RI P2 MS

2012 RI H2 Mathematics Preliminary Examination Paper 2 Qn Solution 1 [5] Let Pn be the statement 1 1sin sin 22cos , 2sin 2 n r n rn        . When 1n , LHS cos  3sin sin22RHS 2sin 2 1 3 1 32cos sin2 2 2 2 2 2 2sin 2 2cos sin 2 cos LHS 2sin 2                                  Hence 1P is true. Assume Pk is true for some k  , i.e. 1 1sin sin 22cos 2sin 2 k r k r       . To prove 1Pk is true, i.e. 1 1 3sin sin 22cos 2sin 2 k r k r        .

2 1 LHS cos cos( 1) 1sin sin 22 cos( 1) 2sin 2 1sin sin 2cos( 1) sin2 2 2 2sin 2 1sin sin sin ( 1) sin ( 1)2 2 2 2 2sin 2 1sin sin sin22 k r rk k k kk k k k kk                                                      31 sin22 2sin 2 3sin sin 22 RHS 2sin 2 k k                  Hence Pk is true implies 1Pk is true. Since 1P is true, and Pk is true implies 1Pk is true, by Mathematical induction, Pn is true for all n  2(i) [6] d ,0d x kx kt   1 ddx k tx   ln , 0x kt c x    e ktxA  When 0t , 80x , thus 80A . When 3t , 20x , 320 80e k 1 1 1ln ln 43 4 3k   . Thus ln 4380e t x   2(ii) [3] 2ln 4 1When 6, 80e (80) 16tx    Just before the (n+1)th injection, the amount of drug present in the blood stream = 1 16 nu

3 1 Immediately after the ( +1)th injection, 1amount of drug present, 80 16 nn n uu  [1] In the long run, the amount present approaches the value 85.3 (3 s.f.). 3a [3] 3 1 0 22 / 7 1 4 9 4 5 3 2 cc                                           223 (1) 7 8 4 4 (2) 5 3 2 (3)cc                   From (1) and (2) we have 22 3 (4)7 4 4 8 (5)         Solve (4) and (5) by GC 13 7,33  Sub into (5)  13 75 3 2 433 c c c                . [3] Let the foot of perpendicular be F . Then     1 40 3 3 1 0 1 1 4 9 4 0 5 3 4 3 3 4 15 1 16 9 36 12 0 1 AF                                                           The coordinates of F are  4, 5, 8 . 3b [2] 1cd 3bi [3] (i) Shortest distance is

4 22 22 0 3 1 31 9 1 4 84 1 5 3 43 261 4 3 40 1 5 26 20 5 8 1 4 45 26 26                                                          3bii [2] 8 0 8 1 9 1 4 1 4 8 15 4                                    r r 3bii i [2]     1 2 : 8 4 5 : 8 4 2 5 8 4 3 8 13 4 p x y z p x y z x y z              r 4(a) [2] 6 3 1 , 2, 1,0,1, 2,3 ki z z e k       4bi [2] 1kkP OP = 2 n  . 2OQ = 2cos n   12PQ = 2sin n   4bii [2] For k = 2, 3, …,n-1 1 1 22cos cos (constant)k kk k OQOQ OQ n OQ n                 Since the ratio of consecutive terms is a constant, 23, ,..., are in GP.nOQ OQ OQ

5 For k = 3, 4, …,n-1 1 1 1 1 2sin 2sin 2cos2sin k k k k kk kk k Q Q OQ n OQQQ n Q Q n OQ n                Since the ratio of consecutive terms is a constant, 2 3 3 4 1, , ..., are in GP.nnQ Q Q Q Q Q  4bii i [4] 1 2 2 3 1 2 1 2 2 2 2 2sin cos sin cos sin 21 cos 2sin 21 cos 2sin 21 cos21 cos nn n n n PQ Q Q Q Q n n n n n n n n n n n                                                                     1 2 1 2sin cos cot 1 1 2sin nn n n                        5 [3] Let X be the monthly rainfall in a region. Since n = 60 is large, by Central Limit Theorem, 245N(159, ) approximately60X P( 150) 0.0607 (3s.f)X  Assume that the rainfall in each month is independent and identically distributed. 6 [4] Consider users from the following 5 strata: below 13 year-olds; 13 – 21 year-olds; 22 – 40 year-olds; 41 – 60 year-olds; above 60 year-olds. Conduct the survey on the first 24 people from each stratum who exits a particular train station (eg City Hall MRT at 12 noon on Weekday). Quota sampling may not give a fair representation of the perception of the general population of MRT users. For example, users who exit from other stations (say along Circle line or NE line) or who travel during peak rush hour in the morning or early evening may have a different travel experience. OR as the surveyor may survey people who appear more approachable and willing to

