Mathematics Grade 12 P2 North West Mock 2019 Memo (English)

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 MATHEMATICS P2/WISKUNDE V2 SEPTEMBER 2019 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 This marking guideline consist of 18 pages./ Hierdie nasienriglyne bestaan uit 18 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 2 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne NOTE: · If a candidate answers a question TWICE, only mark the FIRST attempt. · If a candidate has crossed out an attempt of a question and has 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 nasienriglyne 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 1.1 k + 70 = 80 ü k + 70 = 80 \ k = 10 (1) 1.2 80 - 7 = p OR / OF 63 + 10 = p ü 73 \ p = 73 \ p = 73 (1) 1.3 75 ´ 2 + 105 ´16 + 135 ´ 28 + 165 ´17 + 195 ´10 + 225 ´ 7 ümidpoints/ x = middelwaardes 80 11940 ü ¸ 80 = 80 Answer only: Full Marks üanswer/antwoord \ x = 149, 25 (3) 1.4.1 x + s = 186,89 ümethod/metode s = 37, 64 Answer only: Full Marks üanswer/antwoord (2) 1.4.2 q = x - sx q = 149, 25 - 37, 64 q = 111, 61 üanswer/antwoord (1) 1.5 value of / waarde van m = 3 seconds / sekondes average time elapsed of day 2 will be 3 seconds less / üm=3 gemiddelde tydsverloop van dag 2 is 3 sekondes min der ü xday / dag 2 = 146, 25 xday/ dag 2 =146, 25 (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 3 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1 (68 ; 19) ü (68 ; 19) (1) 2.2 yˆ = a + bx ü a = 7,35 a = 7,35 and / en b = 0,98 ü b = 0,98 ü equation/ \ yˆ = 7,35 + 0,98 x vergelyking (3) 2.3 yˆ = 7,35 + 0,98(83) ü substitute 83 into yˆ = 88, 69% eq. / vervang 83 in vgl. ü 88,69 (2) OR / OF yˆ = 88,35% (calculator / sakrekenaar ) üü 88,35% (2) 2.4 r = 0.96 ü r = 0.96 \ Very strong positive correlation / Baie Sterk positiewe korrelasie üVery strong / Baie sterk (2) 2.5 If the data points were to be plotted on a scatter plot diagram, the new plotted points will be shifted downwards with the üNew data points and same constant value. Hence the line of best fit will also shift line of best fit shifted downwards with the new plotted points, but the slope of the downwards/Nuwe line will remain unchanged. Therefore it has NO influence on datapunte en paslyn the gradient/ word afwaarts geskuif As die datapunte op ‘n spreidiagram geplot moet word, sal die nuwe datapunte afwaarts geskuif word met dieselfde konstante üNo influence on waarde. Vervolgens sal die paslyn ook saam met die nuwe gradient/Geen invloed datapunte afwaarts skuif, maar die helling van die lyn sal op helling onveranderd bly. Dus sal dit GEEN invloed op die helling hê nie. (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 4 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne y R(1; 6) QUESTION/VRAAG 3 S P(–5; 1) O L D x Q(7; –2) 3.1 x = –5 ü x = –5 (1) 3.2 -2 - 6 üMethod/Metode mQR = 7 -1 -8 = 6 4 üAnswer/Antwoord =- 3 (2) 3.3 PS ^ QR ü \ mPS ´ mQR = - 1 mPS ´ mQR = - 1 [ PS ^ QR ] 3 \ mPS = 3 4 ü mPS = Through the point / Deur die punt ( -5;1) 4 3 y -1 = ( x + 5) üSubst./Vervang ( -5;1) 4 3 15 y = x + +1 4 4 3 19 3 19 \y = x + üy = x+ 4 4 4 4 (4) 3.4 1 - (-2) 1 mPQ = ü mPQ = - -5 - 7 4 3 = -12 1 =- ˆ = -1 4 ü tan PDO ˆ = -1 tan PDO 4 4 ˆ \ PDO = - 14,04° +180° = 165,96° ü165,96° (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 5 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne 3.5 -4 -4 tan a = ü tan a = 3 3 \ a = - 53,13° +180° = 126,87° 165,96° ü a = 126,87° ˆ 165,96° = 126,87° + PQR 126,87° ü \ ˆ PQR = 39.09° ˆ 165,96° = 126,87° + PQR Q ü39.09° (4) 3.6 area D PRS RS 4 1 area D PRS 1 = = = üü = area D PQS QS 16 4 area D PQS 4 [same height ;same base/dieselfde hoogte ; dieselfdebasis ] \ RS = a and / en QS = 4a Length of QR / Lengte van QR : (1 - 7 ) + ( 6 - (-2) ) 2 2 d QR = üMethod/Metode d QR = 100 = 10 units / eenhede ü d QR = 10 a + 4 a = 10 \ 5a = 10 \a = 2 Length of QS / Lengte vanQS = 4(2) = 8 units / eenhede ü QS = 8 (5) OR/OF QR 2 = (1 - 7)2 + (6 - (-2)) 2 üMethod/Metode üQR = 10 QR = 10 1 PQ.QS.sinQ area D PQS = 2 area D PQR 1 PQ.QR.sinQ 2 16 x 2 area D PQS 4 = QS üü = 20 x 2 10 area D PQR 5 4 QS = 5 10 4 SQ = (10) 5 = 8 units / eenhede üAnswer/Antwoord (5) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 6 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 y A C(6 ; 3) O x E B(2p ; –p) 4.1 x2 + y2 = r 2 ( 6 ) + ( 3) = r 2 2 2 r 2 = 45 ür2 = 45 \ x 2 + y 2 = 45 ü answer/antwoord (2) 4.2 ( A 0; 45 ) ü0 ü 45 (2) 4.3 3 - 45 üMethod/Metode mAC = 6-0 3-3 5 = 6 1- 5 1- 5 = ü mAC = 2 2 Through the point / Deur die punt 0; 45 ( ) 1- 5 y = x+ 45 üEquation/Vergelyking 2 (3) 4.4 - 45 < k < 45 or / of - 3 5 < k < 3 5 ü - 45 and / en 45 alternative notation / alternatiewe notasie üCorrect notation/ Korrekte notasie k Î (- 45 ; 45 ) or / of k Î (-3 5 ;3 5 ) (2) 4.5.1 ( x - 2 p)2 + ( y + p)2 = 20 ü ( x - 2 p) 2 + ( y + p)2 ü 20 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 7 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne 4.5.2 d OB = 45 + 20 ü dOB = 45 + 20 = 5 5 or / of 125 ü dOB = 5 5 OB2 = (2 p ) 2 + ( p ) 2 [Pythagoras] (5 5 ) = 4 p + p 2 2 2 ü OB2 = (2 p)2 + ( p)2 125 = 5 p 2 25 = p 2 \p =± 5 ü p 2 = 25 But according / Maar vo lg ens diagram p > 0 \p = 5 üp = 5 (5) 4.6 x 2 + 4 x cos q + y 2 + 8 y sin q = - 3 ü Complete the square/Voltooi die vierkant x 2 + 4 x cos q + ( 2 cos q ) + y 2 + 8 y sin q + ( 4sin q ) = 2 2 - 3 + ( 2 cos q ) + ( 4sin q ) 2 2 ü \ r = - 3 + 4 cos q + 16sin q 2 2 2 r 2 = -3 + 4 cos2 q + 16sin 2 q r 2 = - 3 + 4 cos 2 q + 16 (1 - cos 2 q ) ü 16 (1 - cos 2 q ) r 2 = - 3 + 4 cos 2 q + 16 - 16 cos 2 q r 2 = 13 - 12 cos 2 q ü r 2 = 13 - 12 cos 2 q For / vir 13 - 12 cos 2 q to be a maximum / om ' n maksimum te wees : cos 2 q needs to be a minimum / moet ' n min imum wees minimum of / van cos 2 q = 0 ü cos 2 = 0 \ r 2 = 13 - 12(0) = 13 \ r = 13 ü r = 13 OR / OF r 2 = - 3 + 4 (1 - sin 2 q ) + 16sin 2 q ü 4 (1 - sin 2 q ) r 2 = - 3 + 4 - 4sin 2 q + 16sin 2 q ü r 2 = -3 + 4 - 4 sin 2 q + 16 sin 2 q ü r 2 = 1 + 12sin 2 q r 2 = 1 + 12sin 2 q For / vir 1 + 12sin 2 q to be a maximum / om ' n maksimum te wees : ü sin 2 q = 1 maximum of / maksimum van sin q =1 2 \ r 2 = 1 + 12(1) = 13 ü r = 13 \ r = 13 (6) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 8 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1.1 sin(180° - 70°).tan 60° ü sin70° cos180°.tan(180° + 70°).sin 20° ü cos180° ü tan70° sin 70°.tan 60° ü sin20° (-1) tan 70°.sin 20° sin 70° ü sin 70°. 3 cos 70° sin 70° ü sin20° = cos70° (-1). .cos 70° cos 70° ü- 3 =- 3 (7) 5.1.2 1 - 2sin 2 22,5° ü 1 - 2sin 2 22,5° cos 2(22,5°) ü cos 2(22, 5°) cos 45° 1 ü cos 45° = 2 2 1 = ü 2 2 (4) 5.2.1 sin x = 0 and where / en waar tan x is undefined / ongedefinieerd 2 \sin x = 0 ü x = 0° \ x = 0° or / of x = 180° or / of x = 90° ü x = 180° ü x = 90° (3) 5.2.2 LHS / LK cos 2 x.tan x ü sin 2 x cos 2 x = cos2 x - sin 2 x sin x ( cos x - sin x ) æçè cos 2 2 sin x ö xø ÷ ü tan x = cos x = sin 2 x æ sin x ö æ 1 ö = ( cos 2 x - sin 2 x ) ç ÷ç 2 ÷ è cos x ø è sin x ø cos 2 x - sin 2 x cos 2 x - sin 2 x = ü sin x cos x sin x cos x cos 2 x sin 2 x ü = - sin x cos x sin x cos x cos 2 x sin 2 x - sin x cos x sin x cos x cos x sin x = - cos x sin x sin x cos x ü - sin x cos x cos x = - tan x = RHS / RK (5) sin x Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 9 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF LHS / LK cos 2 x.tan x sin 2 x ü ( cos x - sin x ) æçè cos 2 2 sin x ö xø ÷ cos 2 x = cos2 x - sin 2 x = sin x sin 2 x ü tan x = cos x cos x.sin x sin 3 x ü cos x.sin x sin 2 x = - cos2 x sin 3 x sin 2 x sin x ü cos2 x sin x cos x sin x cos x sin x = - ü - sin x cos x sin x cos x cos x = - tan x = RHS / RK (5) sin x [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 10 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 y f 1 x 0° 90° 180° g P(180° ; –3) 6.1 a = 3 and/en b = 2 üa = 3 üb = 2 (2) 6.2 360° ü360° (1) 6.3 2 £ y £ 4 or / of y Î [2; 4] ü values/waardes 2 and/en 4 ü notation / notasie. (2) 6.4 0° < x < 45° or / of 90° < x < 135° ü 0° and/en 45° alternative notation / alternatiewe notasie ü 90° and/en 135° ü notation / notasie x Î (0° ; 45°) È (90° ;135°) (3) 6.5 y = cos x - sin x 2 2 ü y = cos 2x y = cos 2 x to coincide with g / om saam te val met g : y = sin (90° + 2 x) ü sin (90 + 2x) y = sin 2 ( x + 45°) \ q = 45° ü 