JUNIE/ JUNE P/ V015 MARKING/ NASIEN MEMORANDUM NOTE: If a candidate answers a question TWICE, only mark the FIRST attempt. If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version. Consistent accuracy applies in ALL aspects of the marking memorandum. Assuming answers/values in order to solve a problem is NOT acceptable.. NOTA: As 'n kandidaat 'n vraag TWEEKEER beantwoord, merk slegs die EERSTE poging. As 'n kandidaat 'n poging om die vraag te beantwoord, doodgetrek het en nie dit oorgedoen het nie, merk die doodgetrekte poging. Volgehoue akkuraatheid word in ALLE aspekte van die nasienmemorandum toegepas. Aanvaarding van antwoorde/waardes om 'n probleem op te los, is ONaanvaarbaar. QUESTION/ VRAAG 1 Q# SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/ BESKRYWERS 1.1 Hourly earnings/ Uurlikse verdienste Midpoint of interval/ Middelpunte van interval ( x ) Frequency/ Frekwensie ( f ) 9,70 x < 9,90 9,80 5 9,90 x < 10,10 10,00 16 10,10 x < 10,30 10,20 25 10,30 x < 10,50 10,40 30 10,50 x < 10,70 10,60 24 midpoints of intervals/ middelpte van intervalle Answer/ Antw If upper/lower limits used -1 Mean/ Gemiddelde = R10,30 1.2 Standard deviation/ standaardafwyking = 0,23 Answer/ Antw 1.3 1.3 Yes, she is correct. The difference in the mean between men and women is only 5 cents and the difference between the standard deviation is 2 cents / Ja, sy is reg. Die verskil in die gemiddelde tussen mans en vroue is slegs 5 sent and die verskil tussen die standaardafwyking is 2 sent. Yes/ Ja Reason/ Rede If NO, reason must be that there s a difference between mean and standard deviation between men and women./ Indien NEE, rede moet wees dat daar n verskil tussen gemiddelde en standaardafwyking tussen mans en vroue is. [7]
QUESTION/ VRAAG 2 2.1.1 Thandi: correct min/ max/ Cindy korrekte min/maks. Q1 = 4 median/ mediaan = 7 Thandi Q2 = 9 box and whisker diagram/ mond-en-snordiagram 2.1.2 Cindy must get award/ Cindy moet toekenning kry. Scored 9 or more in 50% of the matches while Thandi only did so in 25% of the matches./ Het 9 of meer in 50% van die wedstryde aangeteken, terwyl Thandi dit in 25% van die wedstryde gedoen het. OR/OF The median number of goals scored by Cindy is higher than that of Thandi./Die mediaan aantal doele van Cindy is hoër as die van Thandi. Cindy Reason/ Rede (Must be related to box and whisker diagram/ Moet verwys na die mond- en- snordiagram) 2.2.1 55 Answer/ Antw (1) 2.2.2 Cum. Freq./ Kum Frekw.: AGES/ OUDERDOM FREQUENCY/ FREKWENSIE ME CUMULATIVE FREQUENCY/ KUMULATIEWE FREKWENSIE 18 x < 23 4 4 23 x < 28 8 12 28 x < 33 13 25 33 x < 38 15 40 38 x < 43 10 50 43 x < 48 4 54 48 x < 53 1 55 2.2.3 Voters/ Kiesers 35 years or older/ jaar of ouer: 55 31 = 24 1 st 4 values correct/ 1ste 4 waardes korrek remaining 3 correct/ oorblywende 3 korrek Frequency/ Frekwensie: (5) 1 st 4 values correct/ 1ste 4 waardes korrek remaining 3 correct/ oorblywende 3 korrek Answer/ Antw (1) [13]
