MATHS P2 GR12 MEMO SEPT 2022 Afr English_hlayiso.com_.pdf

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2022 MATHEMATICS P2/WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 16 pages./ Hierdie nasienriglyn bestaan uit 16 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2022) QUESTION 1/VRAAG 1 1.1 Distance of Jumps Number of CF ✓ for cf [6 to 92] Afstand van Spronge athletes KF vir kf [6 tot 92] (in cm) Aantal atlete 420  d ≤ 460 6 6 ✓ for cf [94 to 100] vir kf [94 tot 100] 460  d ≤ 500 14 20 500  d ≤ 540 16 36 540  d ≤ 580 42 78 580  d ≤ 620 14 92 620  d ≤ 660 2 94 660  d ≤ 700 3 97 700  d ≤ 740 2 99 740  d ≤ 780 1 100 (2) 1.2 ✓ anchor point LONG JUMPERS' BEST JUMPS ankerpunt VERSPRINGERS SE BESTE SPRONGE ✓ upper limits 120 boonste limiete ✓ sf / ✓ smooth shape NUMBER OF ATHLETES / AANTAL ATLETE 100 egalige vorm 80 60 40 20 0 400 500 600 700 800 900 DISTANCE OF JUMPS / AFSTAND VAN SPRONGE (IN CM) (4) 1.3 The median jump is 553. Accept between (551 – 555) ✓✓ for answer Die mediaan sprong is 553. Aanvaar tussen (551 – 555) vir antwoord (2) 1.4 Number jumped over 560 cm = 100 – 57 = 43 athletes ✓ for subtraction Therefore, it is 43% of the athletes. vir aftrekking Aantal wat oor 560 cm gespring het = 100 – 57 = 43 atlete ✓ for the answer Dit is daarom 43% van die atlete. vir die antwoord (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2022)) MATHEMATICS P2/WISKUNDE V2 3 QUESTION 2/VRAAG 2 Long jumper / Verspringer 1 2 3 4 5 6 x: Hours practised / Ure geoefen 4,5 2 3,5 4 8 3 y: Distance jumped / Afstand gespring (cm) 650 420 580 490 780 525 2.1 a = 336,699 b = 56,992 ŷ = 336,699 + 56,992x ✓ for/vir a ✓ for/vir b ✓ for/vir a + bx (3) 2.2 ŷ = 336,699 + 56,992 (5.4) = 644, 46 cm ✓ for substitution vir vervanging ✓ for the answer vir die antwoord (2) 2.3 The more they practiced, the further they jumped. ✓✓ for the answer Strong positive correlation. vir die antwoord Hoe meer hulle geoefen het, hoe verder het hulle gespring. Sterk positiewe korrelasie. (2) 2.4.1 The mean will decrease by 13 cm.  for the answer Die gemiddelde sal met 13 cm verminder. vir die antwoord (1) 2.4.2 The range will remain the same / No influence on range.  for the answer Die omvang sal dieselfde bly / Geen invloed op die vir die antwoord omvang. (1) 2.4.3 The standard deviation remains the same.  for the answer Die standaardafwyking bly dieselfde. vir die antwoord (1) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2022) QUESTION 3/VRAAG 3 y E(p ; 8) H(8 ; 7) M x R K F(–2 ; –3) G 3.1.1 M (3; 2) ✓ for/vir x ✓for/vir y (2) 3.1.2 7 − ( −3) ✓ for subst. / vir vervanging mFH = =1 8 − ( −2) ✓ for answer / vir antwoord (2) 3.1.3 mEG = −1 (diagonals bisect at 90) ✓S (hoeklyne halveer loodreg/by 90˚) tan MK̂ X = -1 ✓S ̂ X = 135°  MK̂ X = 135° MK ̂ R = 45° ✓ for answer / vir antwoord ∴MK (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2022)) MATHEMATICS P2/WISKUNDE V2 5 3.2 FE = EH (sides of a rhombus = ) (sye van ʼn rombus = ) ✓ for equating / gelykstel FE = EH 2 2 FE 2 = EH 2 ✓ for squaring / kwadrering ( p + 2 ) + ( 8 + 3) = ( p − 8 ) + ( 8 − 7 ) 2 2 2 2 p 2 + 4 p + 4 + 121 = p 2 − 16 p + 64 + 1 ✓ for simplification 20 p = −120 vir vereenvoudiging ✓ for the answer p = −3 vir die antwoord OR/OF OR/OF ✓ for gradient of E to FH E(p; 8) and Midpoint of HF / en Middelpunt van HF = (3; 2) vir gradiënt van E na FH m FH = 1 gradient from E to midpoint of FH / ✓ statement / stelling gradiënt vanaf E na middelpunt van FH 8−2 6 = = p −3 p −3 FH is perpendicular to EG / FH is loodreg op EG ✓ for the product 6 vir die produk  1 = −1 ✓ for the answer p −3 vir die antwoord  p = −3 (4) 3.3 G(9; – 4) ✓ for/vir x ✓for/vir y (2) 3.4 M(3 ;2)  for coordinates of N N(-9 ;2) vir die koördinate van N MN = 12 units/eenhede  for answer / vir antwoord (3) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2022) QUESTION/VRAAG 4 y P N(6 ; k) R(4 ; 2) x O S 4.1 r 2 = (4 − 0) 2 + (2 − 0) 2 ✓ substitution / vervanging r 2 = 20 ✓ for/vir r 2  ( x − 4 ) + ( y − 2) 2 = 20 2 ✓ for the equation/vir die vergelyking (3) 4.2 2 (6 − 4) + (k − 2) = 20 2  substitution of / vervanging van N (k − 2)2 = 16  simplification / vereenvoudiging k − 2 = ±4  both answers for k / 𝑘 = 6 or/of 𝑘 = -2 beide antwoorde vir k 𝑘=6  selection of 𝑘 = 6 keuse van k = 6 OR/OF OR/OF Sub: N(6; y) into the equation of the circle. Verv. N(6; y ) in die vergelyking van die sirkel. ✓ for substitution / vir vervanging (6 − 4) 2 + ( y − 2) 2 = 20 4 + y 2 − 4 y + 4 − 20 = 0 ✓ for standard form / vir standaardvorm y 2 − 4 y − 12 = 0 ✓ for the factors / vir die faktore ( y − 6)( y + 2) = 0 y = 6 or/of y = −2 ✓ for the answer / vir die antwoord y =6 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2022)) MATHEMATICS P2/WISKUNDE V2 7 4.3 6−2 ✓ for gradient of RN mRN = =2 6−4 vir gradiënt van RN 1 ✓ for gradient of NP mNP = − vir gradiënt van NP 2 ✓ for substitution of N Equation of NP / Vergelyking van NP: vir vervanging van N 1 y − 6 = − ( x − 6) 2 ✓ for/vir 𝑐 = 9 1 y = − x+9 2 ✓ for answer / vir antwoord (5) 4.4.1 1 ✓ for equating / −2 x = − x + 9 vir gelykstelling 2 3 − x=9 ✓ for the simplification 2 vir die vereenvoudiging −3x = 18  x = −6 and/en y = 12 ✓ for the answer /  P( − 6; 12) vir die antwoord (3) 4.4.2 RO = RN = √42 + 22 = 2√5 (radii/radiusse) ✓ use of distance formula gebruik van afstand formule ✓ for RO / RN answer PO = PN = √(-6)2 + 122 = 6√5 (tangents from same pt) vir RO / RN antwoord (raaklyne vanaf dieselfde punt) ✓ for PO / PN answer vir PO / PN antwoord ∴ Perimeter of /Omtrek van PNRO = 2(2√5) + 2(6√5) ✓ for final answer = 16√5 or/of 35,78 units/eenhede vir finale antwoord (4) 4.5 S(8 ;0) ✓✓ coordinates of S koördinate van S T(12 ; -2) ✓✓ coordinates of T koördinate van T (4) [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2022) QUESTION 5/VRAAG 5 64° 1 √1 – p2 26° p 5.1.1 √1 – p2 sin 26° = ✓✓ for the answer 1 vir die antwoord (2) ° 5.1.2 tan 154 = - tan 26° ✓ for the reduction vir die reduksie √1 – p2 =- ✓ for the answer p vir die antwoord (3) ° ° 5.1.3 sin 13 . cos 13 sin 26° = 2 sin 13° . cos 13° ° ✓ for the reduction ° ° sin 26 sin 13 . cos 13 = vir die reduksie 2 √1 − p2 ✓ for the answer sin 13° . cos 13° = 2 vir die antwoord (2) ° 5.2.1 cos(-θ) . tan ( 180 +θ) 2 cos (90° + θ) cos θ . tan θ ✓ cos θ = ✓ tan θ -2 sin θ sin θ ✓ -2 sin θ cos θ . cos θ sin 𝜃 ✓ cos 𝜃 = -2 sin θ ✓ for the answer 1 vir die antwoord =- (5) 2 5.2.2 1 + 2 