Mathematics Grade 11 Paper 2 Gauteng June 2019 Memo

Downloaded from hlayiso.com testpapers.co.za GAUTENG DEPARTMENT OF EDUCATION / GAUTENGSE DEPARTMENT VAN ONDERWYS PROVINCIAL EXAMINATION / PROVINSIALE EKSAMEN JUNE 2019 / JUNIE 2019 GRADE 11 / GRAAD 11 MATHEMATICS / WISKUNDE PAPER 2 / VRAESTEL 2 MARKING GUIDLINE / NASIENRIGLYNE 10 Pages / 10 bladsye 1

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM Paper 2 / Vraestel 2 Grade / Graad 11 GAUTENG DEPARTMENT OF EDUCATION / GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION / PROVINSIALE EKSAMEN MATHEMATICS Paper 2 / WISKUNDE Vraestel 2 MARKING GUIDLINE / NASIENRIGLYNE INSTRUCTIONS AND INFORMATION  Consistent accuracy applies as a general rule.  If a candidate answers a question twice and does not delete either, mark the FIRST attempt.  If a candidate answers a question, crosses it out and does not re-do it, mark the deleted attempt. Marks are awarded as per the guideline, and the following symbols are used: A – Accuracy C.A – Continued Accuracy S – Statement R – Reason S. R – Statement and Reason INSTRUKSIES EN INLIGTING  Deurlopende akkuraatheid geld as ʼn algemene reël.  Indien ʼn kandidaat ʼn vraag twee keer beantwoord en nie een van die antwoorde kanselleer nie, merk die EERSTE poging.  Indien ʼn kandidaat ʼn vraag beantwoord en dit doodtrek maar nie oordoen nie, merk die doodgetrekte poging. Punte word toegeken volgens die riglyn en die volgende simbole word gebruik: A – Akkuraatheid D.A. – Deurlopende Akkuraatheid S – Stelling R – Rede S. R – Stelling en Rede 2

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM Paper 2 / Vraestel 2 Grade / Graad 11 QUESTION 1 / VRAAG 1 1.1  correct formula / 1.1.1 =√ – korrekte form.  correct subst. /  (2  (2))2  (1  (2))2 korrekte verv.  25  answer / 5 antwoord (3) 1.1.2 x1  x2 y y ; 1 2 2 2 correct formula / korrekte form. 2= ; 0= correct subst. / 4=1+ ; 0=2+ korrekte verv. 3= ; –2 = D (3 ; –2) answer / antwoord (3) 1.1.3 mAC  mCE – –  equate m. / = gelykstelling m – – – – – =  correct subst. / – – – korrekte verv. 2 1  4 p2  simplification / 2p 4  4 vereenvoudiging 2p  8 (2p – 4 = 4) p4  (4) 1.1.4 quadrilateral ABDE is a kite when: .  vierhoek ABDE is ʼn vlieër wanneer: – – . – – – – –  correct subst. / . korrekte verv. – – – . ( )  –1 –1 since the product of the gradients = –1 / produk van van gradiënte = –1  explanation / AC BC verduideliking ABDE is a kite (diagonals intersect 90 / ABDE is ʼn vlieër (hoeklyne deursny 90 (4) 3

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM Paper 2 / Vraestel 2 Grade / Graad 11 1.2 1.2.1 P(0 ; 4) answer / antwoord (1) 1.2.2 R(4 ; 0) answer / antwoord (1) 1.2.3 = tan 45 =1  =1 y = mx + c 4 = 1(0) + c 4=c g(x) = x + 4 answer / antwoord (2) 1.2.4.1 = –1  = –1 = tan (–1) = 180 – 45  answer / = 135 antwoord (2) 1.2.4.2 PR̂Q  45 (supplementary angles) (suppl. hoeke)  S.R = 180 – 90 (sum of angles of a / (som van hoeke in driehoek)  answer / = 90 antwoord (2) [22] 4

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM Paper 2 / Vraestel 2 Grade / Graad 11 QUESTION 2 / VRAAG 2 2.1 2.1.1 y correct sketch / korrekte skets 2 5 x θ x2  y 2  r 2 application Pyth. / x 2  (5)2  (2)2 Pythagoras x 2  16 x  4 ……..restriction / beperking (90;270) choose / kies  x  4 x  4 x 4 answer / cos   r 5 antwoord (4) 2.2 2.2.1 cos2 45  sin330.tan 2 210  cos2 45  sin(360  30).tan 2 (180  30) correct expansion / korrekte  cos2 45  sin30.tan 2 30 uitbreiding. 2  2 1  1  2 correct reduction /     .  korrekte reduksie  2  2  3 substitution / 1  vervanging 3 answer / antwoord (6) 2.2.2   cos x cos x.( tan x)( cos x)   tan x  1.( sin x)   sin x  sin x  sin x cos x    (cos x) cos x   cos x   sin x simplification / sin x.cos x  vereenvoudiging  sin x answer / antwoord   cos x (6) 5

