Mathematics Grade 12 Paper 2 Western Cape Mock 2016 Memo

testpapers.co.za SEPTEMBER P/ V2 2016 MARKING/ NASIEN MEMORANDUM 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 the question, mark the crossed out version. • Consistent accuracy applies in ALL aspects of the marking memorandum. • Assuming answers/values in order to solve a problem is NOT acceptable.. NOTA: • As 'n kandidaat 'n vraag TWEEKEER beantwoord, merk slegs die EERSTE poging. • As 'n kandidaat 'n poging om die vraagte beantwoord, doodgetrek het en nie dit oorgedoen het nie, merk die doodgetrekte poging. • Volgehoue akkuraatheid word in ALLE aspekte van die Nasienmemorandum toegepas. • Aanvaarding van antwoorde/waardes om 'n probleem op te los, is ONaanvaarbaar. QUESTION/ VRAAG1 Q1 DESCRIPTORS/ BESKRYWERS MARK/ PUNT 1.1 Aantal leerders Kum f Tyd in minute Number of Cum f Time in minutes learners ✓✓ A/CA 5 < t  10 160 160 Cum frequency values/ Kum.frek. 10 < t  15 150 310 waardes 15 < t  20 110 420 20 < t  25 60 480 25 < t  30 45 525 30 < t  35 15 540 (2) 1.2 550 500 ✓ anchor point (5;0)/ 450 Begin punt 400 ✓ use cum freq/ Kummulatiewe frekwensie 350 gebruik kum.frek. 300 ✓ use upper limits/ 250 boonste limiete 200 ✓ smooth curve/ 150 gladde kurwe 100 50 0 0 5 10 15 20 25 30 35 40 Tyd in minute (4) 1.3 average time = 14,95 min ✓✓ ans/ antw (2) 1.4 SD = 6,89 ✓✓ ans/ antw (2) 1.5 > 14,95 + 6,89 = 21,84 ✓ 21,84  540 – 442 = 98 learners ✓ accept/ aanvaar 440 – 445 ✓ accept/aanvaar (3) 95 – 100 [13]

QUESTION/ VRAAG 2 Q2 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ BESKRYWERS MARK/ PUNT 2.1 a = –22,70544202 ✓ a- value/waarde b = 0,1989294404 ✓ b-value/waarde 𝑦 = −22,71 + 0,20𝑥 ✓Equation/Vergelyking (3) 2.2 r = 0.8496991 = 0,85 ✓ 𝑟 (1) 2.3 𝑦 = −22,71 + 0,20(350) ✓ Substitute/Instel y = 47,29 = 47 ✓ Ans/Antw (2) 2.4 a positive strong correlation/ sterk positief ✓ Ans/Antw (1) [7] QUESTION/ VRAAG 3 Q3 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ BESKRYWERS MARK/ PUNT 3.1 𝐴𝐵 = √(4 + 2)2 + (9 − 1)2 ✓ Subst in correct formula/Instel in korrekte = √62 + 82 formule = √100 ✓ Ans/Antw (2) = 10 3.2 𝐴𝐵 = 𝐵𝐶 = 10 ✓𝑥 value/waarde 𝑦 𝑣𝑎𝑙𝑢𝑒 = 9 + 10 BC ∥ 𝑦 − 𝑎𝑥𝑖𝑠 ✓𝑦 − value/waarde (2) 𝐶(4; 19) 3.3 𝑥1 + 𝑥2 𝑦1 + 𝑦2 ✓Subst in correct formula/ 𝐾( ; ) 2 2 Instel in korrekte formule 4 − 2 19 + 1 ✓Ans/ Antw 𝐾( ; ) 2 2 𝐾( 1 ; 10) (2) 3.4 𝑦2 − 𝑦1 19 − 1 18 ✓𝑚 𝑚= = = =3 𝑥2 − 𝑥1 4+2 6 ✓Subst A or C point into 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) correct formula/ Stel A of C 𝑦 − 1 = 3(𝑥 + 2) in korrekte formule 𝑦 = 3𝑥 + 6 + 1 ✓Ans/Antw 𝑦 = 3𝑥 + 7 (3) 3.5 𝑚𝐴𝐶 = 3 ✓ tan  = m 𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑖𝑛𝑐𝑙𝑖𝑛𝑎𝑡𝑖𝑜𝑛 = tan−1(3) ✓ of inclination/ = 71,57° Inklinasie ∴ 𝜃 = 90° − 71,57° ✓ Comple ’s/ Kompl  = 18,43° ✓Ants/Antw OR OR ✓correct trig ratio/ In the right-angled triangle: korrekte trig verhouding 6 ✓✓ correct values /korrekte tan 𝜃 = 18 waardes 𝜃 = 18.43° ✓Ans/ Antw (4) 3.6 1 ✓correct formula/ korrekte Area ABC = . BC . h formule 2 ✓base/basis = 10 1 = (10)(6) ✓ height/hoogte = 6 2 ✓ ans/ antw = 30 sq units (4) [17]

