NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12

Transcription

1 NATIONAL SENI CERTIFICATE NASIONALE SENI SERTIFIKAAT GRADE/GRAAD MATHEMATICS P/WISKUNDE V NOVEMBER 0 MEMANDUM MARKS/PUNTE: 50 This memorandum consists of 3 pages. Hierdie memorandum bestaan uit 3 bladsye.

2 Mathematics P/Wiskunde V DBE/November 03 NOTE: If a candidate answered a question TWICE, mark only the first attempt. If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out question. Consistent accuracy applies in ALL aspects of the marking memorandum. Assuming values/answers in order to solve a problem is unacceptable. LET WEL: As 'n kandidaat 'n vraag TWEE keer beantwoord het, merk slegs die eerste poging. As 'n kandidaat 'n antwoord deurgehaal en nie oorgedoen het nie, merk die deurgehaalde antwoord. Volgehoue akkuraatheid is DEURGAANS in ALLE aspekte van die memorandum van toepassing. Aanvaarding van waardes/antwoorde om 'n problem op te los, is onaanvaarbaar. QUESTION/VRAAG Interquartile range/interkwartielvariasiewydte 53 4 Answer only: Full marks No CA critical values (4 ; 53) (). 5% of trees have a height in excess of 53 cm. 5% van bome het n hoogte van meer as 53 cm..3 Between Q (50) and Q 3 (53)/Tussen Q en Q 3 REASON / REDE The distance between these two quartiles is the smallest/die afstand tussen hierdie twee kwartiele is die kleinste. The third quarter has smallest length / Die derde kwart het die kortste lengte 5% Q and Q 3 reason () () [6]

3 Mathematics P/Wiskunde V 3 DBE/November 03 QUESTION/VRAAG. and/en points correctly 90 plotted (3 marks) Relative risk of having accident Blood alcohol level % 4 points correctly plotted ( marks) points correctly plotted ( mark) exponential curve/ eksponensiale kromme ().3 The trend shows that as the blood alcohol levels increase, the risk of having an accident increases rapidly. Die tendens(neiging) toon dat indien die bloed-alkoholvlakke toeneem, die risiko van n motorongeluk neem vinnig toe. reason.4 Approximately 47% (Accept 44% - 5%) 47% () () [7] QUESTION/VRAAG 3 3. more than 5 minutes: people Approximately 36 people Answer only: Full marks (Accept 34 37) 3. At 8 minutes approximately 7 people and at minutes approximately 6 people left the auditorium/by 8 minute het ongeveer 7 mense en by minute ongeveer 6 mense die ouditorium verlaat Approximately 35 people left the auditorium between 8 and minutes/ongeveer 35 mense het tussen 8 en minute die ouditorium verlaat. 3.3 Modal class/modale klas: < x 6 x < 6 Mark for critical values and 6 35 (Accept 33 36) < x 6 x < 6 () () () [5]

4 Mathematics P/Wiskunde V 4 DBE/November 03 QUESTION/VRAAG 4 SCHOOL A SCHOOL B SCHOOL C Mean 9,8 9,8 4,8 Standard deviation,3 3,,3 4. School B, because the standard deviation of B is the largest. Skool B, want die standaardafwyking is die grootste. 4. There is no difference in the spread of the marks. Daar is geen verskil in die verspreiding van punte nie. 4.3 Add/increase each score in School A by 5 marks. Vermeerder(tel by) elke punt in Skool A met 5 punte. 4.4 The mean will decrease ( by 0% ) Die gemiddelde sal verminder (met 0%). The standard deviation will also decrease (by 0% ) Die standaardafwyking sal verminder (ook met 0%). School B reason no difference / the same () () increase each mark vermeerder elke punt 5 marks () mean decreased /gemiddeld verminder SD decreased/ SD verminder () [7] Explanation why values decrease by 0%: mean 0,9x n i 0,9 n x i 0,9x SD ( x x) i (0,9x 0,9x) i n 0,9 ( x x) n i 0,9 ( x x) i n

5 Mathematics P/Wiskunde V 5 DBE/November 03 QUESTION/VRAAG 5 y T S 63,43 P( 5 ; 0) O R x 5.. m tan 63,43 tan 63,43 PT 5.. Coordinates of P( 5 ; 0) y y m( x x ) y 0 ( x + 5) c 0 y x + 0 y mx + c 0 ()( 5) + c y x + 0 m PT tan 63,43 tan 63,43 OS 0 y x + 0 OS OP OS 5 Answer only: full marks () substitution of P( 5 ; 0) and m into equation equation () substitution of P( 5 ; 0) and m into equation equation () OS 5 equation ()

