EJC_H2_MATH_P1

2018 JC2 H2 Mathematics Preliminary Examination [Turn over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2018 General Certificate of Education Advanced Level Higher 2 MATHEMATICS Paper 1 [100 marks] 9758/01 12 September 2018 3 hours Additional Materials: Answer Paper List of Formulae (MF26) READ THESE INSTRUCTIONS FIRST Write your name, civics group and question number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all questions. Give non-exact numerical answers correct to 3 signifi cant 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. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 7 printed pages.

2 2018 JC2 H2 Mathematics Preliminary Examination 1 The shaded region R bounded by the curve yx=− , the line yx=− and the x-axis is rotated about the x-axis through O360 . Find the volume of the solid fo rmed, leaving your answer to 2 decimal places. [4] Suggested Solution yx=− 15yx=± − 15yx=+ − intersect yx=− at (4 , 4 )− 15yx=− − cuts the x-axis at (4,0) Volume required () () () 2255 2 44 115 d 15 d 4 ( 4 ) 3 201.06 (to 2 d.p.) xx xxππ π − =+ − − − − − =  2 (i) Solve the inequality 2xa x a axa −− ≥− , where a is a positive real constant, leaving your answer in terms of a. [4] (ii) Hence, by using a suitable value for a, solve the inequality 24e e 1 1 4e 1 4 xx x −− ≥− leaving your answer in exact form. [3] y x O (5,1) R

3 2018 JC2 H2 Mathematics Preliminary Examination Suggested Solution (i) 2xa x a xa −− − ≥ a ( xa≠ ) () () 2 22 0 2 0 xa x aa x a xa xa x a a xa −−− − ≥− −+ − ≥− Consider () 22 2xa x a a−+ − = 0 x = () () () 2 222 4 ( 1 ) 2 aa a a−− ± − − − = 24 2 aa± = aa± Method 1 (test critical points) 22 2 0xa x a a xa −+ − ≥− can be rewritten as ()() ()() 0 xa a xa a xa −+ −− ≥− Using sign test, we can check whether each factor is positive or negative for the different range of values of x Thus, we have or a a xa xa a−≤ < ≥ + Method 2 () () 2xa x aa x a xa −−− − − ≥ 0 () () 22 2xa x a x a a−− + − ≥ 0 ( xa≠ )

4 2018 JC2 H2 Mathematics Preliminary Examina