DHS EJC RVHS H3 Math 2024 Prelim Solutions

2024 JC2 DHS-EJC-RVHS H3 Math Prelim Suggested Solutions Qn Solution 1a d gd xx yy =  dd dd xvx vy v y yy=  = + Substituting into the DE: d d vvy y+ ( )g v= d d vy y ( )g vv=− 11 d d() vyg v v y =− b ( ) 2 2 2 2 2 2 2 11 ed d e e e x y x y x y x y y y xx y xy y y x xy x x x x y y y       −− − ++= = + +    = + +       c Let 211 ge x yx x x x y y y y −− −     = + +           ( ) 2 2 2 2 2 11 11 d dg( ) 11 d d e 1 d d e1 e1 d d 1e 1 ln 1 e ln2 v v v v v vyv v y vy yv v v v v vy y v vy y yc − − − − =− = + + − = + = + + = +    

2 1 2 2 2 22 2 ln 1 e ln ln 1 e e 1 e e e 1 e 1 e 1e v v c vc c v v x y yc y yA yA +− − −   = + − = = + = + = + =+ When ,2ln 9x y= =− ( ) ln9 1 2 2 222 21 e 1 9A A A− = + − = +  =− 2 2 2 2e1e22 2 x x y y y    +=− + =− . Qn Solution 2a 1 ( 1)! 1 ( 1 1)!( 1)! ! ( )! ! ( 1)! ( )! ! ( )!( 1)! ! (shown) n n k n k k nn n k kk n n k k n k k n k n − − − − − + − =  − −−= −− = b Since ,s n and 121 si i i n     , we need to choose s integers from  1, 2,...,n for each term 12 si i ia a a in the sum. Thus, there will be n s   terms in the sum. c Similarly, the number of terms will be 1 1 n s −  − . d ( )( ) ( ) 12 12 12 11 12 1 1 1 1 1 ... ... ... s s n i i j i j k i j n i j k n i i i n i i i n a a a a a a a a a a a a a a a             + + + = + + + + + + +    

( ) ( ) 1 1 1 11 1 1 1 2 1 2 1 2 1 1 1 1 1 1 1 1 1 2 1 2 1 12 ... ... ... (by AM-GM and (b), n n n n nnn n n n n n s s s s nnn nn a a a a a a n a a a a a as − − −                   − − −            − − −            + +          + + + +    ( ) 11 11 2 2 (c)) 1 ... ...12 1 ... ... (by (a)) 12 1 nn snn nn s n snn nnn n n n ng g g g s n n ng g g g s g −−        −                                = + + + + + +                = + + + + + +           =+ Qn Solution 3(a) Observe that the first time the nth line enters the circle, it will create a new region. In addition, each time the nth line intersects with each of the previous 1n lines, a new region is created. Additionally, as there are only 1n− lines previously, and lines cuts other lines at most once, there can only be at most 1n− intersections between the new line and the previous 1n− lines, giving rise to at most n new regions. Since the total possible maximum can only arise when the maximum number of new regions is created with the addition of each line, we have 1 1 ( 1)nn nnmm − += −− = .

2 11 2 2 () 2 122 2 1 ( 2)2 n r n n r r r m m m m r n n nn (b) Note that for each white and black sector on the smaller circle, it will match with exactly 1n sectors and n sectors respectively on the bigger circle.