Mathematics Grade 12 Paper 2 Gauteng Mock 2015 Memo

Downloaded from hlayiso.com GAUTENG PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA PREPARATORY EXAMINATION 2015 MEMORANDUM SUBJECT: MATHEMATICS P2/ WISKUNDE V2 (10612)

Downloaded from hlayiso.com 10612/15 GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION ~ 2015 MATHEMATICS/WISKUNDE (Second Paper/Tweede Vraestel) MEMORANDUM NOTE: If a candidate answered a QUESTION TWICE, mark the FIRST attempt ONLY. Consistent accuracy applies in ALL aspects of the memorandum. Penalise for rounding only in QUESTION 3.6 Assuming answers/values in order to solve a problem is NOT ACCEPTABLE. S/R refers to STATEMENT as well as REASON NOTA: As n kandidaat n vraag TWEEKEER beantwoord, merk slegs die EERSTE poging. Volgehoue akkuraatheid word in ALLE aspekte van die nasienmemorandum toegepas. Penalisering vir afronding word slegs in vraag 3.6 toegepas. Aanvaarding van antwoorde/waardes om n probleem op te los, is ONAANVAARBAAR. S/R verwys na BEWERING sowel as die REDE.

Downloaded from hlayiso.com 10612/15 QUESTION / VRAAG 1 ; nie — ia i a Y (es) (22,6 : 44,5) 70 = v line of best fit = 60 beste paslyn/lyn van beste 3 passing 50 « 6 i @ | 3 _ & 30 os i v | E 20 + / 10 i) 10 20 30 40 | / Age in years i a Q) 1.2.1 25 years /25 jaar v answer / antwoord () 1.2.2 The reaction time according to the line of best fit for age v answer / antwoord (ld) 25 should be 51,46. The reaction time of this patient is Any other valid explanation. £65. Enige ander geldige Die reaksietyd volgens die lyn van beste passing vir die verduideliking ouderdom van 25 moet 51,46 wees. Die reaksietyd vir hierdie pasiént is + 65. on 3 A=-21,03 ¥ A=-21,03 B=2,90 ¥ B=2,90 y = 2,90x — 21,03 Y y= 2,90x — 21,03 (3) 1.4 r=0,95 Vv r=0,95 v very strong / strong very strong correlation / baie sterk korrelasie baie sterk / sterk (2) 15 The 30 year olds have a higher reaction time on average. [ The interquartile range of this group is bigger YIOR (IQR = 15)than that of the 15 year olds(IQR=5). The v for comments / reaction time of the 30 year olds varies more. vir kommentaar Die 30-jariges het gemiddeld n hoér reaksietyd. Die interkwartielvariasiewydte van hierdie groep is groter((IQR=15). as dié van die 15-jariges(IQR=5).. Die (2) reaksietyd van die 30-jariges varieer meer. (11)

Downloaded from hlayiso.com 10612/15 QUESTION / VRAAG 2 Age in years Number of Cumulative frequency Ouderdom in people Kumulatiewe jare Aantal mense Srekwensie Yall weave 20 au answers/ 10<A<20 130 150 alle : 205A <30 152 302 antwoorde 30<A<40 92 394 40<A<60 86 480 60S A<80 18 498 (1) 805A <100 2 500 = _| “plotting : _| — of points iu / plot van / = || punte g 400 v Joining 2 _| of points 2 verbind = _| van punte . _| “grounding 100 | point at | @50) 3 os eee |] grond : i) 20 40 60 0 100 190 | punt by : Age in years (0; 0) / ; we Se ( 3) 2.3.1 | 26 years old (accept 25 ~ 27) v answer 26 jaar oud (aanvaar 25-27) antwoord dg) vw 232 mall, 100 90 500 L 18% = 18% ‘© The percentage of people 16 and above 18 is 82% Y82% Die persentasie mense 16 jaar en ouer is 82% (G3) [3]

