MATHS P2 MEMO GR12 JUNE 2024_Afr+English_hlayiso.com_.pdf

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIORSERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2024 MATHEMATICS P2/WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 19 pages./ Hierdie nasienriglyn bestaan uit 19 bladsye. Hosted by www.ecexams.co.za

Downloaded from hlayiso.com 2 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2024) 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 not redone a question, mark the crossed-out version. • Consistency accuracy applies in ALL aspects of the marking guideline. Stop marking at the second calculation error. • Assuming answers/values in order to solve a problem is NOT acceptable. GEOMETRY S A mark for a correct statement. (A statement mark is independent of a reason). R A mark for the correct reason. (A reason mark may only be awarded only if the statement is correct. S/R Award a mark if a statement and a reason are both correct. NEEM KENNIS: • Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, merk slegs die EERSTE poging. • Indien ʼn kandidaat ʼn poging van ʼn vraag deurgetrek het en dit nie oorgedoen het nie, merk die deurgetrekte weergawe. • Volgehoue akkuraatheid geld in ALLE aspekte van die nasienriglyn. Hou op merk by tweede berekenings fout. • Om antwoorde/waardes te aanvaar om ʼn probleem op te los is NIE aanvaarbaar NIE. MEETKUNDE S ʼn Punt vir korrekte stelling. (ʼn Stelling punt is onafhanklik van die rede) R ʼn Punt vir die korrekte rede. (ʼn Rede punt mag net toegeken word as die stelling korrek is). S/R ʼn Punt word toegeken as die stelling en die rede beide korrek is. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2024) MATHEMATICS P2/WISKUNDE V2 3 QUESTION/VRAAG 1 17 26 27 27 30 32 34 35 36 37 40 1.1  answer / antwoord x = 31 OR/OF OR/OF 341  341 x= 11  answer / antwoord = 31 (2) 1.2 𝛿 = 6,19  answer / antwoord (1) 1.3 x +  = 31 + 6,19  x +  = 31 + 6,19  37,19 = 37 ,19 Temperatures were more than one standard deviation for 1 day. Temperature was meer as een standaardafwyking vir 1 dag.  conclusion / gevolgtrekking (3) 1.4 IQR/IKW = 36 − 27  Q1 =9  Q3 answer / antwoord (3) 1.5  min. and/en max. / maks.  Q1 and/en Q2  correct diagram korrekte diagram (3) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2024) QUESTION/VRAAG 2 2.1 Percentage Frequency / Cumulative obtained / Frekwensie Frequency / Persentasie Kumulatiewe behaal Frekwensie  14 and / en 18 0  x  20 4 4 20  x  40 14 18 14 and / en 4 40  x  60 18 36 60  x  80 14 50 80  x  100 4 54 (2) 2.2 54 matriculants / matriekulante  answer / antwoord (1) 2.3 40  x  60  answer / antwoord (1) 2.4 50 %  reading from the graph lees van grafiek af  answer / antwoord (2) 2.5 12 learners / leerders  reading from the graph lees van grafiek af  answer / antwoord (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2024) MATHEMATICS P2/WISKUNDE V2 5 QUESTION/VRAAG 3 y A (–2 ; p) G θ F B (–4 ; 0) α x D (4 ; –1) C (–1 ; –3) 3.1 BC = (− 4 + 1)2 + (0 + 3)2  correct substitution/ korrekte vervanging =3 2  answer / antwoord (2) 3.2 0+3  correct substitution/ mBC = − 4 +1 korrekte vervanging = −1  answer / antwoord (2) 3.3 mAD = mBC = −1 [ AD || BC ]  mAD = −1 y + 1 = −( x − 4)  correct substitution y = −x + 3 korrekte vervanging  answer / antwoord OR/OF OR/OF m AD = mBC = −1 [ AD || BC ]  mAD = −1 − 1 = −(4) + c  correct substitution/ c=3 korrekte vervanging  answer / antwoord y = −x + 3 