SRJC_H2_MATH_P1_Solution

1 [TURN OVER] SERANGOON JUNIOR COLLEGE 2018 JC2 PRELIMINARY EXAMINATION MATHEMATICS Higher 2 9758/1 11 Sept 2018 3 hours Additional materials: Writing paper List of Formulae (MF 26) TIME : 3 hours READ THESE INSTRUCTIONS FIRST Write your name and class on the cover page and on all the work you hand in. Write in blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. 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 a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calcul ator are not allowed in a question, you are required to present the mathematical steps us ing mathematical notatio ns 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. At the end of the examination, fasten all your work securely together. Total marks for this paper is 100 marks. This question paper consists of 6 printed pages (inclusive of this page) and 2 blank pages.

2 Answer all questions [100 marks]. 1 Find 2 2 d25 x xxx++ . [3] Solution 2 22 25 d1 d25 25 xx x xx xx x +=−++ ++ () 22 22 11 d d 3 d 25 14 xx xxxx x +=− − ++ ++  () 21 31ln 2 5 tan 22 xx xx c − +=− + + − +  2 The complex numbers z and w satisfy the equations *2 1 5 izw z+= and 23 1 1wz+= . Find the complex numbers z and w. [6] Solution ()*2 1 5 i 1zw z+= LL () ( )1 11 3 22wz=− LL Subst (2) into (1) gives ()1 11 3 * 2 15i2 zz z −+ = 211 3 21 5 i22zz z−+ = 215 3 15i22zz−= Let izxy=+ () () 2215 3 i1 5 i22 xy x y+= + + Comparing the real and Imaginary parts, we have 15 152 y = and () 2215 3 3 322 2xx y=+ LL 2y = Subst into (3) gives 2 54 0xx −+ = () ( )41 0xx−− = 1x∴= or 4 When 12 iz =+ , ()1 11 3 6i 4 3i2w =− − = − When 42 iz =+ , ()1111 12 6i 3i22w =− − = − −

3 [TURN OVER] 3 Without the use of a graphing calculator, find the range of values of x for which 218 492 xxx ≥++− . [3] Hence find the exact range of values of x for which 218 e4 e 92e xx x ≥+ +− . [3] Solution 218 492 xxx ≥++− () () 218 2 4 9 02 xx x x +− + + ≥− 3218 2 18 02 xx x x ++ + − ≥− () 2 1 02 xx x + ≥− 1x =− or 02 x≤< Replace x by ex So e1x =− or 0e 2 x≤< No solution or ()0 e and e 2xx≤< No solution or () and ln 2xx∈<¡ ln 2x∴<