SENIOR CERTIFICATE EXAMINATIONS/ SENIORSERTIFIKAAT-EKSAMEN MATHEMATICS P/WISKUNDE V JUNE 06 MEMORANDUM MARKS/PUNTE: 50 This memorandum consists of pages./ Hierdie memorandum bestaan uit bladsye.
Mathematics P/Wiskunde V DBE/06 NOTE: If a candidate answered a question TWICE, mark only the FIRST attempt. If a candidate has crossed out an attempt to answer a question and did not redo it, mark the crossed-out version. Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the second calculation error. Assuming answers/values in order to solve a problem is NOT acceptable. LET WEL: Indien 'n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na. As 'n kandidaat 'n poging om 'n vraag te beantwoord, doodgetrek en nie oorgedoen het nie, sien die doodgetrekte poging na. Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing. Staak nasien by die tweede berekeningsfout. Om antwoorde/waardes om 'n probleem op te los, te veronderstel, word NIE toegelaat NIE. QUESTION/VRAAG 8 8 0 6 9 0 4 5 6. Mean/Gemiddelde 89 7,8. Min 8, max 6 Median/Mediaan 9 Q 0, Q 4 (8 ; 0 ; 9 ; 4 ; 6). Answer only: Full marks Slegs antwoord: Volpunte 89 min, max median Q & Q ().4 The data is skewed to the left/die data is skeef na links. Negatively skewed/negatief skeef box/boks/mond whiskers/snor.5 SD/SA 6,46.6 7,8 + 6,46,64 destinations/bestemmings 8 9 4 6 0 4 8 6 interval () () () () () []
Mathematics P/Wiskunde V DBE/06 QUESTION/VRAAG Temperature at midday (in C) Middaguurtemperatuur (in C) Number of bottles of water (500 ml) Getal bottels water (500 ml) 8 9 6 5 6 40 8 0 5 5 46 5 57 70 6 5. (0 ; 5) answer. a 8,5 b,68 ŷ,68x 8,5. ŷ 6,5 bottles ŷ,68(8) 8,5 6,5 bottles.4 Strong/Sterk The majority of the points lie close to the regression line./die meerderheid punte lê naby die regressielyn. value a value b equation substitution strong/sterk reason/rede () () () () Strong/Sterk r 0,98.5 Temperature cannot rise beyond a certain point as this would be life threatening OR there is only so much water one can consume before it becomes a risk to your health (hyponatremia)./temperatuur kan nie hoёr as 'n sekere punt styg nie, anders raak dit lewensgevaarlik. OF n persoon kan net n sekere hoeveelheid water inneem, anders raak dit n gesondheidsrisiko strong/sterk reason/rede reason/rede () () [9]
Mathematics P/Wiskunde V 4 DBE/06 QUESTION/VRAAG y B(8 ; 0) A( 8 ; 6) T C F 0,54 D( ; 0) O 0 M x.. y y mad x x 0 6 + 8 6 6 mbc y x + c 0 8 + c c 8 y x + 8 mbc y y m( x x ) y 0 ( x 8) y x + 8 [BC AD] [BC AD] substitution gradient () substitute m and (8 ; 0) equation gradient substitute m and (8 ; 0) equation
Mathematics P/Wiskunde V 5 DBE/06. m m BD BD y x 0 0 8 + m AD y x DB AD substitution m BD m AD.4 OR AD 7 or AD 7 or 6 AB 7 or AB 7 or 4 7 BD 00 or BD 00 or 0 AB AD + BD A Dˆ B 90 [converse Pyth th/ omgekeerde Pyth st] tan BDˆ M mbd BDˆ M 45 calculating all sides AB AD + BD tan BDˆ M m BD () OR sin BDˆ M BDˆ M 45 BM BD 0 0 s in BDˆ M ().5 x + x y + y T( x ; y) ; + 8 0 + 0 ; ( ; 5) T symmetrical about BM/T is simmetries om BM distance of T to BM 5 units distance from BM to C C( ; 5) T( ; 5) value of x value of y
