Mathematics Grade 11 Paper 2 KZN November 2018 Paper & Memo

basic education fi J y » Department: (GO) Basic Education AS fis ON“ REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 11 MATHEMATICS P2 i a NOVEMBER 2018 5 Se ey MARKS: 150 TIME: 3 hours This question paper consists of 16 pages and a 24-page answer book. Copyright reserved IA TAA Please turn over

Mathematics/P2 Downloaded from hlayiSO.COM,...ccme-201s CAPS ~ Grade 11 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. Ls 2. a 2. ‘This question paper consists of 1] questions. Answer ALL the questions in the SPECIAL ANSWER BOOK provided. Clearly show ALL calculations, diagrams. graphs, etc. that you used to determine the answers. Answers only will NOT necessarily be awarded full marks. If necessary, round off answers to TWO decimal places, unless stated otherwise. Diagrams are NOT necessarily drawn to scale. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. Write neatly and legibly. Copia seed NA He MO

Downloaded from hlayiso.com Mathematics/P2 3 DBE/November 2018 CAPS — Grade 11 QUESTION 1 A school held a sports day. One of the items on the programme was an obstacle race. Teams of 10 parents and learners participated in this race. The table below shows the time taken, in minutes, by each member of a particular team to complete the race. 1.1 How long, in minutes, did it take for the fastest member of this team to complete the race? Q) 1.2 Determine the mean time taken by this team. (2) 13 Calculate the standard deviation for the data. (1) 1.4 How many members of the team completed the obstacle race outside of two standard deviations of the mean? (3) 1.5 It took another team a total time of x+5 minutes to complete the race. Calculate the value of x if the overall mean of the two teams combined was 18 minutes. (3) [10] Copia seve CL ee

Mathematics/P2 QUESTION 2 Downloaded from hlayiso.com DBE/November 2018 CAPS — Grade 11 2:1 A survey was conducted of the ages of players at a soccer tournament. The results are shown in the cumulative frequency graph (ogive) below. Cumulative frequency graph of the ages of soccer players 50 7 sae = St od T 40 > t FA = 5 30 > = = 3 a Fi : a eo & 20 oS = = | Se E = aa 5 10 = 0 a oe ae ce 0 5 10 Ages of soccer players 2.1.1 How many players took part in the soccer tournament? (1) 212 Determine the number of players between the ages of 24 and 31 years old. (2) 2.1.3 Complete the frequency column of the table below in the ANSWER BOOK. CLASS FREQUENCY | CUMULATIVE INTERVAL FREQUENCY 1S<x<20 4 20<x<25 13 25<x<30 a7 30<x<35 47 35<x<40 50 (3) 2.1.4 Use the grid provided in the ANSWER BOOK to draw a frequency Copyright reserved polygon for the data. (4) IIT ITE A Pease um over

Mathematics/P2 2.2 Downloaded from hlayiso.com DBE/November 2018 CAPS ~ Grade 11 Two Grade 11 Mathematics classes have the same number of learners. The five- number summaries of the marks obtained by these classes for a test are shown below. CLASS A (30; 48 ; 65 ; 82 ; 90) CLASS B (50 ; 58; 65 ; 75 ; 90) The parents of learners in CLASS A and CLASS B observe that both classes have the same median and the same maximum mark and therefore claim that there is no difference in the performance between these classes. Do you agree with this claim? Use at least TWO different arguments to justify your answer. (3) {13} Copynight reserved UATE TOV ETAE Hossam exes

Downloaded from hlayiso.com Mathematics/P2 6 DBE/November 2018 CAPS — Grade [1 QUESTION 3 In the diagram, A(-2 ; 2), B(6; 5), C(0;-3) and D(x; y) are the vertices of a quadrilateral having ADIIBC. BA produced has an x-intercept at E. BD and AC intersect at M. N(¢; 0) isa pointon AD. y D(x; y) 3.1 Calculate the gradient of BC. (2) 3:2 Determine the equation of AD. (3) 3.3 Determine the value of /. (2) 3.4 Calculate the length of AN. (2) 3.5 If DC is defined by y= 3 ~3, determine the coordinates of D. (4) 3.6 Prove that ABCD is a parallelogram. (3) 3.7 Calculate the coordinates of M. @) {19] Comihsrt TNA I—58 9