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2024) 3.4 p = −(−2) + 3  correct substitution/ =5 korrekte vervanging  answer / antwoord OR/OF OR/OF p−0 m AB = −2+4 p = 2 p y − 0 = (x + 4) 2 px y= +2p 2 p (−2)  correct substitution/ + 2 p = − ( −2 ) + 3 2 korrekte vervanging − p + 2p = 5 p=5  answer / antwoord (2) 3.5 1 −3− mCF = 2 5 −1− 2 =1  mCF = 1  mAD  mCF = −11 = −1  m AD  mCF (2) 3.6 tan  = mAD = −1  tan  = mAD = −1  = 135 0  = 135 0  = 450 [ext  of a ] / [ buite van ]  = 45 0 (3) 3.7 AD = (− 2 − 4)2 + (5 + 1)2  AD =6 2 2 2  5  1 CF =  − 1 −  +  4 −   2  2 7 2  CF = 2 Area of trapezium / Oppervlakte van trapezium 1 = ( AD + BC )  CF 2  correct substitution/ 1 ( = 6 2 +3 2  2 ) 7 2 2 korrekte vervanging  answer / antwoord = 31,50 (4) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2024) MATHEMATICS P2/WISKUNDE V2 7 QUESTION/VRAAG 4 y R (k ; 21) Q x C (3; − 1) P T 4.1 ˆ = 900 CQR [tan ⊥ chord] / [raaklyn ⊥ koord] S (1) 4.2 RC2 = QC2 + QR 2 [Pyth. theorem/stelling ]  correct substitution/ RC2 = 102 + 202 korrekte vervanging RC = 500 or/of 10 5  answer / antwoord (2) 4.3 ( k − 3) + ( 21 − ( −1) ) = (10 5 )  RC or application of Pyth./ 2 2 2 RC of toepassing van Pyth. ( k − 3) = 500 − 484 2  simplification/ vereenvoudiging ( k − 3) = 16 2  factors / faktore k − 3 = 4  correct value of k/ korrekte waarde van k k = 7 or / of k  −1 OR/OF OR/OF (k − 3)2 + (21 − (− 1))2 = (10 5 )  RC or application of Pyth./ 2 RC of toepassing van Pyth. k 2 − 6k + 9 + 484 = 500  simplification/ k 2 − 6k − 7 = 0 vereenvoudiging  factors / faktore (k − 7 )(k + 1) = 0  correct value of k/ k = 7 or k  −1 korrekte waarde van k (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2024) 4.4 (x − 3)2 + ( y + 1)2 = 100  LHS / LK  RHS / RK (2) 4.5 TC = 10 and/en TC ⊥ PT T ( 3; − 11) T(3; − 11) y = −11  y = −11 (2) 4.6.1 T(3 ; − 11) 3(-11) − 4 x = 35  correct substitution/ korrekte vervanging  x = -17  x-value/ P(- 17 ; - 11) x-waarde (2) 4.6.2 PQ = PT [tangents from same point are equal in length]  PQ = 20 [raaklyne vanaf dieselfde punt is gelyk ] R = 17 + 3 = 20 (2) 4.6.3 Yes / Ja  Yes / Ja ΔQRC  ΔQCP SS S R (3) 4.7.1 M (3 ; − 16)  answer / antwoord (1) 4.7.2 r=4  answer / antwoord (1) 4.7.3 r1 + r2 = 4 + 10 = 14  r1 + r2 and / en CM 2 = ( 3 − 3) + ( −16 + 1) 2 2 = 152  CM = 15 CM = 15  CM  r1 + r2  The 2 circles do not intersect or touch  conclusion/ Die 2 sirkels sny of raak nie. gevolgtrekking (3) [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2024) MATHEMATICS P2/WISKUNDE V2 9 QUESTION/VRAAG 5 5.1.1 760 1 p  correct sketch/ korrekte skets 14 0 cos 76 0 = p  answer / antwoord OR/OF OR/OF cos 76 0 = sin 14 0  co-ratio / ko-verhouding =p  answer / antwoord (2) 5.1.2 x = 1 − p 2 Pyth.theorem/stelling  x-value / x-waarde ( cos 44 0 = cos 30 0 + 14 0 ) (  cos 30 0 + 14 0 )  expanding compound angle = cos 30 0. cos 14 0 − sin 30 0. sin 14 0 uitbrei van saamgestelde  3 1  answer / antwoord = . 1 − p2 − .p 2 2 (4) 5.1.3 ( ) 2 sin 218 . cos 38 = 2 − sin 38 0 . cos 38 0 0 0  − sin 38 0 = − sin 76 0  − sin 760  answer / antwoord = − 1− p2 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2024) 5.2.1 sin (90 0 +  ). cos ( − 360 0 ) 1+ sin ( − 30 0 −  ) = 1+ (cos  )(cos  )  cos   cos  (− sin 30 0 ) (  sin  − 30 −  0 ) cos  2 = 1− 1 1  2 2 = 1 − 2 cos 2   simplification/ = −(2 cos 2  − 1) vereenvoudiging = − cos 2  answer/antwoord OR/OF OR/OF sin (90 0 +  ). cos ( − 360 0 ) 1+ sin ( − 30 0 )cos  − sin  cos ( − 30 0 ) cos  . cos   cos   cos  = 1+ (sin  cos 30 − sin 30 cos  ) cos  − sin  (cos  cos 30 + sin  sin 30 )  expansion of 0 0 0 0 compound angle/ cos 2  = 1+ uitbrei van 3 1 2 3 1 2 saamgestelde  sin  cos  − cos  − sin  cos  − sin  2 2 2 2 cos 2 1 = 1+  − (cos 2  + sin 2  ) 1 2 2 cos 2  = 1− 1 (1)  simplification/ 2 vereenvoudiging = 1 − 2 cos 2  = −(2 cos 2  − 1) = − cos 2  answer/ antwoord (6) 5.2.2 Max value/Maks. waarde = 1 OR/OF y = 1  answer/ antwoord (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2024) MATHEMATICS P2/WISKUNDE V2 11 5.3 sin 3 x LHS / LK = sin x sin ( 2 x + x )  sin (2 x + x ) = sin x  expansion / uitbreiding sin 2 x cos x + sin x cos 2 x = sin x  sin 2 x = 2 sin x cos x 2sin x cos x.cos x + sin x cos 2 x = sin x sin x ( 2 cos x + 2 cos 2 x − 1) 2  factors / faktore = sin x = 4 cos x − 1 2 = 4 (1 − sin 2 x ) − 1  expression in terms of sin 2 x / uitdrukking in terme van sin 2 x = 4 − 4sin 2 x − 1 = 3 − 4sin 2 x OR / OF OR / OF LHS / LK = sin 3x  sin (2 x + x ) sin x sin ( 2 x + x )  expansion / uitbreiding = sin x sin 2 x cos x + sin x cos 2 x  sin 2 x = 2 sin x cos x = sin x 2sin x cos x.cos x + sin x cos 2 x  factors / faktore = sin x sin x ( cos x + cos 2 x ) 2  expression in terms of sin 2 x / = sin x uitdrukking in terme van sin 2 x = 2 − 2sin x + 1 − 2sin 2 x 2 = 3 − 4sin 2 x (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2024) 5.4.1 sin 2 x + sin 2 x − 3cos 2 x = 0  2 sin x cos x sin 2 x + 2sin x cos x − 3cos 2 x = 0 ( sin x − cos x )( sin x + 3cos x ) = 0  factors / faktore sin x = cos x or / of sin x = −3cos x  both equations in terms of tan x tan x = 1 or / of tan x = −3 beide vergelykings i.t.v tan x  x = 450 + 1800.k x = 45 + 1800.k or / of x = 108, 440 + 1800.k , k    x = 108 ,44 0 + 180 0.k , k  Z OR / OF OR / OF  both equations / x = 450 + 3600.k or / of x = 2250 + 3600.k beide vergelykings or  both equations and k   / x = 108, 44 + 360 .k or / of 0 0 x = 288, 44 + 360 .k , k   0 0 beide vergelykings en k   (5) 5.4.2 x = −71, 44 or / of x = 45 or / of x = 108, 44 0 0 0  each x-value/ elke x-waarde (3) [29] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2024) MATHEMATICS P2/WISKUNDE V2 13 QUESTION/VRAAG 6 6.1 f:  intercepts with the axes/ afsnitte met die asse  turning points/ draaipunte  shape / vorm g:  intercepts with the axes/ afsnitte met die asse  asymptotes/ asimptote  shape / vorm (6) 6.2.1 3600  answer / antwoord (1) 6.2.2 x = −1800 or / of x =1800  x = −180 0  x = 180 0 (2) 6.2.3 −5  y  1 or / of y −5 ; 1  both cv’s correct/ beide kw’s korrek  correct notation/ korrekte notasie (2) 6.2.4 3 solutions / 3 oplossings  answer / antwoord (1) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 14 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2024) QUESTION/VRAAG 7 P 2x 30° 3h h Q R T 7.1 PT̂Q = 30 0  answer / antwoord (1) 7.2 PQ  correct trig. ratio/ = sin PTˆQ korrekte trig. verhouding PT h PT =  correct substitution/ sin 30 0 korrekte vervanging h = 1 2  answer / antwoord = 2h (3) 7.3 RT 2 = PT 2 + PR 2 − 2.PT.PR.cos Pˆ  cosine rule of ∆PRT/ cosinusreël van ∆PRT ( 7h ) = ( 2h ) + (3h ) − 2 ( 2h )(3h ).cos 2 x 2 2 2  correct substitution/ 7h 2 = 4h 2 + 9h 2 − 12h 2 cos 2 x korrekte vervanging 12h 2 cos 2 x = 6h 2 1  simplification/ cos 2 x = vereenvoudiging 2 2 x = 600  correct ratio / x = 30 0 korrekte verhouding  answer / antwoord (5) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2024) MATHEMATICS P2/WISKUNDE V2 15 QUESTION/VRAAG 8 T 