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   