Mathematics Grade 12 Paper 2 Eastern Cape Mock 2016 Memo

NATIONAL SENIOR CERTIFICATE GRADE/GRADE 12 SEPTEMBER 2016 MATHEMATICS P2 / WISKUNDE V2 MEMORANDUM MARKS / PUNTE: 150 This memorandum consists of 16 pages. Hierdie memorandum bestaan uit 16 bladsye.

2 MATHEMATICS P2 (EC/SEPTEMBER 2016) QUESTION 1 / VRAAG 1 Day/Dag 0 1 4 6 9 12 17 19 Weight / Gewig 124 121 103 90 71 50 27 16 1.1 2-4 correct points 5-7 correct points plotting all points 2 – 4 punte korrek 5 – 7 punte korrek alle punte afgesteek (3) 1.2 a 124,84 A b 5,83 B equation / vergelyking y 124,84 5,83x (3) 1.3 x ; y 8;5 ; 75,25 8;5 ; 75,25 and/en y-int 124,84 y-int / y-afsnit 124,84 regression line / regressielyn (2) 1.4 124,84 5,83x 80 substitution / vervanging 5,83x 44,84 answer / antwoord x 7,69 On the morning of the 8th day the bar of soap will be less than 80 grams. Op die oggend van die 8ste dag sal steen seep minder as 80 gram wees. (2) 1.5 r 0,998 answer / antwoord (1) 1.6 Very strong negative correlation. Baie sterk negatiewe answer / antwoord korrelasie. (1) [12] Kopiereg voorbehou Blaai om asseblief

(EC/SEPTEMBER 2016) WISKUNDE V2 3 QUESTION 2 / VRAAG 2 5 7 4 344 3 111468 2 24899 1 5567 0 35 2.1 21 learners / leerders answer / antwoord (1) 2.2 3 pages / bladsye answer / antwoord (1) 2.3 x 28,19 answer / antwoord (2) 2.4 13,12 answer / antwoord (1) 2.5 28,19 13,12; 28,19 13,12 interval / interval 15,07;41,31 8 learners / leerders 8 learners are outside one standard deviation 8 leerders is buite een standaardafwyking 28,57% 8 100 38,10% 21 (3) [8] Copyright reserved Please turn over

4 MATHEMATICS P2 (EC/SEPTEMBER 2016) QUESTION 3 / VRAAG 3 3.1 5 1 subst. P and R into correct formula mPR verv. van P en R in korrekte formule 4 4 1 1 mPR 2 2 (2) 3.2 mSQ 2 SP PT mSQ 2 y 4 2x 1 subst. m and Q into correct formula verv. van m en Q in korrekte formule y 2x 2 y 2x 2 (3) 3.3 Equation of PR / Vergelyking van PR 1 y 1 x 4 substituting m and P into equation of a 2 1 str. line / vervanging van m en P in die y x 3 vergelyking van reguitlyn 2 equation of PR / vergelyking van PR 1 2x 2 x 3 2 equating PR and SQ / gelykstel van PR 4x 4 x 6 en SQ 5 x 10 x-value / x-waarde x 2 y-value / y-waarde y 2 2 2 2 OR/OF substituting m and R into equation of a Kopiereg voorbehou Blaai om asseblief

(EC/SEPTEMBER 2016) WISKUNDE V2 5 1 str. line / vervanging van m en R in die y 5 x 4 2 vergelyking van reguitlyn 2 y 10 x 4 equation of PR / vergelyking van PR 2y x 6 1 y x 3 2 equating PR and SQ / gelykstel van PR 1 en SQ 2x 2 x 3 2 4x 4 x 6 x-value / x-waarde 5 x 10 y-value / y-waarde x 2 (5) y 2 2 2 2 3.4 x 1 y 4 substituting into correct formula 2 2 2 2 vervanging in korrekte formule x 5 and/en y 8 x-value / x-waarde S (−5 ; 8) y-value / y-waarde (3) 3.5 2 2 subt. into correct form / verv. in SQ 5 1 8 4 korrekte formule 6 5 SQ = 6 5 2 2 PT 2 4 2 1 PT = 5 5 1 Area PQS 6 5 5 2 Subt into correct form. / verv. in 15 unit 2 / eenhede2 korrekte formule 15 units2 / eenhede2 OR/OF SQ = 6 5 area of ∆ PQS = ST PT ST = 3 5 3 5 5 PT = 5 15 unit 2 / eenhede2 subst into form / verv. in formule 15 units2 / eenhede2 (5) [18] Copyright reserved Please turn over

6 MATHEMATICS P2 (EC/SEPTEMBER 2016) QUESTION 4 / VRAAG 4 4.1 x2 8 x 16 y2 6y 9 20 completing square 2 2 kwadraatsvoltooiing x 4 y 3 20 2 2 x 4 y 3 20 M 4;3 x-coordinate / x-koördinaat y-coordinate / y-koördinaat (4) 4.2 0 4 2 y 3 2 20 subst. x = 0 into circle equation 2 verv. x = 0 in sirkel vergelyking y 3 4 2 y 3 4 y 3 2 y 1 y 1 Q (0; 1) (3) 4.3 3 1 1 1 mradius mrad 4 0 2 2 mtan/ rkl 2 mtan 2 / mraaklyn = 2 y 1 2 x 0 [tangent radius] subst. mtan/ rkl 2 and Q into [raaklyn correct form. / verv. mtan/ rkl 2 radius] y 2x 1 en Q in korrekte formule. equation / vergelyking (4) 4.4 y 6 answer / antwoord (1) Kopiereg voorbehou Blaai om asseblief