Downloaded from hlayiso.com 10612/15 QUESTION / VRAAG 3 A(=3 39) y B (e D(-5; 1) x 10 3.1 =3-5 9+] MC 4:5) a: (2) 3.2 AM’ = (443) + (5-9) OR/OF DM? =(-5 +4)" + (1-5) | “correct substitution =1+16 =1+16 into distance formula =17 =17 korrekte vervanging in srt 417 afstand formule Yr=Vi7 OR/OF AD? = (—543)* + (1-9) ¥ correct substitution =4+64 into distance formula = 68 korrekte vervanging in “AD = V68 afstand formule : V68 radius = = = Yre= V7 ad Q) ers) Yes, the circle will pass through point C vyes /ja Ja, die sirkel gaan deur punt C B=C=90°, AD is the diameter __ line subtends equal 2°. y 7 B=C=90°, AD die middellyn lynstuk onderspan gelyke ste iced (2) Ze

Downloaded from hlayiso.com 10612/15 BO; y) My XMgpy = —1 VY Mg XMgp =~ 1 -3-0/A-5-0 (25) (=) GC»), » -3-0 eT) 9-10y+y? =-15 ¥ standard form y? -10y+24=0 standaardvorm (y-O(y-4) =0 Y factors / faktore “ y=6ory=4 ~. B(O;6) Y BOO; 6) OR/OF 6) AB’ + BD? = AD? (9— y)? + (3-0)? + (5-0)? +(1— y)? = (5 +3)? +19)? Ons +(-3-0)7 4 81-18y+y? +9+25+1-2y+y? =44+64 (-5-0)2 +(1- y)? 2y* ~20y +48 =0 V (-5+3)? + (1-9)? e -10y+24=0 ¥ standard form/ Standaardvorm (y~- 6)(y-4) =0 ¥ factors / faktore y=6ory=4 -. B(O;6) ¥ BO; 6) . (5) 9-6 AB —3-0 =-1 l my =] my =) yr y, =m(x-x,) OR/OF y=-x+c y-1l =-1(x+5) -5=-l+c ¥ substitution of (—1 ; 5) y-1l =-x-5 4 vervanging van (~1 ; 5) y =-x-4 yerx-4 Vy (3)

Downloaded from hlayiso.com 10612/15 [3.6 _ 1-9 | 1-9 Map = ae Y mp = =4 ~54+3 -~5+3 tanO=4 A(43:9) an O=4 b= 76° BOO ; 6) 1-6 Mp, = = 7 tana=1 ¥ tana =I . v45° = 45 2 BDA= 76° 45° be [ v 31° (-1 if NOT rounded) = ext Zof A 31° (-1 as NIE afgerond) OR/OF AB’ = (~3—0)’ +(9~6)? v (-3-0)? +(9-6)? AB = 18 YAB =418 BD? =(-5-0)? +(1-6)? ¥(-5-0)' +(1-6)? BD = 50 ¥BD = 50 ; a0 Yusing a correct trig ratio tanBDA = 18 =0,6 gebruik korrekte trig verh 50 A : ¥ 31° (-1 if NOT rounded) BDA=31 31° (-1 as NIE afgerond) (6) [20] QUESTION / VRAAG 4 41.1 x+y? -~2x+6y =0 : ' V(@- 1-1 (x-1)?-1 eo. (x—1)’ -1+(y+3)’-9=0 V(y+3"-9 (x-1)? +(y +3)? =149 (x-1)? +(y +3)’ =10 “. centre : (1;—3) ¥ (3-3) ns v i radius : ¥10 vi0 @ 4.1.2 ~34+4 ae 14+2 ot 3 Ym. m, Xm, =—1 : m,=-3 equation of tangent / vergelyking van raaklyn ¥m=-3 p+ 4a — - 2, OR y=-3. ae x +2) 2 a ¥ substitution of (~2; -4) yt4=-3x-6 ~4=-3(-2) +e vervanging van (—2; -4) y =-3x—-10 “c=-l0 y=-3x-10 v y=-3x-10 (4) 6

Downloaded from hlayiso.com —__— 10612/15 BCI) hy : : AG 53) R@s4 Bs 2x+5y+1=0 4.2.1 |RA=RB RA’ = RB” Vv RA’= RB? (p-2) +(q-3)* = (p +1)’ +(q-6)" Y (p-2) +(q-3) p -4p+4+q -6q+9= p’+2ptl+q’ -12q4+36 Y (ptt)? +(q-6)’ —O6p + 6q = 24 a aa v—6p+6q = 24 oe ad (4) 4.2.2 R(p;q) lies onthe line /lé opdielyn 2x+5y+1=0 2p+5q=-1 .....() Pada p=q-4......2) OR sub(1)into(2) / verv (1) in (2) 2p—2q=-8.....(2) 2p—2q=-8....(2) 2(q-4)+5q=-1 2p+5q=-H.....0) 2q-8+5q=-1 (Q-M 7q= 7 Tq=7 g=l1 q=i subginto equation(2) /vervginvgl(2) p=l-4 p=-3 R(-3;1) radius=RA OR/OF _ radius=RB RA? =(-3-2)? +(1—3)* RB’ =(-3+1)? +(1-6)” =25+4 =44+25 RA=29 RB= 29 (x+3) +(5-1)? =29 ¥ 2pt+5q=-1 uv 2(q—-4)+5q=-1/ 2p-2q=-8 Vq= ¥ p=-3 Vv RA?/ RB? ¥ RA/RB=29 ¥ (x43) +(y-1F =29 7) {19}

Downloaded from hlayiso.com 10612/15 QUESTION / VRAAG 5 Ol) oy 271-0 3y =2x oe 2 eee a tan~=m tno = Y tana=m ; (2) S12 / y Pp aN Cc x —_ OP =25532 Byn op= 13 ¥ OP=¥/13 ie or/of a (with rational denominator) Ye a (met rasionale noemer) (2) 5.2.1 2 y iN 5 | 0 QOP =B -a v¥ B-a oO) 5.2.2) sin OOP = sin(B—a) va 2) ae ee a = sin Bcos a —cos B sin a =sin Scosa—cos Ssina Q 3 3 (2) 2) (ae 2) a ig 5 VB) (5 V3 2 a a ou) 5 13 5V13 ai 513 (4) [9]

Downloaded from hlayiso.com QUESTION / VRAAG 6 10612/15 6.1 cos(40° — x). cos x — sin(40° — x). sin x sin 205°. cos 25° _ cos(40° — x + x) ~ —sin 25°. cos 25° ¥ cos(40° — x +x) 2 cose a v —sin 25° — 5 (2sin 25°, cos.25°) _ cos 40° 1 sin 50° v—1 sin soe 2 2 cos 40° sin 50° = ~ OR/OF 1 ¥ sin50°=cos40° or/of 5 cos 40° a sin 50° cos 40° = sin 50° in numerator in noemer =-2 v2 | 6) 6.2.1 LHS /LK = £08 X=sim’ x ¥ cos2x = cos? x~sin? x cos xX +sinx _ (cos x —sin x)(cos x +sin x) i. con one ¥ factorising /faktorisering =cosx—sinx =RHS/RK (2) 622 | coe [es a cosx+sinx 2 : 1 Y cosx—sinx cos x (cos x—sin x)= 6 cos’ x—cosxsinx = a 1 72 ¥ cos’ x—cos xsin x => 2cos* x—2cosxsinx=1 2cos’ x-1=2cos xsin x ¥ 2cos*x—-1 = cos2x cos 2x = sin2x ¥ 2cosxsinx = sin2x (4) 6.2.3 | cosx (=) al cosx+sinx) 2 cos2x =sin2x tan2x=1 oo 2x =45°+k.180° ; keZ V¥ 2x = 45° + k.180° Pyne = ee, ¥ x=22,5°+k.90° 3; keZ OR/OF GB)

Downloaded from hlayiso.com 10612/15 cos2x =sin2x cos2x = cos(90° — 2x) ¥ cos 2x = cos(90° — 2.x) 2x = 90° —2x+k.360° or 2x=2x-90°+k.360° ; keZ [ 4x = 90° + k.360° 4x=90°+K.360° ; keZ Y x= 225° + 90° x= 22,5°+Kk.90° ; keZ (3) 6.3.1 i . il Area AABC = 5m sin4x v Area AABC = —mnsin4x 2 (2) 6.3.2 | Max area if sin4x =1 ¥sin 4x=1 4x =90° Vx = 22,5° (2) E 6 Pirate 6.3.3 | Right-angled triangle/Reghoekige driehoek ¥ answer [18] QUESTION / VRAAG 7 7A sin(60°~ @) =1 60°-@=90° os @=-30° 7 (1) ed 0° period / periode = 36 = 180° ¥ 180° qd) 73 |-2<y<0 of/or [250] ¥-2<y<0 (qd) TA 45° <x <135° or/or v 45° and 135° 150° < x <180° ¥ 150° and 180° ¥ ALL inequalities correct ALLE ongelykhede korrek Subtract I mark for extra intervals Trek 1 punt af vir ekstra intervalle (3) te) O0e = x 1502 ¥ 90° and/en 150° v inequalities / ongelykhede (2) [8] 10

Downloaded from hlayiso.com 10612/15 QUESTION / VRAAG 8 8.1 ——=tana ul v ptana (dy DE= ptana pe 8.2.1 E, =180°-f (opp Z of cyclic quad v E, =180°—B with reason/met rede teenoorst. Z van kvh) F = A, (Z opp equal sides; EH = EF ve BE = A, with reason/met rede Z teenoor gelyke sye) £ +A, =180°-(80°- 8) (of W/Z van A) | VY F,+8,=8 Soe (3) : K=H,= oF 8.2.2 In AEFH po FS sin 1p sin(180°— 8) ¥ substitution in sine rule 2 vervanging in sinus formule sal 7 FEL sin; B FLsind B sin B Vv p= —_—2_ 1 sin B FH. sin 3h - 2sin£ pcos! p ¥ sin B= 2sin4 B ee B 2 ”) i 7 yo) - FH eos; (3) 623 FH? =p + p’ —2p* cos(180°— 8) ¥ substitution in cosine rule =2p?+2p*cosf ees in kosinus reél al ¥2p'+2p’*cosf = 2p (1+ cos @) ¥ 2p2(1+cos 8) FH = p/2(1+cos f) (3) [10] 11

QUESTION / VRAAG 9 Downloaded from hlayiso.com 10612/15 S/R Statement and reason _S/R__Bewering en rede ol (a) AB || CD corr. Z equal / ooreenk Z gelyk (b) line from centre 1 to chord /Midpt. O ; Midpt. koord (c) Radius vreason / rede v reason / rede ¥ reason / rede (3) Join OB / Verbind OB HB = 12 and/en GD = 5 (line from centre .L to chord / Midpt. O ; Midpt. koord) In AOHB (OHY =(OBY -(BHY — 1 =90"(Pyth) = (13) ~(12) =25 OH =5 In AOGD (0G) = (op) -(Gpy = (13) -(5) =144 OG =12 GH = 0G —-OH =12~-5 =7 v S/R v Pythagoras YOH=5 ¥OG = 12 ¥ GH=7 G)

