u x y reduces the differential equation

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1 CATHOLIC JUNIOR COLLEGE H MATHEMATICS 06 JC PRELIM Paper (i) Prove that the substitution (ii) (i) Given u x y, du dy x y dx dx du dy x y (I) dx dx Substitute (I) & u x y and into D.E: we get d u u dx du u dx (shown) Solving, du u dx du u dx Integrate both sides with respect to x: d u u dx dx dx u du dx u u x y reduces the differential equation dy y x x y dx to du u dx. dy Hence, show that the general solution of the differential equation y x x y dx is given by x y x D, where D is an arbitrary constant. dy y x x y dx x C, where C is an arbitrary constant The result in part (i) represents a family of curves. On a single diagram, sketch a nonlinear member of the family which passes through the point, 0. You should state the equation of the graph and axial intercepts clearly on the diagram. [] (iii) State an equation of the line of symmetry for the curve in part (ii). [] [4] Page of 9

2 u x C C u x C x y x D, where D (ii) Since curve passes through point ( -, 0), Using x y x D When x, y 0 ( ) 0 ( ) D 4 D D 4 x y x 4 x y ( x 4) x y x x 8 6 y 8x 6 y 0,4 C y x : 8 6,0 O 0, 4 x (iii) The equation of line of symmetry is y 0. The three distinct roots of the equation x 0 are denoted by, and. (a) Without first finding explicitly, show that 0. [] (b) Given now that 0 arg( ), sketch, on a single Argand diagram, the loci given by (i) z and (ii) arg( z ) arg( ). [] Hence, find the complex number that satisfies both loci, expressing your answer exactly in the form a + ib, where a and b are real numbers. [] [] a Since is a root of x 0, 0 ( )( ) 0 [By long-division since is a root.] Since, so 0 Alternative Method Page of 9

3 Using Geometric series, 0 (as ) b(i) (ii) z arg( z ) arg( ) In this case we need to find w. x 0 x i (0 + kπ) = = e x = e i (kπ/), k = 0, x =, e i (π/), e - i (π/) As 0 arg( ), so w = e i (π/) and w = e - i (π/) ( /) ( /) Thus, z i i z e e ( /) i z e [circle radius, centre e i (π/) = cos π And arg( z ) arg( ) i( /) arg( z ( )) arg( e ) ( ) = π + i sin π = - + i Locus of (i) Im Locus of (ii) M (pt. of intersection) ω - - O Re Using Pythagoras Theorem, OM. The complex number is i. The diagram shows the graph of curve C represented by y f ( x), with oblique asymptotes y x and y x. Page of 9

4 (a) On a separate diagram, sketch a graph of y f ( x), clearly indicating the equation(s) of the asymptote(s) and axial-intercept(s). [] (b) The above curve C is represented by the parametric equations x tan, y sec, where. (i) Show that the normal to the curve at point P, with coordinates (tan, sec ), for 0, is given by y x cosec sec. [] (ii) The normal to the curve at point P intersects the x-axis at point N. Find the coordinates of the mid-point M of PN, in terms of. Hence find a cartesian equation of the locus of M, as varies. [4] (iii) Taking O as the origin, show that the area of triangle OPN is tan sec. Point P moves along the curve such that the rate of change of its parameter with respect to time t is given by d cos. Find the exact rate of change of dt the area of triangle OPN when. [4] 6 (a) [Note: horizontal asymptotes y = ] [Consider what happens to the value of f (x) as x varies] Page 4 of 9

5 b)(i) b)(ii) Given x tan, y sec, where. dx sec d, dy dy dy dx d d dy d sec tan, sec tan tan sin sec sec At point P, gradient of normal = sin. Equation of the normal to the curve at P : y sec ( x tan ), sin x cos sin cos y, y x cosec sec (shown) [Note: Graph is just useful to know but not necessary.] Since equation of normal from (i) is y x cosec sec At point N, y = 0 So, 0 x cosec sec, sec P x tan cosec N Point N is ( tan, 0) Since point P is (tan, sec ) tan tan sec 0 Locus of point M is given by the parametric equations Mid-point of PN is M, tan, sec Page 5 of 9

6 , x tan y sec, where. Using sec tan, ( y) ( x), a cartesian equation for the locus of M. (b) (iii) [Note: Detailed graph is just useful to know but not necessary.] A, area of OPN (base)(height) ( tan ) (sec ) [since 0 ] tan sec (shown) O P N da d sec sec tan sec tan sec sec tan Rate of change of area of da da d dt d dt OPN, (sec sec tan ) cos sec tan [Given d cos ] dt Alternative Method Differentiating A tan sec implicitly with respect to time t, da d sec d sec tan sec tan t dt (sec sec tan ) cos sec tan When 6, So, when 6, tan cos da 4 5 d t sec tan sec, 4 Adam and Gregory signed up for a marathon. In preparation for this marathon, Adam and Gregory Page 6 of 9

7 each planned a 5-week personalised training programme. Adam runs.4 km on the first day of Week, and on the first day of each subsequent week, the distance covered is increased by 0% of the previous week. Gregory also runs.4 km on the first day of Week, but on the first day of each subsequent week, the distance covered is increased by d km, where d is a constant. Assume Adam and Gregory only run on the first day of each week. (i) Find, in terms of d, the total distance covered by Gregory in these 5 weeks. [] (ii) Adam targets to cover a total distance of 70 km in these 5 weeks. Can Adam [] achieve this target? You must show sufficient working to justify your answer. (iii) It is given that Adam covers a longer distance than Gregory on the first day of the 5 th week. Find the maximum value of d, correct to decimal places. Using this value of d, show that the difference in the distance covered by Adam and Gregory for their 5 th week training is 0.4 km correct to significant figures. [4] (iv) Due to unforeseen circumstances, Adam has to end his training programme early. In order for Adam to cover a total distance of 70 km by the end of the weeks, the distance covered has to be increased by x % of the previous week on each subsequent week from Week. Find x. [] 4(i) [Gregory s is a A.P. with a =.4 and d and n = 5] 5 S5 of Gregory's plan (.4) 4d 6 05d 4(ii) [Adam s is a G.P. with a =.4 and r =. and n = 5] S 5.4. of Adam's plan Yes, Adam can achieve his target. 4(iii) U5 of Adam's plan > U 5 of Gregory's plan U of Adam's plan U of Gregory's plan > (.) (.4 4 d) 0 5 [Un of GP = ar n, U n of AP = a + (n )d ] d.0957 Hence maximum value of d is.0 (to dp) Difference in distance covered is = Adam s distance Gregory s distance 4.4(.) (.4 4(.0)) 0.4 (shown) 4(iv) x.4 00 New S of A 70 x 00 From GC, Page 7 of 9

8 x = 5.4% Section B: Statistics [60 marks] 5 A bag contains four red and eight blue balls of which two of the red balls and six of the blue balls have the number 0 printed on them. The remaining balls have the number printed on them. Three balls are randomly drawn from the bag without replacement. (i) Show that the probability that at least one blue ball is drawn is [] Find the probability that (ii) at least one ball of each colour is drawn, [] (iii) the sum of the numbers on the balls drawn is at least two. [] 5(i) 4 red ( 0 and ) and 8 blue (6 0 and ) P(at least one blue ball) = P(no blue balls) = P( red balls) 4 C = - C = (Shown) 55 Alternative Method P(at least one blue ball) = P( red balls) (Shown) 55 Alternative Method (Direct Method) P(at least one blue ball) Page 8 of 9

9 = P( blue and red) + P( blue and red) + P( blue) C C C C C = C C C = (Shown) 55 Alternative Method P(at least one blue ball) = P( blue and red) + P( blue and red) + P( blue) 8 4! 8 7 4! ! 0! = (Shown) 55 5(ii) P(at least one of each colour drawn) = P( red) P( blue) = 4 C C 8 C C = = 8 or 0.77 Alternative Method P(at least one of each colour drawn) = P( blue and red) + P( blue and red) C C C C C C = 8 or (iii) There are 8 balls with 0 and 4 balls with P(sum is at least two) = P(,, 0) + P(,, ) C C C C C = or Alternative Method Page 9 of 9

10 P(sum is at least two) = P(,, 0) + P(,, ) 4 8! 4 0! 0 = or Packets of a particular brand of potato chips are delivered to a supermarket in boxes of 60. On average,.8 packets in a box are underweight. The number of underweight packets from a randomly chosen box is the random variable X. Assume that X has a binomial distribution. (i) Use a suitable approximation to find the probability that two randomly chosen boxes of potato chips contain more than 6 packets of underweight potato chips. State the parameter(s) of the distribution that you use. [4] A batch of 50 boxes of potato chips is delivered to the supermarket. (ii) Use a suitable approximation to find the probability that the mean number of underweight packets per box is more than. [] 6(i) 6(ii) Given that X is the random variable denoting the number of underweight packets from a randomly chosen box i.e. out of 60 packets. Then X ~ B(60, p) Given average, np =.8 p =.8 60 = 0.0 Since n is sufficiently large, [Must state conditions for approx.] p is small and np.8 5 So, X ~ Po.8 approximately For two randomly chosen boxes, X X ~ Po.6 approximately [Note: can add Poisson random variables but not subtract.] [Note: To state the parameter(s) of the distribution that you use is to state X X ~ Po.6 approximately.] that P X X 6 P X X (s.f.) [Key phrase: probability that the mean number of underweight packets per box ] Given X ~ B(60, 0.0) [Should start from Binomial case again] Since sample size = 50 is sufficiently large, by Central Limit Theorem npq X ~ N np, approximately 50 Page 0 of 9

11 .746 X ~ N.8, approximately 50 P X (s.f.) 7 A group of 9 friends, including Albert and Ben, are having dinner at Albert s house. They sit in two groups: a row of 4 on a couch and a group of 5 at a round dining table with 5 identical seats. Find the number of ways they can sit if (i) there are no restrictions, (ii) Albert and Ben sit beside each other, [] (iii) Albert and Ben both sit on the couch or both sit at the round table, but they do not sit beside each other. [] [] 7(i) 7(ii) No. of ways with no restriction = (no. of ways to separate 9 people into a group of 4 and a group of 5) (no. of ways to arrange the row of 4 people) (no. of ways to arrange the circle of 5 people) = 9 C4 4! 5 C5 (5 )! = Alternative Method No. of ways with no restriction = (no. of ways to separate 9 people into a group of 5 and a group of 4) (no. of ways to arrange the circle of 5 people) (no. of ways to arrange the row of 4 people) = 9 C5 (5 )! 4 C4 4! [Note: 9 C4 = 9 C5] = No. of ways if Albert and Berta sit together = (Albert and Berta in the row) + (Albert and Berta in the circle) = (no. of ways to pick remaining people for the row arrange row people A and B swap arrange circle people) + (no. of ways to pick remaining people for the circle arrange circle people A and B swap arrange row people) = ( 7 C! (5 )! ) + ( 7 C (4 )!! 4! ) = = 68 7(iii) No. of ways if Albert and Berta both sit on the couch or both sit at the table = (Albert and Berta in the row) + (Albert and Berta in the circle) = (no. of ways to pick remaining people for the row arrange row people arrange circle people) + (no. of ways to pick remaining people for the circle arrange row people arrange circle people) ( 7 C 4! (5 )! ) + ( 7 C 4! (5 )!) = 56 No. of ways if Albert and Berta sit separately Page of 9

12 = = 68 Alternative Method No. of ways if Albert and Berta sit separately = (Albert and Berta in the row) + (Albert and Berta in the circle) = (no. of ways to pick remaining people for the row arrangement in row arrangement in circle) + (no. of ways to pick remaining people for the circle arrangement in circle arrangement in row) = (no. of ways to pick remaining people for the row [arrange remaining people in row slot in Albert and Berta] arrangement in circle) + (no. of ways to pick remaining people for the circle [arrange remaining people in circle slot in Albert and Bert] arrangement in row) = ( 7 C [! C! ] (5 )! ) + ( 7 C [ ( )! C! ) 4! ) = = 68 8 A factory manufactures a certain product for sale. The following table gives the quantity of product manufactured, x units in thousands, and its corresponding cost of production, y dollars in thousands. The data is recorded during different months of a certain year. Quantity of product, x Cost of production, y (i) Draw a scatter diagram for the data. [] One of the values of y appears to be incorrect. (ii) Indicate the corresponding point in your diagram by labelling it P. [] Remove P from the set of data. (iii) By using the scatter diagram for the remaining points, explain whether y = a + bx or y = a + b ln x is the better model for the relationship between x and y. (iv) Using the better model chosen in part (iii), find the product moment correlation coefficient and the equation of a suitable regression line. Explain what happens to the product moment correlation coefficient and the equation of the regression line if the factory decides to include a fixed cost of M thousand dollars for purchasing a packing machine to the cost of production, y. [4] (v) Use the regression line found in part (iv) to estimate the cost of production when the quantity produced is 6000 units and comment on its reliability. [] [] 8(i) GC screenshot : Page of 9