45° (3) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 11 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 7.1 y C (bcosA ; bsinA) a b x A(0 ; 0) c B(c ; 0) Coordinates of /Koördinate van B(c ; 0) ü coordinates of / koördinate van B ü correct subst. into distance ( b cos A - c ) - ( b sin A - 0 ) 2 2 BC = a = formula / korrekte vervanging in afstandsformule \ a 2 = b 2 cos 2 A - 2bc cos A + c 2 + b 2 sin 2 A ü b2 cos 2 A - 2bc cos A + c 2 + b 2 sin 2 A = b 2 ( cos 2 A + sin 2 A) - 2bc cos A + c 2 ü b 2 ( cos 2 A + sin 2 A ) = b 2 (1) + c 2 - 2bc cos A = b 2 + c 2 - 2bc cos A (4) 7.2 R 60° 50 60 Q 10 60° 60° T P Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 12 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne 7.2.1 Rˆ = Tˆ = Pˆ = 60° ü 60° ü 10 or/of 50 QT = 10 and / en QR = 50 PQ2 = 60 2 + 502 - 2(50)(60) cos 60° ü correct subst in cosine-rule/ korrekte vervanging in cosreël = 3100 ü 3100 OR / OF PQ2 = 102 + 60 2 - 2(10)60) cos 60° = 3100 \ PQ = 55, 68 (4) 7.2.2 R 50 60 Q x 10 S T P ü Diagram S is the nearest to R where RS is the perpendicular heigt / S is die naaste aan R waar RS die loodregte hoogte is. Let / Laat QS = x \ PS = 55, 68 - x ( 50 - x ) = ( 60 - [55, 68 - x] ) = RS 2 2 2 2 2 ü Pythagoras 2500 - x 2 = 3600 - 3100, 26 + 111,36 x - x 2 ü Simplify / vereenvoudig 2000, 26 = 111,36 x \ x = 17,96 units / eenhede ü answer / antwoord (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 13 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne GEOMETRY/MEETKUNDE Please read carefully through the following table before marking QUESTION 8 – 10 / Lees asseblief sorgvuldig deur die volgende tabel alvorens VRAAG 8 –10 nagesien word. The order in which the candidate answer a geometry question must follow logically/ Die volgorde waarin ‘n kandidaat ‘n meetkundevraag beantwoord moet logies volg. Example/Voorbeeld Given/Gegee AB || CD and/en ACˆ E = 105° A D 65° x y B C E The candidate first need to calculate x BEFORE he/she can calculate y/Die kandidaat moet eerste vir x bereken VOORDAT hy/sy vir y kan bereken. A mark for a correct statement (A statement mark is independent of a reason) S ‘n Punt vir ‘n korrekte bewering (‘n Punt vir ‘n bewering is onafhanklik van die rede) A mark for the correct reason (A reason mark may only be awarded if the statement is correct) R ‘n Punt vir ‘n korrekte rede (‘n Punt word slegs vir die rede toegeken as die bewering korrek is) Award a mark if the statement AND reason are both correct S/R Ken ‘n punt toe as die bewering EN rede beide korrek is Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 14 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 D E 1 1 2 2 68o 2 1 3 C B A 8.1.1 ˆ = 68° [ ext.