YJC H2 MATH P2 ANS

1 YISHUN JUNIOR COLLEGE 2015 JC2 PRELIMINARY EXAM PAPER 2 H2 MATHEMATICS SOLUTION Qn Solution 1a cosar  and sinbr  Therefore    i cos sin i cos isina b r r r         Let P be the point on the Argand diagram representing the complex number iab . r is the distance of the point P to the origin.  is the directed angle made by the line segment joining origin to point P, with the positive real axis. Cartesian equation of locus: 22 9xy 1b Given cos isinz  By De Moivre’s Theorem   111 cos isin cos isinzz          OR 1i1 e cos isinzz      OR 1 1 cos isin cos isin cos isinz        2 2 2 cos isin cos i sin     22 cos isin cos sin     cos isin Im( )z iP a b  a b r O Re( )z

2 OR *2* 1 cos i1 sinzzz z z     1 cos isin cos isinz z         2cos (Shown) 1 cos isin (cos isin )z z         2isin (Shown) 2 2 1 1 1 1 zzzz z z z z                Hence 2 2 1 2cos icos or icot1 2isin sin z z       2i 24yx and 28( )y k x 4 8( ) 2( )y y k y y k     2yk 2 4(2 ) 8x k k 22xk Diameter of the rim of the bowl  2 2 2 2kk   or 2 2 2k 42 k (Shown) 2ii Capacity of the bowl 2 8( ) d k k y k y 2 218 2 k k y ky   2 2 2 24182 22k k k k     24 k Volume of material used for the bowl 2 2 0 4 d 4 k y y k 222 0 24 k yk  228 0 4kk   24 k  capacity of the bowl 2iii Given 4k  Required area 22 22 0 112 4 d 84 k x x x  

3 42 2 0 12 4 d 8 xx 30.16988933 30.170 (3 d.p.) 3i C: 2 7 2 x qxy x   , 0q 232 2 qy x q x      or const2 2y x q x     Asymptotes: 2x and 2y x q   3ii If yx is an asymptote of C, then 2x q x   Thus 2 0 2qq    (Shown) 3iii   2 2 27 1 2 16 xx xx    ---- (*) 2 227 162 xx xx   yx y x O 2x Graph of C (0,3.5) ( 1.83,0) (3.83,0)

4 x-coordinates of points of intersection: 0.170, 3.98 Solving (*), 4 0.170x   or 2 3.98x 4i g( ) lnxx , 0x 1h( )x x , 0x Any line yk where k , cuts the graph of g at most once g is a 11 function and therefore function 1g exists. Let g( ) lny x x 1g ( ) e yyx  1g ( ) e xx , 1 ggDR  and 1 ggRD  4ii Since gh RD     , composite function hg does not exist. x g( )yx O (1,0) yk y 216yx yx y x O 2x Graph of C (0,3.5) ( 1.83,0) (3.83,0)

5 4iii 11gh( ) g(h( )) g ln lnx x x xx      gh hDD  gh : lnxx  , 0x ghR  4iv 2 11h ( ) h 1xx xx     9 2 2 2 2 1h ( ) h(h h h h ( )) h( )x x x x   4v(a) g( ) lnxx and gh( ) lnxx Graphs of g and gh are reflections of each other in the x axis. 4v(b) Graphs of g and 1g are reflections of each other in the line yx . 5i Cohort 600 Males 400 Females Sample size 600. 305 00 20 Randomly select 30 males and 20 females from 600 males and 400 females respectively to be surveyed. 5ii A stratified sample is preferable in this context as the population is fairly represented. 6 22.2 26.2 24.22  2 ~ N , 4X    P 22.2 0.0128X  22.2 24.2P 0.0128 2 Z     4 2.2322  1.79 (3 sf) 

6 7 B – Black; R – Red 7i Req. probability 3 2 3 5 2 3 2 7 6 8 7 6 8 7 2 7 5 8         Alternatively: Req. probability 3 2 3! 3 2 1 7 6 2! 5 8 8 7 6 2 7         7ii   4 3 3 2 3 5 2 3 2 1 7 6 7 P 3rd ball red 5 6 8 7 5 88 6 8 7 6 3 8              B R 5 7 5 8 3 8 2 7 4 7 3 7 R R R R B B B B B B R R 3 6 4 6 2 6 3 6 4 6 2 6 1 6 5 6 1st 2nd 3rd

7  P 1st and 3rd ball r 3 5 2 3 2 1 8 7 6 8 7 6 3 2 e 8 d              P 1st ball red | 3rd ball red P 1st and 3rd ball red P 3rd ball red 3 228 3 78  7iii     P at least 2 red | 1st ball red 5 2 2 11 6 7 21 2 P at least 2 red7 7     Hence events are not independent. 8i 0 0: 5H  1 0: 5H  Under 0H , the test statistic is ~ N(0,1)XZ s n  approximately (by CLT) where  = 50, 55x , n = 60,   22 60 19.359s  p- value = 0.0233 Since the p-value < 0.05 (the significance level), we reject H0 and conclude that at the 5% level, there is sufficient evidence that the mean time required by a person to complete the sum has increased. 8ii Since the sample size (60) is l arge, CLT can be used to approximate the sample mean time required to complete the sum to a normal distribution. Hence any assumption is not necessary. 8iii p-value = 2(0.0233) = 0.0466 < 0.05 Since the p-value < 0.05 (the significance level), we reject H0 and conclude that at the 5% level, there is sufficient evidence that the mean time required by a person to complete the sum has changed. 9a M A T H E I C S M A T No repeated letters : 8 5 6720P 

8 1 pair of repeated letters: 73 1 3 5! 63002!C C   2 pairs of repeated letters: 1 3 2 6 5! 5402!2!CC    Number of codewords = 6720 + 6300 + 540 = 13560 9bi 6 1 5! 2!2 Required probability 3 or 0.0133 (3 sf)13560 226 !C    9bii The codeword contains 4 vowels, hence only 1 consonant is contained in the codeword. There are 5 choices of consonants. Required probability 54 12 5 1 3! 5! 2! 3 5 CC C      10i 10ii (a) y on x: 0.980 (3 sf)r (b) y on x : 0.991 (3 sf)r 10iii From (i), the scatter diagra m shows that the rate of increase of y decreases as x increases. Hence the model y c d x should be a better model. This is confirmed by the r value in (ii), which shows that y c d x has 0.991r which is closer to 1. 10iv 0.12102 2.0112 0.121 2.01 (3 sf) yx x   1645 3.yx  Since 1r and estimate is obtained via interpolation, hence estimate is reliable. 10v From GC,  8.6834, 17.585xy Hence a possible point is:  75.4,17.6 (3 sf) x x x x x x x y (200, 28.31) (10, 7.05)

9 11i Assume that the average rate at which people arrive at the bus stop is constant throughout, and that the arrivals at the bus stop are independent of other arrivals. The average rate may not be constant because for example, no on e will take the bus during the wee hours where there are no bus services. The arrivals may not be independent because for example, family members will arrive at the bus stop at the same time for a family trip. 11ii X: no. of people who arrive at the bus stop in a period of 2 minutes.  ~ Po 4X  P 5 0.1562934519 0.156 (3 sf) X   11iii Y: no. of people who arrive at the bus stop in a period of 1 minute.  ~ Po 2Y             1 2 1 2 P at most 1 in 1st minute | 5 in 2 minute s P 0 P 5 P 1 P 4 P5 0.0293050222 0.1562934519 0.1875 (4 sf) Y Y Y Y X         Alternatively: A person who arrived in the 2 minute period has an equal chance to arrive in the 1 st or 2nd min. Req. probability 55 "0 in 1st min" "1 in 1st min" 1 5 3 2 2 16 0.1875     11iv A: no. of people who arrive at the bus stop in a period of 10 min  ~ Po 20A 20 10 . Hence  ~ N 20, 20A approx.    ..22 22.5PP 0.712 (3 sf) ccAA    12i X: no. of successful two point shots out of n attempts.  ~ B , 0.6Xn     P 0.99 P 0 0 1 1 .0 X X        5, P 0 0.0102 (3 sf) 6, P 0 0.00410 (3 sf) nX nX       Hence he needs to attempt at least 6 shots. 12ii Y: no. of successful three point shots out of 60 attempts.  ~ B 60, 0.4Y 60n large, 24 5np , 36 5nq

10  ~ N 24,14.4Y approx    ..P 21 P 20.5 0.704 28 28. (3 s ) 5 f ccYY  