6 Mathematics P/Wiskunde V 6 DBE/November OS 0 units PS 5 + (0) PS P( 5 ; 0) ; OS 0 units PS ( 5 0) + (0 0) Accept PS, PS Accept PS,8 PS 5 cos 63,43 5 PS cos 63,43 PS,8 OS 0 substitution of correct distances into Pythagoras 5 OS 0 substitution of correct distances into Pythagoras 5 ratio 5 PS cos 63,43,8 PS 0 sin 63,43 0 PS sin 63,43 PS, Let T be (x ; y). Then 5 + x 0 + y 0 and x 5 y T(5 ; 0) by inspection: T(5 ; 0) ,5 5 R ; If only x-coordinate : marks ratio 0 PS sin 63,43, x 7,5 / y 0 () () ()

7 Mathematics P/Wiskunde V 7 DBE/November Area ΔPTR (base PR) (height ) 5 (5 + ) 0 5 square units area formula 5 5 +,5 0 5 Area PTR PT.PR.sin TPˆR 5 ( 0 5) sin 63,43 4,99 square units area formula ,99 [5]

8 Mathematics P/Wiskunde V 8 DBE/November 03 QUESTION/VRAAG 6 y O M x R(0 ; 4) N(p ; q) P 6. x x + y 6x y ( x 3) M(3 ; ) 6x + y y ( y + ) 8 If only ( x 3) + ( y + ) r ( r 8), then marks x 6x + 9 y + y + (x 3) (y + ) xm (coefficient of x) xm ( 6 ) xm 3 ym (coefficient of y) ym ( ) ym M(3 ; ) x M ( 6 ) x 3 M y M ( ) y M

9 Mathematics P/Wiskunde V 9 DBE/November ( 4) m RM 3 0 y-intercept is 4 y x 4 MR RP (radius tangent/ raaklyn) mmn mpr q ( ) p 3 p + 3 q + q p substitution into gradient formula m RM equation m MN substitution into gradient formula p + 3 q + MR RP (radius tangent/ raaklyn) mmn mpr y ( ) ( x 3) y + x + 3 y x + q p m MN y x + substitution into equation of line 6.4 ( x 3) + ( y + ) 8 ( p 3) + ( q + ) 8 ( q 3) + ( q + ) 8 q + q + + q + q q + 4q 6 0 method q + q 8 0 ( q + 4)( q ) 0 q 4 or p 6 q q 4 p 6 MRPN is a square/vierkant (rectangle with/reghoek met MN MR) M PˆN 45 But MR has a slope/gradient of, so RN x-axis q 4 and p ( 4) 6 method q 4 p 6

10 Mathematics P/Wiskunde V 0 DBE/November 03 q p ( p 3) + ( p + ) ( p 3) 8 9 p 3 3 p 6 q 4 ( p > 0) Using symmetry: q 4 (since y M y R ) 4 p p 6 p 3 (since x M x N ) method p 6 q 4 method q p 6.5 r² (6)² + ( 4)² x² + y² 5 substitution equation () 6.6 p² + q² (6)² + ( 4)² x² + y² p² + q² 5 area of circle M π r π ( 8) 8π square units 56,55square units 6.7 MRPN is a square (all angles equals 90, adj sides equal) N Mˆ P 45 (diagonals of a square bisect the angles/hoeklyne van vierkant halveer hoeke) NP sin NMP ˆ MP sin 45 or substitution equation r 8 area of circle N Mˆ P 45 NP MP sin NMˆ P () () MRPN is a square (all angles equals 90, adj sides equal) MP MP 6 NP MP 8 6 or MN 8 MP 36 6

11 Mathematics P/Wiskunde V DBE/November 03 By inspection: P(3 ; 7) P(3 ; 7) NP MP (6 3) + (4 7) (3 3) + ( 7 + ) 8 6 or NP 8 MP 6 [4]

