Geylang Methodist Sec 2024_AM_P2_PRELIM_Teacher

Marking Scheme AM P2 (4049/02) Setters : J Joseph(1,2,4,5 and 6 ) Ng SL (3,8,9 and 10) Qn Answer Ma rks Partial Marks Guidance 1 ( ) 422sin 7 1 sin 4xx+ − = 2 1sin 2x= Basic angle 45= 45 ,135 ,225 ,315x= 2(a) 50120 150 ke−= 50 150 120 ke = 150ln 120 50k = 0.00446287= 120 0.00446287150Me −= 87.8 (3 s.f.)g= 2(b) 2When M 150 1003 g=  = 0.00446287100 150 te−= 0.00446287 150 100 te = 150ln 100 0.00446287t = 90.9 hours (3 s.f.)= 3(a) ( ) ( ) 22 2 22 2 xxe x edy dx x −−= − 3(b) ( ) ( ) ( ) ( ) 22 2 2 5 5 22 0 2 2 4 1 0 5 2.52 2 2.5, 2 xx x e x e x ex x ye e −− = − − − = == = 3(c) (2.5, 2e5) is a minimum point.

Qn Answer Ma rks Partial Marks Guidance 4(a) A( ,0) and B(0, )ab gradient of AB b a=− gradient of perpendicular bisector =a b midpoint of AB is = , 22 ab  72gradient of perpendicular bisector = 32 b a + + 14 6 ab ba += + 22 6 14 (shown)a a b b+ = + 4(b) 2ab= ( ) ( ) 2 22 6 2 14b b b b+ = + 23 2 0bb−= ( )3 2 0bb −= 2 3b= a 4 3= 5(a) 2 10, 2 4gf=− =− ( )centre 5,2 ( ) ( ) 22 5 2 25r = − + − − 2units= 5(b) The centre of the circle is 2 units above the x- axis and the radius of the circle is 2 units. Hence the x-axis is a tangent to the circle. 5(c) When 3x= 2y= ( ) ( )3,2 and 5,2 gradient 0= Equation of tangent is 3x= ( )P 3,0

Qn Answer Ma rks Partial Marks Guidance 6(a) 12cosPT = 5sinSR = 12cos 5sind =+ 6(b) 2212 5R=+ 13= 1 5tan 12 − =  22.6= ( )13cos 22.6d =− 6(c) ( )13cos 22.6 10−= 62.3 = 7(a) ( 1) 0f −= → 5ab− + = ( 3) 60f − =− → 33ab− + = 1a= , 6b= 7(b) 32 4 6 0x x x− + + = ( )( ) 21 5 6 0x x x+ − + = 1, 2, 3x=− 7(c) ( ) ( ) 2 63 1 4 3 3 xx x+ = − ( ) ( ) 32 3 4 3 3 6x x x − + + = 0 The equation can be solved by taking each solution in part b 3x .

Qn Answer Ma rks Partial Marks Guidance 8 (a) ( ) ( ) 3 12 8 2 2 1 2 c c =− = 8(b) ( ) 2 2 2 38 2 382 2 2/ dv atdt ms = = − =− = 8(c) 3 2 2 180 2 180 2 16 4 tt tt t t −=  −= = = ( ) ( ) ( ) ( ) 3 24 24 24 24 18 2 8 28 0, 0 0 14 8 4, 14 4 4 8 32 m 5, 14 5 5 8 21.875m Distance travelled = 32 32 21.875 42.125 Average speed = 42.125 5 = 8.425 m/s s t t dt tt c s t c s t t t s t s =− = − + = =  = =− = =− = = =− = = + − =  

Qn Answer Ma rks Partial Marks Guidance 9(a) ( ) ( ) ( ) ( ) 1 2 1 2 1 10 6 62 3 10 6 31, , 4 4 4Gradient of normal = 3 4Equation of normal : 4 1 3 3 4 16 dy xdx x dyx dx y yx yx − − =− − − =− =− = =− − + =− + =− − 9(b) ( ) ( ) 3 2 2 5 3 1 3 3 2 At , = 0, = - 4 5, 0, 3 1Area of triangle = 3 4 6 unit2 Area under curve = 10 6 d 5 310 - 6 x 403 9× -62 -1 4Area of shaded region = 6 + 9 113 or 13.1 unit9 C y x At A y x xx − ==   = − − −   = =