Mathematics Grade 11 P2 Eastern Cape November 2023 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIORSERTIFIKAAT GRADE/GRAAD 11 NOVEMBER 2023 MATHEMATICS P2/WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 13 pages. Hierdie nasienriglyn bestaan uit 13 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2023) QUESTION 1/VRAAG 1 1.1 6 boys / seuns ✓ answer / antwoord (1) 1.2 Skewed to the right / positively skewed ✓ answer / antwoord Skeef na regs / Positief skeef (1) 1.3 127 – 68 = 59 ✓ working out / bewerking ✓ answer / antwoord (2) 1.4.1 Median/Mediaan = 84 / Remains unchanged / ✓ answer / antwoord Bly onveranderd (1) 1.4.2 (87 × 24 + 20) ÷ 24 = 87,8 OR/OF 88 ✓ working out / bewerking ✓ answer / antwoord (2) [7] QUESTION 2/VRAAG 2 2.1 Cumulative Time waiting / Number of frequency Wagtyd learners Kumulatiewe (minutes / minute) Aantal leerders frekwensie ✓ for / vir 6; 16; 35 0<𝑥≤4 6 6 ✓ for / vir 50; 55 4< 𝑥 ≤ 8 10 16 8 < 𝑥 ≤ 12 19 35 12 < 𝑥 ≤ 16 15 50 16 < 𝑥 ≤ 20 5 55 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P2/WISKUNDE V2 3 2.2 ✓ for grounding vir anker ✓ for endpoint vir eindpunt ✓✓ for any 2 correct points vir enige 2 korrekte punte (4) 2.3 55 – 43 = 12 learners / leerders (accept/aanvaar 11 en 13) ✓ 55 ✓43 ✓12 (3) 2.4 8 < 𝑥 ≤ 12 ✓ answer / antwoord (1) 2.5 IQR / IKW = 14 – 7 = 7 ✓ UQ ✓ LQ ✓ answer / antwoord (3) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2023) QUESTION 3 / VRAAG 3 3.1 𝑚𝐴𝐵 = 𝑚𝐵𝐶 −8−0 −14−0 -8 − 0 = ✓ 𝑥 −5 + 8 𝑥−5 + 8 𝑥+8 −8 −14 −14−0 = ✓ 𝑥+8 𝑥+3 𝑥+8 −8𝑥 − 64 = −14𝑥 − 42 ✓ −8𝑥 − 64 6𝑥 = 22 ✓ −14𝑥 − 42 11 11 ✓ 𝑥= 3 𝑥= (5) 3 3.2.1 6+0 -2−6 ✓ substitution / vervanging M( ; ) 2 2 ✓ answer / antwoord = M(3; -4) (2) 3.2.2 2 AC = √(2 − 0)2 + (10 − (−6)) ✓ substitution / vervanging AC = √4 + 256 AC = 2√65 ✓ answer / antwoord (2) 3.2.3 F(−4; 6) ✓ x ✓y (2) 3.2.4 𝑚𝐴𝐵 = 10−(−2) = 12 = −3 ✓ 𝑚𝐴𝐵 2−6 −4 ✓ 𝑚𝐵𝐶 −2+6 4 2 𝑚𝐵𝐶 = = = 6−0 6 3 ✓ 𝑚𝐴𝐵 × 𝑚𝐵𝐶 ≠ −1 𝑚𝐴𝐵 × 𝑚𝐵𝐶 ≠ −1 ∴ABCF is not a rectangle / is nie ʼn reghoek nie OR/OF ✓𝐴𝐵 = √160 = 4√10 𝐴𝐶 = √260 = 2√65 ✓ 𝐵𝐶 = √50 = 5√2 𝐴𝐵 = √160 = 4√10 ✓ 260 ≠ 160 + 50 𝐵𝐶 = √50 = 5√2 260 ≠ 160 + 50 𝐴𝐶 2 ≠ 𝐴𝐵2 + 𝐵𝐶 2 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P2/WISKUNDE V2 5 3.2.5 𝑚𝐴𝐵 = −3 ✓ 𝑚𝐴𝐵 ∴ 𝑦 = −3𝑥 + 𝑐 10 = −3(2) + 𝑐 ✓ substitution / vervanging 𝑐 = 16 ∴ 𝑦 = −3𝑥 + 16 ✓ equation of AB / 10−(−6) vergelyking van AB 𝑚𝐴𝐶 = =8 2−0 ✓ gradient of AC/gradiënt van AC ∴ 𝑦 = 8𝑥 − 6 ✓ equation of /vergelyking van AC (5) 3.2.6 tan AÊX = −3 ̂X = −3 ✓ tan AE ∴ AÊX = 108,43° ✓ AÊX tan AD̂E = 8 ✓ tan AD̂E = 8 ∴ AD̂ E = 82,87° ✓ AD̂E ̂ A = 108,43° − 82,87° = 25,56° ✓✓ answer / antwoord (6) 3.2.7 3 D (4 ; 0) ✓ coordinates of D / koördinate van D 16 E ( 3 ; 0) ✓ coordinates of E / koördinate van E 55 ∴ DE = 12 units / eenhede ✓ length of DE / lengte van DE Perpendicular height / Loodregte hoogte = 10 units / eenhede ✓ height / hoogte 1 55 ∴ A = 2 × 12 × 10 = 22, 92 units2/ eenhede2 ✓ answer / antwoord OR/OF ✓ 𝐴𝐷 = 10,08 𝐴𝐷 = 10,08 ✓𝐴𝐸 = 10,54 𝐴𝐸 = 10,54 ✓ Formula/formule 1 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐴𝐷𝐸 = 𝐴𝐷. 𝐴𝐸 sin 𝐴 2 1 ✓ substitution/vervanging 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐷𝐸 = (10,08)(10,54) sin 25,56 2 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐷𝐸 = 22,92 𝑢𝑛𝑖𝑡𝑠/𝑒𝑒𝑛ℎ𝑒𝑑𝑒 2 ✓answer/antwoord (5) [30] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2023) QUESTION 4 / VRAAG 4 4.1 tan 𝜃 = 7 3 ✓ correct quadrant / korrekte kwadrant 3 𝑟 2 = (−7)2 + (−3)2 ✓ tan 𝜃 = 7 𝑟 2 = 58 𝑟 = √58 ✓ for/vir sin θ + cos θ 2 sin θ −3 −7 +( ) ✓ substitution / vervanging = √58 −3√58 2( ) √58 ✓ simplification / vereenvoudiging −10 √58 = × √58 −6 ✓ answer / antwoord 5 = (6) 3 4.2.1 tan(−32°) ✓ method used / triangle = − tan 32° metode gebruik / driehoek 𝑝 =− ✓ − tan 32° 2 √1−𝑝 ✓ answer / antwoord (3) 4.2.2 sin 418° = sin 58° = sin(90° − 32) ✓ sin 58° ✓ cos 32° = √1 − 𝑝2 = cos 32° ✓ answer / antwoord ✓ answer = √1 − 𝑝2 antwoord (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P2/WISKUNDE V2 7 4.3 sin 120°.tan 300° ✓ sin 60° cos(−60°).tan 225° ✓ − tan 60° sin 60° . −tan 60° ✓ cos 60° = ✓ tan 45° cos 60°.tan 45° ✓ for numerator S/A / vir teller √3 √3 .− ✓ for denominator S/A / vir noemer 2 1 = 1 . 1 2 ✓ answer / antwoord 3 2 = − × = −3 (7) 2 1 4.4 cos2 x−cos 𝑥−sin2 x 1 1 = − 2 sin 𝑥 cos 𝑥+sin 𝑥 tan 𝑥 sin 𝑥 cos2 x−cos 𝑥−(1−cos2 x) ✓ 1 − cos2 x LHS/LK = 2 sin 𝑥 cos 𝑥+sin 𝑥 ✓ simplifying / vereenvoudiging 2𝑐𝑜𝑠 2 𝑥−cos 𝑥−1 ✓ factoring the top / faktorisering teller LHS = 2 sin 𝑥 cos 𝑥+sin 𝑥 ✓ factoring the bottom / (2 cos 𝑥+1)(cos 𝑥−1) faktorisering noemer LHS = sin 𝑥(2 cos 𝑥+1) cos 𝑥−1 cos 𝑥 1 LHS = ✓ sin 𝑥 − sin 𝑥 sin 𝑥 cos 𝑥 1 LHS = − sin 𝑥 sin 𝑥 1 1 LHS = − = RHS/𝑅𝐾 (5) tan 𝑥 sin 𝑥 4.5 sin(5𝜃) = cos(𝜃 − 40°) sin(5𝜃) = sin[90° − (𝜃 − 40°)] ✓ co – function / ko-funksie sin(5𝜃) = sin(130° − 𝜃) 5𝜃 = 130° − 𝜃 + 360°. 