Downloaded from hlayiso.com Mathematics/P2 DBE/November 2018 CAPS — Grade I QUESTION 4 In the diagram, F(2 ;-2), G(5;2) and C(6;y) are the vertices of AFGC. FGLFC. N isapointon FG such that the inclination of NC is 104,04°. The angle of inclination of FG is # and FNC = 8. » v G (5:2) fo) x F(;-2) y C(6;y) 41 Calculate the gradient of I'G. (2) 42 Calculate the value of y. (3) 43 Calculate the size of 6. G) 4.4 Calculate the length of NC. (4) Conia see AL) PSC

Mathematics/P2 Downloaded from hlayiSO.COM), re: 201s CAPS — Grade 11 QUESTION 5 5.1 5.2 5.4 In the diagram below, T(x ;15) is a point in the Cartesian plane such that OT=17 units. P(-2;a) lies on OT. K is a point on the positive x-axis and TOK =0. fy P(-2;a) |e O k * x Determine, with the aid of the diagram, the following: 5.1.1 The value of x (2) 5.1.2 tan Q) $1.3 cos(180° — A) (2) 5.1.4 sin’O 2) 5.1.5 The value of a (3) Simplify WITHOUT using a calculator: sin ee en eee (1) sin 63°.cos 540° Prove the identity: I cos0 cosO 1+sind and ©) Determine the general solution of 3sinx =2tanx (6) [28] Copa reve AA)

Mathematics/P2 CAPS ~ Grade 11 QUESTION 6 The graphs of the functions /(x)=acosb@ and g(x)=ctan@ for xe [- 180° F 180°] are sketched below. The graphs intersect at P(58°; 1,6) and Q. & PA (58° 31.6 Downloaded from HlayiSO.COM oince 01s 6.1 6.2 6.3 6.4 6.5 6.6 Copyright reserved HAAR AA AA Please nm over Write down the range of f- If M(-45° ;-1) lies on g, determine the value of c. Write down the values of a and 6. Determine the coordinates of Q. K lies on f such that KM is parallel to the y-axis. Calculate the length of KM . If the system of axes is shifted 45° to the left and the graphs remain fixed, write down the equation that is now represented by graph /. (2) (1) (2) Q) (2) (2) fy

Downloaded from hlayiso.com Mathematics/P2 DBE/November 2018 CAPS — Grade 11 QUESTION 7 TA In the diagram, p= 67°, PQ=3 cmand PR =9,2 cm. Determine the length of QR. P 67° 3em 9,2 cm Q R (3) Tea In the diagram below, DCB =a, AC=h units and ACB = 6. A 7.2.1 Determine size of ACD in terms of @ and a. (1) 7.2.2 Prove that AD = hsin(O ~ a) (4) cosa Te8 Determine the length of AD if h =17 units, @ =58° and a = 23°. (2) 7.24 Calculate the area of AADC. (3) {13] == HITE A A A Please mm over

Downloaded from hlayiso.com Mathematics/P2 DBE/November 2018 CAPS ~ Grade 11 QUESTION 8 The diagram below shows a cone with a perpendicular height of / cm, a radius of r cm and a slant height of 13 cm. Volume of cone = mh 13cm Total surface area of the cone 5 = EP ES 169xh—nh? 8.1 Show that the volume of the cone is given by a (4) 8.2 If /= 12cm, determine the total surface area of the cone. () [7] Coniston AP

Downloaded from hlayiso.com Mathematics/P2 DBE/November 2018 CAPS — Grade 11 Give reasons for your statements and calculations in QUESTIONS 9, 10 and 11. QUESTION 9 9.1 In the figure, O is the centre of the circle. A, B, C, D and E lie on the circle such that chord AB and chord DC are equal in length and AEB = 39°. 9.1.1 Determine the size of O,. (2) 9.1.2 Determine the size of 0, : (2) Commi esewes TA TTT

Mathematics/P2 Downloaded from hlayiso.com DBE/November 2018 CAPS — Grade 11 9.2 In the diagram, S is the centre of circle PQRT. PT is a diameter. RST =x-8° and PQR=2x—40°, ~ T Q <a R Determine the value of x. (4) 9.3 In the diagram, O is the centre of the circle. Chord AC is perpendicular to radius OD at B. OB = 2x units and AC = 8x units. Cc = B A D Show that the length of BD is 2x(y5-1) units, (5) [13] Copyright reserved NTU AEAAUM A Please mim over

Downloaded from hlayiso.com Mathematics/P2 14 DBE/November 2018 CAPS — Grade 11 QUESTION 10 10.1 In the diagram below, O is the centre of the circle and PTSR is a cyclic quadrilateral. P T S Prove the theorem that states that P +$ =180°. (5) Coors AN LAUT.

Downloaded from hlayiso.com Mathematics/P2 15 DBE/November 2018 CAPS — Grade 11 10.2 In the figure, QS and PR = are diameters of the circle with centre O such that PQ TSR. PS is produced to T. N_ is a point on the circle such that Q, = Q,. SN is drawn. RS intersects QN at M. 5, = 48° 10.2.1 Determine, with reasons, the size of: @ Q G3) (b) R (2) ©) M, (2) 10.2.2 Prove that ST is a tangent to the circle passing through M, N and S. (2) [14] iia III AAA Please tum over

Downloaded from hlayiso.com Mathematics/P2 16 DBE/November 2018 CAPS — Grade 11 QUESTION 11 O is the centre of the larger circle RTAP. OP is the diameter of the smaller circle PSO. NP isa tangent to both circles at P. RN L NP. Let P, =u 111 Prove that PR bisects ORN. (5) 11.2 Prove that ROS = PAT. (5) [10] TOTAL: 150 Copihesered NI 1 0

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Downloaded from hlayiso.com Mathematics/P2/Wiskerde 12 3 CAPS/KABY ~ Grade/Graad 11 - Memorandum QUESTION/FRAAG 2 50 DBE/November 2018 Cumulative frequency graph of the ages of soccer players! Kumulatiewe frekwensie -grafiek van sokkerspelers se ouderdomme : Cumilative frequeney/ Kumulatiewe frekwensie 0 5 10 1s 20 25 30 35 40 Ages of soccer players/ Ouderdomme van sokkerspelers mM 30 players/ spelers ~ DEPARTMENT OF BASIC | / answer anhvoord | PRIVATE Bao x35, PRETORIAGOOT a) 2.1.2 | 40-10 =30 players/ spelers : 2018 -11- 24 40 and/ en 10 | Yanswerl antwoord | APPROVED MARKING GUIDELINE 2) — LPUBHISEXAMINATION™ | ~ { Classinterval’ | Frequency/ Cumulative ' | Klas interval Frekwensie frequency ¥ two correct | Kumulatiewe values/ twee ! frekwensie korrelte waardes 15<x<20 4 I 4 Ww 25 5 a 1 ¥ three correct values/ | 25<x<30 i 24 37 drie korrekte waardes | | 30sx<35 | 10 47 ¥ all correct values! (35sx<40 | 3 50 | al die waardes korrek ¢ 4 CAPS/KABY — Gracle/Gruad 11 - Memorandum DBE/November 2018 INE DN. gh 214 ¢ 9 / Class interval | Class midpoint’ Frequeney/ | jo, b= 8 & Klas- interval | Klas- middelpunt __, Frekwensie | ook ~ H [1s <x<20 cas 4 i bile & El | 20<.x<25 | 22,5 9 i wie Eat 25<x<30 [27,5 24 zal S ale 305x<35 1325 10 ele sla 355x<40 [37.5 3 a |s &|2 é ela a |g & _ | | Frequency polygon of the ages of soccer players/ | Frekweusie- veethaek van sokkerspelers se oudermme ¥ using midpoint | 25 J gebruik | : middelpunte | : 20 + | “plotting the @ b \ } points correctly/ Pa korrekte punte 31s * p > : | geplot a points joined by — 10 straight line/ punte = . a] u verbind met ‘n z 5 reguitlyn a | © grounding 2t/ 9 LETT ROOD CO OTS EE Be geanker by 0 5 0 15 20 25 30 35 40 (12,5;0) and/ en Ages of soccer players/ Ouderdontme van sokkerspelers (42,5 ; 0) (4) 2.2 | The claim is not valid. / Die bewering is nie geldig nie Range of class/ Omvang van klas A= 60 Range of class/ Omvang van klas B = 40 9 | The range of class A is bigger than the range of class B. Therefore the marks of class A are more spread out than the class B./ in klas A meer verspreid as klas B At least 25% of class A have lower marks than any learner in class B/ ten minste 25% van klas A het laer punte as enige leerder in klas B. Class A performed worse at the bottom end. / | Klas A het slegter gevorder aan die onderste groep Die omvang van Klas A is groter as die omvang van klas B. Dus is die punte | ¥ claim not valid/ | bewering nie geldig nie Y comment on the overall spread/ kommentaar oor die algehele verspreiding ¥ comparison of the lower marks/ vergelyk laer punte (3) 03)

