NYJC-TJC-VJC 2025 H3 Mathematics Prelim Solutions

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 16 84 2 0 2 2 2c c c −  −+ + =  = =−  Thus the solutions are 2 2 and 2 2y x y x= − + = − − (b) 2 2 2d 2 4 4 3 ( ) 4( ) 3 0d y x y xy x y y x y xx = + − − + + = − + − + = 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 d d2 2 2 4 4 2 2 4 since 0 at stationary pointsd d d d d y x y xy x yx y y y y yx y y x x yx x x x x = + − − + + = + − + − + = − − = If the stationary point lies on 3yx=− , then 30xy− − = Then 2 2 d 2 2 4 2( 3 1) 2 0d y x y x yx = − − = − − + =  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 0d y x y x yx = − − = − − − =−  Thus the stationary point is a maximum point. (d) x y -2

Q2 Suggested Answers (a) 2 2 f ( ) ( 3) 2 g( ) ( 2) 2 nn nn = + + = − + 2h( ) ( 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) 2 2 2 2 13 024a ab b a b b + + = + +  (b)(ii) 22 3 2 2 2 2 3 33 ( 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 1n n n n n n= − + = − + + To find common integers in the range of p and q, consider 33 33 22 22 ( 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 − + + = + − − − = − − − − − − + − + = − + − − − − + − + + − − =− 22 4 ( 1 ) ( 1) ( 1) 4 11n s n n s s  − − − + − + + =− Thus ( 1 ) 1ns− − =− and 22( 1) ( 1) 4 11n n s s− + − + + = Thus ns= . 22 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 f ( )d f ( )d 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 ( ) d2 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 = − − = − = − =− = =      c(i) ( ) π 2 0 π 2 1 0 π π 2 21 2 2 00 π 2 22 0 π 2 2 0 π 2 2 0 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 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 − −− − − − = = = − − − = − −  = − − = − − −       π 2 0 )dxx

2 2 (1 1) ( 1) ( 1) nn nn n I n I nI n I − − + − = − =− c(ii) ( )( ) π 8 0 π 8 88 0 π 2 88 0 π 2 8 0 8 0 π 2 0 0 cos (2 ) d by part (b)π cos (2 ) d cos 2 π cos (2 )2 ππ 2 cos (2 ) d cos (2 ) symmetrical along 22 π cos (2 ) d π 7 5 3 1π 8642 35π cos (2 ) d128 35π 12 x x x xx xx x x x x xx I I xx  =  −=  ==   = = =     = =      2 π 82 35 π256   = 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 2Re( )z 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 1 4 rruu −  = So 2022 222023 1 2023 1 1 1( ) ( ) 4 rr r u u u u+ =  − −   2022 2023 22 1 2023 1 1 1( ) ( ) 4 rr r W u u u u + =   = − −   2023 2025 2025 14 4 16 4W   = 

Qn. Suggested Solutions 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. 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. 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, 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+= and consequently ,7mn  . 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. 5b (i) The 18 boys can split themselves up according to 6444+++ or 5 5 4 4+ + + . Number of ways 44 12 18 12 8 4 18 13 8 4 C C 72043171206 4 4 4 5 5 4 4          =  +  =                   5b (ii) Each of the 6 players takes 5 dice, leaving 50 dice to be distributed. Number of ways 50 6 1 347876161 +−== − 5c Let A, B and C be the sets of questions that Alex, Ben and Charlie answered correctly respectively. Let a, b, c be the difficult questions that Alex, Ben and Charlie answered correctly respectively.

( ) ( ) ( ) ( ) ( ) ( ) 23 2 2 2 50 40 30 20 10 abc A A B A C A B C B A B B C A B C C A C B C A B C A B C A B A C B C A B C A B C A B A C B C A B C A B C A B C A B C A B C A B C A B C A B C ++ = − − + + − − + + − − + = + + − + + + = + + − + + + − + + + = − + + + = − + + + =+ Hence, there were 10a b c A B C+ + − = more difficult questions than easy questions. 5d Let iF be the event where Alex met the ith friend, 1,2,...,5i= . Number of times Alex participated ( ) ( ) ( ) ( ) ( ) 5 1 5 1 3 3 5 5 5 5 511 5 3 2 1 31 2 3 4 5 29 ii