Mathematics Grade 11 Paper 2 Eastern Cape November 2012 Memo

Downloaded from hlayiso.com Province of the EASTERN CAPE EDUCATION NATIONAL SENIOR CERTIFICATE GRADE 11 GRAAD 11 NOVEMBER 2012 MATHEMATICS P2/WISKUNDE V2 MEMORANDUM MARKS: 150 PUNTE: This memo randum consists of 8 pages. Hierdie memorandum bestaan uit 8 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P2/WISKUNDE V2 (Memo) (NOVEMBER 2012) QUESTION/VRAAG 1 1.1 Quadratic / Kwadraties answer/antwoord (1) 1.2 Yes. The shape of the graph shows it./  Yes/Ja Ja. Die vorm van die grafiek toon dit  reason/rede (2) 1.3 Drivers should drive at a speed between 80 km/h and answer/antwoord 120 km/h. Bestuurders behoort teen ’n spoed tussen 80 km/h en 120 km/h te bestuur. (2) [5] QUESTION/VRAAG 2 2.1.1 11;18;22;25;31;35;36;42;44;49 Method/Metode Median/Mediaan = = 33  Answer/Antwoord Answer only: 2/2 Slegs antwoord :2/2 (2) 2.1.2 Q1 = 22  Q1 Q3 = 42  Q3 Semi IQR= = 10 answer/antwoord (3) 2.2 Min=11, Q1=22, Q2=33, Q3= 42, Max/Maks =49  min  Q1 and/en Q3  Q2  max/maks x 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 22 33 42 49 (4) 2.3 Data is skewed to the left./Data is skeef na links OR/OF  answer/ Data is more widely distributed below the median./ antwoord Data is wyer verspreid onder die mediaan (1) [10]

Downloaded from hlayiso.com (NOVEMBER 2012) MATHEMATICS P2/WISKUNDE V2 (Memo) 3 QUESTION/VRAAG 3 3.1 20+32+25+14+x+38+22+30+19+28+34+40+25 = 27  method/metode 13  answer/antwoord x = 24 (2) 3.2 X x–x (x – x)2  sum/som 20 20 – 27 = – 7 49  32 32 – 27 = 5 25 25 25 – 27 = –2 4  Answer/antwoord 14 14 – 27 = –13 169 Answer only: 3/3 24 24 – 27 = –3 9 Slegs antwoord: 3/3 38 38 – 27 = 11 121 22 22 – 27 = –5 25 30 30 – 27 = 3 9 19 19 – 27 = –8 64 28 28 – 27 = 1 1 34 34 – 27 = 7 49 40 40 – 27 = 13 169 SD = 25 25 – 27 = –2 4 Sum/Som 698 SD = 7,33 (3) 3.3 One standard deviation from the mean/Een standaardafwyking vanaf die gemiddelde:  interval 19,67 to/tot 34,33 9 spectators/toeskouers  answer/antwoord (2) [7] QUESTION/VRAAG 4 4.1 Interval Frequency Cumulative frequency Frekwensie Kumulatiewe First 4 correct/eerste 4 frekwensie korrek 5 10 5 5 Last 3 correct/laaste 3 10 15 9 14 korrek 15 20 14 28 20 25 17 45 25 30 11 56 30 35 7 63 35 40 2 65 (2) 4.2 70  (5 ; 0) 60  All points correct/Alle 50 punte korrek  Shape/vorm 40 30 20 10 0 0 10 20 30 40 50 (3) 4.3 80% of/van 40 = 32  method/metode Approximately 5 learners/Ongeveer 5 leerders answer/antwoord (Accept 4, 5 or 6/Aanvaar 4, 5 of 6) (3) [8]

