2025 Prelim AM Bedok View P2 MS

Bedok View Secondary School Department of Mathematics Mark Scheme 1 Year 2025 Level & Stream 4E Type of Assessment Preliminary Exam Subject & Paper Additional Math P2 Qns Working 1a b 2 2 2 2N( ) ln 2let and ln d 2 ln 2 lnd d1 d 12N ( ) . ln .2 2 2 ln stationary point N ( ) 0 22 ln 0 12 xx x u x v x u x v xx v xx x x x xx xx x x xx x x =  ==  = = − =−     = − +       =− +   = − + =  −+ 0. 2ln 0 20 (rej) or 1 2ln 0 2 2ln 1 21 ln 2 x x x x x e   =  = − + =   =  = 5 0.5 2 2 1.21306 x xe x = = = x 1.1 1.21306 1.3 Sign of N’(x) +ve 0 − ve Sketch of tangent  max absorption rate at x = 1.21 (3 s.f.)

Bedok View Secondary School Department of Mathematics Mark Scheme 2 Qns Working 2a b c d (0) 22 when 0, = 90 90 22 90 22 68 kt k T Ae tT Ae A A =+ = =+ =+ = (3.5) (3.5) (3.5) when 3.5, = 79 79 22 68 57 68 57 68 57ln (3.5)68 57ln 3.568 0.050416 0.0504 (3 s.f.) k k k tT e e e k k k k = =+ = =  = =  =− =− 0.050416 0.050416(10) 22 68 when 10, 22 68 = 63.0728 tTe t Te − − =+ = =+ Since 63.0728 C is between 55 C and 70 C, the coffee is suitable to be served. After a long time, T approaches 22 C.

Bedok View Secondary School Department of Mathematics Mark Scheme 3 Qns Working 3a cos 14 14cos Let be the point on such that sin 3 3sin 14cos 3sin 6 (shown) ED ED G FD BG FD EG EG FD      = = ⊥ = = = + + b c d 2214 3 = 205 3tan 14 12.0947 205 cos( 12.1 ) 6 (3 s.f.) R FD    =+ = = = −  + 205 cos( 12.0947 ) 6 12 6cos( 12.0947 ) 205 12.0947 65.2248 77.3195 77.3 (1 d.p.)      −  + = −  = − =  = = Maximum 205 cos( 12.0947 ) 6 205 6 20.317 it is possible for to be 15 cmFD  −  + =+ =  Or 205 cos( 12.0947 ) 6 15 9 cos( 12.0947 ) 205 12.0947 51.054 63.1 (1 d.p.)     −  + = −  = − =  = FD is 15 cm when 63.1 =

Bedok View Secondary School Department of Mathematics Mark Scheme 4 Qns Working 4a

Bedok View Secondary School Department of Mathematics Mark Scheme 5 Qns Working b 2.5 10 lg lg 10 lg lg lg10 lg lg lg lg gradient = 0.232 (3 s.f.) 0.1 lg vertical axis intercept lg 2.5 0.05 10 =316 (3 s.f.) kt kt kt Na Na Na N a kt N kt a k a a a = = =+ =+ =+ =  =− = = 5a b 2 3 2 2 32 2( 1)( 2)( 1.5) ( 2)( 2 3) 2 3 2 3 4 6 2 5 6 x x x x x x x x x x x x x x − + − − = − − − + =− + + − + − =− + + − ( ) ( ) 2 3 2 32 2 2 3 0 1 3 2 5 0 3 5 3 0 3 2 x x x x x x x x x xx xx x − + − − − + − + − − − + − − + + −+ Quotient 3 x = −

Bedok View Secondary School Department of Mathematics Mark Scheme 6 Qns Working c ( ) ( )( ) ( ) ( ) ( ) 3 2 2 32 32 32 7 2 3 5 2 Quotient 7 2 3 1 2 Quotient 1 1 1 1sub , 7 2 03 3 3 3 1 1 25 ------ ----(1) 27 3 9 sub 2, 2 7 2 2 2 0 hx x kx x x hx x kx x x x h k hk x h k − + − = − −  − + − = + −       =− − − − + − − =           − − = = − + − = ( ) 8 2 30- ---------(2) From (2), 4 15 15 4 ----------(3) 1 1 25sub (3) into (1), 15 427 3 9 1 4 25 5 27 3 9 hk hk kh hh hh += += =− − − − = − − + = 35 70 27 9 6 sub 6 into (3), 15 4(6) 9 h h hk k = = = = − =−

Bedok View Secondary School Department of Mathematics Mark Scheme 7 Qns Working 6a b c (tangent from external point) (tangent from external point) (ratio of corresponding sides are equal) (common angle) is similar to (ratio of 2 pairs of cor respondi AB AE AC AD AB AE AC AD BAE CAD ABE ACD = = =  =  ng sides and a pair of included s are equal) ( is similar to ) (base of isosceles ) 180 (adjacent s on a straight line) Since 180 , ABE ACD ABE ACD ACD ADC EBC ABE ADC EBC  =    =    = −   + =  by the converse of angles in opposite segments, a circle can be drawn to pass through BEDC. Let , (alternate segment thm) // since is similar to (alternate angles, // ) (alternate segment thm) (proven) CBF x BEF x BE CD ABE ACD ECD x BE CD EDF x CBF EDF = =  = =  =

Bedok View Secondary School Department of Mathematics Mark Scheme 8 Qns Working 7a 8 8 8 32 32 3 General term of 1 5 8 (1) 5 8 5 (7 2 ) 1 5 88(7 2 ) ... ... 32 55 88 term = 7 232 55 r r r x x r x r xx xxx xxxx − −  =−   =−  +−         = + + − + − +                     − + −             ( ) 32 33 3 = 7 56 2 (28)125 25 392 56 = 125 25 112 Coefficient of = 125 xx x xx x    −+       −+ − b 2 2 2 2 3 3 3 3 23 22 113322 24 113333 28 113 3 323 48 3 .2 323 .223 ( 1) ( 1)( 2).22 2 3 62 8 nn nn nn nn nn xx nn xx nn nn nn n n n n n n n −− −− −− −−     =           =           =           =          =       − − −=  =− =

Bedok View Secondary School Department of Mathematics Mark Scheme 9 Qns Working 8a b 2 2 1cos 2 3 11 2sin 3 1sin 3 11sin or (rej) 33 1cosec sin 1 = 1 3 = 3 A A A A A A = −= = =− =  ( ) sin105 sin 45 60 sin 45 cos 60 cos 45 sin 60 1 1 1 3..2222 13 2.2 c osec 2 sin105 133 2. 2.2 133 2 31 2 A  = +  =  +   =+ += − +=− +=− −=

Bedok View Secondary School Department of Mathematics Mark Scheme 10 Qns Working 9a b 12cos d2 12sin 2 = 1 2 1 = 4sin 2 Given 0 and 3, find 13 = 4sin (0)2 3 1 = 4sin 3 2 find when 0, 14sin 3 02 13sin 24 1basic of 0.848062 1.6961 1.70 (3 s.f.) v t t t c tc t v c c c vt tv t t t t t =− − + −+ == −+ = −+ = − + = = = = =  0 π t − 2 2 2π a 12cos 2at=−