RI 2024 H3 Math Prelim (Solutions)

2024 Raffles Institution H3 Mathematics Preliminary Examinations (Solutions) Question 1 (a) [2] 0( ) 1Tx = 1()T x x = ( ) 2 2(cos ) cos 2 2cos 1T   = = − and thus 2 2( ) 2 1T x x =− . (b) [3] ( ) ( ) ( ) ( ) 11 11 11 ( ) ( ) cos ( 1)cos cos ( 1)cos 2cos cos cos cos 2 ( ) nn n T x T x n x n x n x x xT x +− −− −− + = + + − = = Alternate Solution ( ) ( ) ( ) ( ) ( )   1 11 11 11 (cos ) cos ( 1) cos cos sin sin 1(cos ) (cos ) cos ( 1) cos ( 1)2 1(cos ) (cos ) (cos ) (cos )2 (cos ) 2(cos ) (cos ) (cos ) ( ) 2 ( ) ( ) n n n n n n n n n n n Tn nn T n n T T T T T T T x xT x T x                  + +− +− +− =+ =− = + + − −  = + − =− =− (c) [3] We have 0(0) 1T = and 1(0) 0T = . Substituting x = 0 into (b) we get 11(0) (0)nnTT+− =− and thus 2 0 if is odd (0) ( 1) if is even nn n T n = − Alternatively, ( ) 2 1 0 if is odd (0) cos cos 0 cos 2 ( 1) if is even nn nnTn n − = = =  −

2024 Raffles Institution H3 Mathematics Preliminary Examinations (Solutions) (d) [2] ( ) 0nTx = ( )0 (cos ) cos ,2 ,2 nTn n k k k knn     = =  = +   = +  Since the polynomial is of degree n (a simple recursion using the relation in (b)) we must have n roots to the equation. The real numbers cos cos , 2 k k knn  = +  are therefore the roots of ( ) 0nTx = . If we restrict k to be 0, 1, …, n – 1, we have n distinct roots as the function cosine is a bijection from ( )0, to ( )1,1− . (e) [4] From (d), we know that ( )( ) ( )0 1 1 21( ) ... , cos 2 n n k kT x a x x x x x x x n − += − − − =  Hence the desired product can be obtained by substituting 0 into the above relation and obtaining ( )( ) ( ) ( ) 0 1 1 0 1 1 (0) ... = 1 ... nn n n T a x x x a x x x − − = − − − − We need to find a. From (b), we can see recursively that 12na −= . Therefore, ( ) 22 1 0 1 1 0 11 21cos ... 2 (0) = 1 0 if is odd = ( 1) ( 1) if is even2 ( 1) 2 nn n n k n n n n n k x x xn T a n n  − − = −− + = −   −− = − 

2024 Raffles Institution H3 Mathematics Preliminary Examinations (Solutions) Question 2 (a) [2] 1 2S = , 2 3S = , 3 4S = (b) [4] The well-spaced subsets of  1,2,3,..., , 1, 2, 3n n n n+ + + contains either the element 3n+ or not. Case 1: 3n+ is an element in the subset. Then elements 1 and 2nn++ are not in the subset and the number of such well- spaced subsets is the number of well-spaced subsets, including the empty set, of the set  1,2,3,..., n = nS Case 2: 3n+ is not an element in the subset. Thus the number of such well-spaced subsets is the number of well-spaced subsets, including the empty set, of the set  1,2,3,..., , 1, 2n n n++ = 2nS + Hence 32n n nS S S++ =+ . 4 3 1 5 4 2 6 5 3 7 6 4 8 7 5 6 9 13 19 28 S S S S S S S S S S S S S S S = + = = + = = + = = + = = + = (c) [4] Let A be the set of k-combinations of  1,2,3,..., n that are well-spaced and B be the set of k-combinations of Y, where Y =  1,2,3,..., 2 2nk−+ . For any a A,  12, ,..., ka a a a= where we assume WLOG 12 ... ka a a   . We define a mapping f: A →B such that  1 2 3f ( ) , 2, 4,..., 2( 1) . ka a a a a k= − − − − Note that f(a) B since 2( 1) 2( 1)kka n a k n k  − −  − − . Clearly, f is injective since if a, b  A and a ≠ b, then iiab for at least one i, so therefore 2( 1) 2( 1)iia i b i− −  − − , so f(a) ≠ f(b). For each  12, ,..., rp p p p=  B, consider  12, 2,..., 2( 1) kq p p p k= + + − .

