NATIONAL SENIOR CERTIFICATE GRADE 12

Transcription

1 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P NOVEMBER 0 MARKS: 50 TIME: hours This questio paper cosists of 8 pages, diagram sheet ad iformatio sheet.

2 Mathematics/P DBE/November 0 NSC INSTRUCTIONS AND INFMATION Read the followig istructios carefully before aswerig the questios This questio paper cosists of questios. Aswer ALL the questios. Clearly show ALL calculatios, diagrams, graphs, et cetera that you have used i determiig your aswers. Aswers oly will ot ecessarily be awarded full marks. You may use a approved scietific calculator (o-programmable ad ographical), uless stated otherwise. If ecessary, roud off aswers to TWO decimal places, uless stated otherwise. Diagrams are NOT ecessarily draw to scale. ONE diagram sheet for aswerig QUESTION. is attached at the ed of this questio paper. Write your cetre umber ad eamiatio umber o this sheet i the spaces provided ad isert the page iside the back cover of your ANSWER BOOK. A iformatio sheet, with formulae, is icluded at the ed of the questio paper. Number the aswers correctly accordig to the umberig system used i this questio paper. Write eatly ad legibly.

3 Mathematics/P DBE/November 0 NSC QUESTION. Solve for :.. ( + ) 6 () Cosider the equatio: + 5y + 6y 0.. Calculate the values of the ratio y. () QUESTION.. Hece, calculate the values of ad y if + y 8. (5) [9]. Give the sequece: 4 ; ; Determie the value(s) of if the sequece is:.. Arithmetic ().. Geometric (). Determie the value of P if P k k 5. Prove that for ay arithmetic sequece of which the first term is a ad the costat differece is d, the sum to terms ca be epressed as S ( a + ( ) d ). [] QUESTION The followig sequece is a combiatio of a arithmetic ad a geometric sequece: ; ; 9 ; 6 ; 5 ; ;. Write dow the et TWO terms. (). Calculate T5 T5. (5). Prove that ALL the terms of this ifiite sequece will be divisible by. () [9]

4 Mathematics/P 4 DBE/November 0 NSC QUESTION 4 A quadratic patter has a secod term equal to, a third term equal to 6 ad a fifth term equal to Calculate the secod differece of this quadratic patter. (5) 4. Hece, or otherwise, calculate the first term of the patter. () [7] QUESTION Cosider the fuctio: f ( ) 5.. Calculate the coordiates of the y-itercept of f. () 5.. Calculate the coordiates of the -itercept of f. () 5.. Sketch the graph of f i your ANSWER BOOK, showig clearly the asymptotes ad the itercepts with the aes For which values of is f ( ) > 0? () 5..5 Calculate the average gradiet of f betwee ad Draw a sketch graph of y a + b + c, where a < 0, b < 0, c < 0 ad a + b + c 0 has oly ONE solutio. [9]

5 Mathematics/P 5 DBE/November 0 NSC QUESTION 6 The graphs of ( ) f 8 ad g ( ) a + b + c are sketched below. B ad C(0 ; 4,5) are the y-itercepts of the graphs of f ad g respectively. The two graphs itersect at A, which is the turig poit of the graph of g ad the -itercept of the graphs of f ad g. y f C(0 ; 4,5) g O A B 6. Determie the coordiates of A ad B. 6. Write dow a equatio of the asymptote of the graph of f. () 6. Determie a equatio of h if h ( ) f () + 8. () 6.4 Determie a equatio of h i the form y... () 6.5 Write dow a equatio of p, if p is the reflectio of h about the -ais. () 6.6 Calculate g ( k) g( k). Show ALL your workig. [4] k 0 5 k 4

6 Mathematics/P 6 DBE/November 0 NSC QUESTION 7 7. How may years will it take for a article to depreciate to half its value accordig to the reducig-balace method at 7% per aum? 7. Two frieds each receive a amout of R6 000 to ivest for a period of 5 years. They ivest the moey as follows: Radesh: 8,5% per aum simple iterest. At the ed of the 5 years, Radesh will receive a bous of eactly 5% of the pricipal amout. Thadi: 8% per aum compouded quarterly. Who will have the bigger ivestmet after 5 years? Justify your aswer with appropriate calculatios. (6) 7. Nicky opeed a savigs accout with a sigle deposit of R 000 o April 0. She the makes 8 mothly deposits of R700 at the ed of every moth. Her first paymet is made o 0 April 0 ad her last paymet o 0 September 0. The accout ears iterest at 5% per aum compouded mothly. QUESTION 8 Determie the amout that should be i her savigs accout immediately after her last deposit is made (that is o 0 September 0). (6) [6] 8. Determie f () from first priciples if f ( ) 4. (5) 8. Evaluate: 8.. dy if y () d 8.. f () if ( ) (7 + ) f []

7 Mathematics/P 7 DBE/November 0 NSC QUESTION 9 The fuctio f ( ) + a + b + c is sketched below. The turig poits of the graph of f are T( ; 9) ad S(5 ; 8). y S(5 ; 8) f O T( ; 9) 9. Show that a, b 60 ad c 4. (7) 9. Determie a equatio of the taget to the graph of f at. (5) 9. Determie the -value at which the graph of f has a poit of iflectio. () [4] QUESTION 0 The graph of y f (), where f is a cubic fuctio, is sketched below. y 4 0 y f / () Use the graph to aswer the followig questios: 0. For which values of is the graph of y f () decreasig? () 0. At which value of does the graph of f have a local miimum? Give reasos for your aswer. () [4]

8 Mathematics/P 8 DBE/November 0 NSC QUESTION Water is flowig ito a tak at a rate of 5 litres per miute. At the same time water flows out of the tak at a rate of k litres per miute. The volume (i litres) of water i the tak at time t (i miutes) is give by the formula V ( t) 00 4t.. What is the iitial volume of the water i the tak? (). Write dow TWO differet epressios for the rate of chage of the volume of water i the tak. (). Determie the value of k (that is, the rate at which water flows out of the tak). () [6] QUESTION A school is plaig a trip for 500 learers. The compay that will be providig the trasport has two types of buses, red buses ad blue buses, available. Each red bus has 50 seats ad each blue bus has 5 seats. The compay has at most 5 bus drivers available. There are at most 8 blue buses available. Let the umber of red buses hired by the school be ad the umber of blue buses hired by the school be y.. Write dow ALL the costraits, i terms of ad y, to represet the above iformatio. (6). Represet the costraits graphically o the attached DIAGRAM SHEET. Clearly idicate the feasible regio.. The cost of hirig a red bus is R600 for the day ad the cost of hirig a blue bus is R00 for the day. Write dow the total trasport cost. ().4.4. Determie ALL possible values of ad y so that the cost will be a miimum. ().4. Calculate the miimum cost of hirig the buses. ().5 If eactly bus drivers are to be used, determie the umber of each type of bus which the school will ow eed to still esure miimum cost. () [7] TOTAL: 50

9 Mathematics/P DBE/November 0 NSC CENTRE NUMBER: EXAMINATION NUMBER: DIAGRAM SHEET QUESTION. y

10 Mathematics/P DBE/November 0 NSC INFMATION SHEET: MATHEMATICS b ± b 4 ac a A P( + i) A P( i) A P( i) A P( + i) i i ( + ) i T ar a( r ) S F f [( + i) ] i f ( + h) f ( ) '( ) lim h 0 h r T a + ( ) d S ( a + ( d ) ; r [ ( + i) ] P i ( ) ( ) + y + y d + y y M ; y m + c y y m ) ( a) + ( y b) r I ΔABC: si a A area Δ ABC ( b c a b + c bc. cos A si B si C ab. si C S ) a ; < r < r y y m m taθ ( α + β ) siα.cosβ cosα. si β si( α β ) siα.cosβ cosα. si β si + cos ( α + β ) cosα.cos β siα. si β cos ( α β ) cosα.cos β + siα. si β cos α si α cos α si α si α siα. cosα cos α ( ; y) ( cosθ + y siθ ; y cosθ siθ ) ( ; y) ( cosθ y siθ ; y cosθ + siθ ) ( i ) σ i f ( A) P( A) P(A or B) P(A) + P(B) P(A ad B) y ˆ a + b ( S ) b ( ) ( ) ( y y)

11 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P NOVEMBER 0 MEMANDUM MARKS: 50 This memoradum cosists of 8 pages.

12 Mathematics/P DBE/November 0 NSC Memoradum NOTE: If a cadidate aswers a questio TWICE, oly mark the FIRST attempt. If a cadidate has crossed out a attempt of a questio ad ot redoe the questio, mark the crossed out versio. Cosistet Accuracy applies i all aspects of the markig memoradum. QUESTION.. ( + ) ( + )( ) 0 or b ± ± b 4ac a () or b ± ( 4) ± 4 ± 4 ± b a 6 ()( 6) ac ± 7,4 or,0 ( 4) 4( )( 8) () Note: Pealise mark for iaccurate roudig off to ANY umber of decimal places if cadidate gives decimal aswers. Note: Aswers by ispectio: award / marks Note: Aswer oly of : award / marks Note: If cadidate coverts equatio to liear: award 0/ marks Note: If cadidate uses icorrect formula: maimum /4 marks (for stadard form) Note: If a error i subs ad 4 ± 80 gets: ad 6 states o solutio : maimum /4 marks If does t coclude with o solutio : maimum /4 marks stadard form factors s () stadard form substitutio ito correct formula s stadard form () substitutio ito correct formula 4 ± or 6 decimal aswer

13 Mathematics/PI DBE/November 0 NSC Memoradum stadard form b ± ( 4) ± b 4ac a,4 or, ( 4 )( ) 0 ( 4) 4( )( 8) () Note: Pealise mark for iaccurate roudig off to ANY umber of decimal places if cadidate gives decimal aswers substitutio ito correct formula factors both critical values of ad 4 or 4 4 ; [ ; ) or 4 4 Note: If cadidate gives either of these correct graphical solutios but writes dow the icorrect itervals or uses AND: ma /4 marks NOTES: If a cadidate gives a aswer of the ma /4 marks. 4 If a cadidate gives a aswer of the ma /4 marks. 4 If a cadidate gives a aswer of ad the ma /4 marks. 4 If the cadidate leaves out the equality of the otatio the pealty of mark. If a cadidate gives a aswer of ; the ma /4 marks. 4 If cadidate gives ad/or, BREAKDOWN: ma /4 marks. 4 If cadidate gives : award /4 marks 4

14 Mathematics/PI 4 DBE/November 0 NSC Memoradum.. + 5y + 6y 0 ( + y)( + y) 0 + y 0 y y Let k y + y 0 y y Note: If a cadidate gives or y y award / marks factors s () k y + 5y + 6y + 5 y + 5k ( k + )( k + ) 0 k or k or y y factors s () + 5y + 6y 5y ± 0 (5y) 4()(6y ) () 5y ± y 5y ± y y y y or + 5y + 6y y 5 + 5y + y 0 6y 5 + y y y ± y 5 y ± 5 + y y substitutes correctly ito correct formula s completig the square ()

15 Mathematics/PI 5 DBE/November 0 NSC Memoradum y y s or y y () Let k y ky + 5y + 6y 0 + 6y 0 k y + 5y k + 6y 0 ( ky) + 5y( ky) y ( k + 5k + 6) 0 ( k + 5k + 6) 0 ( k + )( k + ) 0 k or k or y y Note: (;y) (0;0) is also a solutio, but i this case is udefied y factors s ().. Let y, ( + )( + ) 0 or or y y + y 8 y + y 8 y 8 + y 8 y + y 8 y 8 y 4 y y 8 y y y 8 y y 8 y y 8 y 8 y y 4 y 8 6 factors s () substitutio y subs y y values both values correct (5) 8 y substitutio y values both correct values (5)

16 Mathematics/PI 6 DBE/November 0 NSC Memoradum + y 8 y ( ) ( ) y 4 ( + y)( + y) + y 8 8 y 0 ( y + 8)( y + 8) 0 y 8 or y y 8 y 8 substitutio values correct both y values correct (5) 8 y substitutio y values correct both values correct (5) 8 y ( 8 y) + 5( 8 y) 64 6y + y y y + 4y y + 0 ( y + 8)( y + 4) 0 y 8 or y 4 6 y + 6y + 40y 5y 0 + 6y 0 8 y substitutio factors both y values correct both values correct (5)

17 Mathematics/PI 7 DBE/November 0 NSC Memoradum 8 y ( 8 y) + 5( 8 y) y + 6y y + y + 40y 5y + 6y y + 4y y + y + 0 ± y () ± 6 y 8 or y 4 6 y ()( ) ( 8 ) + 6( 8 ) ( 6)( ) 0 y 4 or y y ( 6 + ) ( 8 ) + 6( 8 ) + 6 ( ) ( 8) ± ( 8) 4( )( 9) () 8 ± 46 y 4 or 6 y 8 Note: If a cadidate uses the formula ad replaces for y ad the aswers are swapped: maimum 4/5 marks 8 y substitutio substitutes ito correct formula both y values correct both values correct (5) y 8 substitutio factors both values correct both y values correct (5) y 8 substitutio substitutes ito correct formula both values correct both correct y values (5) [9]

18 Mathematics/PI 8 DBE/November 0 NSC Memoradum QUESTION a 4 a + d d 8 d ± 8 7 ±8 ±, ± a 4 r 4 ar ± 8 ±8 or ±, or ± ± 4 ± 8 or ±8 or ±, or ± 7 Note: If aswer oly: award / marks Note: If cadidate writes 4 oly (i.e. omits equality) : 0/ marks Note: If cadidate writes oly 4 (i.e. omits equality) : 0/ marks Note: If oly 8 the pealty mark 7 T T T T () a + d ad a 4 () substitutes correctly ito arithmetic mea 4 + formula i.e. s () T T T T 8 both aswers (surd or decimal or epoetial form) both aswers (surd or decimal or epoetial form) () () substitutes correctly ito geometric mea formula i.e. ± 4 both aswers (surd or decimal or epoetial form) ()

19 Mathematics/PI 9 DBE/November 0 NSC Memoradum. k 5 P k P , 49 ( ) 4 k k , 49 or or or 984 or 8 d + a + d S a + [ a + ] [ ] [ a ( ) d] + [ a + ( ) d] S [ a + ( ) d] + [ a + ( ) d] [ a + d] + a S [ a + ( ) d] + [ a + ( ) d] [ a + ( ) d] + [ a + ( ) d] [ a + ( ) d] S [ a + ( ) d] Note: Correct aswer oly: /4 marks oly Note: If the cadidate rouds off ad gets 984,46 (i.e. correct to oe decimal place): DO NOT pealise for the roudig off. a 4 or 8 r subs ito correct formula epad the sum terms i epasio writig out S reversig S epressig S groupig to get S a + d [ ( ) ] S S S S [ a + d] + [ a + d] ( T d) [ a + d] a + ( T d) + T T + + a a + T + a + T + a + T a + T [ a + a + ( ) d] [ a + ( ) d] [ a + ( ) d] Note: If a cadidate uses a circular argumet (eg S + + S T ): ma /4 marks (for writig out S ) writig out S reversig S epressig S groupig to get S [ a + a + ( ) d] Note: If a cadidate uses a specific liear sequece, the NO marks. []

