Mathematics Grade 12 Paper 2 NSC June 2015 Memo

Downloaded from hlayiso.com SENIOR CERTIFICATE EXAMINATION/ SENIORSERTIFIKAAT-EKSAMEN MATHEMATICS P2/WISKUNDE V2 2015 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 18 pages./ Hierdie memorandum bestaan uit 18 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 2 DBE/2015 NSCNSS – Memorandum NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent accuracy applies in ALL aspects of the marking memorandum. LET WEL: • As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na. • Volgehoue akkuraatheid word in ALLE aspekte van die memorandum toegepas. QUESTION/VRAAG 1 3 4 4 5 23 29 32 36 40 47 56 66 68 76 82 1.1.1 571 571 Mean/gemiddelde =  15 15 = 38,07 years/jaar  answer/antw (2) 1.1.2 Median/mediaan = 36 years/jaar  answer/antw (1) 1.1.3 IQR/IKV = 66 – 5  Q 3 = 66 = 61 years/jaar  Q1= 5  answer/antw (3) 1.1.4 Standard deviation/standaardafwyking = 26,388… = 26,39 years/jaar  answer/antw (2) 1.2  max & min  median/ mediaan 3 5 36 82 66 70  quartiles/ 10 20 30 40 50 60 80 kwartiele (3) 1.3 The data is skewed to the right/Die data is skeef na regs OR/OF positively skewed/positief skeef  answer/antw (1) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 3 DBE/2015 NSCNSS – Memorandum QUESTION/VRAAG 2 Number of Saturdays attended 12 11 10 10 9 9 7 6 5 4 12 11 6 Mark (as a %) 96 91 78 83 75 62 70 68 56 34 88 90 59 2.1 a = 22,26252159… 22,26 b = 5,898100173... 5,90 ∴ ŷ = 5,90x + 22,26 equation/vgl (3) 2.2 r = 0,92 (0,9205276443...)  answer/antw (2) 2.3 There is a very strong relationship between the variables/  very strong/ Daar is 'n baie sterk verband tussen die veranderlikes. baie sterk (1) 2.4 ∴ ŷ ≈ 69,447 = 69,45 ≈ 69% (accept 70%)  answer/antw OR/OF (2) ∴ ŷ ≈ 5,90(8) + 22,26  substitution/ substitusie ≈ 69,46 %  answer/antw ≈ 69% (accept 70%) (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 4 DBE/2015 NSCNSS – Memorandum QUESTION/VRAAG 3 y Q M(0 ; 4) P R(4 ; 2) N S(2 ; 1) 1 2 x O T 3.1 2 −1 mRS =  correct subst/ 4−2 1 korrekte subst =  answer/antw 2 (2) 3.2 1 PQ is y = x+6 2  1 ∴PQ | | RS  mPQ = mRS =  S  2 But/maar PS | | QR ∴PQRS = parallelogram (opp sides of quad are | | / teenoorst S/R sye v vh | |) ∴PQ2 = RS2 = (4 – 2)2 + (2 – 1)2  subst of/subst v = 2 2 + 12 R(4 ; 2) and/en S(2 ; 1) ∴PQ = RS = 5 = 2,24 (opp sides of ||m /teenoorst sye v ||m) answer/antw (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 5 DBE/2015 NSCNSS – Memorandum 3.3 4−2 mQR = 0−4 1 =− m QR 2 1 m PT mPT = mQR = − (PS | | QR) 2 Equation of/Vgl van PT: 1 1 y − y1 = − ( x − x1 ) y= − x+c 2 2 1 1 y − 1 = − ( x − 2) 1 = – (2) + c subst of/subst v 2 OR/OF 2 m and/en S(2 ; 1) 1 y −1 = − x +1 2= c 2 1 1 y =− x+2 y= − x+2 2 2  equation/vgl (4) 3.4 N(0 ; 2) coordinates (1) 3.5 1 tan T2 = mPT = − 1 2  tan T2 = − 2 T2 = 153,4° Equation of/Vgl van NR: y = 2  T2 = 153,4° y = 2 ∴ RN̂S = NT̂O (alt ∠s; NR || OT) S RN̂S = NT̂O = 180° – 153,4°  RN̂S = 26,6° = 26,6° (5) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 