2 P 1 1 G 2 1 750 S O 1 4 3 5 2 U 1 V K 8.1.1 Ô1 = 150 0 [ at centre = 2   at circumfere nce ] S R [middelpunts = 2  omtreks] (2) 8.1.2 Û5 = 75 0 [tan chord theorem] / [raaklyn-koord stelling] S R (2) 8.1.3 Tˆ = Uˆ [s opp = sides] / [e teenoor = sye]  S/R 1 4 2Tˆ1 = 1800 − 1500 [s in a ] /[e in ' n ]  S/R T̂1 = 15 0  answer / antwoord (3) 8.1.4 ˆ = Tˆ = 750 U [alt. s, TS PK] / [verw.e, TS PK]  S/R 5 V̂ = 1050 [opp. s of a cyclic quad] S R [teenoorst.e van ' n koordevierhoek ] (3) 8.1.5 ˆ +U U ˆ +U ˆ =Vˆ [tan chord theorem] / [raaklyn - koord stelling ]  S R 3 4 5 Û 3 = 150 (2) 8.1.6 U5 + ˆ ˆ U 4 = 90 0 [tan ⊥ rad] / [raaklyn ⊥ radius] S R Ĝ 2 = 900 [alt. s, TS PK] / [verw.e, TS PK]  answer antwoord (3) 8.2 1 S R TG = GS =  80 = 2 5 [line from centre ⊥ to the chord] 2 [lyn vanaf middelpunt ⊥ op die koord ] (2) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 16 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2024) QUESTION 9 / VRAAG 9 A l h M N B C 9.1 Construction: Draw ⊥ height (h) to AM and ⊥ height (l) to AN.  constructions/ Join BN and MC konstruksies Konstruksie: Teken ⊥ hoogte(h) na AM en ⊥ hoogte(l) na AN. Verbind BN en MC S 1  AM  h Area AMN 2 = R Area  MNB 1  MB  h 2 AM = [same height] / [dieselfde hoogte] MB S 1  AN  l Area AMN 2 = Area  NBC 1  NC  l 2 AN R = [same height] / [dieselfde hoogte] NC Area  MNB = Area NBC [same base, same parallel lines] [dieselfde basis, dieselfde ewewydige lyne] AM AN  = MB NC (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2024) MATHEMATICS P2/WISKUNDE V2 17 9.2 Q E P F 10 G R 49 9.2.1 RG ER S R = [line || to oneside of a Δ]/[lyn || aan een sy van ' n ] FG QE  correct substitution/ RG 5 = korrekte vervanging 10 2 RG = 25  answer / antwoord OR/OF OR/OF QR FR S R = [line || to oneside of a Δ]/[lyn || aan een sy van ' n ] ER GR 7p 10 + GR  correct substitution/ = korrekte vervanging 5p GR 7GR = 50 + 5GR 2GR = 50 GR = 25  answer / antwoord OR/OF OR/OF QE FG = [line || to oneside of a Δ]/[lyn || aan een sy van ' n ]  S R QR FR 2p 10  correct substitution = 7p FG + GR korrekte vervanging 2 10 = 7 10 + GR 20 + 2GR = 70  answer / antwoord 2GR = 50 GR = 25 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 18 MATHEMATICS P2/WISKUNDE V2 (EC/JUNE/JUNIE 2024) 9.2.2 RF 35 RF =  correct value of RP 49 RP 5 RF = korrekte waarde van 7 RP and / en RE 5 RE =  correct value of QR 7 QR RF RE  5 RE  = korrekte waarde van RP QR both / beide = 7  QR ∴ PQ||FE [line divides 2 sides of Δ in proportion]/ [lyn deel 2 sye van Δ eweredig] R [converse prop theorem]/ [omgekeerde eweredigheidstelling] [converse line||to one side of a Δ]/ [omgekeerde lyn || aan een sy van ʼn Δ] (3) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2024) MATHEMATICS P2/WISKUNDE V2 19 QUESTION 10 / VRAAG 10 D P A 2 1 T 2 1 C B Q 10.1 CT BC S R = [line|| to one side of a Δ]/ PC QC [lyn || aan een sy van 'n ] [ prop theorem, BT QP]  QC in terms of BC/ [eweredigheid stelling , BT || QP] QC in terme van BC BC = 6BC 1 = 6 (3) 10.2 ˆ =Q Q ˆ [common] / [gemeen] S S R ˆ =D C ˆ [ s in sameseg] / [e in dies. segment ] ˆ =B A ˆ [3rd s] / [3de e] 1 2  R for/vir  ΔQAC|||ΔQBD [] OR/OF OR / OF ˆ Q=Qˆ [common] / [gemeen] S ˆ =D ˆ S R C [ s in sameseg] / [e in dies. segment ] ˆ A1 = ˆB2 [3rd s] / [3de e]  S for 3rd angles ΔQAC|||ΔQBD [] vir 3de hoeke (4) 10.3 QC QA S R = [||| s ] QD QB QD  QA = QC  QB  6BC  5BC = 6BC  5BC QD.QA = 30BC 2 (3) [10] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief