2024 ZHSS AMATH P2 PRELIM MS

[Turn over ZHONGHUA SECONDARY SCHOOL PRELIMINARY EXAMINATION 2024 SECONDARY 4 EXPRESS/ 5 NORMAL (ACADEMIC) Candidate’s Name Class Register Number STUDENT SOLUTIONS ADDITIONAL MATHEMATICS 4049/02 PAPER 2 9 September 2024 Candidates answer on the Question Paper. No Additional Materials are required. 2 hours 15 minutes READ THESE INSTRUCTIONS FIRST Write your name, class and index number 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 paper clips, glue or correction fluid. Answer all 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. Setter: Mr Francis Tan Vetter: Mr Lionel Ang ___________________________________________________________________ For Examiner’s Use 90

3 1 (a) When 0t = , 0170000 150000ke=+ 170000 150000k=+ 20000k = (b) 140000 20000 150000 ne=+ 120000 150000 ne= 4 5 ne= 4ln 5 n= 0.223144n=− 0.223n=− (c) 0.22370000 20000 150000 te−+ 0.22350000 150000 te− 0.2231 3 te− 1ln 0.2233 t− 4.92t 4t = David must sell the car in 2026

4 2 (a) General term 7 2px x  + = ( ) 77 2 r r pxr x −    Consider the power of x = 7 rr−− = 72 r− If there is an independent term, 7 2 0r−= 7 2r = Since r is not a whole number, there is no term independent of x and as such, every term is dependent on x (b) ( ) 7 2 52px x x  +− Since there is no independent term in 7 2px x  + , the only term to form the independent term is the 1 x in the expansion of ( ) 7 2 52px x x  +− . Find the 1 x term Power of x : 7 2 1r− =− 4r= 1 x term = ( ) 4 37 2 4 px x     = 335 16p x  = 3560p x

5 Thus, the term independent of x 3560 2p xx − = 241920− 3 216p = 6p= 3 (a) Consider ABD ADC = (by tangent chord theorem) ADC BAD = (alternate angles) Thus, ABD BAD = . Since there are 2 equal angles in triangle ABD , triangle ABD is isosceles. (b) Let ABD x= , Since triangle ABD is isosceles (part(a)), 180 2BDA x = − (angle sum of triangle) by tangent chord theorem, ABD ADC x = = Or 180 [ 180 2 ]ADC x x = − + − x= (interior angles) Since tangents from an external point are equal, triangle CDA is also isosceles. Thus, ADC CAD x = = And 180 2DCA x BDA = − = (shown) 4 (a) (i) 10 cos 180x− or 10 cos x − (ii) 190 tan 90 x−−   or 1tan22 x −−   (b) (i) 20 6 72a −=− =− 2 12 6b == 13c= (ii) 14 16x

6 5 (a) cos105 = ( )cos 60 45+ = cos 60cos 45 sin 60sin 45− = 1 2 3 2 2 2 2 2 −  = 26 44− = 26 4 − sec105 1 cos105= 4 26 = − 4 2 6 2 6 2 6 += −+ = 26−− (b) (i) 3 f ( ) 1 xex x= − ( ) 33 ' 2 3 ( 1)f ( ) 1 xxe x ex x −−= − ( ) 3 ' 2 (3 4)f ( ) 1 xexx x −= − 3a= 4b=− (ii) At y axis, 0x= . When x =0, 1y=− f '(0) 4=− Thus, gradient of normal 1 4= Equation of normal: 1 4y x c=+ Sub x =0, 1y=− 1c=− 1 14yx=− When 0y= , 4x= (4,0)P

