Dunman Sec 2025 Prelims 4EXP Add Math Paper 2 Solutions

DUNMAN SECONDARY SCHOOL CANDIDATE NAME CLASS INDEX NUMBER PRELIMINARY EXAMINATION 2025 SECONDARY 4 EXPRESS / 5 NORMAL ACADEMIC MATHEMATICS 4052/02 Paper 2 25 August 2025 2 hours 15 minutes Marking Scheme (Updated 1 Dec 2022 by HOD/Math aligned to Marking Scheme for Specimen Papers 4049)

Question Answer 1a 27 20 5h t t= + − ( ) 25 4 7tt=− − + ( ) 2 5 2 4 7t=− − − + ( ) 2 5 2 27t=− − + maximum height = 27 OR d 10 20d h tt =− + 10 20 0t− + = 2t = When t = 2, ( ) ( ) 2 7 20 2 5 2 27h= + − = m 1b ( ) 2 61x x k k x+ + = − ( ) 2 6 2 0x k x k+ − + = ( ) ( )( ) 2 6 4 1 2 0kk− −  236 12 8 0k k k− + −  2 20 32 0kk− +  ( )( )2 18 0kk− −  2 18x   2a 4tan 3A= 2 42 3tan 2 41 3 A = − 24 7=− Question Answer 2b 12 5sin , cos13 13BB== 3 3 3sin sin cos sin cos2 2 2B B B   + = + ( )12 50113 13=  + −  5 13=− OR 3sin sin 2 22BB     + = − +      

Page 3 of 12 4049/02/PRELIM/4E/2025/SOLUTIONS sin sin22 BB   = − + =− −       5cos 13B=− =− 2c 2cos 2cos 1 2 BB=− 25 2cos 113 2 B=− 2 182cos 2 13 B = 2 9cos 2 13 B = 33cos or (rej)2 13 13 B =−

Page 4 of 12 4049/02/PRELIM/4E/2025/SOLUTIONS Question Answer 3a PQC PSQ = (Alternate segment theorem) PQC PBC = (angles in same segment) Hence PBC PSQ = By alternate angles of parallel lines, BC is parallel to QS 3b PQC PSQ = (Alternate segment theorem) PMQ QMS = (Common angle) PQM is similar to QSM (AA Similarity Test) 3c Since PQM is similar to QSM , QM PM SM QM= ( ) 2 QM PM SM=

Page 5 of 12 4049/02/PRELIM/4E/2025/SOLUTIONS 4a ( )d d coscotd d sin xxx x x =  ( ) ( ) 2 sin sin cos cos sin x x x x x −−= 22 2 sin cos sin xx x −−= 2 1 sin x=− 2cosec (Shown)x=− OR ( ) ( ) 1dd cot tandd xxxx − = ( ) ( ) 2 2tan secxx − =− 2 2 sin 1 cos cos x xx −    =−        2 22 cos 1 sin cos x xx  =−   2 1 sin x=− 2cosec (Shown)x=− 4b 2233 44 cot d 1 cosec dx x x x   − = −   3 4 cotxx  =+ cot cot3 3 4 4       = + − +       1 134 3 = + − − 1 1 or 0.16112 3 = + − −

Page 6 of 12 4049/02/PRELIM/4E/2025/SOLUTIONS Question Answer 5a General term 8 3 8 r r kxr x − =−  ( ) ( ) 838 1 rrrrx k xr −−=− ( ) 848 1 r rrkxr −=− When 8 4 0r−= , 2r= ( ) ( )2 8 4 228 172 kx − −= 228 7k = 2 1 4k = 1 2k = 5b When 8 4 4r− =− , 3r= ( ) 3 3 4 4 8 113 2Tx − =−  47x−=− ( ) ( )( ) 8 4 4 4 31 1 ... 7 7 ... kx x x x x −+ − = + + − + ... 7 7 ... ... 0 ...= + − + = + + There is no constant term in the expansion. (Shown)

Page 7 of 12 4049/02/PRELIM/4E/2025/SOLUTIONS Question Answer 6a AM = 2 cos θ BM = 2 sin θ BC = 4 sin θ Total area = (4 sin θ)2 + 1 2 (2 cos θ)(4 sin θ) = 16 sin2 θ + 4 sin θ cos θ 6b 16 sin2 θ + 4 sin θ cos θ = 8(2 sin2 θ) + 2(2 sin θ cos θ) = 8(1 – cos 2θ) + 2 (sin 2θ) = 2 sin 2θ – 8 cos 2θ + 8 6c R = √22 + 82 = √68 tan α = 8 2 α = 76.0° 2 sin 2θ – 8 cos 2θ = √68 sin (2θ – 76.0°) 7a 22cos sin 1 2sin cos xxLHS xx −= + 22 22 cos sin sin cos 2sin cos xx x x x x −= ++ ( )( ) ( ) 2 cos sin cos sin cos sin x x x x xx +−= + cos sin cos sin xx xx −= + 1 tan 1 tan x RHSx −==+

Page 8 of 12 4049/02/PRELIM/4E/2025/SOLUTIONS Question Answer 7b 1 tan 2 tan1 tan 3 x xx − =+ ( ) 23 1 tan 2 tan tanx x x− = + 22 tan 5tan 3 0xx+ − = ( )( )2 tan 1 tan 3 0xx− + = 1tan or 32x=− 0.464 or 3.61x= 1.89 or 5.03x= 8a

Page 9 of 12 4049/02/PRELIM/4E/2025/SOLUTIONS Question Answer 8b 322 7 2 3 0x x x− + + = Let ( ) 32f 2 7 2 3x x x x= − + + When x = 3, ( ) ( ) ( ) ( ) 32 f 3 2 3 7 3 2 3 3 0= − + + = By factor theorem, ( )3x− is a factor of f(x). ( )( ) 23 2 1 0x x x− − − = ( )( )( )3 2 1 1 0x x x− + − = 1 , 1 or 32x=− 8c 6 4 23 2 7 2 0y y y+ − + = 2 4 6 2 7 230 y y y+ − + = 2 4 6 1 1 13 2 7 2 0y y y      + − + =           23 2 2 2 1 1 13 2 7 2 0y y y      + − + =           2 1x y= , Kun Ye is correct

Page 10 of 12 4049/02/PRELIM/4E/2025/SOLUTIONS Question Answer 9a Height = 3x2 Base = 2x Area = ( )( ) 231 2 3 32 x x x = 9b d 6d y xx = d d d d d d y y x t x t= d36 d xx t= d36 d xx t= d1 d2 x tx= When x = 4, d1 d8 x t = Rate of change of PQ = 1 4 unit/s 9c 2d 9d A xx = d d d d d d A A x t x t= 2d1 9d8 A xt = When x = 4, d 18d A t = unit2/s