Mathematics Grade 11 Paper 2 Eastern Cape November 2015 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 11 NOVEMBER 2015 MATHEMATICS P2/WISKUNDE V2 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 8 pages. Hierdie memorandum bestaan uit 8 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015) 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. • Assuming answers/values in order to solve a problem is NOT acceptable. LET WEL: • Indien ʼn kandidaat ʼn vraag twee keer beantwoord, merk slegs die eerste poging. • Indien ʼn kandidaat ʼn antwoord doodgetrek het, maar nie oorgedoen het nie, merk die doodgetrekte antwoord. • Volgehoue akkuraatheid geld in ALLE aspekte van die memorandum. • Aanname van antwoorde/waardes om ʼn probleem op te los, is Onaanvaarbaar. QUESTION/VRAAG 1: [12] 1.1 Height (cm) Frequency Cumulative Hoogte (cm) Frekwensie frequency Kumulatiewe frekwensie 26 150 55 4 4 155 160 22 26 160 165 56 82 120 165 170 32 114 (2) 6 120 1.2 grounding at(150;0)/anker by (150;0) Ogive/Ogief  plotting (155;4)/plot van 150 (155;4) Cumlative Frequency / 100 plotting with upper limits/plot by boonste 50 limiete 0  joining the points to form Kumulatiewe Frekwensie 140 150 160 170 180 a smooth curve/verbind Heights/Hoogte (in cm) punte om glade kurwe te vorm (4) 1.3 (150, 160,5, 163, 166, 175)  150 OR/OF 160,5 Min = 150  163 Q1 = 160,5 166 Q2 = 163 175 Q3 = 166 Max = 175 (5) 1.4 Skewed to the left / Skeefgetrek na links.  answer/antwoord (1) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2015) MATHEMATICS P2/WISKUNDE V2 3 QUESTION/VRAAG 2 [7] 2.1 ̅  ̅ ANSWER ONLY : FULL MARKS = R3 555 SLEGS ANTWOORD : VOLPUNTE  answer/antwoord (2) 2.2  answer/antwoord (1) 2.3 (3555 – 900,12 ; 3555 + 900,12) = (2654,88 ; 4455,12)  interval 7 werkers lê binne een standaardafwyking  answer/antwoord of workers lie within one standard deviation. (4) van die werkers lê binne een standaardafwyking. [7] QUESTION/VRAAG 3 [13] 3.1 mAB = mBC = mAC  mAB = mBC = mAC  substitute into equation/vervang in vgl ( ) ( )  substituting into mAC/vervang in vgl mAC  answer/antw (4) 3.2 ( )  equation/vgl  subst of m and(-5;5) into ( ) form/vervang van m en (4;-1) in formule  equation/vgl (3) 3.3  subst into correct [ ] formula/vervang in korrekte formule [ ]  coordinates/coordinate =[ ] (3) 3.4 mCD = 0  correct m = 0  substitution into eqn/vervang in vgl  answer/antw OR/OF  equating/ p – 7 = -1  answer p=6 (3) [13] Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com 4 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015) QUESTION/VRAAG 4: [13] 4.1  answer/antw =2 (1) 4.2 MNQ × MMP = -1 [NQ ┴ MP]  answer/antw (1) 4.3   value of /waarde van (3) 4.4 ( )  equation/vgl ( )  subst of m = and (0;6) into eqn/vervang m= en (0;6) in vgl answer/antw (4) 4.5    subst into eqn/vervang in vgl P (2;5)  P (2;5) (4) 4.6 MR = NR R=[ ]  Substitution/vervang. R=[ ]  x value/waarde  y value/waarde R = [-3;-3] (3) [16] QUESTION/VRAAG 5: [23] 5.1.1 tan θ answer/antwoord (2) 5.1.2 ( ) ( ) OP2 √ answer/antwoord (2) 5.1.3a answer/antwoord √ (1) 5.1.3b ( )  -cosθ  answer/antwoord √ ANSWER ONLY FULL MARKS/ (2) SLEGS ANTWOORD:VOLPUNTE Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2015) MATHEMATICS P2/WISKUNDE V2 5 5.2  standard form/st vorm ( )  1 – sin2 x ( )( )= 0 sin x = ½ or  no soln (sin x = 2)  30° ; 150°  k.360, (6) 5.3  sin x  -sin x  cos2 x =  = -tan x  tan x (5) 5.4 ( )  numerator / teller ( )  denominator / noemer  expansion / uitbreiding ( )  simplification / vereenvoudiging ( )  factorisation / ( ) faktorisering (5) ( ) [23] QUESTION/VRAAG 6: [11] 6.1 b2 = a2 + c2 – 2acCos B  a2 + c 2  2acCos B (2) 6.2.1 ̂  Ratio for tan/verhou. vir tan  substitution 65°/verv van 65° PS = 338,83  answer/antwoord (3) 6.2.2 In ∆PQS use of cos = ( )( ) formula/gebruik van cos = 1484499,606 formule SQ = 1218,40 m2 Substitution/vervang  SQ (3) Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com 6 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015) 6.2.3 In ∆RSQ:    value of /waarde van (3) 6.2.4 Area of ΔSPQ = ½ SP.PQ.sin̂  correct formula / korrekte = ½ (338,83)(1 500)sin 30° formule substitute / vervang =127061,25 m2 (338,83), (1500)  Sin 300 answer / antwoord (4) [15] QUESTION/VRAAG 7: [12] 7.1 f  asymtotes / asimptote min value / min waarde  max value / maks waarde g  (-45°; -1)  (45°; 1) (135° ; -1) (6) 7.2 ( ) ( ) ( ) -45 45 0 90 135 180 (6) 7.3 90°  answer / antwoord (1) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2015) MATHEMATICS P2/WISKUNDE V2 7 QUESTION/VRAAG 8: [8] 8.1 bisects the chord.  answer/antwoord (1) 8.2.1 (Pythagoras)  (substitution/vervang)  method/metode  answer/antwoord (3) 8.2.2 (Pythagoras)  (substitution/vervang) method/metode cm  cm But AB = 2AE (OE ┴ AB) AB = 2(3)  S/R = 6 cm  answer/antwoord (4) [8] QUESTION/VRAAG 9: [13] 9.1  construction/konstr C  S/R .O  S/R  S/R  S/R  conclusion / gevolgtrekking B (6) A D CONSTR: Join CO, extend to D PROOF: In ∆AOC i) ̂ ̂ ̂ (ext  of ∆/buitehoek van ∆) ii) ̂ ̂ ̂ (ext  of ∆/buitehoek van ∆) iii) ̂ ̂ (AO = OC) iv) ̂ ̂ (BO = OC) ̂ ̂ ̂ ̂ ̂ = 2( ̂ ̂ ) ̂ 9.2.1 ̂ ( radii equal/radiusse gelyk)  S R (2) 9.2.2 ̂ (ext  of ∆/buitehoek van ∆)  S R (2) 9.2.3 ̂ (angles in same segment/  S R hoeke in dieselfde segment) (2) 9.2.4 ̂ (opp angles of cyclic quad/  S R teenoorstaande hoeke van ʼn koordevierhoek) (2) [14] Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com 8 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015) QUESTION/VRAAG 10: [6] 10.1 ̂ ̂ =x (angles in the same segment/  S R hoeke in dieselfde segment) ̂ ̂ =x (tan chord/tan koord)  S R (4) 10.2 ̂ ̂ (both equal to x/albei gelyk aan x) S/R ̂  conclusion / gevolgtrekking (2) [6] QUESTION/VRAAG 11 [11] 11.1 Are supplementary OR add to 180°.  answer/antwoord (1) 11.2.1 ̂ ̂ (Ext  of cyclic quad/buitehoek S  R van koordevhk) ̂=̂ (corresponding angles, AB ║ DC/ S  R Ooreenkomstige hoeke AB ║ DC) (base angles of ∆ equal/basis hoeke R van ∆ gelyk) (5) 11.2.2 ̂ (angles of ∆/hoeke van ∆) S R (2) 11.2.3 ̂ (  at centre = 2  at circumference/ S  R  by middle = 2  by omtreks) ̂ ̂ S (3) [11] QUESTION/VRAAG 12 [11] 12.1 Sh2 = (Pythagoras) S 2 Sh = 64 + 9 (substitution/vervang)  method/metode Sh = √  answer/antwoord = 8,54 cm (3) 12.2 Area of ∆ face =  formula/formule  substitution/vervang = ( )(√ ) answer/antwoord = 25,63 cm2 (3) 12.3 TSA = area of slanted faces + area of right prism  TSA TBO = oppervlakte van skuinsvlakke + oppervlakte van reghoekige  substitute/vervang prisma  answer/antwoord = 3(25,63) + 62 + 4(6 x 12) (5) = 76,89 + 36 + 288 = 400,89 cm2 [11] TOTAL/TOTAAL: 150 Copyright reserved Please turn over