𝑘 ✓ 5𝜃 = 130° − 𝜃 + 360°. 𝑘 6𝜃 = 130° + 360. 𝑘 𝜃 = 21,67° + 60°. 𝑘 ✓ 𝜃 = 21,67° + 60°. 𝑘 OR / OF 5𝜃 = 180° − (130° − 𝜃) + 360°. 𝑘 ✓ 5𝜃 = 180° − (130° − 𝜃) + 360°. 𝑘 4𝜃 = 50° + 360. 𝑘 ✓ 4𝜃 = 50° + 360. 𝑘 𝜃 = 12,5° + 90°. 𝑘 where/waar 𝑘𝜖𝑍 ✓ 𝜃 = 12,5° + 90°. 𝑘 OR / OF ✓ co-function /ko-funksie 0 − 5𝜃 = 360 − (𝜃 − 40) + 360𝑘 ✓ −6𝜃 = −130 + 360𝑘 −4𝜃 = 310 + 360𝑘 ✓ 𝜃 = 21,67 − 60𝑘 𝜃 = −77,5 − 90𝑘 ✓90 − 5𝜃 = 360 − (𝜃 − 40) + 360𝑘 ✓−4𝜃 = 310 + 360𝑘 ✓ 𝜃 = −77,5 − 90𝑘 (6) [29] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2023) QUESTION 5/VRAAG 5 𝑥 𝑥 5.1 tan 𝜃 = 𝑄𝑅 ✓ tan 𝜃 = 𝑄𝑅 𝑥 𝑄𝑅 = tan 𝜃 ✓ answer / antwoord (2) 2 5.2 𝑥 𝑥 ✓ for the cosine rule 𝑄𝑆 2 = 𝑥 2 + (tan 𝜃) − 2𝑥 (tan 𝜃) cos 120° vir die cosinusreël 𝑥2 2𝑥 2 1 𝑄𝑆 2 = 𝑥 2 + 𝑡𝑎𝑛2 𝜃 − tan 𝜃 (− 2) ✓− 1 2 𝑥2 𝑥2 ✓ simplification 𝑄𝑆 2 = 𝑥 2 + 𝑡𝑎𝑛2 𝜃 + tan 𝜃 vereenvoudiging 1 𝑄𝑆 = √𝑥 2 (1 + 𝑡𝑎𝑛2 𝜃 + tan 𝜃) 1 ✓ for square rooting vir vierkantswortel 1 1 𝑄𝑆 = 𝑥. √𝑡𝑎𝑛2 𝜃 + tan 𝜃 + 1 (4) 5.3.1 1 1 ✓ substitution / vervanging 𝑄𝑆 = 15√𝑡𝑎𝑛2 22° + tan 22° + 1 𝑄𝑆 = 46, 48 m ✓ answer / antwoord (2) 5.3.2 tan 𝑄𝑃̂𝑆 = 46,48 ✓ for using tan / 15 vir gebruik van tan 𝑄𝑃̂𝑆 = 72,11° ✓ answer / antwoord (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P2/WISKUNDE V2 9 QUESTION 6 / VRAAG 6 6.1 For f : ✓ shape ✓ endpoints ✓ TP and/en x-intercepts Vir f : ✓ vorm ✓ eindpunte ✓ DP and/en x-afsnitte For g: ✓ asymptote ✓ shape ✓ y-intercept Vir g: ✓ asimptoot ✓ vorm ✓ y-afsnit (6) 6.2.1 0° < 𝑥 < 90° ✓ answer / antwoord (1) 6.2.2 −45° ≤ 𝑥 ≤ 0° or/of 45° ≤ 𝑥 < 90° ✓✓✓ for each answer or/of 90° < 𝑥 ≤ 180° vir elke antwoord (3) 6.3 1 ≤ 𝑦 ≤ 3 or/of 𝑦𝜖[1; 3] ✓✓ for the answer vir die antwoord (2) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2023) QUESTION 7 / VRAAG 7 7.1 7.1.1 𝑂𝑅 = 3𝑥 ✓ answer / antwoord (1) 7.1.2 UR = 20 cm (line from centre perpendicular…) ✓S✓R (lyn vanaf middelpunt loodreg …) ✓ S and/en R (3𝑥)2 = (2𝑥)2 + 202 (Pythagoras) 9𝑥 2 = 4𝑥 2 + 400 ✓ simplification 5𝑥 2 = 400 vereenvoudiging 𝑥 2 = 80 ✓ answer / antwoord 𝑥 = √80 = 4√5 (5) 7.2 7.2.1 𝑂̂1 = 54° (alt s / verw. e, OD ∥ AB) ✓S✓R 𝐴̂2 = 27° ( at centre) / (middelpunts ) ✓S✓R (4) 7.2.2 𝐶̂ = 153° (opp. s of a cq) / (teenoorst. e van k.v) ✓S✓R (2) 7.2.3 𝐴̂ = 54° (s opp = sides) / (e teenoor = sye) ✓ S and/en R 𝐴̂2 = 27° (proven) / (reeds bewys) ✓S ∴ 𝐴̂1 = 27° ✓S ∴DA bisects OÂ B / DA halveer OA ̂B (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P2/WISKUNDE V2 11 QUESTION 8 / VRAAG 8 8.1 Let /Laat 𝐴̂1 = 𝑥 ∴ 𝐴̂2 = 90° − 𝑥 (tan ⊥ rad) / (raaklyn ⊥ radius) ✓S✓R 𝐹𝐶̂ 𝐴 = 90° ( in a semi-circle) / ( in semi-sirkel) ✓ S and/en R 𝐶̂2 = 90° − 𝑥 (s in the same segment) / (e in dies. segment) ✓ S and/en R ∴ 𝐶̂1 = 𝑥 ̂B = BĈ A ✓S ∴ DA (5) 8.2 8.2.1 ̂ R1 = 39° ( subtended by diameter)/ ( onderspan deur middellyn) ✓S✓R (2) 8.2.2 ̂1 = 78° ( at centre) / (Middelpunts ) O ✓S✓R (2) 8.2.3 ̂ 1 = 51° (s opp = sides) / (e teenoor gelyke sye) Q ✓S✓R (2) 8.2.4 ̂1 = 68° (s in the same segment) / (e in dieselfde segment) U ✓S✓R (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2023) QUESTION 9 / VRAAG 9 9.1.1 ̂F = 𝑥 AB (tan-chord theorem) / (raaklyn-koord stelling) ✓S✓R (2) 9.1.2 BĈD = 𝑥 (corresponding angles, BF∥CD) ✓S✓R (ooreenkomstige hoeke ; BF || CD) (2) 9.2 Exterior angle = interior opp. angle ✓R Buitehoek = teenoorstaande binnehoek. (1) 9.3 ̂ C = 𝑥 (angles in the same segment) / BD ✓S✓R (hoeke in dieselfde segment) ̂ D = 𝑥 (alternate angles BF∥CD) FB (verwissellende hoeke BF || CD) ✓S✓R (4) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P2/WISKUNDE V2 13 QUESTION 10 / VRAAG 10 10.1 𝐷̂ = 𝐶̂3 (s opp. = sides) / (e teenoor = sye) ✓ S and/en R 𝐷̂ = 𝐵̂1 (ext.  of a cq) / (buite  van k.v) ✓S✓R 𝐶̂3 = 𝐵̂2 (s in the same segment) / (e in dieselfde segment) ✓ S and/en R 𝐵̂1 = 𝐵̂2 (4) 10.2 𝐶̂1 + 𝐶̂2 = 𝐷 ̂ (tan-chord theorem) / (raaklyn-koord stelling) ✓ S and/en R But/Maar 𝐷 ̂ = 𝐵̂2 (proven) / (reeds bewys) ∴ 𝐵2 = 𝐶1 + 𝐶̂2 (ext  = int opp.angle) / ̂ ̂ ✓ S and/en R (buite  = teenoorst. binnehoek) ✓R BECH is a c.q. / BECH is ʼn kv (converse. ext.  of a cq) / (omgekeerde buite  van kv) (3) 10.3 𝐵̂2 = 𝐴̂3 + 𝑇̂1 (ext  of a triangle) / (buite  van driehoek) ✓ S and/en R But/Maar 𝐵̂2 = 𝐶̂1 + 𝐶̂2 (proven) / (reeeds bewys) ✓ S and/en R And/ En AT = CT (tangents from the same point) ✓ S and/en R (raaklyne vanaf dieselfde punt) ✓ S and/en R ∴ 𝐶1 + 𝐶2 = 𝐴3 + 𝐴̂2 (s opp. = sides) / (e teenoor = sye) ̂ ̂ ̂ ✓ S and/en R 𝐴̂3 + 𝐴̂2 = 𝐴̂3 + 𝑇̂1 (conv. tan-chord) / (omgekeerde raaklyn-koord) ∴ 𝐴̂2 = 𝑇̂1 CA is a tangent to circle ABT / CA is ʼn raaklyn aan sirkel ABT) (5) [12] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief