Mathematics Grade 12 Paper 2 North West Mock 2016 Memo

Downloaded from hlayiso.com Education and Sport Development Department of Education and Sport Development Departement van Onderwys en Sportontwikkeling Lefapha la Thuto le Tlhabololo ya Metshameko NORTH WEST PROVINCE NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT MARKS: 150 PUNTE: 150 This memorandum consists of 17 pages. Hierdie memorandum bestaan uit 17 bladsye.

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 2 NW/September 2016 NSC/NSS — Memorandum QUESTION 1/ VRAAG 1 11 y= 12,01 + 0,88x Y value of a/waarde van a v value of b/waarde van b Y equation / vergelyking (3) 1.2 y= 12,01 +0,88(46) v sub. 46 into the = 52% equation / vervang 46 in die vergelyking v answer / antwoord (2) 13 No, the preparatory exam mark is the independent v answer / antwoord variable. Hence we cannot determine the prep. exam v reason /rede marks using the final exam./Nee,die voorbereidende (2) eksamenpunt is die onafhanklike veranderlike. Dus kan ons nie die voorbereidende eksamenpunt met behulp van die finale eksmenpunt bepaal nie. 14 x =60,58 Y x = 60,58 y = 65,33 Vv y =65,33 LHS/ ILK = y = 65,33 ¥ sub. into RHS / RHS/RK = 12,01 + 0,88(60,58) = 65,32 vervang in RK LHS/ILK = RHS / RK (x ; y) lies on the regression line v LHS = RHS and conclusion / LK = RK en gevolgtrekking (4) 1.5 r = 0,98 Y value of r/ waarde van r (1) 1.6 There is a very strong positive correlation between ¥ very strong / baie prep. marks and final marks./ Daar is’n sterk positiewe korrelasie tussen die voorbereidende punte en die finale punte. sterk Y positive / positief (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 3 NW/September 2016 NSC/NSS — Memorandum QUESTION 2 / VRAAG 2 21|/a=6 Y answer / antwoord () 22 Expenditure (in rand) Frequency Cumulative ' frequency 50 < x <100 24 24 100 < x <150 ae 76 150 < x <200 14 90 v 76; 90 200 < 25 x <250 6 96 ¥96:100 250 < x <300 4 100 2) 231] 120 3 sg 2 100 8 & g = 804 3 3 3 S69 SI Fs x 40 & £ 3 20 | =| 0 it 0 50 100 150 200 250 300 350 Expenditure/Uitgawes ¥ Correct points Y Shape V Grounding /¥ Korrekte punte ~Vorm “grondvlak @) 2.4 | 100 < x <150 ¥ answer / antwoord () (7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 4 NSC/NSS — Memorandum QUESTION 3 / VRAAG 3 NW/September 2016 y 7 A(455) WA, —> x BC-3 ; -2 D C(6 ; -5) 31) Bc = Yl—x) +O.-9) J(6+3)? + (-5+2) 90 or/of 3V10 or /of 9,49 I ¥ sub. into the distance formula / Vervang in die afstandformule ¥ answer / antwoord (2) 32 _ Msc peTerae —5+2 64+3 1 3 +5 = eG) or/ of yr2= Le 43) 2 3 3 y= -$x-3 or 3y=-x-9 Y sub. into gradient formula / vervang in die gradiént formule gos 3 v equation / vergelyking @) ee ¥ Map = 3 ¥ sub. of the point / vervanging van die punt ¥ equation / vergelyking GB) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 5 NSC/NSS —- Memorandum NW/September 2016 3.4 1 a . Y equating the two equations / stel die _ twee vergelykings gelyk ee aan mekaar -10x = -12 6 x= 5 6 v x- value / x-waarde y = 3(2)-7 5 ul a : 6 17 y-value / y-waarde DG : -=) @) 3.5 _ 5+2 AB = ——— 4 +3 V map = 1 =1 ¥ tana = 1 