ASRJC 9758 2023 Prelim P2 Solution

[Turn over ANDERSON SERANGOON JUNIOR COLLEGE H2 MATHEMATICS JC2 Preliminary Examination Paper 2 (100 marks) 9758/02 3 hours Additional Material(s): List of Formulae (MF26) CANDIDATE NAME CLASS / READ THESE INSTRUCTIONS FIRST Write your name and class in the boxes above. Please write clearly and use capital letters. Write in dark blue or black pen. HB pencil may be used for graphs and diagrams only. Do not use staples, paper clips, glue or correction fluid. Answer all the questions and write your answers in this booklet. Do not tear out any part of this 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. 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. All work must be handed in at the end of the examination. If you have used any additional paper, please insert them inside this booklet. The number of marks is given in brackets [ ] at the end of each question or part question. Question number Marks 1 2 3 4 5 6 7 8 9 10 11 Total This document consists of 21 printed and 3 blank pages.

2 © ASRJC 2023 Section A: Pure Mathematics [40 marks] 1(a) Solution ( ) ( ) p a1 pa pa λλ λλ λ = +− = +− −= − b bb bb BP BAλ=       and common point B ∴the points A, B and P are collinear. (b) Solution Let F be the foot of perpendicular from O to AB ( ) for someOF µµ= +−a ba   0OF AB =      ( ) ( )[ ]0 µ+ − −=a ba ba  2 0µ+ −−=ab b a aa 2 2µ = − a ba OF   = ( ) 2 22+− + aa ba ab 22 2 += −a b ba

3 © ASRJC 2023 [Turn over 2 Solution Let 432P( ) 4 78z z z az bz=− + ++ Since 3 2i+ is a root of P( ) 0z = , P(3 2i) 0+= . ( ) ( ) ( ) ( ) 432 3 2i 4 3 2i 3 2i 3 2i 78 0ab+−++++++ = ( )( 119 120i) 4( 9 46i) (5 12i) 3 2i 78 0ab− + − −+ + + + + + = Equating real parts: 535ab+= ----- (1) Equating imaginary parts: 12 2 64ab+= ----- (2) Solving (1) and (2), we get 7a= and 10b=− Now 432P( ) 4 7 10 78zz z z z= −+−+ Since coefficients of P( )z are all real, 3 2i− is also a root of P( ) 0z = . A quadratic factor of P( )z ( )( )3 2i 3 2izz= −+ −− 2 6 13zz=−+ Consider ( )( ) 432 2 2P( ) 4 7 10 78 6 13z z z z z z z z cz d=− + − += −+ ++ Comparing constants: 78 13 6dd= ⇒= Comparing coefficient of z: 10 6 13 2d cc− = − + ⇒= Equating 2 2 60zz+ += , we have 22 2 4(1)(6) 2(1)z −± −= 2 4( 5) 2 −± −= 1 5i= −± Therefore, the other roots are 3 2i− , 1 5i−+ and 1 5i−− .

4 © ASRJC 2023 3 (a) Solution 2 22 11e sin d e sin e cos d22 x xx xx x xx= −∫∫ 2 2 dsin ed de cos d2 x x vux x u xvx