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7 CAPS/K ABV — Grade/Graad 11 —Memorandi Math ¥ simplifcation/ vereeny. |v xe value/ waarde Y y value/ wacrde | (4) V equating/ vergelyk | ABCD isa parallelogram —[oppsidesare||/ceenoorst syeis|l] OR/OF : ‘ 2oppsidesare =and|| . ABCD isa parallelogram [ echnonst Geico Downloaded from hlayiso.com : - g TOF BASIC EDUCATION 2018 -11- 21 APPROVED MARKING GUIDELINE PART MEN PUBLIC EXAMINATIO Del | [-pavare onc Xess, preTORIA coo" | “Teason/ rede | 3 wa tay 3 iy AB|/DC | reason rede Y length of AD/ lengte van AD ¥ length of BC/ fengte van BC (3) tics! PL/Wiskende V2 8 DBE/November 2018 CAPS/KABV - Grade/Graad 11 - Mer Misthemidpointof AC 1 ~ — Misdie middelpunt van AC al (2)+0 2+(-3) | 2 2 | u- 1-3) Y midpoint of AC/ | | 2 middelpunt van AC | Mis the midpoint of BD M is die middelpunt van BD | w( Giles, Cos8) 2 2 Y midpoint of BD/ u{-urs) | midkdelpunt van AC diagonals bisect each other | | “teason/ rede i 8) hoeklyne halveer mekad Misthemidpointof AC [diagonals bisect] | V Substitution into | M is diemiddelpunt van AC [hoelyne halveer| | the correct formula/ | - a Ve korrekte fo a 2)+0 246 3) | 2 2 F | (- 1-4) | “x= value / sraarde | “p-value / vwaarde @)} | [19] P DEFARTMENT OF BASIS EDUCATION PRIVATE BAG X835, PRETORIA G0! 2018 -11- 24 APPROVED MARKING GUNDELING | PUBLIC EXAMINATION

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Downloaded from hlayiso.com Mathematics/P2/Iiskunde 2 1 CAPS/KABY — Grade/Graad QUESTION/VRAAG 5 T(x315) DBE/November 2018 5.1.1 x’+yl=r? [Pythagoras] (x) +5 S17? x? =64 (P isin quadrant 2/ is in Invadrant 2) ee my | ¥ subst in pyth/ very | ¥ answer! antwoord @) S12 15 1S, un tand=— ry | ¥ answer anhvoord @) 5.1.3 | cos(i80°—4) | =-coso --(3) a answey only of Furl lances li ~cosg Mi | up ¥ answer! anhvoord | Q) S.L4 sin °@ | V substitution/ @) vervanging A ¥ answer! antwoord DEPARTMENT OF BASIC { EDUGATION PRIVATE BAG X095, PRETORIA 0007 2018 -11- 21 APPROVED MARKING GUIDELINE « sin60°.(~cos30°)(-ta "ging" = § ) 12 DBE/November 2018 CAPS/KABV — Grade/Graad 11 -Memorat dy yp r er ce Y tang=4 \ Ace -2°-8 -2 R lef a as il d ; as as Y equating/ stel gelyk on 3 A | OCF e [345/35 Vanswer! antwoord | 4 f 7) t @)| | | | j (-2) | | | v answer/ antwoord | (3) 52 ] sin 120°.cos 210°.tan 315°/Cos27° ) i oA LHS = ane = | ¥ sin 60°/cos 30° | a 2 | ¥ -cos307 A v ~tan45? AV ¥ sin63°ACos279 Be ¥ c0s!80° A go | | 7 L ¥ special angle ¢y 3 —_ ratios/ spesiale hocke _e verhoudings ‘ aN Wey ply > ¥ answer! antwoord cA Lo 1 7 V4 avy ” ASIC % OEP ATION [Craate aac evs, preTonIn cor? } 2018 -1- 24 INE | APPROVED MARKING GUIDE! ME ~