QUESTION/ VRAAG4 Q4 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ BESKRYWERS MARK/ PUNT 4.1 OC  tangent [r  tangent] ✓ both/ beide S’s CE  tangent [r  tangent] (2)  O, C and E straight line [adjacent s = 180°] ✓R 4.2 −2 − 0 ✓ Substitute/ Vervang C 𝑚𝑂𝐶 = = −2 1−0 ✓Ans/ Antw (2) 4.3 𝑚𝐶𝐷 = −2 𝑦2 − 𝑦1 ✓ 𝑚𝑡𝑎𝑛𝑔𝑒𝑛𝑡/𝑟𝑎𝑎𝑘𝑙𝑦𝑛 = −2 𝑥2 − 𝑥1 −6 + 2 ✓ Substitute( any point) = −2 (2) 𝑡−1 correct equation/ Vervang −2(𝑡 − 1) = −4 enige punt 𝑡−1=2 𝑡=3 4.4 1 𝑚𝑡𝑎𝑛𝑔𝑒𝑛𝑡 = 2 ✓ 𝑚𝑡𝑎𝑛𝑔𝑒𝑛𝑡/𝑟𝑎𝑎𝑘𝑙𝑦𝑛 𝐴𝐶: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) 1 ✓ Substitute( any point) 𝑦 + 2 = (𝑥 − 1) 2 correct equation/ Vervang 1 1 enige punt 𝑦 = 𝑥− −2 2 2 1 5 ✓ Answer/ Antw 𝑦= 𝑥− 2 2 (3) 4.5 D is the midpoint of circle 1+𝑥 𝑥 − 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒: =3 2 ✓ 𝑥 value/ waarde 1+𝑥 =6 𝑥=5 ✓ 𝑦 value/ waarde (2) 𝑦−2 𝑦 − 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒: − −6 2 𝑦 − 2 = −12 𝑦 = −10 𝐸(5; −10) 4.6 𝐴𝐶̂ 𝐷 = 90° [𝑟𝑎𝑑𝑖𝑢𝑠 ⊥ 𝑡𝑎𝑛𝑔𝑒𝑛𝑡] ✓AE diameter/ middellyn ∴ 𝐴𝐸 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟  ACE ✓AE = 10 𝐴𝐸 = √(5 − 5)2 + (0 + 10)2 ✓Radius = 5 𝐴𝐸 = √0 + 100 𝐴𝐸 = 10 ✓Midpnt 1 ∴ 𝑟𝑎𝑑𝑖𝑢𝑠 = (10) = 5 2 Equation of Circle/Vegel van 5 + 5 −10 𝑀𝑖𝑑𝑝𝑛𝑡𝐴𝐸 ( ; ) sirkel 2 2 ✓Point/ Punt ∴ 𝐶𝑒𝑛𝑡𝑟𝑒 𝑜𝑓𝑐𝑖𝑟𝑐𝑙𝑒 (5;-5) ✓𝑟 2 (6) Equation of circle ACE: (𝑥 − 5)2 + (𝑦 + 5)2 = 25 4.7 𝐶𝑖𝑟𝑐𝑙𝑒 𝑐𝑒𝑛𝑡𝑟𝑒 𝑂: 𝑥 2 + 𝑦 2 = 5 ✓ 𝑟2 = 5 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 2√5 ✓Diameter / Middellyn (4) √20 < 𝑟 < √20 + 2√5 ✓✓Endpoints/ Eindpunte 2√5 < 𝑟 < 4√5 [21]

