EJC 2025 H3 Mathematics Prelim Question Paper

[Turn over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2025 General Certificate of Education Advanced Level Higher 3 CANDIDATE NAME CIVICS GROUP INDEX NO. MATHEMATICS Paper 1 [80 marks] 9820/01 24 September 2025 3 hours Additional Materials: Printed Answer Booklet List of Formulae and Results (MF27) READ THESE INSTRUCTIONS FIRST Answer all questions. Write your answers on the Printed Answer Booklet. Follow the instructions on the front cover of the answer booklet. 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 must present the mathematical steps using mathematical notations and not calculator commands. You must show all necessary working clearly. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 5 printed pages and 3 blank pages.

2 1 Let a, b, c be positive real numbers such that 36abc= . (a) Show that 2 3 18a b c+ +  . [2] (b) Deduce the minimum value of 2 2 249 3 3 2 2 a b c b c c a a b+++ + + and find the values of a, b and c where this minimum is attained. [5] (c) Hence, or otherwise, prove that if p, q and r are positive real numbers such that 1 36pqr= , then ( ) ( ) ( ) 3 3 3 1 1 1 99 3 36 3 2 4 2 prp r q q r q p+ + + + + . [4] 2 (a) The variables x and y are related by the differential equation 22 d2 d2 y xy x x y= − , 2x y for all ,xy  . Using the substitution y ux= , find the general solution of the differential equation. [7] (b) Describe geometrically the set of points that satisfy the general solution obtained in part (a). [4] 3 Consider 4-by-4 unit-square grids where each unit-square is either shaded or white. For example, the following diagram illustrates an example of such a grid. (a) (i) How many possible 4-by-4 unit-square grids are there? [1] (ii) How many 4-by-4 unit-square grids are there which contain at least one entirely shaded 3 -by-3 square? [4] (b) Determine the minimum number of shaded unit-squares in a 4-by-4 unit-square grid to guarantee that it contains at least one entirely shaded 2-by-2 square. Justify your answer. [3] (c) Define ic to be the number of shaded unit squares in the ith column from the left of a given 4-by-4 unit- square grid. For example, for the 4-by-4 unit-square above, 1 2 3 42, 0, 3, 2c c c c= = = = . How many distinct possible tuples ( )1 2 3 4, , ,c c c c are there if the given 4-by-4 unit-square grid has (i) exactly 4 shaded unit-squares, [1] (ii) at least 12 shaded unit-squares? [3]

3 4 Let S be the set of all integers from 1 to n and T be the set of all subsets of S, including the empty set  and S itself. (a) Explain why 2nT = . [1] The sets 12 and AA , which are not necessarily distinct, are chosen randomly and independently from the 2n subsets in T. (b) In this part of the question, we will calculate ( )12P AA = in 2 ways. (i) Let i be an integer such that 0 in . Show that ( )12 3P 4i A A  = . [2] (ii) Hence, find ( )12P A A = . [1] (iii) Let k be an integer such that 0 kn . Find ( )1 2 1P| A A A k = = . [1] (iv) Deduce that 0 13 24 nn nk k n k + =  =  . [3] (c) (i) Let k be an integer such that 0 kn . Find ( )12P A A k= . [2] (ii) Find ( )12E AA . [2] 5 (a) Let ( )f x be a polynomial with integer coefficients. (i) For integers  and  show that ( ) ( )ff x − has a factor of x − and hence, or otherwise, show that ( ) ( )f f 0− (modulo − ). [2] (ii) Deduce that there does not exist a polynomial f with integer coefficients such that ( ) ( )f 1 2, f 3 5== . [1] (iii) Explain why there does not exist a polynomial f with integer coefficients such that ( ) ( )f 1 0, 0 f 2025 2024=   . [2] (b) A monic degree n polynomial is a polynomial where the coefficient of nx is 1. Let p be an odd prime. Let ( )h x be a monic degree n polynomial with integer coefficients. We call an integer  a root modulo p of h(x) if ( )h0  (modulo p). Moreover, if  (modulo p), we call integers , incongruent. (i) Find two incongruent roots modulo 7 for 2 1xx++ . [1] (ii) For 1n , show that if  is a root modulo p of h( x), then there exist s a monic degree 1n− polynomial ( )0h x with integer coefficients, such that ( ) ( ) ( )0hhx x x − (modulo p). [3] (iii) Prove, by induction, that any monic degree n polynomial ( )h x has at most n incongruent roots modulo p for 0n . [5]

4 Use the information in the mathematical text to answer Question 6. You should read the whole mathematical text before you start answering the questions. On the irrationality of π We denote ( )( )f k x to be the kth derivative of ( )f x with respective to x and ( )( ) ( ) 0 ff xx= . A number is called rational if it can be simplified to the form a b where a is an integer and b is a positive integer where a and b do not share any prime factors. A number is called irrational if it is not a rational number. For example, the numbers 4 390.1, , 33 13− =− are rational numbers but 2 and π are not. However, showing that π is irrational, is not trivial. In 1947, the Canadian-American mathematician Ivan Morton Niven produced an elementary yet elegant proof that π is irrational. His methods could be traced back to French mathematician Charles Hermite’s 1873 paper on proving irrationality of er , where r is a rational number. We will investigate a modified version of Niven’s proof of the irrationality of π . We suppose that π a b= for positive integers a, b, for contradiction. In his proof, for each positive integer n, Niven considered the polynomial ( ) ( ) ( )π f !! n nn nb x xx a bxx nn  − − == . Using binomial expansion, we can write this polynomial in the form ( ) 21f ! n r r rn x c xn = =  , where rc are integers for all r. By considering the expansion, we can show that ( )( )f0 k is an integer, for 0 2kn . This would then imply that ( )( )f π k is an integer, for 0 2kn , as well. Niven also define another function ( )F x as ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) 2 4 2 F f f f ... 1 f n n x x x x x= − + − + − . This is carefully chosen so that the identity ( ) ( ) ( ) ( ) π 0 πf sin d F 0 Fx x x =+ holds for all n. For 0 πx , we have ( ) ! π0f nn x n m for some positive real constant m. We can then develop the bound ( ) ( ) π 0 2π0 f sin d ! nnmx x x n for all 1n . It is a known fact that if lim 0nn u → = , this implies that for sufficiently large n, we have 1nu  . By selecting a sufficiently large n, we will attain the required contradiction to prove that π is irrational.

5 6 In this question you are asked to prove that π is irrational. For the entirety of the question, we will assume that π a b= , where a, b are positive integers, for contradiction. (a) (i) Consider any rc in the sum 21 ! n r r rn cxn =  . Briefly explain why rc is an integer. [2] (ii) Hence, explain why ( )( )f0 k is an integer for 0 2kn . [2] (b) By showing that ( ) ( )ff πxx=− , prove that ( )( )f π k is an integer for 0 2kn . [3] (c) (i) Prove that ( ) ( ) ( )F f Fx x x =− . [2] (ii) Deduce that ( ) ( ) ( ) ( ) π 0 f sin d F 0 F πx x x =+ . [3] (d) (i) Prove that for some positive real constant m, ( ) ! π0f nn x n m for 0 πx . [2] (ii) Hence, show that ( ) ( ) π 0 2π0 f sin d ! nnmx x x n . [1] (iii) By considering the series expansion of ex , or otherwise, explain why 2πlim 0 ! nn n m n→ = . [2]