20 Mathematics/PI 0 DBE/November 0 NSC Memoradum QUESTION. ; 4 Note: 4 If cadidate writes T 8 T 7 4 : award / marks (). k T. k 6 5 ad so T ( k ) 6k T k ad so T 6( 6) 5 5 Note: If cadidate writes out all 5 terms ad gets correct aswer: award 5/5 marks k. T 5 6k T 5 T5 T Note: If cadidate used k 5: ma /5 (5) Cosider sequece P: ; 6 ; P. 6 P Cosider sequece Q: ; 9 ; 5 6 Q ( 6) 5 Q 6 6 T T P Q Note: if cadidate iterchages order i.e. does T5 T5 : ma 4/5 marks Note: writes out all 5 terms ad subtracts T5 T5 : ma 4/5 marks P P 6. 6 Q Q 6 (5)

21 Mathematics/PI DBE/November 0 NSC Memoradum. For all N, k or k for some k N If T k : k Tk. If k : T Tk 6k ( k ) I either case, T has a factor of, so is divisible by. Note: If a cadidate oly illustrates divisibility by with a specific fiite part of the sequece, ot the geeral term: 0/ marks k factors. factors ( k ) () P. Which is a multiple of factors. Q 6 ( ) Which is also a multiple of factors ( ) () Sice T Qk or T P k for all N, T is always divisible by The odd terms are odd multiples of ad the eve terms are times a power of. This meas that all the terms are multiples of ad are therefore divisible by. odd multiples of times a power of () [9]

22 Mathematics/PI DBE/November 0 NSC Memoradum QUESTION 4 4. The secod, third, fourth ad fifth terms are ; 6 ; T 4 ad 4 First differeces are: 7 ; T ; 4 T 4 So T T 4 6 T 4 d or Note: Aswer oly (i.e. d ) with o workig: marks Note: Cadidate gives T 4 ad d oly: award 5/5 marks 7 T T 4 settig up equatio T T ( T T ) + ( T T ) + ( T ) T (5) T T T 4 T d -7+d d 7 + d T 5 T 5 ( T5 T4 ) + ( T4 T ) + ( T T ) ( 7 + d ) + ( 7 + d ) d 6 d d 4a + b + c 9a + b + c 6 5a + b 7 5a + 5b + c 4 6a + b 8 0a + b 4 6a 6 a d a d d Note: Cadidate uses trial ad error ad shows this: award 5/5 marks settig up equatio T T ( T T ) + ( T T ) + ( T ) T (5) 4 a + b + c 9a + b + c 6 5a + 5b + c 4 solved simultaeously (5) T 4 T T T -6 T 4-4 T -7 T T 4 T - 8 T T T 4 d + d 4 7 T T 4 settig up equatio (5)

23 Mathematics/PI DBE/November 0 NSC Memoradum T T T T 4 T 5-6 y -4-7 y y y y 7 y y y + 0 y y y settig up equatio (5) 4. Secod differece y + + T 6 method -9-7 T 0 a 5a + b 7 5() + b 7 b a + b + c 4() + ( ) + c c Note: Aswer oly: award / marks Note: If icorrect d i 4., / CA marks for T d + 8 (sice T 7 d ) T 0 method T 0 () T T () 0 + () + () T T T T 0 y method T 0 () [7]

24 Mathematics/PI 4 DBE/November 0 NSC Memoradum QUESTION y f (0) ; 0 ad y ( ) ; 0 ( ) ( ; 0) (0 ; ) 0 y y Note: Mark 5.. ad 5.. as a sigle questio. If the itercepts are iterchaged: ma /5 marks Note: The graph must ted towards the asymptotes i order to be awarded the shape mark y 0 y 0 6 shape () () both itercepts correct horizotal asymptote vertical asymptote Note: A cadidate who draws oly oe arm of the hyperbola loses the shape mark i.e. ma /4 marks 5..4 < < ( ; ) Note: if cadidate writes < oly: / marks < ad < ad Note: if cadidate writes < oly: / marks iequality iterval otatio ()

25 Mathematics/PI 5 DBE/November 0 NSC Memoradum y 5 5 m 0 ( ) 5 5 formula substitutio f (0) f ( ) m 0 ( ) formula f ( ) 5 substitutio 5. b a < 0 sice b < 0 ad a < 0 y 0 y-itercept egative turig poit o the ais turig poit o the left of the y ais maimum TP ad quadratic shape [9]

26 Mathematics/PI 6 DBE/November 0 NSC Memoradum QUESTION 6 y f C(0 ; 4,5) g O A B f (0) A( ; 0) B(0 ; 7) 6. y 8 y h( ) f () + 8 ( 8) or Note: o CA marks Note: aswer oly: award / marks y 0 for A 0 for B ( 8) of h( ) 4 or () () 6.4 y y 4 y log4 y log y log y log log y log 4 Note: aswer oly award / marks Note: cadidate works out f - ad gets y log ( + 8) award / marks switch ad y i the form y () 6.5 p( ) log 4 p( ) log p( ) log 4 p( ) log 4 aswer () y log

27 Mathematics/PI 7 DBE/November 0 NSC Memoradum 6.6 k 0 g ( k) 5 k 4 g( k) g( 0) + g() + g() + g() g g(5) is the ais of symmetry of g by symmetry g ( ) g ad g ( ) g(5) Aswer g ( 0) + g() 4, ,5 g( 0) + g() + g() + g() g g(5) g ( ) g ad g ( ) g(5) g ( 0) + g() k 0 k 0 5 k 4 g ( k) 5 k 4 g( k) g( k) g(0) + g() + g() + g() g( k) g + g(5) is the ais of symmetry of g by symmetry g g() g(5) g() g( k) k 0 5 g(0) + g() 4, ,5 g( ) a 4,5 9a a g( ) g( k) k 4 ( ) 4,5 a(0 ) k 0 k 0 g ( k) + 0 ( ) 5 k g( k) g( k) g(0) + g() + g() + g() 4, , epasio g ( ) g ad g ( ) g(5) g ( 0) + g() g ( ) ( ) epasio

28 Mathematics/PI 8 DBE/November 0 NSC Memoradum,5 0,5 (5) ) ( g g k g k 4,5,5 7 ) ( ) ( k k k g k g c b a g c b a g c b a g c g c bk ak k g c b a g () 4 () () (0) ) ( ) ( c b a k g k ) ( c b a g c b a g (5) ) ( k c b a k g c b a k g k g k k 7 ) ( ) ( ( ) ( ) 9 ) ( 9 4,5 0 ) (0 4,5 0 ) ( g a a a a g ( ) 4, ) ( ) ( c b a k g k g k k 5, 7 c b a 7 + ( ) ) ( g [4]

29 Mathematics/PI 9 DBE/November 0 NSC Memoradum QUESTION 7 7. A P( i) P P 0,9 log log 0,9 log log 0,9 ( 0,07) 9,55 years Note: If cadidate iterchages A ad P A i.e. uses P : ma /4 marks log 0,9 A P ( i) P P 0,9 9,55 years ( 0,07) Note: If cadidate uses icorrect formula: ma /4 marks P for A P A subs ito correct formula log

30 Mathematics/PI 0 DBE/November 0 NSC Memoradum 7. Radesh: ( + ) ( + 0,085 5) A P i A ,5% of Bous 0, Received R8 850 R Thadi: A P + i ( ) 0, R8 95,68 Thadi's ivestmet is bigger. 7. Fv iitial deposit with iterest + auity Fv R5 8,9 0 0, , ,5 50, , 8 iitial deposit with iterest + auity R5 8,9 8 0, , , ,58 + 0, , , 0 0, 08 i 4 choice made (6) 0,5 i or or 0, , , ,5 (6) 0,5 i or or 0, , , ,5 (6)

