2023 JSS AMATH P2 PRELIM MS

[Turn Over] O JURONG SECONDARY SCHOOL 2023 GRADUATION EXAMINATION SECONDARY 4 EXPRESS/ SECONDARY 5 NORMAL ACADEMIC CANDIDATE NAME CLASS INDEX NUMBER ADDITIONAL MATHEMATICS 4049/02 PAPER 2 2023 Candidates answer on the Question Paper. 2 hours 15 minutes Additional Materials : Writing Paper (1 sheet) READ THESE INSTRUCTIONS FIRST Write your name, class and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. 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. At the end of the examination, fasten all your work securely together. 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 17 printed pages including this page. 90

2 4049/2/GE/23 1. ALGEBRA Quadratic Equation For the equation Binomial expansion , where n is a positive integer, and 2. TRIGONOMETRY Identities Formulae for 02 =++ cbxax a acbbx 2 42−−= nrrnnnnn bbar nbanbanaba ++    ++     +     +=+ −−− ......21)( 221 ! )1)...(1( )!(! ! r rnnn rnr n r n +−−=−=      1cossin 22 =+ AA 22sec 1 tanAA=+ 22cosec 1 cotAA=+ ( ) BABABA sincoscossinsin = ( ) BABABA sinsincoscoscos = ( ) BA BABA tantan1 tantantan  = AAA cossin22sin = AAAAA 2222 sincossin211cos22cos −=−=−= A AA 2tan1 tan22tan − = ABC C c B b A a sinsinsin == Abccba cos2222 −+= Abcsin2 1=

3 4049/2/GE/23 [Turn Over] 1 At the beginning of a virus outbreak, the number of cases of infected people increased with time. After t days, the number of recorded cases was N. It was observed that N can be modelled by the equation 1200 ktNe= . (a) Write down the initial number of cases recorded. [1] The number of cases recorded after 6 days rose to 4800. (b) Estimate the number of cases recorded after 10 days. [4] A pandemic is declared if the number reaches 20 000 cases. (c) Assuming the trend continues, estimate after how many days will it take for a pandemic to be declared. [2] Solution Marks 1200 B1 Solution Marks 6 6 10(0.231049) 4800 1200 4800 1200 48006 ln 1200 0.231049 1200 12095.2 12100 k k e e k k Ne N = = = = = =  M1 – substitute values correctly M1 – value of k – accept rounded off values. M1 A1 Solution Marks 0.23104920000 1200 12.1766 13 days te t = =  M1 – substitute values correctly A1 – Do not accept 12 days

4 4049/2/GE/23 2 The expression 32x px qx r+ + + is divisible by both x and 2x− and it leaves a remainder of 8 when divided by 2x+ . (a) Find the values of p, q and r. [4] (b) Hence, find the remainder when it is divided by 2 23xx+− . [2] Solution Marks (0) 0 0 f r = = (2) 0 8 4 2 0......(1) ( 2) 8 8 4 2 8 16 4 2 0......(2) (1) (2) 88 1 6 f pq f pq pq p p q = + + = −= − + − = − + − = + = = =− [B1] [M1] – Forming correct equations (either one) [A1] [A1] Solution Marks Long division 1x− 2 23xx+− 32 6x x x+− 32( 2 3 )x x x− + − 2 3xx−− 2( 2 3)xx− − − + 3x−− Remainder = 3x−− [M1 – Long division] A1 – Writing out remainder

5 4049/2/GE/23 [Turn Over] Alternative Method 2 3 2 2 3 2 2 2 3 ( 3)( 1) 6 ( )( 2 3) 6 ( )( 2 3) ( ) When 1, 1 1 6 4 ......(1) When 3, 27 9 6( 3) 3 0 3 ......(2) Sub (1) into (2), 0 3( 4 ) 0 12 3 3 1 x x x x x x x Q x x x R x x x Q x x x ax b x ab ab x ab ab bb bb b a + − = + − + − = + − + + − = + − + + = + − = + =− − =− − + − − =− + =− + =− − − + =++ =− =− Remainder = 3x−− [M1] – Forming of equations [A1]

6 4049/2/GE/23 3 (a) Show that 2cos cot 1 2cos cot   + − = can be written as ( )( )2cos 1 sin cos 0  − − = . [3] (b) Hence, solve the equation 2cos 2 cot 2 1 2cos 2 cot 2x x x x+ − = for 0 < x < 180. [4] Solution Marks ( ) ( ) ( ) ( )( ) ( ) ( )( ) 2cos cot 1 2cos cot 2cos cot 1 2cos cot 0 2cos 1 cot 1 2cos 0 2cos 1 cot 2cos 1 0 2cos 1 1 cot 0 cos2cos 1 1 0 sin 2cos 1 sin cos 0                     + − = + − − = − + − = − − − = − − = − − = − − = Alternatively 2 cos cos2cos 2cos 1 0sin sin 2sin cos cos 2cos sin 0 2cos (sin cos ) (sin cos ) 0 (sin cos )(2cos 1) 0                + −  − = + − − = − − − = − − = M1 – factorisation M1 – conversion of cot A1 M1 – conversion of cot M1 – simplification A1 – factorise by grouping Solution Marks ( )( ) cot 2 2cos 2 1 2cos 2 cot 2 2cos 2 1 sin 2 cos 2 0 1cos 2 sin 2 cos 22 tan 2 1 Basic 60 Basic 45 2 60 ,300 2 45 , 225 30 ,150 22.5 ,112.5 x x x x x x x x or x x x xx xx + − = − − = == = = = == == 30 , 22.5 ,112.5 ,150x= M1 – With 2x M1, M1 – Correct Equations A1

7 4049/2/GE/23 [Turn Over] 4 The diagram below shows a mould made of a cylinder and a right circular cone. The diameter of the cylinder is 12x cm and its height is h cm. The vertical height of the cone is 8x cm. (a) Find an expression, in terms of x, for the slant height l of the cone. [1] (b) Given that the entire mould is covered with a plastic sheet whose area is 240π cm2, express h in terms of x. [2] Solution Marks 22(8 ) (6 ) 10 l x x x =+ = B1 Solution Marks 2 2 22 2 2 2 2 240 (6 )(10 ) 2 (6 ) (6 ) 240 60 12 36 240 12 240 96 240 96 12 20 8 rl rh r x x x h x x xh x xh x xh x xh x         + + = + + = + + = =− −= −= M1 A1 8x 12x h l

8 4049/2/GE/23 (c) Show that the volume, V cm3, of the mould is given by 3720 192V x x =− . [3] (d) Hence find the value of x for which the volume has a stationary value and determine whether this value for the volume is a maximum or minimum. [4] Solution Marks Volume 22 2 22 23 3 1 3 1 20 8(6 ) (8 ) (6 ) ( )3 96 720 288 720 192 (shown) coner h r h xx x x x x x x xx       =+ −=+ = + − =− M2 – Cone, cylinder A1 Solution Marks Volume 3 2 2 2 720 192 720 576 0 720 576 0 720 576 1.11803, 1.11803(rej) 1.12 V x x dV xdh dV dh x x x x      =− =− = −= = =− = 2 2 2 2 1152 0 since is positive. dV xdh dV xdh =−  Hence, maximum volume. M1 A1 M1 A1

9 4049/2/GE/23 [Turn Over] 5 (a) Without using a calculator, show that 26cos15 4 += . [3] (b) Hence, state the value of cos( 15 )− . [1] (c) Using your answer from part (i), find the exact value of sec(15 ) [3] Solution Marks cos15 cos(60 45 ) cos 60cos 45 sin 60sin 45 1 2 3 2( ) ( )2 2 2 2 26 (shown)4  = −  =+ =+ += M1 – use of correct formulae with appropriate values M1 – either term A1 Solution Marks 26cos( 15 ) 4 +−  = [B1] Solution M