Downloaded from hlayiso.com 4 MATHEMATICS P2/WISKUNDE V2 (Memo) (NOVEMBER 2012) QUESTION/VRAAG 5 5.1 S(–6 ; 4), T(–1 ; 3), R(–7 ; –1) ST 1 6 3 4 Substitution/instelling ST √26  ST √26 SR 7 6 1 4 SR √26 SR √26 ST = SR. Conclusion/gevolgtrekking ΔSTR is isosceles/gelykbenig. (4) 5.2 B( ; Substitution/instelling  –4 1 B(– 4 ; 1) (3) 5.3 S(–6 ; 4), B(–4 ; 1) SB  Substitution/instelling SB  Gradient/gradiënt 3 4 6 2  Substitution/instelling 5  Answer/antwoord (4) 5.4 5 and/en A(p ; –17)  Substitution of/instelling 3 van – 17 17 5  Substitution of/instelling 2 van p 8  Answer/antwoord (3) 5.5 T(–1 ; 3), R(–7; –1) and/en AS TR TR  TR 3 2 2 3  TR AS 1 Hence AS is perpendicular to TR AS perpendicular to TR/ AS is loodreg op TR AS is loodreg op TR (3) 5.6 STAR is a kite/vlieër  Kite/vlieër Diagonal AS bisects TR at right angles Reason/rede Skuinslyn AS halveer TR en is loodreg op TR OR/OF ΔSTR and ΔRAT are both isosceles. ΔSTR en ΔRAT is albei gelykbenig. (3) [20] QUESTION/VRAAG 6 6.1 x + y + 2 = 0 and/en Q(7;6)  m=–1 y=–x–2  substitution/instelling m=–1  answer/antwoord y – 6 = – (x – 7) y = – x + 13 (3)

Downloaded from hlayiso.com (NOVEMBER 2012) MATHEMATICS P2/WISKUNDE V2 (Memo) 5 6.2 Q(7 ; 6), R(4; –6) 6 6  substitution/instelling QR 4 7 QR 4  answer/antwoord (2) 6.3 P(– 2; 0)  –2 0 (2) 6.4 12  Method/metode QR 1 3 T(1 ; 12)  12 (3) 6.5 tan = 4  tan = 4 = 75,96°  75,96° QR tan =  PQ = 33,69°  = 33,69° PQR = 75,96° – 33,69° = 42,27°  Answer/antwoord (5) [15] QUESTION/VRAAG 7 7.1.1 R/(– 1 ; – 6)  – 1 – 6 (2) 7.1.2 R/(6 ; 1)  6 1 (2) 7.2.1 (x ; y)→ (x ; – y) → (x – 3 ; – y)  x – y (x ; y) → (x – 3 ; – y)  x–3 –y (4) 7.2.2  D//(1 ; – 3) y  E//(– 3 ; 1) 5 4  F//(2 ; 2) 3  Diagram 2 F// E// 1 x -5 -4 -3 -2 -1 1 2 3 4 5 -1 -2 -3 D// -4 -5 (4) 7.2.3 Rigid. The size and shape do not change  Rigid/rigied Rigied. Die grootte en vorm verander nie  reason/rede (2) 7.3.1 N/( ; 6)   6 (2) 7.3.2 Perimeter of/omtrek van KLMN = 10 units/eenhede Method/metode Perimeter of/omtrek van K/L/M/N/= 3 × 10 Answer/antwoord = 30 units/eenhede Answer only/Slegs antwoord: 2/2 (2) 7.4 Rotation about the origin through 90° anticlockwise  90° Rotasie rondom die oorsprong deur 90° anti-kloksgewys  anticlockwise/ anti-kloksgewys (2) [20]

