RVHS 9758 2023 Prelim P1

1 ©RIVER VALLEY HIGH SCHOOL 9758/01/2023 RIVER VALLEY HIGH SCHOOL 2023 JC2 Preliminary Examination Higher 2 NAME CLASS INDEX NUMBER MATHEMATICS Paper 1 Candidates answer on the Question Paper Additional Materials: List of Formulae (MF26) 9758/01 15 Sep 2023 3 hours READ THESE INSTRUCTIONS FIRST This document consists of 25 printed pages and 3 blank pages. For examiner’s use only Question number Mark 1 2 3 4 5 6 7 8 9 10 11 12 Total Calculator Model: Write your class, index number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all the questions. Write your answers in the spaces provided in the question paper. 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. The use of an approved graphing calculator is expected, where appropriate. 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. The total number of marks for this paper is 100.

2 ©RIVER VALLEY HIGH SCHOOL 9758/01/2023 1 (i) Without using a calculator, solve 2 2 4 41 012 xx xx −+ <+− . [3] (ii) Hence solve ( ) 21 21 21 012 2 x xx + + − ≤+− . [3] 2 (i) The complex number iza= + , where a is a real constant, is such that the modulus of ( ) 1 *1 i z z − + is 1 2. Find the value of a where 2a< . [3] (ii) It is given that a is the value found in part (i). The complex number w has argument 4 π. Find the 3 smallest positive integer of n such that the complex number 1 * n z zw −  is purely imaginary with the imaginary part negative. [3] 3 A sequence 1 23, , , ...uuu is such that 1 nu n= for 1n≥ . (i) Show that for 2n≥ , 11 32n nn Au uu nn −+ −+= − , where A is a constant to be found. [2] (ii) Hence find 3 2 1N n nn= −∑ . (You need not express the answer as a single fraction.) [2] (iii) Explain why 3 2 1N n n