NATIONAL SENIOR CERTIFICATE GRADE 12

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1 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P NOVEMBER 0 MARKS: 0 TIME: hours This questio paper cosists of 9 pages 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 o-graphical), uless stated otherwise. If ecessary, roud off aswers to TWO decimal places, uless stated otherwise. Diagrams are NOT ecessarily draw to scale. 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 i each of the followig:.. ( )( + ) 0 ().. (Leave your aswer correct to TWO decimal places.) () < 0 (). Give: y ad y 8.. Solve for ad y simultaeously. (6).. The graph of y is reflected across the lie havig equatio y. What is the equatio of the reflected lie? (). The solutios of a quadratic equatio are give by For which value(s) of p will this equatio have: ± p + 7 QUESTION.. Two equal solutios ().. No real solutios () []. + ; ; 7 are the first three terms of a arithmetic sequece. Calculate the value of. (). The first ad secod terms of a arithmetic sequece are 0 ad 6 respectively... Calculate the th term of the sequece. ().. The sum of the first terms of this sequece is 60. Calculate. (6) [0]

4 Mathematics/P DBE/November 0 NSC QUESTION. Give the geometric sequece: 7 ; 9 ;.. Determie a formula for T, the th term of the sequece. ().. Why does the sum to ifiity for this sequece eist? ().. Determie S. (). Twety water taks are decreasig i size i such a way that the volume of each tak is the volume of the previous tak. The first tak is empty, but the other 9 taks are full of water. Would it be possible for the first water tak to hold all the water from the other 9 taks? Motivate your aswer. (). The th term of a sequece is give by T ( ) Write dow the first THREE terms of the sequece. ().. Which term of the sequece will have the greatest value? ().. What is the secod differece of this quadratic sequece? ().. Determie ALL values of for which the terms of the sequece will be less tha 0. (6) []

5 Mathematics/P DBE/November 0 NSC QUESTION. Cosider the fuctio f ( ) Calculate the coordiates of the y-itercept of the graph of f. ().. Calculate the coordiates of the -itercept of the graph of f. ().. Sketch the graph of f i your ANSWER BOOK. Clearly show ALL asymptotes ad itercepts with the aes. ().. Write dow the rage of f. (). S( ; 0) ad T(6 ; 0) are the -itercepts of the graph of f ( ) a + b + c ad R is the y-itercept. The straight lie through R ad T represets the graph of g ( ) + d. y R f g S( ; 0) T(6 ; 0) 0.. Determie the value of d. ().. Determie the equatio of f i the form f ( ) a + b + c. ().. If f ( ) + +, calculate the coordiates of the turig poit of f. ().. For which values of k will f ( ) k have two distict roots? ().. Determie the maimum value of f ( ) h ( ). () [0]

6 Mathematics/P 6 DBE/November 0 NSC QUESTION The graph of f ( ) 7 for 0 is sketched below. The poit P( ; 9) lies o the graph of f. y 0 P( ; 9) 9) f. Use your graph to determie the values of for which f ( ) 9. (). Write dow the equatio of f i the form... y Iclude ALL restrictios. (). Sketch f, the iverse of f, i your ANSWER BOOK. Idicate the itercept(s) with the aes ad the coordiates of ONE other poit. (). Describe the trasformatio from f to g if g() 7, where 0. () [9] QUESTION 6 The graph of a hyperbola with equatio y f () has the followig properties: Domai: R, Rage: y R, y Passes through the poit ( ; 0) Determie f ( ). []

7 Mathematics/P 7 DBE/November 0 NSC QUESTION 7 7. A busiess buys a machie that costs R The value of the machie depreciates at 9% per aum accordig to the dimiishig-balace method. 7.. Determie the scrap value of the machie at the ed of years. () 7.. After five years the machie eeds to be replaced. Durig this time, iflatio remaied costat at 7% per aum. Determie the cost of the ew machie at the ed of years. () 7.. The busiess estimates that it will eed R by the ed of five years. A sikig fud for R90 000, ito which equal mothly istalmets must be paid, is set up. Iterest o this fud is 8,% per aum, compouded mothly. The first paymet will be made immediately ad the last paymet will be made at the ed of the -year period. Calculate the value of the mothly paymet ito the sikig fud. () 7. Lorraie receives a amout of R upo her retiremet. She ivests this amout immediately at a iterest rate of 0,% per aum, compouded mothly. QUESTION 8 She eeds a amout of R8 000 per moth to maitai her curret lifestyle. She plas to withdraw the first amout at the ed of the first moth. For how may moths will she be able to live from her ivestmet? (6) [7] 8. Determie f () from first priciples if f ( ). () dy 8. Evaluate if y +. () d 8. Give: g ( ) Calculate g () for. () 8.. Eplai why it is ot possible to determie g (). () []

8 Mathematics/P 8 DBE/November 0 NSC QUESTION 9 9. The graph of the fuctio f ( ) is sketched below. y f Calculate the -coordiates of the turig poits of f. () 9.. Calculate the -coordiate of the poit at which f () is a maimum. () 9. Cosider the graph of g ( ) Determie the equatio of the taget to the graph of g at. () 9.. For which values of q will the lie y + q ot itersect the parabola? () 9. Give: h( ) + QUESTION 0 Eplai if it is possible to draw a taget to the graph of h that has a egative gradiet. Show ALL your calculatios. () [7] A particle moves alog a straight lie. The distace, s, (i metres) of the particle from a fied poit o the lie at time t secods ( t 0 ) is give by s ( t) t 8t Calculate the particle's iitial velocity. (Velocity is the rate of chage of distace.) () 0. Determie the rate at which the velocity of the particle is chagig at t secods. () 0. After how may secods will the particle be closest to the fied poit? () [6]

9 Mathematics/P 9 DBE/November 0 NSC QUESTION A calculator compay maufactures two kids of calculators: scietific ad basic. The compay is able to sell all the calculators that it produces. A system of costraits has bee developed for the productio of the calculators. The feasible regio is shaded below. Let ad y respectively be the umber of scietific ad basic calculators produced each day. 0 y 0 Basic Calculators 0 A 0 B 0 FEASIBLE D REGION C Scietific Calculators P. Is it possible for the compay to maufacture scietific calculators ad basic calculators i oe day accordig to their system of costraits? Motivate your aswer. (). Write dow all the algebraic iequalities which describe the costraits related to the maufacturig of the calculators. (6). The profit Q (i hudreds of rads) is give by Q + y. The dotted lie o the graph is a search lie associated with the profit fuctio... Idetify the poit i the regio where the profit is a maimum. Use oly A, B, C or D. ().. Write dow the coordiates of a poit o the dotted lie (if the poit eists) at which the profit is greater tha the profit at P. ().. Give that the profit, whe give by Q a + by ( a > 0 ; b > 0 ), is a maimum at B, determie the maimum value of b a. () [] TOTAL: 0

10 Mathematics/P DBE/November 0 NSC INFMATION SHEET: MATHEMATICS b ± b 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 ) ( 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: 0 This memoradum cosists of 0 pages.

