Mathematics Grade 12 Paper 2 NSC June 2019 Memo

Downloaded from hlayiso.com SENIOR CERTIFICATE EXAMINATIONS/ SENIORSERTIFIKAAT-EKSAMEN NATIONAL SENIOR CERTIFICATE EXAMINATIONS/ NASIONALE SENIORSERTIFIKAAT-EKSAMEN MATHEMATICS P2/ WISKUNDE V2 MARKING GUIDELINES/NASIENRIGLYNE 2019 MARKS: 150 PUNTE: 150 These marking guidelines consist of 20 pages. Hierdie nasienriglyne bestaan uit 20 bladsye.. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 2 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 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. Stop marking at the second calculation error. • Assuming answers/values in order to solve a problem is NOT acceptable. NOTA: • As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na. • As 'n kandidaat 'n antwoord van 'n vraag doodtrek en nie oordoen nie, sien die doodgetrekte poging na. • Volgehoue akkuraatheid word in ALLE aspekte van die nasienriglyne toegepas. Hou op nasien by die tweede berekeningsfout. • Om antwoorde/waardes te aanvaar om 'n probleem op te los, word NIE toegelaat NIE. GEOMETRY • MEETKUNDE A mark for a correct statement (A statement mark is independent of a reason) S 'n Punt vir 'n korrekte bewering ('n Punt vir 'n bewering is onafhanklik van die rede) A mark for the correct reason (A reason mark may only be awarded if the statement is correct) R 'n Punt vir 'n korrekte rede ('n Punt word slegs vir die rede toegeken as die bewering korrek is) Award a mark if statement AND reason are both correct S/R Ken 'n punt toe as die bewering EN rede beide korrek is Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 3 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 1.1 45 children  answer (1) 1.2 x= ∑ fx = (4 × 2) + (8 × 10) + (12 × 9) + (16 × 7) + ( 20 × 8) + (24 × 7) + (28 × 2) n 45 692  692 x= OR x = 15,38 minutes Answer only: full marks 45  answer (2) 1.3 Time taken (t) Number of Cumulative (in minutes) children frequency 2<t≤6 2 2  first 4 cum freq 6 < t ≤ 10 10 12 correct  last 3 cum freq 10 < t ≤ 14 9 21 correct 14 < t ≤ 18 7 28 18 < t ≤ 22 8 36 22 < t ≤ 26 7 43 26 < t ≤ 30 2 45 (2) 1.4 CUMULATIVE FREQUENCY GRAPH (OGIVE) 50 40  plotting cum Cumulative Frequency freq at upper 30 limits correctly (all points)  shape (smooth) 20  grounding (2;0) 10 0 0 5 10 15 20 25 30 Time in minutes (3) 1.5 On graph at the y-value of 22,5 or 23  graph Median = ± 15 minutes. Answer only: full marks  answer (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 4 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1 a = 12,44  value of a b = 0,98 Answer only: full marks  value of b y = 12,44 + 0 ,98 x  equation (3) 2.2.1 15 Percentage = × 100 50 = 30%  answer (1) 2.2.2 yˆ = 12,44 + 0 ,98 x yˆ = 12,4 4 + 0 ,98(30)  substitution of 30 yˆ = 41,84 = 42 Answer only: full marks  answer as integer OR (2) ŷ = 41,87 (if using calculator)  value of y ŷ = 42  answer as integer (2) OR 21 yˆ =   answer 50 (2) 2.3.1 standard deviation =13,88   answer (2) 2.3.2 x = 50,67 − 45,67  50,67 − 45,67 = 5% Answer only: full marks  answer (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 5 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 y K B 1 x E(–2; 0) 0 A C(2 ; –3) D(–2 ; –5) 3.1.1 Midpoint of EC:  − 2 + 2 0 + (−3)   −3 = ;  = 0 ;   x value  y value  2 2   2  (2) 3.1.2 − 3 − (−5) − 5 − (−3) mDC = OR  substitution 2 − (−2) −2−2 2 1 Answer only: full marks = =  answer 4 2 (2) 3.1.3 1 1 mAB = [AB || DC]  mAB = 2 2 1 1 y= x+c y − y1 = ( x − x1 ) 2 2 1 1 0 = (−2) + c OR y − 0 = ( x − (−2))  substitution of ( − 2 ;0) 2 2 c =1 1 equation ∴ y = x +1 2 (3) 3.1.4 1 1 tan α = mAB =  tan α = 2 2 α = 26,57°  value of α θ = 90° + 26,57° [ext ∠of Δ] =116,57°  value of θ (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 6 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 3.2 B(0 ; 1)  coordinates of B 1 − (−3) (−3) − 1 mBC = OR mBC = 0− 2 2 −0 = −2 = −2  mBC = − 2 1 mAB × mBC = × −2 2  product of gradients = –1 = −1 ∴ AB ⊥ BC (3) 3.3.1 AB̂C = 90° ∴ EC is diameter [converse : ∠ in semi circle]  3 ∴ centre of circle =  0 ; −   answer  2 (1) 2 3.3.2 (x − 0)2 +  y + 3  = r 2  substitution of centre  2 2 2 (− 2 − 0)2 +  0 + 3  = r 2 OR (2 − 0)2 +  − 3 −  − 3   = r 2  correct substitution of  2   2  E(–1 ; 0), B(0 ; 1) or   −3  2 C(2 ; –3) to calculate OR (0 − 0 ) + 1 −  2 2   = r r 2 or r   2  EC (−2 − 2) 2 + (0 − (−3))2 OR r = = 2 2 3 OR r = 1 − (− )  value of r 2 or r 2 25 5 ∴ r2 = or r = 4 2 2 2  3 25 x +y+  =  equation  2 4 (4) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 7 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 y L T(–2 ; b) M(2 ; 1) x K P 4.1 ( x − 2) 2 + ( y −1) 2 = 25  equation of the circle substitution of point T (−2 − 2) 2 + (b − 1) 2 = 25 (b − 1) 2 = 9 OF simplification b −1 = ± 3 ∴ b = 4 or b ≠ − 2 ∴  answer (4) 4.2.1 K(2 ; 1 – 5) ∴ K(2 ; –4) Answer only: full marks  x value  y value (2) 4.2.2 4 −1 3 mMT = =−  mMT −2−2 4 4 4 m PL = [radius ⊥ tangent]  mPL = 3 3 4 y= x+ c 3 4 4 = (−2) + c  substitution of mPL 3 and the point T 20 c= 3 4 20 y= x+ 3 3  equation (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 8 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne OR 4 −1 3 mMT = =−  mMT −2−2 4 4 4 m PL = [radius ⊥ tangent]  mPL = 3 3 4 y − y1 = ( x − x1) 3 4 y − 4 = ( x + 2) 3  substitution of mPL and 4 20 the point T y= x+ 3 3  equation OR (4) P(–11 ; –8) 4 − (−8)  coordinates of P mPL = − 2 − (−11) 4 = 3 4 4  mPL = y= x+ c 3 3 4 − 8 = (−11) + c 3 20  substitution of mPL and c= 3 the point P or T 4 20 y= x+ 3 3  equation (4) 4.2.3 4 20 28 28 yL = (2) + =  yL = 3 3 3 3  28  28 40 L 2 ;  and K (2 ; − 4 ) : LK = − (−4) =  length of LK  3  3 3 Coordinates of P: x+2 1 y−4 L(2 ; ) = −4 and = −6 2 2 2 ∴ x = −11 y = −8  xP  y P ∴ P( −11; − 8) ⊥ height (PH) = 2 − (−11) = 13  length of ⊥ height 1 K(2 ; –4) Area ∆ PKL = (LK)(PH) 2 1  40  P(–11 ; –8) =  (13) H(2 ; –8)  substitution into 2 3  the area formula 260 = OR 86,67 square units  answer 3 (7) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 9 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 4.2.3 OR 4 20 28 28 yL = (2) + =  yL = 3 3 3 3  28  28 40 L 2 ;  and K (2 ; − 4 ) : LK = − (−4) =  length of LK  3  3 3 Coordinates of P: L(2; ) x+2 1 y−4 = −4 and = −6 2 2 2 ∴ x = −11 y = −8  xP  y P ∴ P( −11; − 8) K(2 ; –4) PK 2 = (− 11 − 2 ) + (− 8 − (−4) ) 2 2 PK = 185 units θ P(–11 ; –8) − 8 − (−4) 4 H(2 ; –8) mPK = = − 11 − 2 13 4 tan θ = ∴ θ = 17,1027...° 13 ∴ PK̂L = 90° + 17,1027...° = 107,1°  PK̂L 1 Area ∆ PKL = (PK)(LK).sin PK̂L 2 = 1 2 ( )  40  185  sin 107,10°  substitution into  3  the area rule = 86,67 square units  answer (7) 4.3 The centres of the two circles lie on the same vertical line  correct method x = 2. and the sum of the radii = 10  sum of radii = 10 n −1 = 10 1 − n = 10 or  n =11  n = − 9 n =11 n = −9 Answer only: full marks (4) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 10 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1.1 sin 191° = − sin 11°  − sin 11° (1) 5.1.2 cos 22° = cos(2 × 11°) = 1 − 2sin 2 11°  answer (1) 5.2 cos( x − 180°) + 2 sin ( x + 45°) = − cos x + 2 (sin x cos 45° + cos x sin 45°)  − cos x  expansion   1   1  = − cos x + 2  sin x  + cos x    special angle ratios   2   2  = − cos x + sin x + cos x simplification of last 2 terms = sin x answer (5) OR cos( x − 180°) + 2 sin ( x + 45°) = − cos x + 2 (sin x cos 45° + cos x sin 45°)  − cos x  expansion   2  2  = − cos x + 2  sin x  + cos x   special angle ratios  2   2       = − cos x + sin x + cos x simplification of last 2 terms = sin x answer (5) 5.3 sin P + sin Q = sin P + cos P  sin Q = cos P 2 (sin P + cos P )2 =  7   squaring 5 49  expansion sin 2 P + 2 sin P cos P + cos 2 P = 25 49 2 sin P cos P =−1  sin 2 P + cos 2 P = 1 25  49 25  sin 2P =  −   25 25  24 answer = 25 (5) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 11 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1 cos( x − 30°) = 2 sin x cos x cos 30° + sin x sin 30° = 2 sin x  expansion 3 1 cos x + sin x = 2 sin x  special ∠ s 2 2 3 3 cos x = sin x 2 2  simplification 3 tan x =  equation in tan 3 x = 30° + k .180° ; k ∈ Z  30°  k.180°; k ∈ Z OR OR x = 30° + k .360° or x = 210° + k .360° ; k ∈ Z  30° and 210°  k .360° ; k ∈ Z (6) y g D –180° x B 0° A 180° f C 6.2.1(a) A(120° ; 0 )  answer (1) 6.2.1(b) C(− 150° ; − 1)  x value  y value (2) 6.2.2(a) x ∈ (−90° ; 30°) OR − 90° < x < 30°  endpoints  correct interval (2) 6.2.2(b) x ∈ (− 160° ; 20°) OR − 160° < x < 20° endpoints correct interval (2) 6.2.3 y = 2 2 sin x +3 Range of y = 2 sin x : y ∈ [− 2 ; 2] OR − 2 ≤ y ≤ 2 Range of y = 2 sin x + 3 : y ∈ [1 ; 5] OR 1 ≤ y ≤ 5  1 5 Range: y = 2 2 sin x + 3 : y ∈ [2 ; 32] OR 2 ≤ y ≤ 32  2  32  correct interval Answer