QUESTION/ VRAAG 5 Q5 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ BESKRYWERS MARK/ PUNT (1)2 = 𝑝2 + (𝑜𝑝𝑝)2 (1)2 − (𝑝)2 = 𝑥 2 ∴ 𝑥 = √1 − 𝑝2 5.1.1 sin 214° ✓−sin 34° = sin(180° + 34°) = −sin 34° ✓Ans/ Antw (2) = −𝑝 5.1.2 cos 34°.cos (-22°) + sin 34°.sin 338° ✓✓ reduction/ reduksie = cos 34° . 𝑐𝑜𝑠22° + sin 34°(−𝑠𝑖𝑛22°) = cos 34° . cos 22 ° − sin 34° . sin 22° ✓ compound angles/ = cos (34° + 22°) saamgestelde hoeke = cos 56° (4) ∴= 𝑝 ✓ √1 − 𝑝2 5.1.3 cos 68° ✓Double angle/ dubbele = cos 2(34°) hoek = 2 cos2 34° − 1 ✓Ans/ Antw (2) 2 = 2 (√1 − 𝑝2 ) − 1 = 2(1 − 𝑝2 ) − 1 = 1 − 𝑝2 5.2.1 cos(90ᵒ − 2𝜃) . sin(𝜃) . 𝑠𝑖𝑛2 (180ᵒ + 𝜃) cos(720° + 𝜃). ✓ 𝑠𝑖𝑛2𝜃 ✓ (−𝑠𝑖𝑛 𝜃)2 sin 2𝜃 . (sin 𝜃) ✓sin2 𝜃 = ✓ cos 𝜃 (−𝑠𝑖𝑛𝜃)2 . (𝑐𝑜𝑠𝜃) 2 sinθ 𝑐𝑜𝑠𝜃. 𝑠𝑖𝑛 𝜃 ✓2 sinθ 𝑐𝑜𝑠𝜃 = 𝑠𝑖𝑛2 𝜃. 𝑐𝑜𝑠𝜃. ✓2 (6) =2 5.2.2 1 1 𝑐𝑜𝑠 2 𝑥 − ✓ 𝑠𝑖𝑛 2𝑥 𝑡𝑎𝑛2 2𝑥 2 𝑠𝑖𝑛2 𝑥 1 𝑐𝑜𝑠 2 2𝑥 = − 𝑠𝑖𝑛2 2𝑥 𝑠𝑖𝑛2 2𝑥 1 − 𝑐𝑜𝑠 2 2𝑥 1 − 𝑐𝑜𝑠 2 2𝑥 ✓ = 𝑠𝑖𝑛2 2𝑥 𝑠𝑖𝑛2 2𝑥 𝑠𝑖𝑛2 2𝑥 ✓ sin2 2𝑥 = =1 ✓Ans/ Antw (4) 𝑠𝑖𝑛2 2𝑥 [18]

QUESTION/ VRAAG 6 Q6 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ MARK/ BESKRYWERS PUNT 6.1 ✓ shape / vorm y ✓ endpoints/ 1 g eindpunte (3) ✓ intercepts with f axes/ snypunte met x asse -90 -60 -30 0 30 60 90 120 150 180 (180° ; -0,5) (-90° ; -0,87) -1 6.2 ✓ equate the − sin 2𝑥 = cos(𝑥 − 60°) equations / − sin 2𝑥 = sin(90° − (𝑥 − 60°)) vergelyking − sin 2𝑥 = sin(150° − 𝑥) ✓ co-functions/ ko- funksies sin(180° + 2𝑥) = sin(150° − 𝑥) 180° + 2𝑥 = 150° − 𝑥 + 𝑛(360°), 𝑛𝜖 𝑍 sin(180° + 2𝑥)✓ 3𝑥 = −30 + 𝑛(360), 𝑛 ∈ 𝑍 ✓ ✓ Ans/ Antw (5) 𝑥 = −10° + 𝑛(120°) 6.4 ✓ Inequality/ (3) −10° ≤ 𝑥 ≤ 110° Ongelykheid ✓✓Ans/ Antw 6.4 𝑦 = −sin(2𝑥 + 30°) ✓ ✓ Ans/ Antw (2) 6.5 𝑔 must shift 30° right ✓ 30° ✓ right/regs (2) [15]

