2025 Prelim AM Bedok View P1 MS

Bedok View Secondary School Department of Mathematics Mark Scheme 1 Year 2025 Level & Stream 4E5N Type of Assessment Preliminary Exam Subject & Paper Additional Math P1 Qns Working 1 24 6 6 4x xy y− − = ----(1) 4yx−= ------(2) Sub (2) into (1): ( )( ) 2 22 2 2 4 6 (4 ) 6(4 ) 4 4 24 6 24 6 4 2 30 28 0 15 14 0 1 14 0 14 or 1 10 or 3 x x x x x x x x xx xx xx xx yy − + − + = − − − − = + + = + + = + + = =− =− =− = 2 9sin 2sec 29sin cos 2sin cos 9 42sin cos 9 4sin 2 9 xx x x xx xx x =− =− =− =− =− Basic angle = 1 4sin 9 −   = 0.46055 ( )2 0.46055, π 0.46055 0.46055 or 2.68104 0.230 or 1.34 x x =− − − =− − =− −

Bedok View Secondary School Department of Mathematics Mark Scheme 2 Qns Working 3a 3b 11 1 Principal range of cos is 0 cos π, 23 πhence principal value of cos is 24 πinstead of . 4 xx−− −  − − L.H.S: 1cot tan tan tanA A A A+ = + = cos sin sin cos AA AA+ = 22cos sin sin cos AA AA + = 1 sin cosAA = sec cosecAA = RHS (shown) OR RHS: 11sec cos ec cos sinAA AA= = 22 22 1 cos sin sin cos cos sin sin cos cos sin cos sin sin cos cos sin tan cot LHS (shown) AA AA AA AA A A A A AA AA AA += =+ =+ = + =

Bedok View Secondary School Department of Mathematics Mark Scheme 3 Qns Working 4 By long division, ( ) ( ) ( ) ( ) ( ) ( ) 33 2 32 2 2 2 22 22 2 4 2 4 442 8 8 42 2 8 8 4 222 8 8 4 2 2 xx x x xxx xx xx x x A B C xxx x x x x A x Bx x Cx ++ = +++ − − +=+ + − − + = + + +++ − − + = + + + + Let 0x= : 44 1 A A = = Let 2x=− : 12 2 6 C C − =− = Let 1x= : 8 8 4 9 3 6 9 B B − − + = + + =− ( ) ( ) 3 22 2 4 1 9 6 2 222 x xxx x x + = + − + +++ 5a ( ) ( ) ( ) ( ) 2 22 2 22 32 f '( ) 3 23 3 x x x a x x x ax x + − + = + − − += + ( ) ( ) 2 22 22 2 2 2 23 0, for 1 3 Since 3 0, 2 3 0 2 3 0 Given that 1 ( )( 1) 0 0 By comparing: 3 21 2 3 1 1 x ax bx x x x ax x ax bx x b x x bx x b b ab a a − − +    + + − − +  + −   − −  − − +  =− =− − =− = 5b

Bedok View Secondary School Department of Mathematics Mark Scheme 4 Qns Working 6 ( ) 2 2 2 68 Discriminant ( 8) 4( 6)( ) 64 4 24 y k x x k kk kk = − − + = − − − = − + For curve that does not intersect the x-axis, Discriminant < 0 2 6 16 0 ( 2)( 8) 0 2 or 8 kk kk kk − −  + −  −  The curve has a minimum point: 60 6 8 k k k −   7a ( ) ( )d 1 e e 1 ed 2e e x x x xx xxx x − − − −−  − = − − =− 7b ( ) ( ) ( ) ( ) 2e e d 1 e 2e d e d 1 e e d 2e d 1 e e d 2e 1 e 2e e e ee x x x x x x x x x x x x x x x xx x x x c x x x x c x x x x c x x x c xc xc − − − − − − − − − − − − − − − −− − = − + − = − + − =− + − + =− − − + =− − + + =− − +    

Bedok View Secondary School Department of Mathematics Mark Scheme 5 Qns Working 8a ( ) ( ) ( ) ( ) 2 2 2 1 d2 d 1 2 d 1 2 1 d 21 2 1 At 4 and 0, 42 2 2 21 r t t rt t tt tc rc t rt c c r t − − =− + =− + =− + = + + =++ == =+ = =++   8b 2 2 2 π d 2πd dAt 2.6, 5.2 π, d 22.6 2 1 20.6 1 7 3 d2 0.18d 7 13 d 5.2π 0.18d d 0.936π cm /min d Ar A rr Ar r t t t r t A t A t = = == =++ = + = =− =−  + = − =−