cos 105° sin 15° = 1 + 2 cos 75° sin 15° ✓ for reduction of cos 105° = 1 + 2 sin 15° sin 15° vir reduksie van cos 105° =1 − sin 30° ✓ for reduction of cos 75° 1 vir reduksie van cos 75° =1 − 2 ✓ for sin 30° 1 vir sin 30° = 2 ✓ for the answer vir die antwoord (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2022)) MATHEMATICS P2/WISKUNDE V2 9 5.3.1 1 − cos 2 x − sin x = tan x sin 2 x − cos x LHS: 1 − (1 − 2sin 2 x) − sin x ✓ expansion of cos 2x 2sin x cos x − cos x uitbreiding van cos 2x 2sin 2 x − sin x ✓ expansion of sin 2x = 2sin x cos x − cos x uitbreiding van sin 2x sin x(2sin x − 1) ✓ for the simplification = cos x(2sin x − 1) vir die vereenvoudiging ✓ taking out HCF = tan x = RHS uithaal van GGD (4) 5.3.2 sin 2x = cos x  for/vir sin 2𝑥 = cos 𝑥  for any 2 answers 𝑥 = -90° ; 30° ; 90° and/en 150° vir enige 2 antwoorde  for any other 2 answers vir enige 2 antwoorde (3) 5.4 sin x + 2sin x cos x = 3cos x 2 2 ✓ for standard form vir standaardvorm sin 2 x + 2sin x cos x − 3cos 2 x = 0 Divide every term by/Deel elke term deur cos 2 x ✓ for dividing by cos2 x tan 2 x + 2 tan x − 3 = 0 vir deling deur cos2 x ( tan + 3)( tan x − 1) = 0 ✓ for the factors vir die faktore tan x = −3 or/of tan x = 1 ✓ for values of tan x x = 108, 43 + 180.k or/of x = 45 + 180.k vir waardes van tan x where/waar k  Z ✓✓ for the answers vir die antwoorde OR/OF sin2 x + 2 sin x cos x = 3 cos2 x ✓ for standard form vir standaardvorm sin2 x + 2 sin x cos x − 3 cos2 x = 0 (sin x + 3 cos x)(sin x − cos x) = 0 ✓ for the factors vir die faktore sin x = -3 cos x or/of sin x = cos x ✓ for isolating sin x 𝑡𝑎𝑛 𝑥 = −3 or/of 𝑡𝑎𝑛 𝑥 = 1 vir isolering van sin x 𝑥 = 108,43° + 180°. 𝑘 ✓ for values of tan x or/of 𝑥 = 45° + 180°. 𝑘 vir waardes van tan x where/waar, 𝑘 ∈ 𝑍 ✓✓ for the answers vir die antwoorde (7) [30] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2022) QUESTION 6/VRAAG 6 6.1 ✓ shape vorm 𝑔 ✓ start / end points begin / eind punte 𝑓 ✓TP at / DP by 90 (3) 6.2.1 Period / Periode = 180° ✓ answer (1) 6.2.2 -3 ≤ y ≤ -1 ✓ for/vir -3 and/en -1 ✓ for the answer in correct notation OR/OF vir die antwoord in korrekte notasie 𝑦 ∈ [-3 ;-1] (2) 6.3 h(x) = - sin x – 1 ✓ for/vir ℎ(𝑥) Maximum distance/Maksimum afstand ✓ answer / antwoord = 2 units/eenhede (2) 6.4 f(x).g' (x)>0 -90° < x < 90° ✓✓ answer / antwoord (2) 6.5 Graph shifted 1 unit down and 15° to the right.  for 1 unit down / vir 1 eenheid af Grafiek het 1 eenheid af en 15° na regs geskuif.  for 15° to the right / vir 15° na regs (2) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2022)) MATHEMATICS P2/WISKUNDE V2 11 QUESTION 7/VRAAG 7 T 60° P x Q 150° R 7.1 ̂ (30° − x) R ✓ for answer / vir antwoord (1) 7.2 PQ 300 = sin(30 − x) sin 150° ° ✓ for sine-rule vir sinusreël PQ = 600 ✓ for/vir 600 sin(30° − x) ✓ for the answer PQ = 600 sin(30° − x) vir die antwoord (3) 7.3 TP tan 60° = PQ ✓ for/vir tan 60° TP = PQ tan 60° TP = √3.600 sin(30° − x) ✓ for/vir √3.600 sin(30° − x) TP = √3 .600. (sin 30° cos x- cos 30° sin x) 1 √3 ✓ for expansion TP = √3 .600 ( cos x − sin x) 2 2 vir uitbreiding TP = √3 .300(cos x − sin x) ✓ for taking out common factor / vir uithaal van gemene faktor (4) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2022) QUESTION 