d y 2016 PU3H2 Prelims 2 Paper 2 Marking Scheme 1i) Alternatively, e (2) When x=0, y e 1, 2, 4, By Maclaurin Series, 1ii) 1

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1 06 PU3H Prelims Paper Marking Scheme S/N i) ln y sin dy yd dy yd 4 dy d 4 y (shown) () SOLUTION d y dy d 4 y 4 ( 8 ) d d d 0 dy d y When =0, y e,, 4, d d By Maclaurin Series, 4 y... y ii) ln y sin y e sin 3 sin e d 5 d 5 Alternatively, = or.35 (3 s.f.) 8 y e sin ( ) dy sin ( ) e () d 4 dy 4 y (shown) d *Allow use of GC after substituting y

2 06 PU3H Prelims Paper Marking Scheme (i) f ( ) for. Let y f ( ) 5 y 4 5 y 4 5 or y (rejected since ) 4 5 f ( ) 4 5 D [, ), f 4 R (, ] f Alternatively, y y 0 Using quadratic formula, 4()( y) 5 4y (rejected since ) 5 4 or 5 4 f ( ), 5 D [, ), f 4 R (, ] f ii) g( ) e for Dg (, ) and Rg (0, ) 5 Df (, ] and Rf [, ) 4 Since Rg (0, ) Df (, ], composite function fg does not eist. y

3 06 PU3H Prelims Paper Marking Scheme iii) gf ( ) g( ) e Dgf Df (, ] gf : e for. To find range, Either use mapping: 5 f 5 g 4 (, ] [, ) [e, ) R gf [e, ) i) Similarly, use R f 5 [, ) as the new domain in 4 w y y g( ) to obtain: dw dy w y Alternatively, y d d dy dw dy dw y 4 w d d d d 3 dy y d 3 dw y y d dw y y d dw (shown) d d w d w ln c y ln c c y ln where c is an arbitrary constant 5 4 R gf [e, )

4 06 PU3H Prelims Paper Marking Scheme 3ii) 3iii) 3iv) y when c c, y ln As, ln 0, y y 0 C0 C y=0.5 C Where from part (ii) : C0: y ln where c 0, C: y ln, where c, C: y ln Note: when c 4, as increases, y increases gradually and looks as though it tend towards 0.5, instead of the shape seen above. It can be accepted.

5 06 PU3H Prelims Paper Marking Scheme 4i) y z l : 3 y z Let = 3 y z 3 r = 0, 3 p : y p : r 0 To show l lies in p, Method : 0 l is parallel to p a point on l also lies in p 0 Hence, l lies in p Method : Substitute line equation into plane equation: ( ) for all values of. Hence all points on l lies in p. Hence, l lies in p.

6 06 PU3H Prelims Paper Marking Scheme 4ii) 5 r 0 0 where 0 t Direction vector of l is sin 30 o 0 t 0 t 5 0 t 4iii) t 5 t 5 Or t.48 (3 s.f.) p : r s 4 3 Since and meets at l, p p l is perpendicular to the normal vectors of Hence, direction vector of l is: 3 s s s 3 s 9 s 4 4iv) OA Let F be the foot of perpendicular from A to p. Line AF: r = c 4, c 3 4 Alternatively, t t 5 5 p and p.

7 06 PU3H Prelims Paper Marking Scheme c c 3 c 3 3 OF 4 3 Hence, 6 AF OF OA.55 (3sf) 3 4v) p y z y : 3 p3 : r Since all planes have no common point of intersection, line l is parallel but does not lie in p

8 06 PU3H Prelims Paper Marking Scheme 5i) Considering the 3 different positions, arrange the employees in alphabetical order within each position and assign each of them with a number. Select the number of employees to from the 3 different positions in proportion to that of the population: No. of employees selected From administrative positions (50) From professional positions 5 (50) From managerial positions (50) Use simple random sampling (or a random number generator) to randomly select 3, 3 and 6 employees undertaking the administrative, professional and managerial positions respectively. 5ii) In comparison with a stratified sample, a systematic sample may not be a representative subset of the population as the employees undertaking the professional position may be under-represented. 6 X ~ N (, ) P( X 85) P( X 0) By symmetry, P( X 85) By standardising, Z ~ N 0, P( Z ) (allow.6000) 5.00 (3 sf) OR, form equations with and and solve simult. By standardising, Z ~ N 0, 85 P( Z ) P( Z ) (allow.6000) (allow.6000) Using GC, 93, 5.00 (3 sf) 7i) The mode of payment made by a customer is independent of other customers.

9 06 PU3H Prelims Paper Marking Scheme 7ii) X ~ B (0, p ) P X p p p (3sf) shown 7iii) X ~ B (0,0.877) P X a Highest Probability a P X a Last part Hence most likely number is 9. Let random variable Y be the number of customers who pay by credit card out of 00 customers. Y ~ B (00,0.877) Since n=00 is large, np and nq.3 5, npq Y ~ N (87.7,0.787) approimately P Y 80 P Y 79.5 by continuity correction = (3 sf)

10 06 PU3H Prelims Paper Marking Scheme 8i) Let S be the event that he wins a strike. 8ii) P SS ' S ' P S ' SS ' P S ' S ' S 0.45 p p p p p 0.5 p iii) P(a strike for last game wins at most a strike) P SSS ' ' Peactly strike P no strike (3 sf)

11 06 PU3H Prelims Paper Marking Scheme 9a) 9bi) or [Also accept if the points indicate a quadratic relationship] 8. 9bii) (A) r-value = (4 dp) (B) r-value = (4dp) 9biii) Case (): y c d is more appropriate when c 0 and d 0 because for < 0, when increases, the value of y increases, and for 0, when increases, the values of y decreases. Whereas for Case (), each increase in, the values of y increases eponentially. Note: Accept other possible reasons like as increases, y increases and then decreases OR scatter plot resembles that of a quadratic relationship y c d where c 0 Altenatively, r value for Model (B) is closer to than for Model (A), which indicates a stronger linear correlation between and y. 9biv) Equation of regression line: y y y When 9 y, or or.5 ( dp) Hence, the temperature will be.5 o C or.5 o C. Since r is close to and y 9 falls within the given data range of y, interpolation gives a reliable estimate.

12 06 PU3H Prelims Paper Marking Scheme 0a) 0bi) 0bii) P(first and last digit the same) No. of ways 0! [leading from P(not all male guests stand net to each other) = P(all male guests stand net to each other) ] 5! 5! 4! or 0.99 (3 s.f.) 6 0biii) 5 5! 0 Required Prob biv) 55 5 Required Prob i) 0 8.8, Unbiased estimate of population mean, Unbiased estimate of population variance, Idea: Group the 5 males, and they can only be placed at the back row. Choose female out of 5 females, order the groups, arrange the 5 males and then the remaining 4 females in st row. s s (or ) (3 sf)

13 06 PU3H Prelims Paper Marking Scheme ii) To test H 0 : 0 against H : 0 Level of significance: Assuming H0 is true,. Since n is small and T X 0 ~ t s 0. is unknown, Using GC, perform -tailed t-test, p-value = (3 sf) Since p-value 0., we reject H0. ) Last part There is sufficient evidence at 0% level of significance to conclude that the mean mass of butter in a packet is not 0g. i.e. There is sufficient evidence to doubt the company s claim. To test H : 0 against H : 0 0 Level of significance, 0.05 X 0 ~ N 0, 0.5 n Assuming H0 is true, Z Value of test-statistic, n z n Critical region: z (allow.9600) Since the company s claim is overstated, H0 is rejected. Test-statistic falls within critical region, 0. n n 57.6 n 58 n : n, n 58

14 06 PU3H Prelims Paper Marking Scheme i) Let random variable X be the number of travel insurance claims received. X ~ Po ii) iii) In one week, Hence, X ~ Po5.6 P X 5 P X 4 = = (3 sf) In a period of week, X ~ Po 5.6 mean 5.6, variance =5.6 Since n = 50 is large, Central Limit Theorem applies and 5.6 X ~ N (5.6, ) approimately 50 P X 5 = (3 sf) Let random variable W be the number of weeks that have at least 5 claims received per week, out of 0. W ~ B (0, ) iv) W W P 3 P = (3 sf) Let random variable Y be the number of non-travel insurance claims received. X Y ~ Po 7(. 0.8) In week, X Y ~ Po3.3 P X Y 0 = (3 sf) v) In a month, X ~ Po.4 vi) Since.4 0, X In a month, Y ~ Po30.8 Since , Y X X Y P X X Y P X X Y 0 P X X Y (3 sf) ~ N.4,.4 approimately ~ N 30.8, 30.8 approimately ~ N 6.8, 68 approimately by continuity correction The average rate of travel insurance claims may not be constant in a year as there are more people who travel overseas during June and December and that may give rise to more travel insurance claims made.