Mathematics P/Wiskunde V 6 DBE/06.6 0 mdf 8 ( ) Equation of DF: y y m( x x) y 0 ( x + ) y x + Equation of BC: y x + 8 x + x + 8 4x 5 x y + 8 5 C( ; 5) area/ opp BDF area/ opp BDM area/ opp DFM 0 ( 0)( 0) (0) 00 or or, square units/ vk eenh area/ opp BDF.BF.DM 0 ( 0) 00 or or, square units/ vk eenh eq of DF value of x value of y formula/method 0 (DM) 0 (BM) 0 or ( h) (5) formula/method BF DM (5)
Mathematics P/Wiskunde V 7 DBE/06 tan FDˆ M m DC 5 0 + or tan FDˆ M FM DM 0 0 gradient/ratio FDˆ M BFˆD 08,4 0 BF or 6 DF 8,4 (0) DF 0,54 + or 000 [Pyth DFM] 0 0 or [ext ] area/ opp BDF.BF.FD.sinBFˆD 0 0 0 (sin08,4) 00 or or, square units/ vk eenh B Fˆ D DF correct substitution into area rule (5) 0 BF or 6 BD (0 0) + (8 + ) 00 or0 area/ opp BDF.BF.BD.sinDBˆ F 0 00 (sin 45 ) 00 or or, square units/ vk eenh BF BD formula/method correct substitution into area rule (5) area/opp BDF area/opp BCD area/opp BCF 0 ( 0 )( 5 ) ( 5) 00 or or, square units/vk eenh formula/method BD 0 BC 5 0 BF (5)
Mathematics P/Wiskunde V 8 DBE/06 tan FDˆ M FDˆ M BDˆ F m 8,4 6,56 area / opp BDF.BD.DF.sin BDˆ ( ). DC 5 0 + 0 0.0.sin 6,56 00 or or, square units/ vk F or eenh tan FDˆ M FM DM 0 0 gradient/ratio B Dˆ F DF BD correct substitution into area rule (5) [8]
Mathematics P/Wiskunde V 9 DBE/06 QUESTION/VRAAG 4 y R(k ; ) T O C( ; ) x P S 4. radius tangent /raaklyn R 4. 4. CR CR CR CR TR 0 500 ( x 500 ( k ) k + CT + 0 500 or 0 x k 6k 7 0 ( k 7)( k + ) 0 k 7 or k 5 + ( y + ( + ) 6k + 9 + 484 500 ) y ) (Pyth) substitution substitution () () standard form factors k 7 (4) CR 500 ( k ) ( k ) ( x 6 x ) + ( y + (+ ) k 4 or k 4 k 7 or k y ) substitution square form square root k 7 (4)
Mathematics P/Wiskunde V 0 DBE/06 4.4 ( x ) + ( y + ) 00 4.5 CS 0 and CS PS S(; ) y 4.6. S(; ) ( ) 4x 5 x 7 P( 7; ) 4 5 x + 4 68 x x 7 P( 7; ) 4.6. PT PS [tangents from common point/rklyne vanaf dies pt] 7 + 0 units OR (; ) substituting equating R () () () () PC ( 7 ) + ( + ) 500 or 0 5 PT PC TC [Pyth th] 500 00 400 PT 0 value of PC using Pyth OR PC ( 7 ) + ( + ) 500 or 0 5 PTC RTC [90 HS] PT TR PT 0 value of PC /R or proved 4.7. M(; 6) answer ()
Mathematics P/Wiskunde V DBE/06 4.7. Radius 4 4.7. r + r distance CM 0 + 4 4 ( ) 5 5 + ( + 6) r + r 5 () CM > r + r Therefore the two circles do not intersect or touch./daarom sny of raak die twee sirkels nie. explanation []
Mathematics P/Wiskunde V DBE/06 QUESTION/VRAAG 5 5. P T R S 5..(a) 5 sin T 0, 45 value 5 5 () 5..(b) 0 cos S 0, 95 value 0 0 () 5.. cos(t + S) cos T coss sin T sin S expansion 5 0 5 0 5 0 6 simplification 50 50 5 or or 50 (5) 5. tan (80 + θ ) cos(60 θ )sin(90 θ ) cos θ tan θ cos θ (cosθ )(cosθ ) tan θ sin θ sin θ cos θ cos θ cos θ sin θ cos θ identity cos θ sin θ OR cos θ sin θ (6)
Mathematics P/Wiskunde V DBE/06 5. (sin x cos x) sin x sin x cos x + cos sin x cos x sin x cos x sin x 7 6 4 7 6 9 6 x 9 6 squaring both sides expanding LHS using identity simplifying answer (5) [8]
Mathematics P/Wiskunde V 4 DBE/06 QUESTION/VRAAG 6 6. 4sin x + cos x sin x + cos x 0 sin x + ( sin x) 0 sin x sin x 0 sin x(sin x ) 0 sin x 0 or sin x 0 sin x 0 sin x using identity standard form factors sin x 0 or sin x x k.80 or x 90 + k.60, k Z k. 80 90 + k.60, k Z (6) 6.. turning point ( 90 ; ) turning point g (90 ; ) ( 80 ; ) & 80 o 90 o O 90 o 80 o (0 ; ) f 6.. ( 90 ;0 ) 90 < x < 0 6.. f ( x) g( x) 80 ;0 ;90 ;80 f ( x + 0 ) g( x + 0 ) x 0 ;60 ;50 () () any ONE correct other correct () []