QUESTION/ VRAAG 3 3.1 AF = (x 2 x 1 ) 2 + (y 2 y) 2 3.2 3.3 = (3 2) 2 + ( 2 + 3) 2 = 1 + 1 = 2 m BC = y 2 y 1 x 2 x 1 = 3 + 1 p + 4 4 = 1 p+4 2 p + 4 = 8 p = 4 T ( 4 + 2 ; 1 3 ) T(-1; -2) 3.4 tan θ = m BC tan θ = 1 2 θ = 26,6 3.5 m AB = 3+3 4 2 = 3 m BD = 3 0 4 3 = 3 m AB = m BD A; B and/ en D are collinear/ is kollineêr 3.6 y = 2x + 6 subst./ vervang:x = 4: y = 2( 4) + 6 = -2 1 C(-4; -1) does not lie on/ lê nie op y = 2x + 6 nie OR/ OF subst./ vervang:y = 1: 1 = 2x + 6 x -3,5 2 C(-4; -1) does not lie on/ lê nie op y = 2x + 6 nie Distance formule/ Afstand formule Correct subst./ Korrekte vervanging Answer/ Antw. Using gradient formula /Gebruik van gradiënt formule Correct subst./ Korrekte vervanging Equate to 1 / stel = 1 / Answer/ Antw. Correct subst./ Korrekte vervanging -1-2 tan θ = m BC tan θ = 1 2 26,6 m AB = 3 m BD = 3 m AB = m BD collinear/ kollineêr subst./ vervang x = 4-2 C(-4; -1) does not lie on/ lê nie op y = 2x + 6 nie subst./ vervang y = 1-3,5 C(-4; -1) does not lie on/ lê nie op y = 2x + 6 nie. [20]
QUESTION/ VRAAG 4 4.1 M ( 4 2 ; 7 + 3 ) M(1; -2) 4.2 radius: HM = (3 + 2) 2 + ( 2 1) 2 = 34 (x 1) 2 + (y + 2) 2 = 34 4.3 4.4 m HM = 3 + 2 2 1 = 5 3 m tan = 3 5 eqn/ vgl: y y 1 = 3 (x x 5 1) OR/ OF y = 3 x + c 5 subst./ verv (-2; 3): y 3 = 3 (x + 2) 3 5 y = 3 21 x + 5 5 OR/ OF 5y = 3x + 21 tan θ = 5 3 Incl. HG: = 180-59 = 121 HĴG = 90 JĜH = 121-90 = 31 = 3 ( 2) + c 5 4 1 5 = 21 5 = c 1 2 (x 1) 2 (y + 2) 2 34 m HM = 5 3 m tan = 3 5 eqn of straight line/ vgl van reguitlyn subst./ vervang (-2; 3) y = 3 21 x + 5 5 tan θ = 5 3 121 HĴG = 90 4.5 (x + 2) 2 + (y + 2) 2 = 34 (x + 2) 2 (6) 31 (y + 2) 2 and/ en 34 [17] QUESTION/ VRAAG 5 5.1.1 c 2 = 25 2 24 2 Pyth = 49 c = -7 5.1.2 (a) sin θ = 24 25 5.2.2 tan(180 + θ) = tan θ = 24 7 = 24 7 = -33 7 5.2.1 cos 63 = sin 27 = k 1 63 k Pythagoras Answer/ Antw. Answer/ Antw. tan θ Answer/ Antw. sketch/ skets k (1) 27 1 k 2 5.2.2 tan 153 = tan(180 27 ) = - tan 27 = k 1 k 2 tan 27 Answer/ Antw
5.2.3 sin 126 OR/ OF sin 126 = sin 54 = sin(2 63 ) = sin(2 27 ) = 2sin 63 cos 63 = 2 sin 27 cos 27 = 2( 1 k2 ). k k = 2 1 k 2 5.3 cos 2θ = 2 cos 2 θ 1 7m = 2(2m) 2 1 = 8m 2 1 8m 2 7m 1 = 0 (8m + 1)(m 1) = 0 m = 1 8 or 1 Only m = 1 8 ( 1 cos 2x 1) ( 1 cos θ 1) sin(2 63 ) OR/ OF sin(2 27 ) expansion/ uitbreiding Answer/ Antw cos 2θ = 2 cos 2 θ 1 substitution/vervanging 8m 2 7m 1 = 0 m = 1 8 m = 1 8 or/ of 1 (5) [17] QUESTION/ VRAAG 6 6.1. (2 sin 22,5 cos 22,5 ) 2 cos 150 = [sin(2 22,5 )]2 cos(180 30 ) = sin 2 45 cos 30 = ( 1 2 ) 2 = 1 3 3 2 6.2. LK = cos A[cos( A) + sin(180 + A). tan A] sin A = cos A [cos A sin A. cos A ] = cos A [cos A sin2 A cos A ] = cos 2 A sin 2 A = cos 2A = RK sin 2 45 -cos 30 subst. / vervang Answer/ Antw cos( A) = cos A sin( 180 + A) = sin A tan A = sin A cos A cos 2 A sin 2 A (5) 6.3.1 PQ sin 70 = PQ = PR PR.sin 70 cos 2A PRQ=70 PQ = (5) PR.sin 70 6.3.2 110 110 (1) 6.3.3 PB sin 35 = PQ sin 110 PB = PQ. sin 35 sin 110 correct sine ratio/ Korrekte sin-verh. PB = PR.sin 70.sin 35 sin 70.
PR. sin 70. sin 35 = sin 70. PB = PR sin 35 6.3.4 PB = = PR sin 35 54,5 sin 35 = 46,72 m Answer/ Antw. (1) [17] QUESTION/ VRAAG 7 7.1 180 Answer/ Antw. (1) 7.2 cos 2x = sin x 1 2 sin 2 x = sin x 2 sin 2 x sin x 1 = 0 (2 sin x + 1)(sin x 1) = 0 sin x = 1 2 or sin x = 1 1 2 sin 2 x st-form/std vorm sin x = 1 or sin x = 1 2 x = 30 or 90 x = 30 or 90 7.3 y- intercepts/ afsnitte x- intercepts/ afsnitte min/max points Min/maks punte 7.4.1 90 x < 45 Each/ elke interval 0 < x < 45 OR/ OF x [ 90 ; 45 ) (0; 45 ) 7.4.2 x = 90 Answer/ Antw (1) [11]
EUCLIDEAN GEOMETRY/ EUKLIDIESE MEETKUNDE S Statement/ Bewering R Reason/ Rede S/R Statement + Reason/ Bewering + Rede QUESTION/ VRAAG 8 8.1.1 perpendicular / loodreg Answer/ Antw (1) 8.1.2 equal to the angle in the opposite segment subtended by the chord / equal to the angle (1) gelyk aan die hoek in die teenoorgestelde segment onderspan deur die koord in the opposite segment/ gelyk aan die hoek in die teenoorgestelde segment onderspan 8.1.3 supplementary / supplementêr Answer/ Antw (1) 8.2.1 Ĉ 1 = 90 ( in semi circle/ in halfsirkel S R 8.2.2 B 1 = 60 (alt./ verw. e; AB//CD) S R 8.2.3 Ê = 120 (opp. s of cycl. quad/ teenoorst. e van kvh) S R 8.2.4 D 1 = 30 ( s in or isosc. / e in of gelykb. ) S R 8.3.1 2 In AOX, AX AO OX ( Pythagoras) Pythagoras Subst./ vervang (41) (9) 1600 In AOX or/of COX AX 40cm In COX CX OC OX Pythagoras 2, ( ) (15) (9) 144 XC 12cm AC AX XC (40 12) cm AC 52cm 8.3.2 AX = XD (line from centre to chord/ lyn uit midpt.ʘ na koord) Also/Netso BX = XC (line from centre to chord/ lyn uit midpt.ʘ na koord) AX BX = XD XC AB = CD OR/ OF AX XD 40 cm ( drawn from centreto chord ) Also BX XC 12 cm but CD XD XC (40 12) cm 28cm Also AB AX BX (40 12) cm 28cm AB CD 28cm AX =40 cm XC= 12 cm AC =52 cm (5) S/R AX BX= XD XC OR/ OF AX = 40 cm R CD = 28 cm AB = 28 cm [20]
QUESTION/ VRAAG 9 9.1 A 2 1 Consider other proofs as well/ Oorweeg ander bewyse ook Constr./ Konstr.: Draw/Trek BOA and connect/ en verbind AD Proof/ Bewys: B B (Radius tangent/ raakl.) 1 2 90 B 1 = 90 0 B 2 ADB= 90 0 ( in a semi circle/ in halfsirkel) A = 90 0 B 2 ( s in / e in ) E = A ( s in the same segment/ e in dies. segment) DB C = E 9.2.1 F 1 = A ( AB = BF or/ of isosc./ gelykb. ) A = C 2 (exterior of a cyclic quad/ buite van kvh) = x constr/ konstr. S/R S/R (6) 9.2.2 F 3 = B 2 (tan chord thm/ raakl- koordstelling) F 3 + F 4 = B 2 + B 1 (ext. of cyclic quad/ buite van kvh) F 4 = B 1 = 180-2x 9.2.3 F 1 + F 2 = C 2 + E (exterior of FCE/ buite van ) But F 1 = C 2 = x (from/ vanaf 9.2.1) F 2 = E BF is a tangent ( between line and chord / tussen lyn en koord OR/OF converse tan-chord thm/ omgekeerde raakl- koordstelling) Answer/ Antw. (5) S R [19] QUESTION/ VRAAG 10 10.1 QAT = 90 (radius tangent/raakl.) and/ en QBT = 90 AOBT is a cyclic quad / is n kvh (radius tangent/raakl.) (opp/ teenoorst. s is suppl) S/R 10.2 A 1 ( TAB) = C (tan- chord thm/ raakl-koordstelling) = K 1 ( TKB) (corresp./ ooreenk. s; CA PT) 10.3 A 1 = K 1 proved(from 10.1.2) AKBT is a cyclic quad (= s on BT/ = e op BT R (1) [9] Total/Totaal: [150] R