31 Mathematics/PI DBE/November 0 NSC Memoradum 9 0,5 0,5 8 + i or or 0, F v QUESTION 8 0, , ,74 R5 8,9 0,5 9 (correspodig to 700) 8 (correspodig to 00) 8 0, , ,5 9 (6) [6] 8. f ( ) lim h 0 h 0 8 f ( + h) f ( ) h ( + h) ( 4 ) 4 lim h 0 h 4( + h + h ) + 4 lim h 0 h 4 8h 4h + 4 lim h 0 h 8h 4h lim h 0 h h( 8 4h) lim h 0 h lim ( 8 4h) Note: Icorrect otatio: o lim writte: pealty marks lim writte before equals sig: pealty mark Note: A cadidate who gives 8 oly: 0/5 marks Note: A cadidate who omits brackets i the lie lim ( 8 4h) : h 0 NO pealty formula substitutio epasio 8 4h (5)

32 Mathematics/PI DBE/November 0 NSC Memoradum f ( ) 4 f ( + h) 4( + h) substitutio 4 8h 4h epasio f ( + h) y dy d f ( ) 8h 4h 8h 4h f ( ) lim h 0 h h( 8 4h) lim h 0 h lim ( 8 4h) h 0 8 f ( ) (7 + ) f ( ) f () 98() + 4 Note: Icorrect otatio i 8.. ad/or 8..: Pealise mark formula 8 4h multiplicatio 4 98 (5) () f ( ) (7 + ) f ( ) ( 7 + )(. 7) By the chai rule f ( ) f () 98() + 4 chai rule []

33 Mathematics/PI DBE/November 0 NSC Memoradum QUESTION 9 9. f ( ) + a + b + c f ( ) 6 + a + b 6( 5)( ) 6 6 a 4 a b 60 ( 7 + 0) Note: A cadidate who substitutes the values of a, b ad c ad the checks (by substitutio) that T ( ; 9) ad S ( 5;8) lie o the curve: award ma /7 marks f ( ) 6 + a + b 6( 5)( ) b 60 a 4 f (5) c c 4 () 5 + () 5 60() 5 a ; b 60 ; c 4 ( ) ( ) f 6 + a + b f 6() + a() + b a + b b 4 4a f (5) 6(5) + a(5) + b a + b a + (4 4a) a 6a 6 a b 60 + c f () c c 4 ( ) + ( ) 60( ) Note: If derivative equal to zero is ot writte: pealize oce oly + c subs (5 ; 8) or ( ; -9) c 4 (7) Note: A cadidate who substitutes the values of a, b ad c ito the fuctio i.e. gets f ( ) ad the shows by ; 9 S 5;8 are o the curve ad works substitutio that T ( ) ad ( ) out the derivative i.e. gets f ( ) ad shows (by substitutio ito the derivative) that the turig poits are at ad 5 (assumig what s/he sets out to prove ad provig what is give): award ma 4/7 marks as follows: from f ( ) 0 subs ito the derivative ad gets 0 5 from f ( ) 0 subs 5 ito the derivative ad gets 0 substitutio of i f ad gets 9 substitutio of 5 i f ad gets 8 f ( ) 0 f ( 5) 0 f ( ) 6 + a + b 6 a 6 b 60 f (5) () 5 + () 5 60() c c 4 a ; b 60 ; c 4 + c f () c c 4 ( ) + ( ) 60( ) + c subs (5 ; 8) or ( ; -9) c 4 (7)

34 Mathematics/PI 4 DBE/November 0 NSC Memoradum f ( ) 9 i.e a + b + c 9 4 a + b + c 7 f ( 5) 8 i.e a + 5b + c 8 5 a + 5b + c 68 a + b 6 ( ) 6 + a b ad f ( ) 0 f ( 5 ) 0 f + 4 a + b 4 0 a + b a + b + c 9 ad a + 5b + c 8 ( f ( ) 6 + a + b f ) 0 or f ( 5 ) 0 a + b 7 9a 89 a a a + b 450 9a a 9 a 9 a 89 ( ) + b 7 b 80 b 60 b a + b + c 7 + c 7 c 4 ( ) + ( 60) 9. f ( ) m f ta 4 () + 4() 6 60 () () + () 60() Poit of cotact is ( ; ) 5( ) + 5( 60) + 4 5a + 5b + c 68 + c 68 c 4 subs (5 ; 8) or ( ; -9) c 4 (7) f ( ) subs f () m 4 ta f() 9. ( ) y 4 y f ( ) f ( ) y 4 + c 4() + c c 6 y y 4( ) y f ( ) (5) ()

35 Mathematics/PI 5 DBE/November 0 NSC Memoradum ( ) 7 7 ( ) () () [4] QUESTION 0 y 4 0 y f / () 0. -value of turig poit: 4 + > ; 0. f has a local miimum at 4 because: (; y) > ; () 4 graph () f ' f 4 4 / f ( ) < 0 for < 4, so f is decreasig for < 4. / f ( ) > 0 for 4 < <, so f is icreasig for 4 < <. i.e. f has a local miimum at / f ( ) < 0 for < 4 / f ( ) > 0 for 4 < < ()

36 Mathematics/PI 6 DBE/November 0 NSC Memoradum 4 Gradiet of f chages from egative to positive at 4 gradiet egative for < 4 gradiet positive for 4 < < () QUESTION f ( 4) 0 f ( 4) > 0 so graph is cocave up at 4, so f has a local miimum at 4. f ( 4) 0 f ( 4) > 0 4. V ( 0) 00 4(0) 00 litres. Rate i rate out 5 k l / mi. V ( t) 4 l / mi 5 k 4 k 9 l / mi Note: Aswer oly: award / marks 5 k 4 uits stated oce 5 k 4 k 9 () [4] () () () Volume at ay time t iitial volume + icomig total outgoig total t kt 00 4t 5t kt 4t 9t kt 0 ( 9 k) 0 t At miute from start, t, 9 k 0, so k t kt 00 4t k 9 () dv Sice 4, the volume of water i the tak is decreasig by 4 dt litres every miute. So k is greater tha 5 by 4, that is, k 9. k 9 () [6]

37 Mathematics/PI 7 DBE/November 0 NSC Memoradum QUESTION Note: If the wrog iequality y 500 is used, cadidate wrogly says that there are more learers tha available seats. Maimum of 0 marks.., y N + y 5. Blue buses y y y 8 + y 5 y + 5 Note: If cadidate y 8 gives y 500 : y y 500 ma 5/6 marks y 8 Note: for the iequality s marks to be awarded, the LHS ad the RHS must be correct (6) Red buses. C y.4. (6 ; 8) ; (7 ; 6) ; (8 ; 4) ; (9 ; ) ad (0 ; 0) NOTE: The gradiet of the search lie is m.4. C 6 (600) + 8(00) R6 000 or C 7 (600) + 6(00) R6 000 or C 8 (600) + 4(00) R6 000 or C 9 (600) + (00) R6 000 or C 0 (600) + 0(00) R6 000 subs.5 8 red ; 4 blue + y y 500 y 8 feasible regio () marks for all correct solutios marks if oly or 4 correct solutios mark if oly or correct solutios () () () [7] TOTAL: 50

38 Mathematics/P 8 NSC Memoradum QUESTION. y Blue Buses Red Buses

39 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P NOVEMBER 0 MARKS: 50 TIME: hours This questio paper cosists of pages, diagram sheet ad iformatio sheet.

40 Mathematics/P DBE/November 0 NSC INSTRUCTIONS AND INFMATION Read the followig istructios carefully before aswerig the questios This questio paper cosists of questios. Aswer ALL the questios. Clearly show ALL calculatios, diagrams, graphs, et cetera which you have used i determiig the aswers. Aswers oly will ot ecessarily be awarded full marks. You may use a approved scietific calculator (o-programmable ad ographical), uless stated otherwise. If ecessary, roud off aswers to TWO decimal places, uless stated otherwise. Diagrams are NOT ecessarily draw to scale. ONE diagram sheet for QUESTION 4. is attached at the ed of this questio paper. Write your cetre umber ad eamiatio umber o this sheet i the spaces provided ad isert the sheet iside the back cover of your ANSWER BOOK. A iformatio sheet, with formulae, is icluded at the ed of this questio paper. Number the aswers correctly accordig to the umberig system used i this questio paper. Write eatly ad legibly. Copyright Reserved

41 Mathematics/P DBE/November 0 NSC QUESTION Fiftee members of a basketball team took part i a touramet. Each player was allowed the same amout of time o the court. The poits scored by each player at the ed of the touramet are show below Determie the media of the give data. (). Determie the iterquartile rage for the data. (). Draw a bo ad whisker diagram to represet the data. ().4 Use the bo ad whisker diagram to commet o the poits scored by the players i this team. () [9] QUESTION The scores for 8 golfers who played a sigle roud of golf o the same golf course are show below Calculate the mea score. (). Calculate the stadard deviatio for the data. (). How may golfers' scores lie outside oe stadard deviatio of the mea? () [6]

42 Mathematics/P 4 DBE/November 0 NSC QUESTION A group of 8 learers was radomly selected from a class. The performace of these learers i a stadardised test (which couted 50 marks) ad the average umber of hours they sped watchig TV each week was recorded. The data is represeted i the scatter plot below. SCATTER PLOT Test scores obtaied Average umber of hours spet watchig TV. What is the lowest test score for this group of learers? (). Does the data display a liear, quadratic or epoetial relatioship? Justify your choice. (). What coclusio ca be reached about the learers' test scores ad the average umber of hours they sped watchig TV? ().4 Aother learer from the class watches 5 hours of TV per week. Usig the give iformatio, predict his/her performace i the test. () [6]

43 Mathematics/P 5 DBE/November 0 NSC QUESTION 4 Thirty learers were asked to aswer a questio i Mathematics. The time take, i miutes, to aswer the questio correctly, is show i the frequecy table below. TIME, T (IN MINUTES) NUMBER OF LEARNERS t < t < t < t < t < 5 t < 4. Costruct a cumulative frequecy table for the data. () 4. Draw a cumulative frequecy graph (ogive) of the above data o the grid provided o DIAGRAM SHEET. 4. If a learer aswers the questio correctly i less tha 4 miutes, the he/she is classified as a 'gifted learer'. Estimate the percetage of 'gifted learers' i this group. () [9]

44 Mathematics/P 6 DBE/November 0 NSC QUESTION 5 I the diagram below, PQRS is a rectagle with vertices P( 4 ; 0), Q(4 ; a), R(6 ; 0) ad S. Q lies i the first quadrat. y Q(4 ; a) P( 4 ; 0) O R(6 ; 0) S 5. Show that a Determie the equatio of the straight lie passig through the poits S ad R i the form y m + c. 5. Calculate the coordiates of S. 5.4 Calculate the legth of PR. () 5.5 Determie the equatio of the circle that has diameter PR. Give the equatio of the circle i the form ( a) + ( y b) r. () 5.6 Show that Q is a poit o the circle i QUESTION 5.5. () 5.7 Rectagle PQRS udergoes the trasformatio ( ; y) ( + k ; y + l) where k ad l are umbers. What is the miimum value of k + l so that the image of PQRS lies i the first quadrat (that is, 0 ad y 0)? () []

45 Mathematics/P 7 DBE/November 0 NSC QUESTION 6 The circle with cetre B( ; ) ad radius 0 is show. BC is parallel to the y-ais ad CB 5. The taget to the circle at A passes through C. A Bˆ C ADˆ O θ y A C F θ D θ B O 6. Determie the coordiates of C. () 6. Calculate the legth of CA. () 6. Write dow the value of ta θ. () 6.4 Show that the gradiet of AB is. () 6.5 Determie the coordiates of A. (6) 6.6 Calculate the ratio of the area of ABC to the area of ODF. Simplify your aswer. (5) [9]

46 Mathematics/P 8 DBE/November 0 NSC QUESTION 7 7. The followig trasformatio is applied to all poits: Firstly, a poit is traslated by 4 uits to the right. The it is rotated through 80 about the origi. Write dow the geeral rule that represets the above trasformatio i the form ( ; y) 7. Reflect the circle with cetre C( 5 ; ) ad radius of 4 uits about the lie y. Give the equatio of the ew circle i the form a + by + c + dy + e 0. [8] QUESTION O I the diagram, A is the poit (0 ; 4), AB ad AD 5. Rectagle ABCD is rotated about the origi to form rectagle A / B / C / D /. After the rotatio the image of poit A is A / (4 ; 0). y 8. Describe the trasformatio fully i words. () 8. Write dow the coordiates of A B / D. () 8. If ABCD is reflected about the lie to form EFGH, write dow the coordiates of G, the image of C. () D C 8.4 If ABCD is elarged by a scale factor of through the origi to form MNPR, determie the value of area ABCD area MNPR. () [9]

47 Mathematics/P 9 DBE/November 0 NSC QUESTION 9 9. If ta A ad 0 < Â < 90, determie the values of the followig with the aid 40 of a sketch ad without usig a calculator. Leave your aswers i surd form, if ecessary. 9.. cos A () 9.. si (80 + A) () 9. Without usig a calculator, determie the value of the followig epressio: cos 00 ta 0 si ( 0 ) (6) 9. P(4 ; ) ad M(a ; b) are poits o a circle with the origi as cetre. Q ad R are -itercepts of the circle. y P(4 ; ) Q O R M(a ; b) 9.. Write dow the umerical value of si RÔP. () 9.. Calculate the size of QO ˆ P. () 9.. If obtuse P ÔM 5, calculate the value of a, the -coordiate of M, correct to TWO decimal places. () [8]

48 Mathematics/P 0 DBE/November 0 NSC QUESTION 0 The graphs of the fuctios f () a ta ad g() b cos for 0 70 are show i the diagram below. The poit (5 ; ) lies o f. The graphs itersect at poits P ad Q. y f f P (5 ; ) Q g 0. Determie the umerical values of a ad b. 0. Determie the miimum value of g() +. () 0. Determie the period of f. () 0.4 Show that, if the -coordiate of P is θ, the the -coordiate of Q is ( 80 θ ). []

49 Mathematics/P DBE/November 0 NSC QUESTION The figure below represets a triagular right prism with BA BC 5 uits, A Bˆ C 50 ad F ÂC 5. B A E D C F. Determie the area of ABC. (). Calculate the legth of AC. (). Hece, determie the height FC of the prism. () [8]

50 Mathematics/P DBE/November 0 NSC QUESTION. Prove that, if cos( α ) 0, si ( α). cos( α ) (). Determie the geeral solutio of cos cos. (7)... Prove that, for agles A ad B, si A cos A si( A B) si B cos B si B.. Hece, or otherwise, without usig a calculator, show that: (a) si 5B cos5b 4cos B () si B cos B (b) 4cos6 si8 () (c) si 8 is a solutio of the cubic equatio [4] TOTAL: 50

51 Mathematics/P DBE/November 0 NSC CENTRE NUMBER: EXAMINATION NUMBER: DIAGRAM SHEET QUESTION 4. 5 CUMULATIVE FREQUENCY GRAPH OF TIME TAKEN TO ANSWER THE QUESTION Cumulative Frequecy Graph of time take to aswer. 0 Cumulative frequecy Time Time (i (i miutes)

52 Mathematics/P DBE/November 0 NSC INFMATION SHEET: MATHEMATICS b ± b 4 ac a A P( + i) A P( i) A P( i) A P( + i) i i ( + ) i T ar a( r ) S F f [( + i) ] i f ( + h) f ( ) '( ) lim h 0 h r T a + ( ) d S ( a + ( d ) ; r [ ( + i) ] P i ( ) ( ) + y + y d + y y M ; y m + c y y m ) ( a) + ( y b) r I ABC: si a A area ABC ( b c a b + c bc. cos A si B si C ab. si C S ) a ; < r < r y y m m taθ ( α + β ) siα.cos β cosα. si β si( α β ) siα.cos β cosα. si β si + cos ( α + β ) cosα.cos β siα. si β cos ( α β ) cosα.cos β + siα. si β cos α si α cos α si α si α siα. cosα cos α ( ; y) ( cosθ + y siθ ; y cosθ siθ ) ( ; y) ( cosθ y siθ ; y cosθ + siθ ) ( i ) σ i f ( A) P( A) P(A or B) P(A) + P(B) P(A ad B) y ˆ a + b ( S ) b ( ) ( ) ( y y)

53 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P NOVEMBER 0 MEMANDUM MARKS: 50 This memoradum cosists of pages.

