Mathematics Grade 12 Paper 2 North West Mock 2018 Memo

Downloaded from hlayiso.com Education and Sport Development Department of Education and Sport Development Departement van Onderwys en Sportontwikkeling Lefapha la Thuto le Tlhabololo ya Metshameko NORTH WEST PROVINCE NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT " MATHEMATICS P2/WISKUNDE V2 " nm . u 1 SEPTEMBER 2018 1 iy n : MEMORANDUM * + SESS SSSR SSS SEU MARKS/PUNTE: 150 This memorandum consists of 22 pages./ Hierdie memorandum bestaan uit 22 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 NOTE: 2 NSC/NSS — Memorandum NW/September 2018 Ifa candidate answers a question TWICE, only mark the FIRST attempt. * Ifa candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version. * Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the second calculation error. ¢ Assuming answers/values in order to solve a problem is NOT acceptable. NOTA: ¢ As ‘n kandidaat ‘n vraag TWEE KEER beantwoord, merk slegs die EERSTE poging. ¢ As ‘n kandidaat ‘n antwoord van ‘n vraag doodtrek en nie oordoen nie, merk die doodgetrekte poging. ¢ Volgehoue akkuraatheid word in ALLE aspekte van die nasienmemorandum toegepas. Hou op nasien by die tweede berekeningsfout. ¢ Aanvaar van antwoorde of waardes om ‘n probleem op te los, word NIE toegelaat nie. QUESTION/VRAAG 1 11 _ 5478 v 5478 11 ll ¥ = 498 NOTE: ANSWER ONLY, FULL MARKS v 498 @ 12° | o, =119,47 v Vanswer/antwoord @) 1.3 ( -o,;X+0,) (498-119,47 ; 498 +119,47) Y 378,53 (378,53 ; 617,47) Y 617,47 «5 distances /afstande v 5 Q) 1.4.1 | new average/nuwe gemiddeld : Y k=505 - 5555 b=—= lly = 5555-5478) y 4 > 77 k = 505 OR/ OF = ir OR/OF the value of y/ die waardevany : Y Ily=5555 — 5478 =k —¥ = 505-498 =7cm =7om y y Y y=7 ; (2) 1.4.2 | 119,47 vv correct answer / korrekte antwoord (2) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 3 NW/September 2018 NSC/NSS - Memorandum QUESTION/VRAAG 2 2.1 puathx Vv a=57,87 a = 57,87 and/en b = 0,05 ¥ b=0,05 a Y equation/ “. D = 57,87 +0,05x vergelyking (3) 2.2 r= 0,93 vv 0,93 (2) 2.3 Strong positive correlation / Sterk positiewe korrelasie ¥ Strong / sterk ¥ Positive / positief (2) 2.4 Jy = 57,87+0,05(465) Y — substitute 465 Dp =81,12% into eq. / vervang 465 in vgl. VY 81,12 (2) OR/ OF J =80,81% (calculator / sakrekenaar) VV 80,81% / 81% = 81% [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 4 NW/September 2018 NSC/NSS — Memorandum QUESTION/VRAAG 3 y B x E(-11 ; 0) F AC3;0) 9 3.1 Length of AD / Lengtevan AD: v —_——— 2 dy = y(-3+8) +(0-5) dip = y(-3+8) +(0-5)° dp = V'50 Y dyy =5V2 or/ of 7,07 diy = 52 or / of 7,07 units / eenhede Length of DE / Lengte vanDE: doy = \(-8 +11) +(5-0)° v = dy, = 34 or / of 5,83 units / eenhede dog = V34 ot / of 5,83 VY AE =8 Length of AE / Lengte van AE = 8units / eenhede Perimeter / Omtrek A ADE : = 5/2 +3448 =20,9units / eenhede v 2 aVi48 4g or / of 7,07+5,83+8= 20,90 ovlof 7,07 + 5, 6) 3.2.1 | Line from mdpt 1* side of A, parallel to 2™ side, bisects | “reason/rede 3" side / Lyn uit midpt van een sy van A, ewewydig aan 2” sy, halveer 3® sy, qd) 3.2.2 —11+(-8) 0+5 ¥ method/metode oD Yom 19 and/en Yo xg=-— 9 oy 25 2 “XG 3 “Ng 3 v ye 5 2 2 (3) 3.2.3 5/2 ¥ answer/antwoord FG =—AD= — orf of 3,54 units / eenhede L 2 (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 5 NSC/NSS — Memorandum NW/September 2018 3.2.4 0-5 My => —3-(-8) a> 5 =-1 Meg = Mgp [AD| FG] Through the point / Deur die punt (-4 | Y Me, = Myy [AD| FG] Y Mg = ~1 ¥ Subst./Vervang (-2:3) 2°2 =~, 19,5 » 2-2 ys-x-7 Vy=-x-7 : (4) 33 | tanDAO = m,, tan DAO = ~1 ¥ tanDAO = -1 . DAO = - 45°+180° DAO = 135° Y DAO =135° 2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 6 NW/September 2018 NSC/NSS — Memorandum 3.4 DAB = 75° [opp.Zs eyelic quad./ teenoorst.Zekvh| ¥ DAB = 75° OAB = 135°-75° yok = 60° OAB = 60 tan OAB = May tan 60° = m,, = V3 Y tty =v3 equation / vergelyking AB through / deur (-3;0): y-0=43 (x+3) y = v3x43V3 vy = V3x4+3V3 at / by B: v SAS, E58 = V3x43V3 ABSA, Mes3 = Vie 3V3 “Bah, ~ Jax =3V3 — esV8 (4 8) _ sp 228 Vez x=2 x=2 and/en oy =¥3(2)+3V3 OR/OF y= SEAS (HSB ny = 53 V y= 53 . (7) £B(2:5V5) 3] QUESTION/VRAAG 4 4.1 v4yer? ¥ subst. into eq./vervang vty? =(23) in vgl. x+y? =12 ¥ answer/antwoord Q) 421 | 3? —6x4(-3P +y? -2y4(—1)? = 6+(-3" +(-1) Y (x-3P +(y-1) : = Y 16 .(x-3P +(y-1? = 16 . (2) 4.2.2 |M@G;1) are v =] (2) 4.2.3 | 4 v answer/antwoord (1) Copyright teserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 7 NW/September 2018 NSC/NSS — Memorandum 4.2.4 Q&3y) YMQLPQ dyp = (6-3) +(-2-1)° v dyp = V18 dyp = V18 or / of 3N2 dup = 3/2 Pythagoras : PQ? = 3V2)?-4? ¥ PQ? = (3V2)?~4? PQ’ =18-16 “PQ= V2 units / eenhede ¥PQ = V2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 8 NW/September 2018 NSC/NSS — Memorandum 43 y B(2k; 10) 20 95 C} —;— 13°13 . . ACK; 5) \ x, 43.1 | AB =6+7=13 V AB=6+7=13 dy, = «f(2k-(-K)) +(10-5) 13° = (3k) +(5)’ 169 = 9k? +25 144 = 9k? 16 =k? k=2+4 k= 4 (seediagram / sien diagram) “.A(~4;5) and / en B(8;10) Gradient of AB / Gradiént van AB ¥ 13? = Bk) +(5)P Vi? =16 Vk=4 v . 