HCI 9820 2024 H3 Prelim Soln for posting

2024 H3 Math Prelim Suggested Solution Solution 1(a) Consider the first term of a special sequence, it must an odd integer and divided into the following mutually exclusive cases. Case 1: 1 1a = Hence the sequence is of the form  231, , ,...., } ma a a where mn . Consider the function   2 3 2 31, , ,..., } 1, 1,...., 1f mma a a a a a ⎯⎯ → − − − . f is a bijection and the number of special sequences of the type  2 3 11, 1,....., 1 mna a a A −− − − = Number of special sequences with 1 1a = is 1 1nA − + (Accounting for 1 ) Case 2: 1 3a  is of the form  1 2 3, , ,...., } ma a a a where mn . Consider the function   1 2 3 1 2 3, , ,..., } 2, 2, 2,...., 2g mma a a a a a a a ⎯⎯ → − − − − . g is a bijection and the number of special sequences of the type  1 2 3 22, 2, 3,....., 2 mna a a a A −− − − − = Hence, 12 1n n nA A A −−= + + 12 1n n nA A A −− = + + 1 1A = (i.e 1 ) 2 2A = (i.e 1 ,  1, 2 ) 1(b) We want 15A By G.C, 15 1596A =

Solution 2(a) ( ) ' ' g'( ) e f ( ) e f( ) e f ( ) f( ) e xx x x x x x xx =+ =+  2(b) Since g'( ) e xx  , 11 00 1 0 g'( )d e d g(1) g(0) e ef(1) f(0) e 1 ef(1) e 1 e1f(1) e x x x x x a a  −  −  −  − + −+  The largest possible value of f(1) is e1 e a−+ , which is attainable when 'f ( ) f( ) 1xx+= . Let f( )yx= , 'f ( ) f( ) 1 d 1d d 1d 1 d1 ln 1 1e x xx y yx y yx y x Cy y x C yA − += += =− =+− − − = + −=  Since 0,x y a== , 1 f( ) 1 (1 ) e x aA xa − −= = − −

3(a) (i) ( ) 2 2f ( ) 1 (1 ) xx x x x − = = −− ( )( ) ( ) ( ) 1 21 ( 2)( 3) ( ( ))1 2 ( ) 1! ( 2)( 3) ( ( ))0 2 ( 1) 1! n nn nx x x n nx x x n − −  − − −= + − − + + − + − − − −= + + + + − + − Note that 0 0u = and 1 1u = For 2n , coefficient of nx in series expansion of f ( )x ( ) ( ) 1 11 ( 2)( 3) ( ( )) ( 1)1! ( 1) !( 1) 1! n nn n n n n n − −− − − −=− − −−= − = Hence 2f ( ) (1 ) xx x= − is the generating function of nun= . ( ) 3 3f ( ) 1 (1 ) xx x x x − = = −− ( )( ) ( ) ( ) 1 21 ( 3)( 4) ( ( 1))1 3 ( ) 1! ( 3)( 4) ( ( 1))0 3 ( 1) 1! n nn nx x x n nx x x n − −  − − − += + − − + + − + − − − − += + + + + − + − For 2n , coefficient of nx in series expansion of f ( )x

( ) ( ) ( ) ( ) 1 11 ( 3)( 4) ( ( 1)) ( 1)1! ( 1) 1 !( 1) 2 1 ! 1 2 n nn n n n n nn − −− − − − +=− − − + −= − += Note that 0 0(1)0 2u == and 1 1(2)1 2u == Hence the sequence with generating function 3f ( ) (1 ) xx x= − is ( 1) 2 n nnu += , 0n .

3(a) (ii) Method 1 2 ( 1)2 2 nnnn +=−  Hence the generating function for 2 nun= is ( ) ( ) ( ) ( ) 32 2 33 f ( ) 2 11 2 (1 ) 11 xxx xx x x x x x xx  =−  −− − − +== −− Method 2 Consider 2 2 02(1 ) nx x x nxx = + + + + +− Differentiate w..r.t. x, we have 2 21 4 2 21 4 21 3 21 3 (1 ) 2(1 )( 1) 0 1 4(1 ) (1 ) 2 (1 ) 0 1 4(1 ) 12 0 1 4(1 ) 1 0 1 4(1 ) n n n n x x x x n xx x x x x n xx xx x n xx x x n xx − − − − − − − − = + + + + +− − + − = + + + + +− −+ = + + + + +− + = + + + + +− Multiplying x throughout, we have 2 22 2 04(1 ) nxx x x n xx + = + + + + +− Hence the generating function for 2 nun= is ( ) 2 3f ( ) 1 xxx x += −

3(b) Let the generating function of the sequence be f ( )x . ( ) 0 1 1 1 1 11 1 1 10 0 f ( ) 1 1 2 1 12 2 122 (1 ) 12 f ( ) (1 ) n n n n n n n n n nn n nn nn n nn n n n x u x ux ux u x x x u x x x u x x xx x  =  =  − =  − ==  − − ==  = = =+ = + + = + + =+ =+ − =+ −       Therefore 1(1 2 )f ( ) (1 ) 1f ( ) . (1 2 )(1 ) xx x x xx −= − = −−

Solution 4(a) ( )11 = , ( )32 = , ( )12 6 = . 4(b) We note that each ip has 1 1 + choices to be included in a factor, since the choices are 0 1 2, , ,..., i i i i ip p p p  . Hence considering all the prime factors, we have ( ) ( )( ) ( )12 1 1 ... 1 kn   = + + + . 4(c) By part (b), we have ( ) ( )( ) ( )12 1 1 ... 1 kn   = + + + . Hence ( ) ( )( ) ( ) ( ) 12 is odd 1 1 ... 1 is odd 1 . 1 is odd 1 . is even k i i n ik ik        + + +    +    Now, ( ) ( ) 1 2 1 2 12 222 1 2 1 2 2 12 is odd ... ... ... is a perfect square kk k kk k n n p p p p p p n p p p n        = = =  4(d) n Divisors of k … … 8k = 1 2 4 8 7k = 1 7 6k = 1 2 3 6 5k = 1 5 4k = 1 2 4 3k = 1 3 2k = 1 2 1k = 1

4(e) Notice that n gives the number of rows of the table in part (c), and that each kn occurs every kth row, for a total of n k   rows. Hence ( ) 1 1 sum of number of members of each row sum of number of members of each column n k n k k n k  = = = = =   

Solution 5(a) 2 0 22 00 2 2 2 000 2 2 2 00 e sin d 11e sin e cos d22 1 1 1 1e sin e cos e ( sin ) d2 2 2 2 1 1 1e sin e cos e sin d2 4 4 Since sin and cos are bounded, x xx x x x x x x xx x x x x x x x x x x x xx  −  −−  − − −  − − − = − − −    = − + − − − −          = − − −     2 2 2 2 2 0 0 2 0 lim e sin 0, lim e cos 0, 5 1 1 1 1e sin d e sin e cos 0 (0 )4 2 4 4 4 1e sin d 5 x x x x x x x x x x x x x x xx − → − →  − − −  − = = = − − = − − = =   5(b) 2 2 2 0 0 0 22 0 0 22 00 e sin d e sin d e d 1 1 1e d e 0 ( ) 2 2 2 11Hence, e sin d and e sin d exists.52 x x x xx xx x x x x x x x x x x   − − −  −− −−  = − = − − =      

5(c) ( ) ( ) 2 0 232 2 2 02 1 2 ( 1) 1 2 ( 1)1 e sin d e sin d e sin d e sin d ... 1 e sin d 1 e sin d n x x x x nn x n n kk x kk xx x x x x x x xx xx          − − − − − − − − − −= = − + + +− =−        2 2 2 2 2 2 5 1 1e sin d e sin e cos4 2 4 21e sin d e sin e cos55 x x x x x x x x x x x x x x C − − − − − − =− − =− − +   2 2 2 ( 1) ( 1) 2 2( 1) 2 ( 2 2) ( 2 2) 2 22 21e sin d e sin e cos55 110 e cos 0 e cos( 1)55 11e e if is odd55 11e e if is even55 1 e (1 e ) if 5 kk x x x k k kk kk kk k x x x x kk k k k         − − − − − − − − − − + − + − − = − −    = − − − −        += −− + =  22 is odd 1 e (1 e ) if is even5 k k−   −+ ( ) ( ) ( )( ) 22 1 2 2 4 2 22 2 2 22 2 1 e (1 e ) 5 1= (1 e ) e e ... e5 e 1 e1 (1 e )5 1e 1 e 1 e1 5 1e n k k n n n           − = − − − −− − −− − =+ + + + + − =+ − +− = − 