2023 CGSS Prelim 4049 P1 Solutions
Uploaded by hima · 8 October 2023
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Text from the first pagesCEDAR GIRLS’ SECONDARY SCHOOL Preliminary Examination Secondary Four CANDIDATE NAME Solutions CLASS 4 INDEX NUMBER CENTRE/ INDEX NO / ADDITIONAL MATHEMATICS 4049/01 Paper 1 30 August 2023 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/2023 [Turn over Answer all the questions. 1 Express 32 32 2 4 18 2 4 8 x x x x x x − + − − + − in partial fractions. [6] 2 32 2 4 8x x x− + − 322 4 18x x x− + − ( ) 322 4 8 16x x x− − + − 72x−− ( )( ) ( )( ) 32 22 2 4 18 7 2 2 2 4 2 4 x x x x x x x x − + − − − =+ − + − + ( ) 27 2 4 ( )( 2)x A x Bx C x− − = + + + − When 2x= , 14 2 8 2 AA− − = =− Comparing coefficients of x2, 02 A B B= + = Comparing constants, 2 4 2 2 8 2 6 3 A C C C − = − =− + =− =− 32 3 2 2 2 4 18 2 2 3 2.2 4 8 2 4 x x x x x x x x x − + − − = − +− + − − +
4 Cedar Girls’ Secondary School 4049/01/S4 Prelim/2023 2 Two vertices of a rhombus ABCD are A ( )2, 5−− and C ( )4, 7 . (a) Find the equation of the diagonal BD. [3] Gradient of AC = 57 224 −− =−− Gradient of BD = 1 2− Midpoint of AC = ( )2 4 5 7, 1,122 − + − + = Equation of BD: ( )1 211yx− =− − 13 22yx=− + If the gradient of the side BC is 3, find (b) the coordinates of B and of D. [4] Equation of AD: ( )5 3 2yx+ = + or Equation of BC : ( )7 3 4yx− = − 31yx=+ 35yx=− At D, 13 3122xx− + = + At B, 13 3522xx− + = − 1 7x= 13 7x= 1 1031 77y = + = 13 435 77y = − = Coordinates of D = 1 10,.77 Coordinates of B = 13 4,.77 Making use of mid-point formula, let B = ( ),xy or D = ( ),xy 1 7 12 x+ = and 10 7 12 y+ = 13 7 12 x+ = and 4 7 12 y+ = 13 7x= and 4 7y= 1 7x= and 10 7y= Coordinates of B = 13 4,.77 Coordinates of D = 1 10,.77
5 Cedar Girls’ Secondary School 4049/01/S4/Prelim/2023 [Turn over 3 The equation of a curve is 32 9,y x hx kx= + + + where h and k are constants. (a) Show that if y increases as x increases, then 23 0.kh− [3] 2d 32d y x hx kx = + + If y increases as x increases, then 2d 32d y x hx kx = + + > 0, As 30 , then 2 40b ac− ( ) 2 2 4(3)( ) 0hk − 24 12 0hk− 23 0.kh− (b) In the case when 5h=− and 3k = , find the x-coordinate of each of the points at which the curve meets the x-axis. [3] Since curve meets x-axis, 32 5 3 9 0x x x− + + = Let 32f( ) 5 3 9x x x x= − + + Since 32f ( 1) ( 1) 5( 1) 3 9 0− = − − − − + = ( )1x+ is a factor of f(x). 2 69xx−+ 1x+ 32 5 3 9x x x− + + Therefore, ( )( ) 2 6 9 1 0x x x− + + = ( ) ( ) 2 3 1 0xx− + = 3x= or 1x=−
6 Cedar Girls’ Secondary School 4049/01/S4 Prelim/2023 4 (a) Given that the constant term in the binomial expansion of 6 kx x + is 160− , find the value of the constant k. [3] General term = ( ) 66 r r kxr x − 6 2 0r−= 3r = Since constant term is 160− , 36 1603 k =− 3 160 220k −= =− (b) Using the value of k found in part (a), show that there is no constant term in the expansion of ( ) 6 22 3 .kxx x ++ [3] ( ) ( )( ) 6 2 2 22 2 3 2 3 ...Term in Constant term +...x x x xx − − + = + + For term in 2x− , 6 2 2r− =− 4r= Term in 2x− = ( ) 4 2 2 6 2 240 4 x xx − = Constant term in expansion = ( )2 240 3( 160) 0+ − = Hence there is no constant term in the expansion.
7 Cedar Girls’ Secondary School 4049/01/S4/Prelim/2023 [Turn over 5 (a) The equation of a quadratic curve is 22 16.y x px= + + Given that 0y only when 2 xk , find the value of p and of k. [3] Since 22 16 0x px+ + when 2 xk , ( )2 2 ( ) 0x x k− − ( ) 22 2 4 4 0x k x k− + + By comparing, 4 16 4.kk= = By comparing, ( )2 4 (8 4) 12pk=− + =− + =− (b) In the case where 14p=− , find the value of m for which the line 2y x m=+ is a tangent to the quadratic curve, 22 16.y x px= + + [3] Since line cuts curve, 22 2 14 16.x m x x+ = − + 22 16 16 0x x m− + − = Since line is a tangent to curve, 2 40b ac−= ( ) ( ) 2 16 4(2) 16 0 m− − − = ( )16 256 8m− = 16m=−
8 Cedar Girls’ Secondary School 4049/01/S4 Prelim/2023 6 Mary and Sally took part in a shot put competition. The heights, in metres, of Mary’s and Sally’s shot put throws can be modelled by the quadratic functions ( ) 27f ( ) 6 3 180xx=− − + and 21 2 8g( ) 35 5 5x x x=− + + respectively, where x m is the horizontal distance of the shot put from the starting line. (a) Express g(x) in the form ( ) 2 g( )x a x b c= + + where a, b and c are constants. [2] ( ) 2 2 218g( ) 14 7 735 5x x x=− − + − + ( )( ) 218g( ) 7 4935 5xx=− − − + ( ) 21g( ) 7 3 35xx=− − + (b) Evaluate f(0) and g(0) and hence interpret the meaning of your answers. [2] ( ) 27f (0) 0 6 3 1.6180=− − + = g(0) 1.6= Both Mary and Sally threw the shot put from a height of 1.6 m.
9 Cedar Girls’ Secondary School 4049/01/S4/Prelim/2023 [Turn over (c) The winner of the comp
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