2025 CJC JC2 H2 Math Prelim Paper 2 answer key

9758/02/J2PRELIM/2025 [Turn Over CATHOLIC JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2 JC2 Preliminary Examination CANDIDATE NAME CLASS INDEX NUMBER MATHEMATICS 9758/02 Paper 2 17 Sep 2025 3 hours Additional Materials: Printed Answer Booklet List of Formulae (MF27) READ THESE INSTRUCTIONS FIRST Answer all the 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.

2 9758/02/J2PRELIM/2025 Section A: Pure Mathematics [40 marks] 1 The diagram shows the graphs of  fy x and  f 'y x . The graph of  fy x has turning point s at  3, 4 and  3, 4 , crosses the x -axis at  2, 0 and  2, 0 , and the equations of asymptotes are 1x , 1x and 2y . The graph of  f 'y x crosses the x -axis at  3, 0 , and the equations of asymptotes are 1x and 0y . On separate diagrams, sketch the graphs of (a)  fy x , [2] (b)   1 ,f 'y x [3] (c)  f 1 ,y x  [3] labelling clearly the equation(s) of any asymptote(s), coordinates of any axial intercept(s) and turning point(s) where applicable. y x O y =2 (2, 0) (, 0) (3, 4) (3, 4) x = 1 x = 1 x y O y =0 (3, 0) x = 1

3 9758/02/J2PRELIM/2025 [Turn Over (a) Solution: Examiners Feedback: Well attempted. Most students were able to achieve full credit. Others were still able to get the portion of the graph, 0x correct. y x O y =2 (2, 0) (, 0) (3, 4) (3, 4) x = 1 x = 1 x y O y =0 (3, 0) x = 1 y x O y =2 (, 0) (3, 4) x = 1

4 9758/02/J2PRELIM/2025 (b) Examiners Feedback: Poorly attempted. Students were able to identify x-intercept (1,0) and vertical asymptote, 3x but unable to sketch the curve. Others were able to go on and identify that there is a maximum point in the curve   1 ,f 'y x but indicated this point in the 1st quadrant instead of the 4th quadrant. x y O y =0 (3, 0) x = 1 y x O x = 3 (1, 0)

5 9758/02/J