(EC/SEPTEMBER 2016) WISKUNDE V2 7 4.5 6 2x 1 6 = 2x + 1 5 5 x x 2 2 5 U ;6 2 (2) 4.6 11 6 mAU 5 0 2 2 mAU 2 6 11 mAD 1 10 0 mAD 2 1 2 1 mAU mDA 2 2 mAU mDA 1 1 AU DA Â 90º Â 90 DQ̂U 90º [tangent radius] / [raaklyn radius] DQˆ U 90 QUAD is a cyclic quad.[opp. s add up to 180º] R QUAD is ʼn koordevierhoek [ teenoorst. e se som is 180º] (6) [20] Copyright reserved Please turn over

8 MATHEMATICS P2 (EC/SEPTEMBER 2016) QUESTION 5 / VRAAG 5 5.1.1 sin 52 sin 52 − sin 52° answer 1 t2 (3) 5.1.2 cos 2.19 cos 38 cos 2.19 cos 38 2 2 cos 19 1 cos 38 cos 2.19 2 cos 2 19 1 cos 38 1 cos 19 simplification / 2 vereenvoudiging 1 t2 1 cos 19 2 answer / antwoord (4) 5.2 2 cos(180 x) .sin(180 x).sin 74 sin( x 360 ).sin 37 .sin 53 .sin( x 90 ) − cos x 2( cos x) .sin x.sin 74 sin x sin x.sin 37 .cos 37 .( cos x) sin x 2 sin 74 cos37° 1 − cos x sin 74 1 2 sin 74 4 2 answer / antwoord OR/OF 2 cos(1800 x).sin(1800 x).sin 740 sin( x 3600 ).sin 37 0 .sin 530 .sin( x 900 ) 2( cos x).sin x.sin 2 37 0 − cos x sin x sin x.sin 37 0 .cos 37 0 .( cos x) sin x 2sin 2 37 0 cos37° − cos x sin 37 0.cos 37 0 2sin 2 370 4sin 37 0.cos 37 0 sin 37 0.cos 37 0 answer / antwoord 4 (7) Kopiereg voorbehou Blaai om asseblief

(EC/SEPTEMBER 2016) WISKUNDE V2 9 5.3.1 1 cos 2 x 0 1 cos 2 x 0 cos 2 x 1 2x 0 360 .k 2 x 0 k .360 x 0 k .180 ; k x 0 180 .k ; k (4) 5.3.2 2sinx L.H.S/LK 2 1 cos2x 2 sin x 2 2 cos 2 x 2 sin x 1 2 sin 2 x 2 2 1 2 sin 2 x 2 sin x 2 2 2 sin 2 x removing brackets 2 sin x verwyder hakies 2 sin 2 x 1 2 sin 2 x sin x (3) [21] Copyright reserved Please turn over

10 MATHEMATICS P2 (EC/SEPTEMBER 2016) QUESTION 6 / VRAAG 6 6.1 sin x 600 sin 900 2 x co-ratio / ko- funksie x 600 900 2 x 3600.k or/of x 600 1800 900 2 x 3600.k both gen. solns 3x 300 3600.k or/of x 300 3600.k beide algemene x 100 1200.k ;k Z oplossings -30° x 300 ; 100 ; 1300 10° 130° OR/OF co-ratio / ko- cos 2 x cos 30 x funksie 2 x 30 x 360 .k or 2x 30 x 360 .k ; k 3x 30 360 .k or x 30 360 .k both gen. solns beide algemene x 30 ; 10 ; 130 oplossings -30° 10° 130° (5) 6.2 g: x-intercept. x-afsnit y-intercept. y-afsnit shape / vorm f: x-intercept. x-afsnit y-intercept. y-afsnit shape / vorm (6) 6.3 240° answer / antwoord (1) 6.4 h x cos 2 x 90 1 sin 2 x 1 substitution / vervanging sin 2x – 1 (2) [14] Kopiereg voorbehou Blaai om asseblief

(EC/SEPTEMBER 2016) WISKUNDE V2 11 QUESTION 7 / VRAAG 7 7.1 NP 2 ON 2 OP 2 103 2 145 2 using Pyth theorem correctly korrekte gebruik van stelling van Pythagoras NP 177,86 answer / antwoord (2) 7.2 subst. into cosine rule PQ 2 103 2 103 2 2 103 103 .cos(120 ) verv. in cosinus formule answer / antwoord PQ 178,40 (2) 7.3 177,86 2 177,86 178.40 2 2 substitution / vervanging cos N̂ 2(177,86)(177,86) N̂ N̂ 60.20 (2) [6] Copyright reserved Please turn over

