GRADE/GRAAD 12 JUNE/JUNIE 2017 …

1 NATIONAL/NASIONALE. SENIOR. CERTIFICATE/SERTIFIKAAT. GRADE/GRAAD 12. JUNE/JUNIE 2017 . mathematics P2/ wiskunde V2. memorandum . MARKS/PUNTE: 150. This memorandum consists of 12 pages. Hierdie memorandum bestaan uit 12 bladsye. 2 mathematics P2 / wiskunde V2 (EC/JUNE 2017 ). QUESTION 1 / VRAAG 1. Percentages / Frequency / Cumulative Frequency / 3, 12. Persentasies Frekwensie Kumulatiewe Frekwensie 30 < 40 1 1 24, 35, 44. 40 < 50 2 3. 50 < 60 9 12 50. 60 < 70 12 24. 70 < 80 11 35. 80 < 90 9 44. 90 < 100 6 50 (3). Ogive 55. upper limits / bo-limiete 50. 45. 40 cum f / kum. f Cummulative Frequency 35. shape / vorm 30 B. grounded / ge-anker 25. 20. 15. 10. 5. 0. 0 10 20 30 40 50 60 70 80 90 100 110. Percentages (4).

2 Approx. 30 [accept between 28 32] answer/ indicated on Ongeveer 30 [aanvaar tussen 28 32] graph. antwoord / op grafiek aangedui (2). [9]. QUESTION 2 / VRAAG 2. 12,4 15,1 18,9 19,7 19,7 20,0. 20,9 23,7 23,8 31,1 33,6 34,5. 34,9 36,5 40,1. Minimum / Minimum = min & max Lower quartile / Onderste kwartiel (Q1) = Q1. Median / Mediaan (Q2) = Q2. Upper quartile / Boonste kwartiel (Q3) = Q3. Maximum / Maksimum = (4). Copyright reserved Please turn over (EC/ junie 2017 ) wiskunde V2 / mathematics P2 3. min / max Q1 / Q3. Q2 (3). Skewed positively to the right. positively skewed /. Skeef positief na regs positief skeef (1). SD/SA = 8,36 answer / antwoord (2). A small standard deviation indicates that the data is clustered answer / antwoord around the mean.

3 OR/OF. A large standard deviation indicates that the data is more spread out. Klein standaardafwyking dui aan dat die data rondom die gemiddelde gegroepeer is. Groot standaardafwyking dui aan dat die data meer versprei is. (1). [11]. QUESTION 3 / VRAAG 3. R S. M=[ 2 ;. 6+8 9 1. ] x- value of M / x-waarde van M. 2 y- value of M / y-waarde van M. M = (7 ; 4). (2). mFM = 7 4 =. 4 3 1 substituting / vervanging 3. value of mFM / waarde van mFM. 1 1. 1 = 3 ( 1 ) m=3. 1. y 4 = 3(x 7) M = (7; 4) substituting M(7; 4) / vervanging M(7; 4). 1 5. y= x+ answer / antwoord 3 3. (4). Kopiereg voorbehou Blaai om asseblief 4 mathematics P2 / wiskunde V2 (EC/JUNE 2017 ). 1 5 substitution into line equation / vervanging in lyn vergelyking 1= +.

4 3 3 answer (as negative) / antwoord (as negatief). t = -2. OR / OF. OR / OF. mAF = mFM. 3 1 1 substitution into grad eqn / vervanging in gradi nt vergelyking =. 4 3 answer as negative / antwoord as negatief (2). 4 t=6. t = -2. mPC =. 3 ( 1). = -1 substitution / vervanging 4 8 answer / antwoord ANSWER ONLY. FULL MARKS/. SLEGS ANTWOORD. (2). tan = -1 tan = -1. = 135 = 135 (2). 2 1 1. tan = 10 = 5 tan = 5. = 180 = = .. = . = = . answer / antwoord (4). [16]. QUESTION 4 / VRAAG 4. 7 1 1 5. C =[ 2 ; 2 ] x-value /waarde ANSWER ONLY / SLEGS ANTWOORD y-value /waarde (2). = [3; 2]. Copyright reserved Please turn over (EC/ junie 2017 ) wiskunde V2 / mathematics P2 5. CA2 =(3 0)2 + (2 + 2)2 substitution / substitusie CA2 = 25.

5 CA = 5 answer for CA. antwoord vir CA. CB2 = (7 3)2 + (1 + 2)2. CB2 = 25. CB = 5 answer for CB. CA = CB antwoord vir CB (3). 2 5 substitution m AD m AD 7. 0 1. 7. 2 1. m AB substitution 0 7. 1 m AB.. 7. 1 m AD m AB 1. m AD m AB 7 1. 7 (5). AD AB [m AD m AB 1]. DA B 900. ( 3)2 + ( + 2)2 = 25 correct centre /. korrek middelpunt correct / korrekte r2 (2). 1 2 . mBC substitution 7 3. 3. m BC (2). 4. 4 mtan mtan= 3. 4. subst m= 3 and B(7;1). 4. 1 = ( 7) 4. verv. m= 3 en B(7;1). 3. 4 31 answer / antwoord (3). = +. 3 3. AE DB [diametersof same circle] AE = DB reason ABED is a rectangle [diagonals ]. reason (3). [20]. QUESTION 5 / VRAAG 5. sin 238 = sin 58 reduction / reduksie = k answer / antwoord (2). cos 58 = sin 32.

