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 − −=