Bedok View Secondary School Department of Mathematics Mark Scheme 6 Qns Working 9a 6y kx=+ ---- (1) 223x xy−= -----(2) Sub (1) into (2): 2 22 2 2 ( 6) 3 2 6 3 0 (2 ) 6 3 0 x x kx x kx x k x x − + = − − − = − − − = Discriminant 2( 6) 4(2 )( 3) 60 12 k k = − − − − =− For tangent, 60 12 0 5 k k −= = 9b ( )4 2 2 2 3 2 1xx− + = − 4 2 2 2 3 2 1x x x− + = − 2 1 3 2 2 2 4x x x+ = + − ( )2 1 5 2 4x+ = − x = 2 1 5 2 4 5 2 4 5 2 4 ++  −+ x = ( ) 2 2 5(2) 4 2 5 2 4 5 2 4 + + + − x = 9 2 14 50 16 + − x = 97 234 17 +

Bedok View Secondary School Department of Mathematics Mark Scheme 7 Qns Working 10a   2 1 2 0 11Area of 0 3 2 02 1 3 4 22 2.5 units PQR= = + − = 10b 30 21 3 1 3 PR QS m m −= − = =− Equation of QS: 12 ( 0)3 1 2 (1)3 5 (2) yx yx xy − =− − =− + −−− − = −−− Sub (1) into (2): 1 253 21 4 1 4 21 1coordinates of , 44 xx x y S + − = = = =

Bedok View Secondary School Department of Mathematics Mark Scheme 8 Qns Working 11a ( )( ) 1 1 2 2 4 24 2 4 4 24 22 24 4 xx xx x x − − += += += Let 2xy= , ( )( ) 2 2 2 244 4 96 4 96 0 8 12 0 yy yy yy yy += += + − = − + = ( ) 3 8 or 12 2 8 or 2 12 Rejected since 2 0 22 3 x x x x yy x = =− = =−  = = 11b 33log ( 1) log ( 5)xx− = + 3 3 3 log ( 1) log ( 5) log 3 x x− =+ 3 31 2 3 log ( 1) log ( 5) log 3 x x− =+ 332 log ( 1) log ( 5)xx− = + 2 33log ( 1) log ( 5)xx− = + (x – 1) 2 = x + 5 x 2 – 2x + 1 – x – 5 = 0 x 2 – 3x – 4 = 0 (x + 1) (x – 4) = 0 x + 1 = 0 x = − 1 [N.A. as 3log ( 1)x− will be undefined] or x – 4 = 0 x = 4

Bedok View Secondary School Department of Mathematics Mark Scheme 9 Qns Working 12a 3 4 4 3 3 4 of normal 1 of tangent 1 1dAt and 1,2d 1161 2 1 2 831 2 (shown) xy yx m m yx x k k k −= =− = =− = =−  − =−  − =− = 12b 3 3 4 4 At stat points, 160 2 16 2 1 12 1 12 0.537 or 0.537 x x x x x x x −= = = = =− 12c 2 24 4 2 24 2 24 d3 6d2 60 3 02 d3 60d2 d3Since 6 0d2 the gradient has no turning point. y xx x y xx y xx =+    = +  = + 

Bedok View Secondary School Department of Mathematics Mark Scheme 10 Qns Working 13a 21.2 14.4 53.7 At 0, $53.7 thousands C n n nC = − + == 13b ( ) ( ) 2 2 2 2 2 1.2 14.4 53.7 1.2 12 44.75 1.2 6 6 44.75 1.2 6 10.5 C n n nn n n = − + = − + = − − + = − + 13c For 600 pairs of running shoes produced, the minimum cost of 10.5 thousand dollars will be incurred. 13d ( ) ( ) ( ) 2 2 2 1.2 6 10.5 50 1.2 6 39.5 39.56 1.2 6 5.7373 11.7373 or 0.2627 n n n n n − + = −= −= − = = Maximum number of pairs of running shoes for which the cost is at most 50 thousand dollars = 11.7373 = 11.7 hundreds