TECH MATHS P2 MEMO JUNE 2023_Afrikaans+English_hlayiso.com_.pdf

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIORSERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2023 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 17 pages./ Hierdie nasienriglyn bestaan uit 17 bladsye.

Downloaded from hlayiso.com 2 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/JUNE/JUNIE 2023) NOTE: • Continuous accuracy (CA) applies only where indicated in this marking guideline. • Assuming values/answers in order to solve a problem is unacceptable. LET WEL: • Volgehoue akkuraatheid (CA) is slegs van toepassing soos aangedui in hierdie nasienriglyn. • Aanvaarding van waardes/antwoorde om ʼn probleem op te los, is onaanvaarbaar. MARKING CODES / NASIENKODES M Method/Metode A Accuracy/Akkuraatheid AO Answer only/Slegs antwoord CA Consistent accuracy/Deurlopende akkuraatheid F Formula/Formule I Identity/Identiteit R Rounding/Afronding S Simplification/Vereenvoudiging ST Statement/Bewering RE Reason/Rede ST RE Statement and correct reason/Bewering en korrekte rede SF Substitution correctly in correct formula/Korrekte vervanging in die korrekte formule NPU No penalty for omitting units/Geen penalisering vir eenhede uitgelaat Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 3 QUESTION/VRAAG 1 1.1 mAB = tan135 = −1 ✓ M ✓ S A AO: Full marks / Volpunte (2) 1.2 p −5 ✓ M mAB = 8+ 2 p −5 ✓ S A  = −1 8+ 2 ✓ S A  p − 5 = −10  p = −5 (3) 1.3  −2 + 8 5 − 5  ✓ x-value A M AB =  ;  = ( 3;0 ) ✓ y-waarde A  2 2  (2) 1.4 y = −5 ✓A (1) 1.5 C ( −2; −5) ✓ x-value A ✓y-waarde A (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/JUNE/JUNIE 2023) 1.6 −5 − 0 ✓ grad CA mCM = =1 −2 − 3 ✓ product/ produk A ✓ conclusion/ mAB × mCM = −1 × 1 = −1 gevolgtrekking  CM ⊥ AB (3) 1.7 grad of line = mCM = 1 ✓ grad. CA y − y1 = m ( x − x1 ) OR/OF y = mx + c y − 5 = 1( x − ( −2) ) 5 = 1( −2) + c ✓ SF A ✓ equation/vergelyking y = x+7 CA (3) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 5 QUESTION/VRAAG 2 2.1 2.1.1 r = 20 = 2 5 ✓A (1) 2.1.2 xx1 + yy1 = r 2 ✓F ✓ SF A  x ( 4) + y ( 2) = 20 ✓S 2 y = −4x + 20 ✓ equation / vergl  y = −2x + 10 CA OR/OF OR/OF 2 1  mradius = = ✓ grad. radius A 4 2  mtangent/raaklyn = −2 ✓ grad. tan / raaklyn y − y1 = m ( x − x1 ) OR/OF y = mx + c CA  y − 2 = −2 ( x − 4) 2 = −2 ( 4) + c ✓ SF A  y = −2x + 10 ✓ equation / vergl CA (4) 2.1.3 ( −4; −2) ✓ x-value A ✓ y-waarde A (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/JUNE/JUNIE 2023) 2.2 ✓ elliptical shape / elliptiese vorm A ✓ x-intercepts/afsnitte A ✓ y-intercepts/afsnitte A (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 7 QUESTION/VRAAG 3 3.1 3.1.1 8 4 ✓A cos  = = 10 5 (1) 3.1.2 k 2 + 82 = 102 Pythagoras ✓M ✓S  k 2 = 36 ✓ value of / waarde van k k = −6 4th quadrant /4de kwadrant (3) 3.1.3 - 6 tan q 8 = ✓ tan ratio / verh. A cosecq 10 - 6 ✓ cosec ratio / verh. A = 9 ✓S CA 20 (3) 3.2 3cos x − 1 = −1,5 3cos x = −0,5 ✓S A cos x = −0,1666... ✓ Ref / Verw  CA Ref / Verw  = 80, 41 ✓ Quadrants / ∴ x = 180° − 80,41° OR/OF x = 180° + 80,41° Kwadrante A ∴ 𝑥 = 99,59° OR/OF 𝑥 = 260,41° ✓ values of x / waardes van x CA (4) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/JUNE/JUNIE 2023) QUESTION/VRAAG 4 4.1 (1 + cos x )(1 − cos x ) = 1 − cos2 x ✓S A ✓I = sin2 x (2) 4.2 cos ( 2 − x ) tan x 2 2 ✓ cos x2 sin (180 + x ) cosec (180 − x ) sin 2 x ✓ cos 2 x sin 2 x ✓ − sin x cos 2 x ✓ cosec x = cos 2 x ( − sin x )( cosec x ) ✓ −1 sin 2 x ✓ − sin 2 x cos 2 x = cos 2 x ( −1) = − sin 2 x (6) 4.3 LHS / LK = cot x + tan x cos x sin x cos x = + ✓ sin x cos x sin x cos2 x + sin 2 x sin x = ✓ ( sin x )( cos x ) ✓S cos x 1 = ( sin x )( cos x ) ✓ cos2 x + sin2 x = 1 = cosec x sec x = RHS/ RK (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 9 QUESTION/VRAAG 5 f ( x ) = cos 2 x and g ( x ) = sin ( x − 30) for x  0;180 5.1 360 ✓A Period f = = 180 2 (1) 5.2 Amplitude g = 1 ✓A (1) 5.3 f: ✓ y-intercept at / y-afsnit by 1 ✓ x-intercepts at 45 and 135 / x-afsnitte by 45 en 135 ✓ turning point at / draaipunt by (90; −1) ✓ End point at / eindpunt by (180; 1) g: ✓ y-intercept at / y-afsnit by −0,5 ✓ x-intercept at 30 / x-afsnit by 30 ✓ turning point at / draaipunt by (120; 1) ✓ End point at / eindpunt by (180; 0,5) (8) 5.4.1 45  x  135 ✓ 45  x CA ✓ x  135 CA (2) 5.4.2 135  x  180 ✓ 135  x CA ✓ x  180 CA (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/JUNE/JUNIE 2023) QUESTION/VRAAG 6 6.1 p 2 = q 2 + r 2 − 2qr cos P ✓ A OR/OF q = p + r − 2 pr cos Q 2 2 2 OR/OF r = p + q − 2 pq cos R 2 2 2 (1) 6.2 6.2.1  ✓ ST A A BC = 90 ( in semi-circle /  in semi-sirkel) ✓ RE (2) 6.2.2 Ù ✓ ST A BC D = 144° (ext.  of  / buite  van ) ✓ RE (2) 6.2.3 BD2 = BC 2 + CD2 − 2BC CD cos BCD ✓F = 62 + 62 − 2  6  6cos144 ✓ SF A = 130,24922… ✓S CA BD  11,41 cm ✓ BD value / waarde (4) 6.2.4 BC ✓ ratio / verh A = sin 54 AC ✓ value / waarde 6 CA AC = sin 54  7,42 cm (2) 6.2.5 1 ✓F Area of/van ABC = AC  BCsin ACB 2 1 =  7, 42  6  sin 36 ✓ SF CA 2  13, 08 cm 2 ✓ Area CA (3) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 11 QUESTION/VRAAG 7 7.1 BD = 5,5 cm (line from centre ⊥ to chord / lyn vanuit midpt ⊥ ✓ ST A koord) ✓ RE (2) 7.2 OB2 = OD2 + BD2 (Pythagoras) OB 2 = 32 + 5,52 = 39, 25 ✓ ST CA ✓ OB length / OB  6,26 cm lengte (2) 7.3 In ABD and/en ACD: BD = DC (line from centre ⊥ to chord / lyn vanuit midpt ⊥ koord) ✓ ST A AD is common / gemeenskaplik   ✓ ST A D1 = D 2 = 90  ABD  ACD ( SS ) ✓ ST A ✓ RE (4) 7.4 ✓ trig ratio / verh. OD 3 ✓ size of angle / sin B1 = = OB 6, 26 grootte van hoek   B1  28, 64 OR/OF any alternative trig ratio/ enige alternatiewe trig. verhouding (2) 7.5  ✓ ST CA O1 = 61,36 (Int  of  / Binne e van )  O2 = 61,36 (Congruency / Kongruensie) ✓ ST CA   A = 61,36 ( at centre = 2 ×  at circumference / ✓ ST CA ✓ RE middelpts  = 2 × Omtreks) (4) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/JUNE/JUNIE 2023) QUESTION/VRAAG 8 8.1 Tangents from a common point / raaklyne vanuit dieselfde punt ✓A (1) 8.2.1  D1 = 65 (tan – chord th / raaklyn – koord st) ✓ ST A ✓ RE (2) 8.2.2  ✓ ST A D 2 = 25 (Radius ⊥ Tangent / Raaklyn) ✓ RE (2) 8.2.3  ✓ ST A D E F = 66 ( on str line / e op reguit lyn) ✓ RE (2) 8.2.4  ✓ ST A G = 114 (opp. s of cyclic quad / teenoorst. e van kdvh) ✓ RE (2) 8.2.5  ✓ ST A F D K = 66 (tan – chord th / raaklyn – koord st) ✓ RE (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 13 8.3  ✓ ST CA H = 50 (Int s of  / Binne e van )   ✓ RE  H + F  180 ✓ RE ∴EHDF is not cyclic/is nie siklies nie (Opp. s NOT supplementary / Teenoorst. e is NIE supplementêr NIE) OR/OF OR / OF  H = 50 (Int s of  / Binne e van ) ✓ ST CA   H  F D K ✓ RE ∴EHDF is not cyclic/is nie siklies nie (Ext  NOT equal to opp. int  ✓ RE Buite  NIE gelyk aan teenoorst. binne  NIE) (3) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 14 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/JUNE/JUNIE 2023) QUESTION/VRAAG 9 9.1 2k + 3k = 20 for some value k / vir enige waarde k. ✓ setup equation / 5k = 20 vergelyking opstel k = 4 ✓ value of k / waarde ∴ AD = 8 cm and/en DB = 12 cm van k ✓AD and/en DB lengths / lengtes OR/OF OR/OF 2 2 AD =  20 = 8 cm ✓ fraction 5 5 3 ✓ AD length / lengte DB =  20 = 12 cm ✓ DB length / lengte 5 (3) 9.2 BE DB ✓ ST A = (Prop th, DE || AC / Ewer st, DE || AC) BC BA ✓ RE BE 12 = ✓ BE length / lengte 17,34 20 (3) BE  10,40 cm 9.3 In ΔBDE and/en ΔBAC:  B is common / gemeenskaplik   ✓ ST A D = A corr. s / ooreenk e; DE || AC ✓ ST RE A   ✓ ST RE A E = C corr. s / ooreenk. e; DE || AC Last mark for last OR/OF Int s of  / Binne e van  statement and reason BDE ||| BAC (,,) OR for final reason  (3) 9.4 DE BA ✓ ST RE A = (BDE ||| BAC) AC BD DE 20 ✓ ST CA = 7 12 ✓ DE length / lengte DE  11,67 cm (3) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 15 QUESTION/VRAAG 10 10.1 108 km 1 000 m 1h ✓ conversion factors / 108 km/h =   = 30 m/s herleidingsfaktore 1h 1 km 3600 s ✓ answer / antwoord (2) 10.2  =  Dn ✓F ✓ conversion / 30m/s =   ( 0, 25m)  n herleiding ✓ SF A 30 ✓S n= ✓ answer / 0, 25 antwoord n  38, 20 rev/s (5) 10.3  = 2 n ✓F ✓ SF CA = 2  38,20 ✓ answer / antwoord  240,02 rad/s (3) 10.4 s = t OR/OF D = S T ✓F = 30  (10 min  60 s ) ✓ SF CA = 18000m ✓ answer / = 18km antwoord (3) 10.5 number of revolutions/aantal revolusies n= time/tyd 20 ✓ SF CA 38, 20 = t t  0,52 sec ✓ answer / antwoord (2) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 16 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/JUNE/JUNIE 2023) QUESTION/VRAAG 11 11.1.1 s = r ✓F  RT = 8  ✓ SF A 2 RT = 4 cm ✓ RT length / lengte  12,57 cm (3) 11.1.2 rs ✓F Area = 2 8  4 ✓ SF A = 2 ✓ Area = 16 cm2  50, 27 cm 2 OR/OF OR/OF ✓F r 2 Area = 2  ✓ SF A 82  = 2 ✓ Area 2 = 16 cm2  50, 27 cm 2 (3) 11.1.3 Shaded area/Gearseerde area = Area square/vierkant − Area sector/sektor ✓M = 8  8 − 16 ✓ area of square/ app van vierkant  13,73 cm2 ✓ shaded area/ gearseerde app Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 17 11.2 4h2 − 4dh + x2 = 0 ✓F 4 ( 4 ) − 4d ( 4 ) + (10 ) = 0 2 2 ✓SF A ✓S 164 − 16d = 0 ✓ diameter / middellyn d = 10, 25 ✓ radius r = 5,125 cm (5) 11.3 o +o  ✓F AT = a  1 n + o2 + o3 + o4 + ... + on−1   2   9, 42 + 7  ✓SF 113,61 = a  + 8,14 + 6, 42 + 7,1 + 8  A  2  113,61 = a ( 37,87 ) ✓S  a = 3 cm ✓ value/waarde of a OR/OF OR/OF AT = a ( m1 + m2 + m3 + . . . + mn−1 ) ✓F  9, 42 + 8,14 8,14 + 6, 42 6, 42 + 7,1 7,1 + 8 8 + 7  113,61 = a  + + + +  ✓SF A  2 2 2 2 2  113,61 = a ( 37,87 ) ✓S  a = 3 cm ✓ value of a (4) [18] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief