ASRJC 2025 H3 Mathematics Prelim Question Paper

1 MATHEMATICS 9820/01 Paper 1 2 September 2025 3 hours Additional Materials: Answer Booklets List of Formulae (MF27) READ THESE INSTRUCTIONS FIRST Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphing calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 7 printed pages and 1 blank page. ANDERSON SERANGOON JUNIOR COLLEGE JC2 Preliminary Examinations 2025 Higher 3

2 Answer all the questions. 1 By considering a suitable scalar product, prove that ( ) ( )( ) 2 2 2 2 2 2 2ax by cz a b c x y z+ +  + + + + for any real numbers a, b, c, x, y and z. Deduce a necessary and sufficient condition on a, b, c, x, y and z for the following equation to hold: ( ) ( )( ) 2 2 2 2 2 2 2ax by cz a b c x y z+ + = + + + + [3] (i) Show that ( ) 22 2 2 2 2 2( ) ( ) ( )3 abc ab bc ca ++  + + for all real numbers a, b and c. [3] (ii) Find real numbers p, q and r that satisfy both 2 2 2 4 729p q r and 8 8 243p q r . [5] 2 The numbers f ( )r satisfy f ( ) f ( 1)rr+ for 1,2,3,.....r= Show that, for any non-negative integer n, ( ) ( ) ( ) ( ) 1 1 11 f f ( ) 1 f n n k n n n n rk k k k r k k k + − + = −   −  where k is an integer greater than 1. [3] (i) By taking 1f ( )r r= , show that 121 1 11 12 N r N Nr + − = +   + Deduce that the sum 1 1 r r  =  does not converge. [5] (ii) Let ()Sn be the set of positive integers less than n which do not contain the digit 2 in their decimal representation and let ()n be the sum of the reciprocals of the numbers in ()Sn , so for example 11(5) 1 . 34 = + + Show that (1000)S contains 728 distinct numbers. Show that ( ) 80n  for all n. [5]

3 3 A train has n seats, where 2.n For a particular journey, all n seats have been sold and each of the n passengers has been allocated a seat. The passengers arrive one at a time and are labelled 12, ,....., nT T T according to the order in which they arrive. 1T arrives first and nT arrives last. The seat allocated to 1,2,.....,rT r n is labelled .rS Passenger 1T ignores his allocation and decides to choose a seat at random (each of the n seats being equally likely). However, for each 2,r passenger rT sits in ,rS if it is available or, if rS is not available, chooses from the available seats at random. (i) Let nP be the probability that, in a train with n seats, nT sits in nS . Write down the value of 2P and find the value of 3P . [2] (ii) Explain why, for 2,....., 1kn , 11 sits in | sits in ,n n k n kP T S T S P and deduce that, for 3,n 1 2 1 1 n nr r PP n . [6] (iii) Find 4.P Make a conjecture on the value of nP in its simplest f orm and prove your result by induction. [5]

4 4 Fermat’s Little Theorem states that if p is prime and gcd( , ) 1,xp then 1 1 mod .pxp Let n and x be positive integers with gcd( , ) 1.xn Show that if n mp where p is prime and gcd( , ) 1,mp then 11 1 mod if and only if 1 mod .nmx p x p [3] Chinese Reminder Theorem states that a system 1 2 1 mod 1 mod ..... 1 mod k xp xp xp where 12, ,..., kp p p are all pairwise coprime, has unique solution 121 mod ... kx p p p . If n is a product of distinct primes. Show that 1 1 modnxn if and only if , for every prime divisor p of n, 1 1 mod . n p xp [4] Deduce that if every prime divisor p of n satisfies 1 | 1 ,pn then for every x with gcd( , ) 1,xn the congruence 1 1 modnxn holds. [3]

5 5 (i) Let Sn be the number of ways to pick n squares from an n  n matrix such that no two squares are from the same row and no two squares are from the same column. Find a recurrence relation for Sn. [1] (ii) Determine a bijection between the ways to pick n squares from an n  n matrix such that no two squares are from the same row and no two squares are from the same column and the permutation of the set of integers N = {1, 2, 3, …, n}. Hence state the value of Sn in terms of n. [2] Let Dn be the number of ways to pick n squares from the same n  n matrix such that no two squares are from the same row and no two squares are from the same column and no diagonal element is picked. (see Figure 1). Figure 1. The diagonal elements in the case of an 8  8 matrix (iii) Show that 12n n nD sD tD where s and t are functions of n to be determined. State the values of D1 and D2. [4] (iv) By considering the Principle of inclusion-exclusion, show that nD can also be expressed as ( )1 1 1 1! 1 ... 11! 2! 3! ! n nDn n = − + − + + − [4] (v) A teacher received a different christmas gift from each of her six friends. She conceived the idea of giving a gift in return to each of her friends using these 6 different gifts received. She just need to ensure she doesn’t give to each of her friends, the same gift she received from them. How many ways can she do this? [2]

6 Use the information in the mathematical text to answer Question 6. You should read the whole mathematical text before you start answering the questions. Investigating the Generalisations of the Product Rule Throughout this question fg denotes the product of the functions f and g, so that fg( ) f ( ) g( ).x x x= The differential operator, D, when applied to a function, returns its derivative. So Df f '= and the product rule for the derivative of the product of the functions f and g becomes ( )D fg f Dg g Df.=+ (*) The product rule may be generalised in at least two different ways. Firstly, it can be generalised to give a formula for finding higher derivatives of the product of two functions. For example, differentiating both sides of (*) and applying the product rule to each term on the right-hand side gives an expression for the second derivative of fg ( ) ( ) ( ) 2 2 2 2 2D fg f D g Df Dg Dg Df g D f f D g 2Df Dg g D f= + + + = + + A second generalisation of the product rule for two functions gives a formula for the derivative of a product of three or more functions. For example, if h is a third function, then ( ) ( )( ) ( )D fgh D f gh =f D gh gh Df.=+ You can then apply the two-term product rule to find the derivative of D(gh) so that ( ) ( )D fgh f g Dh h Dg gh Df fg Dh fh Dg gh Df= + + = + + This method generalises