6 DBE/2015 NSCNSS – Memorandum QUESTION/VRAAG 4 y A C x O D B(3 ; –4)  1  E− 3; − 8   2  4.1 A(–3 ; 4) x = –3 y = 4 (2) 2 2 2 2 2 2 4.2 r = (3) + (–4) OR r = (–3) + (4)  substitution/ substitusie r2 = 25 r2 = 25 ∴Equation of the circle through A, B and C/ Vgl vd sirkel deur A, B en C: x2 + y2 = 25 answer/antw (3) 4.3 r=5 r = 5 ∴ AB = 10 units/eenhede  answer/antw (2) 4.4 AB ⊥ ED OR AB̂D = 90° (radius ⊥ tangent/raaklyn) S/R BD 2 = AD 2 − AB 2 (Theorem of Pythagoras/st v Pythagoras) BD 2 = ( 125 ) 2 − (10) 2  subst into/in 2 BD = 25 Pyth th/stelling BD = 5 units/eenhede  answer/antw (3) 4.5 1 area of/oppervlakte van ∆ABD = base/basis × ⊥ height/hoogte 2  formula/formule 1  substitution/ = (5)(10) substitusie 2 = 25 square units/vk eenhede  answer/antw (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 7 DBE/2015 NSCNSS – Memorandum 4.6 AE is a diameter/middellyn (line subtending a right angle/  S/R lyn onderspan ‘n regte hoek) AE= 4 – (–8,5) = 12,5 units/eenhede  AE = 12,5 1 25 1 25 ∴r = 6 or/of r = 6 or/of 4 4 4 4 Centre of new circle/middelpunt v nuwe sirkel: 4 + −8,5 = (−3 ; ) x = –3 2 1 9 1 = (−3 ; − 2 ) or (−3 ; − ) or (−3 ; − 2,25) y = − 2 4 4 4 ∴Equation of new circle/Vgl v nuwe sirkel: 9 625 (x + 3)2 + (y + )2 = = 39,06  equation/vgl 4 16 (6) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 8 DBE/2015 NSCNSS – Memorandum QUESTION/VRAAG 5 5.1 1 sketch/skets: cos β = − and/en 180° < β < 360° 5  correct quad/ korrekte kwadr  x = –1 (−1) 2 + y 2 = ( 5 ) 2 β subst into Pyth/ 2 1+ y = 5 subst in Pyth y2 = 4 5 y = −2  value of/waarde van y (–1 ;y) 2 ∴sin β = −  value of/waarde 5 van sin β (5) 5.2 (− tan x).(− sin(90° − x) – tan x 4 sin x  –sin(90° – x) (− tan x).(− cos x)  sin x – cos x 4 sin x sin x (− ).(− cos x) sin x cos x  = cos x 4 sin x 1  answer/antw = 4 (6) 5.3.1 sin A p  answer/antw tan A = = (1) cos A q 5.3.2 p 4 − q 4 = ( p 2 + q 2 )( p 2 − q 2 )  factors/faktore = (sin 2 A + cos 2 A)(sin 2 A − cos 2 A) = (1)(sin 2 A − cos 2 A) identity/identiteit = −1(cos 2 A − sin 2 A)  –1 as CF/GF = −cos2A  answer/antw (4) 5.4.1  writing as single cos 2 θ − cos 2θ LHS/LK = term/skryf as sin θ . cos θ enkelterm cos 2 θ − (cos 2 θ − sin 2 θ )  expansion/ = uitbreiding sin θ . cos θ cos 2 θ − cos 2 θ + sin 2 θ  simplify/vereenv = sin θ . cos θ sin 2 θ  simplify/vereenv = sin θ . cos θ sin θ  simplify/vereenv = = tan θ = RHS/RK cos θ (5) OR Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 9 DBE/2015 NSCNSS – Memorandum cos 2 θ − cos 2θ  writing as single LHS/LK = term/skryf as sin θ . cos θ enkelterm cos 2 θ − ( 2 cos 2 θ − 1 )  expansion/ = sin θ . cos θ uitbreiding 1 − cos 2 θ =  simplify/vereenv sin θ . cos θ sin 2 θ = sin θ . cos θ  identity/identiteit sin θ = cos θ  simplify/vereenv = tan θ = RHS/RK (5) 5.4.2 Undefined when/Ongedefinieerd as: cos θ = 0, sin θ = 0 ∴ θ = 90°  answer/antw (2) 5.5 2(2sin x. cos x) + 3 sin x = 0  expansion/ 4sin x. cos x + 3 sin x = 0 uitbreiding sin x (4cos x + 3) = 