Dunman Sec 2025 Prelims 4EXP Add Math Paper 1 Solutions

s DUNMAN SECONDARY SCHOOL CANDIDATE NAME CLASS INDEX NUMBER PRELIMINARY EXAMINATION 2025 SECONDARY 4 EXPRESS / 5 NORMAL ACADEMIC MATHEMATICS 4052/01 Paper 1 20 August 2025 2 hours 15 minutes Solutions (Updated 1 Dec 2022 by HOD/Math aligned to Marking Scheme for Specimen Papers 4049)

Question Answer 1 24 3x x−= − ( )( )4 3 2xx− − = 2 7 12 2xx− + = 2 7 10 0xx− + = ( )( )2 5 0xx− − = 2 or 5x= When x = 2, y = 2− When x = 5, y = 1 Length = ( ) ( ) 22 2 5 2 1− + − − = 4.24 unit 2a 190 tan 90 p−−    or 1tan22 p −−   2b x − 3 23 15 20xx++ = ( ) 23 5 20xx++ = ( ) ( ) 22 3 2.5 2.5 20x + − + = ( ) 2 3 2.5 1.25x++ Since ( ) 2 3 2.5 1.25 1.25x+ +  , 23 15 20xx++ cannot be smaller than 1.

Page 3 of 12 4049/01/PRELIM/4E/2025/SOLUTIONS Question Answer 4a 2 2 32 (2 1) xx xx ++ + = 2 21 A B C x x x++ + ( ) ( ) 22 3 2 2 1 2 1x x Ax x B x Cx+ + = + + + + When 1 2x=− , 1 3 1 24 2 4 C− + =  C = 3 When x = 0, B = 2 Comparing coefficients of x2, A = 1− 2 2 32 (2 1) xx xx ++ + = 2 1 2 3 21x x x− + + + 4b 2 22 3 2 1 2 3 d d(2 1) 2 1 xx xxx x x x x ++ = − + +++ = ( )23ln ln 2 1 2x x cx− − + + + 5 53log log 5 2 yy−= 5 5 5 log 53log 2 logy y−= 5 5 13log 2 logy y−= ( ) 2 553 log 2log 1 0yy − − = ( )( )553log 1 log 1 0yy+ − = 5 1log 1 or 3y=− y = 5 or 1 35 0.585 (3s.f.)y − == Question Answer 6a ( ) ( )sin 105 sin 60 45 = +  = sin 60 cos45 cos60 sin 45 +   = 3 1 1 1 22 22  +  = 3 1 2 2 2 2 +  = 62 4 + 6b ( )( )1 4 sin105 10 2 32 BC = +

Page 4 of 12 4049/01/PRELIM/4E/2025/SOLUTIONS ( )( )1 6 24 10 2 324 BC  + =+ ( )2 10 2 3 62 6 2 6 2 BC + −= +− ( )( )4 5 3 6 2 62 +− = − 5 6 18 5 2 6= + − − 4 6 3 2 5 2= + − 4 6 2 2=− ( )2 4 3 2=−

Page 5 of 12 4049/01/PRELIM/4E/2025/SOLUTIONS Question Answer 7a ( )9 6 6 4x x x+= 96 644 xx xx+= 2 2 36 6024 xx x + − = 2 33 6022 xx    + − =       When 3 2 x u = , 2 60uu+ − = 7b ( )( )3 2 0uu+ − = 2 or 3 (n.a.)u=− 3 22 x  = ln 2 lg 2 or 33ln lg22 x= = 1.71 8 When x = 2− , ( ) ( ) ( ) 32 2 2 2 2 10 0ab− + − + − + = 4 2 6ab−= 23ab−= - eqn 1 When x = 3, ( ) ( ) ( ) 32 2 3 3 3 10 10ab+ + + = 9 3 54ab+ =− 3 18ab+ =− - eqn 2 eqn 1 + eqn 2: 5 15a=− 3a=− 9b =− Question Answer 9a 2312 0 2tt−= 312 0 2tt −= t = 0 or 8 8 s 9b 8 2 0 3distance 12 d 2t t x=− 8 23 0 16 2tt=−

Page 6 of 12 4049/01/PRELIM/4E/2025/SOLUTIONS    128 0=− = 128 m 9c 12 3at=− When t = 5, a = 3− m/s2

Page 7 of 12 4049/01/PRELIM/4E/2025/SOLUTIONS Question Answer 10 3xy xe −= ( ) ( ) 33d 13d xxy e x ex −−= + − = ( ) 3 13xex− − ( ) 3 1 3 0xex− −= 1 3x= 1 or 0.1233y e= ( )( ) ( ) 2 33 2 d 1 3 3 3d xxy x e ex −−= − − + − When 1 3x= , 2 2 d 0d y x  11,33 e  − is a maximum point.

Page 8 of 12 4049/01/PRELIM/4E/2025/SOLUTIONS Question Answer 11a 11b The 22cos3 xx = is equivalent to 22cos3 1 1xx − + =− + and so its solutions can be found from the interception points of the two graphs.

Page 9 of 12 4049/01/PRELIM/4E/2025/SOLUTIONS Question Answer 12a Midpoint of PQ = 1 9 2 10,22 + − +  = ( )5,4 Gradient of PQ = 2 10 19 −− − = 3 2 Gradient of perpendicular bisector = 2 3− ( )245 3yx− =− − 2 22 33yx=− + 12b 2 22 8 1033xx− + = − 26 52 33x− =− 2x= ( )2,6R 12c Area = 9 2 1 91 10 6 2 102 − = ( ) ( )1 54 4 10 20 6 182 − + − + − = 26 unit2

Page 10 of 12 4049/01/PRELIM/4E/2025/SOLUTIONS Question Answer 13a t 1 2 3 4 5 m 6.55 5.36 4.40 3.60 2.94 ln m 1.88 1.68 1.48 1.28 1.08 13b ln 2.08M = M = 8.00 13c Gradient = 0.2− 0.2k =