Technical Mathematics Grade 12 P2 Gauteng Mock 2020 Memo (Afr)

PREPARATORY EXAMINATION/ VOORBEREIDENDE EKSAMEN 2020 MARKING GUIDELINES/ NASIENRIGLYNE TECHNICAL MATHEMATICS P2/ TEGNIESE WISKUNDE V2 11092 18 pages/bladsye Marking Codes / Nasienkodes A Accuracy / Akkuraatheid AO Answer only / Slegs antwoord CA Consistent Accuracy / Volgehoue akkuraatheid F Substitution in correct formula / Vervanging in korrekte formule M Method / Metode NPR No Penalty for Rounding / Geen Penalisering vir Afronding NPU No Penalty for Units omitted / Geen Penalisering vir Eenhede weggelaat R Rounding / Afronding S Statement / Stelling S/R Statement and Reason / Stelling en Rede SF Simplification / Vereenvoudiging 1

11092/20 NOTES / AANTEKENINGE • If a candidate answers a question TWICE, only mark the FIRST attempt. / Indien 'n kandidaat 'n vraag TWEE keer beantwoord, merk slegs die EERSTE poging. • If a candidate has crossed out an attempt of a question and did not redo it, mark the crossed out version. / Indien 'n kandidaat 'n antwoord doodgetrek het, maar dit nie probeer oordoen het nie, merk die doodgetrekte weergawe. • Consistent Accuracy applies in all aspects of the Marking Guidelines. / Volgehoue akkuraatheid is in al die nasienriglyne van toepassing. QUESTION / VRAAG 1 𝑦 Aሺ−1; 3ሻ F B O E G 𝑥 Dሺ1; −1ሻ Cሺ0; −2ሻ 1.1 𝑦 = −𝑥 + 2 𝑥-intercept / x-afsnit ∴ 𝑦 = 0 ✓A 𝑦=0 0 = −𝑥 + 2 ✓A 𝑥=2 𝑥=2 𝐸ሺ2; 0ሻ (2) 1.2 𝑥𝐴 +𝑥𝐶 𝑦𝐴 +𝑦𝐶 𝐵( 2 ; 2 ) −1 + 0 3 − 2 = 𝐵( ; ) ✓ A x-value / waarde 2 2 1 1 = 𝐵 (− ; ) ✓ A y-value / waarde 2 2 (2) 2

11092/20 1 2 +1 ✓ CA 𝑚𝐵𝐷 = −1 1.3 𝑚𝐵𝐷 = 1 = −1 − −1 2 𝑚𝐴𝐸 = −1 ✓ CA 𝑚𝐴𝐸 = −1 𝑚𝐵𝐷 = 𝑚𝐴𝐸 ∴ 𝐵𝐷|| 𝐴𝐸 ✓ conclusion / gevolgtrekking 𝑚𝐵𝐷 = 𝑚𝐴𝐸 (3) 1.4 3+2 −2−3 𝑚𝐴𝐶 = −1−0 = −5 OR/ OF 𝑚𝐴𝐶 = = –5 ✓ CA 𝑚𝐴𝐶 = −5 0 +1 𝑡𝑎𝑛𝑂𝐺̂ 𝐵 = −5 Ref. angle/Verw.hoek = 78,69° ̂ B = 101,31° ✓ A OG 𝑂𝐺̂ 𝐵 = 180° − 78,69° = 101,31° ̂ C = 78,69° ✓CA OG 𝑂𝐺̂ 𝐶 = 180 − 101,31° = 78,69° (3) 1.5 𝑂𝐵 = √ሺ𝑥𝐵 − 𝑥𝑂 ሻ2 + ሺ𝑦𝐵 − 𝑦𝑂 ሻ2 1 1 1 2 1 2 ✓ CA for/vir 𝐵 (− 2 ; 2) = √(− 2 − 0) + (2 − 0) OR / OF 2 2  1  1 ✓ CA 𝑂𝐵 = 2 √2 = 0 +  + 0 −   2  2 √2 = 2 ✓ A AF = √2 (Answer given / Antwoord 𝐴𝐹 = √ሺ𝑥𝐴 − 𝑥𝐹 ሻ2 + ሺ𝑦𝐴 − 𝑦𝐹 ሻ2 gegee) = √ሺ−1 − 0ሻ2 + ሺ3 − 2ሻ2 = √2 1 √2 ∴ 𝑂𝐵 = 𝐴𝐹 = (3) 2 2 [13] 3