0  factorise/ 3 faktoriseer sin x = 0 or/of cos x = –  both equations/ 4 beide vgls x = 0° + k.360° or 180° + k.360°; k∈Z  x = 0° + k.360° OR/OF or 180° + k.360° x = k.180° ; k∈Z OR/OF x = k.180° or/of x = 138,59° + k.360° or/of 221,41° + k.360° ; k∈Z 138,59+ k.360° or OR/OF 221,41°+ k.360 x = ±138,59° + k.360° ; k∈Z OR/OF ±138,59°+ k.360  k∈Z (6) [29] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 10 DBE/2015 NSCNSS – Memorandum QUESTION/VRAAG 6 1 f -90 ° -60 ° -30 ° 0° 30 ° 60 ° 90° g –1 6.1 Period of/Periode van f = 120°  120° (1) 6.2 b= 3 b = 3 (1) 6.3 x = –45° or/of x = –22,5° or/of x = 67,5° x = –45° x = –22,5° x = 67,5° (3) 6.4 x ∈ (–45° ; –22,5°) ∪ (67,5° ; 90°]  critical values  notation  critical values  notation OR/OF (4) − 45° < x < −22,5° or/of 67,5° < x ≤ 90°  kritieke waardes  notasie  kritieke waardes  notasie (4) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 11 DBE/2015 NSCNSS – Memorandum QUESTION/VRAAG 7 3x x R 2 S P 3x x Q 7.1 QR 2 = PQ 2 + RP 2 − 2.PQ.RP.cos P̂  correct subst into ( 3 x) 2 = x 2 + x 2 − 2.x.x.cos P̂ cosine rule/korrek x 2 + x 2 − ( 3 x) 2 subst in cos-reël cos P̂ = 2 x.x  cos P̂ as subj/ −x 2 onderw cos P̂ = 2x 2 1 cos P̂ = − simplify/vereenv 2 P̂ = 120°  answer/antw (4) 7.2 PR̂Q = PQ̂R = 30° (∠s opp equal sides/∠e teenoor gelyke sye) S QR̂S = 150° (∠s on a str line/∠e op reguitlyn) S 1 Area of/Opp van ∆ QRS = (QR)(RS)(sin QR̂S) 2 1 3  correct subst into = ( 3 x)( x)(sin 150°) 2 2 area rule/korrek 3 3 2 1 subst in opp-reël =( x )( ) simplify/vereenv 4 2 3 3 2 2 = x = 0,65 x  answer/antw 8 (5) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 12 DBE/2015 NSCNSS – Memorandum QUESTION/VRAAG 8 A 1 2 1 T 65° 2 P C 25° 3 2 1 1 2 B D 8.1.1 P̂2 = 65° (∠s opp equal sides/∠e teenoor gelyke sye) S  R (2) 8.1.2 D̂ = 40° (ext∠ of ∆CDP/buite∠ v ∆CDP )  S R OR/OF (∠s on a str line; sum of ∠s in ∆/ ∠e op regt lyn; som v ∠e in ∆) (2) 8.1.3 Â 1 = 40° (ext∠ of ∆CAT/buite∠ v ∆ CAT ) S  R OR/OF (∠s on a str line; sum of ∠s in ∆/ (2) ∠e op regt lyn; som v ∠e in ∆) 8.2 Â 1 = D̂ = 40° S ∴CA is a tangent to the circle (∠ between line and chord)/ R CA is 'n raaklyn aan die sirkel (∠ tussen lyn en koord) (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 13 DBE/2015 NSCNSS – Memorandum QUESTION/VRAAG 9 D E A 2 2 1 1 1 O 2 3 1 B x C 9.1.1 ext∠ of cyclic quad/buite ∠ van koordevh R (1) 9.1.2 ∠ at centre = 2 ×∠ at circumference / midpts∠ = 2 × omtreks∠ R (1)  9.2.1 CD̂A = E1 = x (corresp∠s/ooreenk ∠e; EB || DC)  S R ∴ CD̂A = Ĉ = x ∴ AC = AD (sides opp equal ∠s/sye teenoor gelyke ∠e)  S (justification) (4) 9.2.2 Â = 180° − 2 x (sum of ∠s in ∆/som van ∠e in ∆) S Ô1 = 2 x OR Â + Ô1 = 180° − 2 x + 2 x = 180°  linking the 2 ∠s ∴ ABOD is a cyclic quad/koordevh (opp∠s quad supp/ R teenoorst ∠e van vh suppl) (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 14 DBE/2015 NSCNSS – Memorandum QUESTION/VRAAG 10 10.1 then the line is parallel to the third side/is die lyn ewewydig aan S die derde sy. (1) B F