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM Paper 2 / Vraestel 2 Grade / Graad 11 2.3 √ – – – –tan √ – –  tan √  identity / identiteit simplification / √ vereenvoudiging answer / antwoord sec (5) 2.4 x  0 answer / antwoord x  90 answer / antwoord answer / antwoord x  180 (3) [24] 6

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM Paper 2 / Vraestel 2 Grade / Graad 11 QUESTION 3 / VRAAG 3 3.1 3.1.1  answer / antwoord  answer / antwoord  answer / antwoord (3) 3.2 3.2.1 answer / antwoord x = 325 answer / antwoord (2) 3.2.2 90 x 325 corr. critical value / korr. kritieke waarde corr. inequality korr. ongelykhede (2) 3.3 4  asymptotes / asimptote 3  shape / vorm 2 x-intercepts / x-afsnitte 1 y-intercept / y-afsnit 0 45° 90° 135° 180° 225° 270° -1 -2 -3 -4 (4) [11] 7

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM Paper 2 / Vraestel 2 Grade / Graad 11 QUESTION 4 / VRAAG 4 4.1 ... twice angle at circumference / twee keer die hoek op die omtrek.  answer / antwoord (1) 4.2 Construction / konstruksie: Join CO and produce it./ lyn CO verleng  construction / konstruksie Proof / bewys: In ACO: Oˆ 2  Aˆ  Cˆ 2 (exterior angle of a triangle is   SR equal to sum of int. opp. angles / buite hoek van ) but / maar Aˆ  Cˆ 2 (OA  OC ) (radii / radius)  Ô2 = Ĉ2+ Ĉ2 Ô2 = 2 Ĉ2 Similarly we can prove that / kan ook bewys word dat: Ô1= 2Ĉ1 S Ô2+ Ô1= 2 Ĉ2+ 2 Ĉ1  equating /  AÔB = 2 AĈB gelykstelling  substitution / vervanging (6) [7] 8

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM Paper 2 / Vraestel 2 Grade / Graad 11 QUESTION 5 / VRAAG 5  answer / 5.1 ... supplementary / suppl. antwoord (1) 5.2 5.2.1 Ŝ  Q̂  x (subt. by same arc / omtrekshoeke onderspan deur dieselfde boog)  S.R  S.R QR̂S  Q̂  x (alternate angles/ verwisselende binnehoeke: PQ || RS)  S.R Ŝ  QP̂S  x (alternate angles: verwisselende binnehoeke PQ || RS) (6) 5.2.2 PT̂R  x  x (ext. angle of sum of int. opp. angles / buitehoek van )   S.R  2x  answer / antwoord (3) 5.2.3 PT̂R  2 x (ext. angle of sum of int. opp. angles / buitehoek van )  SR PÔR  2 x (angle at centre th. / middelpunt stelling)  SR Line PR subtends equal angles at PT̂R and PÔR conclusion / gelyke omtrekshoeke onderspan deur dieselfde koord gevolgtrekking quadrilateral PTOR is concyclic / (3) PTOR is ʼn koordevierhoek [13] QUESTION 6 / VRAAG 6 6.1 6.1. SR = (midpoint Th. / middelpunt stelling) substitution / = vervanging = conclusion / AE = EC gevolgtrekking (4) 6.2 Eˆ 1  90 (line from centre of a circle bisecting the chord is to chord) S  R (lyn van middelpunt van koord na middelpunt van sirkel loodreg) (3) 6.3 OE 2  102  82 (Theorem of Pythagoras / Pythagoras Stelling) S R OE  6cm OE = 6cm ED  4cm answer / antwoord (4) [11] 9

Downloaded from hlayiso.com MATHEMATICS / WISKUNDE MEMORANDUM Paper 2 / Vraestel 2 Grade / Graad 11 QUESTION 7 / VRAAG 7 7.1. Let: = x = = x (tan / chord / raaklyn / koord) S R = = x (alternate angles / verwisellende hoeke, DR║ PS) S R but / maar = (given / gegee) = = x (v.o.a / teenoorstande hoeke) S R hence / dus = =x TS is a tangent (angle between tang. and chord is to angle in alter. seg.) R TS is ʼn raaklyn(tussen lyn en koord) (6) 7.1.2 T̂2  D̂ 2 (angles subt. by the same chord / omtrekshoeke onderspan  S R deur dieselfde koord)  S R but / maar D2  PŜT (tan / chord / raaklyn / koord)  S R PŜT  T2 (alternate angles / verwisselende binnehoeke ) TW || PS (6) [12] TOTAL / TOTAAL: 100 10