Mathematics Grade 12 Paper 2 Limpopo Mock 2020 Memo

Downloaded from hlayiso.com LIMPOPO PROVINCIAL GOVERNMENT REPUBLIC OF SOUTH AFRICA OEPARTMENT OF EDUCATION NATIONAL SENIOR CERTIFICATE GRADE/GRAAD 12 MARKS/PUNTE: 150 This memorandum consists of 15 pages / Hierdie memo bestaan uit 15 bladsye.

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 2 LimpopoDoE/September 2020 NSC/NSS/Memorandum QUESTION /VRAAG 1 11.1 A | { v¥ box /mond /¥ whiskers, on B —— —— scale/ stert, en op skaal 10 20 30 40 50 60 70 80 90 (2) 1.1.2. | School B/skool B: ¥School /skool B In B 50% of learners got above 60% whilst 50% of learners of A above 55% / 50% van B het meer as 60%; by A het 50% meer as 55% J Jany two median of B > median of A/ reasons mediaan van 8 > mediaan van A : 75% of B above 30% and at A 75% above 23,5% / Ce eds 75% van B bo 30%; By A75% bo 23,5% OR/OF VS Skewed to the right/ skeef na Regs OR/ OF vv Q) 1.2 135 24 + 225x 32 +200 x = J numerator/ 560 teller 3240 + 7200+ 200x = 560x 560 360x = 10440 x= 29 v 360x = 10440 v answer/antwoord (4) [9] QUESTION/VRAAG 2 21 a=2319 b=0,768.. Vavb y =23,19+ 0,77x v y =23,194+-0,77x 3) 2.2 | y=23,19+,77(30) v = 46,29 substitution/vervanging = 46% ¥answer/antwoord Q)

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 3 LimpopoDoE/September 2020 NSC/NSS/Memorandum 23.1) 6=10,99 ¥/answer /antwoord (2) 2.3.2 | 57,4% —49,4%=8% L8% eee 50 100 x =4 marks ¥4 marks/ punte (2) 9} QUESTION /VRAAG 3 P po S x 0 2 IT (0,-5— (0,-5. > 3.1 | X+ay=a ¥x=0 otay=a y=l vy=l - P(Q;1) v¥ P(0;1) (3) 3.2 | OS:OP=3:1 ¥ S(3;0) 8(3;0):x+ay=a 34a(0) =a ¥ substitution/vervang a=3 Yva=3 3) 3.3 1 Mps =— > V tips => 3 Mp =3 o iy, =3 RT: y=3x- st Jequation/ ver gelyking (3) 34 xX+3y=3 > ys--x4t1 ao x+1=3x al 3 ~x43=9x-17 ¥ equating /stel gelyk

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 4 NSC/NSS/Memorandum LimpopoDoE/September 2020 Substitute: y = -32) +1 v simplification/vereenvaudig y=3 R24) 3 v¥x=2 1 Vy=- an) 4 a areaAPRT = 2x PT xh v PT =67 ab 62x Yh=2 2 3 20 - : =| rie units/ vierkante eenhede J¥answer/ antwoord (3) 3.6 | PT is the diameter(converse: Z in semi ©)/ YSIR OPT is die middellyn (omgekeerde: 2 in semi ©) . 10... radius = 3 units. ZYanswer/antwoord (2) [18]

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 5 NSC/NSS/Memorandum LimpopoDoE/September 2020 QUESTION/VRAAG 4 Vv 4.1.1 | x? ~6x494y? —4y 442124944 Completing the square (x3)? + (y-2)? =25 vierkantsvoltooing C@G;2) ¥LHS/ LK r=5 /RHS/RK v¥ C(3;2) vr=5 (4) 4.1.2 | Y-intercept: /Y as:x =0 vx =0 2 y -4y-12=0 (y-6)(y +2) =0 v “entree y=—2or y=6~. A(0;6) standaardvorm 6-2 4 ¥ factors/faktore Mac => TZ 0-3 3 ¥ A(0;6) 3 : 4 Map = ai (tan 1 radius) ¥ yc = 5 -3 3 yy* 6 V tay = a YR J equation/ vergelyking (8) 413) pet A=O: tano => J tang = 3 4 4 0 =36,87° v¥ 0 =36,87° a= 90° + 36,87° > 7 a =12687° ¥ a = 90° + 36,87 ¥answer / antonard fAN

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 6 NSC/NSS/Memorandum LimpopoDoE/September 2020 42 | x?42x4¢]4+y?-4y44=444144 (x+1) +(y—2)9 = 49 Centre / middelpunt D(—1;2) ; radius = 7 cD=3--1=4 ro try =5+7=12> 4 «. the two circles intercept internally/ twee sirkels sny inwendig “Centre D(-1;2) vt=7 ¥CD=4 Vr thy =12 w>4 ¥ conclusion / gevolgtrekking (6) [22] QUESTION/V RAAG 5 5. tanA= EI 2 5 R diagram vr=13 r=12° +5? = 169 r=13 13sinA —3tanB 5. 412 5 = 13(—) -3(— in A =— G) ZF) ¥sin A ic “3 7 12 ul ¥tanB = a 5 Yanswer /antwoord 6) 5.2 —tan60°—sin x J¥ —tan60° sin x + 2.¢0s30° . . v-sinx _ v3 -sin x a 3 v¥ 2c0s30° sin x +20 2 ) ¥ special angles/ _ —(sinx + V3) bekende hoeke (sin x + ¥3) =-] ~ factorisation/ faktoriseer v¥ answer/

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 7 NSC/NSS/Memorandum LimpopoDoE/September 2020 5.3.1 cos(x + y) =sin(90°-(x+ y) =sin((90° — x) + (—y)) = sin(90° — x) cos(—y) + cos(90° — x) sin(—y) =cosxcos y—sin xsin y “Correct co-fn/ korrekte ko — funksie J regrouping/ herrangskik ¥ Correct expansion/ uitbreiding ¥simplification/ vereenvoudig (4) 5.3.2 | cos(x— y)—cos(x + y) /Expansion/ =cosxcos y +sin xsin y—(cosxcos y~sin xsin y) brei uit - snces +sin xsin y—cosxcos y+sin xsin y Jsimpilifieation / =2sin xsin y vereenvoudig - - (2) 5.3.3 | 2sin195°sin 45° =cos(195° — 45°) — cos(195° + 45°) a =c0s150°—cos240° ¥ Substitution/ =—cos30° —(—cas60°) vervang 31 Roa ta ¥—cos30° _-v3+1 v¥—cos60° 2 ¥ Special angles/ bekende hoeke fanswer/antwoord (5) [22]

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 8 LimpopoDoE/September 2020 NSC/NSS/Memorandum QUESTION/V RAAG 6 ff 4 _ L | 4 i i | | LS H — _ +\} | a k: 1 — | 6.1 g: ¥x-intercepts/x—asse J y-intercepts/y — as ¥ turningpoints / draaipunte Q) 6.2 | 2sinx =cosxcos30°-sin xsin30° “Expansion of g/ sine cose sine brei uit 3 gin y= YB cose ¥ Special angles/ 2 2 bekende hoeke V3 sinx 2 5. 3 ae v¥ =sin x = —cosx cosx 5 2 2 2 tanx = 3 i 5 y sinx cosx (4) 6.3 V3 tanx = rs 419,12 x=19,1° or x =180°+19,1° = 199,1° /199,1° (2) 6.4.1 | x €[60°180°] W[240%;270°] ¥ 60°and /en180° ¥ 240° and / en270° “notation / notasie G3) 6.4.2 | x [19,1°3199,1°] ¥ criticalvalues / kritieke waardes notation / notasie (2)

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 9 LimpopoDoE/September 2020 NSC/NSS/Memorandum QUESTION /VRAAG 7 InAABC :tanx = eee CB v CB h CB= vcp=—" tanx tanx InABCD:CB = BD CD’ = BC’ + BD’ —2BCx BDxcosy ~ Correct application of cos rule/ h h h h korrekte gebruik van cosreél + -2 x x COS Ba tanx’ Canx? tanx tanx i ¥ substitution __2h* 2h" cosy ¥ simplify/ tan*x tan’ x vereenvoudig = 2H" (1-cos y) tan? x uM “factorising/ faktoriseer CD = ——./2(1-cos ae 20 ») CD= a fH Teosy an x [6]

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 10 NSC/NSS/Memorandum QUESTION/VRAAG 8 LimpopoDoE/September 2020 8.1.1 on = R, (tangents from same point)/ raaklyne uit dsde punt) | /SVR 2R, +70° =180° (2'sof APQR/ Z'e van APQR) VS 2R, =110° R, = 55° /answer/ antwoord (4) 8.12) 6 +0, =90° (tangent | radius)/(raaklyn 1 op radius) YSYR O, +55° =90° 0, = 35° J answer / antwoord Q) 8.1.3 | 6 =R, =35° (r=) YS/IR M,=110 (Z'sof AMQR) M, =M,=110 (=chords subtend =2’s at centre)/ YM, =110 (= koorde = middelpunts z'e) vR G3)

8.2 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 11 NSC/NSS/Memorandum LimpopoDoE/September 2020 8.2.1) ¥, =90° (line from centre to midpoint of chord)/ YSYR (van midpunt van ©, mdpt van koord) vs U,+U,=90° (2 insemi ©) +X, =U, +U, =90° VS => RU || SV (corresponding 2's =)/ VR (ooreenkomstige 2’e =) 6) 8.2.2) R, =y — (tan-chord theorem)/ VSVR (4 tussen raaklyn en koord) 0, =y {alternate 2’s, RU || SV y verwisselende 2Z’e, RU || SV ) vs Oo, = ar (4 at centre = 22at circumference) 4 (middelpunts< = 2xomtreksz) SSSR = 2” (3) (20)

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 12 NSC/NSS/Memorandum QUESTION/VRAAG 9 9.1 A LimpopoDoE/September 2020 9.1 Construction: Draw altitudes A and k. Konstruksie: trek hoogtelyne hen k. 1 areabADE 9% 4%4 ay areaABDE 1 Bpxh DB 2 VOAE k areab\ADE 9° *°™" 4g areaADEC 1 ecxk EC But the areaABDE = areaADEC (same base and same height/ dsde basis en hoogte ) _areahADE _ area\ADE “@reaABDE area\DEC _ AD _AE “DB EC /¥ Construction/ konstruksie vS vS YSVR vS (6)

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 13 LimpopeDoE/September 2020 NSC/NSS/Memorandum 9.2 A . J D M a c eee) Lees (line IIto one side of AAMDY | SVR EM FD 3 (lyn || aan een sy van AAMD) = = : J answer G) 9.2.2 | AM:MC=7:7 (diagonals of parm) YSVR CM 7 WE = 3 ¥ answer G) 9.2.3 | areaABDC = areaAABD (diagonals of parm) vS areaAFDC _ areaAFDC . areaABDC areahABD UT) 1 vervang i x FDx height = 1... ¥ area formula / 2 * AD x height oppervlak formule FD ; = a (between same parallel lines)/ (tussen dieselfde I Lyne) YR _ 3 71 (4) [16]

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 14 NSC/NSS/Memorandum QUESTION /VRAAG 10 LimpopoDoE/September 2020 10.1 | InAXYCand ADYX : X, =D, — (tan-chord theorem/ VSVR 4 tussen raaklyn en koord) on =X, (tan-chord theorem/ 4 tussen raaklyn en koord VS ¥,+¥,=¥,+¥,(z'sofa) + AXYC [| | ADYX (222) XY YC >= DY YX vS “XY? = YC.DY vs (5) 10.2 | 4,=2%, (2 at centre is twice 2 at circumference) | YSYR (middelpuntsz = 2x omtreksz) Similarly/netso B, = 2D, y But ¥, = 5, (tan-chord theorem/ ¢ tussen raaklyn en koord) 4, =B, vS (4) 103 | CACY cacay | | | YBX) vs YB YX R_cY ad ¥Substitute/ vervang R cy? rn? yx? . _CYcY v Square both sides/ CY. DY kwadreer albei kante of

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 15 LimpopoDoE/September 2020 NSC/NSS/Memorandum ae _ DY /Replace/vervang YX" R’ CY “answer /antwoord (3) [14]