11092/20 QUESTION / VRAAG 2 2.1 𝑦 Kሺ0; 𝑎ሻ O 𝑥 L Nሺ12; −9ሻ M 2.1.1 𝑟2 = 𝑥2 + 𝑦2 𝑟 2 = ሺ12ሻ2 + ሺ−9ሻ2 ✓ SF 𝑟 2 = 225 𝑥 2 + 𝑦 2 = 225 ✓ A 𝑥 2 + 𝑦 2 = 225 (2) 2.1.2 𝑟 = √225 = 15 𝑎 = 15 ✓A 𝑎 = 15 (1) 𝑦𝑂 − 𝑦𝑁 2.1.3 𝑚𝑂𝑁 = 𝑥 −𝑥 𝑂 𝑁 0−ሺ−9ሻ −9−0 = 0−12 OR / OF = 12 − 0 3 = −4 𝑚𝑂𝑁 × 𝑚𝑀𝑁 = −1 4 ✓ M 𝑚𝑀𝑁 = 3 radius ⊥ tangent /radius ⊥ raak lyn 3 − × 𝑚𝑀𝑁 = −1 4 4 ✓ SF 𝑚𝑀𝑁 = 3 𝑦 − 𝑦𝑁 = 𝑚ሺ𝑥 − 𝑥𝑁 ሻ ✓ CA 4 𝑦 − ሺ−9ሻ = 3 ሺ𝑥 − 12ሻ 4 𝑦 + 9 = 3 𝑥 − 16 OR / OF 4 𝑦 = 3 𝑥 − 25 4

11092/20 OR / OF ✓S 𝑥. 𝑥1 + 𝑦. 𝑦1 = 𝑟² ✓ SF ሺ12ሻ𝑥 + ሺ−9ሻ𝑦 = 225 ✓A 12𝑥 − 9𝑦 = 225 4 y = x − 25 3 OR / OF 4 𝑚𝑂𝑁 × 𝑚𝑀𝑁 = −1 ✓ M 𝑚𝑀𝑁 = 3 radius ⊥ tangent /radius ⊥ raak lyn 3 − 4 × 𝑚𝑀𝑁 = −1 𝑚𝑀𝑁 = 3 4 ✓ SF 𝑁ሺ12; −9ሻ : 𝑦 = 𝑚𝑥 + 𝑐 4 −9 = 3 ሺ12ሻ + 𝑐 ✓ CA 𝑐 = −25 4 (3) ∴ 𝑦 = 3 𝑥 − 25 2.2.1 𝑥2 𝑦2 + =4 9 16 𝑥² 𝑦² ✓A + =1 36 64 (1) 2.2.2 ✓ A x-intercepts / deursnitte ✓ A y-intercepts / deursnitte ✓ CA shape / vorm (3) 2.2.3 Dሺ−6; 0ሻ ✓ A 𝑥D = −6 DE = 2 = FG ✓M Fሺ0; −6ሻ ✓ A 𝑦F = −6 (3) [13] 5