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Downloaded from hlayiso.com nde V2 DBE/November 2018 CAPS/KABV — Grade!Graad 11 —Memorandum, 3siny= 2 tanx. , sin x | 3sinx=2 x | cos x 3sin veos.x=2sinx | 3sin ycos.x-2sinx=0 in. x(3cos x-2)=0 sinx= 0 x=180°K EZ 2 | cosx= = 3 =£48,19°+360°k, keZ ¥ factors/faktore ¥ both equations/ beide vergelykings ¥ general solution/ algemene optossing | | v both general | | solutions! beide | algemene oplossings VkeZ © 28] tS OSPARTMENT OF BASIC 7 ; EDUCATION ea PRIVATE GAG X085, PRETORIA GOOT 2018 -11- 21 APPROVED MARKING GUIDELINE { PUBLIC EXAMINATION} | t ide V2 16 CAPS/KABY — Grade/Gi DBE/November 2018 QUESTION/VRAAG 6 i cae A 3 LY 2 1 M “T 3. [61 | -3S¥S3 gyop | end points eindpunte | yel-3:3] Y notation/ notasie @| | ¥ answer! antwoord | Qs Ya=3 ivb=1 i (2)) 64 | Q(122° 3-16) |v x- value/ waarde Y y- value/ waarde | Q) 65 3 ¥ coordinates off | K eas _ a | 2 " DEPARTMENT GF SxS j kodrdinate van M(~48° ;-1) Pree enucation’ | ; P PRIVATE BAG Kans, PRETORIA SOT. xm? 4, | UB 2 | | re APPROVED MARKING GUIDELINE | © length off lengte Er PUBLIC EXAMINATION van KM | | =312 | | 6.6 | f(x)=3cos(0 - 45°) | 43 } v 45° | @ | [1]

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20 DBE/Noven er 2018, Mathematics/P2/Miskunde 2 19 DBE/November 2018 CAPS/KABV ~ Grade/Graad 11 — Memorandum CAPS/KABV — Grade/Graad 1] — Memorandum | sinsge = 88 7 QUESTION/VRAAG 8 | AB =17sin58° = 2. 14,41682.... Viength of BD / | BD = 14,41682....~10,59289... = 382393... | waste wanceD ase - 382393. i SES 8 sin23° = SO | 3 i SOU ATO BASICS 2393... ° | Giz Be “length of CD/ ro) | sin 23 o = 9, 78660, lengte van CD —_ a | Areno? SADC =3xCD x ACsIN3* \ < | = 49,78660. x17 xsin35° | | | = 47,7 Luni |v answev/antu. (3) BE |? = 13°10 Pythagoras) | Yuting theorem of 3 | | eh wigick? pythagoras/ gebruik 128] | stelling van [o) | | | I prthagoras c | V==Ah — ‘" Yr? = 169-h? =o? | xe) 3 Y substitution/ (<5) + x(169—h?)h vervanging xe} “SEPARTMERT OF SKEET? i 1663 3 ¥ simplification/ ' 5: j eh — mh oO SeaeRBUSATION ASIC | —— vereenvoudig } PRIVATE BAG Xu9s, PRETORIA G00] a {eo} i ® a ; 208 -11- 24 | i c APPROVED MARKING GUIDELINE | = PUBLIC EXAMINATION | ‘BUBLIC EXAMINATION | ro} a2 r=v13'=12? (Pythagoras) | V value off waarde Q =5 j van or Total surface areal buite oppervlakte= ar? tars = a(57)+2(5)03) 90. | ¥ subst. / very. | =900 | =282,74em? | ¥ answer! antwoord i (3) i 7