2024 Raffles Institution H3 Mathematics Preliminary Examinations (Solutions) We need to show that q is well-spaced, which is clear since the difference between any 2 consecutive terms is at least 3 (2 + 1). Hence q A and f(q) = p which implies that f is surjective. Thus, f is a bijection and |A| = |B| = 22 . (shown)nk k −+  Alternative Solution From the set  1,2,3,..., n , if the number is selected, it is represented by ‘0’, otherwise, it is represented by ‘1’. That is, if the subset formed is  1,4,8 , its corresponding binary string representation of length n is 011011101111….111. For the subset to be well-spaced, the binary string representation for such a subset must have at least two ‘1’s between any two consecutive ‘0’s. Thus, to obtain a well-spaced subset of size k, there are k ‘0’s and ()nk− ‘1’s. For the ()nk− ‘1’s , 2( 1)k− of them are placed as a pair between the ‘0’s as shown below. 0 1 1 0 1 1 0 1 1 0 …0 1 1 0 The remaining ( 2( 1)) 3 2n k k n k− − − = − + of the ‘1’s can be placed in any of the ( )1k+ positions (i.e. before/after/in between the ‘0’s). 0 1 1 0 1 1 0 1 1 0 …0 1 1 0        The number of ways this can be done is the same as the number of ways to distribute ( 3 2)nk−+ identical objects into ( 1)k+ distinct boxes. Hence , 3 2 ( 1) 1 2 2 ( 1) 1 nk n k k n kT kk − + + + − − +   ==    +−    . (d) [2] To have a well-spaced subset, we need 222 3 nn k k k +− +    22 3 3 3 , 0 0 0 2 2 2 2 (or equivalently ). nnn n n k k k k n k n kST kk ++          = = = − + − +   ==         

2024 Raffles Institution H3 Mathematics Preliminary Examinations (Solutions) Question 3 (a) [3] Let y = –x. Then ( ) g( )h( ) d g( )h( ) d g( )h( ) d a a a a a a x x x y y y y y y − − − = − − − = − −    Therefore, ( ) ( ) ( )( ) 0 g( )h( ) d 1 g( )h( ) d g( )h( ) d2 1 g( )h( ) d g( )h( ) d2 1 h( ) g( ) g( ) d2 1 h( ) d2 h( ) d since h is even a a aa aa aa aa a a a a a x x x x x x x x x x x x x x x x x x x xx xx − −− −− − − = + − − = + − = + − = =       (b) [4] Let 2h( ) 1xx=− which is even and 1g( ) 12 xx = + . Then 1 1 1 2g( ) g( ) 1. 1 2 1 2 1 2 2 1 x x x x xxx −+ − = + = + =+ + + + Hence 11 2 2 10 1 d 1 d1 2 4x x x x x  − − = − =+ since the integral represents the area of the quadrant of the circle with radius 1 centered at the origin.

2024 Raffles Institution H3 Mathematics Preliminary Examinations (Solutions) (c) [8] Let us first evaluate 2 0 12 01 1212 01 01 1211 01 1 2 1 2 2 f ( )e d e d (2 )e d e e d (2 )e e d e e e e e 1 e e 2 1 111 e e e x xx x x x x xx xx x x x x x x x x − −− − − − − − − − − − − − = + −    = − + + − − −       =− + − + − −    =− + + − = − + = −     Then 2 00 2 1 2( 1) f ( )e d lim f ( )e d lim f ( )e d n xx n kn x n k k x x x x xx  −− → − → = − = =    To evaluate each integral we perform a substitution 2( 1)x k t= − + to bring each integral back to the interval [0, 2]. We thus obtain 2 2( 1) 1 0 2 2( 1) 1 0 2 2( 1) 1 0 12 2 1 2 2 lim f (2( 1) )e d lim f ( )e d since f is 2-periodic = lim e f ( )e d 11= lim 1 ee 111 1 e 1 e1 11e e 111 ee n kt n k n kt n k n kt n k kn n k k t t tt tt − − + → = − − − → = − − − → = − → = −+ =     −       − −= − = = + −+     

2024 Raffles Institution H3 Mathematics Preliminary Examinations (Solutions) Student Solution:

2024 Raffles Institution H3 Mathematics Preliminary Examinations (Solutions) Question 4 (a) [6] 1ua= 21 ()bbu b a u aa= = = ( ) ( )2 3 1 2 2 1 2 2 2u bbu u u a b uu a a = − = − = −  2 bc a = − Hence if we let nP be the statements 2 2 1nn buu a −= and 2 1 2 2nn buu a + =− for all n + , the above shows that 1P . Assuming kP for some k + , 2 2 1kk buu a −= and 2 1 2 2kk buu a + =− . Then ( ) ( ) 2( 1) 2 2 21 2 2 1 2 21 21 2 21 2 1 2( 1) 1 2 2 2 Note: 2 2 2 kk k kk k k k k k kk uu u uuu uu u bbu c c aa bbuuaa ++ + + + + + + + − = =− =−   = − − = − − =  == and similarly ( ) 2( 1) 1 2 3 22 2 1 2 2 21 22 22 21 22 2 2 2( 1) 2 2 2 Note: From induction hypothesis kk k kk k k k k k kk uu u uuu uu u bu a cu cu + + + + ++ + + + + + ++ = =− =−   =−  == Thus kP 1kP +