tana = 1 Ya = 45° a = 45° tanB = 3 B =71,57° ¥B =71,57° BAD =71,57°—45° = 26,57° ¥ answer / antwoord GS) 3.6 | Equation of a line AE || BC/ vergelyking van AE || BC 1 —~5=-—(x-4 y 3% ) 3y-15 = -x+4 AE: 3y +x =19 x-intercept/x-afsnit is 3(0) +x =19 x =19 E (19: 0) 1 Y sub. of 2 and point into the equation / 7 1 vervanging van — 3 en die punt in die vergelyking Y equation of AE/ vergelyking van AE v x-intercepts / x-afsnit Yanswer / antwoord (4) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 6 NSC/NSS — Memorandum NW/September 2016 QUESTION 4 / VRAAG 4 ys NQ; R »M(4 ; 2) oO 41 («-4" +(y-2" =P ¥ sub. of M into equation of a circle / vervang M (0-47 +(0-2P =r in die vergelyking van die sirkel 0~=r vsub. of O(0; 0) / vervang (0;0) 2 2 Y value of r’/ waarde (x-4)°+(y-2)° = 20 ae 3) 4.2 (x-4) + (y—-2)" = 20 Subst/Verv(2 ; y) ¥ sub of N(2; y)/ 2 _4y +(y- 2 = 20 vervang N(2 ; y) 4+y-4y+4=20 vy —4y-12 =0 y- 4y-12 =0 Y(- 6) +2) = 0 (y-6)(y +2) = 0 vy =6 4) y =6 or/of y =-2 N/A OR / OF OR/OF (x-4) +(y-2)° = 20 ¥ sub of N(2; y)/ Subst/Verv(2 ; y) vervang N(2 ; y) (2-4) +(y-2) = 20 2 Vv (y-2) =16 (y-2) =16 Voy, 2 = 4 y-2 =+4 y = 6 ot/of y =—2 N/A/INVT vy =6 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 u NSC/NSS — Memorandum NW/September 2016 43 2 tl 1 Moy “42 ¥ Mou a5 Mog =—2 Y Mop = —2 Equation of OR is /Vergelyking van OR ya-2 Vy =-2x @) 44 -2 Ym, = —-2 m, —— =_ 2) ‘MN MN 2-4 Myr =; Y Myx => 7 =+(-2) ee 2 Vv y-6= 1-2) 2y-12 = x-2 2 2(-2x) - x- 10 =0 -5x = 10 x=-2 y=2(2)=4 RC 2; 4) OR RO? = NR? vty = (x-2P + (W-6" ty =x -4x4+44 y? —12y +36 4x +12y =40 x+3y= 10 and y =-2x x +3(-2x)= 10 x-6x= 10 -5x= 10 x=-2 y=-2(-2) =4 RC 2;4) v 2(-2x) - x- 10 =0 (6) Ve +y = (x29 + (0-67 v4x +12y =40 Y x+3y= 10 ¥ x +3(-2x)= 10 v x=-2 vy=4 © Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 8 NW/September 2016 NSC/NSS — Memorandum 4.5 MNRO is a kite because/vlieér omdat OR =RN and /en ¥ Kite/Vlieér MN = OM ¥ adjacent sides equal /aangr.sye gelyk 2) [18] QUESTION 5 / VRAAG 5 5.1 ‘o Oy) ein 2(Q(0 _ cos(180°+ x) TELL x).sin“ (90° — x) Aghbe: sin(180° —x) (—cos x).(—tan x).cos? x Aer Y¥—cos x sin x v— tanx ¥cos*x sinx 4 vsin x cos x. COS” X = snk +sin’ x sinx = cos’ x + sin? x ¥ cos’ x + sin? x =1 Yanswer/antwoord (6) 5.2.1 | cos(A — B) — cos(A + B) ¥ expansion of cos(A-B) / = cos A cos B+ sin A sin B — [cos A cos B — sin A sin B] | uitbreiding van cos(A-B) = cos A cos B + sin A sin B—cos A cos B + sin A sin B | Yexpansion of cos(A+B) / = 2sin A sinB uitbreiding van cos(A+B) ¥ simplification / vereenvoudig (3) 22 cos15° — cos 75° = cos(45° — 30°) — cos(45° + 30°) v 45° — 30° and /en : 45° + 30° = 2sin 45°.sin 30° v 2sin 45°.sin 30 _ axr2yt or/of 2x-Lyb yy Lyd 272 v2 2 yO ap Db V2 1 ¥ answer / antwoord =— oof —~= 2 v2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 9 NSC/NSS — Memorandum NW/September 2016 OR cos15° — cos 75° = cos(45° — 30°) — cos(45° + 30°) = cos 45°cos30°+sin45°sin 30°— [cos 45°cos 30° sin45°sin30°] = 2sin45° sin 30° = axd2 yh or/of 2x-Lxt D V2 2 v2 1 =— orof — 2 oT v 45° — 30° and /en 45° + 30° v 2sin 45°.sin 30 ya2 1b 1d Me yx 22 2 ¥ answer / antwoord 4) 5.3 AB? = (cos 0-6)" (sin0-7) 86 = cos’0 — 12cos 0+ 36 +sin’0 — 14sin 6 +49 86 = 1+ 36+ 49 - 12cos 0 -14sin 6 0 = - 12cos0 -14sin 0 14sin8 = — 12cos 0 sin® _ -12 cosO 14 tan® =-©/.0,86 7 Y sub. into the distance formula / vervang in die afstandformule ¥ simplification / vereenvoudig v14sin0 = —12cos 0 ¥ tand =-$/0,86 (4) a7) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com NW/September 2016 10 NSC/NSS — Memorandum Mathematics/P2/Wiskunde/V2 QUESTION 6 / VRAAG 6 = Tae ae Cat Q eS @ ial » o g ’ AR fh ~7 ToincccpessqeceneeesPooe —————— . leis § \ 4 H ' A H a] S|] s8 ~ 1 1 ' ' 1 =o ° lo 1 4 t q d ' ' ww S I~ |~].- ¢ ' 7 t 1 ' ' 7 ' S isa wala fala” WO & H 0 ' 0 5 0 4 ' of SCT O/DlOonm gs 1 1 1 1 ' 1 ' " 4 B/Blo- H 4 H H i H ' " i ~..8 Suge - eee ba------- La------ de------ phe ee eee ee Fee — " Qala se SD H H H fe) t H ! t a RR x X88] 1 oO 8 ' H H : i H H ! H . 8 oS > op hs fh > : D ' p ' ' : 1 2 a ' ; : ‘ 1 ' ! ' 5s 5 ! H ‘ i t H 3 1 a ¢ ee ee ee Sey ae ee een ee ee Deere eres || on ek r iH g : ' ' sag H ' ' ' ' i Bo H 4 H H t H ' H Sox ' : \ ' ' ' ! ' 32s 4 ' ' a) eS wn-deneee--- b--------f------- N.----- ha------ Je------- L------- Ae fee Ege ' H H 1 ao ' 1 1 1 mn > 2° 1 ' 1 ete ha} H ! ’ : Ea | ' i t H sas 5 7 a a eos Kay = 4 : S wa — R 4 : 2 ot ah v7 ' j sas ° 1 1 3 oO 2. eas : > 2 a a DD» COUR, CS eee as S al 1 ' a on a a= a iS} ay 1 q £8 e r {| j= 3S ' ; 2 7. f oS isa} no {Ml Ss 8 ee Aco = R bal 1 H £8 " i] ° ' H go R > Isl Io & ' i iS) iS} + ' ‘ Sw — = > wm | ' ' So fs} ke 6 + ce i q ans) 5 iS) | Vv 2 ° 5° prc te cence en penne ne | ie hel ~ Yay onl roy R ! ! ee t Shel ie ' 4 ion I par 7 can) Vv i H r ° ' ' 8 " jes 2 US le x |uj_|y as SoooosS beee-f os a R O77 gl NN jes a ee N N |Qais Ne} Neo} So jololo Please turn over/Blaai om asseblief Copyright reserved/Kopiereg voorbehou

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 ll NW/September 2016 NSC/NSS — Memorandum QUESTION 7 / VRAAG 7 A AC’ = 10? + 6? — 2(10)(6) cos100° ¥ sub. into cosine rule / vervang in die kosinus- = 156,8377813 reél AC = 12,5 units/eenhede Y 156,8377813 B = 80°( opp angles of a cyclic ¥ 12,5 units / eenhede quad/teenoorst<e van koordev ) A vB = 80° In AABC, ue = ae sin40° sinB BC 12,5 ¥ sub. into sine rule / , ea vervang in die sinus- sin40 sin80 reél pow l2oxsin 4d sin 80° Apne Y answer / antwoord = 8,2 units (6) 1 1 re) 72 Area of AABC = Fey Y 60° 1 : ¥ sub. into the area = 298.2) sin 60° tule / vervang in die = 44,4 square units opperviakte-reél v answer / antwoord G3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 12 NSC/NSS — Memorandum NW/September 2016 QUESTION 8 / VRAAG 8 8.1.1 | 90° v ans /antwoord qd) 8.1.2 | Angle in the alternate segment ¥ ans / antwoord Die hoek in die teenoorstaande segment dd) 8.2 5 A o D oe x . VS VR 8.2.1 | Bs =E=x (tan chord theorem / hoek tussen raaklyn en koord) YS YR B, =A=x ( corresponding angles / vS VR (6) ooreenkomstige hoeke) B, =E=x (radii / radiusse OE = OB) A A VS YR 8.2.2 | B2 +B3 =90° (subtended by a diameter / onderspan deur middellyn) | V ans /antwoord CBE = 90°+x (3) 8.2.3 | In ADBE, An = = (line | to one side of a A/ lyn | aaneen |VS VR syvanA ) But/maar ee) =1 (radii / radiusse) vs OD EF = FB v EF = FB (4) F is the midpoint of EB / F is die middelpunt van EB Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 13 NW/September 2016 NSC/NSS — Memorandum OR In A EOF and ABOF E =B2 (Proven in 8.2.1/Bewys in 8.2.1)) A OK VE =B2 EO = OB (radii/radiusse) Di = B3 (Zs opp = sides/ Z e teenoor gelyke sye) A i A A ¥D1 =B3 Di = 03 (corresp Z s/ooreenkomstige Ze BD | AO) A a A A ¥Di = 03 -. B3 = O03 “. B3 =O2 (alt Z s/verwisselende Ze BD | AO) “03 =O2 AEOF = ABOF (AAS/HHS) YAEOF = ABOF (AAS/HHS) EF = FB (4) 8.2.4 OF 1 EB (line from centre to a midpoint / lyn uit v SIR middelpunt van sirkel na middelpunt van koord) g EF = 4 (Fis the midpoint / F is die middelpunt) | ¥ EF = 4 OE’ = OF +EF = 3 442 = 25 VYOE =5 OE =5 Y ans / antwoord ED = 10cm OR/OF OR/ OF F 3 = 90° (corresponding angles / v S/R ooreenkomstige hoeke) EF =4 (Fis the mid point / F is die VEF =4 middelpunt) OE’ = OF +EF = 3° +4? = 25 VYOE =5 OE =5 Vans/antwoord (4) ED = 10cm Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 14 NW/September 2016 NSC/NSS — Memorandum OR/ OF OR/ OF OF = spp (midpoint theorem/middelpunt 1 2 ¥Y OF = —DB formule ) 2 v DB = 6 DB = 6cm In AEDB, Y Applicati f ED’ = 6° +8” (Pythagoras thm/ Pythagoras 6 Ppiication 0 ythagoras formule) thm/Toepassing van formule = 100 Y ans/antwoord (4) ED =10 [19] QUESTION 9 /VRAAG 9 A A 91 A =C (tan CD and chord CE / raaklyn CD YA = Co en koord CE) | “reason / rede = 3 (tan AEH and chord ED / raaklyn AEH | ¥ C = E3 en koord ED) | V reason / rede But they are corresponding angles / Maar hulle is ooreenkomstige hoeke AB ||ED ¥ corresponding angles / ooreenkomstige hoeke GS) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 1S NSC/NSS — Memorandum NW/September 2016 OR/OF A+ + Ei =180° (sum of angles in a A/som van die binne Z° vanA ) Cc stE2 +Di =180° (Sum of angles ina A/ som van OR/OF VY A+Ci +E: = 180° and C +E +Di = 180° die binne Z° vanA ) VA =C v reason / rede But A=C2 (tan CD and chord CE/ raaklyn cD | ¥ Ei = Ds en koord CE) E.=Di (tan AEH and chord CE/ Y altangles / raaklyn AEH en koord CE) | verwisselende hoeke but they are alt.angles/ Maar hulle is verwisselende hoeke AB||ED 9.2 ACDE is a parallelogram because one pair of opposite | V answer / antwoord sides (AC and ED) are equal and parallel / ACDE is ‘n_ | Vreason / rede parallelogram omdat een paar teenoorstaande sye (AC (2) en ED) gelyk en ewewydig is 9.3 In AABH, bce ( proportionality thm or AH || CD/ | ¥ AC _ HD CB DB CB DB eweredigheidstelling of AH | CD ) | “reason / rede HE = HD (proportionality thm or AB | ED/ HE = BD EA DB EA DB eweredigheidstelling AB | ED) | “reason / rede AC _ HE “ CB EA [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 16 NW/September 2016 NSC/NSS — Memorandum QUESTION 10 E F B 2 10.1 Const: On AB ,mark off AP = DE and on AC, mark ¥ Construction/ off AQ = DF/ Op AB, merk op AP = DE en op AC, mark of AQ = DF. Proof/Bewys: In AAPQ and A DEF: AP = DE (const/konstr) AQ = DF (const/ konstr) A=D ( given/ gegee) AAPQ = ADEF (SAS/SHS ) Pi =B ( = B) PQ || BC (corresp. angles =/ ooreenkomstige hoeke) AB = AC (line // oneside of a A/lyn// aan een sy van A AP AQ AB _ AC (AP = DE and AQ = DF) DE DF konstruksie v AAPQ = ADEF (SAS) (SHS) VP =E v PL =B ¥ PQ || BC y AB_AC AP AQ Vine // to one side of a triangle / lyn // aan een sy van ‘n driehoek (7) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 17 NSC/NSS — Memorandum NW/September 2016 10.2.1 Q =R2 (tan-chord theorem/ tan-koord) VYSYR =S2 (alt angles QS//RE/verwisselende) vS QR = RS (side opp. of equal angles/ sye VR (4) teenoor gelyke hoeke) 10.2.2 | In ARST and APQT =T (common/gemeenskaplik) ¥T=T R2+R3=P (ext. angle of a c.q PQRS/buite ¥ R2+R3=P hoek van kdvh PQRS ) | VReason/Rede $3 =Q,+Q; (ext. angle of c.q or 3“ angle inA/ v3" angle or buite hoek van kdvh PQORS ) Reason /3° hoek of rede (4) ARST || APQT (AAA/ HHH) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 18 NW/September 2016 NSC/NSS — Memorandum v 10.2.3 | RS _ ST _ RT (ARST || APOT) R(ARST ||| A PQT) PQ QT PT RS _ RT eee PQ” PP wrest (1) PQ PT In AQST, QS || RE _SE_ QR “ET RT (line drawn parallel to one side of a A/ SE_ QR lyn parallel sy van ‘nA) | ¥ 2. —= = ——— lyn pi lel aan een sy ) Er RT SE RS v Reason/rede “Er "Rr (QR =RS proved/bewys in 10.2.1) _SE _ RS “ET RT ER _ PQ ( from equation/uit vergelyking (1)) ve. Be = = (5) PT [20] Copyright reserved/Kopiereg voorbehou TOTAL / TOTAAL: 150 Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 19 NW/September 2016 NSC/NSS — Memorandum ANALYSIS GRID Q.no Concepts Level 1 | Level2 | Level3 | Level4 | Total 11 Equation of Regression line 3 TZ Prediction 2 1.3 Estimation and Analysis 2 14 @; y) 4 1.5 Correlation coefficient i 1.6 Relationship between the 2 two variables Total 5 ul 2 14 21 Finding the value of 1 unknown 2.2 Cumulative frequency table | 2 2.3 Ogive 3 24 Modal class 1 Total 4 3 7 3.1 Distance 2 | Equation of st. line 3 3.3 Altitude 3 3.4 Points of intersection 3 3.5 Angle between two lines 5) 3.6 Analysis 4 Total 5 6 5 4 20 4.1 Equation of a circle 3 4.2 Distance 4 43 Equation of a tangent 3 44 Point of intersection 6 4.5 Analysis 2 Total 5 4 9 18 5.1 Reduction and identity 6 5.2.1 Compound angle 3 5.2.2 Deduction and special 4 angle Bis) Trig and distance 4 Total 9 4 4 17 6.1 Sketching of graphs 6 6.2.1 Solutions 2 6.2.2 Asympote 2 6.2.3 Range 1 6.2.4 Solutions 1 6.2.4 Inequality 3 Total 4 2 9 15 qi Application of cosine rule 3 3 and sine rule ie Area rule 3 Total 6 3 9 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 20 NW/September 2016 NSC/NSS — Memorandum 8.1.1 Diameter 1 8.1.2 Tan-chord 1 8.2.1 Geo. Reasoning 6 8.2.2 App.tan-chord 3 8.2.3 Proving mid point 4 8.2.4 App. of thm 4 | Total 2 7 10 19 9.1 Tan- chord 3) 9.2 Parallellogram 2 9.3 Proportinality 2 2 Total 5 4 2 11 10.1 Similarity theorem 7 : 10.2.1 Side opp. of equal angles 4 7 Cyclic quad. 4 10.2.3 Application of Similarity 5 Total 4 11 5 20 Gr.Total 29 60 44 17 150 Expected 30 52,5 45 22,5 150 marks Actual 19,3 % 40% 29,3 % 11.3% 100 % Expected 20% 35% 30% 15% 100 % Copyright reserved/Kopiereg voorbehou