2024 CGSS Prelim 4049 P1 Worked Solutions

CEDAR GIRLS’ SECONDARY SCHOOL Preliminary Examination Secondary Four CANDIDATE NAME Worked Solutions CLASS 4 INDEX NUMBER CENTRE/ INDEX NO / ADDITIONAL MATHEMATICS 4049/01 Paper 1 22 August 2024 2 hours 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your Centre number, index number and name in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB 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. The use of an approved scientific calculator is expected, where appropriate. 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. The total number of marks for this paper is 90. For Examiner’s Use This document consists of 21 printed pages and 1 blank page. [Turn over 90

Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 0 2 =++ cbxax , a acbbx 2 4 2 −−= Binomial expansion nrrnnnnn bbar nbanbanaba ++     ++     +     +=+ −−− ......21)( 221 , where n is a positive integer and ! )1(...)1( )!(! ! r rnnn rnr n r n +−−=−=      2. TRIGONOMETRY Identities 1cossin 22 =+ AA AA 22 tan1sec += 22cosec 1 cotAA=+ BABABA sincoscossin)sin( = BABABA sinsincoscos)cos( = BA BABA tantan1 tantan)tan(  = AAA cossin22sin = AAAAA 2222 sin211cos2sincos2cos −=−=−= A AA 2 tan1 tan22tan −= Formulae for ABC C c B b A a sinsinsin == Abccba cos2 222 −+= 1 sin2 bc A=

3 Cedar Girls’ Secondary School 4049/01/S4/Prelim/2024 [Turn over Answer all the questions. 1 Two cylinders are such that the ratio of their heights is 7 : 1 . The height of the smaller cylinder is ( ) 2 2 7 1 27 − − cm. Without using a calculator, find the height of the larger cylinder, expressing your answer in the form ( )7ab+ cm, where a and b are integers. [4] Height of larger cylinder = ( ) 2 2 7 1 7 27 −  − ( ) 2 14 7 27 −= − = 14 7 4 4 7 7 − −+ 14 7 11 4 7 11 4 7 11 4 7 −+= −+ = 154 56 7 11 7 28 9 + − − = 126 45 7 9 +