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. Hence there will be in total 1 1 2 1n n n n n n such matches across all possible rotations. Since there are 21n possible rotations of the smaller circle, there will be at least 2 ( 1) 12 1 2 1 n n n nnnn matches for one of the rotations by Pigeonhole Principle. Qn Solution 4(a) Now, 2 1 2 2 1 2 2 1 2 1n n n n n nF F F F F F+ − + −= +  = − for 1n . ( ) 2 2 1 2 1 11 31 53 2 1 2 3 2 1 2 1 2 1 1 ... Shown nn i i i ii nn nn n F F F FF FF FF FF FF +− == −− +− + =− =− +− + +− +− =−  Similarly, 2 1 2 2 2 1 n in i F F F++ = =− . (b) 1055 SF=  7 9 12191 SF F F= + + 

Note: We can see this from; ( ) ( ) 1 3 5 7 9 1 10 2 10 55 1 54 1 2 5 13 34 F F F F F F F F F =+ = + + + + = + + + + = + − = ( ) ( ) 1 2 4 6 9 12 1 7 1 9 12 7 9 12 191 190 1 1 1 3 8 34 144 . F F F F F F F F F F F F F F =+ = + + + + + = + + + + + = + − + + = + + (ci) ( ) ( ) ( ) ( ) 23 23 34 45 2 32 53 74 21 1 1 ... ... if 4 are are done ... else if 6 are are done ... else if 8 are are done ... else if 2 , for all s s s s u u u u u u u u u u u u si k F F F F F F F F u F F F F u F F F F u F u i i+ + = + + + + + = + + + +  = + + + +  = + + + +  = == Alternatively If 2iui= for all i. Then, 2 4 6 2 ... sk F F F F= + + + + and ( )2 1 2 11 1 1 ss Sk F F +++ = − = + . If 2iui for some i, consider the smallest i where this happens, call this m + 1. (Or consider the largest m such that 1 2iiuu −=+ for all im ). So, 1 22mum+ + . This implies, 23 1 2 2 4 2 ... ... ... s ms u u u m u u k F F F F F F F F F + = + + + + = + + + + + + Thus from (a), ( ) ( ) 1 1 1 1 2 4 2 1 2 1 1 21 1 ... ... ... ... ms ms ms m u u m u u m u u k F F F F F F F F F F F F F F + + + + + + = + + + + + + + = + − + + + = + + +

Since ( )1 2 2 2 1 1mu m m+  + = + + we have that, 1211 ... msm u u Sk F F F +++ = + + + . (cii) Let P(n) be the statement that n S for 1n . Base Case: 21 SF=  So, P(1) is true. Inductive Step: Assume that P(k) is true for some 1k  , i.e. k S , 12 ... su u uk F F F= + + + for some appropriate sequence 12, ,..., su u u . If 1 2u = , from (ci) we are done. If 1 3u = , we can use a similar argument from (ci). Hence it is true. [For more details: If 21iui=+ for all i. Then, 3 5 7 2 1 ... sk F F F F += + + + + and ( )2 2 2 2 211 ssk SF F F+++ = + = − . If 21iui+ for some i, then there exist 1m where 1 23mum+ + . This implies, 23 1 3 3 5 2 1 ... ... ... s ms u u u m u u k F F F F F F F F F ++ = + + + + = + + + + + + Thus, ( ) ( ) 1 1 1 1 3 5 2 1 1 2 2 2 22 1 ... ... ... ... ms ms ms m u u m u u m u u k F F F F F F F F F F F F F F + + + + + + + = + + + + + + + = + − + + + = + + + Since ( )2 3 2 2 1mu m m + = + + we have that, 1221 ... msm u u Sk F F F +++ = + + + .] If 1 4u  , 1221 ... su u u Sk F F F F+ = + + + + as 1 4 2 1u  + . Hence P(k+1) is also true.

Conclusion: As P(1) is true, as well as P(k) is true ( )1Pk+ is true, by Principle of Mathematical Induction, P(n) is true for 1n . Qn Solution 5(a) 11 22( ) 1 ( 2) 11 n k k n k kk Both n and k must be of the same parity. (b) Without restrictions, each kid has k flavours to choose from so there are nk possible orders. Each order can include exactly r flavours where 1 .rk There are ( , )S n r ways to divide the n kids into r disjoint, non-empty subsets. There are k flavours to assign to the first subset, 1k flavours to assign to the second subset and so on. Thus there are 1( ()1) r kk krkP ways to assign r flavours to the subsets. (ci) Note that the total number of possible orders is ! ( , )k S n k (a special case from part (b)) Let iA be the set of orders where flavour i is excluded, 1 .ik Then 1 n iAk and there are 1 k such '.iAs Similarly, we have 2 n i jA A k and there are 2 k such pairs ,. jiAA In general, we have 1 j nr i j A k r and there are k r such sets of 21 , , . ri i iA A A Thus number of orders of n single scoop cones where no flavour is excluded is 0 ( 1) ( 1) ( ( ( 1) ( ) )1 ()1) k k r n n r n n rn k k k k k k r k k rk k krr Thus combining this result with the answer in (b), we have 0 ( 1)1( , ) ( ) ! rn k r k S n k k r rk where ( 1) r rc k r

Qn Solution 6a  2,nan + b + Bi bii ci 22g f ( ) g (2 ) 22 f ( 1) fg( ) (shown) xx x x x = =+ =+ = 2 2 2 4 2f g( ) 4 4 fg f ( ) g fgf ( ) g f ( )x x x x x= + = = = From (b)(i), ( ) ( ) ( ) 2 2 2 2 2f g( ) f g f ( ) g f gf ( ) g g f f ( )x x x x= = = The arrangements are 2 2 2 4 2f g,fg f ,g fgf and g f . g fg fg ( ) 4 4 2i j k x x k j i= + + + cii As the coefficient of x is 4, the function f must be applied twice. Therefore g fg fg ( ) 4 4 2i j k x x k j i= + + + where i, j, k are non-negative integers describes all possible arrangements of f and g which are equal to a function with a coefficient of x of 4. The number of arrangements that satisfy the con