QUESTION/VRAAG 7 Q7 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ BESKRYWERS MARK/ PUNT 7.1 𝐷𝐶 cosθ = 𝑝 p. cosθ = 𝐷𝐶 ✓ DC 𝐵𝐷 𝐷𝐶 = ✓ Subst in Sine Rule/ sin(90° − 𝜃) sin 2𝜃 Sinus reël 𝐵𝐷 𝐷𝐶 = ✓Simplification/ cos 𝜃 2sin 𝜃. cos 𝜃 Vereenvoudig 𝐵𝐷 𝐷𝐶.cos 𝜃 = but𝑝 𝑐𝑜𝑠𝜃 = 𝐷𝐶 1 2sin 𝜃.cos 𝜃 ✓Subst/ vervang DC 𝐵𝐷 𝑝𝑐𝑜𝑠 𝜃. cos 𝜃 = 1 2sin 𝜃. cos 𝜃 ✓ Simplification/ 𝑝𝑐𝑜𝑠𝜃 Vereenvoudig (5) 𝐵𝐷 = 2𝑠𝑖𝑛𝜃 7.2 𝐴𝐶 ✓Substitution /Instel sin 30° = 𝑝 3 sin 30° = 𝐴𝐶 ✓Ans/Antw ∴ 𝐴𝐶 = 1,5 𝑚 (2) 7.3 3 cos 30° 3 1 √3 𝐵𝐷 = = ( )= 2 sin 30° 2 √3 2 ✓BD 2 2 2 𝐴𝐵 = 𝐴𝐷 + 𝐵𝐷 − 2𝐴𝐷. 𝐵𝐷𝑐𝑜𝑠 𝐴𝐷𝐵 2 ✓Cos rule/reël √3 √3 ✓Subst / vervang = 32 + ( ) − 2(3) ( ) cos 70° 2 2 3 ✓AB2 = 9 + − 3√3 cos 70 ° 4 = 7.97 ✓AB (5) 𝐴𝐵 = 2.82 metres [12]

EUCLIDEAN GEOMETRY/ EUKLIDIESE MEETKUNDE ✓S – Statement/ Bewering ✓R – Reason/ Rede ✓S/R – Statement + Reason/ Bewering + Rede QUESTION/ VRAAG 8 8.1 Q8 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ MARK/ BESKRYWERS PUNT 8.1.1 BÂE = 90°  in semi circle/ in halwesirkel ✓ S ✓R (2) 8.1.2 Ê1 = 80 ° opp angles cyclic quad/ teenoorst e koordevierhoek ✓ S ✓R (2) 8.1.3 D1 =45° ext of ∆ FED/ buitevan ∆ FED ✓ S ✓R (2) 8.2 B̂1 = 35° Interior  of ∆ ✓ S✓R ✓S 𝐹̂ = 35° given ✓R (4) ∴ 𝐴𝐵 ∥ 𝐶𝐹 Alternate angles = [10]

QUESTION/ VRAAG 9 Q9 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ MARK/ BESKRYWERS PUNT 9.1 Consider other proofs as well/ Oorweeg ander G 1 H bewyse ook Constr./Konstr.: Measure/Meet AG = DE on/op AB and/en AH = DF ✓constr/ konstr. on/op AC . Draw/Trek GH Proof/ Bewys: 𝐴̂ = D ̂ …given/gegee AG = DE …Constr./ konstr AH = DF …Constr./ konstr ∴ ∆𝐺𝐴𝐻 ≡ ∆𝐸𝐷𝐹 (𝑠;  ; 𝑆) ✓S✓R ̂1 = 𝐸̂ ∴ 𝐺 But / maar ̂𝐵 = 𝐸̂ …given/gegee ̂1 =𝐵̂ ̂1 =𝐵̂ ✓𝐺 ∴ 𝐺 ∴ GHll BC … corresp∠𝑠 = / ooreenk.∠𝑒 = ✓S&R AG AH ∴  ✓S AB AC DE DF ✓S&R ∴  … AG = DE ; AH = DF AB AC (7)

