Geylang Methodist 2024 AM P1 PRELIM Teacher

Marking Scheme AM P1 (4049/01)- Prelim 2024 Qn Answer Ma rks Partial Marks Guidance 1 2 40b ac−   2 2( 2) 4 ( 7) 0p p p+ − +  224( 4 4) 4 28 0p p p p+ + − −  224 16 16 4 28 0p p p p+ + − −  12 16 0p− +  12 16 0p− 4 3p 2 232 xxy e e −=+ 262 xxdy eedx −=− 2 2 2 12 2 xxdy eedx −=+ 2 2 2 2 2 12 2 6 2 3 2x x x x x xd y dy y e e e e e edx dx − − −+ + = + + − + + 221 2xxee −=+ 21A= 2B= 3(a) Period 180 ( ) or = Amplitude 3=

3(b) 4 2 2 d 32d y xx =− d (3 2)d y x dxx =− 23 22 x xc= − + 13c=− 2d3 2 13d2 yx xx = − − 23 2 132 xy x dx= − − 3 2 132 x x x d= − − + 3 2(4)40 (4) 13(4)2 d− = − − + 4d =− 3 2 13 42 xy x x= − − −

5(a) 1 16 7 10(3 2 5)22 BC ++= 16 7 10 3 2 5 BC += + ( )( ) ( )( ) 16 7 10 3 2 5 3 2 5 3 2 5 +− = +− 48 2 16 5 21 20 7 50 13 − + −= 48 2 16 5 42 5 35 2 13 − + −= 13 2 26 5 13 += 2 2 5=+ 5(b) ( ) ( ) 222 3 2 5 2 2 5AC = + + + 18 6 10 5 2 4 10 20= + + + + + 45 10 10=+ 6(a) '( ) 3cos 8sin 2f x x x=+ cos 0 and sin 2 0 for 0 2x x x     3cos 8sin 2 0 for 0 2x x x  +    '( ) 0 for 0 2f x x    Hence f is an increasing function 6(b) 3cos 8sin 266 dy dx  = +   3338 22=  +  311 2= 11 3 22 dy dt = 11 3=

7(a) ( ) 10 2 3 10 2General term r r xr x −  =−  ( ) 10 5 3010 21 rrr xr −−=− when 5 30 6r−= 36 5r = 6Hence there is no term in x 7(b) 5 30 0r−= 6r= ( ) 610 610 2 1 33606 − −= ( ) 6 .... 3360 ... 3 8 x+ + −  6 1Coefficient of 3360 8x −= 420=−

8(a) 4 (similar triangles)9 x yx= + 4 36xy x=+ 364y x=+ 8(b) 1 ( 9)2A x y=+ 1 36( 9) 42 x x = + +  1622 36x x= + + 2 1622dA dx x=− 2 162when 0, 2 0dA dx x= − = 2 1622 x= 2 81x = 9x= 162Minimum Area = 2 9 36 9 + + 272 m= 9(a) (given )BCA ABC AB AC = = (Alternate segment theorem)ABC CAF = BCA CAF = 9(b) 90 (Radius perpendicular to tangent)OAE OAF = = (Alternate segment theorem)BAE BCA = (Using part a)BAE CAF = (Both 90 Equal angles)OAB OAC = − Hence OA bisects angle BAC 10 33log ( 8) log 2xx− + =

(a) 3log ( 8) 2xx −= 2( 8) 3xx −= 2 8 9 0xx− − = ( 9)( 1) 0xx− + = 9 1 (NA)x or x= =− 9x= 10 (b) 1 1log 1logb a a b =  1 log 1 logaa b= − 1 loga b= − 1 1 logb a −=    1 1 c −=  c=− 11 (a) ( ) 2216 16ax x x ax− − = − +

2 2 16 24 aax = − + − 22 16 42 aa x= + − +  2 16 254 a+= 2 94 a = 2 36a = 6a= 2 ab= 3= 11 (b) ( ) 2 25 3yx= − + Max value of 25y= Corresponding value of 3x=− 11 (c) 0y 216 6 0xx− −  2 6 16 0xx+ −  ( 8)( 2) 0xx+ −  82 x−   12 (a) sin sin33 xx   + − −       Alternate solution ( ) 2 2 216 25 2ax x b bx x− − = − + + 2 2 216 25 2ax x b bx x− − = − − − B2 216 25 b=− 2 9b = 3b= M1A1

sin cos cos sin sin cos cos sin3 3 3 3x x x x    = + − −  sin cos cos sin sin cos cos sin3 3 3 3x x x x   = + − + 2cos sin3 x= 12 sin2 x=   sin x= 12 (b) sin 2 sin 2 sin 233 x x x   + − − =       sin 2 2 4sin 2xx−= 3sin 2 2x=− 2sin 2 3x=− Basic angle 0.72973= 2 0.72973, 2 0.72973x = + − 1.94, 2.78x= 13 (a) lg (lg ) lgy b x a=+ 13 (b) lg y 0.792 0.672 0.568 0.462 0.342 0.230 Straight line graph 13 (c) lg 0.9a 7.94a 0.3 0.5lg 0.2222.7 1.8 0.599 b b − −− = 13 (d) when 0.8x= lg 0.72y= 5.23y=