6 participate in the survey such as tourists. No, as the sampling frame, a list of all users of the train system is not easily obtainable. 7i [2] Number of ways in which 1 man and 2 women are chosen = 6 12 6 ( 1) 3 ( 1)2 n nnC C n n    Number of ways in which 3 women are chosen = 3 ( 1)( 2) ( 1)( 2) 3! 6 n n n n n n nC     Total number of ways = ( 1)( 2) ( 1)( 16)3 ( 1)66 n n n n n n nn       1So, 6a 7ii [2] ( 1)( 16)f ( ) is an increasing function for 1.6 n n nnn  From the GC, when n = 9, f(n) = 300 when n = 10, f(n) = 390 Thus, , 9 (or 10)n n n   [1] X does not follow a binomial distribution. Reason: The event that a committee member is a man is not independent of the event another committee member is a man. OR The gender of a committee member is not independent of that of another. 8i [2] 0.929r (3 s.f) r is close to 1, which indicates a strong linear correlation between x and t. So it suggests that a linear model is appropriate. 8ii [2] The scatter diagram shows that as t increases, x increases at an increasing rate, which suggests that the data points may fit better with ln x a bt . 8iii [2] ln 3.02 0.654xt (3 s.f) r = 0.999 8iv [2] When 200x , ln 200 3.0210 0.65432 3.5 (1 d.p) t t   Since x = 200 lies within the given range of data of x between 20 to 510 and there is a strong linear correlation between ln x and t as the product moment correlation coefficient is 0.999 (close to 1), the estimate is reliable. 9i Let R be the mass of a box of red grapes in grams.

7 [2] Then 2(300, 40 )RN Required probability 98 1 P( 310) P( 310) P( 310) 0.0239 ( 3 s.f )C R R R      9ii [3] Let G be the mass of a box of green grapes in grams. Then 2(150, 20 )GN 1 2 3Let 3X R R R G    2 E( ) 3E( ) 3E( ) 450 Var( ) 3Var( ) 3 Var( ) 8400 X R G X R G       Then (450,8400) ( 500) 1 ( 500) 1 ( 500 500) 0.293 (3 s.f ) XN P X P X P X          9iii [4]    1 2 10 1 2 10Let 0.005 ... 0.006 ...Y R R R G G G        22 E( ) 0.005 10E( ) 0.006 10E( ) 6 Var( ) 0.005 10Var( ) 0.006 10Var( ) 0.544 Y R G Y R G           Then (6,0.544)YN ( 7) 0.912 ( 3 s.f )PY  10i [2] P(A  B) P(rise, rise, fall) + P(rise, fall,fall) (0.1 0.7 0.3) (0.1 0.3 0.9)      0.048 10ii [3] P(B) P(A B) P(A' B)    0.048 P(fall, fall, fall) + P(fall, rise, fall) 0.048 (0.9 0.9 0.9) (0.9 0.1 0.3) 0.804        10ii i [2] P(A’  B) 1 [P(A) P(A B)]    1 [0.1 0.048] 0.948    10i v [2] P( B A ) P(A B) 0.048 P(A) 0.1  0.48 [1] Since [P(B) = 0.804] ≠ [ P( B A )= 0.48]  A and B are not independent. 11i [2] Let X be the number of prizes given out in 2 days, out of 40. (40,0.1) ( 2) 1 ( 2) XB P X P X    1 0.22281 0.777 (3s.f)   (ii) [4] Let Y be the number of prizes given out in 7 days, out of 140 (140,0.1)YB Since n = 140 is large such that np = 14 > 5, n(1-p) = 126 > 5,