[HGS] [2023] 4E AM 4049 Prelims P2 MS

HGV Sec 4E Preliminary Examination Additional Mathematics Paper 2 2023 [MARK SCHEME] 1 HILLGROVE SECONDARY SCHOOL PRELIMINARY EXAMINATION 2023 SECONDARY FOUR (EXPRESS) [MARK SCHEME] CANDIDATE NAME ( ) CLASS - CENTRE NUMBER S INDEX NUMBER Additional Mathematics Paper 2 Candidates answer on the Question Paper. No Additional Materials are required. 4049/02 29 August 2023 2 hours 15 minutes 10.05 a.m. – 12.20 p.m. 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, 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 Parent’s/ Guardian’s Signature: ___________________ TOTAL 90 Setters: Mdm Lee Li Lian This document consists of 24 printed pages, including this page.

HGV Sec 4E Preliminary Examination Additional Mathematics Paper 2 2023 [MARK SCHEME] 2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 2 0ax bx c+ + = , 2 4 2 b b acx a −  −= Binomial expansion ( ) 1 2 2 ... ... ,12 n n n n n r r nn n na b a a b a b a b b r − − −     + = + + + + + +           where n is a positive integer and ( ) ! ( 1)...( 1) ! ! ! n n n n n r r r n r r  − − +== − 2. TRIGONOMETRY Identities 22sin cos 1AA+= 22sec 1 tanAA=+ 22cosec 1 cotAA=+ sin( ) sin cos cos sinA B A B A B =  cos( ) cos cos sin sinA B A B A B= tan tantan( ) 1 tan tan ABAB AB = sin 2 2sin cosA A A= 2 2 2 2cos 2 os sin 2 os 1 1 2sinA c A A c A A= − = − = − 2 2 tantan 2 1 tan AA A= − Formulae for ABC sin sin sin a b c A B C== 2 2 2 2 cosa b c bc A= + − 1 sin2 ab C=

HGV Sec 4E Preliminary Examination Additional Mathematics Paper 2 2023 [MARK SCHEME] 3 1 A calculator must not be used in this question. (a) Show that tan15 2 3= − . [4] tan15 tan(45 30 ) tan 45 tan 30 1 tan 45 tan 30 31 3 [A1] 311 3 31 3 31 3 3 3 3 3 33 3 3 3 3 33 3 3 3 3 [M1] 3 3 3 3 9 6 3 3 [ 93 = −  − = +   − = + − = + −+= −= + −−= +− −+= − M1] 12 6 3 6 2 3 [A1] −= =− Alternative Method tan15 tan(60 45 ) tan 60 tan 45 1 tan 60 tan 45 31 [A1] 1 3 1 3 1 3 1 [M1] 3 1 3 1 3 2 3 1 = [M1]2 2 3 [A1] = −  − = +   −= + −−= +− −+ =−

HGV Sec 4E Preliminary Examination Additional Mathematics Paper 2 2023 [MARK SCHEME] 4 (b) Use the result from part (a) to find an expression for 2sec 15 , in the form 3ab+ where a and b are integers. [2] ( ) 22 2 sec 15 1 tan 15 1 2 3 1 4 4 3 3 [A1] 8 4 3 [A1] = +  = + − = + − + =−

HGV Sec 4E Preliminary Examination Additional Mathematics Paper 2 2023 [MARK SCHEME] 5 2 (a) Given that 23 52 f ( ) f ( ) 5x dx x dx − == , find   3 5 3 f ( ) x x dx − − . [4]   3 5 33 55 2 3 3 5 2 5 32 5 3 f ( ) 3 f ( ) 3 3 f ( ) f ( ) 3 [M1] 3(5 5) 3 [M1]2 330 (9 25) [M1]2 54 [A1] x x dx x dx x dx x dx x dx x dx x − −− −− − − =− = + − = + −  = − − =     

HGV Sec 4E Preliminary Examination Additional Mathematics Paper 2 2023 [MARK SCHEME] 6 (b) Differentiate 25 lnxx with respect to x. Hence, find the value of 3 1 5 ln x x dx , giving your answer correct to 2 decimal places. [4] 22 3 32 11 33 32 111 33 32 111 3 32 11 1(5 ln ) 5 ln 10 5 10 ln [A1] (5 10 ln ) 5 ln [M1] 5 10 ln 5 ln 10 ln 5 ln 5 1 1 110 ln 5 ln 5 2 2 2 d x x x x xdx x x x x x x x dx x x x dx x x dx x x x x dx x x x dx x x dx x x x =  +  =+ +=  +=  =− =−      ( ) ( ) ( ) 3 1 323 1 1 1 1 55 ln 45ln 3 5ln1 [A1]2 2 2 1 1 45 5 45ln 3 5ln12 2 2 2 1 1 45 5 45ln 3 5ln12 2 2 2 14.7187765 dx xx x dx  = − −  = − − −  = − − −  =   14.72 (to 2 d.p.) [A1]

HGV Sec 4E Preliminary Examination Additional Mathematics Paper 2 2023 [MARK SCHEME] 7 3 (a) Solve the equation 15 25 6 0xx −− − = . [5] 1 22 2 2 2 2 5 25 6 0 5 5 6 0 (5 )5 6 0 [M1]25 25 5 (5 ) 150 0 Let 5 , 25 150 0 25 150 0 ( 10)( 15) 0 [M1] 10 or 1 5 5 10 xx xx x x xx x x y yy yy yy yy − − − − = − − = − − =  − − = = − − = − + = − − = == = 5 15 [A1] lg 5 lg10 lg 5 lg15 lg 5 lg10 lg 5 lg15 lg10 lg15 [M1 for either shown]lg 5 lg 5 1.43067 x xx xx xx = == == == = 6558 1.68260 6194 1.43 (to 3 s.f.) 1.68 (to 3 s.f.) [A1 for both] = 

HGV Sec 4E Preliminary Examination Additional Mathematics Paper 2 2023 [MARK SCHEME] 8 (b) (i) Given that 3 343 49log logxy= , express y in terms of x. [3] 3 343 49 3 77 77 77 77 77 7 2 log log log log [M1]log 343 log 49 3log log 3log 7 2log 7 1log log [A1]2 log [A1] xy xy xy xy y xy yx = = = = = = = 77 77 2 Alternative Method 1log log [A1]2 2log log [A1] xy xy yx = = = (ii) Find the value of x for which 23 49 343 49 1log ( 11 ) log log 7x x x+ − = . [3] 23 49 343 49 22 49 49 49 22 49 49 7 7 2 49 2 2 2 2 2 2 22 2 1log ( 11 ) log log 7 1log ( 11 ) log [M1] log 7 1log ( 11 ) log log 7 log 49 11log 2 [M1] 11 49 11 2401 2401 11 2400 11 0 (2400 11) 0 110 (N.A.) or 2 x x x x x x x x x xx x xx x xx x x x x xx xx xx + − = + − = + − = + = + = + = =+ −= −= == [A1]400

HGV Sec 4E Preliminary Examination Additional Mathematics Paper 2 2023 [MARK SCHEME] 9 4 A particle travels in a straight line so that, t seconds after leaving fixed point, O, its velocity is, v ms-1, is given by 2 86v t kt k= − + , where k is a constant. The minimum velocity of the particle occurs when 12t = . (a) Show that 3k = . [2] 2 86 Acceeration, 2 8 [A1] When 12, 0 24 8 0 3 (shown) [A1] v t kt k dva dt tk dvt dt k k = − + = =− == −= = (b) Determine whether the particle will return to O during its journey. [4] 2 32 3 2 3 2 32 2 2 24 18 24Displacement, 18 , where is a constant32 When 0, 0, 0 12 18 [A1]3 When 0, 12 18 0 [M1]3 36 54 0 ( 36 54) 0 0 or 36 54 0 v t t tts t c c ts c ts t t s t tt t t t t t t t t t = − + = − + + == =  = − + = − + = − + = − + = = − + = 2( 36) ( 36) 4(1)(54) 2(1) 36 1080 2 34.43167 673 or 1.568323275 [A1] Yes, the particle will return to at t Ot − −  − −= = =  1.57s and 34.4s. [A1]

HGV Sec 4E Preliminary Examination Additional Mathematics Paper 2 2023 [MARK SCHEME] 10 (c) Find the total distance travelled by the particle in the first 2 seconds. [3] 2 2 3 2 When 0, 24 18 0 24 24 4 1 18 21 24 504 = 2 23.22497216 or 0.7750278397 23.2 or 0.775 [M1] 12 183 When 0, 0 m When 0.7750278397, 6.897661467 m [Either this or below M1] When v tt t ts t t ts ts t = − + =  −  =   =  = − + == == = 12, 9 m 3 Total distance travelled by the pa