9.2.1 Tangents from same point / Tans from common pt v reason /rede Raaklyne vanuit dies. punt () Q 110° A T M 9.2.1 | B.+B,=90° (tan _L rad /raaklyn radius) v S/R A, + A, = 90° (tan L rad / raaklyn |. radius) «. AOBT is acyclic quadrilateral/is koordevierhoek Y SIR (opposite Z's sapplementary/oorstaande Z'e suppl) YR @) 9.2.2 | OA=OB (radii) ¥ OA=OB teat i : (radii) r=T, (equal chords subtend equal angels/ YR gelyke hoeke onderspan deur gelyke koorde (2) 923 | A, +B, =110 (ext Z of A/buite Z van A) non But Ay =B, (As opp equal sides / 2° teenoor gelyke sye) Y G=55° ee v SIR AS B, =C=55° (tan chord / Z tussen raaklyn en koord 4 OR/OF a v SIR O, +O, =110° (ext Z of cycl quad/buite Z van kvh) vs ae A v O, +0, =2C (Aat centre = 2x Zat circ /midptsZ = 2xomtreksZ) ‘ e=55° . ] (4) [47] 13

QUESTION / VRAAG 10 Downloaded from hlayiso.com 10612/15 10.1 Construction: Join OD and OB Konstruksie: Verbind OD en OB 6, =2A (Zat centre = 2x Z at circumference / MidptsZ = 2 x OmtreksZ) O, = 2¢ (Zat centre = 2x Z at circumference / MidptsZ = 2 x OmtreksZ) 0, +0, =360° (Z rounda ptOR Z’ ina rev Z°om 'n punt OF omwenteling 2A + 2C = 360° 2A+C) = 360° A+ C=180° ¥ Construction / Konstruksie v SIR SIR v SIR ¥ 2A + 2C = 360° (5) 14

Bowntoaded-from-htiayiso-com 10612/15 : 10.2.1 A 1800" i u ¥ all 3 statements B=180°-x (opp Z of cyclic quad / al 3 bewerings C=180°-z teenoorst. Z° van kvh ) v reason rede (2) 10.2.2 (180° = y) * (180° = x) a (180° -z) ¥ substitution of angles : vervanging van hoeke =540°-(x+ y +z) ¥ simplification = 540° — 180° (sum of Z's of A= 180°) vereenvoudiging = 360° v_ 180° v answer / antw (4) (14) T QUESTION / VRAAG 11 G E M to | Ro =M, (altZ's RE||TM / verw. Z°;RE||TM) y (Sv ®) A ss S R R,=T corr Z's RE||TM /ooreenk. Z° RE||TM) —- (4) 12 | EM RE : ~(S) —~—=—— (line |]one side of A OR prop theorem; RE ||TM oF EG RG (lyn || een sy van A) : eo. YS) YR) But/Maar RT=RM_ (Z‘ oppequalsides ; M, =T/ Z teenoor gelyke sye;M = T) EM _RM ty EG RG

Downloaded from hlayiso.com 10612/15 Hales) In AGYE en AGER G=6 (common /gemeenskaplik) ¥ G=G(SIR) E,=R, (given /gegee) VE, =R, (S/R) ¥,=6, +8, (rd Zof A/3e Z van A) Aina ee equiangular AGYE|JAGER (2,2,4) Igelykhoekige A Le or/of Z,2,Z (R) (4) 114] EG RG_ YG —= = equiang A's / AGYE||| AGER EY RE EG (@iang Ill ) u EG RG (R) equiang A's / AGYE||| AGER an gelykh A’ AGYE ||| AGER (D) 5.1) (RG) =(GM)’-(RM) — GRM =90° Pythag. . ; 3 Y substitute into Pyth =(10)° -(6) vervang in Pyth =64 lea VY RG=8 (2) Ce eee ‘4 GE _GR =— (1 de of A OR th ;RE || TM ¥——=—— fp iG (line || one side o prop theorem; RE || an /reason / rede (lyn || een sy van A) “Gh 14 GE 8 Y¥ substitution of values 10 Vi vervanging van waardes GE= 52 or/of pee or/of 5,7 1units/eenhede ¥ answer /antwoord (4) [19] ee s TOTAL/TOTAAL: 150 16

Downloaded from hlayiso.com