D C A 12 E 8 10.2.1 AE 12 3 S = = AC 20 5 AD 3 = AF 5 AE AD ∴ = S AC AF ∴ DE || FC (line divides two sides of ∆ in prop/ R lyn verdeel twee sye v ∆ in dieselfde verh) (3) 10.2.2 BF 8 = (prop theorem/eweredigh st; BC || FE)  S/R BA 20 8 ∴ BF = (14)  substitute 14/ 20 stel 14 in 28 3 ∴ BF = OR/OF FB = 5 OR/OF FB = 5,6  answer/antw 5 5 (3) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 15 DBE/2015 NSCNSS – Memorandum QUESTION/VRAAG 11 11.1 Draw diameter AD and join DC.  construction/ Trek middellyn AD en verbind DC. konstruksie D C B O A P Proof/Bewys: BÂP + BÂD = 90° (tangent/raaklyn ⊥ radius)  S R DĈB + AĈB = 90° (∠ in semi circle/halfsirkel)  S R but BÂD = DĈB (∠s in same segment/∠e in dies segm)  S/R ∴ BÂP = AĈB (6) OR/OF  construction/ Draw diameter AD and join DB. konstruksie Trek middellyn AD en verbind DB. D C B O A P Proof/Bewys:  S R PÂB + BÂD = 90° (tangent/raaklyn ⊥ radius)  S R DB̂A = 90° (∠ in semi circle/halfsirkel) BÂD + AD̂B = 90° (sum of ∠s in ∆/som van ∠e in ∆) AD̂B = AĈB (∠s in same segment/∠e in dies segm)  S/R ∴ BÂP = AĈB (6) Copyright reserved Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 16 DBE/2015 NSCNSS – Memorandum OR/OF Draw radii OA and OB.  construction/ Trek radii OA en OB. konstruksie C B O A P Proof/Bewys: OÂB + BÂP = 90° (tangent/raaklyn ⊥ radius) ∴ BÂP = 90° − OÂB  S R OÂB = OB̂A (∠s opp equal sides/∠e to gelyke sye)  S AÔB = 180° − 2OÂB (sum of ∠s in ∆/som van ∠e in ∆)  S/R ∴ AĈB = 90° − OÂB (∠ at centre = 2 ×∠ at circumference/  S midpts∠ = 2 × omtreks∠) ∴ BÂP = AĈB (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 17 DBE/2015 NSCNSS – Memorandum B D 1 2 1 C 2 1 2 E 1 2 A P 11.2.1 DĈA = 2 x (EC bisector) P̂ = x (∠ at centre = 2 ×∠ at circumference/ SR midpts∠ = 2 × omtreks∠) Â 1 = P̂ = x (tangent-chord theorem/rkl-kd st) SR In ∆BAD and ∆BCE: B̂ = B̂ (common/gemeen) S Â1 = Ĉ1 (proven above)  S(with justification) ∴∆BAD | | | ∆BCE (∠∠∠) R (7) OR/OF DĈA = 2 x (EC bisector) P̂ = x (∠ at centre = 2 ×∠ at circumference/ SR midpts∠ = 2 × omtreks∠) Â 1 = P̂ = x (tangent-chord theorem/rkl-kd st) SR In ∆BAD and ∆BCE: B̂ = B̂ (common/gemeen) S Â1 = Ĉ1 (proven above)  S(with justification) D̂1 = Ê 1 S ∴∆BAD | | | ∆BCE (7) Copyright reserved Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde V2 18 DBE/2015 NSCNSS – Memorandum 11.2.2(a) BÂC = 90° (tangent/raakl ⊥ radius) R ∴BC2 = 82+ 62 = 100 (Pythagoras theorem/stelling)  substitution into BC = 10 Pyth theorem AC = DC = 6 (radii)  BC = 10 ∴ BD = 10 – 6 = 4 units/eenhede  DC = 6  BD = 4 (5) 11.2.2(b) BA BD S = (∆BAD | | | ∆BCE) BC BE 8 4  substitution/ ∴ = 10 BE substitusie ∴ BE = 5 units/eenhede  BE = 5 (3) 11.2.2(c) AE = 3 In ∆ACE: 3  correct trig ratio/ tan x = korrekte trigvh 6  correct trig eq/ ∴x = 26,57° korrekte trigvgl  answer/antw OR/OF (3) 8  correct trig ratio/ sin 2x = korrekte trigvh 10  correct trig eq/ korrekte trigvgl ∴ 2x = 53,1301... (2x < 90°)  answer/antw ∴ x = 26,57° (3) [24] TOTAL/TOTAAL: 150 Copyright reserved