7 6 (a) Let 32f ( ) 10 9 3 2x x x x= − − + 32 1 1 1 1f ( ) 10 9 3 22 2 2 2      − = − − − − − +           1f ( ) 02−= By factor theorem, 21x+ is a factor. 32f ( ) 10 9 3 2x x x x= − − + 2 32 32 2 2 5 7 2 2 1 10 9 3 2 10 5 ______________________ 14 3 2 14 7 ______________________ 42 42 ______________________ 0 xx x x x x xx xx xx x x −+ + − − + −+ − − + − − − + −+ ( )( ) 2f ( ) 2 1 5 7 2x x x x= + − + 2 1 5 5 5 2 2 2 x x x x x − − −− ( )( )( )f ( ) 2 1 5 2 1x x x x= + − − (b) Let 3y a= . ( ) ( ) 2 135 3 1 2aa a+ = + 2 210 9 3aa a+ = + 3210 2 9 3a a a+ = + 3210 9 3 2 0a a a− − + = Comparing with (a) ( )( )( )2 1 5 2 1 0a a a+ − − = 1 2a=− or 2 5a= or 1a=

8 13 2 y =− or 23 5 y = or 31y = (reject as 30y  ) or 2ln 3 ln 5y = 0y= 0.834y=− 7 (a) 10cos(5 2 ) 50vt= − + Acceleration = d d v t . ( )d 10 sin(5 2 ) 2d v tt = − − − d 20sin(5 2 )d v tt =− When 0t = , 20sin(5) = 19.2− m/s2 7 (b) ( ) 2 24 d 2 cvt t −= +   ( ) 2 24 2 dcv t t − = − + ( ) 1 24 2cv t c − = + + When 0t = , 5cv = Thus, 5 12 c=+ 7c=− ( ) 1 24 2 7cvt − = + − cyclist is instantaneously at rest 0cv= ( ) 1 24 2 7 0t − + − = ( ) 24 72t =+ 24 7( 2)t=+ 24 7 14t=+ 10 7t = (c) Displacement, dcs v t= ( ) 1 24 2 7 ds t t − = + − 24ln( 2) 7s t t c= + − +

9 When 0t = , 0s= 24ln 2c=− 24ln( 2) 7 24ln 2s t t= + − − At 10 7t = , 10 1024ln 2 7 24ln 277s    = + − −       2.93591s= m At 10t = , ( ) ( )24ln 10 2 7 10 24ln 2s= + − − 26.99777s=− m Total distance = 2.93591 + (2.93591+26.99777) =32.9m 8 (a) PQ PD DQ=+ Consider triangle APD , sin3 PD = 3sinPD = Consider triangle DCQ , cos2 DQ = 2cosDQ = Thus, 3sin 2cosPQ =+ (c) ( )13sin 33.7PQ =+ Maximum value = 13 ( )13sin 33.7 13+= ( )sin 33.7 1 += 33.7 90 += 56.3 = 9 (a) At time t, Distance from NEX to P = 10t Distance from NEX to Q = 1000 5 t− ( ) 222 (10 ) 1000 5PQ t t= + − -26.99777 2.93591

10 2 2 2100 1000000 10000 25PQ t t t= + − + 2125 10000 1000000PQ t t= − + 21000000 10000 125s t t= − + (b) Least distance occurs at minimum point i.e. d 0d s t = ( ) ( ) 1 2 2d1 1000000 10000 125 250 10000d2 s t t tt − = − + − ( ) ( ) 1 2 21 1000000 10000 125 250 10000 02 t t t − − + − = 250 10000 0t−= 40t = When 40t = , 21000000 10000(40) 125(40)s= − + 894.4s= 894s= m 10 (a) Year 1980 1990 2000 2010 2020 x 0 1 2 3 4 P 2.45 3.09 3.94 4.95 6.30 ln P 0.90 1.13 1.37 1.60 1.84 (b) From the graph, ln P

11 m = 0.235 ( 0.2 ) c = 0.90 ( 0.2 ) ln 0.235 0.9Px=+ 0.235 0.9xPe += 0.2352.45 xPe= (c) When P = 13 0.23513 2.45 xe= 7.10t = First year in the interval would be 2050. End of paper