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Downloaded from hlayiso.com natics/P2/ ide ¥2 23 DBE/November 2018 24 DBE/November 2018 CAPS/KABV — Grade/Graad 11 - Men CAPS/KABV — Grade/Graad 11 — Mem QUESTION/VRAAG 10 8 iE Fiz 7 Slo 4q |s Ble as Sig Ze Biz 6S c™ off ele Sle sae tele Ole 7 B2le 2 =I, alt a a5 ojo = |g =12 a te = 93 - — | | | apepenay {line fomcentre 1 to chord YSYR | lynvanmiddelpunt L aankoord OA? = (4x)? +(2x}? [Pythagoras] | | {10.1 Construction: Draw radii OR and OT | ¥ construction/ = | Y Substitution’ | Konstruksie: teken raduise OR en OT Konsiruksie vervanging | A . rvanging | | Let/ leat: §=x and/en B= y 6, =28 | angleateentre=2 times angleat cireumferenes/] | | OD=0A=2V3x fradii) | “length of OA / : middelpuntshoek = 2keer omtrekshoek S ovr | spade oa lente van OA | = x -2y - v 6, =2x | =2x(¥5-1) | | BD=2¥5 x -2x , : 5 Similarly? a die selfide manier: O, = 2y vs 6) [13] 2x+2y=360° [anlgesarounda pt/hoeke om'n punt] v SIR x+y =180° SEP TOF BASE + $4 P=180° | DEPARTMENT ON | 1 S4P=180 PRIVATE BAO X395, PRETORIAGO™ * 6 2018 -11- 21 APPROVED MARKING GUIDELINE PUBLIC EXAMINATION. moe

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Downloaded from hlayiso.com maties/PI/Wiskunde V2 27 CAPS/KABY — Grade/Graad (1 - Memorandum DBE/November 2018 Mathematies/P2/N CAPS/KABY = Grade/Gi 28 DBE/November 2018 ] | | | RPT =90° [Zin the semi circle/ in dieselfile segment] vS | 7, =x [tan -chord theorem/ raaktyn-koord st] 17 SIR [A= 90°—xfsum int Z* of A// som binne £! van dl vs | | R, =90°—~ [sum int Z* of A /som binne 2° van] us | iets , | | PRbiscets/ halveer ORN vs | { | OR/ OF | () | I 7, =x [tan—chord theorem raakiyn-koord st] | “sm | p, = x [2 opp.=sides! 2° eeoor gelvke sye | vs 3, = 2x lextZ of A/buiteZ van A] “s R=R= = [2 opp.=sides/ 2° ieeoor geiyke sve] | vg R, = 90° R, =90°—x [sum int 2 of A/ som binne 2° van 4] “S a=’, ©); | PRbiscets/ halveer ORN | BAT 20x fase ricnionta| vs vk §, =90° [Linefiom centre | tochord/fyn van mdpt Laan kon "S “® ROS=90" +x [ext Zof A /buite Z van A] YSR | ROS =PAT | | On OF | (5) ——— DEPARTMENT OF BASIC i EDUCATION | PRRATE BAS xt05, privomAaoeT | 2018 -11- 24 APPROVED MARKING GUIDELINE | PUBLIC EXAMINATION _| | a j | RPT =90° [Zin the semi circle/ in dieselfile seginent] lvs | NET =90° + x | | PAT =NBT =90° +x [tan — chord theorem /raaklyn-koord s a i | | B,=x[Z* opp.=sides / Z* teeoor gelyke sye] | 5 vw | §,=90°[Line fiom centre L to chord / fyn van mdpe Laan tee SR | | 2.0, =90°—x[sum of int 2 of A] | | | | ROS= 6, +6, | | =2x4900—4 | =90° + y ivs | on | | ROS =PAT ©) j \ | | | | | 110) r | TOTAL/TOTAAL: 150 —— DEPARTMENT OF BASIC EDUCATION Se ERESATION __ PRIVATE BAG x895, PRETORIA OO" 2018 -f1- 21 APPROVED MARKING GUIDELINE. PUBLIC EXAMINATION as

