RI H3 Math 2024 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