only: full marks (5) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 12 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 E A θ D B 2θ 60° x+2 x C H F G 7.1.1 x sin θ =  trig ratio AC OR x AC =  simplification sin θ (2) 7.1.2 x +2 cos 60° =  trig ratio CE OR x+2 CE = cos 60°  making CE the x+2 subject = = 2( x + 2 ) 1 2 (2) 7.2 1  use area rule Area ∆ ACE = AC.EC. sin AĈE correctly 2 1 x   substitution of =  (2(x + 2 ))sin 2θ x 2  sin θ  (2( x + 2)) x( x + 2) × 2 sin θ cos θ sin θ =  substitution of sin θ sin 2θ = 2 x(x + 2 ) cos θ (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 13 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 7.3 EC = 2(12 + 2) = 28  EC AE 2 = AC 2 + EC 2 − 2(AC)(EC )cosAĈE  use cosine rule 2 correctly  12   12  (28) cos110° 2 =  + 28 − 2   substitution  sin 55°   sin 55°  AE = 35,77 m  answer (4) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 14 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 8.1 P N M 2 1 O L S 31° T 8.1.1(a) MÔS = 62° [∠ at centre = 2× ∠ at circumf/middelpts∠ =2omtreks∠]  SR (2) 8.1.1(b) L̂ = 31° [equal chords; equal ∠s /= koorde; = ∠e]  SR (2) 8.1.2 LN = NP and LO = OM  LO = OM 1 ∴ ON = 2 PM [ midpoint theorem/middelpuntstelling]  SR 1 ∴ ON = 2 MS [PM = MS] S (4) OR N̂1 = 90° [line from centre to midpt chord/lyn v midpt na midpt kd] SR P̂ = 90° [∠ in semi-circle/∠ in halfsirkel] L̂ is common/gemeen ∴ ∆NLO ||| ∆ PLM (∠∠∠)  S/R NL NO 1 = =  S PL PM 2 1 ∴ ON = 2 PM 1 ∴ ON = 2 MS [PM = MS] S (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 15 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 8.2 B 2y K L 3y A N M C 10x 23x 8.2.1 AN AK = [line || one side of ∆ OR prop theorem; KN ||BM/ R AM AB lyn || sy van ∆ OR eweredigheidst; KN||BM] AN 3 y 3 = = AM 5 y 5 S (2) 8.2.2 AM 10 x = [given] MC 23 x AM = 5 y =10 x ∴ y = 2 x  S LC MC = [line || one side of ∆ OR prop theorem; KN ||LM/  R KL NM lyn || sy van ∆ OR eweredigheidst; KN||BM] 23 x 23 x 23  S = = = 2y 4x 4 (3) OR AM 10 x = [given] MC 23 x AN 3 y 6 x = =  S MN 2 y 4 x LC MC = [line || one side of ∆ OR prop theorem; KN ||LM/  R KL NM lyn || sy van ∆ OR eweredigheidst; KN||BM] 23 x 23 x 23  S = = = 2y 4x 4 (3) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 16 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 K x M 1 2 N 1 2 1 A 2 3 B 9.1 B̂1 = x [ ∠' s opp = sides/∠e teenoor = sye] S M̂ 2 = 2 x [ext ∠ of ∆] OR M̂1 = 180° − 2 x [∠s of ∆] S R BM = MN [ 2 tans from a common point/raaklyne vanuit dieselfde punt] S R 180° − 2 x N̂ 1 = = 90° − x [ ∠' s opp = sides/∠e teenoor = sye] answer 2 (6) OR NM = BM [ 2 tans from a common point/raaklyne vanuit S R dieselfde punt] B̂ 2 = N̂ 1 [ ∠' s opp = sides/∠e teenoor = sye] S R B̂1 = x [ ∠' s opp = sides/∠e teenoor = sye] S In ∆ KBN: x + x + B̂ 2 + N̂ 1 =180° [sum of ∠' s of ∆] 2 x + 2 N̂ 1 =180° x + N̂ 1 = 90° N̂ 1 = 90° − x answer (6) 9.2 MB̂A = B̂ 2 + B̂ 3 = 