12 MATHEMATICS P2 (EC/SEPTEMBER 2016) QUESTION 8 / VRAAG 8 8.1 Bisects the chord / Halveer die koord answer / antwoord (1) 8.2 8.2.1 Q̂ 90 [ in semi circle.] / [ in halwe sirkel] S R S/R L̂1 90 [corresp. s , QN║LO] Q̂ [ooreenk. e , QN║LO] QL = LP [line from centre perp. to chord] [lynstuk vanaf middelpt is loodreg op koord] R (4) 8.2.2 MF 8ML (1) 8.2.3 OP 2 OL2 LP 2 using Pyth correctly 2 2 korrekte gebruik van 4M L 3M L 72 Pythagoras 16M L2 9M L2 49 simplification / 7M L2 49 vereenvoudiging ML 7 ML (3) [9] Kopiereg voorbehou Blaai om asseblief

(EC/SEPTEMBER 2016) WISKUNDE V2 13 QUESTION 9 / VRAAG 9 9.1 D̂3 30 [ s opp equal sides] /[ e teenoor gelyke sye] S R e S/R Ô1 120 [sum of s of a ∆.] / [som v/d van ʼn ] (3) 9.2 Â 60 0 [ at centre = 2 at circumf.] / S R [middelpunts = 2 omtrekshoek] (2) 9.3 Ĉ 120 [opp. s of cyclic quad.] / [teenoorst. e van k.v] S R (2) 9.4 AD̂B 70 [tan chord theorem.] / [raaklyn koord stelling] S R (2) [9] Copyright reserved Please turn over

14 MATHEMATICS P2 (EC/SEPTEMBER 2016) QUESTION 10 / VRAAG 10 10.1 constr. konstr. Û 2 90 Û1 [tan radius.] / [raaklyn radius] S/R Ẑ 90 [ in semi circle] / [ in halwe sirkel] S/R Ŝ 180 (90 Û1 ) 90 [sum of s of a ∆] S/R [som van die e van ʼn ] Ŝ Û1 e Ŝ Ŷ [ s in same segment] / [ in dieselfde segment] S/R Û1 Ŷ (5) Kopiereg voorbehou Blaai om asseblief

(EC/SEPTEMBER 2016) WISKUNDE V2 15 10.2 10.2.1 Â 2 x [ s opp. = sides] / [ e teenoor gelyke sye] S/R S/R Ĉ 2 Â 2 x [tan chord theo.] / [raaklyn koord stelling] S/R Â1 Ĉ1 [tan chord theo.] / [raaklyn koord stelling] Ĉ 2 Ŝ1 [alt. s , CB ║TS] / [verw. e , CB ║TS] S/R e Ĉ1 T̂ [corresp s , CB║TS] / [ooreenk. , CB║TS] S/R (5) 10.2.2 Ŝ1 T̂ x [proven in 10.2.1] / [bewys in 10.2.1] S CS = CT [sides opp. = s ] / [sye teenoor gelyke hoeke] R (2) 10.2.3 AR AT S/R [line ║to one side of a ∆] BR CT [lynstuk ║ aan een sy van ʼn ] CS = CT CS = CT [proved in 10.2.2] / [bewys in 10.2.2] substitution 3 AT 4 vervanging 2 AT AT 6 cm (4) [16] Copyright reserved Please turn over

16 MATHEMATICS P2 (EC/SEPTEMBER 2016) QUESTION 11 / VRAAG 11 F R Z C B P D A N 11.1 Â1 Ĉ 2 [tan chord theo.] / [raaklyn koord stelling] S R Ẑ [tan chord theo] / [raaklyn koord stelling] S/R Ĉ 2 Ẑ [both = x] / [beide = x] BC ║ RZ [corresp. s =] / [ooreenk. e =] R (4) 11.2 Ẑ P̂ [ s in same segment] / [ e in dieselfde segment] S/R S Ĉ 2 [corresp. s ; BC ║ RZ] / [ooreenk. e ; BC ║RZ] BC is a tangent to circle ACP [conv. of tan chord theorem] BC is ʼn raaklyn aan sirkel ACP [ omgekeerde van raaklyn koord stelling] R (3) 11.3 B̂1 D̂ 2 [ext. of a cyclic quad.] S R [buite van koordevierhoek] S/R R̂ B̂1 [corresp. s , BC║RZ] [ooreenk. e , BC║RZ] S/R R̂ D̂ 2 R Ẑ P̂ [ s in same segment] [ e in dieselfde segment] Â 2 Ĉ 4 [3rd ] / [3de ] ∆ RZA ||| ∆ DPC [equiangular or ] [gelykhoekig of ] (5) 11.4 ZA RA [similar ∆s] / [gelykvormige e] S/R PC DC ZA DC ZA DC RA RA . . . .(1) PC PC AR AZ [line ║ to one side of a ∆] S/R AB AC [lynstuk ║ aan een sy van ʼn ] AZ AB AR . . .(2) AC AZ AB RA ZA DC AZ AB AC PC AC DC AC simplification 1 PC AB (5) [17] TOTAL/TOTAAL: 150 Kopiereg voorbehou Blaai om asseblief