12 Mathematics P/Wiskunde V DBE/November 03 QUESTION/VRAAG 7 7. ( x ; y) ( y ; x) ( y ; x) (y ; x) ( y ; x) () 7. ( x ; y) ( x ; y) ( y ; x) ( x ; y) ( y ; x) () 7.3 Mo s claim is correct/mo se bewering is korrek. If order was unimportant then the image of P would be the same in Mo reason both cases. This is not so./as volgorde onbelangrik is, sal die beeld () van P in beide gevalle dieselfde wees. Wat nie so is nie. Choose any point (0 ; 0) and show that their images both cases are not the same. For example: ( 3 ; 4) (4 ; 3) ( 4 ; 3) ( 3 ; 4) ( 3 ; 4) (4 ; 3) Mo is correct calculation of both images Mo () [6] QUESTION/VRAAG 8 8. ABC is translated by 4 units to the left and 4 units up/ ABC word getransleer met 4 eenhede na links en 4 eenhede opwaarts. Accept ( x ; y) ( x 4; y + 4) 8. R / (3 ; 4) Area A / B / C / 6 Area of ABC Scale factor/skaalfaktor AC 0 A / C / EF AC C ( s 0) + ( t ) 0 s s s s + ( t ) 0 + ( t ) 0 + t t t t 9 0 A 0 F(0 ; ) 0 E(s ; t) translation/ translasie 4 left/links and 4 up/opwaarts () recognising that EF AC equation in terms of s and t () () () [0]

13 Mathematics P/Wiskunde V 3 DBE/November 03 QUESTION/VRAAG 9 9. Anti-clockwise / Anti-kloksgewys: 6 6 cos( θ ) sin( θ ) 8...() 6 6 cos( θ ) sin( θ ) () 3 () +(): sin( θ ) sin( θ ) sinθ 0, θ or θ cos( θ ) sin( θ ) 8...() 6 6 cos( θ ) sin( θ ) () 3 () (): cos( θ ) cosθ 0, θ or θ Clockwise /Kloksgewys: 6 6 cos( θ ) + sin( θ ) 8...() 6 6 cos( θ ) + sin( θ ) () () +(): 3 sin( θ ) sin( θ ) sinθ 0, θ or θ substitution into x image of rotation substitution into y image of rotation addition of equations value of sin θ substitution into x image of rotation substitution into y image of rotation subtraction of equations value of cos θ substitution into x image of rotation substitution into y image of rotation addition of equations value of sin θ

14 Mathematics P/Wiskunde V 4 DBE/November cos( θ ) + sin( θ ) 8...() 6 6 cos( θ ) + sin( θ ) () 3 () (): cos( θ ) cosθ 0, θ or θ substitution into x image of rotation substitution into y image of rotation subtraction of equations value of cos θ tanα 6 α 35 T α β W tan α tan β 3 8 β 60 θ tan β T tan α α 45 α β tan β 3 β 60 θ (80 45 ) W tan α 45 tan β

15 Mathematics P/Wiskunde V 5 DBE/November in,3 secs revolution/omwenteling in 360, 3 secs,4 secs/sek 95 minute 60 sec: 60 5 revolutions/omwentelings,4 5 rev/min or 5 omw/min Answer only: mark 95 speed/sec 50 / sec,3 speed/minute / min 9000 no of revolutions 5 rev/min , 3 95,4 secs 60,4 5 rev/min 95 or 50, rev/min [0]

16 Mathematics P/Wiskunde V 6 DBE/November 03 QUESTION/VRAAG 0 0. OP OP OP OP 3 answer r x r y Answer only: full marks r r 3 answer tan(80 α ) tanα 5 sin(30 α ) sin 30 cos30 sinα tan α answer expansion () [8]

17 Mathematics P/Wiskunde V 7 DBE/November 03 QUESTION/VRAAG.. cos (90 + θ ) LHS cos( θ ) + sin(90 θ ).cosθ ( sinθ ) cosθ + cosθ.cosθ cos θ cosθ + cos θ ( cosθ )( + cosθ ) cosθ ( + cosθ ) cosθ cosθ cosθ RHS tan x.sin x + cos x.tan x 0 tan x(sin x + cos x) 0 tan x 0 or x 0 + k.80 ; k Z or sin x + cos x 0 sin x cos x tan x x 35 + k.80 ; k Z sin x sin x.sin x + cos x. 0 cos x cos x sin x cos x.sin x (sin x + cos x) cos x cos x + sin x 0 sin x + cos x.sin x 0 sin x sin x(sin x + cos x) 0 x 35 + k.80 sin x + cos x 0 sin x 0 or tan x x 0 + k.80 ; k Z or x 35 + k.80 ; k Z tan x.sin x + cos x.tan x 0 (cos x 0) tan x + tan x 0 tan x(tan x + ) 0 tan x 0 or tan x + 0 tan x x 0 + k.80 ; k Z or x 35 + k.80 ; k Z cos (90 +θ) sin θ sin(90 θ) cos θ cos( θ ) cos θ cos²θ factors cosθ cosθ factorising tan x 0 and sin x + cos x 0 (6) tan x x 0 x 35 or 45 k.80 k Z (7) factorising sin x 0 and tan x x 0 x 35 or 45 k.80 k Z (7) factorising tan x 0 and tan x + 0 tan x x 0 x 35 or 45 k.80 k Z (7)

