NYJC 2025 Prelim (Answers)

Q1 Suggested Answers (a) y mx c=+ d d y mx = Substitute into the DE: ( ) ( ) ( ) 22 2 4 4 3m x mx c x mx c x mx c= + + − + − + + + Comparing coefficients: 2 :x ( ) 220 1 2 1 0 1m m m m= + −  − =  = Constant: 2 43m c c= + + 2 4 2 0 4 16 8 222 cc c + + = −  −= =−  Thus the solutions are 2 2 and 2 2y x y x= − + = − − . (b) ( ) 22 2 d 2 4 4 3d 4( ) 3 0 y x y xy x yx y x y x = + − − + + = − + − + = Thus ( )( )3 1 0y x y x− + − + = The equations of the two lines are 3yx=− and 1yx=− . (c) 22 2 2 d 2 4 4 3d d d d d2 2 2 4 4d d dd d2 2 4 since 0 at stationary pointsd y x y xy x yx y y y yx y y x x x xx yxy x = + − − + + = + − + − +  = − − = If the stationary point lies on 3yx=− , then 30xy− − = Then 2 2 d 2 2 4 2( 3 1) 2 0 d y x y x y x = − − = − − + =  . Thus the stationary point is a minimum point. If the stationary point lies on 1yx=− , then 10xy− − = Then 2 2 d 2 2 4 2( 1 1) 2 0 d y x y x y x = − − = − − − =−  . Thus the stationary point is a maximum point. (d)

Q2 Suggested Answers (a) 2 2 f ( ) ( 3) 2 g( ) ( 2) 2 nn nn = + + = − + ( ) 2 h( ) 1 4nn= − + Note that f ( ) g( 5)nn=+ for every n . i.e. 5nn + is a bijection on . Hence the functions have the same range. To find common values in the range of f and h, consider ( ) ( ) 22 22 ( 3) 2 1 4 ( 3) 1 2 ns ns + + = − + + − − = Since there are no perfect squares that differ by 2, there are no integer solutions. There are no common values in the ranges of f and h. (b)(i) ( ) 22 2 2 31 24 0a ab b a b b+ + = + +  (b)(ii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 2 2 2 2 3 3 3 ( 1 )( 1 1 ) 1 1 1 1 1 1 a b a a b b a b a a b a b a b b ab − − − + − + = − − − + − − − + − − = − − (c) 3 2 3p( ) 3 7 ( 1) 4 1= − + = − + +n n n n n n To find common integers in the range of p and q, consider ( ) ( )( ) ( ) ( ) ( )( ) ( ) 33 33 2 2 2 2 ( 1) 4 1 4 6 ( 1) 4 6 4 1 ( 1 ) 1 1 4 1 11 ( 1 ) 1 1 4 1 11 n n s s n s s n n s n n s s s n n s n n s s n s − + + = + − − − = − − − − − − + − + = − + − − − − + − + + − − =− x y -2

( ) ( )( ) 2 2 4 ( 1 ) 1 1 4 11n s n n s s  − − − + − + + =− Thus ( 1 ) 1ns− − =− and ( ) ( ) 2 21 1 4 11n n s s− + − + + = Thus ns= . ( ) ( ) 2 2 2 1 1 4 11 3 3 6 0 ( 1)( 2) 0 1 or 2 s s s s ss ss s − + − + + = − − = + − = =− Thus the values common to both ranges are q( 1) 11− =− and q(2) 10= . Q3 Suggested Solutions a(i) ( ) ( )( ) ( ) ( ) ( ) 0 0 0 0 f d f 1 d with fd fd a a a a x x a u u x a u a u u a x x = − − = − =− =−    a(ii) Let π 2 0 cos dsin cos n nn xIx xx= +  . ( ) π 2 0 π 2 0 πcos 2 d by part (a)(i)ππsin cos22 πsin cos2sin dcos sin π& cos sin2 n nn n nn x Ix xx xx x xxx xx  −=    − + −         −= = +  −=       ππ 22 00 π 2 0 cos sin2 d d sin cos cos sin cos sin dsin cos π 2 nn n n n n nn nn xxI x x x x x x xx xxx =+ ++ += + =   π 4I =

b ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 00 0 00 00 00 f d f d by part (a)(i) f d f f f d f d 2 f d f d f d f d 2 aa a aa aa aa x x x a x a x x a x x x x a x a x x x x x x x x a x x ax x x x x = − − = − = − =− = =      applying result from (a)(i) and applying given property of f c(i) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) π 2 0 π 12 0 π π21 2 2 2 0 0 π 222 0 π 22 0 ππ 222 00 cos 2 d cos 2 cos 2 d 1 sin 2 cos 2 1 sin 2 cos 2 d2 1 1 cos 2 cos 2 d 1 cos 2 cos 2 d 1 cos 2 d 1 cos 2 d n n n nn n nn nn I x x x x x x x n x x x n x x x n x x x n x x n x x − −− − − − = = = − − − = − − = − − = − − −       ( ) ( ) ( ) 2 2 1 1 1 1 nn nn n I n I nI n I − − + − = − =− c(ii) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) π 8 0 π 8 880 π 882 0 π 82 0 8 0 π 02 0 cos 2 d by part (b)π cos 2 d cos 2 π cos 22 ππ 2 cos 2 d cos 2 symmetrical along 22 π cos 2 d π 7 5 3 1π 8642 35π cos 2 d128 35ππ 128 2 35 π256 x x x xx xx x x x x xx I I xx  =  −=  ==   = = =     = =  =      2

Q4 (a) ( )( ) 2 ** 1 2 1 2 1 2z z z z z z+ = + + 22 ** 1 2 1 2 1 2z z z z z z= + + + ( ) 22 * 1 2 1 2 2 Rez z z z= + + 22 * 1 2 1 2 2z z z z + + 22 1 2 1 2 2z z z z= + + ( ) 2 12zz=+ Hence 1 2 1 2z z z z+  + . Let Pn be the proposition that 1 2 1 2 nnz z z z z z+ + +  + + + for all n + . Clearly, 1P and 2P is true. Assume Pk is true for some k. Need to prove 1Pk+ is true. i.e. 1 2 1 1 2 1 kkz z z z z z +++ + +  + + + 1 2 1 1 2 1 k k kz z z z z z z +++ + +  + + + + 1 2 1 kz z z + + + + Pk is true 1Pk+ is true. Since 1P and 2P is true and Pk is true 1Pk+ is true, hence by mathematical induction, Pn is true for all n + . (b) Given 2 1 0rruu +− and 2 2023 1 0uu − , 2 12uu− , 2 23uu− , …, 2 2022 2023uu − , 2 2023 1uu − is a sequence of positive numbers. Using AM-GM, 2022 2022 2 2 2 22023 1 2023 1 1 2023 1 11 1( ) ( ) ( ) ( ) 2023 r r r r rr u u u u u u u u++ ==  − −  − + −   Now, 2022 22 1 2023 1 1 2 2 2 2 1 2 2 3 2022 2023 2023 1 2 2 2 2 1 1 2 2 2022 2022 2023 2023 2023 2 1 1 ( ) ( )2023 1 ( ) ( ) ... ( ) ( )2023 1 ( ) ( ) ... ( ) ( )2023 1 ( ) 2023 + = =  − + − = − + − + + − + − = − + − + + − + − =−   rr r rr r u u u u u u u u u u u u u u u u u u u u uu 21 1 1 (2023) since ( )2023 4 4  −  rruu 1 4=

So 2022 222023 1 2023 1 1 1( ) ( ) 4 rr r u u u u+ =  − −  20232022 22 1 2023 1 1 1( ) ( ) 4 rr r W u u u u + =   = − −    2023 2025 2025 14 4 16 4W   =  Qn. Suggested Solutions Comments 5a Let 1A ,...,A m represent the different surnames and 1B ,...,B n represent the different birth months. For each boy (rows 1 to 18 in table below), put a tick under column Ai and Bj if his surname is Ai and birth month is Bj. Surnames Birth months Boy A1 … Am B1 … Bn 1 ✓ ✓ 2 ✓ ✓ 18 ✓ ✓ Let 11,..., , ,...,mna a b b be the number of ticks in the columns 11A ,...,A ,B ,...,Bmn respectively. 111 ,..., , ,..., 8mna a b b since each column has 1 to 8 ticks. Since each row has 2 ticks, the total number of ticks is 11 ... ... 18 2 36mna a b b+ + + + + =  = . Also, since all integers from 0 to 7 were seen in the boys’ responses, and 11 ... ... 1 2 ... 8 36mna a b b+ + + + +  + + + = . Hence, 11,..., , ,...,mna a b b is an arrangement of the integers 1, 2, …, 8, which implies that 8mn+= (there are 8 columns in total) and consequently 1 , 7mn . There is one column which has 8 ticks. WLOG, suppose 1 8a = , i.e. 8 boys have the same surname. Since the number of birth months, n, is at most 7, by pigeonhole principle, there are at least 2 of these 8 boys with the same surname, that have the same birth month. Note: the integers 0 to 7 appear at least once for surname and/or birth months. Method to count and tabulate the data, or to present the data in a meaningful way Count total number of ticks in 2 different ways and compare Apply pigeonhole principle to show answer [AG] 5b (i) Note that the 4 groups are distinguishable because 4 different challenge questions.

The 18 boys can split themselves up according to 6444+++ or 5 5 4 4+ + + . Number of ways 44 12 Which group Which 2 groups has 6 boys have 5 boys each 18 12 8 4 18 13 8 4CC 6 4 4 4 5 5 4 4 7204317120          =  +                     = 5b (ii) Each of the 6 players