# Mathematics Grade 11 Paper 2 Eastern Cape November 2013 Memo > Grade 11 Mathematics Paper 2 memorandum from the Eastern Cape November 2013, 9 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/maths-p2-memo/467f30f0-13c4-4290-a5d3-7c5d2f35167b - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Document type: Memorandum - Year: 2013 - Exam period: Eastern Cape November - Paper: 2 - Curriculum: CAPS - Pages: 9 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Mathematics](https://www.hlayiso.com/library/grade-11/mathematics) › [Mathematics Grade 11 Paper 2 Eastern Cape November 2013 Memo](https://www.hlayiso.com/library/maths-p2-memo/467f30f0-13c4-4290-a5d3-7c5d2f35167b) ## About this document Mathematics · Grade 11 · Eastern Cape November · 2013. Memorandum, 9 pages. Read online or download the PDF. 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Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE / GRAAD 11 NOVEMBER 2013 MATHEMATICS P2 / WISKUNDE V2 MEMORANDUM MARKS / PUNTE: 150 This memorandum consists of 9 pages./ Hierdie memorandum bestaan uit 9 bladsye. ### Page 2 Downloaded from hlayiso.com 2 MATHEMATICS P2 (NOVEMBER 2013) QUESTION / VRAAG 1 1.1 100;143;150;155;164;171;171;180;182;188;190 min and max / min en maks min = 100, Q1=150, Q2=171, Q3= 182, max / maks=190 Q1 Q2 Q3 100 150 171 182 190 (4) 1.2 Skewed to the left / skuins na links answer / antwoord (1) 1.3 100 answer / antwoord (1) [6] QUESTION / VRAAG 2 2.1 Mean / Gemiddelde answer / antwoord (1) 2.2 equating / gelykstel x = 36 answer / antwoord (2) 2.3 8,88 answer / antwoord (2) 2.4 outside / buite [36 – 8,88 ; 36 + 8,88] method / metode = [27,12 ; 44,88] ∴ 3 people / persone answer / antwoord (2) [7] QUESTION / VRAAG 3 3.1 Interval Frequency Cumulative frequency Frekwensie Kumulatiewe frekwensie 2 2 first three correct / 7 9 eerste drie korrek 14 23 remaining two correct / 12 35 oorblywende twee korrek 5 40 (2) 3.2 45 40 35 30 25 20 15 10 5  (0;0) points accurate / punte 0 0 10 20 30 40 50 60 akkuraat shape / vorm (3) ### Page 3 Downloaded from hlayiso.com (NOVEMBER 2013) WISKUNDE V2 3 3.3 35 learners / leerders answer / antwoord Accept 36 or 34 learners / Aanvaar 36 of 34 leerders (2) [7] QUESTION / VRAAG 4 4.1 4.1.1 A(-4 ; 1), B(-1 ; -2), F(k ; p)  method / metode  k=2  p= 5 k=2 (3) p= 5 4.1.2  formula / formule  substitution / instelling  answer / antwoord OR / OF  formula / formule  substitution / instelling  answer / antwoord (3) 4.1.3 B(-1 ; -2) and / en  formula / formule substitution / instelling equation in any form / vergelyking in enige vorm (4) 4.2 C(2 ; 5), A(-4 ; 1), F(2 ; -5) √ formula / formule substitution / instelling √ √  CA √ √ √ √  CF = 10 √ √ AF = √ √ ΔCAF is scalene / ongelyksydig scalene / ongelyksydig (6) 4.3 tan𝛳 = m  tan𝛳 = -1 tan𝛳 = -1  𝛳 = 135° 𝛳 = 135°  ̂ ̂  ̂ (4) ̂ ### Page 4 Downloaded from hlayiso.com 4 MATHEMATICS P2 (NOVEMBER 2013) 4.4 If BC were perpendicular to AF, the triangle would have to explanation / be isosceles. / As BC loodreg op AF was, sou die driehoek verduideliking gelykbenig moes wees. OR / OF (2) C does not satisfy the equation of the perpendicular bisector. / C maak nie die vergelyking van die middelloodlyn waar nie. 4.5 y = 0 as D must be the midpoint of FC / aangesien D die y = 0 (2) middelpunt van FC is. [24] QUESTION / VRAAG 5 5.1 x + 2y – 6 = 0 y = -½ x + 3  y = -½ x + 3 m = -½ m = -½ OR / OF y – 5 = -½ (x +2) substitution / instelling answer / antwoord (4) y = -½ x + 4 5.2 K(-3 ; 5), L(2 ; -3), N(5 ; -9) substitution / instelling  OR / OF  OR / OF K, L, M are NOT collinear / is NIE saamlynig NIE. not collinear / nie saamlynig nie (4) [8] QUESTION 6 / VRAAG 6 6.1 6.1.1 5tan𝛼 + 4 = 0 √  x = 5, y = -4 in y= -5 √ diagram  √ 2cos (180o – 𝛼) = – 2cos 𝛼  – 2cos 𝛼  answer / antwoord √ (4) √ 6.1.2 sin2 (𝛼 – 90o) – sin2𝛼 = cos2 𝛼 – sin2𝛼 cos2 𝛼 substitution / instelling =( ) ( ) √ √ = answer / antwoord (3) ### Page 5 Downloaded from hlayiso.com (NOVEMBER 2013) WISKUNDE V2 5 6.2 4cos2x – tan 45° = 0 4cos2x = 1  tan 45° = 1  any two correct angles / x= 60° or / of x=120° or / of x=240° or / of x=300° enige twee korrekte hoeke remaining two correct angles / oorblywende twee korrekte hoeke (4) [11] QUESTION / VRAAG 7 7.1     =  =  (6) 7.2 7.2.1      (5) 7.2.2 x = -90o  -90o x = 90o  90o (2) 7.3 √ if ( )  ( )  (  𝛳 = 71,57° + k. 180° 𝛳 = 71,57° +  k 180°  (6) [19] ### Page 6 Downloaded from hlayiso.com 6 MATHEMATICS P2 (NOVEMBER 2013) QUESTION / VRAAG 8 8.1 a = -2  a = -2 b=1  b=1 p=1  p=1 r = 45°  r = 45° (4)  8.2 f(x) – g(x) = 0  -75° f(x) = g(x)  105° x = -75° or / of 105° (2) 8.3 360°  360° (1) 8.4 g(x) = sin (x + 45o) h(x) =   ∴ answer only: full marks / slegs antwoord: vol punte (2) [9] QUESTION / VRAAG 9 9.1   Sine rule / sinus reël But  ( 𝛼 ) 𝛼 𝛼 𝛼  𝛼 ∴ answer / antwoord (6) 𝛼 9.2 9.2.1 𝛼 = 76°, β = 72° and / en x = 48 metres / meter 𝛼 𝛼 substitution / instelling MN = 27,16 metres / meter answer / antwoord (2) 9.2.2 Area of ΔKLN = ½KL × LN sin ̂  formula / formule = ½(48)(88)sin 72o  substitution / instelling = 2 008,63 m2  answer / antwoord (3) [11] ### Page 7 Downloaded from hlayiso.com (NOVEMBER 2013) WISKUNDE V2 7 QUESTION / VRAAG 10 10.1 10.1.1 Equal to angle in the alternate segment answer / antwoord Gelyk aan die hoek in die teenoorstaande (1) segment 10.1.2 Interior opposite angle / teenoorstaande answer / antwoord (1) binnehoek 10.2 P 2O 1 S Q R Constr.: Join OQ and OS / Teken OQ en OS construction / konstruksie Proof / Bewys: ̂ ̂ (angle at centre / hoek by middelpunt ) ̂ ̂ reason / rede ̂ ̂ (angle at centre / hoek by middelpunt ) ̂ ̂ But / Maar ̂ ̂ ̂ ̂  ̂ ̂ Hence / dus ̂ ̂ (5) 10.3 10.3.1 P 3 Q 2 1 O 1 R 43 2 5 1 ̂ reason / A S T rede ̂ ̂ (tan-chord / raaklyn-koord) ̂ reason / ̂ ̂ (QR = RS) rede (4) 10.3.2 ̂ (sum of angles of a triangle) reason / rede ̂ ̂ (2) 10.3.3 ̂ ̂ (cyclic quad. / koordevierhoek) reason / rede ∴̂ ̂ (2) ### Page 8 Downloaded from hlayiso.com 8 MATHEMATICS P2 (NOVEMBER 2013) 10.3.4 ̂ ̂ (angle at centre / middelpunthoek)  reason / rede ∴̂ ̂ (2) [17] QUESTION / VRAAG 11 11.1 OD = 25 cm ∴ OC = 25 cm – 18cm = 7 cm OC = 7 cm Pythagoras AC = 24 cm ∴  reason / rede (OD AB) (OD AB) ∴ answer / antwoord(5) 11.2 11.2.1 ̂ (PR||QB, alt angles / verw. hoeke)  ̂ reason / ̂ (Subtended by / onderspan deur RB) rede ̂ (Subtended by / onderspan deur PQ OR /  ̂ reason / OF alt angles / verw hoeke) rede  ̂ reason / rede (6) 11.2.2 ̂ ̂ (angle at centre / middelpunthoek) reason / rede (a) ∴ ̂  ̂ (2) 11.2.2 ̂ ̂ ̂ (angles of / hoeke van )  ̂ (b) ̂  reason / rede (2) ∴ ̂ 11.2.2 ̂  reason / rede (c) (2) 11.2.2 ̂ ̂ angle in semi-circle / hoek in  answer / antwoord (d) half-sirkel)  reason / rede (2) [19] QUESTION / VRAAG 12 12.1 12.1.1 F (a) E D O . x A B C ̂ (tangent is perp. to a diameter)  12.1.1 ̂  (1) (b) ### Page 9 Downloaded from hlayiso.com (NOVEMBER 2013) WISKUNDE V2 9 12.1.2 ̂ (angle at centre / middelpunthoek)  ̂ reason / ∴ ̂ ̂ rede ∴ ̂ ̂ (3) 12.2 ̂ ̂ (alt angles / verw hoeke, AC || DB)  ̂ ̂ reason ̂ ̂ (subtended by / onderspan deur CR) / rede ∴ ̂ ̂ (both / albei = ̂ )  ̂ ̂ reason Hence PDBR is a cyclic quadrilateral / rede (Ext. angle = int. opp. angle / buitehoek = teenoor. binnehoek)  ̂ ̂ conclusion / slotsom  reason / rede (7) [12] TOTAL: / TOTAAL: 150 ## Related documents - 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