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 )