54 Mathematics/P DBE/November 0 NSC Memoradum NOTE: If a cadidate aswers a questio TWICE, oly mark the FIRST attempt. If a cadidate has crossed out a attempt of a questio ad ot redoe the questio, mark the crossed out versio. Cosistet accuracy applies i ALL aspects of the markig memoradum. Assumig aswers/values i order to solve a problem is ot acceptable. QUESTION. Media 4 aswer. Lower quartile Upper quartile 46 Iter quartile rage 46 4 Aswer oly: FULL MARKS () lower quartile upper quartile (). bo-adwhisker with a media skewess idicatig 5 umber summary 7; ; 4; 46; 6 or correct scale ().4 There is a greater spread of scores to the right of the media. There is a greater spread of scores i the top 50%. The spread of the scores o the left had side of the media is closer to each other. The greatest spread of scores lies betwee Q ad the maimum value. Note: Descriptio about the spread based o the bo-ad-whisker diagram must be accepted. If it is idicated that it is skewed to the left because the mea is less tha the media: full marks greater spread right of media () greater spread top 50% () spread closer left of media () greater spread betwee Q ad ma () [9]

55 Mathematics/P DBE/November 0 NSC Memoradum QUESTION. Mea i i 580 7,5 8 Aswer oly: FULL MARKS 580 () Note: If rouded off to 7: mark. Stadard deviatio (σ),78 (, ) Note: If rouded off to,8: mark. golfers scores lie outside stadard deviatio of the mea. The iterval for stadard deviatio of the mea is (7,5,78 ; 7,5 +,78) (69,7 ; 75,8) Aswer oly: FULL MARKS iterval umber () () [6] QUESTION Liear, the poits seem to form a straight lie. liear reaso. The greater the umber of hours spet watchig TV, the lower the test scores The less time a perso speds watchig TV, the higher the test score. Negative correlatio betwee the variables Idirect relatioship betwee the variables deductio.4 60 marks. (Accept marks) deductio () () () () [6]

56 Mathematics/P 4 DBE/November 0 NSC Memoradum QUESTION TIME FREQUENCY CUMULATIVE FREQUENCY t < t < t < t < t < 5 9 t < 0 Note: Oly cumulative frequecy colum full marks Cumulative Frequecy Graph of time take to aswer Oe mark for every two correct cumulative frequecy values () 5 Cumulative Frequecy Time (i miutes) upper limit cumulative frequecy (at least 4 of 6 y- values correctly plotted) groudig at ( ; 0) shape (ot joied by a ruler; smooth curve) 4. Estimated umber of learers that took less tha 4 miutes: approimately 5 learers (Accept 6) Approimate percetage 6,67% (Accept 0%) Note: If usig 9 learers ad approimate percetage 0%: mark If usig 5,5 learers ad approimate percetage 8,%: mark 5 learers 6,67% () [9]

57 Mathematics/P 5 DBE/November 0 NSC Memoradum QUESTION 5 y Q(4 ; a) P( 4 ; 0) O R(6 ; 0) S 5. m PQ m a 0 a a a 8 a 6 a 6 a ± 4 QR a 4; sice a > 0 PQ + QR PR (8 + a ) + (a + ) 0 a a 6 a 4 a 0 a or a 0 a or 4 6 usig gradiet of perpedicular lies a² 6 usig Pythagoras (8 + a ) + (a + ) 0 a² 6 Let A be the midpoit of diagoal PR The A ( ; ) A(; 0). AQ AR (diagoals equal ad bisect each other) AQ AR ( 4 ) + (0 a ) a 5 a 6 a 4 ( ; 0) is cetre AQ AR + a 5 a 6 Note: If cadidate uses a 4 at the begiig, the zero marks.

58 Mathematics/P 6 DBE/November 0 NSC Memoradum 5. Equatio of lie SR: 4 0 m PQ 4 ( 4) m SR mpq PQ SR y y m( ) y 0 y ( 6) m PQ m SR substitutio of m ad (6 ; 0) stadard form m PQ m PQ msr PQ SR y + c c c y S( ; 4) (traslatio) m RS 6 + y + 4 ( + ) y m PQ m SR substitutio of m ad (6 ; 0) stadard form S( ; 4) m SR substitutio of m ad ( ; 4) stadard form 5. Eq. of RS: y Eq. of SP: y 0 ( + 4) ( + 4) y 4 Aswer oly: FULL MARKS m eq. of SP value of value of y

59 Mathematics/P 7 DBE/November 0 NSC Memoradum Midpoit PR M ; ( ; 0) Let S ( ; y). The sice M( ; 0) is this, the midpoit of QS is: + y + y y + 4 ad y y y value of value of y The traslatio that seds Q(4 ; 4) to R(6 ; 0) also seds P( 4 ; 0) to S. (6 ; 0) (4 + ; 4 4) S ( 4 + ; 0 4) ( ; 4) The traslatio that seds Q(4 ; 4) to P( 4 ; 0) also seds R(6 ; 0) to S. ( 4 ; 0) (4 8 ; 4 4) S (6 8 ; 0 4) ( ; 4) m PQ m SR y 6 y 6 m PS m SR y y () + () : 0 () () Substitute : y 6 8 y 4 method or + 4 or y 4 method 8 or 8 4 or y 4 equatios usig the gradiet addig the equatios value of value of y 5.4 PR 6 ( 4) 0 Aswer oly: FULL MARKS 6 ( 4) 0 () PR PR 0 (6 + 4) + (0 0) substitutio i correct formula 0 ()

60 Mathematics/P 8 DBE/November 0 NSC Memoradum ( 4) midpoit PR ( ; ) (; 0) midpoit radius of circle PR 5 uits ( ) + (y 0) 5 ( ) + y 5 Aswer oly: FULL MARKS 5.6 ( ) + y 5 substitute Q(4 ; 4): LHS (4 ) RHS Q is a poit o the circle Note: If substitute poit ito equatio resultig i 5 5: mark No coclusio: mark radius eq. of circle i correct form () substitute Q(4;4) LHS RHS () Distace from cetre ( ; 0) to Q(4 ; 4) Q is a poit o circle, r 5 PR is the diameter of circle PQR therefore Q lies o circle ( P QR ˆ 90 ) (4 ) + y 5 y 6 y 4 Q is a poit o the circle 5 coclusio () diameter P QR ˆ 90 ) () substitute 4 coclusio () ( ) ( ) 9 4 Q is a poit o the circle substitute y 4 coclusio () 5.7 P eeds to shift at least 4 uits to the right ad S eeds to shift at least 4 uits up for the image of PQRS i first quadrat. miimum value of k is 4 ad miimum value of l is 4 miimum value of k + l is 8 Aswer oly: FULL MARKS Note: No CA mark applies i 5.7 if k ad l are ot miimums. k 4 l 4 k + l 8 () []