13 y, Cost of production / $,000 8 Scatter diagram of y against x 0 P 9.4 x, Quantity produced /,000 8(ii) P = (6.0, 5). 8(iii) The better model is y = a + b ln x since the set of data points in the scatter diagram exhibits a non-linear trend in which y increases at a decreasing rate as x increases, rather than a linear trend in which y increases at a constant rate with x. 8(iv) r(ln x), y Line of regression y on ln x. y = ( ) ln x + ( ) y = 47.7 ln x 9.9 If a fixed cost is included in y, the value of r (ln x ), y 0.99 is Page of 9

14 unchanged, as there is no change in the strength of the linear relationship between (ln x) and y, under translation. However, as a constant value M is added to the y value of each data point, the regression line would be also translated in the direction of the positive y-axis by M units, i.e. with resultant regression line equation being y = 47.7 ln x M or y M = 47.7 ln x 9.9 8(v) When x = 6, y = ( ) ln(6) = Hence, the estimated cost of production is $ 65,600 (to s.f). This value is reliable, since r = 0.99 is close to indicating a strong positive linear relationship between y and ln x, and estimating y at x = 6 is an interpolation as x = 6 is within the range of values of x in the data used to construct the regression line (i.e..0 x 9.4). 9 The finishing times in a 0km race with a large number of runners follow a normal distribution. After 40 minutes, 0% of the runners have completed the race. After one hour, 5% of the runners have yet to complete the race. The first 0% of runners who finish the race receive a medal. (i) (ii) Show that, correct to decimal place, the runners have running times with mean 55.4 minutes and standard deviation.0 minutes. Find the maximum time a runner can take to finish the race in order to receive a medal. [] [] A random sample of runners is selected. (iii) Find the probability that more than four runners receive a medal. [] (iv) Given that none of the runners receives a medal, find the probability that the slowest runner completes the race in under one hour. [] 9(i) Let X minutes be the random variable denoting the finishing time of a randomly selected runner in the race. [Must define yr random variable.] X ~ N, Then P X PZ () Page 4 of 9

15 P X PZ () Solving () and (), ( s.f.) and (s.f.) 9(ii) Let the maximum time be a. Since only first 0% will receive a medal, P X a So, 0. a ( s.f.) [ using invnorm(0., 55.77,.998) ] Hence, maximal timing is 45. minutes. [Note: should not use GC table as the time is not an integer value] 9(iii) Let Y be the number of runners, out of, who receive a medal. Y ~ B,0. PY P Y (iv) [Keyword given implies conditional probability] P slowest runner finishes within hour all runners do not receive medal P slowest runner finishes within hour and all runners do not receive medal P all runners do not receive medal P all runners finish within hour and all runners do not receive medal 0. [Since the first 0% of runners receive a medal, so ( - 0.) did not receive a medal] [Since after one hour, 5% of runners have yet to complete the race. So probability of running within one hour is ( 0.5) = Also as first 0% receives a medal, so probability of running within one hour and yet did not get a medal = = 0.45 ] (s. f.) Alternative Method Let A be the random variable denoting the number of runners who finish in an hour and do not receive medal, out of runners. A ~ B, A ~ B, 0.45 Page 5 of 9

16 Required probability P( A ) P( Y 0) ( s. f.) [Recall from (iii) Y ~ B,0. ] 0 (a) A Physical Education teacher wants to plan a volleyball training programme for all students in a secondary school, where each student has exactly one CCA. In order to check on the current fitness level of students in the school, he selects a sample of students by choosing the Captains of every sports team and the Presidents of every Club and Society in the school. (i) (ii) Explain briefly why this may not provide a representative random sample of the student population. Name a more appropriate sampling method which would provide a representative random sample and explain how it can be carried out in this context. [] [] (b) 0a(i) 0a(ii) 0b(i) The vertical jump heights of players from a volleyball team are normally distributed with mean 40 cm. The coach claims that a particular training regime is effective in improving the players jump heights. After the regime is implemented for a period of time, a random sample of 7 players is taken and their jump heights are recorded. The sample mean is 4. cm and the sample standard deviation is k cm. A test is to be carried out at the 0% level of significance to determine whether the training regime has been effective. (i) State appropriate hypotheses for the test. [] (ii) Find the set of values of k for which the result of the test would be to reject the null hypothesis. [] (iii) State the conclusion of the test in the case where the sample variance is 5. [] This method is quota sampling and is not representative of the student population as there is no consideration of the size of each sports team and each club and society relative to the student population, and the student from each group is selected non-randomly i.e. only Captains and Presidents are specifically selected. Stratified sampling. Divide the school population by year or by level (strata), calculate the proportion of each year or level relative to the school population and select the respective number of students from each year or level randomly. Let X cm be the vertical jump height of a randomly selected volleyball player. H : 40 0 H : 40 Page 6 of 9

17 0b (ii) [Since X is normally distributed, population variance,, unknown and sample size, n = 7 is small, use t-test] Under 0 where 40 x 4. X H, test statistics: T ~ t n s / n n 7 7 s k [Note that k is sample standard deviation] 6 [unbiased est. of population variance, s n = (sample variance)] n - To reject H 0, at 0% significance level Test-statistic > invt( 0.,6) x 0 t >.498 s / n >.498 7k 6 7. k > k.498 k.578 Required set is k : 0 k b(iii) Since sample variance is 5 and 5.57 we do not reject H0 at 0% level of significance and conclude that there is insufficient evidence to support the claim that the training regime is effective. The average number of calls per hour received by telephone operators at the Call Centre of bank ECBC is being reviewed. (i) State, in context, two assumptions that need to be made for the number of calls received by a telephone operator to be well modelled by a Poisson distribution. [] The Call Centre has only three telephone operators at any point in time. One handles calls pertaining to credit card queries, another handles calls pertaining to business banking queries while the last operator handles calls pertaining to personal banking queries, with the numbers of calls received in one hour assumed to have the independent distributions Po( ), Po(6) and Po(7) respectively. (ii) It is given that the probability of receiving two calls pertaining to credit card queries within an hour is eight times that of receiving two calls pertaining to credit card Page 7 of 9

18 queries within four hours. Find the exact value of, expressing your answer in the form a ln where a and b b are two positive integers to be found. [] (iii) On a certain day, the Call Centre receives more than 50 calls from 00 to 400 hours. Find the probability that there are no calls pertaining to credit card queries during this period. [] (iv) Using suitable approximations, find the probability that there are more calls pertaining to business banking queries than personal banking queries within a twohour period. [] (i) Any two of the following 4 assumptions: () The calls are made independently of each other. () The average number of calls received over any time interval of the same duration within the day is constant. () The calls received occur randomly. (4) The probability of receiving two or more calls within a very short interval of time is negligible. OR Calls are received one at a time. (ii) Let Cn, B n and P n be the random variables denoting the number of calls received in n hours pertaining to credit card queries, business banking queries and personal banking queries respectively. C ~ Po n,b ~ Po 6n P ~ Po 7n Then and n n n (iii) C ~ Po and C 4 ~ Po4 So, PC 8PC e 4 e!! e 8 7 ln8 ln [Conditional probability] Since C, B, and P are independent, C 4 ~ Po ln and B P ~ Po 67 4 C B P ~ Po ln 6 P(no calls for credit card more than 50 calls in hours) P C 0 C B P 50 P C 0 and C B P 50 P C B P 50 P C 0 and B P 50 P C B P 50 Page 8 of 9

19 0 50 P C B P 50 P C P B P (s.f.) (iv) In hours, B and P ~ Po 6 ~ Po 7 Since 0 and 4 0, So B and P ~ N, approximately ~ N 4,4 approximately Since B and P are independent, B P ~ N( 4, + 4) approximately B P ~ N, 6 approximately 0 P X Y P X Y P X Y 0.5 by continuity correction (s.f.) End Page 9 of 9