18 Mathematics P/Wiskunde V 8 DBE/November sin sin sin 3x sin cos6x 3x (sin 3x x cos x x + cos x) (sin x + cos x) sin 3x sin (x + x) sin (sin x.cos x + cos x.sin x) ( sin x.cos ( sin x( sin ( sin x sin ( 4sin (6 sin x + 3sin x) x 4sin x cos x + 9sin ((sin x.cos x) cos x + ( sin 6 3 x + sin x sin 3 x) x) + sin x sin x + sin x sin 4 x 3 (sin 3 x) x) x) x)sin x) 3 x + cos 6 4 3sin x 48sin x + 8sin x answer.3. Max value ().4. p + sinα (a) q sinα cos α cos α sin α expansion ( + sinα )( sinα ) factorise pq answer p q p q sinα expansion sin α answer p q ( ) x) (sin x + cos x) p + q cos α p + q ( ) p + q expansion answer cos α sin α p + q p q ( ) ( )

19 Mathematics P/Wiskunde V 9 DBE/November 03 expansion p q p + q.4. (b) p + q p + q p q p q sinα sinα sinα tanα sinα p q p + q + sinα p sinα q p + q p + q y ( p + q) y tanα 4 ( p + q) 4 ( p + q) p + q α p + q p + q p q identity answer p + q sketch y 4 ( p + q) answer + sinα p sinα q sinα p q sinα p q x x tanα ( p q) 4 ( p q) p q α 4 ( p q) p q sinα p q sketch x answer 4 ( p q)

20 Mathematics P/Wiskunde V 0 DBE/November cos α sin α pq sin α cos α cos α cos pq + α sinα (tanα) pq tan α + pq tanα + pq pq p q q p p q pq ( p + q)( p q) pq ()(sinα) cos α 4sinα cos α sin α cos α tan α sin α pq + cos α pq (tanα) sinα pq tanα + pq p q pq factorising substituting from.4.(a) and.4.(b) sin α tan α (6) p q q p + sinα sinα ( sinα) ( + sinα) ( + sinα) ( sinα) ( sinα)( + sinα) cos α + sinα + sin α (cos α sinα + sin α) (cos α sin α) 4sinα sin α tan α substitution single fraction 4 sinα sinα tanα (6) [30]

21 Mathematics P/Wiskunde V DBE/November 03 QUESTION/VRAAG. f y O x y tan x + asymptotes and shape for whole domain y intercept x intercepts g - y cos x x intercepts y intercept/tp minimum values (6). Period of g is ().3 Reflected about the x-axis and then translated by 0 to the left/ refleksie om die x-as en dan n translasie van 0 links. reflected about x axis/refleksie om x-as 0 to the left 0 na links () Translated by 0 to the left and then reflected about the x-axis/ Translasie van 0 links en dan n refleksie om die x-as..4 f is always increasing / f ( x) > 0 always g(x) > 0 0 < x < 45 or 35 < x 80 x ( 0 ; 45 ) or (35 ; 80 ] 0 to the left 0 na links reflected about x axis/refleksie om x-as () critical values 0 and 45 inequality critical values 35 and 80 inequality [3]

22 Mathematics P/Wiskunde V DBE/November 03 QUESTION/VRAAG 3 3. In CEB : BC EC + EB (EC)(EB)cosCÊB (3,6) (,) + (,) (,)(,)cos CÊB (,) (3,6) cos CÊB (,) 0, CÊB 63,44 substitution into correct formula 0, ,44 6,3, α 58,8 θ 80 (58,8 ) 63,44 C E h θ, α 6,3 B substitution into correct definition α 58,8 63,44 E 6,3 sinθ, θ 3,7 AEB ˆ θ (3,7 ) 63,44 θ θ, C 6,3 B substitution into correct formula θ 3,7 63,44 3. EF EB BF (,) (6,3) 35403,75 EF 88,6 m GF cos EFˆG EF 6,3 88,6 0,68... using Pythagoras correctly BF 6,3 88,6 using cos EFˆ G 6,3 88,6 EFˆG 5,8 5,8 (6)

23 Mathematics P/Wiskunde V 3 DBE/November 03 sinα h h,,sin 58,8 88,58... In ΔEFG : 6,3 cos EFˆG 88, , EFˆG 5,8 E θ, h α C F 6,3 B h sinα, h subject h 88,58... using cos EFˆ G 6,3 88,6 5,8 (6) E (apex) D 47,9, 88,6 A 64,47 G F 6,3 C 3,6 B [9] TOTAL/TOTAAL: 50