AJC H2Maths 2012Prelim P2 soln

AJC Preliminary Examination 2012 H2 Mathematics Paper 2 (9740/02) Solution Section A: Pure Mathematics (40 marks) 1 DE = 2 2sin 2   = 4sin 2  Perimeter = 4x2 + 2 [ 4sin 2  ] = 8[1 4sin ]2  Unit cost = 2 28000 8[1 sin ] =   1 23500 1 sin    1 2 2 3500 1 3500(1 ...) 3500 1750        2 (i) f(4) = f(-2) = -f(2) = -1. (ii) (iii) 7 5 f ( ) dxx   =   3 2 1 14 4 1 32 x dx       = 33 1 (3 )42 3 x      = 48 [0 8]3 = 56 3 3 line l : 21 32 2 3 3 3 20 yx , z r                          (i) 3 2 1 1 3 3 3 2 2 0 CD k                                            Hence k = -1,  = 0 and  = 2 (ii) Equation of y-z plane is x = 0. At the point of intersection F of the line l and the y-z plane, 2-=0  =2. Hence F (0, 9, 2) (iii) By observation, 0 3 2 OE   and hence 2 '3 2 OC   E D /2 2 2 Alternative, use cosine rule - 4 4 3 6 9 1

Alternative method: Let E be the foot of  from C to y-z plane 1 00 0 r    Line CE: 2 1 2 3 0 3 2 0 2 r                              . When line CE meets y-z plane: 21 3 0 0 2 20 r                     . 2 1 0 3 ( 2) 0 3 2 0 2 OE                                By mid-point theorem, ' 2 OC OCOE   0 2 2 ' 2 2 3 3 3 2 2 2 OC OE OC                                 4 (a) 2 2 24( ) cos sindy x y x xdx   Using u x y , 11du dy dy du dx dx dx dx     2 2 21 4 cos sindu u x xdx   2 2 24 cos 1 sindu u x xdx      221 4 cosdu uxdx   2 11 1 cos 21 4 2 du x dxu   214 1 1 1 1 cos 242 du x dxu  11 1 12tan 2 sin 24 2 2 u x x c        1 1tan 2 sin 2 2x y x x c      (b)(i) 2dx R kxdt  , k is a positive constant At 2 , 0 dxxR dt 20 (2 )R k R 0 (1 4 )R kR   1 4k R 2 4 dx x Rdt R 22 4 14 R dx dtRx  2ln 2 Rx tCRx   22 ,22 C t t CR x R x e e Ae A eR x R x      

At 0, 0 1t x A    Hence 21 221 t t t R x e e x RR x e       (b)(ii) 11 2 1 1 t t exR e     , 1As , 0 2 tt x R e    [or use graphical method] The amount of drug in the body will not exceed 2R mg regardless of the period of treatment. 5 3 12 1i 212 z a    3 21arg( ) 1 4 2 1 2za a      (i)   i233 41 1 i e 2 k zz       , k = 0, +1, -1 2i 12 3e , 0, 1 k zk        12 12cos sinzi   ,    33 44cos sinzi     and    77 12 12cos sinzi  4 33 1i2 1 i1 22 i ww i e          4 4 2 2 i i w z w e z e      The points representing w is an anti-clockwise rotation of the points representing z about the origin by 4  , followed by an enlargement by factor 2 about the origin. Section B: Statistics (60 marks) 6 PP EEE C R T I V i) number of ways = 6! 7 6 5 4 504003! 2!      ii) P _ _ P _ _ _ _ _ _ Case 1: 2 E 7! 5040 Case 2: 1 E   5 1 7!2! 252002!C    Total = 30240 7. (a) (i) ( ) 1P A B and ( ) ( ) ( ) ( )P A B P A P B P A B     0.7 0.8 1   = 0.5 (i) ( ' ) ( ) ( ) 1P A B P A P A B     (Use Venn Diagram) ( ) 0.85 0.7 1 0.55P A B     ( ) 0.55 11( ) 0.6875 ( ) 0.8 16 P A BP A B PB     A & B are not independent as P(AB)  P(A) . or ( ) ( ). ( )P A B P A P B (ii) If ( ) 0.5P A B , then ( ) 1P A B , event A & event B are exhaustive.

(b) Total no. of points = 1 2 3 ... 9 10 55      OR 11 2 55C  Prob. = 11 1 10 9 8 8 0.050355 54 53 159C      OR 10 11 3 1 55 3 8 159 CC C 8 i) Let X = number of hot drinks sold in five minutes. Let Y = number of cold drinks sold in five minutes  2X Po  2.5Y Po  4.5X Y Po    4 1 4P X Y P X Y      = 1- 0.532103… = 0.467896….= 0.468 ii) Required probability =  0 / 4P X X Y       04 4 P X X Y P X Y            04 4 P X P Y P X Y   0.13534 0.89118 0.2270.53210  (iii) Let W= number of periods with more than 4 drinks sold out of the twenty periods.  20,0.46790WB   0.4P W n  1 1 0.4P W n     1 0.6P W n    From GC:  9 0.527 0.6PW     10 0.696 0.6PW    Least n-1 =10  least n = 11 (iv) Number of hot drinks sold in 30 minutes =  ~ 12H Po Number of cold drinks sold in 30 minutes =  ~ 15C Po Since both  >10,  ~ 12,12HN and  ~ 15,15CN  ~ 3,27H C N  0P H C =  0.5 0.25029... 0.250P H C    9 (a) 16n 2848 17816x , 2 2 1 2848509884 19615 16s    , 14s Hypothesis: 00:H  against 10:H  As 2 is unknown, n is small, conduct a 2-tailed t –test, assuming 2( , )X Normal Distribution N  , test statistics: 0 (15) 16 XTt S   At 5% significance level, do not reject 0H if 2.131 2.131t   .

01782.131 2.13114 16    07.4585 178 7.4585    0170.5415 185.4585  0171 185 (b) Combine sample: 16 36 52n   , 2 221 6660120 123642036 36 cc    2848 185(36) 9508y   and 2 509884 1236420 1746304y    . Unbiased estimate for  is = 9508 182.846153852y  Unbiased estimate for 2 is = 2 2 1 95081746304 152.995475151 52s    Hypothesis: 0 : 180H   against 1 : 180H   As 2 is unknown, n is large, conduct a 1-tailed z –test, test statistics: 180 (0,1) 52 XZN s  [ by CLT] From GC, 1.65928407z , 0.04852928p ( 1.65928407)p value P Z   when  = 180) = prob that the sample mean cholesterol attains a value of more than 182.8461538 if the population mean cholesterol content in eggs is 180 Or The p-value is the lowest level of significance for which the null hypothesis of mean cholesterol level being 180, will be rejected. Or The p-value is the probability of obtaining a test statistic more than 1.659, assuming that H0 is true. 10 Let L = waiting time of a patient at Lee’s Clinic and H = waiting time of a patient at Hope Clinic.   225,8LN   237,4HN i) Let 2 1 2 3 4 5 825,55 L L L L LXN      2 2825 37, 4 5X H N     =  12,28.8N  Required probability =  5P X H    55P X H P X H      = 0.000768 + 0.90395 = 0.905 (to 3 sf) Assumption: the waiting times of all patients are independent. ii) Let Y = number of patients with waiting time more than 25 minutes  ,0.5Y B n

Since n > 40,  np > 5 and n(1 – p) > 5   0.5 ,0.25Y N n n approximately.  40 0.95PY   40.5 0.95PY   40.5 0.5 0.95 0.5 nPZ n    40.5 0.5 1.64485 0.5 n n   0.5 1.64485 0.5 40.5 0nn    By GC, 67.487.n Least n = 68 iii) Let W = number of patients treated for influenza in a sample of 20.  20,0.2WB    20 0.2 4EW  and     20 0.2 0.8 3.2Var W  Since n = 60, by CLT, 3.24, 60WN    3.5 0.9848... 0.985PW    iv) Normal model is not likely to be appropriate as