YJC P2 MS

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/2 Pg 1 of 10 Qn Solution 1 (i) Let  be the acute angle between the 2 planes. 12 21 11 3cos 666                   3   (ii) 41 12 21 OB                     for some R 4 12 2         Since B is on 1 , (4 ) 2( 1 2 ) (2 ) 2         62  1 3  11 3 11 51 533 55 3 OB            Thus the coordinates of B is 11 5 5,,3 3 3   . (iii) Normal of 3 1 2 3 1 2 1 3 3 1 1 1 3 1                                            Cartesian equation of 3 : 1 4 1 1 1 1 1 2 1                               r

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/2 Pg 2 of 10 7x y z    7x y z   From G.C. the position vector of the point of intersection of the three planes is 10 3 53 2     2 OPQR forms a rectangle with length OR which is 3 times the length of OP. (a) Let the position vector of complex number z2 be OS 3z x xi   22 2 2 3 8 6z x xi x x i    2 2 8 6 xOS x     3 3 3 3 86 z iz x xi i x xi x xi        8 6 xOQ x   8 6 xOS x xOQx     Since x is a constant and OS xOQ ,  S is collinear with the origin and the point Q. Im(z) Re(z) –3 Q R P O

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/2 Pg 3 of 10 (b) Area of region P =   2 11113 tan 3 tan 4.1723   3 (i)   2f 4 4x x x          22 2 2 2 2 4 2 4 4 x x           Since horizontal line y = 0 cuts the graph  fyx twice, f is not a one-one function.  f does not have an inverse. (ii) Largest value of k = 2 Let   2 22 4 4yx     22 4 4xy     Since 2x  , 22 4 4xy     1 2 2f : 2 4 4 , 4 4x x x          (iii)  1f ,2R      g ,1D   Given 1 2  ,    1 gf , 2 ,1RD        1gf  exists   1gf ln 1 2 ,R       22 , 4 4 Im(z) Re(z) 3 3 –3 –3 P O

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/2 Pg 4 of 10 4 (i) 2 3 4 2 3 4 5 7 9,,333u u u (ii) 21 3 n n nu  (iii) Let Pn be the statement “ 21 ,3 n n nun  ” When n = 1, LHS = 1 1u  , RHS = 21 13   Since LHS = RHS, P1 is true. Assume Pk is true for some values of k  ie 21 3 k k ku  When n = k + 1 Want to show: 1 1 23 3 k k ku      1 1 1 1 LHS 1 323 1 2 13233 23 3 RHS k k kk k kk k u u k k              1P is truek By Mathematical Induction, Pn is true for all n  . 5 (a) dx dt x dx , < 0dt k x k 1 2 d dx x k t   1 22x kt c When t = 0, x=200 2 200C When x = 100, dx 1dt  1 10k  Hence, 1 2 12 2 200 10xt   (shown)

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/2 Pg 5 of 10 When x = 100, 12 100 2 20010 t  82.84t The container has been leaking for 83 min. (b) v x y dd 1dd vy xx     2d 1 ( ) cos 1 d y x y xx      2d1 1 cos 1 d v vxx      2d 1 cosd v vxx   2 1 d cos d1 v x xv  1tan sinv x C  1tan ( ) sinx y x C    ( ) tan(sin )x y x C   tan(sin )y x x C   6 (i) Simple random sample might not be representative if no manager is chosen. (ii) To obtain a sample of 40 staff members, we draw random samples from each category with sample size in the same proportion as the size of each category in the company. Managers Technicians Factory workers Sample size 40 =4.8 5  40 =11.2 11  40 =24 Advantage: Each staff category is represented proportionately. 7 (i) Let be the population mean decrease in cholesterol level H0 :  = 25 H1  > 25 (one-tailed test) Under H0 , (since 2 is unknown and n is small,) the test st atistic is )1(~  nt n S XT 

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/2 Pg 6 of 10 where  = 25, s = 14.51645, n = 12, i.e. T ~ t(11) Level of significance : 0.05 From G.C., the p-value = 0.0180475 Conclusion: Since p value= 0.0180475 < 0.05 (significance level), we reject H0 and conclude that at the 5% level, there is significant evidence to conclude the mean decrease in LDL level is more than 25. (ii) The decrease in LDL level in the underlying population follows a normal distribution. (iii) H0 :  = 25 H1  > 25 (one-tailed test) Under H0 , the test statistic is )1 ,0(N~ n S XZ  approximately (by CLT) where  = 25, n = 100 , s = 14.51645 Level of significance: 5% = 0.05 Critical Region (or Rejection Region): Z > 1.64485 Coffee company not promoting the new coffee product  H0 is not rejected , 25 14.51645 100 x <   < 27.3877 8a b (i) P(Score is 4) = = (ii) P(Score is 6 given that one of them is 2) = = = (i) Using M1 method: N mber of digits Number of integers 1 3C1 = 3 2 3P2 = 6 3 3P3 = 6 Using M2 Method: Number of integers = 4 + 42+ 43 = 84 Hence, total number of integers in set A =3+6+6+84 = 99

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/2 Pg 7 of 10 (ii) P( sum of 2 integers is even) = P(both integer is even) + P(both integer is odd) = + = 9 P(Player wins a prize in a game) = p4 (i) Let X be r.v. “number of winners within the first ten games”. i.e. X ~ B(10, p4) P(at least one winner within the first ten games) ≥ 0.9 P(X ≥ 1) ≥ 0.9 1  P(X = 0) ≥ 0.9 P(X= 0) ≤ 0.1 (1 p4) 10 ≤ 0.1 1 4 101 0.1p 4 0.205672p  0.673 1 p (correct to 3 s.f) (ii) Let Y be the r.v “no. of prizes won out of 100 games”. 4B(100,(0.7) )Y E(Y) =   4 100 0.7 24.01 Var(Y) = 44100(0.7) (1 (0.7) ) 18.245199 Let 1 2 60 ... 60 Y Y YY    Since n = 60 is large, by Central Limit Theorem, 18.245199~ N 24.01, 60Y   approximately.   P 24 0.493Y   Alternative solution Let Y be the r.v “no. of prizes won out of 100 games”. 4B(100,(0.7) )Y np=   4 100 0.7 24.01 >5 nq = 4100(1 (0.7) ) 75.99>5  ~ N 24.01 , 18.24511Y approximately Let 1 2 60 ... 60 Y Y YY    18.245199~ N 24.01, 60Y   approximately.

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/2 Pg 8 of 10   P 24Y 0.5P 24 0.499 60Y    10 (i) Meteors are seen singly, randomly and independently. There is a uniform (mean) rate of occurrence of meteor sightings. (ii) Let X be the r.v. “number of meteors seen by Jess in 5 minutes’. X ~ Po(6.5) P(X >5) = 1  P(X ≤ 5) = 1  0.36904068 = 0.631 (iii) Let Y be the r.v. “number of meteors seen by Jess in 1 hour”. Y ~ P(78) Since  =78 > 10, Y ~ N(78, 78) approximately P(Y  100) P(Y < 100.5) = 0.995 (iv) Let W be the r.v. number of meteors seen by Jess in t minutes W ~ Po(1.3t) P(W > 2) ≥ 0.96 1 P(W ≤ 2) ≥ 0.96 1 ≥ 0.96  0.04 ≤ 0 From GC, t ≥ 5.076. Hence, smallest possible integer t = 6 mins. 11 (i) = 52.75, 356 8y  From the given regression line,  356 0.9978 52.75 4.01048   Therefore,  = 32.9884 33 (ii) (a) r = 0.866 (b) r = 0.907 (iii) From the scatter diagram of y on x, t