HCI H2 MATH P2 ANSWERS

1   2015 H2 Mathematics Prelim Paper 2 Solutions Qn Solutions 1. (a) Locus of R is a straight line passing through the origin and parallel to AB  . 1. (b) a and b are parallel  2 2 2c o s 1 2     aa b b aa b b

1. (c) Using ratio theorem, 12 33ON  ab   and 1 2OM  bc  Since O is the midpoint of MN, ON OM    12 1 33 2 24 33 273        ab b c ab bc abc 0 2. (i)      2f3 2 3 1 9 3 2rr r r r           2 11 2 11 1 32 2 f9 3 2 93 2 33 12 1 1 222 332 33 2 nn rr nn n rr r rr r rr nn n nn n nnn nn n                 Hence, 3a , 3b and 2c . 2. (ii)      1 2 11 3231 4 11 2 qed46 2 6 17 S S        3 21 3 79 2 9 11 0S      4 31 4 1 0 1 221 21 1 3S   

2   Qn Solutions 2. (iii) Conjecture: 31 n nS n  Let P n be the proposition that “ 31 n nS n  for all n  ”. Consider P1 , 1 1 RHSLHS 4S  . Therefore P1 is true. Assume P k is true for some k  , i.e., 31 k kS k  . Want to show P1k  is true, i.e.,  1 1 31 1 k kS k     . Consider P1k  ,             1 1 1 1 f 1 31 31 2 31 1 34 1 31 34 31 1 31 34 1 31 1 k k r S r k k kk kk kk kk kk k k                P1 k  is true. Since P1 is true and   Pi s t r u e P 1kk  is true, by mathematical induction, P n is true for all n  . 3. (i) 3 arg 2 2i4 z       3 arg 1 arg 2 2i4 z      arg 2 2i 04 z     -----(*)

3   Qn Solutions 3. (ii) max arg 3z angle of  0, 2 from  3, 0 1 2tan 3     1 2arg 3 tan43 z     3. (iii) Area = triangle A + trapezium B =     1112 224 722      Re Im O Re Im O   max arg min arg Re Im O A B

4   Qn Solutions 4. (i)  ln 2 1 dx x =  2ln 2 1 d 21x xxx x  =  1ln 2 1 1 d 21x xx x  =  1ln 2 1 ln 2 1 2x xx xc  = 1 ln 2 12x xx c    4. (ii) 4. (iii) Area of shaded region 13 / 2 3/4 1 ln( 2 1)d ln( 2 1)dx xx x    = 1 3/4 1ln( 2 1) ln( 2 1)2xx x x   + 3/2 1 1ln( 2 1) ln( 2 1)2xx x x     x y x x 1 y x

Qn 4. (iv) 5(i) 5(ii) 6 (ai) Solutions =( 0 1 +   = 3 1ln 24 4 Volume = = 2(3) (ln = 26.927 = In order to the partici difficult to informatio We assign Let the t o interval = participant with numb P( | ) P( P( P( A B AB AB AB       30) l n4   33ln 222     1 4 square un 2(3) (ln 5) 2 35) 2  = 26.9 cubic o use strati ipants accor o have this on, etc. Hen n an index n otal numbe r 250.04 x x  t thereafter. bers 5, 30, 5  P(0.675 ) 0.675 ) 0.675 P )0 . 2 7 B B B            131nl242 1 ln 2 ( 02   nits 2 3) ln2  2 ln5 ln 2 ln 2   c units (3s.f fied sampli rding to st r complete in nce it would number to e r of partici p 5 . Pick a r . E.g. If the 55, … until  ) P() P( ) P() 0 AB B AB AB     5 1ln 2   0 10 )   ln 5 ln 2 n 2 x  2 5 e1 d2 y   f.) ing, we ne e rata, e.g. ra c nformation d be difficult each partici pants to a r random num e number 5 4% of the p 0.13 0.08775    2dy dy ed to know ce, gender o due to abse t to use a str ipant when rrive at the mber from was selecte participants A the comple or age-grou entees, incom ratified sam they arrive conference 1 to 25, th ed, we samp are sampled 0.1 ete compos ups. Howev mplete regi mple. at the con f e be x. S a hen pick ev e ple the parti d. B 13 ition of ver, it is istration ference. ampling ery 25th icipants

