RI 2025 Prelim

2025 Raffles Institution H3 Mathematics Preliminary Examinations 1 The function f satisfies f ( ) f (1 )x k x x− − = for all real values of x, where 1.k  (a) Find f ( ).x [2] Consider the equation f ( ) f (1 ) g( ). ( )x x x− − =  (b) Explain why g( ) g(1 ) 0xx+ − = is a necessary condition for there to be a solution f to () . [2] It is now given that there is a solution f to () . (c) If g( )x is a polynomial with degree ,k + explain why k cannot be even. [2] (d) If g is a linear function, give a possible g( )x . [2] (e) Find a solution f to () when 3g( ) (2 1) .xx=− [2] 2 Let z be a complex number such that 3 1, 1.zz= (a) Show that 210 zz+ + = and hence find all the solution to 3 1, 1.zz= [2] Let w deote the solution to 3 1z = with Im( ) 0.w  The complex numbers a, b and c are represented by the points A, B and C respectively in the Argand diagram such that ABC is an equilateral triangle described in the anticlockwise sense. Let m be the complex number representing the point M, the midpoint of AB. (b) Show that ()c m k b m− = − , where k is a complex number to be determined. [2] (c) Hence show that ( ).c b w b a= + − [2] (d) Deduce that 2 0.a bw cw+ + = [2] Let PQR be a triangle, and construct equilateral triangles XQP, YRQ, ZPR on the exterior of triangle PQR. Let 1 2 3,,G G G be the centroids of triangles XQP, YRQ, ZPR respectively. (e) Using the fact that the complex number representing the centroid of a triangle is given by the arithmetic mean of the complex numbers representing the vertices of the triangle, show that triangle 1 2 3G G G is equilateral. [4]

2025 Raffles Institution H3 Mathematics Preliminary Examinations 3 Let m and n be nonnegative integers and , 210 ( ) d . (1 ) mx mn n tI x t t += + (a) Prove that for 2m and 1n , 1 , 2, 1 2 1( ) ( ). 2 (1 ) 2 m m n m n n xmI x I x n x n − −− −=− + + [3] (b) State an inequality in terms of m and n such that 1 2lim 0(1 ) m nx x x − → =+ . [1] (c) (i) Show that 0,1lim . 4x I  → = [3] (ii) Hence find the exact volume of revolution when the region bounded by the curve 2 21 xy x = + and the x-axis is completely rotated about the x- axis. [4] 4 In this question, we will prove the following identity using various methods. 1 ( 1)01 nn m n m n m n m n k k n k k n − − −          − + + − =           −           ----- (*) for positive integers .m k n (a) (i) Use a combinatorial argument to explain why for positive integers 1ni , 11 .1 n n n i i i −−     =+      −      [2] (ii) Prove (*) using mathematical induction on n. [4] (b) By considering the number of ways to select a team of k players from a group of m players such that all n star players are included in the team of k players and the principle of inclusion and exclusion, show (*). [4] (c) (i) Write down the binomial expansion of ( )1 n y− in the form 0 g( , ) n nr r r n y − =  , where g is a function of r and n to be determined. [1] (ii) By considering the coefficient of the term kx in the expansion of ( )( ) ( )1 1 1 , n mn xx − + − + prove (*). [3]

2025 Raffles Institution H3 Mathematics Preliminary Examinations 5 (a) Let 12, ,..., ,...nx x x be a sequence of positive real numbers and let ( )12 12 1 1 1... ...nn n y x x x x x x = + + + + + +   and nnzy = . Note: For any real number x, x denotes the greatest integer less than or equal to x. For example, 2 2, 3.1 3 ==       . (i) Show that for all positive integers n, 1 21n n ny y y+  + + . [3] (ii) Deduce that no two terms in the sequence 12, ,..., ,...nz z z are the same. [2] (b) Let 12, ,..., na a a be real numbers such that 120 ... na a a    . (i) Show that for 2,3,...,in= ( ) 1 2 1 11 .111 ii iii aa aaa − − − −+++ [2] (ii) Hence, using the Cauchy-Schwarz inequality or otherwise, show that 2 1 1 2 1 1 1 1 1 1 1... ... .11 n n na a a a a a a −  + +  + + + + + − − [4] (iii) Using 12 mi ia − + −= , show that for positive integers m, ( ) 2 11 3 2 2 .2 mmm − +  − [3]

2025 Raffles Institution H3 Mathematics Preliminary Examinations Use the information in the mathematical text to answer this Question. You should read the whole mathematical text before you start answering the questions. A Dynamical Systems Proof of Fermat’s Little Theorem We will prove an important and foundational result in number theory using techniques from dynamical systems. The result, known as Fermat’s Little Theorem, states that: Theorem. Let p be a prime number, and a any integer. Then (modulo )pa a p . Usually, the theorem is proved using algebraic ideas. In this note, we will prove it with a simple argument from dynamical systems on the unit interval [0, 1]. The basic idea is to count points of minimum period p of a particular dynamical system, and note that this number must be divisible by p. For every integer n, we define the function :[0,1] [0,1]nT → as follows:   if 1,() 1 if 1. n nx xTx x = = Here the braces denote the fractional part of a real number. For example, {3.1415} = 0.1415, and {3} = 0. We say that x is a fixed point of nT if ()nT x x = . If we apply this operation nT repeatedly, some points x in [0, 1] will be periodic, in the sense that ( )( )( )... ( )n n nT T T x x = for some number of iterations of nT . We call a point k-periodic if it is mapped back to itself after k iterations, that is ( )( )( ) iterations ... ( ) .n n n k T T T x x = We say k is the minimal period if x is not mapped back to itself for any fewer number of iterations than k. So the 1-periodic points are fixed points. We also say that the points 12, ,..., px x x lie in an orbit of size p if 1 2 2 3 1 1( ) , ( ) ,..., ( ) , ( ) .a a a p p a pT x x T x x T x x T x x −= = = = We now determine two crucial properties of the family of functions nT : 1. Let n be an integer greater than 1. The function nT has n fixed points in [0, 1]. 2. Let a and b be positive integers. Then for all ( )( )[0,1], ( ).a b abx T T x T x= Now if p is a prime, and a is any integer, consider the p-periodic points of aT . These points are the fixed points of aT iterated p times. It follows that there are paa− points that have minimal period p. Now consider the orbit of these points. Since each point with minimal period p lies in an orbit of size p, there are ( ) /pa a p− orbits of size p. Since this is an integer, we have the desired result. Adapted from Kevin Iga (2003) A Dynamical Systems Proof of Fermat's Little Theorem, Mathematics Magazine, 76:1, 48-51, DOI: 10.1080/0025570X.2003.11953946.

2025 Raffles Institution H3 Mathematics Preliminary Examinations 6 (a) (i) Sketch the graph of 3()y T x= for 01 x . [2] (ii) Give a graphical explanation why the function nT has n fixed points. [1] (iii) Find the fixed points of nT . [2] (b) (i) Write down an orbit of size 2 for 2T . [1] (ii) Write down an orbit of size 3 for 3T . [2] (c) Show that if a and b are positive integers, then for all [0,1],x ( )( ) ( ).a b abT T x T x = [2] (d) Let m and k be positive integers. Show that for any m-periodic point of aT , the minimal period k must divide m. [3] (e) Let a be a positive integer and p be a prime number. (i) Explain why there are paa− points that have minimal period p amongst the fixed points of aT iterated p times. [2] (ii) Hence show that for all positive integers a, 15 5 3 (modulo 15).a a a a + − [3]