12 Mathematics/P DBE/November 0 NSC Memoradum NOTE: If a cadidate aswered a questio TWICE, mark the FIRST attempt ONLY. If a cadidate crossed out a attempt of a questio ad did ot redo the questio, mark the crossed out questio. Cosistet accuracy applies i ALL aspects of the memoradum. QUESTION.. ( )( + ) 0.. or 0 b ± ( ) ± ( ) ( )( ) ( ) ± 6 6,7 b ac a or, Note: if a cadidate uses icorrect formula award ma mark (for stadard form) Note: if a cadidate has ot rouded off correctly, pealise mark aswer aswer stadard form () subs ito correct formula aswer () + 6 ± 6 ± 6, or, divisio by ± 6 aswer 6 6 ()

13 Mathematics/P DBE/November 0 NSC Memoradum b ± ±, < 0 ( + 8)( ) < 0 b ac a ( ) ( ) ( )( ) ± ( ) 6 9 or, stadard form subs ito correct formula aswer () factors , ( + 8)( ) Therefore the solutio is: 8 < < (8 ; ) -8 aswer () < 0 ( + 8)( ) < 0 factors 8, + 8 < 0 ad > 0 or + 8 > 0 ad < 0 < 8 ad > > 8 ad < No solutio Therefore the solutio is: 8 < < (8 ; ) -8 aswer ()

14 Mathematics/P DBE/November 0 NSC Memoradum NOTE: I this alterative, award ma / marks sice there is o coclusio < 0 ( + 8)( ) < 0 factors 8, 8 graph with bolded lie.. y ad y 8 y (y ) y 8 ( y ) y y y 0 ( y + )( y ) 0 y or y 8 or ( ; y) ( 8 ; ) or ( ; ) Note: If cadidate makes a mistake which leads to both equatios beig LINEAR award maimum /6 marks y substitutio factors y-values -values (6) y ad y 8 y (y ) y 8 ( y ) y By ispectio y or y 8 or ( ; y) ( 8 ; ) or ( ; ) y substitutio y-values -values (6) y ad y 8 y (y ) y 8 ( y ) y y y 0 y ( ) ± ( ) ( )( ) ( ) y or y 8 or ( ; y) ( 8 ; ) or ( ; ) y substitutio subs ito correct formula y-values -values (6)

15 Mathematics/P DBE/November 0 NSC Memoradum y ad y 8 y ( + 8)( ) 0 8 or y or y ( ; y) ( 8 ; ) or ( ; ) y + substitutio factors -values y-values (6) y ad y 8 y ± 8 y y y y 8 0 y y 0 0 ( y )( y + ) ( )( ) ( ) 8 or y or y ( ; y) ( 8 ; ) or ( ; ) y 8 ad y 8 y y or y 8 or ( ; y) ( 8 ; ) or ( ; ) y + substitutio subs ito correct formula -values y-values (6) 8 y substitutio factors y-values -values (6)

16 Mathematics/P 6 DBE/November 0 NSC Memoradum y 8 ad y 8 y 8 y y y y 8 0 y y y ( ) ± ( ) ( )( ) ( ) y or y 8 or ( ; y) ( 8 ; ) or ( ; ) y 8 ad y 8 y ( + 8)( ) 8 or y or y ( ; y) ( 8 ; ) or ( ; ) y 8 ad y 8 y 8 ± + ( )( ) ( ) 8 or y or y ( ; y) ( 8 ; ) or ( ; ) 8 y substitutio subs ito correct formula y-values -values (6) 8 y substitutio factors -values y-values (6) 8 y substitutio subs ito correct formula -values y-values (6)

17 Mathematics/P 7 DBE/November 0 NSC Memoradum.. y iterchages ad y () y y + y y + p + 0 p + 0 p p p + < 0 p < p + 0 or p + 0 or aswer aswer ± 7 0 () () []

18 Mathematics/P 8 DBE/November 0 NSC Memoradum QUESTION. ( ) ( ) T T T T ( ) ( ) T T T ( ) ( ) ( ) ( ) ( ) T T T T T T T T or ( ) ( ) 7 + aswer () T T T + or ( ) ( ) aswer () ( ) T T T T or ( ) ( ) ( ) ( ) aswer ().. ( ) ( )( ) T d a T 0; 6; ; ; 6; 0; ; 8; ; 6; 0 0 T d aswer () epads sequece aswer ()

19 Mathematics/P 9 DBE/November 0 NSC Memoradum.. S ( 0)( + ) oly S ( 0)( + ) 0 oly 0 0 oly [ a + ( ) d] [ ( 0) + ( )( ) ] + or [ a + ( ) d] [ ( 0) + ( )( ) ] + ( 6) ± ( 6) ( )( 80) ( ) or S Note: if cadidate substitutes ito icorrect formula, award 0/6 [ a + ( ) d] [ ( 0) + ( )( ) ] or + Note: if cadidate writes aswer oly, award /6 marks correct formula substitutio of a ad d subs S or 0 0 or factors selects 0 oly correct formula substitutio of a ad d subs S or or subs ito correct formula selects 0 oly correct formula substitutio of a ad d subs S or or factors selects 0 oly (6) (6) (6)

20 Mathematics/P 0 DBE/November 0 NSC Memoradum S T S S 0 sequece epaded series calculated aswer (6) [0] QUESTION.. T ar 7.. < r < or r < Note: The fial aswer ca also be writte as or Note: If cadidate cocludes series is ot coverget, award 0 marks. The commo ratio (r) is which is betwee ad. a 7 ad r substitute ito correct formula aswer aswer () () ().. < < a S r 7 8 or 0, or Note: If r > or r < is substituted the 0/ marks. aswer substitutio aswer () ()

21 Mathematics/P DBE/November 0 NSC Memoradum. Let V be the volume of the first tak. V V V ; ;... 8 Note: If cadidate lets the volume of the first tak be a specific value 9 V (istead of a variable) ad his/her argumet follows correctly, award S / marks 9 Note: If cadidate aswers Yes 87 oly with o justificatio: / marks V 88 0, V < V Yes, the water will fill the first tak without spillig over. Let V be the volume of the first tak. V V V ; ; V S 9 9 V < V V Yes, the water will fill the first tak without spillig over. Let V be the volume of the first tak. V V V ; ;... 8 V S V Sice the first tak will hold the water from ifiitely may taks without spillig over, certaily: Yes, the first tak will hold the water from the other 9 taks without spillig over. V substitute ito correct formula aswer coclusio V substitute ito correct formula observes that 9 < coclusio V substitute ito correct formula correct argumet () () ()

22 Mathematics/P DBE/November 0 NSC Memoradum If the taks are emptied oe by oe, startig from the secod, each tak will fill oly half the remaiig space, so the first tak ca hold all the water from the other 9 taks... T ( ) + 8 Term Term 0 Term 0 Yes (eplicit or uderstood from the argumet.) argumet () 0 0 ().. Term T aswer ().. Secod differece a Secod differece ( ) Secod differece subs ito a aswer () Secod differece 0 first differeces secod differece ().. ( ) + 8 < 0 ( ) + 8 < < < 0 Note: Aswer oly award /6 marks T < 0 stadard form factors 0 9 > 0 ( )( + ) > < or > ; Ν > ; Ν critical values iequalities > (accept: ) (6)

23 Mathematics/P DBE/November 0 NSC Memoradum ( ) ( ) ( ) + 8 < 0 6 > 0 [( ) 8] [( ) + 8] + 8 < 0 > 0 ( )( + ) > 0 T < 0 ( ) 6 > 0 factors critical values < or > ; Ν > ; Ν iequalities > (accept: ) (6) ( ) + 8 < 0 ( ) < 8 ( ) > 6 < 8 or < or > 8 > ; Ν > ; Ν T < 0 ( ) > 8 8 ad 8 > 8 < 8 > (accept: ) (6) T T + 0 < 0 ( ) T < < > 0 ( )( + ) > 0 stadard form factors critical values < or > ; Ν > ; Ν iequalities > (accept: ) (6) ; 0 ; 0 ; 6 ; 8 ; 6 ; 0 ; 0 ; ; ; ; 80 ; 0 ; Ν epasio coclusio of (accept > ) (6) []