6   Qn Solutions Alternative 1           P PPP P0.62 0.49 P P| P 0.27 AB A B AB AB ABAB AB         Alternative 2           P'P' | P 0 . 4 P P P P | 0.27 ABAB B B AB B A B     6 (aii) P( | ) 0.675 P( ) 1 0.13 0.38 0.49 P( | ) P( ) AB A AB A      Therefore the two events A and B are not independent Or P( ) 0.27 P( ) P( ) 0.49 0.4 0.196 P( ) P( ) P( ) AB AB AB A B      Therefore the two events A and B are not independent 6 (b) Required Probability = P( ) P( ) P( ) P( ) P( P( ) .. . . ). . DCC DCDCC CC CDC DCDC C CDCDCC DCC     =    2 24 4 22 2 6 .... 11 11 1 ....p pp p pp p p p pp            22 22 2 2 2 11 1 11 11 1 2 1 2 pp p pp pp p pp pp p               (shown) 7(a) 10 39! 3! 261273600C 

7   Qn Solutions 7(b) [All Universities]=[Any six-person] – [University A&B] – [University B&C] – [University A&B] Case 1: University A&B 9 6C Case 2: University B&C 8 6C Case 3: University A&C 7 6C So the answer is 12 9 8 7 6666 805CCCC Alternate Method [Not advisable because of too many cases] Consider number of candidates from each university. If we have x from A, y from B and z from C, the number of possible ways is 453 xyzCCC A(4) 1 1 1 2 2 2 3 3 4 B(5) 2 3 4 1 2 3 1 2 1 C(3) 3 2 1 3 2 1 2 1 1 Ans 40 120 60 30 180 180 60 120 15 So the total number of possible way is 805. 7(c) All possible groupings: 12C4 8C4 4C4 3!  5775 Method 1 (Direct Method) Case 1: Candidates from University A grouped in 2,2,0 (4 from A versus 8 from ‘the rest’) Number of ways = 48 26 22 22 2! 1260CC CC  Case 2: Candidates from University A grouped in 2,1,1 Number of ways = 48 2 6 22 13 2! 3360CC CC  So the probability is 1260 3360 5775  0.8 Method 2 (Method of Complementation –cases where a group has more than 2 candidates from University A) Case 1: Candidates from University A grouped in 3,1,0 Number of ways = 48 8 31 4 2! 1120CC C  Case 2: Candidates from University A grouped in 4,0,0 Number of ways = 48 44 2! 35CC  So the probability is 1 112035 5775  0.8

8   8(a)     22 2 12 2 12 2N 2 , 2 4 9 i.e. 2 N 2 , 25 XY XY        

     2222 12 2 12 98N 98 , 9 4 8 9 i.e. 9 8 N 9 8 ,900 XY XY        

                12 12 22 12 12 22 12 1 2 22 21 21 P2 P9 8 P2 0 P9 8 0 22 0 2P 25 25 98 98 0 98P 900 900 02 09 8PP 25 900 28 9PP 53 0 XY X Y XY X Y XY XY ZZ ZZ                                                     By symmetry of the standard normal distribution, So 21 2128 9 53 0        21 2 1 21 61 2 89 14 21         1 2 14 2 21 3    8 (bi) N1 0 , 9Y P 16 0.97725Y  = 0.977 (3 sf) 8 (bii) Let S be the number of observations with 16Y  Hence  B 100,0.02275S Since 100n is sufficiently large, 2.275 5np  , Po 2.275S approximately   P 100 95 P 5 0.971SS     (3 s.f.) b a

9   9(i) 17.21 ,3 . 56 kxn   ,nx lies on the regression line, so   0.943 0.484 17.21 0.943 3.5 0.4846 5.50 2 dp x n k k     9(ii)  9 (iii) From the scatter plot, x and n have a curvilinear relationship. Therefore a linear model is inappropriate even though the produ ct moment correlation coefficient is relatively high (i.e. 0.915). 9 (iv) The graph of 2x ab n is concave upwards (or increase at increasing rate) similar to the scatter plot The graph of lnx ab n is concave downwards (or increase at decreasing rate) . From GC, 0.8212061688 0.821 3 s.f.Br  0.9859197289 0.986 3 s.f.Ar  which is closer to 1 and hence suggested a relatively stronger linear relationship between x and ln n as compared to x and 2n . Therefore Model (A) is more appropriate.  6