90° [tangent⊥diameter/raaklyn⊥middellyn] S  R B̂3 = 90° − B̂2 = 90° − (90° − x ) = x S B̂ 3 = K̂ = x S ∴ AB is a tangent/raaklyn converse tan-chord theorem/ R omgekeerde raakl koordst]] (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 17 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne OR B̂ 2 = N̂ 1 B̂1 + B̂2 = x + (90° − x) = 90° S ∴ KN is diameter/middellyn [converse ∠ in semi-circle/ omgekeerde ∠ in halfsirkel] R MB̂A = B̂ 2 + B̂ 3 = 90° [tangent ⊥ diameter] ∴ AB is a tangent/raaklyn converse tan-chord theorem/ S R omgekeerde raakl koordst]] R (5) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 18 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 10.1 A D M N  E F  B C 10.1 Constr: Let M and N lie on AB and AC respectively such that Constr / Konstr AM = DE and AN = DF. Draw MN. Konst: Merk M en N op AB en AC onderskeidelik af sodanig dat AM = DE en AN = DF. Verbind MN. Proof: In ∆ AMN and ∆ DEF AM = DE [Constr] AN = DF [Constr] Â = D̂ [Given]  Δ AMN ≡ Δ DEF ∴ Δ AMN ≡ Δ DEF (SAS) SAS ∴AM̂N = Ê = B̂ MN || BC [corresp ∠' s are equal/ooreenkomstige ∠e =] MN || BC and R AB AC = [line || one side of ∆ OR prop theorem; MN ||BC] AB AC AM AN  = R AM AN AB AC ∴ = [AM = DE and AN = DF] DE DF (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 19 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 10.2 D 4 G 3 2 1 2 1 2 1 A O F B C E 10.2.1(a) DÔB = 90° DĜF = Ĝ 3 + Ĝ 4 = 90° [∠ in semi-circle/∠ in halfsirkel]  S R DÔB + DĜF = 180° ∴ DGFO is a cyclic quad. [converse: opp ∠s of cyclic quad/ R omgekeerde teenoorst ∠e v koordevh] OR ∠s of quad = 180°/∠e van koordevh = 180°] (3) OR EÔB = 90° DĜF = Ĝ 3 + Ĝ 4 = 90° [∠ in semi-circle/∠ in halfsirkel]  S R EÔB = DĜF ∴ DGFO is a cyclic quad. . [converse: ext ∠ = opp int ∠/ R omgekeerde buite∠ = teenoorst ∠] OR ext∠ of quad = opp int ∠/buite∠ v vh = teenoorst ∠] (3) 10.2.1(b) F̂1 = D̂ SR [ext ∠of cyclic quad/buite∠ v koordevh] SR Ĝ 1 + Ĝ 2 = D̂ [tan-chord theorem/raakl koordst] ∴ F̂1 = Ĝ 1 + Ĝ 2 R ∴ GC = CF [ sides opp equal ∠s/sye teenoor = ∠e] (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 20 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 10.2.2(a) AB = DE = 14 [diameters/middellyne] S ∴ OB = 7 units S ∴ BC = OC – OB = 11 – 7 Answer only: full marks = 4 units S (3) 10.2.2(b) In Δ CGB and Δ CAG Ĝ 1 = Â = x [tan-chord theorem/raakl koordst]  S/R Ĉ = Ĉ [common] ∆CGB ||| ∆CAG [∠, ∠, ∠ ] S CG CB = S CA CG CG 4 = 18 CG  CA = 18 CG 2 = 72 CG = 72 or 6 2 or 8,49 units answer (5) 10.2.2(c) OF = OC – FC = 11 – 72  OF OF tan E =  trig ratio OE 11 − 72 = = 0,36 7  substitution Ê = 19,76°  answer (4) OR OF = OC – FC = 11 – 72  OF FE = OE 2 + OF 2 2 = 7 2 + (11 − 72 ) 2 FE = 7,437.. = 7,44 OE OF cos E = OR sin E =  trig ratio FE FE 7 11 − 72 = = 0,94 = = 0,338  substitution 7,44 7,44 Ê = 19,76° Ê = 19,76°  answer (4) [26] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou