2025 JVSS Sec 4 AM Prelim P2 QP

JURONGVILLE SECONDARY SCHOOL PRELIMINARY EXAMINATION 2025 Secondary 4 Express STUDENT NAME CLASS INDEX NUMBER ADDITIONAL MATHEMATICS 4049/02 Paper 2 20 AUGUST 2025 2 hours 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. This document consists of 19 printed pages. 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 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. DO NOT OPEN THE BOOKLET UNTIL YOU ARE TOLD TO DO SO O For Examiner’s Use 90

2 Jurongville Secondary School 4049/02/2025 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 n n 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=+ 22cos ec 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 cos sin 2cos 1 1 2sinA A A 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 bc A

3 Jurongville Secondary School 4049/02/2025 [Turn Over 1 (a) Find the first 3 terms in the expansion of 8(1 3 )x+ in ascending powers of x. [2] 82 2 8(1 3 ) 1 8(3 ) (3 ) ... 2 1 24 252 ... x x x xx + = + + +  = + + + (b) Hence determine the first 3 terms in the expansion of 28(2 6 3 )xx++ . [3] ( ) ( ) 8 2 8 8 2 8 2 2 22 22 2 2 (2 6 3 ) 2 1 3 1256 1 3 11256 1 24 252 ... 256 1 24 12 252 ... 256 1 24 264 ... 256 6144 6758 3 2 2 22 4 ... xx x x x xx x x x xx xx x xx + + = + +   = + +       = + + + + +         = + + + + = + + + = + + + (c) Explain why there is no constant term in the expansion of 2 82 1 133 () xx  − + . [3] 22 82 2 2 11 1 3 1 24 252 ) 2 ( ) 2 ( ...) 44( 33 1 .1 24 2523 ..9 x x xxx xxxx− + + + +    − = − +       =+ ++  0 constant term 4(1) 41(24) (252)39= + = − Hence there is no constant term in the expansion of 2 82 1 133 () xx  − + . M1 A1 M1 M1 A1 M1 M1 A1

4 Jurongville Secondary School 4049/02/2025 2 ADE is a semicircle with centre O. It is given that D and E are midpoints of AB and AC respectively and AB = BC. (a) Prove that ABE CBE  . [3] In ABE and CBE , AB BC= (given) AE EC= (E is the midpoint of AC) BE is a common height ABE is congruent to CBE . (SSS) (b) Prove that 90BEA =  . [1] Since ABE is congruent to CBE , BEA BEC = . (corresponding angles of congruent triangles) 180 (angles on a straight line) 180 90 BEA BEC BEA BEA BEA  + =   + =   =  O F E D CA B B1 M2 for 3 conditions M1 for 2 conditions A1

5 Jurongville Secondary School 4049/02/2025 [Turn Over (c) Determine, with explanation, whether ADE is similar to ABC . [3] Since D and E are midpoints of AB and AC respectively, by midpoint theorem, DE // BC and 1 2DE BC= . In ADE and ABC , DAE BAC = (common angle) ADE ABC = (corresponding angle, DE // BC) ADE is similar to ABC . (AA similarity test) (d) Show that 1 2 DF FC = . [3] In DFE and CFB , EFD BFC = (vertically opposite angle) FDE FCB = (alternate angle, DE // BC) DFE is similar to CFB . By midpoint theorem, 1 2DE BC= . 1 (proven)2 DF DE FC BC= = B1 M1 M1 M1 M1 A1

6 Jurongville Secondary School 4049/02/2025 3 A radioactive substance decays according to the equation 15 btAe −= , where A is the mass in grams of radioactive substance remaining and t is the time in hours after the radioactive substance starts to decay. (a) Given that there are 10 grams of radioactive substance remaining after 30 hours, verify that the value of b is 0.013516. [2] 30 30 10 15 1030 ln 15 1 10ln30 15 0.013516 15 10 15 bt b b A b e e b e− − − = = = = − − = = (b) Find the mass of radioactive substance left at the end of 1 week. [1] 0.013516(7 24)15 1.55g Ae −= = (c) Find the number of hours for the radioactive substance to decay to half of its original mass. [3] When 0t = , 15A= . 0.013516 0.013516 0.013516 15 2 1 2 10.013516 ln 2 51.3 hours 15 15 t t te t A e t e− − − − = = = = = M1 A1 B1 M1 M1 A1

7 Jurongville Secondary School 4049/02/2025 [Turn Over (d) Find the rate at which the substance decays at 50t = hours. [2] 0.013516 0.0135160.20274 15 t t A A e ed dt − −− = = When 50t = , 0.013516(50) 0.103 0.20274d eA dt − =− =− The rate of radioactive decay at 50t = hours is 0.103g/h. M1 A1

8 Jurongville Secondary School 4049/02/2025 4 (a) Prove that cot 1cot(45 ) cot 1 xx x +− = − . [3] 1cot(45 ) tan(45 ) 1 tan 45 tan tan 45 tan 1 tan 1 tan 11 cot 11 cot cot 1 (proven)cot 1 x x x x x x x x x x − = − += − += − + = − += − cos(45 )cot(45 ) sin(45 ) cos 45 cos sin 45 sin sin 45 cos cos 45 sin cos sin cos sin cos 1sin cos 1sin cot 1 (proven)cot 1 xx x xx xx xx xx x x x x x x −− = −  + =  −  += − + = − += − M1 M1 A1 M1 M1 A1

9 Jurongville Secondary School 4049/02/2025 [Turn Over (b) Hence express cot15 in the form 3pq+ , where p and q are integers. [3] ( )cot15 cot 45 30 cot 30 1 cot 30 1 1 1tan 30 1 1tan 30 31 31 3 1 3 1 3 1 3 1 3 2 3 1 31 4 2 3 2 23 = −  += − += − += − ++= −+ ++= − += =+ M1 - sub x = 30o M1 A1

10 Jurongville Secondary School 4049/02/2025 5 It is given that 5cos 13 = , where 0 360   . (a) Show that the value of 22 sin − is 194 169 . [2] 22 2 2 2 sin 2 (1 cos ) 1 cos 51 13 194 169   − = − − =+ =+  = (b) By finding the value of 2sin  , find the exact value of 2tan (90 ) − . [4] ( ) 2 2 2 2 2 2 2 1942 sin 169 144sin 169 1tan 90 tan cos sin 5 13 144 169 25 144       −= = − = = = = M1 A1 – with correct substitution B1 M1 M1 A1