11092/20 QUESTION / VRAAG 3 𝑦 O 𝑥 𝛼 P ሺ√11; 𝑏ሻ 3.1 OP 2 = 𝑥 2 + 𝑦² 2 62 = (√11) + ሺ𝑏ሻ2 ✓M 36 = 11 + 𝑏² ✓ A 𝑏 2 = 25 𝑏 2 = 25 OR /OF b = 5 𝑏 = −5 Answer only / Slegs antwoord 0/2 (2) 3.2.1 6 −5 −30 6 sec 𝛼. tan𝛼 = × = 11 ✓ A sec𝛼 = √11 √11 √11 −5 ✓ A tan𝛼 = √11 ✓ CA Answer (3) Answer only /Slegs antwoord : 1/3 3.2.2 sinሺ180° − αሻ = sinα ✓ M sinα −5 −5 = 6 ✓ CA 6 (2) 2 3.3 LHS = cosec²𝛼 − cot 𝛼 2 2 6 6 = (−5) − ( −5 ) √11 ✓ CA cosecα = −5 √11 36 11 = 25 − 25 = 25 = 1 25 ✓ CA cotα = −5 36 11 25 ✓A − 25 = 25 = 1 25 (3) [10] 6

11092/20 QUESTION / VRAAG 4 4.1.1 √1 − sin2 𝑥. tan𝑥. sin ሺ180° + 𝑥ሻ ✓ S 𝑐𝑜𝑠²𝑥 sin𝑥 sin𝑥 ✓ S cos𝑥 = √cos2 𝑥. ሺ−sin𝑥ሻ cos𝑥 ✓ A −sin𝑥 sin𝑥 = cos𝑥. ሺ−sin𝑥ሻ ✓CA cos𝑥 cos𝑥 = −sin²𝑥 ✓ A −sin²𝑥 (5) 4.1.2 sin240°. sec30° ✓ A −𝑠𝑖𝑛60° cos𝜋. tan315° ✓ A 𝑐𝑜𝑠𝜋 = −1 −𝑠𝑖𝑛60° ✓ A −𝑡𝑎𝑛45° = ሺ−1ሻሺ−𝑡𝑎𝑛45°ሻሺ𝑐𝑜𝑠30°ሻ ✓ A 𝑐𝑜𝑠30° √3 − 2 = = −1 ✓ CA ratios / verhoudings 1 √3 ✓ CA (1) ( 2 ) (6) 4.2 2cos3𝑥 = cot 65,7° 1 2cos3𝑥 = = 0,451.. 𝑡𝑎𝑛65,7° ✓ A cot 65,7° = 0,451.. cos3𝑥 = 0,225.. ✓ M cos3𝑥 = 0,225.. Ref.∠ / 𝑉𝑒𝑟𝑤. ∠ = 76,952. . ° ✓ CA Ref.∠ = 76,952 Quadrant / Kwadrant 1 Quadrant / Kwadrant 4 ✓ A 25,65° 3𝑥 = 76,952. . ° 3𝑥 = 360 − 76,952.. ✓ A 94,35° 𝑥 = 25,65° 3𝑥 = 283,047.. 𝑥 = 94,35° (5) [16] 7

11092/20 QUESTION / VRAAG 5 5.1 𝑎 = −2 ✓A (1) 5.2 0° ሺ360°; −1ሻ ✓ shape / vorm ✓ x-intercepts / x-afsnitte ሺ45°; 0ሻ; ሺ225°; 0ሻ ✓ y-intercepts / y-afsnitte ሺ0; −1ሻ and endpoint / en eindpunt ሺ360°; −1ሻ ✓ Asymptotes / Asimptote 135° and 315° (4) 5.3.1 Period / Periode = 180° ✓A (1) 5.3.2 𝑥 = 135° and/𝑒𝑛 𝑥 = 315° ✓ A 𝑥 = 135° ✓ A 𝑥 = 315° (2) 5.3.3 𝑓ሺ270°ሻ − 𝑔ሺ270°ሻ = 2 − 1 = 1 ✓A (1) 5.3.4 𝑥 ∈ [45° ; 135°] ∪ [225° ; 315°] ✓ CA 45°; 135° values / waardes OR / OF ✓ CA 225°; 315° values / waardes 45° ≤ 𝑥 < 135° or 225° ≤ 𝑥 < 315° ✓ A brackets / hakies OR / OF ✓ CA 45°; 135° values / waardes ✓ CA 225°; 315° values / waardes ✓ inequalities / ongelykhede ≤𝑥< (3) 5.3.5 𝑔ሺ𝑥ሻ − 𝑓ሺ𝑥ሻ = 3 ✓A 𝑥 = 90° (1) [13] 8