10-5 5 . : aa tary Correct subst. into gradient / vervang korrek in gradiént (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 NW/September 2018 NSC/NSS — Memorandum 5 5 = Y¥M,,=— Mas = 35 @ 3k 95 B™ 95 Mc 20 B 5 +k YM ys = 3 13 4k 13 _ 95-65 20+13k 5 30 v = - 95 —65 3k 204+13k 3k = 20+13k 100 + 65k = 90k 100 = 25k k=4 Vk=a v iM. => a> Correct subst. into gradient / @3(4) 12 vervang korrek in gradiént G) 4.3.2 radius | tangent / radius Lraaklyn o Meangent/raaklyn --2 5 Equation of common tangent / Vergelyking van gemeenskaplike raaklyn 95 12 *) Pa t4{ , 48 13. °=«5 13 12 “ysz-sxtll y 5 12 Y Menges /rcakiyn = 7 Y Subst. / vervang (32) G) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 10 NW/September 2018 NSC/NSS — Memorandum QUESTION/VRAAG 5 5.1.1. | sin106° = sin (180°-74°) = sin 74° = cosl6° ¥ cosl6° _ vk vik (2) =o 7 S12 | 5 \e-k make use of diagram/ sinl6° = ; maak van diagram gebruik y Pk vv ye ~k OR/OF 6 | | sin?16°+cos? 16° = 1 v Vek t 2 OR/OF sorrers[) =1 k ¥ sin? 16°+cos?16° =1 in? 16° = 1-— k si a v 5 : ek 216° = a sin* 16 fa v ‘ 6) 2 sinl6° = rok t 5.1.3. | cos16° = cos 2(8°) = 2.cos?(8°)—1 ¥ cos2(8°) vk = 2c0s?(8°)—1 v cos16° = 2cos?(8°)~1 wea = 20s? (8°) v VE 1 vk = 2cos?(8°)~1 op ty = 8B) t cos8? = vk 1 = Vie +t ve 1 or/of 22° V2 V2 2 Vk +t 2t (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 11 NW/September 2018 NSC/NSS — Memorandum 5.2 LHS/ LK yA(L= cos @)(1+ cos) (1—cos6)(1+ cos 6) sin20 v ; =1~-cos’é _ /4(1—cos? 6) ~~ Qsin 8 cosO ¥ 1-cos* =sin? 6 +2 = 2ysin’ VY sin26 =2sin@cosd 2sin@cosO 2sin@cos@ 2sin 0 cosO 4 =| _pns/rx 4) cos@ 5.3.1 | sin p+V3cosp =1 1. 3B 1 1 gin +33. cos =i (+2) Yo sin p +508 p “2 ger PET 2 cos 60°sin p+ sin 60° cos p = 1 (special angles / spesiale hoeke) ¥ cos 60°sin p 1 2 ¥ sin 60° cos p sin (60°+ p) = 3 (compound angles / saamgestelde hoeke) (3) 532; 1 v sin (60°+ p) = > ref.Z/verw.Z = 30° ref.Z/verw.Z = 30° v 30°4m360°/ 60°+ p =30°+n,360° or/of 60°+p = (180°-30°)+n.360° | 2 = 330 oe eo p= ern, = - oO hea = Oo Oo p= —30°+n,360° or/of p=90°+n360° (neZ) Y p= 90°+n360° Y (neZ)if used in correct context / as dit in korrekte konteks gebruik is (4) 5.4 cos 0° + cos1°+cos 2° +... + cos(180 — 2)° + cos(180~1)° + cos 180° + 2 ¥ cos(180—2)° + cos(180~1)° 1+cos1°+cos2°...+ (—cos2°) +(—cos1°)+ (1) +2 1+(-)+2 =2 v ...(— cos 2°) +(~cos1°) v2 6) [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 Mathematics P2/Wiskunde V2 QUESTION/VRAAG 6 NSC/NSS ~ Memorandum NW/September 2018 VV p =30° 6 p = 30° and/en g =1 pP * 2 V¥qa-4 2 (4) 6.2 x =—150° or/of x = 30° v¥~150° v¥ 30° (2) 6.3 The graph of f must be translated 60° to the left / Die grafiek | YV60° to the lef/na van f moet 60° na links getransleer word. links (2) The graph of f must be translated 120° to the right and then be reflected about the x-axis/ Die grafiek van f moet 120° na regs getransleer word en dan in die x-as gereflekteer word. OR/OF v¥ 120° right/regs v — reflect about x- axis/reflekteer in die x-as. {8} Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 13 NW/September 2018 NSC/NSS— Memorandum ‘D QUESTION/VRAAG 7 E 1 1 1 H 1 1 ' tT i 1 A , 71 | B=90° BC x cosa = — = — AC AC x “ AC = COS a @) 72 | ECA =90°-6 YECA = 90°-6 AEC = 180° - 6-(90°—6) ¥ AEC = 180°~8-(90°-6) AEC = 90°- 8 +0 or/of 90°-(B-0) y AEC = 90°-8+6 or/ of 90°-(B-8) or/ of 90°+(6-f) or/ of 90°+(8—-f) (3) 73 InA ABC AC =— (see / sien 7.1) cos@ InAACE . ares Y Correct application of sine-rule/ sing = SMAEC korrekte toepassing van sinus-reél EC x OS a . * _xsin B xsin B VEC = OS ____ = COS sin[90° -(2-6)] sin[90°-(f-6)] xsin B = COS a cos(—A) ¥ sin[90°—(2 —0)] = cos(8—-@) _ xsin B cosa.cos(f—6) G3) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 14 NW/September 2018 NSC/NSS — Memorandum GEOMETRY/MEETKUNDE Ss A mark for a correct statement (A statement mark is independent of a reason.) ‘n Punt vir ‘n korrekte bewering (‘n Punt vir ‘n bewering is onafhanklik van die rede. ) R A mark for a correct reason (A reason mark may only be awarded if the statement is correct.) ‘n Punt vir ‘n korrekte rede (‘n Punt word slegs vir die rede toegeken as die bewering korrek is.) S/R Award a mark if the statement AND reason are both correct. Ken ‘n punt toe as beide die bewering EN rede korrek is. QUESTION/VRAAG 8 8.1 8.1 6, = G +A [ext.Z of A/buite Z van Al YSIR But /maar G =A [Zs opp. equal radii / Ze teenoor gelykeradiusse] | VS/R “0, = 2C, VS In the same manner / op dieselde wyse : 6, =2¢, vs 6, +6, = 26, + 2, .. AOB = 2¢, +2, = 2( C+ ¢,) vS = 2ACB (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 15 NW/September 2018 NSC/NSS — Memorandum 8.2 8.2.1 | D, = 65° [Zat centre = 2x Zat circumference / midpt.Z = 2x omtreksZ] | ¥SYR (2) 8.2.2 B, = 65° [tangent chord theorem /raaklyn — koordstelling | YSYR Q) 8.2.3 | BAD = 65° [alt Zs/verwiss Ze; AD / /EF] yYSR (1) 8.2.4 | 6 =115° loppZs of cyclic quad / teenoorst.Ze van kvh] YSYR (2) 8.2.5 | GBF = 90° [rad 1 tangent / rad raaklyn| VSIR DBF = 65° [alts / verwiss Ze; AD | /EF] YSIR B, = 90°-65° = 25° vs OR/OF _ B, + B, =90° [rad 1 tangent /rad L raaklyn] VSR A v G, = 90° [alts /verwiss Ze; AD //EF | eR B, = 180°— 65°-90° = 25°[ Zs of A/binne Ze van A) @) 8.2.6 | G, = 90° [co-interior Zs / ko - binneZe; AD | /EF]/ [Proved in 8.2.5] ¥ en AG = GD [line from centre | chord; Loodlyn vanuit midpt sirkel na koord ] .op=¥7 vi 4 4 3) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 16 NSC/NSS — Memorandum QUESTION/VRAAG 9 Let/Laat é, =X, NW/September 2018 911 G = A, =x [tangent chord theorem /raaklyn — koordstelling | VSvR @uhe equal circles, equal chords, equal “s | /gelyke sirkels, gelyke koorde, gelykeZe