2014 H2 Maths Prelims 2 P2 Soln

Prelim 2 Paper 2 Suggested Solutions  Qn Suggested Solutions  1(i)     21 1 2 21 2 2 2 22 2 2 2 cos d1 2c o s .d 1 d12 c o sd diff wrt , d2 d 112 dd 21 1 dd12 ( s h o w n )dd yx y xx x yxx x x yx yx xx x x yyxx xx                    Alternative                 21 1 2 32 12 2 2 22 2 1 2 2 22 2 21 2 2 2 2 2 2 2 2 cos d1 2c o s .d 1 diff wrt , d1 1 12. . 2 . cos . 1 2d2 11 d2 1 2. c o s .d 1 11 d12 2 . c o s .d 1 dd12 dd dd1 2 (shown)dd yx y xx x x y x xxx xx y xxx x xx yxxx x x yyxx xx yyxx xx                           (ii)     32 2 2 32 2 32 2 32 diff wrt , dd d d12 0 dd d d dd d13 0 dd d x yy y yxx x xx x x yy yxx xx x       

 2 23 23 2 23 2 23 dd dw h e n 0 , , , 2 , ,2d d d 2 ...22 ! 3 ! ... 46 yy yxy xx x yx x x xx x               (iii)   2 22 23 231 2 2 2 2 cos 46 6 4 2 23 234 2234 232 24 224 ee e . e 1e . 1 ... 62 6 1e . 1 ... 62 1e . 1 ... 62 1e. 1 1 2 xx x xx xx xx x xx x xx x x xx x x xx                                                  2 ... 1 and 1 2ab       

Qn Suggested Solutions  2 dd dd yv x yv vxx x        d d d d d d d 1d d 1d yx yx xyx vx v x vx x x vxx vvx vxv xx vx xvx v vx                1 dd1 ln 1 1e 1e e 1e 1e s h o w n xC Cx x x vxv vx C v v vA yx A                  At the stationary point, d 0.d y x  Hence, equation of locus is:   0x yx xy   y xx y  2y xx y  i.e., 2 1 xy x 

Qn Suggested Solutions  3(i)     2 2 1 5Let . 4 54 54 5 4 reject positive sq root 4 54 5f4 y x x y x y xx y x y x x                ff 1 0,DR  (iii)                               (iv) Solve  f x x .     2 2 2 32 5 4 54 58 1 6 8 16 5 0 shown x x xx xx x xx x          32 2 81 6 5 0 5 310 39 45, 2 35 355, or 22 (rej as 4) xx x xx x xx xx x x            x y

4(i) (ii) 66 0 =6 i 3 j 3 , = 3 , 0 00 6 OA OC OA OV                                 3 2 60 3 1=302 06 3 OM             39 22 36 9 6 3=3 3 230 3 2 AM                  Hence, equation of the line AM is 66 33 , 02 tt         r+   12 0 0 AB     Length of projection of AB  onto the line AM, 222 6 3 2 632 AB AN        = 12 6 03 02 72 77                  Perpendicular distance from P to the line AM =  22 ABA N   =  2 2 7212 6.18 (3 s.f.)7 

Alternative Method Perpendicular distance from P to the line AM  222 61 2 6 30 3 20 2 49632 0 12 2 3 12 4 9 77 6.18 3 s.f. AB                  (iii) A normal vector to the plane AMB = 12 6 0 03 1 2 2 02 3                    06 6 2 03 3 3 1 60 6 2 DV                             Let  be that angle betw. the line DV and plane AMB. 20 12 23 sin 41449 26 = 31 3           

 = 47.7o   (1 d.p.) (iv) If the 3 planes AMB, AMD and  do not have a common point, the line AM is parallel to  but does not lie in  . 61 340 2 61 22 0 3 a a a                            Note that  61 346 1 2 4 0 a           ,    Therefore point A does not lie in  . Hence the line AM does not lie in  .

Qn Suggested Solutions  5 Let X be the random variable denoting the volume of hot chocolate. E(X)=55, Var( X)= 210     By Central Limit Theorem, 210~5 5 ,XN n   approximately.      54 0.77 54 55 0.77 10 / 0.2310 0.738810 10 0.7388 7.388 54.59 the least 55 PX PZ n nPZ n n n n           Alternative Method     Qn Suggested Solutions  6(i)  Obtain a list of households, in order of surnames in that particular constituency.  Randomly select a starting point in the first 20 households on the list and thereafter select every 20th household on the list to be interviewed to get their responses. 6(iii) A better sampling method is stratified sampling. Obtain a list of all households and divide the households according to the different types of housing. Select a random sample from each stratum such that the sample size is proportional to the relative size of each type of housing. Thus, the sample obtained is more representative of the population.   Qn Suggested Solutions  7(i) Let X be the random variable denoting the number of visitors to the blog in two days. X ~ Po (6.4)     P5 1 0 P9 P4 0.650729 0.651 3sf X XX    

(ii) Let Y be the random variable denoting the number of visitors to the blog in a week. Y ~ Po (22.4) Since 22.4 10  Y ~ N(22.4, 22.4) approx P(Y > 30) = P(Y > 30.5) with cc = 0.043500 = 0.0435 (3 sf) (iii) The mean no. of visitors may not be the same for each day, e.g. weekends or during school holidays, mean no. of visitors may be higher.   Qn Suggested Solutions  8 (i) (ii) (iii) (iv) no. of ways the married couple seated together = 2! x 6! = 1440 Total no. of arrangement, without restriction = 7! P(no. of ways the married couple seated together) = 1440/7!=2/7 P(all women sat together | couple sat together) P(all women sat together and couple sat together)= P(couple sat together) (6)3!2! 6!2! 1 20   no. of ways men and women alternate = 4! 3! =144 P(men and women alternate)= 144 1 7! 35 no. of ways to arrange around a table with no restriction = (7-1)! = 720 no. of ways to arrange around a table with one particular woman must sit between two men = 3 2P x (5-1)! = 144 P(no. of ways to arrange around a table with one particular woman must sit between two men) = 144/720=1/5

Qn Suggested Solutions  9(i) Let X be the random variable denoting the number of students who are awarded distinction out of 23 students. X ~ B(23, 0.2) E(X) = np = 4.6 (ii) Let Y be the random variable denoting the number of students who are awarded distinction out of n students. Y ~ B(n ,0.2)       P9 0 . 7 1P 8 0 . 7 P8 0 . 3 From GC, when 50, P 8 0.30733 51, P 8 0.28395 smallest 51 Y Y Y nY nY n           (iii) Let S be the random variable denoting the number of students who pass out of 60 students. S ~ B(60, 0.93) n = 60 is large, np = 55.8 > 5, nq = 4.2 < 5 S’ ~ B(60, 0.07) Since n = 60 is large, np = 4.2 < 5, S’ ~ Po (4.2) approx. P(S > 55) = P(S’ < 5) =  P' 4S  = 0.58982 = 0.590

Qn Suggested Solutions  10 (i) Let A and C be the random variables denoting the mass of a randomly chosen almond cookie and chocolate chip cookie respectively.   2 2 ~N 3 2 , 3 . 0 ~N 2 8 , 1 . 8 A C    1234 2 1234 1234 1234 E( ) (4)(28) 112 Var( ) (4)(1.8 ) ~N 1 1 2 , 1 2 . 9 6 100 120 0.9864368 0.986 3 CCCC CCCC CCCC PC C C C sf          (ii)    1234 22 2 2 1234 1234 1234 E( 2 ) 4 ( ) 2 ( ) (4)(28) 2(32) 48 Var( 2 ) 4 ( ) 2 ( ) (4)(1.8 ) 2 (3.0 ) 48.96 2 ~ N 48, 48.96 P5 0 2 5 0 0.6124961464 0.612 3sf CCCC A E C E A C C C C A Var C Var A CCCC A CCCC A                 (iii)     1234 1234 11 12 20 22 65Let ( ) 100 100 5 ...100 65E( ) (4)E( ) (6)E( ) 0.72100 100 65Var( ) (4)Var( ) ( 14)Va