11092/20 QUESTION / VRAAG 6 6.1 7,5 7,5 ̂E = 𝑠𝑖𝑛AD ̂E = ✓ M 𝑠𝑖𝑛AD 9,6 9,6 7,5 ̂ E = si𝑛−1 ( AD ) 9,6 = 51,38° ✓A OR /OF OR/OF 𝑠𝑖𝑛AD ̂ E 𝑠𝑖𝑛90° ̂E 𝑠𝑖𝑛AD 𝑠𝑖𝑛90° = ✓ M = 7,5 9,6 7,5 9,6 7,5sin90° 𝑠𝑖𝑛AD ̂E = = 0,78.. 9,6 ̂ E = si𝑛−1 ሺ0,78. . ሻ = 51,38° ✓A AD (2) ̂ C = 180° − 61° − 51,38° = 67,62° 6.2 BD ✓A (1) 9

11092/20 6.3 In ∆BDC: 𝐵𝐷 BC ✓ M sine rule / reël = ̂C sin𝐶̂ sinBD BD 2,3 = ✓ SF sin78° sin67,62° 2,3sin78° BD = = 2,43 m ✓ CA BD = 2,43 m sin67,62° (3) 2 2 6.4 AB = AD + BD² − 2AD. BDcosAD ̂B ✓ M cos rule / reël 2 2 2 AB = 9,6 + 2,43 − 2ሺ9,6ሻሺ2,43ሻcos61° ✓ SF AB2 = 75,44 𝐴𝐵 = 8,69 𝑚 ✓ CA AB = 8,69 𝑚 (3) [9] 10

11092/20 QUESTION / VRAAG 7 A 1 2 1 2 E 𝑥 C D 6 O B 7.1 AD = 8 line from centre ⊥ to chord / ✓S ✓R Loodlyn uit midpt. ʘ na koord 𝑟 = AO = √AD2 + OD² Pythagoras ✓ S/R = √82 + 6² = 10 cm ✓ CA (4) 7.2 EO = AO radii /radiusse 𝑥 + 6 = 10 𝑥 = 4 cm ✓ CA (1) 7.3 𝐷𝐶 = 6 + 10 = 16 ✓S AC = √AD2 + DC² Pythagoras ✓ S/R = √82 + 16² = 8√5 cm or /of 17,89 cm ✓CA (3) [8] 11

11092/20 QUESTION / VRAAG 8 8.1 … equal to the interior opposite angle. / ✓S … gelyk aan die teenoorstaande binnehoek. (1) A P 1 2 1 2 3 D 2 3 1 C 80 1 2 B 8.2.1 𝐶̂1 = A𝐵̂ C = 80° s opp equal sides / ✓ S/R e teenoor gelyke sye 𝑃̂1 = 𝐶̂1 = 80° s in the same segment / ✓S ✓R e in dies. segm. ✓S ✓R 𝑃̂3 = A𝐵̂ C = 80° ext.  of cyclic quad / buite  van kvh (5) 8.2.2 𝐵̂2 = 𝐴̂2 = 20° s in the same segment / ✓S ✓R ✓A 𝐵̂1 = 80° − 20° = 60° e in dies. segm. (3) 8.3 ̂ = 𝑃̂1 − 𝐵̂2 𝐷 ext.  of ∆ / buite  van ∆ ✓ S/R = 80° − 20° = 60° 𝐷 = 𝐵̂1 ̂ ✓S AB is a tangent to a circle through points B, D and P / AB is 'n raaklyn aan die sirkel deur punte B, D en P converse tan chord theorem / omgekeerde raaklynstelling ✓R OR / OF  between line and chord /  tussen lyn en koord OR / OF (3) 12

