The shaded region above represents the region in which z lies.


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1 GCE A Level H Maths Solutio Paper SECTION A (PURE MATHEMATICS) (i) Im 3 Note: Uless required i the questio, it would be sufficiet to just idicate the cetre ad radius of the circle i such a locus drawig. Re The shaded regio above represets the regio i which z lies. (ii) Im C Re Miimum z is give by OA OC AC (iii) Maximum z is give by OB OA AB Im P C P D Re Remark : The legth DP i this case, is always the same regardless of which coordiates of P we are takig. This is because it happes that the lie through CD is perpedicular to the lie represetig arg z. This ca be easily verified by fidig the gradiet of the lie through CD. MATHEMATICS DEPARTMENT PAGE
2 Combiig the restrictio give by arg z, the resultig locus of z is give as the shaded regio above (see diagram). The poit D represets the complex umber 6 i. (i) (ii) The maximum value of z 6i is give by DP (there are two possible such positios that give this maximum value). Observe that, ad 5, 5 both lie o the circle which has Cartesia equatio x y 5 3. Thus, these two poits are the possible positios of P. This meas the maximum value required is DP 6 7. From the questio s descriptio, the box will have base area PS SR ad height x. Thus the volume is give by V x x x x x x x Fid 3 x 6x x. (Show!) dv x x dx ad let d V dx x 6 3 x 6 From the diagram (for example, from the side of AD ), 3 we deduce that x, ad thus x 6 is the oly aswer. f x l x 3. 3(i) Write The, x e x f 3 Thus, x f x e 3 fx3 x e.. The domai of, or we ca write as. The rage of f f is is,. Remark : This part ca also be solved by chagig the equatios of the circle ad the halflie to Cartesia form ad do simultaeous equatios solvig. The equatio of the circle is already give while the equatio of the halflie is y x. f Remark : Domai of equals to the rage of f ad rage of f equals to the domai of f. Remark : Geerally, to fid the rage of a fuctio, we will eed a sketch of the graph. MATHEMATICS DEPARTMENT PAGE
3 3(ii) y x 3(iii) The coordiates where the curves itersect the axes are: A, 3, 3 e B,, 3 C, e ad D 3,. Whe the two curves itersect, they also itersect at the lie y x f x f x is equivalet to f x. That is, x. Thus we have x x x l 3. l 3 x Usig the graphic calculator, the values of x are.87 ad 5.8, correct to s.f. Note: TI8 Plus screeshots to get these values (based o the itersectios of the two graphs y x 3 yl x ) are ad show: (a)(i) Use itegratio by parts twice: x x x x x e dx x e x e dx e xe dx x x e x e e dx x e e e e e e e. Note: Usually whe the itegrad cosists of two differet families of terms, we should be prepared to apply itegratio by parts. This is oly a guidelie, as there are l x exceptios such as dx x where it is ot ecessary to employ the techique of itegratio by parts. Aother familiar exceptio is sec 3 xdx where the usual method is to use itegratio by parts to solve it. MATHEMATICS DEPARTMENT PAGE 3
4 (a)(ii) (b) x e x e dx lim. e Note that as. From the diagram, the shaded regio will to udergo a rotatio of oe full roud about the axis. y Note : Itegrals which have limits edig with ifiity such as x x e dx are called improper itegrals. Note : A way to uderstad why e as is by maipulatig the expressio e e ad thik of 3 e...! 3! (accordig to MF5) which ca be cosidered a ifiite sum cosistig of terms i of much higher degrees. x Volume of solid obtaied is give by. 6x V y dx dx. x Usig the substitutio x ta : dx sec d. Whe x ad whe x. ta Thus, V 6 sec d ta MATHEMATICS DEPARTMENT PAGE
5 ta (sice ta sec ta ta sec si 6 d 6 d. ). (Show!) 6 d 6 si d cos sec Thus, V 6 si d cos 6 d (by cosie double agle formula) 8 si 8. GCE A Level H Maths Solutio Paper SECTION B (STATISTICS) 5 From P X..5, we stadardise X : 6(i) X. P.5. PZ () P X , we have Similarly, for 7. PZ () Solvig () ad (): 53.7, 8.3, correct to 3 s.f. Quota samplig might be carried out by Note : To use the IvNorm fuctio i the graphic calculator, we must esure the probability is P X or of the form P X, with the iequality sig beig less tha or less tha or equal. Note : The method of stadardisatio is very efficiet i solvig ormal distributio problems ivolvig ukow mea ad/or ukow variace (or stadard deviatio). First divide the rage of ages ito several groups ( 6, 7, 6 etc). Next, assig a quota (required umber of residets to be iterviewed) for each age group. Iterviewer will the pick residets at his/her ow discretio for iterviews, util all the quotas are fulfilled. MATHEMATICS DEPARTMENT PAGE 5
6 6(ii) 6(iii) 7(i) Oe disadvatage is that the iterviewer may likely have collected a biased sample of residets due to his/her oradom selectio process of residets (selectio is based solely o his/her judgemet). Stratified samplig. This method is ot realistic i this cotext because it will be extremely difficult to gai access to iformatio about our samplig frame. For example, it will be difficult to get a whole list of residets with iformatio such as their ages from a govermet or public agecy. Assumptio : All the frieds should ot have ay kowledge of this experimet to be carried out, before or durig the experimet, to esure that each trial of cotactig a fried is idepedet of oe aother. Remark: Despite the obvious disadvatage of quota samplig, it is still used widely owadays maily due to its big advatage of ot eedig to have a samplig frame to carry out the samplig process. Note: Other aswers are possible such as radomly selected residets may ot be willig to participate i the iterview. Assumptio : The experimet should be coducted i a realistic spa of time i the eveig to esure that time does ot become a factor to affect the costat success probability of cotactig a fried. 7(ii) No matter how quickly the experimet ca be carried out, all the calls are boud to be doe at differet times i the eveig, ad i tur will make it difficult for the probability of success i cotactig a fried to stay costat. 7(iii) For 8 ~ B 8,.7 P R 6, R ad R P 5.55 correct to 3 s.f. TI8 Plus screeshots: MATHEMATICS DEPARTMENT PAGE 6
7 7(iv) For, ~ B,.7 R. TI8 Plus screeshots: Sice is cosidered large, p 8 5 ad p, thus, ~ N8, 8. 5 R approximately. The parameters i this ew distributio are the mea (8) ad variace (8.). cc. Therefore PR 5 PR.5. correct to 3 s.f. Note: c.c stads for cotiuity correctio. 8(i) y Note: The poits should follow a quadratic curve of the form (i the first quadrat): x 8(ii) From the graphic calculator, the value of the product momet correlatio coefficiet, r is.99 correct to 3 s.f. If we observe the differeces of cosecutive y values such as etc: D We would otice that differeces follow a icreasig tred, which would ot be the case if the data have followed a liear model (the differeces would be more or less kept costat). TI8 Plus screeshots to calculatig the value of r : Note: Below is a screeshot from TI8 Plus showig the seve observed poits ad the best fit lie: MATHEMATICS DEPARTMENT PAGE 7
8 8(iii) Let r ad r deote the product momet correlatio coefficiets based o the liear model ad the quadratic model respectively. If r r, the we would deem the 8(iv) quadratic model to be the better model. If r r, the the liear model would be better. From the graphic calculator, the equatio of the leastsquares regressio lie of y o x is y.856 x.39. Though the liear model is ot the best, it is ot obvious from the diagram to coclude that the poits ecessary follow a oliear model. Note: We ca of course fid the value of r for the quadratic model ad do a compariso with that obtaied for the liear model. However, sice the questio asked us to explai how to use, it will be more ideal to just propose a geeral way to help decide which model is better. Some TI8 Plus screeshots: Whe x 3., the estimate value of y is y correct to 3 s.f. MATHEMATICS DEPARTMENT PAGE 8
9 9 A probability tree is draw for the situatio: Remark: A probability tree is for the purpose of clearer illustratio..5 Faulty If the questio did ot ask, it is ot ecessary to draw a A probability tree ad use it to solve.6 probability problems..95 Not faulty. B.7 Faulty.93 Not faulty 9(i)(a) Pa les is faulty P a les is made by A give that the les is faulty 9(i)(b) Ples is made by A les is faulty Ples is made by A ad les is faulty Ples is faulty correct to 3 s.f. P exactly oe of the leses is faulty 9(ii)(a) P st les faulty ad d les ot faulty P st les ot faulty ad d les faulty 9(ii)(b) Pboth were made by A, give that exactly oe is faulty Pboth were made by A exactly oe is faulty Pboth were made by A ad exactly oe is faulty Pexactly oe is faulty (i) correct to 3 s.f. Let deote the mea time take to istall a compoet. Note : We ca also use the complemetary method to obtai the probability: P both faulty P both ot faulty Note : The aswer i this case is exact, thus it is ot ecessary to roud it off to 3 sigificat figures. Null hypothesis H is 38.. Alterative hypothesis H is 38.. MATHEMATICS DEPARTMENT PAGE 9
10 (ii) To reject ull hypothesis, pvalue level of sigificace. (iii) (i) That is T t P.5 t 38. PZ t t Sice t deotes the sample mea time, therefore the set of values required is t : t If the ull hypothesis is ot rejected, pvalue level of sigificace. That is, T t P PZ Sice is a positive iteger, the set of values required ca be writte as : PR.9 correct to 3 s.f. 3 P R r P R r, we have (ii) Startig with 8 8 r r r 9 r r r r 9 r 8!! 8!! r! 8 r! r! r! r! 7 r! 9 r! 3 r! r! 7 r! r! r! r! 8 r! 9 r! 3 r! r r r r r r r r! 7! 9! 3!! 8!!! (Show!) Note :! r r! r! Note : Most probable umber here refers to the mode of R. Note 3: We ca use TI8 Plus to verify the correctess of r : MATHEMATICS DEPARTMENT PAGE
11 r r 8 r 3r r 3 r 8 r r r 5.53 Thus, the required value of r is 6. (i) (ii) Let X deote the umber of people joiig the queue i X ~ Po.8. a period of miutes. The Required probability P X 8 P X 7.3 correct to 3 s.f. Let Y deote the umber of people joiig the queue i a period of t secods.. The Y ~ Po t Y ~ Po.t 6 P Y.7 Give PY PY.7..t.t.t.t Thus e e.7.!!.t.t e.te.7. From the graphic calculator, t Thus t 55 secods (correct to the earest whole umber). TI8 Plus screeshots that display the solvig of the equatio: (iii) (iv) Let J deote the umber of people joiig the queue i a period of 5 miutes ad let L deote the umber of people joiig the queue i a period of 5 miutes. ~ Po 5. L ~ Po 5.8 The J ad J ~ Po8 ad L ~ Po7. Sice i both cases, the mea umber is larger tha. ~ N 8,8 L ~ N 7, 7 approximately. Thus, J ad This meas J L~ N8 7,8 7 approximately. J L~ N 9, 5 By graphic calculator, the required probability is P J L.67 correct to 3 s.f. If the time period is i terms of several hours, the rate of people joiig the checki queue would probably ot be costat throughout, which violates oe of the coditios for a Poisso model to be valid. Remark: I the process of P J L, we computig did ot do a cotiuity correctio eve though J L is a discrete radom variable (it follows a Skellam distributio). Remark: The rate of people joiig the queue would certaily be higher at a certai period of time just before the departure of a flight. MATHEMATICS DEPARTMENT PAGE
12 MATHEMATICS DEPARTMENT PAGE
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