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 + = 14 5 7+ cm

4 Cedar Girls’ Secondary School 4049/01/S4 Prelim/2024 2 The profit, $y of a company can be modelled by the equation 2()y a x h k= − + , where x is the number of goods sold and a, h and k are constants. The company obtained the maximum profit of $24 500 when 800 goods were sold. The company incurred a loss of $7 500 when no goods were sold. (a) State the value of h and of k. [2] Since maximum profit of $24500 occurs when 800 goods were sold, 2( 800) 24500y a x= − + Therefore, 800h= 24500k = (b) Using the values of h and k found in part (a), find the value of a. [2] 2( 800) 24500y a x= − + When 0x= , 7500y=− 27500 ( 800) 24500a− = − + 640000 32000a=− 1 or 0.0520a=− − (c) Find the range of the number of goods the company needed to sell to earn a profit. [2] 21 ( 800) 2450020yx=− − + For the company to be profitable, 0y 21 ( 800) 24500 020 x− − +  2( 800) 490000 0x− −  ( 100)( 1500) 0xx− −  100 1500x

5 Cedar Girls’ Secondary School 4049/01/S4/Prelim/2024 [Turn over 3 Solve the equation 33 27 12log 1 log (8 1)log 3pp+ − = + . [5] 33 27 12log 1 log (8 1)log 3pp+ − = + 2 3 3 3 3log log 27 log 3 log (8 1)pp+ − = + 2 33 27log log (8 1)3 p p=+ 2 33log 9 log (8 1)pp=+ 29 8 1pp=+ 29 8 1 0pp− − = ( )( )1 9 1 0pp− + = 11 or 9pp= =− (reject) x 4 Solve the equation 1 2 15 5 30xx+−=− . [5]

6 Cedar Girls’ Secondary School 4049/01/S4 Prelim/2024 1 2 15 5 30xx+−=− ( ) ( ) 215 5 5 305 xx =− ( ) ( ) 21 5 5 5 30 05 xx − − = ( ) 25 25 5 150 0xx− − = Let 5x be y 2 25 150 0yy− − = ( )( )30 5 0yy− + = 30y= or 5y=− 5 30x = or 55x =− (reject) lg5 lg30x = lg30 2.11 (3 s.f)lg5x== 5 Given that 32 45 10y x px x= − − + is decreasing for 3 xq−   , find the value of the constants p and q. [5]

7 Cedar Girls’ Secondary School 4049/01/S4/Prelim/2024 [Turn over 2d 3 2 45d y x pxx = − − Since y is decreasing, 23 2 45 0x px− −  For y is decreasing for 3 xq−   , 3( 3)( ) 0x x q+ −  23 3 9 9 0x qx x q− + −  23 (9 3 ) 9 0x q x q+ − −  Comparing coefficients, 9 45q− =− 5q= 2 9 3pq− = − 2 9 15p− = − 3p=

8 Cedar Girls’ Secondary School 4049/01/S4 Prelim/2024 6 The graph cos( )y a b cx=+ is defined for 02 x  , where a, b and c are constants. The graph has a period of 4 3  and a minimum value of 1 when 4 3x = . The graph also passes through the point ,33   . (a) Show that 3a= , 2b=− and 3 2c= . [3] Period = 24 3c = ➔ 3 2c= When 3x = , 3y= ➔ 3a= When 4 3x = , 1y= , 13 b=+ ➔ 2b=− (b) Sketch the graph of cos( )y a b cx=+ for 02 x  . [2] (c) Given that there are exactly 3 solutions for the equation cos( )b cx k a =− , for 02 x  , state the range of values of the constant k. [1] 32cos 32 xk− + = For 3 points of intersections, 15 k 5 1

9 Cedar Girls’ Secondary School 4049/01/S4/Prelim/2024 [Turn over 7 The graph of d d y x of a function f ( )yx= is shown in the diagram, passing through the x-axis at ( ) ( )2,0 , 0,0− and ( )2,0 . (a) State the number of stationary points of the graph of f ( )yx= . [1] 3 (b) State the x-coordinate of the minimum point of f ( )yx= and explain why it is the minimum point. [2] 2x= The gradient is negative before the stationary point and is positive after the stationary point. Hence, the point is a minimum point.

10 Cedar Girls’ Secondary School 4049/01/S4 Prelim/2024