YSYR , =C, (4) 92 1) C40, =A, +A, [ext. Zs of cyclic quad / buiteZe van kvh] vSYR But/Maar C, =A, = x [in9.1] “Ss AGA, OR/OF E, =180°-2x [sum of sof A /somvanbinneZevan A] “SIR B =180°-(1 80°—2x) =2x [opps of cyclicquad / teenoorst.Ze vankvh] | YS/R 6,+6,4+6, = B+A, [ext.Zof A / buiteZ van A] But / Maar 6, +¢, =x+x = 2x=B YS u, =A, (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 “417 NW/September 2018 NSC/NSS —- Memorandum 9316 =A, [see/ sien 9.1] “CE = AE [sides opp. = Zs/sye teenoor = Ze| VSR A, =D [tan-chord theorem / raaklyn — koord — stelling] VSR A, =¢, [from / van 9.2] ude, “.CE=DE [sides opp. = Zs/sye teenoor = Ze| YSIR CE = AE =DE “8 . Eis the centre of the circle/is die middelpunt van die sirkel (4) 9.4 | AE = EC = ED [radii / radiusse] vs But / Maar CD = AE [Given / gegee] “. CD = DE = CE vs ~ ..AECD is equilateral / gelyksydig (13) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

QUESTION/VRAAG 10 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 18 NSC/NSS — Memorandum NW/September 2018 R 10.1 Parallel to the third side of the A/ Ewewydig aan die derde sy van | YR qd) die A 10.2.1 | R = 90° [tangent radius /raaklyn Lradius| YSVYR Q, =90° [tangent _L radius /raaklyn Lradius] YS/R Q, =R =90° . PQ//SR [corresponding /s are equal / ooreenk. Ze is gelyk] YR OR/ OF ‘ R =90° [tangent 1 radius /raaklyn Lradius| VYSYR Q, = 90° [tangent L radius /raaklyn Lradius| YSIR “R+Q, =180° “.PQ{|SR [co-interior Zs = 180°; ko —binneZe saam180°] vR (4) 10.2.2 = = oR [prop.theorem; PQ |[SR /eweredigheidst;PQ||SR] | vg vp Tp = FQPS QR PC = BP and/en SC = SR [radii / radiusse] vS/R PS = PC+CS = BP+SR vs TP = LQBP+SR) (4) QR Copyright reserved/Kopiereg voorbehou Please turn over/Blaai'om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 19 NW/September 2018 NSC/NSS — Memorandum 10.2.3 | InATQP and/en ATRS A on vw T =T [common/ gemeenskaplik] s Q, =R [corresp.2s / coreenk.Ze;PQ || SR] vs B =§ [sum of Zs A/som van Ze ny vs - ATQP ||| ATRS OR/ OF T=T [common/ gemeenskaplik | vs Q, =R or /of 8, =§, [corresp.Zs / ooreenk.Ze;PS || QR] vS - ATQPIATRS [2,4,2] VR (3) 10.2.4 | TS’-SR? = TR? [Pythagoras] vs ~ CS= SR [radii / radiusse] .. TS? -CS? = TR? TQ_ QP LT Ihhas] VSIR TR RS_ TS rR = QRS vs QP But / Maar QP = BP and/enRS=CS [radii/radiusse] “TR= TQ.CS BP TQ? = TP? +PQ? -2TP.PQ.cosP, But/Maar PQ = BPand/en §, = P, TQ = (TP? +BP? —2TP.BP cosS) _ TACs? TR? - BP TO’Cs? BP? . ee - (te? +BP? —2TP.BP cosS) CS BP “. TS? -CS? = Vuse cosine-rule to find TQ/ Gebruik cosinus- reél om TQ te vind 2 2 / TQS BP? v Ts cs? = 1YCs" BP* (6) [18] Copyright reserved/Kopiereg voorbehou TOTAL/TOTAAL: 150 Please turn over/Blaai om asseblief