24 Mathematics/P DBE/November 0 NSC Memoradum QUESTION.. y. 0 6 y 6 y (0 ; ) aswer () (; 0) Note: If a cadidate iterchages questio.. ad..: 0/ marks Note: If a cadidate says that. 6 (i.e. wrog mathematics ) s/he will arrive at correct aswer BUT award ma / y 0 -value ().. y ( ; 0) itercepts (0 ; -) f asymptote shape () - -6 y y > 6 ( 6 ; ) aswer ()

25 Mathematics/P DBE/November 0 NSC Memoradum. y R f S( ; 0) T(6 ; 0) 0 g.. y + d 0 ( )(6) + d d substitutio aswer () y y m y 0 ( ) ( 6) y + d substitutio aswer ().. Sice m ad d 6 d y a( 6)( + ) a(0 6)(0 + ) a y ( ) + + m d 6 Note: No marks for aswer oly. substitutio aswer () y a( 6)( + ) subs R(0 ; ) a-value y + + () y a + b + ( ) + ( ) + ( 6) + ( 6) + 0 a b i.e. 0 a b + 0 a b i.e. 0 6a + 6b + 0 b 96 0 a a y ( ) b y a + b + subs S( ;0) ad T ( 6;0) b-value y + + ()

26 Mathematics/P 6 DBE/November 0 NSC Memoradum.. y a + ( ) q 0 a ( ) + q or a ( 6 ) + q a ( 0 ) + q 0 i.e. 0 6a + q i.e. a + q y ( ) + 6 ( + ) + + ( 6)( + ) + 6 y a a( ) ( + + dy 0 d + 0 y () + () + 6 TP of f is ( ; 6) b a ( ) y () + () + 6 TP of f is ( ; 6) f ( ) ( ) TP of f is ( ; 6) + 6 a a q 6 y a( ) + q subs R(0 ; ) ad S( ; 0) (or T(6 ; 0) ) a-value y + + () y a( 6)( + ) epad a-value y + + () -value y-value -value y-value -value y-value () () ()

27 Mathematics/P 7 DBE/November 0 NSC Memoradum + 6 y () 6 + () + -value y-value () TP of f is ( ; 6).. < 6 ;6 aswer k ( ) ().. f ( ) Maimum value of h ( ) occurs at ma value of f() 6 Maimum value 8 Maimum value of h() f () 6 or 8 f ( ) h ( ) occurs at ma value of f() f ( ) + ( ) which has a maimum value of f ( ) Maimum value of h ( ) is or 8 subs 6 for f() or 8 () subs 6 for f() or 8 () subs f() or 8 () [0]

28 Mathematics/P 8 DBE/November 0 NSC Memoradum QUESTION. 0 0 Note: if the cadidate gives 0 0 < <, award / marks (). f - : 7y iterchage - ad y- values 7y y y 0 ( ;0] or ( ;0) () y. [ ;] f. Reflectio about the -ais ( ; y) ( ; y) ; 0 P(9 ; ) 0 shape ed at origi ay other poit o the graph () aswer () aswer () [9]

29 Mathematics/P 9 DBE/November 0 NSC Memoradum QUESTION 6 a f ( ) + a 0 + () a a f ( ) QUESTION 7 ( )( y ) ( )( 0 ) ( )( y ) + k k k y + NOTE: f ( ) as a alterative simplified form. + substitutio of ( ; 0) a ( ) ( y ) () substitutio of ( ; 0) k () [] A P( i) i, ad P idetified 0 000( 0,09) subs ito R7 88,86 NOTE: correct formula Icorrect formula aswer (i 7.. or 7..) A P( + i) award ma / i, ad P idetified 0 000( + 0,07) marks subs ito R68 06, correct formula aswer 7.. Sikig fud eeded: F v R [( + i) ] Fv i 6 0, ,08 R 8,68 NOTE: Icorrect formula award ma / marks Fv R i 0,08 () () 7 00 i auity formula 6 subs ito correct formula aswer ()

30 Mathematics/P 0 DBE/November 0 NSC Memoradum Cosider the sceario as moey deposited at the begiig of every moth, but i the last moth a additioal paymet was made at the ed of the moth: [ ] ( + i) ( + i) Fv + i ( + i) [( + i) ] + i 60 0,08 0, ,08 0, ,08 0,08 0, R8, 68 Preset value of sikig fud eeded: 6 0, Pv + P R8,0 v Usig the preset value formula: [ ( + i) ] Pv i 6 0,08 + 8,0 0,08 R 8,68 0,08 7 i 00 i auity formula 60 i auity formula Fv R subs ito correct formula aswer i 0,08 () 7 00 i auity formula 6 P R8, 0 v subs ito correct formula aswer ()

31 Mathematics/P DBE/November 0 NSC Memoradum 7. [ ( + i) ] Pv i 0, ,0 0, , log 0, ,0 moths She will be able to maitai her curret lifestyle for a little more tha 66 moths usig her pesio moey. Note: If cadidate rouds off early i Questio 7. (ad [ ( + i) ] obtai 8 moths), Pv i pealise mark 0, ,0 0, , ,0 9 log + log 6 66,0 moths She will be able to maitai her curret lifestyle for a little more tha 66 moths usig her pesio moey. Note: If Fv formula used, possibly award oe each for, i, use of logs: ma /6 marks If ay other icorrect formula is used, award 0/6 marks i 0,0 i auity formula subs ito correct formula simplificatio use of logs aswer i moths (6) i 0,0 i auity formula subs ito correct formula simplificatio use of logs aswer i moths (6)

32 Mathematics/P DBE/November 0 NSC Memoradum ( + i) 0, ,0 0, A F v [( + i) ] i 0, ,0 0, , , ,0 0, P 6 0, log 0, ,0 moths ,0 0, , She will be able to maitai her curret lifestyle for a little more tha 66 moths usig her pesio moey ,0 i i auity formula subs ito correct formula simplificatio use of logs aswer i moths (6) [7] QUESTION 8 8. f ( ) f ( + h) ( + h) f ( + h) f ( ) h + h + h + h h + h f ( ) lim h 0 h h( + h) lim h 0 h lim( + h) h 0 Note: If cadidate makes a otatio error Pealise mark Note: If cadidate uses differetiatio rules Award 0/ marks substitutio of of + h simplificatio to h + h formula lim( + h) h 0 aswer ()

33 Mathematics/P DBE/November 0 NSC Memoradum f dy d ( ) lim h 0 lim h 0 f ( + h) f ( ) h [ ( + h) ] ( ) [ ( + h + h ) ] lim h 0 h [ + h + h lim h 0 h h + h lim h 0 h h( + h) lim h 0 h lim h ( + h) g ( ) ( + )( ) + g ( ) ( ) ( ) h ] 8.. The fuctio is udefied at. Divisio by zero is udefied. The deomiator caot be zero. I the defiitio of the derivative, g ( ) does ot eist. + + Note: otatio error pealise mark g lim h 0 Note: cadidates do NOT eed to give their aswer with positive epoets ( + h) g( ) h, but g ( ) formula substitutio of + h simplificatio h + h to h lim( + h) h 0 aswer () 6 () simplificatio aswer aswer () () []

34 Mathematics/P DBE/November 0 NSC Memoradum QUESTION f ( ) f ( ) ( + 8)( ) 0 8 or f ( ) f ( ) ± 8 or ( )( 6) ( ) Note: if either f ( ) 0 or eplicitly stated, award maimum / marks f ( ) + 6 f or ( ) factors values () f ( ) + 6 f or ( ) subs ito formula values () 9.. f ( ) f ( ) aswer () 8 + f ( ) + 6 ( ) ( ) 8 + aswer ( ) ( ) aswer () ()

35 Mathematics/P DBE/November 0 NSC Memoradum f ( ) ( ) ( ) g( ) g( ) ( ) g ( ) 9 m ta ( ) 9 y + c ( ) ( ) + c 7 y + 7 g( ) 9 + g( ) ( ) g ( ) 9 m ta + c ( ) 9 y ( + ) y + 7 y 9( ) + ( ) ( ) aswer ( ) g g ( ) 9 m ta () aswer () ( ) g g ( ) 9 m ta aswer () 7 sketch q > 7 7 correct iequality () y + q ad y q 9 + q q > 7 ( + ) + 7 method 7 correct iequality ()

36 Mathematics/P 6 DBE/November 0 NSC Memoradum y + q ad y q + + q 0 ± 6 ()( q ) () ± 6 8q 6 8q < 0 q > Sice g ( ) ad at, taget equatio is y + 7, y + q ot itersectig g < ( ) + q < q 7 < q 9. h ( ) + For all values of : > 0 For all values of : h ( ) > 0 All tagets draw to h will have a positive gradiet. It will ever be possible to draw a taget with a egative gradiet to the graph of h. h ( ) + Suppose h ( ) < 0 ad try to solve for : + < 0 < but is always positive o solutio for h ( ) 0 for all R i.e. there are o tagets with egative slopes method 7 correct iequality () method 7 correct iequality () h ( ) + clearly argues that h ( ) > 0 coclusio h ( ) + () clearly argues that h ( ) < 0 is impossible coclusio ()

37 Mathematics/P 7 DBE/November 0 NSC Memoradum h ( ) + y h ( ) + h' Sice clearly h ( ) > 0 for all R, it will ever be possible to draw a taget with a egative gradiet to the graph of h. argues h ( ) > 0 by drawig a sketch coclusio () [7]

38 Mathematics/P 8 DBE/November 0 NSC Memoradum QUESTION 0 0. s ( t) t 8t + s ( t) t 8 s ( ) ( ) 8 m / s Note: aswer oly award 0/ marks s (t) subs t 0 ito s ( t) formula aswer () 0. ( t) 0. s m/s aswer t 8 0 t 8 9 t secods or,secods s '( t) 0 aswer () () s 9 9 t + 9 t secods or ( t),secods ( ) 9 9 s t t + aswer () s ( t) t 8t + 8 t ( ) 9 t secods or,secods 8 t ( ) aswer () [6]

39 Mathematics/P 9 DBE/November 0 NSC Memoradum QUESTION 0 y 0 Basic Calculators 0 A 0 B 0 FEASIBLE REGION D C Scietific Calculators P. No, because ( ; ) does ot lie withi the feasible regio. No, because accordig to the costraits, the -value (umber of scietific calculators) must be at least y 0 + y 0 + y y y 0 y + 0y 0 y y y 0 y 0 y 0 aswer (with motivatio) () 0 + y 0 + y 0 y 0 (6).. A aswer.. All poits o the search lie yield the same profit. Hece o such poit eists. No poit eists () () If such a ( ; y) eists, Q + y ad y + so y + Q Q 00 Hece, there is o such poit. No poit eists ()

40 Mathematics/P 0 DBE/November 0 NSC Memoradum.. Q a + by y a b a b a b Q + b The maimum value of b a is. a Q y + b b a b a b aswer () [] TOTAL: 0

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

42 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. ad QUESTION 7. 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.

43 Mathematics/P DBE/November 0 NSC QUESTION The scatter plot below shows the age (i years) ad the average height (i cetimetres) of boys betwee ad years. Scatter Plot Average height (i cm) Age of boys (i years) [Source: Use the scatter plot to determie the average height of a 7-year-old boy. (). Describe the tred i the scatter plot. (). What is the approimate icrease i the average height per aum betwee the ages of ad years? (). Eplai why the observed tred CANNOT cotiue idefiitely. () [6]

44 Mathematics/P DBE/November 0 NSC QUESTION Abe plays for his school's cricket team. The umber of rus scored by Abe i the eight games that he batted i, is show below. (Abe was give out i all of the games.) Determie the average rus scored by Abe i the eight games. (). Determie the stadard deviatio of the data set. (). Abe's scores for the first three of the et eight games were, ad respectively. Describe the effect of his performace o the stadard deviatio of this larger set havig data poits. (). Abe hopes to score a average of 0 rus i the first 6 games. What should his average i the last five games be so that he may reach his goal? () [9] QUESTION I a certai school 60 learers wrote eamiatios i Mathematics ad Physical Scieces. The bo-ad-whisker diagram below shows the marks (out of 00) that these learers scored i the Physical Scieces eamiatio. Physical Scieces. Write dow the rage of the marks scored i the Physical Scieces eamiatio. (). Use the iformatio below to draw the bo-ad-whisker diagram for the Mathematics results o DIAGRAM SHEET. Miimum mark 0 Rage Upper quartile 70 Iterquartile rage 0 Media (). How may learers scored less tha 70% i the Mathematics eamiatio? (). Joe claims that the umber of learers who scored betwee 0 ad i Physical Scieces is smaller tha the umber of learers who scored betwee 0 ad i Mathematics. Is Joe's claim valid? Justify your aswer. () [9]

45 Mathematics/P DBE/November 0 NSC QUESTION As part of a evirometal awareess iitiative, learers of Greeside High School were requested to collect ewspapers for recyclig. The cumulative frequecy graph (ogive) below shows the total weight of the ewspapers (i kilograms) collected over a period of 6 moths by 0 learers. Weight of ewspaper collected (i kilograms). Determie the modal class of the weight of the ewspapers collected. (). Determie the media weight of the ewspapers collected by this group of learers. (). How may learers collected more tha 60 kilograms of ewspaper? () []

46 Mathematics/P 6 DBE/November 0 NSC QUESTION ABCD is a rhombus with A( ; 8) ad C( ; ). The diagoals of ABCD bisect each other at M. The poit E(6 ; ) lies o BC. A( ; 8) θ y B M E(6 ; ) P Q R S T D C( ; ). Calculate the coordiates of M. (). Calculate the gradiet of BC. (). Determie the equatio of the lie AD i the form y m + c. (). Determie the size of θ, that is B ÂC. Show ALL calculatios. (6) []

47 Mathematics/P 7 DBE/November 0 NSC QUESTION 6 A circle cetred at N( ; ) touches the -ais at poit L. The lie PQ, defied by the equatio y +, is a taget to the same circle at poit A. y Q A N( ; ) K O L B P 6. Why is NL perpedicular to OL? () 6. Determie the coordiates of L. () 6. Determie the equatio of the circle with cetre N i the form ( a) + (y b) r () 6. Calculate the legth of KL. () 6. Determie the equatio of the diameter AB i the form y m + c. () Show that the coordiates of A are ;. () 6.7 Calculate the legth of KA. () 6.8 Why is KLNA a kite? () 6.9 Show that A Bˆ K. () 6.0 If the give circle is reflected about the -ais, give the coordiates of the cetre of the ew circle. () []

48 Mathematics/P 8 DBE/November 0 NSC QUESTION 7 Cosider the diagram below where A( ; ), B( ; ) ad C( ; ) are the vertices of ABC. y 0 8 A C 6 B / A / C / B Describe the sigle trasformatio of ABC to A / B / C /. () 7. Write dow the geeral rule of the trasformatio i QUESTION 7.. () 7. A / B / C / is elarged by a scale factor of to form A // B // C //. Draw the elargemet o DIAGRAM SHEET. () 7. Write dow the geeral rule of the trasformatio i QUESTION 7.. () 7. ABC is reflected about the -ais ad the it is reflected about the y-ais to form DEF. 7.. Write dow the coordiates of D, where D is the image of A after the trasformatio described above. () 7.. Write dow the geeral rule of this trasformatio i the form: ( ; y) ( ; ) ( ; ). () 7.. Describe a sigle trasformatio that ABC udergoes to form DEF. () []

