sec set D (sp 2014): PaPer 1

Transcription

1 sec set D (sp 4): PPe Question (5 mks) Question () (i) w + i w ( ) + ( ) + 4 w Im( z) tnθ tn α, wee α is te elted ngle in te fist qudnt. p α tn 6 o p p θ p p + + p + w cos np isin np in genel pol fom. p+ 6np + p+ 6np w cos i sin Question () (ii) θ Re( z) [(cos isin )] n n q + q (cos nq + isin nq) (De Moive s Teoem) z + i z ( + i) z π + 6nπ 6n i + π + π cos sin π + 6nπ + π + 6nπ cos i sin 6 6 [Tee e oots wic you find by putting n nd ten n.] n : z π + π cos i sin ( + i) 8π n z i + : cos sin 6 4π cos i + sin ( i) 8π 6 4π 78 Mtemtics Leving Cetificte

2 Question (b) (i) Let z + bi. z iz i( + bi) i+ bi b+ i Notice tt te el nd imginy pts e swpped between z nd z. Cn you pick out two points on te digm wee tis migt be te sitution? Te two igligted points seem to stisfy tis condition. At tis stge I don t know wic point is z nd z. b Im( z) b Re( z) z kz k( + bi) k+ kbi z nd z e in stigt line wit te oigin. z 4 Im( z) z4 z + z ( b+ i) + ( k + kbi) ( k b) + ( + kb) i z You e now in position to mk in ll te points. b z z b Re( z) (ii) k ª Question (5 mks) Question () (i) steps to Poof by induction. Pove esult is tue fo some stting vlue of n N.. Assume esult is tue fo n k.. Pove esult is tue fo n (k +). nn ( + ) RequiRed to PRove: n step : Pove te esult is tue fo n. ( + ) ( ) [Teefoe, tue fo n.] step : Assume it is tue fo n k. kk ( + ) k Set D Ppe Stte Emintions Commission Hige Level, Set D, Ppe 79

3 step : Pove it is tue fo n k +. Pove ( k+ )( k+ ) ( k) + ( k+ ) Use te esult in Step to pove Step. ( k) + ( k+ ) kk ( + ) + ( k + ) k ( k + ) + k + ( k + ) ( k+ )( k+ ) Teefoe, ssuming tue fo n k mens it is tue fo n k +. So tue fo n nd tue fo n k mens it is tue fo n k +. Tis implies it is tue fo ll n N. Question () (ii) nn S ( + ) n ( ) 5( 5) S S 775 [Tis te te sum of te numbes between 5 nd.] 5 Question (b) log + log ( c) c log + log ( c) c log + log c+ log logc + logc c p + c c c c c [Using log ules &.] Log RuLes log ( y) log + log y...( ) log log log y... y...( ) q log ( ) qlog...( ) log...( 4) log...( 5) log b log log...( 6) b 8 Mtemtics Leving Cetificte

4 Question (5 mks) Question () f( ) + ( k ) + k f( k) ( k) + ( k )( k) + k k k+ k + k If ( - k) is fcto of f () ten k is oot of f (), i.e. f (k) nd vice ves. If f (-k), ten -k is oot of f (). Question (b) If k is oot, ( + k) is line fcto. + ( k ) + k ( + k)( + + ) [A cubic is line multiplied by qudtic.] + + ( k ) + k + ( + k) + ( k + ) + k + k k + ( k ) + k ( + k)( k + ), b k, c k k 4 ( ) ± ( ) ()() () k k + [Solve te qudtic eqution using te fomul.] k k 4 ± Question (c) 4< ( k+ )( k ) < k 4 k 4 k ± b b c ± 4 [To ve ectly one el oot, te qudtic eqution will give comple oots. Tis mens te epession inside te sque oot must be negtive.] Solve te inequlity bove. Fist, solve te equlity to locte te oots. ( k+ )( k ) < ( k + )( k ) < ( k + )( k ) < ( )( 5) < ()( ) < (5)() < Flse Tue Flse nswe: - < k < Set D Ppe Stte Emintions Commission Hige Level, Set D, Ppe 8

5 Question 4 (5 mks) Question 4 () + 8y z...( ) y+ z...( ) + y+ z 5...( ) [Eliminte te s by subtcting pis of equtions.] () ( ): y 5z...( 4)( 4) [Eliminte te z s.] () ( ): 7y 4z 6...( 5)( 5) 44y z 5y+ z 9y 8 y ( ) 5z...( 4) 5z 5z 5 z 5 [Substitute te vlue of y into eqution (4) to find z.] + 8( ) () 5...( ) [Substitute te vlues of y nd z into eqution () to find.] nswe: (,, 5) 8 Mtemtics Leving Cetificte

6 Question 4 (b) (i) f( ) g ( ) ± 5, Cuts -is: Put f () f( ) Cuts y-is: Put [Find te points wee te gps of te functions intesect.] Find wee te f () cuts te es. f ( ) nswes: A (, ), B (5, ), C (, ), D (, ) y D A C B 5 y f ( ) y g() Question 4 (b) (ii) y D A B y f ( ) y g() C 5 < < 5 You cn see fom te gp te vlues of wee f () < g(). nswe: < < 5 Set D Ppe Stte Emintions Commission Hige Level, Set D, Ppe 8

7 Question 5 (5 mks) Question 5 () Sketc te function by finding its tuning points nd end points. f( ) f ( ) + Tuning points: f ( ) + ( )( ), f() 6 8 (5, ) (, -4) f () () 5() + () + 5 (, 48 ) f () () 5 () + () + 5 (, 4) 7 is tuning point. is tuning point. : f( ) ( ) 5( ) + ( ) (, 5) is te stting point. 5: f( 5) ( 5) 5( 5) + ( 5) + 5 (, 5 ) is te finising point. AnsweR: Mimum vlue of f Minimum vlue of f 4 Question 5 (b) Te function f is not injective. An injective function neve mps distinct elements of its domin to te sme element of its nge. Ec element in te domin must mp on to unique element in te nge. Look t te sketc of digm of te function f f() You cn see fom te sketc tt te function is not injective s two points e sown on te -is wic mp on to te sme point on te f () is Mtemtics Leving Cetificte

8 Question 6 (5 mks) Question 6 () (i) 4. I I I n+ n d + c n + (ii) (iii) f( ) d ( ) () is n indefinite integl wic mens tt wen you diffeentite it wit espect to you get () o ( ) f( ). I c 4 Question 6 (b) (i) Let ( ) ln, fo R, >. Find ( ). (ii) Hence, find Ú ln d. solution (i) (ii) ( ) ln ( ) + (ln ) + ln ( ) + ln ln ( ) ln d ( ( ) ) d ( ) + c ln + c PRoduct RuLe y dy uv u dv d d + dy y ln d v du d Set D Ppe Stte Emintions Commission Hige Level, Set D, Ppe 85

9 Question 7 (5 mks) Question 7 () Side l l Top Side Bottom w Side cm Side cm + l+ l+ l+ 5 l ( 5 ) cm + w+ + w w ( ) cm Question 7 (b) V l b ( 5 )( ) Question 7 (c) Sque bottom: l w 5 5 5cm V ( 5 ( 5))( ( 5))( 5) ( )( )() 5 5 cm 86 Mtemtics Leving Cetificte

10 Question 7 (d) ( 5 )( ) 5 [Fom cubic eqution by putting te volume of te bo equl to 5.] ( 5+ ) is solution of tis cubic. Teefoe, ( - 5) is line fcto. Te ote fcto is qudtic. Find te qudtic by lining up ( 5)( + k + 5) ( k 5) + ( 5 5k ) 5 5 k 5 k ( 5)( + ) + 5 [Solve te qudtic using te fomul.], b, c 5 ± ( ) ( ) 4 ()( 5) ( ) b b c ± 4 ± 4 ± ± ± 5 7.,. 9cm [Discd te solution of 7. cm. Te lengt l of te bo is equl to (5 - ). Te lengt is not long enoug to ccomodte vlue of 7. cm.] nswe:.9 cm Set D Ppe Stte Emintions Commission Hige Level, Set D, Ppe 87

11 Question 7 (e) Cpcity in cm Eplntion: % et volume: cm Go to 55 cm on te y-is nd ed off te vlue cm A eigt 7.4 cm is too long to mke bo fom tis piece of cdbod. Te lengt l of te bo is equl to (5 ). Te lengt is not long enoug to ccomodte vlue of 7.4 cm. Question 8 (5 mks) Question 8 () P? F t ye i. F P ( + i) t ( +. ) F P( + i) t Question 8 (b) P t (. ) 88 Mtemtics Leving Cetificte

12 Question 8 (c) I need to clculte te etiement fund tt e s sved fo nd is vilble on te dy of is etiement. is dwn down immeditely. Te net will not be dwn down fo note ye so its pesent vlue on te dy of etiement is (. ). Te net will not be dwn fom te etiement fund fo yes so its pesent vlue is. An so on. (. ) (. ) (. ) (. ) ,, n 5. S n 5 ( (. ) ) S n n ( ) Question 8 (d) (i) ( + i). + i i % (ii) F P(. 466) n His fist pyment P will be compounded 48 times t n inteest te of.466%. His second pyment P will be compounded 479 times t n inteest te of.466%. And so on. (iii) P(. 466) P(. 466) P P(. 466)( (. 466) ) Question 8 (e) yes less: 6 monts 6 P(. 466)( (. 466) ) P Set D Ppe Stte Emintions Commission Hige Level, Set D, Ppe 89

13 Question 9 (5 mks) Question 9 () (i) f( ) f (. ) 5. (. ) + 5 (.) 98. (ii) f( ) f ( ) f () 5 55.() + 55 ( ) ( 55,. ) f ( ) < (, 55. ) is locl mimum. is tuning point. Question 9 (b) (i) f( ) f ().() ( ) f ( ).( 5) + 5 ( ) f () 5.() + 5 ( ) f ( 4).() ( ) f ().() ( ).. [Wok out numbe of points fo te middle pt of te gp nd ten plot it.] velocity (m/s) time (seconds) Question 9 (b) (ii) In te fist. s te spinte does not move. He/se is still in te blocks. To find te distnce tvelled ove te fist 5 s find te e unde te velocity cuve fom. s to 5 s. s vdt ds v s vdt dt (. 5t + 5t 98. ) dt t t t+ c To find te distnce s tvelled eplce t by 5 nd ten t by. nd subtcts te nswes. s + + c 55.() 55 ( ) 5.(.) 5 (. ) 985. ( ) +.98(. ) c m 9 Mtemtics Leving Cetificte

14 Question 9 (b) (iii) Te spinte uns 6.9 m ove te fist 5 seconds. He/se uns te est of te ce (6. m) t te mimum speed of.5 m/s t 548. s t. 5 nswe: Totl time.48 s v s t Question 9 (c) (i) dv dt dv A ka dt 4 d( π ) k( 4π ) dt d 4 π k( 4π ) dt d k dt V 4 π Sufce Ae 4π (ii) d k d k dt dt kt+ c t : c kt+ 4 t : π π k() k + t T : kt + T k ous 9 minutes Set D Ppe Stte Emintions Commission Hige Level, Set D, Ppe 9

15 Question (5 mks) Question () sec set D (sp 4): PPe P(X ) EX ( ) P( ) Question (b) E(X) epesents te men vlue of te ge of ll te second ye students on Jnuy. Question (c) n 6 p(4).575 q(not 4).45 BeRnouLLi trials p P(Success), q P(Filue) P( successes) n Cpq - n 6 4 P( 6 out of e 4 yes of ge ) C (. 575)(. 45).48 6 Question (5 mks) Question () stratified sampling: A pobbility smpling tecnique wee te entie popultion is divided into non-ovelpping subgoups (stt) nd te finl subjects e ndomly selected popotionlly fom te diffeent stt. An dvntge of tis metod ove simple ndom smpling is gete pecision using smples of te sme size. cluster sampling: A pobblity smpling tecnique in wic te smple tkes sevel steps in coosing te smple popultion. Fistly, te popultion is divided into clustes. A simple ndom smple of clustes is ten selected fom ll of tese clustes. Finlly, individuls e selected ndomly fom ec cluste. An dvntge ove simple ndom smple is tt it is muc cepe. Question (b) (i) Mgin of eo n. 9 Mtemtics Leving Cetificte

16 Question (b) (ii) Null ypotesis H o : P. Altentive ypotesis H : P. 4 Smple popotion P. 6 Confidence intevl [.6 -.,.6 +.] [.86,.46] Tee is evidence to suppot te pty s clim tt it s te suppot of % of te electote becuse, bsed on te smple dt, ny vlues in te nge 8% - 4% e possible vlues fo te popotion of te electote wo suppot te pty. % is in tis confidence intevl. Teefoe, you cnnot eject te null ypotesis. Question (5 mks) Question () Is P( 4k, k+ ) l : 4y+? 4 ( k ) 4( k+ ) + k 6 k 4+ [To see if point is on line, substitute te point into te eqution of te line nd sow it stisfies te eqution.] Question (b) l: 4y+ m l : m 4 Eqution of l : Point P(4k, k + ), 4 m 4 m m [Te poduct of te slopes of pependicul lines is -.] y y m ( ) 4 y ( k+ ) ( ( 4k )) ( y k ) 4( 4k+ ) y 9k 4+ 6k 8 4+ y 5k + 5 y l Q(, ) P l Question (c) Q(, ) l : 4+ y 5k () + ( ) 5k k k k Question (d) P is te point t te foot of te pependicul. P( 4k, k+ ) P( 4 ( ), ( ) + ) P(, 6 7) Set D Ppe Stte Emintions Commission Hige Level, Set D, Ppe 9

17 Question 4 (5 mks) y c centre And RAdius of A circle (, -f) (-g, -f) + y - 6 c: + y + g+ fy+ c Cente: ( g, f) g + f c (, ) (-g, ) Sketc te sitution. Te cicle c touces te -is t (-g, ) nd touces te y-is t (, -f ). Te line psses toug te cente of te cicle c. ( g, f) + y 6 ( g) + ( f) 6 g f 6 g+ f 6...( ) [If point is on line, substitute te point into te eqution of te line nd it stisfies te eqution.] Te points (-g, ) nd (, -f ) stisfy te eqution of te cicle. ( g, ) c: + y + g + fy + c (, f) c: + y + g + fy + c ( g) + ( ) + g( g) + f( ) + c ( ) + ( f) + g( ) + f( f) + c g g c g + c c g f f f + c c f...( ) g f g ± f...( ) To find te eqution of ec cicle, coose g +f fo te fist cicle eqution nd g -f fo te second cicle eqution. g + f [Substitute into Eqn. ()] g+ g 6 g 6 g f [Substitute into Eqn. ()] c g ( ) 4 g f f + f 6 f 6 g 6 c g ( 6) 6 equation of circles + y + ( ) + ( ) y+ 4 + y 4 4y+ 4 + y + 6 ( ) + ( 6) y+ 6 + y + y Mtemtics Leving Cetificte

18 Question 5 (5 mks) Question 5 () g: sin P π, R [, ] π : sin P π, R [, ] y sin n π R [, ], P n y () y g() P y f ( ) y - - Question 5 (b) y f( ) sin y y sin - o π sin ( ) 6 π 5π π 7π,,, 5 o S T A C o π π, [ Fist qudnt] 6 6 5π 7π, [ Second qudnt] 6 6 π π, 5π 7π, [Tese e ll te solutions between nd p. ] P y f ( ) y AnsweR: P 7 π, Set D Ppe Stte Emintions Commission Hige Level, Set D, Ppe 95

19 Question 6 (5 mks) Eplntion: Poof by contdiction is fom of poof tt estblises te tut o vlidity of poposition by sowing tt te poposition being flse would imply contdiction. Emple: Pove + fo ll >, R. To pove tis let s ssume it is flse, i.e + < fo ll >, R. + < + < + < ( ) < [Tis sttement is flse fo ll vlues of.] Tis is contdiction. Question 6b (5 mks) OECD is cyclic qudiltel becuse its opposite ngles dd up to 8 o. o o o OEC + ODC It follows tt te ote pi of opposite ngles lso dd up to 8 o s te fou ngles in qudiltel dd up to 6 o. cyclic quadrilaterals A cyclic qudiltel is fou sided figue wose vetices lie on cicle. Opposite ngles of cyclic qudiltel dd up to 8 o. B A A+ C 8 B+ D 8 Convesely, if te opposite ngles of qudiltel dd up to 8 o ten it is cyclic qudiltel. o o C D A E Dw cicle ound te qudiltel. O B D C 96 Mtemtics Leving Cetificte

20 AngLes standing on te same ARc in A circle Angles nd e stnding on te sme c [AB]. Angles stnding on te sme c e equl.. X Y A E O B D C DOC DEC [Bot ngles e stnding on c [DC]] Set D Ppe Stte Emintions Commission Hige Level, Set D, Ppe 97

21 Competitos Competitos Competitos Competitos Question 7 (75 mks) Question 7 () (i) Swim Cycle Run Men ? Medin ? Mode #N/A #N/A #N/A Stndd Devition? Smple Vince Skewness Rnge Minimum Mimum Count Swim 45 Cycle 7 Run Time (minutes) Time (minutes) Time (minutes) You cn mtc te istogms to te events by looking t te mimum nd minimum times to complete ec event (ii) Te medin is line on te istogm tt bisects its e. Te e to te left of it is equl to te e to te igt of it. Using you eye tis ppes to lie long te clss intevl of 4 6 minutes. men ª medin ª5 minutes Run (iii) Swim: Smple Vince s.7 s σ Time (minutes) σ s minutes (iv) Tee ws pobbly no discete modl esult s ll times wee diffeent o tee wee lots of te sme times. Teefoe, tee is no modl time. Tee is modl clss ligt but no modl time. 98 Mtemtics Leving Cetificte

22 Question 7 (b) Cycle vs. Swim: Modete positive coeltion Run vs. Swim: Modetely stong positive coeltion Run vs. Cycle: Stong positive coeltion Question 7 (c) Run/Swim: y Bin: 7.6 mins y.5(7.6) mins Run/Cycle: y Bin: 5.7 mins y.58(5.7) mins Tke te vege of te un times: Te men finising time fo te ovell event ws 88 minutes nd te stndd devition ws minutes. Question 7 (d) mins In ny noml distibution wit men nd stndd devition σ % of te dt flls witin σ of te men % of te dt flls witin σ of te men % of te dt flls witin σ of te men. 88. mins, σ. mins σ 88. (.) mins + σ (. ) 8. 7 mins 95% of te tletes took between 67.5 nd 8.7 minutes to complete te ce. Question 7 (e) P( < ) P(z <.55).877 Numbe of tletes z σ 88. z. 55. Question 7 (f) Let p Pobbilty of completing te ce in less tn minutes p(.877) Let q Pobbilty of completing te ce in moe tn minutes q(.877) q(.) Tis is te ode in wic se inteviews te tletes: Fied q p p p p q 5! No. of pemuttions 5 4! BeRnouLLi trials p P(Success), q P(Filue) P( successes) n Cpq - n 4 P (. ) (. 877) Te pobbility tt te second peson se inteviews will be te sit peson se ppoces is bout 4.5%. Set D Ppe Stte Emintions Commission Hige Level, Set D, Ppe 99

23 Question 8 (5 mks) Question 8 () C C 5 8 A 6 o F D E Cll b te mesue of CFA. Use te Sine Rule to find tis ngle. sin A sin B b [Use te Sine Rule nytime you e given sides nd non-included ngle.] C o sinβ sin 6 5 o 5sin 6 sin β 5sin 6 β sin o 79. Te ngles of tingle dd up to 8 o. 8 o Cll γ, te mesue of ACF. A 6 o 5 C β F γ o o o γ o o o o 5 Use te Cosine Rule to find DE. b + c bccos A [Use te Cosine Rule nytime you e given sides nd n included ngle.] A 6 o 79.8 o F 4. o C 8 DE + 8 ( )( 8)cos 4. o DE o + 8 ( )( 8)cos 4.. cm D C E Question 8 (b) sinα α sin ( ) o Te mimum of te ngle b is 9 o. If it goes beyond 9 o te sol pnel will topple ove. 5 A α β F Mtemtics Leving Cetificte

24 Question 9 (5 mks) o o o o O o 6 o An equiltel tingle fits into te bse of te cylinde ectly. Te cicle t te bse of te cylinde is cicumcicle wit cente O. Te cente of cicumcicle O is found by intesecting te pependicul bisectos of te sides of te tingle. Lift out te igligted tingle to find n epession fo. cos A cos o Adjcent Hypotenuse o o o o O o o Set D Ppe Stte Emintions Commission Hige Level, Set D, Ppe

25 Lift out te igligted tingle to find n epession fo. It is igt-ngled tingle. o o o o O o 6 o + b c Use Pytgos. O + ( ) V π V π Volume of Cylinde π 4 π 8π 8π π Mtemtics Leving Cetificte