Downloaded from hlayiso.com 6 MATHEMATICS P2/WISKUNDE V2 (Memo) (NOVEMBER 2012) QUESTION/VRAAG 8 8.1 V = π r2h and/en V = 4 000 cm3 Substitution of/instelling van π r2 (15) = 4 000 15 Substitution of/instelling van r = 9,21 cm 4 000 answer/antwoord (3) 8.2 Vcone = area of base/opp van basis × H H = 10 9,21 = 3,89 cm  H = 3,89 [or/of 3,9] Vcone/kegel = π (9,21)2 (3,89)  Substitution/Vervanging = 345,54 cm3  V = 345,54 [or/of 346,4] Vcontainer/houer = 4 000 cm3 + 345,54 cm3 = 4 345,54 cm3 [accept/aanvaar 4 346,4]  answer/antwoord (4) 8.3 SA = π r2 + 2π r h + π r s substitution/instelling : = π (9,21)2 + 2π (9,21)(15) + π (9,21)(10)  in π r2  in 2π r h  in π r s = 1 423,85 cm2 answer/antwoord (4) [11] QUESTION/VRAAG 9 9.1.1 sin 29° 1  diagram 1 1 61°  1 p  answer/antwoord 29° 1 cos 29° = 1 OR/OF OR/OF sin2 29° + cos2 29° = 1  identity/identiteit p2 + cos2 29° = 1  substitution/instelling  answer/antwoord cos 29° = 1 (3) 9.1.2 tan (– 569°) = – tan 29°  – tan 29° =  answer/antwoord OR/OF OR/OF tan (– 569°) = – tan 29°  – tan 29° ° = ° =  answer/antwoord (2) 9.1.3 1 – cos 2 61° = 1 – p2 answer/antwoord OR/OF OR/OF 1 – cos 2 61° = sin 261° = cos2 29° =( 1 )2  sin 261° = 1 – p2  answer/antwoord (2)

Downloaded from hlayiso.com (NOVEMBER 2012) MATHEMATICS P2/WISKUNDE V2 (Memo) 7 9.2 LHS = =  = =  = =  1 =  factors/faktore = = RHS  division/deel (5) [12] QUESTION/VRAAG 10 10.1 sin . tan 360° . sin 450°  cos 180°  tan x cos 180° sin . tan . cos  cos x cos 1  sin  –1 sin . cos . cos cos  1 sin cos  1 answer/antwoord (8) 10.2 sin x –3cos x = 0 sin x = 3cos x  sin x = 3cos x tan x = 3  tan x = 3 x = 71,57° + 180°k (k )  71,57° + 180°k  k (4) 10.3 2. √sin α = 1 √sin α = sin =  sin = = 180° – 14,48°  14,48° = 165,52°  165,52° (3) [15]

Downloaded from hlayiso.com 8 MATHEMATICS P2/WISKUNDE V2 (Memo) (NOVEMBER 2012) QUESTION/VRAAG 11 11.1 3  3 (1) 11.2 y f(x)=-sin x f(x)=cos(x - 30) f: 1 x –intercepts/ x-afsnitte  turning points/ draaipunte shape/vorm x -180 -150 -120 -90 -60 -30 30 60 90 120 150 180 g:  y-intercept/ y-afsnit  x-intercepts/ x-afsnitte -1 shape/vorm (6) 11.3 g(x) – f(x) ≤ 0  Values/waardes g(x) ≤ f(x)  Notation/notasie x [–180° ; –30°] or/of x [150° ; 180°]  Values/waardes OR/OF  Notation/notasie –180° ≤ x ≤ –30° or/of 150° ≤ x ≤ 180° per interval (4) 11.4.1 g(x) = cos (x –30°)  –90° h(x) = cos (x – 30° – 60°) + 1  +1 = cos (x – 90°) + 1 (2) 11.4.2 Maximum value of/maksimum waarde van h(x) – f(x) = 3  3 (2) 11.5 – sin x = sin (– x)  sin x  sin (–x) (2) [17] QUESTION/VRAAG 12 12.1 40 sin 30°  sin 30° QS QS QS = 80 m  answer/antwoord (2) 12.2 PSQ 65° / QPS 85°  PSQ / QPS PQ 80  sine formula/sinus sin 65° sin 85° formule PQ = 72,78 m  answer/antwoord (3) 12.3 1 substitution into area 80 72,78 sin 30° formula/instelling in area 2 = 1 455,60 m2 formule  1 Substitution/instelling 40 80 sin 60° 2  = 1 385,64 m2 2 × 1 455,50 + 1 385,64 m2 answer/antwoord = 4 296,84 m2 (5) [10] TOTAL/TOTAAL: 150