Ð cyclic quad / buiteÐ van kvh ] DEA üSüR (2) 8.1.2 Â = 112° [ co-interior Ðs / ko - binne Ðe ; DE / / CA] üS/R (1) 8.1.3 D̂ 2 = 68° [ oppÐs of cyclic quad / teenoorst.Ðe van kvh ] üSüR OR/OF D̂ 2 = 68° [ alt Ðs / verwiss Ðe ; DE / /CA] üSüR (2) 8.1.4 B̂2 = 68° [ Ðs opp. = sides / Ðe teenoor = sye] üS/R (1) 8.1.5 Ê1 = 44° [ sum of Ðs of D /som van binneÐe van D ] üS/R D̂1 = 44° [ tangent chord theorem / raaklyn - koordstelling ] üSüR (3) 8.2 Ĉ = 68° [ sum of Ðs of D /som van binneÐe van D ] üS/R But / maar ˆB1 = 68° [given / gegee] DC = DB [ sides opp.= Ðs / sye teenoor = Ðe ] üS/R \D BDC is isosceles / gelykbenig (2) 8.3 ˆ = 68° D̂ = C üS 2 éconverse tangent chord theorem / ù \ DE is a tangent / 'n raaklyn ê ú ëomgekeerde raaklyn - koordstelling û üR (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 15 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 E 1 2 G 2 1 1 2F 2 D 1 2 1 C 2 A 1 2 1 B 9.1 tangent-chord theorem / raaklyn-koordstelling PR (1) 9.2 In DABC and/en DADB: Â1 = Â1 [common/gemeenskaplik] PS B̂ = D̂ [proven/bewys in 9.1] PS 1 1 \ DABC | | | DADB [Ð ; Ð ; Ð] PR OR/OF In DABC and/en DADB: Â = Â [common/gemeenskaplik] PS 1 1 B̂1 = D̂1 [proven/bewys in 9.1] PS ˆ BCA = ˆ ABD [Ðs of D = 180°/Ðe van D = 180°] PS \ DABC | | | DADB (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 16 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne 9.3 Ê = F̂ 2 1 [alternate Ðs/ verwiss Ðe ; EA | | GF] P S PR F̂1 = D̂ 2 [ext Ð of cyc quad DGFC/buite Ð v kdvh DGFC] P S PR \ Ê 2 = D̂ 2 (4) 9.4 In DAEC and/en DADE: Â 2 = Â 2 [common/gemeenskaplik] PS Ê 2 = D̂ 2 [proven/bewys in 9.3] PS \ DAEC | | | DADE [Ð ; Ð ; Ð] PR AE AC \ = [from ||| Ds / uit ||| De] AD AE PS 2 \ AE = AD ´ AC PS \ AE = AD×AC OR/OF In DAEC and/en DADE: Â 2 = Â 2 [common/gemeenskaplik] PS Ê 2 = D̂ 2 [proven/bewys in 9.3] PS ˆ = AED ACE ˆ [Ðs of D = 180°/Ðe van D = 180°] PS \ DAEC | | | DADE AE AC \ = [from ||| Ds / uit ||| De] PS AD AE \ AE 2 = AD ´ AC PS \ AE = AD×AC (5) 9.5 AB AC = [from |||Ds / uit ||| De] from/van 9.2 PS AD AB 2 AB = AD ´ AC PS 2 = AE [from/van 9.4] PS \AB = AE (3) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 17 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 10.1 A h k D E B C Construction: Connect DC and BE and draw the altitudes k and h P constr/ Konstruksie: Verbind DC en BE en trek hoogtelyne k en h. konstr Area ΔADE 2 ´ AD ´ k AD 1 = = Area ΔBDE 1 ´ BD ´ k BD PS 2 Area ΔADE 2 ´ AE ´ h AE 1 = = Area ΔDEC 1 ´ EC ´ h EC PS 2 but/maar: Area DBDE = Area DDEC PS PR [DE common base/ gemeenskaplike basis with/met DE | | BC] Area ΔADE Area ΔADE Pconclusion/ \ = Area ΔBDE Area ΔDEC afleiding AD AE \ = BD EC (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 18 NW/September 2019 NSC/NSS – Marking Guidelines/Nasienriglyne A 3b 3a S 2b P R 5b 2a T B C 10.2.1 AS AR = [line | | side of D/ lyn | | sy van D] PS PR SP RB OR/OF [ RS | | BP] 3 3 = P 2 2 AS 3 3 = [AP = PC] P SC 7 7 (4) 10.2.2 RT SP = [line | | side of D/ lyn | | sy van D] PS PR TC PC OR/OF [RS | | TP] 2 = Panswer/antwoord 5 (3) 10.2.3 Area of ΔRAS AS 3 = = P 3 Area of ΔRSC SC 7 7 [common height / gemeenskaplike hoogte] PR (2) 10.2.4 Area ΔTPC 1 ˆ TC.PC.sinTCP PCorrect subst = 2 into area rule/ ˆ Area ΔRSC 1 RC.SC.sinTCP 2 Korrekte vervang. TC PC in area-reël = ´ RC SC 5 5 5 = ´ P 7 7 7 25 Panswer/antwoord = (3) 49 [18] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 19 NW/September 2019 NSC/NSS – Memorandum GRID-Analysis (According to BLOOMS TAXONOMY) KNOWLEDGE ROUTINE COMPLEXED PROBLEM PROCEDURES PROCEDURES SOLVING Question Description Low Mod High Low Mod High Low Mod High Low Mod High Statistics(Gr.11) 1.1 Frequency 1 1.2 Cumulative Frequency 1 1.3 Mean average 3 1.4.1 Standard deviation 2 1.4.2 Within standard deviation 1 1.5 Adjusted mean average 2 2 Regression(Gr.12) 2.1 Outlier 1 2.2 Regression line 3 2.3 Application regr. Line 2 2.4 Correlation Coefficient 2 2.5 Application slope 2 3 Analytical (Gr.11) 3.1 Equation vertical line 1 3.2 Gradient of line 2 3.3 Equation of line 4 3.4 Inclination 3 3.5 Angle between straight ln 4 3.6 Integrating geometry 5 4 Analytical(Gr.12) 4.1 Eq circle (0;0) 2 4.2. y-intercept 2 4.3 Equation of line 3 4.4 Roots opposite signs 2 Copyright reserved / Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 20 NW/September 2019 NSC/NSS – Memorandum 4.5.1 Eq. of circle midpt(a;b) 2 4.5.2 Value of p 5 4.6 Max length of radius 6 ROUTINE COMPLEX KNOWLEDGE PROCEDURES PROCEDURES PROBLEM SOLVING Question Description Low Mod High Low Mod High Low Mod High Low Mod High 5 Trigonometry 5.1.1 Reduction 7 5.1.2 Double angles 4 5.2.1 Undefined 3 5.2.2 Identity 5 6 Trig Graphs 6.1 Parameters 2 6.2 Period 1 6.3 Range 2 6.4 Integration interval 3 6.5 Integration of parameters 3 7 2D/3D Trig 7.1 Formulae 4 7.2.1 Application cos-rule 4 7.2.3 Integration 4 8 Geometry 8.1 Ext angle cyclic quad 2 8.2 Co-interior 1 8.3 Opp. Angle Cyclic quads 2 8.4 Angles of triangle 1 8.5 Tangent-chord 3 8.6 Isoceles 2 8.7 Prove tangent 2 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 21 NW/September 2019 NSC/NSS – Memorandum ROUTINE COMPLEX PROBLEM KNOWLEDGE PROCEDURES PROCEDURES SOLVING Question Description Low Mod High Low Mod High Low Mod High Low Mod High 9 Geometry 9.1 Tangent problems 1 9.2 Similar triangles 3 9.3 From similar triangles 4 9.4 Proportionality 5 9.5 Conclusion 3 10 Geometry 10.1 Bookwork 6 10.2.1 Ratio 4 10.2.2 Ratio 3 10.2.3 Ratio of areas 2 10.2.4 Ratio of areas 3 6 15 9 13 21 22 22 23 0 6 7 6 % Breakdown of cognitive 30 20% 56 37,3% 45 30% 19 12,7% levels Expected % 20% 35% 30% 15% Copyright reserved / Kopiereg voorbehou