49 Mathematics/P 9 DBE/November 0 NSC QUESTION 8 Aswer this questio WITHOUT usig a calculator. 8. The poit P(k ; 8) lies i the first quadrat such that OP 7 uits ad T ÔP α as show i the diagram alogside. y 7 P(k ; 8) O α T 8.. Determie the value of k. () 8.. Write dow the value of cosα. () 8.. If it is further give that α + β 80, determie cos β. () 8.. Hece, determie the value of si( β α). () 8. Cosider the epressio: cos si si cos cos si 8.. Prove that: ta si cos () 8.. The above epressio is udefied if si cos 0. Solve this equatio i the iterval () [7] QUESTION 9 9. Simplify as far as possible: si θ si(80 θ ).cos(90 + θ ) + ta () 9. Simplify without the use of a calculator: si0 (cos ta 8.si ) (8) []

50 Mathematics/P 0 DBE/November 0 NSC QUESTION 0 The graphs of f ( ) si( + 0 ) ad g( ) cos for are give below. The graphs itersect at poit P ad poit Q. y g f Q P Calculate f (0) g(0). () 0. Calculate the -coordiates of poit P ad poit Q. (7) 0. For which values of will f ( ) g( )? () 0. Graph h is obtaied by the followig trasformatio of f: h ( ) f ( + 60 ). Describe the relatioship betwee g ad h. () []

51 Mathematics/P DBE/November 0 NSC QUESTION ABCD is a parallelogram with AB uits, BC uits ad A Bˆ C θ for 0 < θ 90. A θ B D C. Prove that the area of parallelogram ABCD is 6siθ. (). Calculate the value of θ for which the area of the parallelogram is square uits. (). Determie the value of θ for which the parallelogram has the maimum area. () [8]

52 Mathematics/P DBE/November 0 NSC QUESTION A hot-air balloo H is directly above poit B o the groud. Two ropes are used to keep the hot-air balloo i positio. The ropes are held by two people o the groud at poit C ad poit D. B, C ad D are i the same horizotal plae. The agle of elevatio from C to H is. C Dˆ B ad C Bˆ D 90. The distace betwee C ad D is k metres. H θ B 90 C k D. Show that CB k si. (). Hece, show that the legth of rope HC is k ta. (). If k 0 m, ad HD,8 m, calculate θ, the agle betwee the two ropes. () []

53 Mathematics/P DBE/November 0 NSC QUESTION The face of a stadard clock is positioed such that the cetre is at the origi. At a certai time, the ed of the miute had is at the poit P( ; ). 7 miutes later, the ed of the miute had is at the poit P / (a ; b). y P D D / O P /. Determie the value of a ad b. (6). OD is the positio of the hour had whe the miute had is at P ad OD / is the positio of the hour had whe the miute had is at P /. Calculate the agle betwee OD ad OD /. () [0] TOTAL: 0

54 Mathematics/P DBE/November 0 NSC CENTRE NUMBER: EXAMINATION NUMBER: DIAGRAM SHEET QUESTION. Physical Sciece Mathematics QUESTION 7. y 0 8 A C 6 B / A / C / B

55 Mathematics/P DBE/November 0 NSC INFMATION SHEET b ± b 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 lim h 0 f ( + h) f ( ) h r S ( a + ( d ) T a + ( ) d ; r ( ) ( ) + 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 ) ( i ) σ i f ( A) P( A) P(A or B) P(A) + P(B) P(A ad B) y ˆ a + b ( S ) [ ( + i) ] P i b ( ) ( ) ( y y)

56 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P NOVEMBER 0 MEMANDUM MARKS: 0 This memoradum cosists of 9 pages.

57 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 uless idicated otherwise QUESTION. Approimately cm (Accept 0 ) aswer (). As the age icreases, the height icreases descriptio () Every year the height icreases by approimately 6, cm Straight lie (liear) with a positive gradiet Strog positive correlatio Icrease i height: icrease i age is a costat Approimate icrease i average height 6, Rage for umerator (87 89 ; 67 70) (Accept ay aswer betwee 6 ad 6, cm) readig off from graph umerator aswer (). Childre stop growig whe they reach adulthood. If the tred cotiues the boys would reach impossible heights The tred will start approachig a costat value. People caot grow idefiitely commet () [6]

58 Mathematics/P DBE/November 0 NSC Memoradum QUESTION. Average umber of rus Stadard deviatio 7, 7, NOTE: Pealty of mark for icorrect roudig off 6 () (). Stadard deviatio 9,7 Stadard deviatio icreases. 9,7 icreases () ad are far from the mea, amely 6. Sice the stadard deviatio depeds o how far data poits are from the mea, the stadard deviatio would be epected to icrease.. Total umber of rus required is Total umber of rus to be scored i last five games Average umber of rus for last five games is 6,6 ad far from mea icrease 0 6,6 () () , ,6 0 6,6 0 6,6 () () [9]

59 Mathematics/P DBE/November 0 NSC Memoradum QUESTION. Rage 8 0. Phy Sc () Maths ma 8 Q 70 Q 0 Media (). From the iformatio give for Mathematics, the value of the third quartile is 70%. Therefore 7% of learers got below 70%. Number of learers below 70% is epected to 7 be learers 00 7% of learers learers (). No, Joe's claim is ivalid. 0% of the learers scored betwee 0% ad % i Physical Scieces. 0% of the learers scored betwee 0% ad % i Mathematics. Therefore the umbers will be equal. No, Joe's claim is ivalid. Same umber of learers (betwee mi ad media) ivalid/o media represets 0% of learers () [9] QUESTION. Modal class is 0 < 60 Correct class () 0< 60 0 to 60. Media positio is learers (grouped data). Approimate weight is about kg. (Accept from kg to kg ). 0 7 learers collected more tha 60 kg. kg () 7 learers () []

60 Mathematics/P DBE/November 0 NSC Memoradum QUESTION A( ; 8) y θ B M E(6 ; ) P Q R S T D C( ; ). Diagoals bisect each other at M: ( ) M ; y M M y M.. M( ; ) m m BC BC m m BC BC y y m( ) y 8 m( + ) m AD m BC y 8 ( + ) y + Lies parallel () substitutio ito gradiet formula m BC 6 () () substitute ( ; 8) gradiets equal equatio ()

61 Mathematics/P 6 DBE/November 0 NSC Memoradum m AD m BC mad y + c 8 ( ) + c c y +. ABCD is a rhombus, therefore AB BC θ BCA ˆ ARS ˆ RSC ˆ ARS ˆ BST ˆ 8 + ta ARS ˆ mac ta ARS ˆ ARS ˆ 80 6, ARS ˆ,69 ta BST ˆ mbc BST ˆ 78,69 θ B Cˆ A,69 78,69 θ Lies parallel gradiets equal substitute ( ; 8) equatio θ BC ˆ A ta A RS ˆ (),69 ta B ST ˆ m BC 78,69 θ (6) 8 + ta ARS ˆ m AC ARS ˆ,69 ta APR ˆ m AD APR ˆ 78,69 PAR ˆ ARS ˆ APR ˆ,69 78,69 θ PAR ˆ Eterior agle of a triagle Diagoals of the rhombus bisect opposite agles ta ARS ˆ,69 ta A PR ˆ m AD 78,69 PA ˆR θ (6)