61 Mathematics/P 9 DBE/November 0 NSC Memoradum QUESTION 6 y A C F θ D θ B O 6. C B y C y B C( ; 6) 6. BA CA (taget radius) CA BC AB (Pythagoras) (5) ( 0 ) 5 CA 5 or,4 uits ta θ m m DC AB m DC m DC AB ta θ m value of value of y () BA CA or B A ˆC 90 substitutio ito Pythagoras () ta ratio ( i ay form) () m m DC AB m DC ta θ ()

62 Mathematics/P 0 DBE/November 0 NSC Memoradum 6.5 Eq. of DC: y 6 ( + ) y + Eq. of AB: y ( + ) y y ( ) y 5 A ( ; 5) Eq. of DC: y 6 ( + ) y + Eq. of AB: y ( + ) y At A: ( ) ad y ( ) 5 A( ; 5) Aswer oly: ( ; 5): mark DC: subst m ad ( ; 6) eq. of DC eq. of AB equatig equatios value of value of y DC: subst m ad ( ; 6) eq. of DC subt m ad ( ;) eq. of AB value of value of y (6) (6) Eq. of DC: y 6 ( + ) y + Eq. of circle: ( + ) + (y ) 0 At A: ( + ) + ( + ) 0 ( + ) + ( + ) ( + ) 0 ad y ( ) + 5 A( ; 5) DC: subst m ad ( ; 6) eq. of DC substitutio value of value of y (6)

63 Mathematics/P DBE/November 0 NSC Memoradum Draw AE BC 5 AE BE cos θ AE BE 4 5 A AE y A + BE A( ; 5) A 5 θ 5 C(-; 6) E θ B(-;) 5 AE 5 5 AE 5 BE 5 5 BE 5 (6) ( + ) + ( y ) 0 () y () ( + ) 5 + ( ) ( + )( ) 0 or subst () i () y subst m ad ( ;) eq of AB eq of circle substatio value of value of y (6)

64 Mathematics/P DBE/November 0 NSC Memoradum Equatio AC : y + 6 C ( ; 6) 5 A ( ; y) taθ θ 6,57 AP AP CP CP y 6 5 A( ; 5) θ 5 cos6,57 5 si 6,57 P θ 6, 57 AP 5 cos 6, 57 AP CP value of value of y (6) 6.6 Area Δ ABC ( 5 )( 0 ) 5 Eq. of DC is y + Therefore OF ad OD. 69 Area Δ ODF ( ) 4 69 Area ΔABC: Area ΔODF 5 : 0 : 69 4 DF ( ) 4. 5 DF (5)( 0) siθ ΔABC ΔODF. 5 ()( )siθ 0 69 ( 5 )( 0 ) OF OD ( ) + ( ) (5). 5 DF (5)( 0) siθ. 5 ()( )siθ (5)

65 Mathematics/P DBE/November 0 NSC Memoradum Δ ODF is a elargemet of Δ ABC area Δ ABC : area Δ ODF AB : OD 0 : OD Equatio of DC is y + D OD area Δ ABC : area Δ ODF AB : OD 0 : 69 elargemet AB :OD 0:OD aswer (5) [9] QUESTION 7 7. ( ; y) ( + 4; y) ( 4; y) ( ; y) ( 4; y) 7. New cetre ( ; 5) ( + ) + ( y + 5) y + y + 0y y y 4 y ( ; 5) ( + )² + (y + 5)² 6 simplificatio [8] QUESTION 8 8. Rotatio of 90 aticlockwise about the origi. Rotatio of 70 clockwise about the origi. Note: if reflectio of 90 aticlockwise: 0 marks 8. D(5 ; 4) D / (4 ; 5 ) rotatio 90 aticlockwise () rotatio 70 clockwise () G ( 7 ; 6) Area ABCD 5 0 square uits 45 Area MNRP 0 () () area ABCD 0 area MNRP 45 Area ABCD Area MNRP (uits) 4 5 ()

66 Mathematics/P 4 DBE/November 0 NSC Memoradum Product ( area ABCD) 9 (5 ) (uits) 5 Note: CA will apply if used i calculatio. () [9] QUESTION A r² r 7 40 cos A si (80 + A) si A 7 or si(80 + A) si80.cos A + cos80.si A 0.cos A.si A si A 7 cos00 ta 0 si ( 0 ) ( cos80 )( ta 60 ) ( si0 ) 40 ( cos80 ) ( ) ) ( cos80 ) A 7 Note: Aswer oly: 0 marks + cos80 Note: If (assume two + si0 egatives cacelled), o pealty ( 40 ; ) sketch r si A 7 () () si A () 7 cos 80 ta 60 or ta 60 si0 si0 cos 80 (6)

67 Mathematics/P 5 DBE/November 0 NSC Memoradum cos00 ta 0 si ( 0 ) ( cos80 )( ta 60 ) ( si0 ) ( si0 ) ( ) ) ( si0 ) cos 80 si0 ta 60 cos 80 si 0 (6) cos00 ta 0 si( 0 ) cos( ) ta si(0 ) si0 si0 ( ) 60 cos( ) si 0 si0 ta 60 (6) y P(4 ; ) Q O R M(a ; b) r 5 si ROP ˆ 0,6 5 5 ratio () 9.. RO ˆ P 6,87 o QOP ˆ 80 6, QOP ˆ 4, Aswer oly: Full Marks 6,869. 4, ()

68 Mathematics/P 6 DBE/November 0 NSC Memoradum 9.. cos + y si m a 4cos5 + si5 a,0 θ θ Note: Pealise mark for roudig icorrectly Note: If icorrect agle is used i the - formula: mark formula substitutio of values a,0 () Rotatio of 5 clockwise 45 aticlockwise cosθ y siθ m a 4cos 45 si 45 a,0 formula substitutio of values a,0 () ta P ˆ 4 P ˆ 6,86... M ˆ 78,... ˆ a cos M 5 a 5 cos 78, a,0 6,86 cos ratio a,0 () [8] QUESTION 0 0. f(5 ) a ta 5 a g(0) 4 b cos 0 4 b 4 0. Miimum value of g() Aswer oly: Full marks 0. Aswer oly: Full marks substitutio a substitutio b Period 60 Aswer oly: Full marks 60 () ()

69 Mathematics/P 7 DBE/November 0 NSC Memoradum 0.4 At P f ( θ ) g( θ ) ta θ 4cos θ ta θ 4cos θ for 80 θ : ta (80 θ) ta θ ta (80 θ) ad 4cos(80 θ) 4cos θ ta θ ta θ 4 cos θ at P 4cos(80 θ) ta θ 4 cos θ 4cos θ ta (80 θ) 4cos (80 θ) at Q ta (80 θ) 4cos (80 θ) taθ 4cosθ siθ cosθ cosθ siθ cos θ ( si θ ) si θ + siθ 0 ± 4()( ) siθ 4 siθ 0, θ 5, or 8,67 the - coordiate of Q is80 - p equatio si θ 0, , 8,67 [] QUESTION. Area ΔABC. AB. BC.si 50 (5)(5) si 50 9,58 uits² Area of ΔABC ()(5) si 5 )(5cos 5 ) 9,58 uits A B C substitutio ito correct formula () base ad height i terms of 5 ad 5 () Area of ΔABC [ (5cos 65 )(5si 65 )]() 9,58 uits base ad height i terms of 5 ad 65 ()

70 Mathematics/P 8 DBE/November 0 NSC Memoradum. AC AC , (5)(5) cos 50 AC 4,uits use of cosie rule substitutio () A ˆ C ˆ 65 (agles opposite equal sides) si 65 si 50 5 AC 5si 50 AC si 65 4,uits use of sie rule substitutio () ( AC) si 5 5 AC (5)si 5 4,uits B sketch/diagram si 5 AC 5 aswer () A C. ( AC) cos 65 5 AC (5) cos 65 AC 4,uits CF ta 5 AC CF 4, ta 5 CF,97 uits sketch/diagram ( AC) cos65 5 aswer () ratio CF as subject () FC 4, si 5 si 65 4,si 5 FC si 65,97 uits sie rule FC as subject () [8]

