ASRJC H3 Math 2024 Prelim Solutions

1 2024 ASRJC H3 Math Prelim Solutions 1 (a) 2d sin sin 0d xxx for (0, ).x π 2 2 d1ln 0d xxx for (0, ).x (i) Given that A , B and C are angles of a triangle, then π0 2A , π0 2B and π0 2C . Applying Jensen inequality, 3 11 11 sin( ) sin33 n kk kk xx ==     ( )11sin sin sin sin33 A B C A B C + +  + +    ( )11sin sin sin sin π33 A B C + +    1 π3sin sin sin sin3 3 2 A B C + +  =  33sin sin sin 2A B C+ +  (Shown) (ii) Let f ( ) lnxx . Then 1 2 1 2 11ln ln ln ln nna a a a a ann 1 12 12ln( ) ln nn n a a aa a a n 1 12 12()n n n a a a a a an (b) (i) 1 1 2 3 5 9 2 1( ) ...... .1 2 4 8 2 n nPn − − +=      Using AM-GM inequality, ( ) ( ) ( ) 1 1 1 1 1 11 11 2 3 5 9 2 1 .....2 3 5 9 2 1 1 2 4 8 2.....1 2 4 8 2 1 1 1 1 11 1 1 1 ..... 11 2 4 8 2() 1 1 1 11 ..... 2 4 8 2() n n n n n n n n n n Pn n n Pn n − − − − − − ++ + + + + +               + + + + + + + + + +                          + + + + + +              

2 ( ) ( ) ( ) 1 1 1 1 111 2 11 2() 12 2() 2() 2( ) 1 n n n n n n n Pn n n Pn n nPn n Pn n −  −  + −  +−  + + (ii) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 23 23 23 2 2( ) 1 ( 1)( 2)... 2 12 ( 1) 2 ( 1)( 2) 2 21 ..... 2! 3! ! 1 1 11 2 2 2 ..... 22! 3! ! 1 1 11 2 2 2 ..... 2 ......2! 3! ! e n n n n Pn n n n nn n n n nn n n n n n n n + −−− − −       = + + + + +                + + + + +  + + + + + +  2 ( )( ) ( ) ( )( ) 1 2 2 2 2 1 1 1 2 1 2 22 2 1 1 2 2 2 1 2 2 2 1 44 44 4 4 nn n n n n n n n n n n n n n n n n n n n n n n n n n n FF w w w w w w w w w w w w w w w w w w w w w w w w w w − − − − − − − − − − − − − − − − − − − = + − − − + = − − + = − + − − = − + − ( )( )1 2 2 1 4n n n n n n nF F w w w w w− − − −− = − + − ----------------------(1) (a) Let nw be nu . Then 21 40n n nu u u −−+ − = . So we have 1 0nnFF −−= for 2n by result above. ie 1nnFF −= for 2n ( )( ) 2 2 2 2 1 1 0 1 0 4 2 1 4 2 1 3F u u u u= + − = + − =− Therefore 3nF =− for all 1.n So we have 22 11 4 3 for 1.n n n nu u u u n−−+ = − 

3 (b)(i) Let nw be nv . ( ) 22 11 2 11 2 1 1 1 4( )(1) 3 4 4 0 20 2 vv vv v v + = − − + = −= = 22 11 4 3 for 1.n n n n nF v v v v n −−= + − =−  From (1), we have 2 0nnvv −−= or 21 40n n nv v v −−+ − = for 2.n . (ii) Since 1,2,1,2,…. satisfies 2 0nnvv −−= for 2.n , then from (1), is constantnF . Also since 0 1v = and 1 2v = , then 3nF =− . So the sequence satisfies (*). (iii) Take the sequence 1,2,7,2,1,2,7,2,….. that satisfies 2 0nnvv −−= for odd