Mathematics Grade 12 P2 KwaZulu Natal November 2015 Memo

2d Trom hlayiso.com Basic Education KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA > MATHEMATICS P2 PREPARATORY EXAMINATION SEPTEMBER 2015 MEMORANDUM NATIONAL SENIOR CERTIFICATE | aBed x84 dH WVEEOL 22 PO GLOZ

Mathematics P] D ow n | fe) a d. 2 d. 3 rs nevAdad AVISO°COnT Examination QUESTION 1 ;id . Scatter plot of Height vs Arm span 200 - 190 | t I E ie = 180 Al “1-4 points S correct S &) v5 —9 points z 170 correct < a} ¥ all points 3 correct 160 E (3) 450 160 170 180 190 200 Height (in em) 12 a= -~36, 58 {a= ~36, 57689...) VV a b= 1,25 (b = 1,25381....) |b p= 36,38 +1,25x Ca! ¥ equation Caer ely Yu Gaya a ie a) 13 f = 36,58 + 1,25(176) eat. |v substitute 176 = 183.42 4. | ¥ answer @) OR yp=184,09 ae) vv answer _ Q) 1.4 There is strong, positive correlation between height and arm span.@~) v” strong, positive @) [10] Copyright reserved Please turn over z abed x84 dH WVEEOL 42 PO GLOZ

Mathematics P1 D 0 wn | oad ed f ro m h | AYISPcG QWMNaory Examination NSC-MEMORANDUM(KZN) QUESTION 2 2.1 : Cumulative Daily Sales Frequency Frequency | ¥ first two 60<5x<70 5 5 ij cumulative 70<x<80 1 16 frequencies correct 80<x<90 22 387 & | ~ next two 90<x<100 13 s1 cumulative frequencies 100 <x <110 7 58) correct 110 <x <120 3 61 | v remainder 7 correct (total = 61) 3) 2.2 . Cumulative frequency graph of Daily Sales 70 -peperrereorpreeerrrarrsr rr rECEIT: perRey : ¥ grounding at 0 E ¥ plotting | et cumulative é Bie frequencies at = 40 usta upper limits ¢ ae Y points correct $s E ¥ smooth shape ki ee of curve E sth 4 5 es «) i He 4 ae arena 40 50 60 70 8&0 90 100 110 130 Daily Sales (in Rands) 2.3 The median for the data is approximately R 87. e&| v reading off g5-#4 from graph Cheeest ) a. | RET 2) 2.4 The upper 25% interval is R96 to R120 4.) ¥ 96 to 120 (Range to accept: 94 to 120) a) [10] Copyright reserved. Please turn over € abed x84 dH WVEEOL 42 PO GLO?

Mathematics P1 D fe) WwW n | fe) ade. df an am, tal AVTSOE Orr?” Examination QUESTION 3 3.1 tartan Yot9_ 5 oe) W x= 5 Xp =5 x, =4 > Yona DG 5 4) Q 3.2 _4-©2) 2% | substitution Into Men = 5-2 gradient formula _ Mey = tan a = 2 tan @ = 2op “= 63,4 as Y answer (3) 33 May = Mcp = 2 ABIICD, equal gradients RLY in gy = 2 yrlxt+e 0=2%-Dte Oa.) v subst (-1 ; 0) c=2 yorxt+2 &) ¥ answer 3) 34 _4-© -2 cay vom = 3 tan (2 of inclination of AD) = : Z of inclination of AD = 33,7° Be | ¥33,7° _ A= 63,4 ~ 33,7° BA cmeue = 29,7° cal v0.7 Copyright reserved Please turn over y abed x84 dH WVYSE0L 42 PO GLOZ

Mathematics P] D Oo wn | oad ed f ro m h | QYISOxG OD aros Examination NSC-MEMORANDUM(K2N) 3.5 3AB=DC “GAB? = DC? of (x +P +(y- 0] =(3—2) +(442) CO) v substitution of (x + iy + ] =45 4nifoc, a) xtl (241) $Y? Soccer (1) Y= IN+Qecrerreransen (2) Swhsiaae (2)in () Sub. 3-8 (x+1+(2x=2) =5 Cal v substitute = y= 2x +2 x? 4241442? +8x+4-5=0 5x7 +10x=0 &.| “ standard form Sx(x+2)=0 x#0 3; x=-2 |) Vv xy=-2 Substitute x =—2 in (1) ya} Blv y= B(-2;-2) (3) [16] Copyright reserved Please turn over g abed x84 dH WVSE0L 42 PO GLOZ

Mathematics P1 D own | oaded. £ rQ mn, layiso-corr Examination QUESTION 4 41 re =(2-47 4G-57 “| ¥ subst into distance formula =8 a) ¥8 x 2 (x= 2) +(y-3¥ =8 “|v 2) ALY (y-3Y z (4) 42 5-3 1m =1 Miyp = rary =] NP Mpp =—l NP PT, product of gradients = - lea} Vv mp, =-1 yo-xte gt xe S=-44e otk ¥ subst (5; 4), e=9 Boive=9 yYa-xt9 | Md Wy aie + 9 Q=-x+9 x=9 “. TY 50) &.| ¥ coordinates of T (6) 43 _ fron ase are mp? cas | ¥ substitution into «« &.% PT = f(9-4)° + (S-0 * v io ? ( ) distance formula =v50 fF & | wv f50 =5¥2 ol Q) 44 Area =x PT? “dé | ~ substitution into gyi yu &, =2x50 area formula =157 thee | YW 1§7 : be @) 45 BY * ig : tan wag = 28. PEL ¥ tan Wer = 3S. Met /50 ¥50 ar MPP saBir = 21,8° Boo ¥21,8° e) 46 NP = NM radii Gv SR PT=T radii @i ¥ SIR MNPT isa kite two pairs of adjacent sides.) reason equal in length 3) Copyright reserved Please turn over 9 abed x84 dH WYSE0L 42 PO GloZ

Mathematics P1 NSC-MEMORANDUM(K2ZN) Downloaded from hla@ydsooG Muy examinsio 47 NTP = NTM = 21,8° diagonal of kite Che ” ot on NPT = NIMT = 90° tangent perpendicular to radius Brew R NMP = 360° - 90? — 90° —43,6° angles in quadilateral cCaly SR = 136,4° a@ |v answer (4) OR NP LPT (rad | tan) ¢alv SR “.TNP = 68,2° (Z sum A) aj“ SR TNP = TNM (prop of kite) aly sR MRP = 2(68,2°) . MNP =136,4° Ou | v answer @ [23] Please turn over Copyright reserved L abed x84 dH WVcri0lL 42 PO GLOZ

Mathematics P1 D own | oaded. £ramolal ayisort BEE” Pemination QUESTION 5 5.1 tan(180° + A). cos (180° — A) . sin (360° - A) cos (90° — A) . : a ~ {tan 4) (cos 4) (sin 4) Rv tan A —cosA sin A an) ¥— sin A“v’ sin A _ sit A os A S y Sin A cos A cos A . = sinA eek. | ¥ answer * “ © 5.2.1 cos 52° = cas 2(26°) &.| ¥ writing 52° in = 2c0s? 26° -1 terms of 26° ; # | “ expansion =2rY-1 =} -] &) @ answer : @) 522 | ni sin 7d? & Vv identity cas71° __ sin(45° + 26°) &| ~ writing in terms of cas(45° + 26°) 26° _ sin 45° cos 26° + cos45°sin 26° & | Vv expansions 00s45* 00s 26°—sin 45° sin 26° ey aR 7 = V2 ; V2 1 v’ substitution = (2 (\i-*) 2 (2 2 (- +f/b- | 8 ¢ -Jjl-r? ) reafb- 7? a a) &\ v answer ) Copyright reserved Please turn over g abed x84 dH WY¢er0lL 42 PO GLO?

Mathematics PJ NSC-MEMORANDUM(KZN) Downloaded from hl ayYiS.xGONaos Examination 5.3 7 sin 2x cos2x + sin? x 2sin xcosx cos* x—sin? x+sin? x __ 28in xcosx 60s? x _ 2sinx "osx = 2tanx = RAS OR tus =—Si02x_ cos2x+sin“ x 2sin xcosx 2cos? x—1+ sin? x 2sin x cosx ” Beos? x —(1-sin? x) _ __ 2sinxcosx ~ 2cos? x —cos’ x _ 2sin xcosx cos’ x _ 2sinx ” “cosx = 2tanx = RHS OR LHS = sin 2x cos2x+sin? x _ ___2sinxcosx ~ 1=2sin? x +sin? x _ 2sinxcosx ~ 1=sin? x _ 2sinxcosx cos’ x _ 2sinx cosx = 2tanx = RHS PP as Ps Y identity for sin2x ¥ identity for cos 2x Y simplification v identity ¥ identity for sin2x Y identity for cos 2x ¥ simplification Y identity (4) v identity for sin2x ¥ identity for cos 2x ~ simplification ¥ identity (4) [19] Copyright reserved 6 abed Please tum over x84 dH WVerOL 22 PO LOZ