71 Mathematics/P 9 DBE/November 0 NSC Memoradum QUESTION. si( α) LHS cos( α ) si(90 + α) cos( α ) cos( α) cos( α ) cos( α ) cos( α ) subtractig 60 cos ( - α) cos(α - ) (). si[90 ( α )] LHS cos( α ) cos( α ) cos( α ) RHS cos cos cos cos (cos )(cos + ) 0 cos cos + cos 0 or cos /a 60 + k.60 ; k Z or 00 + k.60 ; k Z ±60 + k.60 ; k Z subtractig 60 writig as 90 - (α - ) cos(α - ) () cos cos factorisatio cos k.60 k Z (7).. LHS: si Acos B cos Asi B si B cos B si( A B) si B cos B si( A B) RHS si B cos B si( A B) si Bcos B LHS writig as sigle fractio comp. agle epasio comp. agle epasio simplificatio

72 Mathematics/P 0 DBE/November 0 NSC Memoradum LHS: si Acos B cos Asi B si B cos B si( A B) si B cos B si( A B) si B cos B si( A B) si B RHS writig as sigle fractio comp. agle epasio mult. by comp. agle epasio si( A B) RHS si B (si Acos B cos Asi B) si Bcos B si Acos B cos Asi B si Bcos B si Acos B cos Asi B si Bcos B si Bcos B si A cos A si B cos B LHS epasio epasio divide by write as separate fractios

73 Mathematics/P DBE/November 0 NSC Memoradum..(a) A 5B si 5B cos5b si(5b B) si B cos B si B si 4B si B 4si BcosB si B 4cosB si 5B cos5b si B cos B si 5Bcos B cos5bsi B si Bcos B si(5b B) si Bcos B si 4B ()si Bcos B si BcosB si B 4cosB recogisig A 5B substitutig A 5B si 4B si B cos B () writig as sigle fractio si 4B si B cos B compoud agle i deomiator..(b) B 8 si 90 cos90 4cos (8) si8 cos8 0 4cos6 si8 4cos6 si8 recogisig B 8 substitutig B 8 simplify () ()..(c) Let si 8 a 4cos6 si8 4( si 8 ) si8 4( a ) a 4a 8a 8a 4a + 0 Hece si8 is a solutio of si 8 a cos 6 si 8 substitutio of a simplificatio

74 Mathematics/P DBE/November 0 NSC Memoradum si8 4cos6 4( si si8 8 ) 4 8si 8 si8 8(si8 ) 4(si8) + 0 Hece si8 is a solutio of Note: substitutig si8 o ito usig a calculator showig equal to 0: 0 marks cos 6 si 8 simplificatio equatio i.t.o si 8 replacig si 8 [4] TOTAL: 50

75 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P NOVEMBER 0 MARKS: 00 TIME: hours This questio paper cosists of 9 pages, diagram sheets ad iformatio sheet.

76 Mathematics/P DBE/November 0 NSC INSTRUCTIONS AND INFMATION Read the followig istructios carefully before aswerig the questios This questio paper cosists of questios. Aswer ALL the questios. Clearly show ALL calculatios, diagrams, graphs, et cetera, that you have used i determiig your aswers. Aswers oly will ot ecessarily be awarded full marks. You may use a approved scietific calculator (o-programmable ad o-graphical), uless stated otherwise. If ecessary, roud your aswers off to TWO decimal places, uless stated otherwise. Diagrams are NOT ecessarily draw to scale. THREE diagram sheets for aswerig QUESTION 7., QUESTION 8., QUESTION 8., QUESTION 9, QUESTION 0 ad QUESTION are attached at the ed of this questio paper. Write your cetre umber ad eamiatio umber o these sheets i the spaces provided ad isert them iside the back cover of your ANSWER BOOK. A iformatio sheet, with formulae, is icluded at the ed of the questio paper. Number the aswers correctly accordig to the umberig system used i this questio paper. Write legibly ad preset your work eatly.

77 Mathematics/P DBE/November 0 NSC QUESTION Cosider the followig recursive formula: T T ; k ; T k + k. Write dow the first FOUR terms of the sequece. (). How may terms of the above sequece must be added to give a sum of 0? () [6] QUESTION The weights of a radom sample of boys i Grade were recorded. The cumulative frequecy graph (ogive) represets the recorded weights. Cumulative frequecy curve showig weight of boys Cumulative frequecy Weight (i kilograms). How may of the boys weighed betwee 90 ad 00 kilograms? (). Estimate the media weight of the boys. (). If there were 50 boys i Grade, estimate how may of them would weigh less tha 80 kilograms? ().4 It was suggested that the first 50 boys i Grade to arrive at school o that day, be selected as a sample. Eplai why this would ot be a radom sample. () [5]

78 Mathematics/P 4 DBE/November 0 NSC QUESTION Let A ad B be two evets i a sample space. Suppose that P(A) 0,4; P(A or B) 0,7 ad P(B) k.. For what value of k are A ad B mutually eclusive? (). For what value of k are A ad B idepedet? [6] QUESTION 4 The time take for a pizza outlet to deliver to a customer is recorded. The data is foud to be ormally distributed with a mea time of 4 miutes ad a stadard deviatio of miutes. ±68% mea σ stadard deviatio ±96% ±00% σ σ σ + σ + σ + σ Aswer the followig questios with referece to the iformatio provided i the graph. 4. What percetage of pizzas are delivered betwee ad 4 miutes? () 4. What percetage of pizzas are delivered betwee 5 ad 7 miutes? () 4. The outlet advertises that they will ot charge for a pizza that takes loger tha a certai time to deliver. If they wat to give away o more tha % of all deliveries, how may miutes should they allow for delivery? () [8]

79 Mathematics/P 5 DBE/November 0 NSC QUESTION 5 The digits 0,,,, 4, 5 ad 6 are used to make digit codes. 5. How may uique codes are possible if digits ca be repeated? () 5. How may uique codes are possible if the digits caot be repeated? () 5. I the case where digits may be repeated, how may codes are umbers that are greater tha 00 ad eactly divisible by 5? () [7] QUESTION 6 Complaits about a restaurat fell ito three mai categories: the meu (M), the food (F) ad the service (S). I total 7 complaits were received i a certai moth. The complaits were as follows: 0 complaied about the meu. 55 complaied about the food. 67 complaied about the service. 0 complaied about the meu ad the food, but ot the service. complaied about the meu ad the service, but ot the food. 6 complaied about the food ad the service, but ot the meu. The umber who complaied about all three is ukow. 6. Draw a Ve diagram to illustrate the above iformatio. (6) 6. Determie the umber of people who complaied about ALL THREE categories. () 6. Determie the probability that a complait selected at radom from those received, complaied about AT LEAST TWO of the categories (that is. meu, food ad service). () []

80 Mathematics/P 6 DBE/November 0 NSC QUESTION 7 The outdoor temperature, i C, at oo o te days ad the umber of uits of electricity used to heat a house o each of those days, are show i the table below. Noo temperature (i C) Uits of electricity used Draw a scatter graph that shows this iformatio o the grid provided o DIAGRAM SHEET. () 7. Determie the equatio of the least squares regressio lie. 7. Determie the correlatio coefficiet. () 7.4 What ca we coclude about the relatioship betwee the oo temperature ad the umber of uits of electricity used for heatig? () 7.5 Estimate the umber of uits of electricity that was used to heat a house o a day whe the outdoor temperature at oo was 8 C. () []