6 = 1 2 sin 32 . answer / antwoord (2). Kopiereg voorbehou Blaai om asseblief 6 mathematics P2 / wiskunde V2 (EC/JUNE 2017 ). tan(180 30 ).sin(360 60 ).sin 10 - tan 30 . =. cos(180 +45 ).sin(180 45 ).cos( 90 10 ) - sin 60 . - cos 45 . =. ( tan 30 )( sin 60 ) sin 10 sin 45 . ( cos 45 )(sin 45 ) sin 10 sin 10 . 1 3.. simplification /. 3 2 vereenvoudiging = 1 1.. 2 2. = -1 answer / antwoord (7). sin( + ) = cos [ 90 ( + )] cos [ 90 ( + )]. = cos[(90 ) ] cos[(90 ) ]. = cos(90 ) cos sin(90 ) cos(90 ) cos sin(90 ) . = sin cos cos sin sin cos cos sin (4). cos 2 + 1 2 cos 2 1 + 1 identity numerator = identiteit teller sin 2 . tan sin . 2 sin cos . identity denominator cos identiteit noemer 2 2 sin . = . cos . 2 2 . 2.

7 1 2 simplification /. =. 2 vereenvoudiging (4). sin sin . identity / identiteit (cos ). = 2 sin . cos . sin = 2 sin cos . sin 2 sin cos = 0. simplification /. vereenvoudiging sin (1 2 cos ) = 0. 1 factors / faktore sin = 0 or/of cos = (3). 2. sin = 0 1. or cos = 2 = 0 . = 0 + 360 , = 180 . OR = 60 + 360 = 60 . = 180 + 360 360 , (4). [26]. Copyright reserved Please turn over (EC/ junie 2017 ) wiskunde V2 / mathematics P2 7. QUESTION 6 / VRAAG 6. Asymptotes / Asimptote ( f ) Shape / Vorm ( g ). (45 ; 1) ( f ) (-45 ; 0) / (135 ; 0) / x-intercepts ( g ). Endpoints / Eindpunte ( f ) Endpoints / Eindpunte ( g ). (6). x = -45 -45 . (2). (-90 ;45 ] OR/OF 90 x 45. 0 0. 90 and 45. 0 0. correct inequalities (2). 90 answer / antwoord (1).)

8 [11]. QUESTION 7 / VRAAG 7. In ABC. AC . ANSWER ONLY/ proportion / verhouding =. sin sin SLEGS ANTWOORD..sin . AC = answer / antwoord (2). Kopiereg voorbehou Blaai om asseblief 8 mathematics P2 / wiskunde V2 (EC/JUNE 2017 ). In ADC. AC . =. sin(90 ) sin proportion / verhouding .cos . AC =. sin answer / antwoord . AC = . OR/OF. AC 1. AC 1. h tan y . h tan y h AC h tan y AC (2). tan y AC. sin . =. cos . sin . sin .sin . = subst/verv. AC =. cos . sin . d. tan . sin . =. sin OR/OF. OR/OF.. = tan .. sin . = equating AC / gelykstel sin ..sin aan AC (1). = sin . tan . sin .. =.. sin .. = substitution / vervanging . 80 . sin 38 . tan 40 . =. sin 125 answer / antwoord =50,45m (2). [7]. Copyright reserved Please turn over (EC/ junie 2017 ) wiskunde V2 / mathematics P2 9.

9 QUESTION 8 / VRAAG 8. Line from centre perpendicular to chord, bisects the chord. / answer / antwoord Lyn vanaf die middelpunt loodreg op die koord, halveer die koord. (1). AC B = 90 [ angle in semi-circle] / [hoek in semi-sirkel] S R. AC B = D 1 [ both = 90o] / [beide = 90o]. OE || AC [corresp ' s equal] / [ooreenkomstige e is gelyk] R (3). = x [ tan chord] / [raaklyn koord]. A S R. EO B = x [corresp ' s ; AC || OE] / [ooreenkomstige e ; AC || OE] S R (4). EO B = EC B [ both = x] / [beide = x] S. OBEC is cyclic quad [converse angles in same segment] R. OBEC is koordevierhoek [hoeke in dieselfde segment] (2). [10]. Kopiereg voorbehou Blaai om asseblief 10 mathematics P2 / wiskunde V2 (EC/JUNE 2017 ). QUESTION 9 / VRAAG 9.

10 = 49 [ at centre=2 at circumf.] / [Middelpunts ]. A S R (2). C 1 = B. 1 [angles opp equal sides] / [hoeke teenoor gelyke sye] R. 180 98 . 1 =. B [ angles of ] / [hoeke van ]. 2 R. 1 = 41 S (3). BC D = 90 [ 's in semi-circle] / [ e in semi-sirkel] S/R. 2 = C 3 = 26 [ 's in same segment] / [ e in dieselfde segment]. B S/R. C 2 = 23 S (3). [8]. QUESTION 10 / VRAAG 10. QR 2 PQ 2 PR 2 subst. in Pyth 6 8 2 2. QR 10 QR = 10. UR 5.. RQ 10. UR 1. 1 (3). RQ 2. 2. Copyright reserved Please turn over (EC/ junie 2017 ) wiskunde V2 / mathematics P2 11. PM 4 [diagonals bisect each other] R. QM 6 4. 2 2 2. [Pyth. Theo]. QM 2 13 QM 2 13. MS QM 2 13 [Diagonals bisect each other]. MV 13 MV 13. VM 13.. MQ 2 13. 1 VM 1. (4). 2 MQ 2.