8/VRAAG 8 Q R 1 O 42 146 T S P 8.1 ˆ = 84 (  at centre) / (Middelpunts  ) POT ✓ S ✓R (2) 8.2 ˆ = 34 (  s on a straight line) QTR ✓ S and/en R ( e op ʼn reguitlyn) ✓ S and/en R R̂1 = 34 (  s opp. = sides) / ( e teenoor = sye) (2) 8.3 RQ̂ T = 112° ✓ S and/en R ˆ = 68 (opp. s of cq) / (teenoorst. e van kv) RPT ✓S ✓R (3) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2022)) MATHEMATICS P2/WISKUNDE V2 13 QUESTION 9/VRAAG 9 K A 1 2 x E B 1 4 1 2 2 3 M 3 1 2 1 2 D C 9.1 𝐵̂1 = 𝑥 (tangents from common point) (raaklyne van gemene punt) ✓ S and/en R 𝐵̂4 = 𝑥 (vertically opposite angles) ✓ S and/en R (regoorstaande hoeke) ̂1 = 𝑥 𝐷 (tan/chord theorem) (raaklyn/koord stelling) ✓S✓R 𝐸𝐶̂ 𝑀 = 𝑥 (tan/chord theorem) (raaklyn/koord stelling) ✓ S and/en R (5) 9.2.1 BD ̂ D = x (proven / bewys) ̂ M = EC ✓S ∴ 𝐵𝐷||𝐸𝐶 (corresponding angles =) ✓R (ooreenkomstige hoeke =) (2) 9.2.2 𝐴̂2 = 𝐸̂ (angles in the same segment) ✓S✓R (hoeke in dieselfde segment) 𝐵̂2 = 𝐸̂ (corresponding angles = , BD ||EC) ✓ S and/en R (ooreenkomstige hoeke =, BD ||EC) 𝐴̂2 = 𝐵̂2 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 14 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2022) 9.2.3 In ΔMEC: CE ∥ DB (proven / bewys) ME MC = (prop. int. Thm, CE ∥ DB) ✓ S and/en R MB MD (eweredigheid st, CE ∥ DB) ME MB ✓S ∴ = MC MD ∴ ME × MD= MC × MB (2) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2022)) MATHEMATICS P2/WISKUNDE V2 15 QUESTION 10/VRAAG 10 A D P Q E F B C 10.1 Construction: Mark off, on AB and AC, P and Q ✓ construction respectively such that AP = DE and AQ = DF. konstruksie Konstruksie: Merk P en Q onderskeidelik op AB en AC af sodat AP = DE en AQ = DF. In  PAQ and/en  EDF: (1) PA = ED (construction / konstruksie) ✓ S and/en R (2) Aˆ =D ˆ (given / gegee) (3) QA = FD (construction / konstruksie)  PAQ   EDF (SAS)  ˆ APQ = ˆE (congruency / kongruensie) ✓S But/Maar ˆB = Eˆ (given/gegee) ✓S✓R ˆ = Eˆ  APQ ✓S  PQ BC (corresponding s = / ooreenkomstige. e =) AP AQ  = ( prop. int. thm / eweredigheid stelling ) AB AC But/Maar : AP = DE and/en AQ = DF (construction/konstruksie) DE DF  = AB AC (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 16 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2022) 10.2 B 1 O 2 M 1 3 2 A T 2 1 S 10.2.1 BOMˆ = Sˆ (s in the same segment) / (e in dies. segment ) ✓ S ✓ R 3 BOMˆ =A ˆ (corresponding <s, AS OM) ✓ S and/en R (ooreenkomstige e, AS OM)  Sˆ 3 = A ˆ ✓R  TS is a tangent (conv. tan - chord thrm) TS is ' n raaklyn (omgekeerde raaklyn - koord stelling ) (4) 10.2.2 Ŝ2 = 90 (s in a semi - circle)/( in semi - sirkel ) ✓S✓R M̂3 = 90 (corr. s / ooreenk . e, AS OM) ✓S✓R  TS is diameter (conv. s in a semi - circle) ✓R TS is ' n middellyn (omgek . ein semi − sirkel ) (5) 10.2.3 In  ABS and/en  STM (1) Aˆ = Sˆ (proven / bewys ) 3 ✓S ˆ (2) B = STMˆ (s in the same segment)/(e in dies. segement ) ✓ S (3) Sˆ 2 = Mˆ (proven / bewys ) 3 ✓ R or/of 3rd ABS ||| STM (    ) angle/ 3de hoek (3) 10.2.4 AS SB  = (similarity / gelykvormigheid ) ✓S SM MT AS . MT = SM . SB But/Maar : ✓S✓R SB = 2SM (Midpoint thrm / prop.int. , OM AS) (Middelpunt stelling / Ewer ., OM AS)  2SM 2 = AS.MT (3) [21] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief