[WGS] [2023] 4E AM 4049 Prelims P1 MS

Name Index Number Class O LEVEL PRELIMINARY EXAMINATION 2023 LEVEL & STREAM : SECONDARY 4 EXPRESS/ 5 NORMAL ACADEMIC SUBJECT (CODE) : ADDITIONAL MATHEMATICS (4049) PAPER NO : 01 DATE (DAY) : 11 SEPTEMBER 2023 (MONDAY) DURATION : 2 HOURS 15 MINUTES READ THESE INSTRUCTIONS FIRST Write your name, index number and class 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, 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 in this paper is 90. DO NOT TURN OVER THE QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO. Student’s Signature Parent’s Signature Date Date This document consists of 20 printed pages including this cover page. Setter : Ms Nicole Ng 90 MARKING SCHEME

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 02 =++ cbxax a acbbx 2 42 −−= 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 22 22 22 sin cos 1 sec 1 tan cosec 1 cot AA AA AA += =+ =+ ( ) ( ) ( ) 2 2 2 2 2 sin sin cos cos sin cos cos cos sin sin tan tan tan 1 tan tan sin 2 2sin cos cos 2 cos sin 2cos 1 1 2sin 2 tan tan 2 1 tan A B A B A B A B A B A B ABAB AB A A A A A A A A AA A  =  = = = = − = − = − = − Formulae for ABC 2 2 2 sin sin sin 2 cos 1 sin2 a b c A B C a b c bc A bc A == = + − =

3 1 Triangle ABC is such that the length of side AB is ( )1 3 2 cm,+ angle ABC is 45° and its area is ( ) 27 4 2 cm+ . Find, without using a calculator, the exact length of BC, in cm. Leave your answer in the form of ( )2,ab+ where a and b are integers. [4] ( )( ) ( )( ) ( ) ( ) ( ) ( ) 17 4 2 1 3 2 sin 45 12 117 4 2 1 3 2 1(for sin45 )2 2 2 2 7 4 2 1 3 2 1 1 3 2 1 3 2 14 2 16 1 3 2 17 14 2 84 16 48 2 17 34 2 68 17 4 2 2 cm 1 BC M BC M BC M A + = +  + = +   + −= +− +  − = − − + −= − −−= − =+ A B C 45° ( )1 3 2 cm+

4 2 Given that 2 3 24 6 24x x x ++= , find the value of 6x without using a calculator. [4] 2 3 2 2 2 3 2 3 2 2 4 3 2 3 6 3 2 2 3 6 3 2 4 3 13 13 3 4 6 24 2 2 3 8 3 2 3 2 3 1 32 32 3 2 1 3 3 2 2 32 123 861 3 x x x x x x x x x x x x xx xx xx xx x x x M M M A ++ + + + + + + + + ++ ++ + − + − − =   =   =  = =  =  = =

5 3 When a polynomial f ( )x is divided by ( )1x+ and ( )2x+ , the remainders are 3 and 5 respectively. Find the remainder when f ( )x is divided by ( )( )12xx++ . [4] ( )( )f ( ) 1 2 Q( ) f ( 1) 3 3 ........................(1) 1 f ( 2) 5 5 2 .......................(2) 1 (1) (2) : 1 2 1 remainder = 2 1. 1 x x x x ax b a b M a b M M a b xA = + + + + −= =− + −= =− + − −= =  − +

6 4 Given that 24 12 f ( ) d f ( ) d 6x x x x − == , find (a) 42 14 2f ( ) d f ( ) dx x x x − + , [2] 42 14 2 4 4 1 2 2 2f ( ) d f ( ) d 2 f ( ) d f ( ) d f ( ) d 1 2(6 6) 6 18 1 x x x x x x x x x x M A − − + = + − = + − =     (b) the value of k for which   2 1 f ( ) d 9x kx x − += . [3]   ( ) ( ) 2 1 22 11 22 1 22 1 22 f ( ) d 9 f ( ) d d 9 6 9 1 2 32 21 3122 23 2 21 x kx x x x kx x kx M kx kk M kk kA − −− − − += += +=  =     −−=           −= =  

7 5 (a) Find the 1 x term in the expansion of 10 2 2x x  + . [3] ( ) ( ) 1021 1 20 3 71 8 10 21 10 2 20 3 1 1 7 10 15360217 rr r rr T x x Mr xr rM r T x A x − − + − − = = − =− = == (b) Hence, find the constant term in the expansion of ( ) 10 2 213 xx x ++  . [2] ( ) ( )( ) ( ) 10 2 2 153601 3 1 0 3 1 46080 1 x x x M xx A    + + = +       =

8 6 A spherical balloon expands at a constant rate of 8 cm3/s. The balloon is initially empty. (a) Find the rate of increase of its radius when the radius is 2.5 cm, leaving your answer in terms of  . [The volume of a sphere of radius r is 34 3 r .] [3] ( ) 2 2 2 d 41dr dr dr d dV dV dr d d d d dr d 1 dr 8 1 8 4 (2.5) 1 d4 2.5 8 cm/s 125 V rM V ORt V t t t MM t A = =  =  =  =    =  (b) When the radius is beyond 5 cm, besides the expansion, air begins to leak out from the balloon at a rate of 2 cm3/s. Find the rate of change of the radius when it is 8 cm. [2] ( ) ( ) 2 2 dr dr d dV dV dr d d d d dr d 1 d dr 6 1(for 6) 6 4 8 dd48 3 cm/s 1128 V ORt V t t t VM tt A =  =  =  = =    =  Accept 0.00746 m (3 s.f)

9 7 Given that 5sin 13x= and x is obtuse, find the exact value of the following. (a) ( )sec x− [3] ( ) ( ) 1sec 1 cos 1 1cos 13 112 xM x Mx A −= − = =− (b) cos 2 x [3] 2 2 12cos 13 12 2cos 1 113 2 1cos 12 26 26 1 26cos accept ( ) 12 26 26 26 x x M x M x or rej A =− − = − = =−  5 x 13 -12

10 8 The number of ants, N, in a colony after t days can be modelled by 1200 atNe= , where a is a constant. There are 10 000 ants after 6 days. (a) Find the initial number of ants in the colony. [1] (0)1200 1200 1aN e B== (b) How many ants are there after 15 days? Give your answer correct to 2 significant figures. [3] ( )( ) 6 6 0.353377 15 10000 1200 1 10000 1200 100006 ln 1 1200 0.353377 1200 240000 1 a a eM e aM a Ne A = = = = = = (c) Sketch the graph of 1200 atNe= for the first 15 days. [2] B1 – for the shape of the graph B1 – for the y-intercept at 1200 and the point at t = 15 days. (15, 240000) 1200