2025 RI Prelim+P1+Qns

© RI 2025 [Turn over RAFFLES INSTITUTION 2025 YEAR 6 PRELIMINARY EXAMINATION Higher 2 MATHEMATICS Paper 1 9758/01 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 8 printed pages. RAFFLES INSTITUTION Mathematics Department

2 @ RI 2025 9758/01/PRELIM 1 A curve has equation 2 ,1 cy ax b x    where , and a b c are real constants. It is given that the curve crosses the x-axis at 2.x The normal to the curve at the point 0, 3 meets the x-axis at 7.5.x Find the values of , and .a b c [4] 2 The point P travels along the curve C with equation 1sin , 1 1.y x x x    Let the gradient of the curve C at the point P be .m If the x-coordinate of P is increasing at the rate of 9 units per second when 1,2x find the exact value of the rate at which m is changing at this instant. [4] 3 Relative to the origin O, the points A and C have position vectors a and 2b such that 5 2 4p p p  a i j k and 2 2 ,  b i j k where p is a positive constant. It is given that 2 .a b (a) Find the exact value of .p [2] (b) Evaluate 2 2 . a b a b [1] (c) Evaluate a b and hence find the area of triangle .OAC [3] (d) Use a geometrical reason to explain why   1 2 2 .4   a b a b a b [2] 4 (a) Find    2 9 d .2 1 1 x xx x   [4] (b) (i) Differentiate 2 1 1x  with respect to .x [1] (ii) Differentiate 2ln 1x  with respect to .x [1] (iii) Hence find   2 22 ln 1 d .1 x x xx    [4]

3 @ RI 2025 9758/01/PRELIM