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 = =