11092/20 ˆ = 20 A Sum of angles of a triangle ✓ S/R 1 P2 = Aˆ1 = 20 ˆ angles in the same ✓ S D̂ = 60 sege in dies. segm..  Pˆ3 = 80 angles on a straight line D̂ = 60 sum of angles of a triangle Dˆ = Bˆ1 ✓R AB is a tangent converse of tan chord theorem omgekeerde raaklynstelling (3) [12] 13

11092/20 QUESTION / VRAAG 9 9.1.1 GC BG = BE line || one side of   ✓S ✓R CF GC 3 =6 lyn || een sy van ∆ 8 GC = 4 ✓A (3) 9.1.2 GA GC = CF line || one side of   ✓S AD GA 4 =8 lyn || een sy van ∆ 4 GA = 2 ✓ CA (2) 9.2 AB 3+2 = 4+2+3+6 DE ✓ CA 5 1 = 15 = 3 (1) 9.3 Height of / Hoogte van ∆DEF = 3 Height of ∆ABC 1 1 Area / Oppv. ∆ABC = × 𝐴𝐵 × ℎ = ሺ5ሻℎ ✓ M Area / Oppv. ∆ABC 2 2 1 1 Area / Oppv. ∆DEF = 2 × 𝐷𝐸 × 𝐻 = 2 ሺ15ሻ3ℎ ✓ M Area / Oppv. ∆DEF 1 Area / 𝑂𝑝𝑝𝑣. ∆ABC ሺ5ሻℎ 5 1 = 2 = = Area / 𝑂𝑝𝑝𝑣. ∆DEF 1 ሺ15ሻ3ℎ 45 9 ✓ CA 2 OR / OF OR / OF 1 Area / 𝑂𝑝𝑝𝑣. ∆DEF 2 ሺ5ሻ3ℎ 45 = = =9 ✓ CA Area / 𝑂𝑝𝑝𝑣. ∆ABC 1 ሺ15ሻℎ 5 2 (3) [9] 14

11092/20 QUESTION / VRAAG 10 C B 1 3 2 1 S 1 2 T A 10.1 s in semi- circle / e in ʼn semi-sirkel ✓R (1) 10.2 tangent ⊥ radius / raaklyn ⊥ radius ✓R (1) 10.3 In ∆ ACS and / 𝑒𝑛 ∆BCA: ̂ S = AB CA ̂ C = 90° From Question 10.1 and 10.2 / ✓S ✓R Van Vraag 10.1 en 10.2 ✓S Cˆ = Cˆ Common Ŝ = A ̂1 sum of s in ∆ / e van ∆ ∆ACS ||| ∆BCA ∠∠∠ ✓R (4) AC CS AS 10.4 BC = AC = BA ||| s / ∆𝑒 ✓S ✓R AC2 = BC. CS (2) 10.5 AC2 = BC. CS ✓M 752 = BC. 125 75² ✓A BC = 125 = 45 mm (2) 10.6 𝑆𝐵 = 𝐶𝑆 − 𝐵𝐶 ✓ A 𝑆𝐵 = 80 = 125 − 45 = 80 ST = BC SB line || one side of   ✓ S/ R AT ST 80 = 45 lyn || een sy van ∆ 36 80 × 36 ST = 45 ✓A ST = 64 mm (3) [13] 15

11092/20 QUESTION / VRAAG 11 M 1,5 m 125° 0,75 m A C B D 11.1 ✓ SF AB2 = AM 2 − MB² Pythagoras ✓ A (One decimal / AB = √1,52 − 0,75² Een desimaal) = 1,3 m (2) 11.2 AC = 2AB = 2,6 line from centre ⊥ to chord / ✓S loodlyn uit midpt. na koord ✓ SF 1 Area of ∆AMC = 2 AC. MB 1 ✓CA = ሺ2,6ሻሺ0,75ሻ 2 = 0,98 m² (3) 11.3 𝜋 25 125° × 180° = 36 𝜋 = 2,18 radians / radiale ✓A (1) 11.4 𝑟 2𝜃 Area of sector / 𝑂𝑝𝑝𝑣. 𝑣𝑎𝑛 𝑠𝑒𝑘𝑡𝑜𝑟 = ✓M 2 1,5² × 2,18 = ✓ SF 2 ✓ CA = 2,45 m² OR / OF OR / OF s = 𝑟𝜃 ✓ A s = 3,27 s = 1,5 × 2,18 = 3,27 𝑟𝑠 ✓ SF Area of sector / 𝑂𝑝𝑝𝑣. 𝑣𝑎𝑛 𝑠𝑒𝑘𝑡𝑜𝑟 = 2 ✓ CA 𝑟𝑠 = 2 (3) 1,5×3,27 = = 2,45 m² 2 16