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Downloaded from hlayiso.com SPECIAL INSTRUCTIONS QUESTION 3 It has come to our attention that there is a typing error in the diagram of Question 3. The axes are not labelled correctly. This poses many possible scenarios. Scenario 1 Candidates may have read in the question that “D(x ; y)" is a vertex of the quadrilateral and that "BA produced has an x-intercept at E.". They may have then ignored the incorrect labels and proceeded to answer the question as if the horizontal axis is labelled as x and the vertical axis labelled as y. From the limited exposure that we have to candidates” scripts, it would seem like this was the most popular response to the situation. In this instance, responses must be marked in accordance with the marking guideline. Scenario 2 Candidates may have ignored “BA produced has an x-intercept at E.” and used the axes labels exactly as provided in the diagram. In this instance, the coordinates of the points in the diagram will be in the form (v;3). For example, the y-coordinate of A will be -2 and the x-coordinate of A will be 2. For this interpretation to make sense, the x and p variables in all formulae must be switched around. Please see memo below. 31] B(6;5) C(0;-3) j fo] eax | Mee Y subst into grad.form Pm¥ ar _ 3-5 /verv in grad form. | "0-6 | -8 28 ¥ answer/ anavoord | —— q @ a4 “3 | Lj? | “pEPARIMENT OF BASE DUCATION | SANATE GAG Koss, PRETORIA 000% 2018 -11- 21 APPROVED MARKING GUIDELINE 3.2 Mo = Mlge = 5 (ADIBC) fuse | 7 2==(2)+e 3 | c -yo4),14 | ar ROE 4 ,™ = 5 (ADIBC) 4 x-2=>(y—(-2)) 4 14 xetyee a" 3 47 4 woes cyte 8 Y My = wel ¥subst of m and point (-2;2) / very. nt en punt (-2;2) | Y answer/ antwoord @) 4 aD = 3 vm Ysubst of mand point (~2;2)/ very. im en put (-2;2)/ Y answet/ annvoord . (3) 33 j | | Y subst/ very. x=0 | 1% answer/ antwoord { (2) 34 ¥ subst. in distance | formula/ verv. in | afstand formule Y answer/ antwoord (2) [35 7 3 4. 14 TY equati 3, 4.4, 4 quating/ vergelyke | ee | Ysimplifcation/ a - 23 | vereenv. rv 3 | ye value/ waarde =o BEPARTMENT OF sane; | | -4¢-3)44 Ebucarion | 7 3 3 PRIVATE BAG Ke35, PRETORIA GODT” PRETORIA 0001 v x- value/ waarde D(8:-6) ; 8 208 tt 24 | @) oe MARKING GUIDELINE PUBLIC EXAMINATION SOMINATION | rat

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Downloaded from hlayiso.com In addition to Q=67°, candidates further indicated that P =67° in the diagram in the answer book. This will result in APQR being isosceles. The candidate’s response: QR =9,2 cm sides opposite equal angles must be awarded 3 marks. Scenario 3 Candidates did not read through the question carefully and only referred to the diagram in the answer book to calculate the length of QR. In this instance, the solution of QR is far more complex than the original question intended. Candidates will have to do a lot more working to arrive at the answer for QR. The first step is to calculate R , then to calculate B and finally to calculate QR. Such candidates are being prejudiced over the ones that answered the question as it intended. In order to give these candidates the same level of access to this question, only the calculation of R must be marked according to the following memo. "sin _ sink | PR PO" ¥ using the sine rule 67° _ sin | “7 ae ae Y substitution , 3 | 9,2 pert 46 Y answer | : . @)! a SENT OF BASIE RIMENT Of DEPARIDUCATION __ PRNATE BAG X#95, PRETORIA 0007 i 2018 -11- 21 APPROVED MARKING GUIDELINE | eo PUBLIC EXAMINATION wa al