Mathematics P1 D own | oa d le af rom, AvESO 206 apppsiory Examination QUESTION 6 6.1 cos 2x = sin(x - 30°) = cos[90° - (x — 30°)] = cos(120° — x) key angle = 120° — x 2x =120° ~x +3603 Z 3x = 120° +:7.360% xe Z x= 40° +,120°ne Z or 2x = 240° +x4n.360% ne Z xX = 240° + 7.360% € Z 2x = 360° ~ (120° ~x) +2.360°;ne Z v using co-ratio ¥ 120°-x ¥ 2x =120°-x+ 360° v x= 40° + 2.120° ¥ -2x =360° - 120° ~ x) + 71.360° v¥ x= 240° + 1.360° ¥neZ (7) SAAS x-intercepts turning points shape = oe intercepts tuming points shape << 408 (6) 6.3 —120° < x < -80° or 40° <x < 90° critical values: —120° ; —80° critical values: 40° ; 90° correct notation SAK G) (16 Copyright reserved. OL abed Please turn over x84 dH WVcri0lL 42 PO GLOZ

vatemiser DOWNIoade|d From HlaYiSOs6OM aioe roninaion NSC-MEMORANDUM(KZN) QUESTION 7 cml AB tanx = BD &| ¥ using tan ratio =. BD Bpe—/- ol v pp- tan x tanx — (2) 72 BC=BD CD? = BC? + BD’ -2BC.BD.cos y &! v using cosine 2 1 formula (2) (4) AA) © \tanx tanx tanx J tanx J} ” C&.| v substitution 77,). a ne 2h Sa ts OS & | “simplification tan’ x tan? x tan’x _ 2h cosy “tan x tan® x” 2H =——(I-cosy) O& “common factor tan’ x 4 _ 2h (1-cos.y) "tan x [6} Copyright reserved LL aBed Please turn over x84 dH WYPr0L 42 PO GLO?

Mathematics Pi D fe) wn | oaded. fram blayise:cornr Examination QUESTION 8 8.1 BE Construction: Draw diameter AD. Join Dio C 3.) ¥ construction Proof: ; EAC+DAC =90° tan L radius & VSR DCA = 90° - Zinsemi circle om ¥ : iR ADC + DAC = 90° sum of Zs in A BAC = ADC ale & But ABC= ADC <s in the same segment ae SYR 6) Copyright reserved Please tum over ZL abed x84 dH WVGrOL 42 PO GLOZ

Mathematics P1 D Oo wn | oad ed f ro m h | AVIS OFA Marory Examination NSC-MEMORANDUM(KZN) 8.2.1 Two tangents drawn from the same external point are equal in &@,| “ answer length. (¢3) 8.2.2 (a) | § =x angles opposite equal sides &l/“ SYR 2) 8.2.2 (b) | R ,=x tan-chord theorem a iY“ SYR 0) B22) |v, =\2x ext Z ofA Riv SVR @) x = cay v v 8.2.3 R, =0, =x Sy -B, alt Zs; PQ || RS a , : ve Q,=S,=~x R,=B sin the same segmentebsra FR oR, =8, 4) 8.2.4 W, =2x ext Z ofA al’ SYR W, =V, “.QVSW is acyclic quad converse: ext Z of cyclicquad * “ SVR (4) 825@ | O,sy tan-chord theorem av SYR By 8250) | P=6,=y corresp Zs; PQ {| RS Ri Y SVR OR ext Z of A @) 8.2.6 Jon MtoR and Meo P é, =x proven in 8.2.3 a PMR =2x Z at centre =2 x/ at circumference ¥ S/R PMR = W, = 2x Ry gs -. PMWR is a cyclic quad converse: angles in same segment iv R (3) [28] Copyright reserved Please turn over €L abed x84 dH WVGHOL 42 PO GLOZ

Mathematics PL Downloaded. framalalayisoxearmr bens QUESTION 9 9.1 6, =¢ vert opp angles Siw S wR eG: ppane wl SYR Q, =A, tan-chord theorem A,=R ext angle of cyclic quad aly SVR 56,=R “PQ=PR sides opp equal angles Rl YR i) 92 In APBO and APQA (i) P, is common aly Ss (ii) Bs Q, tan-chord theorem aly oR (i) ~=PQB=A, remaining angles intriangle “P & ¥ ware ~. APB APQA equiangular ai i QI APQ Joi @ a3 PA _PQ from 9.2 @ ¥ deduction PQ PB But PQ= PR &)¥ PQ=PR PA_PR “PR PB .. PA, PR and PB form a oy geometric sequence the ratio is constant me ee in @) (14) Copyright reserved Please turn over +L aBed x84 dH WYL70L 42 PO GLO?

GL abed vatenwiser DOWNloaded From Nlayi$OnCOM ry carina NSC-MEMORANDUM(KZN) QUESTION 10 10.1 In AHKG: | = = = (prop theorem; RD || KG) a/v S/R DG _? A, ¥ substitution 2 3 “DG = 6 &@) “ answer @) 10.2 LetFD =» “EG =6-y) Co| + siatement D&E aoa GF GE G2 K (prop theorem; FE || HK) miv S/R — - -yo DEeMFD |G <> 1 £a.| ¥ substitution yt+2 2 26-y)=y +2 12 - 2p sp +2 & | v simplification —3y = -10 10 & | ~ answer = i. =FD yrs (5) [8] TOTAL: 150 Copyright reserved x84 dH WYL70L 42 PO GLO?