9.2 C F B E D A 9.2.1 ̂1 = 𝐸 1) 𝐵 ̂1 …tan-chord thm/ raakl- koordstelling ✓S&R ̂ ̂1 …equal chords; equal ’s /gelyke koorde gelykee 𝐸2 = 𝐸 (3) ✓S ̂2 = 𝐵 ∴ 𝐸 ̂1 9.2.2 In ∆𝐵𝐷𝐴 and / en ∆𝐸𝐹𝐵 : ̂ A = 𝐹̂ …ext of cyclic quad/ buitevan koordevierhoek BD ✓✓S&R ̂2 = 𝐵 ∴ 𝐸 ̂1 Proven ✓S ∴ ∆𝐵𝐷𝐴lll∆𝐸𝐹𝐵 (;;) (4) ✓S&R OR In ∆𝐵𝐷𝐴 and / en ∆𝐸𝐹𝐵 : ̂ A = 𝐹̂ 1) BD … ext of cyclic quad/ buitevan koordevierhoek ✓S&R ̂1 = 𝐸 2) 𝐵 ̂1 … tan-chord thm/ raakl- koordstelling ̂ ̂1 𝐸2 = 𝐸 … equal chords; equal ’s /gelyke koorde gelykee ✓S&R ̂2 = 𝐵 ∴ 𝐸 ̂1 ✓S &R 𝐴̂ = 𝐵 ̂3 … sum of ’s in ∆/ e van ∆ ∴ ∆𝐵𝐷𝐴lll∆𝐸𝐹𝐵 ✓S &R 9.2.2 𝐵𝐷 𝐷𝐴 ✓S = 𝐸𝐹 𝐹𝐵 ∴ 𝐵𝐷. 𝐹𝐵 = 𝐸𝐹. 𝐷𝐴 ✓S&R (2) ∴ BD² = DA.EF … BD = FB [16]

QUESTION/ VRAAG 10 P S D F G E A B C Q10 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ MARK/ BESKRYWERS PUNT 10.1. alt ∠s / verwisselende ∠e; YT ∥ RQ ✓R (1) 10.2.1 BP 𝐵𝑆  … Prop. Thm; EF ll PS/ Ewer. Stelling; EF ll PS BE 𝐵𝐹 ✓S✓R (2) 𝐵𝑃. 𝐵𝐹 𝐵𝐸 = 𝐵𝑆 10.2.2 In ΔBGP and/ en ΔBEG: (4) 1) Ĝ1 = P̂1 … Tan chord thm/ Raaklyn koordstelling ✓✓ S&R 2) Bˆ  Bˆ … common ∠ / gemene∠ ✓S &R ∴ ∆𝐵𝐺𝑃lll∆𝐵𝐸𝐺 (;;) ✓S &R OR In ΔBGP and/ enΔBEG 1) Ĝ1 = P̂1 … Tan chord ✓✓ S&R 2) Bˆ  Bˆ … common ∠ / gemene∠ ✓S &R 3) BGˆ P  BEˆ G … sum of ’s in ∆/ e van ∆ ∴ ∆𝐵𝐺𝑃lll∆𝐵𝐸 ✓S 𝐵𝐺 𝐵𝑃 = …∆𝐵𝐺𝑃lll∆𝐵𝐸𝐺 10.2.3 𝐵𝐸 𝐵𝐺 ✓S ∴ BG² = BP.BE 𝐵𝑃. 𝐵𝐹 ✓S 𝐵𝐺 2 = 𝐵𝑃. (3) 𝐵𝑆 ✓ Subst 2 𝐵𝑃 . 𝐵𝐹 𝐵𝐺 2 = 𝐵𝑆 𝐵𝐺 2 𝐵𝐹 ∴ = 𝐵𝑃2 𝐵𝑆 [10]

QUESTION/ VRAAG 11 20 cm Q11 SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ MARK/ BESKRYWERS PUNT 11.1.1 FC 4 ✓✓S&R  … EF llAD 20 5 ✓answer/antw (3) ∴ FC=16 11.1.2 36 4 ✓DC = 36  … DE llAB DB 5 ✓✓S&R (4) ∴ DB= 45 ✓answer/antw 1 11.2 𝐴𝑟𝑒𝑎 ∆ 𝐸𝐶𝐹 2 .4𝑘. 8. sin 𝐶 ✓½. 4𝑘. 8. 𝑠𝑖𝑛𝐶 = 𝐴𝑟𝑒𝑎 ∆𝐴𝐵𝐶 1 . 9𝑘. 81. sin 𝐶 ✓½. 9𝑘. 40.5. 𝑠𝑖𝑛𝐶 2 𝐴𝑟𝑒𝑎 ∆ 𝐸𝐶𝐹 32 = ✓ ✓ Answer 𝐴𝑟𝑒𝑎 ∆𝐴𝐵𝐶 81 (4) TOTAL / TOTAAL: [150] [11]