e van QPR)  Pˆ = Rˆ 4 2 (4) 10.2.4 In ASP and/en PQR ˆ = Pˆ (proven / bewys) statement and reason A 2 stelling en rede Pˆ4 = Rˆ 2 (proven / bewys) statement / stelling reason for similarity ASP ||| PQR (AAA) / () rede vir gelykvormigheid AS SP AP = = (similar triangles) PQ QR PR (gelykvormige driehoeke)  AP . QR = SP . PR AP PR AP PR  =  = PS RQ PS RQ AP BR AP BR But / Maar: = (from / vanaf 10.2)  = PS RS PS RS PR BR PR BR  =  = RQ RS RQ RS  BR.RQ = RS.RP (6) [19] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief