AJC H2 MATH P2 soln

Anderson Junior College Preliminary Examination 2015 H2 Mathematics Paper 2 (9740/02) 1a (i) 2 2 2d 2(e ) ( 2e )d xxw Ax      2 22 4e (e ) x x A    22 22 d4 LHS = e = RHS (verified)d (e ) x x w wx A    (ii) 2 d2 d e x y x A  2 2 d e xyx A  2 2 1 2 e d 1e x x A xA A   21 ln 1 e , xABA   where B is an arbitrary constant. 1b Let S be the amount of salt (in grams) at time t minutes.   d rate of salt entering tank rate of salt leaving tankd 2(5) 5100 200 g/min20 S t S S    1 20 1 20 1 20 1 d 1 200 d 20 1 d d 200 1ln 200 20 200 e 200 e t t S St StS S t c SA SA            When 0,t 50S  50 200 A 150A 1 20200 150e t S     When concentration is 1 g/litre, 100.S  1 20100 200 150e t    ⇒ 1 20e 2 3 t  ∴ 20ln(2 3) 8.11 min (3 s.f.)t  2 5 5 ( 2 ) 2()55 33 5 5 5 5 32 0 32 32 2 , 0, 1, 2 2 , 2 , 2 , 2 , 2 ki k i i i i i i z ze z e k e e e e e                   2i Method 1 45 51 3 2 5 2 * 2 n in nn i n in z e ez e          

For 1 2 * n z z    to be real and positive, smallest n = 5 Method 2     1 12 2 12 arg arg( ) arg( *)* arg( ) arg( ) 3 55 4 5 n z n z zz n z z n n            For 1 2 * n z z    to be real and positive, 4 2 , 5 n kk  5 , 2n k k , so smallest 5n 2iii Let the complex number represented by A’ be x+iy BA rotates 90o about B to get BA’: 5( ) 2 ( ) 2 2 iiix iy e i e e          1 5( 2) 2 ( 2) i x iy i e         since ei = -1 2 2cos 2 sin 2)55x iy i i      Real part = 22 2 sin 2 2sin55i     3(i) 4 6 2 1 2 0 2 2 1 4 0 4 2 AB                                         and 3 6 3 2 0 2 6 0 6 AC                              Normal to plane ABC = 1 3 2 1 2 0 2 6 1 AB                              Equation of plane ABC is 2 2 6 r 0 0 0 12 1 1 0                               5 2 i Ae   2 i Be   Re(z) Im(z) 2 A’

3(ii) Since S lies on the perpendicular from (3,-1, 4) to plane 1x y z   , 31 1 1 41 OS                       (Note S lies along perpendicular from D to p1. Line SD: r = 1 + pOD n ) S lies on plane ABC  32 1 0 12 41                        6 2 4 12 2 3         11 331 25 1 1 3341 14 3 OS                               . So S = ( 11 3 , 5 3 , 14 3 ). 3(iii) (iii) Let M be the mid-point of DS. (Note a pt. on plane required is the mid pt of D and S) 2 OS ODOM  11 10 33 3 1 5 4 12 3 3 414 13 33                                     Normal to p2 = 1 1 1   Equation of p2 : 10 311 4r 1 1 9 311 13 3                               1 r 1 9 1      4(i) Let n be the number of intervals between the first hook to the last.  2(50) ( 1)( 2) 5002 n n    2 51 500 0nn   13.2n or 37.8n Thus smallest 13n 37.8n is rejected since any additional intervals beyond 13 will give a total length greater than 500. Since the smallest number of intervals is 13, number of hooks = 13+1 = 14 Notice OM is NOT 1 2 SD .

4(ii)  2 80 1 0.9 80 80(0.9) 80(0.9) ... 80(0.9) 1 0.9 n n         80 1 0.9 5001 0.9 n  0.9 0.375n  ln(0.9) ln(0.375)n  9.31n Smallest n = 10 4(iii) Length of the ribbon between the nth and (n+1)th hook must satisfy the condition:     1 80 0.9 50 2 1 n n     4(iv) Using GC to solve the inequality in (iii), 9.28n or 21.3n (N.A since there are only 14 hooks from (i)) Thus the number of hooks with ribbons will be 10. 5(i) The interviewer will station beside the auditorium exit and interview the first 4 Chinese, 4 Malays, 4 Indians, and 3 from other races that exits the auditorium after the film screening has ended. [Appropriate strata must be suggested (male/female, race, age group, etc.), and quota must be set for each strata.] 5(ii) To obtain a stratified sample of size 30 from a population of size 650, draw random samples (using simple random sampling) from each mutually exclusive subgroup, divided based on age group, with sample sizes as follows: Subgroup (age group) below 21 years old 21 – 35 years old 36 – 50 years old above 50 years old Total Sample size 30 650 130 6   30 650 195 9   30 650 260 12   30 650 65 3   30 After selecting the 30 people, the surveyor will contact them through telephone using the contact numbers in the registration list. Difficulty could be encountered in contacting some of the selected people after the event. For stratified sampling, the person selected that is hard to contact cannot be simply replaced by another person (this is allowed if it is quota sampling). 6 No of ways   3 (3 1)! 3!   = 432 6 last part Method 1 (Direct) Case 1 : Charles is with Father & his mother is separated.   2 3! 2! 2! 2 96    Case 2 : Charles is with Mother & his father is separated. No of ways = 96 (same as case 1) Case 3: Charles ‘s family is together:   2 3! 2! 3! 144   Total no of ways = 96 x 2 + 144 = 336 FaAMa FbBMb C Fc Mc Mc 3 units Andy & Ben C & Fc Mc FaAMa FbBMb CFcMc 3 units Andy & Ben C family

Method 2 (Complement method) Total no of ways if Andy and Ben stand between their parents =   2 5! 2! 480 No. of ways where Andy and Ben stand between their parents and charles family all separated =   2 2! 2! 3 2 1 48     No. of ways where Andy and Ben stand between their parents, charles parents are together but not with Charles =   2 3! 2! 2! 2 96    Total = 480 – 48 – 96 = 336 Method 3: Case 1: Charlie’s family are together No of ways =   2 3! 2! 3! 144   Case 2 : Charlie’s parents are separated No of ways =   2 2! 2! 3 2 4 192     Total number of ways = 144+192 = 336 7 X: Heights of plants supplied.   2~ N ,X    25 2525 0.06 0.06 1.55477P X P Z          -- (1)   91 9191 0.04 0.96 1.750686P X P Z          ---- (2) Solving (1) and (2), 56.044 56.0, 19.967 20.0    Let m be the required median height of plants accepted by Evergreen.   1 0.06 0.040.06 0.51 2P X m     56.5m cm Y: number of plants supplied, out of 20, that are rejected.  ~ B 20 , 0.1Y  4 0.957PY    1 2 3 12 1 2 3 232 2 2 0 0.00254 X X X XXP P X X X            (Answer uses sigma = 20,   5sf, Use 5 sf for both parameters = 0.00250, use mu = 168, 20  , get 0.00256) 8i 0.9550961661 0 0.955 (to 3 sig.fig)r   Since r value is close to 1 , it suggests a strong negative linear correlation between x and y, hence a linear model is appropriate.    2 1 2 32 2 ~ N 168.132 , 9 20X X X FaAMa FbBMb C Fc Mc 3 units Charlie family FaAMa FbBMb Andy & Ben 2 units Andy & Ben Slot in Fc, Mc first, then Charlie has 4 choices to be beside his parents FaAMa FbBMb C Mc Fc FaAMa FbBMb FaAMa FbBMb FcMc

8ii 8iii With point