62 Mathematics/P 7 DBE/November 0 NSC Memoradum 8 + ta ARS ˆ m AC ARS ˆ,69 ta APR ˆ APR ˆ 78,69 θ PAR ˆ θ ARS ˆ APR ˆ θ,69 78,69 θ Diagoals of the rhombus bisect opposite agles Eterior agle of a triagle ta A RS ˆ,69 ta A PR ˆ m AD 78,69 θ PA ˆ R θ (6) 8 + ta ARS ˆ m AC ARS ˆ,69 ta BST ˆ BST ˆ 78,69 θ RCS ˆ RCS ˆ + BST ˆ RCS ˆ + RSC ˆ ARS ˆ θ ARS ˆ BSˆ T,69 78,69 BABC ta A RS ˆ,69 ta B ST ˆ 78,69 θ RC ˆ S θ (6) ABCD is a rhombus, therefore AB BC ACB ˆ BAC ˆ taθ ta ACB ˆ ta( ARS ˆ BST ˆ ) ta ARS ˆ ta BST ˆ + ta ARS ˆ.ta BST ˆ θ ACB ˆ BAˆ C ta θ ta AC ˆ B formula substitutio ta θ θ (6)

63 Mathematics/P 8 DBE/November 0 NSC Memoradum From., M has coordiates ( ; ) Joi ME m ME 6 From., mbc mme. mbc MEC ˆ 90 ME ( ) ( ) ( 6) + ( ) 6 EC MEC is a right-agled triagle. E C ˆM gradiet of ME gradiet of BC M E ˆC 90 ME 6 EC 6 E C ˆM ABCD is a rhombus, therefore AB BC θ BC ˆM (6) ( ) + ( 8 ) AM Now to calculate the coordiates of B: m m m AC BD BD 8 + mac diagoals bisect at right agles Equatio of BD is y + Equatio of BC is y 9 BD ad BC itersect at B. Solve equatios simultaeously to get B(7 ; 6). AM y + y 9 B(7 ; 6) ( 7 ) + ( 6 ) BM BM AM Sice AMB ˆ 90 BM taθ AM taθ θ BM (6) []

64 Mathematics/P 9 DBE/November 0 NSC Memoradum QUESTION 6 y Q A N( ; ) K O L B P 6. The radius (NL) of a circle is perpedicular to the taget (OL) at the poit of cotact. 6. L( ; 0) ( ; 0) 6. Cetre N ( ; ) ad r NL Equatio of the circle N: ( a) + ( y b) r ( ) + ( y ) 6. Coordiates of K. K is the -itercept of the taget. y K( ;0) KL ( ) KL + KL radius taget r () () ( ) + ( y ) () substitute y 0 ito equatio of taget KL ()

65 Mathematics/P 0 DBE/November 0 NSC Memoradum y K( ;0) substitute y 0 ito equatio of taget KL ( ) + ( y y ) KL KL KL ( + ) 6 + (0 0) KL () For AK, m, c ta AKO ˆ OK OK KL OK OK KL () y K( ;0) KN NL + KL ( ) + (0 ) + KL 0 + KL 6 KL KL Theorem of Pythagoras KN NL + KL KL ()

66 Mathematics/P DBE/November 0 NSC Memoradum 6. m m AB AK m y y m AB y y y AK m( ) ( ) taget radius m AK m AB substitutio of poit (;) ito equatio equatio () m m AB AK m m AB AK y + c ( ) + c 8 9 c + 7 c 7 y + taget radius m AK m AB substitutio of poit (;) ito equatio equatio ()

67 Mathematics/P DBE/November 0 NSC Memoradum 6.6 Poit A lies o PQ ad AB. Therefore y + 6 y 7 6 A ; equatio substitutio of () Poit A lies o PQ ad the circle. Therefore ( ) + ( + ) ( ) + ( ) ( 7) y + 6 y equatio equatio ( )( 7) 0 ( 7) 0 substitutio of substitutio of ()

68 Mathematics/P DBE/November 0 NSC Memoradum Poit A lies o the circle ad lie AB ( ) + ( y ) () y Subs () i () : ( )( 7) ( ( + ) y + 6 y Usig rotatio: Let θ AKN ˆ LKˆ N () 7 + ) equatio ( )( 7) 0 substitutio of () Move diagram uit to the right. The A / is L / rotated through θ. taθ AN KA si θ siθ cosθ ( cos θ cos θ si θ ( )( ) ) ( ) values of siθ ad cosθ A L cos θ yl si θ ( ) (0)( ) 6 y A L si θ + yl cos θ ( ) (0)( ) 6 A ( ; ) Now to get back to A, move back uit to the left. 7 6 A( ; ) substitutio ito rotatio formulae 6 A ( ; ) ()

69 Mathematics/P DBE/November 0 NSC Memoradum y A N K θ θ M L Let ˆ NL N KL θ. So, ta θ. KN Hece si θ ad cos θ ta θ Let AM ais with M o - ais NAK NLK AKN ˆ NKL ˆ θ AKL ˆ θ y A AM AK si θ KLsi θ si θ si θ siθ cosθ 6 y A A OL NAsi MAN ˆ si si θ 8 7 ( 90 MAK ˆ ) si θ solve for ad y ()

70 Mathematics/P DBE/November 0 NSC Memoradum 6.7 KA ( ) + ( y y ) distace formula substitutio () KN KA KN KA AN 0 KN 0 KA KN AN () KA KL Tagets from a commo poit are equal KA 6.8 AN NL Radii are equal KA KL KLNA is a kite two pairs of adjacet sides are equal. 6.9 AB AN + NB + AK AB KAB ˆ 90 taget radius ΔAKB is a right agled isosceles triagle AKB ˆ + ABK ˆ 90 ABK ˆ 90 ABK ˆ KAKL reaso AN NL KA KL AB AK AB K A ˆB 90 () () ()

71 Mathematics/P 6 DBE/November 0 NSC Memoradum N is midpoit of AB Let B be ( ; y B B ) 7 B + B B ; ta β mab β 80 6,87 y B y β, 0 taα m KB 7 + α 8, ABK ˆ α + (80 β ) 8, + 6, B K α A B β, 8, A BK ˆ α + (80 β ) () N is midpoit of AB Let B be ( B ; y B ) 7 B + B B ; KB ( + ) + ( ) cosθ θ y B y 6 + B + ( ) ()( ) cosθ K A B substitutio ito cosie formula cos θ () 6.0 / N (; ) / N (; ) () []

72 Mathematics/P 7 DBE/November 0 NSC Memoradum QUESTION 7 NOTE: CA ot applicable i this questio 7. Rotatio about the origi through 90 i a clockwise directio. Rotatio about the origi through 70 i a ati-clockwise directio. Rotatio about the origi through -90. rotatio of 90 clockwise directio () rotatio of 70 ati-clockwise directio () statemet () 7. ( ; y) ( y; ) 7. 0 y A // (both) ( ; y) ( y; ) () 9 A C B / B // A / C // C / oe poit correct all poits correct ad triagle draw () B ( ; y) (;y) ( ;y) () 7.. A ( ;) ( ; ) D(; ) () 7.. ( ; y) ( ; y) ( ; y) ( ; y) ( ; y) 7.. Rotatio of 80 through the origi i either directio. Reflectio about the origi. rotatio 80 () reflectio origi () () []

73 Mathematics/P 8 DBE/November 0 NSC Memoradum QUESTION 8 No calculator allowed i this questio 8.. OT k, PT 8 ad OP 7 k k k k 896 ± k > 0 k k 7 8 k (7 8)(7 + 8) 9 k ± k > 0 k cos α 7 7 α + β 80 β 80 α cos β cos(80 α) cosα 7 substitutio ito Pythagoras k substitutio ito Pythagoras k cos( 80 α) or cos α 7 () () () () 8 7 α β cos β cos(80 α) cosα 7 cos( 80 α) or cos α 7 ()

74 Mathematics/P 9 DBE/November 0 NSC Memoradum 8.. si( β α) si β cosα cos β siα epasio 8 si β 7 8 si α () 8.. β α ( 80 α ) α 80 α si( β α) si(80 α ) siα siα.cosα cos si LHS si cos ( si ) si si cos cos si si si cos cos si (si ) cos (si ) si cos ta RHS substitute β siαcosα si α si si cos () either si (si ) or cos (si ) si cos ()

75 Mathematics/P 0 DBE/November 0 NSC Memoradum RHS LHS ta cos si ) (si cos ) (si si cos cos si si si cos cos si si ) cos ( cos cos si si cos cos cos si si ) ( cos cos si si cos RHS LHS + + ta cos si ) (si cos ) (si si cos cos si si si cos cos si si si si cos cos si si si cos cos cos si si ) si (cos cos si si cos cos si cos either ) (si si or ) (si cos cos si () si cos si cos either ) (si si or ) (si cos cos si ()

76 Mathematics/P DBE/November 0 NSC Memoradum 8.. si cos 0 si cos cos 0 cos (si ) 0 cos k si k or or k or k 90 or 70 or 0 or 0 k Z si cos cos 0 ad si for two correct aswers for four correct aswers () si cos si si(90 ) k ; k Z or 80 ( 90 ) k k k k 0 or 0 or 70 or k si( 90 ) k ad k for two correct aswers for four correct aswers () [7]

77 Mathematics/P DBE/November 0 NSC Memoradum QUESTION si θ si(80 θ ).cos(90 + θ ) + ta si θ (siθ )( siθ ) + si θ si θ + si θ cos θ ta θ si0 (cos ) ta 8 si si 76.cos 0 ta 8.( si ) si 8 cos8 si 8 ( cos8 ) cos8 si 8 cos8 si 8 cos8 si0 (cos ) ta 8 si si ( ).(cos ) si 8.( si ) cos8 si cos.cos 0 cos ( si ) si cos0. NOTE: If cos 0 is missig: deduct mark Aswer oly: 0/8 siθ siθ cos²θ ta θ si 76 cos0 si 8 cos8 si () si8 cos8 si cos8 (8) si( ) si 8 cos8 si si cos cos0 cos si8 ad si cos8 (8)

78 Mathematics/P DBE/November 0 NSC Memoradum si0 (cos ) ta 8 si si0.cos 0 si 8 ( si ) cos8 (si0 ) si 8 ( cos 8) cos8 si0 si 8 cos8 si0 si 76 si 76 cos or si 76 cos cos0 si 8 cos8 si cos 8 si76 (8) si0 (cos ) ta 8 si si0.cos 0 si 8.( si ) cos8 si0. cos si si0. cos ( si ) ( si ) si0. si0 cos0 si 8 cos8 si cos si8 ad si cos8 cos 0.si 0 si0 (8) []

79 Mathematics/P DBE/November 0 NSC Memoradum QUESTION 0 0. f ( 0) g(0) 0, ( ),, () 0. si( + 0 ) cos equatio si.cos 0 + cos.si 0 cos epasio of si(+0 ) si cos cos + si + cos cos si cos ta 09, + 80 k ; k Z P 70,89 ad Q 09, si( + 0 ) cos cos(90 0 ) cos cos(60 ) cos cos60 cos + si 60 si cos cos + si cos cos + si cos si cos ta 09, + 80.k ; k Z P 70,89 ad Q 09, 0. 70,89 09, [ 70,89 ; 09, ] P Q substitutio of special agles simplificatio ta P 70, 89 Q 09, (7) equatio epasio of cos(60 ) substitutio of special agles simplificatio ta P 70, 89 Q 09, (7) agles correct iterval () 0. h( ) si( ) si( + 90 ) cos g( ) h is the reflectio of g about the -ais. f is shifted to the left through 60 ad the doubled. h is the reflectio of g about the -ais. reflectio about the -ais or lie y 0 () reflectio about the -ais or lie y 0 () []

80 Mathematics/P DBE/November 0 NSC Memoradum QUESTION. Area parallelogram ABCD Area ΔABC 6siθ ( )( ) siθ area ABC substitutio ito area rule () h siθ h siθ Area ABCD base height h. siθ 6siθ h θ NOTE: If o workig is show, the 0/ h θ si h siθ b.h () Area of parallelogram ABCD area of ΔABC + area of ΔADC ( )( ) siθ si 6 si θ Area (sum of // sides) h ( + ) si θ 6 si θ. Area of parallelogram ABCD 6 siθ siθ θ 60 + ( )( ) θ NOTE: Deduct mark if both 60 ad 0 are give as aswers 6 si 60 θ 60. Maimum area of parallelogram occurs whe si θ, that is whe θ 90 sum of areas equal sides ad equal agles () formula h siθ substitutio 6 siθ si θ 60 () () 6 siθ 60 () si θ θ 90 () [8]

81 Mathematics/P 6 DBE/November 0 NSC Memoradum QUESTION. CB si BDˆ C CB k si si(90 ) CB CB CD si CBˆ D k.si si(90 ) k.si cos cos k si DCB ˆ 80 (90 + ) 90 DC DB k D Usig the sie rule i triagle CBD CB k si si(90 ) k.si si(90 ) si.cos cos () DCB ˆ DBC ˆ 90 DC DB k k k B F C Draw DF BC CF si CD CF k si CB CF CB k si DCB ˆ 80 (90 + ) 90 DC DB k CB CD +BD.CD.BD.cos CB k + k k cos k ( cos ) k ( ( si )) k (si ) k si (k si ) CB k si C DF ˆ CF k si CB CF () DCB ˆ DBC ˆ 90 DC DB k usig cosie rule i triagle CDB factors simplificatio ()

82 Mathematics/P 7 DBE/November 0 NSC Memoradum. BC cos HC BC HC cos k si cos k ta HC BC si90 si(90 ) BC HC si(90 ) k si cos k ta cos HC BC HC BC cos substitutio of BC () BC HC si( 90 ) substitutio of BC si(90 ) cos (). HC k ta (0).ta( ), I HCD: CD HC + HD HC. HD.cosθ HC + HD CD cosθ HC. HD (,979...) +,8 0 (,979...)(,8) cosθ 0,6... θ 7,8 value of HC substitutio ito cos formula cos θ 0,6... 7,8 () []

83 Mathematics/P 8 DBE/November 0 NSC Memoradum QUESTION. Agle that miute had moves is: mi : mi : 6 7 mi : 7 6 P is rotated by 60-8 i a ati-clockwise directio: a cos8 si8 b cos8 + si8 ad,6, substitutio of 8 0 ito formula for ad y, 6, 6 (6) Agle that miute had moves is: P is rotated by i a clockwise directio: a cos + si b cos si ad,6,6 b a α β 80 - β substitutio of 0 ito formula for ad y, 6, 6 (6) taα α 6, α + 80 β β 6, + 80, a 0 cos,,6 b 0 si,,6 ta α α 6, α + 80 β β,, 6, 6 (6)

84 Mathematics/P 9 DBE/November 0 NSC Memoradum. The miute had moves through 60 i 60 miutes. The hour had moves through 0 i 60 miutes, that is, that of the miute had. So whe the miute had moves through, the hour had moves through 8, 60 The hour had moves through 0 i 60 miutes 7 it moves through 0 8, i 7 miutes , , () () [0] TOTAL : 0

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

86 Mathematics/P DBE/November 0 NSC INSTRUCTIONS AND INFMATION Read the followig istructios carefully before aswerig the questios This questio paper cosists of 0 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, aswers should be rouded off to TWO decimal places, uless stated otherwise. Diagrams are NOT ecessarily draw to scale. FIVE diagram sheets for aswerig QUESTION., QUESTION., QUESTION 7., QUESTION 7., QUESTION 8., QUESTION 9 ad QUESTION 0 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 eatly ad legibly.

87 Mathematics/P DBE/November 0 NSC QUESTION A recordig compay ivestigates the relatioship betwee the umber of times a CD is played by a atioal radio statio ad the atioal sales of the same CD i the followig week. The data below was collected for a radom sample of 0 CDs. The sales figures are rouded to the earest 0. Number of times CD is played Weekly sales of the CD Idetify the idepedet variable. (). Draw a scatter plot of this data o the grid provided o DIAGRAM SHEET. (). Determie the equatio of the least squares regressio lie. (). Calculate the correlatio coefficiet. (). Predict, correct to the earest 0, the weekly sales for a CD that was played times by the radio statio i the previous week. ().6 Commet o the stregth of the relatioship betwee the variables. () [] QUESTION Each of the 00 employees of a compay wrote a competecy test. The results are idicated i the table below: PASS FAIL TOTAL Males 6 78 Females 7 0 Total Are the evets PASS ad FAIL mutually eclusive? Eplai your aswer. (). Is passig the competecy test idepedet of geder? Substatiate your aswer with the ecessary calculatios. () [6]

88 Mathematics/P DBE/November 0 NSC QUESTION A compay producig televisio sets decided to check the lifespa (i years) of their most popular model. They selected 0 sets of the most popular model at radom for this test. The lifespa of each set was recorded. The iformatio is represeted i the table below. LIFESPAN FREQUENCY (IN YEARS),9 <,6,6 < 6, 6 6, < 7,0 8 7,0 < 7,7 7 7,7 < 8, 8, < 9,. Costruct a histogram to represet the data. Use the grid provided o DIAGRAM SHEET. (). Calculate the estimated mea lifespa of the most popular model of televisio set. (). The data represetig the lifespa of this batch of televisio sets is ormally distributed. This implies that approimately 68% of the data lies withi oe stadard deviatio of the mea, approimately 98% of the data lies withi two stadard deviatios of the mea ad approimately 00% of the data lies withi three stadard deviatios of the mea. The stadard deviatio of this data set is 0,76 years. Calculate the lifespa of the most popular model of televisio set such that 98% of the lifespa of all the sets will eceed this value. (). The compay wats to issue a -year guaratee with this model of televisio set. What would you recommed? Justify your recommedatio. () []

89 Mathematics/P DBE/November 0 NSC QUESTION Durig summer i a certai city i South Africa the probability of a suy day is 7 ad the probability of a raiy day is 7. If it is a suy day, the the probability that Vusi cycles to work is 0 7, the probability that Vusi drives to work is ad the probability that Vusi takes the trai to work is 0. If it is a raiy day, the the probability that Vusi cycles to work is 9, the probability that Vusi drives to work is 9 ad the probability that Vusi takes the trai to work is.. Draw a tree diagram to represet the above iformatio. Idicate o your diagram the probabilities associated with each brach as well as all the outcomes. (). For a day selected at radom, what is the probability that:.. It is raiy ad Vusi will cycle to work ().. Vusi takes the trai to work (). If Vusi works days i a year, o approimately how may occasios does he drive to work? () [] QUESTION Every cliet of CASHSAVE Bak has a persoal idetity umber (PIN) which is made up of digits chose from the digits 0 to 9.. How may persoal idetity umbers (PINs) ca be made if:.. Digits ca be repeated ().. Digits caot be repeated (). Suppose that a PIN ca be made up by selectig digits at radom ad that the digits ca be repeated. What is the probability that such a PIN will cotai at least oe 9? () [8] QUESTION 6 6. Write dow a recursive formula for the sequece: ; ; ; 9 ; 6;... () 6. Write dow the et term of the give recursive sequece: ; 7 ; ; ; ;... () [6]

90 Mathematics/P 6 DBE/November 0 NSC NOTE: Give reasos for all statemets made i QUESTION 7, QUESTION 8, QUESTION 9 ad QUESTION 0. QUESTION 7 7. If i LMN ad FGH it is give that Lˆ Fˆ ad Mˆ Ĝ, prove the theorem that LM LN states. FG FH L M F N G H (7) 7. I the diagram below, VRK has P o VR ad T o VK such that PT RK. VT uits, PR 9 uits, TK 6 uits ad VP 0 uits. Calculate the value of. V P T 6 K R () []

91 Mathematics/P 7 DBE/November 0 NSC QUESTION 8 8. Complete the followig statemet: The agle betwee the taget ad the chord is equal... () 8. I the diagram poits P, Q, R ad T lie o the circumferece of a circle. MW ad TW are tagets to the circle at P ad T respectively. PT is produced to meet RU at U. M PˆR 7 P Qˆ T 9 Q Tˆ R Let T PˆW a, R PˆT b, M PˆQ c ad R Tˆ U d, calculate the values of a, b, c ad d. R Q M 9 7 c P b a T d U W (9) [0]

92 Mathematics/P 8 DBE/November 0 NSC QUESTION 9 O is the cetre of the circle CAKB. AK produced itersects circle AOBT at T. AĈB C B T O K A 9. Prove that Tˆ 80. () 9. Prove AC KB. () 9. Prove BKT CAT () AC 9. If AK : KT :, determie the value of KB () []

93 Mathematics/P 9 DBE/November 0 NSC QUESTION 0 I the diagram below, O is the cetre of the circle. Chord AB is perpedicular to diameter DC. CM : MD : 9 ad AB uits. D B O M C A 0. Determie a epressio for DC i terms of if CM uits. () 0. Determie a epressio for OM i terms of. () 0. Hece, or otherwise, calculate the legth of the radius. () [7] TOTAL: 00

94 Mathematics/P DBE/November 0 NSC CENTRE NUMBER: EXAMINATION NUMBER: DIAGRAM SHEET QUESTION. 000 Scatter plot showig the umber of times a CD was played vs the CD sales i the followig week

95 Mathematics/P DBE/November 0 NSC CENTRE NUMBER: EXAMINATION NUMBER: DIAGRAM SHEET QUESTION. 0 Histogram showig the frequecy of the lifespa of the most popular televisio set (years) Frequecy Lifespa (years)

96 Mathematics/P DBE/November 0 NSC CENTRE NUMBER: EXAMINATION NUMBER: DIAGRAM SHEET QUESTION 7. L M F N G H QUESTION 7. V P T 6 K R

97 Mathematics/P DBE/November 0 NSC CENTRE NUMBER: EXAMINATION NUMBER: DIAGRAM SHEET QUESTION 8. R Q M 9 7 c P b a T d U QUESTION 9 W C B T O K A

98 Mathematics/P DBE/November 0 NSC CENTRE NUMBER: EXAMINATION NUMBER: DIAGRAM SHEET QUESTION 0 D B O M C A

99 Mathematics/P DBE/November 0 NSC INFMATION SHEET: MATHEMATICS b ± b 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 lim h 0 f ( + h) f ( ) h r S ( a + ( d ) T 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 ( m y S y b c a b + c bc. cos A si B si C ab. si C ) a ; < r < r 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 ) ( 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) ( )

100 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P NOVEMBER 0 MEMANDUM MARKS: 00 This memoradum cosists of 6 pages.