Mathematics P/Wiskunde V 5 DBE/06 QUESTION/VRAAG 7 θ A 8 87 64 D B C 7. 7. ABˆ D θ [alternate s; lines] BD 64 cosθ AB 8 θ 8 64 sin BÂD 8 BÂD 5,8 θ 8 correct trig ratio substitution into correct ratio (to the nearest degree) correct trig ratio substitution into correct ratio (to the nearest degree) BC AB + AC (AB)(AC) cos BÂC use cosine rule correct substitution 8 + 87 (8)(87) cos8,6 into cosine rule 4,754... BC 0,97 m answer
Mathematics P/Wiskunde V 6 DBE/06 7. sind ĈB sinbdˆ C BD BC BD.sinBDˆ C sindĉb BC 64.sin0 sin DĈB 0,97 DĈB,8 use sine rule substitution [9]
Mathematics P/Wiskunde V 7 DBE/06 QUESTION/VRAAG 8 8. P M 8 O 48 T R 4 o S 8.. Pˆ [opp s of cyclic quad/ teenoorst e v koordevh] R 8.. ( ) 64 [ centre at circum/ midpts omtreks ] Ô R () 8.. 8..4 reflex Ô 96 [ centre at circum/ midpts omtreks ] Ô 64 [ s around a point/ e om ' n punt] OMˆ S 80 ( + 8 + 4 ) [sum s / som e ] 87 Rˆ TMˆ P 87 + 8 6 99 [ext cyclicquad/ buite koordevh] and R R () () () ()
Mathematics P/Wiskunde V 8 DBE/06 8. A B D E x C 8.. corres s/ ooreenk e; AB DC R () 8.. Ê x [tan - chord theorem/ raakl koordst] S R Bˆ x [ s opp sides/ e teenoor sye] Any s correct R Ê x [alt s/ verwiss e; AB DC] R D ÂB x /teenoor e m alternate/verwiss s/e; BC AD] (6) 8.. Dˆ 80 x [co - int s suppl/ ko binne e suppl; AD BC] S R Bˆ + Dˆ 80 ABED a cyc quad/ kdvh [converse opp s of cyclic quad/ omgek teenoorst e koordevh] R DÂB x Ê DÂB x m [opp s/ teenoor e ] OR/ OF [alt s/ verwiss e;bc AD] S R ABED a cyc quad/ kdvh [converse ext of cyc quad/ omgek buite v koordevh] R [8]
Mathematics P/Wiskunde V 9 DBE/06 QUESTION/VRAAG 9 9. in the alternate segment/ in die( teen)oorstaande segment () 9. A D B 4 C G F 9.. 9.. 9.. Â Bˆ 4 Dˆ Â Dˆ + Dˆ + Dˆ [tan chord theorem/ raakl koordst] [ext Δ/ buite ] S R R Bˆ 4 Bˆ [vert opp s/ regoorst e] Dˆ + Dˆ Bˆ [proven/ bewys] Ĝ [ s in same segment/ ein dies segment] R R AGCD is cyc quad/ kvh [converse ext cyc quad/ omgek buite kvh] Dˆ Â [ s in same segment/ e in dies segment ] R Â Fˆ [ s in same segment/ e in dies segment ] Dˆ Fˆ DC CF [sides opp s/ sye teenoor e] R (4) (4) (4) []
Mathematics P/Wiskunde V 0 DBE/06 QUESTION/VRAAG 0 M A B C K F V T 0. Constr/ Konstr : Draw line BC such that MB AK and MC AF Trek lyn BC sodat MB AK en MC AF Proof/ Bewys : In BMC and/ en KAF MB AK [constr/ konstr] Mˆ Â [given/ gegee] MC AF [constr/ konstr] BMC KAF [s s] MBˆ C AKˆ F or MĈB AFˆK [ ] but / maar Vˆ Kˆ or Tˆ Fˆ [given/ gegee] MBˆ C Vˆ or MĈB Tˆ But these are corresponding s/ maar hulleis ooreenk e BC [corr s / ooreenk e ] MV MB but / maar MV AK VT MT MC MB MT AF [prop theorem/ eweredighst; AK and MC AF [constr/ konstr] BC VT] constr/konstr / R / R S R (7)
Mathematics P/Wiskunde V DBE/06 0. E M H F G K 0..(a) 0..(b) 0..(c) 0.. In KGH and KEF Kˆ is common/ gemeen Ĥ Fˆ [ext cyclicquad/ buite koordevh] Ĝ Ê [sum s OR ext cyclicquad/ som e OR buite koordevh] KGH KEF [ ] S R naming third angle OR (4) EF KE GH KG [ s] EF KE GH EF [KG EF] EF KE.GH () KG EM KF EF [prop theorem/ eweredighst; MG EK] R but EF KG [given/ gegee] KG EM KF KG KG EM.KF KE.GH EM.KF 0 4 KE.GH EM.KF EM 6 substitution 5 units [9] TOTAL/TOTAAL: 50