11092/20 11.5 Area / Oppv. ABCD = AMCD - ∆AMC ✓M = 2,45 − 0,98 = 1,47 m² ✓A (2) 11.6 Volume = area of the base x height / basis oppv. × hoogte = 1,47 × 5 ✓SF = 7,35 m³ ✓A (2) [13] QUESTION / VRAAG 12 12.1.1 13 n =13 rev / minute = 60 rev / s = 0,22 rev/s 13 ✓A n = 60 rev /s 𝑣 = 𝜋𝐷𝑛 13 ✓SF 10,3 = 𝜋𝐷 (60) 𝐷 = 15,13 𝑚 ✓CA (3) 12.1.2 𝜔 = 2𝜋𝑛 ✓M 13 = 2𝜋 ( ) ✓SF 60 = 1,36 m/s ✓CA (3) 12.2.1 DE = 20 cm ✓A (1) 2 2 12.2.2 BC = AC − AB² Pythagoras AB = √22 − 1² ✓ SF ✓A = √3 = 1,73 cm (2) 12.2.3 Surface area / Buite oppv. = π . DE. CE − π. BC. AC ✓M ✓ SF = π ሺ20ሻሺ25ሻ − πሺ1,73ሻሺ2ሻ ✓ CA (3) 24 827 = 𝜋 = 1 549,06 cm² 50 12.2.4 AD = √252 − 20² = 15 cm ✓ A AD = 15 cm 1 1 ✓M Volume = 3 πDE 2 . AD − 3 πBC. AB ✓ SF ✓ CA 1 1 (4) = πሺ20ሻ2 ሺ15ሻ − πሺ1,73ሻ2 ሺ1ሻ 3 3 = 6 280,05 cm³ [16] 17

11092/20 QUESTION / VRAAG 13 25 𝑎 = 5 = 5 cm ✓ A 𝑎 = 5 cm 𝑦1 + 𝑦6 𝑦 +𝑦 ✓ SF 1 2 6 = 5,5 𝐴𝑇 = 𝑎 [ + 𝑦2 + 𝑦3 + 𝑦4 + 𝑦5 ] 2 ✓ M other ordinates / 10+ 1 = 5 [ 2 + 10 + 7,3 + 4,1 + 6] ander ordinate = 5[5,5 + 27,4] ✓ M 5[5,5 + 27,4] = 164,5 cm² ✓ CA OR /OF OR/ OF 𝐴𝑇 = 𝑎[𝑚1 + 𝑚2 + 𝑚3 + 𝑚4 + 𝑚5 ] ✓M 10+10 𝑚1 = = 10 ✓ A 𝑚1 𝑎𝑛𝑑 / 𝑒𝑛 𝑚2 2 10 + 7,3 𝑚2 = = 8,65 2 ✓ A 𝑚3 , 𝑚4 , 𝑚 5 7,3 + 4,1 𝑚3 = = 5,7 2 4,1 + 6 𝑚4 = = 5,05 2 6+1 𝑚1 = = 3,5 2 ✓ SF ✓ CA 𝐴𝑇 = 5[10 + 8,65 + 5,7 + 5,05 + 3,5] 𝐴𝑇 = 5[32,9] 𝐴𝑇 = 164,5 cm² (5) [5] TOTAL / TOTAAL: 150 18